Maths 10a Teacher Edition.pdf

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First published 2012 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12pt Times LT © John Wiley & Sons Australia, Ltd 2012 The moral rights of the authors have been asserted. National Library of Australia Cataloguing-in-Publication data Title: Maths quest 10+10A for the Australian curriculum/ Kylie Boucher [et al.] ISBN: 978 0 7303 4178 9 (student ed. : pbk) 978 0 7303 4179 6 (student ed. : ebook) 978 0 7303 4181 9 (teacher ed. : pbk) 978 0 7303 4182 6 (teacher ed. : ebook) Series: Maths quest series. Notes: Includes index Target Audience: For secondary school age. Subjects: Mathematics—Textbooks. Mathematics—Study and teaching (Secondary) Other Authors/ Contributors: Boucher, Kylie. Dewey Number: 510 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Internal design images: © Shutterstock/Nikolai Bird, 2010 Illustrated by Aptara and the Wiley Art Studio Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10  9  8  7  6  5  4  3  2  1

Contents 3B Determining linear equations  64

Introduction  viii About eBookPLUS  x Acknowledgements  xi Chapter 1

Exercise 3B  67 3C The distance between two points on a Number and algebra Patterns and algebra

Indices  1 Are you ready?  2

1A Review of index laws  3

Exercise 1A  5 1B Negative indices  7 Exercise 1B  10 1C Fractional indices  12 Exercise 1C  14 1D Combining index laws  17 Exercise 1D  20 Summary  23 Chapter review  24 eBookPLUS activities  26

Chapter 2

straight line  68 Exercise 3C  70 3D The midpoint of a line segment  71 Exercise 3D  73 3E Parallel and perpendicular lines  74 Exercise 3E  79 Summary  83 Chapter review  85 eBookPLUS activities  88

Chapter 4

Simultaneous linear equations and inequations  89 Are you ready?  90

4A Graphical solution of simultaneous linear Number and algebra Patterns and algebra

Linear algebra  27 Are you ready?  28

4B

4C

2A Substitution  29

Exercise 2A  31 2B Adding and subtracting algebraic

fractions  33 Exercise 2B  36 2C Multiplying and dividing algebraic fractions  37 Exercise 2C  39 2D Solving linear equations  40 Exercise 2D  43 2E Solving equations with algebraic fractions and multiple brackets  45 Exercise 2E  48 Summary  50 Chapter review  52 eBookPLUS activities  54

Chapter 3

Number and algebra LINEAR AND NON-LINEAR RELATIONSHIPS

Number and algebra

4D

4E 4F 4G

equations  91 Exercise 4A  94 Solving simultaneous linear equations using substitution  96 Exercise 4B  98 Solving simultaneous linear equations using elimination  99 Exercise 4C  101 Problem solving using simultaneous linear equations  103 Exercise 4D  105 Solving linear inequations  106 Exercise 4E  108 Sketching linear inequations  110 Exercise 4F  113 Solving simultaneous linear inequations  116 Exercise 4G  118

Summary  123 Chapter review  125 eBookPLUS activities  130

Chapter 5

Measurement and geometry PYTHAGORAS AND TRIGONOMETRY

LINEAR AND NON-LINEAR RELATIONSHIPS

Coordinate geometry  55

Trigonometry I  131

Are you ready?  56

Are you ready?  132

3A Sketching linear graphs  57

5A Pythagoras’ theorem  133

Exercise 3A  62

Exercise 5A  137

5B Pythagoras’ theorem in three

7E Mixed factorisation  240

5C

Summary  242 Chapter review  244 eBookPLUS activities  246

5D

5E

5F 5G 5H

dimensions  140 Exercise 5B  143 Trigonometric ratios  145 Exercise 5C  149 Using trigonometry to calculate side lengths  151 Exercise 5D  154 Using trigonometry to calculate angle size  156 Exercise 5E  158 Angles of elevation and depression  161 Exercise 5F  163 Bearings and compass directions  165 Exercise 5G  169 Applications  172 Exercise 5H  173

Summary  177 Chapter review  179 eBookPLUS activities  182

Chapter 6

Measurement and geometry USING UNITS OF MEASUREMENT

Surface area and volume  183 Are you ready?  184

6A Area  185

Exercise 6A  189 6B Total surface area  193 Exercise 6B  199 6C Volume  203 Exercise 6C  208

Chapter 8

Quadratic equations  247 Are you ready?  248

8A Solving quadratic equations  249

Exercise 8A  253 8B The quadratic formula  255

Exercise 8B  257 8C Solving quadratic equations by inspecting

graphs  258 Exercise 8C  261 8D Finding solutions to quadratic equations by interpolation and using the discriminant  263 Exercise 8D  267 8E Solving a quadratic equation and a linear equation simultaneously  269 Exercise 8E  272 Summary  274 Chapter review  276 eBookPLUS activities  278

Functions  279 Are you ready?  280

9A Plotting parabolas  281

Exercise 9A  284 9B Sketching parabolas using the basic graph of Number and algebra

9C

Quadratic expressions  219 Are you ready?  220

9D

7A Expanding algebraic expressions  221

Exercise 7A  225 7B Factorising expressions with three terms  227 Exercise 7B  229 7C Factorising expressions with two or four terms  231 Exercise 7C  234 7D Factorising by completing the square  236 Exercise 7D  239 Contents

Number and algebra LINEAR AND NON-LINEAR RELATIONSHIPS

patterns and algebra

iv

Number and algebra LINEAR AND NON-LINEAR RELATIONSHIPS

Chapter 9

Summary  213 Chapter review  214 eBookPLUS activities  218

Chapter 7

Exercise 7E  240

9E 9F 9G

y = x2  287 Exercise 9B  291 Sketching parabolas in turning point form  292 Exercise 9C  296 Sketching parabolas of the form y = ax2 + bx + c  298 Exercise 9D  302 Exponential functions and their graphs  306 Exercise 9E  309 The hyperbola  312 Exercise 9F  314 The circle  315 Exercise 9G  317

Summary  319 Chapter review  321 eBookPLUS activities  324

Chapter 10

Measurement and geometry

Chapter 13

Statistics and probability data representation and interpretation

GEOMETRIC REASONING

Deductive geometry  325

Univariate data  429

Are you ready?  326

Are you ready?  430

13A Measures of central tendency  431

10A Congruence review  327 10B 10C 10D

10E

Exercise 13A  435

Exercise 10A  329 Similarity review  332 Exercise 10B  335 Congruence and proof  336 Exercise 10C  338 Quadrilaterals: definitions and properties  340 Exercise 10D  341 Quadrilaterals and proof  344 Exercise 10E  345

13B Measures of spread  439

Exercise 13B  442 13C Box-and-whisker plots  444

Exercise 13C  447 13D The standard deviation  449 Exercise 13D  451 13E Comparing data sets  454 Exercise 13E  455 13F Skewness  459 Exercise 13F  461

Summary  347 Chapter review  349 eBookPLUS activities  351

Summary  464 Chapter review  466 eBookPLUS activities  470

projects plus 

ICT activity

Chapter 14

pro-0099 Backyard flood  352

Statistics and probability DATA REPRESENTATION AND INTERPRETATION

Chapter 11

Problem solving

Are you ready?  472

Problem solving I  355 Chapter 12

14A Identifying related pairs of

Statistics and probability CHANCE

Probability  379 Are you ready?  380

12A Review of probability  381 12B

12C 12D

12E 12F

Exercise 12A  392 Complementary and mutually exclusive events  396 Exercise 12B  400 Two-way tables and tree diagrams  403 Exercise 12C  410 Independent and dependent events  413 Exercise 12D  415 Conditional probability  417 Exercise 12E  419 Subjective probability  420 Exercise 12F  421

Summary  423 Chapter review  425 eBookPLUS activities  428

Bivariate data  471 variables  474 Exercise 14A  476 14B Graphing bivariate data  477 Exercise 14B  481 14C Scatterplots  483 Exercise 14C  488 Summary  491 Chapter review  492 eBookPLUS activities  496

Chapter 15

Statistics and probability DATA REPRESENTATION AND INTERPRETATION

Statistics in the media  497 Are you ready?  498

15A Populations and samples  499

Exercise 15A  502 15B Primary and secondary

data  503 Exercise 15B  508 15C Evaluating inquiry methods and statistical reports  511 Exercise 15C  518 Contents

v

15D Statistical investigations

Exercise 15D

521

18F Logarithms

525

Summary 527 Chapter review 529 eBookPLUS activities 533

prOjeCts plus

ICt aCtIvIty

pro-0100 Climate change

534

Chapter 16

Financial maths

537

16C 16D 16E 16F

Chapter 19

polynomials

539

Are you ready?

Exercise 16A 540 Buying on terms 542 Exercise 16B 543 Successive discounts 546 Exercise 16C 547 Compound interest 549 Exercise 16D 551 Depreciation 553 Exercise 16E 554 Loan repayments 556 Exercise 16F 558

19B

19C 19D 19E

19F 19G

Chapter 17

prOblem sOlvINg

637 Exercise 19A 638 Adding, subtracting and multiplying polynomials 639 Exercise 19B 640 Long division of polynomials 641 Exercise 19C 646 Polynomial values 647 Exercise 19D 648 The remainder and factor theorems 649 Exercise 19E 650 Factorising polynomials 651 Exercise 19F 654 Solving polynomial equations 655 Exercise 19G 657

Number aNd algebra

Chapter 20

Are you ready?

vi

Contents

Are you ready?

590

Exercise 18A 594 595 Exercise 18B 597 18C Operations with surds 599 Exercise 18C 607 18D Fractional indices 609 Exercise 18D 612 18E Negative indices 614 Exercise 18E 616 18B Surds

Functions and relations

589

18A Number classification review

Number aNd algebra lINear aNd NON-lINear relatIONshIps

real Numbers

real numbers

636

Summary 659 Chapter review 661 eBookPLUS activities 662

565

10a Chapter 18

635

19A Polynomials

Summary 560 Chapter review 562 eBookPLUS activities 564

problem solving II

Number aNd algebra patterNs aNd algebra

538

16A Purchasing goods 16B

Summary 630 Chapter review 632 eBookPLUS activities 634

Number aNd algebra mONey aNd FINaNCIal mathematICs

Are you ready?

617 Exercise 18F 618 18G Logarithm laws 619 Exercise 18G 622 18H Solving equations 624 Exercise 18H 628

591

664

20A Functions and relations 20B 20C 20D 20E

663

665 Exercise 20A 669 Exponential functions 671 Exercise 20B 675 Cubic functions 679 Exercise 20C 682 Quartic functions 683 Exercise 20D 685 Transformations 686 Exercise 20E 692

22C  Area of triangles  745

Summary  694 Chapter review  696 eBookPLUS activities  698

Chapter 21

Measurement and geometry GEOMETRIC REASONING

Circle geometry  699 Are you ready?  700

21A Angles in a circle  701

Exercise 21A  706 21B Intersecting chords, secants and

tangents  708 Exercise 21B  713 21C Cyclic quadrilaterals  715 Exercise 21C  717 21D Tangents, secants and chords  718 Exercise 21D  720 Summary  724 Chapter review  726 eBookPLUS activities  730

Exercise 22C  747 22D The unit circle  749 Exercise 22D  752 22E Trigonometric functions  755 Exercise 22E  757 22F Solving trigonometric equations  759 Exercise 22F  761 Summary  762 Chapter review  764 eBookPLUS activities  766

Chapter 23

Statistics and probability DATA REPRESENTATION AND INTERPRETATION

Interpreting data  767 Are you ready?  768

23A Bivariate data  769

Exercise 23A  773 23B Lines of best fit  776

Exercise 23B  784 23C Time series  786

Chapter 22

Measurement and geometry PYTHAGORAS & TRIGONOMETRY

Trigonometry II  731 Are you ready?  732

22A The sine rule  733

Exercise 22A  739 22B The cosine rule  741

Exercise 22B  744

Exercise 23C  790 Summary  794 Chapter review  795 eBookPLUS activities  798

Answers  799 Glossary  889 Index  901

Contents

vii

Introduction Australian Mathematics education is entering a historic phase. A new curriculum offers new opportunities to engage future generations of students in the exciting and challenging world of Mathematics. The Australian Mathematics Curriculum provides students with essential mathematical skills and knowledge through the content strands of Number and algebra, Measurement and geometry and Statistics and probability. The Curriculum focuses on students becoming proficient in mathematical understanding, fluency, reasoning and problem solving. Maths Quest 10 + 10A for the Australian Curriculum is specifically written and designed to meet the requirements and aspirations of the Australian Mathematics Curriculum. This resource contains: a student textbook with accompanying eBookPLUS ■■ a teacher edition with accompanying eGuidePLUS ■■ a TI-Nspire CAS Calculator companion ■■ a Casio ClassPad CAS Calculator companion. ■■

Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos, and to make navigation through the text easier. Are you ready? sections at the start of each chapter provide introductory questions to establish students’ current levels of understanding. Each question is supported by a SkillSHEET that explains the concept involved and provides extra practice if needed. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities and eLessons are available online to help with the teaching and learning of particular concepts. Worked examples in a Think/Write format provide clear explanation of key steps and suggest presentation of solutions. Exercises contain many carefully graded skills and application problems, including multiplechoice questions. Cross-references to relevant worked examples appear with the first ‘matching’ question throughout the exercises. Each chapter concludes with a summary and chapter review exercise containing examinationstyle questions (multiple-choice, short-answer and extended-response), which help consolidate students’ learning of new concepts. A glossary is provided to enhance students’ mathematical literacy. There are two problem-solving chapters designed to encourage students to apply their mathematical skills in non-routine situations.

Student website — eBookPLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, tablets, school or home computers, and cut and paste material for revision, assignments or the creation of notes for exams.

viii

Introduction

WorkSHEET icons link to editable Word documents, and may be completed on-screen, or printed and completed by hand. Individual pathway activity icons link to online activity sheets below, at and above the level presented in the text, for each exercise. These activities allow students to work at their own pace and to engage with the concepts being taught at an appropriate differentiated level. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations, which help students to understand difficult concepts. eLesson icons link to videos or animations designed to elucidate concepts in ways that are more than what the teacher can achieve in the classroom. Hungry brain activities provide engaging, whole-class activities to introduce each chapter. Test yourself tests are also available. Answers are provided for students to receive instant feedback. Word searches and crosswords are available for each chapter. Two ProjectsPLUS activities provide students with the opportunity to work collaboratively and creatively, online, on a mathematics project.

Teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. Maths Quest 10 + 10A for the Australian Curriculum provides ample material, such as exercises, problem-solving questions, projects, worksheets and technology files, from which teachers can assess their students.

Introduction

ix

About eBookPLUS

Next generation teaching and learning This book features eBookPLUS: an electronic version of the entire textbook and supporting multimedia resources. It is available for you online at the JacarandaPLUS website (www.jacplus.com.au).

Using the JacarandaPLUS website To access your eBookPLUS resources, simply log on to www.jacplus.com.au using your existing JacarandaPLUS login and enter the registration code. If you are new to JacarandaPLUS, follow the three easy steps below. Step 1. Create a user account The first time you use the JacarandaPLUS system, you will need to create a user account. Go to the JacarandaPLUS home page (www.jacplus.com.au), click on the button to create a new account and follow the instructions on screen. You can then use your nominated email address and password to log in to the JacarandaPLUS system. Step 2. Enter your registration code Once you have logged in, enter your unique registration code for this book, which is printed on the inside front cover of your textbook. The title of your textbook will appear in your bookshelf. Click on the link to open your eBookPLUS.

Using eBookPLUS references eBookPLUS logos are used throughout the printed books to inform you that a multimedia resource is available for the content you are studying. Searchlight IDs (e.g. INT-0001) give you instant access to multimedia resources. Once you are logged in, simply enter the searchlight ID for that resource and it will open immediately.

Minimum requirements • A modern internet browser such as Internet Explorer 7+, Mozilla Firefox 3+, Google Chrome 8+, Safari 3+ or Opera 9+ • Adobe Flash Player 10+ • Javascript must be enabled (most browsers are enabled by default).

Step 3. View or download eBookPLUS resources Your eBookPLUS and supporting resources are provided in a chapter-by-chapter format. Simply select the desired Troubleshooting • Go to the JacarandaPLUS help page at chapter from the drop-down list. Your eBookPLUS www.jacplus.com.au/jsp/help.jsp. contains the entire textbook’s content in easy-to-use HTML. The student resources panel contains supporting • Contact John Wiley & Sons Australia, Ltd. Email: [email protected] multimedia resources for each chapter. Phone: 1800 JAC PLUS (1800 522 7587) Once you have created your account, you can use the same email address and password in the future to register any JacarandaPLUS titles you own.

x

About eBookPLUS

Acknowledgements The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book.

Images • Coral Connor: 353 (top left)/© Coral Connor • Copyright Agency Limited: 516/The Sunday Mail, 5 September, 2010, p. 15 • Corbis Australia: 55/Corbis/Ladislav Janicek • Corbis Royalty Free: 219, 268/© Corbis Corporation • Creatas Images: 494/© Creatas Images • Cultura: 131/ © Cultura • Digital Stock: 247 (bottom), 281 (left), 556/© Digital Stock/Corbis Corporation; 545 (centre right)/© Digital Stock • Digital Vision: 140/© Digital Vision; 207/© Digital Vision/Stephen Frink • Image Disk Photography: 164 (bottom)/© Image Disk Photography • iStockphoto: 32/© iStockphoto.com/Steve O’Connor; 171; 231; 484/© Vladimir Popovic/ iStockphoto.com; 671/© iStockphoto.com/Sebastian Kaulitzki; 767/© iStockphoto.com/technotr • John Wiley & Sons Australia: 311, 403, 458, 489, 530 (right)/© John Wiley & Sons Australia/ Photo by Kari-Ann Tapp; 366, 507/© John Wiley & Sons Australia/Photo by Renee Bryon • NASA: 535/© NASA • Photodisc: 105 (bottom), 138 (bottom), 138 (top), 139, 144 (bottom), 144 (centre), 144 (top), 164 (top), 172, 180, 198, 247 (top), 254 (bottom), 254 (top), 286 (bottom), 286 (top), 318, 322, 394, 411, 415, 416 (bottom), 416 (top), 492, 537, 545 (bottom left), 547, 555, 563, 637, 748, 749, 774, 785, 791 • Photolibrary: 490/Photolibrary/Science Photo Library/ROBERTO DE GUGLIEMO; 493/Photolibrary/Index Stock Imagery/Vohra Aneal • Photolibrary Royalty Free: 546/Photolibrary • Shutterstock (all images used under licence from Shutterstock): 1/© mattasbestos; 7/© ARTSILENSEcom, 2009; 16/© arbit , 2009; 89/ © Christina Richards/Shutterstock.com; 105 (top left)/© Angelo Gilardelli, 2009; 105 (top right)/© Hydromet, 2009; 160/© Aleix Ventayol Farrés; 174/© Jim Parkin/Shutterstock. com; 183/© Steven Frame/Shutterstock.com; 226/© Lord_Ghost, 2009; 235/© krechet, 2009; 236/© Benis Arapovic, 2009; 245; 279/© Alistair Michael Thomas, 2009; 281 (right)/© Samot/ Shutterstock.com; 305/© robcocquyt, 2009; 325/Tomasz Trojanowski; 352 (bottom)/© sevenke/ Shutterstock.com; 352 (top)/© sevenke/Shutterstock.com; 353 (bottom left)/© Dan Breckwoldt/ Shutterstock.com; 353 (right)/© Creative Illus/Shutterstock.com; 355/© Benis Arapovic; 374/ © Ronald Sumners; 375/© Steve Heap; 379/© Rafal Olkis, 2009; 413/© Neale Cousland/2009; 429/© Darren Baker; 435/© Omer N Raja; 436/© iofoto; 438/© Brasiliao; 439/© Lisa F. Young, 2010; 449/© Givaga, 2011; 452/© Mark Herreid; 456/© Brad Remy; 463/© Laurence Gough; 466/© Blend Images; 467/© Inger Anne Hulbækdal; 468/© Konstantin L; 471/ © Andresr, 2010; 477/© photobank.kiev.ua, 2010; 482/© AVAVA, 2010; 495/© Nagy-Bagoly Arpad, 2010; 497/© l i g h t p o e t/Shutterstock.com; 530 (left)/© Sonja Foos/Shutterstock.com; 531 (left)/© Nejron Photo/Shutterstock.com; 531 (right)/© Zina Seletskaya/Shutterstock.com; 534 (bottom)/© Mariia Sats/Shutterstock.com; 534 (top)/© Lysithee/Shutterstock.com; 539/ © Milevski Petar, 2009; 545 (bottom right)/© Sasha Radosavljevich, 2009 page 027/ © KOUNADEAS IOANNHS; 545 (top left)/© Marc Dietrich, 2009; 545 (top right)/© Losevsky Pavel, 2009; 565/© Kinetic Imagery; 566/© Harald Høiland Tjøstheim; 573/Kosarev Alexander; 575/© Julian W; 580/© Chris Hellyar; 583/© Simon Krzic; 588/© Serg64; 589/Markus Gebauer; 591 (left)/© Johan Larson; 591 (right)/© R. Gino Santa Maria; 614/Markus Gebauer; 629 (bottom)/© gmwnz; 629 (top)/EmiliaU; 635/© StudioSmart; 639/© BrunoRosa; 663/ © Neale Cousland; 699/© Egolii; 731/© Olivier Le Queinec • Viewfinder Australia Photo Library: 22/© Viewfinder Australia Photo Library • Jennifer Wright: 740/Creative Cohesions

Text • AFP, page 524/ ‘Single women earn more’, The Weekend Australian, 4–5 September 2010, p. 20 • Copyright Agency Limited: 516/The Sunday Mail, September 2010; 517/‘Sponges are Acknowledgements

xi

toxic’, The Sunday Mail, 5 September 2010, p. 36; 523/‘Word limit’, by Professor Emeritus Roland Sussex taken from The Courier Mail, 14–15 August 2010; 524/‘Egg Shortage’, by Paddy Hintz, The Courier Mail, 28–29 August 2010 • News Limited: 532/‘Taste Test’, The Sunday Mail, 4 April 2010, p. 26 Every effort has been made to trace the ownership of copyright material. Information that will enable the publisher to rectify any error or omission in subsequent editions will be welcome. In such cases, please contact the Permissions Section of John Wiley & Sons Australia, Ltd.

xii

Acknowledgements

number AnD AlgebrA • pAtterns AnD AlgebrA

1

       

1A  1B  1C  1D 

Review of index laws Negative indices Fractional indices Combining index laws

WhAt Do you knoW ?

Indices

1 List what you know about indices. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of indices. eBook plus

Digital doc

Hungry brain activity Chapter 1 doc-5167

opening Question

If you open a new social networking account with a single friend and double the number of friends each day, how long would it take for you to have 1000 friends?

number AnD AlgebrA • pAtterns AnD AlgebrA

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 1.1 doc-5168

eBook plus

Digital doc

SkillSHEET 1.2 doc-5169

eBook plus

Digital doc

SkillSHEET 1.3 doc-5170

eBook plus

Digital doc

SkillSHEET 1.4 doc-5171

eBook plus

Digital doc

SkillSHEET 1.5 doc-5172

eBook plus

Digital doc

SkillSHEET 1.6 doc-5173

eBook plus

Digital doc

SkillSHEET 1.7 doc-5174

Index form   1  State■the■base■and■power■for■each■of■the■following. a  34 Base■is■3,■power■is■4 b  25 Base■is■2,■power■is■5

c  157

Using a calculator to evaluate numbers given in index form   2  Calculate■each■of■the■following. a  24 16 b  53 125

c  46

Linking between squares and square roots   3  Complete■the■following■statements. 3 a  If■32■=■9,■then■ 9 ■=■.■■.■■.

Digital doc

SkillSHEET 1.8 doc-5175

2

4096

11 b  If■112■=■121,■then■ 121 ■=■.■■.■■.

17 c  If■172■=■289,■then■ 289 ■=■.■■.■■.

Calculating square roots   4  Find■each■of■the■following.

64

a 

8

100 10

b 

Linking between cubes and cube roots   5  Complete■the■following■statements. 3 2 a  If■23■=■8,■then■ 8 ■=■.■■.■■.

25 5

c 

3 5 b  If■53■=■125,■then■ 125 ■=■.■■.■■.

3 9 c  If■93■=■729,■then■ 729 ■=■.■■.■■.

Calculating cube roots   6  Find■each■of■the■following. a 

3

64 4

b 

3

216

6

c 

3

1 1

Estimating square roots and cube roots   7  Estimate,■to■the■nearest■whole■number,■the■value■of■each■of■the■following.■(Do■not■use■a■

calculator.) 23

a  d 

eBook plus

Base■is■15,■power■is■7

3

60

102 10

b 

5

e 

4

3

11

2

40

c  f 

3

120

6 5

Using a calculator to evaluate square roots and cube roots   8  Use■a■calculator■to■fi■nd■the■value,■correct■to■4■decimal■places,■of■each■square■root■or■cube■root■ in■question■7. a■ 4.7958     d  3.9149

b  10.0995 e  2.2240

c  6.3246 f  4.9324

maths Quest 10 for the Australian Curriculum

number AND algebra • patterns and algebra

Review of index laws

1A

■■

Index laws are the basis for exponential functions, which we will cover in a later chapter. A number in index form has two parts; the base and the index, and is written as: base

■■ ■■

ax

index (power or exponent)

Another name for an index is an exponent or a power. The first two index laws relate to multiplication and division of index expressions. First Index Law: When terms with the same base are multiplied, the indices are added. am ì an = am + n Second Index Law: When terms with the same base are divided, the indices are subtracted. am ó an = am - n

■■

Note: Constants or normal numbers should be treated normally when solving equations. Only apply the index laws to the indices themselves. This will become clearer in the following worked examples.

Worked Example 1

Simplify each of the following. a  m4n3p ì m2n5p3 b  2 a2b3 ì 3 a b4 2 x 5 y4 c 

10 x 2 y3

Think a

b

c

Write

1

Write the expression.

2

Multiply the terms with the same base by adding the indices. Note: p = p1.

1

Write the expression.

2

Simplify by multiplying the coefficients, then multiply the terms with the same base by adding the indices.

1

Write the expression.

2

Simplify by dividing both of the coefficients by the same factor, then divide terms with the same base by subtracting the indices.

■■

a m4n3p ì m2n5p3

= m4 + 2 n3 + 5 p1 + 3 = m6n8p4 b 2a2b3 ì 3ab4

= 2 ì 3 ì a2 + 1 ì b3 + 4 = 6a3b7

c

2x5 y4 10 x 2 y 3 1x 5 − 2 y 4 − 3 5 3 x y = 5

=

The Third Index Law is used in calculations when a zero index is involved. Third Index Law: Any term (excluding 0) with an index of 0, is equal to 1. a0 = 1 Chapter 1 Indices

3

number AND algebra • patterns and algebra

Worked Example 2

Simplify each of the following. a  (2  b3)0

b  -4(a2b5)0

Think a

b

Write a (2b3)0

1

Write the expression.

2

Apply the Third Index Law, which states that any term (excluding 0) with an index of 0, is equal to 1.

1

Write the expression.

2

The term inside the brackets has an index ■ of 0 so the bracket is equal to 1.

= -4 ì 1

3

Simplify.

= -4

■■

=1

b -4(a2b5)0

The Fourth, Fifth and Sixth Index Laws involve removing brackets from an index expression. Fourth Index Law: To remove brackets, multiply the indices inside the brackets by the index outside the brackets. Where no index is shown, assume that it is 1. (am)n = amn  ifth Index Law: To remove brackets containing a product, raise every part of the product to F the index outside the brackets. (ab)m = ambm  ixth Index Law: To remove brackets containing a fraction, multiply the indices of both the S numerator and denominator by the index outside the brackets. m

am  a  b  = m b ■■

Note: Do not forget to raise constants to the correct power as well.

Worked Example 3

Simplify each of the following. a  (2n4)3 b  (3 a2b7)3

 2 x3  c     y4 

Think a

b

4

4

d  (-4)3

Write a (2n4)3

1

Write the expression.

2

Apply the Fourth Index Law by multiplying the indices inside the brackets by the index outside the brackets. Simplify any constants raised to a power. Note: 2 = 21.

1

Write the expression.

2

Apply the Fifth Index Law by multiplying the indices inside the brackets by the index outside the brackets. Note: 3 = 31.

= 31 ì 3 ì a2 ì 3 ì b7 ì 3 = 33a6b21

3

Simplify.

= 27a6b21

Maths Quest 10 for the Australian Curriculum

= 21 ì 3n4 ì 3 = 23n12 = 8n12 b (3a2b7)3

number AnD AlgebrA • pAtterns AnD AlgebrA

c

Write■the■expression.■

2

Apply■the■Sixth■Index■Law■by■multiplying■ the■indices■of■both■the■numerator■and■ denominator■by■the■index■outside■the■ brackets.

3

d

 2x3    y4 

4

c 

1

=

Simplify.

=

21 × 4 × x 3 × 4 y4 × 4 16 x12 y16

d (-4)3

1

Write■the■expression.

2

Expand■the■brackets.

=■-4■ì■-4■ì■-4

3

Simplify,■taking■careful■note■of■the■sign.

= -64

Hint:■A■negative■number■raised■to■an■odd■power■will■ always■remain■negative;■a■negative■number■raised■to■ an■even■power■will■always■become■positive.■Why?

remember

To■simplify■expressions■with■constants■and/or■pronumerals■in■index■form,■the■following■ index■laws■are■used. 1.■ am■ì■an■=■am■+■n 2.■ am■ó■an■=■am■-■n 3.■ a0■=■1■(when■a■≠■0) 4.■ (am)n■=■amn 5.■ (ab)m■=■ambm m

am  a 6.■   =  b bm exerCise

1A inDiViDuAl pAthWAys eBook plus

Activity 1-A-1

Reviewing index operations doc-4948 Activity 1-A-2

Practising the index laws doc-4949

review of index laws FluenCy   1  We 1a, b ■Simplify■each■of■the■following. a  a3■ì■a4 a7 b  a2■ì a3■ì■a a6 2 3 5 4 7 d  ab ■ì■a b e  m2n6■ì■m3n7 m5n13 ab 5 3 4 6 4 5 g  mnp■ì■m n p m n p h  2a■ì■3ab 6a2b j 

  2  We 1c ■Simplify■each■of■the■following. a  a4■ó■a3 a b  a7■ó■a2 d 

Activity 1-A-3

Applying the index laws doc-4950

3m3■ì■2mn2■ì■6m4n5 36m8n7 k  4x2■ì■ 1 xy3■ì■6x3y3 12x6y6 2

g  j 

4a

7

3a

3

m 7 n3 4 2

m n

4 4 ■a 3

m3n

7ab5c4■ó■ab2c4 7b3

e  h  k 

21b 7b

6

2

2x 4 y3 4

4x y

3b4

f 

1 2 ■y 2

16m3 n3 p2

2

l 

2x3y2■ì■4x■ì■ 12 x4y4

c  b6■ó■b3

a5

20 m 5 n3 p4

c  b■ì■b5■ì■b2 b8 f  a2b5c■ì■a3b2c2 a5b7c3 1 i  4a2b3■ì■5a2b■ì■ b5 10a4b9

i  5 2 2 ■m p 4

l 

8

48m

12m3

4x8y6

b3 4m5

6x7y■ó■8x4 14 x 3 y 4 z 2 28 x 2 y 2 z 2

3 3 ■x y 4 1 ■xy2 2

Chapter 1 Indices

5

number AND algebra • patterns and algebra 3   WE2  Simplify each of the following. a a0 1 d 3x0 3

b (2b)0 1 e 4b0 4

 a 4

0

g 4a0 -    

c (3m2)0 1 f -3 ì (2n)0

h 5y0 - 12

3

-7

4   WE3  Simplify each of the following. a (a2)3 a6

 2 n4    3 

4 8  n 9

g

(2m3n5)4

 5m3  j    n2 

4

16m12n20

e (a2b)3

a6b3

f (3a3b2)2

 3m 2 n  h    4 

3

 a2  i    b3 

n8

m (-3)5

3

n (-7)2

-243

27 6 3  m n 64

343 x 3 8 y15 49

l

1 8 m 81

2

 3a   3  5b

o (-2)5

9a6b4 a4 b6

4

81a 4 625b12 -32

5   MC  a   2m10n5 is the simplified form of: A m5n3 ì 2m4n2 ✔ D

2n(m5)2

ì

B

6m10 n4 3n

 2m 5  E    n3 

n4

b The value of 4 - (5a)0 is: A -1 ✔ D 3

B 9 E 5

6   MC  a  4 a3b ì b4 ì 5a2b3 simplifies to: A 9a5b8 B 20a5b7 5 7 D 9a b E 21a5b8 b

c

15 x 9 × 3 x 6 9 x10 × x 4 A 5x9 D 9x9 3 p7 × 8q 9 12 p3 × 4 q 4

d

B 9x ✔ E 5x

p4 q 4 2 4 q E 24

5a6 b 2 A

✔ D

6

✔ B

÷

C 1

20a5b8

C 5x29

simplifies to:

p4 q 4 24

7a 5 b 3

2

simplifies to:

A 2q4 D

C (2m5n2)2

✔ C

7b 3 a 2 5b 5a 4

C

q4 2

simplifies to:

49a3 b 25 ab3

Maths Quest 10 for the Australian Curriculum

25a3 b 49 25ab3 E 49 B

C a3b

4

4

 m2  c    3 

 7x  k  5   2y 

625m12

5x0 - (5xy2)0

b (2a5)4 16a20

2

d 

i

-3

number AnD AlgebrA • pAtterns AnD AlgebrA unDerstAnDing   7  Evaluate■each■of■the■following. a  23■ì■22■ì■2 64 b  2■ì■32■ì■22 d  g 

5

6

4

4

3 ×4

3 ×4 4 4 × 56 4 3 × 55

c  (52)2

72

48

e  (23■ì■5)2

20

h  (33■ì■24)0 1

 3

1600

f     5 i 

  8  Simplify■each■of■the■following. a  (xy)3z x 3yz b  ab■ì■(pq)0 ab

 a2    b3 

d  

x

a2 x b

e 

3x

n3 m 2 p

n m

q

625

3 27 125

4(52■ì■35)0

4

c  ma■ì■nb■ì■(mn)0

manb

f  (am■+■n)■p amp + np

n3■- pm2■- q

reAsoning 2

  9  Find■algebraically■the■exact■value■of■x■if■4x■+■1■=■2x .■Justify■your■answer. 1■ê 3  10  Binary■numbers■(base■2■numbers)■are■used■in■computer■operations.■As■the■name■implies,■

binary■uses■only■two■types■of■numbers,■0■and■1,■to■express■all■numbers.■A■binary■ number■such■as■101■(read■one,■zero,■one)■means■ (1■ì■22)■+■(0■ì■21)■+■(1■ì■20)■=■4■+■0■+■1■=■5■ (in■base■10,■the■base■we■are■most■familiar■with). ■ The■number■1010■(read■one,■zero,■one,■zero)■ means■(1■ì■23)■+■(0■ì■22)■+■(1■ì■21)■+■(0■ì■20)■=■8■ +■0■+■2■+■0■=■10.■ ■ If■reading■the■binary■number■from■right■to■ left,■the■index■of■2■increases■by■one■each■time,■ beginning■with■a■power■of■zero. ■ Using■this■information,■write■out■the■numbers■ 1■to■10■in■binary■(base■2)■form. reFleCtion 

 

Why are these laws called index laws?

1 ô 1     6 ô 110

1b eBook plus

Interactivity Negative indices

int-2777

negative indices ■■

2   ô 10  7 ô 111

3   ô 11  8 ô 1000

4   ô 100  9 ô 1001

5   ô 101  10 ô 1010

So■far■we■have■dealt■only■with■indices■that■are■positive■whole■numbers■or■zero.■To■extend■this,■ we■need■to■consider■the■meaning■of■an■index■that■is■a■negative■whole■number.■Consider■the a3 a3 expression■ 5 .■Using■the■Second■Index■Law,■ 5 ■=■a3■-■5 a a ■ =■a-2 a3 a×a×a Writing■terms■in■the■expanded■notation■we■have:■ 5 ■=■ a × a×a×a×a a 1 ■ =■ a×a 1 ■ =■ 2 a 1 By■equating■the■results■of■simplifi■cation,■using■the■two■methods,■we■have:■a-2■=■ 2 . a Chapter 1 Indices

7

number AND algebra • patterns and algebra

■■

In general terms,

1 a

n

=

a0 n

(1 = a0)

a = a0 - n (using the Second Index Law) = a-n 1 Seventh Index Law: a-n = n a The convention is that an expression should be written using positive indices and so we use the Seventh Index Law to do this.



■■

Worked Example 4

Express each of the following with positive indices. a  x-3 b  2m-4n2

c 

Think a

b

c

4 a−3

Write

1

Write the expression.

2

Apply the Seventh Index Law.

1

Write the expression.

2

Apply the Seventh Index Law to write the expression with positive indices.

1

Copy the expression and rewrite the fraction, using a division sign.

2

Apply the Seventh Index Law to write the expression with positive indices.

3

To divide the fraction, use the ‘multiply and flip’ method.

■■ ■■

a x-3

=

1 x3

b 2m-4n2

=

2 n2 m4

4

c

a −3

= 4 ó a-3 =4ó



=4ì



= 4a3

1 a3 a3 1

Worked Example 5

Simplify each of the following, expressing the answers with positive indices. −2  2 m3  2 x 4 y2 2 -3 -5 a  a b ì a b b  c   −2   n  3 xy5 Think a

8

1

= an. a −n All laws discussed in the previous section are applicable to the terms with negative indices. Part c from Worked example 4 demonstrates the converse of the Seventh Index Law

Write

1

Write the expression.

2

Apply the First Index Law. Multiply terms with the same base by adding the indices.

3

Express the answer with positive indices.

Maths Quest 10 for the Australian Curriculum

a a2 b-3 ì a-5b

= a2 + -5b-3 + 1 = a-3b-2 1 = 3 2 a b

number AND algebra • patterns and algebra

b

Write the expression.

2

Apply the Second Index Law. Divide terms with the same base by subtracting the indices.

=

2x 4 − 1y 2 − 5 3

=

2 x 3 y −3 3

Express the answer with positive indices.

=

3

c

4 2 b 2x y

1

3 xy 5

3y3

 2m3  c  −2   n 

1

Write the expression.

2

Apply the Sixth Index Law. Multiply the indices of both the numerator and denominator by the index outside the brackets. Remember that 2 = 21.

=

3

Express all terms with positive indices.

=

4

Simplify.

=

■■

2x3 −2

2 −2 m−6 n4

1 2

2 m6 n 4 1 4 m6 n 4

Numbers in index form can be easily evaluated if they are expressed with positive indices first. Consider the following example.

Worked Example 6

Evaluate 6 ì 3-3 without using a calculator. Think

Write

1

Write the multiplication.

6 ì 3-3

2

Express 3-3 with a positive index.

=6ì

3

Multiply the numerator of the fraction by the whole number.

=

4

Evaluate the denominator.

=

6 27

5

Cancel by dividing both the numerator and denominator by the same number.

=

2 9

1 33

6 33

Chapter 1 Indices

9

number AnD AlgebrA • pAtterns AnD AlgebrA a■     c      e 

1

  b 

5

x 2

  d 

a9 3x 2

  f 

y3 6a 3

1 y4 4

remember

5a3 1

1.■ A■term■with■a■negative■index■can■be■expressed■with■a■positive■index■using■the■Seventh■ Index■Law. 1 (a)■ a-n■=■ n ■ a 1 (b)■ −n ■=■an a 2.■ All■index■laws■apply■to■terms■with■negative■indices. 3.■ Always■express■answers■with■positive■indices■unless■otherwise■instructed. 4.■ Numbers■and■pronumerals■without■an■index■are■understood■to■have■an■index■of■1.

4 m3 n 4

  h  a6 bc 5 2a 4     i    j  2ab2 3     g 

    k 

7b 3

  l 

2a 4

2 m3 a 2 3b 4 n5

exerCise

1b

negative indices FluenCy

inDiViDuAl pAthWAys eBook plus

  1  We4 ■Express■each■of■the■following■with■positive■indices. a  x-5 b  y-4 d 

Activity 1-B-1

g 

Negative indices doc-4951 Activity 1-B-2

j 

Harder negative indices doc-4952 Tricky negative indices doc-4953

    c 

a 2b3 3 n8

    e  2 y 3x     g      i      k      m      o 

  b    d    f 

3 m 2 n2 1 3 3

3m n 4 q8 p14 27q 9 6

8p 1

  h    j    l 

6

j 

x6 y 4 a2b5 5y 6x3

6a

k 

3b −2

1

i 

−6

a 7a −4

l 

2b −3

2 3a −4 2m3 n −5 3a−2b 4

(2a3m4)-5

4x2 y9

i 

x 7 y −3

k  4(p7q-4)-2

−3

 a− 4  n     2b −3 

l 

2

20

a8b12

g  j 

6 −3

2

8 9

2m 2 n− 4 6m 5 n−1 3(a-2b-3)4

 6a 2  o     3b −2 

h 

16 × 2

4

−4

8 ×2

k 

4

c  3-4

1 16

e  4-3■ì■22

48

0

2

1 36

b  6-2

1 8

d  3-2■ì■23

1 32a m 3

h 

2 n3 m 6

a  2-3

x5 15

f  5x-2y3■ó■6xy2

−3

  3  We6 ■Evaluate■each■of■the■following■without■using■a■calculator.

4 × 3−3

32 27

−3

2

5 × 250

i 

3

2

25 × 5

−4

6 6

d   

 1  −8   2

e 

256

 3   4

−7

l 

125

B  -5x

maths Quest 10 for the Australian Curriculum

✔ D 

1 x

5

5 36

× 5−2 × 34 34 × 4 2

27 ■=■1 2 25 25

3 4

123 × 150

0.000■■059■■499

f  (0.045)-5

7.491■■540■■923

C  5x

1 3

c  7-5

5  mC ■a■ ■x-5■is■the■same■as: A  -x5

1 81

f  5■ì■6-2

  4  Evaluate■each■of■the■following,■using■a■calculator. a  3-6 0.001■■371■■742 b  12-4 0.000■■048■■225

8a b

10

f  2-2m-3n-4

e  2xy6■ó■3x2y5

6m 4 n

 2 p2  m     3q3 

4 y12

6   n  b 4 a8

h 

d  4a3b2■ó■a5b7 g 

a■

e 

c  2a-9

3x2y-3

  2  We5 ■Simplify■each■of■the■following,■expressing■the■answers■with■positive■indices. a  a3b-2■ì■a-5b-1 b  2x-2y■ì■3x-4y-2 c  3m2n-5■ì■m-2n-3

Activity 1-B-3

1

4 -3 a 5 6a3b-1c-5

5■■419■■228.099

E 

1 x −5

number AND algebra • patterns and algebra

b

1

is the same as: a−4 a 4a 1

d c

1 8

b -4a

✔ c

a4

e -a4

a4

is the same as:

A 23

✔ B

D 3-2

E

2-3

C 32

1 2−3

6   MC  a  Which of the following, when simplified, gives A D

3m − 4n−2 4 22 n−2

✔ B

✔ D

4 n2

?

3 × 2−2 × m 4 × n−2

C

3n−2 2−2 m − 4

E 3m 4 × 22 n−2

3−1 m − 4

b When simplified, 3a-2b-7 ó A

3m 4

4

3 -4 6 a b is equal to: 4 9b B

a6 b13 4a2

E

b13

C

4 a6

9a 2 4b

4a2 b

c When (2x6y-4)-3 is simplified, it is equal to: A

D

2 x18

B

y12 8 y12

E

x18

x18

✔ C

8 y12

y12 8 x18

x18 6 y12

3

 2a x  8b 9 , then x and y (in that order) are: is equal to  a6  by 

d If 

A -3 and -6 D -3 and -2

B -6 and -3 ✔ E

C -3 and 2

-2 and -3

Understanding 7 Simplify, expressing your answer with positive indices. a

m−3 n−2 m−5 n6 3 −3 2

c

5(a b ) − 4 −1

(ab )

m2

b

n8

÷

−2

−1

(5a b)

25

−4

3

a 7b6

(a b)

(m3 n−2 )−7

n2 m

−5 3 4

(m n )

8 Simplify, expanding any expressions in brackets. a (r3 + s3) (r3 - s3) r6 - s6 b (m5 + n5)2 c

( x a + 1 )b × x a + b x a ( b + 1) × x 2 b

1

 px + 1  d    px − 1 

m10 + 2m5n5 + n10

−4

×

p8( x + 1) ( p2 x )4

×

p2 ( p12 x )0

p2

Chapter 1 Indices

11

number AND algebra • patterns and algebra

 2r × 8r  ar + b. 22r - 4  in the form 2  22r × 16 

9 Write 

10 Write 2-m ì 3-m ì 62m ì 32m ì 22m as a power of 6. 11 Solve for x if 4x - 4x - 1 = 48.

63m

x=3

Reasoning 12 Look at the following pattern:

22 = 4 21 = 2 20 = 1 a What is changing on the left hand side of the equation each time? The power is reduced by 1. b What is the pattern shown on the right hand side of the equation? Each answer is divided by 2 to get

reflection 

Do any of the index laws from exercise 1A not apply to negative indices?

the next answer.

c How can this pattern be used to help display the

rule a-n =

1c

1

an

? If the pattern continues we will get 2-1 =

Fractional indices ■■

 

2-2 =

1 , 2 1

1 1 = , etc. which illustrates a-n = n . 4 22 a

Terms with fractional indices can be written as surds, using the following laws: 1

1. a n = n a m

n 2. a n = a m

■■

=

( a) n

m

.

To understand how these laws are formed, consider the following numerical examples. Using 1

1

4 2 × 4 2 = 41 the First Index Law and we also know that 4 × 4 = 16 =4 1

If these two identities are true, then 4 2 = 4 . Similarly: 1

1

1

Using the First Index Law 8 3 × 8 3 × 8 3 = 81 3

and we also know that

8 × 3 8 × 3 8 = 3 512 =8 1

If these two identities are true, then 8 3 = 3 8 . 1

This can be generalised to a n = n a . m ■■

Now consider: a n = a



m×

1 n

m

=

n

×m

a m = ( n a )m m n

n Eighth Index Law:  a = a m = ( n a )m .

12

1

a n = an m 1  1 m n = (a ) =  a n    or

Maths Quest 10 for the Australian Curriculum

number AND algebra • patterns and algebra

■■

As can be seen from the above identities, the denominator of a fraction (n) indicates the power or type of root. That is, n = 3 implies cube root, n = 4 implies fourth root, and so on. Note that when n = 2 (square root), it is the convention not to write 2 at the square root sign.

Worked Example 7

Evaluate each of the following without using a calculator. 3

1

b  16 2

a  9 2 Think

Write 1

a

1

Rewrite the number using the Eighth Index Law.

2

Evaluate.

9

=3 m

b

a 92 = 3 16 2

3 = ( 16 )

1

Rewrite the number using a n = ( n a )m.

2

Evaluate the square root.



= 43

3

Evaluate the result.



= 64

b

Worked Example 8 1

Simplify each of the following. 1

2

a  m 5 × m 5

1 2 3 b  ( a b ) 6

 2 2  x3  c   3  y 4 

Think

Write 1

a

1

Write the expression.

2

a m5 × m5

=

3 m5

2

Apply the First Index Law to multiply terms with the same base by adding the indices.

1

Write the expression.

2

Use the Fourth Index Law to multiply each index inside the brackets by the index outside the brackets.

= a6 b6

3

Simplify the fractions.

= a3b2

1

b

b (a 2 b 3 ) 6 2 3

1 1

1

c

1

Write the expression.

 22 3 c x   3  y 4  1

2

Use the Sixth Index Law to multiply the index in both the numerator and denominator by the index outside the brackets.

=

x3 3

y8 Chapter 1 Indices

13

number AnD AlgebrA • pAtterns AnD AlgebrA

remember

1.■ Fractional■indices■are■those■which■are■expressed■as■fractions. 2.■ Terms■with■fractional■indices■can■be■written■as■surds,■using■the■following■identities: 1

an = n a m

a n = n a m = ( n a )m . 3.■ All■index■laws■are■applicable■to■fractional■indices. exerCise

1C inDiViDuAl pAthWAys

Fractional indices FluenCy   1  We7 ■Evaluate■each■of■the■following■without■using■a■calculator.

eBook plus

Activity 1-C-1

Fractional indices doc-4954 Activity 1-C-2

Harder fractional indices doc-4955

a 

1 16 2

g 

1 16 4

eBook plus

Digital doc

SkillSHEET 1.9 doc-5176

2

b  h 

3 25 2

5 125

c 

1 812

i 

3 36 2

9 216

d 

1 83

j 

7 100 2

10■■000■■000 1

1

2

4

f  814

3

8

2 27 3

9

e  64 3 k 

3 4 16

l 

  2  Using■a■calculator,■evaluate■each■of■the■following.■Give■the■answer■correct■to■2■decimal■places. 1

a  3 3

Activity 1-C-3

Tricky fractional indices doc-4956

4

1 25 2

d  g 

1.44

1 89

3  2 2   3

1.26

b 

1 52

e 

3 8 12

h 

0.54

1

c  7 5 1.48

2.24

4

f 

2.54

3  3 4   4

i 

0.81

(0.6) 5 2  4 3  5 

0.66 0.86

  3  We8a ■Simplify■each■of■the■following. 3

1

d  g 

3 x4 −

×

1

4

a  4 5 × 4 5 2 x5

23 x 20

2

−4 y 9

1

1

1

5

c  a 2 × a 3

22

e 

1 5m 3

×

h 

3 2 8 a 5

× 0.05a 4

20

4y2 × y 9

3

b  2 8 × 2 8

45

1 2m 5

8 10 m15

3

f 

3 1 7 b 2

i 

5x3 × x 2

×

2 4b 7

2b

1

9

0.02a 8

a6 5 7

7

5x 2

  4  Simplify■each■of■the■following. 2

3

1

3

3

a  a 3 b 4 × a 3 b 4 3

1

ab 2

3 2

1 1

4 5

b  x 5 y 9 × x 5 y 3

x5y9

1 1

2

Digital doc

SkillSHEET 1.10 doc-5177

1

1

a  3 2 ÷ 3 3 d 

6 a7

÷

3 a7

1

36 3 a7

2

1

b  5 3 ÷ 5 4 e 

3 x2

÷

1 x4

1

8 17

6a 5 b 15

1 1 1

19 2 19 5 5 1 4 5 3 e  x y 2 z 3 × x 6 y 3 z 2 x 6 y 6 z 6 m n 2m 28 n 5 3   5  Simplify■each■of■the■following.

d  6m 7 ×

eBook plus

3 4

c  2ab 3 × 3a 5 b 5

5

512

2 3 1

f 

3 3

2a 5 b 8 c 4 × 4 b 4 c 4

3 2 c  12 ÷ 12 2

1

12 2

4 5 x4

f 

m5

11

m 45

5

m9 3

g 

2x 4 3

4x 5 14

3 3

1 20 x 2

maths Quest 10 for the Australian Curriculum

h 

7 n2 4

21n 3

2

1 3 n 3

i 

25b 5 1

20 b 4

7

5 20 b 4

2 9

8a 5 b 8 c

number AnD AlgebrA • pAtterns AnD AlgebrA   6  Simplify■each■of■the■following. 4

3

3 2 a  x y ÷ x 3 y 5

d 

4 10 x 5 y

2 1 5x 3 y 4

÷

5 2

5 7

2

2

2 3 2 x 15 y 4

e 

1

Digital doc

SkillSHEET 1.11 doc-5178

3 4

4

1 a 4

1

3

3 11

1 8 56 m n 3

c  m 8 n 7 ÷ 3n 8

a 45 b 15

7

3 3 4 5a b 5

20 a 5 b 4 eBook plus

7

b  a 9 b 3 ÷ a 5 b 5

x3y5

11 7 20 b 20

1 5

p8 q 4

f 

1

1 24 12 p q 7

2 1

7 p3 q6

  7  Simplify■each■of■the■following. 3

1

 35 a   2 4    d 

 24 1 b   5 3  56  

9 2 20

 1 c   7 5   

3

1 (a3 )10

f 

 13 1 1  2b 2  2 3 b 6  

i 

 a c b a  3m b  3 c m c  

n

14

b

 m p m h   x n  xp  

2  3  15 g  4  p 7  4 p 5  

6

75 1

1  48 e   m 9  m6  

3 a 10

6

  8  We8b, c ■Simplify■each■of■the■following. 1

1 1  1 12 a   a 2 b 3  a4b6  

1  1 3 33 3a 3 b 5 c 4

d  

 



 4 5 g   m   7  n 8 

2

1

b  (a

1

1

1

33a9b5c4

e 

4

3 b) 4

1  1 2 22 5 x2y3z5

 

 3 7 c   x 5 y 8   

3 a 3b 4

 

1

1

f 

 33 1  a4  a2 2  b   b 3

i 

1 7  72  4x  2 2 x 2 3  3  2 y 4  y8

1

5x 4 y 3 z 5

2

7 n4

6 7

x5y4

2

1

 33 2  b5  b 5 h   4  8  c 9  c 27

8 m5

2

2

  9    mC ■a■

y 5 ■is■equal■to:

 1 A   y 2   

5

2 B  y × 5

C 

1 (y5 ) 2

D  2 5 y

✔

 1 5 E   y   

2

2

b  k 3 ■is■not■equal■to:

 1 A   k 3    1

c  5

g

2

2

B 

3

k

2

✔

 1 C   k 2   

3

D 

( ) 3

k

−

5 2

2

1 2 E  (k ) 3

■is■equal■to: 2

5 A  g

✔ B 

g

−

2 5

5

C  g 2

D  g

1

E  2 g 5 Chapter 1 Indices

15

number AnD AlgebrA • pAtterns AnD AlgebrA m

1  3 n  10  mC  a■ ■If■ a 4  ■is■equal■to■a 4 ,■then m and n could■not■be:   A  1■and■3 B  2■and■6 C  3■and■8 D  4■and■9 ✔ E  both■C■and■D p

 mm an  ■is■equal■to: b  When■simplifi■ed,■   n  p  b m

A 

D 

eBook plus

Digital doc

SkillSHEET 1.12 doc-5179

ap

✔ B 

n bm

a b

p

C 

n bm

a

E 

m

p n a

b

a

mp n n

bm

m2 np nm p2

 11  Simplify■each■of■the■following. a 

a8

d 

16 x 4

g 

3

a4 4x2

27m 9 n15

3m3■n5

b 

3

b9

e 

3

8y 9

h 

5

32 p5 q10

b3 2y3 2pq2

c 

4

m16

f 

4

16 x8 y12 2x2■y3

i 

3

216a6 b18 6a2■b6

m4

unDerstAnDing  12  The■relationship■between■the■length■of■a■pendulum■(L)■in■a■grandfather■

clock■and■the■time■it■takes■to■complete■one■swing■(T)■in■seconds■is■ given■by■the■following■rule.■Note■that■g■is■the■acceleration■due■to■ gravity■and■will■be■taken■as■9.8. 1

 L 2 T = 2π    g

2.007■s a  Calculate■the■time■it■takes■a■1m■long■pendulum■to■complete■one■swing. b  Calculate■the■time■it■takes■the■pendulum■to■complete■10■swings. 20.07■s c  How■many■swings■will■be■completed■after■10■seconds? 4.98■swings

reAsoning eBook plus

Digital doc

WorkSHEET 1.1 doc-5180

16

( 7 )7

 13  Consider■the■term■ a . Check■with■your■teacher. a  Use■the■Eighth■Index■Law■combined■with■the■

First■Index■Law■to■show■that■ ( 7 a ) ■=■a. b  Use■the■Eighth■Index■Law■combined■with■the■ 7

Fourth■Index■Law■to■show■that■ ( 7 a ) ■=■a.

maths Quest 10 for the Australian Curriculum

7

reFleCtion 

 

Why is it easier to perform operations with fractional indices than with expressions using surds?

number AND algebra • patterns and algebra

Combining index laws

1d

■■ ■■

In most practical situations, more than one index law is needed to simplify the expression. The following examples show simplification of expressions with indices, using several index laws.

Worked Example 9

Simplify. a  b 

( 2 a ) 4 b4 6 a 3 b2 3n − 2 × 9 n + 1 81n − 1

Think a

b

Write 4 4 a (2a) b

1

Write the expression.

2

Apply the Fourth Index Law to remove the bracket.

=

3

Apply the Second Index Law for each number and pronumeral to simplify.

=

4

Write the answer.

1

Write the expression.

2

Rewrite each term in the expression so that it has a base of 3.

=

3

Apply the Fourth Index Law to expand the brackets.

=

4

Apply the First and Second Index Laws to simplify.

=

6a 3 b 2

Write your answer.

6a 3 b 2 8ab 2 3

8ab 2 3 n −2 × 9 n +1 b 3

81n −1

= 5

16a 4 b 4

3n −2 × (32 ) n +1 (34 ) n −1 3n −2 × 32 n + 2 34 n − 4 33 n 34 n − 4 1 3n − 4 1

3

n−4

Chapter 1 Indices

17

number AND algebra • patterns and algebra

Worked Example 10

Simplify each of the following. a  (2a3b)4 ì 4a2b3

b 

7 xy3 ( 3 x 3 y2 )

2

c 

Think a

b

18

7 m3 n3 × mn2

Write a (2a3b)4 ì 4a2b3

1

Write the expression.

2

Apply the Fourth Index Law. Multiply each index inside the brackets by the index outside the brackets.

= 24a12b4 ì 4a2b3

3

Evaluate the number.

= 16a12b4 ì 4a2b3

4

Multiply coefficients and multiply pronumerals. Apply the First Index Law to multiply terms with the same base by adding the indices.

= 16 ì 4 ì a12 + 2b4 + 3 = 64a14b7

1

Write the expression.

2

Apply the Fourth Index Law in the denominator. Multiply each index inside the brackets by the index outside the brackets.

=

3

Apply the Second Index Law. Divide terms with the same base by subtracting the indices.

=

4

c

2 m 5 n × 3 m 7 n4

1 to express the answer am with positive indices. −m Use a =

b

7 xy 3 (3 x 3 y 2 )2

=

7 xy 3 9x6 y 4

7 x −5y −1 9 7 9x5 y

5 7 4 c 2m n × 3m n

1

Write the expression.

2

Simplify each numerator and denominator by multiplying coefficients and then terms with the same base.

=

3

Apply the Second Index Law. Divide terms with the same base by subtracting the indices.

=

6m8 n0 7

4

Simplify the numerator using a0 = 1.

=

6m8 × 1 7

=

6m8 7

Maths Quest 10 for the Australian Curriculum

7m3 n3 × mn2 6m12 n5 7m 4 n 5

number AND algebra • patterns and algebra

■■

■■

When more than one index law is used to simplify an expression, the following steps can be taken. Step 1:  If an expression contains brackets, expand them first. Step 2: If an expression is a fraction, simplify each numerator and denominator, then divide (simplify across then down). Step 3:  Express the final answer with positive indices. The following example illustrates the use of index laws for multiplication and division of fractions.

Worked Example 11

Simplify each of the following. a 

( 5 a 2 b3 ) 2 a10

×

a 2 b5 ( a 3 b)7



b 

8 m3 n−4 ( 6 mn2 ) 3

÷

4 m−2 n− 4 6 m−5 n

Think a

b

Write a

(5a 2 b3 )2

×

1

Write the expression.

2

Remove the brackets in the numerator of the first fraction and in the denominator of the second fraction.

=

3

Multiply the numerators and then multiply the denominators of the fractions. (Simplify across.)

=

4

Divide terms with the same base by subtracting the indices. (Simplify down.)

= 25a-25b4

5

Express the answer with positive indices.

=

1

Write the expression.

2

Remove the brackets.

=

3

Change the division sign to multiplication and flip the second fraction (multiply and flip).

=

4

Multiply the numerators and then multiply the denominators. (Simplify across.)

­=

5

Cancel common factors and divide pronumerals with the same base. (Simplify down.)

­=

6

Simplify and express the answer with positive indices.

­=

b

a10 25a 4 b6 a10

×

a2b5 (a 3 b) 7 a2b5 a 21b 7

25a6 b11 a31b 7

25b 4 a 25

8m3 n −4 (6mn2 )3

÷

8m3 n− 4 216m3 n6 8m3 n− 4 216m3 n6

4 m −2n −4 6m −5 n ÷ ×

4 m −2n− 4 6m −5n 6m −5n 4 m−2n − 4

48m −2n−3 864 mn2 m −3n−5 18 1 18m3 n5

Note that the whole numbers in part b of Worked example 11 could be cancelled in step 3. Chapter 1 Indices

19

number AnD AlgebrA • pAtterns AnD AlgebrA

remember

1.■ Simplifi■cation■of■expressions■with■indices■often■involves■application■of■more■than■one■ index■law. 2.■ If■an■expression■contains■brackets,■they■should■be■removed■fi■rst. 3.■ If■the■expression■contains■fractions,■simplify■across■then■down. 4.■ When■dividing■fractions,■change■ó■to■ì■and■fl■ip■the■second■fraction■(multiply■ and■fl■ip). 5.■ Express■the■fi■nal■answer■with■positive■indices. exerCise

1D inDiViDuAl pAthWAys

Combining index laws FluenCy   1  We10a ■Simplify■each■of■the■following. 2 2 3 4 3 a  (3a b ) × 2a b

eBook plus

Activity 1-D-1

c  2m3 n−5 × (m 2 n−3)

Review of indices doc-4957

7 2 2 3 3 2 e  (2a b ) × (3a b )

Activity 1-D-2

Indices practice doc-4958

36a

20b10

g  6 x 2 y 3 × 

 



Activity 1-D-3

Tricky indices doc-4959

1  3 42 4x 4 y 5

1 1

 i  2  

3 − 2 1 4 p3 q3

 

 × 3 

5 2 3 6 b  (4 ab ) × 3a b

54a10b9 13 −6 2 n

48a5b16

3 2 2 4 3 d  (2 pq ) × (5 p q ) 500p8q18

m9

15b 2

2 −2 3 5 −4 f  5(b c ) × 3(bc ) 3

7 11

12 x 8 y 15

1 − 1 3 −  3 p4 q 4

 

7

1

1

(16m3 n4 ) 4 ×  m 2 n 4 

j 

1 −  1 2 3 8p5q3



6 p12



h 

 

 

c 26

3



15 15

8m 4 n 4 2

1 33 7 5  ×  64 p 3 q 4  8 p 45 q 18  

  2  We10b ■Simplify■each■of■the■following. a 

5a 2 b3 3

(2a b)

5

3

8a

 4 x 3 y10  d     2x7 y4  g 

(

64 y 36

) ( )

1 5 p6 q 3

25

6

b 

7

e 

x 24

4 x 5 y6

x

3 4

(2 xy )

4y

3a3 b −5

2 1 1 3 p2 q4

35 1 p 3 q2

 3b 2 c3  h     5b −3c − 4 

−4

(3m 2 n3 )3

625 81b 20 c 28

i 

27

5 5 7

128m29 n26

(2m n )

 3g 2 h 5  f     2g4 h 

24a24b7

(2a 7 b 4 )−3

2

c 

6

(

(

3

1 1 1 x3 y4 z2

27h12 8g 6

) )

2 5 1 3

x3 y8z 2

3 − 2 1 1 2 − x3 y 4z3

  3  We10c ■Simplify■each■of■the■following. 2 3 4 a  2a b × 3a b

4a b

3a 2 2

6x3 y 2 × 4 x6 y

4 x5

9 xy 5 × 2 x 3 y 6

3y8

3 5

d 

g 

a3 b 2 × 2(ab 5 )3

b7

6(a 2 b3 )3 × a 4 b

3a 4

6 3 5 b  4 m n × 12mn

8n2

c 

36 x 6 y

f 

7 6

6m n

e 

h 

(6 x 3 y 2 )4 9 x 5 y 2 × 4 xy 7

2 p− 4q −2 × (5 pq 4 )−2

m 2 n4 3

2 3

12m n × 5m n 5 x 2 y 3 × 2 xy 5

y2

10 x 3 y 4 × x 4 y 2

x4

3 1

4

3

6x 2 y 2 × x 5 y 5

i  2 p11

maths Quest 10 for the Australian Curriculum

4

( p6 q 2 )−3 × 3 pq 75q 5

20

10 m6 n5 × 2m 2 n3

2

( )

1 1 5 x2y

1 1

× 3x 2 y 5

17

7

x 10 y 10

number AND algebra • patterns and algebra 4   WE 11a  Simplify each of the following. a b c d e

2 5a 4 b 7 4 a 3b 3 15 n9 4 m9 4 m5 9 n15 4 81x 2 y14

f 48x11y6 g

h

3 p4 5q

9

1 2b 12 17 3a 24 1

i

a

a3 b 2

×

5a 4 b 7

2a 6 b

b

a 9 b3 3

5p q g 3q − 4

10 a b

×

4 ab6 6a

c

3

 2 xy 2   x3 y 9  e  3 5  ×  10   3x y   2y 

−2

 5 p6 q 4  ×   3 p5 

7 3

4

 2m3 n 2  6m 2 n 4 × d   4 m3 n10  3mn5  6 −5

(2a 6 ) 2

h

1 1 2a 2 b 3 1 1 6a 3 b 2

6

(m n)

2

f

( ) ×

1 1 2 4a 4 b

i

1 b4a

4

4 x −5y−3 ( x 2 y 2 )−2

×

( m 3 n3 )3

×

(2mn)2 3x 5 y6 2−2 x −7y

2 1

1

3x 3 y 5

4x 2

1 1 9x 3 y 4

×

3 4 x y

5   WE 11b  Simplify each of the following. a

5a 2 b3 7 5

6a b

÷

a9b4 3ab

5 13

6

2a

3

 4a9   3a 7  c  6  ÷  5   b   2b   x 5 y −3  e    2 xy 5 

−4

÷

b

4

1024 b 2 81a

7a 2 b 4

 3ab  ÷  3a6 b 7  2a6 b 4 

d

4 x 6 y −10

4 y 36

(3 x −2y 2 )−3

27 x16

(2x y ) 4 5

3m3 n4

f

2m−6 n−5

21

g

1 3 4m 2 n 4

÷

1 1 6m 3 n 4

3

2

56a11b6 81

(4x y) ÷

3

6

5x 2 y6

4 x 12 3 y 20

( m 4 n3 ) 2

10 xy

25

3

 2 m 4 n6  ÷  −1   m n 

128 x 23 y 4 −2

6m19n19

1

3 12  1 −2 3 3  − h  4 b c  ÷ 2b 3 c 5    1     6c 5 b 

11 16m12 n

3 1 8m 4 n 2

3

11

4b 2 1

7

3 2 c 30

Understanding 6 Evaluate each of the following. 2 0 −3 0 5 6 −1 −3 a (5 × 2) × (5 × 2 ) ÷ (5 × 2 ) 3 3 −2 b (2 × 3 ) ÷

( 2 6 × 39 ) 0

125 8

1

26 × (3−2)−3

7 Evaluate the following for x = 8. (Hint: Simplify first.) 2

2x  x (2 x )−3 ×   ÷ 3 4  2 (2 )

1

8 a  Simplify the following fraction.

a 2 y × 9b y × (5ab) y (a y )3 × 5(3b y )2

5y - 1

b Find the value of y if the fraction is equal to 125. 9   MC  Which of the following is not the same as 3 3

3 B ( 4 xy )

A 8 x 2 y 2 1

D

(2 x 3 y 3 ) 2 ( 32 )−1

1

✔ E

y=4

3 (4 xy ) 2

? C

64 x 3 y 3

1

4 xy 2 × (2 xy 2 ) 2 Chapter 1 Indices

21

number AnD AlgebrA • pAtterns AnD AlgebrA

 10  The■expression■ ✔ A  1

a■ m 6 n     b 

−

7 6

or

6

m n7

3 g − 6h3 n 2

    c  3

−

7 3

× 5

−

7 6

a

2 3

(2 xy )

÷

xy 16 x 0

2

B 

x 2 y6 2 xy

■is■equal■to:

E 

6

2x2

C  2x2y6

b6 1 128 xy 5

 11  Simplify■the■following. a 

1

    d  2-2■or■ 4     e  a6b-8■or■

D 

x2 y

3

2

m n ÷ mn

3

b 

(

1

)

1

 1 2 g h ×  −3  n  −2

3

c 

14

15 14     f  d 15 or d

d 

3 22

×4

−

1 4

× 16

−

3 4

 a3 b −2  e   −3 −3  3 b 

−2

3

3

9 4 × 15 2

6

b8

45 3

 3−3 a −2 b  ÷  4 −2   a b 

2

f  (

5

3 2 2 d )

×(

3

1 5 5 d )

reAsoning  12  In■a■controlled■breeding■program■at■the■Melbourne■Zoo,■

the■population■(P)■of■koalas■at■t■years■is■modelled■by■ P■=■P0■ì■10kt.■The■initial■number■of■koalas■is■20■and■the■ population■of■koalas■after■1■year■is■40.■ a  Determine■the■value■of■P0■and■k.■ P0 = 20, k = 0.3 b  Calculate■the■number■of■koalas■after■2■years. 79■koalas c  When■will■the■population■be■equal■to■1000? During■the■6th■year. eBook plus

Digital doc

WorkSHEET 1.2 doc-5181

 13  The■decay■of■uranium■is■modeled■by■D■=■D0■ì■2-kt.■If■it■takes■

6■years■for■the■mass■of■uranium■to■halve,■find■the■percentage■ remaining■after: a  2■years 79% b  5■years 56% c  10■years. 31%

reFleCtion 

 

Do index laws need to be performed in a certain order?

22

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • pAtterns AnD AlgebrA

summary Review of index laws 

To■simplify■expressions■with■constants■and/or■pronumerals■in■index■form,■the■following■index■ laws■are■used. ■■ am■ì■an■=■am■+■n ■■ am■ó■an■=■am■-■n ■■ a0■=■1■(when■a■ò■0) ■■ (am)n■=■amn ■■ (ab)m■=■ambm m am  a ■■  b  = m b Negative indices ■■

■■ ■■ ■■

A■term■with■a■negative■index■can■be■expressed■with■a■positive■index■using■the■Seventh■Index■ Law. 1 1 (a)■ a-n■=■ n ■■ (b)■ −n ■=■an a a All■index■laws■apply■to■terms■with■negative■indices. Always■express■answers■with■positive■indices■unless■otherwise■instructed. Numbers■and■pronumerals■without■an■index■are■understood■to■have■an■index■of■1. Fractional indices

■■ ■■

Fractional■indices■are■those■which■are■expressed■as■fractions. Terms■with■fractional■indices■can■be■written■as■surds,■using■the■following■identities: ■ 1 an = n a m

■■

a n = n am = ( n a ) . All■index■laws■are■applicable■to■fractional■indices. m

Combining index laws ■■ ■■ ■■ ■■ ■■

Simplifi■cation■of■expressions■with■indices■often■involves■application■of■more■than■one■ index■law. If■an■expression■contains■brackets,■they■should■be■removed■fi■rst. If■the■expression■contains■fractions,■simplify■across■then■down. When■dividing■fractions,■change■ó■to■ì■and■fl■ip■the■second■fraction■(multiply■and■fl■ip). Express■the■fi■nal■answer■with■positive■indices.

MAPPING YOUR UNDERSTANDING

Homework  Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■1. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

Chapter 1 Indices

23

number AND algebra • patterns and algebra

Chapter review Fluency 1 3d10e4 is the simplified form of: 8

A d  6e2 ì 3d 4e3 B

6d 10 e 5

2e 2 C (3d 5e 2)2 ✔ D

3e

A

2

4

4x-3

ó A 2 D 2x-1

D

9

E

12 x8 × 2 x 7

2

✔ C

2x6

✔ C

4x

✔ A

D

7 16

1 16

4p q 1 4 p16 q −

3 4

A 2 D 3 3 8 24

÷

3 92

16m 7

E

5

B 2a6b13

C

a3 b6 2

22 16

p q

m7 7

10

5 1

2

✔ B

2 1 2 2i 7 j 11 k 5 2

50 25 2i 7 j 11 k10

1

2

10

5

32i 7 j 11 k 2 C 5

2i 7 j 11 k 5 E D 5 10 Simplify each of the following.

b

26a 4 b6 c 5

13ab3c 2 6

12a3 b3 c3

 14 p7  d    21q3 

4

9x10y10

3

1000m15n6 16 p28 81q12

11 Evaluate each of the following. 0

a

( p5 q 2 )2 can be simplified to: 2 pq 5 B

8

l3

32i 7 j 11 k 2 can be simplified to:

 20 m 5 n2  6  

a3 b6 E 4 ÷

7

4m7

2l 3

c 

(2a 2 b) 2

a6 b13 4 a6 b13 D 2

5

✔ C

l3

3 5

simplifies to:

✔ A

2m 7

a 5x3 ì 3x5y4 ì  x2y6

2 3 5 5 The expression (a b ) is equal to:

( 2 p 5 q 2 )3

11

2

x9

B 8x E 4x29

( p 2 q) 4

B

32i 7 j 11 k 5 A 5

is equal to:

A 4x5 D 8x5

6

8m 7

l3

B 2x0

6x9 × x5

can be simplified to:

2

l3

2 8m3n ì n4 ì 2m2n3 simplifies to: A 10m5n8 B 16m5n7 5 8 D 10m5n7 ✔ C 16m n 5 8 E 17m n

8x3

−3

 1  8  lm −2  16 

(d  5)2 ì e3

 d5  E 3  2  e 

3

 2   2l 9 m −1   

C

1 4 p8

E 22p16q

5a0

 2a  -   + 12 16  3

b -(3b)0 -

(4 b) 0 2

3

-2

12 Simplify each of the following and express your

answer with positive indices. a 2a-5b2 ì 4a-6b-4 b 4x-5y-3 ó 20x12y-5 c (2m-3n2)-4

a c

8

b

11 2

a b m12

16 n8

13 Evaluate each of the following without using a

calculator.

can be simplified to: ✔ B

E

1 216 1 2

C

8 27

Maths Quest 10 for the Australian Curriculum

a

 1   2 −3

c 4

−3

−3 b 2 × (3) ×

8

×

5 8− 2

−5 0

 9   2

2

3 2

y2 5 x17

number AnD AlgebrA • pAtterns AnD AlgebrA  14  Simplify■each■of■the■following. a  b 

4 1 2a 5 b 2

×

1 3 3a 2 b 4

3 1 43 x 4 y 9

4

4 1 16 x 5 y 3

1 2 x 20 y 9

×

3 2 5a 4 b 5

 18  Simplify■each■of■the■following■and■then■evaluate. 41 33 30a 20 b 20

a  (3 ×

1

6×



2

b   125 3 −

1

 16  Simplify. a 

3

a 9 + 4 16a8 b 2 − 3 ( 5 a )

b 

5

32 x 5 y10 +

64 x 3 y 6

b 

2a13

2a 2 b3 × 5−2 a −3 b − 6

5b 2

2 x 4 y −5 6 −2

3y x

 4 xy −2  ×  −6 3   3x y  1

46

6 1

- 18

−3 0

−6 × (3 )

1 (9a 4 m ) 2

−3

4

1 36

(4 b) m ×   2  Answer■the■following■and■explain■your■reasoning 3 a  What■is■the■ten’s■digit■of■33 ? 8 309 b  What■is■the■one’s■digit■of■6 ? 6 c  What■is■the■one’s■digit■of■81007? 2 eBook plus

Interactivities

6xy2

(5a −2b) −3 × 4 a6 b −2

1 22 27 3 

− 2a 3 + 2a 2 b 2

Test yourself Chapter 1 int-2828

 17  Simplify■each■of■the■following. a 

2

0

1

15

3

1  1  32 × 63 

6a3m × 2b 2 m × (3ab)− m

calculator.■Show■all■working.

1 16 2

1  −  +  36 × 5 2 

  1  If m =■2,■determine■the■value■of:

 15  Evaluate■each■of■the■following■without■using■a■

a 

×5

−2

problem solVing

b

1 × 814

×

3 32

b  (6 × 3−2 )−1 ÷

1  12 4 a 3  2a 6 c   3  b3  2

3 16 4

1 56 ) 2

Word search Chapter 1 int-2826 Crossword Chapter 1 int-2827

9y4 32 x15

1   3 43  3 n −2 2 m n 4 m     c  ÷ 1  2    5− 3   5m 2 n 

−

1 2 4

23 m

Chapter 1 Indices

25

eBook plus

ACtiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■1■(doc-5167)■ (page 1) Are you ready? Digital docs (page 2) •■ SkillSHEET■1.1■(doc-5168):■Index■form •■ SkillSHEET■1.2■(doc-5169):■Using■a■calculator■to■ evaluate■numbers■given■in■index■form •■ SkillSHEET■1.3■(doc-5170):■Linking■between■ squares■and■square■roots •■ SkillSHEET■1.4■(doc-5171):■Calculating■square■ roots •■ SkillSHEET■1.5■(doc-5172):■Linking■between■cubes■ and■cube■roots •■ SkillSHEET■1.6■(doc-5173):■Calculating■cube■ roots •■ SkillSHEET■1.7■(doc-5174):■Estimating■square■roots■ and■cube■roots •■ SkillSHEET■1.8■(doc-5175):■Using■a■calculator■to■ evaluate■square■roots■and■cube■roots

1A   Review of index laws Digital docs (page 5) •■ Activity■1-A-1■(doc-4948):■Reviewing■index■ operations •■ Activity■1-A-2■(doc-4949):■Practising■the■index■ laws •■ Activity■1-A-3■(doc-4950):■Applying■the■index■ laws

1B   Negative indices

(page 10) •■ Activity■1-B-1■(doc-4951):■Negative■indices •■ Activity■1-B-2■(doc-4952):■Harder■negative■indices •■ Activity■1-B-3■(doc-4953):■Tricky■negative■indices Digital docs

Interactivity

•■ Negative■indices■(int-2777)■(page 7)

26

maths Quest 10 for the Australian Curriculum

1C   Fractional indices Digital docs

•■ Activity■1-C-1■(doc-4954):■Fractional■indices■ (page 14) •■ Activity■1-C-2■(doc-4955):■Harder■fractional■indices■ (page 14) •■ Activity■1-C-3■(doc-4956):■Tricky■fractional■indices■ (page 14) •■ SkillSHEET■1.9■(doc-5176):■Addition■of■fractions■ (page 14) •■ SkillSHEET■1.10■(doc-5177):■Subtraction■of■ fractions■(page 14) •■ SkillSHEET■1.11■(doc-5178):■Multiplication■of■ fractions■(page 15) •■ SkillSHEET■1.12■(doc-5179):■Writing■roots■as■ fractional■indices■(page 16) •■ WorkSHEET■1.1■(doc-5180):■Fractional■indices■ (page 16) 1D   Combining index laws Digital docs

•■ Activity■1-D-1■(doc-4957):■Review■of■indices■ (page 20) •■ Activity■1-D-2■(doc-4958):■Indices■practice■(page 20) •■ Activity■1-D-3■(doc-4959):■Tricky■indices■(page 20) •■ WorkSHEET■1.2■(doc-5181):■Combining■index■laws■ (page 22) Chapter review

(page 25) •■ Test■Yourself■Chapter■1■(int-2828):■Take■the■end-ofchapter■test■to■test■your■progress■ •■ Word■search■Chapter■1■(int-2826):■an■interactive■word■ search■involving■words■associated■with■this■chapter •■ Crossword■Chapter■1■(int-2827):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • PAtterns AnD AlgebrA

2

2A Substitution 2B Adding and subtracting algebraic fractions 2C Multiplying and dividing algebraic fractions 2D Solving linear equations 2E Solving equations with algebraic fractions and multiple brackets WhAt Do you knoW ?

linear algebra

1 List what you know about linear equations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of linear equations. eBook plus

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oPening Question

Tod thinks of a number. If 5 is added to this number and then quadrupled, Tod’s answer is 224. What number did Tod start with?

number AnD AlgebrA • PAtterns AnD AlgebrA

Are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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SkillSHEET 2.1 doc-5183

Like terms 1 Select the like terms from each of the following lists. 1

a abc, 3acb, ab, 2 bc b x2y, -3y,

1 2 yx , 4

xy

abc and 3acb x2y and 14 yx 2

c pq, -2q2p, 2pq2, p 2q2 -2q2p and 2pq2 eBook plus

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SkillSHEET 2.6 doc-5188

28

Collecting like terms 2 Simplify each of the following expressions. a 2x - 5 + 7 - 5x -3x + 2 b -3a - 4 - 2a - 5 -5a - 9 c 4p - 2q + 8 - 6p -2p - 2q + 8 Finding the highest common factor 3 Find the highest common factor for each of the following pairs of terms. a 6x and 24y 6 b 6ab and 9abc 3ab c -12pq and -20pqr -4pq Addition and subtraction of fractions 4 Calculate each of the following. a

2 3

+

3 4

15

12

b

7 8

−

5 12

11 24



c

4 15

+

3 20



5 12

Multiplication of fractions 5 Perform the following multiplications. a

3 1 × 4 3

1 4

b

5 12

×

4 15



b

14 15

÷

21 25

1 19

1 9

1

1

c 2 × 1 2 2 2 15 3

Division of fractions 6 Calculate each of the following. a

5 8

÷

3 4

5 6

maths Quest 10 for the Australian Curriculum

c 2 1 ÷ 3 3 4 8

2 3

number AND algebra • Patterns and algebra

2A

Substitution When the numerical values of pronumerals are known, they can be substituted into an algebraic expression and the expression can then be evaluated. It can be useful to place any substituted values in brackets when evaluating an expression.

Worked Example 1

If a = 4, b = 2 and c = -7, evaluate the following expressions. a  a - b b  a3 + 9b - c Think a

b

Write a a-b

1

Write the expression.

2

Substitute a = 4 and b = 2 into the expression.

=4-2

3

Simplify.

=2

1

Write the expression.

2

Substitute a = 4, b = 2 and c = -7 into the expression.

= (4)3 + 9(2) - (-7)

3

Simplify.

= 64 + 18 + 7 = 89

b a3 + 9b - c

Worked Example 2

If c = a2 + b2 , calculate c if a = 12 and b = -5. Think

Write

c=

a2 + b2

1

Write the expression.

2

Substitute a = 12 and b = -5 into the expression.

=

(12)2 + ( −5)2

3

Simplify.

=

144 + 25

= 169 = 13

Number laws ■■

■■

Recall from previous studies that when dealing with numbers and pronumerals, particular rules must be obeyed. Before progressing further, let us briefly review the Commutative, Associative, Identity and Inverse Laws. Consider any three pronumerals x, y and z, where x, y and z are elements of the set of Real numbers.

Commutative Law 1. x + y = y + x  (example: 3 + 2 = 5 and 2 + 3 = 5) 2. x - y ò y - x  (example: 3 - 2 = 1 but 2 - 3 = -1) Chapter 2 Linear algebra

29

number AND algebra • Patterns and algebra

3. x ì y = y ì x  (example: 3 ì 2 = 6 and 2 ì 3 = 6) 3

2

4. x ó y ò y ó x  (example: 3 ó 2 = 2 , but 2 ó 3 = 3 ) Therefore, the Commutative Law holds true for addition and multiplication, since the order in which two numbers or pronumerals are added or multiplied does not affect the result. However, the Commutative Law does not hold true for subtraction or division.

Associative Law 1. x + ( y + z) = (x + y) + z  [example: 2 + (3 + 4) = 2 + 7 = 9 and (2 + 3) + 4 = 5 + 4 = 9] 2. x - ( y - z) ò (x - y) - z  [example: 2 - (3 - 4) = 2 - -1 = 3 and (2 - 3) - 4 = -1 - 4 = -5] 3. x ì ( y ì z) = (x ì y) ì z  [example: 2 ì (3 ì 4) = 2 ì 12 = 24 and (2 ì 3) ì 4 = 6 ì 4 = 24] 4. x ó ( y ó z) ò (x ó y) ó z 3

4

8

2

2

1

2

1

[example: 2 ó (3 ó 4) = 2 ó 4 = 2 ì 3 = 3 but (2 ó 3) ó 4 = 3 ó 4 = 3 ì 4 = 12 = 6 ] The Associative Law holds true for addition and multiplication since grouping two or more numbers or pronumerals and calculating them in a different order does not affect the result. However, the Associative Law does not hold true for subtraction or division.

Identity Law The Identity Law states that in general: x+0=0+x=x xì1=1ìx=x In both of the examples above, x has not been changed (that is, it has kept its identity) when zero is added to it or it is multiplied by 1.

Inverse Law x + -x = -x + x = 0 1 1 xì = ìx=1 x x That is, when the additive inverse of a number or pronumeral is added to itself, it equals 0. When the multiplicative inverse of a number or pronumeral is multiplied by itself, it equals 1. The Inverse Law states that in general:

Closure Law A law that you may not yet have encountered is the Closure Law. The Closure Law states that, when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. For example, addition is closed on natural numbers (that is, positive integers: 1, 2, 3,  .  .  .) since adding a pair of natural numbers produces a natural number. Subtraction is not closed on natural numbers. For example, 5 and 7 are natural numbers and the result of adding them is 12, a natural number. However, the result of subtracting 7 from 5 is -2, which is not a natural number.

Worked Example 3

Find the value of the following expressions, given the integer values x = 4 and y = -12. Comment on whether the Closure Law for integers holds for each of the expressions when these values are substituted. a  x + y      b  x - y      c  x ì y      d  x ó y Think a

30

Write a x + y = 4 + -12

1

Substitute each pronumeral into the expression.

2

Evaluate and write the answer.



3

Determine whether the Closure Law holds; that is, is the result an integer?

The Closure Law holds for these substituted values.

Maths Quest 10 for the Australian Curriculum

= -8

number AnD AlgebrA • PAtterns AnD AlgebrA b Repeat steps 1–3 of part a.

b x - y = 4 - -12

c Repeat steps 1–3 of part a.

c

d Repeat steps 1–3 of part a.

d x ó y = 4 ó -12

= 16 The Closure Law holds for these substituted values. x ì y = 4 ì -12 = -48 The Closure Law holds for these substituted values. 1

= -3 The Closure Law does not hold for these substituted values since the answer obtained is a fraction, not an integer. ■

It is important to note that, although a particular set of numbers may be closed under a given operation, for example multiplication, another set of numbers may not be closed under that same operation. For example, in part c of Worked example 3, integers were closed under multiplication. However, in some cases, the set of irrational numbers is not closed under multiplication, since 3 ì 3 = 9 = 3. In this example, two irrational numbers produced a rational number under multiplication. remember

1. When the numerical values of pronumerals are known, they can be substituted them into an algebraic expression and the expression can then be evaluated. 2. It is sometimes useful to place any substituted values in brackets when evaluating an expression. 3. When dealing with numbers and pronumerals, particular rules must be obeyed. (a) The Commutative Law holds true for addition and multiplication. (b) The Associative Law holds true for addition and multiplication. (c) The Identity Law states that, in general: x + 0 = x and x ì 1 = x. 1 (d) The Inverse Law states that, in general: x + -x = 0 and x ì = 1. x (e) The Closure Law states that, when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. exerCise

2A inDiViDuAl PAthWAys eBook plus

Activity 2-A-1

Substitution doc-4960 Activity 2-A-2

Harder substitution doc-4961 Activity 2-A-3

Tricky substitution doc-4962

substitution fluenCy 1 We1 If a = 2, b = 3 and c = 5, evaluate the following expressions. a a + b 5 b c - b 2 c c - a - b 0 d c - (a - b) g abc 30 j c2 + a 27

6

e 7a + 8b - 11c -17 h ab(c - b) 12 k -a ì b ì -c 30

2 If d = -6 and k = -5, evaluate the following. a d + k -11 b d - k -1 d kd 30 e -d(k + 1) g k3 -125

h

k −1 d

1

-24

a b c + + 3 2 3 5 i a2 + b2 - c2 -12 l 2.3a - 3.2b -5 f

c k - d 1 f d 2 36 i

3k - 5d 15

Chapter 2 linear algebra

31

number AnD AlgebrA • PAtterns AnD AlgebrA

eBook plus

1

1

3 If x = 3 and y = 4 , evaluate the following.

Digital doc

a x + y

SkillSHEET 2.7 doc-5189

c xy

b y - x - 121

7 12

x

d 1 1 y

1 12

e x2y3

3

1 576

f

9x y2

48

4 We2 Calculate the unknown variable in the following real-life mathematical formulas.

a 2 + b 2 , calculate c if a = 8 and b = 15.

a If c = b If A = c d e f g h i

17

1 bh, 2

determine the value of A if b = 12 and h = 5. 30 The perimeter, P, of a rectangle is given by P = 2L + 2W. Calculate the perimeter, P, of a rectangle, given L = 1.6 and W = 2.4. 8 C If T = , determine the value of T if C = 20.4 and L = 5.1. 4 L n +1 If K = , determine the value of K if n = 5. 1.5 n −1 9C Given F = + 32, calculate F if C = 20. 68 5 If v = u + at, evaluate v if u = 16, a = 5, t = 6. 46 The area, A, of a circle is given by the formula A = p r2. Calculate the area of a circle, correct to 1 decimal place, if r = 6. 113.1 1 If E = 2 mv2, calculate m if E = 40, v = 4. 5

A , evaluate A to 1 decimal place if r = 14.1. 624.6 π 5 mC a If p = -5 and q = 4, then pq is equal to: A 20 B 1 C -1 j

Given r =

✔ D

-20

5

E - 4

b If c2 = a2 + b2, and a = 6 and b = 8, then c is equal to: A 28 B 100 D 14 E 44 c Given h = 6 and k = 7, then kh2 is equal to: ✔ B 252 A 294 D 5776 E 85

✔ C

10

C 1764

unDerstAnDing 6 Knowing the length of two sides of a right-angled triangle, the third side can be calculated

using Pythagoras’ theorem. If the two shorter sides have lengths of 1.5 cm and 3.6 cm, calculate the length of the hypotenuse. 3.9 cm 4 7 The volume of a sphere can be calculated using the formula 3 pr3. What is the volume of a sphere with a radius of 2.5 cm? Give your answer correct to 2 decimal places. 65.45 cm3 2.5 cm

8 A rectangular park is 200 m by 300 m. If Blake runs along the diagonal of the park, how far will he run? Give your answer to the nearest metre. 361 m 32

maths Quest 10 for the Australian Curriculum

Reasoning 9   WE 3  Determine the value of the following expressions, given the integer values x = 1, y = -2

9

and z = -1. Comment on whether the Closure Law for integers holds true for each of the expressions when these values are substituted. a x + y b y - z c y ì z d x ó z e z - x f x ó y 10 Find the value of the following expressions, given the natural number values x = 8, y = 2 and z = 6. Comment on whether the Closure Law for natural numbers holds true for each of the expressions. a x + y b y - z c y ì z d x ó z e z - x f x ó y 11 For each of the following, complete the relationship to illustrate the stated law. Justify your answer. a + (2b + 4c) a (a + 2b) + 4c = _______________ Associative Law x ì (3y ì 5c) b (x ì 3y) ì 5c = _______________ Associative Law ó q ò _______________ q ó 2p c 2p Commutative Law + q = _______________ q + 5d d 5d Commutative Law 0 + 3z = 3z reflection    e 3z + 0 = _______________ Identity Law 1 1 f 2x ì _______ = _______________ Inverse Law = × 2x = 1 Why is knowledge of the 2x 2x 4x ó (3y ó 5z) g (4x ó 3y) ó 5z ò _______________ Associative Law Commulative Law useful? - 4y ò _______________ 4y - 3d h 3d Commutative Law

2B

Adding and subtracting algebraic fractions

10 e -2 — in this case, subtraction is not closed on natural numbers. f 4 — in this case, division is closed on natural numbers.

■■

To add or subtract algebraic fractions, we perform the following steps. 1. Find the lowest common denominator (LCD) by finding the lowest common multiple (LCM) of the denominators. 2. Rewrite each fraction as an equivalent fraction with this common denominator. 3. Express as a single fraction. 4. Simplify the numerator.

Worked Example 4

Simplify the following expressions. 2x x a   - 2 3

b 

x+1 x+ 4 + 4 6

Think a

Write a

c 1 2 — in this case, multiplication is closed on natural numbers. 10 a 10 — in this case, addition is closed on natural numbers. b  4 — in this case, subtraction is not closed on natural numbers. d 43 — in this case, division is not closed on natural numbers.

a -1 — in this case, addition is closed on integers. b -1 — in this case, subtraction is closed on integers. c 2 — in this case, multiplication is closed on integers. d -1 — in this case, division is closed on integers. e -2 — in this case, subtraction is closed on integers. f - 12 — in this case, division is not closed on integers.

number AND algebra • Patterns and algebra

2x x − 3 2

1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of 3 and 2 is 6.

=

3

Express as a single fraction.

=

4 x − 3x 6

4

Simplify the numerator.

=

x 6

2x 2 x 3 × − × 3 2 2 3 4 x 3x = − 6 6

Chapter 2 Linear algebra

33

number AND algebra • Patterns and algebra

b

x +1 x + 4 + 6 4

1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of 6 and 4 is 12, not 24. Note that 24 is a common multiple but not the lowest common multiple. If 24 is used as the common denominator, then additional calculations will need to be performed to arrive at the final simplified answer.

=

3

Express as a single fraction.

=

4

Simplify the numerator by expanding brackets and collecting like terms.

=

■■

b

x +1 2 x + 4 3 × + × 6 2 4 3 2( x + 1) 3( x + 4) = + 12 12

2( x + 1) + 3( x + 4) 12

2 x + 2 + 3 x + 12 12 5 x + 14 = 12

If pronumerals appear in the denominator, we can treat these separately from their coefficients. This is demonstrated in the following worked example.

Worked Example 5

Simplify

2 1 − . 3x 4x

Think

Write

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of 3 and 4 is 12. The LCM of x and x is x. So the LCD is 12x, not 12x2. If you take care to ensure that you have found the LCM at this step, then the subsequent mathematics will be simpler.

3

Express as a single fraction.

=

8−3 12 x

4

Simplify the numerator.

=

5 12x

■■

34

2 1 − 3x 4 x

1

2 4 1 3 × − × 3x 4 4 x 3 8 3 = − 12 x 12 x =

When there is an algebraic expression in the denominator of each fraction, we can obtain a common denominator by writing the product of the denominators. For example, if x + 3 and 2x - 5 are in the denominator of each fraction, then a common denominator of the two fractions will be (x + 3)(2x - 5).

Maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra

Worked Example 6

Simplify

x + 1 2x − 1 + by writing it first as a single fraction. x+3 x+2

Think

Write

1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of x + 3 and x + 2 is the product (x + 3)(x + 2).

x + 1 2x − 1 + x+3 x+2 =

( x + 1) ( x + 2) (2 x − 1) ( x + 3) × + × ( x + 3) ( x + 2) ( x + 2) ( x + 3))

=

( x + 1)( x + 2) (2 x − 1)( x + 3) + ( x + 3)( x + 2) ( x + 3)( x + 2)

3

Express as a single fraction.

=

( x + 1)( x + 2) + (2 x − 1)( x + 3) ( x + 3)( x + 2)

4

Simplify the numerator by expanding brackets and collecting like terms. Note: The denominator is generally kept in factorised form. That is, it is not expanded.

=

( x 2 + 2 x + x + 2) + (2 x 2 + 6 x − x − 3) ( x + 3)( x + 2)

=

x 2 + 3x + 2 + 2 x 2 + 5x − 3 ( x + 3)( x + 2)

=

3x 2 + 8x − 1 ( x + 3)( x + 2)

Worked Example 7

Simplify

x+2 x−1 by writing it first as a single fraction. + x − 3 ( x − 3)2

Think 1

Write the expression.

2

Rewrite each fraction as an equivalent fraction using the LCD. The LCM of x - 3 and (x - 3)2 is (x - 3)2 not (x - 3)3.

Write

x+2 x −1 + x − 3 ( x − 3)2 = = =

3

Express as a single fraction.

=

4

Simplify the numerator.

=

x+2 x−3 x −1 × + x − 3 x − 3 ( x − 3)2 ( x + 2)( x − 3) ( x − 3)

2

x2 − x − 6 ( x − 3)2

+

+

x −1 ( x − 3)2

x −1 ( x − 3)2

x2 − x − 6 + x − 1 ( x − 3)2 x2 − 7 ( x − 3)2

Chapter 2 Linear algebra

35

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1. Algebraic fractions contain pronumerals that may represent particular numbers or changing values. 2. To add or subtract algebraic fractions we perform the following steps. (a) Find the lowest common denominator (LCD) by finding the lowest common multiple (LCM) of the denominators. (b) Rewrite each fraction as an equivalent fraction with this common denominator. (c) Express as a single fraction. (d) Simplify the numerator.

exerCise

2b inDiViDuAl PAthWAys eBook plus

Activity 2-B-1

Introducing algebraic fractions doc-4963 Activity 2-B-2

Working with algebraic fractions doc-4964 Activity 2-B-3

Advanced algebraic fractions doc-4965

eBook plus

Digital doc

SkillSHEET 2.4 doc-5186

eBook plus

Digital doc

SkillSHEET 2.8 doc-5190

2

Adding and subtracting algebraic fractions fluenCy 1 Simplify each of the following.

( )

4 2 26 5 1 5 49 1 b + + 21 21 7 3 8 9 72 4 3 17 3 2 1 d e − − 9 11 99 7 5 35 5x 4 15x − 4 3 2 x 15 − 16 x g − h − 40 27 9 27 8 5 2 We4 Simplify the following expressions. 2y y 5y y y 3y a − b − 40 3 4 12 8 5 8 x 2 x 14 x 2w w 3w d e + − 9 9 3 14 28 28 12 y y 89 y 10 x 2 x 32 x g h + + 35 5 7 5 15 15 x + 2 x + 6 7 x + 30 2 x − 1 2 x + 1 2 x − 11 j k + − 12 30 4 3 5 6 a

3 We5 Simplify the following. 2 1 5 a + 4 x 8 x 8x

12 + 5x 2 g 100 x d

4 8 15 x 3x 7 37 + 100x 20 x

3 1 5 − 4 x 3 x 12x 1 1 7 e + 6 x 8 x 24x 1 5 51 h + 10 x x 10x b

3 6 1 + 5 15 1 x 6 − 5x f − 30 5 6 5 2 15 − 2 x i − 3x x 3 c

4 x x 13x − 3 4 12 y y y f − 5 20 4 x + 1 x + 3 7 x + 17 i + 10 5 2 3 x + 1 5 x + 2 19 x + 7 l + 6 2 3 c

5 1 38 + 3 x 7 x 21x 9 9 9 f − 4 x 5 x 20x 4 3 1 − i − 6x 3x 2 x c

4 We6, 7 Simplify the following by writing as single fractions. 2 3x 2x 5 2 x 2 + 3 x + 25 + + a b ( x + 4) ( x − 2) ( x + 5) ( x − 1) ( x + 5)( x − 1)



3 x + 14 x − 4 ( x + 4)( x − 2)



2 x 2 + 6 x − 10 (2 x + 1)( x − 2)

c

5 x + (2 x + 1) ( x − 2)

d

2x 3 4 x 2 − 17 x − 3 − ( x + 1) (2 x − 7) ( x + 1)(2 x − 7)



7x2 + x ( x + 7)( x − 5)

e

4x 3x + ( x + 7) ( x − 5)

f

x+2 x − 1 2x2 + 6x + 7 + x + 1 x + 4 ( x + 1)( x + 4)

36

maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra 2



− x + 7 x + 15 ( x + 1)( x + 2)



x 2 + 3x + 9 ( x + 2)(3 x − 1)



x + 8 2x + 1 − x +1 x+2 x + 1 2x − 5 i − x + 2 3x − 1 4 3 + k 2 x +1 ( x + 1)

3x + 7 ( x + 1)2



2c

h

g

j l

x −7 ( x + 3)( x − 2)

x+5 x −1 − x+3 x−2 2 3 − x −1 1− x 3 1 3x − 4 − x − 1 ( x − 1)2 ( x − 1)2

reflection 

 

Why can't we just add the numerators and the denominators of fractions: a c a +c + = ? b d b +d

5 − 5x 5 = ( x − 1)(1 − x ) x − 1

Multiplying and dividing algebraic fractions ■■

■■

When multiplying algebraic fractions, first cancel any common factors if possible, then multiply the numerators together and finally multiply the denominators together. Simplify the expression further if necessary. 2 x 4 xy 2 x × 4 xy For example, = × 3y × 7 3y 7 8x 2 (Cancel y from the numerator and denominator.) 21 When dividing algebraic fractions, change the division sign to a multiplication sign and write the following fraction as its reciprocal (swap the numerator and the denominator). 8x 4 x 8x 5 For example, ÷ = × (The process then follows that for multiplication.) 3 5 3 4x =



=

10 3

(Cancel 4x from the numerator and denominator.)

Worked Example 8

Simplify each of the following. 5 y 6z a  × 3x 7 y b 

2x x+1 × ( x + 1)(2 x − 3) x

Think a

Write a

5y 6z × 3x 7 y

1

Write the expression.

2

Examine the fractions and see if you can cancel any common factors in the numerator and denominator. The y can be cancelled in the denominator and the numerator. Also the 3 in the denominator can divide into the 6 in the numerator.

=

5 2z × x 7

3

Multiply the numerators, then multiply the denominators.

=

10 z 7x Chapter 2 Linear algebra

37

number AND algebra • Patterns and algebra

b

b

2x x +1 × ( x + 1)(2 x − 3) x

1

Write the expression.

2

Check for common factors in the numerator and the denominator. (x + 1) and the x are common in the numerator and the denominator and can therefore be cancelled.

=

2 1 × (2 x − 3) 1

3

Multiply the numerators, then multiply the denominators.

=

2 2x − 3

Worked Example 9

Simplify the following expressions. 3 xy 4 x a  ÷ 2 9y

b 

4 x−7 ÷ ( x + 1)(3 x − 5) x + 1

Think a

b

38

Write a

3 xy 4 x ÷ 2 9y

1

Write the expression.

2

Change the division sign to a multiplication sign and write the second fraction as its reciprocal.

=

3 xy 9 y × 2 4x

3

Check for common factors in the numerator and denominator and cancel. The pronumeral x is common to both the numerator and denominator and can therefore be cancelled.

=

3y 9 y × 2 4

4

Multiply the numerators, then multiply the denominators.

=

27 y 2 8

1

Write the expression.

2

Change the division sign to a multiplication sign and write the second fraction as its reciprocal.

=

4 x +1 × ( x + 1)(3 x − 5) x − 7

3

Check for common factors in the numerator and denominator and cancel. (x + 1) is common to both the numerator and denominator and can therefore be cancelled.

=

4 1 × 3x − 5 x − 7

4

Multiply the numerators, then multiply the denominators.

=

4 (3 x − 5)( x − 7)

Maths Quest 10 for the Australian Curriculum

b

4 x−7 ÷ ( x + 1)(3 x − 5) x + 1

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1. When multiplying algebraic fractions, first cancel any common factors if possible, then multiply the numerators together and finally multiply the denominators together. Simplify the expression further if necessary. 2. When dividing algebraic fractions, change the division sign to a multiplication sign and write the following fraction as its reciprocal (swap the numerator and the denominator). The process then follows that for multiplication.

exerCise

2C inDiViDuAl PAthWAys eBook plus

Activity 2-C-1

Learning operations with algebraic fractions doc-4966 Activity 2-C-2

Operations with algebraic fractions doc-4967

multiplying and dividing algebraic fractions fluenCy 1 We8a Simplify each of the following.

x 20 × y 5 x 9 × d 2 2y 3 y 8z g × 4 x 7y j

Activity 2-C-3

Advanced operations with algebraic fractions doc-4968

eBook plus

eBook plus

eBook plus

Digital doc

5y x × 3x 8y

k

a

2x x −1 × ( x − 1)(3 x − 2) x

c

−20 y −21z 12z × x 7x 5y

l

y x −x × −3w 2 y 6 w

c

b

5x 4x + 7 5 × x −3 ( x − 3)(4 x + 7) x

9x 5x + 1 9 × 2( x − 6) (5 x + 1)( x − 6) 2x

d

( x + 4) x +1 1 × ( x + 1)( x + 3) x + 4 x + 3

e

2x x −1 2x × x + 1 ( x + 1)( x − 1) ( x + 1)2

f

2 x ( x + 1) × x (2 x − 3) 4

g

2x 3a a × 4(a + 3) 15 x 10(a + 3)

h

15c 21d 35d × 12(d − 3) 6c 8(d − 3)

j

7 x 2 ( x − 3) 3( x − 3)( x + 1) 3x × 2 10 ( x − 1) 5 x ( x + 1) 14( x − 3) ( x − 1)

Digital doc

SkillSHEET 2.9 doc-5191

5 24

9x 4y



y 16 4y × x 4 x 3w −7 3w f × −14 x 2x x −9 z −3x × i 2y 3z 2y

b

2 We8b Simplify the following expressions.

Digital doc

SkillSHEET 2.5 doc-5187

6z 7x

x 12 3x × y 4 y x −25 −5x e × 4y 10 2y −y 6z 2z × h 3 x −7 y 7 x

4x y

a

i

6x2 20( x − 2)2

×

15( x − 2)

4

16 x

2 3x − 2

9 32 x 2 ( x − 2)

x +1 2(2 x − 3)

3 We9a Simplify the following expressions. a

3 5 ÷ x x

d

20 20 ÷ 3 y 3y

g

3 xy 3 x 4 y 2 ÷ 7 7 4y

j

8wx 3w 32 xy ÷ 15 5 4y

SkillSHEET 2.10 doc-5192

3 5

b

2 9 ÷ x x

2 9

c

1 5 1 ÷ 5w w 25 2 xy 5 x 2 y 2 h ÷ 25 5 y e

k

2 xy 3 xy ÷ 5 5

2 3

4 12 ÷ x x

1 3

7 3 35 or 5 5 ÷ 6 2 x 5x 6 6 y 3x 8y 2 ÷ i 9 9 4 xy f

l

10 xy 20 x ÷ y2 7 14 y

Chapter 2 linear algebra

39

number AnD AlgebrA • PAtterns AnD AlgebrA 4 We9b Simplify the following expressions. a

9 x+3 ÷ ( x − 1)(3 x − 7) x − 1

b

1 x−9 1 ÷ ( x + 2)(2 x − 5) 2 x − 5 ( x + 2)( x − 9)

9 (3 x − 7)( x + 3)

refleCtion 

12( x − 3)2 4( x − 3) 21( x − 3) c ÷ x+5 ( x + 5)( x − 9) 7( x − 9) 13 3( x + 1) d ÷ 2 2 ( x − 4)( x − 1) 6( x − 4) ( x − 1) 13

eBook plus

Digital doc WorkSHEET 2.1 doc-5193

Is

3

x +2 Explain.

the same as

■ ■

Solve the following equations. a a + 27 = 71

b

e = 0.87

40

1

Write the equation.

2

27 has been added to a to result in 71. The addition of 27 has to be undone. Therefore, subtract 27 from both sides of the equation to obtain the solution.

1

Write the equation.

2

Express 3 14 as an improper fraction.

3

The pronumeral d has been divided by 16 to result in 13 . Therefore 4 the division has to be undone by multiplying both sides of the equation by 16 to obtain d.

1

Write the equation.

2

The square root of e has been taken to result in 0.87. Therefore, the square root has to be undone by squaring both sides of the equation to obtain e.

maths Quest 10 for the Australian Curriculum

d 1 = 34 16

d f2 =

think

c

1

x +2

+

1

x +2

?

Equations are algebraic sentences that can be solved to give a numerical solution. An equation consists of two algebraic expressions joined by an equals sign. Remember, to solve any equation we need to isolate the pronumeral we wish to find; that is, we must undo all the operations that have been performed on the pronumeral.

WorkeD exAmPle 10

b

x +2

+

solving linear equations ■

a

1

9( x − 4)( x + 1)

2D

c

 

4 25

Write a

a + 27 = 71 a + 27 - 27 = 71 - 27 a = 44

b

d 1 = 34 16 d 13 = 16 4 d 13 ì 16 = ì 16 16 4 d = 52

c

e = 0.87

( e)

2

= 0.872 e = 0.7569

number AND algebra • Patterns and algebra

d

1

Write the equation.

2

The pronumeral f has been squared 4 to result in 25 . Therefore the squaring has to be undone by taking the square root of both sides of the equation to obtain f. Note that there are two possible solutions, one positive and one negative, since two negative numbers can also be multiplied together to produce a positive one.

■■ ■■

d

4 25

f2=

4 25

f=± 2

f = ê 5

Each of the equations in Worked example 10 was a one-step equation. Remember that in two-step equations, the reverse order of operations must be applied; that is, address addition and subtraction first, then multiplication and division, then exponents and roots and, lastly, any bracketed numbers.

Worked Example 11

Solve the following. a  5y - 6 = 79

b 

Think a

b

Write

1

Write the equation.

2

Add 6 to both sides of the equation.

3

Divide both sides of the equation by 5 to obtain y.

1

Write the equation.

2

Multiply both sides of the equation by 9.

3

Divide both sides of the equation by 4 to obtain x.

5y - 6 = 79

a

5y - 6 + 6 = 79 + 6 5y = 85

b



Express the improper fraction as a mixed number fraction.

5 y 85 = 5 5 y = 17 4x =5 9

4x ì9=5ì9 9 4x = 45

4

4x =5 9



4 x 45 = 4 4 45 x= 4 1

x = 11 4

Equations where the pronumeral appears on both sides ■■

In solving equations where the pronumeral appears on both sides add or subtract one of the pronumeral terms so that it is eliminated from one side of the equation. Chapter 2 Linear algebra

41

number AND algebra • Patterns and algebra

Worked Example 12

Solve the following equations. a  5h + 13 = 2h - 2 c  2(x - 3) = 5(2x + 4)

b  14 - 4 d = 27 - d

Think a

b

c

42

Write a 5h + 13 = 2h - 2

1

Write the equation.

2

Eliminate the pronumeral from the right-hand side by subtracting 2h from both sides of the equation. Note that it is also possible to instead subtract 5h from both sides, leaving -3h on the right-hand side. However, it simpler to work with positive pronumerals.

3

Subtract 13 from both sides of the equation.

3h = -15

4

Divide both sides of the equation by 3 and write your answer.

h = -5

1

Write the equation.

2

Create a single pronumeral term by adding 4d to both sides of the equation.

3

Subtract 27 from both sides of the equation.

-13 = 3d

4

Divide both sides of the equation by 3.

−

5

Express the improper fraction as a mixed number fraction.

-4 3 = d

6

Write your answer so that d is on the left-hand side.

1

Write the equation.

2

Expand the brackets on both sides of the equation.

2x - 6 = 10x + 20

3

Isolate the pronumeral on the righthand side by subtracting 2x from both sides of the equation.

2x - 2x - 6 = 10x - 2x + 20

4

Subtract 20 from both sides of the equation.

5

Divide both sides of the equation by 8.

6

Simplify and write your answer with the pronumeral on the left.

Maths Quest 10 for the Australian Curriculum

3h + 13 = -2

b 14 - 4d = 27 - d

14 = 27 + 3d

13 =d 3 1

1

d = -4 3 c

2(x - 3) = 5(2x + 4)

-6 - 20 = 8x + 20 - 20 -26 = 8x −

26 =x 8 x=−

13 4

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1. Equations are algebraic sentences that can be solved to give a numerical solution. 2. Equations are solved by undoing any operation that has been performed on the pronumeral. 3. When solving two-step equations, the reverse order of operations must be applied. exerCise

2D inDiViDuAl PAthWAys eBook plus

Activity 2-D-1

Simple puzzling equations doc-4969 Activity 2-D-2

Puzzling equations doc-4970 Activity 2-D-3

Advanced puzzling equations doc-4971

solving linear equations fluenCy 1 We10a Solve the following equations. a a + 61 = 85 a = 24 b k - 75 = 46 k = 121 d r - 2.3 = 0.7 r = 3 e h + 0.84 = 1.1 h = 0.26 1 1 1 g t - 12 = -7 t = 5 h q + = q = 6 3

2

c g + 9.3 = 12.2 g = 2.9 f i + 5 = 3 i = -2 i

x - 2 = -2 x = 0

2 We10b Solve the following equations.

f =3 4

f = 12

b

d 9v = 63

v=7

e 6w = -32 w = -5 13

a

i = -6 i = -60 10

m 7 5 = m = 16 8 19 8 3 We10c, d Solve the following equations. g 4a = 1.7 a = 0.425

a

t = 10

t = 100

d f 2 = 1.44 f = ê1.2 g

15 225 g = 22 g = 484

4 We11a Solve the following. a 5a + 6 = 26 a = 4 d 7f - 18 = 45 f = 9 s = 4 65 g 6s + 46 = 75 5 Solve the following.

h

b y2 = 289 e

h j2 =

f + 6 = 16 f = 40 4 r c + 6 = 5 r = -10 10 n e + 5 = 8.5 n = 28 8

h=

16 49

j = ê14 31

k 5 = k = 10 12 6 y 3 i = 5 8 y = 21 12 4 f

c

q = 2.5 q = 6.25

f p2 = i

9 64

p = ê 83

7

a2 = 2 9 a = ê1 23

c 8i - 9 = 15 i = 3 r = 5 25 f 10r - 21 = 33 i 8a + 88 = 28 a = -7 12

g + 4 = 9 g = 30 6 m d - 12 = -10 m = 18 9 p f - 1.8 = 3.4 p = 62.4 12 b

6 Solve the following. 1 a 6(x + 8) = 56 x = 1 3 c 5(m - 3) = 7 m = 4 25 2 e 5(3n - 1) = 80 n = 5 3 7 We11b Solve the following.

3k = 15 k = 25 5 8u d = -3 u = -418 11

196 961

y = ê17

b 6b + 8 = 44 b = 6 q = 118 e 8q + 17 = 26 h 5t - 28 = 21 t = 9 45

a

a

h=

4 7

c 6z = -42 z = -7

b 7( y - 4) = 35 y = 9 k = 1 12 d 3(2k + 5) = 24 f 6(2c + 7) = 58 c = 1 13

9m = 18 m = 16 8 11x e = 2 x = 118 4 b

7p = -8 p = -11 37 10 4v f = 0.8 v = 3 15 c

Chapter 2 linear algebra

43

number AND algebra • Patterns and algebra

p + 2 = 7 is: 5 ✔ B p = 25 A p = 5 D p = 10 E p = 1 b If 5h + 8 = 53, then h is equal to: 1 A 5 B 12.2

8   MC  a  The solution to the equation

C p = 45

C 225

✔ E 9 D 10 c The exact solution to the equation 14x = 75 is:

A x = 5.357 142 857

B x = 5.357 (to 3 decimal places)

D x = 5.4

E x = 5.5

9 Solve the following equations. a -x = 5 x = -5 d -7 - x = 4 x = -11

v 5 10 Solve the following equations. a 6 - 2x = 8 x = -1 g − = 4 v = -20

d -3 - 2g = 1 g = -2

g = -1 13

12

13

14

c 5 - p = -2 p = 7 f -6t = -30 t = 5

h −

i

r 1 = r = -3 12 4

-4g = 3.2

g = -0.8

b 10 - 3v = 7 v = 1

c 9 - 6l = -3 l = 2

e -5 - 4t = -17 t = 3

f −

3e = 14 e = -23 13 5 4f i − + 1 = 8 f = -12 14 7

8j k = 9 j = -3 83 h − - 3 = 6 k = -36 3 4   WE 12a  Solve the following equations. a 6x + 5 = 5x + 7 x = 2 b 7b + 9 = 6b + 14 b = 5 c 11w + 17 = 6w + 27 w = 2 d 8f - 2 = 7f + 5 f = 7 e 10t - 11 = 5t + 4 t = 3 f 12r - 16 = 3r + 5 r = 2 13 1 g 12g - 19 = 3g - 31 h 7h + 5 = 2h - 6 h = -2 5 i 5a - 2 = 3a - 2 a = 0   WE 12b  Solve the following equations. 2 a 5 - 2x = 6 - x x = -1 b 10 - 3c = 8 - 2c c = 2 c 3r + 13 = 9r - 3 r = 2 3 1 d k - 5 = 2k - 6 k = 1 e 5y + 8 = 13y + 17 y = -18 f 17 - 3g = 3 - g g = 7 g 14 - 5w = w + 8 w = 1 h 4m + 7 = 8 - m m = 15 i 14 - 5p = 9 - 2p p = 1 23   WE 12c  Solve the following equations. 4 a 3(x + 5) = 2x x = -15 b 8( y + 3) = 3y y = -4 5 3) u = -2 75 c 6(t - 5) = 4(t + 3) t = 21 d 10(u + 1) = 3(u 1 e 12( f - 10) = 4( f - 5) f = 12 2 f 2(4r + 3) = 3(2r + 7) r = 7 12 g 5(2d + 9) = 3(3d + 13) d = -6 h 5(h - 3) = 3(2h - 1) h = -12 i 2(4x + 1) = 5(3 - x) x = 1   MC  a  The solution to 8 - 4k = -2 is: 1 1 1 B k = -2 2 C k = 1 2 ✔ A k = 2 2 1

D k = -1 2 b The solution to − 1

A n = 3 3 ✔ D

E k =

2 5

6n + 3 = -7 is: 5

1

n = 83

1

B n = -3 3

A p =

2 5

2

E p =

4 5

Maths Quest 10 for the Australian Curriculum

1 3

1

✔ B

C p = 4 3

C n =

E n = -8 3

c The solution to p - 6 = 8 - 4p is:

44

5

x = 514

b 2 - d = 3 d = -1 e -5h = 10 h = -2

g − 11

✔ C

4

p = 25

D p =

2 3

reflection 

 

Describe in one sentence what it means to solve linear equations.

number AnD AlgebrA • PAtterns AnD AlgebrA

solving equations with algebraic fractions and multiple brackets

2e eBook plus

Interactivity Solving euqations

int-2778

equations with multiple brackets Many equations need to be simplified by expanding brackets and collecting like terms before they are solved. Doing this reduces the equation to one of the basic types covered in the previous exercise.

WorkeD exAmPle 13

Solve each of the following linear equations. a 6(x + 1) - 4(x - 2) = 0 b 7(5 - x) = 3(x + 1) - 10 think a

b

Write a 6(x + 1) - 4(x - 2) = 0

1

Write the equation.

2

Expand all the brackets. (Be careful with the -4.)

3

Collect like terms.

4

Subtract 14 from both sides of the equation.

5

Divide both sides of the equation by 2 to find the value of x.

1

Write the equation.

2

Expand all the brackets.

35 - 7x = 3x + 3 - 10

3

Collect like terms.

35 - 7x = 3x - 7

4

Create a single pronumeral term by adding 7x to both sides of the equation.

35 = 10x - 7

5

Add 7 to both sides of the equation.

42 = 10x

6

Divide both sides of the equation by 10 to solve for x and simplify.

42 10

=x

21 5

=x

7

Express the improper fraction as a mixed number fraction.

45 = x

8

Rewrite the equation so that x is on the left-hand side.

6x + 6 - 4x + 8 = 0 2x + 14 = 0 2x = -14 x = -7 b 7(5 - x) = 3(x + 1) - 10

1

1

x = 45

equations involving algebraic fractions ■

■

To solve equations involving algebraic fractions, write every term in the equation as a fraction with the same lowest common denominator. Every term can then be multiplied by this common denominator. This has the effect of eliminating the fraction from the equation. Chapter 2 linear algebra

45

number AND algebra • Patterns and algebra

Worked Example 14

Solve each of the following linear equations. x 3x 1 a  − = 2 5 4 3 4 b  =1− 2x x Think a

b

46

Write

1

Write the equation.

2

The lowest common denominator of 2, 5, and 4 is 20. Write each term as an equivalent fraction with a denominator of 20.

3

Multiply both sides of the equation by 20. This is the same as multiplying each term by 20, which cancels out the 20 in the denominator and effectively removes it.

x 3x 1 − = 2 5 4

a

x 10 3 x 4 × − × = 2 10 5 4 10 x 12 x = − 20 20

1 5 × 4 5 5 20

 10 x 12 x  5  20 − 20  ì 20 = 20 ì 20 10 x 12 x 5 ì 20 ì 20 = ì 20 20 20 20 10x - 12x = 5 -2x = 5

4

Simplify the left-hand side of the equation by collecting like terms.

5

Divide both sides of the equation by -2 to solve for x.

x = - 5

6

Express the improper fraction as a mixed number fraction.

x = -2 12

1

Write the equation.

2

The lowest common denominator of 2x and x is 2x. Write each term as an equivalent fraction with a denominator of 2x.

3

Multiply each term by 2x. This effectively removes the denominator.

3 = 2x - 8

4

Add 8 to both sides of the equation.

11 = 2x

5

Divide both sides of the equation by 2 to solve for x.



6

Express the improper fraction as a mixed number.

7

Rewrite the equation so that x is on the lefthand side.

Maths Quest 10 for the Australian Curriculum

2

b



3 4 =1x 2x 3 1 2x 4 2 = × − × 2x 1 2 x x 2 3 2x 8 = − 2x 2 x 2 x

11 2

=x

1

52 = x 1

x = 52

number AND algebra • Patterns and algebra

Worked Example 15

Solve each of the following linear equations. 4 1 5( x + 3) 3( x − 1) = a  b  =4+ 3 ( x − 1 ) x +1 6 5 Think a

b

Write

5( x + 3) 3( x − 1) =4+ 6 5

a

1

Write the equation.

2

The lowest common denominator of 5 and 6 is 30. Write each term as an equivalent fraction with a common denominator of 30.

3

Multiply each term by 30. This effectively removes the denominator.

25(x + 3) = 120 + 18(x - 1)

4

Expand the brackets and collect like terms.

25x + 75 = 120 + 18x - 18 25x + 75 = 102 + 18x

5

Subtract 18x from both sides of the equation.

6

Subtract 75 from both sides of the equation.

7

Divide both sides of the equation by 7 to solve for x.

x=

8

Express the improper fraction as a mixed number.

x=

1

Write the equation.

2

The lowest common denominator of 3, x + 1 and x - 1 is 3(x - 1)(x + 1). Write each term as an equivalent fraction with a common denominator of 3(x - 1)(x + 1).

3

Multiply each term by the common denominator.



4(x + 1) = 3(x - 1)

4

Expand the brackets.



4x + 4 = 3x - 3

5

Subtract 3x from both sides of the equation.

6

Subtract 4 from both sides of the equation to solve for x.

25( x + 3) 120 18( x − 1) = + 30 30 30

7x + 75 = 102

7x = 27 27 7 6 37

4 1 = 3( x − 1) x + 1

b

4( x + 1) 3( x − 1) = 3( x − 1)( x + 1) 3( x − 1)( x + 1)

x + 4 = -3

x + 4 - 4 = -3 - 4 x = -7

remember

1. For equations involving brackets, expand the brackets and collect like terms. This will reduce the equation to a more basic type. 2. For complicated algebraic fraction equations, the following steps may be used. (a) Write each term in the equation as an equivalent fraction with the lowest common denominator. (b) Multiply each term by the common denominator. This has the effect of removing the fraction from the equation. (c) Continue to solve the equation using the same methods as for a basic algebraic equation. Chapter 2 Linear algebra

47

number AnD AlgebrA • PAtterns AnD AlgebrA

exerCise

2e inDiViDuAl PAthWAys eBook plus

Activity 2-E-1

Algebraic equations with fractions doc-4972 Activity 2-E-2

Harder algebraic equations with fractions doc-4973 Activity 2-E-3

Tridy algebraic equations with fractions doc-4974

solving equations with algebraic fractions and multiple brackets fluenCy 1 We13 Solve each of the following linear equations. a 6(4x - 3) + 7(x + 1) = 9 x = 20 31 x = 35 b 9(3 - 2x) + 2(5x + 1) = 0 8 c 8(5 - 3x) - 4(2 + 3x) = 3 x = 29 36 d 9(1 + x) - 8(x + 2) = 2x x = -7 8 e 6(4 + 3x) = 7(x - 1) + 1 x = -211 f 10(4x + 2) = 3(8 - x) + 6 x = 10 43 g 8(x + 4) + 2(x - 3) = 6(x + 1) x = -5 h 7.2(3x - 1) + 2.3(5 - x) = -34.3 x = -2 i 6(2x - 3) - 2(6 - 3x) = 7(2x - 1) x = 5 43 11 j 9(2x - 5) + 5(6x + 1) = 100 x = 212 k 5(2x - 1) - 3(6x + 1) = 8 x = -2 l 7(2x + 7) - 5(2x + 1) = 2(4 - x) x = -6 2 Solve each of the following linear equations.

7x + 5 12 − 5 x = 11 x = 4 b = -13 x = 18 3 6 3x − 2 8x + 3 c = 5x x = -172 d = 2x x = 32 5 4 2x − 1 x − 3 4x + 1 x + 2 x= 2 e f = x = -113 or x = -3 23 = 13 5 4 3 4 6 − x 2x − 1 8 − x 2x + 1 x=3 x= 5 g h = = 7 3 5 9 3 5x − 3 1 i - = 0 x = 13 20 2 8 3 We14 Solve each of the following linear equations. x 4x 1 x x 3 x= 5 a b − = x = 15 + = 17 3 5 3 4 5 4 x 4x −3 x x 1 10 c d − = 2 x = -6 29 + = x = -19 4 7 5 8 4 2 x x −3 5x 2x x = -192 e - = x = -1 12 f -8= 3 6 4 8 3 2 x 3x 4 1 2 x= 4 g h − = − = x = 12 7 x 6 x 7 8 8 15 2 1 4 5 i j + = x = 3 - 4 = x = 3 14 x x 3 x x 2x − 4 x 4 x − 1 2x + 5 k l + 6 = x = 52 = 0 x = 1 85 5 2 2 3 4 We15 Solve each of the following linear equations. 3( x + 1) 5( x + 1) 2( x + 1) 3(2 x − 5) 31 a + = 4 x = 195 b + = 0 x = 1 58 2 3 7 8 2(4 x + 3) 6( x − 2) 1 8( x + 3) 3( x + 2) x = -315 11 c d = x = 414 = 17 5 2 2 5 4 a

48

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

e g eBook plus

Digital doc WorkSHEET 2.2 doc-5194

i k

5(7 − x ) 2(2 x − 1) = + 1 x = 5 20 43 2 7 −5( x − 2) 6(2 x − 1) 1 x = 1 2 = 61 3 5 3 1 3 8 + = x = 1.5 x −1 x +1 x +1 1 3 −1 x=3 - = x − 1 x x −1

f h j l

2(6 − x ) 9( x + 5) 1 = + x = -110 13 3 6 3 9(2 x − 1) 4( x − 5) x = -4 9 = 26 7 3 3 5 5 x = -4 1 + = 3 x +1 x − 4 x +1 4 5 −1 - = x=1 x 2x − 1 x

refleCtion 

 

Do the rules for the order of operations apply to algebraic functions? Explain.

Chapter 2 linear algebra

49

number AND algebra • Patterns and algebra

Summary Substitution ■■ ■■ ■■

When the numerical values of pronumerals are known, they can be substituted them into an algebraic expression and evaluated. It is sometimes useful to place any substituted values in brackets when evaluating an expression. When dealing with numbers and pronumerals, particular rules must be obeyed. (a) The Commutative Law holds true for addition and multiplication. (b) The Associative Law holds true for addition and multiplication. (c) The Identity Law states that, in general: x + 0 = x and x ì 1 = x. 1 (d) The Inverse Law states that, in general: x + -x = 0 and x ì = 1. x (e) The Closure Law states that, when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. Adding and subtracting algebraic fractions

■■ ■■

Algebraic fractions contain pronumerals that may represent particular numbers or changing values. To add or subtract algebraic fractions we perform the following steps. (a) Find the lowest common denominator (LCD) by finding the lowest common multiple (LCM) of the denominators. (b) Rewrite each fraction as an equivalent fraction with this common denominator. (c) Express as a single fraction. (d) Simplify the numerator. Multiplying and dividing algebraic fractions

■■

■■

When multiplying algebraic fractions, first cancel any common factors if possible, then multiply the numerators together and finally multiply the denominators together. Simplify the expression further if necessary. When dividing algebraic fractions, change the division sign to a multiplication sign and write the following fraction as its reciprocal (swap the numerator and the denominator). The process then follows that for multiplication. Solving linear equations

■■ ■■ ■■

Equations are algebraic sentences that can be solved to give a numerical solution. Equations are solved by undoing any operation that has been performed on the pronumeral. When solving two-step equations, the reverse order of operations must be applied. Solving equations with algebraic fractions and multiple brackets

■■ ■■

50

For more complicated equations involving brackets, expand the brackets and collect like terms. This will reduce the equation to a more basic type. For complicated algebraic fraction equations, the following steps may be used. (a) Write each term in the equation as an equivalent fraction with the lowest common denominator. (b) Multiply each term by the common denominator. This has the effect of removing the fraction from the equation. (c) Continue to solve the equation using the same methods as for a basic algebraic equation.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

MAPPING YOUR UNDERSTANDING

Homework Book

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 27. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

Chapter 2 linear algebra

51

number AND algebra • Patterns and algebra

Chapter review Fluency

10 Simplify the following. 5y y 7 y a − 3 2 6

1

1 Given E = 2 mv2 where m = 0.2 and v = 0.5, the

value of E is: A 0.000  625 C 0.005 E 0.0025

5 1 22 − 3 x 5 x 15x 11 Simplify the following. y 32 8y a × x 4 x 20 y 35z 25z b × 7 x 16 y 4 x x+6 5( x + 1) 5 × c x+3 ( x + 1)( x + 3) x+6

B 0.1 ✔ D 0.025

c

2 The expression -6d + 3r - 4d - r simplifies to: A 2d + 2r ✔ B -10d + 2r C -10d - 4r D 2d + 4r E -8dr 3 The expression 5(2f + 3) + 6(4f - 7) simplifies to: A 34f + 2 B 34f - 4 D 34f + 14 ✔ C 34f - 27 E 116f -14

12 Solve the following equations. a p - 20 = 68 p = 88 b s - 0.56 = 2.45 s = 3.01

6 Simplify the following by collecting like terms. a 3c - 5 + 4c - 8 7c - 13 b -3k + 12m - 4k - 9m -7k + 3m c -d + 3c - 8c - 4d -5d - 5c d 6y2 + 2y + y2 - 7y 7y2 - 5y 1

1 1 = × 7p = 1 3p ó (5q ó 7r) 7 p 7 p

8 For each of the following, complete the relationship

Associative Law Commutative Law Inverse Law Associative Law Identity Law Associative Law Commutative Law Commutative Law

d

x = 12 x = 144 y g - 3 = 12 y = 60 4 i 5 - k = -7 k = 12

h a2 = 36 a = ê6

13 Solve the following. a 42 - 7b = 14 b = 4 b 12t - 11 = 4t + 5 t = 2 c 2(4p - 3) = 2(3p - 5) p = -2 14 Solve each of the following linear equations. 1 a 5(x - 2) + 3(x + 2) = 0 x = 2 1 b 7(5 - 2x) - 3(1 - 3x) = 1 x = 6 5 3 c 5(x + 1) - 6(2x - 1) = 7(x + 2) x = - 14 d 8(3x - 2) + (4x - 5) = 7x x = 1 2 e 7(2x - 5) - 4(x + 20) = x - 5 x = 12 9 1 x = 16 f 3(x + 1) + 6(x + 5) = 3x + 40 15 Solve each of the following equations.

x x 3 x x 6 + = x=7 b − = 3 x = 22 12 2 5 5 3 5 1 x x 3 2 5 = − x=2 + = x = 5 c − d 21 7 6 x 5 x 2x − 3 3 x + 3 3 x = 38 e − = 2 5 5 2( x + 2) 3 5( x + 1) x = -16 = + f 21 3 7 3 a

the natural number values x = 12, y = 8 and z = 4. Comment on whether the Closure Law holds for each of the expressions when the values are substituted. 96 — in this case, multiplication is closed on natural numbers. a x ì y b z ó x c y - x

Maths Quest 10 for the Australian Curriculum

r = -5 r = -35 7 f 2(x + 5) = -3 x = -132

c 3b = 48 b = 16 e

7 If A = 2 bh, determine the value of A if b = 10 and 35 h = 7.

52

3 x 2 + 2 x − 17 ( x + 3)( x + 2)

25 30 5 ÷ x x 6 xy 10 x y 2 e ÷ y 50 5 2x 9x +1 2x ÷ f ( x − 1 )( 9 x + 1) ( x + 8)( x − 1) x + 8

5 If 14p - 23 = 6p - 7 then p equals: A -3 b -1 ✔ d 2 c 1 e 4

9 Find the value of the following expressions given



d

4 The expression 7(b - 1) - (8 - b) simplifies to: A 8b - 9 ✔ B 8b - 15 C 6b - 9 D 6b - 15 E 8b + 1

to illustrate the stated law. a + (3b + 6c) a (a + 3b) + 6c = _______ 3b - 12a b 12a - 3b ò _______ c 7p ì _______ = _______ x ì (5y ì 7z) d (x ì 5y) ì 7z = _______ 0 + 12p = 12p e 12p + 0 = _______ f (3p ó 5q) ó 7r ò _______ 11e + 9d g 9d + 11e = _______ b ó 4a h 4a ó b ò _______

x + 4 x + 2 7 x + 18 + 10 5 2 x − 1 2x − 5 + d x+3 x+2 b

9 b 1 — in this case, division is not closed on natural numbers. 3 c  4 — in this case, subtraction is not closed on natural numbers.

number AnD AlgebrA • PAtterns AnD AlgebrA Problem solVing

2 You are investigating prices for having business

1 A production is in town and many parents are

cards printed for your new games store. A local printing company charges a flat rate of $250 for the materials used and $40 per hour for labour. a If h is the number of hours of labour required to print the cards, construct an equation for the cost of the cards, C. C = 250 + 40h b You have budgeted $1000 for the printing job. How many hours of labour can you afford? Give your answer to the nearest minute. c The printer estimates that it can print 1000 cards per hour of labour. How many cards will 18750 be printed with your current budget? d An alternative to printing is photocopying. The company charges 15 cents per side for the first 10 000 cards and then 10 cents per side for the remaining cards. Which is the cheaper option for 18 750 single-sided cards and by how Printing is the cheaper option by $1375. much?

18 hours 45 minutes

taking their children. An adult ticket costs $15 and a child’s ticket costs $8. Every child must be accompanied by an adult and each adult can have no more than 4 children with them. It costs the company $12 per adult and $3 per child to run the production. There is a seating limit of 300 people and all tickets are sold. a Determine how much profit the company makes on each adult ticket and on each child’s ticket. $3 per adult ticket; $5 per child’s ticket b To maximise profit, the company should sell as many children’s tickets as possible. Of the 300 available seats, determine how many should be allocated to children if there is a maximum of 4 children per adult. 240 c Using your answer to part b, determine how many adults would make up the remaining seats. 60 d Construct an equation to represent the profit that the company can make depending on the number of children and adults attending the production. P = 3a + 5c, where a = number of adults and c = number of children e Substitute your values to calculate the maximum profit the company can make. $1380

eBook plus

Interactivities

Test yourself Chapter 2 int-2831 Word search Chapter 2 int-2829 Crossword Chapter 2 int-2830

Chapter 2 linear algebra

53

eBook plus

ACtiVities

Chapter opener Digital doc

• Hungry brain activity Chapter 2 (doc-5182) (page 27) Are you ready?

(page 28) • SkillSHEET 2.1 (doc-5183): Like terms • SkillSHEET 2.2 (doc-5184): Collecting like terms • SkillSHEET 2.3 (doc-5185): Finding the highest common denominator • SkillSHEET 2.4 (doc-5186): Addition and subtraction of fractions • SkillSHEET 2.5 (doc-5187): Multiplication of fractions • SkillSHEET 2.6 (doc-5188): Division of fractions Digital docs

2A Substitution Digital docs

• Activity 2-A-1 (doc-4960): Substitution (page 31) • Activity 2-A-2 (doc-4961): Harder substitution (page 31) • Activity 2-A-3 (doc-4962): Tricky substitution (page 31) • SkillSHEET 2.7 (doc-5189): Order of operations (page 32) 2B Adding and subtracting algebraic fractions

(page 36) • Activity 2-B-1 (doc-4963): Introducing algebraic fractions • Activity 2-B-2 (doc-4964): Working with algebraic fractions • Activity 2-B-3 (doc-4965): Advanced algebraic fractions • SkillSHEET 2.4 (doc-5186): Addition and subtraction of fractions • SkillSHEET 2.8 (doc-5190): Writing equivalent algebraic fractions with the lowest common denominator Digital docs

2C Multiplying and dividing algebraic fractions Digital docs

• Activity 2-C-1 (doc-4966): Learning operations with algebraic fractions (page 39) • Activity 2-C-2 (doc-4967): Operations with algebraic fractions (page 39)

54

maths Quest 10 for the Australian Curriculum

• Activity 2-C-3 (doc-4968): Advanced operations with algebraic fractions (page 39) • SkillSHEET 2.5 (doc-5187): Multiplication of fractions • SkillSHEET 2.9 (doc-5191): Simplification of algebraic fractions (page 39) • SkillSHEET 2.10 (doc-5192): Division of fractions (page 39) • WorkSHEET 2.1 (doc-5193): Algebraic fractions (page 40) 2D Solving linear equations

(page 43) • Activity 2-D-1 (doc-4969): Simple puzzling equations • Activity 2-D-2 (doc-4970): Puzzling equations • Activity 2-D-3 (doc-4971): Advanced puzzling equations Digital docs

2E Solving equations with algebraic fractions and multiple brackets Digital docs

• Activity 2-E-1 (doc-4972): Algebraic equations with fractions (page 48) • Activity 2-E-2 (doc-4973): Harder algebraic equations with fractions (page 48) • Activity 2-E-3 (doc-4974): Tricky algebraic equations with fractions (page 48) • WorkSHEET 2.2 (doc-5194): Solving equations with fractions (page 49) Interactivity

• Solving equations (int-2778) (page 45) Chapter review

(page 53) • Test Yourself Chapter 2 (int-2831): Take the end-ofchapter test to test your progress. • Word search Chapter 2 (int-2829): an interactive word search involving words associated with this chapter • Crossword Chapter 2 (int-2830): an interactive crossword using the definitions associated with the chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

  3a  Sketching linear graphs   3B  Determining linear equations   3C  The distance between two points on a straight line   3d  The midpoint of a line segment   3E  Parallel and perpendicular lines WhAt Do you knoW ?

Coordinate geometry

1 List what you know about linear graphs and their equations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of linear graphs and their equations. eBook plus

Digital doc

Hungry brain activity Chapter 3 doc-5195

opening Question

How can a knowledge of coordinate geometry help us design structures like this one?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■click■on■the■SkillSHEET■icon■next■to■the■question■ on■your■eBookPLUS■or■ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 3.1 doc-5196

Measuring the rise and run   1  State■the■rise■and■the■run■for■each■of■the■following■straight■line■graphs. y y Rise■=■6,■run■=■2 a  b  4 2 0 -2

eBook plus

Digital doc

Rise■=■-2,■run■=■5

2 2

x

0

x

5

Describing the gradient of a line   2  State■whether■each■line■in■question■1■has■a■positive,■zero,■negative■or■undefi■ned■gradient. a■ Positive     b  Negative

SkillSHEET 3.2 doc-5197

eBook plus

Digital doc

SkillSHEET 3.3 doc-5198

eBook plus

Digital doc

Plotting a line using a table of values   3  Draw■up■a■table■of■values■and■plot■the■graph■for■each■of■the■following■rules. a  y■=■x■+■3 b  y■=■x■-■2 c  y■=■2x   3  a■ y■=■x■+■3 Stating the y-intercept from a graph   4  State■the■y-intercept■for■each■graph■shown■in■question■1.

x -2 -1 0 1 2 y ■1 ■2 3 4 5

a■ -2     b  2

y 6

SkillSHEET 3.4 doc-5199

y=x+3

4 2

eBook plus

Digital doc

SkillSHEET 3.5 doc-5200

eBook plus

Digital doc

SkillSHEET 3.6 doc-5201

Solving linear equations that arise when finding x- and y-intercepts -4 -2 0 2 4x -2   5  Consider■the■equation■3y■+■4x■=■12. a  Substitute■x■=■0■and■solve■to■fi■nd■the■value■of■y. ■ y■=■4     b  y■=■x■-■2     x■=■3 b  Substitute■y■=■0■and■solve■to■fi■nd■the■value■of■x. x -2 -1 ■ 0 ■ 1 2 Using Pythagoras’ theorem   6  Find■the■length■of■side■AB. a  B

    a■ 10■m■

y -4 -3 -2 -1 0

b■ 5■cm

y 4

b  B

2

13 cm

-4

8m A

6m

C

A

0 -2

2

4x

-4

C

12 cm

-2

y=x-2

    c  y■=■2x x

-2

-1

0

1

2

y

-4

-2

0

2

4

y 4

-4

-2

0 -2 -4

56

maths Quest 10 for the Australian Curriculum

y = 2x

2 2

4x

number AND algebra • Linear and non-linear relationships

3A

Sketching linear graphs ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

A linear graph has an equation that can be written in the standard form y = mx + c, where m is the gradient of the line, and c is the y-intercept. An alternative form of the linear equation is ax + by = k, where a, b and k are constants. The gradient (or slope) is a measure of the steepness of a graph. If the gradient, m, of the line is positive, the graph will have an upward slope to the right. If the gradient, m, of the line is negative, the graph will have a downward slope to the right. The greater the magnitude of the gradient, the steeper the linear graph will be. A linear graph is drawn on a Cartesian plane, with two axes (x and y) meeting at the origin (0, 0). The axes divide the plane into four regions, or quadrants. A point is specified by its x- and y-coordinates. A graph of the line y = 2x + 5 is shown in the figure at right. y y = 2x + 5 One method which can be used to draw a graph of an equation 10 is to simply plot the points on graph paper. Quadrant 2 Quadrant 1 5 The points can be plotted manually, or a graphing calculator can be used to plot the points. -10

0

-5

Quadrant 3

10 x

5

-5

Quadrant 4

-10

Worked Example 1

Plot the linear graph defined by the rule y = 2 x - 5 for the x-values -3, -2, -1, 0, 1, 2, 3. Think 1

2

3

Create a table of values using the given x-values. Find the corresponding y-values by substituting each x-value into the rule. Plot the points on a Cartesian plane and rule a straight line through them. Since the x-values have been specified, the line should only be drawn between the x-values of -3 and 3.

Write

x

-3

-2

-1

0

1

2

y

3  

x

-3

-2

-1

 0

 1

 2

  3 

y

-11

-9

-7

-5

-3

-1

1

y 2 1 0 -3 -2 -1 -1

(3, 1) 1 2 3x (2, -1)

-2 -3 (1, -3) -4 -5 (0, -5) -6 (-1, -7) -7 -8 (-2, -9) -9 -10 y = 2x - 5 (-3, -11) -11 -12 4

Label the graph.

Chapter 3 Coordinate geometry

57

number AND algebra • Linear and non-linear relationships

■■ ■■

A minimum of two points are necessary to plot a straight line. Two methods can be used to plot a straight line: •• Method 1: The x- and y-intercept method. •• Method 2: The gradient–intercept method.

Sketching a straight line using the x- and y-intercept method ■■

■■

As the name implies, this method involves finding the x- and y-intercepts, then joining them to sketch the straight line. If the equation is in the form y = mx + c, the value of c gives the y-intercept.

Worked Example 2

Sketch graphs of the following linear equations by finding the x- and y-intercepts. a  2x + y = 6 b  y = -3x - 12 Think a

Write/draw a 2x + y = 6

1

Write the equation.

2

Find the x-intercept by substituting y = 0.

x-intercept: when y = 0, 2x + 0 = 6 2x = 6 x=3 x-intercept is (3, 0).

3

Find the y-intercept by substituting x = 0.

y-intercept: when x = 0, 2(0) + y = 6 y=6 y-intercept is (0, 6).

4

Rule a straight line passing through both points that have been plotted.

y 2x + y = 6 (0, 6) 0 (3, 0)

b

58

x

5

Label the graph.

1

Write the equation.

2

Find the x-intercept by substituting y = 0.   i  Add 12 to both sides of the equation. ii Divide both sides of the equation by -3.

x-intercept: when y = 0, -3x - 12 = 0 -3x = 12 x = -4 x-intercept is (-4, 0).

3

Find the y-intercept. The equation is in the form y = mx + c, so compare this with our equation to find the y-intercept, c.

c = -12 y-intercept is (0, -12).

Maths Quest 10 for the Australian Curriculum

b y = -3x - 12

number AND algebra • Linear and non-linear relationships

4

y

Rule a straight line passing through both points that have been plotted. (-4, 0)

x

0

(0, -12) y = -3x - 12

5

Label the graph.

The gradient–intercept method ■■

■■

This method is often used if the equation is in the form y = mx + c, where m represents the gradient (slope) of the straight line, and c represents the y-intercept. The steps below outline how to use the gradient–intercept method to sketch a linear graph. Step 1:  Plot a point at the y-intercept. rise . (To write a whole number as a fraction, place Step 2: Write the gradient in the form m = run it over a denominator of 1.) Step 3: Starting from the y-intercept, move up the number of units suggested by the rise (move down if the gradient is negative). Step 4: Move to the right the number of units suggested by the run and plot the ■ second point. Step 5:  Rule a straight line through the two points.

Worked Example 3 2

Plot the graph of y = 5 x - 3 using the gradient-intercept method. Think

Write/DRAW 2

1

Write the equation of the line.

y = 5x - 3

2

Identify the value of c (that is, the y-intercept) and plot this point.

c = -3, so y-intercept: (0, -3).

3

Write the gradient, m, as a fraction. (In this case, it is a fraction already.)

m=

4

5

rise , interpret the numerator of the run fraction as the rise and the denominator as ■ the run. Since m =

Starting from the y-intercept at -3, move 2 units up and 5 units to the right to find the second point. We have still not found the x-intercept.

2 5

So, rise = 2; run = 5.

y x 0 -1 1 2 3 4 5 6 7 8 (5, -1) -2 -3 (0, -3) -4

Chapter 3 Coordinate geometry

59

number AND algebra • Linear and non-linear relationships

6

2

y = 5x - 3

To find the x-intercept, let y = 0.

2

0 = 5x - 3 2

3 = 5x 5

3ì2 =x x=

15 2

15

( 2 , 0) is the x-intercept. 7

Label the graph and draw a line through all the points found.

(152 , 0)

y

x 0 -1 1 2 3 4 5 6 7 8 (5, -1) -2 -3 (0, -3) y = 25 x - 3 -4

Sketching linear graphs of the form y = mx ■■ ■■ ■■

These graphs have the value of c = 0. They pass through the origin (0, 0), providing only one point to plot. This means that a second point must be determined by either: •• choosing any x-value, then calculating the corresponding y-value •• using the gradient–intercept method.

Worked Example 4

Sketch the graph of y = 3x. Think

Write/draw

1

Write the equation.

y = 3x

2

Find the x- and y-intercepts. Note: By recognising the form of this linear equation, y = mx you can simply state that the line passes through the origin, (0, 0).

x-intercept: when y = 0, 0 = 3x x=0 y-intercept: when x = 0, y = 0 Both the x- and y-intercepts are at (0, 0).

3

Find another point to plot by finding the y-value when x = 1.

When x = 1,  y = 3 ì 1         = 3 Another point on the line is (1, 3).

4

Plot the two points (0, 0) and (1, 3) and rule a straight line through them.

y

y = 3x

3 (0, 0)

5

60

Label the graph.

Maths Quest 10 for the Australian Curriculum

(1, 3) 1

x

number AND algebra • Linear and non-linear relationships

Sketching linear graphs of the form y = c and x = a ■■

■■ ■■

■■

It is possible to have an equation for a straight line that contains only an x-term or only a y-term. These equations can be written in the form y = c or x = a, where c and a are both constants. Lines of the form y = c are parallel to the x-axis, having a gradient of zero and a y-intercept of c. Lines of the form x = a are parallel to the y-axis, having an undefined (infinite) gradient and no y-intercept (since they do not cross the y-axis).

Worked Example 5

Sketch graphs of the following linear equations. a  y = -3 b  x = 4 Think a

1

Write the equation.

2

The y-intercept is -3. As x does not appear in the equation, the line is parallel to the x-axis, such that all points on the line have a y-coordinate equal to -3. That is, this line is the set of points (x, -3) where x is an element of the set of real numbers.

3

Sketch a horizontal line through (0, -3).

Write/draw a y = -3

y-intercept = -3

y

x

0

y = -3

(0, -3)

b

4

Label the graph.

1

Write the equation.

2

The x-intercept is 4. As y does not appear in the equation, the line is parallel to the y-axis, such that all points on the line have an x-coordinate equal to 4. That is, this line is the set of points (4, y) where y is an element of the set of real numbers.

3

Sketch a vertical line through (4, 0).

4

b x=4

x-intercept = 4

y

x=4

0

(4, 0)

x

Label the graph.

Chapter 3 Coordinate geometry

61

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

  1  a x -5 -4 -3 -2 -1 0 1

y -25 -15 -5 5 15 25 35

y 35 30 y = 10x + 25 25 20 15 10 5 -5 -4 -2 -1-5 -3 -10 -15 -20 -25

1 2

exerCise

3A inDiviDuAl pAthWAys eBook plus

Activity 3-A-1

Sketching linear graphs doc-4975 Activity 3-A-2

Graphs of linear equations doc-4976 Activity 3-A-3

More graphs of linear equations doc-4977

x

1.■ The■Cartesian■plane■is■a■grid,■consisting■of■two■axes■(x■and■y),■meeting■at■the■origin■ (0,■0). 2.■ A■location■(point)■is■specifi■ed■by■its■x-■and■y-coordinates. 3.■ A■linear■graph■consists■of■an■infi■nite■set■of■points■that■can■be■joined■to■form■a■straight■ line,■but■to■sketch■a■linear■graph,■the■coordinates■of■only■two■points■are■needed. 4.■ A■linear■rule■or■equation■can■be■used■to■obtain■the■coordinates■of■points■that■belong■to■ its■graph. 5.■ Linear■equations■may■be■written■in■several■different■forms.■The■two■most■common■ forms■are■y■=■mx■+■c■and■ax■+■by■=■k. 6.■ When■a■linear■equation■is■expressed■in■the■form■y■=■mx■+■c,■then■m■represents■the■ gradient■(slope)■of■the■straight■line■and■c■represents■the■y-intercept. 7.■ A■straight■line■with■a■positive■gradient■slopes■upward■to■the■right■and■a■straight■line■ with■a■negative■gradient■slopes■downward■to■the■right. 8.■ The■x-■and■y-intercept■method■allows■us■to■sketch■the■graph■of■any■linear■equation■by■ fi■nding■two■specifi■c■points:■the■x-intercept■and■y-intercept.■An■exception■is■the■case■of■ lines■passing■through■the■origin. 9.■ Graphs■of■equations■in■the■form■y■=■mx■pass■through■the■origin.■To■fi■nd■the■second■ point,■substitute■a■chosen■x-value■into■the■equation■to■fi■nd■the■corresponding■y-value. 10.■ Graphs■of■equations■in■the■form■y■=■c■have■a■gradient■of■zero■and■are■parallel■to■the■ x-axis. 11.■ Graphs■of■equations■in■the■form■x■=■a■have■an■undefi■ned■(infi■nite)■gradient■and■are■ parallel■to■the■y-axis.

sketching linear graphs FluenCy   1  We 1 ■Using■a■graphing■calculator,■generate■a■table■of■values■and■then■plot■the■linear■graphs■

defi■ned■by■the■following■rules■for■the■given■range■of x-values. y 20 Rule x-values 15 y = -3x + 2 10 -5,■-4,■-3,■-2,■-1,■0,■1 a  y■=■10x■+■25 5 -1,■0,■1,■2,■3,■4 b  y■=■5x■-■12 -10 -5 0 5 10 x -6,■-4,■-2,■0,■2,■4 c  y■=■-0.5x■+■10 -5 -10 ■ 0,■1,■2,■3,■4,■5 d  y■=■100x■-■240 -15 -3,■-2,■-1,■0,■1,■2 e  y■=■-5x■+■3 -20 -3,■-2,■-1,■0,■1,■2 f  y■=■7■-■4x   2  Plot■the■linear■graphs■defi■ned■by■the■following■rules■for■the■given■range■of x-values. Rule x-values a  y =■-3x■+■2 x -6 -4 -2 0 2 4 6 b  y■=■-x■+■3

eBook plus

Digital doc

SkillSHEET 3.7 doc-5202

62

c  y■=■-2x■+■3

maths Quest 10 for the Australian Curriculum

y

20

14

8

x

-3

-2

-1

y

6

5

x

-6

y

15

-4

-10

-16

0

1

2

3

4

3

2

1

0

-4

-2

0

2

4

6

11

7

3

-1

-5

-9

2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

eBook plus

Digital doc

SkillSHEET 3.5 doc-5200

  4  a y

y = 4x + 1 (1, 5)

5

1 01

x

  b  y

y = 3x - 7 01

x (1, -4)

4

  3  We2 ■Sketch■graphs■of■the■following■linear■equations■by■fi■nding■the■x-■and■y-intercepts. a  5x■-■3y■=■10 b  5x■+■3y■=■10   c  y y 5x - 3y = 10 y   3  a   b  4 4 4 c  -5x■+■3y■=■10 2 2 d  -5x■-■3y■=■10 2 e  2x■-■8y■=■20 -4 -2 0 -2 0 2 4x 2 4x -2 0 2 4x -2 -2 f  4x■+■4y■=■40 -2 -5x + 3y = 10 -4 g  -x■+■6y■=■120 5x + 3y = 10 h  -2x■+■8y■=■-20   f  y y y    d   e 4 i  10x +■30y■=■-150 10 4x + 4y = 40 5 2 j  5x■+■30y■=■-150 5 -10 -5 0 5 10 x k  -9x■+■4y■=■36 x 0 -5 2x - 8y = 20 -4 -2 2 4 -5 0 5 10 x -2 l  6x■-■4y■=■-24 -5 m  y■=■2x■-■10 -4 -5x - 3y = 10 n  y■=■-5x■+■20 1 o  y■=■− 2 x■-■4   4  We3 ■Sketch■graphs■of■the■following■linear■equations■using■the■gradient-intercept■method. a  y =■4x■+■1 b  y =■3x■-■7 c  y =■-2x■+■3 1 2

d  y =■-5x■-■4

e  y =■ x■-■2

g  y =■0.6x■+■0.5

h  y =■8x

f  y =■- 27 x■+■3 i 

  5  We4 ■Sketch■the■graphs■of■the■following■linear■equations. a  y =■2x b  y =■5x

7

  c  y

c  y =■-3x 2

3

e  y =■ 3 x

(1, 1)

1 01

x

y = -2x + 3

eBook plus

Digital doc

SkillSHEET 3.8 doc-5203

  6  a y 10

y = 10

5 -10 -5 0 -5

5 10 x

  b 

d  y f  y

  5  a

  b 

y y = 2x 2

1 =■ 2 x 5 =■− 2 x

  6  We5 ■Sketch■the■graphs■of■the■following■linear■equations. a  y =■10 b  y =■-10 c  x =■10 d  x =■-10 e  y =■100 f  y =■0 g  x =■0 h  x =■-100 i  y■=■-12

y = x -■7

0 1

   c

y

y = 5x

5 x

 d

y 0 1

x

0 1 y

y = 1–2 x

1– 2

x

0

-3

1

x

y = -3x

  7  Transpose■each■of■the■equations■to■standard■form■(that■is,■y■=■mx■+■c). State■the■x-■and■

y-intercept■for■each. a  5(y■+■2)■=■4(x■+■3) ■ x-intercept:■-0.5;■y-intercept:■0.4 b  5(y■-■2)■=■4(x■-■3)     x-intercept:■0.5;■y-intercept:■-0.4 c  2(y■+■3)■=■3(x■+■2)     x-intercept:■0;■y-intercept:■0 d  10(y■-■20)■=■40(x■-■2)   x-intercept:■-3;■y-intercept:■12     x-intercept:■-4;■y-intercept:■-4 e  4(y■+■2)■=■-4(x■+■2) f  2(y■-■2)■=■-(x■+■5)     x-intercept:■-1;■y-intercept:■-0.5     x-intercept:■2.75;■y-intercept:■2.2 g  -5(y■+■1)■=■4(x■-■4)     x-intercept:■7;■y-intercept:■3.5 h  8(y■-■5)■=■-4(x■+■3)     x-intercept:■9.75;■y-intercept:■-3.9 i  5(y■+■2.5)■=■2(x■-■3.5) j  2.5(y■-■2)■=■-6.5(x■-■1)     x-intercept:■ 23■ö■1.77;■y-intercept:■4.6 13

   e

y

y = 2–3 x

2

0

  f

3

x

y 0 - –25

x

1 y=

- –25 x

unDerstAnDing

y 5 -10 -5 0 -5

5 10 x

-10

y = -10

  8  Find■the■x-■and■y-intercepts■of■the■following■lines. a  -y■=■8■-■4x ■ (2,■0),■(0,■-8)     (- 12,■0),■(0,■3) b  6x■-■y■+■3■=■0     (-5,■0),■(0,■25) c  2y■-■10x■=■50

reFleCtion 

 

What types of straight lines have an x- and y-intercept of the same value?

Chapter 3 Coordinate geometry

63

number AND algebra • Linear and non-linear relationships

3B

Determining linear equations The equation of a straight line can be determined if given: •• two points through which the line passes •• the gradient of the line and one other point through which the line passes. ■■ The gradient of a straight line can be calculated from the coordinates of two points (x1, y1) and (x2, y2) which lie on the line. y rise y2 − y1 B = •• Gradient = m = run x2 − x1 (x2, y2) ■■

■■

■■

■■

The equation of the straight line can then be ■ found in the form y = mx + c, where c is the y-intercept. Once the gradient has been found, substitute ■ one pair of known x- and y-values into the standard linear equation y = mx + c to determine the value of c. If the y-intercept has already been given, then this is the value of c and no further calculation is required.

rise = y2 - y1 A (x1, y1) x-intercept y-intercept

Worked Example 6

Find the equation of the straight line shown in the graph. y 6

0

3

Think

x

Write

1

There are two points given on the straight line: the x-intercept (3, 0) and the y-intercept (0, 6). Therefore, c is 6.

c=6

2

We can now find the gradient of the line by using the rise y2 − y1 = , where (x1, y1) = (3, 0) formula m = run x2 − x1 and (x2, y2) = (0, 6).

m=

rise run y −y = 2 1 x2 − x1

6−0 0−3 6 = −3 = -2 The gradient m = -2. =

3

64

Substitute m = -2, and c = 6 into y = mx + c to find the equation.

Maths Quest 10 for the Australian Curriculum

y = mx + c y = -2x + 6

run = x2 - x1 x

number AND algebra • Linear and non-linear relationships

Worked Example 7 y

Find the equation of the straight line shown in the graph.  

1

(2, 1) 0

Think

2

x

Write

1

There are two points given on the straight line: the x- and y-intercept (0, 0) and another point (2, 1). The y-intercept, c, is 0.

c=0

2

We can now find the gradient of the line by using the rise y2 − y1 , where (x1, y1) = (0, 0) and = formula m = run x2 − x1 (x2, y2) = (2, 1).

m=

rise run y −y = 2 1 x2 − x1

1− 0 2−0 1 = 2 1 The gradient m = 2 . =

3

Substitute m = 12 and c = 0 into y = mx + c to determine the equation.



y = mx + c 1 y = 2x + 0



y = 12 x

Worked Example 8

Find the equation of the straight line passing through (-2, 5) and (1, -1). Think

Write

y = mx + c y −y m= 2 1 x2 − x1

1

Write the general equation of a straight line.

2

Write the formula for calculating the gradient of a line between two points.

3

Let (x1, y1) and (x2, y2) be the two points (-2, 5) and (1, -1) respectively. Substitute the values of the pronumerals into the formula to calculate the gradient.

m=

4

Substitute the value of the gradient into the general rule.

y = -2x + c

5

Select either of the two points, say (1, -1), and substitute its coordinates into y = -2x + c.

Point (1, -1): -1 = -2 ì 1 + c

6

Solve for c; that is, add 2 to both sides of the equation.

-1 = -2 + c 1=c

7

State the equation by substituting the value of c into y = -2x + c.

The equation of the line is y = -2x + 1.

−1 − 5 1 − −2 −6 = 3 = -2

Chapter 3 Coordinate geometry

65

number AND algebra • Linear and non-linear relationships

Finding the equation of a straight line using the gradient and another point (point–gradient method) Worked Example 9

Find the equation of the straight line with gradient of 2 and y-intercept of -5. Think

Write

1

Write the known information. In this instance, the other point is the y-intercept, which makes the calculation of c straightforward.

Gradient = 2, y-intercept = -5

2

State the values of m and c.

m = 2, c = -5

3

Substitute these values into y = mx + c to find the equation.

y = mx + c y = 2x - 5

■■ ■■

Sometimes the gradient and another point which is not the y-intercept is given. The value of c can then be found by substituting the coordinates of this point into y = mx + c.

Worked Example 10

Find the equation of the straight line passing through the point (5, -1) with a gradient of 3. Think

Write

1

Write the known information.

Gradient = 3, point (5, -1).

2

State the values of m, x and y.

m = 3, (x, y) = (5, -1)

3

Substitute these values into y = mx + c and solve to find c.

y = mx + c -1 = 3(5) + c -1 = 15 + c -16 = c

4

Substitute m = 3 and c = -16 into y = mx + c to determine the equation.

The equation of the line is y = 3x - 16.

remember

y2 − y1 rise or m = x − x . run 2 1 2. An equation of a straight line can be found if you are given either:■ (i)  two points that lie on the line or■ (ii)  the gradient of the line and another point (the point–gradient method).■ Note that alternative (i) can reduce to alternative (ii) since the gradient can be calculated using the two given points. 3. The equation of a straight line can be found by substituting the values of the gradient, m, into y = mx + c. The value of c can then be found by substituting the x- and y-values of a given point into y = mx + c. If one of the points given is the y-intercept then it is simply a matter of letting c = y-intercept. 1. The gradient of a straight line is equal to m =

66

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

3b inDiviDuAl pAthWAys eBook plus

Determining linear equations FluenCy   1  We 6 ■Determine■the■equation■for■each■of■the■straight■lines■shown. y a  b c y

Activity 3-B-1

x

-2 0

Activity 3-B-2

Linear equations doc-4979

5

12

4

Determining linear equations doc-4978

y

0

0

x

4

5

x

Activity 3-B-3

More complex linear equations doc-4980

d

y

e

y

f

y 3

eBook plus

0

x

4

0

-16

x

0

-6

-4

Digital doc

SkillSHEET 3.1 doc-5196

x

-8 g

h

y

0

-5

x

–5 7

       

y

x

0

a  b  c  d 

y■=■2x■+■4 y■=■-3x■+■12 y■=■-x■+■5 y■=■2x■-■8

  e  y■=■ 12 x■+■3 1

  f  y■= 4 x■-■4 -5

  a■ y■=■2x     b  y■=■-3x

  g  y■=■7x■-■5   h  y■=■-3x■-■15

-15

  2  We 7 ■Determine■the■equation■of■each■of■the■straight■lines■shown. y a  b y (-4, 12) 12 6 (3, 6)

    c  y■=■ 1 x

x

0 3

2

c

eBook plus

Digital doc

SkillSHEET 3.9 doc-5204

0 -2

y

d

y

-4 (-4, -2)

x

-4 0

    d  y■=■− 43 x

x

(-8, 6) -8

6 0

x

  3  We 8 ■Find■the■equation■of■the■straight■line■that■passes■through■each■pair■of■points. a  (1,■4)■and■(3,■6) y■=■x■+■3 b  (0,■-1)■and■(3,■5)   y■=■2x■-■1   y■=■− 12 x + 72     y■=■ 12 x + 12 ■ c  (-1,■4)■and■(3,■2) d  (3,■2)■and■(-1,■0) e  (-4,■6)■and■(2,■-6)   y■=■-2x■-■2 f  (-3,■-5)■and■(-1,■-7)   y■=■-x■-■8■ Chapter 3 Coordinate geometry

67

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

 

  4  We 9 ■Find■the■linear■equation■given■the■information■in■each■case■below. a  Gradient■=■3,■y-intercept■=■3 ■ y■=■3x■+■3 b  Gradient■=■-3,■y-intercept■=■4   y■=■-3x■+■4     y■=■-4x■+■2 d  Gradient■=■4,■y-intercept■=■2   y■=■4x■+■2 c  Gradient■=■-4,■y-intercept■=■2 e  Gradient■=■-1,■y-intercept■=■-4     y■=■-x■-■4 f  Gradient■=■0.5,■y-intercept■=■-4   y■=■0.5x■-■4     y■=■5x■+■2.5 h  Gradient■=■-6,■y-intercept■=■3   y■=■-6x■+■3 g  Gradient■=■5,■y-intercept■=■2.5 i  Gradient■=■-2.5,■y-intercept■=■1.5 j  Gradient■=■3.5,■y-intercept■=■6.5  y■=■3.5x■+■6.5   y■=■-2.5x■+■1.5   5  We 10 ■For■each■of■the■following,■fi■nd■the■equation■of■the■straight■line■with■the■given■gradient■

and■passing■through■the■given■point. a  Gradient■=■5,■point■=■(5,■6) ■ y■=■5x■-■19 b  Gradient■=■-5,■point■=■(5,■6)   y■=■-5x■+■31 c  Gradient■=■-4,■point■=■(-2,■7)   y■=■-4x■-■1 d  Gradient■=■4,■point■=■(8,■-2)   y■=■4x■-■34■ e  Gradient■=■3,■point■=■(10,■-5) y■=■3x■-■35 f  Gradient■=■-3,■point■=■(3,■-3)   y■=■-3x■+■6 g  Gradient■=■-2,■point■=■(20,■-10)     y■=■-2x■+■30 h  Gradient■=■2,■point■=■(2,■-0.5)   y■=■2x■-■4.5     y■=■0.5x■-■19 i  Gradient■=■0.5,■point■=■(6,■-16) j  Gradient■=■- 0.5,■point■=■(5,■3)   y■=■-0.5x■+■5.5

eBook plus

Digital doc

WorkSHEET 3.1 doc-5205

3C

reFleCtion 

 

What problems might you encounter when calculating the equation of a line whose graph is actually parallel to one of the axes?

the distance between two points on a straight line Distance between two points ■■ ■■

The■distance■between■two■points■can■be■calculated■using■Pythagoras’■theorem. Consider■two■points■A(x1,■y1)■and■B(x2,■y2)■on■the■Cartesian■plane■as■shown. AC■=■x2■-■x1 y BC■=■y2■-■y1 y2

By■Pythagoras’■theorem: AB2■=■AC2■+■BC2 =■(x2■-■x1)2■+■(y2■-■y1)2

y1

A

B(x2, y2) C

(x1, y1)

Hence■ AB = ( x2 − x1 )2 + ( y 2 − y1 )2

x1

x2

x

The■distance■between■two■points■A(x1,■y1)■and■B(x2,■y2)■is: AB = ( x2 − x1 )2 + ( y 2 − y1 )2 ■■

This■distance■formula■can■be■used■to■calculate■the■distance■between■any■two■points■on■the■ Cartesian■plane.

WorkeD exAmple 11 y

Find the distance between the points A and B in the figure at right.

4 A think

68

Write

1

From■the■graph■fi■nd■points■A■and■B.

A(-3,■1)■and■B(3,■4)

2

Let■A■have■coordinates■(x1,■y1).

Let■(x1,■y1)■=■(-3,■1)

maths Quest 10 for the Australian Curriculum

-3

B

1 3

x

number AND algebra • Linear and non-linear relationships

3

Let B have coordinates (x2, y2).

Let (x2, y2) = (3, 4)

4

Find the length AB by applying the formula for the distance between two points.

AB = ( x2 − x1 )2 + ( y2 − y1 )2 = [3 − (−3)]2 + (4 − 1)2 = (6)2 + (3)2 = 36 + 9 = 45 =3 5 = 6.71 (correct to 2 decimal places)

Worked Example 12

Find the distance between the points P(-1, 5) and Q(3, -2). Think

Write

1

Let P have coordinates (x1, y1).

Let (x1, y1) = (-1, 5)

2

Let Q have coordinates (x2, y2).

Let (x2, y2) = (3, -2)

3

Find the length PQ by applying the formula for the distance between two points.

PQ = ( x2 − x1 )2 + ( y2 − y1 )2 = [3 − (−1)]2 + (−2 − 5)2 = (4)2 + (−7)2 = 16 + 49 = 65 = 8.06 (correct to 2 decimal places)

■■

The distance formula is useful in proving geometric properties of polygons.

Worked Example 13

Prove that the points A(1, 1), B(3, -1) and C(-1, -3) are the vertices of an isosceles triangle. Think 1

Plot the points. Note: For triangle ABC to be isosceles, two sides must have the same magnitude.

Write/draw y -1 C

2

Find the length AC. A(1, 1) = (x2, y2) C(-1, -3) = (x1, y1)

A

1 1

3 B

x

From the diagram, AC appears to have the same length as BC.

-3

AC = [1 − (−1)]2 + [1 − (−3)]2 = (2)2 + (4)2 = 20 Chapter 3 Coordinate geometry

69

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

4

5

Find■the■length■BC. B(3,■-1)■=■(x2,■y2) C(-1,■-3)■=■(x1,■y1)

BC = [3 − (−1)]2 + [−1 − (−3)]2

Find■the■length■AB. A(1,■1)■=■(x1,■y1) B(3,■-1)■=■(x2,■y2)

AB = [3 − (1)]2 + [−1 − (1)]2

State■your■proof.

Since■AC■=■BC,■triangle■ABC■is■an■isosceles■triangle.

= (4)2 + (2)2 = 20

= (2)2 + (−2)2 = 4+4 = 8 =2 2

remember

The■distance■between■two■points■A(x1,■y1)■and■B(x2,■y2)■is: AB = ( x2 − x1 )2 + ( y2 − y1 )2 exerCise

3C inDiviDuAl pAthWAys eBook plus

Activity 3-C-1

Finding the distance between two points on a straight line doc-4981 Activity 3-C-2

Calculations of distance between two points doc-4982 Activity 3-C-3

Applications of distance between two points doc-4983

the distance between two points on a straight line FluenCy   1  We11 ■Find■the■distance■between■each■pair■of■points■     shown at■right.   2  We12 ■Find■the■distance■between■the■following■pairs■of■

points. (2,■5),■(6,■8) ■ 5 (-1,■2),■(4,■14)   13 (-1,■3),■(-7,■-5)   10 (5,■-1),■(10,■4) 7.07 (4,■-5),■(1,■1)   6.71 (-3,■1),■(5,■13)   14.42     13 (5,■0),■(-8,■0) (1,■7),■(1,■-6)   13 a2 + 4b2 (a,■b),■(2a,■-b)  (-a,■2b),■(2a,■-b)  3 a 2 + b 2

a b c d e f g h i j

y

O 6 B 5 4 P 3 C 2A E H N L 1 -6 -5 -4-3-2-1 0 1 2 3 4 5 6 x -1 F -2 M -3 I J D -4 -5 -6 G

K

  1  AB■=■5,■CD■=■2 10■or■6.32,■EF■=■3 2■or■4.24, ■

■ GH■=■2 5■or■4.47,■IJ■=■5,■KL■=■ 26■or■5.10,

■

■ MN■=■4 2 ■or■5.66,■OP■=■ 10 ■or■3.16

  3  We13 ■Prove■that■the■points■A(0,■-3),■B(-2,■-1)■and■C(4,■3)■are■the■vertices■of■an     isosceles■triangle.

Answers■will■vary.

unDerstAnDing eBook plus

Digital doc

Spreadsheet 021 doc-5206

70

  4  The■points■P(2,■-1),■Q(-4,■-1)■and■R(-1,■3 3 − 1)■are■joined■to■form■a■triangle.■     Prove■that■triangle■PQR■is■equilateral.

Answers■will■vary.

Answers■will■vary.

  5  Prove■that■the■triangle■with■vertices■D(5,■6),■E(9,■3)■and■F(5,■3)■is■a■right-angled■triangle.

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships AB = 4.47, BC = 2.24, CD = 4.47, DA = 2.24

6 The vertices of a quadrilateral are A(1, 4), B(-1, 8), C(1, 9) and D(3, 5). a Find the lengths of the sides. c What type of quadrilateral is it?

b Find the lengths of the diagonals. Rectangle



AC = 5, BD = 5

Reasoning   7   MC  If the distance between the points (3, b) and (-5, 2) is 10 units, then the value of b is: a -8 B -4 C 4 ✔ d 0 E 2   8   MC  A rhombus has vertices A(1, 6), B(6, 6), C(-2, 2) and D(x, y). The coordinates of D are: a (2, -3) B (2, 3) C (-2, 3) d E (3, 2) (3, -2) ✔   9 A rectangle has vertices A(1, 5), B(10.6, z), C(7.6, -6.2) and D(-2, 1). Find: a the length of CD 12 reflection    b the length of AD 5 c the length of the diagonal AC 13 How could you use the distance d the value of z. formula to show that a series of -2.2

10 Show that the triangle ABC with coordinates

A(a, a), B(m, -a) and C(-a, m) is isosceles.

points lay on the circumference of a circle with centre C ?

Answers will vary.

3D

The midpoint of a line segment Midpoint of a line segment ■■ ■■

The midpoint of a line segment is the half-way point. The x- and y-coordinates of the midpoint are half-way between those of the coordinates of the end points.

Midpoint formula y Consider the line segment connecting the points B(x2, y2) A(x1, y1) and B(x2, y2). (y2 - y) Let P(x, y) be the midpoint of AB. AC is parallel to PD. P(x, y) (x2 - x) D PC is parallel to BD. (y - y1) AP is parallel to PB (collinear). A Hence triangle APC is similar to triangle PBD. (x1, y1) (x - x1) C But AP = PB (since P is the midpoint of AB). x Hence, triangle APC is congruent to triangle PBD. Therefore x - x1 = x2 - x 2x = x1 + x2 x + x2 x= 1 2 y +y Similarly it can be shown that y = 1 2 . 2 In general, the coordinates of the midpoint of a line segment joining y (x2, y2) the points (x1, y1) and (x2, y2) can be found by averaging the x- and y-coordinates of the end points, respectively. M x_____, y1 + y2 1 + x2 _____ The coordinates of the midpoint of the line segment joining

x +x y +y  (x1, y1) and (x2, y2) are:  1 2 , 1 2   2 2 

(x1, y1)

(

2

2

)

Chapter 3 Coordinate geometry

x 71

number AND algebra • Linear and non-linear relationships

Worked Example 14

Find the coordinates of the midpoint of the line segment joining (-2, 5) and (7, 1). Think 1

Label the given points (x1, y1) and (x2, y2).

2

Find the x-coordinate of the midpoint.

Write

Let (x1, y1) = (-2, 5) and (x2, y2) = (7, 1) x +x x= 1 2 2 −2 + 7 = 2 5 =2 1

3

Find the y-coordinate of the midpoint.

=22 y +y y= 1 2 2 5+1 = 2 6 =2 =3

4

Give the coordinates of the midpoint.

1

Hence, the coordinates of the midpoint are (2 2 , 3).

Worked Example 15

The coordinates of the midpoint, M, of the line segment AB are (7, 2). If the coordinates of A are (1, -4), find the coordinates of B. Think

2

Label the start of the line segment (x1, y1) and the midpoint (x, y). Find the x-coordinate of the end point.

3

Find the y-coordinate of the end point.

4

Give the coordinates of the end point.

5

Check that the coordinates are feasible.

1

Write/DRAW

Let (x1, y1) = (1, -4) and (x, y) = (7, 2) x1 + x2 2 1 + x2 7= 2 14 = 1 + x2 x2 = 13 y +y y= 1 2 2 − 4 + y2 2= 2 4 = -4 + y2 y2 = 8 Hence, the coordinates of the point B are (13, 8). x=

y 8 2

B (13, 8) M (7, 2)

1 7 -4 A (1, -4)

72

Maths Quest 10 for the Australian Curriculum

13

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

The■coordinates■of■the■midpoint■of■the■line■segment■ joining■(x1,■y1)■and■(x2,■y2)■are: ■

■

y (x2, y2)

 x1 + x2 y1 + y2   2 , 2 

M

1

(x1, y1)

exerCise

3D inDiviDuAl pAthWAys eBook plus

Activity 3-D-1

Finding the midpoint of a line segment doc-4984 Activity 3-D-2

Calculations — midpoint of a segment doc-4985 Activity 3-D-3

Applications — midpoint of a segment doc-4986

eBook plus

Digital doc

Spreadsheet 075 doc-5207

+ x _____ y +y (x_____, 2 2 ) 2

1

2

x

the midpoint of a line segment FluenCy   1  We14 ■Use■the■formula■method■to■fi■nd■the■coordinates■of■the■midpoint■of■the■line     segment joining■the■following■pairs■of■points. a  (-5,■1),■(-1,■-8) (-3,■-3 12 ) c  (0,■4),■(-2,■-2)       (-1,■1)       (2a,■ 12b) e  (a,■2b),■(3a,■-b)

1 b  (4,■2),■(11,■-2)   (7 2 ,■0)   (0,■1 12 ) d  (3,■4),■(-3,■-1) f  (a■+■3b,■b),■(a■-■b,■a■-■b) (a + b,■ 1 a) 2

  2  We15 ■The■coordinates■of■the■midpoint,■M,■of■the■line■segment■AB■are■(2,■-3).■If■the■ coordinates■of■A■are■(7,■4),■fi■nd■the■coordinates■of■B. (-3,■-10) unDerstAnDing   3  Find: a  the■coordinates■of■the■centre■of■a■square■with■vertices■A(0,■0),■B(2,■4),■C(6,■2)■and■ D(4,■-2) ■ (3,■1) b  the■side■length   4.47 c  the■length■of■the■diagonals.   6.32   4  mC ■The■midpoint■of■the■line■segment■joining■the■points■(-2,■1)■and■(8,■-3)■is: a  (6,■-2) B  (5,■2) C  (6,■2) E  (5,■-2) ✔ d  (3,■-1)   5  mC ■If■the■midpoint■of■AB■is■(-1,■5)■and■the■coordinates■of■B■are■(3,■8),■then■A■has■

coordinates: a  (1,■6.5) d  (4,■3)

B  (2,■13) E  (7,■11)

✔ C 

(-5,■2)

  6  a■ The■vertices■of■a■triangle■are■A(2,■5),■B(1,■-3)■and■C(-4,■3).■Find:   i  the■coordinates■of■P,■the■midpoint■of■AC (-1,■4)   ii  the■coordinates■of■Q,■the■midpoint■of■AB (1 12,■1)   iii  the■length■of■PQ   3.9   iv  the■length■of■BC. 7.8 b  Hence■show■that■BC■=■2PQ.   Answers■will■vary.   7  a■ ■A■quadrilateral■has■vertices■A(6,■2),■B(4,■-3),■C(-4,■-3)■and■D(-2,■2).■Find:   i  the■midpoint■of■the■diagonal■AC (1,■-0.5)   ii  the■midpoint■of■the■diagonal■BD. (1,■-0.5) b  Comment■on■your■fi■nding.   Answers■will■vary. Chapter 3 Coordinate geometry

73

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships   8  a■ ■■The■points■A(-5,■3.5),■B(1,■0.5)■and■C(-6,■-6)■are■the■vertices■of■a■triangle.■Find:   i  the■midpoint,■P,■of■AB   (-2,■2)   ii  the■length■of■PC 8.94   iii  the■length■of■AC 9.55   iv  the■length■of■BC. 9.55 b  Describe■the■triangle.■What■could■PC■represent?    Isosceles.■PC■could■be■the■perpendicular■ height■of■the■triangle. reAsoning eBook plus

  9  Find■the■equation■of■the■straight■line■that■passes■through■the■midpoint■of■A(-2,■5)■and■   y■=■-3x■-■2 B(-2,■3),■and■has■a■gradient■of■-3.

Digital doc

 10  Find■the■equation■of■the■straight■line■that■

3e

reFleCtion 

passes■through■the■midpoint■of■A(-1,■-3)■and■ 2 B(3,■-5),■and■has■a■gradient■of■ 3.   3y■-■2x■+■14■=■0

WorkSHEET 3.2 doc-5208

If the midpoint of a line segment is the origin, what are the possible values of the x- and y-coordinates of the end points?

parallel and perpendicular lines parallel lines ■■

 

eBook plus

Interactivity Parallel and perpendicular lines

Lines■which■have■the■same■gradient■are■parallel■lines.

int-2779

WorkeD exAmple 16

Show that AB is parallel to CD given that A has coordinates (-1, -5), B has coordinates (5, 7), C has coordinates (-3, 1) and D has coordinates (4, 15). think 1

Write

Find■the■gradient■of■AB.

Let■A(-1,■-5)■=■(x1,■y1)■and■B(5,■7)■=■(x2,■y2) y −y Since■ ■ m = 2 1 x2 − x1 7 − (−5) mAB = 5 − (−1) 12

= 6 =■2 2

Find■the■gradient■of■CD.

Let■C(-3,■1)■=■(x1,■y1)■and■D(4,■15)■=■(x2,■y2) 15 − 1 mCD = 4 − (−3) =

14 7

=■2 3

Compare■the■gradients■to■determine■ if■they■are■parallel.■(Note:■||■means■ ‘is■parallel■to’.)

■■

74

Since■parallel■lines■have■the■same■gradient■and■ mAB■=■mCD■=■2,■then■AB||CD.

Collinear■points■are■points■which■all■lie■on■the■same■straight■line.

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 17

Show that the points A(2, 0), B(4, 1) and C(10, 4) are collinear. Think 1

Write

Let A(2, 0) = (x1, y1) and B(4, 1) = (x2, y2) y −y Since m= 2 1 x2 − x1 1− 0 mAB = 4−2

Find the gradient of AB.

2

1 2

Let B(4, 1) = (x1, y1) and C(10, 4) = (x2, y2)

Find the gradient of BC.



3

=

Show that A, B and C are collinear.

mBC = 4 − 1 10 − 4



=



=

3 6 1 2 1

Since mAB = mBC = 2 then AB||BC Since B is common to both line segments, A, B and C must lie on the same straight line. That is, A, B and C are collinear.

Perpendicular lines ■■ ■■

There is a special relationship between the gradients of two perpendicular lines. Consider the diagram shown below where the line segment AB is perpendicular to the line segment BC, AC is parallel to the x-axis, and BD is the perpendicular height of the resulting triangle ABC. Let mAB = m1 y B a = b a q = tan (q  ) a Let mBC = m2 a C A q a c b =− D c x = -tan (a ) b =− a −1 = m1 Hence or

m2 = −

1 m1

m1m2 = -1 Chapter 3 Coordinate geometry

75

number AND algebra • Linear and non-linear relationships

Hence, if two lines are perpendicular to each other, then the product of their gradients is -1. Two lines are perpendicular if and only if:                   m1m2 = -1 −1 or               m2 = m1 Worked Example 18

Show that the lines y = -5x + 2 and 5y - x + 15 = 0 are perpendicular to one another. Think

Write

y = -5x + 2 m1 = -5

1

Find the gradient of equation 1.

Hence

2

Find the gradient of equation 2.

5y - x + 15 = 0 Rewrite in the form y = mx + c 5y = x - 15 x y= -3 5 m2 =

Hence 3

Test for perpendicularity. (The two lines are perpendicular if the product of their gradients is -1.)

1 5 1

m1m2 = -5 ì 5 = -1 Hence, the two lines are perpendicular to each other.

Determining the equation of a straight line parallel or perpendicular to another straight line ■■

The gradient properties of parallel and perpendicular straight lines can be used to determine the equations of other lines with particular attributes.

Worked Example 19

Find the equation of the straight line that passes through the point (3, -1) and is parallel to the straight line with equation y = 2x + 1. Think

Write

Write the general equation.



2

Find the gradient of the given line.

y = 2x + 1 has a gradient of 2 Hence m = 2

3

Substitute for m in the general equation.

so

4

Substitute the given point to find c.

5

76

y = mx + c

1

Substitute for c in the general equation.

Maths Quest 10 for the Australian Curriculum

y = 2x + c (x, y) = (3, -1) \ -1 = 2(3) + c =6+c c = -7

y = 2x - 7 or 2x - y - 7 = 0

number AND algebra • Linear and non-linear relationships

Worked Example 20

Find the equation of the line that passes through the point (0, 3) and is perpendicular to a straight line with a gradient of 5. Think

Write

m=5 1 m1 = − 5

1

Find the gradient of the perpendicular line.

Given

2

Substitute for m and (x1, y1) in the general equation.

Since y - y1 = m(x - x1) and (x1, y1) = (0, 3) 1

then y - 3 = − 5 (x - 0) x =− 5 5(y - 3) = -x 5y - 15 = -x x + 5y - 15 = 0

Horizontal and vertical lines ■■

Recall the following. •• Horizontal lines are parallel to the x-axis, have a gradient of zero, are expressed in the form y = c and have no x-intercept. •• Vertical lines are parallel to the y-axis, have an undefined (infinite) gradient, are expressed in the form x = a and have no y-intercept.

Worked Example 21

Find the equation of: a  the vertical line that passes through the point (2, -3) b  the horizontal line that passes through the point (-2, 6). Think

Write

a For a vertical line, there is no y-intercept so y does not

a x=2

b For a horizontal line, there is no x-intercept so x does not

b y=6

appear in the equation. The x-coordinate of the point is 2. appear in the equation. The y-coordinate of the point is 6.

Worked Example 22

Find the equation of the perpendicular bisector of the line joining the points (0, -4) and (6, 5). Think 1

Find the gradient of the line joining the given points using the general equation.

Write/draw

Let (0, -4) = (x1, y1) Let       (6, 5) = (x2, y2) y − y1 m= 2 x2 − x1 5 − (−4) m= 6−0 9 =6 =

3 2

Chapter 3 Coordinate geometry

77

number AND algebra • Linear and non-linear relationships

2

Find the gradient of the perpendicular line.

For lines to be perpendicular, m2 = − 2

m1 = − 3 3

1 m1

x1 + x2 2 0+6 = 2 =3 y + y2 y= 1 2 −4 + 5 = 2 1 = 2 x=

Find the midpoint of the line joining the given points.

1

Hence (3, 2 ) are the coordinates of the midpoint. 4

Substitute for m and (x1, y1) in the general equation.

Since y - y1 = m(x - x1) 2 1 and (x1, y1) = (3, 2 ) and m1 = − 3

5

Simplify by removing the fractions.

then y -

= − 3 (x - 3)

(a)  Multiply both sides by 3.



= -2(x - 3)

(b)  Multiply both sides by 2.

= -2x + 6 6y - 3 = -4x + 12 4x + 6y - 15 = 0

1 2 1 3(y - 2 ) 3 3y - 2

y 5

Note: The diagram at right shows the geometric situation.

2

(6, 5)

1 2 –2 1– 2

-4

3

6 x

-4

remember

1. The equation of a straight line may be expressed in the form: y = mx + c where m is the gradient of the line and c is the y-intercept, or y - y1 = m(x - x1) where m is the gradient and (x1, y1) is a point on the line. 2. The gradient can be calculated if two points, (x1, y1) and (x2, y2), are given by using y −y m= 2 1 x2 − x1 3. Parallel lines have the same gradient. 4. Collinear points lie on the same straight line. Two lines are perpendicular if and only if: m1m2 = -1 1 or m2 = − . m1 The equation of a straight line can be determined by two methods: 78

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

5.■ The■y■=■mx■+■c■method. This■requires■the■gradient,■m,■and■a■given■point■to■be■known,■in■order■to■establish■the■ value■of■c. If■the■y-intercept■is■known,■then■this■can■be■directly■substituted■for■c. 6.■ Alternative■method:■y■-■y1■=■m(x■-■x1) This■requires■the■gradient,■m,■and■a■given■point■(x1,■y1)■to■be■known. 7.■ The■general■equation■for■a■vertical■line■is■given■by■x■=■a■and■a■horizontal■line■is■given■ by■y■=■c.

exerCise

3e inDiviDuAl pAthWAys eBook plus

Activity 3-E-1

Parallel and perpendicular lines doc-4987 Activity 3-E-2

More difficult parallel and perpendicular lines doc-4988 Activity 3-E-3

Complex parallel and perpendicular lines doc-4989

parallel and perpendicular lines FluenCy   1  We 16 ■Find■if■AB■is■parallel■to■CD■given■the■following■coordinates. a  b  c  d  e  f 

A(4,■13),■B(2,■9),■C(0,■-10),■D(15,■0). ■ No A(2,■4),■B(8,■1),■C(-6,■-2),■D(2,■-6).   Yes A(-3,■-10),■B(1,■2),■C(1,■10),■D(8,■16).   No A(1,■-1),■B(4,■11),■C(2,■10),■D(-1,■-5).       No A(1,■0),■B(2,■5),■C(3,■15),■D(7,■35).   Yes A(1,■-6),■B(-5,■0),■C(0,■0),■D(5,■-4).   No

  2  Which■pairs■of■the■following■straight■lines■are■parallel? a  2x■+■y■+■1■=■0■ c  2y■-■x■=■3■

    b,■f ;■c ,■e

b■ y■=■3x■-■1 d■ y■=■4x■+■3

x −1 ■ 2 g  3y■=■x■+■4■

e  y =

■

f■ 6x■-■2y■=■0 h■ 2y■=■5■-■x

 

Answers■will■vary.

  3  We 17 ■Show■that■the■points■A(0,■-2),■B(5,■1)■and■C(-5,■-5)■are■collinear.   4  Show■that■the■line■that■passes■through■the■points■(-4,■9)■and■(0,■3)■also■passes■through■ the■point■(6,■-6).   Answers■will■vary.     Answers■will■vary.   5  We 18 ■Show■that■the■lines■y■=■6x■-■3■and■x■+■6y■-■6■=■0■are■perpendicular■to■one■another. eBook plus

Digital doc

Spreadsheet 085 doc-5209

  6  Determine■if■AB■is■perpendicular■to■CD,■given■the■following■coordinates. a  A(1,■6),■B(3,■8),■C(4,■-6),■D(-3,■1)■ ■ Yes b■ A(2,■12),■B(-1,■-9),■C(0,■2),■D(7,■1)   Yes     Yes c  A(1,■3),■B(4,■18),■C(-5,■4),■D(5,■0)■  No d■ A(1,■-5),■B(0,■0),■C(5,■11),■D(-10,■8) e  A(-4,■9),■B(2,■-6),■C(-5,■8),■D(10,■14)■  Yes f■ A(4,■4),■B(-8,■5),■C(-6,■2),■D(3,■11)   No   7  We 19 ■Find■the■equation■of■the■straight■line■that■passes■through■the■point■(4,■-1)■and■is■     y■=■2x■-■9 parallel■to■the■straight■line■with■equation■y■=■2x■-■5.

eBook plus

Digital doc

Spreadsheet 029 doc-5210

  8  We 20 ■Find■the■equation■of■the■line■that■passes■through■the■point■(-2,■7)■and■is■perpendicular 2

to■a■line■with■a■gradient■of■ 3.   3x■+■2y■-■8■=■0   9  Find■the■equations■of■the■following■straight■lines. a  Gradient■3■and■passing■through■the■point■(1,■5). y■=■3x■+■2 b  Gradient■-4■and■passing■through■the■point■(2,■1).   y■=■-4x■+■9 c  Passing■through■the■points■(2,■-1)■and■(4,■2).   3x■-■2y■-■8■=■0 d  Passing■through■the■points■(1,■-3)■and■(6,■-5).   5y■+■2x■+■13■=■0 e  Passing■through■the■point■(5,■-2)■and■parallel■to■x■+■5y■+■5■=■0.     x■+■5y■+■5■=■0 f  Passing■through■the■point■(1,■6)■and■parallel■to■x■-■3y■-■2■=■0.   x■-■3y■+■17■=■0 g  Passing■through■the■point■(-1,■-5)■and■perpendicular■to■3x■+■y■+■2■=■0.       x■-■3y■-■14■=■0 Chapter 3 Coordinate geometry

79

number AND algebra • Linear and non-linear relationships 10 Find the equation of the line which passes through the point (-2, 1) and is: a parallel to the straight line with equation 2x - y - 3 = 0 2x - y + 5 = 0 b perpendicular to the straight line with equation 2x - y - 3 = 0. x + 2y = 0 11 Find the equation of the line that contains the point (1, 1) and is: a parallel to the straight line with equation 3x - 5y = 0 3x - 5y + 2 = 0 b perpendicular to the straight line with equation 3x - 5y = 0. 5x + 3y - 8 = 0 12   WE 21  Find the equation of: a the vertical line that passes through the point (1, -8) x = 1 b the horizontal line that passes through the point (-5, -7). y = -7 13   MC  a  The vertical line passing through the point (3, -4) is given by: ✔ B x = 3 A y = -4 C y = 3x - 4 D y = -4x + 3 E x = -4 b Which of the following points does the horizontal line given by the equation y = -5 pass

14

15 16 17

through? A (-5, 4) B (4, 5) D (5, -4) ✔ C (3, -5) E (5, 5) c Which of the following statements is true? A Vertical lines have a gradient of zero. B The y-coordinates of all points on a vertical line are the same. C Horizontal lines have an undefined gradient. ✔ D The x-coordinates of all points on a vertical line are the same. E A horizontal line has the general equation x = a. d Which of the following statements is false? A Horizontal lines have a gradient of zero. ✔ B The straight line joining the points (1, -1) and (-7, -1) is vertical. C Vertical lines have an undefined gradient. D The straight line joining the points (1, 1) and (-7, 1) is horizontal. E A horizontal line has the general equation y = c. The triangle ABC has vertices A(9, -2), B(3, 6) and C(1, 4). a Find the midpoint, M, of BC. (2, 5) b Find the gradient of BC. 1 c Show that AM is the perpendicular bisector of BC. Answers will vary. d Describe triangle ABC. Isosceles triangle   WE 22  Find the equation of the perpendicular bisector of the line joining the points (1, 2) and (-5, -4). y = -x - 3 Find the equation of the perpendicular bisector of the line joining the points (-2, 9) and (4, 0). 4x - 6y + 23 = 0 ABCD is a parallelogram. The coordinates of A, B and C are (4, 1), (1, -2) and (-2, 1) respectively. Find: a the equation of AD y = -x + 5 b the equation of DC y = x + 3 c the coordinates of D. (1, 4)

understanding Answers will vary. 18 In each of the following, show that ABCD is a parallelogram. a A(2, 0), B(4, -3), C(2, -4), D(0, -1) b A(2, 2), B(0, -2), C(-2, -3), D(0, 1) c A(2.5, 3.5), B(10, -4), C(2.5, -2.5), D(-5, 5) 80

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships 19 In each of the following, show that ABCD is a trapezium. Answers will vary. a A(0, 6), B(2, 2), C(0, -4), D(-5, -9) b A(26, 32), B(18, 16), C(1, -1), D(-3, 3) c A(2, 7), B(1, -1), C(-0.6, -2.6), D(-2, 3) 20   MC  The line that passes through the points (0, -6) and (7, 8) also passes through: A (4, 3) D (1, -8)

✔ B

(5, 4)

C (-2, 10)

E (1, 4)

21   MC  The point (-1, 5) lies on a line parallel to 4x + y + 5 = 0. Another point on the same line

as (-1, 5) is: A (2, 9) B (4, 2) C (4, 0) ✔ E (3, -11) D (-2, 3) 22 Find the equation of the straight line given the following conditions: a passes through the point (-1, 3) and parallel to y = -2x + 5 y = -2x + 1 b passes through the point (4, -3) and parallel to 3y + 2x = -3. 3y + 2x + 1 = 0 23 Determine which pairs of the following straight lines are perpendicular. a x + 3y - 5 = 0 d 2y = x + 1 g 2x + y = 6

b y = 4x - 7 e y = 3x + 2 h x + y = 0

a, e; b, f ; c , h; d, g

c y = x f x + 4y - 9 = 0

24 Find the equation of the straight line that cuts the x-axis at 3 and is perpendicular to the line with equation 3y - 6x = 12. y = − 12 x + 32

25 Calculate the value of m for which lines with the following pairs of equations are perpendicular

to each other. 8 a 2y - 5x = 7 and 4y + 12 = mx m = − 5

m = 185 b 5x - 6y = -27 and 15 + mx = -3y

26   MC  The gradient of the line perpendicular to the line with equation 3x - 6y = 2 is: A 3 B -6 C 2 D

1 2

✔ E

-2

27   MC  Triangle ABC has a right angle at B. The vertices are A(-2, 9), B(2, 8) and C(1, z). The

value of z is: 1

A 8 4 3

D 7 4

✔ B

4

C 12

E -4

Reasoning 28 The map shows the proposed course for a yacht race.

y

Scale: 1 unit «1 km

N

Buoys have been positioned at A(1, 5), B(8, 8) and 11 10 C(12, 6), but the last buoy’s placement, D(10, w), is yet 9 Buoy B to be finalised. 8 a How far is the first stage of the race, that is, from the 7 Buoy 6 A start, O, to buoy A? 5.10 km Buoy C M 5 b The race marshall boat, M, is situated halfway 4 between buoys A and C. What are the coordinates of E 3 Buoy D 2 H the boat? (6.5, 5.5) 1 O c Stage 4 of the race (from C to D) is perpendicular to stage 3 (from B to C). What is the gradient of CD? 2 (Start) 1 2 3 4 5 6 7 8 9 10 11 12 x d Find the linear equation that describes stage 4. y = 2x - 18 e Hence determine the exact position of buoy D. (10, 2) 2 f An emergency boat is to be placed at point E, 3 of the way from buoy A to buoy D. The coordinates of E are (7, 3). How far is the emergency boat from the hospital, located at H, 2 km North of the start? 7.071 km Chapter 3 Coordinate geometry

81

number AND algebra • Linear and non-linear relationships 29 Show that the following sets of points form the vertices of a right-angled triangle.   Answers will vary. a A(1, -4), B(2, -3), C(4, -7) b A(3, 13), B(1, 3), C(-4, 4) c A(0, 5), B(9, 12), C(3, 14) 30 Prove that the quadrilateral ABCD is a rectangle when A is (2, 5), B(6, 1), C(3, -2) and D(-1, 2).   Answers will vary. 31 Prove that the quadrilateral ABCD is a rhombus, given A(2, 3), B(3, 5), C(5, 6) and D(4, 4). Hint: The diagonals of a rhombus intersect at right angles. Answers will vary. reflection 

 

How could you use coordinate geometry to design a logo for an organisation?

82

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Summary Sketching linear graphs ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

■■ ■■ ■■

The Cartesian plane is a grid, consisting of two axes (x and y), meeting at the origin (0, 0). A location (point) is specified by its x- and y-coordinates. A linear graph consists of an infinite set of points that can be joined to form a straight line, but to sketch a linear graph, the coordinates of only two points are needed. A linear rule or equation can be used to obtain the coordinates of points that belong to ■ its graph. Linear equations may be written in several different forms. The two most common forms are y = mx + c and ax + by = k. When a linear equation is expressed in the form y = mx + c, then m represents the gradient (slope) of the straight line and c represents the y-intercept. A straight line with a positive gradient slopes upward to the right and a straight line with a negative gradient slopes downward to the right. The x- and y-intercept method allows us to sketch the graph of any linear equation by finding two specific points: the x-intercept and y-intercept. An exception is the case of lines passing through the origin. Graphs of equations in the form y = mx pass through the origin. To find the second point, substitute a chosen x-value into the equation to find the corresponding y-value. Graphs of equations in the form y = c have a gradient of zero and are parallel to the x-axis. Graphs of equations in the form x = a have an undefined (infinite) gradient and are parallel to the y-axis. Determining linear equations

■■ ■■

■■

y2 − y1 rise or m = x − x . run 2 1 An equation of a straight line can be found if you are given either:■ (i)  two points that lie on the line or■ (ii)  the gradient of the line and another point (the point–gradient method).■ Note that alternative (i) can reduce to alternative (ii) since the gradient can be calculated using the two given points. The equation of a straight line can be found by substituting the values of the gradient, m, into y = mx + c. The value of c can then be found by substituting the x- and y-values of a given point into y = mx + c. If one of the points given is the y-intercept then it is simply a matter of letting c = y-intercept. The gradient of a straight line is equal to m =

The distance between two points on a straight line

The distance between two points A(x1, y1) and B(x2, y2) is: AB = ( x2 − x1 )2 + ( y2 − y1 )2 The midpoint of a line segment

The coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are:

 x1 + x2 y1 + y2   2 , 2 

y (x2, y2) M

+ x _____ y +y (x_____, 2 2 ) 1

(x1, y1)

2

1

2

x

Chapter 3 Coordinate geometry

83

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships Parallel and perpendicular lines ■■

■■

■■ ■■

■■

■■

The■equation■of■a■straight■line■may■be■expressed■in■the■form: y■=■mx■+■c where■m■is■the■gradient■of■the■line■and■c■is■the■y-intercept,■or■ y■-■y1■=■m(x■-■x1)■ where■m■is■the■gradient■and■(x1,■y1)■is■a■point■on■the■line. The■gradient■can■be■calculated■if■two■points,■(x1,■y1)■and■(x2,■y2),■are■given■by■using y −y m= 2 1 x2 − x1 Parallel■lines■have■the■same■gradient. Collinear■points■lie■on■the■same■straight■line. Two■lines■are■perpendicular■if■and■only■if: m1m2■=■-1 1 or m2■=■− . m1 The■equation■of■a■straight■line■can■be■determined■by■two■methods: •■ The■y■=■mx■+■c■method. This■requires■the■gradient,■m,■and■a■given■point■to■be■known,■in■order■to■establish■the■value■ of■c. If■the■y-intercept■is■known,■then■this■can■be■directly■substituted■for■c. •■ Alternative■method:■y■-■y1■=■m(x■-■x1) This■requires■the■gradient,■m,■and■a■given■point■(x1,■y1)■to■be■known. The■general■equation■for■a■vertical■line■is■given■by■x■=■a■and■a■horizontal■line■is■given■ by■y■=■c.

MaPPING YOUR UNdERSTaNdING

Homework  Book

84

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■55. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships 10 x -10 -8 -6 -4 -2 0

Chapter review Fluency

3 ✔

2 x

0

3x + 2y = 6 C 2x + 3y = 6 E 2x - 3y = -6

x-intercept of 4 is: A y = -3x - 12 C y = -3x - 4 E y = 4x - 3

labelling the x- and y-intercepts.

B y = -3x + 4 ✔ D y = -3x + 12

straight lines. a y = -7x + 6 c y =

✔ C

4 −3

D 3

E -4

1

(13 3 ), y-intercept c = -5 3

21 (1 5 ), y-intercept c = − 4 - 43 x-intercept = 16 16

13 Sketch graphs of the following linear equations by

finding the x- and y-intercepts. a 2x - 3y = 6 b 3x + y = 0 c 5x + y = -3 d x + y + 3 = 0

2

c x = -2

0 -2

x

3

y 3 -1 0

15 Sketch the graph of the equation

3(y - 5) = 6(x + 1).

x y = -3x

16 Find the equations of the straight lines in the

following graphs.

16 a y = 2x - 2 b

y

0 0

y 2x - 3y = 6

d y = 7

y 7 (0, 7) - –27

a

14 Sketch the graph of each of the following. b 1 a y = x b y = -4x

a

3x - 4y + 7 = 0 is:

B

40 3

d y = 0.5x + 2.8 x-intercept = -5.6, y-intercept c = 2.8

13

8 The gradient of the line perpendicular to 4 3

3 8 4 x 7

6

x-intercept = 7 , y-intercept c = 6

b y = x - 5 x-intercept =

5 16

3 4

x

12 Find the x- and y-intercepts of the following

193 E 12   5 The midpoint of the line segment joining the points (-4, 3) and (2, 7) is: B (-2, 10) C (-6, 4) ✔ A (-1, 5) D (-2, 4) E (-1, 2)   6 If the midpoint of the line segment joining the points A(3, 7) and B(x, y) has coordinates (6, 2), then the coordinates of B are: A (15, 3) B (0, -6) ✔ C (9, -3) D (4.5, 4.5) E (-9, 3) 7 If the points (-6, -11), (2, 1) and (x, 4) are collinear, then the value of x is: B 3.2 ✔ A 4 E 3

y = 3x - 2 (1, -1)

1

0 1 -2 (0, -2)

7 5

D

D

3

d y = x - 3

(6, - 7) is:

✔ C

( 2– , 0)

2

  4 The distance between the points (1, 5) and

29

2 4 6 8 10 x

y

c y = − 3 x + 1

through (2, -7) and (-2, -2) is: ✔ B 5x + 4y + 18 = 0 d 5x - 4y - 18 = 0

B

2x + y - 1 = 0 and passing through the point y (1, 4) is: 80 60 A 2x + y - 6 = 0 40 20 B 2x + y - 2 = 0 -10 -8 -6 -4 -2 0 C x - 2y + 7 = 0 -20 -40 D x + 2y + 9 = 0 -60 -80 E x - 2y = 0

a y = 3x - 2 b y = -5x + 15

A 4x - 5y + 18 = 0 c 5x + 4y - 18 = 0 e 4x + 5y + 18 = 0

A

10

11 Sketch the graph of the following linear equations,

3 The equation of a linear graph which passes

1 4

8

the equation y = -5x + 15 for values of x between -10 and +10.

2 The equation of a linear graph with gradient -3 and

C

6

10 Produce a table of values, and sketch the graph of

B 3x - 2y = 6 D 2x - 3y = 6

✔ A

53

4

9 The equation of the line perpendicular to

1 The equation of the line drawn below is: y

A

2

65 55 45 35 25 15 5 -5 -15 -25 -35

y

x

3(y - 5) = 6(x + 1)

-2

1

x

b y = -x - 4 y

-4

0

x

-4

Chapter 3 Coordinate geometry

85

1

16 c y = − 3 x + 2

d y = 4x

e y = - 4

3

f x = 5

number AND algebra • Linear and non-linear relationships c

y

d

y

• (2, 8)

3x - 2y + 16 = 0 28 Find the equation of the straight line joining the point

(-2, 5) and the point of intersection of the straight lines with equations y = 3x - 1 and y = 2x + 5. 2 29 Using the information given in the diagram. x x 0 0 6 a Find: 4 y i the gradient of AD   − 5 B(4, 9) 5 9 ii the gradient of AB   4   4x + 5y - 61 = 0 iii the equation of BC C y e f y 4 A iv the equation of DC O D   (9, 5) v the coordinates of C. x x 0 45 9 b Describe quadrilateral ABCD. Square   5x - 4y - 25 = 0 - –43 x 0 5 30 In triangle ABC, A is (1, 5), B is (-2, -3) and C is (8, -2). a Find: 1 17 Find the linear equation given the information in i the gradient of BC   10 1 each case below. (− 2 , 1) ii the midpoint, P, of AB 1 1 a gradient = 3, y-intercept = -4 y = 3x - 4 iii the midpoint, Q, of AC. (4 2 , 1 2 ) b gradient = -2, y-intercept = -5 y = -2x - 5 b Hence show that: Answers will vary. 1 c gradient = , y-intercept = 5 y = 12 x + 5 i PQ is parallel to BC 2 ii PQ is half the length of BC. d gradient = 0, y-intercept = 6 y = 6 18 For each of the following, find the equation of the

3 5

d gradient = , point (1, -3)

problem solving 1 John has a part-time job working as a gardener and

is paid $13.50 per hour. a Complete the following table of values relating the amount of money received to the number of hours worked.

y = 35 x − 18 15

19 Find the distance between the points (1, 3) and (7, -2). 61 20 Prove that triangle ABC is isosceles given A(3, 1), B(-3, 7) and C(-1, 3). Answers will vary.

23 Show that the points A(3, 1), B(5, 2) and C(11, 5) are collinear. Answers will vary.

2

4

6

8

10

Pay ($)

0

27

54

81

108

135

money received to the number of hours worked. c Sketch the linear equation on a Cartesian plane

over a suitable domain. d Using algebra, calculate the pay that John will 3 $91.13 receive if he works for 6 4 hours. 2 A fun park charges a $12.50 entry fee and an

24 Show that the lines y = 2x - 4 and x + 2y - 10 = 0 are perpendicular to one another. Answers will vary. 25 Find the equation of the straight line passing

through the point (6, -2) and parallel to the line x + 2y - 1 = 0. x + 2y - 2 = 0 2x + 3y - 9 = 0

26 Find the equation of the line perpendicular to

3x - 2y + 6 = 0 and having the same y-intercept.

27 Find the equation of the perpendicular bisector of

the line joining the points (-2, 7) and (4, 11). 86

0

b Find a linear equation relating the amount of

21 Show that the points A(1, 1), B(2, 3) and C(8, 0) are the vertices of a right-angled triangle. Answers will vary. 22 The midpoint of the line segment AB is (6, -4).

If B has coordinates (12, 10), find the coordinates of A. (0, -18)

Number of hours

Pay = $13.50 ì (number of hours worked)

straight line with the given gradient and passing through the given point. a gradient = 7, point (2, 1) y = 7x - 13 b gradient = -3, point (1, 1) y = -3x + 4 1 1 c gradient = , point (-2, 5) y = 2 x + 6 2

3x + 2y - 21 = 0 Maths Quest 10 for the Australian Curriculum

additional $2.50 per ride. a Complete the following table of values relating the total cost to the number of rides. Number of rides Cost ($)

0

2

4

6

8

10

12.50 17.50 22.50 27.50 32.50 37.50

b Find a linear equation relating total cost to the number of rides. Cost = $2.50 ì number of rides + $12.50 c Sketch the linear equation on a Cartesian plane

over a suitable domain. d Using algebra, calculate the cost for 7 rides. $30

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships   3  The■cost■of■hiring■a■boat■is■$160■plus■

$22.50■per■hour. a  Sketch■a■graph■showing■the■total■cost■for■ C■=■22.50h■+■160 between■0■and■12■hours. b  State■the■equation■relating■cost■to■time■rented. c  Predict■the■cost■of■hiring■a■boat■for■12■hours■ and■15■minutes.   Approx■$436       4  ABCD■is■a■quadrilateral■with■vertices■A(4,■9),■ B(7,■4),■C(1,■2)■and■D(a,■10).■ Given■that■the■diagonals■are■perpendicular■to■each■ other,■fi■nd: a  the■equation■of■the■diagonal■AC ■ 7x■-■3y■-■1■=■0 b  the■equation■of■the■diagonal■BD   3x■+■7y■-■49■=■0   -7 c  the■value■of■a.   5  An■architect■decides■to■design■a■building■with■a■ 14-metre-square■base■such■that■the■external■walls■ are■initially■vertical■to■a■height■of■50■metres,■but■ taper■so■that■their■separation■is■8■metres■at■its■peak■ height■of■90■metres.■A■profi■le■of■the■building■is■ shown■with■the■point■(0,■0)■marked■as■a■reference■at■ the■centre■of■the■base. y

500 •

Cost ($)

400

C

300

0

2

24

S(-30, 24)

—) B(- 1–2 , 57 40

A(-1, 4–5 ) -30

57 – 40

4– 5

-1 -1–2

x

If■so,■which■one? b  Due■to■bias,■the■displaced■guard■ball■is■

B 0

y

a  Will■player■Y■displace■one■of■the■guard■balls?■

8m

•

100

1 57

B(- 2 ,■ 40 ).■Player■Y■bowls■a■ball■so■that■it■travels in■a■straight■line■toward■the■jack.■The■ball■is■ bowled■from■the■position■S,■with■the■coordinates■ (-30,■24).

(Not to scale)

•

200

so■that■it■travels■in■a■straight■line■in■order■to■ displace■an■opponents■‘guard■balls’.■In■a■particular■ game,■player■X■has■2■guard■balls■close■to■the■ jack.■The■coordinates■of■the■jack■are■(0,■0)■and■the■ 4 coordinates■of■the■guard■balls■are■A(-1,■ 5 )■and■

90 m

4 6 8 10 12 Time (hours)

50 m

A

0 14 m

x

knocked■so■that■it■begins■to■travel■in■a■straight■ line■(at■right■angles■to■the■path■found■in■part■a).■ Find■the■equation■of■the■line■of■the■guard■ball. c  Show■that■guard■ball■A■is■initially■heading■ directly■toward■guard■ball■B. d  Given■its■initial■velocity,■guard■ball■A■can■ travel■in■a■straight■line■for■1■metre■before■ its■bias■affects■it■path.■Calculate■and■explain■ whether■guard■ball■A■will■collide■with■guard■ ball■B.

a  Write■the■equation■of■the■vertical■line■ connecting■A■and■B. ■ a■=■-7     B(-7,■50),■C(-4,■90) b  Write■the■coordinates■of■B■and■C. c  Find■the■length■of■the■tapered■section■of■wall■ from■B■to■C.       40.1■metres   6  a■ ■Since■the■gradient■of■SA■=■the■ gradient■of■SO■=■-0.8,■the■points■S,■   6  In■a■game■of■lawn■bowls,■the■object■is■to■bowl■a■ A■and■O■are■collinear.■Player■Y■will■ biased■ball■so■that■it■gets■as■close■as■possible■to■ displace■guard■ball■A.

a■smaller■white■ball■called■a■jack.■During■a■game,■ 41     b  y■=■ 45x■+■ 20 ■or■25x■-■20y■+■41■=■0 a■player■will■sometimes■bowl■a■ball■quite■quickly■

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Interactivities

Test yourself Chapter 3 int-2834 Word search Chapter 3 int-2832 Crossword Chapter 3 int-2833

    c   Since■the■gradient■of■the■path■AB■is■ 5 ,■which■is■the■same■as■the■gradient■ 4 of■the■known■path■of■travel■from■ the■common■point■A,■the■direction■ of■travel■is■toward■B.     d   dAB■=■0.8■metres.■Yes,■guard■ball■A■ will■collide■with■guard■ball■B■as■it■ will■not■be■deviated■from■its■linear■ path■under■1■metre■of■travel.

Chapter 3 Coordinate geometry

87

eBook plus

ACtivities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■3■(doc-5195)■(page 55) are you ready? Digital docs (page 56) •■ SkillSHEET■3.1■(doc-5196):■Measuring■the■rise■and■ the■run •■ SkillSHEET■3.2■(doc-5197):■Describing■the■gradient■ of■a■line •■ SkillSHEET■3.3■(doc-5198):■Plotting■a■line■using■a■ table■of■values •■ SkillSHEET■3.4■(doc-5199):■Stating■the■y-intercept■ from■a■graph •■ SkillSHEET■3.5■(doc-5200):■Solving■linear■ equations■that■arise■when■fi■nding■x-■and■y-intercepts •■ SkillSHEET■3.6■(doc-5201):■Using■Pythagoras’■ theorem

3a   Sketching linear graphs Digital docs

•■ Activity■3-A-1■(doc-4975):■Sketching■linear■graphs■ (page 62) •■ Activity■3-A-2■(doc-4976):■Graphs■of■linear■ equations■(page 62) •■ Activity■3-A-3■(doc-4977):■More■graphs■of■linear■ equations■(page 62) •■ SkillSHEET■3.7■(doc-5202):■Substitution■into■a■ linear■rule■(page 62) •■ SkillSHEET■3.5■(doc-5200):■Solving■linear■ equations■that■arise■when■fi■nding■x-■and■y-intercepts■ (page 63) •■ SkillSHEET■3.8■(doc-5203):■Transposing■linear■ equations■to■standard■form■(page 63) 3B   determining linear equations Digital docs

•■ Activity■3-B-1■(doc-4978):■Determining■linear■ equations■(page 67) •■ Activity■3-B-2■(doc-4979):■Linear■equations■(page 67) •■ Activity■3-B-3■(doc-4980):■More■complex■linear■ equations■(page 67) •■ SkillSHEET■3.1■(doc-5196):■Measuring■the■rise■and■ the■run■(page 67) •■ SkillSHEET■3.9■(doc-5204):■Finding■the■gradient■ given■two■points■(page 67) •■ WorkSHEET■3.1■(doc-5205):■Gradient■(page 68) 3C   The distance between two points on a straight line

(page 70) •■ Activity■3-C-1■(doc-4981):■Finding■the■distance■ between■two■points■on■a■straight■line Digital docs

88

maths Quest 10 for the Australian Curriculum

•■ Activity■3-C-2■(doc-4982):■Calculation■of■distance■ between■two■points •■ Activity■3-C-3■(doc-4983):■Applications■of■distance■ between■two■points •■ Spreadsheet■021■(doc-5206):■Distance■between■two■ points 3d   The midpoint of a line segment Digital docs

•■ Activity■3-D-1■(doc-4984):■Finding■the■midpoint■of■ a■line■segment■(page 73) •■ Activity■3-D-2■(doc-4985):■Calculations■—■ midpoint■of■a■segment■(page 73) •■ Activity■3-D-3■(doc-4986):■Applications■—■ midpoint■of■a■segment■(page 73) •■ Spreadsheet■075■(doc-5207):■Midpoint■of■a■segment■ (page 73) •■ WorkSHEET■3.2■(doc-5208):■Midpoint■of■a■line■ segment■(page 74) 3E   Parallel and perpendicular lines

(page 79) •■ Activity■3-E-1■(doc-4987):■Parallel■and■ perpendicular■lines •■ Activity■3-E-2■(doc-4988):■More■diffi■cult■parallel■ and■perpendicular■lines •■ Activity■3-E-3■(doc-4989):■Complex■parallel■and■ perpendicular■lines■ •■ Spreadsheet■085■(doc-5209):■Perpendicular■ checker •■ Spreadsheet■029■(doc-5210):■Equation■of■a■straight■ line Digital docs

Interactivity

•■ Parallel■and■perpendicular■lines■(int-2779)■ (page 74) Chapter review

(page 87) •■ Test■Yourself■Chapter■3■(int-2834):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■3■(int-2832):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■3■(int-2833):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

4 simultaneous linear equations and inequations

4a Graphical solution of simultaneous linear equations 4b Solving simultaneous linear equations using substitution 4c Solving simultaneous linear equations using elimination 4d Problem solving using simultaneous linear equations 4e Solving linear inequations 4F Sketching linear inequations 4G Solving simultaneous linear inequations WhAt Do you knoW ? 1 List what you know about linear equations and linear inequations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of linear equations and linear inequations. eBook plus

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Hungry brain activity Chapter 4 doc-5211

opening Question

How could John decide which of the two concreting companies he should use — Angelico’s Concrete ($700 plus $20 per m2 of concrete) and Baux Cementing ($1200 plus $15 per m2 of concrete)?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 4.1 doc-5212

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SkillSHEET 4.4 doc-5215

Substitution into a linear rule 1 Substitute■-1■for■x■in■each■of■the■following■equations■to■calculate■the■value■of■y. a y■=■4x■-■2 ■ -6 b y■=■3■-■x 4 c y■=■-2■-■5x 3

Solving linear equations that arise when finding x- and y-intercepts 2 a■ i■ y=■2 ii x■=■3 2 For■each■of■the■following■equations,■substitute: b i■ y=■-3 ii x■=■9 i x■=■0■to■fi■nd■the■corresponding■value■of■y c i■ y= - 32 ii x■=■2 ii y■=■0■to■fi■nd■the■corresponding■value■of■x a 2x■+■3y■=■6 b x■-■3y■=■9 c 4y■=■3x■-■6 Transposing linear equations to standard form 3 Write■the■following■equations■in■the■form■y■=■mx■+■c. a 2y■+■4x■=■8 ■ y=■-2x■+■4 b 8x■-■2y■=■10 y=■4x■-■5

Measuring the rise and the run 4 a■ 1 4 Find■the■gradient■for■each■of■the■following■straight■lines. a b c y y 10 20 5 10 -10 -5 0 -5

5 10 x

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SkillSHEET 4.7 doc-5218

b 2

c -1 y 10 5

-10 -5 0 5 10 x -10 -20

-10

eBook plus

c 2x■+■3y■+■5■=■0 y= − 23 x −

5 10 x

-10 -5 0 -5 -10

Finding the gradient given two points 5 Find■the■gradient■of■the■line■passing■through■each■of■the■following■pairs■of■points. 7 1 a (1,■2)■and■(3,■7) ■ 25 b (-1,■-4)■and■(2,■3) 3 c (6,■-1)■and■(-2,■1) - 4

Graphing linear equations using the x- and y-intercept method 6 Graph■each■line■with■the■following■equations■using■the■x-■and■y-intercept■method. a 5y■-■4x■=■20 b 4y■- 2x■=■5 c 3y■+■4x■=■-12

Checking whether a given point makes the inequation a true statement 7 For■each■of■the■following,■use■substitution■to■check■if■the■given■point■makes■the■inequality■a■

true■statement. a 3x■-■2y■<■12■(5,■1) ■ False

b y■Ç■5x■+■1■(-2,■1) False

6 a

b

y 5y - 4x = 20 -5

4 0

c x■-■y■> -8■(2,■-12) True c

y 4y - 2x = 5

y

1 1–4 x

-2 1–2

0

x

-3

0

x

-4 3y + 4x = -12

90

maths Quest 10 for the Australian Curriculum

5 3

number AND algebra • Linear and non-linear relationships

4A

Graphical solution of simultaneous linear equations Simultaneous linear equations ■■ ■■ ■■ ■■

Any two linear graphs will meet at a point, unless they are parallel. At this point, the two equations simultaneously share the same x- and y-coordinates. This point is referred to as the solution to the two simultaneous linear equations. Simultaneous equations can be solved graphically or algebraically.

Graphical solution ■■ ■■ ■■ ■■

This method involves drawing the graph of each equation on the same set of axes. The intersection point is the simultaneous solution to the two equations. An accurate solution depends on drawing an accurate graph. Graph paper or graphing software can be used.

Worked Example 1

Use the graph of the given simultaneous equations below to determine the point of intersection and, hence, the solution of the simultaneous equations. x + 2y = 4 y = 2x - 3

y 3

y = 2x - 3

2 1 -1 0 -1

x + 2y = 4 1

2

3

4

5 x

-2 -3

Think

Write

1

Write the equations and number them.

x + 2y = 4 y = 2x - 3

2

Locate the point of intersection of the two lines. This gives the solution.

Point of intersection (2, 1) Solution: x = 2 and y = 1

3

Check the solution by substituting x = 2 and y = 1 into the given equations. Comment on the results obtained.

Check equation [1]: LHS = x + 2y RHS = 4 = 2 + 2(1) =4 LHS = RHS Check equation [2]: LHS = y RHS = 2x - 3 = 1 = 2(2) - 3 =4-3 =1 LHS = RHS In both cases LHS = RHS, therefore the solution set (2, 1) is correct.

[1] [2]

Chapter 4 Simultaneous linear equations and inequations

91

number AND algebra • Linear and non-linear relationships

■■

It is always important to check the solution.

Worked Example 2

For the following simultaneous equations, use substitution to check if the given pair of coordinates, (5, -2), is a solution. 3x - 2y = 19 [1] 4y + x = -3 [2] Think

Write

1

Write the equations and number them.

3x - 2y = 19   4y + x = -3

2

Check by substituting x = 5 and y = -2 into equation [1].

Check equation [1]: LHS = 3x - 2y = 3(5) - 2(-2) = 15 + 4 = 19 LHS = RHS

Check by substituting x = 5 and y = -2 into equation [2].

Check equation [2]: LHS = 4y + x RHS = -3 = 4(-2) + 5 = -8 + 5 = -3 LHS = RHS In both cases, LHS = RHS. Therefore, the solution set (5, -2) is correct.

3

■■

[1] [2] RHS = 19

In order to obtain an accurate solution to a pair of simultaneous equations it is important to draw an accurate graph. This is demonstrated in the example below.

Worked Example 3

Solve the following pair of simultaneous equations using a graphical method. x + y = 6 2x + 4y = 20 Think 1

Write the equations, one under the other and ■ number them.

x + y = 6 2x + 4y = 20

2

Calculate the x- and y-intercepts for equation [1]. For the x-intercept, substitute y = 0 into equation [1].

Equation [1] x-intercept: when y = 0, x+0=6 x=6 The x-intercept is at (6, 0). y-intercept: when x = 0, 0+y=6 y=6 The y-intercept is at (0, 6).

For the y-intercept, substitute x = 0 into equation [1].

92

Write/draw

Maths Quest 10 for the Australian Curriculum

[1] [2]

number AND algebra • Linear and non-linear relationships

3

Calculate the x- and y-intercepts for equation [2]. For the x-intercept, substitute y = 0 into equation [2]. Divide both sides by 2.

For the y-intercept, substitute x = 0 into equation [2]. Divide both sides by 4. 4

Use graph paper to rule up a set of axes and label the x-axis from 0 to 10 and the y-axis from 0 to 6.

Equation [2] x-intercept: when y = 0, 2x + 0 = 20 2x = 20 x = 10 The x-intercept is at (10, 0). y-intercept: when x = 0, 0 + 4y = 20 4y = 20 y=5 The y-intercept is at (0, 5). y 6 5 4 3 2 1

(2, 4)

5

Plot the x- and y-intercepts for each equation.

6

Produce a graph of each equation by ruling a straight line through its intercepts.

7

Label each graph.

8

Locate the point of intersection of the lines.

The point of intersection is (2, 4).

9

Check the solution by substituting x = 2 and y = 4 into each equation.

Check [1]: LHS = x + y =2+4 =6 LHS = RHS Check [2]: LHS = 2x + 4y = 2(2) + 4(4) = 4 + 16 = 20 LHS = RHS

10

State the solution.

■■

x 0 1 2 3 4 5 6 7 8 910 -3-2-1 -1 -2 x+y=6 -3

RHS = 6

RHS = 20

In both cases, LHS = RHS. Therefore, the solution set (2, 4) is correct. The solution is x = 2, y = 4.

A CAS calculator can be used to obtain a graphical (as well as an algebraic) solution to simultaneous linear equations.

Parallel lines It is possible for two simultaneous linear equations to have no solution. ■■ This occurs when the graphs of the two equations do not cross because they have the same gradient. ■■ In other words, the two graphs are parallel. ■■ Consider the following pair of simultaneous equations. 2x - y = 5 [1] 4x - 2y = 6 [2] They can be graphed to show two parallel lines. ■■

2x + 4y = 20

6 5 4 3 2 1 -1 -10 -2 -3 -4 -5 -6

y 4x - 2y = 6 x 1

2

3

4

5

2x - y = 5

Chapter 4 Simultaneous linear equations and inequations

93

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

■■

■■

We■can■confi■rm■that■the■two■lines■are■in■fact■parallel■by■rearranging■each■equation■into■the■ form■y■=■mx■+■c■and■checking■the■gradient■of■each. ■ 2x■-y■=■5■ [1] -y■=■5■-■2x■ -y=■-2x■+■5■  y=■2x■-■5■ gradient■m■=■2■ ■ 4x■-■2y=■6■■ [2] ■ -2y=■6■-■4x ■ -2y=■-4x■+■6 ■ y=■2x■-■3 ■gradient■m■=■2 It■is■also■possible■for■two■simultaneous■linear■equations■to■have■many■solutions.■This■occurs■ when■the■two■linear■equations■are,■in■fact■the■same■equation,■simply■expressed■in■a■different■ form.■For■example, y■=■2x■-■5■ ■ [1] 6x■-■3y■=■15■ ■ [2] Simplifying■equation■[2]■by■dividing■by■3■gives■2x■-■y■=■5. Rearranging■it■in■the■same■form■as■equation■[1]■gives■y■=■2x■-■5. A■word■of■caution■here:■Make■sure■that■the■signs■are■exactly■the■same■in■both■equations.■They■ will■not■represent■the■same■equation■if■this■is■not■the■case. remember

1.■ When■solving■simultaneous■equations■graphically,■obtaining■an■accurate■solution■ depends■on■drawing■accurate■graphs. 2.■ The■solution■to■linear■simultaneous■equations■is■the■point■where■their■graphs■ intersect. 3.■ Lines■that■have■the■same■gradient■are■parallel. 4.■ If■the■graphs■of■the■two■simultaneous■equations■are■parallel■lines,■then■the■simultaneous■ equations■have■no■solution,■as■they■have■no■point■of■intersection. exerCise

4A inDiviDuAl pAthWAys eBook plus

Activity 4-A-1

Investigating graphs of simultaneous equations doc-4990 Activity 4-A-2

Graphing simultaneous equations doc-4991

94

graphical solution of simultaneous linear equations fluenCy 1 We1 ■Use■the■graphs■below■of■the■given■simultaneous■equations■to■write■the■point■of■

intersection■and,■hence,■the■solution■of■the■simultaneous■equations. a x■+y=■3 b x■+y=■2 x■-y=■1 ■ (2,■1) 3x■-y=■2 (1,■1)

y 5 4 3 2 1 -3 -2-1 0 -1 -2 -3

y

x-y=1

1 2 3 4 5

x

x+y=3

maths Quest 10 for the Australian Curriculum



6 5 4 3 2 1 -0.5 -10 -2 -3 -4

3x - y = 2

x 0.5

1.0

1.5

2.0

2.5

y+x=2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships c y■-x■=■4

inDiviDuAl pAthWAys

d y■+■2x■=■3 (2,■-1) 2y■+x■=■0

3x■+■2y■=■8 (0,■4)

eBook plus

3x + 2y = 8

Activity 4-A-3

Further graphing of simultaneous equations doc-4992

6

y

4

-3

y + 2x = 3

2

x

x

-1 0 -2

-2

y

1

2 -4

3

y-x=4

1

2

-1 0 -1

3

-4

-2

-6

-3

■

1

2

3

4

5

2y + x = 0

e y■-■3x■=■2 x■-y=■2 (-2,■-4) 6

f 2y■-■4x■=■5

4y■+■2x■=■5 (-0.5,■1.5)

y

y

y - 3x = 2

6

4

-3

-2

-1

0

1

2

2y - 4x = 5

4

x-y=2 x

2

2

3

-1.0 -0.5

-2

0

4y + 2x = 5 0.5

1.0

1.5 2.0

x

-2

-4

-4

-6

■

-6



2 We2 ■For■the■following■simultaneous■equations,■use■substitution■to■check■if■the■given■pair■of■

coordinates■is■a■solution. a (7,■5)■ 3x■+■2y■=■31 ■ ■ 2x■+■3y■=■28 ■ No c (9,■1)■ x■+■3y■=■12 ■ ■ 5x■-■2y■=■43 Yes e (4,■-3)■ y■=■3x■-■15 ■ ■ 4x■+■7y■=■-5 Yes g (4,■-2)■ 2x■+y=■6 ■ ■ x■-■3y■=■8 No i (-2,■-5)■ 3x■-■2y■=■-4 ■ ■ 2x■-■3y■=■11 No eBook plus

Digital doc

SkillSHEET 4.6 doc-5217

b (3,■7)■ d f h j

■ ■ (2,■5)■ ■ ■ (6,■-2)■ ■ ■ (5,■1)■ ■ ■ (-3,■-1)■ ■ ■

y■-x■=■4 2y■+x■=■17 Yes x■+y=■7 2x■+■3y■=■18 No x■-■2y■=■2 3x■+y=■16 No y■-■5x■=■-24 3y■+■4x■=■23 Yes y■-x■=■2 2y■-■3x■=■7 Yes

3 We3 ■Solve■each■of■the■following■pairs■of■simultaneous■equations■using■a■graphical■method. a x■+y=■5 b x■+■2y■=■10 2x■+y=■8 (3,■2) 3x■+y=■15 (4,■3) c 2x■+■3y■=■6 d x■-■3y■=■-8 2x■-y=■-10 (-3,■4) 2x■+y=■-2 (-2,■2) e 6x■+■5y■=■12 f y+■2x■=■6 (2,■0) 5x■+■3y■=■10 2y■+■3x■=■9 (3,■0) g y=■3x■+■10 h y=■8 y=■2x■+■8 (-2,■4) 3x■+y=■17 (3,■8) i 4x■-■2y■=■-5 j 3x■+y=■11 (- 12 ,■■112) x■+■3y■=■4 4x■-y=■3 (2,■5) k 3x■+■4y■=■27 l 3y■+■3x■=■8 x■+■2y■=■11 (5,■3) 3y■+■2x■=■6 (2,■ 2) 3

Chapter 4 simultaneous linear equations and inequations

95

number AND algebra • Linear and non-linear relationships 4 Solve each of the following pairs of simultaneous equations. a y = 8 - x b y = 3x + 10 y = x + 2 (3, 5) y = 2x + 8 (-2, 4) d y = 3 + 4x

e y = 16 - 3x y = 11 - 2x (5, 1)

g y = 7

h y =

y = 1 + 3x (-2, -5) y = 2x + 15 (-4, 7)

c y = 2x - 3 x = 5 (5, 7) f 3y + x = 0

2y = 3x - 22 (6, -2)

2x +2 3 y = 2x - 2 (3, 4)

Understanding 5 Using technology, determine which of the following pairs of simultaneous equations have no

solutions. Confirm by finding the gradient of each line. a y = 2x - 4 b 5x - 3y = 13 3y - 6x = 10 No solution 4x - 2y = 10 (2, -1) d y = 4x + 5 e 3y + 2x = 9 2y - 10x = 8 (1, 9) 6x + 4y = 22 (3, 1) g 4y + 3x = 7 h 2y - x = 0 12y + 9x = 22 No solution 14y - 6x = 2 (2, 1)

c x + 2y = 8

5x + 10y = 45 No solution

f y = 5 - 3x

3y = -9x + 18 No solution

Reasoning 6 Two straight lines intersect at the point (3, -4). One of the lines has a y-intercept of 8. The

120

Cost

100 80 60 40 20 0

1

second line is a mirror image of the first in the line x = 3. Determine the equation of the second line. (Hint: Draw a graph of both lines) y = 4x - 16 7 At a well-known beach resort it is possible to hire a jet-ski by the hour in two different locations. On the Northern beach the cost is $20 plus $12 per hour, while on the Southern D beach the cost is $8 plus $18 per hour. The jet-skis can be rented for up to 5 hours. C a Write the rules relating cost to the length of rental. Northern beach C = 20 + 12t Southern beach D = 8 + 18t b On the same set of axes sketch a graph of cost (y-axis) against length of rental (x-axis) C = 20 + 12t for 0–5 hours. reflection    D = 8 + 18t c For what rental times, if any, is the Northern beach What do you think is the major 2 3 4 5 rental cheaper than the Southern beach rental? Use error made when solving Time (hours) your graph to justify your answer. Time > 2 hours simultaneous equations d For what length of rental time are the two rental graphically? schemes identical? Use the graph and your rules ■ to justify your answer. T  ime = 2 hours, cost = $44 N  orthern beaches in red, southern beaches in blue

4b

Solving simultaneous linear equations using substitution ■■ ■■

There are two algebraic methods which can be used to solve simultaneous equations. They are the substitution method and the elimination method.

Substitution method ■■ ■■ ■■

96

This method is particularly useful when one (or both) of the equations is in a form where ■ one of the two variables is the subject. This variable is then substituted into the other equation, producing a third equation with only one variable. This third equation can then be used to determine the value of the variable.

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 4

Solve the following simultaneous equations using the substitution method. y = 2x - 1 and 3x + 4y = 29 Think

Write

1

Write the equations, one under the other and number them.

y = 2x - 1 3x + 4y = 29

2

Substitute the expression (2x - 1) for y from equation [1] into equation [2].

Substituting (2x - 1) into [2]: 3x + 4(2x - 1) = 29

3

Solve for x. (i) Expand the brackets on the LHS of the equation. (ii) Collect like terms. (iii) Add 4 to both sides of the equation. (iv) Divide both sides by 11.

3x + 8x - 4 = 29

4

Substitute the value of x into any of the equations, say [1], to find the value of y.

Substituting x = 3 into [1]: y = 2(3) - 1 =6-1 =5

5

Write your answer.

Solution: x = 3, y = 5 or (3, 5)

6

Check the answer by substituting the point of intersection into equation [2].

Check: Substitute into 3x + 4y = 29. LHS = 3(3) + 4(5) RHS = 29 = 9 + 20 = 29 As LHS = RHS, the solution is correct.

■■ ■■ ■■

[1] [2]

[3]

11x - 4 = 29 11x = 33 x=3

In some cases, both equations may be written with the same variable as the subject. They can then be made equal to each other. This produces a third equation with only one variable.

Worked Example 5

Solve the following pair of simultaneous equations using the substitution method. y = 5x - 8 and y = -3x + 16 Think

Write

y = 5x - 8 y = -3x + 16

1

Write the equations, one under the other and number them.



2

Both equations are written with y as the subject, so equate them.

5x - 8 = -3x + 16

3

Solve for x. (i) Add 3x to both sides of the equation. (ii) Add 8 to both sides of the equation. (iii) Divide both sides of the equation by 8.

4

Substitute the value of x into either of the original equations, say [1], and solve for y.

[1] [2]

8x - 8 = 16 8x = 24 x=3 Substituting x = 3 into [1]: y = 5(3) - 8 = 15 - 8 =7

Chapter 4 Simultaneous linear equations and inequations

97

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 5

Write■your■answer.

Solution:■x■=■3,■y■=■7■or■(3,■7)

6

Check■the■answer■by■substituting■the■point■of■ intersection■into■equation■[2].

Check:■Substitute■into■y■=■-3x■+■16. LHS■=■y ■ =■7■ RHS■=■-3x■+■16 ■ =■-3(3)■+■16 ■ =■-9■+■16 ■ =■7 As■LHS■=■RHS,■the■solution■is■correct.

remember

When■using■the■substitution■method■to■solve■simultaneous■equations: 1.■ choose■the■equation■in■which■one■of■the■variables■is■the■subject 2.■ substitute■this■expression■for■the■variable■into■the■other■equation■and■solve 3.■ substitute■the■value■you■have■found■into■the■rearranged■equation■to■solve■for■the■ other■variable 4.■ check■your■solution. exerCise

4b inDiviDuAl pAthWAys eBook plus

Activity 4-B-1

Learning substitution doc-4993 Activity 4-B-2

Practising substitution doc-4994 Activity 4-B-3

Tricky substitution doc-4995

98

solving simultaneous linear equations using substitution fluenCy 1 We4 ■Solve■the■following■simultaneous■equations■using■the■substitution■method.■Check■your■

solutions■using■technology. a x■=■-10■+■4y

b 3x■+■4y■=■2 3x■+■5y■=■21 ■ (2,■3) x■=■7■+■5y (2,■-1) c 3x■+y=■7 d 3x■+■2y■=■33 x■=■-3■-■3y (3,■-2) y=■41■-■5x (7,■6) e y=■3x■-■3 f 4x■+y=■9 -5x■+■3y■=■3 (3,■6)■ y=■11■-■5x (2,■1) g x■=■-5■-■2y h x=■-4■-■3y 5y■+■x=■-11 (-1,■-2) -3x-■4y=■12 (-4,■0) i x■=7■+■4y j x■=■14■+■4y 2x■+■y■=-4 (-1,■-2) -2x■+■3y■=■-18 (6,■-2) k 3x+■2y■=■12 l y■=2x■+■1 x■=■9■-■4y (3,■1 12 ) -5x■-■4y■=■35 (-3,■-5) 2 We5 ■Solve■the■following■pairs■of■simultaneous■equations■using■the■substitution■method.■ Check■your■solutions■using■technology. a y■=■2x■-■11■and■y■=■4x■+■1 ■ (-6,■-23) b y■=■3x■+■8■and■y■=■7x■- 12 (5,■23) c y■=■2x■-■10■and■y■=■-3x (2,■-6) d y■=■x■-■9■and■y■=■-5x  32 , − 152  e y■=■-4x■-■3■and■y■=■x■-■8 (1,■-7) f y■=■-2x■-■5■and■y■=■10x■+■1 (- 12 ,■-4)  3 1  − 2, − 2  g y■=■-x■-■2■and■y■=■x■+■1  1 4 h y■=■6x■+■2■and■y■=■-4x  − 5, 5  i y■=■0.5x■and■y■=■0.8x■+■0.9 (-3,■-1.5)

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships (1,■0.3) j y■=■0.3x■and■y■=■0.2x■+■0.1 2 4 7 7 3 1 y■=■-x■and■y■=■- 4 x■- 4

k y■=■-x■and■y■=■- x■+

 4 4 ,  − 5 5 

l

(1,■-1)

unDerstAnDing 3 A■small■farm■has■sheep■and■chickens.■There■are■twice■

as■many■chicken■as■sheep,■and■there■are■104■legs■ between■the■sheep■and■the■chickens.■How■many■ chickens■are■there? 26■chickens

4C eBook plus

Interactivity Simultaneous equations

int-2780

refleCtion 

 

When would you choose the substitution method in solving simultaneous equations?

solving simultaneous linear equations using elimination ■■ ■■ ■■

Elimination■is■best■used■when■the■two■equations■are■given■in■the■form■ax■+■by■=■k. The■method■involves■combining■the■two■equations■so■that■one■of■the■variables■is■eliminated. Addition■or■subtraction■can■be■used■to■reduce■the■two■equations■with■two■variables■into■ one■equation■with■only■one■variable.

WorkeD exAmple 6

Solve the following pair of simultaneous equations using the elimination method. -2x - 3y = -9 2x + y = 7 think

Write

1

Write■the■equations,■one■under■the■other■and■number■ them.

-2x■-■3y■=■-9■ 2x■+y=■7■

2

Look■for■an■addition■or■subtraction■that■will■eliminate■ eitherx■or■y. Note:■Adding■equations■[1]■and■[2]■in■order■will■ eliminatex.

[1]■+■[2]: -2x■-■3y■+■(2x■+■y)■=■-9■+■7 -2x■-■3y■+■2x■+■y■=■-2 -2y■=■-2

3

Solve■foryby■dividing■both■sides■of■the■equation■by■-2.

4

Substitute■the■value■ofyinto■equation■[2].■ Note:y=■1■may■be■substituted■into■either■equation.

5

Solve■forx. ■(i)■ Subtract■1■from■both■sides■of■the■equation. (ii)■ Divide■both■sides■of■the■equation■by■2.

6

Answer■the■question.

Solution:x■=■3,y=■1■or■(3,■1)

7

Check■the■answer■by■substituting■the■point■of■ intersection■into■equation■[1]■since■equation■[2]■was■ used■to■fi■nd■the■value■ofx.

Check:■Substitute■into■-2x■-■3y■=■-9. LHS■=■-2(3)■-■3(1) =■-6■-■3 =■-9 RHS■=■-9 LHS■=■RHS,■so■the■solution■is■correct.

■■

[1] [2]

y=■1 Substitutingy=■1■into■[2]: 2x■+■1■=■7 2x■=■6 x■=■3

When■the■like■terms■do■not■have■the■same■coeffi■cient,■multiply■one■or■both■equations■by■a■ constant■so■as■to■create■the■same■coeffi■cient. Chapter 4 simultaneous linear equations and inequations

99

number AND algebra • Linear and non-linear relationships

Worked Example 7

Solve the following pair of simultaneous equations using the elimination method. x - 5y = -17    2 x + 3y = 5 Think

Write

1

Write the equations, one under the other and number them.

  x - 5y = -17 2x + 3y = 5

[1] [2]

2

Look for a single multiplication that will create the same coefficient of either x or y. Multiply equation [1] by 2 and call the new equation [3].

[1] ì 2:   2x - 10y = -34

[3]

3

Subtract equation [2] from [3] in order to eliminate x.

[3] - [2]: 2x - 10y - (2x + 3y) = ‑34 - 5 2x - 10y - 2x - 3y = ‑39 -13y = -39

4

Solve for y by dividing both sides of the equation by -13.

5

Substitute the value of y into equation [2].

6

Solve for x. (i) Subtract 9 from both sides of the equation. (ii)  Divide both sides of the equation by 2.

y=3 Substituting y = 3 into [2]: 2x + 3(3) = 5 2x + 9 = 5 2x = -4 x = -2

7

Write your answer.

Solution: x = -2, y = 3 or (-2, 3)

8

Check the answer by substituting into equation [1].

Check: Substitute into x - 5y = -17. LHS = (-2) - 5(3) = -2 - 15 = -17 RHS = -17 LHS = RHS, so the solution is correct.

Note: In this example, equation [1] could have been multiplied by -2 (instead of by 2), then the two equations added (instead of subtracted) to eliminate x. ■■ Sometimes it is necessary to multiply both equations by a constant in order to achieve the same coefficient for one of the variables. Worked Example 8

Solve the following pair of simultaneous equations using the elimination method. 6x + 5y = 3    5x + 4y = 2 Think

100

Write

1

Write the equations, one under the other and number them.

6x + 5y = 3 5x + 4y = 2

[1] [2]

2

Decide which variable to eliminate, say y. Multiply equation [1] by 4 and call the new equation [3]. Multiply equation [2] by 5 and call the new equation [4].

Eliminate y. [1] ì 4:   24x + 20y = 12 [2] ì 5:   25x + 20y = 10

[3] [4]

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

Subtract■equation■[3]■from■[4]■in■order■to■eliminate■y.■

[4]■-■[3]: 25x■+■20y■-■(24x■+■20y)■=■10■-■12 25x■+■20y■-■24x■-■20y■=■-2 x■=■-2

4

Substitute■the■value■ofx■into■equation■[1].

Substitutingx■=■-2■into■[1]: 6(-2)■+■5y■=■3 -12■+■5y■=■3

5

Solve■for■y. ■(i)■ Add■12■to■both■sides■of■the■equation (ii)■ Divide■both■sides■of■the■equation■by■5.

5y■=■15 y■=■3

6

Write■your■answer.

Solution:x■=■-2,y=■3■or■(-2,■3)

7

Check■the■answer■by■substituting■the■solution■into■ equation■[2].

Check:■Substitute■into■5x■+■4y■=■2. LHS■=■5(-2)■+■4(3) =■-10■+■12 =■2 RHS■=■2 LHS■=■RHS,■so■the■solution■is■correct.

Note:■Equation■[1]■could■have■been■multiplied■by■-4■(instead■of■by■4),■then■the■two■equations■ added■(instead■of■subtracted)■to■eliminate■y. remember

1.■ Simultaneous■equations■of■the■form■ax■+■by■=■k■can■be■solved■by■the■elimination■ method■by■looking■for■an■addition■or■subtraction■of■the■equations■that■will■eliminate■ one■of■the■variables. 2.■ For■like■terms■with■the■same■coeffi■cient■but■opposite■signs,■add■the■equations.■For■like■ terms■with■the■same■coeffi■cient■and■the■same■sign,■subtract■the■equations. 3.■ If■the■terms■do■not■have■the■same■coeffi■cient,■multiply■one■or■both■equations■by■ a■constant■to■create■the■same■coeffi■cient.■Remember■to■multiply■both■sides■of■the■ equation■to■keep■it■balanced. 4.■ Once■one■variable■has■been■eliminated,■solve■the■single■variable■equation■formed.■ Substitute■the■solution■back■into■one■of■the■original■equations■to■fi■nd■the■value■of■the■ variable■that■was■originally■eliminated. 5.■ Check■your■solution■by■substitution. exerCise

4C inDiviDuAl pAthWAys eBook plus

Activity 4-C-1

Elimination practice doc-4996

solving simultaneous linear equations using elimination fluenCy 1 We 6 ■Solve■the■following■pairs■of■simultaneous■equations■by■adding■equations■to■eliminate■

eitherx■or■y.

b 5x■+■4y■=■2 c -2x■+y=■10 -x■+■4y■=■1 ■ (3,■1) 5x■-■4y■=■-22 (-2,■3) 2x■+■3y■=■14 (-2,■6)■ 2 Solve■the■following■pairs■of■equations■by■subtracting■equations■to■eliminate■eitherx■or■y. a 3x■+■2y■=■13 b 2x■-■5y■=■-11 c -3x■-y=■8 5x■+■2y■=■23 (5,■-1) 2x■+y=■7 (2,■3) -3x■+■4y■=■13 (-3,■1) a x■+■2y■=■5

Chapter 4 simultaneous linear equations and inequations

101

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

inDiviDuAl pAthWAys eBook plus

Activity 4-C-2

Let’s eliminate doc-4997 Activity 4-C-3

More elimination doc-4998

3 Solve■each■of■the■following■equations■using■the■elimination■method. a x■+■2y■=■12 b 3x■+■2y■=■-23 3x■-■2y■=■12 ■ (6,■3) 5x■+■2y■=■-29 (-3,■-7) c 6x■+■5y■=■-13

-2x■+■5y■=■-29 (2,■-5) e x■-■4y■=■27 3x■-■4y■=■17 (-5,■-8) g -5x■+■3y■=■3 (1 12 ,■3 12 ) -5x■+y=■-4 i 4x■-■3y■-■1■=■0 4x■+■7y■-■11■=■0 (1,■1)

d 6x■-■5y■=■-43 (-3,■5) 6x■-y=■-23 f -4x■+y=■-10 4x■-■3y■=■14 (2,■-2) h 5x■-■5y■=■1

2x■-■5y■=■-5 (2,■145)

4 We7 ■Solve■the■following■pairs■of■simultaneous■equations. a x■+■2y■=■4 b 3x■+■2y■=■19 3x■-■4y■=■2■ (2,■1) 6x■-■5y■=■-7 (3,■5) c -2x■+■3y■=■3

d 6x■+y=■9 (1,■3) -3x■+■2y■=■3

e

f 5x■+y=■27

g i k

5x■-■6y■=■-3 (3,■3) x■+■3y■=■14 3x■+y=■10■ (2,■4) -6x■+■5y■=■-14 (4,■2) -2x■+y=■-6 -3x■+■2y■=■6 1 x■+■4y■=■-9 (-3,■-1 2 ) 2x■+■3y■=■9 4x■+y=■-7 (-3,■5)

4x■+■3y■=■26 (5,■2) h 2x■+■5y■=■14 3x■+y=■-5 (-3,■4) j 3x■-■5y■=■7 (-6,■-5) x■+y=■-11 l -x■+■5y■=■7 5x■+■5y■=■19 (2,■1.8)

5 We8 ■Solve■the■following■pairs■of■simultaneous■equations. a 2x■+■3y■=■16 b 5x■-■3y■=■6 3x■+■2y■=■19 ■ (5,■2) 3x■-■2y■=■3■ (3,■3) c 3x■+■2y■=■6 e g i k

4x■+■3y■=■10 (-2,■6) 2x■-■3y■=■14 3x■-■5y■=■21 (7,■0) -4x■+■5y■=■-9 2x■+■3y■=■21■ (6,■3) 2x■-■2y■=■-4 5x■+■4y■=■17 (1,■3) x y + =2 2 3 x y + = 4 (-8,■18) 4 3

d 2x■+■7y■=■3 3x■+■2y■=■13 (5,■-1) f -3x■+■7y■=■-2 4x■+■2y■=■14 (3,■1) h 2x■+■5y■=■-6 3x■+■2y■=■2■ (2,■-2) j 2x■-■3y■=■6

4x■-■5y■=■9 (-1.5,■-3) x y 3 l + = 3 2 2 x y 1 + = − (-3,■5)■ 2 5 2

6 Solve■the■following■simultaneous■equations■using■an■appropriate■method.■Check■your■answer■

using■technology. a 7x■+■3y■=■16 y■=■4x■-■1 ■ (1,■3) c -3x■+■2y■=■19 4x■+■5y■=■13 (-3,■5) e -4x■+■5y■=■-7 x■=■23■-■3y (8,■5) 102

maths Quest 10 for the Australian Curriculum

b 2x■+■y■=■8

4x■+■3y■=■16 (4,■0) d -3x■+■7y■=■9 4x■-■3y■=■7 (4,■3) f y■=■-x 1 2   y■= - 5 x - 5  13, − 13 

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships reAsoning 7 Ann,■Beth■and■Celine■wanted■to■weigh■themselves■on■a■coin■weighing■machine.■The■problem■

was■they■only■had■enough■money■for■one■weighing.■They■decided■to■weigh■themselves■in■ pairs,■one■stepping■off■as■another■stepped■on. • Ann■and■Beth■weighed■119■kg refleCtion    • Beth■and■Celine■weighed■112■kg How does eliminating • Celine■and■Ann■weighed■115■kg one variable help to solve How■much■did■each■of■the■girls■weigh? simultaneous equations?

eBook plus

Digital doc

WorkSHEET 4.1 doc-5220

Ann■61■kg,■Beth■58■kg,■Celine■54■kg

4D

problem solving using simultaneous linear equations ■■ ■■

Many■word■problems■can■be■solved■using■simultaneous■linear■equations. Follow■these■steps. •■ Defi■ne■the■unknown■quantities■using■appropriate■pronumerals. •■ Use■the■information■given■in■the■problem■to■form■two■equations■in■terms■of■these■ pronumerals. •■ Solve■these■equations■using■an■appropriate■method. •■ Write■the■solution■in■words. •■ Check■the■solution.

WorkeD exAmple 9

Ashley received better results for his Maths test than for his English test. If the sum of the two marks is 164 and the difference is 22, calculate the mark he received for each subject. think

Write

1

Defi■ne■the■two■variables.

Letx■=■the■maths■mark. Lety=■the■English■mark.

2

Formulate■two■equations■from■the■information■given■and■ number■them. Note:■Sum■means■to■add■and■difference■means■to■subtract.

x■+y=■164 x■-y=■22

3

Use■the■elimination■method■by■adding■equations■[1]■and■ [2]■to■eliminate■y.■

[1]■+■[2]:■ ■ 2x■=■186

4

Solve■forx■by■dividing■both■sides■of■the■equation■by■2.

5

Substitute■the■value■ofx■into■equation■[1].

6

Solve■foryby■subtracting■93■from■both■sides■of■the■ equation.

7

Answer■the■question.

Solution: Maths■mark■(x)■=■93 English■mark■(y)■=■71

8

Check■the■answer■by■substituting■x■=■93■andy=■71■into■ equation■[1].

Check:■Substitute■intox■+y=■164. LHS■=■93■+■71■ RHS■=■164 =■164 As■LHS■=■RHS,■the■solution■is■correct.

[1] [2]

x■=■93 Substitutingx■=■93■into■[1]: x■+y=■164 93■+y=■164 y=■71

Chapter 4 simultaneous linear equations and inequations

103

number AND algebra • Linear and non-linear relationships

Worked Example 10

To finish a project, Genevieve buys a total of 25 nuts and bolts from a hardware store. If each nut costs 12 cents, each bolt costs 25 cents and the total purchase price is $4.30, how many nuts and how many bolts does Genevieve buy? Think

Write

1

Define the two variables.

Let x = the number of nuts. Let y = the number of bolts.

2

Formulate two equations from the information given and number them. Note: The total number of nuts and bolts is 25. Each nut cost 12 cents, each bolt cost 25 cents and the total cost is 430 cents ($4.30).

x + y = 25 12x + 25y = 430

3

Solve simultaneously using the substitution method since equation [1] is easy to rearrange.

4

Rearrange equation [1] to make x the subject by subtracting y from both sides of equation [1].

Rearrange equation [1]: x + y = 25 x = 25 - y

5

Substitute the expression (25 - y) for x into equation [2].

Substituting (25 - y) into [2]: 12(25 - y) + 25y = 430

6

Solve for y.

300 - 12y + 25y = 430 300 + 13y = 430 13y + 300 = 430 13y = 130 y = 10

7

Substitute the value of y into the rearranged equation x = 25 - y from step 4.

Substituting y = 10 into x = 25 - y: x = 25 - 10 x = 15

8

Answer the question.

Solution: The number of nuts (x) = 15. The number of bolts (y) = 10.

9

Check the answer by substituting x = 15 and y = 10 into equation [1].

Check: Substitute into x + y = 25. LHS = 15 + 10    RHS = 25 = 25 As LHS = RHS, the solution is correct.

[1] [2]

remember

1. To solve worded problems, read the question carefully and define the two variables using appropriate pronumerals. 2. Formulate two equations from the information given and number them. 3. Use either the elimination method or the substitution method to solve the two equations simultaneously. 4. Check your answer by substituting the values obtained for each variable into the original equations. 104

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

4D inDiviDuAl pAthWAys eBook plus

Activity 4-D-1

Problem solving doc-4999

problem solving using simultaneous linear equations Maths■mark■=■97,■English■mark■=■66 fluenCy 1 We 9 ■Rick■received■better■results■for■his■Maths■test■than■for■his■English■test.■If■the■sum■of■his■

two■marks■is■163■and■the■difference■is■31,■fi■nd■the■mark■for■each■subject. 2 We 10 ■Rachael■buys■some■nuts■and■bolts■to■fi■nish■a■project.■She■does■not■buy■the■same■number■

of■nuts■and■bolts,■but■buys■30■items■in■total.■If■each■nut■costs■10■cents,■each■bolt■costs■20■cents■ and■the■total■purchase■price■is■$4.20,■how■many■nuts■and■how■many■bolts■does■she■buy? 18■nuts,■12■bolts

Activity 4-D-2

Harder problem solving doc-5000 Activity 4-D-3

Tricky problem solving doc-5001

unDerstAnDing 8■and■3 3 Find■two■numbers■whose■difference■is■5■and■whose■sum■is■11. 4 The■difference■between■two■numbers■is■2.■If■three■times■the■larger■number■minus■double■the■ smaller■number■is■13,■fi■nd■the■two■numbers. 9■and■7 5 One■number■is■9■less■than■three■times■a■second■number.■If■the■fi■rst■number■plus■twice■the■ second■number■is■16,■fi■nd■the■two■numbers. 6■and■5 6 A■rectangular■house■has■a■perimeter■of■40■metres■and■the■length■is■4■metres■more■than■the■ width.■What■are■the■dimensions■of■the■house? Length■=■12■m■and■width■=■8■m 7 Mike■has■5■lemons■and■3■oranges■in■his■shopping■basket.■The■cost■of■the■fruit■is■$3.50.■Voula,■

with■2■lemons■and■4■oranges,■pays■$2.10■for■her■fruit.■How■much■does■each■type■of■fruit■cost? Lemons■cost■ 55■cents■and■ oranges■cost■ 25■cents.

■ Length■60■■m■and■width■20■■m■ Eight■20-cent■coins■and■three■50-cent■coins 8 A■surveyor■measuring■the■dimensions■of■a■block■of■land■fi■nds■that■the■length■of■the■block■is■ three■times■the■width.■If■the■perimeter■is■160■metres,■what■are■the■dimensions■of■the■block? 9 Julie■has■$3.10■in■change■in■her■pocket.■If■she■has■only■50■cent■and■20centpieces■and■the■total■ Twelve■$1■coins■ and■nine■$2■coins

number■of■coins■is■11,■how■many■coins■of■each■type■does■she■have? 10 Mr■Yang’s■son■has■a■total■of■twenty-one■$1■and■$2■coins■in■his■moneybox.■When■he■counts■his■ money,■he■fi■nds■that■its■total■value■is■$30.■How■many■coins■of■each■type■does■he■have? 11 If■three■Magnums■and■two■Paddlepops■cost■$8.70■and■the■difference■in■price■between■a■ Magnum■and■a■Paddlepop■is■90■cents,■how■much■does■each■type■of■ice-cream■cost? 12 If■one■Redskin■and■4■Golden■roughs■cost■$1.65,■whereas■2■Redskins■and■3■Golden■roughs■cost■ $1.55,■how■much■does■each■type■of■sweet■cost? Cost■of■the■Golden■rough■=■35■cents■

Paddlepops■cost■$1.20■and■a■Magnum■costs■$2.10.

and■cost■of■the■Redskin■=■25■cents■

reAsoning 13 A■catering■fi■rm■works■out■its■pricing■based■on■a■fi■xed■cost■

for■overheads■and■a■charge■per■person.■It■is■known■that■a■party■ Fixed■costs■=■$87,■ cost■per■person■=■$23.50 for■20■people■costs■$557,■whereas■a■party■for■35■people■costs■ $909.50.■Use■this■information■to■work■out■the■fi■xed■cost■and■the■ cost■per■person■charged■by■the■company. 14 The■difference■between■Sally’s■PE■mark■and■Science■mark■is■ 12,■and■the■sum■of■the■marks■is■154.■If■the■PE■mark■is■the■higher■ mark,■what■did■Sally■get■for■each■subject? PE■mark■is■83■and■Science■mark■is■71.

Chapter 4 simultaneous linear equations and inequations

105

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 15 Mozza’s■cheese■supplies■sells■six■Mozzarella■cheeses■and■eight■Swiss■cheeses■to■Munga’s■deli■

for■$83.60,■and■four■Mozzarella■cheeses■and■four■Swiss■cheeses■to■Mina’s■deli■for■$48.■How■ much■does■each■type■of■cheese■cost? Mozzarella■costs■$6.20,■Swiss■cheese■costs■$5.80. 16 If■the■perimeter■of■the■triangle■in■the■diagram■is■12■cm■and■the■length■of■the■rectangle■is■1■cm■ more■than■the■width,■fi■nd■the■value■ofx■and■y. x■=■3■and■y■=■4

Fixed■costs■=■$60,■ cost■per■person■=■$25

2x cm

y cm

x cm m

5c

(y + 3) cm

17 Mr■and■Mrs■Waugh■want■to■use■a■caterer■for■a■birthday■party■for■their■twin■sons.■The■manager■

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Digital doc

WorkSHEET 4.2 doc-5221

says■the■cost■for■a■family■of■four■would■be■$160.■However,■the■sons■want■to■invite■8■friends,■ making■12■people■in■all.■The■cost■for■this■would■be■$360.■If■the■total■cost■in■each■case■is■made■up■ of■the■same■cost■per■person■and■the■same■fi■xed■cost,■fi■nd■the■cost■per■person■and■the■fi■xed■cost. 18 Joel■needs■to■buy■some■blank■DVDs■and■zip■disks■to■ back■up■a■large■amount■of■data■that■has■been■generated■ refleCtion    by■an■accounting■fi■rm.■He■buys■6■DVDs■and■3■zip■disks■ How do you decide which for■$96.■He■later■realises■these■are■not■suffi■cient■and■so■ method to use when solving buys■another■5■DVDs■and■4■zip■disks■for■$140.■How■ word problems using much■did■each■DVD■and■each■zip■disk■cost?■(Assume■ simultaneous linear equations? the■same■rate■per■item■was■charged■for■each■visit.) $4■each■for■DVDs■and■$24■each■for■zip■disks

4e

solving linear inequations ■■ ■■

■■ ■■

An■equation■is■a■statement■of■equality■such■as■x■=■2;■an■inequation■is■a■statement■of■ inequality■such■as■x■<■2■(x■is■less■than■2). The■solution■to■a■linear■equation■is■a■single■point■on■a■number■line,■but■the■solution■to■an■ inequation■is■a■portion■of■the■number■line.■That■is,■the■solution■to■the■inequation■has■many■ values. The■following■table■shows■four■types■of■simple■inequations■and■their■corresponding■ representation■on■a■number■line. Note■that■an■open■circle■placed■over■the■2■indicates■that■2■is■not■included;■that■is,■2■does■not■ satisfy■the■inequality■statement.■A■closed■or■solid■circle■indicates■that■2■is■included;■that■is,■it■ does■satisfy■the■inequality■statement. Mathematical statement

106

English statement

x■>■2

x■is■greater■than■2

x■í■2

x■is■greater■than■or■equal■to■2

x■<■2

x■is■less■than■2

x■Ç■2

x■is■less■than■or■equal■to■2

maths Quest 10 for the Australian Curriculum

Number line diagram 0

2

4

6

8 10 x

-10 -8 -6 -4 -2 0

2

4

6

8 10

-10 -8 -6 -4 -2 0

2

4

6

8 10

-10 -8 -6 -4 -2 0

2

4

6

8 10

-10 -8 -6 -4 -2

x

x

x

number AND algebra • Linear and non-linear relationships

■■

The basic technique for solving inequations is to: 1. imagine that in place of the inequality sign, there is an equals sign 2. solve the inequation as if it were a linear equation, except that in place of the equals sign keep the original inequality sign unless the special case outlined below applies.

Worked Example 11

Solve each of the following linear inequations. a x + 3 Ç 4 b 4x - 1 < -2 c 6x - 7 í 3x + 5 Think a

b

c

Write a

1

Write the inequation.

2

Obtain x by subtracting 3 from both sides of the inequation. Keep the inequality sign the same throughout.

1

Write the inequation.

2

Add 1 to both sides of the inequation.

3

Obtain x by dividing both sides of the inequation by 4.

1

Write the inequation.

2

Combine the pronumeral terms by subtracting 3x from both sides of the inequation.

3

Add 7 to both sides of the inequation.

4

Obtain x by dividing both sides of the inequation by 3.

x+3Ç4 x+3-3Ç4-3 xÇ1

b

4x - 1 < -2 4x - 1 + 1 < -2 + 1 4x < -1 1 4x <− 4 4 1 x<− 4

c

6x - 7 í 3x + 5 6x - 7 - 3x í 3x + 5 - 3x 3x - 7 í 5 3x - 7 + 7 í 5 + 7 3x í 12 3 x 12 í 3 3 xí4

The special case — multiplying or dividing both sides of the inequation by a negative number ■■ ■■

■■

Consider the inequation 6 > 5 (6 is greater than 5). If we multiply both sides of the inequation by -1 we get: -6 > -5, which is not correct. In fact -6 < -5. Applying this to inequations generally, when we multiply or divide an inequation by a negative number, the direction of the inequality sign must change. When multiplying or dividing by a negative number, change the direction of the inequality sign; that is, change: < to > > to < Ç to í í to Ç Chapter 4 Simultaneous linear equations and inequations

107

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

WorkeD exAmple 12

Solve each of the following linear inequations. a -3m + 5 < -7 b 5(x - 2) í 7(x + 3) think a

b

Write

1

Write■the■inequation.

2

Subtract■5■from■both■sides■of■the■inequation.■ (No■change■to■the■inequality■sign.)

3

Obtain■m■by■dividing■both■sides■of■the■ inequation■by■-3.■Reverse■the■inequality■sign,■ since■you■are■dividing■by■a■negative■number.

1

Write■the■inequation.

2

Expand■both■brackets.

3

Combine■the■pronumeral■terms■by■subtracting■ 7x■from■both■sides■of■the■inequation.

4

Add■10■to■both■sides■of■the■inequation.

5

Obtain■x■by■dividing■both■sides■of■the■ inequation■by■-2.■Since■we■need■to■divide■by■ a■negative■number,■reverse■the■direction■of■the■ inequality■sign.

-3m■+■5■<■-7

a

-3m■+■5■-■5■<■-7■-■5 -3m■<■-12 −3m −12 ■>■ −3 −3 m■>■4 b

5(x■-■2)■í■7(x■+■3) 5x■-■10■í■7x■+■21 5x■-■10■-■7x■í■7x■+■21■-■7x -2x■-■10■í■21 -2x■-■10■+■10■í■21■+■10 -2x■í■31 −2 x 31 ■Ç■ −2 −2 −31  x■Ç■ 2 

1

x■Ç■-15 2

remember

1.■ The■solution■to■an■inequation■is■a■portion■of■the■number■line.■(That■is,■there■are■an■ infi■nite■number■of■solutions■to■any■given■inequation.) 2.■ When■solving■an■inequation,■imagine■an■equals■sign■in■place■of■the■inequality■sign■and■ solve■as■if■it■was■a■linear■equation.■Remember■to■keep■writing■the■original■inequality■ sign■between■the■two■sides■of■each■step. 3.■ Special■case:■if■in■the■process■of■the■solution■you■need■to■multiply■or■divide■both■sides■ of■the■inequation■by■a■negative■number,■reverse■the■inequality■sign.■That■is,■change■ <■to■>,■>■to■<,■Ç■to■í■and■í■to■Ç. exerCise

4e inDiviDuAl pAthWAys eBook plus

Activity 4-E-1

Puzzling inequations 1 doc-5002 Activity 4-E-2

Puzzling inequations 2 doc-5003

108

solving linear inequations fluenCy 1 We11a ■Solve■each■of■the■following■inequations. a x■+■1■>■3■ x■>■2 b■ a■+■2■>■1■ a■>■-1 m■í■4 d m■-■1■í■3■ e■ p■+■4■<■5■ p■<■1 m■Ç■9 g m■-■5■Ç■4■ h■ a■-■2■Ç■5■ a■Ç■7 m■í■2 j 5■+■m■í■7■ k■ 6■+■q■í■2■ q■í■-4

y■-■3■í■4 y■í■7 x■+■2■<■9 x■<■7 x■-■4■>■-1 x■>■3 5■+■a■>■-3 a■>■-8 2 Solve■each■of■the■following■inequations.■Check■your■solutions■by■substitution. a 3m■>■9■ m■>■3 b■ 5p■Ç■10■ p■Ç■2 c■ 2a■<■8 a■<■4 d 4x■í■20■ x■í■5 e■ 5p■>■-25■ p■>■-5 f■ 3x■Ç■-21 x■Ç■-7

maths Quest 10 for the Australian Curriculum

c■ f■ i■ l■

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

m■í■-0.5 g 2m■í■-1■

inDiviDuAl pAthWAys

Activity 4-E-3

Puzzling inequations 3 doc-5004

3

4

eBook plus

5

Digital doc

SkillSHEET 4.7 doc-5218

6

7

8

9

m ■>■6 m■>■18 3 m l■ ■í■5 m■í■25 5 i■

x a x■<■8 k■ Ç■-2■ a■Ç■-14 ■<■4■ 2 7 We11b ■Solve■each■of■the■following■inequations. a 2m■+■3■<■12■ m■<■4.5 b■ 3x■+■4■í■13■ x■í■3 c■ 5p■-■9■>■11 p■>■4 n■Ç■2 d 4n■-■1■Ç■7■ e■ 2b■-■6■<■4■ b■<■5 f■ 8y■-■2■>■14 y■>■2 m■Ç■-1 g 10m■+■4■Ç■-6■ h■ 2a■+■5■í■-5■ a■í■-5 i■ 3b■+■2■<■-11 b■<■-4 1 3 c■Ç■-1 j 6c■+■7■Ç■1■ k■ 4p■-■2■>■-10■ p■>■-2 l■ 3a■-■7■í■-28 a■í■-7 We11c ■Solve■each■of■the■following■inequations. ■ m■>■3 a 2m■+■1■>■m■+■4■ b■ 2a■-■3■í■a■-1■ a■í■2 c■ 5a■-■3■<■a■-■7 a■<■-1 a■Ç■-3 d 3a■+■4■Ç■a■- 2■ e■ 5x■-■2■>■40■-■2x■ x■>■6 f■ 7x■-■5■Ç■11■-■x x■Ç■2 b■<■4 g 7b■+■5■<■2b■+■25■ h■ 2(a■+■4)■>■a■+■13■ a■>■5 i■ 3(m■-■1)■<■m■+■1 m■<■2 m■Ç■3 k■ 3(5b■+■2)■Ç■-10■+■4b■ j 5(2m■-■3)■Ç■3m■+■6■ l■ 5(3m■+■1)■í■2(m■+■9)■ m■í■1 16 Solve■each■of■the■following■inequations. b■Ç■- 11 x−2 x+7 x +1 ■í■-4■ x■í■-18 ■<■-1 x■<■-10 a b■ c■ ■Ç■4■ ■ x■Ç■7 2 5 3 2x + 3 3x − 1 5x + 9 ■>■6■ ■í■2■ x■í■5 ■<■0 x■<■-1 45 d e■ f■ x■>■10 1 2 4 7 6 We12 ■Solve■each■of■the■following■inequations. ■ m■<■-2 a -2m■>■4■ b■ -5p■Ç■15■ p■í■-3 c■ -2a■í■-10 a■Ç■5 p■í■-5 d -p■-■3■Ç■2■ e■ 10■-■y■í■13■ y■Ç■-3 f■ 14■-■x■<■7 x■>■7 1 g 1■-■6p■>■1■ h■ 2■-■10a■Ç■0■ a■í■ 5 i■ 2(3■-■x)■<■12 x■>■-3 p■<■0 a■Ç■-11 j -4(a■+■9)■í■8■■ k■ -15■Ç■-3(2■+■b)■■ b■Ç■3 l■ 2x■-■3■>■5x■+■6 x■<■-3 k■>■8 m k■+■5■<■2k■-■3■ n■ 3(x■-■4)■<■5(x■+■5)■ o■ 7(a■+■4)■í■4(2a■-■3) a■Ç■40 1 x■>■-18 mC ■When■solving■the■inequation■-2x■>■-7■we■need■to: 2 ✔ b■ change■the■sign■to■<■ a change■the■sign■to■í■ c■ change■the■sign■to■= d change■the■sign■to■Ç e■ keep■the■sign■unchanged Solve■each■of■the■following■inequations. 5− m 2− x −3 − x ■í■2■ m■Ç■-3 ■<■-4 x■>■17 a b■ c■ ■>■1■ ■ x■<■-1 3 4 5 3 − 8a 4 − 3m −2m + 6 a■> 5 d e■ f■ ■<■-1■ ■Ç■0■ m■í■1 13 ■Ç■3 m■í■-12 8 2 2 10 Solve■each■of■the■following■inequations. 2 a 3k■>■6■ ■ k■>■2 b■ -a■-■7■<■-2■ a■>■-5 c■ 5■-■3m■í■0 m■Ç■1 3 d x■+■4■>■9■ e■ 10■-■y■Ç■3■ y■í■7 f■ 5■+■3d■<■-1 d■<■-2 x■>■5 7p 1− x − 4 − 2m −6 p■í■ 7 ■>■0 m■<■-2 g h■ i■ ■í -2■ ■Ç■2■ x■í■-5 3 3 5 a■<■9 j 5a■-■2■<■4a■+■7■ k■ 6p■+■2■Ç■7p■-■1■ p■í■3 l■ 2(3x■+■1)■>■2x■-■16 x■>■-4 1 2 j

eBook plus

h■ 4b■>■-2■ b■>■-0.5

unDerstAnDing eBook plus

10 Write■linear■inequations■for■the■following■statements,■using■x■to■represent■the■unknown.■

(Do■not■attempt■to■solve■the■equations.) a The■product■of■5■and■a■certain■number■is■greater■than■10. ■ 5x■>■10 SkillSHEET 4.8 b When■three■is■subtracted■from■a■certain■number■the■result■is■less■than■or■equal■to■5. x■-■3■Ç■5 doc-5219 c The■sum■of■seven■and■three■times■a■certain■number■is■less■than■42. 7■+■3x■<■42 11 Given■the■positive■numbers■a,■b,■c■and■d■and■the■variable■x,■there■is■the■following■relationship:■ -c■<■ax■+■b■<■-d. a Find■the■possible■range■of■values■of■x■if■a■=■2,■b■=■3,■c■=■10■and■d■=■1 –6.5■<■x■<■–2 −d − b −c − b b Rewrite■this■relationship■in■terms■of■x■only■(x■by■itself■between■the■<■signs). ■<■x■<■ Digital doc



a

a

Chapter 4 simultaneous linear equations and inequations

109

number AND algebra • Linear and non-linear relationships REASONING 12 Two speed boats are racing along a section of Lake Quikalong. The speed limit along this



section of the lake is 50 km/h. Ross is travelling 6 km/h faster than Steven and together they are travelling at a speed greater than 100 km/h. a Write an inequation and solve it to describe all possible speeds that Steven could be travelling at. S > 47 b At Steve’s lowest possible speed, is he over the speed limit? No c The water police issue a warning to Ross for exceeding the speed limit on the lake. Show that the police were justified in issuing a warning to Ross. Answers will vary. 13 At the beginning of this chapter we looked at the decision about which of two companies John should use when pouring different volumes of concrete. Angelico’s Concrete charges $700 plus $20 per cubic metre of concrete. Baux Cementing charges $1200 plus $15 per cubic metre of concrete. a Write an algebraic equation for the cost of using Angelico’s Concrete. CA = 700 + 20x b Write an algebraic equation for the cost of using Baux Cementing. CB = 1200 + 15x c Write an inequation that, when solved, will tell you the volume of concrete for which it is cheaper 7 00 + 20x < 1200 + 15x, reflection    to use Angelico’s Concrete. x < 100 d For what volume of concrete will it be cheaper to What is similar and different when solving linear inequations use Baux Cementing? x > 100 to linear equations? e For what volume of concrete will the cost be the same (if any)? x = 100

4f

Sketching linear inequations ■■ ■■

■■

■■

110

Linear inequations replace the equality sign with an inequality sign, namely, >, í, < and Ç. The graph of linear inequations is a half plane and it is related to the graph of the corresponding linear equation in that the line forms the boundary between the two half ■ planes. Consider the linear inequation y í x + 2. There are many points (x, y) that satisfy this inequation.   All the points that lie on the line y = x + 2 satisfy this y (2, 5) inequation, as well as many other points on the Cartesian plane for which the y-coordinate is greater than two more than the x-coordinate. (0, 2)   For example, (2, 5), (2, 6), (-4, 0) are some points that (-4, 0) satisfy this inequation. x   The graph of an inequation is a half plane; in this (-2, 0) example, it is the region of the Cartesian plane above the line.   The region that is required has been shaded, but sometimes it is helpful to shade the region that isn’t ■ required. Now consider the inequation y < x + 2. All the coordinates y that have a y-coordinate less than 2 more than their (2, 5) x-coordinate will satisfy this inequation.   Some solution points would be (2, 1), (3, 2) and (4, 2).   To sketch the graph of this inequation we need to shade (0, 2) (-4, 0) a region of the Cartesian plane that is below the line. x   In this example the points that lie on the line are not (-2, 0) part of the solution to the inequation, so the line is dotted, indicating that it is not included in the solution.

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Summary 1 If the inequation is of the form of y > mx + c, then the region above the line is shaded

and the line is dotted, indicating that the points that lie on the line are not part of the ■ solution. y

y

x

m>0

Required region

x

m<0

Required region

2 If the inequation is of the form of y í mx + c, then the region above the line is shaded

and the line is solid, indicating that the points that lie on the line are part of the ■ solution. y

y

x

m>0

Required region

x

m<0

Required region

3 If the inequation is of the form of y < mx + c, then the region below the line is shaded

and the line is dotted, indicating that the points that lie on the line are not part of the ■ solution. y

y

x

m>0

Required region

x

m<0

Required region

Chapter 4 Simultaneous linear equations and inequations

111

number AND algebra • Linear and non-linear relationships 4 If the inequation is of the form of y Ç mx + c, then the region below the line is shaded and the

line is solid, indicating that the points that lie on the line are part of the solution. y

y

x

m>0

x

m<0

Required region

Required region

Worked Example 13

Sketch the half plane given by each of the following inequations. a  y í x + 2 b  y < 3x - 1 Think a

Write a yíx+2

1

Write the inequation.

2

Sketch the linear equation, showing the x- and y-intercepts.

y=x+2 Let y = 0, x = -2.

3

To find the x-intercepts, let y = 0.

Therefore (-2, 0) is the x-intercept.

4

To find the y-intercept, let x = 0.

The y-intercept can be read from the equation, y-intercept is (0, 2).

5

Sketch the line labelling the x- and y-intercepts.

y (0, 2) (-2, 0)

6

112

Since the inequation is of the form ■ y í mx + c, then the region above the line is shaded and the line is solid. The region of the plane where the points always have the y-coordinate greater than or equal to the x-coordinate plus two will be above the line. Therefore the region that is required is above the line. Shade this region.

Maths Quest 10 for the Australian Curriculum

x

y (0, 2) (-2, 0)

x

Required region

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships b

b y■<■3x■-■1

1

Write■the■inequation.

2

Sketch■the■linear■equation,■showing■ the■x-■and■y-intercepts.

y■=■3x■-■1■

3

To■fi■nd■the■x-intercepts,■let■y■=■0.

Let■y■=■0,■x■= 3 .

4

To■fi■nd■the■y-intercept,■let■x■=■0.

5

Sketch■the■line■labelling■the■x-■and■ y-intercepts.■Since■the■inequality■is■of■ the■form■<■then■this■line■should■be■a■ broken■line.

1

Therefore■( 1,■0)■is■the■x-intercept. 3 The■y-intercept■can■be■read■from■the■equation. The■y-intercept■is■(0,■-1). y y = 3x -1 (1–3, 0)

x

(0, -1)

6

Since■the■inequation■is■of■the■form y■< mx+ c,■then■the■region■below■the■ line■is■shaded■and■the■line■is■dotted.

y

y < 3x - 1 (1–3, 0) (0, -1)

x

Required region

remember y

(-1, 0 )

(0, 1) 0

x

exerCise

4f inDiviDuAl pAthWAys eBook plus

Activity 4-F-1

Understanding linear inequations doc-5005

1.■ Inequations■involve■the■inequality■sign■> (greater■than),■í (greater■than■and■equal■to),■ < (less■than)■and■Ç■(less■than■and■equal■to). 2.■ The■graph■of■a■linear■inequation■is■a■half■plane. 3.■ A■broken■line■is■used■for■>■or■<■signs,■and■a■solid■line■is■used■for■í■and■Ç■signs. 4.■ If■the■inequation■is■of■the■form■y■í■mx■+■c■or■y■>■mx■+■c,■then■the■region■above■the■line■ is■the■required■region. 5.■ If■the■inequation■is■of■the■form■y■Ç■mx■+■c■or■y■<■mx■+■c,then■the■region■below■the■line■ is■the■required■region.

sketching linear inequations fluenCy 1 We13 ■Sketch■the■half■plane■given■by■each■of■the■following■inequations. y a y■í■x■+■1 b y■<■x■-■6 c y■>■-x■-■2 d y■<■3■-■x e y■>■x■-■2 f y■<■4 (-2, 0) 0 x g x■í -5 h y■Ç■x■-■7 (0, -2) i y■í -2 j y■<■x■+■7 k x■<■6 l y■Ç■3x

y

(6, 0) 0

x

(0, -6)

Chapter 4 simultaneous linear equations and inequations

113

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

inDiviDuAl pAthWAys eBook plus

2 Verify■your■solutions■to■question■1■using■technology. 3 mC ■a■ ■The■shaded■region■satisfying■the■inequation■y■>■2x■-■1■is: y y ✔ b a (1–2, 0)

Activity 4-F-2

Graphing linear inequations doc-5006

(0, -1)

(1–2, 0)

x

(0, -1)

y

c

(1–2, 0)

x

(0, -1)

Activity 4-F-3

Interpreting linear inequation graphs doc-5007

d

y

e (1–2, 0)

(0, -1)

y (- –21 , 0)

x

x

(0, -1)

b The■shaded■region■satisfying■the■inequation■y■Ç■x■+■4■is: y a b (0, 4)

x

(-4, 0)

y

c

x

(-4, 0)

e

(0, 4)

x

(-4, 0)

y

✔ d

(0, 4)

y

(0, 4)

x

(-4, 0)

y (0, 4) (4, 0) x

c The■region■satisfying■the■inequation■y■<■-3x■is: y b ✔ a

y (1, 3)

(0, 0) x (1, -3)

114

maths Quest 10 for the Australian Curriculum

(0, 0)

x

x

number AND algebra • Linear and non-linear relationships y

c

y

d

(-1, 3)

(0, 3) x

(0, 0)

x

(-1, 0)

y

e

(-1, 0)

(0, 1) x

understanding 4

y 10 9 8 7 6 5 4 3 2 1 0

a Find the equation of the line l shown in the diagram at left. y = 12 x + 3 b Write down three inequations which define the region R. y í 1 x + 3, x > 2, y Ç 7 2

l R

1 2 3 4 5 6 7 8 9 10

x

5 Happy Yaps Dog Kennels charges $35 per day for large dogs (dogs over 20 kg) and $20 per

day for small dogs (less than 20 kg). On any day, Happy Yaps Kennels can only accommodate a maximum of 30 dogs. a If l represent the number of large dogs and s represents the number of small dogs. Write down an inequation, in terms of l and s, that represents the total number of dogs at Happy Yaps. l + s Ç 30 b Another inequation can be written as s í 12. In the context of this problem, write down what this inequation represents. At least 12 small dogs c The inequation l Ç 15 represents the number of large dogs that Happy Yaps can accommodate on any day. This inequation is shown as a bold line on the graph below, clearly shade in the area that is not within the region for this inequation. l

l 30

30

15

15 0

0

12

30

12

30

s

s

d Explore the maximum number of small and large dogs Happy Yaps Kennels can accommodate to receive the maximum amount in fees. 15 large and 15 small dogs Chapter 4 Simultaneous linear equations and inequations

115

number AND algebra • Linear and non-linear relationships Reasoning y>x+2

6 a Given the following graph, state the inequation it describes. b Prove, by choosing a point on the graph, that the inequation is correct. Answers will vary.

reflection 

4 2 -4

 

y

0

-2

2

4

x

-2

How are the graphs for linear equations and inequations similar and different?

-4 Region required

4G

Solving simultaneous linear inequations ■■ ■■ ■■ ■■ ■■ ■■

The graph of a linear inequation represents a region of the Cartesian plane and not simply a line. This means that two linear inequations will have two regions as graphs. If these regions intersect, they have an infinite number of points in common. It is easier to solve simultaneous inequations graphically rather than algebraically. The process involves drawing each half plane on a Cartesian plane, shading the regions that are required for each inequation. The region that is shaded by both inequations is the solution to the simultaneous inequations.

Worked Example 14

Use a graphical technique to solve the following simultaneous inequations. y í x + 1 2 x - y > 4 Think

116

Write/draw

1

Write the inequations, one under the other and number them.

y í x + 1 2x - y > 4

2

Find the x- and y-intercepts for the boundary equation of inequation [1].

For [1], the boundary is y = x + 1. x-intercept: when y = 0, 0=x+1 x+1=0 x = -1 The x-intercept is at (-1, 0). y-intercept: when x = 0, y=0+1 y=1 The y-intercept is at (0, 1).

3

Rule a pair of coordinate axes and choose a suitable scale to allow all the intercepts to be marked.

4

For inequation [1] (í), sketch a solid line through both intercepts.

5

Label the inequation.

6

For inequation [1] test the point (0, 0) to see if a TRUE or FALSE statement is generated.

Maths Quest 10 for the Australian Curriculum

Test point: (0, 0) Is 0 í 0 + 1? Is 0 í 1?     FALSE

[1] [2]

number AND algebra • Linear and non-linear relationships

7

y

As the statement is FALSE, the opposite side of the line is required. Shade the region required. Note: Since the equation is in the form y í x + 1, we would expect the required region to be above the line. y≥x+

14 12 10 8 6 4 12

−8 −6 −4 −2−20 −4 −6 −8 Region required −10

8

Find the x- and y-intercepts for the boundary equation to inequation [2].

9

For inequation [2] (>), sketch a broken line through both intercepts.

10

Label the inequation.

11

For inequation [2] test the point (0, 0) to see if a TRUE or FALSE statement is generated.

12

As the statement is FALSE, the opposite side of the line is required. Shade the region required (below the dotted line).

13

Indicate the solution region on the graph. This is the overlapping of the two shaded regions.

For [2], the boundary is 2x - y = 4. x-intercept: when y = 0, 2x - 0 = 4 2x = 4 x=2 The x-intercept is at (2, 0). y-intercept: when x = 0, 2(0) - y = 4 -y = 4 y = -4 The y-intercept is at (0, -4).

Test point: (0, 0) Is 2(0) - 0 > 4?     FALSE

y

y-x+

Check your solution by substituting a point from the solution region, say (7, 9), into each of the original inequations.

l So

n

io

ut

14 12 10 8 6 4 12

-8 -6 -4 -2-20 -4 -6 -8 -10

14

x

2 4 6 8 10 12

2 4 6 8 10 12

x

2x - y > 4

Region required

Check: Substituting (7, 9) into [1]: yíx+1 9í7+1 9 í 8 is true. Substituting (7, 9) into [2]: 2x - y > 4 2(7) - 9 > 4 14 - 9 > 4 5 > 4 is true. The solution region is correct.

Chapter 4 Simultaneous linear equations and inequations

117

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

1.■ The■graph■of■a■linear■inequation■represents■a■region■of■the■Cartesian■plane. 2.■ A■graphical■technique■can■be■used■to■solve■simultaneous■inequations. 3.■ The■graph■of■two■simultaneous■inequations■consists■of■the■intersection■of■two■regions■ and■therefore■has■an■infi■nite■number■of■solutions. 4.■ It■is■usual■to■shade■the■wanted■region■on■the■graph■of■an■inequation. exerCise

solving simultaneous linear inequations

4g

fluenCy

inDiviDuAl pAthWAys

Activity 4-G-1

Introducing simultaneous linear inequations doc-5008

3

Activity 4-G-3

1

y

-6

-4

-2

2

4

6 x

-2 -4



2

3

4

5

-3 c 3x■-■2y■>■12 y 6

-3 d 4x■+yí■-8■

2 4

-1 0 -2

x

x 1

2

4

6

y■í■x■+■4■ y

y=x+4

5

-6

-6

-4

-2

-2 -4

118

2

4

-4

y



-2

-2

2

1

2

4

6

-4

maths Quest 10 for the Australian Curriculum



10

4x■+■y■í■-8■ y 2 1 -4

x 1

2

3

-2 -10 -2 -3 -4 -5 -6 -7 -8 -9

2

x

4x + y = -8 y■<■3■-■3x

2 x

8

y

0 -1 -2 -4 -6 -8 -10 -12

4

0

6



3

-2

x

-3

f y■<■3■-■3x

4 2

-4

3x - 2y = 12

-6 e y■íx■+■4■

6

3

4

12 10 8 6 4 2

4

-4

0

2

-2

3x■-■2y■>■12■ y 2 1 -4 -2 -10 -2 -3 -4 -5 -6 -7

2

x

-2 0 -1

-2

4 2

x 1

x + 2y = 6

1

x

-1 0 -1

6

0

2

x+y=3

6

x■+■y■>■3

2

y

4



3

2

Further simultaneous linear inequations doc-5010

y

x■+■2y■Ç■6■

Practising simultaneous linear inequations doc-5009



Activity 4-G-2

y 3 2 1 -4 -2-10 -2

Note:■Questions■1■and■2■revise■the■skills■used■when■working■with■inequations. 1 For■each■of■the■following,■use■substitution■to■check■if■the■given■pair■of■coordinates■makes■the■ inequation■true■or■false. a (2,■4)■ x■+■3y■>■13 y■-■2x■<■7 False c (0,■5)■ y■Ç■5x■+■4 False True b (-3,■2)■ d (1,■-4)■ 5x■+y<■6 True e (7,■1)■ 2x■+■5y■Ç■19 f (2,■3)■ 2x■-y>■6 False g (-2,■-3)■ 2x■-■3y■>■3 True h (-5,■-4)■ y■>■7■+■2x False i (3,■0)■ y■Ç■-3x False j (0,■4)■ y■+■2x■>■4 False True 2 Use■the■graphs■of■the■equations■given■below■to■sketch■the■graph■of■the■given■inequations.■ Note:■the■shaded■region■is■the■region■required. (Remember■to■shade■the■region■required.) a x■+y>■3 b x■+■2y■Ç■6■

eBook plus

-3 -2 -1 0 -1

y

y 3 2

y = 3 - 3x x 1

2

3

1 -1

0

-1 -2

1

2

3

x

number AND algebra • Linear and non-linear relationships



g y - 3x < 9

y - 3x < 9  y 10 9 8 7 6 5 4 3 2 1 -3 -2 -1 0

h 2x + y í 8 y

9 6 y - 3x = 9 3

1

x

-9 -6 -3 0 -3

2 x

3

9

6

y

12 10 8 6 4 2 -6

-2 -20 -4 -6

-4



–6

2x + y = 8 x 2

4

6

3   MC  For each of the following pairs of simultaneous inequations, choose the graph which

gives the correct solution. (Remember the required region is shaded.) a y > x + 3

x+yÇ4

✔ a

6 5 4 3 2 1

y

6 5 4 3 2 1

3

4

d

2

3

4

6 5 4 3 2 1

1

3

4

5

y

x 2

1

-2

3

4

5

Region required

–4

y

h 2x + y í 8 y Region required

10 9 8 7 6 5 4 3 2 1

x

0

-1

2

Region required

0

-1

5

Region required

–4

E

6 5 4 3 2 1

x 1

x

-2

y

-2

y

0

-1

5

Region required

0

-1

2

1

-2

c

6 5 4 3 2 1

x

0

-1

b

2

1

3

4

5

-2 –4

0

-1

1

2

3

4

x

b x Ç 5

2y + x > 2 a

3

y

b

Region required

2

y

Region required

2

1 -1 0 -1

3

x 1

2

3

4 5

1 -1 0 -1

-2

-2

-3

-3

x 1

2

3

4 5

Chapter 4 Simultaneous linear equations and inequations

119

number AND algebra • Linear and non-linear relationships ✔ c

3

d

y

1

1

x

-1 0 -1

1

2

x

-1 0 -1

3 4 5

1

2

3 4 5

-2

-2 -3 3

y

2

2

E

3

-3

Region required

y

Region required

Region required

2 1

x

-1 0 -1

1

2

3

4 5

-2 -3 c y í 3 - x

2x + 3y Ç 6 a

3

✔ b

y

1 1

2

3

4

y

d

Region required

3

4

5

3

y

Region required

2

1

x

-1 0 -1

1

2

3

4

1 -1 0 -1

5

-2

-2

-3

-3 y

Region required

2 1

x 1

2

3

4

-2 -3 120

2

-3

Region required

2

-1 0 -1

1

-2

-3

3

x

-1 0 -1

5

-2

E

Region required

1

x

-1 0 -1

3

y

2

2

c

3

Maths Quest 10 for the Australian Curriculum

5

x 1

2

3

4

5

number AND algebra • Linear and non-linear relationships 4 Note: The shaded region is the region required. a

y

6

d x - y < 3

x - 2y í 4 a

2x - y í 4

3

2

1

–2

-1 0 -1

x+y<3

–4 –6

c

x

0

6

4

2

-2

10

8

x + 5y Ç 10

-4 -6

c

2y > x - 2

2 1 0

3

2

1

-1 -2

x

5

4

✔ E

y<3-x

-3

d

6

y > 2x + 4

-2

-1

0

-2 -4

2

1

y < 4 - 2x

y

x yÇx+2

4

y í 4 - 2x

0

2

x

y yÇx+2

4 2 2

d

Region required

3

1

1 1

2

3

4

5

x

-2

-3

-3 y

4 e

Region required

y

–1 0 -1

-2

y

6

4

5

Region required

1

2

3

4

5

x

-6

-4

-2

0

-2 -4

-2

-6

4

x

5

y

x+y>4

2

1

3

2

f y - 2x Ç 5

4

2

4

6 x

20 18 16 14 12 10 8 6 4 2 0

-2 -4

3x + y > 17

y<8 x 1

2

3

4

5

5 2–3

6

understanding 4   WE 14  Use a graphical technique to solve the following simultaneous inequations. a x + y < 3 b 3x + 2y > 12 c 2y > x - 2

2x - y í 4

y

-2 0

y

2

d y > 2x + 4

2 –2 0

3

2

-3

2

3

1

-3

5 Note: The shaded region is the region required. a   i 

3

-1 0 -1

3 x

-6

b

5

1

2

ii 

4

2

y

4

-3

3

Region required

-1 0 -1

y

3

-1

2

-2

-3

2 -2

1

3x + 2y > 12

4

Region required

x

-1 0 -1

-2 y

6

y

1

x

x

5

4

3 2

1

0

b

b

y

2

4 2 –1

3

x y > 4 - 2x

y < 4 - 2x g x + 2y í 10 3x + y > 15 j y - x > 4 2x + 3y Ç 6 m x + y > 7 2x - 3y í 18

x + 5y Ç 10

e y - 2x Ç 5

x+y>4 h y > 2x - 3 x<5 k y + 2x > 3 y < 2x n y > 4 y í 2x

y<3-x

f y < 8

3x + y > 17 i 3y - 2x < 6 y í 2x - 2 l y - 2x í 9 x + y Ç 4 Note: The shaded region is the region required. y 2

2x + y < 0 1 Sketch the half plane represented by the region:  i y Ç x + 2 -1 0 i i y í 4 - 2x. Show the region where both the inequations y Ç x + 2 and y í 4 - 2x hold true. 6 Show the region where the inequations 2x + y < 0 and x - 2y > 0 simultaneously hold true.

5 a b

x - 2y > 0 x

1 2

Chapter 4 Simultaneous linear equations and inequations

121

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 7 Natsuko■is■starting■to■plan■a■monthly■budget■by■classifying■expenditures■such■as■rent■and■other■

expenses■(r)■and■savings■(x).■Her■total■net■income■is■$2000■per■month.■She■can■spend■no■more■ Note: The■shaded■region■ than■30■per■cent■of■her■income■on■rent. a Write■an■inequation■to■express■the■constraint■that■Natsuko■can■spend■no■more■than■ is■the■region■required. x $2000■per■month. r■+■x■Ç■2000 r■Ç■600 b Write■an■inequation■to■express■the■30■per■cent■rent■and■other■expenses■limitation. 2000 r■í■0,■x■í■0.■Amount■of■money■ c Do■any■other■inequations■apply■to■this■situation?■Explain. d Sketch■a■graph■of■the■region■that■applies■for■all■your■inequations. cannot■be■negative. e State■three■possible■solutions■of■allocating■rent■and■other■expenses/savings. Answers■will■vary. 8 Monica■wants■to■take■a■minimum■of■450■units■of■vitamin■C■and■300■units■of■vitamin■E■per■ day.■Each■brand■A■tablet■provides■100■units■of■vitamin■C■and■50■units■of■vitamin■E,■while■each■ 0 600 2000 r brand■B■tablet■provides■75■units■of■vitamin■C■and■75■units■of■vitamin■E. a Write■an■inequation■which■indicates■the■combination■needed■of■each■brand■of■vitamin■ tablet■to■meet■the■daily■requirement■of■vitamin■C. 100a■+■75b■í■450 (Hint:■Let■a■=■the■number■of■brand■A■tablets■and■b=■the■number■of■brandBtablets.) Note: The■shaded■region■ b Write■an■inequation■which■indicates■the■combination■needed■of■each■brand■of■vitamin■ is■the■region■required. 50a■+■75b■í■300 tablet■to■meet■the■daily■requirement■of■vitamin■E. b c Graph■the■two■inequations■and■indicate■the■region■which■provides■a■solution■to■both■the■ 6 vitamin■C■and■vitamin■E■requirements. d Recommend■to■Monica■two■different■vitamin■plans■that■fi■t■the■restrictions. Answers■will■vary. 4 9 Maria■is■making■some■high-energy■sweets■using■peanuts■and■chocolate■chips.■She■wanted■ 2 to■make■a■maximum■of■400■g■of■the■sweets,■but■wanted■them■to■contain■at■least■180■g■of■ carbohydrates. 0 a 1 2 3 4 5 6 a Let■the■mass■of■peanuts■be■p■and■the■mass■of■chocolate■chips■be■c.■Write■an■inequation■to■ represent■the■constraint■that■the■total■mass■must■be■at■most■400■■g. ■ p■+■c■Ç■400 b On■a■Cartesian■plane,■sketch■the■region■defi■ned■by■the■inequation■obtained■in■part■a. (Hint:■Consider■only■the■positive■axes■as■the■values■of■both■p■and■c■must■be■positive.) c The■peanuts■provide■30%■of■their■mass■in■carbohydrates■and■the■chocolate■chips■provide■ 60%■of■their■mass■in■carbohydrates.■Write■an■inequation■that■represents■the■constraint■that■ g. 0.3p■+■0.6c■í 180 the■mass■of■carbohydrates■must■be■at■least■180■■ d On■a■Cartesian■plane,■sketch■the■region■defi■ned■by■the■inequation■obtained■in■part■c . e On■a■Cartesian■plane,■show■the■region■where■the■inequations■sketched■in■parts■b■and■d■ both■hold■true. eBook plus f The■region■obtained■in■part■e■shows■all■possible■ Digital doc refleCtion    masses■of■peanuts■and■chocolate■chips■that■meet■ WorkSHEET 4.3 Maria’s■requirements.■List■fi ■ ve■sets■of■possible■ How do the solutions from a doc-5222 masses■of■peanuts■and■chocolate■chips■that■would■ system of equations differ from a system of inequations? meet■her■requirements. Answers■will■vary.

122

600 500 400 300 200 100 p + c Ç 400 0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

e 700 600 500 400 300 200

0.3p + 0.6c í 180

100

maths Quest 10 for the Australian Curriculum

0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

Mass of chocolate chips in grams (c)

700

d Mass of chocolate chips in grams (c)

Mass of chocolate chips in grams (c)

9 b■ Note: The■shaded■region■ is■the■region■required.

700 600 500 400 300 200 100 0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

number AND algebra • Linear and non-linear relationships

Summary Graphical solution of simultaneous linear equations ■■

■■ ■■ ■■

When solving simultaneous equations graphically, obtaining an accurate solution depends on drawing accurate graphs. The solution to linear simultaneous equations is the point where their graphs intersect. Lines that have the same gradient are parallel. If the graphs of the two simultaneous equations are parallel lines, then the simultaneous equations have no solution, as they have no point of intersection. Solving simultaneous linear equations using substitution

When using the substitution method to solve simultaneous equations: choose the equation in which one of the variables is the subject ■■ substitute this expression for the variable into the other equation and solve ■■ substitute the value you have found into the rearranged equation to solve for the other ■ variable ■■ check your solution. ■■

Solving simultaneous linear equations using elimination ■■

■■

■■

■■

■■

Simultaneous equations of the form ax + by = k can be solved by the elimination method by looking for an addition or subtraction of the equations that will eliminate one of the variables. For like terms with the same coefficient but opposite signs, add the equations. For like terms with the same coefficient and the same sign, subtract the equations. If the terms do not have the same coefficient, multiply one or both equations by a constant to create the same coefficient. Remember to multiply both sides of the equation to keep it balanced. Once one variable has been eliminated, solve the single variable equation formed. Substitute the solution back into one of the original equations to find the value of the variable that was originally eliminated. Check your solution by substitution. Problem solving using simultaneous linear equations

■■

■■ ■■

■■

To solve worded problems, read the question carefully and define the two variables using appropriate pronumerals. Formulate two equations from the information given and number them. Use either the elimination method or the substitution method to solve the two equations simultaneously. Check your answer by substituting the values obtained for each variable into the original equations. Solving linear inequations

■■

■■

■■

The solution to an inequation is a portion of the number line. (That is, there are an infinite number of solutions to any given inequation.) When solving an inequation, imagine an equals sign in place of the inequality sign and solve as if it was a linear equation. Remember to keep writing the original inequality sign between the two sides of each step unless the special case applies. Special case: if in the process of the solution you need to multiply or divide both sides of the inequation by a negative number, reverse the inequality sign. That is, change < to >, > to <, Ç to í and í to Ç. Chapter 4 Simultaneous linear equations and inequations

123

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships Sketching linear inequations ■■ ■■ ■■ ■■ ■■

Inequations■involve■the■inequality■sign■>■(greater■than),■í■(greater■than■and■equal■to),■ <■(less■than)■and■Ç■(less■than■and■equal■to). The■graph■of■a■linear■inequation■is■a■half■plane. A■broken■line■is■used■for■>■or■<■signs,■and■a■solid■line■is■used■for■í■and■Ç■signs. If■the■inequation■is■of■the■form■y■í■mx■+■c■or■y■>■mx■+■c,■then■the■region■above■the■line■is■the■ required■region. If■the■inequation■is■of■the■form■y■Ç■mx■+■c■or■y■<■mx■+■c,■then■the■region■below■the■line■is■the■ required■region. Solving simultaneous linear inequations

■■ ■■ ■■ ■■

The■graph■of■a■linear■inequation■represents■a■region■of■the■Cartesian■plane. A■graphical■technique■can■be■used■to■solve■simultaneous■inequations. The■graph■of■two■simultaneous■inequations■consists■of■the■intersection■of■two■regions■and■ therefore■has■an■infi■nite■number■of■solutions. It■is■usual■to■shade■the■wanted■region■on■the■graph■of■an■inequation.

MaPPING YOUR UNdeRSTaNdING

Homework book

124

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■Whatdoyouknow?■on■page■89. Have■you■completed■the■two■Homeworksheets,■the■Richtask■and■two■Codepuzzles■in■ your■MathsQuest10HomeworkBook?

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Chapter review Fluency

7 During a walk-a-thon, Sarah receives $4 plus

$3 per kilometre. The graph which best represents Sarah walking up to 5 kilometres is:

1 The equation of the line drawn below is: y

A

3

2 x

0 ✔ A

3x + 2y = 6

0

B 3x - 2y = 6 D 2x - 3y = 6

C 2x + 3y = 6 E 2x - 3y = -6

B

2 The equation of a linear graph with gradient -3 and

x-intercept of 4 is: A y = -3x - 12 C y = -3x - 4 E y = 4x - 3

B y = -3x + 4 ✔ D

y = -3x + 12

(2, -7) and (-2, -2) is: 4x - 5y + 18 = 0 5x + 4y + 18 = 0 5x + 4y - 18 = 0 5x - 4y - 18 = 0 4x + 5y + 18 = 0

✔ C

A ✔ B c d e

4 The inequation that is represented by the region D

y 2 2

x

Region required

yí2-x C y Ç 2 - x E y í 2x

✔ A

E

5 The equation of a linear graph which passes

✔

through the origin with gradient -3 is: A y = -3 B x = -3 C y = -3x D y = 3 - 3x E y = 3x - 3

0

6 An online music shop charges $5 postage for

✔

$ 24 20 16 12 8 4

2 CDs and $11 for 5 CDs. The equation that best represents this, if C is the cost and n is the number of CDs, is: A C = 5n + 11 B C = 6n + 5 C C = n + 2 D C = 5n + 1 E C = 2n + 1

d (km) (5, 24)

1 2 3 4 5

d (km)

(5, 19)

1 2 3 4 5

$ 18 15 12 9 6 3 0

b y í x - 2 D y Ç x - 2

1 2 3 4 5

$ 24 20 16 12 8 4 0

shown below is:

(5, 18)

$ 24 20 16 12 8 4 0

3 The equation of a linear graph which passes through

0

$ 18 15 12 9 6 3

d (km) (5, 19)

1 2 3 4 5

d (km)

(5, 18)

1 2 3 4 5 d (km)

8 Which of the following pairs of coordinates is the

✔

solution to the given simultaneous equations? 2x + 3y = 18 5x - y = 11 A (6, 2) B (3, -4) C (3, 9) D (3, 4) E (5, 11)

Chapter 4 Simultaneous linear equations and inequations

125

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

-4 -6 -8 -10

b

1 -3 3

1

22

1 2 3 4 5x (3, -1)

y 10 8 6 5 4 2

0 -6 -5 -4 -3 -2 -1 -2 (-3, -1) -4

11 Sketch■the■graph■of■the■following■linear■equations,■

labelling■the■x-■and■y-intercepts. a y■=■3x■-■2 b y■=■-5x■+■15 2

16

7

d y■=■ 5 x■-■3■

1 -3 3

4

3

b y=■ x■-■5 8 x-intercept■=■-5.6,■ d y=■0.5x■+■2.8 y-intercept■c■=■2.8

3

c y=■ 7 x■-■ 4

a y■=■ 2 x

b y=■-4x

c x= -2

d y=■7

15 Sketch■the■graph■of■the■equation■

3(y■-■5)■=■6(x■+■1).

y 2x - 3y = 6 0 -2

3

x

y 7 (0, 7) x

0

- –27

3(y - 5) = 6(x + 1)

16 Find■the■equations■of■the■straight■lines■having■the■

following■graphs. a (3, 1)

y 14 a

a■ y■=■2x■-■2

0

1– 2

b

y x

0 -4

1 -2 2 -4 -5 -6 -8 -10

maths Quest 10 for the Australian Curriculum

c

(1, 1–2 ) 0

-2

-4

1 2 3 4 5x

y

x

1

b

e none■of■the■above 126

3

a y= -7x■+■6

14 Sketch■the■graph■of■each■of■the■following.

y 10 8 6 4 2

0 -6 -5 -4 -3 -2 -1 -2

x

12 Find■the■x-■and■y-intercepts■of■the■following■ 40 straight■lines. x-intercept■=■ 3 ■(131),■y-intercept■c■=■-5

1 2 3 4 5 x b y■=■-x■-■4 1 1 3 -4 3 2 2 c y■=■− 13 x■+■2 -6 -5 -8 -10

(-3, -1)

0 1

–2

1

0 -6 -5 -4 -3 -2 -1 -2

d

y = 3x - 2 (1, 1)

1

c y■=■- 3 x■+■1

fi■nding■the■x-■and■y-intercepts. a 2x■-■3y■=■6 b 3x■+y=■0 c 5x■+y= -3 d x+y+■3■=■0

1 2 3 4 5x

y 10 8 6 4 2

y

13 Sketch■graphs■of■the■following■linear■equations■by■

-6 -8 -10

c

the■equationy=■-5x■+■15■for■values■ofxbetween■ -10■and■+10.

21

0 -6 -5 -4 -3 -2 -1 -2

1

1 22 33

6

y 10 8 6 5 4 2

✔ a

10 Produce■a■table■of■values,■and■sketch■the■graph■of■

12 a■ x-intercept■=■ 7 ,■y-intercept■c■=■6

simultaneous■equations■is: y■=■5■-■2x y■=■3x■-■10

c x-intercept■=■16 ■(1 5 ),■y-intercept■c■=■- 43

9 The■graphical■solution■to■the■following■pair■of■

y

x 0 1 y = -4x -4

c

y

2 0

y

x = -2

d x

x

0

-2

6

x

1 y = 1–2 x

y 7 y=7 0

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships d

y

16 d■ y■=■4x

• (2, 8)

e

21 Use■the■graphs■below,■showing■the■given■

3 y■=■- 4

f x■=■5 x

0

20 Note:■The■shaded■region■ is■the■region■required. a

e

■ (3,■1)

1 yÇx+1

y

y c

b 3x■+■2y■=■12

x

2y■=■3x (2,■3)

y > 3x - 12

x

0

17 Find■the■linear■equation■given■the■information■in■

a■ y■=■-x■+■8

2

b y■=■- 32 x■+■12 c y■=■ 5 x +

27 5

1

■

c gradient■= 2 ,■y-intercept■=■5 y■=■ 12 x■+■5 d gradient■=■0,■y-intercept■=■6 y■=■6 18 For■each■of■the■following,■fi■nd■the■equation■of■the■

straight■line■with■the■given■gradient■and■passing■ through■the■given■point. y■=■7x■-■13 a gradient■=■7,■point■(2,■1) b gradient■= -3,■point■(1,■1) y■=■-3x■+■4 1

y■=■ 2 x■+■6

3

19 Find■the■equation■of■the■straight■line■that■passes■

through■each■pair■of■points. a (1,■7)■and■(3,■5) b (8,■0)■and■(6,■3) c (-1,■5)■and■(4,■7)

y = 5x

0 1

x y < 5x

y

0

x=7

7

xí7

x

2

4

6

8

x

coordinates■is■a■solution■to■the■given■simultaneous■ equations. a (7,■1)■ x■-■2y■=■5 ■ 5y■+■2x■=■18 ■ No b (4,■3)■ y■=■7■-x ■ 5y■-■2x■=■7 Yes

23 Solve■each■of■the■following■pairs■of■simultaneous■

equations■using■a■graphical■method. a 4y■-■2x■=■8 b y■=■2x■-■2 x■+■2y■=■0 ■ (-2,■1) x■-■4y■=■8 (0,■-2) c 2x■+■5y■=■20 y■=■2 (5,■2)

24 Solve■the■following■simultaneous■equations■using■

18

y 5

■



d gradient■= 5,■point■(1,■-3) y■=■ 35 x■-■ 5

20 Sketch■the■half■plane■given■by■each■of■the■

y

22 Use■substitution■to■check■if■the■given■pair■of■

each■case■below. a gradient■=■3,■y-intercept■=■-4 ■ y■=■3x■-■4 b gradient■= -2,■y-intercept■=■-5 y■=■-2x■-■5

following■inequations. a y■Ç■x■+■1 d b y■í■2x■+■10 c y>■3x■-■12 d y<■5x e xí7 1 f y■Ç■ 2 x■+■1 e g 2x■+■y■í■9 h 4x■-■3y■í■48 i y> -12

x

6 5 4 3 2 1 -2 -10 -2 -3 -4 -5

x

4

-12

1

7

-4

y

c gradient■= 2 ,■point■(-2,■5)

6

10

-5 0

5

4

y y í 2x + 10

0

2

-6

x b

- –43

-4 -2 0 -2

x

-1 0

0

f

y

simultaneous■equations,■to■write■the■point■of■ intersection■of■the■graphs■and,■hence,■the■solution■ of■the■simultaneous■equations. y a x■+■3y■=■6 3 y=■2x-■5 2

the■substitution■method. a y■= 3x■+■1 b x■+■2y■=■16 (2,■7) c 2x■+■5y■=■6 d 3 y■=■ 2 x■+■5 (-2,■2)

y■=■2x■+■7 3y■-■4x■=■11 (-5,■-3) y■= -x  7 7  − 3, 3  y■=■8x■+■21

e y■=■3x■-■11

f y■=■4x■-■17 5 y■=■5x■+■17 y■=■6x■-■22 ( 2,■-7) (-14,■-53) 25 Solve■the■following■simultaneous■equations■using■ the■elimination■method. a 3x■+y=■17 b 4x■+■3y■=■1 7x■-y=■33 ■ (5,■2) -4x +y=■11 (-2,■3) c 3x■-■7y■=■-2 d 4y■-■3x=■9 (1,■3) -2x -■7y■=■13 y■+■3x■=■6 (-3,■-1) e 5x■+■2y■=■6 f x■-■4y■=■-4 4x■+■3y■=■2 (2,■-2) 4x■-■2y■=■12 (4,■2) Chapter 4 simultaneous linear equations and inequations

127

number AND algebra • Linear and non-linear relationships 26 Solve the following simultaneous equations using

4 Write the following as a pair of simultaneous

an appropriate method. a 3x + 2y = 6 b 6x - 4y = -6 3y + 5x = 9 (0, 3) 7x + 3y = -30 (-3, -3) c 6x + 2y = 14 x = -3 + 5y (2, 1)

equations and solve. a Find two numbers whose difference is 5 and whose sum is 23. Numbers are 9 and 14. b A rectangular house has a total perimeter of 34 metres and the width is 5 metres less than the length. What are the dimensions of the 27 Solve the following simultaneous inequations. house? Length = 11 metres, width = 6 metres a y Ç x + 4 b 2y - 3x í 12 c If two Chupa Chups and three Wizz Fizzes cost yí3 y + 3x > 0 $2.55, but five Chupa Chups and seven Wizz c 5x + y < 10 C  hupa-chups cost 45 cents and Fizzes cost $6.10, find the price of each type of x + 2y < 11 Whizz fizzes cost 55 cents. lolly. 5 Laurie buys milk and bread for his family on the

problem solving 1 John has a part-time job working as a gardener and

Number of hours

0

2

4

6

8

10

Pay ($)

0

27

54

81

108

135

b Find a linear equation relating the amount

and kangaroos (4-legged). The total number of animals is 21 and they have 68 legs in total. Using simultaneous equations, determine how many cockatoos and kangaroos there are in the paddock. 7 At a fun park, the cost of a rollercoaster ride and

a Ferris wheel ride is $10. The cost of a Gravitron ride and a Ferris wheel ride is $12. The cost of a Rollercoaster ride and a Gravitron ride is $14. What is the cost of each ride? 8 There are two sections to a concert hall. Seats in the

‘Dress circle’ are arranged in rows of 40 and cost $140 each. Seats in the ‘Bleachers’ are arranged in rows of 70 and cost $60 each. There are 10 more rows in the ‘Dress circle’ than in the ‘Bleachers’ and the capacity of the hall is 7000. Number a If d represent the number of rows in the ‘Dress 0 2 4 6 8 10 of rides circle’ and b represents the number of rows in the ‘Bleachers’ then write an equation in terms Cost ($) 12.50 17.50 22.50 27.50 32.50 37.50 of these two variables based on the fact that b Find a linear equation relating total cost to the there are 10 more rows in the ‘Dress circle’ Cost = $2.50 ì number of rides + $12.50 than in the ‘Bleachers’. d = b + 10 number of rides. c Sketch the linear equation on a Cartesian plane b Write an equation in terms of these two over a suitable domain. variables based on the fact that the capacity of d Using algebra, calculate the cost of entry and the hall is 7000 seats. 7000 = 70b + 40d $30 7 rides. c Solve the two equations from a and b simultaneously using the method of your choice 3 The cost of hiring a boat is $160 plus b = 60 and d = 70 to find the number of rows in each section. $22.50 per hour. d Now that you have the number of rows in each a Sketch a graph showing the total cost for 0 to section, calculate the number of seats in each 12 hours. section. b State the equation relating cost to time rented. e Hence, calculate the total receipts for a concert c Predict the cost of hiring a boat for 12 hours and 15 minutes. where all tickets are sold. $435.63 $644  000 2 A fun park charges a $12.50 entry fee and an

Cost = 22.5 ì time + 160

additional $2.50 per ride. a Complete the following table of values relating the total cost to the number of rides.

128

N  umber of seats in ‘Bleachers’ is 4200; the number Maths Quest 10 for the Australian Curriculum of seats in the ‘Dress circle’ is 2800.

Rollercoaster ride $6, Ferris wheel ride $4, Gravitron ride $8

of money received to the number of hours Pay = $13.50 ì (number of hours worked) worked. c Sketch the linear equation on a Cartesian plane over a suitable domain. d Using algebra, calculate the pay that John will 3 $91.13 receive if he works for 6 4 hours.

6 A paddock contains some cockatoos (2-legged)

13 kangaroos and 8 cockatoos

is paid $13.50 per hour. a Complete the following table of values relating the amount of money received to the number of hours worked.

way home from school each day, paying with a $10 note. If he buys three cartons of milk and two loaves of bread, he receives 5 cents in change. If he buys two cartons of milk and one loaf of bread, he receives $4.15 in change. How much does each item cost? Milk $1.75, bread $2.35

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 9 John■is■comparing■two■car■rental■companies,■

10 Frederika■has■$24■■000■saved■for■a■holiday■and■a■ Golden■Ace■Rental■Company■and■Silver■Diamond■ new■stereo.■Her■travel■expenses■are■$5400■and■her■ Rental■Company.■Golden■Ace■Rental■Company■ daily■expenses■are■$260. charges■a■fl■at■rate■of■$38■per■day■and■$0.20■per■ a Write■down■an■equation■for■the■cost■of■her■ kilometre.■The■Silver■Diamond■Rental■Company■ holiday■if■she■stays■for■d■days. charges■a■fl■at■rate■of■$30■per■day■plus■$0.32■per■ ■■ Upon■her■return■from■holidays■Frederika■wants■ kilometre. to■purchase■a■new■stereo■system■that■will■cost■ a Write■an■algebraic■equation■for■the■cost■of■ her■$2500. 5400■+■260d■=■CH renting■a■car■for■three■days■from■the■Golden■ b How■many■days■can■she■spend■on■her■holiday■ Ace■Rental■Company■in■terms■of■the■number■of■ if■she■wishes■to■purchase■a■new■stereo■upon■her■ kilometres■travelled,■k. ■ CG■=■114■+■0.20k return? 61■days b Write■an■algebraic■equation■for■the■cost■of■ 11 Mick■the■painter■has■fi■xed■costs■(e.g.■insurance,■ renting■a■car■for■three■days■from■the■Silver■ equipment,■etc)■of■$3400■per■year.■His■running■cost■ Diamond■Rental■Company■in■terms■of■the■ to■travel■to■jobs■is■based■on■$0.75■per■kilometre.■ number■of■kilometres■travelled,■k. CS■=■90■+■0.32k Last■year■Mick■had■costs■that■were■less■than■ c How■many■kilometres■would■John■have■to■ $16■■000. travel■so■that■the■cost■of■hiring■from■each■ a Write■an■inequality■to■show■this■information■ company■for■three■days■is■the■same? 200■km and■solve■it■to■fi■nd■how■many■kilometres■Mick■ d Write■an■inequation■that,■when■solved,■will■tell■ travelled■for■the■year.■ ■ n■<■16■800■km■ you■the■number■of■kilometres■for■which■it■is■ b Explain■the■information■you■have■found. cheaper■to■use■Golden■Ace■Rental■Company■ M ick■travelled■less■than■16■■800■km■ when■renting■for■three■days. eBook plus for■the■year■and■his■costs■stayed■ below■$16■■000. e For■what■number■of■kilometres■will■it■be■ Interactivities cheaper■to■use■Silver■Diamond■Rental■ Test yourself Chapter 4 Company■for■three■days’■hire? k■<■200 int-2837 114■+■0.20k■<■90■+■0.32k■\■k■>■200

Word search Chapter 4 int-2835 Crossword Chapter 4 int-2836

Chapter 4 simultaneous linear equations and inequations

129

eBook plus

ACtivities

chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■4■(doc-5211)■(page89) are you ready? Digital docs

(page90)

•■ SkillSHEET■4.1■(doc-5212):■Substitution■into■a■linear■rule •■ SkillSHEET■4.2■(doc-5213):■Solving■linear■equations■ that■arise■when■fi■nding■x-■and■ y-intercepts •■ SkillSHEET■4.3■(doc-5214):■Transposing■linear■ equations■to■standard■form •■ SkillSHEET■4.4■(doc-5215):■Measuring■the■rise■and■ the■run •■ SkillSHEET■4.5■(doc-5216):■Finding■the■gradient■given■ two■points •■ SkillSHEET■4.6■(doc-5217):■Graphing■linear■equations■ using■the■x-■and■y-intercept■method •■ SkillSHEET■4.7■(doc-5218):■Checking■whether■ a■given■point■makes■the■inequation■a■true■statement 4a Graphical solution of simultaneous linear equations

•■ Activity■4-A-1■(doc-4990):■Investigating■graphs■of■ simultaneous■equations■(page94) •■ Activity■4-A-2■(doc-4991):■Graphing■simultaneous■ equations■(page94) •■ Activity■4-A-3■(doc-4992):■Further■graphing■of■ simultaneous■equations■(page95) •■ SkillSHEET■4.6■(doc-5217):■Graphing■linear■equations■ using■the■x-■and■y-intercept■method■(page95) 4b Solving simultaneous linear equations using substitution

(page98)

•■ Activity■4-B-1■(doc-4993):■Learning■substitution■ •■ Activity■4-B-2■(doc-4994):■Practising■substitution■ •■ Activity■4-B-3■(doc-4995):■Tricky■substitution■ 4c Solving simultaneous linear equations using elimination Digital docs

•■ Activity■4-C-1■(doc-4996):■Elimination■practice■ (page101) •■ Activity■4-C-2■(doc-4997):■Let’s■eliminate■(page102) •■ Activity■4-C-3■(doc-4998):■More■elimination■(page102) •■ WorkSHEET■4.1■(doc-5220):■Simultaneous■equations■ I■(page103) Interactivity

•■ Simultaneous■linear■equations■(int-2780)■(page99) 4d Problem solving using simultaneous linear equations Digital docs

•■ Activity■4-D-1■(doc-4999):■Problem■solving■ (page105) 130

4e Solving linear inequations Digital docs

•■ Activity■4-E-1■(doc-5002):■Puzzling■inequations■1■ (page108) •■ Activity■4-E-2■(doc-5003):■Puzzling■inequations■2■ (page108) •■ Activity■4-E-3■(doc-5004):■Puzzling■inequations■3■ (page109) •■ SkillSHEET■4.7■(doc-5218):■Checking■whether■a■ given■point■makes■the■inequation■a■true■statement■ (page109) •■ SkillSHEET■4.8■(doc-5219):■Writing■equations■from■ worded■statements■(page109) 4F Sketching linear inequations Digital docs

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•■ Activity■4-D-2■(doc-5000):■Harder■problem■solving■ (page105) •■ Activity■4-D-3■(doc-5001):■Tricky■problem■solving■ (page105) •■ WorkSHEET■4.2■(doc-5221):■Simultaneous■ equations■II■(page106)

maths Quest 10 for the Australian Curriculum

•■ Activity■4-F-1■(doc-5005):■Understanding■linear■ inequations■(page113) •■ Activity■4-F-2■(doc-5006):■Graphing■linear■ inequations■(page114) •■ Activity■4-F-3■(doc-5007):■Interpreting■linear■ inequation■graphs■(page114) 4G Solving simultaneous linear inequations Digital docs

•■ Activity■4-G-1■(doc-5008):■Introducing■ simultaneous■linear■inequations■(page118) •■ Activity■4-G-2■(doc-5009):■Practising■simultaneous■ linear■inequations■(page118) •■ Activity■4-G-3■(doc-5010):■Further■simultaneous■ linear■inequations■(page118) •■ WorkSHEET■4.3■(doc-5222):■Simultaneous■ equations■III■(page122) chapter review

(page129) •■ Test■Yourself■Chapter■4■(int-2837):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■4■(int-2835):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■4■(int-2836):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

5 Trigonometry I

5A Pythagoras’ theorem 5B Pythagoras’ theorem in three dimensions 5C Trigonometric ratios 5D Using trigonometry to calculate side lengths 5E Using trigonometry to calculate angle size 5F Angles of elevation and depression 5G Bearings and compass directions 5H Applications WhaT Do you knoW ? 1 List what you know about trigonometry. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of trigonometry. eBook plus

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Hungry brain activity Chapter 5 doc-5223

oPening QuesTion

How can Raylene find her way to the finish line of the orienteering course?

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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SkillSHEET 5.1 doc-5224

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SkillSHEET 5.2 doc-5225

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Rounding to a given number of decimal places 1 Round the following numbers to 3 decimal places. a 0.6845 b 1.3996 1.400 0.685

c 0.7487 0.749

Rounding the size of an angle to the nearest minute and second 2 Round the following angles: a i 15è33Å ii 15è32Å41ë b i 63è16Å ii 63è15Å32ë i to the nearest minute c i 27è10Å ii 27è10Å16ë ii to the nearest second. a 15è32Å40.5ë b 63è15Å32.4ë c 27è10Å15.8ë Labelling the sides of a right-angled triangle 3 Label the sides of the following right-angled triangles using the letters H (for hypotenuse),

O (for opposite) and A (for adjacent) with respect to angle q.

SkillSHEET 5.3 doc-5226

a

b

H

O

A

q



c

A q

H

H

q O

q

O

q

A

q

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SkillSHEET 5.5 doc-5227

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Digital doc

SkillSHEET 5.6 doc-5228

Rearranging formulas 4 Rearrange each of the following formulas to make x the subject. 4.2 x 4.2 a tan 15è = b tan 28è = x = x tan(28°) 30 x = 30 ì tan (15è)

x = 5.3 x = 5.3 ì tan (64è) tan 64°

Drawing a diagram from given directions 5 Draw a diagram for each of the following situations. a Kate’s bushwalking route took her from A to B, a distance of 5 km at a bearing of 25èT

then to C, a further distance of 7.5 km at a bearing of 120èT. b A ship steamed S20èE for a distance of 180 km, then the ship travelled N60èW for a

further 70 km. b

N

km

N

S

B 120è

5k

m

25è A

20è

180

a

132

c

7.

5

70 km

N

km 60è

C

maths Quest 10 for the australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

5A

Pythagoras’ theorem Similar right-angled triangles In the two similar right-angled triangles shown below, the angles are the same and the corresponding sides are in the same ratio. D

A

6 cm

3 cm

B

10 cm

5 cm

4 cm

C

E

8 cm

F

The corresponding sides are in the same ratio. AB AC BC = = . DE DF EF To write this using the side lengths of the triangles gives: AB 3 1 = = DE 6 2 AC 5 1 = = DF 10 2 BC 4 1 = = EF 8 2 This means that for right-angled triangles, when the angles are fixed, the ratios of the sides in the triangle are constant. We can examine this idea further by completing the following activity. Using a protractor and ruler, draw an angle of 70è, measuring horizontal distances of 3 cm, 7 cm and 10 cm as demonstrated in the diagram below.

c b a 70è 3 cm 7 cm 10 cm Note: Diagram not drawn to scale.

Measure the perpendicular heights a, b and c. a ö 8.24 cm    b ö 19.23 cm    c ö 27.47 cm Chapter 5 Trigonometry I

133

measurement AND geometry • Pythagoras and trigonometry

To test if the theory that for right-angled triangles, when the angles are fixed, the ratios of the sides in the triangle are constant is correct, calculate the ratios of the side lengths. a 8.24 ≈ ≈ 2.75 3 3 b 19.23 ≈ ≈ 2.75 7 7 c 27.47 ≈ ≈ 2.75 10 10 The ratios are the same because the triangles are similar. This important concept forms the basis of trigonometry.

Review of Pythagoras’ theorem ■■

■■ ■■

Pythagoras’ theorem states that in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The rule is written as c2 = a2 + b2 where a and b are the two shorter sides and c is the hypotenuse. The hypotenuse is the longest side of a right-angled triangle and is always the side that is opposite the right angle. Pythagoras’ theorem gives us a way of finding the length of the third side in a triangle, if we know the lengths of the two other sides.

c

a

b x

4 7

Finding the hypotenuse ■■

To calculate the length of the hypotenuse when given the length of the two shorter sides, substitute the known values into the formula c2 = a2 + b2.

Worked Example 1

For the triangle at right, calculate the length of the hypotenuse, x, correct to 1 decimal place.

x

4 7 Think 1

Write/draw

Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse as c.

c=x

a=4

b=7 3

Substitute the values of a, b and c into this rule and simplify.

4

Calculate x by taking the square root of 65. Round the answer correct to 1 decimal place.

x = 65 x = 8.1

Maths Quest 10 for the Australian Curriculum

a2

+

x2

2

134

=

Write Pythagoras’ theorem.

c2

b2

= + 72 = 16 + 49 = 65 42

measurement AND geometry • Pythagoras and trigonometry

Finding a shorter side Sometimes a question will give you the length of the hypotenuse and ask you to find one of the shorter sides. In such examples, we need to rearrange Pythagoras’ formula. Given that c2 = a2 + b2, we can rewrite this as: a2 = c2 - b2 or b2 = c2 - a2. ■■

Worked Example 2

Calculate the length, correct to 1 decimal place, of the unmarked side of the triangle at right. 14 cm 8 cm Think 1

Write/draw

Copy the diagram and label the sides a, b and c. Remember to label the hypotenuse as c; it does not matter which side is a and which side is b.

a c = 14 b=8

2

Write Pythagoras’ theorem.

3

Substitute the values of a, b and c into this rule and simplify.

4

Find a by taking the square root of 132. Round to 1 decimal place.

■■

c2 = a2 + b2 142 = a2 + 82 196 = a2 + 64 a2 = 196 - 64 = 132 a = 132 = 11.5  cm

In many cases we are able to use Pythagoras’ theorem to solve practical problems.   First model the problem by drawing a diagram, then use Pythagoras’ theorem to solve the right-angled triangle. Use the result to give a worded answer.

Worked Example 3

A ladder that is 4.5  m long leans up against a vertical wall. The foot of the ladder is 1.2  m from the wall. How far up the wall does the ladder reach? Give your answer correct to 1 decimal place. Think 1

Write/draw

Draw a diagram and label the sides a, b and c. Remember to label the hypotenuse as c.

c = 4.5 m

a

b = 1.2 m Chapter 5 Trigonometry I

135

measurement AND geometry • Pythagoras and trigonometry

2

Write Pythagoras’ theorem.

c2 = a2 + b2

3

Substitute the values of a, b and c into this rule and simplify.

4.52 = a2 + 1.22 20.25 = a2 + 1.44 a2 = 20.25 - 1.44 = 18.81

4

Find a by taking the square root of 18.81. Round to 1 decimal place and include the unit of measurement (m).

5

Answer the question in a sentence.

■■

a = 18.81 = 4.3  m The ladder will reach a height of 4.3  m up the wall.

Sometimes the unknown length involves more than one side.   In these cases, substitute into Pythagoras’ theorem, then solve the following equation for the unknown.

Worked Example 4

Calculate the value of the pronumeral, correct to 2 decimal places, in the triangle at right.

3x 78 2x

Think 1

Copy the diagram and label the sides a, b and c.

Write/draw b = 3x c = 78 a = 2x

136

c2 = a2 + b2

2

Write Pythagoras’ theorem.

3

Substitute the values of a, b and c into this rule and simplify.

782 = (3x)2 + (2x)2 6084 = 9x2 + 4x2 6084 = 13x2

4

Rearrange the equation so that the pronumeral is on the left-hand side of the equation.

13x2 = 6084

5

Divide both sides of the equation by 13.

6

Find x by taking the square root. Round the answer correct to 2 decimal places.

Maths Quest 10 for the Australian Curriculum

13 x 2 6084 = 13 13 x2 = 468 x = 468 = 21.63

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

rememBer

1. The hypotenuse is the longest side of the triangle and is opposite the right angle. 2. On your diagram, check whether you are finding the length of the hypotenuse or one of the shorter sides. 3. The length of a side can be found if we are given the length of the other sides by using the formula c2 = a2 + b2. 4. When using Pythagoras’ theorem, always check the units given for each measurement. 5. If necessary, convert all measurements to the same units before using the rule. 6. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. 7. Worded problems should be answered in a sentence. exerCise

5a inDiViDual PaThWays eBook plus

Activity 5-A-1

c 980.95 f 175.14

1 We1 For each of the following triangles, calculate the length of the hypotenuse, giving

answers correct to 2 decimal places. 4.7

a

b

c

19.3

804 6.3

Activity 5-A-2

27.1

Practising Pythagoras’ theorem doc-5012 Activity 5-A-3

b 33.27 e 2.85

FluenCy

Review of Pythagoras’ theorem doc-5011

More of Pythagoras’ theorem doc-5013

1 a 7.86 d 12.68

Pythagoras’ theorem

562 0.9

e

d

152

f

7.4

87 10.3

eBook plus

2.7

Digital doc

SkillSHEET 5.1 doc-5224



2 We2 Find the value of the pronumeral, correct to 2 decimal places. a b s c 1.98 a 36.36 30.1 b 1.62 c 15.37 47.2 2.56 d 0.61 e 2133.19 t f 453.90 d

e

0.28

0.67

w

17.52

f

2870

v

u

8.4

1920

468 x

114 Chapter 5 Trigonometry I

137

measurement AND geometry • Pythagoras and trigonometry

23.04  cm

3   WE3  The diagonal of the rectangular sign at right is 4

a 14.14  cm b 24.04  cm c 4.53  cm

5

6

a 6.06 b 4.24 c 4.74

7

34  cm. If the height of this sign is 25 cm, find the width. A right-angled triangle has a base of 4  cm and a height of 12  cm. Calculate the length of the hypotenuse to 2 decimal places. 12.65  cm Calculate the lengths of the diagonals (to 2 decimal places) of squares that have side lengths of: a 10  cm b 17  cm c 3.2  cm. The diagonal of a rectangle is 120  cm. One side has a length of 70  cm. Determine: a the length of the other side b the perimeter of the rectangle 334.94  cm 97.47  cm c the area of the rectangle. 6822.90  cm2   WE4  Find the value of the pronumeral, correct to 2 decimal places for each of the following. a

c

b 25

3x

2x

3x

4x 18 x

30

6x

understanding 8 An isosceles triangle has a base of 30  cm and a height of 10  cm. Calculate the length of the two equal sides. 18.03  cm 9 An equilateral triangle has sides of length 20  cm. Find the height of the triangle. 17.32  cm 10 A right-angled triangle has a height of 17.2  cm, and a base that is half the height. Calculate the length of the hypotenuse, correct to 2 decimal places. 19.23  cm 11 The road sign shown below is in the form of an equilateral triangle. Find the height of the sign and, hence, find its area. 65.82  cm; 2501.16  cm2

76 cm

12 A flagpole, 12  m high, is supported by three wires, attached from the top of the pole to the

ground. Each wire is pegged into the ground 5  m from the pole. How much wire is needed to support the pole? 39  m 13 Ben’s dog ‘Macca’ has wandered onto a frozen pond, and is too frightened to walk back. Ben estimates that the dog is 3.5  m from the edge of the pond. He finds a plank, 4  m long, and thinks he can use it to rescue Macca. The pond is surrounded by a bank that is 1  m high. Ben uses the plank to make a ramp for Macca to walk up. Will he be able to rescue his dog? Yes 14 Sarah goes canoeing in a large lake. She paddles 2.1  km to the 3.8 km north, then 3.8  km to the west. Use the triangle at right to find out how far she must then paddle to get back to her starting point in 2.1 km the shortest possible way. 4.34  km Starting point 138

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry 15 A baseball diamond is a square of side length 27  m. When a runner on first base tries to steal

second base, the catcher has to throw the ball from home base to second base. How far is that throw? 38.2  m

Second base

27 m First base

Home base Catcher 16 Penny, a carpenter, is building a roof for a new house. The roof has

17 18

19 20

21 22

a gable end in the form of an isosceles triangle, with a base of 6  m and sloping sides of 7.5  m. She decides to put 5 evenly spaced vertical 7.5 m 7.5 m strips of wood as decoration on the gable as shown at right. How many metres of this decorative wood does she need? 20.61  m Calculate the length, in mm, of the hypotenuse of a right-angled 6m triangle, if the two shorter sides are 5  cm and 12  cm. 130  mm The hypotenuse and one other side of a right-angled triangle are given for each case below. Find the length of the third side in the units specified. Give your answers correct to 2 decimal places. 386.13  mm a Sides 46  cm and 25  cm, third side in mm 62.09  cm b Sides 843  mm and 1047  mm, third side in cm c Sides 4500  m and 3850  m, third side in  km 2.33  km 16.15  cm d Sides 20.3  cm and 123  mm, third side in cm e Sides 6420  mm and 8.4  m, third side in cm 541.70  cm f Sides 0.358  km and 2640  m, third side in m 2615.61  m g Sides 491  mm and 10.8  cm, third side in mm 478.97  mm h Sides 379  000  m and 82  700  m, third side in  km 369.87  km A rectangle measures 35  mm by 4.2  cm. Calculate the length of its diagonal in millimetres to 2 decimal places. 54.67  mm A rectangular envelope has a length of 21  cm and a diagonal measuring 35  cm. Calculate: a the width of the envelope 28  cm b the area of the envelope. 588  cm2 A sheet of A4 paper measures 210  mm by 297  mm. Calculate the length of the diagonal in centimetres to 2 decimal places. 36.37  cm A right-angled triangle has a hypotenuse of 47.3  cm and one other side of 30.8  cm. Calculate 552.86  cm2 the area of the triangle. Chapter 5 Trigonometry I

139

measurement AND geometry • Pythagoras and trigonometry 23 A swimming pool is 50  m by 25  m. Peter is bored by his usual training routine, and decides

24

25 26 27

28

to swim the diagonal of the pool. How many diagonals must he swim to complete his normal 21.46 diagonals, so would need to complete 22 distance of 1200  m? Give your answer to 2 decimal places. Sarah is making a gate that has to be 1200  mm wide. It must be braced with a diagonal strut made of a different type of timber. She has only 2  m of this kind of timber available. What is the maximum height of the gate that she can make? 1600  mm A hiker walks 4.5  km west, then 3.8  km south. How far in metres is she from her starting point? Give your answer to 2 decimal places. 5889.82  m 7.07  cm A square has a diagonal of 10  cm. What is the length of each side? Wally is installing a watering system in his garden. The pipe is to go all around the edge of the rectangular garden, and have a branch diagonally across the garden. The garden measures 5  m by 7.2  m. If the pipe costs $2.40 per metre (or part thereof), what will be the total cost of the pipe? $81.60 The size of a rectangular television screen is given by the length of its diagonal. What is the size of the screen at right to the nearest centimetre if its dimensions are 158  cm 185  cm wide and 96  cm deep?

Reasoning 29 During a recent earthquake, Helen’s large

bookshelf fell over. The bookshelf is x metres wide and 2.5 metres high. The ceiling is 3 metres high. Show that if the bookshelf is lying on its side next to the wall and is able to be stood up directly, then x is less than 1.658 metres Students own working.

5B

 

The square root of a number usually gives us both a positive and negative answer. Why do we only take the positive answer when using Pythagoras’ theorem?

Pythagoras’ theorem in three dimensions ■■

Many real-life situations involve 3-dimensional (3-D) shapes: shapes with length, width and height. Some common 3-D shapes used in this section include boxes, pyramids and rightangled wedges.

Box ■■

140

reflection 

Pyramid

Right-angled wedge

The important thing about 3-D shapes is that in a diagram, right angles may not look like right angles, so it is important to redraw sections of the diagram in two dimensions, where the right angles can be seen accurately.

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

Worked Example 5

Determine the length AG in this box.

A

B 6 cm

C

D

F

E

5 cm 10 cm

H Think 1

G

Write/draw A

Draw the diagram in three dimensions.

B 6 cm

C

D

F

E

5 cm H 2

3

Draw in two dimensions, a right-angled triangle that contains AG and label the sides. Only 1 side is known, so we need to find another right-angled triangle to use. Draw EFGH in two dimensions and show the diagonal EG. Label the side EG as x. We have two of the three side lengths so we can calculate the unknown.

10 cm

A

6 E

G

E

F

5

x

5

H

10

G

4

Use Pythagoras’ theorem to calculate EG.

c2 = a2 + b2 x2 = 52 + 102 = 25 + 100 = 125 x = 125 = 11.18  cm

5

Place this information on triangle AEG. Label the side AG as y. Now we have two of the three side lengths.

A 6 E

6

Use Pythagoras’ theorem to find AG.

G

y

11.18

G

c2 = a2 + b2 y2 = 62 + ( 125 )2 = 36 + 125 = 161 y = 161 = 12.69

7

Answer the question in a sentence.

The length of AG is 12.69  cm.

Chapter 5 Trigonometry I

141

measurement AND geometry • Pythagoras and trigonometry

Worked Example 6

A piece of cheese in the shape of a right-angled wedge sits on a table. It has a rectangular base measuring 14  cm by 8  cm, and is 4  cm high at the thickest point. An ant crawls diagonally across the sloping face. How far, to the nearest millimetre, does the ant walk? Think 1

Draw a diagram in three dimensions and label the vertices. Mark BD, the path taken by the ant, with a dotted line.

Write/draw B E A

14 cm

D

C 4 cm F 8 cm

x 2

Draw in two dimensions a right-angled triangle that contains BD, and label the sides. Only one side is known, so we need to find another rightangled triangle to use.

B 4 D

E 3

Draw EFDA in two dimensions, and show the diagonal ED. Label the side ED as x.

4

Use Pythagoras’ theorem to calculate ED.

5

Place this information on triangle BED. Label the side BD as y.

E

F

8

x

8

A

14

D

c2 = a2 + b2 x2 = 82 + 142 = 64 + 196 = 260 x = 260 = 16.12  cm B y

4 E

142

6

Solve this triangle for BD.

7

Check the answer’s units. We need to convert cm to mm, so multiply by 10.

8

Answer the question in a sentence.

Maths Quest 10 for the Australian Curriculum

16.12

D

c2 = a2 + b2 2 y2 = 42 + ( 260 ) = 16 + 260 = 276 y = 276 = 16.61  cm = 166.1  mm The ant walks 166  mm, correct to the nearest millimetre.

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

rememBer

1. Pythagoras’ theorem can be used to solve problems in three dimensions. 2. Some common 3-D shapes include boxes, pyramids and right-angled wedges. 3. To solve problems in three dimensions it is helpful to draw sections of the original shape in two dimensions. exerCise

5B inDiViDual PaThWays eBook plus

Activity 5-B-1

Pythagoras in 3-dimensions doc-5014

Pythagoras’ theorem in three dimensions Where appropriate in this exercise, give answers correct to 2 decimal places. FluenCy 1 We5 Calculate the length, AG. a b A B C

D

10

a 17.32 A

B

b 12.25 c 15.12 c A

C

D

B

C

D

10.4

Activity 5-B-2

Pythagoras in 3-D figures doc-5015

E

eBook plus

Digital doc

E

F 7.3

10

Activity 5-B-3

Investigating triangles in 3-D figures doc-5016

10

F

H

10

G

H

F

E

G

8.2

5 H

5

G

2 Calculate the length of CE in the wedge at right and, hence, obtain AC. 12.21, 12.85

A

E

SkillSHEET 5.4 doc-5229

D 3 If DC = 3.2 m, AC = 5.8 m, and CF = 4.5 m in the figure at right, calculate the length of AD and BF. 4.84 m, 1.77 m

7

C

10

B 4 F

B

A

F D

C

4 Calculate the length of BD and, hence, the height of the pyramid 11.31, 5.66 at right.

V 8 A 8

D 5 The pyramid ABCDE has a square base. The pyramid is 20 cm high.

B

8

C

E

Each sloping edge measures 30 cm. Calculate the length of the sides of the base. 31.62 cm

EM = 20 cm

A D

B M C

Chapter 5 Trigonometry I

143

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 6 The sloping side of a cone is 10 cm and the height is 8 cm. What is the length of the radius of the base? 6 cm 10 cm

8 cm 7 An ice-cream cone has a diameter across the top of 6 cm, and 12.65 cm sloping side of 13 cm. How deep is the cone?

r

8 We6 A piece of cheese in the shape of a right-angled wedge

B

sits on a table. It has a base measuring 20 mm by 10 mm, and is 4 mm high at the thickest point, as shown in the figure. A fly crawls diagonally across the sloping face. How far, to the nearest millimetre, does the fly walk? 23 mm

E A

20 mm

C 4 mm F D 10 mm

unDersTanDing 9 Jodie travels to Bolivia, taking with her a suitcase as shown in the photo. She buys a carved

walking stick 1.2 m long. Will she be able to fit it in her suitcase for the flight home? 30

cm

No: maximum stick can be only 115 cm long.

65 cm

90 cm

10 A desk tidy is shaped like a cylinder, height 18 cm and diameter

10 cm. Pencils that are 24 cm long rest inside. What lengths of the pencils are above the top of the cylinder? 3.41 cm

11 A 10-m high flagpole is in the corner of a rectangular park 10 m that measures 240 m by 150 m. A 240 m a Calculate: i the length of the diagonal of the park 283.02 m 150 m 240.21 m ii the distance from A to the top of the pole B iii the distance from B to the top of the pole. 150.33 m b A bird flies from the top of the pole to the centre of the park. How far does it fly? 144

maths Quest 10 for the australian Curriculum

141.86 m

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 12 A candlestick is in the shape of two cones, joined at the vertices

as shown. The smaller cone has a diameter and sloping side of 7 cm, and the larger one has a diameter and sloping side of 10 cm. How tall 14.72 cm is the candlestick? 13 The total height of the shape below is 15 cm. Calculate the length of the sloping side of the pyramid. 13.38 cm 15 cm 6 cm 14 cm 14 cm 14 A sandcastle is in the shape of a truncated cone as shown. Calculate the length of the diameter of the base. 42.27 cm 20 cm

30 cm

eBook plus

32 cm

15 A tent is in the shape of a triangular prism, with a height of

120 cm

120 cm as shown at right. The width across the base of the door is 1 m and the tent is 2.3 m long. Calculate the length of each sloping side, in metres. Then calculate the area of fabric used in the construction of the sloping 1.3 m, 5.98 m2 rectangles which form the sides.

Digital doc

WorkSHEET 5.1 doc-5230

2.3 m 1m

reasoning 16 Stephano is renovating his apartment, which is at the end of two corridors. The corridors of the

apartment building are 2 m wide with 2 m high ceilings, and the first corridor is at right angles to the second. Show that he can carry lengths of timber up to 6 m long to his apartment. Students’ own working reFleCTion 

 

The diagonal distance across a rectangle of 2

2

dimensions x by y is x + y . What would be the rule to find the length of a diagonal across a cuboid of dimensions x by y by z ? Use your rule to check your answers to question 1.

5C

Trigonometric ratios angles and the calculator ■ ■

Last year you were shown that each angle has specific values for its sine, cosine and tangent. These values are needed for practically every trigonometry problem and can be obtained with the aid of a calculator. Chapter 5 Trigonometry I

145

measurement AND geometry • Pythagoras and trigonometry

Worked Example 7

Calculate the value of each of the following, correct to 4 decimal places, using a calculator. a  cos 65è57Å b  tan 56è45Å30ë Think

Write

a Write your answer to the correct number

of decimal places. b Write your answer to the correct number

of decimal places.

a cos 65è57Å = 0.4075 b tan 56è45Å30ë = 1.5257

Worked Example 8

Calculate the size of angle q, correct to the nearest degree, given sin q  = 0.6583. Think

Write

1

Write the given information.

sin q = 0.6583

2

To find the size of the angle, we need to undo sine with its inverse, sin-1. Ensure your calculator is in degrees mode.

q = sin-1 (0.6583)

3

Write your answer to the nearest degree.

q = 41è

■■ ■■

Sometimes, we need to be able to find an angle correct to either the nearest minute or nearest second. When we use an inverse trigonometric function, the angle is expressed in degrees as a decimal. It should be converted to degrees, minutes and seconds (DMS).

Worked Example 9

Calculate the value of q : a  correct to the nearest minute, given that cos q  = 0.2547 b  correct to the nearest second, given that tan q  = 2.364. Think a

b

146

1

Write the equation.

2

Write your answer, rounding to the nearest minute. Remember there are 60 minutes in 1 degree and 60 seconds in 1 minute. Hence, for the nearest minute, we round up at 30ë or higher.

1

Write the equation.

2

Write your answer, rounding to the nearest second.

Maths Quest 10 for the Australian Curriculum

Write a cos q = 0.2547

cos-1 0.2547 = 75è15Å

b tan q = 2.364

tan-1 2.364 = 67è4Å16ë

measurement AND geometry • Pythagoras and trigonometry

Review of SOH CAH TOA We are able to find a side length in a right-angled triangle if we are given one other side length and the size of one of the acute angles. These sides and angle are related using one of the three trigonometric ratios. ■■ The sine ratio The sine ratio is defined as the ratio of the length of the side opposite angle q (O) to the length of the hypotenuse (H). This is O written as sin q  = . H The sine of an angle is not dependent on the size of the right-angled triangle as all these triangles are similar in shape. q ot en

yp ot en us e

H yp

The cosine ratio The cosine ratio is defined as the ratio of the length of the adjacent side (A) to the length of the hypotenuse (H) and is written as A cos q  = . H The cosine of an angle also does not depend on the size of the right-angled triangle.

H

■■

Opposite

us e

■■

q Adjacent

■■

Opposite

The tangent ratio O The tangent ratio is defined as tan q  = , where O is the length of A the side opposite angle q and A is the length of the side adjacent to it. Again, the tangent ratio does not depend on the size of the right-angled triangle. q Adjacent

■■

■■

Having defined the three trigonometric ratios, we need to decide in each case which of the three to use. We do this by labelling the sides relative to the angle we have been given. We then select the ratio that contains both the side we are finding and the side we have been given. The three ratios can be remembered easily by using the mnemonic or abbreviation SOH CAH TOA: •  SOH stands for ‘Sine, Opposite, Hypotenuse’. •  CAH stands for ‘Cosine, Adjacent, Hypotenuse’. •  TOA stands for ‘Tangent, Opposite, Adjacent’.

Worked Example 10

For the triangle shown, write the expressions for the sine, cosine and tangent ratios of the given angle.

c a q b Chapter 5 Trigonometry I

147

measurement AND geometry • Pythagoras and trigonometry

Think

Write/draw

Label the diagram using the symbols O, A, H with respect to the given angle (angle q ).

1

c=H a=O q

b=A 2

From the diagram, identify the values of O (opposite side), A (adjacent side) and H (the hypotenuse).

O = a, A = b, H = c

3

Write the formula for each of the sine, cosine and tangent ratios.

sin q  =

4

Substitute the values of A, O and H into each formula.

O A O , cos q  = , tan q  = H H A

a b a sin q  = , cos q  = , tan q  = c c b

Worked Example 11

Write the trigonometric ratio which must be used in order to find the value of the pronumeral in each of the following triangles. Set up a suitable equation. a               b  18 15

x

6

50è b

Think a

1

Label the sides of the triangle whose lengths are given, using the appropriate symbols.

Write/draw a 15 = H

6=O

b

b

148

O H

2

We are given the lengths of the opposite side (O) and the hypotenuse (H). Write the ratio that contains both of these sides.

sin q  =

3

Identify the values of the pronumerals.

O = 6, H = 15

4

Substitute the values of the pronumerals into the ratio. (Since the given angle is denoted with the letter b, replace q with b.)

sin b =

1

Label the sides of the triangle whose lengths are either given, or need to be found, using the appropriate symbols.

Maths Quest 10 for the Australian Curriculum

b

6 2 = 15 5

18 = A 50è

x=O

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

O A

2

The length of the adjacent side (A) is given and the length of the opposite side (O) needs to be found. Write the ratio that contains these sides.

tan q  =

3

Identify the values of the pronumerals.

4

Substitute the values of the pronumerals into the ratio.

O = x, A = 18, q = 50è x tan 50è = 18

rememBer

1. When using the calculator to find values of sine, cosine and tangent, make sure the calculator is in Degree mode. 2. To find the size of an angle whose sine, cosine or tangent is given, perform an inverse operation; that is, sin-1, cos-1 or tan-1. 3. Use the calculator’s conversion function to convert between decimal degrees and degrees, minutes and seconds. 4. There are 60 minutes in 1 degree and 60 seconds in 1 minute. 5. The three trigonometric ratios, sine, cosine and tangent, are defined as: O A O sin q  = , cos q  = and tan q  = , H H A where H is the hypotenuse, O is the opposite side and A is the adjacent side. 6. The three ratios are abbreviated to the useful mnemonic SOH CAH TOA. 7. To determine which trigonometric ratio to use, follow these steps. (a) Label the sides of the right-angled triangle that are either given, or need to be found, using the symbols O, A, H with respect to the angle in question. (b) Consider the sides that are involved and write the trigonometric ratio containing both of these sides. (Use SOH CAH TOA to assist you.) (c) Identify the values of the pronumerals in the ratio. (d) Substitute the given values into the ratio.

exerCise

5C inDiViDual PaThWays eBook plus

Activity 5-C-1

Review of trigonometry doc-5017 Activity 5-C-2

Using trigonometry doc-5018 Activity 5-C-3

Applying trigonometry doc-5019

Trigonometric ratios FluenCy 1 Calculate each of the following, correct to 4 decimal places. a sin 30è 0.5000 b cos 45è 0.7071 d 0.8387 d sin 57è e tan 83è 8.1443

c tan 25è 0.4663 f cos 44è 0.7193

2 We7 Calculate each of the following, correct to 4 decimal places. a sin 40è30Å 0.6944 b cos 53è57Å 0.5885 c -1.5013 d tan 123è40Å e sin 92è32Å 0.9990 f 0.8120 g cos 35è42Å35ë h tan 27è42Å50ë 0.5253 i 0.4063 j sin 23è58Å21ë k cos 8è54Å2ë 0.9880 l 1.7321 m tan 420è n cos 845è -0.5736 o

tan 27è34Å 0.5220 sin 42è8Å 0.6709 cos 143è25Å23ë -0.8031 sin 286è -0.9613 sin 367è35Å 0.1320 3 We8 Find the size of angle q, correct to the nearest degree, for each of the following. 50è a sin q  = 0.763 b cos q  = 0.912 24è c tan q  = 1.351 53è d cos q  = 0.321 71è e tan q  = 12.86 86è f cos q  = 0.756 41è Chapter 5 Trigonometry I

149

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry e f d 4 We9a Find the size of the angle q in each of the following, correct to the nearest minute. ii cos (q) = 54è29Å a sin q  = 0.814 b sin q  = 0.110 6è19Å c tan q  = 0.015 0è52Å f 72è47Å d cos q  = 0.296 e tan q  = 0.993 44è48Å f sin q  = 0.450 26è45Å e iii tan (q) = d 5 We9b Find the size of the angle q in each of the following, correct to the nearest second. i a tan q  = 0.5 26è33Å54ë b cos q  = 0.438 64è1Å25ë c sin q  = 0.9047 64è46Å59ë b i sin (a) = g 48è5Å22ë d tan q  = 1.1141 e cos q  = 0.8 36è52Å12ë f tan q  = 43.76 88è41Å27ë h ii cos (a) = 6 Find the value for each of the following, correct to 3 decimal places. g a 3.8 cos 42è 2.824 b 118 sin 37è 71.014 c 2.5 tan 83è 20.361 i iii tan (a) = 220 226.735 2 cos 23° 2 h 2.828 d e 1.192 f l 5 sin 18° sin 45° cos 14° c i sin (b ) = k 12.8 18.7 55.7 g 32.259 h i 4909.913 j 7.232 ii cos (b ) = tan 60è32Å sin 35è25Å42ÅÅ cos 89è21Å k 2.5 sin 27è8Å 3.8 tan 1è51Å44ÅÅ 3.2 cos 34è52Å l j k l 0.904 14.814 0.063 iii tan (b ) = 10.4 cos 83è2Å j 4.5 sin 25è45Å 0.8 sin 12è48Å

7 a i sin (q) =

eBook plus

Digital doc

SkillSHEET 5.3 doc-5226

d e f

n i sin (g ) = m o ii cos (g ) = m n iii tan (g ) = o b i sin (b ) = c a ii cos (b ) = c b iii tan (b ) = a v i sin (g ) = u t ii cos (g ) = u v iii tan (g ) = t eBook plus

Digital doc

SkillSHEET 5.7 doc-5231

8 a sin (q ) =

angles: i sine iii tangent a d q

25 30

c tan (q ) =

4 5

2.7 p 17 e sin (35è) = t

d tan (q ) =

f sin (a ) =

ii cosine b

c b

h

i

k

j

a

e

f

g l

d

e

o g

f a

n

b

b

m

u

v

c

g t 8 We11 Write the trigonometric ratio which must be used in order to find the value of the

pronumeral in each of the following triangles. a

b

12

12 15

b cos (q ) =

150

7 We10 For each of the following triangles, write the expressions for ratios of each of the given

c

25

5

q

15

4

30 q

d

2.7

e

p q

14.3 17.5

maths Quest 10 for the australian Curriculum

t

17

14.3

f

35è

17.5 a

q

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

g

9.8

31

a

20 31

i cos (a ) =

i

q 20

7 x

g sin (15è) = h tan (q ) =

h

7

3.1

x

3.1 9.8

15è

reasoning 9 Consider the right-angled triangle shown at right. a Label each of the sides using the letters O, A, H with O = 34 mm, A = 39 mm, H = 51 mm

b c

d e

f T he sin of an angle is equal to the cos of its complement g angle. h

eBook plus

Interactivity Using trigonometry

int-1146

a

O

using trigonometry to calculate side lengths ■

Consider this right-angled triangle. Labelling the sides with respect to the 42è angle, we can see that the unknown side is opposite and we are given the hypotenuse. x From the diagram at right, sin 42è = . Using a 24 calculator, we know that the sine ratio of a 42è angle is approximately 0.6691. We can now solve this equation to find the value of x. We are therefore able to calculate a side length if we are given the size of an angle and one other side.

x

24 m 42è

Opposite

5D

a

H

41è respect to the 41è angle. A Measure the side lengths (to the nearest millimetre). Determine the value of each trigonometric ratio. (Where applicable, answers should be given correct to 2 decimal places.) i sin 41è sin (41è) = 0.67 41è ii cos 41è cos (41è) = 0.76 iii tan 41è tan (41è) = 0.87 What is the value of the unknown angle, a ? a = 49è Determine the value of each of these trigonometric ratios, correct to 2 decimal places. i sin a sin (49è) = 0.76 ii cos a cos (49è) = 0.67 iii tan a tan (49è) = 1.15 (Hint: First relabel the sides of the triangle with respect to angle a.) reFleCTion    What do you notice about the relationship between sin 41è and cos a ? They are equal. How do we determine which of sin, What do you notice about the relationship cos or tan to use in a trigonometry question? between sin a and cos 41è? They are equal. Make a general statement about the two angles.

x

Hy po 24 tenus m e 42è Adjacent

Chapter 5 Trigonometry I

151

measurement AND geometry • Pythagoras and trigonometry

■■

■■

The solution to the above problem is: O the sine ratio formula sin q  = H x the result of substituting into the formula sin 42è = 24 x = 24 ì sin 42è rearranging the formula to make x the subject x ö 16.06  m the result of the calculation. We need to apply this method using any of the three trigonometric ratios to find a side length. The steps used in solving the problem are as follows. Step 1. Label the sides of the triangle, which are either given, or need to be found, with respect to the given angle. Step 2. Consider the sides involved and determine which of the trigonometric ratios is required. (Use the mnemonic SOH  CAH  TOA to help you.) (a) Use the sine ratio if the hypotenuse (H) and the opposite side (O) are involved. (b) Use the cosine ratio if the hypotenuse (H) and the adjacent side (A) are involved. (c) Use the tangent ratio if the opposite (O) and the adjacent (A) sides are involved. Step 3. Substitute the values of the pronumerals into the ratio. Step 4. Solve the resultant equation for the unknown side length.

Worked Example 12

Find the value of the pronumeral for each of the following. Give answers correct to 3 decimal places. a         b  6 cm

a

32è 0.346 cm

f

35è

Think a

1

Label the sides of the triangle, which are either given, or need to be found.

Write/draw a 6 cm

H

a

35è

152

2

Identify the appropriate trigonometric ratio to use.

3

Substitute O = a, H = 6, q = 35è.

4

Make a the subject of the equation.

5

Calculate and round the answer, correct to 3 decimal places.

Maths Quest 10 for the Australian Curriculum

O

sin q  =

O H

sin 35è =

a 6

6 sin 35è = a a = 6 sin 35è a ö 3.441  cm

measurement AND geometry • Pythagoras and trigonometry b

1

Label the sides of the triangle, which are either given, or need to be found.

b H

32è

0.346 cm

2

Identify the appropriate trigonometric ratio to use.

3

Substitute A = f, H = 0.346 and q = 32è.

4

Make f the subject of the equation.

5

Calculate and round the answer, correct to 3 decimal places.

A f

A H

cos q  =

f 0.346 0.346 cos 32è = f cos 32è =

f = 0.346 cos 32è ö 0.293  cm



Worked Example 13

Find the value of the pronumeral in the triangle shown. Give the answer correct to 2 decimal places. 120 m 5è P Think 1

Write/draw

Label the sides of the triangle, which are either given, or need to be found.

5è

A

2

Identify the appropriate trigonometric ratio to use.

tan q  =

O A

3

Substitute O = 120, A = P and q = 5è.

tan 5è =

120 P

4

Make P the subject of the equation. (i)  Multiply both sides of the equation by P. (ii)  Divide both sides of the equation by tan 5è.

5

Calculate and round the answer, correct to 2 decimal places.

O 120 m

H P

P ì tan 5è = 120 120 P= tan 5° P ö 1371.61  m

remember

The trigonometric ratios can be used to find a side length in a right-angled triangle when we are given one other side length and one of the acute angles. Chapter 5 Trigonometry I

153

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

exerCise

5D inDiViDual PaThWays eBook plus

Activity 5-D-1

Calculating side lengths using trigonometry doc-5020

using trigonometry to calculate side lengths FluenCy 1 We12 Find the length of the unknown side in each of the following, correct to 3 decimal places. a 8.660 b 7.250 c 8.412 a

c

b 8

x 10 cm

Activity 5-D-2

Applying trigonometry to simple figures doc-5021

25è

a

31è

a

14

60è

Activity 5-D-3

Practical applications of trigonometry doc-5022

2 We13 Find the length of the unknown side in each of the following triangles, correct to

2 decimal places. a 0.79 a

b 4.72

c 101.38 b

71è

4.6 m 13è

m

n 2.3 m

c 94 mm 68è t 3 Find the length of the unknown side in each of the following, correct to 2 decimal places. b

a

8è5

3 a 33.45 m b 74.89 m c 44.82 m d 7.76 mm e 80.82 km f 9.04 cm

c

P

2'4

14 m

5''

11.7 m

43.95 m

2'

t

1 8è

1

x

40è26'

d

f

e

80.9 cm

x 75.23 km ' 42 11.2 mm

154

maths Quest 10 for the australian Curriculum

21è25'34"

x

è 34

6è25' x

measurement AND geometry • Pythagoras and trigonometry 4 Find the value of the pronumeral in each of the following, correct to 2 decimal places. a a x = 31.58  cm b y = 17.67  m c z = 14.87  m d p = 67.00  m e p = 21.38  km, q = 42.29  km f a = 0.70  km, b = 0.21  km

b

x



23.7 m 36è42' y

43.9 cm

46è

c

d

34

z

2 è1

p

15.3 m

'

12.3 m

13è12' e

f

0.732 km

q p

a

73è5'

63è11' 47.385 km

b

understanding 5 Given that the angle q is 42è and the length of the hypotenuse is 8.95  m in a right-angled

triangle, find the length of: a the opposite side 6.0  m b the adjacent side. 6.7  m Give each answer correct to 1 decimal point. 6 A ladder rests against a wall. If the angle between the ladder and the ground is 35è and the foot of the ladder is 1.5  m from the wall, how high up the wall does the ladder reach? 1.05  m Reasoning 7 Tran is going to construct an enclosed rectangular desktop that is at an incline of 15è. The

diagonal length of the desktop is 50 cm. At one end, the desktop will be raised 8 cm. The desktop will be made of wood. The diagram below represents this information. Side view of the desktop x 15è

x = 30.91 cm, y = 29.86 cm, z = 39.30 cm

Top view of the desktop 8 cm

y

a Determine the values (in centimetres) of x, y

and z of the desktop. Write your answers correct to 2 decimal places. b Using your answer from part a determine the minimum area of wood, in cm2, Tran needs to construct his desktop. Write your answer 2941.54 cm2 correct to 2 decimal places.

z 50 cm

reflection 

 

How does solving a trigonometric equation differ when we are finding the length of the hypotenuse side compared to when finding the length of a shorter side?

Chapter 5 Trigonometry I

155

measurement AND geometry • Pythagoras and trigonometry

Using trigonometry to calculate angle size

5E

■■

To find the size of an angle using the trigonometric ratios, we need to be given the length of any two sides.

Worked Example 14

For each of the following, find the size of the angle, q, correct to the nearest degree. a  b  5m 5 cm

3.5 cm

q 11 m

q Think a

1

Label the sides of the triangle, which are either given, or need to be found.

Write/draw a H

O

5 cm

3.5 cm

q

b

2

Identify the appropriate trigonometric ratio to use. We are given O and H, so choose the sine ratio.

sin q  =

3

Substitute O = 3.5 and H = 5 and evaluate the expression.

sin q  =

4

Make q the subject of the equation using inverse sine.

5

Evaluate q and round the answer, correct to the nearest degree.

1

Label the sides of the triangle, which are either given, or need to be found.

O H

3.5 5 = 0.7

q = sin-1 0.7 = 44.427  004è

q ö 44è

b

O 5m q

2

156

Identify the appropriate trigonometric ratio to use. We are given O and A, so choose the tangent ratio.

Maths Quest 10 for the Australian Curriculum

tan q  =

11 m

O A

A

measurement AND geometry • Pythagoras and trigonometry

3

Substitute O = 5 and A = 11. As the value of tan (q ) is a simple fraction, we do not need to evaluate the expression.

4

Make q the subject of the equation using inverse tangent.

5

5

tan q  = 11 5

q = tan-1 11 = 24.443  954  78è

Evaluate q and round the answer, correct to the nearest degree.

■■

q ö 24è



When asked for a more accurate measurement of an angle, we are able to use the calculator to find an angle correct to the nearest minute or nearest second.

Worked Example 15

Find the size of angle q  in each of the triangles shown below. a  b  3.1 m q 55 cm

7.2 m

q 42 cm

(Answer correct to the nearest minute.)

(Answer correct to the nearest second.)

Think a

1

Label the sides of the triangle, which are either given, or need to be found.

Write/draw a

3.1 m A q O 7.2 m

O A

2

Identify the appropriate trigonometric ratio to use.

tan q  =

3

Substitute O = 7.2 and A = 3.1 and evaluate the expression.

tan q  =

4

Make q the subject of the equation using inverse tangent.

q = tan-1 2.322  580  645

5

Evaluate q and write the calculator display.

q = 66.705  436  75è

6

Use the calculator to convert the answer to degrees, minutes and seconds and round the answer to the nearest minute.

7.2 3.1 = 2.322  580  645



= 66è42Å19.572ë q ö 66è42Å Chapter 5 Trigonometry I

157

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry b

1

b

Label the sides of the triangle, which are either given, or need to be found.

H 55 cm

q 42 cm A 2

Identify the appropriate trigonometric ratio to use.

cos q  =

A H

3

Substitute A = 42 and H = 55.

cos q  =

42 55

4

Make q the subject of the equation using inverse cosine.

q = cos-1

5

Evaluate q and write the calculator display.

q = 40.214 171 02è

6

Use the calculator to convert the answer to degrees, minutes and seconds and round the answer to the nearest second.

= 40è12Å51.016ë q ö 40è12Å51ë

42 55

rememBer

1. The trigonometric ratios can be used to find the size of the acute angles in a right-angled triangle when we are given the length of two sides. 2. To find an angle size we need to use the inverse trigonometric functions. 3. Answers may be given correct to the nearest degree, minute or second, or as decimal degrees.

exerCise

5e inDiViDual PaThWays eBook plus

Activity 5-E-1

using trigonometry to calculate angle size FluenCy 1 We14 Find the size of the angle, q, in each of the following. Give your answer correct to the

nearest degree.

5.2

Activity 5-E-2

158

b 47è

c 69è c

b

Review of angle calculations doc-5023 Calculation angles using trigonometry doc-5024

a 67è

a

4.8

q

maths Quest 10 for the australian Curriculum

4.7 q 3.2

8

q 3

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

inDiViDual PaThWays eBook plus

2 We15a Find the size of the angle marked with the pronumeral in each of the following. Give

your answer correct to the nearest minute. b

a

c

7.2 m

b

12

Activity 5-E-3

17

Applying trigonometry to angle calculations doc-5025

4m

q

2 a 54è47Å

12

q b 33è45Å

10

c 33è33Å

3 We15b Find the size of the angle marked with the pronumeral in each of the following. Give

your answer correct to the nearest second. a

b

a

5m

8 3m q eBook plus

Digital doc

SkillSHEET 5.8 doc-5232

2 c 2.7 3 a 75è31Å21ë

a

b 36è52Å12ëÅ

c 37è38Å51ë

3.5 4 Find the size of the angle marked with the pronumeral in each of the following, giving your

answer correct to the nearest degree.

a 41è b 30è c 49è d 65è e 48è f 37è

a

b a

13.5

89.4

15.3 c

c

77.3

d

106.4

d 43.7

18.7

92.7 b

e

f

12.36 13.85

e

7.3 cm

12.2 cm

18.56 9.8 cm

a

Chapter 5 Trigonometry I

159

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

eBook plus

Digital doc

WorkSHEET 5.2 doc-5233

5 Find the size of each of the angles in the following, giving your answers correct to the nearest

minute. b

a

c

d

a

x

0.798

5.7

2.3

56.3

y 0.342

b

e 5 a a = 25è47Å, b = 64è13Å b d = 25è23Å, e = 64è37Å c x = 66è12Å, y = 23è48Å

27.2

unDersTanDing 6 a Calculate the length of the sides r, l and h. Write r = 57.58, l = 34.87, h = 28.56

your answers correct to 2 decimal places. b Calculate the area of ABC, correct to the nearest square centimetre. 714 cm2 c Calculate ±BCA. 29.7è

A h D

l 20 cm

r 125è 30 cm

B

C

reasoning 7 In the sport of air racing, small planes have to travel between two large towers (or pylons). The

gap between a pair of pylons is smaller than the wing-span of the plane, so the plane has to go through on an angle with one wing ‘above’ the other. The wing-span of a competition airplane is 8 metres.

a Determine the angle, correct to 1 decimal place, that the plane has to tilt if the gap

between pylons is: i 124.42 km/h ii 136.57 km/h iii 146.27 km/h

160

i 7 metres ii 6 metres 41.4è 29.0è b Because the plane has rolled away from the

horizontal as it travels between the pylons it loses speed. If the plane’s speed is below 96 km/h it will stall and possibly crash. For each degree of ‘tilt’ the speed of the plane is reduced by 0.98 km/h. What is the minimum speed the plane must go through each of the pylons in part a? Write your answer correct to 2 decimal places.

maths Quest 10 for the australian Curriculum



iii 5 metres. 51.3è

reFleCTion 

 

How is finding the angle of a right-angled triangle different to finding a side length?

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

5F eBook plus

eLesson Height of a satellite

angles of elevation and depression ■

■

eles-0173

Many people use trigonometry at work. It is particularly important in careers such as the building trades, surveying, architecture, navigation and engineering. Trigonometric ratios have a variety of applications, some of which will be discussed in this section. Trigonometric ratios can be used to solve problems. When solving a problem, the following steps can be of assistance. 1. Sketch a diagram to represent the situation described in the problem. 2. Label the sides of the right-angled triangle with respect to the angle involved. 3. Identify what is given and what needs to be found. 4. Select an appropriate trigonometric ratio and use it to find the unknown measurement. 5. Interpret your result by writing a worded answer.

angles of elevation and depression ■

■

When we need to look up or down in order to see a certain object, our line of vision (that is, the straight line from the observer’s eye to the object) is inclined. The angle of inclination of the line of vision to the horizontal when looking up is referred to as the angle of elevation, and when looking down it is referred to as the angle of depression. The angle of elevation is measured up from the horizontal line to the line of vision.

q

Angle of elevation Horizontal

■

The angle of depression is measured down from the horizontal line to the line of vision. Horizontal q

■

Angle of depression

For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B. Angle of depression q of A from B

q

B

Angle of elevation of B from A

A Chapter 5 Trigonometry I

161

measurement AND geometry • Pythagoras and trigonometry

Worked Example 16

From an observer, the angle of elevation of the top of a tree is 50è. If the observer is 8 metres from the tree, find the height of the tree. Think 1

Write/draw

Sketch a diagram and label the sides of the triangle with respect to the given angle. Let the height of the tree be h. h O 50è 8m

A

2

Identify the appropriate trigonometric ratio. We are given A and need to find O, so choose the tangent ratio.

tan q  =

O A

3

Substitute O = h, A = 8 and q = 50è.

tan 50è =

h 8

4

Rearrange to make h the subject.



5

Calculate and round the answer to 2 decimal places.



6

Give a worded answer.

The height of the tree is 9.53  m.

h = 8 tan 50è ö 9.53

remember

1. To solve a problem involving trigonometric ratios, follow these steps: (a)  Draw a diagram to represent the situation. (b) Label the diagram with respect to the angle involved (either given or that needs to be found). (c)  Identify what is given and what needs to be found. (d) Select an appropriate trigonometric ratio and use it to find the unknown side or angle. (e)  Interpret the result by writing a worded answer. 2. The angle of elevation is measured up and the angle of depression is measured down from the horizontal line to the line of vision. Horizontal q

q

Angle of elevation Horizontal

3. For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B.

Angle of depression q of A from B

A

162

Angle of depression

Maths Quest 10 for the Australian Curriculum

q

Angle of elevation of B from A

B

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

exerCise

5F inDiViDual PaThWays eBook plus

Activity 5-F-1

Identifying elevation and depression doc-5026 Activity 5-F-2

Calculating elevation and depression doc-5027 Activity 5-F-3

Applications of elevation and depression doc-5028

eBook plus

Digital doc

SkillSHEET 5.6 doc-5228

angles of elevation and depression FluenCy 1 We16 The angle of elevation from an observer to the top of a tree is 54è22Å. If the tree is known to be 12.19 m high, how far is the observer from it? 8.74 m 2 From the top of a cliff 112 m high, the angle of depression to a boat is 9è15Å. How far is the boat from the foot of the cliff? 687.7 m 3 A person on a ship observes a lighthouse on the cliff, which is 830 metres away from the ship.

The angle of elevation of the top of the lighthouse is 12è. 176.42 m a How far above sea level is the top of the lighthouse? b If the height of the lighthouse is 24 m, how high is the cliff? 152.42 m 4 At a certain time of the day a post, 4 m tall, casts a shadow of 1.8 m. What is the angle of elevation of the sun at that time? 65è46Å 5 An observer, who is standing 47 m from a building, measures the angle of elevation of the top of the building as 17è. If the observer’s eye is 167 cm from the ground, what is the height of the building? 16.04 m unDersTanDing 6 A surveyor needs to determine the height of a building. She measures the angle of elevation of

the top of the building from two points, 38 m apart. The surveyor’s eye level is 180 cm above the ground.

h 47è12

'

x

35è5

0' 38 m

180 cm

a Find two expressions for the height of the building, h, in terms of x using the two angles. h = x tan (47è12Å) m; h = (x + 38) tan (32è50Å) m a . x = 76.69 m b Solve for x by equating the two expressions obtained in c Find the height of the building. 84.62 m 7 The height of another building needs to be determined but cannot be found directly. The

surveyor decides to measure the angle of elevation of the top of the building from different sites, which are 75 m apart. The surveyor’s eye level is 189 cm above the ground.

h 43è35

'

x

32è1

8'

75 m

189 cm

a Find two expressions for the height of the building, h, in terms of x using the two angles. h = x tan (43è35Å) m; h = (x + 75) tan (32è18Å) m b Solve for x. 148.40 m c Find the height of the building. 141.24 m Chapter 5 Trigonometry I

163

measurement AND geometry • Pythagoras and trigonometry 8 A lookout tower has been erected on top of a cliff. At a distance of 5.8  km from the foot of the

cliff, the angle of elevation to the base of the tower is 15.7è and to the observation deck at the top of the tower is 16è respectively as shown in the figure below. How high from the top of the cliff is the observation deck? 0.033  km or 33  m

  

16è 15.7è 5.8 km

9 Elena and Sonja were on a camping trip to the

Angle of depression

1.3 km Grampians, where they spent their first day 20è hiking. They first walked 1.5  km along a path 1.5 km 150 m inclined at an angle of 10è to the horizontal. 10è Then they had to follow another path, which 1.4 km was at an angle of 20è to the horizontal. They walked along this path for 1.3  km, which brought them to the edge of the cliff. Here Elena spotted a large gum tree 1.4  km away. If the gum tree is 150  m high, what is the angle of depression from the top of the cliff to the top of the gum tree? 21è

10 a Find the height of a telegraph

pole in the photograph at right if the angle of elevation to the top of the pole is 8è from a point at the ground level 60  m from the base of the pole. 8.43  m b Find the height of the light pole in the figure below. 56.54  m

43.3è

8è

60 m 60 m

11 From a point on top of a cliff, two boats are observed.

If the angles of depression are 58è and 32è and the cliff is 46  m above sea level, how far apart are the boats? 44.88  m

32è 58è 46 m

164

Maths Quest 10 for the Australian Curriculum

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 12 Joseph is asked to obtain an estimate of the height

42è 1.76 m

15 m

eBook plus

Digital doc

WorkSHEET 5.3 doc-5234

of his house using any mathematical technique. He decides to use an inclinometer and basic trigonometry. Using the inclinometer, Joseph determines the angle of elevation, q, from his eye level to the top of his house to be 42è. The point h x from which Joseph measures the angle of elevation is 15 m away from his house and the distance from Joseph’s eyes to the ground is 1.76 m. a Fill in the given information on the diagram provided (substitute values for the pronumerals). b Determine the height of Joseph’s house. 15.27 m 13 The competitors of a cross-country run are nearing the finish line. From a lookout 100 m above the track, the angles of depression to the two leaders, Nathan and Rachel, are 40è and 62è respectively. How far apart are the two competitors? 66 m 14 A 2.05 m tall man, standing in front of a street light 3.08 m high, casts a 1.5 m shadow. a What is the angle of elevation from the ground to 54è the source of light? b How far is the man from the bottom of the light pole? 0.75 m

x q d

40è 62è 100 m

3.08 m

2.05 m 1.5 m

reasoning 15 The angle of elevation of a hot air balloon changes

reFleCTion    from 27è at 7.00 am to 61è at 7.03 am, according to an observer who is 300 m away from the take-off point. What is the difference between an angle of elevation and an a Assuming a constant speed, calculate that speed angle of depression? (in m/s and km/h) at which the balloon is rising, 2.16 m/s, 7.77 km/h correct to 2 decimal places. b The balloon then falls 120 metres. What is the angle of elevation now? Write your answer correct to 1 decimal place. 54.5è

5g

Bearings and compass directions ■

A bearing can be expressed as either a true bearing or a compass direction.

Compass directions ■

■

Compass (conventional) bearings are directions measured from the north–south line in either a clockwise or anticlockwise direction. To identify the compass direction of an object we need to state: 1. whether the angle is measured from north (N) or south (S) 2. the size of the angle and 3. whether the angle is measured in the direction of west (W) or east (E). Chapter 5 Trigonometry I

165

measurement AND geometry • Pythagoras and trigonometry

■■

For example, the compass direction of S20èE means the direction is 20è from south towards east, while the compass direction N40èW means the direction that is 40è from north towards west. N

N

N40èW

40è

W

E

W

E

20è

S

S20èE

S

      

True bearings ■■ ■■

True bearings are measured from north in a clockwise direction. They are always expressed in 3 digits. The diagrams below show the bearings of 025è true and 250è true respectively. (These true bearings are more commonly written as 025èT and 250èT.) N

N

025è true

25è

W

E

W

E

250è

250èT S

S

      

Worked Example 17

A boat travels a distance of 5  km from P to Q in a direction of 035èT. a  How far east of P is Q? b  How far north of P is Q? c  What is the true bearing of P from Q?

a

1

Draw a diagram to represent the situation. Label the hypotenuse and the opposite and adjacent sides.

Write/draw a

N

O x

5k m

Think

Ay

35è P 2

166

To determine how far Q is east of P, we need to find the value of x. We are given the length of the hypotenuse (H) and need to find the length of the opposite side (O). Choose the sine ratio.

Maths Quest 10 for the Australian Curriculum

sin q  =

O H

Q H

measurement AND geometry • Pythagoras and trigonometry

b

c

3

Substitute O = x, H = 5 and q = 35è.

4

Make x the subject of the equation.

5

Evaluate and round the answer, correct to 2 decimal places.

6

Write the answer in words.

1

To determine how far Q is north of P, we need to find the value of y. This can be done in several ways, namely: using the cosine ratio, the tangent ratio, or Pythagoras’ theorem. Let’s use the cosine ratio.

2

Substitute P = y, H = 5 and q = 35è.

3

Make y the subject of the equation.

4

Evaluate and round the answer, correct to 2 decimal places.

5

Write the answer in words.

1

To find the bearing of P from Q, we need to draw the compass directions through Q and then measure the angle in the clockwise direction from the north line through Q to the line PQ. Show the required angle on the diagram.

sin 35è =

x 5

x = 5 sin 35è = 2.87 Point Q is 2.87  km east of P. b

cos q  =

A H

cos 35è =

y 5

y = 5 cos 35è = 4.10 Point B is 4.10  km north of A. c

N

N Q

a 35è P

2

Study the diagram. The angle that represents the true bearing is the sum of 180è (from north to south) and the angle, labelled a. Now the north lines through P and Q are parallel and so the line PQ is a transversal. Therefore angle 35è and angle a are equal (being alternate angles). Calculate the true bearing.

True bearing = 180è + a a = 35è True bearing = 180è + 35è = 215è

3

Write the answer in words.

The bearing of P from Q is 215èT.

■■

Sometimes a person or an object (for example, a ship) changes direction during their journey. (This can even happen more than once.) In situations like this we are usually interested in the total distance the object has moved and its final bearing from the starting point.   The following worked example shows how to deal with such situations.

Worked Example 18

A boy walks 2  km on a true bearing of 090è and then 3  km on a true bearing of 130è. a How far east of the starting point is the boy at the completion of his walk? (Answer correct to 1 decimal place.) b How far south of the starting point is the boy at the completion of his walk? (Answer correct to 1 decimal place.) c What is the bearing of the boy (from the starting point), in degrees and minutes, at the completion of his walk? Chapter 5 Trigonometry I

167

measurement AND geometry • Pythagoras and trigonometry

Think

Write/draw

Draw a diagram of the boy’s journey.

N

N 130è

2 km

3 km

a

1

The first leg of the journey is due east so we find the eastern component of the second leg. Construct a triangle about the second leg of the journey. We can calculate one of the missing angles by using the rule of supplementary angles: 180è - 130è = 50è.

a

N

N 130è

2 km Ay

50è H

E

3 km

x O

b

2

We need to find the eastern component of the journey, x, which is the opposite side and have been given the hypotenuse. Choose the sine ratio.

3

Substitute O = x, H = 3 and q = 50è.

4

Make x the subject of the equation.

5

Evaluate and round correct to 1 decimal place.

6

Add to this the 2  km east that was walked in the first leg of the journey and give a worded answer.

1

In the first part of the journey the boy has not moved south at all. Thus the distance that he moved south of the starting point is the southern component of the second leg, labelled y. (See the diagram in part a.)

2

To find y we can use Pythagoras’ theorem, as we know the lengths of two out of three sides in the right-angled triangle. Note that the hypotenuse, c, is 3 and one of the sides is 2.3, as found in part a. Round the answer correct to 1 decimal place. Note: Alternatively, the cosine ratio could have been used.

3

168

Write the answer in words.

Maths Quest 10 for the Australian Curriculum

sin q  =

O H

x 3 x = 3 sin 50è

sin 50è =

= 2.3  km Total distance east = 2 + 2.3 = 4.3  km The boy walked a total of 4.3  km east of the starting point. b Distance south = y  km

a2 = c2 - b2 y2 = 32 - 2.32 = 9 - 5.29 = 3.71 y = 3.71 = 1.9  km

y 3 y = 3 cos 50è = 1.9  km

cos 50è =

The boy walked a total of 1.9 km south of the starting point.

measurement AND geometry • Pythagoras and trigonometry c

1

Draw a diagram of the journey and write in the distances found in parts a and b. The bearing of the boy from the starting point is represented by the angle a (that is, the angle measured in a clockwise direction from north to the line joining the starting and the finishing points of the journey).

c

N

a

N 130è

2 km

q 3 km

A 1.9 km 4.3 km O

2

The size of angle a cannot be found directly. Find the size of the supplementary angle labelled q.

3

We have the lengths of the opposite side and the adjacent side, so choose the tangent ratio.

tan q  =

4

Substitute O = 4.3 and A = 1.9 and evaluate.

tan q  =

5

Make q the subject of the equation using the inverse tangent function.

q = tan-1 2.263  157  895

6

Evaluate and round to the nearest minute.

7

Find the angle a.

a = 180è - 66è10Å = 113è50Å

8

Write the answer in words.

The bearing of the boy from his starting point is 113è50Å T.

O A

4.3 1.9 = 2.263  157  895

= 66.161  259  82è = 66è9Å40.535ë = 66è10Å

remember

1. To identify the compass direction of an object we need to state (in this order): (a) whether the angle is measured from north (N) or south (S) (b) the size of the angle and (c) whether the angle is measured in the direction of west (W) or east (E). 2. True bearings are measured from north in a clockwise direction and expressed as 3 digits and with a T. 3. When solving problems involving bearings or compass directions, always draw a clear diagram prior to attempting the problem. Exercise

5G

Bearings and compass directions fluency 1

Change each of the following compass directions to true bearings. a N20èE 020èT d S28èE 152èT

b N20èW 340èT e N34èE 034èT

c S35èW 215èT f S42èW 222èT Chapter 5 Trigonometry I

169

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 2 Change each of the following true bearings to compass directions.

inDiViDual PaThWays

b 132èT S48èE e 086èT N86èE

c 267èT S87èW f 234èT S54èW



a 049èT N49èE N30èW d 330èT

eBook plus

Bearings doc-5029

a

b

N

N

Activity 5-G-2

3k

Calculations involving bearings doc-5030

m

35è W

Activity 5-G-3

22è 2.5 km

Applications involving bearings doc-5031

S

a

N

c

100è 30 km

N 35è 2.5 km

40

40è

d

N

km

N

W N 135è

35è m 8k

0

23

N

km

S

m

0k

14

e

240è

12 km 65è

N

N 50è

m 7k

km

N N 260è 120è 0.8 km N 32è 1.3 km

f

N

40è

30 0m

N 50è

2.1

c

4 km

E

50 0m

b

E

a 3 km 325èT b 2.5 km 112èT c 8 km 235èT d 4 km 090èT, then 2.5 km 035èT e 12 km 115èT, then 7 km 050èT f 300 m 310èT, then 500 m 220èT

3 Describe the following paths using true bearings.

Activity 5-G-1

30è

50è

7k

S

a A ship travels 040èT for 40 km and then 100èT for 30 km. b A plane flies for 230 km in a direction 135èT and a further 140 km in a direction 240èT. c A bushwalker travels in a direction 260èT for 0.8 km, then changes direction to 120èT for

N

m

5k

m

8k

e

4 Show each of the following by drawing the paths.

N

d

40è

1.3 km, and finally travels in a direction of 32è for 2.1 km. d A boat travels N40èW for 8 km, then changes direction to S30èW for 5 km and then

m

S50èE for 7 km. e A plane travels N20èE for 320 km, N70èE for 180 km and S30èE for 220 km.

N

5 We17 a You are planning a trip on your yacht. If you travel 20 km from A to B on a bearing

S

of 042èT: i how far east of A is B? 13.38 km ii how far north of A is B? 14.86 km 222èT iii what is the bearing of A from B? b In the next part of the journey you decide to travel 80 km from B to C on a bearing of 130èT. N i Show the journey to be travelled using a diagram. 130è B N ii How far south of B is C? 51.42 km 80 iii How far east of B is C? 61.28 km 42è km iv What is the bearing of B from C? 310èT A 20

km

320 km

30è

m

20è

0k

N

22

70è180 km

170

maths Quest 10 for the australian Curriculum

C

measurement AND geometry • Pythagoras and trigonometry

In the next part of the journey you decide to travel 45  km from C to D on a bearing of 210èT. Show the journey to be travelled using a diagram. How far south of C is D? 38.97  km How far west of C is D? 22.5  km What is the bearing of C from D? 030èT N 6 If a farmhouse is situated 220  m N35èE from a shed, what is the true bearing of the shed from the house? 215èT C c

N B 130è

N

80

km

45

km

A

20 km

42è

D

a 9.135  km b 2.305  km c 104è10Å T

  i  ii iii iv

210è

Understanding 7 A pair of hikers travel 0.7  km on a true bearing of 240è and then 1.3  km on a true bearing of 300è. How far west have they travelled from their starting point? 1.732  km 8   WE 18  A boat travels 6  km on a true bearing of 120è and then 4  km on a true bearing of 080è. a How far east is the boat from the starting point on the completion of its journey? b How far south is the boat from the starting point on the completion of its journey? c What is the bearing of the boat from the starting point on the completion of its journey? 9 A plane flies on a true bearing of 320è for 450  km. It then flies on a true bearing of 350è for

130  km and finally on a true bearing of 050è for 330  km. How far north of its starting point is 684.86  km the plane? 10 Find the final bearing for each of the following. Express your answer in true bearings, correct to the nearest minute. a A boat travels due east for 4  km and then travels N20èE for 3  km. What is the final bearing of the boat from the starting point? 60è43Å T b A bushwalker travels due north for 3  km, then due east for 8  km. What is the final bearing of the bushwalker from the starting point? 69è27Å T c A car travels due south for 80  km, then travels due west for 50  km, and finally due south for a further 30  km. What is the final bearing of the car from the starting point? 204è27Å T Reasoning 11 A yacht is sailing around islands in the Pacific

y

Ocean. The sailor sees a mountain range on an island that is on a bearing of 330èT. The 29è yacht sails at a rate of 5 km/h for 30 minutes N N x due West, such that the mountain range is 30è now on a bearing of 050èT. From this new 50è a position, the sailor determines that the angle q of elevation to the highest point on the d km 330è mountain range is 29è, as shown in the Note: Diagram is not drawn to scale. diagram at right. a Determine the exact values of the angles labelled q and a. q = 60è, a = 40è b Determine the distance d, in kilometres, the yacht sailed on a bearing of due West before another bearing of the mountain range was taken. 2.5 km c Using any suitable method, determine the value of x, in kilometres. Write your answer correct to 3 decimal places. 2.198 km reflection    d Using the value of x from part b determine the What is the difference between value of y, the height above sea level of highest true bearings and compass point on the mountain range, in kilometres. Write directions? your answer correct to 2 decimal places. 1.22 km Chapter 5 Trigonometry I

171

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

applications

5h eBook plus

Interactivity Applying trigonometry to drafting problems

■

■

int-2781

■

Many people use trigonometry at work. It is particularly important in careers such as the building trades, surveying, architecture and engineering. Trigonometric ratios can be used to find angles of elevation and depression, as well as to calculate distances which we could not otherwise easily measure. When solving a problem, remember the following steps. 1. Sketch a diagram to represent the situation described in the problem. 2. Label the sides of the right-angled triangle with respect to the angle involved. 3. Identify what is given and what needs to be found. 4. Select an appropriate trigonometric ratio and use it to find the unknown measurement. 5. Interpret your result by writing a worded answer.

WorkeD examPle 19

A ladder of length 3 m makes an angle of 32è with the wall. a How far is the foot of the ladder from the wall? b How far up the wall does the ladder reach? c What angle does the ladder make with the ground? Think

WriTe/DraW

Sketch a diagram and label the sides of the right-angled triangle with respect to the given angle.

(wall)

32è y

3m H a

a

172

1

We need to find the distance of the foot of the ladder from the wall (O) and are given the length of the ladder (H). Choose the sine ratio.

2

Substitute O = x, H = 3 and q = 32è.

3

Make x the subject of the equation.

4

Evaluate and round the answer to 2 decimal places.

5

Write the answer in words.

maths Quest 10 for the australian Curriculum

A

x O

a sin q =

O H

sin 32è =

x 3

x = 3 sin 32è ö 1.59 m The foot of the ladder is 1.59 m from the wall.

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry

b

c

b cos q =

A H

1

We need to find the height the ladder reaches up the wall (A) and are given the hypotenuse (H). Choose the cosine ratio.

2

Substitute A = y, H = 3 and q = 32è.

3

Make y the subject of the equation.

y = 3 cos 32è

4

Evaluate and round the answer to 2 decimal places.

y ö 2.54 m

5

Write the answer in words.

1

To find the angle that the ladder makes with the ground, we could use any of the trigonometric ratios, as the lengths of all three sides are known. However, it is quicker to use the angle sum of a triangle.

2

cos 32è =

y 3

The ladder reaches 2.54 m up the wall. c a  + 90è + 32è = 180è

a + 122è = 180è a  = 180è - 122è a  = 58è

Write the answer in words.

The ladder makes a 58è angle with the ground.

rememBer

To solve a problem involving trigonometric ratios, follow these steps: (a) Draw a diagram to represent the situation. (b) Label the diagram with respect to the angle involved (either given or that needs to be found). (c) Identify what is given and what needs to be found. (d) Select an appropriate trigonometric ratio and use it to find the unknown side or angle. (e) Interpret the result by writing a worded answer. exerCise

5h inDiViDual PaThWays eBook plus

Activity 5-H-1

Trigonometry applications 1 doc-5032 Activity 5-H-2

Trigonometry applications 2 doc-5033

applications FluenCy 1 We19 A 3 m-long ladder is placed against a wall so that it reaches 1.8 m up the wall. 36è52Å a What angle does the ladder make with the ground? b What angle does the ladder make with the wall? 53è8Å c How far from the wall is the foot of the ladder? 2.4 m 2 Jamie decides to build a wooden pencil box. He wants his ruler to be able to lie across

the bottom of the box, so he allows 32 cm along the diagonal. The width of the box is to be 8 cm. 32 cm q

Activity 5-H-3

Trigonometry applications 3 doc-5034

8 cm

Calculate: a the size of angle q 14è29Å b the length of the box. 31 cm Chapter 5 Trigonometry I

173

measurement AND geometry • Pythagoras and trigonometry 3 A carpenter wants to make a roof pitched at 29è30Å, as shown in the diagram. How long should he cut the beam, PR? 6.09  m R

P

29è30'

Q

10.6 m

4 The sloping sides of a gable roof are each 7.2  m long. They rise to a height of 2.4  m in the centre. What angle do the sloping sides make with the horizontal? 19è28Å

5 The mast of a boat is 7.7  m high. A guy wire from the top of the mast is fixed to the deck 4  m from the base of the mast. Determine the angle the wire makes with the horizontal. 62è33Å understanding 6 A desk top of length 1.2  m and width 0.5  m rises to 10  cm.

0.5 m

E

F 10 cm

C

D

A

B

1.2 m

Calculate: a ±DBF 11è32Å b ±CBE. 4è25Å 7 A cuboid has a square end. H

G X

D

C O

E

F

45 cm

A 25 cm B a If the length of the cuboid is 45  cm and its height and width are 25  cm each, calculate:   35.36  cm 51.48  cm i the length of BD    ii  the length of BG   57.23  cm iii the length of BE 51.48  cm iv  the length of BH v ±FBG vi  ±EBH. 29è3Å 25è54Å 174

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

If the midpoint of FG is X and the centre of the rectangle ABFE is O calculate: i the length OF 25.74  cm    ii the length FX 12.5  cm iii ±FOX 25è54Å iv the length OX.   28.61  cm 8 In a right square-based pyramid, the length of the side of the base is 12  cm and the height is 26  cm.

26 cm

b

12 cm

Determine: a the angle the triangular face makes with the base 77è b the angle the sloping edge makes with the base 71è56Å c the length of the sloping edge. 27.35  cm 9 In a right square-based pyramid, the length of the side of the square base is 5.7  cm.

68è 5.7 cm

If the angle between the triangular face and the base is 68è, determine: a the height of the pyramid 7.05  cm b the angle the sloping edge makes with the base 60è15Å c the length of the sloping edge. 8.12  cm 10 In a right square-based pyramid, the height is 47  cm. If the angle between a triangular face and

the base is 73è, calculate: a the length of the side of the square base 28.74  cm b the length of the diagonal of the base 40.64  cm c the angle the sloping edge makes with the base. 66è37Å 11 The height of a vertical cone is 24.5  cm. 48è37'10"

24.5 cm

If the angle at the apex is 48è37Å10ë, determine: a the length of the slant edge of the cone 26.88  cm b the radius of the cone. 11.07  cm Chapter 5 Trigonometry I

175

measurement AND geometry • Pythagoras and trigonometry Reasoning 12 Aldo the carpenter is lost in a rainforest. He comes across a large river and he knows that he

can not swim across it. Aldo intends to build a bridge across the river. He draws some plans to calculate the distance across the river as shown in the diagram below.

72è Tree

River

4.5 cm 88è

a Aldo used a scale of 1 cm to represent 20 m. Find the real-life distance represented by 4.5 cm in Aldo’s plans. 90 m b Use the diagram below to write an equation for h in terms of d and the two angles. h=

d tan θ ì tan q2 tan θ1 + tan θ 2

h J1

J2

d-x

x d

c Use your equation from b to find the distance across the river, correct to the nearest metre. 250 m reflection 

 

What are some real-life applications of trigonometry?

176

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

Summary Pythagoras’ theorem ■■ ■■ ■■ ■■ ■■ ■■ ■■

The hypotenuse is the longest side of the triangle and is opposite the right angle. On your diagram, check whether you are finding the length of the hypotenuse or one of the shorter sides. The length of a side can be found if we are given the length of the other sides by using the formula c2 = a2 + b2. When using Pythagoras’ theorem, always check the units given for each measurement. If necessary, convert all measurements to the same units before using the rule. Worded problems can be solved by drawing a diagram and using Pythagoras’ theorem to solve the problem. Worded problems should be answered in a sentence. Pythagoras’ theorem in three dimensions

■■ ■■ ■■

Pythagoras’ theorem can be used to solve problems in three dimensions. Some common 3-D shapes include boxes, pyramids and right-angled wedges. To solve problems in three dimensions it is helpful to draw sections of the original shape in two dimensions. Trigonometric ratios

When using the calculator to find values of sine, cosine and tangent, make sure the calculator is in Degree mode. ■■ To find the size of an angle whose sine, cosine or tangent is given, perform an inverse operation; that is, sin-1, cos-1 or tan-1. ■■ Use the calculator’s conversion function to convert between decimal degrees and degrees, minutes and seconds. ■■ There are 60 minutes in 1 degree and 60 seconds in 1 minute. ■■ The three trigonometric ratios, sine, cosine and tangent, are defined as: O A O sin q  = , cos q  = and tan q  = , H H A where H is the hypotenuse, O is the opposite side and A is the adjacent side. ■■ The three ratios are abbreviated to the useful mnemonic SOH  CAH  TOA. ■■ To determine which trigonometric ratio to use, follow these steps. (a) Label the sides of the right-angled triangle that are either given, or need to be found, using the symbols O, A, H with respect to the angle in question. (b) Consider the sides that are involved and write the trigonometric ratio containing both of these sides. (Use SOH  CAH  TOA to assist you.) (c) Identify the values of the pronumerals in the ratio. (d) Substitute the given values into the ratio. ■■

Using trigonometry to calculate side lengths ■■

The trigonometric ratios can be used to find a side length in a right-angled triangle when we are given other side length and one of the acute angles. Using trigonometry to calculate angle size

■■ ■■ ■■

The trigonometric ratios can be used to find the size of the acute angles in a right-angled triangle when we are given the length of two sides. To find an angle size we need to use the inverse trigonometric functions. Answers may be given correct to the nearest degree, minute or second, or as decimal degrees. Chapter 5 Trigonometry I

177

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry Angles of elevation and depression ■

The angle of elevation is measured up and the angle of depression is measured down from the horizontal line to the line of vision. Horizontal q

q

■

Angle of depression

Angle of elevation Horizontal

For any two objects, A and B, the angle of elevation of B, as seen from A, is equal to the angle of depression of A as seen from B. Angle of depression q of A from B

q

B

Angle of elevation of B from A

A Bearings and compass directions ■

■

■

To identify the compass direction of an object we need to state (in this order): (a) whether the angle is measured from north (N) or south (S) (b) the size of the angle (c) whether the angle is measured in the direction of west (W) or east (E). True bearings are measured from north in a clockwise direction and expressed as 3 digits and with a T. When solving problems involving bearings or compass directions, always draw a clear diagram prior to attempting the problem. Applications

■

To solve a problem involving trigonometric ratios, follow these steps: (a) Draw a diagram to represent the situation. (b) Label the diagram with respect to the angle involved (either given or that needs to be found). (c) Identify what is given and what needs to be found. (d) Select an appropriate trigonometric ratio and use it to find the unknown side or angle. (e) Interpret the result by writing a worded answer.

MAPPING YOUR UNDERSTANDING

Homework Book

178

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 131. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

maths Quest 10 for the australian Curriculum

measurement AND geometry • Pythagoras and trigonometry

Chapter review Fluency

7 Which of the following statements is correct? A sin 55è = cos 55è B sin 45è = cos 35è C cos 15è = sin 85è D sin 30è = sin 60è ✔ E sin 42è = cos 48è

1 The most accurate measure

✔

for the length of the third side in the triangle at right is: A 4.83  m B 23.3  cm C 3.94  m D 2330  mm E 4826  mm

5.6 m 2840 mm

8 Which of the following can be used to find the

value of x in the diagram below?

2 What is the value of x in

28.7

this figure? a 5.4 5 b 7.5 c 10.1 ✔ d 10.3 e 4 3 What is the closest length of AG of the cube at right? D A 10 B 30 C 20 D 14 ✔ E 17

x

2

35è 7

A 28.7 sin 35è

A

B C

✔ B

F 10

D

28.7 sin 35°

E

28.7 cos 35°

9 Which of the following expressions can be used to

find the value of a in the triangle shown?

G

10

28.7 cos 35è

C 28.7 tan 35è

10

E

H

x

4 If sin 38è = 0.6157, which of the following will

also give this result? sin 218è sin 322è sin 578è sin 682è sin 142è 5 The angle 118è52Å34ë is also equal to: 52 A 118.5234è B 118 ° A B C D ✔ E

75

a

34

C 118.861è E 118.786è

✔ D

A 35 sin 75è

118.876è

C sin-1

6 Which trigonometric ratio for the triangle shown at

✔

right is incorrect? b A sin a = c a B sin a = c a C cos a = c b D tan a = a a E tan q  = b

35

75 35

✔ B

35

sin-1 75 35

D cos-1 75

75

E cos-1 35

10 If a school is 320  m S42èW from the police station, a a

b c

q

✔

what is the true bearing of the police station from the school? A 042èT B 048èT C 222èT D 228èT E 312èT Chapter 5 Trigonometry I

179

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 11 Calculate x, correct to 2 decimal places. a

x = 113.06 cm



ProBlem solVing 1 A surveyor needs to determine the height of a

building. She measures the angle of elevation of the top of the building from two points, 64 m apart. The surveyor’s eye level is 195 cm above the ground.

x 123.1 cm 48.7 cm b

117 mm 82 mm

x = 83.46 mm

x

12 Calculate the value of the pronumeral, correct to 2 decimal places. 9.48 cm 13.4 cm x

x

8.25 mm 13 Calculate the height of this pyramid.

10 mm

8 mm

8 mm

14 A person standing 23 m away from a tree observes

15

17

18

180

maths Quest 10 for the australian Curriculum

h 47è48

x

'

36è2

4' 64 m

195 cm

a Find the expressions for the height of the

h = tan (47è48Å)x m h = tan (36è24Å) (x + 64) m

16

the top of the tree at an angle of elevation of 35è. If the person is 1.5 m tall, what is the height of 17.6 m the tree? A man of height 1.8 m stands at the window of a tall building. He observes his young daughter in the playground below. If the angle of depression from the man to the girl is 47è and the floor on which the man stands is 27 m above the ground, how far from the bottom of the building is the 26.86 m child? A plane flies 780 km in a direction of 185èT. How far west has it travelled from the starting 67.98 km point? A hiker travels 3.2 km on a bearing of 250èT and then 1.8 km on a bearing of 320èT. How far west has she travelled from the starting point? 4.16 km If a 4 m ladder is placed against a wall and the foot of the ladder is 2.6 m from the wall, what angle does the ladder make with the wall? 40è32Å

building, h, in terms of x using the two angles. b Solve for x by equating the two expressions obtained in part a. 129.10 m 144.32 m c Find the height of the building. 2 The height of a right square-based pyramid is 13 cm. If the angle the face makes with the base is 67è, find: 11.04 cm a the length of the edge of the square base b the length of the diagonal of the base 15.6 cm c the angle the slanted edge makes with the base. 59è2Å

measuremenT anD geomeTry • PyThagoras anD TrigonomeTry 3 A boat sails on a compass direction of E12èS for

10 km then changes direction to S27èE for another 20 km. The boat then decides to return to its starting point. A 12è

10 km

B

27è

20 km

C a How far, correct to 2 decimal places, is the boat from its starting point? 27.42 km b On what bearing should the boat travel to

return to its starting point? Write the angle N43èW or 227èT correct to the nearest degree.

4 A car is travelling northwards on an elevated

expressway 6 m above ground at a speed of 72 km/h. At noon another car passes under the expressway, at ground level, travelling west, at a speed of 90 km/h. a How far apart, in metres, are the two cars 40 seconds after noon? 1280.6 m b At this time the first car stops, while the second car keeps going. At what time will they be 3.5 km apart? Write your answer correct to the 12:02:16.3 pm nearest tenth of a second. 5 Two towers face each other separated by a distance, d, of 20 metres. As seen from the top of the first tower, the angle of depression of the second tower’s base is 59è and that of the top is 31è. What is the height, in metres correct to 2 decimal places, of each of the towers? 33.29 m, 21.27 m

eBook plus

Interactivities

Test yourself Chapter 5 int-2840 Word search Chapter 5 int-2838 Crossword Chapter 5 int-2839

Chapter 5 Trigonometry I

181

eBook plus

aCTiViTies

Chapter opener Digital doc

• Hungry brain activity Chapter 5 (doc-5223) (page 131) Are you ready? Digital docs

(page 132)

• SkillSHEET 5.1 (doc-5224): Rounding to a given number of decimal places • SkillSHEET 5.2 (doc-5225): Rounding the size of an angle to the nearest minute and second • SkillSHEET 5.3 (doc-5226): Labelling the sides of a right-angled triangle • SkillSHEET 5.5 (doc-5227): Rearranging formulas • SkillSHEET 5.6 (doc-5228): Drawing a diagram from given directions 5A Pythagoras’ Theorem Digital docs

(page 137)

• Activity 5-A-1 (doc-5011): Review of Pythagoras’ theorem • Activity 5-A-2 (doc-5012): Practising Pythagoras’ theorem • Activity 5-A-3 (doc-5013): More of Pythagoras’ theorem • SkillSHEET 5.1 (doc-5224): Rounding to a given number of decimal places 5B Pythagoras’ theorem in three dimensions Digital docs

• Activity 5-B-1 (doc-5014): Pythagoras in 3-dimensions (page 143) • Activity 5-B-2 (doc-5015): Pythagoras in 3-D figures (page 143) • Activity 5-B-3 (doc-5016): Investigating triangles in 3-D figures (page 143) • SkillSHEET 5.4 (doc-5229): Drawing 3-D shapes (page 143) • WorkSHEET 5.1 (doc-5230): Pythagoras’ theorem (page 145) 5C Trigonometric ratios Digital docs

• Activity 5-C-1 (doc-5017): Review of trigonometry (page 149) • Activity 5-C-2 (doc-5018): Using trigonometry (page 149) • Activity 5-C-3 (doc-5019): Applying trigonometry (page 149) • SkillSHEET 5.3 (doc-5226): Labelling the sides of a right-angled triangle (page 150) • SkillSHEET 5.7 (doc-5231): Selecting an appropriate trigonometric ratio based on the given information (page 150) 5D Using trigonometry to calculate side lengths Interactivity

• Using trigonometry (int-1146) (page 151) Digital docs

(page 154)

• Activity 5-D-1 (doc-5020): Calculating side lengths using trigonometry • Activity 5-D-2 (doc-5021): Applying trigonometry to simple figures 182

maths Quest 10 for the australian Curriculum

• Activity 5-D-3 (doc-5022): Practical applications of trigonometry 5E Using trigonometry to calculate angle size Digital docs

• Activity 5-E-1 (doc-5023): Review of angle calculations (page 158) • Activity 5-E-2 (doc-5024): Calculating angles using trigonometry (page 158) • Activity 5-E-3 (doc-5025): Applying trigonometry to angle calculations (page 159) • SkillSHEET 5.8 (doc-5232): Rounding angles to the nearest degree (page 159) • WorkSHEET 5.2 (doc-5233): Using trigonometry (page 160) 5F Angles of elevation and depression eLesson

• Height of a satellite (eles-0173) (page 161) Digital docs

• Activity 5-F-1 (doc-5026): Identifying elevation and depression (page 163) • Activity 5-F-2 (doc-5027): Calculating elevation and depression (page 163) • Activity 5-F-3 (doc-5028): Applications of elevation and depression (page 163) • SkillSHEET 5.6 (doc-5228): Drawing a diagram from given directions (page 163) • WorkSHEET 5.3 (doc-5234): Elevation and depression (page 165) 5G Bearings and compass directions Digital docs

(page 170)

• Activity 5-G-1 (doc-5029): Bearings • Activity 5-G-2 (doc-5030): Calculations involving bearings • Activity 5-G-3 (doc-5031): Applications involving bearings 5H Applications Interactivity

(page 172)

Digital docs

(page 173)

• Applying trigonometry to drafting problems (int-2781) • Activity 5-H-1 (doc-5032): Trigonometry calculations 1 • Activity 5-H-2 (doc-5033): Trigonometry calculations 2 • Activity 5-H-3 (doc-5034): Trigonometry calculations 3 Chapter review Interactivities

(page 181)

• Test Yourself Chapter 5 (int-2840): Take the end-ofchapter test to test your progress • Word search Chapter 5 (int-2838): an interactive word search involving words associated with this chapter • Crossword Chapter 5 (int-2839): an interactive crossword using the definitions associated with the chapter To access eBookPLUS activities, log on to www.jacplus.com.au

measurement anD geometry • using units of measurement

6 surface area and volume

6a Area 6b Total surface area 6c Volume What Do you knoW ? 1 List what you know about measurement. Create a sunshine wheel to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large sunshine wheel that shows your class’s knowledge of measurement. eBook plus

Digital doc

Hungry brain activity Chapter 6 doc-5235

opening Question

How can you determine the most efficient size of a cylindrical water tank of a particular volume?

measurement anD geometry • using units of measurement

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

Digital doc

SkillSHEET 6.1 doc-5236 eBook plus

Digital doc

SkillSHEET 6.2 doc-5237

Conversion of area units 1 Convert each of the following to the units given in brackets. Write the answer using scientific

notation (standard form). a 3.6 m2 (mm2)

b 20 000 cm2 (km2)

3.6 ì 106 mm2 2 ì 10-6 km2 Using a formula to find the area of a common shape a 24 m2 b 30 cm2 c 4.9 cm2 2 Find the area of each of the following plane figures.

a

b

Digital doc

SkillSHEET 6.3 doc-5238

c

8 cm

3m

3 cm

8m

eBook plus

c 5.2 ha (m 2 ) 5.2 ì 104 m2

2.5 cm

12 cm

a 150 cm2 b 232 cm2 c 1.22 m2 Total surface area of cubes and rectangular prisms 3 Find the total surface area of each of the following prisms. a b c 10 cm 10 cm 5 cm

3m

10 cm

2 cm 8 cm

eBook plus

Digital doc

SkillSHEET 6.4 doc-5239

eBook plus

Digital doc

Conversion of volume units 4 Convert each of the following to the units given in brackets. Give the answers in scientific

notation (standard form). a 3.4 m3 (cm3)

b 250 000 mm3 (m3)

3.4 ì 106 cm3 2.5 ì 10-4 m3 Volume of cubes and rectangular prisms 5 Find the volume of each of the prisms in question 3.

SkillSHEET 6.5 doc-5240

184

3 ) 6.5 ì 103 mm3 c 6.5 cm3 (mm

maths Quest 10 for the australian Curriculum

a 125 cm3 b 160 cm3 c 0.03 m3

measurement anD geometry • using units of measurement

6a

area ■ ■

eBook plus

■

Digital doc

SkillSHEET 6.1 doc-5236

■

The area of a figure is the amount of surface or flat space within the boundaries of the figure. The units used for area are mm2, cm2, m2, km2 or ha (hectares), depending upon the size of the figure. l ha = 10 000 (or 104) m2 There are many real-life situations that require an understanding of the area concept. Some are, ‘the area to be painted’, ‘the floor area of a room or house’, and ‘how much land one has’, ‘how many tiles are needed for a wall’. It is important that you are familiar with converting units of area. This can be revised by completing SkillSHEET 6.1.

using area formulas ■

The area of many plane figures can be found by using a formula. The table below shows the formula for the area of some common shapes. Shape

Formula A = l 2, where l is a side length.

1. Square l

2. Rectangle

A = lw, where l is the length and w is the width.

l w

1

A = 2 bh, where b is the base length and h the height.

3. Triangle h b a

4. Trapezium h

1

A = 2 (a + b)h, where a and b are lengths of parallel sides and h the height.

b

A = p r2, where r is the radius.

5. Circle r

A = bh, where b is the base length and h the height.

6. Parallelogram h b

(continued) Chapter 6 surface area and volume

185

measurement anD geometry • using units of measurement

Shape

Formula

θ ì p r2, where q is the sector angle 360° in degrees and r is the radius.

A=

7. Sector

q

r

A=

8. Rhombus

1 2

xy, where x and y are diagonals.

x

y

9. Ellipse

A = p ab, where a and b are the lengths of the semi-major and semi-minor axes respectively.

b a

■

Measurements must be in the same unit of length before substituting into an area formula.

alternative way to find the area of a triangle eBook plus

eLesson Heron’s formula

eles-0177

■

If the lengths of all three sides of a triangle are known, its area, A, can be found by using Heron’s formula: A = s( s − a )( s − b )( s − c ) where a, b and c are the lengths of a+b+c the three sides and s is the semi-perimeter or s = . 2

c

WorkeD example 1

Find the areas of the following plane figures, correct to 2 decimal places. a

b 3 cm

5 cm

c

2 cm

15 cm

5 cm 40è

6 cm think a

186

Write

1

The shape is a triangle with three side lengths given, but not the height. In this case Heron’s formula is used.

2

Identify the values of a, b and c.

3

Calculate the value of s, the semi-perimeter of the triangle.

maths Quest 10 for the australian Curriculum

a

a

b

A = s(s − a)(s − b)(s − c)

a = 3, b = 5, c = 6 a+b+c s= 2 3+ 5+6 = 2 14 = 2 =7

measurement AND geometry • Using units of measurement

b

c

A = 7(7 − 3)(7 − 5)(7 − 6)

Substitute the values of a, b, c and s into Heron’s formula and evaluate, correct to 2 decimal places.

4

= 7 × 4 × 2 ×1 = 56 = 7.48  cm2 b A = p ab

1

The shape shown is an ellipse. Write the appropriate area formula.

2

Identify the values of a and b (the semimajor and semi-minor axes).

a = 5, b = 2

3

Substitute the values of a and b into the formula and evaluate, correct to 2 decimal places.

A=pì5ì2 = 31.42  cm2

1

The shape shown is a sector. Write the formula for finding the area of a sector.

2

Write the value of q and r.

q = 40è, r = 15

3

Substitute and evaluate the expression, correct to 2 decimal places.

A=

c

A=

θ ì p r2 360°

40° ì p ì 152 360° = 78.54  cm2

Areas of composite figures ■■ ■■

A composite figure is a figure made up of a combination of simple figures. The area of a composite figure can be calculated by: •• calculating the sum of the areas of the simple figures that make up the composite figure •• calculating the area of a larger shape and then subtracting the extra area involved.

Worked Example 2

Find the area of each of the following composite shapes. C

a 

AB = 8 cm EC = 6 cm FD = 2 cm

A

E

F

Think

B 9 cm D

C

2 cm

B

D

a

b  A

E 5 cm H

10 cm

F G

Write a Area ACBD = Area ABC + Area ABD

1

ACBD is a quadrilateral that can be split into two triangles: ABC and ABD.

2

Write the formula for the area of a triangle containing base and height.

Atriangle = 2 bh

3

Identify the values of b and h for ABC.

ABC: b = AB = 8, h = EC = 6

1

Chapter 6 Surface area and volume

187

measurement AND geometry • Using units of measurement

1 2 1 2

Substitute the values of the pronumerals into the formula and, hence, calculate the area of ABC.

Area of ABC =

5

Identify the values of b and h for ABD.

ABD: b = AB = 8, h = FD = 2

6

Calculate the area of ABD.

Area of ABD =

4

=

ì AB ì EC ì8ì6

= 24  cm2

=

1 2 1 2

AB ì FD ì8ì2

= 8  cm2

b

7

Add the areas of the two triangles together to find the area of the quadrilateral ACBD.

Area of ACBD = 24  cm2 + 8  cm2 = 32  cm2

1

One way to find the area of the shape shown is to find the total area of the rectangle ABGH and then subtract the area of the smaller rectangle DEFC.

b Area = Area ABGH - Area DEFC

2

Write the formula for the area of a rectangle.

Arectangle = l ì w

3

Identify the values of the pronumerals for the rectangle ABGH.

Rectangle ABGH: l = 9 + 2 + 9 = 20 w = 10

4

Substitute the values of the pronumerals into the formula to find the area of the rectangle ABGH.

Area of ABGH = 20 ì 10 = 200  cm2

5

Identify the values of the pronumerals for the rectangle DEFC.

Rectangle DEFC: l = 5, w = 2

6

Substitute the values of the pronumerals into the formula to find the area of the rectangle DEFC.

Area of DEFC = 5 ì 2 = 10  cm2

7

Subtract the area of the rectangle DEFC from the area of the rectangle ABGH to find the area of the given shape.

Area = 200 - 10 = 190  cm2

remember

1. Area is a measure of the amount of surface within the boundaries of a figure. 2. The units for measuring area are mm2, cm2, m2 and km2. 3. Land area is usually measured in hectares (ha) where 1 ha = 10  000 (or 104) m2. 4. Areas can be calculated by using formulas that are specific to the given plane figure. 5. Areas of composite figures can be calculated by adding the areas of the simple figures making the composite figure or by calculating the area of an extended figure and subtracting the extra area covered.

188

Maths Quest 10 for the Australian Curriculum

measurement anD geometry • using units of measurement

exerCise

6a inDiViDual pathWays eBook plus

Activity 6-A-1

Review of area doc-5035

area Where appropriate, give answers correct to 2 decimal places. fluenCy 1 Find the areas of the following shapes. a

4 cm

Activity 6-A-2

4 cm

Area problems doc-5036

15 cm

12 cm

Activity 6-A-3

Tricky area problems doc-5037

eBook plus

c

b

a 16 cm2 b 48 cm2 d 120 cm2 e 706.86 cm2 2 g 254.47 cm h 21 m2 d

c 75 cm2 f 73.5 mm2 i 75 cm2

10 cm

e

12 cm

f

Digital doc

SkillSHEET 6.2 doc-5237

8 cm

15 cm

13 mm

8 mm

18 cm 7 mm

g

h

i 15 cm

6m

10 cm

7m

18 cm

2 Express the area in questions 1e and 1g in terms of p. Part e = 225p cm2; part g = 81p cm2 3 We 1a Use Heron’s formula to find the area of the following triangles. a

b 3 cm

8 cm

5 cm

16 cm 6 cm 12 cm

a 20.7 cm2

b 7.64 cm2

Chapter 6 surface area and volume

189

measurement anD geometry • using units of measurement 4 We1b Find the area of the following ellipses. Answer correct to 1 decimal place. a

b

a 113.1 mm2 b 188.5 mm2

9 mm

12 mm

4 mm 5 mm

5 We1c Find the area of the following shapes, i stating the answer exactly; that is in terms of p and ii correct to 2 decimal places. a i 12p cm2 ii 37.70 cm2 b i 69π mm2 2 ii 108.38 mm2

a

b

c

30è

70è

6 mm

18 cm

12 cm

c i 261p cm2 ii 819.96 cm2

345è

6 mC A figure has an area of about 64 cm2. Which of the following cannot possibly represent

the figure? a A triangle with base length 16 cm and height 8 cm b A circle with radius 4.51 cm c A rectangle with dimensions 16 cm and 4 cm d A square with side length 8 cm ✔ e A rhombus with diagonals 16 cm and 4 cm 7 mC The area of the quadrilateral shown below right is to be calculated. Which of the following lists all the lengths required to calculate the area? a AB, BC, CD and AD b AB, BE, AC and CD B c BC, BE, AD and CD ✔ d AC, BE and FD e AC, CD and AB

C F

E D

A 8 We2 Find the area of the following composite shapes. a

b

20 cm

15 cm

190

maths Quest 10 for the australian Curriculum

a 123.29 cm2 b 1427.88 m 40 m

28 m

measurement anD geometry • using units of measurement c

d

8 cm 2 cm

3 cm

8 c 52 cm2 d 30.4 m2 e 78 cm2 f 2015.5 cm2

4 cm 3.8 m

e

2.1 m

f 28 cm 18 cm

5 cm 12 cm 9 Find the shaded area in each of the following. a 125.66 cm2 b 102.87 cm2 c 13.73 m2 d 153.59 m2 e 13.86 m2 f 37.5 m2

a

b

16 m 8m

2m

2m

3 cm r = 7 cm



d

c

8m

e

3m 40è

f

8m

5m

15 m 5m

2m

7.5 m

13 m 7 m

5m Chapter 6 surface area and volume

191

measurement AND geometry • Using units of measurement understanding 10 A sheet of cardboard is 1.6  m by 0.8  m. The following shapes are cut from the cardboard:

•  a circular piece with radius 12  cm •  a rectangular piece 20  cm by 15  cm •  2 triangular pieces with base 30  cm and height 10  cm •  a triangular piece with side length 12  cm, 10  cm and 8  cm. What is the area of the remaining piece of cardboard? 11  707.93  cm2 11 A rectangular block of land, 12  m by 8  m, is surrounded by a concrete path 0.5  m wide. Find the area of the path. 21  m2 12 Concrete slabs 1  m by 0.5  m are used to cover a footpath 20  m by 1.5  m. How many slabs are needed? 60 13 A city council builds a 0.5  m wide concrete path around the garden as shown below. 12 m 5m 8m

3m

Find the cost of the job if the workman charges $40.00 per m2 . $840 14 A game of tennis can be played with 4 people using the whole court or it can be played with 2 people using the singles court, which excludes the edge on either side, as shown in the diagram.

1.8 m

x

8.23 m

6.40 m

10.97 m

11.89 m

a b c d e f

What is the total area of the whole tennis court? 260.87  m2 What is the area of the singles court? 195.71  m2 What area can one person use when playing doubles? 130.43  m2 97.85  m2 What area can one person use when playing singles? 37.5% What percentage of the total tennis court area is used by one person for singles? How far is the ball served before it bounces if it follows the path indicated in the 18.75  m diagram? That is, calculate x.

15 Ron the excavator operator has 100 metres of barricade mesh and needs to enclose an area to

work in safely. He chooses to make a rectangular region with dimensions x and y. a Write an equation that connects x, y and the perimeter. 50 = x + y y = 50 – x b Write y in terms of x. x. Area = 50x − x2 c Write an equation for the area of the region in terms of d Fill in the table for different values of x.

192

x

0

5

10

15

20

25

30

35

40

45

50

Area

0

225

400

525

600

625

600

525

400

225

0

Maths Quest 10 for the Australian Curriculum

eBook plus

Digital doc

WorkSHEET 6.1 doc-5241

6b eBook plus

Digital doc

SkillSHEET 6.3 doc-5238

Can x have a value more than 50? Why? No, impossible to make a rectangle. Sketch a graph of area against x. 600 500 x = 25 Determine the value of x that makes the area a maximum. 400 What is the value of y for maximum area? y = 25 300 200 What shape encloses the maximum area? Square 100 Calculate the maximum area. 625 m2 0 10 20 30 40 50 60 70 x Ron decides to choose to make a circular area with the barricade mesh. r = 15.915 m k What is the radius of this circular region? l What area is enclosed in this circular region? 795.77 m2 170.77 m2 m How much extra area does Ron now have compared to his rectangular region? e f g h i j

Area

 1 2  1  n m2; rectangular (square) area,  n2  m2. Circular area,   16   4π    1 1 4 ÷ Circular area is always or 1.27  times larger.  4π 16  π

measurement anD geometry • using units of measurement

reasoning 16 Dan has purchased a country property with layout and dimensions

as shown in the diagram. a Show that the property has a total area of 987.5 ha. Students’ work b Dan wants to split the property in half (in terms of area) 1500 m 5000 m by building a straight-lined fence either vertically or 2000 m horizontally through the property. Assuming the cost of the fencing is a fixed amount per linear metre, justify where the 1000 m fence should be built (that is, how many metres from the top left-hand corner and in which direction), to minimise 2020.83 m; horizontal the cost. 17 In question 15, Ron the excavator operator could choose to enclose a rectangular or circular area with 100 m of barricade mesh. In this case, the circular region resulted in a larger safe Circular area, 1790.49 m2; rectangular area, 1406.25 m2 work area. a Show that for 150 m of barricade mesh, a circular refleCtion region again results in a larger safe work area as opposed to a rectangular region. How are perimeter and area b Show that for n metres of barricade mesh, a circular different but fundamentally related? region will result in a larger safe work area as opposed to a rectangular region.

total surface area ■

The total surface area (TSA) of a solid is the sum of the areas of all the faces (outside surfaces) of that solid. It can be found by calculating the area of the net of the solid.

tsa of rectangular prisms and cubes ■

The following formulas have been introduced in previous years.

h

Rectangular prism (cuboid): w

TSA = 2(lh + lw + wh) l ■

A special case of the rectangular prism is the cube where all sides are equal (l = w = h). Cube: TSA = 6l2 l

To see a worked example and revise finding the total surface area of cubes and rectangular prisms, complete the SkillSHEET shown. Chapter 6 surface area and volume

193

measurement anD geometry • using units of measurement

tsa of spheres and cylinders

eBook plus

Sphere:

Interactivity TSA-sphere

TSA = 4p r2

int-2782

r

Note: The mathematics required to obtain the rule for the total surface area of a sphere is beyond the scope of Year 10. Cylinder: h

TSA = 2p r(r + h) or 2p r2 + 2p rh r

■

The formula for the TSA of the cylinder is found from the area of the net as shown. TSA = p r 2 + p r 2 + 2p rh r = 2p r 2 + 2p rh A = pr 2 = 2p r(r + h) 2p r

A = 2prh

h

r A = pr 2

WorkeD example 3

Find the total surface area of the solids below, correct to 1 decimal place. a

r = 7 cm

b

r

think a

b

194

50 cm 1.5 m

Write a TSA = 4p r2

1

Write the formula for the TSA of a sphere.

2

Identify the value for r.

3

Substitute and evaluate.

1

Write the formula for the TSA of a cylinder.

2

Identify the values for r and h. Note that the units will need to be the same.

r = 50 cm, h = 1.5 m = 150 cm

3

Substitute and evaluate.

TSA = 2 ì p ì 50 ì (50 + 150) = 62 831.9 cm2

maths Quest 10 for the australian Curriculum

r=7 TSA = 4 ì p ì 72 = 615.8 cm2 b TSA = 2p r(r + h)

measurement AND geometry • Using units of measurement

TSA of cones ■■

The total surface area of a cone can be found by considering its net. SA = Acircular base + Acurved surface, = p r2 + Asector of radius, s

s s

r

r

The sector is a fraction of the full circle of radius, s, with circumference, 2p s. The sector has arc length, equivalent to the circumference of the base of the cone, 2p r. ■■ The fraction of the full circle represented by the sector can be found by writing the arc length as a fraction of the circumference of the full circle, 2π r = r . 2π s s Area of a sector = fraction of the circle ì p r2 r = × π s2 s = p rs ■■ ■■

Therefore, SA = Acircular base + Acurved surface = p r2 + p rs



= p r(r + s) Cone: TSA = p r(r + s) or p r2 + p rs

Worked Example 4

Find the total surface area of the cone shown.

15 cm 12 cm

Think

Write

1

Write the formula for the TSA of a cone.

TSA = p r (r + s)

2

State the values of r and s.

r = 12, s = 15

3

Substitute and evaluate.

TSA = p  ì 12 ì (12 + 15) = 1017.9  cm2

TSA of other solids ■■ ■■ ■■

TSA can be found by summing the areas of each face. The areas of each face may need to be calculated separately. Always consider top, bottom, front, back, left and right faces. Chapter 6 Surface area and volume

195

measurement AND geometry • Using units of measurement

Worked Example 5

Find the total surface area of the square-based pyramid shown. 5 cm 6 cm

Think

Write/draw

1

There is no formula, so write the components of the TSA. These are the square base and four identical triangles.

TSA = Area of square base + Area of four triangular faces

2

Find the area of the square base.

Area of base = l2, where l = 6 Area of base = 62 = 36  cm2

3

The four side faces are isosceles triangles. Draw one face and write the formula for finding its area.

h

5 cm

3 cm 1

Area of a triangular face = 2 bh; b = 6 4

Find the height of the triangle using Pythagoras’ theorem.

a2 = c2 - b2, where a = h, b = 3, c = 5 h2 = 52 - 32 h2 = 25 - 9 h2 = 16 h = 4  cm

5

Calculate the area of the triangular face by substituting b = 6 and h = 4.

Area of triangular face =

Calculate the TSA by adding the area of the square base and the area of four triangular faces together.

TSA = 36 + 4 ì 12 = 36 + 48 = 84  cm2

6

■■

1 2

ì6ì4

= 12  cm2

Note that the area of a triangular face of the square-based pyramid in the previous example could also be calculated using Heron’s formula, as the lengths of all three sides were given.

TSA of composite solids ■■ ■■

196

The TSA of a composite solid is calculated by summing the areas of the solid’s faces. When two smaller solids are joined, the surfaces involved in the join will not be part of the surface of the composite solid. For example, if a square-based pyramid is stacked on top of a cube of the same base dimensions, the two faces in the join (one of the cube’s faces and the pyramid’s base) are inside and therefore not part of the surface of the solid.

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement

Worked Example 6

Find the total surface area of the object shown. 6 cm

10 cm 10 cm

Think

Write/draw

1

The solid shown is made up of a cube and a square pyramid. Only five faces of the cube are on the surface. Likewise, only triangular faces of the pyramid are on the surface. Thus, the TSA of the solid consists of five identical squares and four identical triangles.

TSA = 5 ì area of a square + 4 ì area of a triangle

2

Find the area of the square face with the side length 10  cm.

Asquare = l2, where l = 10 A = 102 A = 100  cm2

3

Draw a triangular face and work out its height using Pythagoras’ theorem.

6 cm

h

5 cm

a2 = c2 - b2, where a = h, b = 5, c = 6 h2 = 62 - 52 h2 = 36 - 25 h2 = 11 h = 3.32  cm 4

5

1 2 1 2

Find the area of the triangular face with the base of 10  cm and the height of 3.32  cm.

Atriangle =

Find the TSA of the solid by adding the area of 5 squares and 4 triangles together.

TSA = 5 ì 100 + 4 ì 16.6 = 500 + 66.4 = 566.4  cm2

■■

bh, where b = 10, h = 3.32

= ì 10 ì 3.32 = 16.6  cm2

Applications of surface area are commonly seen when calculating the amount of material needed for building structures such as silos, tanks, swimming pools, or painting or tiling surfaces. Chapter 6 Surface area and volume

197

measurement AND geometry • Using units of measurement

Worked Example 7

The silo shown at right is to be built from metal. The top portion of the silo is a cylinder of diameter 4  m and height 8  m. The bottom part of the silo is a cone of slant height 3  m. The silo has a circular opening of radius 30  cm on the top. a What area of metal (to the nearest m2) is required to build the silo? b If it costs $12.50 per m2 to cover the surface with an anti-rust material, how much will it cost to cover the silo completely?

Think a

b

198

1

8m

4m

3m

Write

The surface area of the silo consists of the circle (the top face), the curved part of the cylinder and the curved part of the cone. The circular opening is cut out from the top face and thus its area must be subtracted.

a TSA = area of a large circle



- area of a small circle + area of curved section of a cylinder + area of curved section of a cone

2

To find the area of the top face, subtract the area of the small circle from the area of the larger circle. Let R = radius of small circle.

Area of top face = Alarge circle - Asmall circle = p r2 - p R2 4 where r = 2 = 2  m and R = 30  cm = 0.3  m. Area of top face = p ì 22 - p ì 0.32 = 12.28  m2

3

The middle part of the silo is the curved part of a cylinder. Find its area. (Note that in the formula TSAcylinder = 2p r 2 + 2p rh, the curved part is represented by 2p rh.)

Area of curved section of cylinder = 2p rh where r = 2, h = 8. Area of curved section of cylinder =2ìpì2ì8 = 100.53  m2

4

The bottom part of the silo is the curved section of a cone. Find its area. (Note that in the formula TSAcone = p r 2 + p rs, the curved part is given by prs.)

Area of curved section of cone = p rs where r = 2, s = 3. Area of curved section of cone = p ì 2 ì 3 = 18.85  m2

5

Find the total surface area of the silo by finding the sum of the surface areas calculated above.

TSA = 12.28 + 100.53 + 18.85 = 131.66  m2

6

Write the worded answer. (Remember that we were asked to give the answer to the nearest square metre.)

The area of metal required is 132  m2.

To find the total cost, multiply the total surface area of the silo by the cost of an anti-rust material per m2 ($12.50).

Maths Quest 10 for the Australian Curriculum

b Cost = 132 ì $12.50

= $1650.00

measurement anD geometry • using units of measurement

remember

1. The total surface area (TSA) of a figure is the sum of the areas of all its outside faces. 2. TSA of a cube with the length of the edge, l, is given by the formula TSA = 6l 2 3. TSA of a rectangular prism with dimensions l, w and h is TSA = 2(lw + lh + wh) 4. TSA of a closed cylinder of radius, r, and height, h, is TSA = 2p rh + 2p r 2 5. TSA of a sphere of radius, r, is TSA = 4p r 2 6. TSA of a closed cone with radius, r, and slant height, s, is TSA = p rs + p r 2 7. TSA of a pyramid = area of base + area of triangular faces 8. TSA of a composite shape can be found by calculating the areas of individual faces that are on the surface and then adding them together

exerCise

6b inDiViDual pathWays eBook plus

Activity 6-B-1

total surface area fluenCy

Note: Where appropriate, give the answers correct to 1 decimal place. 1 Find the total surface areas of the solids shown.



b

a

Introducing surface area doc-5038

a 600 cm2 b 384 cm2 c 1440 cm2 d 27 m2

Activity 6-B-2

Surface area problems doc-5039 Activity 6-B-3

Tricky surface area problems doc-5040

10 cm c

8 cm d

12 cm

2m 1.5 m

15 cm

3m

20 cm eBook plus

Digital doc

SkillSHEET 6.3 doc-5238

2 We3 Find the total surface area of the solids shown below. a b 21 cm r=3m

r

c



a 113.1 m2 b 6729.3 cm2 c 8.2 m2 d 452.4 cm2

30 cm

d

0.5 m

12 cm 2.1 m

Chapter 6 surface area and volume

199

measurement anD geometry • using units of measurement 3 We4 Find the total surface area of the cones below. a b

a 1495.4 cm2 b 502.7 cm2 8 cm

20 cm 12 cm

14 cm

4 We5 Find the total surface area of the solids below. a

2.5 m

b 12 cm

15 cm c

1.5 m d

9.1 cm 8 cm

14 cm

6 cm

7.2 cm

5.1 cm

a 506.0 cm2 b 9.4 m2 c 340.4 cm2 d 224.1 cm2

10 cm 7 cm

5 Find the surface areas of the following. a A cube of side length 1.5 m 13.5 m2 b A rectangular prism 6 m ì 4 m ì 2.1 m 90 m2 c A cylinder of radius 30 cm and height 45 cm, open at one end 11 309.7 cm2 9852.0 mm2 d A sphere of radius 28 mm e An open cone of radius 4 cm and slant height 10 cm 125.6 cm2 f A square pyramid of base length 20 cm and slant edge 30 cm 1531.4 cm2 6 We6 Find the total surface area of the objects shown.

a 880 cm2 b 3072.8 cm2 c 75 cm2 d 70.4 cm2

a

10 cm

b

8 cm 5 cm

12 cm 5 cm

20 cm

20 cm 35 cm 12 cm

c

d 2 cm

m

2.5 c

5 cm

3 cm

3 cm

200

maths Quest 10 for the australian Curriculum

measurement AND geometry • Using units of measurement e 6 e 193.5  cm2 f 1547.2  cm2

f

5 cm

3.5 cm

20 cm

10 cm 12 cm

15 cm 7   MC  A cube has a total surface area of 384  cm2. The length of the edge of the cube is: a 9  cm c 7  cm ✔ b 8  cm d 6  cm e 5  cm understanding 8 Open cones are made from nets cut from a large sheet of paper 1.2  m ì 1.0  m. If a cone has

a radius of 6  cm and a slant height of 10  cm, how many cones can be made from the sheet? 63 (Assume there is 5% wastage of paper.) 9 A steel girder is to be painted. Calculate the area of the surface to be painted. 11  216  cm2 2 cm

2 cm



5 cm 20 cm 2 cm

120 cm

12 cm

10   WE 7  The greenhouse shown below is to be built using shade cloth. It has a wooden door of

dimensions 1.2  m ì 0.5  m. a Find the total area of shade cloth needed to complete the greenhouse. 70.0  m2 b Find the cost of the shade cloth at $6.50 per m2 . $455

5m

2.5 m 3m

11 A cylinder is joined to a hemisphere to make a cake holder, as shown below. The surface of the

cake holder is to be chromed at 5.5 cents per cm2. a Find the total surface area to be chromed. 3063.1  cm2 b Find the cost of chroming the cake holder. $168.47

15 cm

10 cm

Chapter 6 Surface area and volume

201

measurement anD geometry • using units of measurement 12 A soccer ball is made up of a number of hexagons sewn together

2 cm

on its surface. Each hexagon can be considered to have dimensions as shown in the diagram. y a Calculate q o. q = 120 è x b Calculate the values of x and y exactly. x = 1; y = 3 c Calculate the area of the trapezium in the diagram. 3 3 cm2 6 3 cm2 d Hence, determine the area of the hexagon. e If the total surface area of the soccer ball is 192 3 cm2, how q many hexagons are on the surface of the soccer ball? 32 13 a Determine the exact total surface area of a sphere with radius 2 metres. An inverted cone with side length 4 metres is placed on top of the sphere so that the centre of its base is 0.5 metres above the centre of the sphere. 8p m2 7 b Find the radius of the cone exactly. m 2 c Find the area of the curved surface of the cone exactly. 2 7π m2 d What are the exact dimensions of a box that could precisely fit the cone connected to 7 × 7 ×1 the sphere? Complete the following question without the aid of a calculator. 14 The table shown below is to be varnished (including the base of each leg). The table top has a

thickness of 180 mm and the cross-sectional dimension of the legs is 50 mm by 50 mm. 80 cm 60 cm

C heapest: 30 cm by 30 cm, $269.50; 20 cm by 20 cm (individually) $270; 20 cm by 20 cm (boxed) $276.50

70 cm



eBook plus

Digital doc

WorkSHEET 6.2 doc-5242

202

A friend completes the calculation as shown. Assume there are no simple calculating errors. Analyse the working presented and justify if the TSA calculated is correct. Calculation is correct.

Table top

0.96

2 ì (0.8 ì 0.6)

Legs

0.416

16 ì (0.52 ì 0.05)

Table top edging

0.504

TSA

0.18 ì (2(0.8 + 0.6)) 2

1.88 m

15 A shower recess with dimensions 1500 mm (back wall) by 900 mm (side wall) needs to have

the back and two side walls tiled to a height of 2 m. a Calculate the area to be tiled in m2. 6.6 m2 b Justify that 180 tiles (including those that need to be cut) of dimension 20 cm by 20 cm will be required. Disregard the grout. c Evaluate the cheapest option of tiling; $1.50/tile or $39.50/box, where a box covers 1 m2 or tiles of dimension 30 cm by 30 cm costing $3.50/tile. refleCtion 16 If the surface area of a sphere to a cylinder is in the Why is calculating the total surface ratio 4 : 3 and the sphere has a radius of 3a, show area of a composite solid more that if the radius of the cylinder is equal to its height, then the radius of the cylinder is 3 3a . 2

maths Quest 10 for the australian Curriculum

r =

difficult than for a simple solid such as a rectangular prism or cylinder?

3 3a 2

measurement anD geometry • using units of measurement

6C eBook plus

Digital doc

SkillSHEET 6.4 doc-5239

Volume ■ ■ ■

Volume of prisms and other shapes ■

■

■

■

eBook plus

The volume of a 3-dimensional figure is the amount of space it takes up. The units for volume are mm3, cm3 and m3. To revise the technique of converting from one unit to another, complete SkillSHEET 6.4.

The volume of any solid with a uniform cross-sectional area is given by the formula: V = AH, where A is the cross-sectional (or base) area and H is the height of the solid. The height of a prism simply means the dimension perpendicular to a solid’s cross-sectional base. This is often the physical height, depth or length. Prisms are the most recognisable solids with uniform cross-sectional areas. A prism is a solid shape with identical opposite ends joined by straight edges, forming a congruent cross-section. In some cases a special formula can be developed from the formula V = AH. Volume = AH = area of a square ì height = l2 ì l = l3

Cube

Interactivity Maximising the volume of a cuboid

l

int-1150

Rectangular prism h

Volume = AH = area of a rectangle ì height = lwh

w l

Cylinder

r h

Volume = AH = area of a triangle ì height 1 = 2 bh ì H

Triangular prism

H

h b

eBook plus

Digital doc

SkillSHEET 6.5 doc-5240

Volume = AH = area of a circle ì height = p r 2h

■

To see a worked example and revise the volume of a cube and rectangular prism, complete SkillSHEET 6.5. Chapter 6 surface area and volume

203

measurement AND geometry • Using units of measurement

Worked Example 8

Find the volumes of the following shapes. a 14 cm

b

4 cm

20 cm

10 cm

Think a

b

5 cm

Write a V = pr 2h

1

Write the formula for the volume of the cylinder.

2

Identify the value of the pronumerals.

r = 14, h = 20

3

Substitute and evaluate.

V = p ì 142 ì 20 ö 12 315.04  cm3

1

Write the formula for the volume of a triangular prism.

2

Identify the value of the pronumerals. (Note h is the height of the triangle and H is the depth of the prism.)

b = 4, h = 5, H = 10

3

Substitute and evaluate.

V = 2 ì 4 ì 5 ì 10 = 100  cm3

b

1

V = 2 bh ì H

1

Worked Example 9 a What effect will doubling each of the side lengths of a cube have on its volume (in comparison

with the original shape)?

b What effect will halving the radius and doubling the height of a cylinder have on its volume

(in comparison with the original shape)?

Think a

204

Write a V = l3

1

Write the formula for the volume of the cube.

2

Identify the value of the pronumeral. Note: Doubling is the same as multiplying by 2.

lnew = 2l

3

Substitute and evaluate.

Vnew = (2l)3

4

Compare the answer obtained in step 3 with the volume of the original shape.

5

Write your answer.

Maths Quest 10 for the Australian Curriculum

= 8l 3 Doubling each side length of a cube will increase the volume by a factor of 8; that is, the new volume will be 8 times as large as the original volume.

measurement AND geometry • Using units of measurement b

1

Write the formula for the volume of the cylinder.

2

Identify the value of the pronumerals. Note: Halving is the same as dividing by 2.

3

Substitute and evaluate.

b V = p r2h

r rnew = , hnew = 2h 2 2

Vnew = p  r  2h  2  =π× =

4

Compare the answer obtained in step 3 with the volume of the original shape.

5

Write your answer.

r2 × 2h 24

π r 2h 2 1

= 2 pr2h Halving the radius and doubling the height of a cylinder will decrease the volume by a factor of 2; that is, the new volume will be half as large as the original volume.

Volume of spheres ■■

4

Volume of a sphere of radius, r, can be calculated using the formula: V = 3 p r 3.

Worked Example 10

Find the volume of a sphere of radius 9  cm. Answer correct to 1 decimal place. Think

Write 4

1

Write the formula for the volume of a sphere.

V = 3 pr3

2

Identify the value of r.

r=9

3

Substitute and evaluate.

V=

4 3

ì p ì 93 = 3053.6  cm3

Volume of pyramids ■■

Pyramids (including cones) are not prisms as the cross-section changes from the base upwards.

■■

It has been found that the volume of a pyramid is one-third the volume of an equivalent prism with the same base area and height.



1

Volume of a pyramid = 3  AH

H Area of base = A base Chapter 6 Surface area and volume

205

measurement AND geometry • Using units of measurement

■■

Since a cone is a pyramid with a circular cross-section, the volume of a cone is one-third the volume of a cylinder with the same base area and height.

1

Volume of a cone = 3  AH



1

= 3 p r 2h



h r

Worked Example 11

Find the volume of each of the following solids. a b  12 cm

10 cm 8 cm

8 cm Think a

b

Write 1

a V = 3 p r2h

1

Write the formula for the volume of a cone.

2

Identify the values of r and h.

r = 8, h = 10

3

Substitute and evaluate.

V = 3 ì p ì 82 ì 10 = 670.21  cm3

1

Write the formula for the volume of a pyramid.

2

Find the area of the square base.

A = l2 where l = 8 A = 82 = 64  cm2

3

Identify the value of H.

H = 12

4

Substitute and evaluate.

V = 3 ì 64 ì 12 = 256  cm3

1

b V=

1 3

AH

1

Volume of composite figures ■■ ■■ ■■

206

A composite solid is made from smaller solids. The volume of each smaller solid component can be calculated separately. The volume of a composite solid is calculated by summing the volumes of each of the smaller solid components.

Maths Quest 10 for the Australian Curriculum

measurement AND geometry • Using units of measurement

Worked Example 12

Calculate the volume of the composite solid shown.

3m 1.5 m

Think

Write

1

The given solid is a composite figure, made up of a cube and a square-based pyramid.

V = Volume of cube + Volume of pyramid

2

Find the volume of the cube.

Vcube = l  3 where l = 3 Vcube = 33 = 27  m3

3

Write the formula for finding the volume of a square-based pyramid.

Vsquare-based pyramid = 3 AH

4

Find the area of the square base.

A = l2 = 32 = 9  m2

5

Identify the value of H.

H = 1.5

6

Substitute and evaluate the volume of the pyramid.

Vsquare-based pyramid = 3 ì 9 ì 1.5 = 4.5  m3

7

Find the total volume by adding the volume of the cube and pyramid.

V = 27 + 4.5 = 31.5  m3

1

1

Capacity ■■ ■■ ■■

Capacity measures the amount of liquid that will fit in a 3-dimensional figure. The units for capacity are: mL, L and kL. Volume and capacity are fundamentally related: 1  cm3 = 1  mL 1000  cm3 = 1  L 1  m3 = 1000  L = 1  kL

Worked Example 13

Find the capacity (in litres) of a rectangular aquarium, which is 50  cm long, 30  cm wide and 40  cm high.

Chapter 6 Surface area and volume

207

measurement anD geometry • using units of measurement

think

Write

1

Write the formula for the volume of a rectangular prism.

V = lwh

2

Identify the values of the pronumerals.

l = 50, w = 30, h = 40

3

Substitute and evaluate.

V = 50 ì 30 ì 40 = 60 000 cm3

4

State the capacity of the container in millilitres, using 1 cm3 = 1 mL.

60 000 cm3 = 60 000 mL

5

Since 1 L = 1000 mL, to convert millilitres to litres divide by 1000.

6

Give a worded answer.

= (60 000 ó1000) L = 60 L The capacity of the fish tank is 60 L.

remember

1. Volume of a 3-dimensional figure is the amount of space it takes up 2. Volume is measured in cubic units 3. Volume of a prism = AH, where A is the cross-sectional area (or base) and H is the height of the prism 4. The height of a prism is the dimension perpendicular to the prism’s cross-section. 4 5. Volume of a sphere = 3 p r3 1

6. Volume of a cone = 3 p r2h 1

7. Volume of a pyramid = 3 AH 8. Capacity of a 3-dimensional figure is the amount of liquid that will fit in that figure. 9. The relationship between the volume of a solid and the capacity (amount of liquid it can hold) is: 1 cm3 = 1 mL, 1000 cm3 = 1 L. 10. 1 m3 = 1000 L = 1 kL exerCise

6C inDiViDual pathWays eBook plus

Volume fluenCy 1 Find the volumes of the following prisms. a



b

Activity 6-C-1

Review of volume and capacity doc-5041 Activity 6-C-2

Volume and capacity problem doc-5042

3 cm c

a 27 cm3 b 74.088 m3 c 3600 cm3 d 94.5 cm3

4.2 m d

12 cm

Activity 6-C-3

Tricky volume and capacity problems doc-5043

15 cm

4.2 cm

20 cm

7.5 cm 3 cm

208

maths Quest 10 for the australian Curriculum

measurement anD geometry • using units of measurement

eBook plus

2 Calculate the volume of each of these solids. a

a 450 mm3 b 360 cm2 b

Digital doc

SkillSHEET 6.5 doc-5240

18 mm

15 cm

[Base area: 25 mm2]

[Base area: 24 cm2]

3 We8 Find the volume of each of the following. Give each answer correct to 1 decimal place

where appropriate. a

14 cm



a 6333.5 cm3 b 19.1 m3 c 280 cm3 d 288 mm3

b

2.7 m

12 cm

1.5 m

c

d

10 cm 7 cm

12 mm

8 mm 6 mm

8 cm

4 We10 Find the volume of a sphere (correct to 1 decimal place) with a radius of: a 1.2 m 7.2 m3 14 137.2 cm3 b 15 cm c 7 mm 1436.8 mm3 d 50 cm. 523 598.8 cm3 5 We11a Find the volume of each of the following cones, correct to 1 decimal place. a b a 377.0 cm3 22 mm b 2303.8 mm3 20 mm 10 cm

6 cm 6 We11b Find the volume of each of the following pyramids. a

b

12 cm

a 400 cm3 b 10 080 cm3 c 576 cm3

42 cm 24 cm 10 cm

30 cm

c 12 cm

m

16 cm

18 c

Chapter 6 surface area and volume

209

measurement anD geometry • using units of measurement 7 We12 Calculate the volume of each of the following composite solids correct to 2 decimal

a 1400 cm3 b 10 379.20 cm3 c 41.31 cm3 d 48.17 cm3 e 218.08 cm3 f 3691.37 cm3

places where appropriate. a

b

8 cm

10 cm

5 cm

12 cm 5 cm





20 cm

20 cm 35 cm 12 cm

d

c

2 cm

m

2.5 c

5 cm

3 cm



a Vnew = 27l3, the volume will be 27 times as large as the original volume. b Vnew = 18l2, the volume will be 18 of the original volume. c Vnew = 2p r2h, the volume will be twice as large as the original volume. d Vnew = p r2h, the volume will remain the same. e Vnew = 3lwh, the volume will be 3 times as large as the original value.

3 cm

210

e

f

5 cm

3.5 cm

20 cm

10 cm



12 cm



15 cm

unDerstanDing 8 We9 a What effect will tripling each of the side lengths of a cube have on its volume

(in comparison with the original shape)? b What effect will halving each of the side lengths of a cube have on its volume

(in comparison with the original shape)? c What effect will doubling the radius and halving the height of a cylinder have

on its volume (in comparison with the original shape)? d What effect will doubling the radius and dividing the height of a cylinder by 4 have on its

volume (in comparison with the original shape)? e What effect will doubling the length, halving the width and tripling the height of a

rectangular prism have on its volume (in comparison with the original shape)? 9 mC A hemispherical bowl has a thickness of 2 cm and an outer 2 cm diameter of 25 cm. If the bowl is filled with water to its full capacity, the volume of the water will be: a 1526.04 cm3 b 1308.33 cm3 c 3052.08 cm3 25 cm d 2616.66 cm3 3 ✔ e 2424.52 cm

maths Quest 10 for the australian Curriculum

measurement anD geometry • using units of measurement 10 Tennis balls of diameter 8 cm are packed in a box 40 cm ì 32 cm ì 10 cm, as shown. What is unused space in the box? 7438.34 cm3

11 We13 A cylindrical water tank has a diameter of 1.5 m and a height of 2.5 m. What is the capacity (in litres) of the tank? 4417.9 L 12 A monument in the shape of a rectangular pyramid (base length of 10 cm, base width of 6 cm,

height of 8 cm), a spherical glass ball (diameter of 17 cm) and conical glassware (radius of 14 cm, height of 10 cm) are packed in a rectangular prism of dimensions 30 cm by 25 cm by 20 cm. The extra space in the box is filled up by a packing material. What volume of packing material is required? 10 215.05 cm3 13 A swimming pool is being constructed so that it is the 8m upper part of an inverted square-based pyramid. a Calculate H. H = 6 m 19 bins b Calculate the volume of the pool. 112 m3 3m 3 c How many 6 m bins will be required to take the dirt away? d How many litres of water are required to fill this pool? 112 000 L 4 m H e How deep is the pool when it is half-filled? 1.95 m from floor 14 A soft drink manufacturer is looking to repackage cans of soft drink to minimise the cost of packaging while keeping the volume constant. Consider a can of soft drink with a capacity of 400 mL. a If the soft drink was packaged in a spherical can: i find the radius of the sphere 4.57 cm ii find the total surface area of this can. 262.5 cm2 b If the soft drink was packaged in a cylindrical can with a radius of 3 cm: i find the height of the cylinder 14.15 cm ii find the total surface area of this can. 323.27 cm2 c If the soft drink was packaged in a square-based pyramid with a base side length of 6 cm: i find the height of the pyramid 33.3 cm ii find the total surface area of this can. 434.28 cm2 d Which can would you recommend the soft drink manufacturer use for its repackaging? Why? Sphere. Costs less for a smaller surface area. reasoning 15 Marion has mixed together ingredients for a cake. The recipe requires a baking tin that is

cylindrical in shape with a diameter of 20 cm and a height of 5 cm. Marion only has a tin as shown and a muffin tray consisting of 24 muffin cups. Each of the muffin cups in the tray is a portion of a cone as shown in the diagram. Should Marion use the tin or muffin tray? Explain. Required volume = 1570.79 cm3; tin volume =1500 cm3; muffin tray volume = 2814.72 cm3. Marion should use the tin with approximately 70 cm3 mixture left over.

8 cm

12 cm

4 cm

4 cm

10 cm

15 cm

8 cm

Chapter 6 surface area and volume

211

measurement anD geometry • using units of measurement 16 Nathaniel and Andrew are going to the snow for

survival camp. They plan to construct an igloo, consisting of an entrance and hemispherical structure, as shown. Nathaniel and Andrew are asked to redraw their plans and increase the size of the liveable region (hemispherical structure) so that the total volume (including the entrance) is doubled. How can this be achieved? Increase radius of hemispherical section to 1.92 m.

1.5 m

1m

1.5 m

17 Sam is having his 16th birthday party and wants to make an ice trough to keep drinks cold. He

eBook plus

Digital doc

WorkSHEET 6.3 doc-6753

has found a square piece of sheet metal with a side length of 2 metres. He cuts squares, of side length x metres, from each corner then bends the sides of the remaining sheet. When four squares of the appropriate side length are cut from the corners the capacity of the trough can be maximised at 588 litres. Explain how Sam should proceed to maximise the capacity of the trough. Cut squares of side length, s = 0.3 m or 0.368 m from the corners. 18 The Hastings family house has a rectangular roof with dimensions 17 m ì 10 m providing water to three water tanks, each with a radius of 1.25 m and a height of 2.1 m. When rain falls it is measured in millimetres. This means that this is the depth to which the water would fill if it were captured. Show that approximately 182 millimetres of rain must fall on the roof to fill the tanks. Volume of water needed; 30.9 m3. refleCtion Volume is measured in cubic units. How is this reflected in the volume formula?

212

maths Quest 10 for the australian Curriculum

measurement anD geometry • using units of measurement

summary area ■ ■ ■ ■ ■

Area is a measure of the amount of surface within the boundaries of a figure. The units for measuring area are mm2, cm2, m2 and km2. Land area is usually measured in hectares (ha) where 1 ha = 10 000 (or 104) m2. Areas can be calculated by using formulas that are specific to the given plane figure. Areas of composite figures can be calculated by adding the areas of the simple figures making the composite figure or by calculating the area of an extended figure and subtracting the extra area covered. Total surface area

■ ■ ■ ■ ■ ■ ■ ■

The total surface area (TSA) of a figure is the sum of the areas of all its outside faces. TSA of a cube with the length of the edge, l, is given by the formula TSA = 6l 2 TSA of a rectangular prism with dimensions l, w and h is TSA = 2(lw + lh + wh) TSA of a closed cylinder of radius, r, and height, h, is TSA = 2p rh + 2p r 2 TSA of a sphere of radius, r, is TSA = 4p r 2 TSA of a closed cone with radius, r, and slant height, s, is TSA = p rs + p r 2 TSA of a pyramid = area of base + area of triangular faces TSA of a composite shape can be found by calculating the areas of individual faces that are on the surface and then adding them together Volume

■ ■ ■

■ ■ ■ ■ ■ ■

■

Volume of a 3-dimensional figure is the amount of space it takes up Volume is measured in cubic units Volume of a prism = AH, where A is the cross-sectional area (or base) and H is the height of the prism The height of a prism is the dimension perpendicular to the prism’s cross-section. Volume of a sphere = 43 πr 3 Volume of a cone = 13 πr 2h

Volume of a pyramid = 13 AH Capacity of a 3-dimensional figure is the amount that will fit in that figure The relationship between the volume of a solid and the capacity (amount of liquid it can hold) is: 1 cm3 = 1 mL, 1000 cm3 = 1 L. 1 m3 = 1000 L = 1 kL

MaPPING YOUR UNdeRSTaNdING

Homework book

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 183. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

Chapter 6 surface area and volume

213

measurement anD geometry • using units of measurement

Chapter review fluenCy

5 Find the areas of the following plane figures. All

1 If all measurements are in cm, the area of the figure

below is: 7

measurements are in cm. a

3 14

a 16.49 cm2 c 9.81 cm2 e 30 cm2

2 If all measurements are in

✔

12

b 39.25 cm2 ✔ d 23.56 cm2

b

10

6

centimetres, the area of the figure at right is: a 50.73 cm2 b 99.82 cm2 c 80.18 cm2 d 90 cm2 e 119.45 cm2

8 5 15 5 c 5

3 7

3 If all measurements are in centimetres, the shaded

area of the figure below is:

5

30è d 3

2 7

a 3.93 cm2 c 388.77 cm2 ✔ e 129.59 cm2

b 11.52 cm2 d 141.11 cm2

6

e 10

4 The total surface area of the solid below is:

12

28 mm

f

40 mm

80è 10

8444.6 mm2 c 14146.5 mm2 e 16609.5 mm2

✔ a

214

b 9221 mm2 d 50271.1 mm2

maths Quest 10 for the australian Curriculum

a 84 cm2 b 100 cm2 c 6.5 cm2 d 56.52 cm2 e 60 cm2 f 244.35 cm2

measurement anD geometry • using units of measurement 6 Find the areas of the following figures. All

measurements are in cm.

8 Find the total surface area of each of the following

solids.

a

a

35 cm

a 300 cm2 b 224.52 cm2 c 160 cm2

15

50 cm

20 b 12

b

a 18 692.48 cm2 b 1495.40 cm2 c 804.25 cm2 d 642 cm2 e 873.36 mm2 f 760 cm2

14 mm

10 20 mm 8 c c

10 6

10

8 cm 5

20 7 Find the shaded area in each of the following. All

measurements are in cm. a

Q

QO = 15 cm SO = 8 cm PR = 18 cm O

P

14 cm 12 cm

d

a 499.86 cm2 b 44.59 cm2 c 128.76 cm2

R

S

18 cm

7 cm

e b

10 cm

10 mm 10 mm

14 mm

12.5

4 mm

[closed at both ends] f 12 cm

c

5 10 cm 10 cm 10 cm Chapter 6 surface area and volume

215

measurement anD geometry • using units of measurement 9 Find the volume of each of the following. a



7 cm

g

a 343 cm3 b 672 cm3 c 153 938.04 cm3 d 1.45 m3 e 1800 cm3 f 1256.64 cm3 g 297 cm3 h 8400 cm3 i 7238.23 mm3

11 cm

9 cm h 30 cm

b 7 cm

20 cm 8 cm

42 cm

12 cm

c

i

35 cm

12 mm 40 cm

problem solVing 1 A rectangular block of land 4 m ì 25 m is

d

surrounded by a concrete path 1 m wide. a Calculate the area of the path. 62 m2 b Calculate the cost of concreting at $45 per square metre. $7290

3.7 m

3 V = π r 2 h, the volume 2 will be 1.5 times as large as the original volume.

1m

the height of a cylinder by 6 have on its volume (in comparison with the original shape)? 3 What effect will halving the length, tripling the

e 10 cm

30 cm

V = 3lwh, the volume will be 3 times as large as (or triple) the original volume. 4

12 cm f 12 cm

10 cm

216

2 What effect will tripling the radius and dividing

maths Quest 10 for the australian Curriculum

width and doubling the height of a rectangular prism have on its volume (in comparison with the original shape)? A cylinder of radius 14 cm and height 20 cm is joined to a hemisphere of radius 14 cm to form a bread holder. a Find the total surface area. 3605.55 cm2 b Find the cost of chroming the bread holder on the outside at $0.05 per cm2 . $180.33 c What is the storage volume of the bread holder? 18062.1 cm3 d How much more space is in this new bread holder than the one it is replacing, which had a quarter circle end with a radius of 18 cm and a length of 35 cm? 9155.65 cm3

measurement anD geometry • using units of measurement 5 Bella Silos has two rows of silos for storing wheat.

d How much wheat can be stored altogether in 303.48 m3 Each row has 16 silos and all the silos are identical, these silos? with a cylindrical base (height of 5 m, diameter e Wheat is pumped from these silos into cartage of 1.5 m) and conical top (diameter of 1.5 m, height trucks with rectangular containers 2.4 m wide, of 1.1 m). 5 m long and 2.5 m high. How many truckloads 11 trucks 1.33 m a What is the slant height of the conical tops? are necessary to empty all the silos? b What is the total surface area of all the silos? f If wheat is pumped out of the silos at c What will it cost to paint the silos if one litre 2.5 m3/min, how long will it take to fill one 12 minutes of paint covers 40 m2 at a bulk order price of truck? $28.95 per litre? $655.85 2 910.81 m

eBook plus

Interactivities

Test yourself Chapter 6 int-2843 Word search Chapter 6 int-2841 Crossword Chapter 6 int-2842

Chapter 6 surface area and volume

217

eBook plus

aCtiVities

chapter opener Digital doc

• Hungry brain activity Chapter 6 (doc-5235) (page 183) are you ready? Digital docs (page 184) • SkillSHEET 6.1 (doc-5236): Conversion of area units • SkillSHEET 6.2 (doc-5237): Using a formula to find the area of a common shape • SkillSHEET 6.3 (doc-5238): Total surface area of cubes and rectangular prisms • SkillSHEET 6.4 (doc-5239): Conversion of volume units • SkillSHEET 6.5 (doc-5240): Volume of cubes and rectangular prisms

6a area Digital docs

• Activity 6-A-1 (doc-5035): Review of area (page 189) • Activity 6-A-2 (doc-5036): Area problems (page 189) • Activity 6-A-3 (doc-5037): Tricky area problems (page 189) • SkillSHEET 6.1 (doc-5236): Conversion of area units (page 185) • SkillSHEET 6.2 (doc-5237): Using a formula to find the area of a common shape (page 189) • WorkSHEET 6.1 (doc-5241): Area (page 193) eLesson

• Heron’s formula (eles-0177) (page 186) 6b Total surface area Interactivity

• TSA-sphere (int-2782) (page 194) Digital docs

• Activity 6-B-1 (doc-5038): Introducing surface area (page 199)

218

maths Quest 10 for the australian Curriculum

• Activity 6-B-2 (doc-5039): Surface area problems (page 199) • Activity 6-B-3 (doc-5040): Tricky surface area problems (page 199) • SkillSHEET 6.3 (doc-5238): Total surface area of cubes and rectangular prisms (page 193, 199) • WorkSHEET 6.2 (doc-5242): Surface area (page 202) 6c Volume Digital docs

• Activity 6-C-1 (doc-5041): Review of volume and capacity (page 208) • Activity 6-C-2 (doc-5042): Volume and capacity problem (page 208) • Activity 6-C-3 (doc-5043): Tricky volume and capacity problems (page 208) • SkillSHEET 6.4 (doc-5239): Conversion of volume units (page 203) • SkillSHEET 6.5 (doc-5240): Volume of cubes and rectangular prisms (page 203, 209) • WorkSHEET 6.3 (doc-6753): Volume (page 212) Interactivity

• Maximising the volume of a cuboid (int-1150) (page 203) chapter review Interactivities (page 217) • Test Yourself Chapter 6 (int-2843): Take the end-ofchapter test to test your progress • Word search Chapter 6 (int-2841): an interactive word search involving words associated with this chapter • Crossword Chapter 6 (int-2842): an interactive crossword using the definitions associated with the chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • PAtterns AnD AlgebrA

7

7A Expanding algebraic expressions 7B Factorising expressions with three terms 7C Factorising expressions with two or four terms 7D Factorising by completing the square 7E Mixed factorisation WhAt Do you knoW ?

Quadratic expressions

1 List what you know about quadratic expressions. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of quadratic expressions. eBook plus

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Hungry brain activity Chapter 7 doc-5243

oPening Question

What distance does the dolphin cover in one leap?

number AnD AlgebrA • PAtterns AnD AlgebrA

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 7.1 doc-5244

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SkillSHEET 7.8 doc-5249

220

Expanding brackets 1 Expand■each■of■the■following. a 4(3x■+■5) ■ 12x■+■20

c -4x(3■-■2x) -12x■+■8x2

10x2■-■15x b 5x(2x■-■3)

Expanding a pair of brackets 2 Expand■each■of■the■following■expressions. a (x■+■2)(x■-■2) b (2x■-■3) 2 4x2■-■12x■+■9 ■ x2■-■4

6x2■-■11x■-■10 c (3x■+■2)(2x■-■5)

Factorising by taking out the highest common factor 3 Factorise■each■of■the■following■expressions. ■ 4x(x■+■2) a 4x2■+■8x b -15x2■-■9x -3x(5x■+■3)

c 6x2■-■x x(6x■-■1)

Factorising by taking out a common binomial factor 4 Factorise■each■of■the■following. a 3x(x■+■2)■+■4(x■+■2) b 4x(x■-■1)■-■(x■-■1)

c -2x(x■+■3)■-■(x■+■3) -(x■+■3)(2x■+■1)

(x■-■1)(4x■-■1)

■ (x■+■2)(3x■+■4)

Simplifying algebraic fractions 5 Write■each■of■the■following■fractions■in■simplest■form. a

( x + 3)( x − 2) ( x − 2)

b

x+7

( x − 3)( x + 7)

5 a■ x■+■3

Simplifying surds 6 Simplify■each■of■the■following. a 24 ■ 2 6

maths Quest 10 for the Australian Curriculum

2

b 3 12 6 3

c b

3 x ( x + 2)2 ( x + 3) 6 x( x + 3)2 ( x + 2)

1 ( x − 3)( x + 7)

c

c 4 243 36 3

x+2 2( x + 3)

number AND algebra • Patterns and algebra

7A

Expanding algebraic expressions Binomial expansion ■■ ■■

When an expression contains two sets of brackets, expansion is known as binomial expansion. Consider the expression (a + b)(c + d). Its expansion can be represented visually by this area model. a

+ b

c +

ac

bc

d

ad

bd

This diagram visually shows that (a + b)(c + d) = ac + ad + bc + bd

■■ ■■

factorised expanded ■ form form Generally, binomial expressions contain only one variable, together with constants. Expansion of the binomial expression (x + 3)(x + 2) can be shown visually by this area model. x x

+

3

x ì x = x2

3ìx = 3x

2 ì x = 2x

3ì2 =6

+ 2

Expressed mathematically this is: (x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6

■■

factorised expanded ■ form form It is not necessary to draw this area model when expanding binomial expressions. The shortcut is to simply multiply each term in the first bracket by each term in the second bracket.

Worked Example 1

Expand each of the following. a (x + 3)(x + 2)

b (x - 7)(6 - x)

Think a

Write a (x + 3)(x + 2)

1

Write the expression.

2

Multiply the terms in the second bracket by the first term in the first bracket and then the second term in the first bracket.

= x(x + 2) + 3(x + 2)

3

Remove the brackets by multiplying each term in the brackets by the term outside the bracket.

= x2 + 2x + 3x + 6

4

Collect like terms.

= x2 + 5x + 6 Chapter 7 Quadratic expressions

221

number AND algebra • Patterns and algebra b

b (x - 7)(6 - x)

1

Write the expression.

2

Multiply the terms in the second bracket by the first term in the first bracket and then the second term in the first bracket. Notice that the minus sign stays with the second term in the first bracket (-7).

= x(6 - x) - 7(6 - x)

3

Remove the brackets by multiplying each term in the brackets by the term outside the bracket. Remember to change the sign when the term outside the bracket is negative.

= 6x - x2 - 42 + 7x

4

Collect like terms.

= -x2 + 13x - 42

FOIL method ■■

The word FOIL provides us with an acronym for the expansion of a binomial product.

■■

First: multiply the first terms in each bracket together

■■

Outer: multiply the two outer terms

■■

Inner: multiply the two inner terms

I (x + a)(x - b)

■■

Last: multiply the last terms in each bracket together

L (x + a)(x - b)

F (x + a)(x - b) O (x + a)(x - b)

Worked Example 2

Use FOIL to expand (x + 2)(x - 5). Think

Write

1

Write the expression.

(x + 2)(x - 5)

2

Multiply the first term in each bracket, then the outer terms, the inner terms and finally the last two terms.

= x ì x + x ì -5 + 2 ì x + 2 ì -5 = x2 - 5x + 2x - 10

3

Collect like terms.

= x2 - 3x - 10

■■

If there is a term outside the pair of brackets, expand the brackets and then multiply each term of the expansion by that term.

Worked Example 3

Expand 3(x + 8)(x + 2). Think

222

Write

1

Write the expression.

3(x + 8)(x + 2)

2

Use FOIL to expand the pair of brackets.

= 3(x2 + 2x + 8x + 16)

3

Collect like terms within the brackets.

= 3(x2 + 10x + 16)

4

Multiply each of the terms inside the brackets by the term outside the brackets.

= 3x2 + 30x + 48

Maths Quest 10 for the Australian Curriculum

number AND algebra • Patterns and algebra

■■

This method can be extended to include three or even more sets of brackets. In such examples, expand two brackets first and then multiply the result by the third bracket.

Expanding expressions that are perfect squares ■■ ■■

A special binomial expansion involves the expansion of a perfect square. The expansion of (a + b)2 can be represented visually by this area model. +

a

a

b

a ì a = a2

aìb = ab

a ì b = ab

bìb = b2

+ b

■■

■■ ■■

(a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2

This result tells us that to expand a perfect square: square the first term multiply the two terms together and then double them square the last term. Similarly (a - b)2 = a2 - 2ab + b2. (Try this expansion for yourself.) Any perfect square can also be expanded using FOIL; however, this method provides a quicker alternative for performing such expansions. (a + b)2 = a2 + 2ab + b2 or (a - b)2 = a2 - 2ab + b2

Worked Example 4

Expand and simplify each of the following. a (2x - 5)2 b -3(2x + 7)2 Think a

b

Write a (2x - 5)2

1

Write the expression.

2

Expand using the rule (a - b)2 = a2 - 2ab + b2.

1

Write the expression.

2

Expand the brackets using the rule (a + b)2 = a2 + 2ab + b2.

= -3[(2x)2 + 2 ì 2x ì 7 + (7)2] = -3(4x2 + 28x + 49)

3

Multiply every term inside the brackets by the term outside the brackets.

= -12x2 - 84x - 147

= (2x)2 - 2 ì 2x ì 5 + (5)2 = 4x2 - 20x + 25 b -3(2x + 7)2

Difference of two squares rule ■■ ■■

This rule results from the expansion of an expression of the form (a + b)(a - b). Note: The brackets can be in any order. The two brackets contain the same terms, but one has the terms added and the other subtracted. Chapter 7 Quadratic expressions

223

number AND algebra • Patterns and algebra

■■ ■■

The area model for the difference of two squares rule shows a large square with a smaller square removed from it. Consider the larger square has a side length of a, while the smaller square has a side length of b. a

a a

b

a2 - b

a

b2

a

a2 - b2

a-b

a-b =

= a-b

b

b

a-b

The final figure shows two rectangles with dimensions a by (a - b) and b by (a - b). So, a(a - b) + b(a - b) = a2 - b2 To factorise, take out a common factor of (a - b) on the left-hand side. (a - b)(a + b) = a2 - b2 Alternatively, the difference of two squares rule is usually written as (a + b)(a - b) = a2 - b2 Worked Example 5

Expand and simplify each of the following. a (3x + 1)(3x - 1) b 4(2 x - 7)(2 x + 7) Think a

b

Write a (3x + 1)(3x - 1)

1

Write the expression.

2

Expand using the rule (a + b)(a - b) =

1

Write the expression.

2

Expand using the difference of two squares rule.

= 4[(2x)2 - (7)2] = 4(4x2 - 49)

3

Multiply every term inside the brackets by the term outside the brackets.

= 16x2 - 196

a2

-

b2.

= (3x)2 - (1)2 = 9x2 - 1 b 4(2x - 7)(2x + 7)

remember

1. When expanding an algebraic expression with: (a) one bracket — multiply each term inside the bracket by the term outside the bracket (b) two brackets — multiply the terms in order: First terms, Outer terms, Inner terms and then Last terms (FOIL) (c) a term outside the two brackets — expand the pair of brackets first, then multiply each term of the expanded expression by the term outside the brackets (d) three brackets — expand any two of the brackets and then multiply the expanded expression by the third bracket. 2. Perfect squares rule: (a + b)2 = a2 + 2ab + b2 or (a - b)2 = a2 - 2ab + b2 3. Difference of two squares rule: (a + b)(a - b) = a2 - b2 224

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

exerCise

7A inDiViDuAl PAthWAys eBook plus

Activity 7-A-1

Review of expansion doc-5044 Activity 7-A-2

Expanding algebraic expressions doc-5045 Activity 7-A-3

Expanding more complex algebraic expressions doc-5046

2x3■-■12x2■+■18x eBook plus

Digital doc

SkillSHEET 7.1 doc-5244

expanding algebraic expressions FluenCy 1 Expand■each■of■the■following. a 2(x■+■3) ■ 2x■+■6 d -(x■+■3) -x■-■3 15x2■-■6x g 3x(5x■-■2) 2 4x3■-■6x2 j 2x (2x■-■3)

b e h k

4x■-■20 4(x■-■5) x(x■+■2) x2■+■2x 10x■-■15x2 5x(2■-■3x) 2 3x (2x■-■1) 6x3■-■3x2

2 We1, 2 ■Expand■each■of■the■following. a (x■+■3)(x■-■4) b (x■+■1)(x■-■3) c ■ x2■-■x■-■12 x2■-■2x■-■3 2 x ■-■6x■+■5 d (x■-■1)(x■-■5) e (2■-■x)(x■+■3) -x2■-■x■+■6 f g (2x■-■3)(x■-■7) 2x2■-■17x■+■21 h (x■-■1)(3x■+■2) 3x2■-■x■-■2 i 2 21■-■17x■+■2x k (5■-■2x)(3■+■4x) j (3■-■2x)(7■-■x) l 15■+■14x■-■8x2 2 3 We3 ■Expand■each■of■the■following. 8x ■-■28x■-■16 ■ 2x2■-■4x■-■6 b 4(2x■+■1)(x■-■4) a 2(x■+■1)(x■-■3) c 3x3■-■75x f d 2x(x■-■1)(x■+■1) e 3x(x■-■5)(x■+■5) 2x3■-■2x g -2x(3■-■x)(x■-■3) h -5x(2■-■x)(x■-■4) i 5x3■-■30x2■+■40x 4 Expand■each■of■the■following. a (x■-■1)(x■+■1)(x■+■2) b (x■-■3)(x■-■1)(x■+■2) c d (x■-■1)(x■-■2)(x■-■3) e (2x■-■1)(x■+■1)(x■-■4) f

Digital doc

SkillSHEET 7.2 doc-5245

(x■-■7)(x■+■2) x2■-■5x■-■14 (x■-■4)(x■-■2) x2■-■6x■+■8 6x2■-■17x■+■5 (3x■-■1)(2x■-■5) (11■-■3x)(10■+■7x) 110■+■47x■-■21x2

-2(x■+■1)(x■-■7) -2x2■+■12x■+■14 6x3■-■54x 6x(x■-■3)(x■+■3) 6x(x■+■5)(4■-■x) -6x3■-■6x2■+■120x (x■-■5)(x■+■1)(x■-■1) (3x■+■1)(2x■-■1)(x■-■1)

6 mC ■a■ ■(3x■-■1)(2x■+■4)■expands■to: B ✔ A 6x2■+■10x■-■4 D 6x2■-■10x■-■4 E b -2x(x■-■1)(x■+■3)■expands■to: A x2■+■2x■-■3 B D -2x3■+■4x2■-■6x E

h ( 2■- 3x)( 3■+■2x)■-■ 5x 6 + 2 2 x − 3 3x − 6 x 2 − 5x

5x2■-■24x■+■3 6x2■-■4

C 3x2■+■2x■-■4

-2x2■-■4x■+■6 -2x3■-■3

✔ C

-2x3■-■4x2■+■6x 4 a■ x3■+■2x2■-■x■-■2 b x3■-■2x2■-■5x■+■6 c x3■-■5x2■-■x■+■5 d x3■-■6x2■+■11x■-■6 e 2x3■-■7x2■-■5x■+■4 f 6x3■-■7x2■+■1

7 mC ■The■expression■(x■-■1)(x■-■3)(x■+■2)■is■not■the■same■as: A (x■–■3)(x■–■1)(x■+■2) ✔ B (x■+■3)(x■–■1)(x■–■2) C (x■–■1)(x■+■2)(x■–■3) D (x■+■2)(x■–■1)(x■–■3) E (x■–■3)(x■+■2)(x■–■1)

-12■+■108x■-■243x2

3(7■-■x) 21■-■3x 2x(x■-■4) 2x2■-■8x 8x2■+■2x 2x(4x■+■1) 2 15x3■+■20x2 5x (3x■+■4)

5 Expand■each■of■the■following■and■simplify. a (x■+■2)(x■-■1)■-■2x b 3x■-■(2x■-■5)(x■+■2) ■ x2■-■x■-■2 -2x2■+■4x■+■10 19x■-■23 c (2x■-■3)(x■+■1)■+■(3x■+■1)(x■-■2) 5x2■-■6x■-■5 d (3■-■2x)(2x■-■1)■+■(4x■-■5)(x■+■4) -5x■-■1 e (x■+■1)(x■-■7)■-■(x■+■2)(x■-■3) f (x■-■2)(x■-■5)■-■(x■-■1)(x■-■4) -2x■+■6 g (x■-■3)(x■+■1)■+■ 3x x2■-■2x■-■3■+■ 3x

eBook plus

c f i l

8 We4a ■Expand■and■simplify■each■of■the■following. 2 ■ x2■-■2x■+■1 a (x■-■1) b (x■+■2) 2 x2■+■4x■+■4 2 2 d (4■+■x) 16■+■8x■+■x e (7■-■x) 2 49■-■14x■+■x2 2 2 g (3x■-■1) h (12x■-■3) 2 144x2■-■72x■+■9 9x ■-■6x■+■1 2 2 2 25■-■40x■+■16x2 j (2■-■3x) 4■-■12x■+■9x k (5■-■4x)

c f i l

9 We4b ■Expand■and■simplify■each■of■the■following. a 2(x■-■3) b 4(x■-■7) 2 4x2■-■56x■+■196 2 ■ 2x2■-■12x■+■18 2 2 2 -49x2■+■14x■-■1 -4x ■-■12x■-■9 d -(2x■+■3) e -(7x■-■1) 2 g -3(2■-■9x) h -5(3■-■11x)2

2 3x2■+■6x■+■3 c 3(x■+■1) f 2(2x■-■3) 2 8x2■-■24x■+■18■ 2 -16x2■-■16x■-■4 i -4(2x■+■1)

2 x2■+■10x■+■25 (x■+■5) (12■-■x) 2 144■-■24x■+■x2 (5x■+■2) 2 25x2■+■20x■+■4 (1■-■5x) 2 1■-■10x■+■25x2

10 We5 ■Expand■and■simplify■each■of■the■following. -45■+■330x■-■605x2 x2■-■25 a (x■+■7)(x■-■7) ■ x2■-■49 b (x■+■9)(x■-■9) c (x■-■5)(x■+■5) x2■-■81 2 2 x ■-■1 4x ■-■9 9x2■-■1 d (x■-■1)(x■+■1) e (2x■-■3)(2x■+■3) f (3x■-■1)(3x■+■1) 2 2 g (7■-■x)(7■+■x) h (8■+■x)(8■-■x) i (3■-■2x)(3■+■2x) 49■-■x 64■-■x 9■-■4x2 Chapter 7 Quadratic expressions

225

number AnD AlgebrA • PAtterns AnD AlgebrA unDerstAnDing 11 The■length■of■the■side■of■a■rectangle■is■(x■+■1)■cm■and■the■width■is■(x■-■3)■■cm. a Find■an■expression■for■the■area■of■the■rectangle. (x■+■1)(x■-■3) x2■-■2x■-■3 b Simplify■the■expression■by■expanding. c If■x■=■5■■cm,■fi■nd■the■dimensions■of■the■rectangle■and,■hence,■its■area. 6■■cm,■2■■cm,■12■■cm2 12 Chickens■are■kept■in■a■square■enclosure■with■sides■measuring■x■m.■The■number■of■chickens■is■ a■

xm

b

(x + 2) m

increasing■and■so■the■size■of■the■enclosure■is■to■have■1■metre■added■to■one■side■and■2■metres■to■ the■adjacent■side. a Draw■a■diagram■of■the■original■enclosure. b Add■to■the■fi■rst■diagram■or■draw■another■one■to■show■the■new■enclosure.■Mark■the■lengths■ on■each■side■on■your■diagram. (x■+■1)(x■+■2) c Find■an■expression■for■the■area■of■the■new■enclosure. (x + 1) m d Simplify■the■expression■by■removing■the■brackets. x2■+■3x■+■2 e If■the■original■enclosure■had■sides■of■2■metres,■fi■nd■the■area■of■the■original■square■and■ then■the■area■of■the■new■enclosure. 4■■m2,■12■■m2 13 A■jewellery■box■has■a■square■base■with■sides■measuring■(x■+■2)■■cm■and■is■5■■cm■high. ■ (x■+■2)2 a Write■an■expression■for■the■area■of■the■base■of■the■box. b Write■an■expression■for■the■volume■of■the■box.■ 5(x■+■2)2 (V■=■area■of■base■ì■height) c Simplify■the■expression■by■expanding■the■brackets. 5x2■+■20x■+■20 d If■x■=■8■■cm,■fi■nd■the■volume■of■the■box■in■cm 3. 500■■cm3 e Find■the■area■of■the■lid■of■the■box■and,■hence,■fi■nd■how■many■ 1-cm■square■tiles■could■be■inlaid■in■the■lid. 100■■cm2,■100■tiles 14 In■redesigning■their■courtyard,■Linda■and■Finn■removed■a■section■of■ paving■as■shown■in■the■diagram■below. x Section of paving removed

x+1

x+3 3x + 5

a■=■4,■b■=■4,■ c■=■-24,■ d■=■0,■e■=■3

a Write■down■an■expression,■in■terms■of■x,■for■the■area■of■the■section■of■paving■that■is■ removed.■Write■your■answer■in■expanded■form.■ ■ x2■+■x b Find■the■area■of■paving■that■remains■in■terms■of■x. 5x2■+■21x +■20 c A■circular■fountain■is■to■be■placed■in■the■section■created■from■removing■the■paving.■ x ■ i Write■down■the■largest■possible■radius,■in■terms■of■x,■for■the■circular■fountain. 2 2 ii If■the■area■of■the■circular■fountain■is■1.77■m ,■determine■the■value■of■x.■Write■your■ 1.50■m answer■to■the■nearest■centimetre. reAsoning 15 Find■the■values■a,■b,■c,■d and■e■that■make■the■following■identity■true.

4x2(x■−■2)(x■+■3)■+■3■=■ax4■+■bx3■+■cx2■+■dx■+ e 16 A■tissue■box■has■the■side■lengths■shown■at■right.■ a Write■an■expression■in■factorised■form■for■the■ volume■of■the■box. ■ (x■+■2)(x■–■1)(x■–■3) b Find■the■volume■when■x is■5■■ cm. 56■■cm3 c What■is■the■volume■when x■=■1 0 226

maths Quest 10 for the Australian Curriculum

(x - 3) cm (x + 2) cm

(x - 1) cm

number AND algebra • Patterns and algebra

Is it possible for x to be zero? Justify your answer. No; you can’t have a negative volume. What is the smallest number x can be so that all the measurements are feasible? x>3 3 Find the volume of the box when x is 4  cm. 18  cm If the volume of the box is 120  cm3, find the value reflection    of x. (Hint: Try substituting values of x into the expression until you find the correct one.) x=6 Why does the Difference of Two h Write the expression for volume in expanded form. Squares rule have that name? Use this to calculate the volume when x = 6  cm. d e f g

x3 - 2x2 – 5x + 6

7B

Factorising expressions with three terms ■■ ■■

An expression with three terms is called a trinomial. Quadratic trinomials can be written in the form ax2 + bx + c where the highest power is a squared term.

Factorising ax2 + bx + c when a = 1 ■■ ■■ ■■ ■■ ■■ ■■ ■■

■■

+ 3 x Factorising is the inverse of expanding, so the area model for expanding a binomial can be considered in reverse. 3x x2 x Looking in reverse, it can be seen that ■ x2 + 3x + 2x + 6 = (x + 3)(x + 2). + This means that the factorised form of x2 + 5x + 6 is (x + 3)(x + 2). 2x 6 2 The technique is to find two factors of 6 which add to 5 (3 and 2). Note that there are other factors of 6 (6 and 1, -3 and -2, -6 and -1), but there is only one pair of factors which adds to 5. Obviously, it is not necessary to draw an area model to factorise every trinomial. The following method works for every possible trinomial (when a = 1) that can be factorised. Step 1.  Place the trinomial in the correct order or standard form x2 + bx + c. Step 2.  Find all the factor pairs of c (the constant term). Step 3.  Identify the factor pair whose sum equals b. Step 4.  Express the trinomial x2 + bx + c in factor form; that is, (x + __)(x + __). Remember to first check for and take out any common factors. As always, you can check your answer by expanding the brackets to re-create the original expression.

Worked Example 6

Factorise each of the following. a x2 - x - 20 b -2x2 + 16x - 14 Think a

Write a x2 - x - 20

1

Write the expression.

2

Check for a common factor (none).

3

Identify the factors of x2 as x and x and find the factors of the last term (-20) which add to equal the coefficient of the middle term (-1).

-20: 5 + -4 = 1, -5 + 4 = -1

4

Write the expression and its factorised form.

x2 - x - 20 = (x - 5)(x + 4)

Chapter 7 Quadratic expressions

227

number AND algebra • Patterns and algebra b

b -2x2 + 16x - 14

1

Write the expression.

2

Check for a common factor. (-2 can be taken out.)

= -2(x2 - 8x + 7)

3

Identify the factors of x2 as x and x and find the factors of the last term (7) which add to equal the coefficient of the middle term (-8).

7: 1 + 7 = 8, -1 + -7 = -8

4

Write the expression and its factorised form.

-2(x2 - 8x + 7) = -2(x - 1)(x - 7)

Factorising ax2 + bx + c when a ò 1 ■■

■■ ■■

If the coefficient of x2 is not 1, and there is not a common factor, we factorise the expression by splitting up the x-term so that the expression can then be factorised by grouping. A quadratic trinomial of the form ax2 + bx + c is broken up into four terms by finding two numbers that multiply to give ac and add to give b. Alternatively, the cross-product method could be used. This is illustrated in the following worked example.

Worked Example 7

Factorise 10x2 - x - 2 by: a grouping b the cross-product method Think a

b

a 10x2 - x - 2

1

Write the expression and check for a common factor (none).

2

Find the factor pair of ac (-20) which gives a sum of b (-1).

-20: 2 + -10 = -8, -4 + 5 = 1, 4 + -5 = -1

3

Rewrite the expression by breaking the x-term into two terms using the factor pair from step 2.

10x2 - x - 2 = 10x2 + 4x - 5x - 2

4

Factorise by grouping terms.

= 2x(5x + 2) - (5x + 2) = (5x + 2)(2x - 1)

1

Write the expression.

2

List the factor pairs of the first term (10x2) and the last term (-2). Note: There are four possible factor pairs for the first term: x and 10x; 2x and 5x; -x and -10x; -2x and -5x. In some cases, all variations will have to be tested to obtain the required middle term.

3

228

Write

Calculate the sum of each crossproduct pair until you find the combination that produces the middle term from the original expression (shown in red).

Maths Quest 10 for the Australian Curriculum

b 10x2 - x - 2

Factors of Sum the 10x2 -2 cross-products 2x 5x

-2 1

2x 5x

Result

2x 5x

-2 1

2x - 10x = -8x

1 -2

2x 5x

1 -2

-4x + 5x = x

2x 5x

2 -1

2x 5x

2 -1

-2x + 10x = 8x

2x 5x

-1 2

2x 5x

-1 2

4x – 5x = -x

number AnD AlgebrA • PAtterns AnD AlgebrA

4

10x2■-■x■-■2■=■(2x■-■1)(5x■+■2)

Express■the■trinomial■in■factor■form. Note:■The■fi■rst■pair■of■brackets■contain■ the■fi■rst■row■entries■and■the■second■pair■ of■brackets■contain■the■second■row■ entries■that■produce■the■middle■term■ from■the■original■expression.

remember

1.■ When■factorising■any■expression,■look■for■a■common■factor■fi■rst. 2.■ To■factorise■a■quadratic■trinomial■when■the■coeffi■cient■of■x2■is■1■(that■is,■x2■+■bx■+■c): (a)■ identify■the■factor■pair■of■c■whose■sum■is■equal■to■b (b)■express■the■trinomial■in■factor■form,■x2■+■bx■+■c■=■(x■+■__)(x■+■__). 3.■ To■factorise■a■quadratic■trinomial■when■the■coeffi■cient■of■x2■is■not■1■(that■is,■ ax2■+■bx■+■c■where■a■ò■1): (a)■ identify■the■factor■pair■of■ac■whose■sum■is■equal■to■b (b)■rewrite■the■expression■by■breaking■the■x-term■into■two■terms■using■the■factor■pair■ from■the■previous■step (c)■ factorise■the■resulting■expression■by■grouping. Alternatively,■the■cross-product■method■could■be■used■to■solve■any■quadratic■trinomial. 4.■ All■factorisations■can■be■checked■by■expanding■to■re-create■the■original■expression. exerCise

7b inDiViDuAl PAthWAys eBook plus

Activity 7-B-1

Introducing quadratic factorisation doc-5050 Activity 7-B-2

Practising quadratic factorisation doc-5051 Activity 7-B-3

Tricky quadratic factorisation doc-5052

eBook plus

Digital doc

SkillSHEET 7.5 doc-5250

Factorising expressions with three terms FluenCy 1 We6a ■Factorise■each■of■the■following. (x■+■3)(x■+■1) a x2■+■3x■+■2 b x2■+■4x■+■3 ■ (x■+■2)(x■+■1) 2■-■2x■-■3 2 d x2■+■8x■+■16 e x (x■+■4) (x■-■3)(x■+■1) g x2■-■11x■-■12 (x■-■12)(x■+■1) h x2■-■4x■-■12 (x■-■6)(x■+■2) (x■+■5)(x■-■1) (x■+■7)(x■-■1) j x2■+■4x■-■5 k x2■+■6x■-■7 m x2■-■4x■+■3 n x2■-■9x■+■20 (x■-■3)(x■-■1) (x■-■4)(x■-■5)

c f i l o

(x■+■8)(x■+■2) x2■+■10x■+■16 x2■-■3x■-■4 (x■-■4)(x■+■1) (x■+■4)(x■-■1) x2■+■3x■-■4 2 (x■+■5)(x■-■2) x ■+■3x■-■10 (x■+■14)(x■-■5) x2■+■9x■-■70

2 We6b ■Factorise■each■of■the■following. a -2x2■-■20x■-■18■ b -3x2■-■9x■-■6 2 d -x ■-■11x■-■10 e -x2■-■7x■-■10■ 2 g -x ■-■7x■-■12 h -x2■-■8x■-■12 2 j 3x ■+■33x■+■30 k 5x2■+■105x■+■100

c f i l

-x2■-■3x■-■2 -x2■-■13x■-■12 2x2■+■14x■+■20 5x2■+■45x■+■100

3 Factorise■each■of■the■following. (a■-■7)(a■+■1) a a2■-■6a■-■7 (m■+■5)(m■-■3) d m2■+■2m■-■15 (k■+■19)(k■+■3) g k2■+■22k■+■57 j v2■-■28v■+■75 (v■-■25)(v■-■3)

c f i l

(b■+■4)(b■+■1) b2■+■5b■+■4 c2■+■13c■-■48 (c■+■16)(c■-■3) g2■-■g■-■72 (g■+■8)(g■-■9) x2■-■19x■+■60 (x■-■15)(x■-■4)

b e h k

(t■-■4)(t■-■2) t2■-■6t■+■8 (p■-■16)(p■+■3) p2■-■13p■-■48 (s■-■19)(s■+■3) s2■-■16s■-■57 (x■+■16)(x■-■2) x2■+■14x■-■32

4 mC ■a■ To■factorise■-14x2■-■49x■+■21,■the■fi■rst■step■is■to: A fi■nd■factors■of■14■and■21■that■will■add■to■-49 B take■out■14■as■a■common■factor ✔ C take■out■-7■as■a■common■factor D fi■nd■factors■of■14■and■-49■that■will■add■to■make■21 E take■out■-14■as■a■common■factor

2 a■ -2(x■+■9)(x■+■1) b -3(x■+■2)(x■+■1) c -(x■+■2)(x■+■1) d -(x■+■10)(x■+■1) e -(x■+■2)(x■+■5) f -(x■+■12)(x■+■1) g -(x■+■3)(x■+■4) h -(x■+■2)(x■+■6) i 2(x■+■2)(x■+■5) j 3(x■+■1)(x■+■10) k 5(x■+■20)(x■+■1) l 5(x■+■4)(x■+■5)

Chapter 7 Quadratic expressions

229

number AND algebra • Patterns and algebra b The expression 42x2 - 9x - 6 can be completely factorised to: ✔ B 3(2x - 1)(7x + 2) A (6x - 3)(7x + 2) C (2x - 1)(21x + 6) D 3(2x + 1)(7x - 2) E 42(x - 3)(x + 2) 5   MC  When factorised, (x + 2)2 – (y + 3)2 equals: A (x + y – 2)(x + y + 2) B (x – y – 1)(x + y – 1) D (x – y + 1)(x + y + 5) E (x + y – 1)(x + y + 2)



a (2x + 1)(x + 2) b (2x - 1)(x - 1) c (4x + 3)(x - 5). d (2x - 1)(2x + 3) e (x - 7)(2x + 5) f (3x + 1)(x + 3) g (3x - 7)(2x - 1) h (4x - 7)(3x + 2) i (5x + 3)(2x - 3) j (4x - 1)(5x + 2) k (3x + 2)(4x - 1) l (3x - 1)(5x + 2)

✔ C

(x – y – 1)(x + y + 5)

6 Which method of factorising is the most appropriate for each of the following expressions? a Factorising using common factors b Factorising using the difference of two squares rule c Factorising by grouping d Factorising quadratic trinomials   i 5x2 + 3x − 2 d ii 25a2 − b 2 b iii x2 + 6x + 9 – y 2 b 2 2 2 a d iv 16x – 25x v 4x – 4y + x – y c vi x2 + 14x - 32 7   WE 7  Factorise each of the following using an appropriate method. a 2x2 + 5x + 2 b 2x2 - 3x + 1 c 2 d 4x + 4x - 3 e 2x2 - 9x - 35 f g 6x2 - 17x + 7 h 12x2 - 13x - 14 i j 20x2 + 3x - 2 k 12x2 + 5x - 2 l

4x2 - 17x - 15 3x2 + 10x + 3 10x2 - 9x - 9 15x2 + x - 2 8 Factorise each of the following, remembering to look for a common factor first. 3(3x + 1)(x - 7) a 4x2 + 2x - 6 2(x - 1)(2x + 3) b 9x2 - 60x - 21 2 c 72x + 12x - 12 12(2x + 1)(3x - 1) d -18x2 + 3x + 3 -3(3x + 1)(2x - 1) 3a(4x - 7)(2x + 5) e -60x2 + 150x + 90 -30(2x + 1)(x - 3) f 24ax2 + 18ax - 105a 2 -(2x - 7)(5x + 2) g -8x + 22x - 12 -2(4x - 3)(x - 2) h -10x2 + 31x + 14 i -24x2 + 35x - 4 -(8x - 1)(3x - 4) j -12x2 - 2xy + 2y 2 -2(3x - y)(2x + y) 2 2 -5(2x - 7y)(3x + 2y) k -30x + 85xy + 70y l -600x2 - 780xy - 252y 2 -12(5x + 3y)(10x + 7y) Understanding 9 Consider the expression (x - 1)2 + 5(x - 1) - 6. a Substitute w = x - 1 in this expression. w2 + 5w - 6 b Factorise the resulting quadratic. (w + 6)(w - 1) c Replace w with x - 1 and simplify each factor. This is the factorised form of the original expression. (x + 5)(x - 2) 10 Use the method outlined in question 9 to factorise each of the following expressions. a (x + 1)2 + 3(x + 1) - 4 x(x + 5) b (x + 2)2 + (x + 2) - 6 x(x + 5) c (x - 3)2 + 4(x - 3) + 4 (x - 1)2 (x + 9)(x + 5) d (x + 3)2 + 8(x + 3) + 12 2 e (x - 7) - 7(x - 7) - 8 (x - 15)(x - 6) f (x - 5)2 - 3(x - 5) - 10 (x - 10)(x - 3) 11 Factorise x2 + x - 0.75. (x - 0.5)(x + 1.5) 12 Students decide to make Valentine’s Day cards. The total area of

each card is equal to (x2 - 4x - 5) cm2. a Factorise the expression to find the dimensions of the cards in x. (x - 5)(x + 1) terms of b Write down the length of the shorter side in terms of x. (x - 5) cm c If the shorter sides of a card are 10  cm in length and the x. x = 15 cm longer sides are 16  cm in length, find the value of c. 160  cm2 d Find the area of the card proposed in part e If the students want to make 3000 Valentine’s Day cards, how much cardboard will be required? Give your answer in terms of x. 3 000(x - 5)(x + 1)  cm2 or (3000x2 - 12 000x - 15 000)  cm2

230

Maths Quest 10 for the Australian Curriculum

Happy Valentine's Day

number AnD AlgebrA • PAtterns AnD AlgebrA 13 The■area■of■a■rectangular■playground■is■given■by■the■general■expression■(6x2■+■11x■+■3)■m2■

where■x■is■a■positive■whole■number. a Find■the■length■and■width■of■the■play■ground■in■terms■of■x. ■ (2x■+■3)(3x■+■1) b Write■an■expression■for■the■perimeter■of■the■playground. P■=■10x■+■8 c If■the■perimeter■of■a■particular■playground■is■88■metres,■fi■nd x. x■=■8■metres reAsoning 14 Cameron■wants■to■build■an■in-ground■‘endless’■pool.■Basic■models■have■a■depth■of■2■metres■





and■a■length■triple■the■width.■A■spa■will■also■be■attached■to■the■end■of■the■pool. a The■pool■needs■to■be■tiled.■Write■an■expression■for■the■surface■area■of■the■empty■pool■ (that■is,■the■fl■oor■and■walls■only). ■ SA■=■3x2■+■16x b The■spa■needs■an■additional■16■m2■of■tiles.■Write■an■expression■for■the■total■area■of■tiles■ needed■for■both■the■pool■and■the■spa. Total■area■=■3x2■+■6x■+■16 c Factorise■this■expression. (3x■+■4)(x■+■4) d Cameron■decides■to■use■ tiles■that■are■selling■at■a■ l■=■21■■m;■w■=■7■■m;■d■=■2■■m discount■price,■but■there■ are■only■280■m2■of■the■tile■ available.■Find■the■maximum■ dimensions■of■the■pool■if■the■ width■is■in■whole■metres. e What■area■of■tiles■is■actually■ needed■to■construct■the■ pool? 275■■m2 f What■volume■of■water■can■the■ pool■hold? 294■■m3 15 A■quilt■is■made■by■repeating■the■patch■at■right. b y y a■ Yellow■=■3■cm■ì■3■cm The■letters■indicate■the■colours■of■fabric■that■make■up■the■patch■—■ ■ Black■=■3■cm■ì■6■cm yellow,■black■and■white.■The■yellow■and■white■pieces■are■square■and■ b b w ■ White■=■6■cm■ì■6■cm the■black■pieces■are■rectangular.■Many■of■these■patches■are■sewn■ b Yellow■=■0.36■m2 together■in■rows■and■columns■to■make■a■pattern.■The■fi■nished■quilt,■ Black■=■0.72■m2 b y y made■from■100■patches,■is■a■square■with■an■area■of■1.44■m2.■ 2 White■=■0.36■m ■ An■interesting■feature■is■created■when■the■blocks■are■sewn■together:■ c each■colour■forms■a■shape.■The■shape■and■its■area■are■exactly■the■same■ for■each■colour.■(The■feature■appears■throughout■the■quilt,■except■at■the■edges.) a Determine■the■size■of■each■yellow,■black■and■white■ fabric■piece■in■a■patch. reFleCtion    eBook plus b How■much■(in■m2)■of■each■of■the■different■colours■ In your own words, describe would■be■needed■to■construct■the■quilt?■(Ignore■ Digital doc how you would factorise a seam■allowances.) WorkSHEET 7.1 quadratic trinomial. doc-5251 c Sketch■a■section■of■the■fi■nished■product.

7C

Factorising expressions with two or four terms ■■ ■■ ■■

Factorising■to■the■inverse■or■opposite■of■expanding. The■factorised■form■shows■the■expression■as■a■product■of■factors,■while■the■expanded■form■ shows■the■expression■as■a■sum■or■difference■of■terms. The■most■straightforward■type■of■factorisation■is■where■a■common■factor■is■removed■from■ the■expression.■Once■this■has■been■done,■we■need■to■consider■the■number■of■terms■in■the■ expression■to■see■whether■other■types■of■factorisation■may■be■possible■to■further■simplify. Chapter 7 Quadratic expressions

231

number AND algebra • Patterns and algebra

Factorising expressions of the type a2 - b2 ■■ ■■

Recall the area model for the difference of two squares rule. When factorising an algebraic expression of the type a2 - b2, follow these steps. Look for a common factor first. If there is one, factorise by taking it out. Rewrite the expression showing the two squares and identifying the a and b parts of the expression. Factorise, using the rule a2 - b2 = (a + b)(a - b).

Worked Example 8

Factorise each of the following. a 4x2 - 9 b 7x2 - 448 c x2 - 17 Think a

b

c

Write a 4x2 - 9

1

Write the expression.

2

Check for a common factor and write as two perfect squares.

= (2x)2 - 32

3

Factorise using a2 - b2 = (a + b)(a - b).

= (2x + 3)(2x - 3)

1

Write the expression.

2

Check for a common factor and take it out.

= 7(x2 - 64)

3

Write the terms in the bracket as two perfect squares.

= 7(x2 - 82)

4

Factorise using a2 - b2 = (a + b)(a - b).

= 7(x + 8)(x - 8)

1

Write the expression.

2

Check for a common factor and write as two perfect squares. In this case, a surd needs to be used to rewrite 17 as a perfect square.

= x2 - ( 17 )2

3

Factorise using a2 - b2 = (a + b)(a - b).

= (x + 17 )(x - 17 )

b 7x2 - 448

c

x2 - 17

Factorising expressions with four terms ■■ ■■

If there are four terms to be factorised, look for a common factor first. Then group the terms in pairs and look for a common factor in each pair. It may be that a new common factor emerges as a bracket (common binomial factor).

Worked Example 9

Factorise each of the following. a x - 4y + mx - 4my b x2 + 3x - y2 + 3y Think a

232

1

Write

Write the expression and look for a common factor. (There isn’t one.)

Maths Quest 10 for the Australian Curriculum

a x - 4y + mx - 4my

number AND algebra • Patterns and algebra

b

2

Group the terms so that those with common factors are next to each other.

= (x - 4y) + (mx - 4my)

3

Take out a common factor from each group (it may be 1).

= 1(x - 4y) + m(x - 4y)

4

Factorise by taking out a common binomial factor. The factor (x - 4y) is common to both groups.

= (x - 4y)(1 + m)

1

Write the expression and look for a common factor.

b x2 + 3x - y2 + 3y

2

Group the terms so that those with common factors are next to each other.

= (x2 - y2) + (3x + 3y)

3

Factorise each group.

= (x + y)(x - y) + 3(x + y)

4

Factorise by taking out a common binomial factor. The factor (x + y) is common to both groups.

= (x + y)(x - y + 3)

■■ ■■

In Worked example 9, grouping occurred in pairs. This is known as grouping ‘two and two’. Now we will look at grouping a different combination, known as grouping ‘three and one’.

Worked Example 10

Factorise the following expression: x2 + 12x + 36 - y2. Think

Write

1

Write the expression and look for a common factor.

x2 + 12x + 36 - y2

2

Group the terms so that those that can be factorised are next to each other.

= (x2 + 12x + 36) - y2

3

Factorise the quadratic trinomial. This is the form of a perfect square.

= (x + 6)(x + 6) - y2 = (x + 6)2 - y2

4

Factorise the expression using a2 - b2 = (a + b)(a - b).

= (x + 6 + y)(x + 6 - y)

remember

1. To factorise an expression with two terms: (a) take out any common factors (b) check whether the difference of two squares rule can be used. 2. To factorise an expression with four terms: (a) take out any common factors (b) check whether they can be grouped using the ‘two and two’ method or the ‘three and one’ method. Chapter 7 Quadratic expressions

233

number AnD AlgebrA • PAtterns AnD AlgebrA

exerCise

7C inDiViDuAl PAthWAys eBook plus

Activity 7-C-1

Factorising expressions with two or four terms doc-5047 Activity 7-C-2

More factorising expressions with two or four terms doc-5048 Activity 7-C-3

Advanced factorising expressions with two or four terms doc-5049

2 a■ (x■-■2)(3x■+■2) b (x■+■3)(5■-■2x) c (x■-■1)(x■+■5) d (x■+■1)(x■-■1) e (x■+■4)(x■-■2) f (x■-■3)(4■-■x)

Factorising expressions with two or four terms FluenCy 1 Factorise■each■of■the■following■by■taking■out■a■common■factor. a x2■+■3x b x2■-■4x ■ x(x■+■3) x(x■-■4) 2 4x(x■+■4) d 4x ■+■16x e 9x2■-■3x 3x(3x■-■1) 2 3x(4■-■x) g 12x■-■3x h 8x■-■12x 2 4x(2■-■3x)

3x(x■-■2) c 3x2■-■6x f 8x■-■8x 2 8x(1■-■x) i 8x2■-■11x x(8x■-■11)

2 Factorise■each■of■the■following■by■taking■out■a■common■binomial■factor. a 3x(x■-■2)■+■2(x■-■2) b 5(x■+■3)■-■2x(x■+■3) c (x■-■1)2■+■6(x■-■1) d (x■+■1)2■-■2(x■+■1) e (x■+■4)(x■-■4)■+■2(x■+■4) f 7(x■-■3)■-■(x■+■3)(x■-■3) 3 We8a ■Factorise■each■of■the■following. (x■+■3)(x■-■3) a x2 ■-■1 ■ (x■+■1)(x■-■1) b x2■-■9 2 (x■+■10)(x■-■10) 2 (■y■+■k)(■y■-■k) d x ■-■100 e y2■-■k (4a■+■7)(4a■-■7) h 25p2■-■36q2 g 16a2■-■49

c x2■-■25 (x■+■5)(x■-■5) f 4x2■-■9y 2 (2x■+■3y)(2x■-■3y)■ 2 (1■+■10d)(1■-■10d) i 1■-■100d

4 We8b ■Factorise■each■of■the■following. (5p■+■6q)(5p■-■6q) a(x■+■3)(x■-■3) a 4x2 ■-■4 ■ 4(x■+■1)(x■-■1) b 5x2■-■80 c ax2■-■9a 5(x■+■4)(x■-■4) 2 2 2 d 2b ■-■8d 2(b■+■2d )(b■-■2d ) e 100x ■-■1600 100(x■+■4)(x■-■4) f 3ax2■-■147a 3a(x■+■7)(x■-■7) 2 3(6■+■x)(6■-■x) g 4px2■-■256p 4p(x■+■8)(x■-■8) h 36x2■-■16 4(3x■+■2)(3x■-■2) i 108■-■3x 5 mC ■a■ ■If■the■factorised■expression■is■(x■+■7)(x■-■7),■then■the■original■expression■must■have■

been:

A x2■-■7 D x2■+■49

eBook plus

Digital doc

6

Digital doc

✔

7

5(x■+■ 3)(x■-■ 3) 8

9

✔

x2 ( 3 ) − C 4 ( )2 5 2

x2 9 B − 16 25

x2 ( 3 ) E − 16 ( )2 5 2 2 c The■factorised■form■of■64x ■-■9y ■is: ✔ B (8x■+■3y)(8x■-■3y) A (64x■+■9y)(64x■-■9y) C (8x■-■3y)(8x■-■3y) D (8x■+■3y)(8x■+■3y) E (16x■+■3y)(16x■-■3y) mC ■Which■of■the■following■expressions■would■be■factorised■by■grouping■‘two■and■two’? A x2■–■a2■+■12a■−■36 B x2■−7x −10 2■ C 2x −■6x■−■xy +■3y D (s■–■5)2■–■25(s■+■3)2 E (r■+■5)■–■(r■+■3)(r■+■5) We8c ■Factorise■each■of■the■following. (x■+■ 15)(x■-■ 15) (x■+■ 7)(x■-■ 7) a x2■-■11 b x2■-■7 c x2■-■15 ■ (x■+■ 11)(x■-■ 11) 2 2 2■-■66 3(x■+■ 22 )(x■-■ 22) d 4x ■-■13 e 9x ■-■19 f 3x (3x■+■ 19)(3x■-■ 19) (2x■+■ 13)(2x■-■ 13) g 5x2■-■15 h 2x2■-■4 i 12x2■-■36 12(x■+■ 3)(x■-■ 3) 2(x■+■ 2)(x■-■ 2) Factorise■each■of■the■following■expressions. (x■-■5)(x■+■1) ■ (x■-■3)(x■+■1) (x■-■4)(x■+■6) c (x■-■2)2 ■-■9 a (x■-■1)2■-■4 b (x■+■1)2■-■25 2 (6■-■x)(x■+■8) f 36■-■(x■-■4)2 d (x■+■3)2■-■16 e 49■-■(x■+■1) (x■-■1)(x■+■7) (10■-■x)(x■+■2) 2 8(x■-■3) h 4(x■+■2)2■-■9(x■-■1)2 g (x■-■1)2■-■(x■-■5) i 25(x■-■2)2■-■16(x■+■3)2 We9a ■Factorise■each■of■the■following. (7■-■x)(5x■+■1) (x■-■22)(9x■+■2) ■ (x■-■2y)(1■+■a) a x■-■2y■+■ax■-■2ay b 2x■+■ax■+■2y■+■ay (x■+■y)(2■+■a) c ax■-■ay■+■bx■-■by d 4x■+■4y■+■xz■+■yz (x■-■y)(a■+■b) (x■+■y)(4■+■z) (f■-■2)(e■+■3) e ef■-■2e■+■3f■-■6 f mn■-■7m■+■n■-■7 (n■-■7)(m■+■1) g 6rt■-■3st■+■6ru■-■3su h 7mn■-■21n■+■35m■-■105 3(2r■-■s)(t■+■u) 7(m■-■3)(n■+■5) x2 9 D − 4 25

234

x2■-■49

 x 3  x 3 ■then■the■original■expression■must■have■ + − 4 5   4 5 

been:

x2 3 A − 4 5

SkillSHEET 7.4 doc-5247

✔ C

b If■the■factorised■expression■is■  

SkillSHEET 7.3 doc-5246

eBook plus

B x2■+■7 E x2■-■14x■+■49

maths Quest 10 for the Australian Curriculum

2

number AND algebra • Patterns and algebra



(5p - 4t + 3t)(5p - 4t - 3t)

i 64 - 8j + 16k - 2jk 2(8 - j)(4 + k) k 5x2 + 10x + x2y + 2xy x(5 + y)(x + 2)

j l

3a2 - a2b + 3ac - abc a(3 - b)(a + c) 2m2 - m2n + 2mn - mn 2 m(m + n)(2 - n)

10 Factorise each of the following. a xy + 7x - 2y - 14 (y + 7)(x - 2) c pq + 5p - 3q - 15 (q + 5)( p - 3) e a2b - cd - bc + a2d (b + d)(a2 - c)

b mn + 2n - 3m - 6 (m + 2)(n - 3) d s2 + 3s - 4st - 12t (s + 3)(s - 4t) f xy - z - 5z2 + 5xyz (1 + 5z)(xy - z)

11   WE 9b  Factorise each of the following. a a2 - b2 + 4a - 4b (a - b)(a + b + 4) c m2 - n2 + lm + ln (m + n)(m - n + l) e 5p - 10pq + 1 - 4q 2 (1 - 2q)(5p + 1 + 2q)

b p2 - q2 - 3p + 3q ( p - q)( p + q - 3) d 7x + 7y + x2 - y 2 (x + y)(7 + x - y) f 49g2 - 36h2 - 28g - 24h (7g + 6h)(7g - 6h - 4)

12   WE 10  Factorise each of the following. a x2 + 14x + 49 - y2 (x + 7 + y)(x + 7 - y) b x2 + 20x + 100 - y2 (x + 10 + y)(x + 10 - y) 2 2 c a - 22a + 121 - b (a - 11 + b)(a - 11 - b) d 9a2 + 12a + 4 - b2 (3a + 2 + b)(3a + 2 - b) e 25p2 - 40p + 16 - 9t2 f 36t2 - 12t + 1 - 5v (6t − 1 + 5v )(6t − 1 − 5v ) 13   MC  a  In the expression 3(x - 2) + 4y(x - 2), the common binomial factor is: A 3 + 4y B 3 - 4y C x ✔ E x - 2 D -x + 2 b Which of the following terms is a perfect square? B (x + 1)(x - 1) C 3x2 ✔ A 9 2 D 5(a + b) E 25x c Which of the following expressions can be factorised using grouping? A x2 - y2 B 1 + 4y - 2xy + 4x2 C 3a2 + 8a + 4 2 2 ✔ D x + x + y - y E 2a + 4b - 6ab + 18 14   MC  When factorised, 6(a + b) − x(a + b) equals: ✔ B (6 – x)(a + b) A 6 – x(a + b) D (6 + x)(a – b) E (6 + x)(a + b)

C 6(a + b – x)

understanding 15 The area of a rectangle is (x2 - 25) cm2. a Factorise the expression. (x - 5)(x + 5) b Using the factors, find a possible length and width of the rectangle. (x - 5)  cm, (x + 5)  cm 2  cm, 12  cm c If x = 7  cm, find the dimensions of the rectangle. d Hence, find the area of the rectangle. 24  cm2 120  cm2 or 6 times bigger e If x = 13  cm, how much bigger would the area of this rectangle be? Reasoning 16 A circular garden of diameter 2r m

A2 = p(r + 1)2  m2

is to have a gravel path laid around it. The path is to be 1  m wide. a Find the radius of the garden. r metres b Find the radius of the circle that includes both the garden and the path. (r + 1) m c Find the area of the garden in terms of r. A1 = pr2  m2 d Find the area of the garden and path together in terms of r, using the formula for the area of a circle. e Write an equation to find the area of the path, then write your equation in fully factorised form. A = p(r + 1)2 - pr2 = p(2r + 1) m2 f If the radius of the garden is 5  m, use the answer to part e to find the area of the path, correct to 2 decimal places. 34.56  m2 Chapter 7 Quadratic expressions

235

number AnD AlgebrA • PAtterns AnD AlgebrA 17 A■roll■of■material■is■(x■+■2)■metres■wide.■Annie■buys■(x■+■3)■metres■and■Bronwyn■buys■

5■metres. a Write■an■expression,■in■terms■of■x,■for■the■area■of■ each■piece■of■material. b If■Annie■has■bought■more■material■than■Bronwyn,■ write■an■expression■for■how■much■more■she■has■than■ Bronwyn. (x■+■3)(x■+■2)■–■5(x■+■2) c Factorise■and■simplify■this■expression. d Find■the■width■of■the■material■if■Annie■has■5■■m2■more■ than■Bronwyn. Width■=■5■■m e How■much■material■does■each■person■have?■explain■ your■answer. Annie■has■30■■cm2■and■Bronwyn■has■25■■cm2.

■ Annie■=■(x■+■3)(x■+■2)■m2 Bronwyn■=■5(x■+■2)■m2

(x■+■2)(x■-■2)■=■x2■–■4

reFleCtion 

 

What do you always check for first when factorising?

7D eBook plus

Interactivity Completing the square

int-2783

Factorising by completing the square A■quadratic■equation■can■be■written■in■general■form:■y■=■ax2■+■bx■+■c■and■in■turning■point■ form:■y■=■a(x■–■h)2■+■k. ■■ A■process■called■‘completing■the■square’■makes■it■possible■to■change■from■the■general■form■ to■the■turning■point■form.■This■process■is■useful■when■a■quadratic■equation■will■not■factorise■ easily■and■the■turning■point■and■the■x-intercepts■are■required. ■■ Completing■the■square■relies■on■the■knowledge■of■perfect■squares■and■the■difference■of■two■ squares.■Recall: ■■

Perfect■squares:■ ■ Difference■of■two■squares:■

(x■+■a)2■=■x2■+■2ax■+■a2 (x■-■a)2■=■x2■–■2ax■+■a2 x2■–■a2■=■(x■-■a)(x■+■a) x

Suppose■we■want■to■factorise■x2■+■4x■+■2.■There■are■no■factors■of■2■which■add■to■ give■4,■so■we■cannot■factorise■using■the■cross-product■method.■Instead,■the■ fi■rst■two■terms■can■be■used■to■‘build’■a■perfect■square.■ Start■with■a■square,■x2,■as■shown■at■right.

x2

x

x

2

x

x2

2x

2

2x

To■show■4x,■add■two■rectangles,■each■with■an■area■of■2x.

We■now■have■two■sides■of■a■square■with■side■length■(x■+■2). Two■units■with■a■value■of■1■are■added■to■make■x2■+■4x■+■2.■ There■are■2■units■missing. This■square,■then,■can■be■expressed■as:■x2■+■4x■+■2■=■(x■+■2)2■-■2 This■can■then■be■factorised■using■surds■and■the■difference■of■two■ squares■rule,■with■2■expressed■as■( 2 )2 .■ ■ ■ 236

(x■+■2)2■-■2■=■(x■+■2)2■-■( 2 )2 =■(x■+■2■-■ 2)(x■+■2■+■ 2)

maths Quest 10 for the Australian Curriculum

x

2

x

x2

2x

2

2x

1

1

number AND algebra • Patterns and algebra

■■

The factors include surds. This means that the expression has been factorised over the Real Number field. If the factors are not surds they have been factorised over the Rational Number field. To complete the square algebraically, we add the square of half the coefficient of x and, to compensate for adding this number, we need to subtract the same number to keep the equation equivalent. For example: x2 + 6x + 1 = x2 + 6x + (3)2 + 1 - (3)2 = (x + 3)2 + 1 - 9 = (x + 3)2 - 8



Using the difference of two squares rule, this factorises to: = (x + 3 - 8)(x + 3 + 8).

■■

Odd coefficients of x are more difficult to deal with; Fractions or decimals will need to be used.

Worked Example 11

Factorise each of the following by first completing the square. a  x2 - 8x + 5 b  x2 + 5x + 1 c  2x2 + 8x - 3 Think a

b

Write a x2 - 8x + 5

1

Write the expression.

2

Identify the coefficient of x, halve it and square the result.

3

Add the result of step 2 to the expression, placing it after the x-term. To balance the expression, we need to subtract the same amount as we have added.

= x2 - 8x + 16 + 5 - 16

4

Insert brackets around the first three terms to group them and then simplify the remaining terms.

= (x2 - 8x + 16) - 11

5

Factorise the first three terms to produce a perfect square.

= (x - 4)2 - 11

6

Rewrite the expression as the difference of two squares.

= (x - 4)2 - ( 11)2

7

Factorise using the difference of two squares rule.

= (x - 4 + 11)(x - 4 - 11)

1

Write the expression.

2

Identify the coefficient of x, halve it and square the result.

3

Add the result of step 2 to the expression, placing it after the x-term. To balance the expression, we need to subtract the same amount as we have added.

(

1 2

)

2

× −8 = (-4)2 = 16

b x2 + 5x + 1

(

1 2

) =( )

×5

2

5 2

= x 2 + 5x +

25 4

2

=

25 4

+1−

25 4

Chapter 7 Quadratic expressions

237

number AND algebra • Patterns and algebra

4

5

Insert brackets around the first three terms to group them and simplify the remaining terms. (Convert them to equivalent fractions.)

( = (x

Factorise the first three terms to produce a perfect square.

= x+ 2

) )−

25 = x 2 + 5 x + 25 + 44 − 4 4

+ 5x +

25 4

( )−

21 4

2

5

2

21 4

2

6

Rewrite the expression as the difference of two squares.

2 5   21   =x+  −  2   2 

7

Factorise using the difference of two squares rule.

 5 21   5 21  = x+ + x+ −   2 2  2 2    5 + 21   5 − 21  or  x + x+   2  2  

c

c

2x2 + 8x − 3

1

Write the expression.

2

Factorise the expression.

3

Identify the coefficient of x, halve it and square the result.

4

Add the result of step 3 to the expression, placing it after the x-term. To balance the expression, we need to subtract the same amount as we have added.

= 2 x 2 + 4 x + 4 − 32 − 4

5

Insert brackets around the first three terms to group them and simplify the remaining terms. (Convert them to equivalent fractions.)

= 2 ( x 2 + 4 x + 4) − 32 − 4

6

Factorise the first three terms to produce a perfect square.

7

Rewrite the expression as the difference of two squares.

8

Factorise using the difference of two squares rule. Leave the factor of 2 outside the difference of two squares expression.

9

Rationalise the surd denominator.

(

= 2 x 2 + 4 x − 32

(

1 2

×4

)

2

)

= (2)2 = 4

(

)

( ) = 2 ( ( x + 4 x + 4) − ) = 2 ( ( x + 2) − ) 11 2

2

2

 = 2  ( x + 2)2 − 

(

= 2 ( x + 2) −

11 2

( )  )((x + 2) + ) 11 2

11 2

2

11 2

 22   22  = 2  ( x + 2) −   ( x + 2) + 2  2     4 − 22   4 + 22  or 2  x + x+   2  2  

■■ ■■

238

Remember that you can expand the brackets to check your answer. If the coefficient of x2 ò 1, factorise the expression before completing the square.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

remember

1.■ If■a■quadratic■trinomial■cannot■be■factorised■by■fi■nding■an■integer■factor■pair,■then■ factorise■using■the■completing■the■square■method: (a)■ if■possible,■take■out■a■common■factor■and■write■it■outside■the■brackets (b)■halve■the■value■of■the■coeffi■cient■of■the■x-term■and■square■the■result (c)■ add■this■number■to■the■expression,■writing■it■after■the■x-term.■Balance■the■ expression■by■also■inserting■the■necessary■subtraction. (d)■factorise■the■fi■rst■three■terms■as■a■perfect■square■and■then■simplify■the■remaining■ terms (e)■ rewrite■the■expression■as■the■difference■of■two■squares (f)■ factorise■using■the■difference■of■two■squares■rule. 2.■ All■factorisations■can■be■checked■by■expanding■to■the■original■expression.

exerCise

7D inDiViDuAl PAthWAys eBook plus

Activity 7-D-1

Introducing completing the square doc-5053

Factorising by completing the square FluenCy 1 Complete■the■square■for■each■of■the■following■expressions. ■ (x■+■5)2■-■25 (x■+■3)2■-■9 a x2■+■10x b x2■+■6x 2 2 (x■-■2) ■-■4 c x ■-■4x d x2■+■16x (x■+■8)2■-■64 (x■-■10)2■-■100 e x2■-■20x f x2■+■8x (x■+■4)2■-■16 g x2■-■14x h x2■+■50x (x■-■7)2■-■49 (x■+■25)2■-■625 2 2 (x■-■1) ■-■1 i x ■-■2x

2 We11a ■Factorise■each■of■the■following■by■fi■rst■completing■the■square. a x2■-■4x■-■7 b x2■+■2x■-■2 ■ (x■-■2■+■ 11)(x■-■2■-■ 11) (x■+■1■+■ 3)(x■+■1■-■ 3) Practising 2 c x ■-■10x■+■12 d x2■+■6x■-■10 completing the (x■-■5■+■ 13)(x■-■5■-■ 13) (x■+■3■+■ 19)(x■+■3■-■ 19) square e x2■+■16x■-■1 f x2■-■14x■+■43 (x■+■8■+■ 65)(x■+■8■-■ 65) (x■-■7■+■ 6)(x■-■7■-■ 6) doc-5054 g x2■+■8x■+■9 h x2■-■4x■-■13 (x■-■2■+■ 17 )(x■-■2■-■ 17) (x■+■4■+■ 7)(x■+■4■-■ 7) Activity 7-D-3 i x2■-■12x■+■25 (x■-■6■+■ 11)(x■-■6■-■ 11) 21 21 Completing the (x■-■ 3 ■+■ 2 )(x■-■ 3 ■-■ 2 ) 2 square 3 We11b ■Factorise■each■of■the■following■by■fi■rst■completing■the■square. 2 5 5 doc-5055 a x2■-■x■-■1 b x2■-■3x■-■3 ■ (x■-■ 1 ■+■ 2 )(x■-■ 1 ■-■ 2 ) 2 2 13 13 2 c x ■+■x■-■5 d x2■+■3x■-■1 (x■+■ 3 ■+■ 2 )(x■+■ 3 ■-■ 2 ) 2 2 17 17 5 5 2 2 21 21 (x■+■ ■+■ 2 )(x■+■ ■-■ 2 ) f x ■+■5x■-■2 33 (x■+■ 1 ■+■ 2 )(x■+■ 1 ■-■ 2 ) e x ■+■5x■+■2 2 2 (x■+■ 5■+■ 2 )(x■+■ 5■-■ 2 2 2 2 g x2■-■7x■-■1 h x2■-■9x■+■13 13 13 2 1 1 29 29 9 9 i x ■-■x■-■3 (x■-■ ■+■ 2 )(x■-■ ■-■ 2 ) (x■-■ ■+■ 2 )(x■-■ ■-■ 2 ) 2 2 Activity 7-D-2

2

33 ) 2

2

4 We11c ■Factorise■each■of■the■following■by■fi■rst■looking■for■a■common■factor■and■then■

(x■-■ 7 ■+■ 2

53 53 )(x■-■ 7 ■-■ 2 ) 2 2

completing■the■square. ■ 2(x■+■1■+■ 3)(x■+■1■-■ 3) a 2x2■+■4x■-■4 5(x■+■3■+■2 2)(x■+■3■-■2 2) c 5x2■+■30x■+■5 5(x■-■3■+■ 7)(x■-■3■-■ 7) e 5x2■-■30x■+■10 g 3x2■+■30x■+■39 3(x■+■5■+■2 3)(x■+■5■-■2 3) i 6x2■+■36x■-■30 6(x■+■3■+■ 14)(x■+■3■-■ 14)

b d f h

4(x■-■1■+■ 6)(x■-■1■-■ 6) 4x2■-■8x■-■20 3(x■-■2■+■ 17)(x■-■2■-■ 17 ) 3x2■-■12x■-■39 6x2■+■24x■-■6 6(x■+■2■+■ 5)(x■+■2■-■ 5) 2x2■-■8x■-■14 2(x■-■2■+■ 11)(x■-■2■-■ 11)

unDerstAnDing 5 Which■method■of■factorising■is■the■most■appropriate■for■each■of■the■following■expressions? a Factorising■using■common■factors b Factorising■using■the■difference■of■two■squares■rule c Factorising■by■grouping Chapter 7 Quadratic expressions

239

number AnD AlgebrA • PAtterns AnD AlgebrA

Factorising■quadratic■trinomials Completing■the■square i 3x2■−■8x■−■3 ■d ii 49m2■−■16n 2 b iii x2■+■8x■+■4■–■y 2 c a d iv 7x2■–■28x v 6a –■6b■+■a2■–■b 2 c vi x2■+■x■–■5 2■–■7x■–■1 vii (x■–■3)2■+■3(x■–■3)■–■10 viii x d e 6 mC ■a■ ■To■complete■the■square,■the■term■which■should■be■added■to■x2■+■4x■is: A 16 D 2 C 4x E 2x ✔ B 4 b To■factorise■the■expression■x2■-■3x■+■1,■the■term■that■must■be■both■added■and■ subtracted■is: d e

A 9

B 3

C 3x

D

7 mC ■The■factorised■form■of■x2■–■6x■+■2■is:

9

3 2

✔ E 4

A (x■+■3■-■ 7)(x■+■3■+■ 7)

B (x■+■3■-■ 7 )(x■-■3■+■ 7 )■

C (x■-■3■-■ 7)(x■-■3■-■ 7)

D (x■-■3■-■ 7 )(x■+■3■+■ 7 )

✔ E

(x -■3■+■ 7)(x■-3■-■ 7)

reAsoning 8 A■square■measuring■x■cm■in■side■length■has■a■cm■added■to■its■

reFleCtion 

length■and■b■cm■added■to■its■width.■The■resulting■rectangle■ has■an■area■of■(x2■+■6x■+■3)■cm2.■Find■the■values■of■a■and■b,■ correct■to■2■decimal■places. a■=■0.55;■b■=■5.45

7e 7e inDiViDuAl PAthWAys eBook plus

Activity 7-E-1

Mixed factorisation doc-5056 Activity 7-E-2

Harder mixed factorisation doc-5057 Activity 7-E-3

Advanced mixed factorisation doc-5058

-4(x■+■6)(x■+■1)

Why is this method called completing the square?

mixed factorisation ■■

exerCise

 

The■following■exercise■will■help■you■to■practise■recognising■the■appropriate■method■of■ factorising■needed■for■a■given■expression. 2 (x■+■2■+■3y)(x■+■2■-■3y)■ 7■ (c■+■e)(5■+■d) 14 5(x■+■10)(x■+■2)■ 20 5(n■+■1)(2m■-■1)■ 31 (u■+■v)(t■-■3)■

mixed factorisation

16■ (x■-■4■+■y)(x■-■4■-■y) 29■ (2■+■r)(■p■-■s) 39■ 4(3■-■x■+■2y)(3■-■x■-■2y)

FluenCy

Factorise■each■of■the■following■expressions■in■questions■1–45. 2 x2■+■4x■+■4■-■9y2

3(x■+■3) 1 3x■+■9

4 x2■-■49 (x■+■7)(x■-■7) 7 5c■+■de■+■dc■+■5e 10

(x■+■4)(x■-■3) x2■+■x■-■12

13 16x2■-■4x 4x(4x■-■1)

5 5x2■-■9x■-■2 (5x■+■1)(x■-■2) 6 15x■-■20y 8

5x2■-■80

5(x■+■4)(x■-■4)

14 5x2■+■60x■+■100

16

17

19

x2■-■5

20 10mn■-■5n■+■10m■-■5

4x2■+■8

4(x2■+■2)

(x■+■2)(x■-■2) 22 x2■-■10x■-■11 (x■+■1)(x■-■11) 23 x2■-■4 25 xy■-■1■+■x■-■y (y■+■1)(■x■-■1) 26 28 -4x2■-■28x■-■24

9

5(3x■-■4y)

-x2■-■6x■-■5

-(x■+■5)(x■+■1)

(m■+■1)(n■+■1) 12 x2■-■7 11 mn■+■1■+■m■+■n (x■+■ 7)(x■-■ 7)

x2■-■8x■+■16■-■y2 (x■+■ 5)(x■-■ 5)

3 x2■-■36 (x■+■6)(x■-■6)

3x2■+■5x■+■2

3(3■-■y)(x■+■2) 15 18■+■9x■-■6y■-■3xy 18 fg■+■2h■+■2g■+■f h (g■+■h)(f■+■2) (x■+■5)(x■+■1) 21 x2■+■6x■+■5 (a■+■b)(c■-■5) 24 -5a■+■bc■+■ac■-■5b

(3x■+■2)(x■+■1) 27 7x2■-■28 7(x■+■2)(x■-■2)

29 2p■-■rs■+■pr■-■2s

30 3x2■-■27 3(x■+■3)(x■-■3)

31 -3u +■tv■+■ut■-■3v

32

34 (x■-■1)2■-■4 (x■+■1)(x■-■3)

35 (x■+■2)2■-■16 (x■+■6)(x■-■2) 36 (2x■+■3)2■-■25 4(x■-■1)(x■+■4)

x2■-■11

(4x■-■1)(3x■-■1) (x■+■ 11 )(x■-■ 11) 33 12x2■-■7x■+■1

37 3(x■+■5)2■-■27 3(x■+■2)(x■+■8) 38 25■-■(x■-■2) 2 (3■+■x)(7■-■x) 39 4(3■-■x)2■-■16y2 3(y■+■x)■(y■-■x)

43 (x■+■3)2■+■5(x■+■3)■+■4

41 (x■+■3)2■-■(x■+■1) 2 4(x■+■2) 42 (2x■-■3y)2■-■(x■-■y) 2 (3x■-■4y)(x■-■2y)

44 (x■-■3)2■+■3(x■-■3)■-■10 45 2(x■+■1)2■+■5(x■+■1)■+■2 (x■+■2)(x■-■5) 2(2x■+■3)(x■+■3) maths Quest 10 for the Australian Curriculum

(x■+■7)(x■+■4) 240

40 (x■+■2y)2■-■(2x■+■y)2

number AnD AlgebrA • PAtterns AnD AlgebrA unDerstAnDing

eBook plus

46 Consider■the■following■product■of■algebraic■fractions.■

Digital doc

x 2 + 3 x − 10

SkillSHEET 7.6 doc-5252

2

x −4 x+5 x−4

eBook plus

SkillSHEET 7.7 doc-5248

e g

Digital doc

WorkSHEET 7.2 doc-5254

i



46 a■

2

x − 2x − 8

( x + 5)( x − 2) ( x + 2)( x + 2) × ■ b■ ( x + 2)( x − 2) ( x − 4)( x + 2)

( x + 5) ( x − 2) ( x + 2) ( x − 2)

×

( x + 2) ( x + 2) ( x − 4) ( x + 2)

47 Use■the■procedure■in■question■46■to■factorise■and■simplify■each■of■the■following.

c

eBook plus

x2 + 4x + 4

a Factorise■the■expression■in■each■numerator■and■denominator. b Cancel■factors■common■to■both■the■numerator■and■the■denominator. c Simplify■the■expression■as■a■single■fraction.

a

Digital doc

×

x2 − 4x + 3 x 2 − 4 x − 12 6 x − 12 2

x −4

×

×

x 2 + 5 x + 6 ■ x − 1 x−6 x2 − 9

3x + 6 18 x ( x − 5) x( x − 5)

x2 + 4x − 5 x2 + x − 2

÷

x 2 + 10 x + 25 x2 + 4x + 4

d

x+2 x+5

4 ab + 8a 5ac + 5 a 4(b + 2) ÷ 2 5 (c − 3) c − 2c − 3 m 2 + 4 m + 4 − n2 2

4 m − 4 m − 15

5(m + 2 + n) 2(2m − 5)

÷

2m 2 + 4 m − 2mn 2

b

10 m + 15m

f

h

j

3 x 2 − 17 x + 10 6 x 2 + 5x − 6 6x2 − x − 2

×

x2 − 1 x2 − 6x + 5



x +1 2x + 3

2 x 2 + x − 1 2 x − 1 2 x 2 + 3 x + 1 3 x 2 + 10 x − 8 x + 4 x2 − 7x + 6 x2 + x − 2 p2 − 7 p 2

p − 49

÷

×

÷

x 2 − x − 12 x − 6

x+3

x2 − 2x − 8

p2 + p − 6 2

p + 14 p + 49

d 2 − 6d + 9 − 25e 2 2

4 d − 5d − 6

reFleCtion 

÷



p( p + 7) ( p + 3)( p − 2)

4 d − 12 − 20 e 5(d − 3 + 5e) 4(4 d + 3) 15d − 10

 

When an expression is fully factorised, what should it look like?

Chapter 7 Quadratic expressions

241

number AND algebra • Patterns and algebra

Summary Expanding algebraic expressions ■■

■■ ■■

When expanding an algebraic expression with: (a) one bracket — multiply each term inside the bracket by the term outside the bracket (b) two brackets — multiply the terms in order: First terms, Outer terms, Inner terms and then Last terms (FOIL) (c) a term outside the two brackets — expand the pair of brackets first, then multiply each term of the expanded expression by the term outside the brackets (d) three brackets — expand any two of the brackets and then multiply the expanded expression by the third bracket. Perfect squares rule: (a + b)2 = a2 + 2ab + b2 or (a - b)2 = a2 - 2ab + b2 Difference of two squares rule: (a + b)(a - b) = a2 - b2 Factorising expressions with three terms

■■ ■■

■■

■■

When factorising any expression, look for a common factor first. To factorise a quadratic trinomial when the coefficient of x2 is 1 (that is, x2 + bx + c): (a) identify the factor pair of c whose sum is equal to b (b) express the trinomial in factor form, x2 + bx + c = (x + __)(x + __). To factorise a quadratic trinomial when the coefficient of x2 is not 1 (that is, ax2 + bx + c where a ò 1): (a) identify the factor pair of ac whose sum is equal to b (b) rewrite the expression by breaking the x-term into two terms using the factor pair from the previous step (c) factorise the resulting expression by grouping. Alternatively, the cross-product method could be used to solve any quadratic trinomial. All factorisations can be checked by expanding to re-create the original expression. Factorising expressions with two or four terms

■■

■■

To factorise an expression with two terms: (a) take out any common factors (b) check whether the difference of two squares rule can be used. To factorise an expression with four terms: (a) take out any common factors (b) check whether they can be grouped using the ‘two and two’ method or the ‘three and one’ method. Factorising by completing the square

■■

■■

242

If a quadratic trinomial cannot be factorised by finding an integer factor pair, then factorise using the completing the square method: (a) if possible, take out a common factor and write it outside the brackets (b) halve the value of the coefficient of the x-term and square the result (c) add this number to the expression, writing it after the x-term. Balance the expression by also inserting the necessary subtraction. (d) factorise the first three terms as a perfect square and then simplify the remaining terms (e) rewrite the expression as the difference of two squares (f) factorise using the difference of two squares rule. All factorisations can be checked by expanding to the original expression.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • PAtterns AnD AlgebrA

■■

■■

This■method■will■not■always■give■a■difference■of■two■squares.■A■sum■of■two■squares■will■ sometimes■be■obtained.■If■the■coeffi■cient■of■x2■ò■1,■factorise■the■expression■before■completing■ the■square. This■method■can■be■used■to■convert■an■expression■into■turning■point■form■to■fi■nd■the■turning■ point■of■a■quadratic■graph.

MAPPING YOUR UNDERSTANDING

Homework Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■219. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

Chapter 7 Quadratic expressions

243

number AND algebra • Patterns and algebra 10 a x2 - 14x + 49 b 4 - 4x + x2 c 9x2 + 6x + 1 d -18x2 + 24x - 8 e -28x2 - 140x - 175 f -160x2 + 400x - 250 g x2 - 81 h 9x2 - 1 i 25 - 4x2

Chapter review Fluency

d e f g h i

3

✔

4

✔

10 Expand and simplify each of the following. a (x - 7)2 b (2 - x)2 2 c (3x + 1) d -2(3x - 2)2 2 e -7(2x + 5) f -10(4x - 5)2 The factorised form of -3d 2 - 9d + 30 is: g (x + 9)(x - 9) h (3x - 1)(3x + 1) A -3(d - 5)(d - 2) B -3(d + 5)(d - 6) i (5 + 2x)(5 - 2x) C -(3d + 5)(d - 2) D -(3d + 5)(d - 6) 13 a (x - y)(a + b) 11 Factorise each of the following. E -3(d + 5)(d - 2) b (x + y)(7 + a) a 2x2 - 8x 2x(x - 4) c (x + 2)( y + 5) If the factorised expression is (2x – 5)(2x + 5), then d (1 + 2q)(mn - q) b -4x2 + 12x -4x(x - 3) the original expression must have been: 2 e (5r + 1)( pq - r) ax(3 - 2x) c 3ax - 2ax f (v - 1)(u + 9) A 2x2 – 5 B 4x2 – 5 2(2x - 5)(4 - x) d (x + 1)2 + (x + 1) (x + 1)(x + 2) g (a - b)(a + b + 5) 2 2 C 4x – 25 D 4x – 20x + 25 e 3(2x - 5) - (2x - 5)2 h (d - 2c)(d + 2c - 3) E 2x2 + 25 f (x - 4)(x + 2) - (x - 4) (x - 4)(x + 1) i (1 + m)(3 - m)

5 To factorise -5x2 - 45x + 100, the first step is to: 12 A find factors of 5 and 100 that sum to -45 B take out 5 as a common factor ✔ C take out -5 as a common factor 4a(x + 2y)(x - 2y) D find factors of 5 and -45 that will add to 13

make 100 E take out -5x as a common factor

6 To complete the square, the term which should be ✔

C -12x

Factorise each of the following by grouping. a ax - ay + bx - by b 7x + ay + ax + 7y c xy + 2y + 5x + 10 d mn - q - 2q2 + 2mnq 2 e pq - 5r - r + 5pqr f uv - u + 9v - 9 g a2 - b2 + 5a - 5b h d2 - 4c2 - 3d + 6c 2 i 2 + 2m + 1 - m

14 Factorise each of the following by grouping. a 4x2 + 12x + 9 - y2 (2x + 3 + y)(2x + 3 - y) b 49a2 - 28a + 4 - 4b 2 (7a - 2 + 2b)(7a - 2 - 2b) c 64s2 - 16s + 1 3t (8s - 1 + 3t )(8s - 1 - 3t )

7 Which of the following is equivalent to 5x2 – 20x – 5? A 5(x – 2)2 B 5(x – 2)2 – 3 2 C 5(x – 2) – 15 D 5(x – 2)2 – 20 2 ✔ E 5(x – 2) – 25

16 Factorise each of the following by completing the

2

53 )(x 2

+72

53 ) 2

2(x + 9 + 2

85 )(x 2

+92

17 ) 2

(x + 7 +

b x2 - 10x - 3 d x2 - 5x + 2 f 2x2 + 18x - 2

2

Maths Quest 10 for the Australian Curriculum

square. a x2 + 6x + 1 c x2 + 4x - 2 e x2 + 7x - 1

-5-

necessary. a 3x(x - 4) 3x2 - 12x -21x2 - 7x b -7x(3x + 1) c (x - 7)(x + 1) x2 - 6x - 7

17 )(x 2

9 Expand each of the following and simplify where

✔

2

following is incorrect? A The value of the constant is -15. B The coefficient of the x term is 2. C The coefficient of the x term is -8. D The coefficient of the x2 term is 1. E The expansion shows this to be a trinomial expression.

15 Factorise each of the following. a x2 + 10x + 9 b x2 - 11x + 18 2 c x - 4x - 21 d x2 + 3x - 28 2 e -x + 6x - 9 f 3x2 + 33x - 78 2 g -2x + 8x + 10 h -3x2 + 24x - 36 2 i 8x + 2x - 1 j 6x2 + x - 1 2 k 8x + 4x - 12 l 105x2 - 10x - 15 m -12x2 + 62x - 70 n -45x2 - 3x + 6 o -60x2 - 270x - 270

8 In the expanded form of (x − 3)(x + 5), which of the

85 ) 2

16 a (x + 3 + 2 2)(x + 3 - 2 2) b (x - 5 + 2 7)(x - 5 - 2 7)

added to x2 – 12x is: A 36 B -12 D -6 E -6x

Factorise each of the following. (x + 5)(x - 5) a x2 - 16 (x + 4)(x - 4) b x2 - 25 c 2x2 - 72 2(x + 6)(x - 6) d 3x2 - 27y 2 3(x + 3y)(x - 3y) e 4ax2 - 16ay2 f (x - 4)2 - 9 (x - 1)(x - 7)

d (x - 5 +

244

2 When expanded, (3x + 7)2 becomes: A 9x2 + 49 B 3x2 + 49 C 3x2 + 21x + 49 ✔ D 9x2 + 42x + 49 E 9x2 + 21x + 49

c (x + 2 + 6)(x + 2 - 6)

15 a (x + 9)(x + 1) b (x - 9)(x - 2) c (x - 7)(x + 3) d (x + 7)(x - 4) e -(x - 3)2 f 3(x + 13)(x - 2) g -2(x - 5)(x + 1) h -3(x - 6)(x - 2) i (4x - 1)(2x + 1) j (3x - 1)(2x + 1) k 4(2x + 3)(x - 1) l 5(7x - 3)(3x + 1) m -2(3x - 5)(2x - 7) n -3(3x - 1)(5x + 2) o -30(2x + 3)(x + 3)

1 When expanded, -3x(x + 4)(5 - x) becomes: A -3x3 - 3x2 – 27x B -3x3 + 3x2 – 27x C 3x3 + 3x2 – 60x D -3x3 + 3x2 – 60x 3 2 ✔ E 3x - 3x – 60x

(2x - 5)(x - 3) 2x2 - 11x + 15 (4x - 1)(3x - 5) 12x2 - 23x + 5 3(x - 4)(2x + 7) 6x2 - 3x - 84 (2x - 5)(x + 3)(x + 7) 2x3 + 15x2 - 8x - 105 3x2 - 5x + 65 (x + 5)(x + 7) + (2x - 5)(x - 6) (x + 3)(5x - 1) - 2x 5x2 + 12x - 3

17 a■ 3x(x■-■4) b (x■+■3■+■ 7 )(x■+■3■-■ 7 ) c (2x■+■5)(2x■-■5) d (2x■+■5)(x■+■2) e (a■+■2)(2x■+■3) f -3(x■-■2)(x■+■3)

number AnD AlgebrA • PAtterns AnD AlgebrA 17 Factorise■each■of■the■following■using■the■most■

2 A■section■of■garden■is■to■have■a■circular■pond■of■

appropriate■method. 2( x + 4) a 3x2■-■12x b x2■+■6x■+■2 18 a 5( x + 1) 2 2 c 4x ■-■25 d 2x ■+■9x■+■10 2 e 2ax■+■4x■+■3a■+■6 f -3x ■-■3x■+■18 18 First■factorise■then■simplify■each■of■the■following. x+4 2 x − 12 3 x + 6 7 x − 42 7 b a × × 5 x − 30 x +1 4 x − 24 6 x + 12 8 x 2 − 4 x 2 + 4 x − 5 ( x − 2)( x − 1) c × x ( x − 4) x 2 + 5x x 2 − 2 x − 8 3 Problem solVing 1 A■large■storage■box■has■a■square■base■with■sides■

measuring■(x■+■2)■cm■and■is■32■■cm■high.

radius■2r■with■a■2■■m■path■around■its■edge.■ ■ 4r a State■the■diameter■of■the■pond.■ 2r■+■2 b State■the■radius■of■the■pond■and■path. 4pr■■2 c State■the■area■of■the■pond. (4pr■■2■+■8r■+■4)p d State■the■area■of■the■pond■and■path. e Write■an■expression■to■fi■nd■the■area■of■the■path■ 4p■■(2r■+■1) only■and■write■it■in■factorised■form. f If■the■radius■of■the■pond■is■3■metres,■fi■nd■the■ 28p■m2 area■of■the■path. In■order■to■make■the■most■of■the■space■available■for■ headlines■and■stories,■the■front■page■of■a■newspaper■ is■given■an■area■of■x2■–■5x■–■14■■cm2.■■ a Factorise■the■expression■to■fi■nd■the■dimensions■ of■the■paper■in■terms■of■x. ■ (x■–■7)(x■+■2) b Write■down■the■length■of■the■shorter■side■in■ x■–■7■cm terms■of x. c If■the■shorter■side■of■the■front■page■is■28■■cm,■ 35 fi■nd■the■value■of■x. 1036■cm2 d Find■the■area■of■this■particular■paper.■ Division■by■zero■in■Step■5 4 Here■is■a■well-known■puzzle. Let■a■=■b■=■1 Step■1:■Write■a■=■b. a■=■b Step■2:■■Multiply■both■ a2 = ab sides■by■a. Step■3:■■Subtract■b2■ a2 - b2 = ab - b2 from■both■sides. Step■4:■■Factorise. (a■+■b)(a■-■b)■=■b(a■-■b) Step■5:■■Simplify■by■ (a + b)■=■b dividing■by■ (a - b). Step■6:■■Substitute■ 1■+■1■=■1 a = b =■1. Where■is■the■error?■ Show■your■thinking.

a Write■an■expression■for■the■area■of■the■base■of■ the■box. ■ (x■+■2)2 b Write■an■expression■for■the■volume■of■the■box■ 32(x■+■2)2 (V■=■area■of■base■ì■height). c Simplify■the■expression■by■expanding■the■ brackets. 32x2■+■128x■+■128 d If■x■=■30■■cm,■fi■nd■the■volume■of■the■box■in■cm 3. 32■■768■■cm3

eBook plus

Interactivities

Test yourself Chapter 7 int-2846 Word search Chapter 7 int-2844 Crossword Chapter 7 int-2845

Chapter 7 Quadratic expressions

245

eBook plus

ACtiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■(doc-5243)■(page 219) Are you ready?

(page 220) •■ SkillSHEET■7.1■(doc-5244):■Expanding■brackets■ •■ SkillSHEET■7.2■(doc-5245):■Expanding■a■pair■of■ brackets■ •■ SkillSHEET■7.3■(doc-5246):■Factorising■by■taking■ out■the■highest■common■factor •■ SkillSHEET■7.4■(doc-5247):■Factorising■by■taking■ out■a■common■binomial■factor •■ SkillSHEET■7.7■(doc-5248):■Simplifying■algebraic■ fractions •■ SkillSHEET■7.8■(doc-5249):■Simplifying■surds Digital docs

7A Expanding algebraic expressions

(page 225) Activity■7-A-1■(doc-5044):■Review■of■expansion Activity■7-A-2■(doc-5045):■Expanding■algebraic■ expressions Activity■7-A-3■(doc-5046):■Expanding■more■ complex■algebraic■expressions SkillSHEET■7.1■(doc-5244):■Expanding■brackets SkillSHEET■7.2■(doc-5245):■Expanding■a■pair■of■ brackets

Digital docs

•■ •■ •■ •■ •■

7B Factorising expressions with three terms Digital docs

•■ Activity■7-B-1■(doc-5050):■Introducing■quadratic■ factorisation■(page 229) •■ Activity■7-B-2■(doc-5051):■Practising■quadratic■ factorisation■(page 229) •■ Activity■7-B-3■(doc-5052):■Tricky■quadratic■ factorisation■(page 229) •■ SkillSHEET■7.5■(doc-5250):■Finding■a■factor■pair■ that■adds■to■a■given■number■(page 229) •■ WorkSHEET■7.1■(doc-5251):■Factorising■and■ expanding■(page 231) 7C Factorising expressions with two or four terms Digital docs (page 234) •■ Activity■7-C-1■(doc-5047):■Factorising■expressions■ with■two■or■four■terms •■ Activity■7-C-2■(doc-5048):■More■factorising■ expressions■with■two■or■four■terms

246

maths Quest 10 for the Australian Curriculum

•■ Activity■7-C-3■(doc-5049):■Advanced■factorising■ expressions■with■two■or■four■terms •■ SkillSHEET■7.3■(doc-5246):■Factorising■by■taking■ out■the■highest■common■factor •■ SkillSHEET■7.4■(doc-5247):■Factorising■by■taking■ out■a■common■binomial■factor 7D Factorising by completing the square Digital docs (page 239) •■ Activity■7-D-1■(doc-5053):■Introducing■completing■ the■square •■ Activity■7-D-2■(doc-5054):■Practising■completing■ the■square •■ Activity■7-D-3■(doc-5055):■Completing■the■ square Interactivity

•■ Completing■the■square■(int-2783)■(page 236) 7E Mixed factorisation Digital docs

•■ Activity■7-E-1■(doc-5056):■Mixed■factorisation■ (page 240) •■ Activity■7-E-2■(doc-5057):■Harder■mixed■ factorisation■(page 240) •■ Activity■7-E-3■(doc-5058):■Advanced■mixed■ factorisation■(page 240) •■ SkillSHEET■7.6■(doc-5252):■Factorising■by■grouping■ three■and■one■(page 241) •■ SkillSHEET■7.7■(doc-5248):■Simplifying■algebraic■ fractions■(page 241) •■ WorkSHEET■7.2■(doc-5254):■Mixed■factorisation■ (page 241) Chapter review Interactivities (page 245) •■ Test■Yourself■Chapter■7■(int-2846):■Take■the■end-ofchapter■test■to■test■your■progress. •■ Word■search■Chapter■7■(int-2844):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■7■(int-2845):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter

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number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

8 Quadratic equations

8a Solving quadratic equations 8B The quadratic formula 8c Solving quadratic equations by inspecting graphs 8D Finding solutions to quadratic equations by interpolation and using the discriminant 8E Solving a quadratic equation and a linear equation simultaneously WhAt Do you knoW ? 1 List what you know about quadratic equations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of quadratic equations. eBook plus

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Hungry brain activity Chapter 8 doc-5255

opening Question

How can you tell that these images show shapes which are represented by quadratic equations?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 8.1 doc-5256

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SkillSHEET 8.5 doc-5260

Factorising by taking out the highest common factor 1 Factorise■each■of■the■following■expressions. a x2■-■3x ■ x(x■-■3) b 4x2■+■12x 4x(x■+■3)

12x(3x■-■1)

Finding a factor pair that adds to a given number 2 For■each■of■the■following,■fi■nd■a■factor■pair■of■the■fi■rst■number■that■adds■to■the■second■number. a -6,■1 ■ -2■and■3 b 6,■-5 c -6,■-1 2■and■-3 -2■and■-3

Simplifying surds 3 Simplify■each■of■the■following. a

b 2 50 10 2

24 ■ 2 6

c 3 288 36 2

Substituting into quadratic equations 4 Substitute■the■x-value■in■brackets■into■each■of■the■following■quadratic■equations■to■determine■

the■y-value.

a y■=■x2■-■4x■+■3■ (x■=■3) ■ 0 b y■=■-3x2■+■2x■-■8■ (x■=■2) -16 c y■=■-8x2■-■3x■-■12■ (x■=■-2) -38 Equation of a vertical line 5 Write■the■equation■for■each■of■the■lines■shown■below. y y a b

-4 -3 -2 -1 0 1 2 3

248

c 36x2■-■12x

x

maths Quest 10 for the Australian Curriculum

x -2-1 0 1 2 3 4

a■ x■=■-2 b x■=■3 c x■=■1.5 c

y

x -1 0 1 2 3 4 5

number AND algebra • Linear and non-linear relationships

8A

Solving quadratic equations ■■ ■■ ■■

The general form of a quadratic equation is ax2 + bx + c = 0. To solve an equation means to find the value of the pronumeral(s) or variables, which when substituted, will make the equation a true statement. This is done by using the Null Factor Law. If a ì b = 0 then a = 0 or b = 0 (or possibly both a and b equal 0).

■■

Equations that are not in factor form need to be factorised before the Null Factor Law can be applied.

Worked Example 1

Solve the equation (x - 7)(x + 11) = 0. Think

Write

1

Write the equation and check that the right-hand side equals zero.

(x - 7)(x + 11) = 0

2

The left-hand side is factorised so use the Null Factor Law to find two linear equations.

x - 7 = 0  or  x + 11 = 0

3

Solve for x.

x = 7

x = -11

Worked Example 2

Solve each of the following equations. a x2 - 3x = 0 c x2 - 13x + 42 = 0 Think a

b

b 3x2 - 27 = 0 d 36x2 - 21x = 2 Write a x2 - 3x = 0

1

Write the equation. Check that the righthand side equals zero.

2

Factorise by taking out any common factors (x).

x(x - 3) = 0

3

Use the Null Factor Law to write two linear equations.

x = 0  or  x - 3 = 0

4

Solve for x.

x = 0 b

x=3 3x2

- 27 = 0

1

Write the equation. Check that the righthand side equals zero.

2

Take out any common factors (3).



3

Look at the number of terms to factorise and identify the appropriate method. Factorise using the difference of two squares rule.

3(x2 - 32) = 0 3(x + 3)(x - 3) = 0

4

Use the Null Factor Law to write two linear equations.

x + 3 = 0  or  x - 3 = 0

5

Solve for x.

   x = -3 x=3   (Alternatively, x = ê3)

3(x2 - 9) = 0

Chapter 8 Quadratic equations

249

number AND algebra • Linear and non-linear relationships c

d

c

x2 - 13x + 42 = 0

1

Write the equation. Check that the right-hand side equals zero. Check for any common factors (none).

2

Look at the number of terms to factorise and identify the appropriate method. Factorise by finding a factor pair of ■ 42 that adds to -13.

   42: -6 + -7 = -13 (x - 6)(x - 7) = 0

3

Use the Null Factor Law to write two linear equations.

x - 6 = 0  or  x - 7 = 0

4

Solve for x.

   x = 6 36x2

d   

x=7

- 21x = 2

1

Write the equation. Check that the righthand side equals zero. (It does not.)

2

Rearrange the equation so the right-hand side of the equation equals zero. Check for any common factors (none).

36x2 - 21x - 2 = 0

3

Recognise that the expression to factorise is a quadratic trinomial. Identify a factor pair of ac (-72) which adds to the coefficient of x (-21).

   -72: 3 + -24 = -21

4

Rewrite the expression by breaking the x-term into two terms using the factor pair from step 3.

36x2 - 24x + 3x - 2 = 0

5

Factorise the expression by grouping.

12x(3x - 2) + (3x - 2) = 0     (3x - 2)(12x + 1) = 0

6

Use the Null Factor Law to write two linear equations.

3x - 2 = 0  or  12x + 1 = 0     3x = 2 12x = -1

7

Solve for x.



2

x = 3

1

x = - 12

Solving quadratic equations by completing the square ■■

■■

If it is not possible to find an integer factor pair when factorising a quadratic trinomial, the completing the square method can be used before applying the Null Factor Law to the equation. This method allows us to find irrational solutions. In other words, the solutions will be surds.

Worked Example 3

Find the solutions to the equation x2 + 2x - 4 = 0. Give exact answers. Think

250

1

Write the equation. Check that the right-hand side equals zero.

2

Identify the coefficient of x, halve it and square the result.

Maths Quest 10 for the Australian Curriculum

Write

x2 + 2x - 4 = 0

( × 2) 1 2

2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

Add■the■result■of■step■2■to■the■equation,■placing■ it■after■the■x-term.■To■balance■the■equation,■we■ need■to■subtract■the■same■amount■as■we■have■ added.

4

Insert■brackets■around■the■fi■rst■three■terms■to■ group■them■and■then■simplify■the■remaining■ terms.

5

Factorise■the■fi■rst■three■terms■to■produce■a■ perfect■square.

6

Express■as■the■difference■of■two■squares■and■ then■factorise.

x2 + 2x +

( × 2) − 4 − ( × 2) ■=■0 2

1 2

1 2

2

x2■+■2x■+■(1)2■-■4■−■(1)2■=■0 x2■+■2x■+■1■-■4■-■1■=■0 (x2■+■2x■+■1)■-■5■=■0

(x■+■1)2■-■5■=■0 (x■+■1)2■-■( 5)2■=■0 (x■+■1■+■ 5)(x■+■1■-■ 5)■=■0

7

Use■the■Null■Factor■Law■to■fi■nd■linear■ equations.

x■+■1■+■ 5■=■0■or■x■+■1■-■ 5■=■0

8

Solve■for■x.■Keep■the■answer■in■surd■form■to■ provide■an■exact■answer.

x■=■-1■-■ 5 ■ x■=■-1■+■ 5

■■ ■■

(Alternatively,■x■=■-1■ê■ 5 .)

There■are■many■problems■that■can■be■modelled■by■a■quadratic■equation.■You■should■fi■rst■form■ the■quadratic■equation■that■represents■the■situation■before■attempting■to■solve■such■problems. Recall■that■worded■problems■should■always■be■answered■with■a■sentence.

WorkeD exAmple 4

eBook plus

When two consecutive numbers are multiplied together, the result is 20. Determine the numbers. think

eLesson Completing the square

eles-0174

Write

1

Defi■ne■the■terms■by■using■a■pronumeral■for■one■ of■the■numbers■and■adding■1■to■it■to■give■the■ second■number.

2

Write■an■equation■multiplying■the■numbers■to■ give■the■answer.

3

Rearrange■the■equation■so■that■the■right-hand■ side■equals■zero.

4

Expand■to■remove■the■brackets.

5

Factorise.

(x■+■5)(x■-■4)■=■0

6

Use■the■Null■Factor■Law■to■solve■for■x.

x■+■5■=■0■ or■ x■-■4■=■0 x■=■-5■ x■=■4

7

Use■the■answer■to■determine■the■second■ number.

If■x■=■-5,■x■+■1■=■-4. If■x■=■4,■x■+■1■=■5.

8

Answer■the■question■in■a■sentence.

The■numbers■are■4■and■5■or■-5■and■-4.

9

Check■the■solutions.

Check:■4■ì■5■=■20■ -5■ì■-4■=■20

Let■the■two■numbers■be■x■and■(x■+■1).

x(x■+■1)■=■20 x(x■+■1)■-■20■=■0 x2■+■x■-■20■=■0

Chapter 8 Quadratic equations

251

number AND algebra • Linear and non-linear relationships

Worked Example 5

The height of a football after being kicked is determined by the formula h = -0.1d 2 + 3d, where d is the horizontal distance from the player. a How far away is the ball from the player when it hits the ground? b What horizontal distance does the ball cover when the height of the ball first reaches 20  m? Think a

b

Write a h = -0.1d 2 + 3d

1

Write the formula.

2

The ball hits the ground when h = 0. Write the formula with h = 0 on the right-hand side.

-0.1d 2 + 3d = 0

3

Factorise the expression.

-0.1d 2 + 3d = 0 d(-0.1d + 3) = 0

4

Use the Null Factor Law and then simplify the expression.

d = 0  or  -0.1d + 3 = 0 -0.1d = -3 −3 d = −0.1 = 30

5

Interpret the solutions.

d = 0 is the origin of the kick. d = 30 is the distance from the origin the ball has travelled when it lands.

6

Answer the question in a sentence.

The ball is 30  m from the player when it hits the ground.

1

To find the height of the ball at 20 m, substitute 20 for h.

2

Rearrange the expression.

0.1d2 – 3d + 20 = 0

3

Multiply both sides of the equation by 10 to remove the decimal from the coefficient.

d2 – 30d + 200 = 0

4

Factorise the expression.

    (d – 20)(d – 10) = 0

5

Apply the Null Factor Law.

d − 20 = 0  or  d – 10 = 0

6

Solve.

   d = 20

7

Interpret the solution. The first time the ball reaches a height of 20  m is the smaller value of d. Answer in a sentence.

The ball first reaches a height of 20  m after it has travelled a distance of 10 m.

b

h = -0.1d 2 + 3d 20 = -0.1d 2 + 3d

d = 10

remember

1. The general form of a quadratic equation is ax2 + bx + c = 0. 2. To solve a quadratic equation: (a) make sure the right-hand side of the equation equals zero (b) take out any common factors (c) factorise the left-hand side if applicable (d) use the Null Factor Law to solve for x. 3. An exact answer is a surd or an answer that has not been rounded or approximated. 252

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

solving quadratic equations

8A

fluenCy

inDiViDuAl pAthWAys

1 We 1 ■Solve■each■of■the■following■equations. a (x■+■7)(x■-■9)■=■0 ■ -7,■9 b (x■-■3)(x■+■2)■=■0 -2,■3■ d x(x■-■3)■=■0 0,■3 e x(x■-■1)■=■0 0,■1

eBook plus

-1.2,■-0.5

Activity 8-A-1

Solving simple quadratics doc-5059



Solving quadratic equations doc-5060 Activity 8-A-3

Solving more complex quadratics doc-5061

a■   c e g   i





b■ d■ f■ h■

h 9x(x■+■2)■=■0 -2,■0

j -(x■+■1.2)(x■+■0.5)■=■0 k 2(x■-■0.1)(2x■-■1.5)■=■0 0.1,■0.75 2 Solve■each■of■the■following■equations. a (2x■-■1)(x■-■1)■=■0 b (3x■+■2)(x■+■2)■=■0 d (7x■+■6)(2x■-■3)■=■0 e (5x■-■3)(3x■-■2)■=■0 g x(x■-■3)(2x■-■1)■=■0 h x(2x■-■1)(5x■+■2)■=■0 3 We 2a ■Solve■each■of■the■following■equations. a x2■-■2x■=■0 0,■2 b x2■+■5x■=■0 -5,■0 2 2 d 3x ■=■-2x e 4x2■-■6x■=■0 - 3 ,■0 0,■1 12

Activity 8-A-2

1 ,■1■ 2 1 ,■7■ 4 3 2 ,■ ■ 5 3 0,■ 12 ,■3■ 0,■-3,■ 25

0,■3 g 2x(x■-■3)■=■0

-2,■- 23 - 67 ,■1 12 - 85,■ 23 0,■ 12 ,■- 25

0,■ g 4x2■-■2 7x■=■0

7 2

h 3x2■+■ 3x■=■0 -  33 ,■0

4 We 2b ■Solve■each■of■the■following■equations. -5,■5 a x2■-■4■=■0 ■ -2,■2 b x2■-■25■=■0 2 d 4x ■-■196■=■0 -7,■7 e 9x2■-■16■=■0 -1 13 ,■1 13 - 23 ,■ 23 g 9x2■=■4 j

1 2 x ■-■ 4 ■=■0 36 9

-4,■4

c (x■-■2)(x■-■3)■=■0 2,■3 f x(x■+■5)■=■0 -5,■0 1

1

- 12 ,■ 12

i

(x■-■ 2 )(x■+■ 2 )■=■0

l

(x■+■ 2)(x■-■ 3)■=■0 - 2 ,■ 3

c (4x■-■1)(x■-■7)■=■0 f (8x■+■5)(3x■-■2)■=■0 i x(x■+■3)(5x■-■2)■=■0 c x2■=■7x 0,■7 1 f 6x2■-■2x■=■0 0,■ 3 2 i 15x■-■12x ■=■0

0,■1 14

-2,■2 c 3x2■-■12■=■0 -2 12 ,■2 12 f 4x2■-■25■=■0 1

h 36x2■=■9 - 12 ,■ 12

i

x2■-■ 25 ■=■0

− 5 ,■ 5 k x2■-■5■=■0

l

9x2■-■11■=■0

- 15 ,■ 15

- 

11 ,■ 311 3

5 We 2c ■Solve■each■of■the■following■equations. -4,■-2 a x2■-■x■-■6■=■0 ■ -2,■3 b x2■+■6x■+■8■=■0 c x2■-■6x■-■7■=■0 -1,■7 2 2 -1,■4 d x ■-■8x■+■15■=■0 3,■5 e x ■-■2x■+■1■=■0 1 f x2■-■3x■-■4■=■0 2 2 a■ 2 + 2 , 2 − 2 2,■6 -2,■5 g x ■-■10x■+■25■=■0 h x ■-■3x■-■10■=■0 i x2■-■8x■+■12■=■0 5 b −1 + 3 , −1 − 3 -3,■7 j x2■-■4x■-■21■=■0 k x2■-■x■-■30■=■0 -5,■6 l x2■-■7x■+■12■=■0 3,■4 c −3 + 10 , −3 − 10 6 mC ■The■solutions■to■the■equation■x2■+■9x■-■10■=■0■are: ✔ B x =■1■and■x =■-10 a x■=■1■and■x■=■10 c x■=■-1■and■x■=■10 d 4 + 2 3 , 4 − 2 3 D x■=■-1■and■x■=■-10 E x■=■1■and■x■=■9 e 5 + 2 6 , 5 − 2 6 7 mC ■The■solutions■to■the■equation■x2■–■100■=■0■are: f 1 + 3 , 1 − 3 a x■=■0■and■x■=■10 B x■=■0■and■x■=■-10 ✔ c x■=■-10■and■x■=■10 D x■=■0■and■x■=■100 E x■=■-100■and■x■=■100 g −1 + 6 , −1 − 6 e - 143 ,■1 h -1 43 ,■-113 8 We 2d ■Solve■each■of■the■following■equations. h −2 + 10 , − 2 − 10 1 ■ - 12 ,■3 23,■-1 a 2x2■-■5x■=■3 b 3x2■+■x■-■2■=■0 c 5x2■+■9x■=■2 c -2,■ 5 i −2 + 15 , − 2 − 15 1 1 1 1 2 2 2 d 6x ■-■11x■+■3■=■0 3,■1 2 e 14x ■-■11x■=■3 f 12x ■-■7x■+■1■=■0 4 ,■ 3 1 1 2 1 2 2 2 -1 3 ,■2 2 g 6x ■-■7x■=■20 h 12x ■+■37x■+■28■=■0 i 10x ■-■x■=■2 - 5 ,■ 2 j 6x2■-■25x■+■24■=■0 l 3x2■-■21x■=■-36 3,■4 1 12,■2 23 k 30x2■+■7x■-■2■=■0 - 25 ,■ 16 3 5 3 5 a■  +  ,■  -  9 We 3 ■Find■the■solutions■for■each■of■the■following■equations.■Give■exact■answers. 2 2 2 2 5 29 5 29 a x2■-■4x■+■2■=■0 b x2■+■2x■-■2■=■0 c x2■+■6x■-■1■=■0 b - 2 ■+■ 2 ,■- 2■-■ 2 2 2 d x ■-■8x■+■4■=■0 e x ■-■10x■+■1■=■0 f x2■-■2x■-■2■=■0 33 7 33 7 2 2 c  +  ,■  -  2 g x ■+■2x■-■5■=■0 h x ■+■4x■-■6■=■0 i x2■+■4x■-■11■=■0 2 2 2

d 12 ■+■

21 1 ,■ ■-■ 221 2 2

e 11  +  2

117 11 ,■ 2  -  117 2 2

f - 12 ■+■ 25 ,■- 12 ■-■ g - 32 ■+■ h 25 ■+■ i

5 2

37 ,■- 32 ■-■ 237 2

37 5 37 ,■ 2 ■-■ 2 2

9  +■ 265 ,■ 92 ■-■ 265 2

10 Find■the■solutions■for■each■of■the■following■equations.■Give■exact■answers. a x2■-■3x■+■1■=■0 b x2■+■5x■-■1■=■0 c x2■-■7x■+■4■=■0 2 2 d x ■-■5■=■x e x ■-■11x■+■1■=■0 f x2■+■x■=■1 g x2■+■3x■-■7■=■0 h x2■-■3■=■5x i x2■-■9x■+■4■=■0

11 Solve■each■of■the■following■equations,■rounding■answers■to■2■decimal■places. -4.24,■0.24 c 5x2■-■10x■-■15■=■0 -1,■3 a 2x2■+■4x■-■6■=■0 ■ -3,■1 b 3x2■+■12x■-■3■=■0 2 2 d 4x ■-■8x■-■8■=■0 -0.73,■2.73 e 2x ■-■6x■+■2■=■0 0.38,■2.62 f 3x2■-■9x■-■3■=■0 -0.30,■3.30 2 -2.30,■1.30 i 4x2■+■8x■-■2■=■0 -2.22,■0.22 g 5x ■-■15x■-■25■=■0 h 7x2■+■7x■-■21■=■0 -1.19,■4.19 Chapter 8 Quadratic equations 253

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships No■real■solutions■—■when■we■complete■the■square■we■get■the■sum■of■two■squares,■not■the■difference■of■two■squares■and■we■cannot■factorise■ the■expression. unDerstAnDing 12 Are■there■real■solutions■to■the■equation■x2■+■4x■+■10■=■0?■Give■reasons■for■your■answer. 6■and■8,■ -6■and■-8

13 We 4 ■When■two■consecutive■numbers■are■multiplied,■the■result■is■72.■Find■the■numbers. 8■and■9■or■ -8■and■-9 14 When■two■consecutive■even■numbers■are■multiplied,■the■result■is■48.■Find■the■numbers. 15 When■a■number■is■added■to■its■square■the■result■is■90.■Find■the■number. 9■or■-10 16 Twice■a■number■is■added■to■three■times■its■square.■If■the■result■is■16,■fi■nd■the■number.

2

2■or■-2 3

17 Five■times■a■number■is■added■to■two■times■its■square.■If■the■result■is■168,■fi■nd■the■number. 8■or■-10 12 18 We 5 ■A■soccer■ball■is■kicked.■The■height,■h,■in■metres,■of■the■soccer■ball■



45

cm

2x cm

t■seconds■after■it■is■kicked■can■be■represented■by■the■equation■h■=■-t(t■-■6).■ Find■how■long■it■takes■for■the■soccer■ball■to■hit■the■ground■again. 6■seconds 19 The■length■of■an■Australian■fl■ag■is■twice■its■width■and■the■diagonal■length■ is■45■■cm. a If■x■cm■is■the■width■of■the■fl■ag,■fi■nd■the■length■in■terms■of■x. ■ l■=■2x b Draw■a■diagram■of■the■fl■ag■marking■in■the■diagonal.■Mark■the■length■and■the■width■in■ x cm terms■of■x. c Use■Pythagoras’■theorem■to■write■an■equation■relating■the■lengths■of■the■sides■to■the■ x2■+■(2x)2■=■452,■5x2■=■2025 length■of■the■diagonal. d Solve■the■equation■to■fi■nd■the■dimensions■of■the■Australian■fl■ag.■Round■your■answer■to■ Length■40■cm,■width■20■cm the■nearest■cm. 20 If■the■length■of■a■paddock■is■2■■m■more■than■its■width■and■the■area■is■48■■m2,■fi■nd■the■length■and■ width■of■the■paddock. 8■m,■6■m 21 Solve■for■x. a

x2 + 4x + 3 2

x + 4x + 4 2

b

c

x − 2x − 3 x2 − 6x + 8

= =

x2 + 2x + 1 2

x + 5x + 6

■ - 73

x 2 + x − 12

x■=■0

( x 2 − 16)( x − 2)

( x − 1)( x 2 + 2 x + 1) 2x2 + 6x + 4

=

3x 2 + 4 x + 1 3 x 2 + 15 x + 18

x■=■ê  11 3

reAsoning 22 H enrietta■is■a■pet■rabbit■who■lives■in■an■



254

enclosure■that■is■2■■m■wide■and■4■■m■ long.■Her■human■family■has■decided■to■ purchase■some■more■rabbits■to■keep■ her■company■and■so■the■size■of■the■ enclosure■must■be■increased. ■ a Draw■a■diagram■of■Henrietta’s■ 2m enclosure,■clearly■marking■the■ lengths■of■the■sides. 4m b If■the■length■and■width■of■the■ enclosure■are■increased■by■x■m,■ (2■+■x)■m,■(4■+■x)■m fi■nd■the■new■dimensions. c If■the■new■area■is■to■be■24■■m2,■ write■an■equation■relating■the■ sides■and■the■area■of■the■enclosure■ (2■+■x)(4■+■x)■=■24 (Area■=■length■ì■width). d Use■the■equation■to■fi■nd■the■value■of■x■and,■hence,■the■length■of■the■sides■of■the■new■ enclosure. x■=■2,■4■m■wide,■6■m■long maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 23 A■student■is■required■to■cover■an■area■of■620■■cm2■with■mosaic■tiles.■The■tile■pattern■is■to■be■

surrounded■by■a■border■2■■cm■wide■to■complete■the■display■page.■The■length■of■the■display■ page■is■l■cm■and■its■width■is■4■■cm■less■than■its■length. a Find■the■width■of■the■display■page■in■terms■of■l. ■ (l■-■4)■cm b Find■the■width■and■length■of■the■tile■pattern■in■terms■of■l. l■-■8,■l■-■4 c Using■the■answers■from■b,■write■an■equation■relating■the■area■of■the■tile■pattern■to■its■ dimensions. (l■-■8)(l■-■4)■=■620 d Use■the■completing■the■square■method■to■solve■the■equation■and,■hence,■fi■nd■the■length,■ l■cm,■of■the■display■page.■Round■your■answer■to■the■nearest■cm. 31■cm e Find■the■area■of■the■display■page.■Round■your■answer■to■the■nearest■cm 2. 836■cm2 24 The■cost■per■hour,■C(s),■in■thousands■of■dollars■of■running■two■cruise■ships,■Annabel■and■Betty,■ travelling■at■a■speed■of■s■knots■is■given■by■the■following■relationships. ■ ■CAnnabel(28)■=■$364■800,■ CAnnabel(s)■=■0.3s2■+■4.2s■+■12■and■CBetty(s)■=■0.4s2■+■3.6s■+■8 CBetty(28)■=■$422■400 a Determine■the■cost■per■hour■for■each■ship■if■they■are■both■ travelling■at■28■knots. eBook plus b Find■the■speed■in■knots■at■which■both■ships■must■travel■ refleCtion    for■them■to■have■the■same■cost. 10■knots Digital doc What does the Null WorkSHEET 8.1 c Explain■why■only■one■of■the■solutions■obtained■in■your■ Factor Law mean? doc-5261 S peed■can■only■be■a■positive■ working■for■part■b■is■valid.■ quantity,■so■the■negative■ solution■is■not■valid.

8b

the quadratic formula ■■ ■■

The■method■of■solving■quadratic■equations■by■completing■the■square■can■be■generalised■to■ produce■what■is■called■the■quadratic formula. Consider■solving■the■general■equation■ax2■+■bx■+■c■=■0.■We■will■fi■rst■follow■the■steps■involved■ in■completing■the■square. b c x2 + x + = 0 a a

1.■ Divide■both■sides■of■the■equation■by■a.■ 2

x2 +

2.■ Complete■the■square.■

2

b c  b  b x+  −  + =0  2a   2a  a a 2

3.■ Factorise■the■fi■rst■three■terms■as■a■perfect■square.■

b b2 c  x + − + =0  2a  4a 2 a

4.■ Add■the■fi■nal■two■terms.■

b b 2 − 4 ac  =0  x + 2a  − 4a2

2

2

2  b 2 − 4 ac  b  x + −   =0    2a  2a

5.■ Write■as■the■difference■of■two■squares.■

6.■ ■Factorise■using■the■difference■of■two■■ squares■rule. 7.■ Solve■the■two■linear■factors.■ ■

x+

 b b 2 − 4 ac   b b 2 − 4 ac  + − x+ x+  =0     2a 2a 2a 2a b b 2 − 4 ac b b 2 − 4 ac + = 0 ■ or■ x + − =0 2a 2a 2a 2a x=

−b b 2 − 4 ac −b b 2 − 4 ac ■ ■■ x= + − 2a 2a 2a 2a Chapter 8 Quadratic equations

255

number AND algebra • Linear and non-linear relationships

■■

■■

− b ± b 2 − 4 ac where a is the coefficient of x2, b is 2a the coefficient of x and c is the constant or the term without an x. This formula can be used to solve any quadratic equation. The solution can be summarised as x =

Worked Example 6

Use the quadratic formula to solve each of the following equations. a 3x2 + 4x + 1 = 0 (exact answer) b -3x2 - 6x - 1 = 0 (round to 2 decimal places) Think a

b

Write a 3x2 + 4x + 1 = 0

1

Write the equation.

2

Write the quadratic formula.

x=

3

State the values for a, b and c.

where a = 3, b = 4, c = 1

4

Substitute the values into the formula.

x=

5

Simplify and solve for x.

1

Write the equation.

2

Write the quadratic formula.

x=

3

State the values for a, b and c.

where a = -3, b = -6, c = -1

4

Substitute the values into the formula.

x=

5

Simplify the fraction.

=

6

Solve for x.

x ö -1.82   or  x ö -0.18

− b ± b 2 − 4 ac 2a − 4 ± (4)2 − (4 × 3 × 1) 2×3

−4 ± 4 6 −4 ± 2 = 6 −4 − 2 −4 + 2 x=   or x = 6 6 1 x = -1 x = - 3 =

b -3x2 - 6x - 1 = 0

− b ± b 2 − 4 ac 2a −( −6) ± 36 − 4 × −3 × −1 2 × −3

6 ± 24 −6 6±2 6 = −6 3± 6 = −3 3+ 6 3− 6 x=   or  −3 −3

Note: When asked to give an answer in exact form, you should simplify any surds as necessary. ■■

256

If the value inside the square root sign is negative, then there are no solutions to the equation.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

The■quadratic■formula■x = form■ax2■+■bx■+■c■=■0. exerCise

8b inDiViDuAl pAthWAys

− b ± b 2 − 4 ac ■can■be■used■to■solve■quadratic■equations■of■the■ 2a

the quadratic formula fluenCy

Introducing the quadratic formula doc-5062

1 State■the■values■for■a,■b■and■c■in■each■of■the■following■equations■of■the■form■ax2■+■bx■+■c■=■0. a 3x2■-■4x■+■1■=■0 a■=■3,■b■=■-4,■c■=■1 b 7x2■-■12x■+■2■=■0 a■=■7,■b■=■-12,■c■=■2 2 a■=■1,■b■=■-5,■c■=■7 c 8x ■-■x■-■3■=■0 d x2■-■5x■+■7■=■0 a■=■8,■b■=■-1,■c■=■-3 2 a■=■5,■b■=■-5,■c■=■-1 e 5x ■-■5x■-■1■=■0 f 4x2■-■9x■-■3■=■0 a■=■4,■b■=■-9,■c■=■-3 a■=■12,■b■=■-29,■c■=■103 h 43x2■-■81x■-■24■=■0 a■=■43,■b■=■-81,■c■=■-24 g 12x2■-■29x■+■103■=■0 i 6x2■-■15x■+■1■=■0 a■=■6,■b■=■-15,■c■=■1

Activity 8-B-2

2 We 6a ■Use■the■quadratic■formula■to■solve■each■of■the■following■equations.■Give■exact■

eBook plus

Activity 8-B-1

Practice using the quadratic formula doc-5063 Activity 8-B-3

Using the quadratic formula doc-5064

eBook plus

3

Digital doc

SkillSHEET 8.6 doc-5262

eBook plus

4

Digital doc

SkillSHEET 8.7 doc-5263

5

6 ✔

7

answers. −3 ± 13 −5 ± 21 a x2■+■5x■+■1■=■0 a■ b x2■+■3x■-■1■=■0 g 9 ± 73 ■ h■ 3 ± 2 3 ■ b■ 2 2 2 2 c x ■-■5x■+■2■=■0 d x2■-■4x■-■9■=■0 2 2 5 ± 17 e x ■+■2x■-■11■=■0 f x ■-■7x■+■1■=■0 i −4 ± 31 ■ j■ 1 ± 21 ■ d■ 2 ± 13   c 2 2 2 g x ■-■9x■+■2■=■0 h x2■-■6x■-■3■=■0 5 ± 33 2 2 7 ± 45 i x ■+■8x■-■15■=■0 j -x ■+■x■+■5■=■0 k ■ l■ −1 ± 2 2 e −1 ± 2 3 ■ f■ 2 2 k -x2■+■5x■+■2■=■0 l -x2■-■2x■+■7■=■0 We 6b ■Use■the■quadratic■formula■to■solve■each■of■the■following■equations.■Give■approximate■ answers■rounded■to■2■decimal■places. a 3x2■-■4x■-■3■=■0 -0.54,■1.87 b 4x2■-■x■-■7■=■0 -1.20,■1.45 2 -4.11,■0.61 -0.61,■0.47 c 2x ■+■7x■-■5■=■0 d 7x2■+■x■-■2■=■0 2 0.16,■6.34 e 5x ■-■8x■+■1■=■0 0.14,■1.46 f 2x2■-■13x■+■2■=■0 g -3x2■+■2x■+■7■=■0 -1.23,■1.90 h -7x2■+■x■+■8■=■0 -1.00,■1.14 2 i -12x ■+■x■+■9■=■0 -0.83,■0.91 j -6x2■+■4x■+■5■=■0 -0.64,■1.31 2 k -11x ■-■x■+■1■=■0 l -4x2■-■x■+■7■=■0 -1.45,■1.20 -0.35,■0.26 2 m -2x ■+■12x■-■1■=■0 n -5x2■+■x■+■3■=■0 -0.68,■0.88 0.08,■5.92 2 0 -3x ■+■5x■+■2■=■0 -0.33,■2.00 mC ■The■solutions■of■the■equation■3x2■-■7x■-■2■=■0■are: a 1,■2 B 1,■-2 ✔ c -0.257,■2.59 D -0.772,■7.772 E -1.544,■15.544 mC ■In■the■expansion■of■(6x■–■5)(3x■+■4),■the■coeffi■cient■of■x■is: a 18 B -15 ✔ c 9 D 6 E -2 mC ■In■the■expanded■form■of■(x■−■2)(x■+■4),■which■of■the■following■is■incorrect? a The■value■of■the■constant■is■-8. B The■coeffi■cient■of■the■x■term■is■-6. c The■coeffi■cient■of■the■x■term■is■2. D The■coeffi■cient■of■the■x2■term■is■1. E The■expansion■shows■this■to■be■a■trinomial■expression. mC ■An■exact■solution■to■the■equation■x2■+■2x■–■5■=■0■is: a -3.449 D

2 + −16 2

B -1■+■ 24 E

✔ c

-1■+■ 6

2 + 24 2 Chapter 8 Quadratic equations

257

number AND algebra • Linear and non-linear relationships

Understanding a 0.5, 3 b 0, 5 8 Solve each of the following equations using any suitable method. Round to 3 decimal places c -1, 3 where appropriate. d 0.382, 2.618 e 0.298, 6.702 a 2x2 - 7x + 3 = 0 b x2 - 5x = 0 c x2 - 2x - 3 = 0 2 2 f 2, 4 d x - 3x + 1 = 0 e x - 7x + 2 = 0 f x2 - 6x + 8 = 0 g No real solution 2 2 g x - 5x + 8 = 0 h x - 7x - 8 = 0 i x2 + 2x - 9 = 0 h -1, 8 2 2 j 3x + 3x - 6 = 0 k 2x + 11x - 21 = 0 l 7x2 - 2x + 1 = 0 i -4.162, 2.162 2 2 m -x + 9x - 14 = 0 n -6x - x + 1 = 0 o -6x2 + x - 5 = 0 j -2, 1 k -7, 1.5 Reasoning l No real solution m 2, 7 9 The surface area of a closed cylinder is given by the formula SA = 2πr(r + h), where r cm is 2p r2 + 14p r - 231 = 0 the radius of the can and h cm is the height. n - 12 , 13 o No real solution

  The height of a can of wood finish is 7  cm and its surface area is 231  cm2. a Substitute values into the formula to form a quadratic equation using the pronumeral, r. b Use the quadratic formula to solve the equation and, hence, find the radius of the can. Round the answer to 1 decimal place. 3.5 cm c Calculate the area of the paper label on the can. Round the answer to the nearest square 154 cm2 centimetre. x 10 To satisfy lighting requirements, a window must have an area x(x + 30) of 1500  cm2. 30 cm a Find an expression for the area of the window in terms of x. b Write an equation so that the window satisfies the lighting x(x + 30) = 1500 requirements. x c Use the quadratic formula to solve the equation and find x to the 265 mm nearest mm. 11 Two competitive neighbours build rectangular pools that cover the same area but are different

shapes. Pool A has a width of (x + 3) m and a length that it 3 m longer than its width. Pool B has a length that is double the width of Pool A. The width of Pool B is 4 m shorter than its length. a Find the exact dimensions of each pool if their areas are the same. b Verify mathematically that the areas are the 2 2 Pool A: 3 3 m by 6 3 m; The area of each is 24 49 m2. same. reflection    1 1 Pool B: 3 3 m by 7 3 m 12 A block of land is in the shape of a rightWhat kind of answer will you get if angled triangle with a perimeter of 150 m and a the value inside the square root sign hypotenuse of 65 m. Determine the lengths of the in the quadratic formula is zero? other two sides. 25 m, 60 m

8c

Solving quadratic equations by inspecting graphs y

■■ ■■

■■

■■

258

The graph of a quadratic equation is called a parabola. To solve quadratic equations graphically means to find the values of x where y = 0 or where the parabola intercepts the x-axis. A quadratic equation written in standard form has solutions when the graph of y = ax2 + bx + c is equal to zero. In this section, we will find solutions (also called roots or zeros) of quadratic equations by inspecting their corresponding graphs.

Maths Quest 10 for the Australian Curriculum

y = ax2 + bx + c

(0, c) 0

x Solutions/roots/zeros to ax2 + bx + c = 0

number AND algebra • Linear and non-linear relationships

Worked Example 7

Determine the solution (or roots) of each of the following quadratic equations by inspecting their corresponding graphs. Round answers to 1 decimal place where appropriate. a x2 + x - 2 = 0 b 2x2 - 4x - 5 = 0 Think

Write/Draw

a The graph of y = x2 + x - 2 is equal to zero

a

when y = 0. Look at the graph to find where y = 0; that is, where it intersects the x-axis.

y 3 2 1 0 -3 -2 -1 -1 -2 -3

1 2 3x y = x2 + x - 2

x2 + x - 2 = 0 From the graph, the solutions are x = 1 and x = -2. b The graph of y = 2x2 - 4x - 5 is equal to zero

b

when y = 0. Look at the graph to see where y = 0; that is, where it intersects the x-axis. By sight, we can only give estimates of the solutions.

y 6 4 2 0 -3 -2 -1 -2

1 2 3x

-4 -6 -8

2x2 - 4x – 5 = 0 From the graph, the solutions are x ö -0.9 and x ö 2.9.

■■

Some quadratic equations have only one solution. For example, the graph of x2 - 4x + 4 = 0 has the one solution of x = 2. That is, the graph of equation touches the x-axis only at x = 2. y y = x2 - 4x + 4 5

-2 ■■

0

2

4

x y

There are also quadratic equations that have no real solutions. For example, the graph of y = 3x2 - 4x + 4 does not intersect the x-axis and so 3x2 - 4x + 4 = 0 has no real solutions (that is, no solutions that are real numbers).

10 5

-2

y = 3x2 - 4x + 4 x 0 2

Chapter 8 Quadratic equations

259

number AND algebra • Linear and non-linear relationships

Confirming solutions ■■

It is possible to confirm the solutions obtained by sight. As we saw with linear equations, this is achieved by substituting the solution or solutions into the original quadratic equation. If both sides of the equation are equal, the solution is correct.

Worked Example 8

Confirm, by substitution, the solutions obtained in Worked example 7. a x2 + x - 2 = 0; solutions: x = 1 and x = -2 b 2x2 - 4x - 5 = 0; solutions: x ö -0.9 and x ö 2.9 Think a

b

Write a

When x = 1, x2 + x - 2 = 12 + 1 - 2

1

Write the left-hand side of the equation and substitute x = 1 into the expression.

2

Simplify to check that the expression is equal to zero.

3

Write the expression and substitute x = -2.

  When x = -2, x2 + x - 2 = (-2)2 + -2 - 2

4

Simplify to check that the expression is equal to zero.



1

Write the left-hand side of the equation and substitute x = -0.9 into the expression.

2

Simplify. As the x-values are only estimates, the results should be reasonably close to zero.

= 1.62 + 3.6 - 5 = -0.22 As -0.9 is only an estimate, the left-hand side expression can be said to be close to zero.

3

Write the expression and substitute x = 2.9 into the expression.

When x = 2.9, 2x2 - 4x - 5 = 2 ì (2.9)2 - 4 ì 2.9 - 5

4

Simplify to check that the expression is reasonably close to zero.

= 16.82 - 11.6 - 5 = -0.22 As 2.9 is only an estimate, the left-hand side expression can be said to be close to zero.

  = 0   Solution is confirmed.

b

=4-2-2 = 0   Solution is confirmed.

When x = -0.9, 2x2 - 4x - 5 = 2 ì (-0.9)2 - 4 ì -0.9 - 5

Worked Example 9

A golf ball hit along a fairway follows the path shown in the graph. The height, h metres after it has 1 travelled x metres horizontally, follows the rule h = - 270 (x2 - 180x). Use the graph to find how far the ball landed from the golfer. h 30

1 2 h = - ––– 270 (x - 180x)

20 10 0 260

Maths Quest 10 for the Australian Curriculum

90

180 x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

think

Write

On■the■graph,■the■ground■is■represented■by■the■x-axis■ since■this■is■where h =■0.■The■golf■ball■lands■when■ the■graph■intersects■the■x-axis.

The■golf■ball■lands■180■■m■from■the■golfer.

remember

1.■ The■solution(s)■(also■known■as■roots■or■zeros)■of■a■quadratic■equation■can■be■found■by■ inspecting■the■graph■of■the■equation.■You■may■need■to■draw■the■graph■of■the■equation■ fi■rst■using■a■calculator■or■graphing■software. 2.■ The■root■of■any■graph■is■the■x-intercept■or■the■x-coordinate■of■the■point■where■the■graph■ crosses■the■x-axis. 3.■ The■roots■or■intercepts■of■the■quadratic■graph y =■ax2■+■bx■+ c are■the■solutions■to■the■ equation■ax2■+■bx■+ c =■0. exerCise

8C inDiViDuAl pAthWAys eBook plus

Activity 8-C-1

solving quadratic equations by inspecting graphs fluenCy 1 We 7 ■Determine■the■roots■of■each■of■the■following■quadratic■equations■by■inspecting■the■

corresponding■graphs.■Round■answers■to■1■decimal■place■where■appropriate. b x2■-■11x■+■10■=■0 x =■1, x =■10 x =■-2, x =■3

a x2■- x -■6■=■0

Finding solutions to quadratic equations by inspecting graphs doc-5065 Activity 8-C-2

Solving quadratic equations by inspecting graphs doc-5066 Activity 8-C-3

Harder solutions to quadratic equations by inspecting graphs doc-5067



y

y

12 8 4

8

-6 -4 -2 0 2 4 6 x -4 2 -8 y = x - x - 6

x =■-5, x =■5 c -x2■+■25■=■0





2 4 6 x

e x2■-■3x■-■4■=■0 x =■-1, x =■4 y 2 15 y = x - 3x - 4 10 5

-1 0 1 2 3 4 5 6 x -10

2 4 6 8 10 12 x y = x2 - 11x + 10

d 2x2■-■8x■+■8■=■0

x =■2

y y = 2x2 - 8x + 8 20 10

y y = -x2 + 25 30 20 10 -6 -4 -2 0 -10

-2 0 -8 -16 -24



-1 0 -10

1 2 3 4 5 x

x ö■-1.4, x ö■4.4 f x2■-■3x■-■6■=■0 y 2 15 y = x - 3x - 6 10 5 -1 0 1 2 3 4 5 6 x -10 Chapter 8 Quadratic equations

261

number AND algebra • Linear and non-linear relationships g x2 + 15x - 250 = 0 x = -25, x = 10 y

h -x2 = 0 y 5 0

200 100 -30 -20 -10 0 -100 -200 -300 -400 y = x2 + 15x - 250

i

10

-5

x



5 x

-5

j

2x2 + x - 3 = 0 x ö -1.5, x = 1

5

-5

y = -x2

-10

x2 + x - 3 = 0 x ö -2.3, x ö 1.3

-4 -2 0

x=0

5

2 4

x

-2 -1 0

y = x2 + x - 3

-5

1 2

x

y = 2x2 + x - 3

Understanding 2   WE 8  Confirm, by substitution, the solutions obtained in question 1. 3   WE9  A golf ball hit along a fairway follows the path shown in the graph.  Confirm by substitution of above values into quadratic equations.

h 28

1 2 h = - ––– 200 (x - 150x)

0

150 x

75

The height, h metres after it has travelled x metres horizontally, follows the rule 1

h = - 200 (x2 - 150x). Use the graph to find how far the ball lands from the golfer. 150 m 4 A ball is thrown upwards from a building and follows the path shown in the graph until it lands on the ground. h 25

h = -x2 + 4x + 21

21

0

2

7

x

The ball is h metres above the ground when it is a horizontal distance of x metres from the building. The path of the ball follows the rule h = -x2 + 4x + 21. Use the graph to find how far from the building the ball lands. 7 m 262

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships reAsoning T he■answer■for■ 5 a■ ■Use■a■graphics■calculator■to■sketch■the■two■functions■ part■b■are■the■ x-coordinates■of■the■ y■=■2x2■-■5x■-■3■and■y■=■-x2■–■3x. intersection■of■the■ b Use■the■calculator■to■fi■nd■the■solution■to■ ■ x■=■-0.72,■1.39 quadratic■in■part■a. 2 2

2x ■-■5x■-■3■=■-x ■-■3x,■correct■to■2■decimal■places. c Comment■on■your■answers■to■parts■a■and■b.

8D

refleCtion 

 

What does ‘the root of a graph’ mean?

finding solutions to quadratic equations by interpolation and using the discriminant interpolation

eBook plus

■■

Interactivity Solving by interpolation

Consider■the■quadratic■equation■x2■-■3x■–■6■=■0. Using■a■graphing■calculator■or■graphing■software■to■sketch■the■graph■of■the■equation y =■x2■-■3x■–■6,■we■can■see■there■is■a■solution■between x =■4■and x =■5. y 20

int-1147

y = x2 - 3x - 6

10 -2 -1 0 -10

1 2 3 4 5 6

x

This■can■be■confi■rmed■using■the■following■logic: Step■1.■ The■value■of y =■x2■-■3x■-■6■when x =■4■can■be■expressed■as y(4)■=■42■-■3■ì■4■-■6■ ■ ■ =■-2 ■ The■value■of■y■=■x2■-■3x■-■6■when■x■=■5■can■be■expressed■as y(5)■=■52■-■3■ì■5■-■6■ ■ ■ =■4 Since■the■graph■moves■from■below■the x-axis■at x =■4,■to■above■the x-axis■at x =■5,■it■is■ reasonable■to■assume■that■there■is■a■solution■somewhere■between x =■4■and x =■5. Step■2.■ Choose■a■value■between x =■4■and x =■5;■for■example,■4.5. ■■

■ ■

y(4)■=■42■-■3■ì■4■-■6■ =■-2 y(4.5)■=■4.52■-■3■ì■4.5■-■6■ =■0.75 Since■the■graph■moves■from■below■the x-axis■at x =■4,■to■above■the x-axis■at x =■4.5,■it■is■ reasonable■to■assume■there■is■a■solution■somewhere■between x =■4■and x =■4.5.

■■

Step■3.■ ■Repeat■step■2,■checking■that■your■equations■are■approaching■zero.■The■solution■is■ approximately x =■4.372. Repeat■the■process■to■fi■nd■the■other■root,■somewhere■between■-2■and■0. Notes 1.■ ■This■process■can■also■be■done■on■a■spreadsheet. 2.■ ■A■CAS■calculator■can■also■help■you■to■fi■nd■the■roots.■Rather■than■using■the■trace■function,■ try■using■the■table■function■after■drawing■the■graph.■Step■up■in■increments■of■0.1■then■0.01■ to■pinpoint■the■solution. Chapter 8 Quadratic equations

263

number AND algebra • Linear and non-linear relationships

Using the discriminant ■■

■■ ■■

−b ± b 2 − 4 ac gives the solutions to the general quadratic equation 2a 2 ax + bx + c = 0. By examining the expression under the square root sign, b2 - 4ac, we can determine the number and type of solutions produced and, hence, the number of x-intercepts to expect when the quadratic equation is graphed. The expression b2 - 4ac is known as the discriminant and is denoted by the symbol D (delta). The formula x =

Case 1: D < 0 If x2 + 2x + 3 = 0, then a = 1, b = 2 and c = 3. x=

D = b2 - 4ac = 22 - (4 ì 1 ì 3) = -8

−b ± b 2 − 4 ac 2a

− 2 ± −8 2 If the discriminant is less than zero, there are no real solutions because the expression under the square root sign is negative. It is not possible to find a real number that is the square root of a negative number. Hence, the graph of y = x2 + 2x + 3 will not intersect the x-axis; i.e., there will be no x-axis intercepts. ■■

=

Case 2: D = 0 If 4x2 + 12x + 9 = 0, then a = 4, b = 12 and c = 9. D = b2 - 4ac = 122 - (4 ì 4 ì 9) = 144 - 144 =0

−b ± b 2 − 4 ac 2a −12 ± 0 = 2×4

x=

12

= - 8 3

= - 2 If the discriminant is equal to zero then the two solutions are the same. That is, if b2 - 4ac = 0, −b −b + 0 −b − 0 then x = and x = . This may be regarded as one rational solution that is equal to . 2a 2a 2a One solution indicates that the quadratic trinomial is a perfect square that can be factorised easily using the perfect squares rule; that is, 4x2 + 12x + 9 = (2x + 3)2. Hence, the graph of y = 4x2 + 12x + 9 will touch the x-axis once. ■■

Case 3: D > 0 If the discriminant is positive, there are two distinct solutions. We can determine more information than this by checking whether the discriminant is also a perfect square. (a)  If 2x2 - 7x - 4 = 0, then a = 2, b = -7 and c = -4. D = b2 - 4ac = (-7)2 - (4 ì 2 ì -4) = 49 + 32 = 81

2 x = − b ± b − 4 ac 2a

7 ± 81 2×2 7±9 = 4 1 x = 4 or x = - 2 =

If the discriminant is positive and a perfect square, the quadratic trinomial will have ■ two rational solutions. This means the quadratic trinomial can be factorised easily; that is,■ 2x2 - 7x - 4 = (2x + 1)(x - 4). 264

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

(b)  If x2 - 5x - 1 = 0 then a = 1, b = -5 and c = -1. D = b2 - 4ac = (-5)2 - (4 ì 1 ì -1) = 25 + 4 = 29

x =

−b ± b 2 − 4 ac 2a

x = 5 ± 29 2 ×1

5 ± 29 2 If the discriminant is positive but not a perfect square, the factors are irrational and the quadratic formula must be used to find the two irrational (surd) solutions. Hence, the graphs of both equations shown in (a) and (b) will each have two x-intercepts. The table below summarises the three cases. x =

■■

D > 0 (positive) D < 0 (negative)

D = 0 (zero)

Not a perfect square

Perfect square

Number of solutions

No solutions

1 rational solution

2 rational solutions

Graph

Graph does not cross or touch the x-axis

Graph touches the x-axis

Graph crosses the x-axis twice

y

y

2 irrational (surd) solutions

y

y

x x

a

b

x

-b -a

a

b

x

 

Worked Example 10

By using the discriminant, determine whether the following equations have:   i two rational solutions ii two irrational solutions iii one rational solution (two equal solutions) iv no real solutions. a x2 - 9x - 10 = 0 b x2 - 2x - 14 = 0 2 c x - 2x + 14 = 0 d x2 + 14x = -49 Think a

Write a x2 - 9x - 10 = 0

1

Write the equation.

2

Identify the coefficients a, b and c.

a = 1, b = -9, c = -10

3

Find the discriminant.

D = b2 - 4ac = (-9)2 - (4 ì 1 ì -10) = 121

4

Identify the number and type of solutions when D > 0 and is a perfect square.

The equation has two rational solutions.

Chapter 8 Quadratic equations

265

number AND algebra • Linear and non-linear relationships b

c

d

b x2 - 2x - 14 = 0

1

Write the equation.

2

Identify the coefficients a, b and c.

a = 1, b = -2, c = -14

3

Find the discriminant.

D = b2 - 4ac = (-2)2 - 4 ì 1 ì -14 = 60

4

Identify the number and type of solutions when D > 0 but not a perfect square.

The equation has two irrational solutions.

1

Write the equation.

2

Identify the coefficients a, b and c.

a = 1, b = -2, c = 14

3

Find the discriminant.

D = b2 - 4ac = (-2)2 - (4 ì 1 ì 14) = -52

4

Identify the number and type of solutions when D < 0.

The equation has no real solutions.

1

Write the equation, then rewrite it so the right side equals zero.

c

x2 - 2x + 14 = 0

d x2 + 14x = -49

x2 + 14x + 49 = 0

2

Identify the coefficients a, b and c.

a = 1, b = 14, c = 49

3

Find the discriminant.

D = b2 - 4ac = 142 - (4 ì 1 ì 49) =0

4

Identify the number and types of solutions when D = 0.

The equation has 1 rational solution.

Remember, the number of solutions of a quadratic equation is the same as the number of x-intercepts obtained when the equation is graphed.

remember

1. Interpolation can be used to find approximate solutions to quadratic equations. 2. The discriminant of a quadratic equation is given by D = b2 - 4ac. 3. If D < 0, there are no real solutions to the equation. 4. If D = 0, there is only one rational solution (or two equal solutions) to the equation. The equation can be factorised easily. 5. If D > 0, there are two distinct solutions to the equation. (a) If the discriminant is a perfect square, the solutions are rational and the equation can be factorised easily. (b) If the discriminant is not a perfect square, the solutions are irrational and the equation can be solved using the quadratic formula or the completing the square method. 6. The number of solutions of a quadratic equation corresponds to the number of x-intercepts obtained when the equation is graphed. 266

Maths Quest 10 for the Australian Curriculum

7.■ This■information■can■be■summarised■in■the■following■table: D > 0 (positive) D < 0 (negative)

3 a■ No■real■solutions■ c 2■rational■solutions■ e 2■irrational■solutions■ g 2■irrational■solutions■ i No■real■solutions■ k 2■irrational■solutions■

b■ d■ f■ h■ j■ l■

1■rational■solution 1■rational■solution 1■rational■solution No■real■solutions 2■irrational■solutions 2■rational■solutions

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Perfect square

Not a perfect square

Number of solutions

No■solutions

1■rational■ solution

2■rational■ solutions

Graph

Graph■does■not■ cross■or■touch■ the■x-axis

Graph■touches■ the■x-axis

Graph■crosses■the■x-axis■twice

2■irrational■ (surd)■solutions



exerCise

D = 0 (zero)

8D inDiViDuAl pAthWAys eBook plus

Activity 8-D-1

Finding solutions to quadratic equations by interpolation doc-5068 Activity 8-D-2

Harder solutions to quadratic equations by interpolation doc-5069 Activity 8-D-3

Difficult solutions to quadratic equations by interpolation doc-5070

finding solutions to quadratic equations by interpolation and using the discriminant fluenCy 1 Use■a■graphing■calculator■or■graphing■software■to■sketch■the■graph■of■each■of■the■following■

equations■and■then■use■the■process■of■interpolation■to■fi■nd■approximate■solutions. a x2■+■3x■-■7■=■0 b 3x2■-■2x■-■4■=■0 -0.87,■1.5 c 2x2■+■7x■-■10■=■0 -4.6,■1.1 ■ -4.5,■1.5 2 Determine■the■discriminant■for■each■of■the■following■equations. a x2■-■3x■+■5 ■ -11 b 4x2■-■20x■+■25■=■0 c x2■+■9x■-■22■=■0 169 0 2 2 0 37 d 9x ■+■12x■+■4 e x ■+■3x■-■7■=■0 f 25x2■-■10x■+■1■=■0 0 2 2 g 3x ■-■2x■-■4■=■0 52 h 2x ■-■5x■+■4■=■0 -7 i x2■-■10x■+■26■=■0 -4 2 2 109 j 3x ■+■5x■-■7■=■0 k 2x ■+■7x■-■10■=■0 129 l x2■-■11x■+■30■=■0 1 3 We10 ■By■using■the■discriminant,■determine■whether■the■equations■in■question■2■have: i two■rational■solutions ii two■irrational■solutions iii one■rational■solution■(two■equal■solutions) iv no■real■solutions. 4 With■the■information■gained■from■the■discriminant,■use■the■most■effi■cient■method■to■solve■each■ equation■in■question■2.■Where■appropriate,■round■answers■to■3■decimal■places. unDerstAnDing

a■ No■real■solutions b

2 12

c -11,■2 d - 23 ■

■



5 Consider■the■equation■3x2■+■2x■+■7■=■0. a What■are■the■values■of■a,■b■and■c? ■ a■=■3,■b■=■2,■c■=■7 b What■is■the■value■of■b2■-■4ac? -80 c How■many■real■solutions,■and■hence■x-intercepts,■are■there■for■this■equation?

No■real■solutions

6 Consider■the■equation■-6x2■+■x■+■3■=■0. a What■are■the■values■of■a,■b■and■c? a■=■-6,■b■=■1,■c■=■3 −3 ± 37  ö -4.541, e b What■is■the■value■of■b2■-■4ac? 73 2 c How■many■real■solutions,■and■hence■x-intercepts,■are■there■for■this■equation? 2■real■solutions ■ ■ 1.541 d With■the■information■gained■from■the■discriminant,■use■the■most■effi■cient■method■to■solve■ f 15 the■equation.■Give■an■exact■answer. 1 ± 73 2 1 ± 13 12  ö -0.869, 7 mC ■The■discriminant■of■the■equation■x ■-■4x■-■5■=■0■is: g 3 B 11 c 4 D 0 E -4 ✔ a 36 ■ ■ 1.535 8 mC ■Which■of■the■following■quadratic■equations■has■two■irrational■solutions? a x2■-■8x■+■16■=■0 B 2x2■-■7x■=■0 ✔ c x2■+■8x■+■9■=■0 h No■real■solutions 2■ 2■ i No■real■solutions D x -■4■=■0 E x -■6x■+■15■=■0 Chapter 8 Quadratic equations

267

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 9 mC ■The■equation■x2■=■2x■-■3■has: a two■rational■solutions ✔ c no■solutions E one■rational■and■one■irrational■solution

B exactly■one■solution D two■irrational■solutions

reAsoning 10 Find■the■value■of■k if■x2■-■2x - k■=■0■has■one■solution. k■=■-1 11 Find■the■values■of■m■for■which■mx2■-■6x■+■5■=■0■has■one■solution. m■=■1,■8 12 Find■the■values■of■n when■x2■-■3x■-■n■=■0■has■two■solutions. n■>■- 49

13 Show■that■3x2■+■px■-■2■=■0■will■have■real■solutions■for■all■values■of■p. p2■can■only■ give■a■positive■ number■which,■ when■added■to■ 24,■is■always■ a■positive■ solution.

eBook plus

Digital doc

WorkSHEET 8.2 doc-5264

268

14 The■path■of■a■dolphin■as■it■leaps■out■of■the■water■can■be■modelled■by■the■equation■

h■=■-0.4d2■+■d,■where■h■is■the■dolphin’s■height■above■water■and■d■is■the■horizontal■distance■ from■its■starting■point.■Both■h■and■d■are■in■metres.

a How■high■above■the■water■is■the■dolphin■when■it■has■travelled■2■■m■horizontally■from■its■ 0.28■m starting■point? ■ 0.4■m b What■horizontal■distance■has■the■dolphin■covered■when■it■fi■rst■reaches■a■height■of■25■■cm? c What■horizontal■distance■has■the■dolphin■covered■when■it■next■reaches■a■height■of■25■■cm?■ 2.20■m Explain■your■answer. d What■horizontal■distance■does■the■dolphin■cover■in■one■leap?■(Hint:■What■is■the■value■of■ h■when■the■dolphin■has■completed■its■leap?) 2.5■m e Can■this■dolphin■reach■a■height■of: i 0.5■■m Yes ii 1■■m■during■a■leap? No refleCtion    ■How■can■you■determine■this■without■actually■solving■

the■equation? f Find■the■greatest■height■the■dolphin■reaches■during■ a■leap. 1.25■m

maths Quest 10 for the Australian Curriculum

What does the discriminant tell us?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

8e

solving a quadratic equation and a linear equation simultaneously There■are■occasions■when■it■is■necessary■to■fi■nd■the■intersections■points■(if■any)■of■a■quadratic■ and■a■linear■equation. ■■ There■are■three■possible■outcomes■in■this■situation. 1.■ The■straight■line■can■intersect■twice■with■the■parabola. y ■■

eBook plus

Interactivity Simultaneous quadratic equations

int-2784

0

2.■ ■The■straight■line■can■be■a■tangent■to■the■parabola.■In■ this■case■the■straight■line■touches■the■parabola■at■ one■point.

y

0

3.■ ■The■straight■line■may■not■intersect■at■all■with the■parabola.

■■ ■■ ■■ ■■

x

y

0

■■

x

x

A■quadratic■equation■can■be■solved■simultaneously■with■a■linear■equation■using■the■ substitution■method. The■x2■term■in■the■quadratic■equation■makes■it■impractical■to■use■the■elimination■method. Each■equation■is■best■written■with■y■as■the■subject■so■that■the■right-hand■side■of■each■equation■ can■then■be■set■equal■to■one■another. The■result■can■then■be■simplifi■ed■to■produce■a■new■quadratic■equation. The■quadratic■equation■will■have■two■solutions■if■the■straight■line■cuts■the■parabola■twice,■ one■solution■if■the■straight■line■is■a■tangent■to■the■parabola,■and■no■solution■if■the■line■does■ not■intersect■the■parabola.

Chapter 8 Quadratic equations

269

number AND algebra • Linear and non-linear relationships

Worked Example 11

Solve the simultaneous equation pair y = x2 + 2x + 2 and y = 7 - 2x. Think

Write

1

Write the equations, one under the other, and number them.

y = x2 + 2x + 2 y = 7 - 2x

2

Both equations are written with y as the subject, so equate them.

x2 + 2x + 2 = 7 - 2x

3

Move every term to the left-hand side to create a new quadratic equation.

x2 + 4x - 5 = 0

4

Factorise and use the Null Factor Law to solve the quadratic equation.

(x + 5)(x - 1) = 0 x + 5 = 0   or  x - 1 = 0 x = -5  or x=1

5

Substitute each answer for x into equation [2] to find the corresponding values of y.

Substituting x = -5 into [2]:   y = 7 - 2(-5)   y = 17 Substituting x = 1 into [2]:   y = 7 - 2(1)   y=5

6

Answer the question.

Solution: x = -5, y = 17  (-5, 17)  or x = 1, y = 5  (1, 5)

7

Check the answers by substituting the point of intersection into equation [1].

Check: Substitute (-5, 17) into y = x2 + 2x + 2 RHS = (-5)2 + 2(-5) + 2 = 25 - 10 + 2 = 17 = LHS Substitute (1, 5) into y = x2 + 2x + 2 RHS = (1)2 + 2(1) + 2 =1+2+2 =5 = LHS Both solutions are correct.

■■ ■■

[1] [2]

When there is only one solution to the simultaneous equation pair, the straight line is a tangent to the parabola. When this situation arises, the new quadratic equation formed will be a perfect square.

Worked Example 12

Solve the simultaneous equation pair y = x2 - 5x + 2 and y = x - 7. Think

270

Write

1

Write the equations, one under the other, and number them.

y = x2 - 5x + 2   [1] y = x - 7 [2]

2

Both equations are written with y as the subject, so equate them.

x2 - 5x + 2 = x - 7

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

3

Move every term to the left-hand side to create a new quadratic equation.

x2 - 6x + 9 = 0

4

Factorise and use the Null Factor Law to solve the quadratic equation.



5

Substitute for x into equation [2] to find the corresponding value of y.

Substituting x = 3 into [2]: y=3-7 = -4

6

Answer the question.

Solution: x = 3, y = -4  (3, -4)

7

Check the answer by substituting the point of intersection into equation [1].

Check: Substitute into y = x2 - 5x + 2 RHS = (3)2 -5(3) + 2 = 9 - 15 + 2 = -4 = LHS The solution is correct.

■■

(x - 3)2 = 0 x-3=0 x=3

When there is no intersection of the parabola and the straight line, the discriminant has a value less than zero.

Worked Example 13

Show that the equations y = x2 + x + 4 and y = 2x - 1 have no solution when solved simultaneously. Think

Write

1

Write the equations, one under the other, and number them.

y = x2 + x + 4 y = 2x - 1

2

Both equations are written with y as the subject, so equate them.

x2 + x + 4 = 2x - 1

3

Move every term to the left-hand side to create a new quadratic equation.

x2 - x + 5 = 0

4

If the equations are to have no solution, ■ then D < 0.

D = b2 - 4ac = (-1)2 - 4 ì 1 ì 5 = 1 - 20 = -19 <0

5

Draw a conclusion.

There is no solution to the pair of simultaneous equations as the discriminant of the resulting quadratic is less than 0.

[1] [2]

remember

1. To solve a quadratic equation with a linear equation we use the substitution method. 2. Make y the subject of both the quadratic equation and the linear equation, and then equate the right-hand side expressions formed. This will leave a new quadratic equation to solve for x. Chapter 8 Quadratic equations

271

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 3. ■ If■the■new■quadratic■equation■has■two■solutions,■both■must■be■substituted■back■into■one■

of■the■original■equations■to■fi■nd■the■corresponding■values■of■y. 4.■ In■some■cases■there■will■only■be■one■value■of■x.■This■occurs■when■the■linear■equation■is■ a■tangent■to■the■parabola. 5.■ In■some■cases■there■will■be■no■solution■to■a■straight■line■and■a■parabola.■If■this■is■the■ case,■the■discriminant■of■the■equation■formed■when■solving■will■be■less■than■0,■that■is,■ b2■-■4ac■<■0. exerCise

solving a quadratic equation and a linear equation simultaneously

8e inDiViDuAl pAthWAys

(-4,■1)■and■(1,■6) 1 We11 ■Solve■the■simultaneous■equation■pair■y■=■x2■+■4x■+■1■and■y■=■x■+■5. 2 Solve■each■of■the■following■pairs■of■simultaneous■equations. a y■=■x2■+■5x■+■16 b y■=■x2■+■x■-■7 c y■=■x2■-■7x■+■10 (-2,■-5)■and■(6,■35) y■=■x■-■5 (3,■-2)■and■(5,■0) y■=■4■-■2x ■ (-4,■12)■and■(-3,■10) y■=■5x■+■5

eBook plus

Activity 8-E-1

Introduction to quadratic and linear solutions doc-5071

3 We12 ■Solve■the■simultaneous■equation■pair■y■=■x2■-■3x■+■6■and■y■=■x■+■2. (2,■4) 4 We13 ■Show■that■the■equations■y■=■x2■-■4x■+■7■and■y■=■2x■-■4■have■no■solution■when■solved■ simultaneously.   D = -8

Activity 8-E-2

5 Solve■each■of■the■following■pairs■of■simultaneous■equations. a y■=■x2■-■x■-■2 b y■=■x2■+■4x■-■5 c y■=■2x■+■8 (-2,■4)■and■(5,■18) y■=■x■-■7 (-2,■-9)■and■(-1,■-8) d y■=■x2■+■6x■+■11 e y■=■x2 f y■=■4■-■2x (-7,■18)■and■(-1,■6) y■=■4x■-■3 (1,■1)■and■(3,■9)

Practising quadratic and linear solutions doc-5072 Activity 8-E-3

Tricky quadratic and linear solutions doc-5073

y■=■x2■-■4x■+■10 y■=■4x■-■6 (4,■10) y■=■x2■-■9x■+■12 y■=■2x■+■2 (1,■4)■and■(10,■22)

(-3,■1)■and■(-2,■1) 6 Solve■the■simultaneous■pair■y■=■x2■+■5x■+■7■and■y■=■1. unDerstAnDing

9 a

7 a■ Find■the■point■of■intersection■of■the■parabola■y■=■x2■+■5x■-■11■with■the■line■x■=■1. ■ (1,■-5) b Is■the■line■a■tangent■to■the■parabola?■If■not,■explain■why■there■is■still■only■one■solution.

y (3, 32)

(2, 21)

5 -3 -2

-1

b

x

1 — 11

y (2, 8)

-6

(1, 0) -6

c

x

-8

y

14 -7

-2

5 x

8 The■parabolas■y■=■x2■-■4■and■y■=■4■-■x2■intersect■in■two■places.■Find■the■coordinates■of■their■ N o,■but■the■straight■line■is■ points■of■intersection. (-2,■0)■and■(2,■0) vertical■and■intersects■at■ 9 For■each■of■the■following■pairs■of■equations: one■point■only. i solve■simultaneously■to■fi■nd■the■points■of■intersection ii illustrate■the■solution■(or■lack■of■solution)■using■a■sketch■graph. a y■=■x2■+■6x■+■5■and■y■=■11x■-■1 y y e 9 d b y■=■x2■+■5x■-■6■and■y■=■8x■-■8 c y■=■x2■+■9x■+■14■and■y■=■3x■+■5 3 d y■=■x2■-■7x■+■10■and■y■=■-11x■+■6 (-2, 28) -1 6 x e y■=■x2■- 2x■-■3■and■y■=■x■-■6 -6 f y■=■x2■+■11x■+■28■and■y■=■10x■+■40 g y■=■x2■+■5x■-■36■and■y■=■15x■-■61 10 h y■=■x2■-■6x■-■16■and■y■=■-4x■-■17 6 6 — 11 i y■=■x2■-■2x■-■24■and■y■=■4x■+■3 x 2 5 j y■=■x2■-■7x■+■10■and■y■=■-4x■+■6 y y k y■=■-x2■+■4x■+■21■and■y■=■x■+■11 g f 2 l y■=■-x ■+■14x■-■48■and■y■=■13x■-■54 (3, 70) (5, 14) m y■=■-x2■+■4x■+■12■and■y■=■9x■+■16 n y■=■x2■+■7x■+■12■and■y■=■20 x 40 4 -9 o y■=■-x2■-■4x■+■5■and■y■=■-4x■+■9 2 p y■=■x ■-■4x■+■4■and■y■=■8x■-■32 28 -36

(-3, -4)

(-4, 0) -7

272

maths Quest 10 for the Australian Curriculum

-4

x

-61

number AND algebra • Linear and non-linear relationships 10 An engineer’s plans for a proposed road through a

mountain are shown at right. At what heights above sea level will the entrance and exit to the tunnel be, given the equations of the mountain profile and road path as shown on the plan? 1.322 km and 2.553 km

y y=-

x–2 +5 2

y = x–4 +2 Proposed road

Entrance of tunnel

x

Sea level

Reasoning 11 A graphic designer draws a logo involving a parabola sitting

in a V shape on a set of axes as shown at right. Find the equation of the parabola, given it is of the form y = kx2 and the points of intersection of the V with the parabola. y=

x2 , (2, 2) and (-2, 2) 2

reflection 

y y=

kx 2

-1 0 -2

1

x

 

If a Cartesian plane showed two upright and two inverted parabolas, how many intersection points would be possible?

Chapter 8 Quadratic equations

273

number AND algebra • Linear and non-linear relationships

Summary Solving quadratic equations ■■ ■■

■■

The general form of a quadratic equation is ax2 + bx + c = 0. To solve a quadratic equation: (a)  make sure the right-hand side of the equation equals zero (b)  take out any common factors (c)  factorise the left-hand side if applicable (d)  use the Null Factor Law to solve for x. An exact answer is a surd or an answer that has not been rounded or approximated. The quadratic formula

The quadratic formula x = ax2 + bx + c = 0.

− b ± b 2 − 4 ac can be used to solve quadratic equations of the form 2a

Solving a quadratic equations by inspecting graphs ■■

■■ ■■

The solution(s) (also known as roots or zeros) of a quadratic equation can be found by inspecting the graph of the equation. You may need to draw the graph of the equation first using a CAS calculator or graphing software. The root of any graph is the x-intercept or the x-coordinate of the point where the graph crosses the x-axis. The roots or intercepts of the quadratic graph y = ax2 + bx + c are the solutions to the equation ax2 + bx + c = 0. Finding solutions to quadratic equations by interpolation

■■ ■■ ■■ ■■

■■ ■■

The discriminant of a quadratic equation is given by D = b2 - 4ac. If D < 0, there are no real solutions to the equation. If D = 0, there is only one rational solution (or two equal solutions) to the equation. The equation can be factorised easily. If D > 0, there are two distinct solutions to the equation. (a) If the discriminant is a perfect square, the solutions are rational and the equation can be factorised easily. (b) If the discriminant is not a perfect square, the solutions are irrational and the equation can be solved using the quadratic formula or the completing the square method. The number of solutions of a quadratic equation corresponds to the number of x-intercepts obtained when the equation is graphed. This information can be summarised in the following table:

D < 0 (negative) Number of No solutions solutions Graph

D = 0 (zero) 1 rational solution

D > 0 (positive) Perfect square Not a perfect square 2 rational ■ solutions

2 irrational (surd) solutions

Graph does not cross■ Graph touches Graph crosses the x-axis twice or touch the x-axis the x-axis

Solving a quadratic equation and a linear equation simultaneously ■■ ■■

274

To solve a quadratic equation with a linear equation we use the substitution method. Make y the subject of both the quadratic equation and the linear equation, and then equate the right-hand side expressions formed. This will leave a new quadratic equation to solve for x.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

■■ ■■ ■■

If■the■new■quadratic■equation■has■two■solutions,■both■must■be■substituted■back■into■one■of■the■ original■equations■to■fi■nd■the■corresponding■values■of■y. In■some■cases■there■will■only■be■one■value■of■x.■This■occurs■when■the■linear■equation■is■a■ tangent■to■the■parabola. In■some■cases■there■will■be■no■solution■to■a■straight■line■and■a■parabola.■If■this■is■the■case,■the■ discriminant■of■the■equation■formed■when■solving■will■be■less■than■0,■that■is,■b2■-■4ac■<■0.

MaPPING YOUR UNDERSTaNDING

Homework Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■247. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

Chapter 8 Quadratic equations

275

number AND algebra • Linear and non-linear relationships

Chapter review Fluency

9 Ten times an integer is added to seven times its

square. If the result is 152, what was the original number? 4

1   MC  The solutions to the equation

✔

x2 + 10x - 11 = 0 are: a x = 1 and x = 11 B x = 1 and x = -11 c x = -1 and x = 11 D x = -1 and x = -11 E x = 1 and x = 10

10 Solve each of the following by using the quadratic

formula, rounding answers to 3 decimal places.

a 4x2 - 2x - 3 = 0 -0.651, 1.151 b 7x2 + 4x - 1 = 0 -0.760, 0.188 c -8x2 - x + 2 = 0 0.441, -0.566

2   MC  The solutions to the equation -5x2 + x + 3 = 0

are: a x = 1 and x = ✔ B

x = -0.68 and x = 0.88

12 The graph of y = x2 - 4x - 21 is shown. y

5

x2

2 4 6

x

10

B 241 D 19

4   MC  Which of the following quadratic equations

has two irrational solutions? a x2 - 6x + 9 = 0 B 4x2 - 11x = 0 c x2 - 25 = 0 D x2 + 8x + 2 = 0 E x2 - 4x + 10 = 0

-21

(2, -25)

Use the graph to find the solutions to the quadratic equation x2 - 4x - 21 = 0. -3, 7 13 Determine the roots of the quadratic graph shown. -3, 1

5 The area of a pool is (6x2 + 11x + 4) m2. Find

the length of the rectangular pool if its width is (3x + 4) m (2x + 1) m 6 Solve each of the following quadratic equation by

y = -2x2 - 4x + 6 y 10

14 a 2 irrational solutions b 2 rational solutions

✔

y = x2 - 4x - 21

-4 -2 0 5

3   MC  The discriminant of the equation

- 11x + 30 = 0 are: 1 c 91 E -11

answers to 3 decimal places.

a 18x2 - 2x - 7 = 0 -0.571, 0.682 b 29x2 - 105x - 24 = 0 -0.216, 3.836 c -5x2 + 2 = 0 -0.632, 0.632

3 5

c x = 3 and x = -5 D x = 0.68 and x = -0.88 3 E x = 1 and x = - 

✔ a

11 Solve each of the following equations, rounding

5 -6 -4 -2 0 -5

2 4 6 x

first factorising the left-hand side of the equation. -10 x2 + 8x + 15 = 0 -5, -3 b x2 + 7x + 6 = 0 -6, -1 2 2 x + 11x + 24 = 0 d x + 4x - 12 = 0 2, -6 -8, -3 x2 - 3x - 10 = 0 5, -2 f x2 + 3x - 28 = 0 4, -7 14 Identify whether each of the equations below has x2 - 4x + 3 = 0 3, 1 h x2 - 11x + 30 = 0 5, 6 no real solutions, one solution or two solutions. x2 - 2x - 35 = 0 7, -5 State whether the solutions are rational or irrational. a x2 + 11x + 9 = 0 b 3x2 + 2x - 5 = 0 7 Solve each of the following quadratic equations. 2 2 2 c x - 3x + 4 = 0 No real solutions a 2x + 16x + 24 = 0 b 3x + 9x + 6 = 0 c 4x2 + 10x - 6 = 0 d 5x2 + 25x - 70 = 0 15 Solve the following pairs of simultaneous e 2x2 - 7x - 4 = 0 f 6x2 - 8x - 8 = 0 equations. (-8, 22) and (2, 2) g 2x2 - 6x + 4 = 0 h 6x2 - 25x + 25 = 0 a y = x2 + 4x - 10 7 a -2, -6 b -2, -1 i 2x2 + 13x - 7 = 0 y = 6 - 2x 1 2 c , -3 d 2, -7 b y = x - 7x + 20 8 Solve each of the following by completing the 2 1 -4 ê 17 square. Give an exact answer for each one. y = 3x - 5 (5, 10) e - 2 , 4 f - 23 , 2 2 2 2 c y = x + 7x + 11 a x + 8x - 1 = 0 b 3x + 6x - 15 = 0 g 2, 1 h 35 , 25 y = x No solution c -4x2 - 3x + 1 = 0 -1, 14 1 a c e g i

-1 ê 6

276

Maths Quest 10 for the Australian Curriculum

i -7,

2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships problem solVing



Use■the■graph■to■determine: a how■far■the■diver■landed■from■the■edge■of■the■ 1 When■a■number■is■added■to■its■square,■the■result■is■ pool■and 6■m 56.■Determine■the■number. -8■and■7 b how■high■the■diving■board■was■above■the■water. 6■m 2 Leroy■measure■his■bedroom■and■fi■nds■that■its■length■ 7 Let■m and■n be■the■solutions■to■the■quadratic■ is■3■metres■more■than■its■width.■If■the■area■of■the■ 2 equation■x2■–■2 5x –■2■=■0.■Determine■the■value■ bedroom■is■18■m ■,■calculate■the■length■and■width■ of■m2■+■n 2. 24 of■the■room. Length■=■6■m,■width■=■3■m 8 While■it■requires■a■minimum■of■2■points■to■ 3 The■surface■area■of■a■cylinder■is■given■by■the■ determine■the■graph■of■a■line,■it■requires■a■ formula■SA■=■2p r(r■+■h),■where■r■cm■is■the■radius■ minimum■of■3■points■to■determine■the■shape■of■a■ of■the■cylinder■and■h■cm■is■the■height. parabola.■The■general■equation■of■a■parabola■is■ ■ ■ ■ The■height■of■a■can■of■soft■drink■is■10■cm■and■its■ y =■ax2■+■bx +■c,■where■a,■b■and■c■are■the■constants■ surface■area■is■245■cm2. ■ 2p r(r■+■10)■=■245 to■be■determined. a Substitute■values■into■the■formula■to■form■a■ a Determine■the■equation■of■the■parabola■that■has■ quadratic■equation■using■the■pronumeral■r. a■y-intercept■of■(0,■–2),■and■passes■though■the■ b Use■the■quadratic■formula■to■solve■the■equation■ points■(1,■–5)■and■(–2,■16).■ y■=■2x2■-■5x■-■2 and,■hence,■fi■nd■the■radius■of■the■can.■Round■ b Determine■the■equation■of■a■parabola■which■ your■answer■to■1■decimal■place. 3.0■cm goes■through■the■points■(0,■0),■(2,■2)■and■(5,■5).■ c Calculate■the■area■of■the■label■on■the■can.■The■ Show■full■working■to■justify■your■answer.■ label■covers■the■entire■curved■surface.■Round■ 188■cm2 the■answer■to■the■nearest■square■centimetre. 9 When■the■radius■of■a■circle■increases■by■6■cm,■its■ area■increases■by■25%.■Use■the■quadratic■formula■ 4 Find■the■values■of■d■when■2x2■-■5x■-■d■=■0■has■one■ to■fi■nd■the■exact■radius■of■the■original■circle. solution. - 25 8

5 For■what■values■of■k■does■

N o■parabola■is■possible.■The■points■

are■on■the■same■straight■line. (k■-■1)x2■-■(k■-■1)x■+■2■=■0■have■two■distinct■ solutions? k■>■9■and■k■<■1 6 A■platform■diver■follows■a■path■determined■by■the■ equation■h■=■-0.5d2■+■2d■+■6,■where■h■represents■ the■height■of■the■diver■above■the■water■and■d■ represents■the■distance■from■the■diving■board.■Both■ pronumerals■are■measured■in■metres.

h 8 6 4 2 -4 -2-20

12( 5■+■2)■cm eBook plus

Interactivities

Test yourself Chapter 8 int-2849 Word search Chapter 8 int-2847 Crossword Chapter 8 int-2848

2 4 6 8 d

Chapter 8 Quadratic equations

277

eBook plus

ACtiVities

chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■8■(doc-5255)■(page 247) are you ready?

(page 248) •■ SkillSHEET■8.1■(doc-5256):■Factorising■by■taking■ out■the■highest■common■factor •■ SkillSHEET■8.2■(doc-5257):■Finding■a■factor■pair■ that■adds■to■a■given■number •■ SkillSHEET■8.3■(doc-5258):■Simplifying■surds •■ SkillSHEET■8.4■(doc-5259):■Substituting■into■ quadratic■equations •■ SkillSHEET■8.5■(doc-5260):■Equation■of■a■vertical■ line Digital docs

8a Solving quadratic equations eLesson

•■ Completing■the■square■(eles-0174)■(page 251) Digital docs

•■ Activity■8-A-1■(doc-5059):■Solving■simple■ quadratics■(page 253) •■ Activity■8-A-2■(doc-5060):■Solving■quadratic■ equations■(page 253) •■ Activity■8-A-3■(doc-5061):■Solving■more■complex■ quadratics■(page 253) •■ WorkSHEET■8.1■(doc-5261):■Solving■quadratic■ equations■(page 255) 8B The quadratic formula

(page 257) •■ Activity■8-B-1■(doc-5062):■Introducing■the■quadratic■ formula■ •■ Activity■8-B-2■(doc-5063):■Practice■using■the■ quadratic■formula■ •■ Activity■8-B-3■(doc-5064):■Using■the■quadratic■ formula■ •■ SkillSHEET■8.6■(doc-5262):■Substituting■into■the■ quadratic■formula •■ SkillSHEET■8.7■(doc-5263):■Simplifying■surds Digital docs

8c Solving quadratic equations by inspecting graphs Digital docs (page 261) •■ Activity■8-C-1■(doc-5065):■Finding■solutions■to■ quadratic■equations■by■inspecting■graphs■

278

maths Quest 10 for the Australian Curriculum

•■ Activity■8-C-2■(doc-5066):■Solving■quadratic■ equations■by■inspecting■graphs■ •■ Activity■8-C-3■(doc-5067):■Harder■solutions■to■ quadratic■equations■by■inspecting■graphs■ 8D Finding solutions to quadratic equations by interpolation and using the discriminant Interactivity

•■ Solving■by■interpolation■(int-1147)■(page 263) Digital docs

•■ Activity■8-D-1■(doc-5068):■Finding■solutions■to■ quadratic■equations■by■interpolation■(page 267) •■ Activity■8-D-2■(doc-5069):■Harder■solutions■to■ quadratic■equations■by■interpolation■(page 267) •■ Activity■8-D-3■(doc-5070):■Diffi■cult■solutions■to■ quadratic■equations■by■interpolation■(page 267) •■ WorkSHEET■8.2■(doc-5264):■Using■the■discriminant■ (page 268) 8E Solving a quadratic equation and a linear equation simultaneously Interactivity

•■ Simultaneous■quadratic■equations■(int-2784)■ (page 269) Digital docs (page 272) •■ Activity■8-E-1■(doc-5071):■Introduction■to■quadratic■ and■linear■solutions •■ Activity■8-E-2■(doc-5072):■Practising■quadratic■and■ linear■solutions■ •■ Activity■8-E-3■(doc-5073):■Tricky■quadratic■and■ linear■solutions■ chapter review

(page 277) •■ Test■yourself■Chapter■8■(int-2849):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■8■(int-2847):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■8■(int-2848):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

9

9a Plotting parabolas 9b Sketching parabolas using the basic graph of y = x 2 9c Sketching parabolas in turning point form 9d Sketching parabolas of the form y = ax 2 + bx + c 9e Exponential functions and their graphs 9F The hyperbola 9G The circle WhAt Do you knoW ?

Functions

1 List what you know about functions. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of functions. eBook plus

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Hungry brain activity Chapter 9 doc-5265

opening Question

The path of this thrown ball follows the quadratic equation y = -0.45x2 + 2.2x + 1.5. What is the maximum height reached by the ball?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■ebookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 9.1 doc-5266

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SkillSHEET 9.2 doc-5267

Substitution into quadratic equations 1 Substitute■the■x-value■in■brackets■into■each■of■the■following■quadratic■equations■to■determine■

the■y-value.

a y■=■x2■-■4x■+■3■ (x■=■3) ■ 0 b y■= -3x2■+■2x■-■8■ (x■=■2) -16 c y■= -8x2■-■3x■-■12■ (x■= -2) -38 Equation of a vertical line 2 Write■the■equation■for■each■of■the■lines■shown■below. a b y y

-4 -3 -2-1 0 1 2 3

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SkillSHEET 9.3 doc-5268

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SkillSHEET 9.5 doc-5270

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SkillSHEET 9.6 doc-5271

280

x

3 2 a■ x■=■-2 b x■=■3 c x■=■ 2 ■or■x■=■1.5

c

x -2 -1 0 1 2 3 4

Completing the square 3 Complete■the■square■for■each■of■the■following. a x2■+■2x■+■2 (x■+■1)2■+■1 b x2■- 3x■+■4

y

-1 0 1 2 3 4 5

x

2

3 7   x − 2  + 4

c 2x2■-■4x■+■6 2(x■-■1)2■+■4

Solving quadratic equations using the quadratic formula 4 Solve■the■following■quadratic■equations■using■the■quadratic■formula.■Leave■your■answer■in■

surd■form. −1 ± 5 a x2■+■x■-■1■=■0 ■ 2

b 2x2■-■4x■+■1■=■0

2± 2 2

c -3x2■-■2x■+■2■=■0

1 ± 7 −1 ∓ 7 = Solving quadratic equations of the type ax 2 + bx + c = 0 where a = 1 −3 3 5 Solve■the■following■quadratic■equations■by■factorising. a x2■+■5x■+■6■=■0 b x2■+■x■-■2■=■0 c x2■-■4■=■0 x■=■2■or■x■=■-2 ■ x■=■-2■or■x■=■-3

x■=■1■or■x■=■-2

Solving quadratic equations of the type ax 2 + bx + c = 0 where a ò 1 6 Solve■the■following■quadratic■equations■by■factorising. a 2x2■+■5x■+■2■=■0 b 3x2■-■5x■-■2■=■0 c 6x2■-■13x■+■6■=■0 1

x■=■- 2 ■or■x■=■-2

maths Quest 10 for the Australian Curriculum

1

x■=■2■or■x■=■- 3

3

2

x■=■ 2 ■or■x■=■ 3

number AND algebra • Linear and non-linear relationships

9A

Plotting parabolas ■■

A parabola is the graphical form of a quadratic equation. Its shape is seen in many everyday situations, some of which are parts of nature and some are made by humans.

■■

If you look at these pictures, you will notice that: 1. the parabolas are symmetrical. For each one, a line could be drawn down the middle to divide the parabola exactly in half. This line is called the axis of symmetry. 2. they have either a maximum turning point ( ) or a minimum turning point ( ) 3. some are wider or narrower than others but they have basically the same shape. If the equation of the parabola is given, a table of values can be produced by substituting x-values into the equation to obtain the corresponding y-values. These x- and y-values provide the coordinates for points which can be plotted and joined to form the shape of the graph. When plotting graphs, use grid or graph paper for accuracy. The graph of y = x 2, shown below, has been produced in this manner. The axis of symmetry is the y-axis, with the equation x = 0. The turning point is (0, 0).

■■

■■

x

-3

-2

-1

0

1

2

3

y

 9

 4

 1

0

1

4

9

y

y = x2

10 8 6 4 2 -4 -3-2-1 0 1 2 3 4 -2 (0, 0)

■■

x

The graph of y = x2 is the basic graph of a quadratic equation or parabola. It has a minimum turning point at (0, 0). This parabola can be dilated (made wider or narrower), translated (moved horizontally and/or vertically) and reflected (turned upside down), to form other parabolas. How some of these transformations take place can be seen in the worked examples that follow.

Worked Example 1

Plot the graph of each of the following equations. In each case, use the values of x shown as the values in your table. State the equation of the axis of symmetry and the coordinates of the turning point. a y = 2x2 for -3 Ç x Ç 3 1 b y = x2 for -3 Ç x Ç 3 2 Chapter 9 Functions

281

number AND algebra • Linear and non-linear relationships

Think a

Write/draw

1

Write the equation.

2

Produce a table of values using x-values from -3 to 3.

3

4

a y = 2x2

x

-3

-2

-1

0

1

2

 3

y

18

 8

 2

0

2

8

18

Draw a set of clearly labelled axes, plot the points and join them with a smooth curve. The scale would be from 20 to -2 on the y-axis and -4 to 4 on the x-axis.

y 20 18

Label the graph.

14

16 12

y = 2x2

10 8 6 4 2

-4-3-2-10 1 2 3 4 -2

b

x

5

Write the equation of the line that divides the parabola exactly in half.

The equation of the axis of symmetry is x = 0.

6

Write the coordinates of the turning point.

The turning point is (0, 0).

1

Write the equation.

2

Produce a table of values using x-values from -3 to 3.

3

4

Draw a set of clearly labelled axes, plot the points and join them with a smooth curve. The scale would be from 6 to -2 on the y-axis and -4 to 4 on the x-axis. Label the graph.

b

y = 12 x 2 x

-3

-2

-1

0

1

2

3

y

4.5

 2

0.5

0

0.5

2

4.5

y 6 5 y = 1–2 x2

4 3 2 1 -4 -3 -2 -1 0 -1

1

2

3

4 x

-2

282

5

Write the equation of the line that divides the parabola exactly in half.

The equation of the axis of symmetry is x = 0.

6

Write the coordinates of the turning point.

The turning point is (0, 0).

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 2

Plot the graph of each of the following equations. In each case, use the values of x shown as the values in your table. State the equation of the axis of symmetry, the coordinates of the turning point and the y-intercept for each one. a  y = x2 + 2 for -3 Ç x Ç 3 b  y = (x + 3)2 for -6 Ç x Ç 0 c  y = -x2 for -3 Ç x Ç 3 Think a

Write/draw

1

Write the equation.

2

Produce a table of values.

3

Draw a set of clearly labelled axes, plot the points and join them with a smooth curve. The scale on the y-axis would be from 0 to 12 and -4 to 4 on the x-axis.

4

a y = x2 + 2

x y

-3 11

-2  6

-1  3

0 2

1 3

2 6

 3 11

y 12 9

Label the graph.

6

y = x2 + 2

3

(0, 2) -4 -2 0

b

2

x

4

5

Write the equation of the line that divides the parabola exactly in half.

The equation of the axis of symmetry is x = 0.

6

Write the coordinates of the turning point.

The turning point is (0, 2).

7

Find the y-coordinate of the point where the graph crosses the y-axis.

The y-intercept is 2.

1

Write the equation.

2

Produce a table of values.

3

Draw a set of clearly labelled axes, plot the points and join them with a smooth curve. The scale on the y-axis would be from 0 to 10 and -7 to 1 on the x-axis.

4

Label the graph.

b y = (x + 3)2

x y

-6  9

-5  4

-4  1

-3  0

-2  1

-1  4

0 9

y 10 8

y = (x + 3)2

(0, 9)

6

(-3, 0)

4 2

-7-6-5-4-3-2-1 0 1

x

5

Write the equation of the line that divides the parabola exactly in half.

The equation of the axis of symmetry is x = -3.

6

Write the coordinates of the turning point.

The turning point is (-3, 0).

7

Find the y-coordinate of the point where the graph crosses the y-axis.

The y-intercept is 9.

Chapter 9 Functions

283

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships c

1

Write■the■equation.

2

Produce■a■table■of■values.

3

Draw■a■set■of■clearly■labelled■axes,■plot■ the■points■and■join■them■with■a■smooth■■ curve.■The■scale■on■the■y-axis■would■be■ from■-10■to■1■and■from■-4■to■4■on■the■ x-axis.

c

y■=■-x2 x y

-3 -9

-2 -4

-4 -2 0

2

-9

2 a

y 30 25 20 15 10 5

x

2 -4

3 -9

y = -x2

Write■the■equation■of■the■line■that■divides■ the■parabola■exactly■in■half.

The■equation■of■the■axis■of■symmetry■is■x■=■0.

6

Write■the■coordinates■of■the■turning■point.

The■turning■point■is■(0,■0).

7

Find■the■y-coordinate■of■the■point■where■ the■graph■crosses■the■y-axis.

The■y-intercept■is■0.

y = 3x2

remember

x■=■0,■(0,■0) y

4

1 -1

5

-3 -2-1 0 1 2 3 x

b

0 0

y

Label■the■graph.

4

-1 -1

1

y = –4 x2

2 1 -3 -2-1 0 1 2 3 x

1.■ Produce■a■table■of■values■by■substituting■each■integer■value■of x into■the■equation. 2.■ Plot■a■graph■by■drawing■and■labelling■a■set■of■axes,■plotting■the■points■from■the■table■ and■joining■the■points■to■form■a■smooth■curve. 3.■ The■axis■of■symmetry■is■the■line■that■divides■the■parabola■exactly■in■half. 4.■ The■turning■point■is■the■point■where■the■graph■changes■direction■or■turns. 5.■ The■turning■point■is■a■maximum■if■it■is■the■highest■point■on■the■graph■and■a■minimum■if■ it■is■the■lowest■point■on■the■graph. 6.■ The■x-intercepts■are■the■x-coordinates■of■the■points■where■the■graph■crosses■the■x-axis. 7.■ The■y-intercept■is■the■y-coordinate■of■the■point■where■the■graph■crosses■the■y-axis.

y

8

y = x2

FluenCy 1 Plot■the■graph■of y =■x2■for■values■of x between■-3■and■3.■State■the■equation■of■the■axis■of■

symmetry■and■the■coordinates■of■the■turning■point.

2 We1 ■Plot■the■graph■of■each■of■the■following■equations.■In■each■case,■use■the■values■of■

x■shown■as■the■values■in■your■table.■State■the■equation■of■the■axis■of■symmetry■and■the■ coordinates■of■the■turning■point. Placing■a■number■greater■than■1■in■front■of■x2■makes■ a y■=■3x2■for■-3■Ç■x■Ç■3 the■graph■thinner.■Placing■a■number■greater■than■0■ 1 4

b y■= x2■for■-3■Ç■x■Ç■3

but■less■than■1■in■front■of■x2■makes■the■graph■wider.

3 Compare■the■graphs■you■have■drawn■for■question■2■with■that■of■y■=■x2■in■question■1.■Explain■

how■placing■a■number■in■front■of■x2■affects■the■graph■obtained.

284

10

Activity 9-A-2

Plotting parabolas doc-5075

6

Review of plotting parabolas doc-5074

4

Activity 9-A-1

x

eBook plus

You■may■wish■to■use■a■graphing■calculator■for■this■exercise.

(0, 0)

inDiviDuAl pAthWAys

plotting parabolas

2

9A

-4 -3–2-1 1 2 3 -2

exerCise

x■=■0,■(0,■0)

x■=■0,■(0,■0)

maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

equation■of■the■axis■of■symmetry,■the■coordinates■of■the■turning■point■and■the■y-intercept■for■ each■one. a y =■x2■+■1 b y =■x2■+■3 c y =■x2■-■3 d y =■x2■-■1

eBook plus

Activity 9-A-3

5 Compare■the■graphs■you■have■drawn■for■question■4■with■that■for y =■x2■in■question■1.■Explain■

Trends in plotting parabolas doc-5076

how■adding■to■or■subtracting■from■x2■affects■the■graph■obtained.

6 We2b ■Plot■the■graph■of■each■of■the■following■equations.■In■each■case,■use■the■values■of

x shown■as■the■values■in■your■table.■State■the■equation■of■the■axis■of■symmetry,■the■coordinates■ of■the■turning■point■and■the■y-intercept■for■each■one. Adding■a■number■moves■the■graph■of■ a y =■(x■+■1)2■ -5■Ç x Ç■3 y■=■x2■horizontally■to■the■left■by■that■ b y =■(x■+■2)2■■ -6■Ç x Ç■2 number■of■units.■Subtracting■a■number■ c y =■(x■-■2)2■ -1■Ç x Ç■5 moves■the■graph■of■y■=■x2■horizontally■ d y =■(x■-■1)2■ -2■Ç x Ç■4 to■the■right■by■that■number■of■units.

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SkillSHEET 9.1 doc-5266

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SkillSHEET 9.2 doc-5267

■

7 Compare■the■graphs■you■have■drawn■for■question■6■with■that■for y =■x2■in■question■1.■Explain■

how■adding■to■or■subtracting■from x before■squaring■affects■the■graph■obtained.

8 We2c ■Plot■the■graph■of■each■of■the■following■equations.■In■each■case,■use■the■values■of

1

9 Compare■the■graphs■you■have■drawn■for■question■8■with■that■for y =■x2■in■question■1.■Explain■

x■=■0,■(0,■1),■1

c

y 4 3 2 1 0 123456 -2 -3 -4 -5

how■a■negative■sign■in■front■of■x2■affects■the■graph■obtained.■Also■compare■the■graphs■obtained■ in■question■8■with■those■in■questions■4■and■6.■Which■graphs■have■the■same■turning■point?■How■ are■they■different?

unDerstAnDing

x

y = -(x - 3)2 + 4

x■=■3,■(3,■4),■max,■-5 y d 5

-2 0 -5 -10 -15 -20 -25

■

2

4

x

y = -3(x - 1)2 + 2

x■=■1,■(1,■2),■max,■-1

10 Plot■the■graph■of■each■of■the■following,■and■state: i the■equation■of■the■axis■of■symmetry ii the■coordinates■of■the■turning■point■and■whether■it■is■a■maximum■or■a■minimum iii the■y-intercept. 10 a y b y y = 2(x + 2)2 - 3 a y =■(x■-■5)2■+■1■ 0■Ç x Ç■6 y = (x - 5)2 + 1 16 26 b y =■2(x■+■2)2■-■3■ -5■Ç x Ç■1 2 12 c y =■-(x■-■3) ■+■4■ 0■Ç x Ç■6 8 d y =■-3(x■-■1)2■+■2■ -2■Ç x Ç■4 4 e y =■x2■+■4x■-■5■ -6■Ç x Ç■2 1 -8 -6 -4 -2 0 x x f y =■-x2■-■2x■+■15■ -6■Ç x Ç■4 0 12 3 4 5 6 -4 2 g y =■-3x ■-■6x■+■24■ -5■Ç x Ç■3 x■=■5,■(5,■1),■min,■26 x■=■-2,■(-2,■-3),■min,■5 h y =■(x■-■2)2■+■1■ -2■Ç x Ç■4

11 a If■the■x2■term■is■ positive,■the■parabola■ has■a■minimum■ turning■point.■If■the■ x2■term■is■negative,■ the■parabola■has■a■ maximum■turning■ point. b If■the■equation■is■of■the■ 12 form■y■=■a(x■-■b)2■+■c,■ the■turning■point■has■ coordinates■(b,■c).

Use■your■graphs■from■question■10 a–d■to■answer■the■following. a Explain■how■you■can■determine■whether■a■parabola■has■a■minimum■or■maximum■turning■ point■by■looking■only■at■its■equation. b Explain■how■you■can■determine■the■coordinates■of■the■turning■point■of■a■parabola■by■ looking■only■at■the■equation. c Explain■how■you■can■obtain■the■equation■of■the■axis■of■symmetry■by■looking■only■at■the■ equation■of■the■parabola. mC ■For■the■graph■of y =■(x■-■2)2■+■5,■the■turning■point■is:

a (5,■2) d (-2,■-5)

✔ c (2,■5) b (2,■-5) e (-2,■5) The■equation■of■the■axis■of■symmetry■can■be■found■from■ the■x-coordinate■of■the■turning■point.■That■is,■x■=■b. Chapter 9 Functions

The■negative■sign■inverts■the■graph■of■y■=■x2.The■graphs■with■the■same■turning■points■are: y■=■x2■+■1■and■y■=■-x2■+■1;■y■=■(x■-■1)2■and■y■=■-(x■-■1)2;■y■=■(x■+■2)■and■y■=■-(x■+■2)2;■y■=■x2■-■3■and■y■=■-x2■-■3. They■differ■in■that■the■fi■rst■graph■is■upright■while■the■second■graph■is■inverted.

x shown■as■the■values■in■your■table.■State■the■equation■of■the■axis■of■symmetry,■the■coordinates■ of■the■turning■point■and■the■y-intercept■for■each■one. a y =■-x2■+■1■ -3■Ç x Ç■3 b y =■-(x■-■1)2■ -2■Ç x Ç■4 c y =■-(x■+■2)2■ -5■Ç x Ç■1 d y =■-x2■-■3■ -3■Ç x Ç■3

y -3-2-1 01 2 3 4 x -2 -3 -4 -5 -6 -7 -8 y = -x2 + 1

Adding■a■number■raises■the■graph■of■y■=■x2■vertically that■number■of■units.■Subtracting■a■number■lowers■the■ graph■of■y■=■x2■vertically■that■number■of■units.

4 We2a ■Plot■the■graph■of■each■of■the■following■for■values■of x between■-3■and■3.■State■the■

inDiviDuAl pAthWAys

285

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 13 mC ■For■the■graph■of y =■3(x■-■1)2■+■12,■the■turning■point■is: ✔ b (1,■12) a (3,■12) d (-3,■12) e (-1,■-12) 14 mC ■For■the■graph■of y =■(x■+■2)2■-■7,■the■y-intercept■is: a -2 b -7 e 7 d -11

■

c (-1,■12)

✔ c

-3

15 mC ■Which■of■the■following■is■true■for■the■graph■of y =■-(x■-■3)2■+■4? ✔ a Turning■point■(3,■4),■y-intercept■-5 b Turning■point■(3,■4),■y-intercept■5 c Turning■point■(-3,■4),■y-intercept■-5 d Turning■point■(-3,■4),■y-intercept■5 e Turning■point■(3,■-4),■y-intercept■13

y

reAsoning 0

x

16 A■ball■is■thrown■into■the■air.■The■height,■h■metres,■

of■the■ball■at■any■time,■t■seconds,■can■be■found■by■ using■the■equation■h■=■-(t■-■4)2■+■16. a Plot■the■graph■for■values■of■t■between■0■and■8. b Use■the■graph■to■fi■nd: i the■maximum■height■of■the■ball ■ 16■■m ii how■long■it■takes■for■the■ball■to■fall■back■to■ the■ground. 8■■s

y

h 0

x

18 16 14 12 10 8 6 4 2

h = -(t - 4)2 + 16

0 1234567 8

x

y

0

286

t

h 18 16 14 12 10 8 6 4 2

0

y

0

■

x

1 2 3

t

17 From■a■crouching■position■in■a■ditch,■an■archer■wants■

to■fi■re■an■arrow■over■a■horizontal■tree■branch,■which■is■ 15■metres■above■the■ground.■The■height,■in■metres■(h),■ of■the■arrow■t■seconds■after■it■has■been■fi■red■is■given■by■ the■equation■h■=■-8t(t■-■3). a Plot■the■graph■for■t■=■0,■1,■1.5,■2,■3. b From■the■graph■fi■nd: i the■maximum■height■the■arrow■reaches ■ 18■■m ii whether■the■arrow■clears■the■branch■and■the■ distance■by■which■it■clears■or■falls■short■of■the■ branch Yes,■by■3■■m iii the■time■it■takes■to■reach■maximum■height 1.5■■s iv how■long■it■takes■for■the■arrow■to■hit■the■ground. 3■■s 18 There■are■0,■1,■2■and■infi■nite■possible■points■of■ intersection■for■two■parabolas.■ a Illustrate■these■on■separate■graphs. b Explain■why■infi■nite■points■of■intersection■are■ reFleCtion    possible.■Give■an■example. What x-values can a parabola c How■many■points■of■intersection■are■possible■for■a■ have? What y-values can a parabola■and■a■straight■line?■Illustrate■these.

An■infi■nite■number■of■points■of■intersection■occur■when■the■two■equations■ represent■the■same■parabola,■with■the■effect■that■the■two■parabolas■ superimpose.■For■example■y■=■x2■+■4x■+■3■and■2y■=■2x2■+■8x■+■6. maths Quest 10 for the Australian Curriculum

parabola have?

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

sketching parabolas using the basic graph of y = x 2

9b eBook plus

■■

Interactivity Dilation of y = x2

int-1148

Identifying■the■type■of■transformation■which■has■occurred■to■the■graph■of y =■x2■allows■us■to■ sketch■rather■than■plot■a■parabola.■A■sketch■graph■is■not■drawn■on■grid■or■graph■paper■and■there■ is■no■scale■shown■along■each■axis.■It■allows■us■to■see■the■relationship■between x and y and■the■ key■features■of■the■graph■such■as■the■coordinates■of■the■turning■point■and■the■x-■and■y-intercepts. y

Dilation ■■

■■

■■

Compare■the■graph■of y =■2x2■with■that■of y =■x2.■This■graph■is■ thinner■or■closer■to■the■y-axis.■As■the■coeffi■cient■of■x2■increases,■the■ graph■becomes■narrower■and■closer■to■the■y-axis. The■turning■point■has■not■changed■under■the■transformation■and■is■ still■(0,■0).■This■is■called■an■invariant■point. 1 Compare■the■graph■of y =■ 4 x2■with■that■of y =■x2. The■graph■is■wider■or■closer■to■the■x-axis. The■turning■point■has■not■changed■and■is■still■(0,■0). As■the■coeffi■cient■of■x2■decreases■(but■remains■positive),■the■ graph■becomes■wider■or■closer■to■the■x-axis.

y = 2x2 y = x2

x

(0, 0) y

y = x2 y = 1–4 x2 x

(0, 0)

WorkeD exAmple 3

State whether each of the following graphs is wider or narrower than the graph of y = x2 and state the coordinates of the turning point of each one. 1 a y = x2 b y = 4x2 5 think a

b

Write 1

a y■=■ 5 x2

1

Write■the■equation.

2

Look■at■the■coeffi■cient■of■x2■and■decide■ whether■it■is■greater■than■or■less■than■1.

3

The■dilation■doesn’t■change■the■turning■point.

1

Write■the■equation.

2

Look■at■the■coeffi■cient■of■x2■and■decide■ whether■it■is■greater■than■or■less■than■1.

4■>■1,■so■the■graph■is■narrower■than■that■ of■y■=■x2.

3

The■dilation■doesn’t■change■the■turning■point.

The■turning■point■is■(0,■0).

eBook plus

Interactivity Vertical translation of y = x2 + c

1 ■<■1,■so■the■graph■is■wider■than■that■of■y■=■x2. 5

The■turning■point■is■(0,■0). b y■=■4x2

vertical translation ■■

y

y = x2 + 2

Compare■the■graph■of y =■x2■+■2■with■that■of y =■x2. ■ The■whole■graph■has■been■moved■or■translated■2■units■ upwards.■The■turning■point■has■become■(0,■2).■

y = x2

int-1192

(0, 2) x

Chapter 9 Functions

287

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

■■

Compare■the■graph■of y =■x2■-■3■with■that■of y =■x2. ■ The■whole■graph■has■been■moved■or■translated■3■units■ downwards.■The■turning■point■has■become■(0,■-3).

y

y = x2 y = x2 - 3

x (0, -3)

WorkeD exAmple 4

State the vertical translation and the coordinates of the turning point for the graphs of each of the following equations. a y = x2 + 5 b y = x2 - 4 think a

b

Write a y■=■x2■+■5

1

Write■the■equation.

2

+5■means■the■graph■is■translated■upwards■5■units.

Vertical■translation■of■5■units■up

3

Translate■the■turning■point■of■y■=■x2,■which■is■ (0,■0).■The■x-coordinate■of■the■turning■point■ remains■0,■and■the■y-coordinate■has■5■added■to■it.

The■turning■point■becomes■(0,■5).

1

Write■the■equation.

2

-4■means■the■graph■is■translated■downwards■ 4■units.

Vertical■translation■of■4■units■down

3

Translate■the■turning■point■of■y■=■x2■which■is■ (0,■0).■The■x-coordinate■of■the■turning■point■ remains■0,■and■the■y-coordinate■has■4■subtracted■ from■it.

The■turning■point■becomes■(0,■-4).

b y■=■x2■-■4

Note: There■are■no■invariant■points■under■a■vertical■translation. eBook plus

Interactivity Horizontal translation of y = (x - h)2

horizontal translation ■■

Compare■the■graph■of y =■(x■-■2)2■with■that■of y =■x2. ■ The■whole■graph■has■been■moved■or■translated■2■units■to■the■ right.■The■turning■point■has■become■(2,■0).■

int-1193

y y = x2 y = (x - 2)2

(0, 4) (2, 0) ■■

Compare■the■graph■of y =■(x■+■1)2■with■that■of y =■x2. ■ The■whole■graph■has■been■moved■or■translated■1■unit■left.■The■ turning■point■has■become■(-1,■0).

y

maths Quest 10 for the Australian Curriculum

y = (x + 1)2 y = x2

(0, 1) (-1, 0)

288

x

x

number AND algebra • Linear and non-linear relationships

Worked Example 5

State the horizontal translation and the coordinates of the turning point for the graphs of each of the following equations. a  y = (x - 3)2 b  y = (x + 2)2 Think a

b

Write a y = (x - 3)2

1

Write the equation.

2

-3 means the graph is translated to the right 3 units.

Horizontal translation of 3 units to the right

3

Translate the turning point of y = x2 which is (0, 0). The y-coordinate of the turning point remains 0, and the x-coordinate has 3 added to it.

The turning point becomes (3, 0).

1

Write the equation.

2

+2 means the graph is translated to the left 2 units.

Horizontal translation of 2 units to the left

3

Translate the turning point of y = which is (0, 0). The y-coordinate of the turning point remains 0, and the x-coordinate has 2 subtracted from it.

The turning point becomes (-2, 0).

b y = (x + 2)2

x2

Note: There are no invariant points under a horizontal translation.

Reflection ■■

■■

y

Compare the graph of y = -x2 with that of y = x2.   In each case the axis of symmetry is the line x = 0 and the turning point is (0, 0). The only difference between the equations is the negative sign in y = -x2 and the difference between the graphs is that y = x2 ‘sits’ on the x-axis and y = -x2 ‘hangs’ from the x-axis. (One is a reflection or mirror image of the other.) y = x2 has a minimum turning point and y = -x2 has a maximum turning point. What is the invariant point? Any quadratic graph where x2 is positive has a shape and is said to be upright. Conversely, if x2 is negative the graph has a shape and is said to be inverted.

y = x2

x

(0, 0)

y = -x2

Worked Example 6

For each of the following graphs, give the coordinates of the turning point and state whether it is a maximum or a minimum. a  y = -(x - 7)2 b  y = 5 - x2 Think a

Write a y = -(x - 7)2

1

Write the equation.

2

It is a horizontal translation of 7 units to the right, so 7 units is added to the x-coordinate of (0, 0).

The turning point is (7, 0).

3

The sign in front of the x2 term is negative, so it is inverted.

Maximum turning point

Chapter 9 Functions

289

number AND algebra • Linear and non-linear relationships b

b y = 5 - x2

1

Write the equation.

2

Rewrite the equation so that the x2 term is first.

y = -x2 + 5

3

The vertical translation is 5 units up, so 5 units is added to the y-coordinate of (0, 0).

The turning point is (0, 5).

4

The sign in front of the x2 term is negative, so the graph is inverted.

Maximum turning point

Worked Example 7

For each of the following quadratic equations:     i state the appropriate dilation, reflection and translation of the graph of y = x2 needed to obtain the graph    ii state the coordinates of the turning point iii hence, sketch the graph. a y = (x + 3)2 b  y = -2x2 Think a

Write/draw a y = (x + 3)2

1

Write the quadratic equation.

2

Identify the transformation needed — horizontal translation only, no dilation or reflection.

i Horizontal translation of 3 units to

3

State the turning point.

ii The turning point is (-3, 0).

4

Sketch the graph of y = (x + 3)2. You may find it helpful to lightly sketch the graph of y = x2 on the same set of axes first.

iii

the left

y

y = (x + 3)2

y = x2

x

(-3, 0) b

b y = -2x2

1

Write the quadratic equation.

2

Identify the transformations needed — dilation (2 in front of x2) and reflection (negative in front of x2 term), no translation.

i Reflection so graph is inverted. As

3

The turning point remains the same as there is no translation.

ii The turning point is (0, 0).

4

Sketch the graph of y = -2x2. You may find it helpful to lightly sketch the graph of y = x2 on the same set of axes first.

iii

2 > 1, the graph is narrower than that of y = x2.

y

y = x2

(0, 0)

y = -2x2

290

Maths Quest 10 for the Australian Curriculum

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

1.■ If■the■graph■of y =■x2■is■translated■k■units■vertically,■the■equation■becomes■y =■x2■+■k. 2.■ If■the■graph■of y =■x2■is■translated■h■units■horizontally,■the■equation■becomes■ y =■(x■-■h)2. 3.■ If■the■graph■of y =■x2■is■dilated■by■factor■a,■the■graph■becomes■narrower■if■a■>■1■and■ wider■if■0■<■a■<■1. 4.■ If■the■x2■term■is■positive,■the■graph■is■upright.■If■there■is■a■negative■sign■in■front■of■the■ x2■term,■the■graph■is■inverted. 5.■ Invariant■points■are■points■that■do■not■change■under■a■transformation. exerCise

9b inDiviDuAl pAthWAys eBook plus

Activity 9-B-1

Review of sketching parabolas doc-5077 Activity 9-B-2

Sketching basic parabolas doc-5078 Activity 9-B-3

Trends in sketching basic parabolas doc-5079

sketching parabolas using the basic graph of y = x 2 FluenCy 1 We3 ■State■whether■each■of■the■following■graphs■is■wider■or■narrower■than■the■graph■of y =■x2■

and■state■the■coordinates■of■the■turning■point■of■each■one. 1 a y =■5x2 ■ Narrower,■TP■(0,■0) b y =■ 3 x2 Wider,■(0,■0) c y =■7x2 Narrower,■TP■(0,■0) d y =■10x2■ Narrower,■TP■(0,■0) 2 2 e y =■ 5 x f y =■0.25x2 Wider,■TP■(0,■0) Wider,■TP■(0,■0) g y =■1.3x2 Narrower,■TP■(0,■0)

h y =■ 3x2

Narrower,■TP■(0,■0) 2 We4 ■State■the■vertical■translation■and■the■coordinates■of■the■turning■point■for■the■graphs■of■

each■of■the■following■equations.

a y =■x2■+■3 c y =■x2■-■7

Vertical■3■up,■TP■(0,■3) Vertical■7■down,■TP■(0,■-7)

b y =■x2■-■1

1 =■x2■- 2

1 Vertical■ 2 ■down,■TP■(0,■- 12 )

f

e y

g y =■x2■+■2.37

d y

y

Vertical■1■down,■TP■(0,■-1)

1 1 1 =■x2■+■ 4 Vertical■ 4 ■up,■TP■(0,■ 4 ) =■x2■-■0.14

Vertical■0.14■down,■TP■(0,■-0.14)

h y =■x2■+

3 Vertical■ 3■up,■TP■(0,■ 3) 3 We5 ■State■the■horizontal■translation■and■the■coordinates■of■the■turning■point■for■the■graphs■of■ each■of■the■following■equations. a y =■(x■-■1)2 ■ Horizontal■1■right,■(1,■0) b y =■(x■-■2)2 Horizontal■2■right,■(2,■0) 2 c y =■(x■+■10) Horizontal■10■left,■(-10,■0) d y =■(x■+■4)2 Horizontal■4■left,■(-4,■0) 1 1 2 1 1 1 e y =■(x■- ) Horizontal■ 2 ■right,■( 2 ,■0) f y =■(x■+ 5 )2 Horizontal■ 15■left,■(- 5 ,■0) 2 g y =■(x■+■0.25)2

Vertical■2.37■up,■TP■(0,■2.37)

Horizontal■0.25■left,■(-0.25,■0) h y =■(x■+

3)2 Horizontal■ 3■left,■(− 3,■0) 4 We6 ■For■each■of■the■following■graphs■give■the■coordinates■of■the■turning■point■and■state■ whether■it■is■a■maximum■or■a■minimum.■ a y =■-x2■+■1 (0,■1),■max b y =■x2■-■3 (0,■-3),■min 2 (-2,■0),■max c y =■-(x■+■2) d y =■3x2 (0,■0),■min 2 e y =■4■-■x (0,■4),■max f y =■-2x2 (0,■0),■max 2 g y =■(x■-■5) (5,■0),■min h y =■1■+■x2 (0,■1)■min 5 In■each■of■the■following■state■whether■the■graph■is■wider■or■narrower■than y =■x2■and■whether■it■ has■a■maximum■or■a■minimum■turning■point. a y =■3x2 Narrower,■min b y =■-3x2 Narrower,■max 1

c y =■ 2 x2 e y g y

=■- 43 x2 2 = 3x

Wider,■min

1 5

d y =■- x2

y

Wider,■max

=■0.25x2

Narrower,■max

f

Narrower,■min

h y =■-0.16x2

Wider,■min Wider,■max

Chapter 9 Functions

291

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 6 a ■■■i■ ■Horizontal■ translation■1■ left ii (-1,■0) b i■ ■Refl■ected,■ narrower■ (dilation) ii (0,■0) c

unDerstAnDing 6 We7 ■■■For■each■of■the■following■quadratic■equations: i state■the■appropriate■dilation,■refl■ection■and■translation■of■the■graph■of y =■x2■needed■

to■obtain■the■graph

i■ ■Vertical■ translation■ 1■up ii (0,■1)

a y =■(x■+■1)2

b y =■-3x2

c y =■x2■+■1

d y =■ x2

e y =■x2■-■3

f

1 3

d i■ ■Wider■ (dilation) ii (0,■0)

g y =■- x2

e

y k y =■-x2■-■4

y =■(x■-■4)2 h y =■5x2 j y =■-(x■-■6)2 l y =■-(x■+■1)2

m y■=■2(x■+■1)2■-■4

n y■=■ 2 (x■-■3)2■+■2

i

2 5 =■-x2■+■2

1

3

7

p y■=■- 4 (x■-■1)2■-■ 2

reAsoning 7 A■vase■25■■cm■tall■is■positioned■on■a■bench■near■a■wall■as■shown.

y

The■shape■of■the■vase■follows■the■curve y =■(x■-■10)2,■where y cm■is■the■height■of■the■vase■and x cm■is■the■distance■of■the■vase■ from■the■wall. a How■far■is■the■base■of■the■vase■from■the■wall? 10■■cm b What■is■the■shortest■distance■from■the■top■of■the■vase■to■the■ wall? 5■■cm c If■the■vase■is■moved■so■that■the■top■just■touches■the■wall,■ fi■nd■the■new■distance■from■the■wall■to■the■base. 5■■cm d Find■the■new■equation■that■follows■the■shape■of■the■vase. y■=■(x■-■5)2

Digital doc

WorkSHEET 9.1 doc-5272

6 m■■■■i■ ■Narrower■(dilation),■horizontal■ translation■1■left,■vertical■ translation■4■down ■ ii (-1,■-4) n i■ ■Wider■(dilation),■horizontal■ translation■3■right,■vertical■ translation■2■up ii (3,■2)

o i■ ■Wider■(dilation),■refl■ected,■ horizontal■translation 1 2■left,■vertical■translation■ 4 ■up ii (-2,■ 14 ) p i■ ■Narrower■(dilation),■refl■ected,■ horizontal■translation■1■right,■ vertical■translation■ 3 ■down 2 ii (1,■- 32 )

reFleCtion 

Bench

x

 

What are the turning points of the graphs y = x2 + k and y = (x - h)2?

sketching parabolas in turning point form ■■ ■■ ■■

292

1

o y■=■- 3 (x■+■2)2■+ 4 ■

eBook plus

9C

1

Wall

i■ ■Vertical■ translation■3■ down ii (0,■-3) f i■ ■Horizontal■ translation■4■ right ii (4,■0) g i■ ■Refl■ected,■ wider■(dilation) ii (0,■0)

6 h■ ■■■i■ Narrower■(dilation) ii (0,■0) i i■ ■Refl■ected,■vertical■ translation■2■up ii (0,■2) j i■ ■Refl■ected,■horizontal■ translation■6■right ii (6,■0) k i■ ■Refl■ected,■vertical■ translation■4■down ii (0,■-4) l i■ ■Refl■ected,■horizontal■ translation■1■left ii (-1,■0)

ii state■the■coordinates■of■the■turning■point iii hence,■sketch■the■graph.

So■far,■to■sketch■quadratic■graphs■we■have■looked■at■transforming■the■graph■of y =■x2■by■ dilation,■refl■ection■or■translation.■ These■transformations■can■be■combined■into■what■is■called■the■turning■point■form■of■a■ quadratic■equation: y =■a(x■-■h)2■+■k.■ We■can■easily■fi■nd■some■of■the■key■features■of■a■parabola■from■the■turning■point■form■of■a■ quadratic■equation.■These■include: ■■(i)■ the■turning■point■(h,■k) ■(ii)■ ■whether■the■graph■is■upright■with■a■minimum■turning■point■(a■>■0)■or■the■graph■is■ inverted■with■a■maximum■turning■point■(a■<■0) (iii)■ ■whether■it■is■wider■(|a| > 1),■narrower■(|a| < 1)■or■the■same■width■(|a| = 1)■as■the■graph■of y =■x2.

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 8

For each of the following equations, state the coordinates of the turning point of the graph and whether it is a maximum or a minimum. a y = (x - 6)2 - 4              b  y = -(x + 3)2 + 2 Think a

b

Write a y = (x - 6)2 - 4

1

Write the equation.

2

Identify the transformations — horizontal translation of 6 units to the right and a vertical translation of 4 units down. State the turning point.

The turning point is (6, -4).

3

As a is positive (a = 1), the graph is upright with a minimum turning point. Write the equation.

Minimum turning point

1

b y = -(x + 3)2 + 2

2

Identify the transformations — horizontal translation of 3 units to the left and a vertical translation of 2 units up. State the turning point.

The turning point is (-3, 2).

3

As a is negative (a = -1), the graph is inverted with a maximum turning point.

Maximum turning point

■■ ■■ ■■

Other key features such as the x- and y-intercepts can also be determined from the equation of a parabola. The point(s) where the graph cuts or touches the x-axis are called the x-intercept(s). At these points, y = 0. The point where the graph cuts the y-axis is called the y-intercept. At this point, x = 0.

Worked Example 9

Determine i the y-intercept and ii the x-intercepts (where they exist) for the parabolas with equations: a  y = (x + 3)2 - 4          b  y = 2(x - 1)2          c  y = -(x + 2)2 - 1. Think a

Write a y = (x + 3)2 - 4

1

Write the equation.

2

Find the y-intercept by substituting x = 0 into the equation.

i y-intercept: when x = 0,

Find the x-intercepts by substituting y = 0 into the equation and solving for x. Add 4 to both sides of the equation. Take the square root of both sides of the equation. Subtract 3 from both sides of the equation. Solve for x.

ii x-intercepts: when y = 0,

3

y = (0 + 3)2 - 4 =9-4 =5 The y-intercept is 5.

(x + 3)2 - 4 = 0 (x + 3)2 = 4 (x + 3) = +2 or -2

x = 2 - 3  or   x = -2 - 3 x = -1     x = -5 The x-intercepts are -5 and -1. Chapter 9 Functions

293

number AND algebra • Linear and non-linear relationships

b

c

1

Write the equation.

2

Find the y-intercept by substituting x = 0 into the equation.

b y = 2(x − 1)2 i y-intercept: when x = 0,

y = 2(0 - 1)2 =2ì1 =2 The y-intercept is 2. ii x-intercepts: when y = 0, 2(x − 1)2 = 0 (x − 1)2 = 0 x−1=0 x=0+1 x=1 The x-intercept is 1.

3

Find the x-intercepts by substituting y = 0 into the equation and solving for x. Note that there is only one solution for x and so there is only one x-intercept. (The graph touches the x-axis.)

1

Write the equation.

2

Find the y-intercept by substituting x = 0 into the equation.

i y-intercept: when x = 0,

Find the x-intercepts by substituting y = 0 into the equation and solving for x. We cannot take the square root of -1 to obtain real solutions; therefore, there are no x-intercepts.

ii x-intercepts: when y = 0,

3

c

y = -(x + 2)2 - 1 y = -(0 + 2)2 - 1 = -4 - 1 = -5 The y-intercept is -5.

-(x + 2)2 - 1 = 0 (x + 2)2 = -1 There are no real solutions, so there are no x-intercepts.

Worked Example 10

For each of the following:    i write the coordinates of the turning point   ii state whether the graph has a maximum or a minimum turning point iii state whether the graph is wider, narrower or the same width as the graph of y = x2 iv find the y-intercept    v find the x-intercepts vi sketch the graph. a  y = (x - 2)2 + 3 b  y = -2(x + 1)2 + 6 Think a

a

y = (x - 2)2 + 3

1

Write the equation.

2

State the coordinates of the turning point from the equation. Use (h, k) since the equation is in the turning point form of y = a(x - h)2 + k where a = 1, h = 2 and k = 3.

i The turning point is (2, 3).

3

State the nature of the turning point by considering the sign of a.

ii The graph has a minimum turning point as

Specify the width of the graph by considering the magnitude of a.

iii The graph has the same width as y = x2 since

4

294

Write/Draw

Maths Quest 10 for the Australian Curriculum

the sign of a is positive.

a = 1.

number AND algebra • Linear and non-linear relationships

Find the y-intercept by substituting x = 0 into the equation.

iv y-intercept: when x = 0,

Find the x-intercepts by substituting y = 0 into the equation and solving for x. As we have to take the square root of a negative number, we cannot solve for x.

v x-intercepts: when y = 0,

7

Sketch the graph, clearly showing the turning point and the y-intercept.

vi

8

Label the graph.

5

6

y = (0 - 2)2 + 3 =4+3 =7 y-intercept is 7.

(x - 2)2 + 3 = 0 (x − 2)2 = -3 There are no real solutions, and hence no x-intercepts. y y = (x - 2)2 + 3 7 3

(2, 3)

0

b

1 2

3 4 5

6

7

8

2

x

b y = -2(x + 1)2 + 6 Write the equation. i The turning point is (-1, 6). State the coordinates of the turning point from the equation. Use (h, k) since the equation is in the turning point form of y = a(x - h)2 + k where a = -2, h = -1 and k = 6.

State the nature of the turning point by considering the sign of a.

ii The graph has a maximum turning point as

Specify the width of the graph by considering the magnitude of a.

iii The graph is narrower than y = x2

Find the y-intercept by substituting x = 0 into the equation.

iv y-intercept: when x = 0,

Find the x-intercepts by substituting y = 0 into the equation and solving for x.

v x-intercepts: when y = 0,

Sketch the graph, clearly showing ■ the turning point and the x- and y-intercepts.

the sign of a is negative. since |a| > 1.

y = -2(0 + 1)2 + 6 = -2 ì 1 + 6 =4 The y-intercept is 4.

-2(x + 1)2 + 6 = 0 2(x + 1)2 = 6 (x + 1)2 = 3 x + 1 = 3  or  x + 1 = - 3 x = -1 + 3      x = -1 - 3 The x-intercepts are -1 - 3 and -1 + 3 (or approximately -2.73 and 0.73).

vi

y (–1, 6) 4

Label the graph. –1 – 3

0

–1 + 3

x

y = –2(x + 1)2 + 6

Chapter 9 Functions

295

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

Unless■otherwise■stated,■exact■values■for■the■intercepts■should■be■shown■on■sketch■graphs.■ remember 1 a■ b c d e f g

1.■ If■the■equation■of■a■parabola■is■in■turning■point■form, y =■a(x■-■h)2■+■k,■then■the■turning■ point■is■(h,■k). 2.■ If■a■is■positive,■the■graph■is■upright■with■a■minimum■turning■point. 3.■ If■a■is■negative,■the■graph■is■inverted■with■a■maximum■turning■point. 4.■ If■the■magnitude■of■a■is■greater■than■1,■the■graph■is■narrower■than■the■graph■of y =■x2. 5.■ If■the■magnitude■of■a■is■between■0■and■1,■the■graph■is■wider■than■the■graph■of y =■x2. 6.■ To■fi■nd■the■y-intercept,■substitute x =■0■into■the■equation. 7.■ To■fi■nd■the■x-intercepts,■substitute y =■0■into■the■equation■and■solve■for■x.■

(1,■2),■min (-2,■-1),■min (-1,■1),■min (2,■3),■max (5,■3),■max (-2,■-6),■min 1

3

(- 2 ,■- 4 ),■min 1 2

h ( 3 ,■ 3 ),■min i (-0.3,■-0.4),■min

exerCise

sketching parabolas in turning point form

9C

FluenCy

inDiviDuAl pAthWAys

1 We8 ■For■each■of■the■following■equations,■state■the■coordinates■of■the■turning■point■of■the■

eBook plus

Activity 9-C-1

Reviewing turning point form doc-5080

Activity 9-C-3

Interpreting turning point form trends doc-5082

b c d e f

g y =■(x■+ 2 )2■-

i

3 4

1

h y =■(x■- 3)2■+

2 3

y =■(x■+■0.3)2■-■0.4

2 For■each■of■the■following■state: i the■coordinates■of■the■turning■point ii whether■the■graph■has■a■maximum■or■a■minimum■turning■point iii whether■the■graph■is■wider,■narrower■or■the■same■width■as■that■of y =■x2. a y =■2(x■+■3)2■-■5 b y =■-(x■-■1)2■+■1 c y =■-5(x■+■2)2■-■4 1 1 1 1 f y =■0.2(x■+ )2■-■ 2 d y =■ (x■-■3)2■+■2 e y =■- (x■+■1)2■+■7

Activity 9-C-2

i■ (-3,■-5) iii Narrower i■ (1,■1) iii Same i■ (-2,■-4) iii Narrower i■ (3,■2) iii Wider i■ (-1,■7) iii Wider 1 1 i■ (- 5,■- 2 )

c y =■(x■+■1)2■+■1 f y =■(x■+■2)2■-■6

1

Turning point form doc-5081

a

graph■and■whether■it■is■a■maximum■or■a■minimum. b y =■(x■+■2)2■-■1 e y =■-(x■-■5)2■+■3

a y =■(x■-■1)2■+■2 d y =■-(x■-■2)2■+■3

4

5

2

3 Select■the■equation■that■best■suits■each■of■the■following■graphs. i ii y y ii Min 3

iii

y

1

ii Max ii Max

0

ii Min ii Max

-3 iv

v

y

ii Min

vi

y

y

3

iii Wider

0 1 3 i■ b■ y■=■-(x■-■2)2■+■3 ii e■ y■=■-x2■+■1 iii f■ y■=■(x■+■1)2■-■3 iv d■ y■=■-(x■+■2)2■+■3 v c■ y■=■x2■-■1 vi a■ y■=■(x■-■1)2■-■3 296

x

-1 0

x

0

x

2

-2

0

x

a y =■(x■-■1)2■-■3 d y =■-(x■+■2)2■+■3

maths Quest 10 for the Australian Curriculum

0 -1

b y =■-(x■-■2)2■+■3 e y =■-x2■+■1

x -3 c y =■x2■-■1 f y =■(x■+■1)2■-■3

x

number AND algebra • Linear and non-linear relationships 6 a     i  (4, 2)   ii  Min iii  Same width iv  18  v  No x-intercepts b     i  (3, -4)   ii  Min iii  Same width iv  5   v  1, 5 c     i  (-1, 2)   ii  Min iii  Same width iv  3  v  No x-intercepts d   i  (-5, -3)   ii  Min iii  Same width iv  22   v -5 - 3, -5 + 3 (approx. -6.73, -3.27) e   i  (1, 2) ii  Max iii  Same width iv  1   v 1 - 2, 1 + 2 (approx. -0.41, 2.41) f   i  (-2, -3) ii  Max iii  Same width iv  -7 v  No x-intercepts g   i  (-3, -2) ii  Max iii  Same width iv  -11 v  No x-intercepts h   i  (1, 3)   ii  Min iii  Narrower iv  5   v  No x-intercepts i   i  (-2, 1) ii  Max iii  Narrower iv  -11 1 1   v -2 , -2 + 3 3 (approx. -2.58, -1.42)

1

1 3

4   MC  a The translations required to change y = x2 into y = (x - 2 )2 + are: ✔ a

1

right 2 , up 13 1 2 1 , 3

c right , down e right

up

1 2 1 , 2

b left , down 1 3

d left

1 3

1

up 3

1 2 1

1 2

1 3

b For the graph 4 (x - )2 + , the effect of the

1 4

on the graph is:

no effect to make the graph narrower to make the graph wider to invert the graph 1 e to translate the graph up of a unit 4

a b ✔ c d

c Compared to the graph of y = x2, y = -2(x + 1)2 - 4 is: a inverted and wider ✔ b inverted and narrower c upright and wider d upright and narrower e inverted and the same width d A graph with minimum turning point (1, 5) and which is narrower than the graph of

y = x2 is: a y = (x - 1)2 + 5

✔ c

y = 2(x - 1)2 + 5

1 2

b y = (x + 1)2 + 5 d y = 2(x + 1)2 + 5

1 2

e y = (x - 1)2 + 5 e Compared to the graph of y = x2, the graph of y = -3(x - 1)2 - 2 has the following

features. 5 a i  -3 ii -3, 1 Maximum TP at (-1, -2), narrower b i  12 ii 2 Maximum TP at (1, -2), narrower c i  -18 ii No x-intercepts Maximum TP at (1, 2), wider d i  -5 ii -1, 5 e i  4 ii No x-intercepts Minimum TP at (1, -2), narrower f i  4 ii -3 - 5, -3 + 5 (approx. -5.24, -0.76) Minimum TP at (-1, -2), wider 5   WE9  Determine i the y-intercept and ii the x-intercepts (where they exist) for the parabolas with equations: a y = (x + 1)2 - 4 b y = 3(x - 2)2 c y = -(x + 4)2 - 2 2 2 d y = (x - 2) - 9 e y = 2x + 4 f y = (x + 3)2 - 5 a ✔ b c d e

Understanding 6   WE10  For each of the following:   i write the coordinates of the turning point  ii state whether the graph has a maximum or a minimum turning point iii state whether the graph is wider, narrower or the same width as the graph of y = x2 iv find the y-intercept  v find the x-intercepts vi sketch the graph. a y = (x - 4)2 + 2 b y = (x - 3)2 - 4 c y = (x + 1)2 + 2 2 2 d y = (x + 5) - 3 e y = -(x - 1) + 2 f y = -(x + 2)2 - 3 2 2 g y = -(x + 3) - 2 h y = 2(x - 1) + 3 i y = -3(x + 2)2 + 1 7 Consider the equation 2x2 - 3x - 8 = 0. a Complete the square. 2(x - 43 )2 - 73 =0 8 x = 43 ê 473 b Use the result to determine the exact solutions to the original equation. c Determine the turning point of y = 2x2 - 3x - 8 and indicate its type. ( 43 , - 73 ), minimum 8 Chapter 9 Functions

297

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 8 a■ ■Find■the■equation■of■a■quadratic■which■has■a■turning■point■of■(-4,■6)■and■has■an■x-intercept■ at■(-1,■0).■ ■ y■=■-23(x■+■4)2■+■6 b State■the■other■x-intercept■(if■any).■ (-7,■0) p ($) 1.9

reAsoning 9 The■price■of■shares■in■fl■edgling■company■‘Lollies’r’us’■plunged■dramatically■one■afternoon,■

1.4 1.0

0

3

following■the■breakout■of■a■small■fi■re■on■the■premises.■However,■Ms■Sarah■Sayva■of■Lollies■ Anonymous■agreed■to■back■the■company,■and■share■prices■began■to■rise. ■ Sarah■noted■at■the■close■of■trade■that■afternoon■that■the■company’s■share■price■followed■ 5 t (Hours after 12 pm.) the■curve:■P■=■0.1(t■-■3)2■+■1■where■$P■is■the■price■of■shares■t■hours■after■noon. a Sketch■a■graph■of■the■relationship■between■time■and■share■price■to■represent■the■ situation. b What■was■the■initial■share■price? $1.90 c What■was■the■lowest■price■of■shares■that■afternoon? $1 d At■what■time■was■the■price■at■its■lowest? 3■■pm e What■was■the■fi■nal■price■of■‘Lollies’r’us’■shares■as■trade■closed■at■5■pm? $1.40 10 Rocky■is■practising■for■a■football■kicking■competition.■After■being■kicked,■the■path■that■the■ball■ follows■can■be■modelled■by■the■quadratic■relationship: h=

−1 (d − 15)2 + 8, 30

where■h■is■the■vertical■distance■the■ball■reaches■(in■metres),■and■d■is■the■horizontal■distance■ (in■metres).■ Maximum■height■ is■8■metres■when■ horizontal■distance■ is■15■metres.

a Determine■the■initial■vertical■height■of■ the■ball. ■ 0.5■m b Determine■the■exact■maximum■horizontal■ distance■the■ball■travels.■ (15■+■4 15)■m c Write■down■both■the■maximum■height■and■

the■horizontal■distance■when■the■maximum■ height■is■reached.■

9D eBook plus

Interactivity Sketching parabolas

int-2785

reFleCtion 

 

Does a in the equation y = a(x - h)2 + k have any impact on the turning point?

sketching parabolas of the form y = ax 2 + bx + c ■■ ■■

The■standard■form■of■a■quadratic■equation■is y =■ax2■+■bx■+■c■where■a,■b■and■c■are■ constants.■ As■seen■in■the■previous■section,■to■sketch■a■parabola■we■need■to■know: 1.■y-intercept 2.■x-intercepts 3.■the■nature■of■the■turning■point;■that■is,■whether■it■is■a■maximum■or■a■minimum■ turning■point 4.■the■coordinates■of■the■turning■point.

Finding the turning point of a parabola when the equation is not in turning point form 1. Changing to turning point form ■■ As■seen■previously,■when■an■equation■is■written■in■turning■point■form■the■coordinates■of■the■ turning■point■can■be■read■from■the■equation.■That■is,■the■coordinates■of■the■turning■point■for y =■a(x■-■h)2■+■k■are■(h,■k). ■■ To■change■to■turning■point■form■we■use■the■completing■the■square■method.■ 298

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Worked Example 11

Change each of the following equations into turning point form and hence state the coordinates of the turning point for each one. a  y = x2 + 6x + 2 b  y = -2x2 + 7x - 3 Think a

b

Write

1

Write the equation.

2

Complete the square:  (i) Halve the coefficient of x and square it. (ii) Add this new term to the equation, placing it after the x-term. (iii) Balance the equation by subtracting the same term from the right-hand side of the equation.

3

Factorise the perfect square and simplify the remaining terms.

4

State the coordinates of the turning point.

1

Write the equation.

2

Take out a common factor.

3

Complete the square:  (i) Halve the coefficient of x and square it. (ii) Add this new term to the equation, placing it after the x-term. (iii) Balance the equation by subtracting the same term from the right-hand side of the equation.

a y = x2 + 6x + 2  6

2

 6

2

= x2 + 6x +  2  −  2  + 2 = x2 + 6x + (3)2 - (3)2 + 2 = (x2 + 6x + 9) - 9 + 2

= (x + 3)2 - 9 + 2 = (x + 3)2 - 7 The turning point is (-3, -7). b y = -2x2 + 7x - 3

= -2  x 2 − 7 x + 2



= −2  x 2 − 72 x + = = =

3  2 1  2

2

1

7 2

× − 72  −  2 × − 2  +

 2 2 3  7 −2  x 2 − 2 x +  − 47  −  − 47  + 2     2 7 49 3 49  8 −2  x − x + 16  − 16 + 2 × 8  2    2 7  −2  x − x + 49  − 49 + 24  2 16 16 16   2   −2  x − 7  − 25  2 16  

4

Factorise the perfect square and simplify the remaining terms.

=

5

Multiply the common factor by each term in the square brackets so that the equation is in turning point form.

 = −2  x − 47  +

6

State the coordinates of the turning point.

2

3  2 

25 8

The turning point is

 7 25  ,  4 8 

or 13 , 31  . 4 8

Worked Example 12

Sketch the graph of y = 2x2 - 4x - 2 using the completing the square method to find the coordinates of the turning point. Show all relevant points. Think

Write/draw

1

Write the equation.

y = 2x2 - 4x - 2

2

Find the y-intercept by substituting x = 0.

y-intercept: when x = 0, y=0-0-2 = -2 The y-intercept is -2. Chapter 9 Functions

299

number AND algebra • Linear and non-linear relationships

3

Find the x-intercepts by substituting y = 0. Factors cannot be easily found, so use the quadratic formula to solve for x.

x-intercepts: when y = 0, 0 = 2x2 - 4x - 2 2x2 - 4x - 2 = 0 2(x2 - 2x - 1) = 0 x2 - 2x - 1 = 0 − b ± b 2 − 4 ac 2a where a = 1, b = -2, c = -1 x=

2 ± (−2)2 − 4(1)(−1) 2 2± 8 x= 2

x=

2±2 2 2 The x-intercepts are 1 - 2 and 1 + 2 (approx. -0.41 and 2.41). x=

4

Find the turning point by taking out a common factor from the original equation.

y = 2x2 - 4x - 2 = 2(x2 - 2x - 1)

5

Complete the square:  (i)  Halve the coefficient of the x-term and square it. (ii) Add this new term to the equation, placing it after the x-term. (iii) Balance the equation by subtracting the same term from the right-hand side of the equation.

  2 2 2 2 = 2  x 2 − 2 x + − 2 − − 2 − 1   = 2[(x2 - 2x + (-1)2) - (-1)2 - 1] = 2[(x2 - 2x + 1) - 1- 1]

6

Factorise the perfect square and simplify the remaining terms.

= 2[(x - 1)2 - 2]

7

Multiply the common factor by each term in the square brackets so that the equation is in turning point form. State the coordinates of the turning point.

= 2(x - 1)2 - 4

State the nature of the turning point. As the sign of a is positive, the parabola has a minimum turning point.

The parabola has a minimum turning point.

8 9

10

Sketch the graph.

( ) ( )

Turning point is (1, -4).

y y = 2x2 - 4x - 2

11

Label the graph. 1- 2 0 1 -2 -4

1+ 2 x

(1, -4)

2. Using the x-intercepts to find the x-coordinate of the turning point ■■ A parabola is symmetrical, so the x-intercepts are the same distance from the axis of symmetry (the line which divides the graph exactly in half ). This means that the x-coordinate of the turning point is halfway between the x-intercepts. 300

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

■■

y

In this graph, the x-intercepts are exactly 1 unit from the y-axis which is the axis of symmetry. One way to find the x-coordinate of the turning point is to calculate the average of the two x-intercepts. In this case, 1 + −1 = 0 is the x-coordinate of the

y = x2 - 1

2

turning point. The y-coordinate of the turning point can then be found by substituting the x-coordinate into the equation.

-1 0 -1

1

x

Worked Example 13

Sketch the graph of y = x2 - 10x + 21, using the x-intercepts to find the coordinates of the turning point. Think

Write/draw

1

Write the equation.

y = x2 - 10x + 21

2

Find the y-intercept by substituting x = 0.

y-intercept: when x = 0, y = 0 - 0 + 21 = 21 The y-intercept is 21.

3

Find the x-intercepts by substituting y = 0.

x-intercepts: when y = 0, x2 - 10x + 21 = 0

4

Factorise and solve for x by using the Null Factor Law.

5

Find the x-coordinate of the turning point by averaging the x-intercepts, x1 + x 2 . (This is halfway between the 2 x-intercepts.)

(x - 7)(x - 3) = 0 x - 7 = 0  or  x - 3 = 0 x = 7      x = 3 The x-intercepts are 3 and 7. 3+7 x-value of the turning point = 2 10 = 2 =5

6

Find the y-coordinate of the turning point by substituting the x-coordinate into the equation and solving for y.

When x = 5, y = 52 - 10(5) + 21 = -4

7

State the coordinates of the turning point.

The turning point is (5, -4).

8

State the nature of the turning point. As the sign of a is positive, the parabola has a minimum turning point.

Parabola has a minimum turning point.

9

Sketch the graph.

10

Label the graph.

y

y = x2 - 10x + 21

21

0 -4

3 5 7 (5, -4)

x

Chapter 9 Functions

301

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 2 a y■=■(x■+■1)2■-■6,■ x-intercepts■ are■-1■ê■ 6■ (ö■-3.4,■1.4)

remember

1.■ If■the■equation■is■in■the■form y =■ax2■+■bx■+■c,■the■coordinates■of■the■turning■point■can■ be■found■by: (a)■ ■using■the■completing■the■square■method■to■change■the■equation■into■turning■point■form (b)■■fi■nding■the■x-coordinate■of■the■point■exactly■halfway■between■the■two■x-intercepts.■ This■is■the■x-coordinate■of■the■turning■point.■Then■substitute■the x-value■into■the■ equation■to■fi■nd■the■y-coordinate. −b .■Then■substitute■the■x-value■into■the■equation■to■fi■nd■the■y-coordinate. (c)■ ■using■ x = 2a 2.■ The■graph■should■also■show■both■the■y-intercept■and■the■x-intercepts■of■the■parabola■if■ they■exist.

y = x2 + 2x - 5 y

-1 - 6

-1 0

x -1 + 6

-5 (-1, -6) -6

exerCise

9D inDiviDuAl pAthWAys

sketching parabolas of the form y = ax 2 + bx + c FluenCy 1 We11 ■Change■each■of■the■following■equations■into■turning■point■form■and■write■the■

eBook plus

Activity 9-D-1

Understanding parabola sketching doc-5083

2

Activity 9-D-2

Parabola sketching doc-5084 Activity 9-D-3

Sketching tricky parabolas doc-5085

3 eBook plus

Digital doc

SkillSHEET 9.3 doc-5268

4 eBook plus

Digital doc

SkillSHEET 9.4 doc-5269

5 y -–1 2

-9

0

y = 2x2 - 17x - 9 9 x

(4 1–4, -451–8 )

✔

coordinates■of■the■turning■point■for■each■one. a y =■x2■+■4x■-■2 b y =■x2■+■12x■-■4 c y =■x2■-■8x■+■6 2 2 d y =■x ■-■2x■+■12 e y =■x ■+■3x■+■1 f y =■x2■+ x■-■2 2 2 g y =■x ■+■7x■+■2 h y =■2x ■+■4x■+■8 i y =■3x2■-■12x■+■6 We12 ■Sketch■the■graph■of■each■of■the■following■using■the■completing■the■square■method■to■ fi■nd■the■coordinates■of■the■turning■point.■Show■all■relevant■points. a y =■x2■+■2x■-■5 b y =■x2■-■4x■+■7 c y =■x2■+ 6x■-■3 d y =■x2■-■5x■+■1 e y =■-x2■-■5x■+■1 f y =■-x2■+ x■-■3 g y =■3x2■+■3x■-■12 h y =■-5x2■+■10x■-■35 2 i y =■-7x ■-■7x■+■49 We13 ■Sketch■the■graph■of■each■of■the■following,■using■the■x-intercepts■to■fi■nd■the■coordinates■ of■the■turning■point.■ 3 a■ y = x2 + x - 12y a y =■x2■+ x -■12 b y =■x2■-■12x■+■32 0 -4 c y =■x2■-■8x■-■9■ d y =■-x2■-■6x■-■8 2 2 e y =■-x ■-■6x■+■27 f y =■-x ■+■2x■+■35 g y =■x2■+■4x■- 5 -12 (- 1–, -12 1–) Sketch■the■graphs■of■each■of■the■following. 2 4 a y =■2x2■-■17x■-■9 b y =■3x2■-■23x■+■14 2 c y =■5x ■+■27x■+■10 d y =■6x2■+■7x■-■3 2 e y =■-2x ■+■7x■+■4 f y =■-2x2■+■11x■+■21 2 g y =■-6x ■+■5x■+■6 h y =■-18x2■+■67x■-■14 2 i y =■2x ■-■7x■+■8 y mC ■a■ The■equation■that■best■suits■the■graph■shown■is: a y =■x2■+■2x■-■24 1 a■ y■=■(x■+■2)2■-■6,■(-2,■-6) b y =■3x2■+■6x■-■72 b y■=■(x■+■6)2■-■40,■(-6,■-40)■ c y =■x2■-■2x■-■24 c y■=■(x■-■4)2■-■10,■(4,■-10) d y =■3x2■-■6x■-■72 d y■=■(x■-■1)2■+■11,■(1,■11) 0 5 5 3 3 (-6, 0) (4, 0) x e y =■2x2■+■4x■-■72 e y■=■(x■+■ 2 )2■-■ 4 ,■(-■2 ,■-■4 )■■ 9

1

9

f y■=■(x■+■ 12 )2■-■ 4 ,■(- 2 ,■- 4 )■ g

41 41 7 7 y■=■(x■+■ 2 )2■-■ 4 ,■(- 2 ,■- 4 ) y■=■2(x■+■1)2■+■6,■(-1,■6)

h i y■=■3(x■-■2)2■-■6,■(2,■-6) 302

maths Quest 10 for the Australian Curriculum

-72

3

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

eBook plus

Digital doc

b The■equation y =■x2■+■5x■-■24■is■best■represented■by: a b y

SkillSHEET 9.5 doc-5270

y 24

0

-3

eBook plus

Digital doc

8

x

-3 0

x

8

-24

SkillSHEET 9.6 doc-5271

✔ c

d

y

y 24

0

-8

3

x

3

x

-8

0

3

x

2

3

x

-24

e

y

0

-8

-12

6 Match■each■of■the■following■graphs■with■the■appropriate■equation. 6 a■ b c d

iv vii vi iii

a

b

y

y 0

0

c

4

x

d

y

1

y 9

-2 0

6 x

-3

0

3

x

Chapter 9 Functions

303

number AND algebra • Linear and non-linear relationships

e

6 e i f viii g ii h v

f

y

y

5 0 1

g

5

x

0

h

y

-1

0

1

3– 2

x

5– 2

y

x

0

1

x

-1

i   y = x2 - 6x + 5 iii   y = -x2 + 9   v   y = 2x - 2x2 vii   y = -x2 + 5x - 6

 ii   y = x2 - 1     iv   y = 3x2 - 12x     vi   y = -x2 + 4x + 12 viii   y = -4x2 + 16x - 15

Understanding 7 a Find the equations for parabolas A and B in both turning point form and standard form. b For the two parabolas, A and B, state the transformations on A to create B. Translated 4 units to the left and 5 units down, reflected in the x-axis, and dilated by 32 in the y-direction.

A

6

y f1(x) =

6·x + 5

2

f2(x) = 2·x + 1 1 0

1

x

-6

0 -4 -2 (-1, -1) -2

(-2, -3)

B

2

4

6x

-4 -6

8 Use a graphical method on your calculator to find the points of intersection of these parabolas, correct to 2 decimal places. (-0.32, 3.18) and (-4.68, -1.18)

y = x2 + 6x + 5 y = -x2 - 4x + 2 9 Use simultaneous equations to show that the parabola y = x2 + 6x + 5 and the straight line y = 2x + 1 intersect at one point only. a Find the coordinates of this point of intersection. (-2, -3) b Verify this by graphing the simultaneous equations on your calculator and finding the point of intersection. 304



(3, 4)

4 x2 +

A: y = 2(x - 3)2 + 4, y = 2x2 - 12x + 22; B: y = -3(x + 1)2 - 1, y = -3x2 - 6x - 4

y 8

Maths Quest 10 for the Australian Curriculum



number AND algebra • Linear and non-linear relationships Reasoning

y 14

10 Consider the family of parabolas y = x2 + px + 5. a Sketch, on the same set of axes, the parabolas for p = -4, -2, 2, 4. b Discuss the effect of p on the graph.

12 10 8

11 The height, h metres, of a model rocket above the ground t seconds after launching is given by

6 4 2 5x

0

-5

-2 y = x2 + 2x + 5

y = x2 - 2x + 5

y = x2 + 4x + 5

y = x2 - 4x + 5

11 a

h 2500

(25, 2500)

50 t

0

y 200







(10, 200)

20 x

0

h 17

-1.72 0

h = -4.9t2 + 1.5t + 17

t

2.02

A 11 250

0

(75, 11 250)

150 x

the equation h = 4t(50 - t) for 0 Ç t Ç 50. If p < 0, the turning point is on the right side of the y-axis. If a Sketch the graph of the rocket’s flight. p > 0 the turning point is on the b Determine the height of the rocket when it is launched. left side of the y-axis. As the h = 0 magnitude of p increases the c What is the greatest height the rocket reaches? 2500  m turning point moves away from d After how long does the rocket reach the greatest height? the y-axis. All graphs have the same y-intercept (0, 5). e How long is the rocket in the air? 50  s 25  s after launching 12 A farmer decides to fence a new rectangular paddock with the greatest possible area, using an existing fence for one side and 40 metres of fencing to make the other three sides. Let the area of the paddock be defined as A  m2. a Write an equation using x and y to describe the area of the paddock. A = xy  m2 b Write an equation relating x and y and the length of fencing Existing fence available. 2x + y = 40  m c Rearrange the equation so that y is the subject. y = (40 - 2x)  m d Substitute this value of y into the equation for the area. A = 2x(20 - x)  m2 xm e Using the intercept method find the coordinates of the New paddock turning point. (10, 200) xm f Sketch the graph. ym g Use the graph to find the maximum area of the paddock and its dimensions. Maximum area is 200  m2, paddock is 10  m wide and 20  m long. 13 A daring feat performed in Acapulco, Mexico, is for a person to dive from a cliff into the ocean. Starting from about 17 metres above the water, ■ the height, h (in metres), of a diver t seconds after he jumps can be represented by the equation ■ h = -4.9t2 + 1.5t + 17. a Sketch a graph to represent the diver’s height after jumping. b How long does it take for the diver to reach the water (to the nearest second)? 2  s c When does the diver reach his maximum height above the water? Give your answer correct to 2 decimal places. 0.15  s d What is the diver’s greatest height above the water? Give your answer correct to the nearest cm. 17.11  m 14 A farmer has 300 metres of fencing with which to fence 3 sides of a rectangular paddock. a Using the method described in question 8, find an equation relating the area and the width of the paddock. A = 2x(150 - x)  m2 b Sketch the graph. 11  250  m2, 75  m and 150  m c Use the graph to find the greatest possible area for the paddock and its dimensions. Chapter 9 Functions

305

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 15 In■preparation■for■an■Archery■Games■opening■ceremony,■an■archer■shoots■a■fi■ery■arrow■that■

follows■a■parabolic■path■into■a■cauldron.■The■cauldron■is■15■metres■high■with■its■centre■a■ distance■of■10■metres■from■a■wall. The■archer■releases■the■arrow■at■a■distance■of■34■metres■horizontally■from■the■wall■as■ represented■by■the■diagram■below. P

Wall

Cauldron

15 m 10 m

34 m

This■event■can■be■represented■on■a■set■of■axes■and■the■path■of■the■arrow■can■be■modelled■by■ a■quadratic■equation■of■the■form■y■=■ax2■+■bx■+■c. a Given■that■if■the■cauldron■was■not■in■the■way,■the■arrow■would■land■2■metres■from■the■ wall.■Show■that■the■path■of■the■arrow■can■be■represented■by■the■equation −5 2 45 85 −5 45 ,■b■=■16 ,■c■=■ −85 y= x + x − . ■ a■=■ 64 16 64 16 16 b Use■your■calculator■to■graph■the■equation■ reFleCtion    and■hence■fi■nd■the■exact■coordinates■ Which feature is most clearly displayed in where■the■arrow■reaches■its■maximum■ an equation of the type y = ax 2 + bx + c: (18,■20) height,■P. the x-intercept(s), the y-intercept or the c Convert■the■equation■in■part■a■to■turning■ turning point? point■form■to■show■that■your■answer■to■b■ −5 (x■-■18)2■+■20 is■correct. y■=■ 64

eBook plus

Digital doc

WorkSHEET 9.2 doc-5273

9e

exponential functions and their graphs ■■ ■■

Relationships■of■the■form y =■ax■are■called■exponential functions■with■base■a,■where■a■is■a■ real■number■not■equal■to■1,■and x is■the■index■power■or■exponent. The■term■‘exponential’■is■used,■as■the■independent■variable x is■the■exponent■(or■index).

WorkeD exAmple 14

Complete the table of values below and use it to plot the graph of y = 2x. x

-4

-3

-2

-1

0

1

2

3

4

y think 1

2

306

Substitute■each■value■of x into■the■function y =■2x■to■ obtain■the■corresponding y-value. Plot■each■point■generated■on■a■set■of■axes.

maths Quest 10 for the Australian Curriculum

Write/DrAW

x

-4 -3 -2 -1

0■

1

2

3

4

y

1 16

1

2

4

8

16

1 8

1 4

1 2

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

3

y

Join■with■a■smooth■curve.

y = 2x

18 16 14 12 10 8 6 4 2 -4 -3 -2 -1 0

4

x

1 2 3 4

Label■the■graph.

eBook plus

Interactivity Exponential graphs

int-1149

The■graph■in■Worked■example■14■has■several■important■features. ■■ The■graph■passes■through■(0,■1).■That■is,■the■y-intercept■is■1.■The■graph■of■any■equation■in■the■ form y =■ax■will■pass■through■this■point. ■■ The■graph■passes■through■the■point■(1,■2).■All■graphs■of■the■form y =■ax■will■pass■through■the■ point■(1,■a). ■■ y >■0■for■all■values■of■x.■You■will■notice■that■for■negative■values■of x, the■graph■gets■very■close■ to■but■will■never■touch■the■x-axis.■When■this■occurs,■the■line■that■the■graph■approaches■is■ called■an■asymptote.■The■equation■of■the■asymptote■for y =■ax■is y =■0;■i.e.■the■x-axis.

WorkeD exAmple 15 a Plot the graph of y = 3 ì 2x for -3 Ç x Ç 3. c Write the equation of the horizontal asymptote.

b State the y-intercept.

think a

1

2

Write/DrAW

Prepare■a■table■of■values■taking x-values■ from■-3■to■3.■Fill■in■the■table■by■substituting■ each■value■of x into■the■given■equation.■

a

Draw■a■set■of■axes■on■graph■paper■to■plot■ the■points■from■the■table■and■join■them■with■ a■smooth■curve.

x

-3

-2

-1

0■

1

2

3

y

3 8

3 4

1 12

3

6

12

24

24 22 20 18 16 14 12 10 8 6 4 2 -3 -2 -1 0

3

y y = 3 ì 2x

y=0 1 2 3

x

Label■the■graph.

b

Locate■where■the■curve■cuts■the■y-axis.■ Alternatively,■fi■nd■the■y-value■for x =■0■in■the■table.

b The■y-intercept■is■3.

c

Find■an■imaginary■line■to■which■the■curve■gets■ closer■and■closer■but■does■not■cross.■As■it■is■a■ horizontal■asymptote,■the■equation■will■be■of■the■ form y =■constant.

c

The■equation■of■the■asymptote■is y =■0.

Chapter 9 Functions

307

number AND algebra • Linear and non-linear relationships

■■

■■

Compare the graphs drawn in Worked examples 14 and 15. When 2x was multiplied by a constant, the graph was dilated; that is, its width changed. Since the constant was a positive number greater than 1, the graph became narrower. If the constant had been a fraction or decimal between 0 and 1, the graph would have become wider. The following worked example considers the effect of a negative exponent.

Worked Example 16

Plot the graph of y = 3-x for -3 Ç x Ç 3, clearly showing the y-intercept and the horizontal asymptote. Think 1

Write/draw

Draw up a table of values.

2

Substitute the values of x into the equation to find the corresponding y-values.

3

Draw a set of axes, plot the points generated from the table and join with a smooth curve.

x

-3

-2

-1

0

1

2

3

y

27

 9

 3

1

1 3

1 9

1 27

y y = 3–x

28 26 24 22 20 18 16 14 12 10 8 6 4 2

-3 -2 -1 0

4

y=1 1 2 3

y=0 x

Label the graph.

Further exponential graphs ■■ ■■ ■■

Recall the dilation, reflection and translation rules for quadratic graphs. These rules also apply to exponential graphs. Adding to or subtracting from the basic function shifts the graph up or down the y-axis. y = ax + k  or  y = ax - k Adding to or subtracting from x shifts the graph left or right along the x-axis. y = ax - h  or  y = ax + h remember

1. Relationships of the form y = ax, where a ò 1 are called exponential functions with base a. 2. To obtain the graph of an exponential function, construct a table of values first and then plot the points from the table and join them with a smooth curve. Alternatively use a graphics calculator, CAS calculator or graphing software. 3. An asymptote is a line which the graph approaches but never cuts or touches. 4. Multiplying by a constant dilates the basic graph — that is, makes it narrower or wider. 5. If x is a negative number, the graph is reflected across the y-axis. 308

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

exponential functions and their graphs

9e

You■may■use■a■graphing■calculator■or■graphing■software■to■assist■you■in■this■exercise.

inDiviDuAl pAthWAys

FluenCy 1 a■ ■ We14 ■Complete■the■table■below■and■use■the■table■to■plot■the■graph■of y =■10x■for■

eBook plus

-2 Ç■x Ç■2.

Activity 9-E-1

Exploring exponential graphs doc-5086 Activity 9-E-2

Features of exponential graphs doc-5087

Digital doc

y = 10x

2 a■

y

2 x

1

y = 4x

80

8 9

60

10

40 20

(1, 4)

-4 -3 -2 -1 0 1 2 3 4 x

b■

y

y = 5x

100 80 60 40 (0, 1)

20

y

y = 6x

100 80 60 40 (0, 1) 20

1

2

3

4

1

10

100

1000

10■000

Providing■a■realistic■scale■is■diffi■cult.

c y =■6x

x

-3

-2

-1

0

1

2

3

2x

0.125

0.25

0.5

1

2

4

8

3■ì■2x

0.375

0.75

1.5

3

6

12

24

1 ■ì■2x 5

0.025

0.05

0.1

0.2

0.4

0.8

1.6

11

equation y =■k■ì■ax. We16 ■Plot■the■graph■of y =■2-x for■-3■Ç x Ç■3,■clearly■showing■the■y-intercept■and■the■ The■coeffi■cient,■k,■affects■the■ horizontal■asymptote. steepness■of■the■graph:■the■larger■the■ x -x On■the■one■set■of■axes,■sketch■the■graphs■of y =■3 ■and■y =■3 . value■of■k,■the■steeper■the■graph. Use■your■answer■to■question■9■to■describe■the■effect■of■a■negative■index■on■the■graph■of y =■ax. The■negative■index■refl■ects■the■graph■in■the■y-axis. a■ ■■Complete■the■table■of■values■below■and■use■the■points■generated■to■sketch■the■graph■of

( ).

y =■

1 2

x

x

-3

-2

-1

0

1

2

3

y

8

4

2

1

0.5

0.25

0.125

(1, 5)

-4 -3 -2 -1 0 1 2 3 4 x

c■

0

1 10

7 Study■the■graphs■in■question■6■and■state■the■effect■that■the■value■of■k■has■on■graphs■with■

100

(0, 1)

-1

1 10 0

y =■ 15■ì■2x on■the■same■set■of■axes.

10 0

-2

6 Complete■the■following■table■of■values■and■then■plot■the■graph■of■y =■2x,■y =■3■ì■2x,■and■

y

-1

1 10 00

4 Use■your■answer■to■question■3■to■describe■the■effect■of■increasing■the■value■of■a■on■the■graph■ Increasing■the■value■of■a■increases■ of y =■ax. the■steepness■of■the■graph■where■x■is■ x 5 We15 ■a■ Plot■the■graph■of y =■2■ì■3 ■for■-3 Ç■x Ç■3. positive■and■fl■attens■the■graph■where■x■ b State■the■y-intercept. 2 is■negative. c Write■the■equation■of■the■horizontal■asymptote. y■=■0

SkillSHEET 9.7 doc-5274

-2

-3

3 On■the■one■set■of■axes,■draw■the■graphs■of y =■2x, y =■2-x,■y =■3x, y =■3-x,■y =■4x and■y =■4-x.

eBook plus

(0, 1)

y

1 10 000

b Why■would■it■be■diffi■cult■to■draw■the■graph■for■-4 Ç■x Ç■4?

Tricky exponential graphs doc-5088

100

-4

2 Plot■the■graph■of■each■of■the■following■exponential■functions. a y =■4x b y =■5x

Activity 9-E-3

1 a

x

(1, 6)

-4 -3 -2 -1 0 1 2 3 4 x

1 2

( ) ■and

b By■writing■ ■with■a■negative■index,■show■algebraically■that■the■functions y =■ x

 1 −1 x −x y =■2-x■are■identical.  2  = (2 ) = 2 x x x 12 Draw■the■graphs■of y =■(1.2) , y =■(1.5) ■and y =■(1.8) . 13 a■ Draw■the■graph■of y =■10■ì■(1.3)x. b State■the■y-intercept. 10 c Write■the■equation■of■the■horizontal■asymptote. y■=■0

1 2

x

V

40 000

15 000 0

V = 40 000 ì (0.85)n 1

2

Chapter 9 Functions

3

4

5 n

309

number AND algebra • Linear and non-linear relationships 14 Use a calculator to draw the graphs of each of the following on the one set of axes. a y = 3x b y = 3x + 2 c y = 3x - 3

15 a y = 0 b y = 2 c y = -3

15 For the graphs drawn in question 14, state the equation of the horizontal asymptote. 16 Use your answers to questions 14 and 15 to state the effect that changing the value of c has on the graph of y = 3x + c. Moves the graph vertically Understanding 17 Match each of the graphs (a–d) with the correct equation below (i–iv). y 10 8 6 4 2 (1, 1.6)

y = 3x

-12 -10 -8 -6 -4 -2 0 -2 y = 3x - 3 -4

2

4

y=

3x

+2

6

8

y

10

2

4

6

x

a

-10

0

-5

5

-4

     i  y = 3 ì 4x   ii  y = 2 ì 8x iii  y = 3 ì 0.25x iv  y = 2 ì 0.8x

x -4

2

4

x

y 10 8 6 4 2 -4

(1, 12)

0

-2

(1, 16)

0

-2

10

d

y 20 16 12 8 4

c

y 12 10 8 6 4 2

b

2

(1, 0.75)

0

-2

2

y 10 8 6 4 2 -4

-2

(1, 3)

0

2

19   MC  The graph of y = -3 ì 4x is best represented by: y

✔ B

y x

0 (0, 1) 0

310

x

4

17 a ç iv b ç i c ç ii d ç iii

18   MC  The equation for the graph at right is: A y = 2x ✔ B y = 3x C y = 2 ì 3x D y = 3 ì 2x -x E y = 2

A

x

4

x

Maths Quest 10 for the Australian Curriculum

(0, -3)

4

x

c

y

d

y

1

(0, 0.5)

2

3

c

-6 -5 -4 -3 -2 -1

1

-1

(0, 3) x

0 (0, -1) x

0

y

e

y

n A

0 1 2 3 4 5 6 1000 1100 1210 1331 1464.10 1610.51 1771.56 A

1000

0

1

V 40 000

15 000 0

10 5

(0, 1) 5

10

10 8

0

-10 -5 0 -5

22 c

y = 2x + 4

(0, 16)

15

y

y = 2x

x

d

(0, 1) 2

y

y = 2x

y = 2x - 1

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

(0, -1)

x

6 y = 4x + 1

4 2

-8 -6 -4 -2 0 -2

20 On■the■same■set■of■axes■draw■the■graphs■of: a y =■4x b y =■4x + 1

y = 4x - 3 y = 4x

2

4

6

8

x

c y =■4x -■3.

21 Use■your■answer■to■question■20■to■state■the■effect■that■changing■the■value■of■b■has■on■the■graph■ of■y =■4x - b. Moves■the■graph■horizontally 22 Sketch■graphs■of■each■of■the■following■on■the■one■set■of■axes,■showing■the■y-intercept■and■the■

horizontal■asymptote.■(Remember■that■a■sketch■graph■shows■the■basic■shape■of■the■relationship■ and■its■key■features.■A■sketch■graph■is■not■drawn■by■plotting■points■from■a■table■of■values.) a y =■2x■and y =■2x■+■4 b y =■2x■and y =■2x■-■1 c y =■2x■and y =■2x■+■4 d y =■2x■and y =■2x■-■1 Check■your■answers■with■a■graphing■calculator■or■graphing■software. reAsoning 23 Myung-Hye■invests■$1000■at■10%■p.a.■interest■compounded■annually.■This■investment■can■be■

represented■by■the■function■A■=■1000■ì■(1.1)n,■where■A■is■the■amount■to■which■the■investment■ grows■and n is■the■number■of■years■of■the■investment. a Prepare■a■table■of■values■for■0■Ç n Ç■6.■Substitute■integer■values■of n into■the■equation■and■ use■a■calculator■to■determine■corresponding■values■of■A. b Plot■the■points■generated■by■the■table,■clearly■labelling■the■axes.■Join■the■points■with■a■ smooth■curve. A = 1000 ì (1.1)n c Use■the■table■of■values■or■the■graph■to■fi■nd■the■value■of■the■investment■after■3■years. $1331 24 Kevin■buys■a■car■for■$40■■000.■The■car■depreciates■at■the■rate■of■15%■p.a.■The■value,■$V,■of■the■ car■after n years■can■be■given■by■the■equation■V■=■40■■000■ì■(0.85)n. a Prepare■a■table■of■values■for■ 0■Ç n Ç■5.■Substitute■integer■ 2 3 4 5 6 n values■of n into■the■equation■ and■use■a■calculator■to■fi■nd■ corresponding■values■of■V.■ Round■answers■to■the■nearest■ whole■number■as■required. b Plot■the■points■generated■by■the■table,■clearly■labelling■the■axes.■Join■the■points■with■a■ V = 40 000 ì (0.85)n smooth■curve. As■n■increases,■the■value■ 1 2 3 4 5 n c Describe■what■is■happening■to■the■value■of■the■car■as n increases. of■the■car■decreases. d Find■the■value■of■the■car■after■5■years.■Give■the■answer■to■the■nearest■dollar.■ $17■748 24 a

n V

0 1 2 3 4 5 40■000 34■000 28■900 24■565 20■880 17■748

Chapter 9 Functions

311

number AND algebra • Linear and non-linear relationships

9F

25 The graph shows the growth rate of two different bacteria. a Determine when each bacteria reaches a

population of 500  000. b Estimate the starting population of each bacteria. c Which bacteria grows at a faster rate? Bacteria A 240 s d When are the populations equal? 26 The rat population in Hamlin is very prolific; the rats double their population every 2 days. An initial count of rats in the town shows 2048 rats. a What is the rat population 10 days after the initial count? 65 536 b Predict the population after 100 days. 2.3 ì 1018 c Write an equation that enables you to predict the x rat population. 2048 ì ( 2 ) d Predict when the rat population will reach: 18 days   i 1 million ii 10 million   25 days iii 1 billion (1000 million).   38 days

Bacteria population ( ì 1000)

190 s (bacteria A); 110 s (bacteria B)

Bacteria A starts at 20 000; bacteria B starts at 260 000.



1400 1200 1000

B A

800 600 400 200 0

50 100 150 200 250 300 Time (s)

reflection 

 

What are two different transformations that can be done to exponential graphs?

The hyperbola ■■

k A hyperbola is a function of the form xy = k or y = . x

Worked Example 17

1 Complete the table of values below and use it to plot the graph of y = . x x

-3

-2

-1

1

-2

1 2

0

1

2

3

y Think 1

2

Write/draw

Substitute each x value into the function 1 y = to obtain the corresponding y value. x

1

x

-3 -2 -1 - 2

y

- 3 - 2 -1 -2 Undefined

1

-3 -2 -1

312

1

1 2

2

1

2

3

1

1 2

1 3

y

Draw a set of axes and plot the points from the table. Join them with a smooth curve.

■■

0

2 1 0

y = —1x 1 2 3 -1 -2

x

The graph in Worked example 17 has several important features. 1. There is no function value (y value) when x = 0. At this point the hyperbola is undefined. When this occurs, the line that the graph approaches (x = 0) is called a vertical asymptote.

Maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

■■

2. As x becomes larger and larger, the graph gets very close to but will never touch the x-axis. The same is true as x becomes smaller and smaller. The hyperbola also has a horizontal asymptote at y = 0. 3. The hyperbola has two separate branches. It cannot be drawn without lifting your pen from the page and is an example of a discontinuous graph. k 1 As with exponential functions, graphs of the form y = are the same basic shape as y = x x with y values dilated by a factor of k.

Worked Example 18 a Plot the graph of y =

4 for -2 Ç x Ç 2. x

b  Write down the equation of each asymptote.

Think a

1

2

Write/draw

Prepare a table of values taking x values from -2 to 2. Fill in the table by substituting each x value into the given equation to find the corresponding y value.

a

1

x

-2 -1 - 2

y

-2 -4 -8 Undefined

Consider any lines that the curve approaches but does not cross.

■■

1 2

1

2

8

4

2

y

Draw a set of axes and plot the points from the table. Join them with a smooth curve. -2 -1

b

0

8 4 0

y = —4x x

1 2 -4 -8

b Vertical asymptote is x = 0.

Horizontal asymptote is y = 0.

Consider the effect of negative values of k.

Worked Example 19

Plot the graph of y =

−3 for -3 Ç x Ç 3. x

Think 1

2

Draw a table of values and substitute each x value into the given equation to find the corresponding y value. Draw a set of axes and plot the points from the table. Join them with a smooth curve.

Write/draw 1

x -3 -2 -1 - 2

0

1

  2   1      2   3

y   1 1.5   3   6 Undefined -6 -3 -1.5 -1 y 6 3 -3 -2 -1 0 -3

— y = –3 x

1 2 3 x

-6

Chapter 9 Functions

313

-10

0

x

k 1.■ A■hyperbola■is■a■function■of■the■form■y = . x 2.■ To■obtain■the■graph■of■a■hyperbola,■construct■a■table■of■values.■Plot■the■points■and■ join■them■with■a■smooth■curve.■Alternatively,■use■a■graphics■calculator■or■a■computer■ graphing■package. 3.■ A■hyperbola■will■often■have■both■a■horizontal■and■a■vertical■asymptote.

-3 -2 -1

10

y

remember

1 2 3

— y = 10

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

exerCise

9F inDiviDuAl pAthWAys

the hyperbola You■may■use■a■graphics■calculator■or■computer■graphing■package■to■assist■you■in■this■exercise. FluenCy

eBook plus

Activity 9-F-1

1 We17 ■Complete■the■table■of■values■below■and■use■it■to■plot■the■graph■of■y =

Hyperbola graphs doc-5089 Activity 9-F-2

Exploring the hyperbola doc-5090 Activity 9-F-3

x

More hyperbola graphs doc-5091

3

0

100

y

1

x

—– y = 100

4

5

x

iii

0

1

x

— y = 20

20

y

0

1

x

5 y=—

x

ii 5

y

-5

-4

-3

-2

-1

0

1

2

3

4

5

y

-2

-2.5

-3.3

-5

-10

Undefi■ned

10

5

3.3

2.5

2 y (1, 4)

(1, 3) 4 20 100 y=— x 3 x■=■0,■y■=■0 ii■ y = ■ x■=■0,■y■=■0 iii y = y=— x 2 (1, 2) x x y=— x 0 x 2 3 4 On■the■same■set■of■axes,■draw■the■graphs■of■y = , y = ■and■y = . x x x Use■your■answer■to■question■3■to■describe■the■effect■of■increasing■the■value■of■k■on■the■ k It■increases■the■y-values■by■a■factor■of■k■ y graph■of■y = . and■hence■dilates■the■curve■by■a■factor■of■k. x −10 10 y = -10 —– x We19 ■Plot■the■graph■of■y = ■for■-5■Ç■x■Ç■5. x 1 2 3 2 3 x 6 −6 -5 -4 -3 -2 -1 0 On■the■same■set■of■axes,■draw■the■graphs■of■y = ■and■y = . x x -10 −k Use■your■answer■to■question■6■to■describe■the■effect■of■the■negative■in■y = . k x The■negative■refl■ects■the■curve■y■=■ ■in■the■x-axis. 1 x Complete■the■table■of■values■below■and■use■the■points■to■plot■y = .■State■the■equation■of■ x −1 the■vertical■asymptote. Equation■of■vertical■asymptote■is■x■=■1.

5 i y = ■ x■=■0,■y■=■0 x

7

2 a■ ■ i■

x

2 We18 ■a■ Plot■the■graph■of■each■hyperbola. b Write■down■the■equation■of■each■asymptote.

6

314

10 . x

8

x

-3

-2

-1

0

1

2

3

4

y

-0.25

-0.33

-0.5

-1

Undefi■ned

1

0.5

0.33

9 Plot■the■graph■of■each■hyperbola■and■label■the■vertical■asymptote. y 1 9 a y b a y=

x−2 1 b y= x−3 1 c y= x +1

1

y = —— x-2

1 0 1 -— 2

maths Quest 10 for the Australian Curriculum

x=2

y 1

1

y = —— x+1

y = —— x-3 1

2 3

■ c

x

1 -— 3

-2 -1

0 34

x=3

x

1 0

x = -1

-1

1

x

number AND algebra • Linear and non-linear relationships y

a

Understanding -4 y = —— x+1

(-2, 4)

-1 0 -4

b

The a translate the graph left 1 . or right, and x = a becomes the x−a vertical asymtote. 11 Sketch each of the following, showing the position of the vertical asymptote. y −4 a y = 1 x +1 2— 2 2 b y = -2 0 x 5 x −1 y = —— x+2 5 reflection    (-3, -5) c y = x+2 How could you summarise the effect of the 12 Give an example of the equation of a transformations dealt with in this exercise on 1 hyperbola that has a vertical asymptote of: the shape of the basic hyperbola y = ? a x = 3 x b x = -10. 10 Use your answers to question 9 to describe the effect of a in y =

x

y 2

y = —— x-1 (2, 2) 0 1 -2

x

9G

The circle ■■ ■■

Check with your teacher. Possible answers: a  y =

1 1    b  y = x−3 x + 10

A circle is the path traced out by a point at a constant distance (the radius) from a fixed point (the centre). Consider the circles shown below right. The first circle has its centre at the origin and radius r.   Let P (x, y) be a point on the circle. y   By Pythagoras: x 2 + y2 = r 2. P(x, y)   This relationship is true for all points, P, on the circle. r y

■■

If the circle is translated h units to the right, parallel to the x-axis, and k units upwards, parallel to the y-axis, then: The equation of a circle, with centre (h, k) and radius r, is: (x - h)2 + (y - k)2 = r 2

x

x

The equation of a circle, with centre (0, 0) and radius r, is: x2 + y2 = r 2 y y k

P(x, y) (y - k) (x - h) h

x x

Worked Example 20

Sketch the graph of 4x2 + 4y2 = 25, stating the centre and radius. Think 1

Express the equation in standard form by dividing both sides by 4.

Write/draw

x2 + y2 = r2 + 4y2 = 25

4x2

x2 + y2 =

25 4

2

State the coordinates of the centre.

Centre (0, 0)

3

Find the length of the radius by taking the square root of both sides. (Ignore the negative results.)

r2 =

25 4

r=

5 2

Radius = 2.5 units Chapter 9 Functions

315

number AND algebra • Linear and non-linear relationships

4

y 2.5

Sketch the graph. -2.5

2.5

x

-2.5

Worked Example 21

Sketch the graph of (x - 2)2 + ( y + 3)2 = 16, clearly showing the centre and radius. Think

Write/draw

1

Express the equation in standard form.

(x - h)2 + ( y - k)2 = r 2 (x - 2)2 + ( y + 3)2 = 16

2

State the coordinates of the centre.

Centre (2, -3)

3

State the length of the radius.

r 2 = 16 r=4 Radius = 4 units

4

Sketch the graph.

y 1 -2 -3

2

6 x

4

-7

Worked Example 22

Sketch the graph of the circle x2 + 2x + y2 - 6y + 6 = 0. Think 1

Express the equation in standard form by completing the square on the x terms and again on the y terms.

3

State the coordinates of the centre. State the length of the radius.

4

Sketch the graph.

2

Write/draw

(x - h)2 + ( y - k)2 = r 2 + 2x + y2 - 6y + 6 = 0 2 (x + 2x + 1) - 1 + ( y2 - 6y + 9) - 9 + 6 = 0 (x + 1)2 + ( y - 3)2 - 4 = 0 (x + 1)2 + ( y - 3)2 = 4 Centre (-1, 3) r2 = 4 r=2 Radius = 2 units x2

y 5 3 1 -3 -1 1

316

Maths Quest 10 for the Australian Curriculum

x

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships

remember

Circle■graphs: x2■+■y 2■=■r 2■ (x■-■h)2■+■(■y■-■k)2■=■r 2■ exerCise

FluenCy

inDiviDuAl pAthWAys eBook plus

Activity 9-G-1

Circle graphs doc-5092 Activity 9-G-2

Exploring the circle doc-5093 More circle graphs doc-5094 y 7 5 (1, 2) 6 x

-3

b

y 3 -2 6 -3

-8

4 x

-9

c

y 8 1 -3

4

x

-6

d

3 We22 ■Sketch■the■graphs■of■the■following■circles. a x2■+■4x■+■y2■+■8y■+■16■= 0 b x2■-■10x■+■y2■-■2y■+■10■= 0 c x2■-■14x■+■y2■+■6y■+■9■= 0 d x2■+■8x■+■y2■-■12y■-■12■= 0 e x2■+■y2■-■18y■-■19■= 0 f 2x2■-■4x■+■2y2■+■8y■-■8■= 0

a

1 d

y 5

3 4

y 9

x

–2

-1 -3

2

2 x

■

6 -2

12

x

4

y

e x2■+■(y■-■9)2■=■102 y

c

y

■ Centre■(0,■0),■radius■9 y 5

x

-2 5 x

✔ d

-10 -1

2

3– 3

1

10

1 3

■

x

-2

-2

1

4 x

-5

1 3– x

-3 –

15 x

2 -5

x

y

y

x ■ ■ Centre■(0,■0),■radius■5 y f 1

4 10

f (x■-■1)2■+■(y■+■2)2■=■32

-5

-5 5

5 9

-5

y

-5

-10

x

-9

y

-5

14

-10

9 x

e

10

7

-3

b

-9

12 x

8

-13

f

y 14

19

-4 -5

-2

y 4

4 mC ■The■graph■of■(x■-■2)2■+■(■y■+■5)2■=■4■is:

y

e

-6

c■ (x■-■7)2■+■(y■+■3)2■=■72■ d■ (x■+■4)2■+■(y■-■6)2■=■82

unDerstAnDing

7 -10

1 We20 ■Sketch■the■graphs■of■the■following,■stating■the■centre■and■radius■of■each. a x2■+■y2■=■49 y c 1 a■ b y y 7 b x2■+■y2■=■42 6 4 c x2■+■y2■=■36 -7 7 x -6 6 x -4 4 x d x2■+■y2■=■81■ 2 2 e 2x ■+■2y ■=■50■ -7 -6 -4 f 9x2■+■9y2■=■100 ■ Centre■(0,■0),■radius■6 ■ ■ Centre■(0,■0),■radius■7■ ■ Centre■(0,■0),■radius■4 2 We21 ■Sketch■the■graphs■of■the■following,■clearly■showing■the■centre■and■the■radius. a (x■-■1)2■+■(■y■-■2)2■=■52 3 a■ (x■+■2)2■+■(y■+■4)2■=■22■ b■ (x■-■5)2■+■(y■-■1)2■=■42 b (x■+■2)2■+■(■y■+■3)2■=■62 y y c (x■+■3)2■+■(■y■-■1)2■=■49 5 -4 -2 -2 x d (x■-■4)2■+■(■y■+■5)2■=■64 1 x 2 2 1 5 9 e x ■+■(■y■+■3) ■=■4■ -3 -4 2 2 f (x■-■5) ■+■y ■=■100

Activity 9-G-3

-4

radius■r radius■r

the circle

9g

a■

centre■(0,■0)■ centre■(h,■k)■

3

1 -3 – 3

■ Centre■(0,■0),■radius■103

Chapter 9 Functions

317

number AND algebra • Linear and non-linear relationships 5   MC  The centre and radius of the circle (x + 1)2 + ( y - 3)2 = 4 is: ✔ b (-1, 3), 2 a (1, -3), 4 c (3, -1), 4 d (1, -3), 2 y 6 Find the equation representing

the outer edge of the galaxy as shown in the photo at right, using the astronomical units provided. (x - 5)2 + (y - 3)2 = 16 3

5

reflection 

9

 

How could you write equations representing a set of concentric circles (circles with the same centre, but different radii)?

318

Maths Quest 10 for the Australian Curriculum

x

number AND algebra • Linear and non-linear relationships

Summary Plotting parabolas ■■ ■■ ■■ ■■ ■■ ■■ ■■

Produce a table of values by substituting each integer value of x into the equation. Plot a graph by drawing and labelling a set of axes, plotting the points from the table and joining the points to form a smooth curve. The axis of symmetry is the line that divides the parabola exactly in half. The turning point is the point where the graph changes direction or turns. The turning point is a maximum if it is the highest point on the graph and a minimum if it is the lowest point on the graph. The x-intercepts are the x-coordinates of the points where the graph crosses the x-axis. The y-intercept is the y-coordinate of the point where the graph crosses the y-axis. Sketching parabolas using the basic graph of y = x 2

■■ ■■ ■■ ■■ ■■

If the graph of y = x2 is translated c units vertically, the equation becomes y = x2 + c. If the graph of y = x2 is translated h units horizontally, the equation becomes y = (x − h)2. If the graph of y = x2 is dilated by factor a, the graph becomes narrower if a > 1 and wider if 0 < a < 1. If the x2 term is positive, the graph is upright. If there is a negative sign in front of the x2 term, the graph is inverted. Invariant points are points that do not change under a transformation. Sketching parabolas in turning point form

■■ ■■ ■■ ■■ ■■ ■■ ■■

If the equation of a parabola is in turning point form, y = a(x - h)2 + k, then the turning point is (h, k). If a is positive, the graph is upright with a minimum turning point. If a is negative, the graph is inverted with a maximum turning point. If the magnitude of a is greater than 1, the graph is narrower than the graph of y = x2. If the magnitude of a is between 0 and 1, the graph is wider than the graph of y = x2. To find the y-intercept, substitute x = 0 into the equation. To find the x-intercepts, substitute y = 0 into the equation and solve for x. Sketching parabolas of the form y = ax2 + bx + c

■■

■■

If the equation is in the form y = ax2 + bx + c, the coordinates of the turning point can be found by: (a) using the completing the square method to change the equation into turning point form (b) finding the x-coordinate of the point exactly halfway between the two x-intercepts. This is the x-coordinate of the turning point. Then substitute the x-value into the equation to find the y-coordinate. −b . Then substitute the x-value into the equation to find the y-coordinate. (c) using x = 2a The graph should also show both the y-intercept and the x-intercepts of the parabola if they exist. Exponential functions and their graphs

■■ ■■

■■ ■■ ■■

Relationships of the form y = ax, where a ò 1 are called exponential functions with base a. To obtain the graph of an exponential function, construct a table of values first and then plot the points from the table and join them with a smooth curve. Alternatively use a graphics calculator, CAS calculator or graphing software. An asymptote is a line which the graph approaches but never cuts or touches. Multiplying by a constant dilates the basic graph — that is, makes it narrower or wider. If x is a negative number, the graph is reflected across the y-axis. Chapter 9 Functions

319

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships The hyperbola ■■ ■■ ■■

k A■hyperbola■is■a■function■of■the■form■y = . x To■obtain■the■graph■of■a■hyperbola,■construct■a■table■of■values.■Plot■the■points■and■join■them■ with■a■smooth■curve.■Alternatively,■use■a■graphics■calculator■or■a■computer■graphing■package. A■hyperbola■will■often■have■both■a■horizontal■and■a■vertical■asymptote.

The circle ■■

Circle■graphs: x2■+■y 2■=■r 2■ (x■-■h)2■+■(■y■-■k)2■=■r 2■

centre■(0,■0)■ centre■(h,■k)■

radius■r radius■r

MaPPING YOUR UNdeRSTaNdING

Homework book

320

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■279. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Chapter review y 10

✔ b

Fluency 1 The turning point for the graph y = 3x2 − 4x + 9 is: A

 1 2  , 1  3 3

B

 1 2 ,  3 3 

C

 1 1  6 , 16  2

-4

-2

C

2 , 2  3 6 3 

x-intercepts closest together? ✔ A y = x2 + 3x + 2 B y = x2 + x - 2 C y = 2x2 + x - 15 D y = 4x2 + 27x - 7 E y = x2 - 2x - 8 3 Which graph of the equations below has the largest y-intercept? A y = 3(x - 2)2 + 9 B y = 5(x - 1)2 + 8 C y = 2(x - 1)2 + 19 ✔ D y = 2(x - 5)2 + 4 E y = 12(x - 1)2 + 10 4 The translation required to change y = x2 into 1 y = (x - 3)2 + 4 is: right 3, up

1 4

4x

C left 3, down D left 3, up

-2

D

4x

y 10 5 -4

-2

1 0 -5 -10 y 10

-2

0 -5

5 The graph of y = -3 ì 2x is best represented by:

2

4x

x -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

y 16  7  0 -5 -8 -9 -8 -5  0  7 16 y 16 14 12 10 8 6 4 2

0 -8 -6-4 -2 -2 2 4 6 8 -4 -6 -8 (-4, -9) -10

x

TP (-4, -9); x-intercepts: -7 and -1

-10 6 Use the completing the square method to find the

y 10

turning point for each of the following graphs. a y = x2 - 8x + 1 (4, -15) b y = x2 + 4x 5 (-2, 9)

5 3

-10

2

-10

-4

-5

4x

5 3

1 4

0

2 -5

1 4

1 4

-2

-4

1 4

E right , up 3

A

5 3

E

B right 3, down

-4

2

y 10

2 Which graph of the following equations has the

✔ A

0 -3 -5 -10

2

✔ D  , 7  3 3

E

5

2

4x

7 For the graph of the equation y = x2 + 8x + 7,

produce a table of values for the x-values between -9 and 1, and then plot the graph. Show the y-intercept and turning point. From your graph, state the x-intercepts.

Chapter 9 Functions

321

number AND algebra • Linear and non-linear relationships 8 For each of the following, find the coordinates of

21 Find the equation of this circle. x2 + y2 = 36 y y 6 6

the turning point and the x- and y-intercepts and sketch the graph. a y = (x - 3)2 + 1 b y = 2(x + 1)2 - 5 9 For the equation y =

4

-x2

- 2x + 15, sketch the graph and determine the x- and y-intercepts and the coordinates of the turning point.

10 For the exponential function y = 5x: a complete the table of values below y

y = 5x

160 140 120 100 80 60 40 (0, 1) 20

Increasing the value of a makes the graph steeper for positive x-values and flatter for negative x-values.

-4 -3 -2 -1 0

(1, 5) 1

2

3 4x

x -3 -2 -1  0  1  2  3

-6

y = -(x -

0.2

Changing the sign of the index reflects the graph in the y-axis.

+3=

-x2

+ 4x - 1

5

h

25

2 The height, h, in metres of a golf ball t seconds

(2, 4)

2

h = 4t - t

125

0

4 t

for -4 Ç x Ç 4.

for -4 Ç x Ç 4.

1

y = 2 ì 3x, y = 5 ì 3x and y = 2 ì 3x. b Use your answer to part a to explain the effect of changing the value of k in the x. Increasing the value of k equation y = ka makes the graph steeper.

15 a On the same set of axes sketch the graphs of

y = (2.5)x and y = (2.5)-x. b Use your answer to part a to explain the effect of a negative index on the equation y = ax.

16 Sketch each of the following. b y = −

2 x

after it is hit is given by the formula h = 4t - t2. a Sketch the graph of the path of the ball. b What is the maximum height the golf ball reaches? 4  m c How long does it take for the ball to reach the maximum height? 2 s d How long is it before the ball lands on the 4s ground after it has been hit? 3 A ball is thrown upwards from a building and follows the path given by the formula h = -x2 + 4x + 21. The ball is h metres above the ground when it is a horizontal distance of x metres from the building. a Sketch the graph of the path of the ball. b What is the maximum height the ball reaches? 25  m c How far is the ball from the wall when it 2  m reaches the maximum height? 7  m d How far from the building does the ball land? 4 A soccer ball is kicked upwards in the air. The height, h, in metres, t seconds after the kick is modelled by the quadratic equation h = -5t2 + 20t.

h 21 2 h = -x + 4x + 21

−3 x−2 -3 18 Give an example of an equation of a hyperbola that has a vertical asymptote at x = -3.

17 Sketch y =

19 Sketch each of these circles. Clearly show the b (x - 5)2 + (y + 3)2 = 64

h

(2, 25)

322

(2, 20) h = -5t2 + 20t

0

0

7

a Sketch the graph of this relationship. b For how many seconds is the ball in the air? 4 s c For how many seconds is the ball above a

height of 15 m? That is, solve the quadratic inequation -5t2 + 20t > 15. 2 s Check with your teacher. 1 d For how many seconds is the ball above a . Possible answer is y =  he ball is never above height of 20   m? T x+3

Maths Quest 10 for the Australian Curriculum

4 t

x

20 Sketch the following circles. Remember to first

complete the square. a x2 + 4x + y2 - 2y = 4 b x2 + 8x + y2 + 8y = 32

-4 y = -x2 + 4x - 1 y = x2 - 4x + 7

b Confirm your result graphically.

14 a On the one set of axes draw the graphs of

centre and the radius. a x2 + y2 = 16

4x

just touches the one above at the turning point.

1

and y = (1.5)x. b Use your answer to part a to explain the effect of changing the value of a in the equation of y = a x.

4 x

2

1 Consider the quadratic equation: y = x2 - 4x + 7. a Determine the equation of the quadratic which

0.04

10-x

0

-2

problem solving

13 a On the same axes draw the graphs of y = (1.2)x

a y =

2)2

0.008

11 Draw the graph of y = 10 ì

6 x

-2

y

3x

0 -6

b plot the graph.

12 Draw the graph of y =

2

a height of 20  m.

number AnD AlgebrA • lineAr AnD non-lineAr relAtionships 5 The■height■of■the■water■level■in■a■cave■is■determined■

d If■the■owners■decide■on■the■fi■rst■design,■P(x),■ by■the■tides.■At■any■time,■t,■in■hours■after■9■am,■ the■percentage■of■area■within■the■courtyard■ the■height,■h(t),■in■metres,■can■be■modelled■by■the■ without■grass■is■40.5%.■By■using■any■method,■ function■h(t)■=■t2■-■12t■+■32,■0■Ç■t■Ç■12.■ fi■nd■the■approximate■percentage■of■area■of■ a What■values■of■t■is■the■model■valid■for?■Write■ courtyard■without■lawn■with■the■new■design,■ your■answer■in■interval■notation.■ ■ [0,■12] N(x)■ 28.6% 32■m b Determine■the■initial■height■of■the■water. 7 A■stone■arch■bridge■has■a■span■of■50■metres.■The■ c Bertha■has■dropped■her■keys■onto■a■ledge■ shape■of■the■curve■AB■can■be■modelled■using■a■ which■is■7■metres■from■the■bottom■of■the■cave.■ quadratic■equation.■ ■ By■using■a■graphics■calculator,■determine■the■ b(x) times■in■which■she■would■be■able■to■climb■ down■to■retrieve■her■keys.■Write■your■answers■ correct■to■the■nearest■minute. 11:41■am■to■6:19■pm 6 A■grassed■area■is■planted■in■a■courtyard■that■has■ 4.5 m a■width■of■5■metres■and■length■of■7■metres.■The■ perimeter■of■the■grassed■area■is■described■by■the■ function■P■=■-x2■+■5x,■where■P■is■the■distance,■ A B x 50 m (0, 0) in■metres,■from■the■house■and■x■is■the■distance,■ ■ Check■with■your■teacher. in■metres■from■the■side■wall.■The■diagram■below■ a Taking■A■as■the■origin■(0,■0)■and■given■that■the■ represents■this■information■on■a■Cartesian■plane. maximum■height■of■the■arch■above■the■water■ 7m level■is■4.5■metres,■show■using■algebra,■that■the■ shape■of■the■arch■can■be■modelled■using■the■ Wall equation■b(x)■=■−0.0072x2■+■0.36x,■where■b(x)■ is■the■vertical■height■of■the■bridge,■in■metres,■ and■x■is■the■horizontal■distance,■in■metres.■ x b A■fl■oating■platform■20■metres■wide■and■ 5m House p■metres■high■is■towed■under■the■bridge.■Given■ that■the■platform■needs■to■have■a■clearance■of■ a In■terms■of■P,■write■down■an■inequality■that■ at■least■30■centimetres■on■each■side,■explain■ describes■the■region■where■the■grass■has■been■ why■the■maximum■value■of■p■is■less■than■ planted. P■<■x2■-■5x 10.7■centimetres. b Determine■the■maximum■distance■the■grass■ area■will■be■from■the■house.■ 6.25■m eBook plus When■x■=■0.3,■b■=■10.7.■ c The■owners■of■the■house■have■decided■that■ Therefore■if■p■is■greater■ Interactivities than■10.7■cm■the■platform■ they■would■prefer■the■grassed■area■to■be■in■ Test yourself Chapter 9 would■hit■the■bridge. a■maximum■distance■of■3.5■metres■from■the■ int-2852 house.■The■perimeter■of■the■lawn■following■this■ Word search Chapter 9 design■can■be■described■by■the■equation int-2850 2■+■bx■+■c■ Crossword Chapter 9 N(x)■=■ax int-2851 i Using■algebra,■show■that■this■new■ design■can■be■described■by■the■function■ N(x)■=■-0.48x(x■-■5) ■Check■with■your■teacher. ii Describe■the■transformation■that■maps■P(x)■ to■N(x)■ Dilation■by■a■factor■of■0.48

Chapter 9 Functions

323

eBook plus

ACtivities

chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■9■(doc-5265)■(page 279) are you ready? Digital docs

(page 280)

•■ SkillSHEET■9.1■(doc-5266):■Substitution■into■quadratic■ equations •■ SkillSHEET■9.2■(doc-5267):■Equation■of■a■vertical■line •■ SkillSHEET■9.3■(doc-5268):■Completing■the■square •■ SkillSHEET■9.4■(doc-5269):■Solving■quadratic■ equations■using■the■quadratic■formula •■ SkillSHEET■9.5■(doc-5270):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■=■1 •■ SkillSHEET■9.6■(doc-5271):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■ò■1 9a

Plotting parabolas

Digital docs

•■ Activity■9-A-1■(doc-5074):■Review■of■plotting■ parabolas■(page 284) •■ Activity■9-A-2■(doc-5075):■Plotting■parabolas■(page 284) •■ Activity■9-A-3■(doc-5076):■Trends■in■plotting■parabolas■ (page 285) •■ SkillSHEET■9.1■(doc-5266):■Substitution■into■quadratic■ equations■(page 285) •■ SkillSHEET■9.2■(doc-5267):■Equation■of■a■vertical■line■ (page 285) 9b

Sketching parabolas using the basic graph of y = x 2

Interactivities

•■ Dilation■of■y■=■x2■(int-1148)■(page 287) •■ Vertical■translation■of■y■=■x2■+■c■(int-1192)■(page 287) •■ Horizontal■translation■of■y■=■(x■-■h)2■(int-1193)■(page 288) Digital docs

•■ Activity■9-B-1■(doc-5077):■Review■of■sketching■basic■ parabolas■(page 291) •■ Activity■9-B-2■(doc-5078):■Sketching■basic■parabolas■ (page 291) •■ Activity■9-B-3■(doc-5079):■Trends■in■sketching■basic■ parabolas■(page 291) •■ WorkSHEET■9.1■(doc-5272):■Quadratic■graphs■ (page 292) 9c

Sketching parabolas in turning point form

Digital docs

(page 296)

•■ Activity■9-C-1■(doc-5080):■Reviewing■turning■point■ form■ •■ Activity■9-C-2■(doc-5081):■Turning■point■form■ •■ Activity■9-C-3■(doc-5082):■Interpreting■turning■point■ form■trends■ 9d

Sketching parabolas of the form y = ax 2 + bx + c

Interactivity

•■ Sketching■parabolas■(int-2785)■(page 298) 324

maths Quest 10 for the Australian Curriculum

Digital docs

•■ Activity■9-D-1■(doc-5083):■Understanding■parabola■ sketching■(page 302) •■ Activity■9-D-2■(doc-5084):■Parabola■sketching■ (page 302) •■ Activity■9-D-3■(doc-5085):■Sketching■tricky■parabolas■ (page 302) •■ SkillSHEET■9.3■(doc-5268):■Completing■the■square■ (page 302) •■ SkillSHEET■9.4■(doc-5269):■Solving■quadratic■ equations■using■the■quadratic■formula■(page 302) •■ SkillSHEET■9.5■(doc-5270):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■=■1■ (page 303) •■ SkillSHEET■9.6■(doc-5271):■Solving■quadratic■ equations■of■the■type■ax2■+■bx■+■c■=■0■where■a■ò■1■ (page 303) •■ WorkSHEET■9.2■(doc-5273):■y■=■ax2■+■bx■+■c■ (page 306) 9e

exponential functions and their graphs

Interactivity

•■ Exponential■graphs■(int-1149)■(page 307) Digital docs

(page 309)

•■ Activity■9-E-1■(doc-5086):■Exploring■exponential■ graphs■ •■ Activity■9-E-2■(doc-5087):■Features■of■exponential■ graphs■ •■ Activity■9-E-3■(doc-5088):■Tricky■exponential■ graphs■ •■ SkillSHEET■9.7■(doc-5274):■Substitution■into■index■ expressions 9F

The hyperbola

Digital docs

(page 314)

•■ Activity■9-F-1■(doc-5089):■Hyperbola■graphs •■ Activity■9-F-2■(doc-5090):■Exploring■the■ hyperbola■ •■ Activity■9-F-3■(doc-5091):■More■hyperbola■graphs■ 9G

The circle

Digital docs

(page 317)

•■ Activity■9-G-1■(doc-5092):■Circle■graphs■ •■ Activity■9-G-2■(doc-5093):■Exploring■the■circle■ •■ Activity■9-G-3■(doc-5094):■More■circle■graphs■ chapter review Interactivities

(page 323)

•■ Test■yourself■Chapter■9■(int-2852):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■9■(int-2850):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■9■(int-2851):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter To access eBookPLUS activities, log on to www.jacplus.com.au

measurement anD geometry • geometric reasoning

10

 10a   10B  10C  10d

Congruence review Similarity review Congruence and proof Quadrilaterals: definitions and properties   10e Quadrilaterals and proof What Do you knoW ?

Deductive geometry

1 List what you know about geometry. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. eBook plus 3 As a class, create a large concept map Digital doc that shows your Hungry brain activity class’s knowledge Chapter 10 of geometry. doc-5275

opening Question

Can you predict the path of a ball after it has been hit by the cue? Will the ball land in a pocket?

measurement anD geometry • geometric reasoning

are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 10.1 doc-5276

eBook plus

Digital doc

SkillSHEET 10.2 doc-5277

Naming angles, lines and figures   1  For■the■triangle■shown■at■right,■use■correct■mathematical■

notation■to■name: a  the■triangle DABC b  the■angle■marked■ñ ±ACB c  the■line■opposite■the■right■angle. AC

C A B

Corresponding sides and angles of congruent triangles   2  The■two■triangles■below■are■congruent. A

Q P

C a  b  c  d  eBook plus

Digital doc

SkillSHEET 10.3 doc-5278

B Triangle 1

Triangle 2

R

Which■side■in■Triangle■1■corresponds■to■side■QR■in■Triangle■2? CA Which■side■in■Triangle■2■corresponds■to■side■BC■in■Triangle■1? PQ Name■the■angle■in■Triangle■1■that■corresponds■to■±PQR■in■Triangle■2. ±BCA Name■the■angle■in■Triangle■2■that■corresponds■to■±ABC■in■Triangle■1. ±QPR

Writing similarity statements   3  For■the■pair■of■similar■triangles■shown■at■right,■write■the■similarity■

A

statement■and■list■the■ratios■of■the■corresponding■sides.■ DADE■~■DABC,■

AD AE DE = = AB AC BC

B eBook plus

Digital doc

SkillSHEET 10.4 doc-5279

Identifying quadrilaterals   4  Name■the■following■shapes. a 

a■ Parallelogram     b  Trapezium     c  Kite

326

maths Quest 10 for the australian curriculum

b 

c 

D

E

C

measurement AND geometry • geometric reasoning

10A

Congruence review ■■ ■■

Congruent figures have the same size and the same shape; that is, they are identical in all respects. It can be said that congruent figures are similar figures with a scale factor of 1. The symbol used for congruency is @. For example, DABC in the diagram below is congruent to DPQR. This is written as DABC @ DPQR. C

A

■■

Q

B

R

Note that the vertices of the two triangles are written in corre­sponding order. Of all the shapes that are being tested for congruency, we are par­ticularly interested in triangles. There are four tests designed to check whether triangles are congruent. Three of these tests are for any type of triangle and one is specifically designed for right-angled triangles. The tests are summarised in the table below. Test

■■

P

Diagram

Abbreviation

All three sides in one triangle are equal in length to the corresponding sides in the other triangle.

SSS

Two corresponding sides and the included angle are the same in both triangles.

SAS

Two corresponding angles and a pair of corresponding sides are the same.

ASA

The hypotenuse and one pair of the other corresponding sides in rightangled triangles are the same.

RHS

In each of the tests we need to show three equal measurements about a pair of ­triangles in order to show they are congruent. Chapter 10 Deductive geometry

327

measurement AND geometry • geometric reasoning

Worked Example 1

Select a pair of congruent triangles from the diagram below, giving a reason for your answer. A

18 cm

Q 50è

15 cm

95è C

95è

P

B

L

35è 15 cm

Think

N

35è

95è

R

M Write

1

In each triangle the length of the side opposite the 95è angle is given. If triangles are to be congruent, the sides opposite the angles of equal size must be equal in length. Draw your conclusion.

AC = PR = 15 cm, LN = 18 cm Since LN ò AC and LN ò PR, DLMN is not congruent to DABC and DPQR.

2

We have a pair of equal sides. For DABC and DPQR to be congruent, 2 pairs of corresponding angles must be shown to be equal.

DABC: ±A = 50è, ±B = 95è, ±C = 180è - 50è - 95è = 35è ±B = ±Q ±C = ±R

3

Triangles ABC and PQR have a pair of corresponding sides equal in length and 2 pairs of angles the same, so draw your conclusion.

DABC @ DPQR (ASA)

■■

■■

Note that in the above worked example the congruent triangles were identified by using the ASA test, which implies that two angles and one pair of corresponding sides must be the same. Note that if two pairs of corresponding angles are equal, the third pair must also be equal.

Worked Example 2

Given that DABD @ DCBD, find the values of the pronumerals in the figure at right.

B

A Think

328

40è

x

z

y D

Write

1

In congruent triangles corresponding sides are equal in length. Side AD (marked x) corresponds to side DC, so state the value of x.

DABD @ DCBD AD = CD, AD = x, CD = 3 So x = 3 cm.

2

Since triangles are congruent, corresponding angles are equal. State the angles corresponding to y and z and hence find the values of these pronumerals.

±A = ±C ±A = 40è, ±C = y So y = 40è. ±BDA = ±BDC ±BDA = z, ±BDC = 90è So z = 90è.

Maths Quest 10 for the Australian Curriculum

3 cm

C

measurement anD geometry • geometric reasoning

WorkeD example 3

Prove that DPQS is congruent to DRSQ.

think

P

Q

S

R

Write

1

Study■the■diagram■and■state■which■sides,■and/or■ angles■are■equal.■

QP■=■SR■(given) ±SPQ■= ±SRQ■=■90°■(given) QS■is■common.

2

Select■the■appropriate■congruency■test.■(In■this■ case■it■is■RHS■because■the■triangles■have■an■equal■ side,■a■right■angle■and■a■common■hypotenuse.)

So■DPQS■@■DRSQ■(RHS).

remember

1.■ Congruent■fi■gures■are■identical■in■all■respects;■that■is,■they■have■the■same■shape■and■the■ same■size. 2.■ Triangles■are■congruent■if■any■one■of■the■following■applies: (a)■ corresponding■sides■are■the■same■(SSS) (b)■two■corresponding■sides■and■the■included■angle■are■the■same■(SAS) (c)■ two■angles■and■a■pair■of■corresponding■sides■are■the■same■(ASA) (d)■the■hypotenuse■and■one■pair■of■the■other■corresponding■sides■are■the■same■in■a■ right-angled■triangle■(RHS). 3.■ The■symbol■used■for■congruency■is■@.

exercise

10a inDiViDual pathWays eBook plus

Activity 10-A-1

Review of congruent shapes doc-5095 Activity 10-A-2

Practice with congruent figures doc-5096

congruence review Fluency   1  We1 ■Select■a■pair■of■congruent■triangles■in■each■of■the■following,■giving■a■reason■for■your■

answer.■All■side■lengths■are■in■cm. I■and■III,■SAS

a 

65è

65è 4

3 I

65è

II

3 70è

70è

4

III

4

3 45è

chapter 10 Deductive geometry

329

measurement anD geometry • geometric reasoning

inDiViDual pathWays

b 

110è

eBook plus

  b  I■and■II,■AAS   c  II■and■III,■RHS   d  I■and■II,■SSS

6 cm

I

40è

6 cm III

II

Activity 10-A-3

Tricky congruent figures doc-5097

110è

110è

40è 6 cm

40è eBook plus

c 

3

Digital doc

SkillSHEET 10.1 doc-5276

4

3

5

II

III

4

I

3

eBook plus

Digital doc

SkillSHEET 10.2 doc-5277

d 

3.5

2

3.5

2

I III

II

4.8

2.5

4.8

3.5

4.8

unDerstanDing   2  We2 ■Find■the■value■of■the■pronumeral■in■each■of■the■following■pairs■of■con■gruent■triangles.■

All■side■lengths■are■in■cm. b 

a  4

c 

80è

3

30è

85è z x

4

e 

d 

7 30è

330

y x

maths Quest 10 for the australian curriculum

x

x

x

y

       

a■ x■=■3■cm   b  x■=■85è   c  x■=■80è,■y■=■30è,■z■=■70è   d  x■=■30è,■y■=■7■cm   e  x■=■40è,■y■=■50è,■z■=■50è,■m■=■90è,■n■=■90è

40è

n m

y

z

measurement anD geometry • geometric reasoning reasoning eBook plus

  3  We3 ■Prove■that■each■of■the■following■pairs■of■triangles■are■congruent. b  a  P P

  Q     

Digital doc

SkillSHEET 10.5 doc-5280

c 

R

S

Q

P

S

Q

d 

S

R

A

B

D

C

a■ Use■SAS   b  Use■SAS.   c  Use■ASA.   d  Use■ASA.   e  Use■SSS.

R e 

Q

P

R

S   4  mc ■Note:■There■may■be■more■than■one■correct■answer.■

Which■of■the■following■is■congruent■to■the■triangle■ shown■at■right?

3 cm

5 cm

35è a 

3 cm

B 

5 cm

3 cm 35è

35è

✔ C 

✔ d 

3 cm 35è

5 cm

3 cm 35è

5 cm 5 cm

chapter 10 Deductive geometry

331

measurement anD geometry • geometric reasoning   5  Prove■that■DABC■@■DADC■and■hence■fi■nd■the■values■of■the■pronumerals■in■each■of■the■

following. b    B

A

a 

30è 30è

D

w

7 cm

x

x

B

70è

A B

c 

65è A

C

y D

30è y

x

C

D

a■ x■=■110è,■y■=■110è,■z■=■4■cm,■w■=■7■cm     b  x■=■70è c  x■=■30è,■y■=■65è

4 cm 40è 40è z C   6  Explain■why■the■triangles■

shown■at■right■are■not■ necessarily■congruent.

40è

5 cm

5 cm

The■third■sides■are■not■ necessarily■the■same.

7 cm

40è 7 cm

  7  Explain■why■the■triangles■

8 cm

shown■at■right■are■not■ congruent.

8 cm

30è

Corresponding■sides■ are■not■the■same.

30è 70è

70è

  8  Show■that■DABO■@■DACO,■if■O■is■the■centre■of■the■circle.

A

B

Use■SSS.

reFlection How can you be certain that two figures are congruent?

10b

O C

similarity review ■■ ■■ ■■

■■

Similar■fi■gures■have■identical■shape■but■different■size. The■corresponding■angles■in■similar■fi■gures■are■equal■in■size■and■the■corresponding■sides■are■ in■the■same■ratio,■called■a■scale■factor. The■sign■used■to■denote■similarity■is■~■which■is■read■as■‘is■similar■to’. •■ Similar■fi■gures■can■be■obtained■as■a■result■of■an■enlargement■or■reduction. •■ If■an■enlargement■(or■a■reduction)■took■place,■the■original■fi■gure■can■be■called■the■object■ and■the■enlarged■(or■reduced)■fi■gure■called■the■image.■It■can■also■be■said■that■the■object■ maps to■the■image. For■any■two■similar■fi■gures,■the■scale■factor■can■be■obtained■using■the■following■formula: length■of■the■image length■of■the■object Note:■The■size■of■the■scale■factor■indicates■whether■the■original■object■has■been■enlarged■or■ reduced.■That■is,■if■the■scale■factor■is■greater■than■1,■an■enlargement■has■occurred.■If■it■is■less■ than■1,■a■reduction■in■size■has■occurred. Scale■factor■=■

332

maths Quest 10 for the australian curriculum

measurement AND geometry • geometric reasoning

■■

Consider the pair of similar triangles below. U A 3 B

10

6

5 C

4

V

8

W

The following can be said about these ­triangles. – Triangle UVW is similar to ­triangle ABC or, using symbols, DUVW ~ DABC. – The corresponding angles of the two triangles are equal in size: ±CAB = ±WUV, ±ABC = ±UVW and ±ACB = ±UWV. UV VW UW = = = 2; – The corresponding sides of the two triangles are in the same ratio. AB BC AC that is, DUVW has each of its sides twice as long as the ­corresponding sides in DABC. – The scale factor is 2.

Testing triangles for similarity ■■

Triangles can be checked for similarity using one of the tests described in the table below. Test

■■

Abbreviation

All corresponding angles are equal in size.

AAA or equiangular (angle–angle–angle)

All corresponding sides are in the same ratio.

SSS (side–side–side)

Two pairs of corresponding sides are in the same ratio and the included angles are equal in size.

SAS (side–angle–side)

Both are right-angled triangles; the hypotenuses and one other pair of corresponding sides are in the same ratio.

RHS (right angle–hypotenuse–side)

Note: When using the equiangular test, only two corresponding angles have to be checked. Since the sum of the interior angles in any tri­angle is a constant number (180è), the third pair of corresponding angles will automatically be equal, provided that the first two pairs match exactly.

Worked Example 4

Find a pair of similar triangles among those shown. Give a reason for your answer. a  b  c  3 cm

140è 2 cm

6 cm

3 cm

140è

5 cm

140è 4 cm

Chapter 10 Deductive geometry

333

measurement AND geometry • geometric reasoning

Think 1

Write

In each triangle we know the size of two sides and the included angle, so the SAS test can be applied. Since all included angles are equal (140è), we need to find ratios of corresponding sides, taking two triangles at a time.

For triangles a and b: 6 4 =2=2 3 For triangles a and c: 5 3 = 1.6 , 2 = 1.5 3 For triangles b and c: 5 3 = 0.83 , 4 = 0.75 6

2

Triangle a ~ triangle b (SAS)

Only triangles a and b have corresponding sides in the same ratio (and included angle of equal size). State your conclusion, specifying the similarity test that has been used.

Worked Example 5

Prove that DABC is similar to DEDC.

A

D C

B Think

E Write

1

AB is parallel to DE. Transversal BD forms two alternate angles: ±ABC and ±EDC.

±ABC = ±EDC (alternate angles)

2

Transversal AE forms two alternate angles: ±BAC and ±DEC.

±BAC = ±DEC (alternate angles)

3

While crossing each other, the two transversals form vertically opposite angles at C.

±BCA = ±DCE (vertically opposite angles)

4

Triangles ABC and EDC have three pairs of corresponding angles of equal size and therefore are similar. State this using appropriate mathematical symbols and specify the similarity test being used.

\ DABC ~ DEDC (equian­gular, AAA)

remember

1. Similar figures have the same shape but different size. 2. Corresponding angles of similar figures are equal in size. 3. Corresponding sides of similar figures are in the same ratio, called the scale factor. 4. Triangles can be tested for similarity using the following requirements: (a) corresponding angles are equal in size (AAA or equiangular) (b) corresponding sides are in the same ratio (SSS) (c) two pairs of corresponding sides are in the same ratio, and angles included between those sides are equal in size (SAS) (d) one angle in each triangle is right (90­è); the hypotenuses and one pair of corresponding sides are in the same ratio (RHS). 5. The symbol for similarity is ~. 334

Maths Quest 10 for the Australian Curriculum

measurement anD geometry • geometric reasoning

exercise

10b inDiViDual pathWays eBook plus

Activity 10-B-1

similarity review Fluency   1  We4 ■Find■a■pair■of■similar■triangles■among■those■shown■in■each■part.■Give■a■reason■for■your■

answer. a  i■

ii 

Review of similar shapes doc-5098

iii 

5

a■ i■and■iii,■RHS   b  i■and■ii,■SAS   c  i■and■iii,■SSS   d  i■and■iii,■AAA   e  i■and■ii,■SSS

    10    

5

Activity 10-B-2

Similarity practice doc-5099 Activity 10-B-3

Tricky similarity problems doc-5100

3

6

4

b  i■

iii 

ii  4 20è

c  i■

20è

5 ii 

2 4

8

2

12

iii 

2 5

3

20è

2.5

6

4.5

4

3 d  i■

iii 

ii 

40è

50è

60è

e  i■

40è

60è

ii 

iii 

4 6

3

60è

2

8

7

5 4

4

unDerstanDing   2  Name■two■similar■triangles■in■each■of■the■following■fi■gures. b  a  Q A        

a■ Triangles■PQR■and■ABC   b  Triangles■ADB■and■ADC   c  Triangles■PQR■and■TSR   d  Triangles■ABC■and■DEC   e  Triangles■ABC■and■DEC

Q

P

B B A

C

D

R

C

P d 

c 

R

A

B

e 

B D

D

S

T

E A

C

E

C chapter 10 Deductive geometry

335

measurement anD geometry • geometric reasoning reasoning eBook plus

Digital doc

  3  We5 ■Prove■that■DABC■is■similar■to■DEDC■in■each■of■the■following. a  b  C D A

SkillSHEET 10.3 doc-5278

E

eBook plus

c 

SkillSHEET 10.6 doc-5281

C

B

B

E d 

E

A

Digital doc

C

D

A

Check■with■your■teacher.

D B

B

D A C

E

AB BC AC = = . AD DE AE b  Find■the■value■of■the■pronumerals. f■=■9,■g■=■8

  4  a■ Complete■this■statement:■

2

A

D

4

B 3

f

4 C

g E

  5  Find■the■value■of■the■pronumeral■in■the■diagram■at■right.■ x■=■4

Q A

x

2 P

4

  6  The■triangles■shown■at■right■are■similar.■Find■the■ value■of■x■and■y.■ x■=■20è,■y■=■2 14

45è

R

4

B

4 45è 1

20è

  7  Find■the■values■of■x■and■y■in■the■diagram■at■right. eBook plus

x■=■3,■y■=■4 P

reFlection

Digital doc WorkSHEET 10.1 doc-5282

How can you be certain that two figures are similar?

9

y S

1.5

3 y Q

x

R

8 6 x T

10c

congruence and proof ■■ ■■

336

In■geometry,■congruence■is■one■of■the■main■tools■we■can■use■to■prove■both■familiar■and■ unfamiliar■properties■of■shapes. In■this■section,■we■will■aim■to■prove■key■deductive■geometry■theorems■about■triangles.■In■a■ proof,■it■is■important■to■give■reasons■for■all■steps.

maths Quest 10 for the australian curriculum

measurement AND geometry • geometric reasoning

Worked Example 6 A

Prove that if two sides of a triangle are equal, then the angles opposite those sides are equal, that is, ±ABC = ±ACB.

C

B Think 1

Write/draw

Draw a diagram. Construct AD so that ■ ±BAD = ±CAD.

A

B

C

D

2

State the known facts about the sides and angles.

AB = AC (given) ±BAD = ±CAD (by construction) AD is common.

3

Summarise the given information.

Two sides and the included angle in DBAD and DCAD are equal.

4

State which congruency test applies.

DBAD @ DCAD (SAS)

5

State the conclusion.

±ABC = ±ACB (corresponding angles in congruent triangles are equal)

Worked Example 7

PQR is an isosceles triangle with PQ = PR. Also, MQ = NR. Prove that MR = NQ.

P

M

N R

Q Think 1

Write/draw P

Draw a diagram, labelling all given information.

M Q

N R Chapter 10 Deductive geometry

337

measurement anD geometry • geometric reasoning

2

State■the■known■facts■about■the■sides■and■angles.

MQ■=■NR■(given) and■PQ■=■PR■(given) \■PM■=■PN also■±QPR■is■common.

3

Summarise■the■given■information.

Two■sides■and■the■included■angle■in■DMPR■and■ DNPQ■are■equal.

4

State■which■congruency■test■applies.

DMPR■@■DNPQ■(SAS)

5

State■the■conclusion.

\■MR■=■NQ■(corresponding■sides■in■congruent■ triangles■are■equal)

remember

1.■ Many■deductive■geometry■proofs■can■be■completed■using■congruent■triangle■tests. 2.■ In■a■proof,■it■is■important■to■give■reasons■for■all■steps. exercise

10c inDiViDual pathWays eBook plus

Activity 10-C-1

Congruent triangles doc-5101

congruence and proof reasoning   1  We6 ■Prove■that■if■two■angles■of■a■triangle■are■equal,■the■sides■

A

opposite■those■angles■are■equal.■(See■the■fi■gure■shown■at■right.) ■ ■ Hint:■Construct■a■line■perpendicular■to■BC■through■A■ and■prove■that■DABD■@■DACD. Use■AAS.

Activity 10-C-2

Matching congruent triangles doc-5102 Activity 10-C-3

Harder congruent triangles doc-5103

B

C

D

  2  Prove■that■each■angle■of■an■equilateral■triangle■is■60è.

Check■with■your■teacher. M   3  Prove■that■the■bisector■of■the■vertical■angle■of■an■isosceles■

triangle■bisects■the■base.■(See■the■fi■gure■shown■lower■right.)   4  Prove■that■the■intervals■joining■the■midpoints■of■

the■three■sides■of■a■triangle■cut■the■original■triangle■

Use■SAS;■then■ corresponding■sides■ into■four■congruent■triangles. Check■with■your■teacher. in■congruent■triangles■ are■equal.■NO■=■OP.

N   5  We7 ■Use■congruence■to■prove■that■AB■||■CD. Use■SAS;■then■alternate■angles■ in■congruent■triangles■are■equal.■ Hence■AB■||■CD.

C

A E D

B   6  Prove■that■DWXY■@■DYZW.

Use■AAS.

W

X

Z 338

maths Quest 10 for the australian curriculum

P

O

Y

measurement anD geometry • geometric reasoning   7  Prove■that■DADO■@■DABO. Use■RHS.

B

A

O C

D   8  Prove■that■DPTS■@■DQTR.

Use■AAS.

Q

P

T

Use■RHS■or■AAS;■then■ corresponding■sides■and■angles■in■ congruent■triangles■are■equal.

S

R

  9  DABC■is■isosceles■with■AB■=■AC.■D■lies■on■BC■so■that■AD■^■BC.■Prove■that■AD■bisects■  10 

■ ■  11  ■ ■ ■ ■ ■

±ADB■=■90è■given ±ABC■=■90è■given ±BAD■=■xè■given ±BAC■=■xè■given À■■±ABD■=■90è■-■x■and■ ±ACB■=■90è■-■x. ■ ■ À■DBAD■~■DCAB.

■■ ■■ ■■ ■■

±BAC■and■D■is■the■midpoint■of■BC. PQR■is■a■triangle.■M■lies■on■PR■so■that■QM■^■PR.■N■lies■on■PQ■so■that■PQ■^■RN.■Also■ RN■=■QM.■Prove■that■±PRQ■=■±PQR■and■hence■that■DPQR■is■an■isosceles■triangle. (Hint:■Prove■DQNR■@■DRMQ.) Use■RHS. B Can■we■prove■Pythagoras’■theorem■using■our■ knowledge■of■similar■triangles?■Consider■the■triangle■ shown.■DABC■is■a■right-angled■triangle■with■ ±ABC■=■90°.■We■can■construct■a■perpendicular■ from■B■to■AC,■meeting■AC■at■D.■Let■±BAD■=■x.■ x A C D Our■aim■is■to■show■that■AB2■+■BC2■=■AC2. a  Copy■the■diagram■and■label■the■size■of■all■other■angles■ B in■the■triangle. y x b  Using■the■equiangular■test,■prove■that■DBAD■~■DCAB. ■ Hint:■Show■that■±ABD■=■90°■-■x■and■±ACB■=■90°■-■x. x y A c  Copy■and■complete: C D AD AB = ■ ■ ±BDC■=■90è■given AB AC ■ ■ ■ ±ABC■=■90è■given ■ ■ \ AB2■= AD■ì■AC ■ ■ ■ ±ACB■=■90è■-■x d  Using■the■equiangular■test,■prove■that■DBCD■~■DACB.■ ACB.■■ ■ ■ ±DCB■=■90è■-■xè ■ ■ ■ À■±DBC■=■xè e  Copy■and■complete: ■ ■ ■ ±BAC■=■xè■given CD BC = ■ ■ ■ ■ ■ ■ \■DBCD■~■DACB BC AC ■ ■ ■ \ BC2■= CD.AC f  Combining■step■c■and■step■e,■copy■and■complete: ■ ■ ■ AB2■+■BC2■=■__________ AD■ì■AC■+■CD■ì■AC AC■(AD■+■CD) (by■factorising) ■ ■ =■__________■ AC■ì■AC ■ ■ =■__________■ (by■simplifying) ■ ■ ■ \ AB2■+■BC2■=■AC2 g  Challenge:■Can■you■prove■the■converse■of■Pythagoras’■theorem?■That■is,■if■the■square■on■ one■side■of■a■triangle■equals■the■sum■of■the■squares■on■ the■other■two■sides,■then■the■angle■between■these■other■ two■sides■is■a■right■angle. Students■to■do. reFlection What is the most important thing to include in a proof?

chapter 10 Deductive geometry

339

measurement anD geometry • geometric reasoning

10D eBook plus

Interactivity Quadrilateral definitions

int-2786

Quadrilaterals: definitions and properties ■■

There■are■many■important■properties■of■quadrilaterals■that■can■be■shown■using■deductive■ geometry. Shape

Definition

Properties

Trapezium

A■trapezium■is■a■ quadrilateral■with■two■pairs■ of■equal■adjacent■angles.

One■pair■of■opposite■sides■is■ parallel.

Parallelogram

A■parallelogram■is■a■ quadrilateral■with■both■pairs■ of■opposite■sides■parallel.

■■

A■rhombus■is■a■ parallelogram■with■four■ equal■sides.

■■

A■rectangle■is■a■ parallelogram■whose■ interior■angles■are■right■ angles.

Diagonals■are■equal■and■ bisect■each■other.

A■square■is■a■parallelogram■ whose■interior■angles■are■ right■angles■with■four■equal■ sides.

■■

Rhombus

Rectangle

Square

■■ ■■

■■

■■ ■■

■■

340

maths Quest 10 for the australian curriculum

Opposite■angles■are■equal. Opposite■sides■are■equal. Diagonals■bisect■each■ other.

Diagonals■bisect■each■ other■at■right■angles. Diagonals■bisect■the■ angles■at■the■vertex■ through■which■they■pass.

All■angles■are■right■angles. All■side■lengths■are■equal. Diagonals■are■equal■in■ length■and■bisect■each■ other■at■right■angles. Diagonals■bisect■the■ vertex■through■which■they■ pass■(45è).

measurement anD geometry • geometric reasoning

WorkeD example 8

Use the definitions and properties of the five special quadrilaterals to answer the following statements as true or false. a A parallelogram is a trapezium. b A trapezium is a rectangle. c A square is a parallelogram. think a

b

Write a A■trapezium■has■one■pair■of■parallel■sides.

1

Consider■the■properties■of■a■trapezium.

2

Consider■the■properties■of■a■ parallelogram.

A■parallelogram■is■a■quadrilateral■with■both■ pairs■of■opposite■sides■parallel.

3

Decide■if■a■parallelogram■fi■ts■the■ defi■nition■of■a■trapezium.

Statement■is■false.

1

Consider■the■defi■nition■of■a■rectangle.

b A■rectangle■is■a■parallelogram■whose■interior■

angles■are■right■angles.

c

2

Consider■the■defi■nition■of■a■trapezium.

A■trapezium■is■a■quadrilateral■two■pairs■of■equal■ adjacent■angles.

3

Decide■if■a■trapezium■fi■ts■the■defi■nition■of■ a■rectangle.

A■trapezium■does■not■necessarily■have■a■right■ angle,■therefore,■the■statement■is■false.

1

Consider■the■defi■nition■of■a■ parallelogram.

2

Consider■the■defi■nition■of■a■square.

A■square■is■a■parallelogram■whose■interior■ angles■are■right■angles■with■four■equal■sides.

3

Decide■if■the■square■fi■ts■the■defi■nition■of■ a■parallelogram.

A■square■is■a■■parallelogram;■therefore,■the■ statement■is■true.

c

A■parallelogram■is■a■quadrilateral■whose■ appropriate■sides■are■parallel.

remember

1.■ A■trapezium■is■a■quadrilateral■with■two■pairs■of■equal■adjacent■angles. 2.■ A■parallelogram■is■a■quadrilateral■with■both■pairs■of■opposite■sides■parallel. 3.■ A■rhombus■is■a■parallelogram■with■four■equal■sides. 4.■ A■rectangle■is■a■parallelogram■whose■interior■angles■are■right■angles. 5.■ A■square■is■a■parallelogram■whose■interior■angles■are■right■angles■with■four■equal■sides. exercise

10D inDiViDual pathWays eBook plus

Activity 10-D-1

Quadrilaterals doc-5104

Quadrilaterals: definitions and properties Fluency   1  We8 ■Use■the■defi■nitions■of■the■fi■ve■special■quadrilaterals■to■decide■if■the■following■■statements■

are■true■or■false. A■square■is■a■rectangle. True A■square■is■a■rhombus. True A■square■is■a■trapezium. True A■trapezium■is■a■rhombus. False

a  c  e  g 

b  d  f  h 

A■rhombus■is■a■parallelogram. True A■rhombus■is■a■square. False A■parallelogram■is■a■rectangle. False A■rectangle■is■a■square. False chapter 10 Deductive geometry

341

measurement anD geometry • geometric reasoning

inDiViDual pathWays eBook plus

Activity 10-D-2

Harder quadrilaterals doc-5105 Activity 10-D-3

Tricky quadrilaterals doc-5106

eBook plus

Digital doc SkillSHEET 10.4 doc-5279

unDerstanDing   2  Draw■three■different■trapeziums.■Using■your■ruler■and■protractor,■decide■which■of■the■ following■properties■are■true■in■a■trapezium. None■are■true. a  Opposite■sides■are■equal. b  All■sides■are■equal. c  Opposite■angles■are■equal. d  All■angles■are■equal. e  Diagonals■are■equal■in■length. f  Diagonals■bisect■each■other. g  Diagonals■are■perpendicular. h  Diagonals■bisect■the■angles■they■pass■through.   3  Draw■three■different■parallelograms.■Using■your■ruler■and■protractor■to■measure,■decide■which■ of■the■following■properties■are■true■in■a■parallelogram. a,■c,■f a  Opposite■sides■are■equal. b  All■sides■are■equal. c  Opposite■angles■are■equal. d  All■angles■are■equal. e  Diagonals■are■equal■in■length. f  Diagonals■bisect■each■other. g  Diagonals■are■perpendicular. h  Diagonals■bisect■the■angles■they■pass■through.   4  Draw■three■different■rhombuses.■Using■your■ruler■and■protractor■to■measure,■decide■which■of■ the■following■properties■are■true■in■a■rhombus. a,■b,■c,■f,■g,■h a  Opposite■sides■are■equal. b  All■sides■are■equal. c  Opposite■angles■are■equal. d  All■angles■are■equal. e  Diagonals■are■equal■in■length. f  Diagonals■bisect■each■other. g  Diagonals■are■perpendicular. h  Diagonals■bisect■the■angles■they■pass■through.   5  Draw■three■different■rectangles.■Using■your■ruler■and■protractor■to■measure,■decide■which■of■ the■following■properties■are■true■in■a■rectangle. a,■c,■d,■e,■f a  Opposite■sides■are■equal. b  All■sides■are■equal. c  Opposite■angles■are■equal. d  All■angles■are■equal. e  Diagonals■are■equal■in■length. f  Diagonals■bisect■each■other. g  Diagonals■are■perpendicular. h  Diagonals■bisect■the■angles■they■pass■through.   6  Draw■three■different■squares.■Using■your■ruler■and■protractor■to■measure,■decide■which■of■the■ following■properties■are■true■in■a■square. a,■b,■c,■d,■e,■f,■g,■h a  Opposite■sides■are■equal. b  All■sides■are■equal. c  Opposite■angles■are■equal. d  All■angles■are■equal. e  Diagonals■are■equal■in■length. f  Diagonals■bisect■each■other. g  Diagonals■are■perpendicular. h  Diagonals■bisect■the■angles■they■pass■through.   7  Name■two■quadrilaterals■that■have■diagonals■that■bisect■each■other■at■right■angles.

Rhombus,■square

  8  Name■two■quadrilaterals■with■all■angles■equal. Rectangle,■square Parallelogram,■rhombus,■   9  Name■four■quadrilaterals■that■have■at■least■one■pair■of■opposite■sides■that■are■parallel■and■equal. rectangle,■square  10  Name■a■quadrilateral■that■has■equal■diagonals■that■bisect■each■other■and■bisect■the■angles■they■ pass■through. Square reasoning  11  Pool■is■played■on■a■rectangular■table.■Balls■are■hit■with■a■cue■

and■bounce■off■the■sides■of■the■table■until■they■land■in■one■of■ the■holes■or■pockets. a  Draw■a■rectangular■pool■table■measuring■5■cm■ by■3■cm■on■graph■paper. ■ Mark■the■four■holes,■one■in■each■corner. 342

maths Quest 10 for the australian curriculum

measurement anD geometry • geometric reasoning b  A■ball■starts■at■A.■It■is■hit■so■that■it■travels■at■

a■45è■diagonal■across■the■grid.■When■it■hits■ the■side■of■the■table,■it■bounces■off■at■a■45è■ diagonal■as■well.■How■many■sides■does■the■ ball■bounce■off■before■it■goes■in■a■hole? 6■sides c  A■different■size■table■is■7■cm■by■2■cm.■How■ many■sides■does■a■ball■bounce■off■before■it■ goes■in■a■hole■when■hit■from■A? 7■sides d  Complete■the■following■table. Table size 11     e   If■the■ratio■of■the■sides■ is■written■in■simplest■ form■then■the■pattern■ is■m■+■n■-■2.     f   There■are■two■routes■ for■the■ball■when■hit■ from■B.■Either■2■or■3■ sides■are■hit.■The■ball■ does■not■end■up■in■the■ same■hole■each■time.■ ■ A■suitable■ justifi■cation■would■be■ a■diagram■—■student■ to■draw.■     g   Isosceles■triangles■ and■parallelograms.■ The■triangles■are■ congruent.     h   The■shapes■formed■are■ parallelograms.■There■ is■only■one■possible■ path■although■the■ball■ could■be■hit■in■either■ of■two■directions■ initially.■     i   Given■m■:■n■is■the■ ration■length■to■width■ in■simplest■form.■ When■m■is■even■and■n■ is■odd■the■destination■ pocket■will■be■the■ upper■left.■When■m■ and■n■are■both■odd,■ the■destination■pocket■ will■be■the■upper■right.■ When■m■is■odd■and■n■ is■even■the■destination■ pocket■will■be■the■ lower■right.■     j  Students■to■investigate.

A

Number of sides hit

5■cm■ì■3■cm

6

7■cm■ì■2■cm

7

4■cm■ì■3■cm

5

4■cm■ì■2■cm

1

6■cm■ì■3■cm

1

9■cm■ì■3■cm

2

12■cm■ì■4■cm

2

e  Can■you■see■a■pattern?■How■many■sides■would■a■

f 

■

■ g 

■

h 

i 

j 

ball■bounce■off■before■going■in■a■hole■when■hit■ from■A■on■an■m■ì■n■table? The■ball■is■now■hit■from■B■on■a■5■cm■ì■3■cm■ pool■table. How■many■different■paths■can■a■ball■take■when■ hit■along■45è■diagonals?■Do■these■paths■all■hit■ the■same■number■of■sides■before■going■in■a■hole?■ B Does■the■ball■end■up■in■the■same■hole■each■time? Justify■your■answer. The■ball■is■now■hit■from■C■along■the■path■shown. What■type■of■triangles■and■quadrilaterals■ are■formed■by■the■path■of■the■ball■with■ itself■and■the■sides■of■the■table?■Are■any■ of■the■triangles■congruent? A■ball■is■hit■from■C■on■a■6■cm■by■3■cm■ table.■What■shapes■are■formed■by■the■path■ of■the■ball■with■itself■and■the■sides■of■the■table?■ C Is■there■only■one■path■possible? Challenge:■A■ball■is■hit■from■A■along■45è■ diagonals.■The■table■is■m■ì■n.■Can■you■fi■nd■a■ formula■to■predict■which■hole■the■ball■will■go■in? Challenge: What■would■happen■if■the■game■was■played■on■a■trapezoidal■table? reFlection

eBook plus

Digital doc WorkSHEET 10.2 doc-5283

What is the difference between the definitions and properties of shapes?

chapter 10 Deductive geometry

343

measurement AND geometry • geometric reasoning

10E

Quadrilaterals and proof ■■ ■■

In the previous exercise, we investigated some of the properties of quadrilaterals by construction and measurement. It is also important to be able to prove these properties from the definitions of the shapes.

Worked Example 9

Use the definition of a parallelogram to prove that the opposite sides of a parallelogram are equal. Think 1

Write/draw

Draw a diagram.

A

B

D 2

State the definition of a parallelogram.

3

Construct diagonal BD on the diagram.

4

Prove that the two triangles formed are congruent.

In DBAD and DDCB, BD is common. ±ADB = ±DBC (alternate angles equal as AD || BC) ±ABD = ±BDC (alternate angles equal as AB || DC) \ DBAD @ DDCB (AAS)

5

State conclusions.

AD = BC and AB = DC (corresponding sides in congruent triangles are equal) \ Opposite sides in a parallelogram are equal.

■■

A parallelogram is a quadrilateral with both pairs of opposite sides par­allel.

It is also often useful to prove that a particular quadrilateral is a parallelogram, for instance. If we know that the opposite sides are parallel, then we can use the definition and show the quadri­lateral is a parallelogram. However, there are also other tests for some of the quadrilaterals as outlined in the table below. Shape

344

C

Tests

Parallelogram

A quadrilateral is a parallelogram if: (a) opposite sides are parallel or (b) opposite sides are equal or (c) opposite angles are equal or (d) one pair of sides is both equal and parallel or (e) the diagonals bisect each other.

Rhombus

A quadrilateral is a rhombus if: (a) all sides are equal or (b) the diagonals bisect each other at right angles or (c) the diagonals bisect the angles they pass through.

Rectangle

A quadrilateral is a rectangle if: (a) all angles are equal or (b) the diagonals are equal and bisect each other.

Maths Quest 10 for the Australian Curriculum

measurement anD geometry • geometric reasoning

WorkeD example 10

Prove that PQRS is a parallelogram.

P

Q

S think

R

Write

1

State■the■given■information.

PS■=■QR ±PSQ■=■±RQS

2

Draw■conclusions■from■the■given■facts.

\■PS || QR■(alternate■angles■are■equal)

3

State■reasons■why■PQRS■is■a■parallelogram.

\■PQRS■is■a■parallelogram■since■PS■and■QR■are■ both■equal■and■parallel.

remember

1.■ A■quadrilateral■is■a■parallelogram■if: (a)■ opposite■sides■are■parallel■or (b)■opposite■sides■are■equal■or (c)■ opposite■angles■are■equal■or (d)■one■pair■of■sides■is■both■equal■and■parallel■or (e)■ the■diagonals■bisect■each■other. 2.■ A■quadrilateral■is■a■rhombus■if: (a)■ all■sides■are■equal■or (b)■the■diagonals■bisect■each■other■at■right■angles■or (c)■ the■diagonals■bisect■the■angles■they■pass■through. 3.■ A■quadrilateral■is■a■rectangle■if: (a)■ all■angles■are■equal (b)■the■diagonals■are■equal■and■bisect■each■other. exercise

10e inDiViDual pathWays eBook plus

Activity 10-E-1

Quadrilaterals and proof reasoning W

  1  We9 ■Use■congruence■to■prove■that■the■opposite■angles■

X

(±ZWX■and■±XYZ)■are■equal■in■a■parallelogram. Use■AAS■to■show■DZWX■@■DZYX.

Quadrilateral proofs doc-5107

Z

Y

Activity 10-E-2

Harder quadrilateral proofs doc-5108 Activity 10-E-3

Tricky quadrilateral proofs doc-5109

  2  Use■congruence■on■DADE■and■DCBE■to■prove■that■

A

B

the■diagonals■of■a■parallelogram■bisect■each■other.

E

Use■AAS■to■show■DAED■@■DCEB■and■ hence■AE■=■EC■and■DE■=■EB.

D chapter 10 Deductive geometry

C 345

measurement anD geometry • geometric reasoning   Use■SAS■to■show■DDAE■@■DBAE.■Hence,■DE■=■EB.■(See■previous■question.)   3  a■ Prove■that■DAED■@■DCED. Use■SAS. b  Hence,■show■that■±AED■=■±CED■=■90è.■(That■is,■

B

A E

the■diagonals■of■a■rhombus■are■perpendicular.) c  Show■that■BD■bisects■±ADC.■(That■is,■the■diagonals■ of■a■rhombus■bisect■the■angles■they■pass■through.)   4  Prove■that■the■diagonals■of■a■rhombus■bisect■each■other.   5  Prove■that■all■angles■in■a■rectangle■are■right■angles.

D

C

P

Q

S

R

A

B

D

C

  Use■co-interior■angles■ and■parallel■lines.

    b   ±AED■=■±CED■(corresponding■angles■in■congruent■ triangles■equal)■and■±AED■+■±CED■=■180è■(angle■ sum■of■straight■lines■is■180è) \■±AED■=■±CED■=■90è     c   Corresponding■angles■in■congruent■triangles■are■equal.

  6  Use■congruence■on■DADC■and■DBCD■to■show■that■

the■diagonals■in■a■rectangle■are■the■same■length.     Use■SAS.■AC■=■BD■(corresponding■ sides■in■congruent■triangles■are■equal).

  7  We10 ■ABCD■is■a■parallelogram.■X■is■the■midpoint■of■AB■

X

A

and■Y■is■the■midpoint■of■DC.■Prove■that■AXYD■is■ also■a■parallelogram.     AX■||■DY■because■ABCD■is■a■parallelogram■AX■=■DY■(given)■ \■AXYD■is■a■parallelogram■since■opposite■sides■are■equal■and■parallel.

D

midpoints■of■their■respective■sides■of■ABCD. a  Prove■DPAS■@■DRCQ. ■ Use■SAS. b  Prove■DSDR■@■DPBQ.     Use■SAS. c  Hence,■prove■that■PQRS■is■also■a■parallelogram.

C

Y P

A

  8  ABCD■is■a■parallelogram.■P,■Q,■R■and■S■are■all■

B

B

S

Q

D

C

R

      Opposite■sides■are■equal.   9  AC■and■BD■are■diameters■of■a■circle■with■centre■O.■

A

Prove■that■ABCD■is■a■rectangle.     AC■=■DB■(diameters■of■the■same■circle■are■equal)■ AO■=■OC■and■OD■=■OB■(radii■of■the■same■circle■are■equal)■ \■ABCD■is■a■rectangle.■(Diagonals■are■equal■and■bisect■each■other.)

D

O

B

 10  The■diagonals■of■a■parallelogram■meet■at■right■angles.■ Prove■that■the■parallelogram■is■a■rhombus.   Check■with■your■teacher.  11  Two■congruent■right-angled■triangles■are■arranged■

as■shown.■Show■that■PQRS■is■a■parallelogram.

P

intersect■at■P■and■Q.■Prove■that■PNQM■is■a■rhombus.

  MP■=■MQ■(radii■of■same■circle)■PN■=■QN■(radii■ of■same■circle)■and■circles■have■equal■radii.■ \■All■sides■are■equal.■ \■PNQM■is■a■rhombus.

reFlection How do you know if a quadrilateral is a rhombus?

maths Quest 10 for the australian curriculum

R

S

 12  Two■circles,■centred■at■M■and■N,■have■equal■radii■and■

Digital doc WorkSHEET 10.3 doc-5284

346

Q

P

  PS■=■QR■(corresponding■sides■in■congruent■triangles■are■equal)■ PS■||■QR■(alternate■angles■are■equal)■ \■PQRS■is■a■parallelogram■since■one■pair■of■opposite■sides■are■parallel■and■equal. eBook plus

C

M

N Q

measurement AND geometry • geometric reasoning

Summary Congruence review ■■

■■

■■

Congruent figures are identical in all respects; that is, they have the same shape and the same size. Triangles are congruent if any one of the following applies: (a) corresponding sides are the same (SSS) (b) two corresponding sides and the included angle are the same (SAS) (c) two angles and a pair of corresponding sides are the same (ASA) (d) the hypotenuse and one pair of the other corresponding sides are the same in a rightangled triangle (RHS). The symbol used for congruency is @. Similarity review

■■ ■■ ■■ ■■

■■

Similar figures have the same shape but different size. Corresponding angles of similar figures are equal in size. Corresponding sides of similar figures are in the same ratio, called the scale factor. Triangles can be tested for similarity using the following requirements: (a) corresponding angles are equal in size (AAA or equiangular) (b) corresponding sides are in the same ratio (SSS) (c) two pairs of corresponding sides are in the same ratio, and angles included between those sides are equal in size (SAS) (d) one angle in each triangle is right (90­è); the hypotenuses and one pair of corresponding sides are in the same ratio (RHS). The symbol for similarity is ~. Congruence and proof

■■ ■■

Many deductive geometry proofs can be completed using congruent triangle tests. In a proof, it is important to give reasons for all steps. Quadrilaterals: definitions and properties

■■ ■■ ■■ ■■ ■■

A trapezium is a quadrilateral with two pairs of equal adjacent angles. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A rhombus is a parallelogram with four equal sides. A rectangle is a parallelogram whose interior angles are right angles. A square is a parallelogram whose interior angles are right angles with four equal sides. tQuadrilaterals and proof

■■

■■

A quadrilateral is a parallelogram if: (a) opposite sides are parallel or (b) opposite sides are equal or (c) opposite angles are equal or (d) one pair of sides is both equal and parallel or (e) the diagonals bisect each other. A quadrilateral is a rhombus if: (a) all sides are equal or (b) the diagonals bisect each other at right angles or (c) the diagonals bisect the angles they pass through. Chapter 10 Deductive geometry

347

measurement anD geometry • geometric reasoning

■■

A■quadrilateral■is■a■rectangle■if: (a)■ all■angles■are■equal (b)■the■diagonals■are■equal■and■bisect■each■other.

MaPPINg Your uNderStaNdINg

Homework  Book

348

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■325. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the australian curriculum

measurement anD geometry • geometric reasoning

chapter review Fluency

  4  Test■whether■the■following■pairs■of■triangles■are■

  1  Select■a■pair■of■congruent■triangles■in■each■of■the■

following■sets■of■triangles,■giving■a■reason■for■your■ answer.■All■angles■are■in■degrees■and■side■lengths■ in■cm.■(The■fi■gures■are■not■drawn■to■scale.) a 

4

75è

4

40è

III

4

II

6

6

I

a 

47è

47è

2 110è

75è 65è

6

similar.■For■similar■triangles■fi■nd■the■scale■factor.■ All■angles■are■in■degrees■and■side■lengths■in■cm. 3

110è 5

7.5

b 

  1  a■ I■and■III,■ASA■or■SAS     b  I■and■II,■RHS

75è

5

3

b  I

6

10

6

8

6

II

50è 1

III

8

50è 2

c 

  2  Find■the■value■of■the■pronumeral■in■each■pair■of■

  4  a■ Similar,■scale■factor■=■1.5     b  Not■similar     c  Similar,■scale■factor■=■2

congruent■triangles.■All■angles■are■given■in■degrees■ and■side■lengths■in■cm. b  a  ■

2

2 70è

8

4

x 2

  5  Find■the■value■of■the■pronumeral■in■each■pair■of■

similar■triangles.■All■angles■are■given■in■degrees■ and■side■lengths■in■cm.

x c 

y

z

x

60è

    a■ x■=■8■cm     b  x■=■70è c  x■=■30è,■y■=■60è,■z■=■90è

a 

A

5 y

30è D x A

3  a■ ■Prove■that■the■two■

B

E 3

C b  ••

A

    a■ x■=■48è,■y■=■4.5■cm     b  x■=■86è,■y■=■50è,■z■=■12■cm

1 50è

D S

2

C

triangles■shown■in■ the■diagram■at■right■ are■congruent. ■ Use■SAS.

b  Prove■that■DPQR■

B

48è

R

B

C 1.5

44è

z

E

x

is■congruent■to■ DQPS.     Use■ASA.

8 y P

Q

D chapter 10 Deductive geometry

349

measurement anD geometry • geometric reasoning     c    x■=■60è,■y■=■15■cm,■z■=■12■cm P x y

9

Q

A

30è

C

    Use■equiangular■test.

Q

Q

  7  Prove■that■

 14  State■three■tests■that■can■be■used■to■show■that■a■

S

quadrilateral■is■a■rhombus.  15  Prove■that■WXYZ■is■a■parallelogram.■

    Use■equiangular■test. P

T

R

W

  8  Prove■that■the■angles■opposite■the■equal■sides■in■an■

problem solVing  11  ABC■is■a■triangle.■D■is■the■midpoint■of■AB,■E■is■

the■midpoint■of■AC■and■F■is■the■midpoint■of■BC.■ DG■^■AB,■EG■^■AC■and■FG■^■BC. A

Z

 16  Prove■that■the■diagonals■in■a■rhombus■bisect■the■

angles■they■pass■through.  17  Explain■why■the■triangles■shown■below■are■not■ congruent.   Corresponding■sides■are■not■the■same. 5 cm 80è

80è

25è

5 cm

E   A■rhombus■is■a■ parallelogram■with■ two■adjacent■sides■ equal■in■length. H

F O

that■have■equal■diagonals.

a  Prove■that■DGDA■@■DGDB. ■ Use■SAS. b  Prove■that■DGAE■@■DGCE.   Use■SAS. c  Prove■that■DGBF■@■DGCF.   Use■SAS. maths Quest 10 for the australian curriculum

25è

 18  Prove■that■DEFO ~ DGHO.

 20  Name■any■quadrilaterals■ C

X

Y

G

 19  State■the■defi■nition■of■a■rhombus.

E G

350

50è

  Rectangle,■square

    ±FEO■=■±OGH■(alternate■angles■equal■as■EF■||■HG) ±EFO■= ±OHG■(alternate■angles■equal■as■EF■||■HG) ±EOF■=■±HOG■(vertically■opposite■angles■equal) \■DEFO■~■DGHO■(equiangular)

isosceles■triangle■are■equal.   9  mc ■Note:■There■may■be■more■than■one■correct■ answer. ■ ■ A■quadrilateral■with■two■adjacent■sides■equal■could■ be■a: ✔ a  rhombus■ ✔ B■ square ✔ C  rectangle■ ✔ d■ parallelogram  10  True■or■false?■ a  A■rhombus■is■a■square. ■ False b  A■square■is■a■rectangle.       True c  A■rectangle■is■a■trapezium.     True

130è

    WZ■||■XY■(co-interior■angles■are■ supplementary)■and■WZ■=■XY■(given)     \ WXYZ■is■a■parallelogram■since■ one■pair■of■sides■is■parallel■and■equal.

DPST ~ DPRQ.

F

S

R

 13  Name■any■quadrilaterals■that■have■diagonals■that■     Rhombus,■square bisect■the■angles■they■pass■through.

E

B

P

C

B

D

 14  A■quadrilateral■is■a■ rhombus■if:     a  all■sides■are■equal     b   the■diagonals■ bisect■each■other■ at■right■angles     c   the■diagonals■ bisect■the■angles■ they■pass■through.

D

A

DABC ~ DEDC.

DPQS■is■isosceles.

B

4

  6  Prove■that■

 12  PR■is■the■perpendicular■bisector■of■QS.■Prove■that■

  Use■SAS. PQ■=■PS■(corresponding■ sides■in■congruent■ 3 triangles■are■equal)

5

R

z

  They■are■all■the■same■length.

d  What■does■this■mean■about■AG,■BG■and■CG? e  A■circle■centred■at■G■is■drawn■through■A.■What■ other■points■must■it■pass■through?     B■and■C

eBook plus

Interactivities

Test yourself Chapter 10 int-2855 Word search Chapter 10 int-2853 Crossword Chapter 10 int-2854

eBook plus

actiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■10■(doc-5275)■ (page 325) are you ready? Digital docs (page 326) •■ SkillSHEET■10.1■(doc-5276):■Naming■angles,■lines■ and■fi■gures •■ SkillSHEET■10.2■(doc-5277):■Corresponding■sides■ and■angles■of■congruent■triangles •■ SkillSHEET■10.3■(doc-5278):■Writing■similarity■ statements •■ SkillSHEET■10.4■(doc-5279):■Identifying■ quadrilaterals

10a   Congruence review Digital docs

•■ Activity■10-A-1■(doc-5095):■Review■of■congruent■ shapes■(page 329) •■ Activity■10-A-2■(doc-5096):■Practice■with■congruent■ fi■gures■(page 329) •■ Activity■10-A-3■(doc-5097):■Tricky■congruent■ fi■gures■(page 330) •■ SkillSHEET■10.1■(doc-5276):■Naming■angles,■lines■ and■fi■gures■(page 330) •■ SkillSHEET■10.2■(doc-5277):■Corresponding■sides■ and■angles■of■congruent■triangles■(page 330) •■ SkillSHEET■10.5■(doc-5280):■Angles■and■parallel■ lines■(page 331) 10B   Similarity review Digital docs

•■ Activity■10-B-1■(doc-5098):■Review■of■similar■ shapes■(page 335) •■ Activity■10-B-2■(doc-5099):■Similarity■practice■ (page 335) •■ Activity■10-B-3■(doc-5100):■Tricky■similarity■ problems■(page 335) •■ SkillSHEET■10.3■(doc-5278):■Writing■similarity■ statements■(page 336) •■ SkillSHEET■10.6■(doc-5281):■Calculating■unknown■ side■lengths■in■a■pair■of■similar■triangles■(page 336) •■ WorkSHEET■10.1■(doc-5282):■Deductive■geometry■I■ (page 336)

10C   Congruence and proof

(page 338) •■ Activity■10-C-1■(doc-5101):■Congruent■triangles •■ Activity■10-C-2■(doc-5102):■Matching■congruent■ triangles •■ Activity■10-C-3■(doc-5103):■Harder■congruent■triangles Digital docs

10d   Quadrilaterals: definitions and properties Interactivity

•■ Quadrilateral■defi■nitions■(int-2786)■(page 340) Digital docs

•■ Activity■10-D-1■(doc-5104):■Quadrilaterals■(page 341) •■ Activity■10-D-2■(doc-5105):■Harder■quadrilaterals■ (page 342) •■ Activity■10-D-3■(doc-5106):■Tricky■quadrilaterals■ (page 342) •■ SkillSHEET■10.4■(doc-5279):■Identifying■ quadrilaterals■(page 342) •■ WorkSHEET■10.2■(doc-5283):■Deductive■ geometry■II■(page 343) 10e   Quadrilaterals and proof Digital docs

•■ Activity■10-E-1■(doc-5107):■Quadrilateral■proofs■ (page 345) •■ Activity■10-E-2■(doc-5108):■Harder■quadrilateral■ proofs■(page 345) •■ Activity■10-E-3■(doc-5109):■Tricky■quadrilateral■ proofs■(page 345) •■ WorkSHEET■10.3■(doc-5284):■Deductive■ geometry■III■(page 346) Chapter review

(page 350) •■ Test■yourself■Chapter■10■(int-2855):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■10■(int-2853):■an■interactive■word■ search■involving■words■associated■with■this■chapter •■ Crossword■Chapter■10■(int-2854):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

chapter 10 Deductive geometry

351

eBook plus

ict actiVity Process ■■

backyard flood SearCHLIgHt Id: Pro-0099

Scenario Mrs■Effi■ciency■is■at■home■busy■preparing■for■ Christmas■and■needs■to■water■the■fruit■trees■in■the■ back■yard.■When■she■goes■to■connect■the■hose■to■the■ tap■she■discovers■that■Mr■Effi■ciency■has■taken■the■one■ she■usually■uses.■Her■only■option■is■to■use■one■twice■ the■diameter.■She■assumes■this■will■deliver■water■at■ twice■the■rate■of■the■smaller■hose■and■so■leaves■it■for■ half■her■usual■time■of■20■minutes.■She■returns■to■a■ fl■ood■in■the■backyard.■ At■the■evening■meal■she■discusses■the■event■with■ the■family.■She■is■really■distressed■because■she■ realises■that■her■action■has■led■to■a■waste■of■water■(a■ very■scarce■resource).■Her■Year■10■daughter■decides■ to■research■the■situation■and■present■a■summary■to■her■ mother■to■explain■what■has■happened.

■■ ■■

■■

■■

task ■■

■■ ■■ ■■

352

You■will■need■to■look■at■the■Interactive■websites■ found■in■your■eBookPLUS■to■summarise■how■ doubling■and■halving■dimensions■affects■length,■ area■and■volume. You■will■use■Google■Sketchup■to■design■Babushka■ solids■to■be■used■as■measures■for■cooking. You■will■also■investigate■if■the■weight■gain■of■a■ baby■fi■ts■this■same■model. You■will■then■write■up■your■procedure■and■ conclusions■in■Word■and■present■it■as■a■book■in■ Calameo.■You■should■assume■the■role■of■Mrs■ Effi■ciency’s■daughter■who■is■preparing■a■report■for■ her■mother.■Remember■to■provide■some■advice■to■ Mrs■Effi■ciency■regarding■conservation■of■water.

maths Quest 10 for the australian curriculum

■■

■■

Open■the■ProjectsPLUS■application■in■your■ eBookPLUS.■Watch■the■introductory■video■lesson,■ click■the■‘Start■Project’■button■and■then■set■up■ your■project■group.■You■can■complete■this■project■ individually■or■invite■other■members■of■your■class■ to■form■a■group.■After■saving■your■settings,■the■ project■will■be■launched. Navigate■to■your■Research Forum.■Here■you■will■fi■nd■ a■series■of■topics■to■help■you■complete■your■task. Research.■Keep■a■journal■of■your■discoveries■as■ you■proceed■with■the■task.■The■journal■entries■ should■be■short■phrases■to■explain■what■you■are■ doing/thinking,■similar■to■using■twitter■or■sending■a■ text■message■to■a■friend■—■do■not■use■abbreviations■ for■your■words.■Start■each■message■with■the■title■ of■the■topic■you■are■reporting■on.■Each■person■in■ your■group■should■report■on■at■least■3■different■ sources■of■information.■You■can■view■and■comment■ on■the■other■group■members’■entries■and■rate■the■ comments■they■have■made.■When■your■research■is■ complete,■print■your■Research■Report■to■hand■to■ your■teacher. Go■to■the■Media Centre■in■your■eBookPLUS■and■ open■the■Enlargements■Teaching■Tool.■Complete■ the■table.■Answer■the■questions■provided■in■the■ Enlargements■Teaching■Tool■fi■le■in■the■Media Centre.■Summarise■your■discoveries■in■200■words■ or■fewer.■Include■screenprints■of■each■enlargement. Wordle■is■a■site■that■creates■an■image■of■the■words■ in■an■article■according■to■the■frequency■of■their■ usage.■Open■the■Wordle■site.■Select■the■create■tab.■ Copy■the■summary■of■your■article■into■the■text■box.■ Keep■selecting■the■Randomise button■until■you■are■ happy with■the■result.■Print■your■fi■nal■choice.■Take■ a■screenprint■of■your■fi■nal■choice.■Take■it■into■Paint for■use■as■an■image■in■your■presentation.■Save■your■ Paint■fi■le■as■a■jpeg■fi■le. Go■to■the■Media Centre■in■your■eBookPLUS■ and■open■the■Area■Calculator■site.■Make■sure■the■ calculations■are■done■in■metres.■For■each■of■the■ shapes,■choose■a■number■to■enter■into■the■box■to■ calculate■the■area.■Recalculate■the■area■for■half■ this■dimension■and■double■it.■How■can■this■be■ summarised?■Repeat■this■process■for■the■Volume■ Calculator■site.■Summarise■your■fi■ndings.■Include■ this■summary■in■your■report. Download■Google Sketchup. Watch■the■ introductory■video■to■get■started.■Use■the■measuring■

tool to help you create a set of nesting solids to be used instead of babushka (matryoshka) measuring cups — look at the photo to see the proportions of the cups. Take a screenprint of your masterpiece to include in your report.

area, surface area and suggested volume in terms of the software scale factor. Discuss • ProjectsPLUS whether or not the • Microsoft Word weight gain of a baby • Internet Explorer fits this same model. • Google Sketchup Write a paragraph • Wordle • Calameo explaining the error of • Enlargements Mrs Efficiency’s Teaching Tool calculations, and why her backyard flooded. Assume the role of Mrs Efficiency’s daughter who is presenting her research findings to her mother.

Remember: 1 cup is equal to 250 mL.

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Use the Australian Icon weblink in your eBookPLUS and choose 3 icons to compare with the real object. Calculate the scale factor (range of numbers). Give measurements and calculations to support your conclusion. Calculate the enlargement factor of the icon from the real object. Calculate how many times larger the surface area and volume of the icon is compared with real object. Go to the Media Centre in your eBookPLUS and open the Sydney Harbour Bridge link. Download the model. Use the measuring tool in Google Sketchup to find the length and height of the model. Compare this with the actual measurements of the Sydney Harbour Bridge. Comment on its accuracy in your report. Take a screen capture, with your measurements visible on the model, to include in your report.

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Use Microsoft Word when preparing your presentation. Your Media Centre includes images that can help to liven up your presentation. As you arrange your images on your Word page make them appear as they would in the pages of a book. Each page of your Word document will be a page in your Calameo production of your report. Summarise the findings of your research. Give formulas for length,

Use Word to develop your presentation. Remember that you are trying to deliver a comprehensive summary of similarity and clarify why there was a flood in the backyard. You are also showing applications of similarity. Make sure you include all the results of your research, and that your presentation will grab the attention of those listening. Try to limit your report to either 4 or 8 pages for the most professional delivery. Use Word to type up your dialogue to present your report (200–500 words). Before you present your findings, use the Calameo weblink in your eBookPLUS and sign up for a free account. You will need an email address so ensure you have your teacher’s or your parents’ permission to do this. Take the guided tour to learn how to prepare a Calameo presentation. Create a publication and upload your Word report. Choose the radio button for a Private Publication Mode. Click the Start Uploading button. To show your report, click the button Read the publication and choose full screen view before you commence your delivery. (If the pagination alters using Word — and this is a problem — delete your Calameo publication and repeat the process by uploading a pdf of your report.)

ICT Activity — projectsplus

353

probleM solving

11 problem solving I

60 cm

m

40 c

opening QUesTion

How far does the centre of the racquet travel if the girl swings through an angle of 300è?

problem solving 1 A cuboid has dimensions 10 cm by 12 cm by 18 cm. Find the length of the diagonal space. 23.83 cm 2 Expand (3x - 2y)4. 81x4 - 216x3y + 216x2y2 - 96xy3 + 16y4 3 Sketch a possible graph of y = 2x2 -3x + g. Determine axial intercepts and the coordinates of 1231.5

cm3

4 5 6 7



the turning point, if any. Find the volume of a right cone with a base diameter of 14 cm and slant height of 25 cm. Consider a right-angled triangle, such that the two shorter sides are 6.4 mm and 8.9 mm in length. Find the angle between the shortest side and the hypotenuse. 54.28è The perimeter of a rectangle is 20 cm and its area is 14 cm2. Calculate the dimensions of the rectangle, correct to 1 decimal place. 8.3 cm by 1.7 cm Solve for x and y. y = ax - 3b dx + ey = f

x=

 f + 3be  f + 3be ; y = a  − 3b ae + d  ae + d 

8 When two algebraic fractions are equal, a method known as ‘cross-multiplying’ makes

finding the value of x a lot quicker. a c =   so, a(x + d) = c(x + b) x+b x+d Expand and solve normally. Use the above method to find x in each of the following. 3 5 a = x−2 x+2 b

3x + 4 5x − 4 = 2 6



y g

x

0 3 - ø 9 - 8g 4

x=8 x-intercepts: x = x = -4

3 + ø 9 - 8g 4

( 43 , g - 98 ) 3 ± 9 − 8g ; 4

y-intercept: y = g; 3 9 turning point:  , g −  8 4

2x + 1 4 x=1 = 2 1 + 3x 5 9 Mary has baked a birthday cake in the shape below. c

16 cm 14 cm 18 cm  L − l 2 4 π  L − l 3 3 π  cm   l+ 3  2   2 

10 11 12

13

She has 60 cm of ribbon which she wants to wrap around the sides of the cake. Does she have enough ribbon? Explain your answer. No, Mary will need 64.5 cm of ribbon. Find the angle of elevation to the top of a 27.3-m high Norfolk Pine tree that is 83.6 m from the observer. Assume that the observer’s eye is 1.667 m above ground level. 17.05è Solve a(x - p)(x + q) í 0 for x if a < 0 and p > -q. -q Ç x Ç p A cylinder of length, l m, has both circular ends removed and replaced with hemispheres. The container now has a length, L m. Determine the volume of the container now in terms of L and l. Marlon substituted numbers into the equation below until he had a true statement. x(x - 3) = 10 Marlon’s answer for the problem was 5. a Is Marlon’s answer right or wrong? Explain. This is a quadratic equation, which means that there

356

Maths Quest 10 for the Australian Curriculum

is a possibility of two different answers. Marlon has one of the two parts of the answer correct.

problem solving No. b A friend of Marlon’s showed him another way to solve the problem. x(x - 3) = 10 x2 - 3x = 10 x(x - 3) = 10 x2 - 3x - 10 = 0 x = 10 x - 3 = 10 (x - 5)(x - 2) = 0 (x - 5) = 0   or   (x - 2) = 0 x = 13 x = 5    x=2 (10, 13)

14

= 4 2b2 cm2.

= 12 ì 2b ì 4 2b

c Area = 12 base ì height

15

16

17

Is Marlon’s friend correct? If the solution is correct explain why, and if the solution is incorrect, provide a correct solution. Sketch the graph of y = 2(4)x - 8, showing axial intercepts and asymptotes, if any. A piece of flat pastry is cut in the shape of a right-angled triangle. The longest side is 6b cm and the shortest is 2b cm a Find the length of the third side. Give your answer in exact form. 4 2b b Find the sizes of the angles in the triangle. 19.5è, 70.5è, 90è. c Prove that the area of the triangle is equal to 4 2b 2 cm2. Cameron purchased 500 tickets for an AFL game so that all of the teachers and students in the school would be able to attend the match. y Teachers AFL tickets: $15/ticket Students AFL tickets: $5/ticket The total cost for the game was $3500. How many students attended the game? 400 students Determine the length of the diagonal x. 11.75 cm 0 1

8 cm

x

x

-6 y = -8

5 cm

7 cm

18 Parallel lines on a Cartesian plain have the same gradient but different y-intercepts. Find the

If radius and height are 19 both halved, the surface area is one-quarter its original value.

y=

2 − 4k  h +k x −   2 h2 

20

21  f 2  e 2   ≤ A≤  4 4

22

pairs of parallel lines from the following list and state the gradient and y-intercept for each. a 3y + 6x = -36 18 a (y-intercept -12) b 4y = -4x + 20 b (y-intercept 5) c 3y + 1 = 9x c (y-intercept - 1) 3 d 12 = 2x + 2y d (y-intercept = 6), gradient = -1; 1 e (y-intercept - 4) e 10 y = − x − 8 5 2 f (y-intercept = 2), gradient = –2; f 12x = -6y + 12 g (y-intercept =  72 ), gradient = 3; g 2y - 6x - 7 = 0 1 h (y-intercept = 1 ), gradient = - 20 ; h 20y = -x + 5 4 2 The formula that can be used to find the surface area, A cm , of a solid cylinder with radius r cm and height h cm is A = 2p r(r + h). a Find an exact value for A when r = 4, h = 6 A = 80p cm2 b Describe in words the changes that will occur to A if r and h are both halved. Justify your reasoning mathematically. A rocket is fired from ground level (from an underground concealed bunker) and lands h kilometres away, across horizontal terrain. If the maximum height the rocket reaches is k kilometres, find the equation of its path in terms of h and k. The perimeter, P, of a square lies in the range e to f, i.e. e  Ç  P  Ç  f. In terms of e and f what is the range of values for its area, A? Find the equation of the straight line going through (-1, 5), parallel to the line which passes through (0, 4) and (5, -3). y = - 7 x + 18 5

5

Chapter 11 Problem solving I

357

problem solving 23 One method of measuring the height of a building (h) is shown in the figure below. This

method is often used when you are not able to measure the distance along the ground (x) because of a lake or some other obstacle.

h q

a D

x

Step 1: From a location on ‘your’ side of the lake, measure the angle a, using an angle measuring device called a transit or an inclinometer. Write an equation involving x, h and a. Express this equation with x on the LHS and all other terms on the RHS. Step 2: Move a further distance, D, away from the building, where D is the known distance between the first location and the second location. Measure the angle, q, to the top of the building from the second location.

Any false statement that occurs during the solving of simultaneous equations indicates the lines are parallel, 24 and have no points of intersection. 25

47 cm for the 26 circle and 53 cm for the square 27

28

Write an equation involving x, h, D and q. Express this formula with x + D on the LHS.

Step 3: Solve the equations from Steps 1 and 2 for x in terms of all other variables. Then since each of the expressions equals x, equate the expressions. Step 4:  Solve this equation for h, in terms of D, a and q. Step 5: Use this equation to determine the height of a building where D = 50 m, a = 34.3è and q = 30.7è. 229.1 m P2 The perimeter of a rectangle is P cm and its area is cm2. Determine the dimensions of the 16 rectangle in terms of P. P by P 4 4 Warwick was solving a pair of simultaneous equations using the elimination method and reached the result that 0 = -5. Suggest a solution to the problem, giving a reason for your answer. A piece of wire 1 m long is cut into two pieces. One piece is used to make a circle, the other a square. Determine where to cut the wire so that the areas of the square and circle are equal. Translate each of these parabolas by the given amounts, then write the new equation in standard form. a y = x2 - 4x + 1 translated 3 units left, 2 units up. y = x2 + 2x b y = -4x2 + 6x - 2 translated 3 units down and 1 unit right. c y = (x - 4)2 - 5 translated 2 units left and 5 units up. y = -4x2 + 14x - 15 A cone has a radius (r) of 8 cm and a height (h) of 16 cm, as shown in the figure below. The ‘top’ is sliced off to leave a ‘frustum’ (shaded area). y = x2 - 4x + 4

Cone Frustum

h r

a Calculate the total volume of the cone. 1072 cm3 b If the volume of the sliced off top is 20% of the total volume, determine the height of the frustum. 9.4 cm 358

Maths Quest 10 for the Australian Curriculum

problem solving 29 A movie projector uses 35 mm film (35 mm wide and 24 mm high) with a light source 60 mm

from the film’s surface. a How far away is the projector’s light source from the screen if the width of the image on the screen is 16.5 m? 28.3 m b If the distance between the film and the light source is halved, what happens to the width of the image on the screen? The image width doubles. 30 A landscape gardener wishes to put a fence around a rectangular lawn. The lawn’s width is 3 m shorter than its length and there is to be allowance for a 2-m wide gate. a Develop a formula for the total length of the fence in terms of the length of the lawn. b The cost of the fence is $23 per metre plus a $100 additional fixed fee. Modify your Total length = 4l - 8, formula to provide the cost of the fence. Cost = 23(4l - 8) + 100 where l is the length of c Use your formula to determine the cost of a fence for a lawn whose width is 12 m. $1296 the lawn. 31 The equation of a quadratic can be determined directly from a table of values and differences. Consider the table below. x

-3

-2

-1

 0

1

 2

 3

y

 3

-3

-5

-3

3

13

27

Now determine the 1st differences x

-3

-2

-1

 0

1

 2

 3

y

 3

-3

-5

-3

3

13

27

-6

1st difference

-2

 2

6

10

14

Add the 2nd differences to the table

The equation of the quadratic is y = 2x2 + 4x - 3. y

x

-3

-2

-1

 0

1

 2

 3

y

 3

-3

-5

-3

3

13

27

1st difference

-6

-2

 2

6

10

14

25

2nd difference

20 15 10 5 -3

-2

-1

0

1

2

-5

V = 130 x − x 2

3

 4

 4

4

 4

 4

Note how the 2nd differences are constant. The theory of differences states that this constant is equal to 2a in the equation y = ax2 + bx + c. a Determine the value of a. a = 2 b Re-do the table to subtract the term ax2 from each y-value, leaving a table for bx + c. x 3 Use this table to determine b. Note that this equation is now linear and the first difference gives the value of b. b = 4 c Use a similar method to determine c. c = -3 d What is the equation of the quadratic? Confirm your result by plotting the graph. 32 A box with a lid is to be constructed with a total surface area of 260 cm2. The box is to have a square base of side length x and height L. 130 − x 2 a Write the equation for total surface area and make L the subject of the equation. L = 2x b Write the formula for the volume in terms of L and x. Substitute in your result from part a  to derive a formula for the volume in terms of x. c By trial-and-error, or another method, determine the values of L and x that make the volume as large as possible. x = 6.6 cm, L = 6.55 cm Chapter 11 Problem solving I

359

problem solving 33 A right-angled triangle is inscribed in a circle of diameter length d cm as shown in the diagram below. Check with your teacher.

d

ø8 4

a Show that d = 4 6 cm b Show that the proportion of the area of the triangle to the area of the circle is

2 6π

34 Develop a formula for the volume of a cone with a radius r and a slant height s. V = 13p r2 s 2 − r 4 35 Given the quadratic equation y = x2 - 4x + 7 and a ‘general’ quadratic y = ax2 + bx + c,

determine the conditions on a, b and c such that: a the two quadratics never intersect b2 + 8b + 28a + 4c - 4ac - 12 < 0 b the two quadratics intersect (or touch) once b2 + 8b + 28a + 4c - 4ac - 12 = 0 c the two quadratics intersect twice. b2 + 8b + 28a + 4c - 4ac - 12 > 0 36 A set of wine bottles is stacked lying down as shown in the diagram below, making a triangular effect. The radius of each bottle is r cm.

55 bottles

a b c d

How many bottles are required if there are 10 bottles on the base and 1 bottle on the top? What is the length of the base of the 10-bottle ‘triangle’? 20r cm What is the total height of the triangle? 17.32r cm Generalise your result for the height of a ‘triangle’ of n bottles on the base. 3nr cm

37 The cost of a return airline ticket to Perth from Sydney varies between airlines. If a ticket

travelling with Virgin Green Airline costs $458 and a ticket travelling with Qintas costs $506, determine the number of people who travelled by air to Perth from Sydney if there were 20% more passengers that flew with Virgin Green and the total price for all airline tickets was $63  336. 132 passengers — 72 Virgin Green passengers, 60 Qintas passengers 38 The diagram below represents the safety ratio for placing ladders against vertical structures.

4 units q 1 unit a Using the values shown in the diagram, determine the value of the angle, q. Write your answer correct to the nearest minute. 75è58Å b A 3-metre ladder is placed against a vertical wall. Determine the horizontal distance,

to the nearest centimetre, that the ladder should be placed so that it satisfies the safety regulations. 73 cm 360

Maths Quest 10 for the Australian Curriculum

problem solving 39 Concentric circles are circles that share a common centre. A circle is drawn with a radius of

x cm. The next circle drawn has a radius of (x + 1) cm. This pattern continues until five circles are drawn. The diagram below shows the 5 concentric circles.

x+1 x

3rd is (x + 2) cm; 4th is (x + 3) cm; 5th is (x + 4) cm Circumference of 4th a Write down the radii of the 3rd, 4th and 5th circles, in terms of x. 100 % larger b Write down an expression, in terms of x, that can be used to determine the ratio of the circle is x+3 area of the 2nd circle to the area of the 5th circle. If this ratio is 94 , determine the value than the 3rd circle’s of x. 5 cm circumference. c What is the percentage increase in circumference, in terms of x, in moving from the

3rd circle to the 4th circle? 40 Carol is celebrating her 16th birthday. In her excitement she cuts the cake into unequal

sections. Her brother takes the largest piece which is twice as much as her mother’s piece. Her sister takes one of the smaller pieces which is 1 the size of her mother’s piece of cake. 3

Carol and her father each take a piece of cake which is 1 12 times as large as Carol’s mother’s

piece. If Carol had cut the cake into 8 equal slices, each slice would have been the same size as her mother’s piece of cake. 19 a What fraction of the cake was eaten by Carol and her family? 24 5 b What fraction of the cake remains? 24 c If the exact amount of cake, in cm2, remaining after Carol and her family have eaten their one piece of cake is 160p, determine the exact diameter of the cake. 32 3 cm 41 Calculate the x-coordinate of the intersection point(s), if any, of:

y = 2x2 - 5x - 3 and y = -x2 - 3x x = Give your answer in exact form.

1 ± 10 3

42 A large advertising banner is to be placed on the side of a building. The banner has a diagonal

7 3 metres 6

of length 4 17 m and a height of 5 5m. a Determine the exact width, in metres, of the banner. 7 3 metres b Determine the exact area, in m2, of the banner. 35 15 m2 c To attach the banner to the side of the building, anchor points are attached around the border. There is an anchor point attached to each corner and anchor points across the width and the height. i There are 7 anchor points across the width at the top and 7 at the bottom of the banner. Determine the exact length, in metres, between these points. ii Anchor points are placed

5 5 3

metres apart along both sides of the height of the

banner. Determine the total number of anchor points needed for the banner. 18 43 Shane is a coach driver who conducts tours in outback Australia. All tours are based on twin share at a cost of $x per passenger. For passengers wishing to have their own room, an additional cost of $385 is added to the overall tour price. Usually on any tour, an average of 75% of passengers choose twin share. a If there are n passengers, write down an equation, in terms of n and x, that can be used to determine the total amount, A, in dollars, collected in tour money. A = nx + 96.25n Chapter 11 Problem solving I

361

problem solving b Shane conducts a tour with 50 passengers. On the next tour there are 45 passengers.

The difference in the total amount between the first and second tour is $17  981.25. Determine the value of x. $3500 44 A section of a stained glass window is shown below. y 2 3

1

D

A

0

B

x

C

The pattern formed can be modelled using three intersecting parabolas labelled 1, 2 and 3 with equations: 2 10 Parabola 1: y = − x 2 + x [1] 15 3 Parabola 2:  y = -0.192x2 + 9.6x - 90 Parabola 3: y = −

[2]

2 2 500 x + 10 x − 15 3

[3]

a By solving the equation [2] for y = 0 using any method, find the coordinates of points B and C. B (12.5, 0) and C (37.5, 0) b      i Set up an equation, in terms of x, that will determine the point of intersection 22 2 94 between parabolas 1 and 2. Write your values in exact form. x - x + 90 = 0 375 15 ii Using your calculator, find the coordinates of point A. (17.1, 18) c      i By finding the turning point format of equations [1] and [3], state the transformation made to equation [1] so that it maps onto equation [3]. Translated 25 units to the right ii Hence, write down the coordinates of point D. ( 75, 125 ) 6 2

45 Chilly Treats want to remodel their ice snacks packages. Currently frozen fruit juice is sold

in containers in the shape of an equilateral triangular-based prism. The sides of the triangular base are x cm and the height of the container is 10 cm.

10 cm

x

x x

362

Maths Quest 10 for the Australian Curriculum

problem solving

49  b Solving 36x2 - 6x - 6 = 97 026 using any method gives x = -51.833 metres and x = 52. Ignore the negative solution because x > 0 for measurement units. Possible dimensions could be: 3(2x - 1) by 2(3x + 1) 2(2x - 1) by 3(3x + 1) 6(2x - 1) by 1(3x + 1) (2x - 1) by 6(3x + 1). Or any possible combination for numbers whose product is 6, such as 1.5 and 4

The total capacity of the container is 250 mL. Chilly Treats’ new design will be in the shape of a cylinder. x 2 = 100 a Find the exact value of x2. 3 10 b If the capacity and vertical height of the container is to remain the same, determine the π exact diameter, in centimetres, of the cylinder. c If the area of one of the side faces of the triangular based prism is approximately 40 cm2, determine the difference in the total surface area, in cm2, between the The cylinder has two different containers. Write your answer correct to 2 decimal places. Clearly larger surface area by 57.25 cm2. indicate which container has the larger surface area. 46 Jacques is test driving a new model Rocket Roadster. The speed of the car can be modelled using the equation S(t) = -3t2 + 12t + 27, where S is the speed in metres per second (m/s) and t is time, in seconds. a What was the initial speed, in km/h, of the car when the testing began? Write your answer correct to 1 decimal place. 97.2 km/h b By converting S(t) into turning point format, determine the maximum speed, in km/h correct to 1 decimal place, that the Rocket Roadster can reach in this road test and the time taken, in seconds, to reach that speed. 140.4 km/h in 2 seconds 47 A circular piece of paper has a 90è sector removed as shown in the diagram below. r 90è h 30 20

The remaining area is carefully folded to make the shape of a cone. 10 a What is the slant height of this cone? r 0 b Develop a formula for the radius of the base of this cone. 3 r 1 2 3 t 4 3 c Show that the volume of the cone is 7 π r 3. Check with your teacher. 64 48 A ball is thrown upwards in the air. The maximum height of the ball is given by the rule h = h0 + V0t - 16t2, where h = height (in metres) and t = time (in seconds). a What is the height at t = 0? At t = 0, h = h0 = the starting height. b The ball is thrown up in the air from the edge of a roof 20 m high. It goes up, reaches a maximum height and returns just past the edge of the roof and hits the ground. The initial speed of the ball, V0, is 12 m/s (about 43 km/h). Sketch a graph of this situation. c What is the maximum height of the ball? Estimate your answer from the graph then calculate the exact answer. 22.25 m d How long does it take to reach the maximum height? Estimate your answer from the graph then calculate the real answer. 0.375 seconds e How long is the ball in the air before it hits the ground? Estimate your answer from the graph then calculate the real answer. 1.55 seconds 49 Farmer Gordon has two paddocks ready for sowing. One paddock has dimensions (2x - 1) metres by (3x + 1) metres, and the second paddock has an area of (36x2 - 6x - 6) m2. (2 x − 1)(3 x + 1) 1 a Show that the area ratio of paddock 1 to paddock 2 is 1 : 6. = 6(2 x − 1)(3 x + 1) 6 b The area of paddock 2 is 97  026 m2. Show that the value of x is 52. Hence find a set of possible dimensions for paddock 2 using your answer from part a. 50 Greg is building a garden shed. He measured out the width and length of the shed to be (3 m ê 5 cm) by (4.5 m ê 5 cm). Width: 2.95 m to 3.05 m, length: 4.45 m to 4.55 m a i Determine the possible range of values for both the width and length. ii D  etermine the percentage error in Greg’s measurement for the width if the actual width was 2.98 m. Write your answer correct to 2 decimal places. 0.67% Chapter 11 Problem solving I

363

probleM solving b The walls of the shed will be 2 metres high and the roof will be a height of 3 metres

from the base. The shed will be constructed entirely from corrugated iron. The diagram below shows the shed and its dimensions. The shed will be built on a concrete slab. Using Greg’s measurements of length of 4.5 m and width of 3 m, determine the minimum amount of iron, in m2, required. Write your answer correct to 2 decimal places. 49.23 m2

1m

2m

4.5 m 3m c Corrugated iron is sold in lineal metres at $11.80 per metre. The effective width of the

corrugated iron sheet is 762 mm (this is allowing for overlapping of sheets). Determine the minimum cost for the corrugated iron. Write your answer to the nearest $10. $760 51 To estimate the length along the side of an inaccessible bushland, a surveyor marks out a circular path around the bushland so that the four corners of the bushland lie on the circumference of the circle. She is able to measure three of the four sides of the bushland as shown in the sketch below. (The diagram is not drawn to scale.) D 790 m A

2sè 3

980 m

pè xm

3 pè 2

C

The square numbers are 1, 4, 9, 16, 25, 36, ………… The difference between these numbers is 3, 5, 7, 9, 11... If this continues to 75, it is the 37th number, so 382 – 372 = 75. So, the two natural numbers are 37 and 38.

sè 850 m ABCD is trapezoidal with AD||BC. B ±BAD = ±CDA = 72è ±ABC = ±BCD = 108è a Using one of the theorems of circle geometry, determine the exact values of p and s, in degrees. p = 72è, s = 108è b Describe the shape of the figure ABCD. 52 Explain how you could develop a pattern to determine two consecutive natural numbers

whose squares differ by 75. 53 A certain type of carpet has a width of (x + 2) metres. Customers can purchase the carpet in any length. Mr Barnes buys (x + 5) m of this carpet for his rumpus room and Mr Snowdon buys a 4 m length for his family room. Mr Barnes has (x + 2)(x + 5) m2; a Write an expression for the area of carpet that each man buys. Mr Snowdon has 4(x + 2) m2. b Write an expression for the difference in area if Mr Barnes has the longer piece of carpet. c Factorise and simplify this expression. (x + 2)(x + 1) b (x + 2)(x + 5) - 4(x + 2) d If Mr Barnes has bought 6 m2 more than Mr Snowdon, find the width of the carpet. e What area of carpet did each man buy? Mr Barnes bought 18 m2 and The carpet has Mr Snowdon bought 12 m2.

364

Maths Quest 10 for the Australian Curriculum

a width of 3 m.

problem solving 54 A square is drawn within a semicircle as shown in this diagram. The area of the square is 32 cm2.

A square is drawn inside a circle with the same radius as the semicircle.

Check with your teacher.

Determine the length of the side of the square in the circle. The side of the square is 4 5 cm. 55 Rebecca and Bethany are participating in a fund raising charity door knock. In the first hour, each girl collected x number of gold coins. In the second hour Rebecca collected (x - 1) gold coins and in the third hour she collected (3x - 4) gold coins. In the fourth hour, the number of gold coins each girl collected was the product of the number of coins collected during the second and third hours. Bethany’s number of gold coins collected in the fourth hour can be expressed as 2x2 + 5x - 7. a Show that the number of coins Rebecca collected during the fourth hour was 3x2 - 7x + 4. b Show that the expression, in terms of x, for the combined number of coins both Rebecca and Bethany collected in the fourth hour is (x - 1)(5x + 3). 56 The shape of this vase be approximated using a truncated cone, as shown in the diagram below. 20 cm

30 cm 15 cm

Check with your teacher.

Show that the total amount of water, in litres, that can be poured into the vase is 2.3p. 57 Consider the diagram below. If the length of AO is one-third the length of AD and the length of AC is 2 units, explain why the length of BD is 4 units. Use similar triangles. C

D

O A

B

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problem solving

58 In an election for Year 10 representatives on the school council,



1 4

1

2 5

of the votes went to James,

1

to Jennifer, 6 to Raoul, 10 to Amy and the remaining 20 votes went to Diana. How many

students were there in year 10? 240 students 59 An Xbox games package comes in a box that has a length 10 cm longer than its width and a height that is 7 cm greater than its width. a Using the variable x to represent the width of the box, write an equation for the box’s volume. V = x(x + 10)(x + 7) b Find the volume of the box if its width is 30 cm. 44 400 cm3 c The manufacturer wants to include another controller and decides that the box should have a volume of 94  000 cm3 but retain the same shape. Using trial and error, find the width for this volume, if x must be an integer. 40 cm

N 163è20Å 16è40Å b km

25 km N 121è40Å a km d km

19 km c km 58è20Å

k = 12 makes the two equations represent the same line, giving an infinite number of solutions. All other values of k generate two parallel lines.

60 Find values of k for which the simultaneous equations 2x + y = 6 and 2y = -4x + k have: a an infinite number of solutions k = 12 b no solutions. k ≠ 12

Explain how you arrived at your answers. 61 A yacht sails for 25 km on a bearing of 163è20ÅT and then for a further 19 km on a bearing of 23.3 km east and 33.9 km 121è40ÅT. south of its starting point a Draw a fully labelled diagram to show the yacht’s course. b Determine how far east and south the yacht is from its starting point. c Calculate the true bearing of the yacht from its starting point. Give your answer correct to the nearest minute. 145è30ÅT 62 Martha decides to redesign the front cover of her diary which has an area equal to (x2 - 3x - 10) cm2. a Factorise this expression to find the dimensions of the diary cover in terms of x. (x - 5)(x + 2) b Write down the shorter length in terms of x. (x - 5) is the shorter length. c If the shorter sides of the diary cover are 12 cm in length, find the value of x. x = 17 d What is the area of the front cover of Martha’s diary? 228 cm2 63 A rectangular hallway rug is five times as long as it is wide. Its diagonal length is 410 cm. How wide and long is the rug? The rug is 80 cm wide and 400 cm long. 366

Maths Quest 10 for the Australian Curriculum

probleM solving 64 A circular dining table made of cedar timber is inlaid with glass as shown in the diagram

65 The factors of 24 are: 1 and 24; 2 and 12, 3 and 8 and 4 and 6. To make the first bracket equal 1, then x must be 7 and to make the second bracket equal 24, then y must be 28. This pattern continues until all possibilities are found. They are:

1 2 3 4 6 8 12 24

Factors

24 12 8 6 4 3 2 1

x 7 8 9 10 12 14 18 30

y 28 16 12 10 8 7 6 5

below. The radius of the glass top is 2r cm with a 20 cm ring of cedar around it.

State the diameter of the glass (in terms of r). 4r cm Give the radius of the glass and wood (the table top). (2r + 20) cm Calculate the area of the glass. 4p r2 cm2 Determine the area of the top of the table (glass and wood) (4p r2 + 80p r + 400p ) cm2 Write an equation to find the area of the wood section only and write it in factorised form. 80p (r + 5) cm2 f If the radius of the glass is 40 cm, find the area of the wood needed to surround the glass. Give your answer in m2. 1.131 m2 g The manufacturers want to make a slightly larger table in the same design using the same width cedar ring. If the area of the table top is to be 2 m2, find the size of r (to the nearest cm). 30 cm 65 How many solutions can you find for the equation (x - 6)(y - 4) = 24 if x and y are positive integers? 66 A cross brace (shown in red) has been placed to support the roof of a garage as shown. Find the length (in mm) of this supporting beam. 5704 mm a b c d e

1800 mm

5200 mm 3000 mm

360 adults and 190 children

67 When the movie The Fellowship of the Ring was shown in the cinema, every seat (550) was

taken. The price of admission for adults was $9.50 and for children $4.50. The takings for one night were $4275. How many adults and children were present at the movie? 68 A coffee table rectangular cloth is to be decorated by sewing lace onto the edge of the The cloth is x cm wide and 4x cm long. material. Its length is four times its width. a If the width of the material is x cm, express the dimensions of the cloth in terms of x. b Give an equation for the perimeter and the area of the cloth in terms of x. P = 10x, A = 4x2 c Find the length and width of the cloth if its perimeter is 3 m. Length = 120 cm, width = 30 cm. d If the width of the lace is 6 cm, what is the outside perimeter of the cloth now, and how much area does it cover? (Answer in terms of x) e Given the original perimeter was 3 m, what increase in area of the cloth was achieved by Perimeter = (10x + 48) cm, adding the lace? The area has increased by 1944 cm2. area = (4x2 + 60x + 144) cm2 69 Solve each of the following. a Find the value of r if x2 - 4x - r = 0 has one solution. r = -4 b Find the value of s if 2x2 - 5x + s = 0 has two solutions. s < 25 8 c Find the value of t when tx2 - 3x - 8 = 0 has one solution. t = - 9 32 Chapter 11 problem solving I

367

probleM solving Dan is 25 years old.

70 Five years ago Dan was twice as old as he was 15 years ago. What is Dan’s age now? 71 A surveyor measures the angle of elevation to the top of a lighthouse from a point on the

ground 130 m from its base as 37è. When he looks further down the lighthouse, he sees a large balcony. The angle of elevation to the balcony from the same point is 31è. What is the distance from the balcony to the top of the lighthouse? 19.85 m 72 Robyn keeps guinea pigs in a small square enclosure with sides measuring x m. The number 1m of guinea pigs is increasing so she wants to increase the size of the enclosure by 1 m on one side and 3 m on the adjacent side. a Draw a labelled diagram of the original square and show the additions to it. b Write an expression for the area of the new enclosure. (x + 1)(x + 3) c To satisfy animal safety requirements, the area of the enclosure must be at least 15 m2. Find the dimensions of the enclosure. 5 m by 3 m d To make sure the enclosure is big enough, Robyn decides to make the area 17 m2. Determine the new dimensions of the enclosure (to the nearest cm). 5.24 m by 3.24 m 73 During an 8-hour period, an experiment is done in which the temperature of a room follows the relationship T = h2 - 8h + 21, where T is the temperature in degrees Celsius h hours after starting the experiment. a Change the equation into turning point form and hence sketch the graph of this quadratic. b What is the initial temperature? 21èC c After three hours, is the temperature increasing or decreasing? Decreasing d After five hours is the temperature increasing or decreasing? Increasing e State the minimum temperature and when it occurred. 5 èC after 4 hours f What is the temperature after 8 hours? 21èC 74 The height of a playground swing above the ground of is 2 3 m. The base of the swing’s pipe supports must be 5 m apart so that the structure is stable once children start to swing.

xm xm

Temperature degrees celsius

3m

T 25 20 15 10 5 0

2

4 6 Hours

8h

5m 2 3m

The struts each need 73 m long. to be 2

a Find the length of the struts (use exact values) that make up the supports for the swing. b If the base of the swing seat is to be 1 m off the ground, how much chain is required for 2 swings? (Use exact values.) The chain length is (2 3 − 1) m, so (8 3 − 4) m of chain is required. c How much pipe is required to build the swing if the length at the top of the swing is to

be 30 m long? (2 73 + 30 ) m of pipe is required. 75 A bike chain is wrapped around 3 gear wheels that are the same size. The radius of each wheel is 8 cm. How long is the chain? Approximately 98.3 cm

368

Maths Quest 10 for the Australian Curriculum

problem solving 76 The parabola y = x2 + bx + c has x intercepts (2, 0) and (6, 0). a Find the values of b and c. b = -8, c = 12 b State the equation. y = x2 - 8x + 12 d Sketch the parabola. (4, -4) c Complete the square and find the turning point. 77 Two guide wires are used to support a flagpole as shown. One reaches the top of the flagpole

and the other part way down the pole.

y 15

y 15

Wire

10

5

9.5 m

5 -15 -10

0

-5

10

5

10

0

3m

15 x

2

4

6

8

10 x

-5

-5

6m

-10

a What is the height of the flagpole (to the nearest metre)? 8 m b What angle (to the nearest degree) does the longer guide wire make with the ground? 57è c The shorter wire is attached to the flagpole 1 m from the top. How long is this wire? 7.6 m

-15

78 When a drop of water hits the flat surface of a pool, circular ripples are made. One ripple is

a  i

6 cm

10 cm

8 cm

10 cm

12 cm

represented by the equation x2 + y2 = 9 and 5 seconds later, the ripple is represented by the equation x2 + y2 = 225, where the lengths of the radii are in cm. a State the radius of each of the ripples. First ripple’s radius is 3 cm, second ripple’s radius is 15 cm. b Sketch these equations. c How fast is the ripple moving outwards? 2.4 cm/s d If the ripple continues to move at the same rate, when will it hit the edge of the pool which is 2 m from its centre? 1 minute 23 seconds after it is dropped 79 Stephie, a tennis player, serves the ball in a tournament. She throws the ball in the air and hits it over the net. Her arm length is 60 cm and it is 40 cm from her grip on the tennis racquet to the centre of the racquet. How far does the centre of the racquet travel if she swings through an angle of 300è? The centre of the racket travels 5.24 m. 80 This 8 cm by 12 cm rectangle is cut into two sections as shown. iii 

  ii 10 cm

6 cm

6 cm

6 cm

8 cm 10 cm

10 cm

8 cm

10 cm

8 cm

10 cm

Area of trapezium = 1(18 + 6) ì 8 = 96 cm2.

8 cm

2

12 cm

12 cm

Area of rectangle = 12 ì 8 = 96 cm2. Area of parallelogram = 12 ì 8 = 96 cm2, 12 × 16 Area of triangle = 2 = 96 cm2,

6 cm

12 cm

0

2

4

6

81 Bridgette is practising her golf drives. The path the golf ball takes is defined by the quadratic

1 equation h = − (d − 6)2 + 6, where h is the height of the ball above the ground for a 6 Perimeter of rectangle = 40 cm. horizontal distance of d. Both h and d are in metres. Perimeter of parallelogram = 44 cm, (6, 6) a Find the value of h when d = 0. 0 Perimeter of triangle = 48 cm, 1 Perimeter of trapezium = 44 cm. 2 b State the turning point of the equation h = − (d − 6) + 6 The triangle has the largest perimeter, 6 while the rectangle has the smallest. c Sketch the graph of this relationship. d What horizontal distance does the golf ball cover in its flight? 12 m

Height (metres)

h 8

2

4 6 8 10 Distance (metres)

12 d

a Draw labelled diagrams to show how the two sections can be rearranged to form a:    i parallelogram ii right-angled triangle iii trapezium. b Show that these figures, as well as the original rectangle, all have the same area. c Comment on the perimeters of the figures.

Chapter 11 Problem solving I

369

problem solving e What is the maximum height the golf ball reaches? 6 m f At what horizontal distance was the golf ball at its maximum height? 6 m 82 A yacht is anchored off an island. It is 2.3 km from the yacht club and 4.6 km from a weather

station. The three points form a right angled triangle at the yacht club. Weather station

Yacht club

2.3 km 4.6 km Yacht a Calculate the angle at the yacht between the yacht club and the weather station. b Calculate the distance between the yacht club and the weather station. 3.98 km

60è

The next day the yacht travels directly towards the yacht club, but is prevented from reaching the club because of dense fog. The weather station notifies the yacht that it is now 4.2 km from the station. c Calculate the new angle at the yacht between the yacht club and the weather station. 71è d Determine how far the yacht is now from the yacht club. 1.34 km 83 The minute hand in Penny’s watch is 1 cm long. Someone told her that the tip of the hand travels more than 30 m in 8 hours. Is this true? Show full working to justify your answer. 84 This cable drum has the measurements shown. True: the tip travels 30.2 m. y 12

10 cm

Height (metres)

10 8

25 cm

30 cm

6 4

10 cm

2 0

1 2 3 4 5 6 Horizontal distance (metres)

x

50 cm

a What volume of wood was used in its construction? 56 941 cm3 b Determine its surface area. 11 938 cm2 85 Jan is practising for the World Diving Championships. The path she takes from the diving board

into the water is given by the quadratic equation y = -0.75x2 + 3x + 8, where y metres is the height above the water level and x metres is the horizontal distance from the edge of the board. a Using a graphics calculator, sketch the graph of y = -0.75x2 + 3x + 8. b What is the height of the diving board above the water? 8 m c What was the maximum height Jan reached during her dive? 11 m above the water d What was the horizontal distance Jan covered before she hit the water? 5.83 m 86 A skip bin for waste has been delivered to a building site. Its shape is in the form of a trapezoidal prism with dimensions as shown in the diagram. 4m 2m

3.5 m 3m 370

Maths Quest 10 for the Australian Curriculum

problem solving a Calculate the volume of material this skip can hold. (Assume that it is not loaded beyond the top rim.) 24.5 m3 b A smaller skip has a volume one-eighth the size of the larger one. If its shape is similar

The dimensions of the smaller skip are half those of the larger one.

to that of the larger one, what would its dimensions be? 87 In a children’s play gym, two cylindrical foam shapes are placed on the ground and a board covered in foam rests on it. The cylinders have radii of 50 cm and 40 cm and their distance apart on the ground is 1.5 m. Calculate the angle the board makes with the ground. 7.6è

50 cm 40 cm 1.5 m y 5

88 The small and large triangles in this figure are similar.

4 3

25

19

2

(1, 2)

1

x -2

15

y

-1

0

1

2

3

4

5x

-1 -2

22

Determine the lengths of the pronumerals. x = 13.75, y = 11.4 89 The equation (x - 1)2 + (y - 2)2 = 4 describes a circle. a State the centre of the circle. Centre is (1, 2). b State the radius of the circle. Radius is 2. c Find the x- and y-intercepts of this circle. d Sketch the circle, clearly marking the centre.

x-intercept 1; y-intercepts (− 3 + 2) and ( 3 + 2)

90 The parabola y = ax2 + bx + c has a turning point (-3, 4) and passes through the point

(1, -28). a Determine its equation. y = -2x2 - 12x - 14 b State the values of a, b and c. a = -2, b = -12, c = -14 c Sketch the parabola. 91 Find the sum of the angles at the tips of this regular star. 180è (-3, 4)

y 4 2 (-1.6, 0)

(-4.4, 0) 0 -5 -4 -3 -2 -1 -2

1 2 3x

-4 -6 -8 y = -2x2 - 12x - 14 -10 -12 (0, -14) -14 -16

Chapter 11 Problem solving I

371

problem solving 92 Tina is re-covering a footstool in the shape of a cylinder with diameter 50 cm and height

30 cm. She also intends to cover the base of the cushion. The area of material required is 1.04 m2. If Tina is careful in placing the pattern pieces, she may be able to cover the footstool.

She has 1 m2 of fabric to make this footstool. When calculating the area of fabric required, allow an extra 20% of the total surface area to cater for seams and pattern placings. Explain whether Tina has enough material to cover the footstool. 93 The Gold Coast City Council has decided to construct a ceremonial arch at the entrance to Surfers Paradise beach. The arch is to be in the shape of a parabola. The maximum height is to be 15 metres and the width at the base is to be 20 metres. A path, 16 metres wide will pass beneath the arch with its centre immediately beneath the highest point of the arch.

15 m 16 m 20 m

The designers decide to use a mathematical model of the arch in the design process. They place an origin of coordinates at ground level immediately beneath the highest point of the arch. y 14 12 10 8 6 4 2 -10

-5

0

5

10 x

a Prove that the equation of this curve is y = 15 - 0.15x2 Prove the equation is y = 15 - 0.15x2. b The designers are concerned that there will not be sufficient clearance for vehicles up to

6 metres in height to pass along the path beneath the arch. Show that the vertical height 372

Maths Quest 10 for the Australian Curriculum

problem solving

of this arch above the road level at the edge of the road would not be sufficient to allow a 6 m high vehicle to pass through. The height at the edge of the road is 5.4 m. c How wide must the road be to allow a 6-m high vehicle to pass through the arch? The road needs to be 15.5 m wide. a−b 94 a Q b is defined as . a b 1 1 Θ What is the value of 2 3 in simplest terms? - 4 9 1 1 Θ 3 2 95 David has calculated the time, in minutes, it takes him to drive to work in the morning as m2 - 10m + 50, where m is the number of minutes after 8 am that he leaves home. a How long does it take David to reach work if he leaves at 8 am. 50 minutes b At what time can he leave so that the trip takes 30 minutes? 8.03 am or 8.07 am c The trip will take one hour if he leaves at what time? 8.11 am d When should he leave to take the smallest amount of time. 8.05 am e How long will it take him if he leaves home at this time? 25 minutes f If he decides that he cannot take longer than 30 minutes to get to work, between what times would he have to leave home? Between 8.03 am and 8.07 am 96 There is a theorem which says: If two distinct numbers are exactly divisible by the difference of the two numbers, the difference is the HCF of the two numbers. Matt is travelling at 45 km/h and Explain what this means, illustrating with an example.

Steve is travelling at 60 km/h.

97 Matt and his brother Steve start from home in their cars. Matt travels directly east, while

Steve travels directly north at a speed 15 km/h faster than Matt’s speed. After travelling for 1 hour 20 minutes, the two cars are 100 km apart. At what speeds are the two cars travelling? 98 A regular octagon is inscribed inside a circle with all its vertices lying on the circumference of the circle. The circle has a radius of 10 cm. Determine the perimeter of the octagon. 99 An obtuse-angled isosceles triangle has equal angles of xè and equal side lengths of y cm. For example: take the two numbers 48 and 60. Their HCF is 12. Difference = 60 - 48 = 12 60 ó 12 = 5 and 48 ó 12 = 4 The two numbers are exactly divisible by 12. This theorem says, then, that the HCF of 48 and 60 is 12, which is the case.

The perimeter of the octagon is 61 cm.

xè y cm xè y cm

There is not enough information about the triangle to use a traditional formula 1 A = ( 2 base ì height) to find its area. Show how you could use trigonometry to develop the following formula to calculate its area. 1 Area = y2sin (2xè) Check with your teacher. 2 100 Farmer Max has a rectangular field 150 m by 100 m. His son offers to help him mow the field, but says he will do only half. The ride-on mower cuts a strip 1 m wide. Max starts mowing at a corner, and mows around the field towards the centre. He stops and hands over to his son when he has done n circuits of the field. Construct an algebraic equation involving n, and solve to find its value. Give your answer correct to 1 decimal place. 17.4 circuits Chapter 11 Problem solving I

373

problem solving 101 Pulsars are rapidly spinning stars. They spin at incredible rates when they are first formed —

about 30 times every second. As they age, they slow down. Astronomers have represented the spinning rates of two pulsars (Crab Nebula and AP 2016 + 28) by two simple equations. Crab Nebula: P = 0.033 + 0.000 013T AP 2016 + 28:  P = 0.558 + 0.000 000 004 7T P is the time (in seconds) it takes for the pulsar to spin once on its axis, and T is the number of years since today. In approximately how many years from now will the two pulsars be spinning at the same rate? 102 The two arches of the Sydney Harbour Bridge can both be modelled as parabolas. Approximately 40 400 years from now

Using the reference point (0, 0) as the bottom of the left side of the lower arch, the two equations are: The 8 different ways are: 1 12

=

1 12

=

1 12 1 12

=

1 12

1 13 1 14

+

1 15 1 16

+

1 60

+

=

1 18

+

1 48 1 36

1 12

=

1 20

+

1 30

1 12

=

1 21

+

1 28

1 12

=

1 24

+

1 24

=

+

1 156 1 84

Lower arch: y = -0.001  92x2 + 0.96x Upper arch: y = -0.001  28x2 + 0.64x + 60 Note: The measurements are in metres. Write a description of the two arches, giving as much information as you can. Remember to support your information with mathematical evidence. Check with your teacher. 1 1 1 1 103 The unit fraction can be expressed as = + where a and b are natural 12 12 12 + a 12 +b numbers. Use this equation to find the product of a and b, then list the 8 different representations for 1 the unit fraction . 12 104 Explain why all perfect squares have an odd number of factors. Give an example to support your explanation. 105 A cone is formed from a sector of a circle with a central angle of 72è. The radius of the base of the cone is 3.18 cm. What is the radius of the circle from which the sector was taken? 15.9 cm 106 This tile pattern is made using congruent triangular tiles. There is 1 tile in row 1, 3 in row 2, and so on.

The factors of a number are generally written in pairs, producing an even number of factors. With a perfect square, one of these factors will be paired with itself, producing an odd number of factors. This occurs for all perfect squares.

Develop a formula to determine the number of tiles needed to complete a pattern of this type with r rows. The total number of tiles needed for r rows is r2. 374

Maths Quest 10 for the Australian Curriculum

problem solving 107 The school cafeteria sells apples at one price, and bananas at a different price. Six apples and

4 bananas cost $7, while 1 apple and 9 bananas cost $4.50. a Show how you can determine how much more an apple costs, compared with a banana, without actually finding the cost of each. b What is this difference in price? An apple costs 50c more than a banana. 108 When children are sick, it’s important they’re given the correct dosage of medicine. If adult medicines are the only ones available, you need to convert the adult dosage into a safe dose for young children. One rule which can be used is: Age in years + 1 ì adult dose Child’s dose = 24 a For an adult dose of 5 mL, how many mL would you give a 5-year old? Another rule commonly used is: 1.25 mL Answers may vary. The two answers are only slightly different. Because of an inability to measure Age in years to that degree of accuracy in the home, they both ì adult dose Child’s dose = age + 12 provide a good guide to a safe amount to administer. b How many mL of a 5-mL adult dose would you administer to a 5-year old using this formula? 1.47 mL c Your two answers should be different. Comment on this difference. d Is there any age for which these two formulae give the same dosage? Approx 1 and 10 years 109 Ben’s teacher shows him a graph of a quadratic equation. It is not labelled, except for three points (-2, 19), (0, 1) and (3, 4) on the curve. His task is to find the equation of the quadratic. What is the equation of the curve? y = 2x2 - 5x + 1 110 Draw two straight lines across the face of this clock, so that the sum of the numbers in each region formed is the same.  et the cost of an apple be a cents, L and the cost of a banana be b cents. 6a + 4b = 700 [1] 1a + 9b = 450 [2] Subtract [2] from [1]. 5a - 5b = 250 Divide through by 5. a - b = 50

11

11 12

1 2

10 9

3

8

4 7

6

12

1 2

10 9

3

8

4 7

6

5

5

The total of the numbers here in each region is 26.

111 A group of four people out bushwalking comes across a suspension bridge as the last obstacle

they need to cross to reach their campsite. They can’t all cross at once, because the bridge can only support a maximum of 2 people at a time. Unfortunately it is approaching dark, and they only have 1 torch among the 4 of them.

Call the people A, B, C and D who take times of 1, 2, 5 and 10 minutes respectively. A and B go over 2 min B returns 2 min C and D go over 10 min A returns 1 min A and B go over 2 min Total time = 2 + 2 + 10 + 1 + 2 = 17 min

When walking alone, the four people would take 1, 2, 5 and 10 minutes to cross one way. With 2 people walking together, because they need the flashlight for safety reasons, they must travel at the speed of the slower person. One person must then travel back across the bridge each time to bring the flashlight back. How can the group arrange themselves for the bridge-crossing to take the minimum time? What is this minimum time? Chapter 11 Problem solving I

375

problem solving

If the cut was vertical, a rectangle (or straight line if the cut was on the very edge) would result. A horizontal cut would result in a circle. A cut which goes through the sides at an angle would produce an ellipse or a parabola. An ellipse results if the cut is simply through the sides. A parabola results if the cut is through the side and the base (or top).

112 Tennis balls are stacked in the shape of a triangular pyramid, with 5 balls on each side of the base. How many balls are in the whole stack? 35 tennis balls 113 Timber railings are manufactured to be 100 cm long, with a possible error of 4%. Dana’s

deck is 55 m long, and she plans to place the railings end-to-end. How many of these railings should she order to ensure she can cover the whole length? 58 railings 114 Suppose you had some ‘blue-tack’ in the shape of a cylinder. How could you cut through the cylinder to expose a surface in the shape of a parabola? Explain any restrictions. Answers may vary. The value of n 115 Prove that the set of numbers represented by: 2n, (n2 - 1) and (n2 + 1)

must be greater than 1 because the second number would be 0 if n were 1.

produces Pythagorean triples for all values of n greater than 1. Explain the restriction on the value of n. 116 The Jackson family wants to put an L-shaped deck on one corner of their house. It is to be symmetrical around the corner, as shown. Deck

House

 here are 5 T different shapes.

  They have 30 m of a specially designed hand railing to use around the perimeter of the deck. What dimensions of the deck will give them the maximum area, using the whole 30 m of the handrail? The deck should be 5 m wide and 10 m long on each side. 117 A palimage of a number is the number that has the same digits as the given number, but in the reverse order. For example, the palimage of 476 is 674. If the sum of a number and its palimage is 968, what could the original number have been? There is more than one answer. See how many you can find. The original number could have been 187, 781, 286, 682, 385, 583 or 484. 118 Consider this question on a test paper. Solve the equation x = −3. Here is Kelly’s solution. Comment on the solution.

( x )2 = (−3)2 x=9

There is no solution to the equation. It is not possible for the square root of a number to be negative. Squaring both sides of the equation has produced an invalid solution.

119 A domino is a 2 dimensional figure formed by two congruent squares that share a common

side.

A tetromino is a 2 dimensional figure formed by four congruent squares that share common sides. They all have a perimeter of 10 units, a Draw the shape of all the different tetrominoes. except for the last one which has a b Compare their perimeters. perimeter of 8 units. 120 A motor boat leaves a ship at sea and travels north at 80 km/h. The ship precedes 30è south of east at 32 km/h. The motor boat only has fuel for 4 hours. How far north can the motor boat travel, so it can safely return to the ship in time? 112 km 376

Maths Quest 10 for the Australian Curriculum

problem solving

121 You are familiar with the quadratic formula x =

− b ± b 2 − 4 ac 2a

An alternative form of the quadratic formula is x =

2c − b ± b 2 − 4 ac

Choose a quadratic equation and show that the two formulae give the same answers. 122 Consider the number 234 written in words.

Check with

your teacher. TWO HUNDRED AND THIRTY FOUR The letters of each word are cycled separately as shown below, and placed in a numbered vertical list. 1 TWO HUNDRED AND THIRTY FOUR 2 WOT UNDREDH NDA HIRTYT OURF 3 OTW NDREDHU DAN IRTYTH URFO .. .. .. n TWO HUNDRED AND THIRTY FOUR

If n > 500, what is its smallest value? 505 123 You will be familiar with the following unit fraction additions.

1 1 1 Fold the left vertical line forward, so = + 2 3 6 that page 3 sits on top of page 4. Fold the entire bottom half backwards, 1 1 1 = + so that page 5 sits behind page 4. 3 4 12 Fold forward the left vertical fold, so that page 3 sits on top of page 2. 1 1 1 = + Finally, fold forward the left vertical 5 6 30 fold, so that page 6 sits on page 7. Write a general equation involving n of the type The pages are now in order from page 1 to page 8. 1 = . . . . . . . . 1 1 1 n + = to represent unit fraction additions of this type. n n + 1 n( n + 1) 124 A puzzle company prides itself on its unique designs. The following design of a map when folded flat is printed on one piece of paper. 3 6

4 5

2 1

7 8

The numbers represent the page numbers of the map. How can the map be folded so that its pages are in the correct order? 125 Trains travel along two straight parallel tracks between Allensville and Bentley, with the journey taking 4 h 15 min each way. The trains leave both towns on the hour every hour. If I leave Allensville at 12 noon and travel towards Bentley, how many trains will pass by me coming in the opposite direction? 9 trains 126 It has been said that if you multiply the y-coordinates for a particular x-value of two straight lines, then plot this y2-value against the particular x-value, a parabola will result. To investigate this claim, consider the two straight lines y = -2x + 4 and y = x - 3. Complete the following tables. y = -2x + 4 x

-2

-1

0

1

2

3

4

5

6

y Chapter 11 Problem solving I

377

problem solving

y=x-3 x The claim appears to be true. Further investigation would need to be conducted to determine if it worked in all cases — for example, if the lines were parallel, at right angles, vertical or horizontal.

127

128 Take the total, subtract 16, then divide by 4 to get the first number in the square. The other numbers are 129 1, 7 and 8 larger.

130

 he value of T a represents the horizontal stretching factor. Positive a, open to right, negative a open to left.

-2

-1

0

1

2

3

4

5

6

-2

-1

0

1

2

3

4

5

6

y x 2

y

What is your conclusion regarding the claim? A small company manufactures and sells muesli bars. The set-up cost to make the bars themselves was $5400, and the cost for ingredients is 45c per bar. The set-up cost to package them was $7500, and the cost of materials for each bar is 35c. The bars are sold in a multipack at 5 for $6. 6450 multi-packs How many multi-packs would the manufacturer need to sell in order to break even? Consider the following puzzle. ■■ Take a calendar displaying the dates row by row of any month in the year. ■■ Choose 4 days that form a square on the calendar. ■■ Find the total of those 4 dates. Explain how, from the total, you can determine which 4 dates have been selected. The hour and the minute hand of a clock each trace a circular path as they sweep around the clock’s face. How many times will they lie on top of each other in a 12-hour period? 11 times You are familiar with the equation of the parabola with its vertical axis of symmetry. It is also possible to get a horizontal parabola. Its shape is similar to that of a vertical parabola, but its axis of symmetry is horizontal. A horizontal parabola can be represented by the following general equation. x - h = a(y - k)2 Where (h, k) is the vertex of the parabola, and a is a constant. Investigate this general equation with particular values for h, k and a, then complete the following. a How can you distinguish a vertical parabola from a horizontal parabola, simply from its equation? x2 but no y2 gives vertical parabola, y2 but no x2 gives a horizontal parabola. b What is the effect of the value of a? c A horizontal parabola has the equation x + 4 = (y + 1)2. Draw a sketch of its shape, labelling its vertex and all x- and y-intercepts. y x + 4 = (y + 1)2

2 1

(-3, 0) -4

-3 (-4, -1)

-2

-1

0

(0, 1) 1

-1 -2 -3 -4

378

Maths Quest 10 for the Australian Curriculum

(0, -3)

2

x

stAtistiCs AND probAbility • ChANCe

12

12A Review of probability 12B Complementary and mutually exclusive events 12C Two-way tables and tree diagrams 12D Independent and dependent events 12E Conditional probability 12F Subjective probability WhAt Do yoU kNoW ?

probability

1 List what you know about chance. Create a concept map to show your list 2 Share what you know with a partner and then with a small group 3 As a class, create a large concept map that shows your class’s knowledge of chance. eBook plus

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Blackjack (or 21) is a popular card game. What is the chance of being dealt two cards whose total is 21?

stAtistiCs AND probAbility • ChANCe

Are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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380

Set notation 1 Three different sets are identified as follows:

A = {1, 4, 9} B = {2, 4, 6, 8} C = {2, 3, 5, 7}. Use these sets to answer the following. a How many numbers appear in each set? Set A: 3, Set B: 4, Set C: 4 b Which number is common to set A and set B? 4 c Which number is common to set B and set C? 2

Simplifying fractions 2 Write each of the following fractions in simplest form. a

13 52

1 4

b

4 36

1 9

c

8 12

2 3

Determining complementary events 3 Two events which have nothing in common, but when added together form the sample

space, are called complementary events. Determine the complementary event for each of the following. a Drawing an ace from a standard deck of playing cards. Not drawing an ace b Drawing a black card from a standard deck of playing cards. Drawing a red card c Obtaining a factor of 6 when a six-sided die is rolled. Obtaining a 4 or a 5 Addition and subtraction of fractions 4 Simplify each of the following. a

1 2

+

1 6

2 3

b

1 52

+

5 6

133 156

c

1 13

c

1 2

+

1 2

−

1 26

7 13

Multiplying fractions for calculating probabilities 5 Simplify each of the following. a

1 2

×

1 6

1 12

Maths Quest 10 for the Australian Curriculum

b

1 52

×

5 6

5 312

×

1 2

×

1 2

1 8

stAtistiCs AND probAbility • ChANCe

12A eBook plus

review of probability ■

Interactivity Random number generator

■

int-0089

■

■

■

■

This chapter, investigates such things as the probability of selecting 3 hearts from a deck of cards and the probability of tossing two heads when a coin has been tossed twice. Probability deals with the likelihood or chance of an event occurring. The probability of an event is represented by a number in the range 0 to 1 inclusive, which can be expressed as a fraction, decimal or percentage. There are times when it is certain that an event will not occur; for example, it is certain that an athlete will not complete a 100 m race in less than 5 seconds. Therefore, the probability for the event ‘an athlete completes 100 m in less than 5 seconds’ is 0. Alternatively, it can be certain that an event will occur. For instance, it is certain that the day following Saturday is Sunday. The probability of such an event is 1. In the range 0 to 1 inclusive there is an infinite set of numbers giving the probabilities of various events, where the chance of an event occurring increases as the probability gets closer to 1. The probability scale shown below displays the range of probabilities in the range 0 to 1 inclusive. Chances decrease Highly unlikely Impossible

0

0.1

Unlikely Very unlikely

0.2

Less than even chance

0.3

Likely

Even chance

0.4

0%

Better than even chance

0.5

0.6

Very likely

0.7

0.8

Highly likely Certain

0.9

50%

1 100%

Chances increase ■

■

■

A probability of 0.5 indicates that there is an equal chance of an event occurring as there is for the event not occurring. The probability of an event can also be described by words and phrases, such as impossible, highly unlikely, very unlikely, less than even chance, even chance, better than even chance, very likely, highly likely, certain and so on. Some terms that are used in the study of probability are defined below.

Definitions ■ ■

■ ■

■

■

Trial: the number of times a probability experiment is conducted. Outcome: the result of an experiment. For example, if a die is rolled, the outcome is a number in the range 1 to 6 inclusive. Event: a desired or favourable outcome. Equally-likely outcomes: outcomes that have the same chance of occurring. For example, if a coin is tossed, then the chance of tossing a Head is equal to the chance of tossing a Tail. Hence, they are equally-likely outcomes. Sample space, S: the set of all possible outcomes for an experiment. For example, in rolling a die, the sample space, S, is S = {1, 2, 3, 4, 5, 6}. Frequency: the number of times an outcome occurs.

experimental probability ■

The experimental probability of an event is based on past experience. Experimental probability =

number of times an event has occurred total number of trials Chapter 12 probability

381

statistics AND probability • Chance

Worked Example 1

A discus thrower has won 7 of her last 10 competitions. a  What is the probability that she will win the next competition? b  What is the probability that she will not win the next competition? Think a

b

1

Write a Number of wins = 7

Write the number of wins and the total number of competitions.

Total number of competitions = 10 number of times this event occurred total number of trials

2

Write the rule for probability.

P(event) =

3

Substitute the known values into the rule.

P(she wins) = 10

4

Write your answer.

The probability she will win the next competition 7 is 10 .

1

Write the number of losses and the total number of competitions.

7

b Number of losses = 3

Total number of competitions = 10 number of times this event occurred total number of trials

2

Write the rule for probability.

P(event) =

3

Substitute the known values into the rule.

P(she loses) = 10

4

Write your answer.

The probability she will lose the next competition 3 is 10 .

■■

3

The event ‘she will win the next competition’ and the event ‘she will not win the next competition’ are called complementary events. Complementary events will be discussed in more detail in the next section.

Relative frequency frequency of the score f or total sum of frequencies S f

■■

Relative frequency of a score =

■■

The symbol S (sigma) means ‘the sum of’. The relative frequency of a score is the same as the experimental probability of that score and is useful when analysing tabulated results.

■■

Worked Example 2

A Year 10 class has the following composition. a Calculate the relative frequency of 16-year-old girls in the class. b If a student is selected at random, determine the probability that the student is a boy. Frequency (  f  )

382

15-year-olds

16-year-olds

Total (S f  )

Boys

 7

 9

16

Girls

 6

 8

14

Total (S f )

13

17

30

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Think a

1

Write

Write the number of 16-year-old girls and the total number of students in the class.

a Number of 16-year-old girls (  f  ) = 8

Total number in the class (S f  ) = 30 f S f

2

Write the rule for relative frequency.

Relative frequency =

3

Substitute the known values into the rule.

Relative frequency = 30

4

Simplify and evaluate.

5

Write your answer.

8

4

= 15 The relative frequency of 16-year-old girls in the class is 4 . 15

b

1

Write the number of boys in the class and the total number of students in the class.

b Number of boys = 16

Total number in the class = 30 number of times this event occurred total number of trials

2

Write the rule for probability.

P(event) =

3

Substitute the known values into the rule.

P(boy selected) = 30

4

Simplify and evaluate.

5

Write your answer.

16 8

= 15 The probability of a boy being chosen is 8 . 15

Theoretical probability ■■ ■■

■■

The theoretical probability of an event, P(E), depends on the number of favourable outcomes and the total number of possible outcomes (that is, the sample space). The theoretical probability of an event is given by the rule: number of favourable outcomes P(event) = number of possible outcomes This may be simplified to: n(E) P(E) = n(S) where n(E) = number of times or ways an event, E, can occur and n(S) = number of elements in the sample space or number of ways all outcomes can occur, given all the outcomes are possible.

Worked Example 3

A card is drawn from a shuffled pack of 52 cards. Determine the probability that the card chosen is: a  a heart b  a king. Think a

1

Write

Define the events and write the number of favourable outcomes and the total number of possible outcomes. Note: There are 13 cards in each of the 4 suits.

a H is the event that a heart is chosen.

S is the sample space. n(H ) = 13 n(S ) = 52 Chapter 12 Probability

383

stAtistiCs AND probAbility • ChANCe

2

b

Write the rule for probability.

3

Substitute the known values into the rule.

4

Simplify and evaluate.

5

Write your answer.

1

Define the event and write the number of favourable outcomes and the total number of possible outcomes.

2

Write the rule for probability.

3

Substitute the known values into the rule.

4

Simplify and evaluate.

5

Write your answer.

Using P(E) =

n(E ) n(S)

P(H) =

n(H) n(S)

13

P(H) =  52 =

1 4

The probability of choosing a heart is 1 . 4

b K is the event that a king is chosen.

S is the sample space. n(K) = 4 n(S) = 52 Using P(E ) =

n(E ) n(S)

P(K) =

n(K) n(S)

P(K) =

4 52 1

= 13 The probability of choosing a king is 1 . 13

venn diagrams ■ ■

■

■

eBook plus

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SkillSHEET 12.1 doc-5286

384

Venn diagrams provide a means of representing outcomes diagrammatically. A common way of drawing Venn diagrams is to use a rectangle which represents the sample space and a series of circles representing other smaller, sorted sets. In Venn diagrams, overlapping circles represent the intersection of, or common elements in, those sets. The sample space is also known as the universal set, x .

Definitions Terminology associated with Venn diagrams is defined below. 1. A set is a collection of similar elements. 2. The universal set, x , is the largest set that contains all the possible outcomes for that experiment and is represented by the rectangle of the Venn diagram. Consider all the outcomes from an experiment where a die is rolled. The sample space, S, for this experiment is also known as the universal set, x   = {1, 2, 3, 4, 5, 6}. 3. The intersection of sets (symbol ¶) is x = Universal set represented by the common elements in two (or more) sets. A B The shaded region is A ¶ B.

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

4. The union (symbol ß) of the sets A and B is given by the combined set of A and B. It is the set of elements that are in set A or set B or in both.   The shaded region is A ß B.   Note: Common elements are written only once.

x = Universal set

A

5. The complement of a set, A (written AÅ), is the set of elements that are in x   but not in A. The shaded region represents the complement of A.

B

x = Universal set

A

B

The definitions of set, universal set, intersection, union and complement are illustrated in the following example. Example 1 Consider when rolling a die the two events: event A: rolling an even number event B: rolling a multiple of 3. The universal set is written as, x  = {1, 2, 3, 4, 5, 6} and sets A and B are, A = {2, 4, 6} and B = {3, 6}. These are represented in the Venn diagram below. x = {1, 2, 3, 4, 5, 6}

A

B 2

6

3

4 1

5

Also, Intersection of sets A and B

Union of sets A and B

x = {1, 2, 3, 4, 5, 6}

A

A

B 2

6

x = {1, 2, 3, 4, 5, 6}

3

B 2

4

6

3

4 1

A ¶ B = {6}

5

1

5

A ß B = {2, 3, 4, 6} Chapter 12 Probability

385

statistics AND probability • Chance

Complement of set A x = {1, 2, 3, 4, 5, 6}

A

B 2

3

6

4 1

5

AÅ = {1, 3, 5} 6. The subset (symbol ´) of a set is a smaller set from within the set. The shaded region in the diagram shows that A is a subset of x ; that is, A ´ x . x

A

The definition, subset, is illustrated in the example below. If M = {2, 3, 4} and N = {2, 3}, then N is a subset of M, written as N ´ M. x

M N 2

3 4

7. Disjoint sets are sets that have nothing in common with each other. That is, A ¶ B = { } = f It can be argued that the intersection of disjoint sets has nothing in it. The set {} or f  is known as the empty set, or null set. x

A B

386

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Worked Example 4 a Draw a Venn diagram representing the relationship between the following sets. Show the position

of all the elements in the Venn diagram. x  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {3, 6, 9, 12, 15, 18} B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} b Determine: i  P(A)    ii  P(B)    iii  P(A ¶ B)    iv  P(A ß B)    v  P(AÅ ¶ BÅ). Think a

Draw a rectangle with two partly intersecting circles labelled A and B.

1 2

b

Write/draw

A

Analyse sets A and B and place any common elements in the central overlap.

3

Place the remaining elements of set A in circle A.

4

Place the remaining elements of set B in circle B.

5

Place the remaining elements of the universal set x  in the rectangle. i

ii

n(x ) = 20

a

B 3 15

11 13 1 5 7 17 19 b

i n(A) = 6, n(x  ) = 20

1

Write the number of elements that belong to set A and the total number of elements.

2

Write the rule for probability.

P(A) =

n(A) n(x  )

3

Substitute the known values into the rule.

P(A) =

6 20

4

Evaluate and simplify.

1

Write the number of elements that belong to set B and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

3

= 10 ii n(B) = 10, n(x  ) = 20

P(B) =

n(B) n(x  )

P(B) =

10 20

= iii

iv

1

Write the number of elements that belong to set A ¶ B and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

1

Write the number of elements that belong to set A ß B and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

2 4 8 10 14 16 20

6 12 18

9

1 2

iii n(A ¶ B) = 3, n(x  ) = 20

P(A ¶ B) =

n(A ¶ B) n(x  )

P(A ¶ B) =

3 20

i v n(A ß B) = 13, n(x  ) = 20

P(A ß B) = P(A ß B) =

n(A ß B) n(x  )

13 20

Chapter 12 Probability

387

statistics AND probability • Chance v

1

Write the number of elements that belong to set AÅ ¶ BÅ and the total number of elements.

2

Repeat steps 2 to 4 of part b i.

v n(AÅ ¶ BÅ) = 7, n(x  ) = 20

P(AÅ ¶ BÅ) =

n(AÅ ¶ BÅ) n(x  ) 7

P(AÅ ¶ BÅ) = 20

Worked Example 5

In a class of 35 students, 6 students like all three subjects: PE, Science and Music. Eight of the students like PE and Science, 10 students like PE and Music, and 12 students like Science and Music. Also, 22 students like PE only, 18 students like Science only and 17 like Music only. Two students don’t like any of the subjects. a Display this information on a Venn diagram. b Determine the probability of selecting a student who: i  likes PE only           ii  does not like Music. c Find P[(Science ß Music) ¶ PEÅ]. Think a

1

Write/draw

Draw a rectangle with three partly intersecting circles, labelled PE, Science and Music.

n(x ) = 35

a PE

Science

Music 2

Extract the information relating to students liking all three subjects. Note: The central overlap is the key to solving these problems. Six students like all three subjects, so place the number 6 into the section corresponding to the intersection of the three circles.

n(x ) = 35 PE

Science

6

Music 3

388

Extract the relevant information from the second sentence and place it into the appropriate position. Note: Eight students like PE and Science; however, 6 of these students have already been accounted for in step 2. Therefore, 2 will fill the intersection of only PE and Science. Similarly, 4 of the 10 who like PE and Music will fill the intersection of only PE and Music, and 6 of the 12 students will fill the intersection of only Science and Music.

Maths Quest 10 for the Australian Curriculum

n(x ) = 35 PE

Science 2 4

6

6

Music

statistics AND probability • Chance

4

5

6

b

i

ii

1

n(x ) = 35

Extract the relevant information from the third sentence and place it into the appropriate position. Note: Twenty-two students like PE and 12 have already been accounted for in the set. Therefore, 10 students are needed to fill the circle corresponding to PE only. Similarly, 4 students are needed to fill the circle corresponding to Science only to make a total of 18 for Science. One student is needed to fill the circle corresponding to Music only to make a total of 17 for Music.

PE 2

10 4

4

6

6

1 Music

n(x ) = 35

Extract the relevant information from the final sentence and place it into the appropriate position. Note: Two students do not like any of the subjects, so they are placed in the rectangle outside the three circles.

PE

Science 2

10 4

Check that the total number in all positions is equal to the number in the universal set. 10 + 2 + 4 + 4 + 6 + 6 + 1 + 2 = 35 Write the number of students who like PE only and the total number of students in the class.

Science

4

6

6

1 Music b

2

i n(students who like PE only) = 10

n(x  ) = 35

2

Write the rule for probability.

P(likes PE only) =

n(likes PE only) n(x  )

3

Substitute the known values into the rule.

P(likes PE only) =

10 35

4

Evaluate and simplify.

5

Write your answer.

1

Write the number of students who do not like Music and the total number of students in the class. Note: Add all the values that do not appear in the Music circle as well as the two that sit in the rectangle outside the circles.

=

2 7

The probability of selecting a student who 2 likes PE only is 7 . ii n(students who do not like Music) = 18

n(x  ) = 35

2

Write the rule for probability.

P(does not like Music) n(does not like Music) = n(x  )

3

Substitute the known values into the rule.

P(does not like Music) = 35

4

Write your answer.

The probability of selecting a student who does not like Music is 18 . 35

18

Chapter 12 Probability

389

statistics AND probability • Chance c

1

Write the number of students who like Science and Music but not PE. Note: Add the values that appear in the Science and Music circles but do not overlap with the PE circle.

2

Repeat steps 2 to 4 of part b ii.

n[(Science ß Music) ¶ PEÅ] = 11 n(x  ) = 35

c

P[(Science ß Music) ¶ PEÅ] n[(Science ß Music) ¶ PEÅ] = n(x  ) 11

P[(Science ß Music) ¶ PEÅ] = 35 The probability of selecting a student who likes Science or Music but not PE is 11 . 35

Odds ■■ ■■ ■■

Probabilities in gambling can be expressed as odds. 5 This is very common in racing, where odds are given as ratios; for example 5–1 (or 1 or 5  :  1). In the odds of a–b, a–b a is the chance against the event

■■

If the odds for a horse to win are given as 5–1, then from 6 races the horse is expected to lose 5 and win 1. The probability that the horse wins or loses can be calculated from the odds given. These calculations are shown below. P(win) =

■■

b is the chance for the event

n(expected wins) n(races) 1

P(lose) = =

= 6

n(expected losses) n(races) 5 6

If given odds of a–b, then: n(E) b = n(x  ) a + b ñ  P(the event does not occur), P(EÅ) = n(EÅ) = a n(x  ) a+b

ñ  P(the event occurs), P(E) =

Payouts ■■ ■■

■■

390

The payout in races is based on the odds given. a If the odds given are a–b, you can win $ for every $1 bet, and alternatively stated, $a for b every $b bet. The bookmaker will pay out your win plus the initial bet. The TAB quotes a whole payout figure for a horse, made up of the winnings and the initial bet. For example: Odds

Bet

Winnings

5–1

$10

$5 for every $1 bet: 1 ì $10 = $50

7–2

$14

$7 for every $2 bet: 2 ì $14 = $49

Maths Quest 10 for the Australian Curriculum

Payout figure

5

$60 ($50 + $10)

7

$63 ($49 + $14)

statistics AND probability • Chance

Worked Example 6

The odds given for the horse Gunnawin to win the Melbourne Cup are 9–4. a Determine the probability of Gunnawin winning the Melbourne Cup. b Tony decides to bet $12 on Gunnawin to win. If the horse does win, what is Tony’s payout? c In the same race, the probability that the horse ‘Can’t Lose’ wins is given as 5 . What are the odds 17 that this horse will win? Think a

b

c

1

Write

Write the number of ways Gunnawin can win (4) and the total number of outcomes (9 + 4 = 13).

a n(Gunnawin wins) = 4

n(x  ) = 13

2

Write the rule for probability.

P(Gunnawin wins) n(Gunnawin wins) = n(x  )

3

Substitute the known values into the rule.

P(Gunnawin wins) = 13

4

Write your answer.

The probability of Gunnawin winning the 4 Melbourne Cup is 13 .

1

Explain what the ratio means and relate it to the bet.

4

9

b In the odds 9–4 the punter can win $ 4 for

every $1 that is bet (or for every $4 bet the punter will win $9). Therefore, if Tony bets 9 $12 he will win 4 ì $12 = $27.

2

Add the original amount invested to the amount returned.

Payout = $27 + $12 = $39

3

Write your answer.

Tony’s payout will be $39.

1

Look at the given fraction. The numerator corresponds to the ‘win’ component (second number) of the ratio.

2

The lose component of the ratio is always the first number.

Therefore the lose–win ratio is 12–5.

3

Write your answer.

The odds of Can’t Lose winning the Melbourne Cup are 12–5.

c

This horse has been given the chance of 5 winning as 17 . Therefore its chance of losing 12 is 17 .

remember

1. Probabilities can be expressed as a percentage, fraction or decimal in the range 0 to 1, inclusive. number of times an event has occurred 2. Experimental probability = total number of trials frequency of the score f or 3. Relative frequency of a score = total sum of frequencies S f n(E) 4. Theoretical probability that an event, E, will occur is P(E) = where n(x  ) n(E) = number of times or ways an event, E, can occur and n(x  ) = the total number of ways all outcomes can occur. Chapter 12 Probability

391

stAtistiCs AND probAbility • ChANCe 5. P(x  ) = 1

6. Venn diagrams provide a diagrammatic representation of sample spaces. 7. If given odds of a – b, then: b n(E) ■ P(the event occurs) P(E) = = a + b n(x  ) n(EÅ) a ■ P(the event does not occur) P(EÅ) = . = n(x  ) a + b exerCise

review of probability

12A

flUeNCy

iNDiviDUAl pAthWAys

1 Complete the relative frequency column in the given table.

eBook plus

Activity 12-A-1

Review of probability doc-5110 Activity 12-A-2

General probability problems doc-5111 Activity 12-A-3

x

f

1

2

2

5

3

6

4

3

5

4

1

eBook plus

Digital doc

a

SkillSHEET 12.6 doc-5291

17 2

1 9

4 a b c d

8

10

A¶B XÅ ¶ Y AÅ ¶ BÅ A ¶ C ¶ BÅ

or 0.2 1.00

1

3

A

B x

A

B x

b X

Y

12

14 18

c

d

A

B

C 392

x

20

4 16

19 6

or 0.15

x

B

15

or 0.3

Show the position of all the elements in the Venn diagram. x  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} 12 3 B = {1, 4, 9, 16} 2 1 20 = 5 20 = 10 b Calculate: 4 1 10 1 i P(A) 20 = 2 ii P(B) 20 = 5 iii P(A ¶ B) iv P(A ß B) v P(AÅ ¶ BÅ). 4 Using the given Venn diagrams, indicate the set each of the following shaded areas represents.

Digital doc

SkillSHEET 12.2 doc-5287

3

or 0.25

a 5 5 b 1 10 c 3 10 3 a We4 Draw a Venn diagram representing the relationship between the following sets.

eBook plus

7 13

or 0.1

2 For the table of values in question 1, what is the probability of selecting the following numbers

if a number is chosen at random?

5 11

1 10 1 4 3 10 3 20 1 5

S f = 20

Tricky probability problems doc-5112

A

Relative frequency

Maths Quest 10 for the Australian Curriculum

x

8 20

=

2 5

statistics AND probability • Chance 5 Write the following odds as probabilities. 1 a 5–1 6 b 13–4

4 17

2 9

c 7–1

1 8

6   MC  The probability of written as odds is: ✔ A

7–2

B 2–7

C 2–9

D 9–2

E  11–9

Understanding 7   WE 1  Terry has kicked 9 goals of the 10 attempts he made during a football match. 9 a What is the probability that he will kick a goal on his next attempt? 10 1 b What is the probability that he will not kick a goal on his next attempt? 10 8 Rachel attended 12 meetings in December. She was elected as the chairperson in 10 of those

meetings. What is the probability that she will be elected as the chairperson in the next meeting 5 she attends? 6 9 For a survey, a student counted the vehicles driving out of a sports complex at the end of day 1 of a sports carnival. She recorded the results in a table as shown below. Vehicle type Number

Bus

Car

Motorbike

4-wheel drive

3

17

4

6

Assuming that on day 2 there is a similar traffic movement, what is the probability that a randomly selected vehicle will be: 4 1 17 a a car 30 b a bus 10 c not a 4-wheel drive? 5 10   WE 2  Visitors to the Queen Victoria market were interviewed. The composition of this survey group is given by the following table. Females

Males

Total

NSW

 7

 9

16

Qld

 5

 7

12

Tasmania

 3

 2

 5

Europe

16

17

33

Asia

10

 4

14

Total

41

39

80

a b

Calculate the relative frequency of: 3

   i visitors from Queensland 20 1   ii European female visitors 5 iii male visitors from New South Wales 7 iv Asian visitors. 40

9 80

If a person is selected at random from this group, find the probability that the person is: 1

   i a Tasmanian visitor 16 17   ii a European male visitor 80 iii a female visitor from Queensland.

1 16

11   MC  Which statement is true for the information given in the table in question 10? ✔ a The probability of selecting a European visitor from this group is higher than that of

selecting a person from any other visitor group. b The probability of selecting a European visitor from this group is the same as that of

selecting a person from any other visitor group. c The probability of selecting a European male visitor from this group is the same as that of

selecting a European female visitor. d The probability of selecting an Asian visitor is the lowest. e The probability of selecting a visitor from New South Wales is the same as the probability

of selecting a visitor from Queensland. Chapter 12 Probability

393

stAtistiCs AND probAbility • ChANCe 12 We3 A card is drawn from a shuffled pack of 52 cards. Find the probability that the card

drawn is: 1 1 a an ace 13 b a club 4 12 1 c a red card 2 d not a jack 13 1 e a green card 0 f not a red card. 2 13 A bag contains 4 blue marbles, 7 red marbles and 9 yellow marbles. All marbles are of the same size. A marble is selected at random. What is the probability that the marble is: 7 1 a blue 5 b red 20 11 c not yellow d black? 0 20 14 MC Fifty Year 10 students on an excursion were asked to indicate their preference for an evening activity. It was concluded that, if a student is selected at random, the probability that he or she has chosen ice-skating is 15. a The number of students who chose ice-skating is: A 5 B 1 ✔ C 10 D 40 E 8 b The probability that a randomly selected student did not choose ice-skating is: A ✔ D

1 5 4 5

B

2 5

C

3 5

E 1

c The probability that a randomly selected student

chose tenpin bowling is: 1 5

B

D 1

✔ E

A

4 5

C 0

not able to be determined

15 A sporting club has members who play different sports, as shown by the given Venn diagram. Volleyball

Walking 10

15 Volleyball

Walking 10

15

x

Tennis

38 17

17 6

Volleyball

Walking 10

15

Tennis

6

Tennis Copy the given Venn diagram and shade the areas that represent: i members playing tennis only ii members walking only iii members playing both tennis and walking but not playing volleyball. How many members belong to the sporting club? 96 Volleyball Determine the probability of members who: 35 i play volleyball 96 10 8 1 15 ii are involved in all three activities. 96 = 12 8 Determine the probability of members who do not: 2 63 21 i play tennis 96 = 32 6 23 ii walk. 96 Tennis

Maths Quest 10 for the Australian Curriculum

38

8 2

a b c d

394

38

8 2

8

2

x

17 6

Walking 38 17

x

x

stAtistiCs AND probAbility • ChANCe 16 We5 Thirty students were asked which lunchtime sports they enjoyed — volleyball, soccer or

tennis. Five students chose all three sports. Six students chose volleyball and soccer, 7 students chose volleyball and tennis while 9 chose soccer and tennis. Fifteen students chose volleyball, 14 students chose soccer and 18 students chose tennis. a Copy the Venn diagram shown and enter the given information. n(x ) = 30

x = 30

Volleyball

Soccer 1

7 2

5 7

Volleyball

Soccer

4 4 Tennis

Tennis b c

If a student is selected at random, determine the probability of selecting a student who: chose volleyball 12 1 chose all three sports 6 x = 35 1 chose both volleyball and soccer but not tennis 30 Calculator Graph book 2 did not choose tennis 5 chose soccer. 157 7 18 5 Determine: 1 i P[(Soccer ß Tennis) ¶ VolleyballÅ] 2 5 8 ii P[(Volleyball ß Tennis) ¶ SoccerÅ]. 15 17 Thirty-five Year 10 students were required to bring a calculator and a graph book to a maths lesson. On checking, it was found that 18 students had brought both, 7 students had the calculator only and 5 students had the graph book only. Five students had neither the calculator nor the graph book. a Show this information on a Venn diagram. b How many students had: i a calculator 25 ii a graph book? 23 c If a student is selected at random, determine the probability that the student: 18 i had both the calculator and the graph book 35 5 ii had a calculator 7 1 iii had neither eBook plus 7 12 iv did not have a graph book. 35 Digital docs d Calculate: SkillSHEET 12.7 i P(calculator only) 15 doc-5292 6 ii P(calculator or graph book or both) 7 SkillSHEET 12.8 1 iii P(graph book only). 7 doc-5293 18 We6 The odds given for the greyhound ‘Dog’s Breakfast’ to win its race are 7–3. 3 a Determine the probability of Dog’s Breakfast winning its race. 10 b Maria decides to bet $15 on Dog’s Breakfast to win the race. If Dog’s Breakfast wins, calculate Maria’s payout. $50 4 c The dog ‘Zoom Top’ is also in the race. If the probability of Zoom Top winning is 13 , what odds should be given for Zoom Top? 9–4 3 19 The probability of a horse winning a race is given as . What are the horse’s chances, given as 7 odds? 4–3 i ii iii iv v

Chapter 12 probability

395

statistics AND probability • Chance Reasoning 20 Azi and Robyn are playing a dice game. Azi has an eight-sided die (faces numbered 1 to 8

1

Yes. P(Azi wins) = 2 and P(Robyn wins) =

1 2

21

b A  nswers may vary, 22 check with your teacher.   i Example — rolling an even number ii Example — rolling a 3 iii Example — rolling a number 23 greater than 3

24 25

inclusive) and Robyn has a six-sided die (faces numbered 1 to 6 inclusive). They both roll their 1 1 No. P(Azi rolls a 5) = 8 and P(Robyn rolls a 5) = 6 die. a The person who rolls the number 5 wins. Is this game fair? b The person who rolls an even number wins. Is this game fair? A six-sided die has three faces numbered 1 and the other three faces numbered 2. Are the events ‘rolling a 1’ and ‘rolling a 2’ equally likely? Yes. Both have a probability of 12. Using a six-sided die, an eight-sided die, a twelve-sided die and a sixteen-sided die (all faces numbered consecutively beginning with 1): a analyse and comment on the fairness of a game that constitutes a win by rolling a The person with the 6-sided die has less chance of winning. multiple of 4. 1 1 For the 8-, 12- & 16-sided dice: P(mult 4) = 4 ; for 6-sided die, P(mult 4) = 6 b devise rolling games where:   i the game is fair regardless of the die used.  ii it is more probable to win using a die with a smaller number of faces. iii it is more probable to win using a die with a larger number of faces. P(Alex wins) = 15 ; P(Rene wins) = 25 Alex places a $5 bet on a horse to win at 4–1 and Rene bets $10 on another horse. The pay-out figure for both bets reflection    is $25. What is the probability that Rene’s horse wins? What basic formula must Are the odds 10–6 the same as 5–3? Explain. be remembered in order to With the use of diagrams, show that calculate simple probabilities? P(AÅ ¶ BÅ) = P(A ß B)Å. 6 3 Yes, equivalent fractions; 16 =

8

Complementary and mutually exclusive events

12B

Complementary events ■■ ■■

The complement of a given set is made up of all the elements that belong to the universal set, x , but not to the particular given set. This is illustrated in the Venn diagram below, where the complement of set M, denoted as MÅ, will constitute all the elements outside set M; that is {5, 6, 7, 8}.

25

A

A

B

n(x ) = 8

B

M Overlaying AÅ and BÅ shows AÅ ¶ BÅ as the area surrounding A and B

The union of A and B is shown in brown, leaving the surrounding area as (A ß B)Å

N

1

5 3

2

4

7

6 8

■■

■■

396

Consider sets M and MÅ They have no elements in common and make up the universal set, x . Using set notation ñ  M ¶ MÅ = f , the null or empty set (indicating no elements) ñ  M ß MÅ = x . If A and AÅ are complementary events then P(A) + P(AÅ) = 1. This may be rearranged to P(AÅ) = 1 - P(A) or P(A) = 1 - P(AÅ).

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Worked Example 7

A card is drawn from a pack of 52 playing cards. Determine: a  the probability of drawing a spade b  the probability of not drawing a spade. Think a

b

1

Write

Write the number of favourable outcomes; that is, the number of ways a spade may be drawn and the total number of possible outcomes.

a n(drawing a spade) = 13

n(x  ) = 52

n(drawing a spade) n(x  )

2

Write the rule for probability.

P(spade) =

3

Substitute the known values into the rule.

P(spade) = 52

4

Simplify and evaluate.

5

Write your answer.

1

Write the rule for obtaining the complement of drawing a spade.

2

Substitute the known values into the given rule.

13

=

1 4 1

The probability of drawing a spade is 4 . b P(AÅ) = 1 - P(A)

P(not a spade) = 1 - P(spade) =1-

3

Evaluate.

4

Write your answer.

=

1 4

3 4

The probability of not drawing a spade is 43 .

Worked Example 8

A player is chosen from a cricket team. Are the events ‘selecting a batsman’ and ‘selecting a bowler’ complementary events if a player can have more than one role? Give a reason for your answer. Think

Write

Explain the composition of a cricket team. Players who can bat and bowl are not necessarily the only players in a cricket team. There is a wicket-keeper as well. Some players (all rounders) can bat and bowl.

No, the events ‘selecting a batsman’ and ‘selecting a bowler’ are not complementary events. These events may have common elements; that is, the all rounders in the team who can bat and bowl. The cricket team also includes a wicket-keeper.

Mutually exclusive events ■■ ■■

Two events that have no common elements and that cannot occur simultaneously are defined as mutually exclusive events. That is A ¶ B = { } or f . Using set notation, if two events A and B are mutually exclusive then P(A ¶ B) = 0 since 0 n(A ¶ B) = P(A ¶ B) = =0 n(x  ) n(x  ) Chapter 12 Probability

397

statistics AND probability • Chance

■■

■■ ■■

Some examples of mutually exclusive events are listed below. 1. A die is rolled. Let event A = obtaining an even number = {2, 4, 6} and event B = obtaining a factor of 3 = {1, 3}. A 2. In the given Venn diagram, event A = {1, 7, 10} and 1 event B = {2, 3}. 3. Two coins are tossed. Let event A = obtaining 7 2 Heads and event B = obtaining 2 Tails. 10 Complementary events are mutually exclusive; their union forms the universal set. The union of mutually exclusive events do not always form the universal set and as such, mutually exclusive events are not necessarily complementary events.

n(x ) = 5 B 2 3

Complementary events, A and AÅ

Mutually exclusive events, A and B

Common elements

A ¶ AÅ = •

A¶B=•

Union of sets

A ß AÅ = x

AßBòx

The Addition Law of probability ■■

■■

Consider the shaded region in the Venn diagram shown. 1. Counting the number of elements in the shaded region gives, n(A ß B) = 12. 2. For sets A and B, n(A) = 7 and n(B) = 8. 3. Sets A and B have three common elements, n(A ¶ B) = 3. 4. Adding together the number of elements in sets A and B, counts the common elements twice and hence does not give the number of elements in the shaded region, n(A) + n(B) = 15 ò 12 = n(A ß B). 5. Adding together the number of elements in sets A and B and subtracting the number of common elements gives the number of elements in the shaded region, n(A) + n(B) - n(A ¶ B) = 7 + 8 - 3 = 12 = n(A ß B) Therefore the number of elements in A ß B is n(A ß B) = n(A) + n(B) - n(A ¶ B) So, n(A ß B) x P(A ß B) = n(x  ) A B n(A) + n(B) - n(A ¶ B) = n(x  ) n(A) + n(B) - n(A ¶ B) = n(x  ) n(x  ) n(x  ) = P(A) + P(B) - P(A ¶ B)

■■

■■

398

This is known as the Addition Law of probability. If events A and B are not mutually exclusive, the Addition Law of probability states that: P(A or B) = P(A) + P(B) - P(A and B)  or  P(A ß B) = P(A) + P(B) - P(A ¶ B) If events A and B are mutually exclusive, the Addition Law of probability states that: P(A or B) = P(A) + P(B)  or  P(A ß B) = P(A) + P(B) since P(A ¶ B) = 0

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Worked Example 9

A card is drawn from a pack of 52 playing cards. What is the probability that the card is a heart or a club? Think

Write

1

Determine whether the given events are mutually exclusive.

The two events are mutually exclusive as they have no common elements.

2

Determine the probability of drawing a heart and of drawing a club.

P(heart) =

3

Write the Addition Law for two mutually exclusive events.

P(A or B) = P(A) + P(B) where A = drawing a heart and B = drawing a club

4

Substitute the known values into the rule.

P(heart or club) = P(heart) + P(club) 1 = 4 + 14

=

13 52 1 4

P(club) = =

= 5

Evaluate and simplify.

6

Write your answer.

Note: Alternatively, we can use the formula for theoretical probability.

13 52 1 4

2 4

= 12 1

The probability of drawing a heart or a club is 2 . P(heart or club) =

n(heart or club) n(x  ) 26

= 52 =

1 2

Worked Example 10

A die is rolled. Determine: a P(an odd number) b P(a number less than 4) c P(an odd number or a number less than 4). Think a

b

1

Write

Determine the probability of obtaining an odd number; that is, {1, 3, 5}.

2

Write your answer.

1

Determine the probability of obtaining a number less than 4; that is, {1, 2, 3}.

2

Write your answer.

a

3

P(odd) = 6 =

1 2 1

The probability of obtaining an odd number is 2 . b P(less than 4) =

=

3 6 1 2

The probability of obtaining a number less 1 than 4 is 2 .

Chapter 12 Probability

399

statistics AND probability • Chance c

c

1

Determine whether the given events are mutually exclusive.

2

Write the Addition Law for two non-mutually exclusive events.

P(A or B) = P(A) + P(B) - P(A and B) where A = selecting an odd number and B = selecting a number less than 4.

3

Substitute the known values into the rule. Note: P(A and B) = 62 = 13 since the

P[odd number ß (number < 4)] = P(odd number) + P[(number < 4)] - P[odd number ¶ (number < 4)] 1 1 1 =2+2−3

( )

events have two elements in common.

The two events are not mutually exclusive as they have common elements; that is, 1 and 3.

2

4

Evaluate and simplify.

=3

5

Write your answer.

The probability of obtaining an odd number or a 2 number less than 4 is 3 . The set that has elements that are odd numbers or numbers less than 4 is {1, 2, 3, 5}. P[odd number ß (number < 4)] n[odd number ß (number less than 4)] = n(x  ) 4 =6

Note: Alternatively, we can use the formula for theoretical probability.

=

2 3

remember

1. Complementary events have no common elements and together make up the universal set. 2. If A and AÅ are complementary events then P(A) + P(AÅ) = 1. This may be rearranged to: P(AÅ) = 1 - P(A) or P(A) = 1 - P(AÅ). 3. Mutually exclusive events have no common elements and cannot occur simultaneously. 4. If events A and B are not mutually exclusive then: P(A or B) = P(A) + P(B) - P(A and B) or P(A ß B) = P(A) + P(B) - P(A ¶ B) where P(A ¶ B) is the probability of the intersection of sets A and B or the common elements in sets A and B. 5. If events A and B are mutually exclusive then: P(A or B) = P(A) + P(B) or P(A ß B) = P(A) + P(B) since P(A ¶ B) = 0. 6. Mutually exclusive events may or may not be complementary events. 7. Complementary events are always mutually exclusive. Exercise

12B

Complementary and mutually exclusive events fluency 1 Consider complementary events, A and B. a 1 - P(B) = ? P(A) c If P(B) = 0.65, then P(A) = ?

400

Maths Quest 10 for the Australian Curriculum

0.35

3

b If P(A) = 7 , then P(B) = ? d P(A ß B) = ? 1

4 7

stAtistiCs AND probAbility • ChANCe

eBook plus

Activity 12-B-1

Activity 12-B-2

Harder complementary and mutually exclusive events doc-5114 Activity 12-B-3

Tricky complementary and mutually exclusive events doc-5115

eBook plus

Digital doc

SkillSHEET 12.3 doc-5288

2 Events M and N are not mutually exclusive events. a Draw a Venn diagram to illustrate events M and N. b P(M ß N) = ? P(M ) + P(N ) - P(M ¶ N ) c State if the following statements are true, false or cannot be determined? i P(M ¶ N) = • False ii P(M ¶ N) > 0 True iii P(M ß N) = 1 Cannot be determined

N

Complementary and mutually exclusive events doc-5113

M

iNDiviDUAl pAthWAys

UNDerstANDiNg 3 We7 A card is drawn from a pack of 52 playing cards. Determine: 1 a the probability of obtaining an ace 13 12 b the probability of not obtaining an ace. 13 4 The weather bureau announced that there is an 80% probability of having a rain shower on Friday. What is the probability of not having a rain shower on that day? 20% or 15 5 A number is selected from the set {1, 2, 3 . . . 20}. Let E1 be the event of selecting an even

number and E2 be the event of selecting an odd number.

a Determine: 1 1 i P(E1) ii P(E2). 2 2 b Are E1 and E2 complementary?

Yes

6 A bag contains 50 balls, of which there are 10 blue balls, 5 red balls and 3 yellow balls. What 16

is the probability of picking a ball that is not blue, red or yellow? 25 Questions 7 and 8 refer to the following information. A number is selected from 1 to 100, inclusive. Let: E1 = a multiple of 10 is picked E2 = a factor of 20 is picked E3 = a multiple of 2 is picked E4 = an odd number is picked. 7 MC Which of the following represents a pair of complementary events? A E1 and E2 B E2 and E3 ✔ C E3 and E4 D E2 and E4 E E1 and E3 8 Calculate: 1 a P(multiple of 10) 10 9 b P(not a multiple of 10) 10 47 c P(not a factor of 20). 50 Questions 9 and 10 refer to the following information. The ages of 50 Year 10 students are shown in the following table. Age (years) 15

16

17

Total

Girls

7

10

9

26

Boys

9

8

7

24

Total

16

18

16

50

9 MC Which of the following represent a pair of complementary events? A Selecting a 15-year-old boy and selecting a 15-year-old girl B Selecting a 15-year-old student and selecting a 16-year-old student C Selecting a 17-year-old student and selecting a 15-year-old student ✔ D Selecting a 15- or 16-year-old student and selecting a 17-year-old student E Selecting a 17-year-old student and selecting a 15- or 16-year-old girl Chapter 12 probability

401

stAtistiCs AND probAbility • ChANCe 10 Calculate: 9 a P(selecting a 15-year-old boy) 50 b P(not selecting a 15-year-old boy) 12 c P(selecting a boy) 25 13 d P(selecting a girl). 25

41 50

1

11 a When a coin is tossed 4 times, the probability of getting 4 Heads is 16 . What is the 15 16

probability of not getting 4 Heads?

2 9

b The probability that a horse will win a race is . What is the probability that one of the

other horses will win the race?

7 9

12 Are the events ‘getting 2 Tails’ and ‘getting 0 Tails’ complementary when a coin is tossed twice? No, getting 1 Tail is possible too. 13 In a school raffle, 200 tickets were sold. Margaret and Julie bought 25 tickets between them. 1 a What is the probability that Margaret or Julie will win? 8 7 b What is the probability that neither of them will win? 8 eBook plus

Digital doc

SkillSHEET 12.4 doc-5289

2 3

14 We8 A die is rolled. What is the probability that the outcome is an even number or a 5? 15 A number is chosen from the set {1, 2, 3 . . . 25}. What is the probability that the number is: 9 a a multiple of 4 or a multiple of 7 25 b a multiple of 4 or an odd number 19 25 9 c less than 5 or more than 20? 25 16 A card is drawn from a well-shuffled pack of 52 playing cards. Calculate: 7 a P(a spade or ace of hearts is drawn) 26 2 b P(a king or a queen is drawn) 13 3 c P(a jack or a king or an ace is drawn). 13 17 MC Which of the following represents a pair of mutually exclusive events when a die is

rolled? Obtaining an even number or obtaining a 4 Obtaining an odd number or obtaining a 3 Obtaining a number less than 3 or obtaining a number more than 5 Obtaining a multiple of 2 or obtaining a multiple of 3 Obtaining a factor of 6 or obtaining a multiple of 6

A B ✔ C D E

18 In a 3-horse race, the probability for each of the horses to win is given as:

Our Lady: 37

Shaka:

4 9

Speedy:

Determine the probability that: a either Our Lady or Speedy wins 4 b either Shaka or Speedy wins. 7

8 . 63

5 9

19 Christine’s teaching timetable for Monday and Tuesday is given below.

Period and class 1

2

Monday

10B

Tuesday

8B

3

8A

4

5

8B

8A

10A

6

7 9A

9B

She is organising a music tuition class for a lesson when she is not teaching, but she cannot use the first lesson on any day because of her responsibility as a senior teacher. Determine the probability that: 8 4 a she cannot take music tuition because she is teaching or 7 14 2 1 b she cannot take music tuition because it is the first lesson or 7 14 10 5 c she cannot take music tuition. 14 or 7 402

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe 20 We9 A card is drawn from a well-shuffled pack of 52 playing cards. Calculate: 1 a P(a king is drawn) 13 1 b P(a heart is drawn) 4 4 c P(a king or a heart is drawn). 13 21 Two coins are tossed. Event 1 is obtaining 2 Heads and event 2 is obtaining 2 Tails. a Are events 1 and 2 mutually exclusive? Yes 1 b What is P(event 1 or event 2)? 2 reAsoNiNg 22 We10 For each of the following pairs of events: i state, giving justification, if the pair are complementary events ii alter the statements, where applicable, so that the events become complementary events. a Having Weet Bix or having Strawberry Pops for breakfast b Walking to a friend’s place or driving there c Watching TV or reading as a leisure activity d Rolling a number less than 5 or rolling a number greater than 5 with a ten-sided die Yes. There are only two possible outcomes; passing or failing. with faces numbered 1 to 10 e Passing a maths test or failing a maths test 23 Pat suggests that for a single roll of a die, getting a factor of 4 and getting a factor of 6 are mutually exclusive. Is he right? Why or why not? No. The number 2 is common to both events. eBook plus

24 Two tetrahedral dice (4-sided) are rolled and the sum of the outcome on each is taken.

(Note: The outcome is the number on the bottom face.) Let: Event 1 = the sum is 6 Event 2 = the sum is 3 Event 3 = the sum is more than 4 Event 4 = the sum is less than 4. a Decide whether the following statements are true or false. i Events 1 and 2 are mutually exclusive. T ii Events 2 and 4 are mutually exclusive. F iii Events 2 and 3 are mutually exclusive. T iv Events 1 and 2 are complementary events. F v Events 2 and 4 are complementary events. F vi Events 2 and 3 are complementary events. F b Determine: 3 i P(event 1) 16 1 ii P(event 2) 8 5 iii P(event 3) refleCtioN    3 8 iv P(event 4). 16 How are the differences between c Determine: mutually exclusive events and 5 i P(event 1 or event 2) 16 complementary events reflected 3 ii P(event 2 or event 4) in the addition law of probability? 16 iii P(event 2 or event 3). 3

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SkillSHEET 12.9 doc-5294

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Digital doc

WorkSHEET 12.1 doc-5295

4

12C

two-way tables and tree diagrams ■

■

When more than one event has to be considered, a diagrammatic representation of the sample space is helpful in calculating the probabilities of various events. Two-way tables and tree diagrams may be used.

two-way tables ■

A two-way table (sometimes referred to as a lattice diagram) is able to represent two events in a 2-dimensional table. Chapter 12 probability

403

statistics AND probability • Chance

■■ ■■

With the help of the information in each row and each column, all the pairs of outcomes are listed and the diagram ensures that none of the pairs is omitted. A two-way table for the experiment of tossing a coin and rolling a die simultaneously is shown in the following table. Die outcomes

Coin outcomes

1

■■

2

H (H, 1) (H, 2) T

(T, 1)

(T, 2)

3

4

5

6

(H, 3)

(H, 4)

(H, 5)

(H, 6)

(T, 3)

(T, 4)

(T, 5)

(T, 6)

Two-way tables can be used to display the combined outcomes of only two events.

Worked Example 11

Two dice are rolled. The outcome is the pair of numbers shown uppermost. a Show the results on a two-way table. b Calculate the probability of obtaining an identical ordered pair; that is, P[(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]. Think

Rule a two-way table.

2

Label the first row as ‘Die 2’ and write all the outcomes.

3

Label the first column as ‘Die 1’ and write all the outcomes.

4

Write each ordered pair in its respective position.

1

Look at the two-way table from part a and highlight the identical pairs.

a

Die 2 outcomes 1

404

2

3

5

6

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Die 2 outcomes 1

2

3

4

5

6

1

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

2

Write the number of identical pairs and the total number of possible pairs.

n(identical pairs) = 6 n(x  ) = 36

3

Since each outcome is equally likely, write the rule for probability.

P(identical pairs) =

Maths Quest 10 for the Australian Curriculum

4

1

b

Die 1 outcomes

b

1

Die 1 outcomes

a

Write

n(identical pairs) n(E) = n(x  ) n(S)

stAtistiCs AND probAbility • ChANCe

4

Substitute the known values into the rule.

5

Simplify and evaluate.

6

Write your answer.

■

6

P(identical ordered pairs) = 36 1

=6 The probability of obtaining an identical ordered pair is 16 .

Two-way tables are limited to displaying two events occurring simultaneously. Alternative representations are used to display more than two events.

tree diagrams ■

■

eBook plus

eLesson Games at Wimbledon

■

eles-1032 ■

■

■

■

Another way of representing the sample space is to construct a tree diagram. This is a branching diagram that helps list all the outcomes. Coin 1 Coin 2 Tree diagrams are very helpful when there are multiple 1– H events; for example, when a coin is tossed twice. Each stage 2 of a multiple event experiment produces a part of a tree. 1– H 2 1– The first stage of the experiment is tossing coin 1. The two T 2 possible results that can be obtained are Heads or Tails and 1– H these are listed at the end of each branch. The probability of 2 1– 2 T obtaining the result listed is written along the branches. 1– The second stage of the experiment is tossing coin 2, for T 2 which the possible results are also Heads or Tails. A pair of branches is attached to each of the ends of the existing branches. Again, the branches are labelled with the appropriate outcomes and probabilities. After the diagram has been completed, the outcomes are listed at the right-hand side of the tree diagram. This is done by beginning at the starting point and following along each set of branches, then listing the combinations. The possible results or outcomes obtained by following along the combined branches are (H, H), (H, T), (T, H) and (T, T). The probability for each outcome is calculated by taking the product of the probabilities associated with the respective branches. For example, the probability of (H, H) is obtained by multiplying the individual probabilities of the two H branches; that is P(H, H) = P(H) ì P(H) =2ì2

 

=

1 4

The completed tree diagram is illustrated below. Coin 1

Coin 2 Outcomes HH H

1– 2 1– 2

1– 2

■

1

1

 

H

T

Probability 1– 1– 1– 2 ì 2 = 4

1– 2

T

HT

1– 2

ì

1– 2

=

1– 4

1– 2

H

TH

1– 2

ì

1– 2

=

1– 4

1– 2

T

TT

1– 2

ì

1– 2

=

1– 4

— 1

When added together, all the probabilities should sum to 1. If more than one outcome is included in a particular event, then the respective probabilities are added. For example: P(1 Head) = P(H, T) + P(T, H) 1

 

=4+

 

=

1 4

1 2

Chapter 12 probability

405

statistics AND probability • Chance

■■

Tree diagrams may be extended to display three or more events occurring simultaneously.

■■

Tree diagrams are useful in working out the sample space and calculating probabilities of various events. On each branch of a tree diagram, the probability associated with the branch is listed. The products of the probabilities given on the branches are taken to calculate the probability for a particular outcome.

Worked Example 12

Three coins are tossed simultaneously. Draw a tree diagram for the experiment. Calculate the following probabilities. a  P(3 Heads) b  P(2 Heads) c  P(at least 1 Head) Think

Write/draw

1

Use branches to show the individual outcomes for the first part of the experiment (tossing the first coin). Place a 1 above the coin toss outcomes. Label the ends of the branches H and T and place the probabilities along the branches.

2

Link each outcome of the first toss with the outcomes of the second part of the experiment (tossing the second coin). Place a 2 above the second toss outcomes. Label the ends of the branches H and T and place the probabilities along the branches.

3

Link each outcome from the second toss with the outcomes of the third part of the experiment (tossing the third coin). Place a 3 above the third toss outcomes. Label the ends of the branches H and T and place the probabilities along the branches.

4

List each of the possible outcomes on the right-hand side of the tree diagram.

5

Determine the probability of each result. Note: The probability of each result is found by multiplying along the branches and in each case this will be 1 1 1 1 ì 2 ì 2 = 8. 2

1

2 1– 2

1– 2

1– 2

H

T

1– 2 1– 2 1– 2

H

1– 2 1– 2

T 1– 2

H

1– 2

T 1– 2

406

Maths Quest 10 for the Australian Curriculum

3 H

HHT

1– 2

H

HTH

T

HTT

H

THH

T

THT

H

TTH

T

TTT

1– 2 1– 2 1– 2 1– 2 1– 2 1– 2

T 1– 2

1– 2

1– 2

Outcomes Probability 1– 1– 1– HHH 2 ì 2 ì 2 = ì

1– 2

ì

1– 2 1– 2 1– 2 1– 2 1– 2 1– 2

ì ì ì ì ì

ì

1– 2

=

ì

1– 2 1– 2 1– 2 1– 2 1– 2 1– 2

=

ì ì ì ì ì

= = = = =

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

statistics AND probability • Chance

6

a

b

c

Place calculations and results next to the respective outcomes. Note that in this example, each of the outcomes has the same probability, therefore each outcome is equally likely.

Refer to the listed outcomes and calculations next to the tree diagram and write your answer. 1

Refer to the listed outcomes and calculations next to the tree diagram. Note: This combination occurs three times.

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram. Note: At least 1 Head means any outcome that contains one or more Head. This is every outcome except 3 Tails. That is, it is the complementary event to obtaining 3 Tails.

2

Write your answer.

■■

a The probability of obtaining 3 Heads is 1 . 8

b P(2 Heads)

= P(H, H, T) + P(H, T, H) + P(T, H, H) 1 1 1 =8+8+8 3

=8

The probability of obtaining exactly 2 Heads is 83 . c

P(at least 1 Head) = 1 - P(T, T, T) 1 =1-8 7

=8

7

The probability of obtaining at least 1 Head is 8 .

As can be seen from the tree diagram in Worked example 12, the probabilities of all outcomes add up to 1.

Worked Example 13

Two dice are rolled simultaneously. Draw a tree diagram for the experiment and find: a  P(two 6s)            b  P(one 6) c  P(no 6s)             d  P(at least one 6). Think

Write/draw

1

Draw the tree diagram using two outcomes — S (getting a 6) and SÅ (not getting a 6). Note: Recall for complementary events, P(S) + P(SÅ) = 1.

2

List each of the possible outcomes on the right-hand side of the tree diagram.

1

2 1– 6

1– 6

5– 6

S

5– 6 1– 6

Probability 1– 1– 1 — 6 ì 6 = 36

S

Outcomes SS

S'

SS'

1– 6

ì

5– 6

5 =— 36

S

S'S

5– 6

ì

1– 6

5 =— 36

S'

S'S'

5– 6

ì

5– 6

25 =— 36 — 1

S' 5– 6

Chapter 12 Probability

407

statistics AND probability • Chance

a

b

3

Determine the probability of each possible result by multiplying along the branches.

4

Place the calculations and results next to the respective outcomes.

1

Refer to the listed outcomes and calculations next to the tree diagram.

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram. Note: This combination occurs twice.

a P(two 6s) = P(S, S) 1 = 36

1 . The probability of obtaining two 6s is 36

b

P(one 6) = P(S, SÅ) + P(SÅ, S) 5 5 = 36 + 36 10

= 36 = 185

c

d

5

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram.

2

Write your answer. Refer to the listed outcomes and calculations next to the tree diagram. Note: P(at least one 6) means any outcome that contains one or more six. This includes every outcome except for the (SÅ, SÅ) combination. That is, it is the complementary event to obtaining the (SÅ, SÅ) combination.

1

2

The probability of obtaining one 6 is 18 . c

P(no 6s) = P(SÅ, SÅ) 25 = 36 25

The probability of obtaining no 6s is 36 . d P(at least one 6) = 1 - P(SÅ, SÅ). 25 = 1 - 36 = 11 36

11

Write your answer.

The probability of obtaining at least one 6 is 36.

Alternatively, part d of Worked example 13 could have been a calculated in the following manner. P(at least one 6) = P(one or more 6s) = P(S, S) + P(S, SÅ) + P(SÅ, S)

=



=

1 36 11 36

+

5 36

5

+ 36

Worked Example 14

The letters A, B, C and D are written on identical pieces of card and placed in a box. A letter is drawn at random from the box. Without replacing the first card, a second one is drawn. Use a tree diagram to find: a P(first letter is A)    b P(second letter is B)    c P(both letters are the same).

408

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Think

Write/draw

1

Draw a tree diagram of the situation. There are four letters to choose as the first letter of the pair of letters. Each has a probability of 14 of being chosen.

2

For each letter chosen as the first letter, there are three letters remaining to choose from. Each has a probability 1 of 3 of being chosen.

3

Write the sample space and calculate the probability of each outcome.

1– 3

A

1– 3

1– 4 1– 4

B 1– 4

1– 3

C

1– 4

1– 3

D

a

1

Refer to the listed outcomes and calculations next to the tree diagram.

1– 3

1– 3

1– 3

1– 3

1– 3

1– 3

1– 3

1– 3

B C D A C D A B D A B C

Outcomes AB AC AD BA BC BD CA CB CD DA DB DC

Probability 1– 1 1 4 ì –3 = — 12 1– 1– 1 — ì = 4 3 12 1– 1– 1 — 4 ì 3 = 12 1– 1– 1 — 4 ì 3 = 12 1– 1– 1 — ì = 4 3 12 1– 1– 1 — 4 ì 3 = 12 1– 1– 1 4 ì 3 = — 12 1– 1– 1 — ì = 4 3 12 1– 1 1 4 ì –3 = — 12 1– 1– 1 4 ì 3 = — 12 1– 1– 1 — 4 ì 3 = 12 1– 1 1 4 ì –3 = — 12 — 1

a P(first letter A) = P(A, B) + P(A, C) + P(A, D) 1

1

1

P(first letter A) = 12 + 12 + 12 3

= 12 = 14

2

b

1

1

The probability that the first letter is A is 4 .

Write your answer. Refer to the listed outcomes and calculations next to the tree diagram.

b

P(second letter B) = P(A, B) + P(C, B) + P(D, B) 1 1 1 P(second letter B) = 12 + 12 + 12 3

= 12 =

c

2

Write your answer.

1

Refer to the listed outcomes and calculations next to the tree diagram.

2

Write your answer.

1 4

The probability that the second letter is B is 14 . c

P(both letters are the same) = 0

As the first card is not replaced before the second is drawn, the probability that both letters are the same is 0.

Chapter 12 Probability

409

stAtistiCs AND probAbility • ChANCe

reMeMber

1. Two-way tables give a clear diagrammatic representation of the sample space; however, they are limited to displaying two events. 2. Tree diagrams are useful in working out the sample space and calculating probabilities of various events, especially if there is more than one event. On each branch of a tree diagram, the probability associated with the branch is listed. The products of the probabilities given on the branches are taken to calculate the probability for an outcome. 3. The probabilities of all outcomes add to 1.

12C iNDiviDUAl pAthWAys

two-way tables and tree diagrams flUeNCy 1 For the tree diagram below, calculate the following probabilities:

eBook plus

Activity 12-C-1

H

0.5

T

0.5

H

0.5

T

0.5

H

0.5

T

Die 2 outcomes 1

R

Review of two-way tables and tree diagrams doc-5116

0.4 0.4

Activity 12-C-2

Practice with two-way tables and tree diagrams doc-5117

G

0.2 B

Activity 12-C-3

Tricky two-way tables and tree diagrams doc-5118

0.5

Die 1 outcomes

exerCise

a P(R, H) 0.2 b P(B, H) 0.1 d P(H) 0.5 e P(R, H or G, T) 2 i Copy and complete the two-way table below.

2

3

4

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Coin outcomes

Card outcomes Club, Ü

Spade, â

Diamond, á

Heart, à

H

H, Ü

H, â

H, á

H, à

T

T, Ü

T, â

T, á

T, à

ii Calculate the following probabilities. 1 a P(T, Ü) 8 b P(T)

0.5

c P(red card)

0.5

UNDerstANDiNg eBook plus

Digital doc

SkillSHEET 12.5 doc-5290

410

3 We11 Two dice are rolled. The outcome is the pair of numbers shown on each die. a Show the results on a two-way table. b Calculate the probability of obtaining an ordered pair where the second digit is half the 1

value of the first. 12 4 A 10-sided die is rolled at the same time that a coin is tossed. a Show the outcomes on a two-way table. b Calculate the probability of event (H, n), where n is a factor of 10. 1 c Calculate P(T, even number). 4

Maths Quest 10 for the Australian Curriculum

6

1

c P(B) 0.2 f P(BÅ) 0.8

0.4

5

1 5

RBR RBB

BRR BRB

BBR BBB

R B

R B 1– 2

1– 2

1– 2

R

B

1– 2

1– 2

1– 2

B

1– 2

B

1– 2

1– 2

6

6

1– 2

1

R

1– 2

2

R

1– 2

1– 2

1– 2

R B

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

5 A green octahedron (a 3-dimensional shape with 8 regular faces) is rolled simultaneously with

3 R B

Outcomes Probability 1– RRR 8 1– RRB 8

statistics AND probability • Chance

— 1

B

BB

BR

BG G

1 — 6 1 — 12 1– 4 1– 2

8

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

9

GGB GGG

1– 2 1– 2

G

10

G

1– 2

B

1– 2

1– 2 1– 2

1– 2

G

B G

GBB GBG B G

BGB BGG B G 1– 2

B

1– 2

B

1– 2

1– 2

2 1

1– 2

1– 2

1– 2

3 B G

Outcomes Probability 1– BBB 8 1– BBG 8

7  a

1– 3

1– 2

1– 6

B

G

1– 2

1– 3

1– 6

B

R

GG G

1– 6 1– 3 1– 2

R

1

GB

GR R

B

RB

1 — 18 1 — 36 1 — 12

1 — 18 1– 6

RG G

1– 3

1– 6

R

2 Outcomes Probability 1– RR 9

7

a yellow octahedron. Both figures have the faces numbered 1, 2, 3  .  .  .  8. a Show the sample space on a two-way table. b On the diagram, highlight the event (n, n), where n is a number in the range 1 to 8 inclusive. 1 c What is the probability of getting (n, n) as described in part b above? 8   WE 12  A circular spinner is divided into two equal halves, coloured red and blue, and spun 3 times. Draw a tree diagram for the experiment. Calculate the following probabilities. a P(3 red sectors) 1 3 3 a 8 b 8 c 8 b P(2 red sectors) c P(1 red sector) 7 1 1 d 8 e 8 f 2 d P(0 red sectors) e P(at least 1 red sector) f P(at least 2 red sectors) A bag contains 6 identical marbles, 2 of which are red, 1 green and 3 blue. A marble is drawn, the colour is noted, the marble is replaced and another marble is drawn. a Show the possible outcomes on a tree diagram. b List the outcomes of the event ‘the first marble is red’. {(R, R), (R, G), (R, B)} 1 c Calculate P(the first marble is red). 3 7 d Calculate P(2 marbles of the same colour are drawn). 18 Assuming that it is equally likely that a boy or a girl will be born, answer the following. a Show the possibilities of a 3-child family on a tree diagram. b In how many ways is it possible to have exactly 2 boys in the family? 3 3 c What is the probability of getting exactly 2 boys in the family? 8 d Which is more likely, 3 boys or 3 girls in the family? They are equally likely. 7 e What is the probability of having at least 1 girl in the family? 8 A tetrahedron (prism with 4 identical triangular faces) is numbered 1, 1, 2, 3. It is rolled twice. 1 2 Outcomes Probability The outcome is the number facing downwards. 11 1 a Show the results on a tree diagram. 12 2 1 b Are the outcomes 1, 2 and 3 equally likely? No 13 3 1 21 c Find the following probabilities: 1 22 2 2   i P(1, 1) 4 23 3 1  ii P(1 is first number) 2 1 31 3 3 iii P(both numbers equal) 8 32 2 9 3 33 iv P(both numbers are odd). 16 — 1   WE 13  A die is rolled twice to check whether a 3 occurs. Draw a tree diagram for the experiment and calculate: 1 5 a P(two 3s) 36 b P(one 3) 18 11 25 c P(no 3s) d P(at least one 3). 36 36 A card is drawn from a pack of 52 playing cards and checked to see whether a spade has been selected. The card is replaced, the pack reshuffled and another card is selected. a Draw a tree diagram for the activity and list the sample space. 1 b What is the probability that both cards are spades? 16 9 c What is the probability that neither of the cards is a spade? 16 3 d What is the probability that one of the cards is a spade? 8   WE 14  The letters X, Y, W and Z are written on identical pieces of card and placed in a box. A letter is drawn at random from the box. Without replacing the first card, a second one is drawn. Use a tree diagram to find: 1 a P(first letter is W) 4 b P(second letter is Z) 14 c P(both letters are different). 1

1– 2

8  a

1– 2

11

12

1– 4

1– 4

1– 4

1– 4

1– 2

1– 4

1– 4

1– 2

1– 4

1– 4

1– 4 1– 8 1– 8

1– 8 1 — 16 1 — 16 1– 8 1 — 16 1 — 16

Chapter 12 Probability

411

stAtistiCs AND probAbility • ChANCe

1–

GG G

GB

5– 9

B 4– 9

G

4 — 10

6 — 10

B

1

13 a

3 — 1

4 — 15

4 — 15 6– 9

3– 9

B

G

BB

BG

2 — 15

2 Outcomes Probability

13 A group of students is made up of 6 girls and 4 boys. Two students are to be selected to

represent the group on the student representative council. They decide to write all names on identical pieces of paper, put them in a hat and choose two names randomly. They want to check the composition (boys or girls) of the two-person team. a Show the selections on a tree diagram (note that the probabilities for the second selection change). 2 b Determine the probability of 2 boys being selected. 15 1 c Determine the probability of 2 girls being selected. 3 d Determine the probability of selection of 1 boy and 1 girl. 158 e Are the events ‘0 boys’, ‘1 boy’ and ‘2 boys’ equally likely? No

reAsoNiNg

1–

6 — 1

GG

1– 3

G

GR

2 3

s the first counter is not replaced, A the probability of drawing the second counter is altered. This is reflected in the probabilities along the branches of the tree diagram; P(2 counters of the same colour) = 13;

P(2 counters of different colours) = .

2– 4

G

1– 3

R 2– 3

2– 3 2– 4

R

1– 3

R

G

RG

1– 3

Outcomes Probability 1– RR 6

14 Robyn is planning to watch 3 footy

eBook plus

Digital doc

WorkSHEET 12.2 doc-5296

412

games on one weekend. She has a choice of two games on Friday night; (A) Carlton vs West Coast and (B) Collingwood vs Western Bulldogs. On Saturday, she can watch one of the three games; (C) Geelong vs Brisbane Lions, (D) Melbourne vs Fremantle and (E) Kangaroos vs Adelaide. On Sunday, she also has a choice of three games; (F) St Kilda vs Sydney, (G) Essendon vs Port Adelaide and (H) Richmond vs Hawthorn. a To determine the different combinations of games Robyn can watch, she draws a tree diagram using codes A, B, . . . H. Suggest a sample space for Robyn’s selections. b Robyn’s favourite team is Carlton. What is the probability that one of the games Robyn watches involves Carlton? 12 c Robyn has a good friend that plays for St Kilda. What is the probability that Robyn watches both the matches involving Carlton and St Kilda? 16 15 A small hospital awaits the arrival of four children showing symptoms of a particular virus. The virus has two different strands, strand A and strand B. A child suffering from strand A cannot share a room with a child suffering from strand B. Given a room can fit at most two children and that there is an equal chance that a child has strand A or B, decide if 2 or 3 rooms need to be made up in order to house the children. There is a 50% chance that a third room will be needed. 16 To pass an exam Susan must answer two of the last three multiple choice questions correctly. As Susan is running out of time she decides to guess the answers to these three questions. Susan notes that two of the questions give six possible answers rather than the usual standard four choices. Analyse and comment on how the inclusion of six possible answers (as opposed to 4) for two of these questions will affect her chances of passing the exam? 17 i Four identical counters, 2 red and 2 green, are placed in a bag. One counter is drawn, its colour refleCtioN    recorded, it is replaced in the bag and a second one is drawn. When calculating probabilities, what is the significance of an item a Show the sample space on a tree diagram. 1 being replaced, or not replaced, b Calculate P(2 counters of the same colour). 2 1 before a second event occurs? c Calculate P(2 counters of different colours). 2 ii Suppose the first counter is not replaced. Analyse Susan would have 16% chance of passing and explain how this will affect the probabilities? the exam if the last three questions had the

Maths Quest 10 for the Australian Curriculum

standard 4 choices. This chance is reduced to 48% with the inclusion of two questions offering 6 possible answers.

stAtistiCs AND probAbility • ChANCe

12D eBook plus

Interactivity A pack of cards

int-2787

Independent and dependent events ■

■ ■

■

If a coin is tossed the outcome is a Head or a Tail. The outcome of the first toss does not affect the outcome of the next toss of the coin. The second toss will still yield a Head or a Tail irrespective of the outcome of the first toss. Similarly, the outcome on the roll of a die will not affect the outcome of the next roll. If successive events have no effect on each other, they are called independent events. If events A and B are independent then the Multiplication Law of probability states that: P(A and B) = P(A) ì P(B) or P(A ¶ B) = P(A) ì P(B) The reverse is also true. If: P(A and B) = P(A) ì P(B) or P(A ¶ B) = P(A) ì P(B) is true then event A and event B are independent events.

WorkeD exAMple 15

Adam is one of the 10 young golfers to represent his state. Paz is one of the 12 netball players to represent her state. All the players in their respective teams have an equal chance of being nominated as captains. a Are the events ‘Adam is nominated as captain’ and ‘Paz is nominated as captain’ independent? b Determine: i P(Adam is nominated as captain) ii P(Paz is nominated as captain). c What is the probability that both Adam and Paz are nominated as captains of their respective teams? thiNk a

b

Write

Determine whether the given events are independent and write your answer.

a Adam’s nomination has nothing to do with Paz’s

i

b

1

Determine the probability of Adam being nominated as captain. He is one of 10 players.

nomination and vice versa. Therefore, the events are independent. i P(Adam is nominated) = P(A)

=

n(Adam is nominated) n(x  ) 1

P(Adam is nominated) = 10

ii

2

Write your answer.

1

Determine the probability of Paz being nominated as captain. She is one of 12 players.

The probability that Adam is nominated as 1 captain is 10 . ii P(Paz is nominated) = P(P)

=

n(Paz is nominated) n(x  ) 1

P(Paz is nominated) = 12

c

2

Write your answer.

1

Write the Multiplication Law of probability for independent events.

The probability that Paz is nominated as 1 captain is 12 . c

P(A and P) = P(A ¶ P) = P(A) ì P(P) P(Adam and Paz are nominated) = P(Adam is nominated) ì P(Paz is nominated)

2

Substitute the known values into the rule.

1 1 = 10 × 12

Chapter 12 probability

413

statistics AND probability • Chance 1

3

Evaluate.

= 120

4

Write your answer.

The probability that both Adam and Paz are 1 nominated as captains is 120 .

■■

■■ ■■

Sometimes one event affects the outcome of another. For example, if a card is drawn from a 13 1 pack of playing cards, the probability that its suit is hearts, P(hearts), is 52 (or 4 ). If this card is not replaced, then this will affect the probability of subsequent draws. The probability that 12 the second card drawn is a heart will be 51 while the probability that the second card is not a 39 heart will be 51 . When one event affects the occurrence of another, the events are called dependent events. If two events are dependent, then the probability of occurrence of one event affects that of the other.

Worked Example 16

A bag contains 5 blue, 6 green and 4 yellow marbles. The marbles are identical in all respects except in their colours. Two marbles are picked in succession without replacement. Determine the probability of picking 2 blue marbles. Think 1

Determine the probability of picking the first blue marble.

Write/draw

P(picking a blue marble) =

n(B) n(x  ) 5

P(picking a blue marble) = 15 1

=3 2

3

Determine the probability of picking the second blue marble. Note: The two events are dependent since marbles are not being replaced. Since we have picked a blue marble this leaves 4 blue marbles remaining out of a total of 14 marbles. Calculate the probability of obtaining 2 blue marbles.

P(picking second blue marble) =

n(B) n(x  ) 4

P(picking second blue marble) = 14 2

=7 P(2 blue marbles) = P(1st blue) ì P(2nd blue) 1 2 = × 3

7

2 = 21

4

Write your answer.

Note: Alternatively, a tree diagram could be used to solve this question.   The probability of selecting 2 blue marbles successively can be read directly from the first branch of the tree diagram.

2

The probability of obtaining 2 blue marbles is 21.

5 — 15

10 — 15

4 — 14

Blue

10 — 14

Not blue

5 — 14

Blue

Blue

Not blue 9 — 14

P(2 blue marbles) = = = 414

Maths Quest 10 for the Australian Curriculum

5 × 4 15 14 1 ×2 3 7 2 21

Not blue

stAtistiCs AND probAbility • ChANCe

reMeMber

1. Events are independent if the occurrence of one event does not affect the occurrence of the other. 2. If events A and B are independent, then P(A ¶ B) = P(A) ì P(B). This is the Multiplication Law of probability. Conversely, if P(A ¶ B) = P(A) ì P(B) then events A and B are independent. 3. Dependent events affect the probability of occurrence of one another. exerCise

12D iNDiviDUAl pAthWAys eBook plus

Activity 12-D-1

Simple independent and dependent events doc-5119 Activity 12-D-2

Independent and dependent events doc-5120 Activity 12-D-3

Tricky independent and dependent events doc-5121

eBook plus

Interactivity Random numbers

Independent and dependent events flUeNCy 1 If A and B are independent events and P(A) = 0.7 and P(B) = 0.4, calculate: a P(A and B) 0.28 b P(AÅ and B) where AÅ is the complement of A 0.12 c P(A and BÅ) where BÅ is the complement of B 0.42 d P(AÅ and BÅ). 0.18 UNDerstANDiNg 2 We15 A die is rolled and a coin is tossed. a Are the outcomes independent? Yes b Determine: 1 i P(Head) on the coin 2 1 ii P(6) on the die. 6 c Determine P(6 on the die and Head on the coin).

1 12

3 A tetrahedron (4-faced) die and a 10-sided die are rolled simultaneously. What is the

probability of getting a 3 on the tetrahedral die and an 8 on the 10-sided die? 401 4 A blue die and a green die are rolled. What is the probability of getting a 5 on the blue die and not a 5 on the green die? 365 4 5 Dean is an archer. The experimental probability that Dean will hit the target is 5 .

int-0085

a b c d

What is the probability that Dean will hit the target on two successive attempts? 16 25 What is the probability that Dean will hit the target on three successive attempts? 1 What is the probability that Dean will not hit the target on two successive attempts? 25 What is the probability that Dean will hit the target on the first attempt but miss on the 4 second attempt? 25 Chapter 12 probability

64 125

415

stAtistiCs AND probAbility • ChANCe 6 MC A bag contains 20 apples, of which 5 are bruised.

Peter picks an apple and realises that it is bruised. He puts the apple back in the bag and picks another one. a The probability that Peter picks 2 bruised apples is: A

1 4

B

1 2

1

✔ C 16

D

3 4

E

15 16

b The probability that Peter picks a bruised apple first but a

good one on his second attempt is: A

1 4

B

1 2

C

3 4

3

✔ D 16

1 16 is 1 and 7

E

7 The probability that John will be late for a meeting

the probability that Phil will be late 3 for a meeting is 11 . What is the probability that: 3 48 a John and Phil are both late 77 b neither of them is late 77 8 18 c John is late but Phil is not late 77 d Phil is late but John is not late? 77 8 On the roulette wheel at the casino there are 37 numbers, 0 to 36 inclusive. Bidesi puts his chip on number 8 in game 20 and on number 13 in game 21. 1 a What is the probability that he will win in game 20? 37 1 b What is the probability that he will win in both games? 1369 73 c What is the probability that he wins at least one of the games? 1369 9 Based on her progress through the year, Karen was given a probability of 0.8 of passing the Physics exam. If the probability of passing both Maths and Physics is 0.72, what is her probability of passing the Maths exam? 0.9 10 Suresh found that, on average, he is delayed 2 times out of 7 at Melbourne airport. Rakesh made similar observations at Brisbane airport, but found he was delayed 1 out of every 4 times. Find the probability that both Suresh and Rakesh will be delayed if they are flying out of their respective airports. 1 14

11 Bronwyn has 3 pairs of Reebok and 2 pairs of Adidas running shoes. She has 2 pairs of

Reebok, 3 pairs of Rio and a pair of Red Robin socks. Preparing for an early morning run, she grabs at random for a pair of socks and a pair of shoes. What is the probability that she chooses: 1 a Reebok shoes and Reebok socks 5 1 b Rio socks and Adidas shoes 5 1 c Reebok shoes and Red Robin socks 10 1 d Adidas shoes and socks that are not Red Robin? 3 12 We16 Two cards are drawn successively and without replacement from a pack of playing cards. Determine the probability of drawing: 25 1 1 a 2 hearts 17 b 2 kings 221 c 2 red cards. 102 416

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

b c

26 145 136 435 221 435

13 In a class of 30 students there are 17 girls. Two students are picked randomly to represent the

class in the Student Representative Council. Determine the probability that: a both students are boys b both students are girls c one of the students is a boy.

Reasoning No. Coin tosses are independent events. 14 Greg has tossed a tail on each of 9 successive coin tosses. He believes that his chances of No one toss affects tossing a Head on his next toss must be very high. Is Greg correct? Justify your answer. the outcome of the next. The probability 15 The multiplication law of probability relates to of a Head or Tail on independent events. Tree diagrams can illustrate the reflection    a fair coin is always sample space of successive dependent events and How are dependent events, 0.5. Greg has a 50% the probability of any one combination of events can independent events and the chance of tossing be calculated by multiplying the stated probabilities multiplication law of probability a Head on the next reflected on a tree diagram? along the branches. Is this a contradiction to the coin toss as was the chance in each of the multiplication law of probability? Explain. previous 9 tosses.

12E

Conditional probability ■■ ■■ ■■

Conditional probability is when the probability of an event is conditional (depends) on another event occurring first. The effect of conditional probability is to reduce the event space and thus increase the probability of the desired outcome. For two events, A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B|  A) and can be calculated using the formula: P ( A ∩ B) P(B|  A) = , P( A) ≠ 0 P( A)

Worked Example 17

A group of students was asked to nominate their favourite food, spaghetti (S) or lasagne (L). The results are illustrated in the Venn diagram at right. Use the Venn diagram to calculate the following probabilities relating to a student’s favourite food. a What is the probability that a randomly selected student prefers spaghetti? b What is the probability that a randomly selected student likes lasagne given that they also like spaghetti? Think a

1

From 40 students surveyed, shown in blue, 20 nominated their favourite food as ‘spaghetti’ or ‘spaghetti and lasagne’ as shown in red.

x

S 11

L 9

15 5

No. As events are illustrated on a tree diagram, the individual probability of each outcome is recorded. The probability of a dependent event is calculated (altered according to the previous event) and can be considered as if it was an independent event. As such, the multiplication law of probability can be applied along the branches to calculate the probability of successive events.

13 a

Write/draw a

x

S

L

11

9

15 5

2

The probability that a randomly selected student prefers spaghetti is found by substituting these values into the probability formula.

number of favourable outcomes total number of possible outcomes 20 P(spaghetti) = 40 P(event) =

=

1 2

Chapter 12 Probability

417

statistics AND probability • Chance b

The condition imposed ‘given they also like spaghetti’ alters the sample space to the 20 students described in part a, as shaded in blue. Of these 20 students, 9 stated their favourite foods as lasagne and spaghetti, as shown in red.

1

The probability that a randomly selected student likes lasagne, given that they like spaghetti, is found by substituting these values into the probability formula for conditional probability.

2

b

x

S 11

L 9

15 5

P(B|  A) = P(L|  S) = =

P ( A ∩ B) P( A) 9 40 1 2 9 20

Worked Example 18

If P(A) = 0.3, P(B) = 0.5 and P(A ß B) = 0.6, calculate: a  P(A ¶ B)              b  P(B | A) Think a

b

Write

1

State the addition law for probability to determine P(A ß B).

a P(A ß B) = P(A) + P(B) - P(A ¶ B)

2

Substitute the values given in the question into this formula and simplify.

0.6 = 0.3 + 0.5 - P(A ¶ B) P(A ¶ B) = 0.3 + 0.5 - 0.6 = 0.2

1

State the formula for conditional probability.

2

Substitute the values given in the question into this formula and simplify.

■■

b P(B|  A) =

P(B|  A) = =

P ( A ∩ B) , P(A) ò 0 P( A) 0.2 0.3 2 3

It is possible to transpose the formula for conditional probability to calculate P(A ¶ B): P ( A ∩ B) P(B |  A) = , P(A) ò 0 P( A) P(A ¶ B) = P(A) ì P(B| A) This is called the multiplication rule for probability. remember

1. Conditional probability is when the probability of an event is conditional (depends) on another event occurring first. 2. For two events, A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B|  A) and can be calculated using the formula: P ( A ∩ B) P(B|  A) = , P( A) ≠ 0 P( A) 3. The multiplication rule for probability gives P(A ¶ B) = P(A) ì P(B| A) 418

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe

exerCise

12e iNDiviDUAl pAthWAys eBook plus

Activity 12-E-1

Conditional probability flUeNCy 1 We17 A group of students was asked to nominate their favourite form of dance, hip hop (H)

or jazz (J ). The results are illustrated in the Venn diagram below. Use the Venn diagram given to calculate the following probabilities relating to a student’s favourite form of dance.

Introducing conditional probability doc-5122

x

H 35

Activity 12-E-2

J 12

29

Practice with conditional probability doc-5123 Activity 12-E-3

Tricky conditional probability problems doc-5124

14 41

a What is the probability that a randomly selected student prefers jazz? P(J) = 90 b What is the probability that a randomly selected student prefers hip hop, given that they prefer jazz? P(H | J) = 12 41

2 A group of students was asked which seats they found most comfortable, the seats in the

computer lab or the science lab. The results are illustrated in the Venn diagram below. Use the Venn diagram given to calculate the following probabilities relating to the most comfortable seats. x

C 15

S 8

5 2

13 a What is the probability that a randomly selected student prefers the science lab? P(S) = 30 b What is the probability that a randomly selected student prefers the science lab, given that they might prefer the computer lab or the science lab? P(S | (C ß S)) = 13 28

3 We18 If P(A) = 0.7, P(B) = 0.5 and P(A ß B) = 0.9, calculate: 3 a P(A ¶ B) 0.3 b P(B | A). 7 4 If P(A) = 0.65, P(B) = 0.75 and P(A ¶ B) = 0.45, calculate: a P(B | A) 139 b P(A | B).

3 5

UNDerstANDiNg 5 A medical degree requires applicants to participate in two tests, an aptitude test and an

emotional maturity test. 52% passed the aptitude test, while 30% passed both tests. Use the conditional probability formula to calculate the probability that a student who passed the aptitude test also passed the emotional maturity test. 0.58 or 15 26 6 At a school classified as a ‘Music school for excellence’ the probability that a student elects to study Music and Physics is 0.2. The probability that a student takes Music is 0.92. What is the probability that a student takes Physics, given that the student is taking Music? 0.22 or 235 7 The probability that a student is well and misses a work shift the night before an exam is 0.045, while the probability that a student misses a work shift is 0.05. What is the probability that a student is well, given they miss a work shift the night before an exam? 0.9 8 Two marbles are chosen, without replacement, from a jar containing only red and green marbles. The probability of selecting a green marble and then a red marble is 0.67. The probability of selecting a green marble on the first draw is 0.8. What is the probability of selecting a red marble on the second draw, given the first marble drawn was green? 0.8375 Chapter 12 probability

419

statistics AND probability • Chance 9 Consider rolling a red and a black die and the probabilities of the following events:

Event A Event B Event C

the red die lands on 5 the black die lands on 2 the sum of the dice is 10. a   MC  The initial probability of each event described is: A P(A) =

1 6

B P(A) =

5 6

C P(A) =

5 6

P(B) =

1 6

P(B) =

2 6

P(B) =

2 6

P(C) =

1 6

P(C) =

7 36

P(C) = 18

P(A) = 6

1

E P(A) =

1 6

1 6

P(B) =

2 6

1

P(C) = 12

✔ D

P(B) =

i  P(A  |  B) = 1 6



P(C) = 12

ii  P(B  |  A) = 16

b Calculate the following probabilities.    i P(A | B) ii P(B | A)

iii  P(C  |  A) = 16 iv  P(C  |  B) = 0

5

1



iii P(C | A)

iv  P(C | B)

10   MC  A group of 80 schoolgirls consists of 54 dancers and 35 singers. Each member of the

✔

group is either a dancer or a singer, or both. The probability that a randomly selected student is a singer given that she is a dancer is: A 0.17 B 0.44 C 0.68 D 0.11 E 0.78

Conditional Reasoning probability 11 Explain how imposing a condition alters probability calculations. is when the probability of one 12 At your neighbouring school, 65% of the students are male and 35% are female. Of the male event depends on students, 10% report that dancing is their favourite activity; of the female students, 25% report the outcome of that dancing is their favourite activity. another event.

Find the probability that: a a student selected at random prefers dancing and is female 0.0875 b a student selected at random prefers dancing and is male. 0.065 13 Using the information presented in Question 12 above, construct a tree diagram. From your diagram, calculate: reflection    a the probability that a student is male and does not prefer dancing 0.585 How does imposing a condition b the overall percentage of students who prefer alter the probability of an event? dancing. 0.1525 or 15.25%.

12F

Subjective probability ■■

■■

420

Consider the following claims: ‘I feel the Australian cricket team will win this year’s Test cricket series because, in my opinion, they have a stronger side than the opposition.’ Claims like this are often made by people who may not have all the facts, and may also be biased.   ‘I think this summer will be a cold one.’ A statement like this will have merit if it comes from an individual with relevant knowledge, such as a meteorologist or a scientist. However, often people make these remarks with limited observation. Subjective probability is usually based on limited mathematical evidence and may involve one or more of the following: judgements, opinions, assessments, estimations and conjectures by individuals. It can also involve beliefs, sentiments and emotions that may result in a certain amount of bias.

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • ChANCe

WorkeD exAMple 19

On Anzac Day Peter plays two-up, which involves tossing two coins. Heads win if both coins land Heads, while Tails win if both coins land Tails. If the coins land with one Head and one Tail they are called ‘odd’, and the coins are tossed again until either Heads or Tails wins. After observing for a while, Peter notices that the last five tosses had either Tails winning or were odd. This leads Peter to believe that Heads will win the next game, so he places $50 on Heads and loses. Peter questions the fairness of the game and states that the game is biased and favours Tails. Discuss the accuracy of Peter’s statement. thiNk

Write

Discuss the statement made and comment on the probability of obtaining Heads or Tails in this particular game.

The outcome depends upon whether it is a Test match or a one-day game and how effective the bowlers and batsmen are; not forgetting the pitch usually favours spin bowling.

exerCise

12f iNDiviDUAl pAthWAys eBook plus

Activity 12-F-1

Subjective probability doc-5125

reMeMber

Subjective probability is based on judgements and opinions. It can also involve beliefs, emotions and bias.

subjective probability

Activity 12-F-3

In-depth subjective probability doc-5127

1 b The outcome depends on which team is better on the day and which team can adjust to the conditions. c No. The third one has an equal chance of being a girl or a boy.

UNDerstANDiNg 1 We19 Discuss the accuracy of these statements. a The team batting last can never win a cricket match at the MCG. b The Australian cricket team is so good that not even bad weather can stop it from winning. c Two children in John’s family are girls so the third one will be a girl, too. d The Wallabies defeated the All Blacks three times last year so they will win the first game

this year. This is not necessarily true. Current position and form of both teams should be used as a gauge.

Activity 12-F-2

Harder subjective probability doc-5126

Each game is independent and so five Tails or odd outcomes in the previous games have no effect on the outcome of the current game. The game is not biased. Peter took a risk and paid for it. He is wrong in suggesting that the game is not fair.

e It rained heavily on the last three consecutive Fridays so do not organise sport on a Friday. It does not mean it will rain again on Friday. f According to the weather report only three in every twenty houses were damaged by the

cyclone, so my house will not be damaged. g New Zealand lost its cricket match against Australia because their team uniform looked There is no certainty about that. boring. It depends upon the condition h This coin is biased because we obtained six Heads in a row. and location of your house. i The USA topped the medal tally in the last Olympics so they will do the same again in

the next Olympics. Australian Rules football is the best sport in the world. No horse is certain to 2 Comment on the accuracy of these statements. You still have a chance. win. Lots of problems can occur on the track. This is not true. a I have bought only one ticket for the raffle, therefore I cannot win. Even though Heads b This particular horse has odds of 1–2. It is certain to win. and Tails have equal c If you keep on betting on Heads, you cannot lose. chances, it does not d If you want to win at all times, bet on the favourites. mean half the results e It is no use betting on the favourites as you cannot win a great deal of money, therefore will show Heads. you should bet on the outsiders. Sometimes outsiders pay well, if you back the right one! j

You can lose more money than you win.

Favourites do not always win.

Chapter 12 probability

421

stAtistiCs AND probAbility • ChANCe 3 Assign a probability to each of the following, based on your experience or judgement. Answers will vary. a The probability that you will be late for a class this week Class discussion b The probability that your favourite sporting team will win its next match required as there c The probability that two traffic lights in a row will be red when you approach successive are many factors intersections to consider. d The probability that you will see a dog some time today There is a contradiction. reAsoNiNg The job was never hers. The team may have had She had to do well to 4 Comment on the contradictions involved in the following statements. a lead but a match is only win the position. a That job was hers but she did not do well in the interview. won when finished. b The team had won the match but they became a little complacent towards the end. eBook plus c ‘Makybe Diva’ was certain to win. I cannot believe she lost the race. No horse is certain to win. Digital doc

WorkSHEET 12.3 doc-5297

5 Compare and contrast experimental probability,

theoretical probability and subjective probability. 6 Describe a situation where subjective probability may

endanger people. Answers will vary. Class discussion may be required. Example only: medical — our town is so far away from any major airports that it is unlikely our residents will need immunisation from swine flu.

422

Maths Quest 10 for the Australian Curriculum

refleCtioN 

 

Is subjective probability reliable?

Experimental probability is based on data collected from trials. The more trials undertaken, the closer the experimental probability will reflect theoretical probability. Theoretical probability is based on mathematical models. A theoretical probability does not guarantee a particular outcome in real life situations. Subjective probability is based on judgements and opinions and hence may be biased. Subjective probability may approach theoretical probability if the assigned probability is based on real experiences and judgements made from an objective and educated position.

statistics AND probability • Chance

Summary Review of probability ■■ ■■

■■

■■

■■ ■■ ■■

Probabilities can be expressed as a percentage, fraction or decimal in the range 0 to 1, inclusive. number of times an event has occurred Experimental probability = total number of trials frequency of the score f or Relative frequency of a score = total sum of frequencies S f n(E ) Theoretical probability that an event, E, will occur is P(E ) = where n(E ) = number of n(x  ) times or ways an event, E, can occur and n(x  ) = the total number of ways all outcomes can occur. P(x  ) = 1 Venn diagrams provide a diagrammatic representation of sample spaces. b If the odds for an event are given as a–b, then P(the event occurs) = and a+b a . P(the event does not occur) = a+b Complementary and mutually exclusive events

■■ ■■ ■■ ■■

■■

■■ ■■

Complementary events have no common elements and together make up the universal set. If A and AÅ are complementary events then P(A) + P(AÅ) = 1. This may be rearranged to: P(AÅ) = 1 - P(A) or P(A) = 1 - P(AÅ). Mutually exclusive events have no common elements and cannot occur simultaneously. If events A and B are not mutually exclusive then: P(A or B) = P(A) + P(B) - P(A and B) or P(A ß B) = P(A) + P(B) - P(A ¶ B) where P(A ¶ B) is the probability of the intersection of sets A and B or the common elements in sets A and B. If events A and B are mutually exclusive then: P(A or B) = P(A) + P(B) or P(A ß B) = P(A) + P(B) since P(A ¶ B) = 0. Mutually exclusive events may or may not be complementary events. Complementary events are always mutually exclusive. Two-way tables and tree diagrams

■■ ■■

■■

Two-way tables give a clear diagrammatic representation of the sample space; however, they are limited to displaying two events. Tree diagrams are useful in working out the sample space and calculating probabilities of various events, especially if there is more than one event. On each branch of a tree diagram, the probability associated with the branch is listed. The products of the probabilities given on the branches are taken to calculate the probability for an outcome. The probabilities of all outcomes add to 1. Independent and dependent events

■■ ■■

■■

Events are independent if the occurrence of one event does not affect the occurrence of the other. If events A and B are independent, then P(A ¶ B) = P(A) ì P(B). This is the Multiplication Law of probability. Conversely, if P(A ¶ B) = P(A) ì P(B) then events A and B are independent. Dependent events affect the probability of occurrence of one another. Chapter 12 Probability

423

stAtistiCs AND probAbility • ChANCe Conditional probability ■

■

■

Conditional probability is when the probability of an event is conditional (depends) on another event occurring first. For two events, A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B| A) and can be calculated using the formula: P ( A ∩ B) P(B| A) = , P( A) ≠ 0 P( A) The multiplication rule for probability gives P(A ¶ B) = P(A) ì P(B| A) Subjective probability

■

Subjective probability is based on judgements and opinions. It can also involve beliefs, emotions and bias.

MAPPING YOUR UNDERSTANDING

Homework Book

424

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 379. Have you completed the two Homework sheets, the Rich task and two Code puzzles in your Maths Quest 10 Homework Book?

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Chance

Chapter review b AÅ ¶ BÅ

Fluency 1 Which of the following is always true for an event,

M, and its complementary event, MÅ? P(M) + P(MÅ) = 1 B P(M) - P(MÅ) = 1 C P(M) + P(MÅ) = 0 D P(M) - P(MÅ) = 0 E P(M) ì P(MÅ) = 1 2 A number is chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Which of the following pairs of events is mutually exclusive? A {2, 4, 6} and {1, 2, 3} ✔ B {1, 2, 3, 5} and {4, 6, 7, 8} C {0, 1, 2, 3} and {3, 4, 5, 6} D {multiples of 2} and {factors of 8} E {even numbers} and {multiples of 3} 3 Which of the following states the Multiplication Law of probability correctly? A P(A ¶ B) = P(A) + P(B) ✔ B P(A ¶ B) = P(A) ì P(B) C P(A ß B) = P(A) ì P(B) D P(A ß B) = P(A) + P(B) E P(A) = P(A ß B) ì P(B) 4 The odds 3-2 expressed as a probability are:

A

B

B x

A

x

✔ A

A c e

1 5 1 2 1 3

B ✔ d

3 5 2 5

The following information relates to questions 5 and 6. x  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 3, 4} and B = {3, 4, 5, 8} 5 A ¶ B equals: A {2, 3, 3, 4, 4, 5, 8} ✔ B {3, 4} C {2, 3, 4} D {2, 3, 4, 5, 8} E {2, 5, 8} 6 A ¶ BÅ equals: ✔ B {2} A {3, 4} C {2, 3, 4, 5, 8} D {2, 3, 4} E {1, 2, 6, 7, 9, 10} 7 Shade the region stated for each of the following Venn diagrams. a AÅ ß B A B x A

B

x

c AÅ ¶ BÅ ¶ C

A

x

A

B

x

B

C

C

8 Convert the following odds to probabilities. a 3–7 107 2 b 5–2 7 5 c 12–5 17 9 Convert the following probabilities to odds. a

7 11

4–7

b

6 7

1–6

25 33

c

8–25

problem solving 1 From past experience, it is concluded that there

is a 99% probability that July will be a wet month in Launceston (it has an average rainfall of approximately 80 mm). The probability that July will not be a wet month next year in Launceston is: A 99%

B 0.99

D 1

E 0

✔ C

1 100

2 A card is drawn from a well-shuffled deck of

52 cards. What is the theoretical probability of not selecting a red card? A C

3 4 1 13

B

1 4

✔ D

1 2

E 0 3 Which of the following events is not equally likely? A Obtaining a 5 or obtaining a 1 when a die is

rolled B Obtaining a club or obtaining a diamond when

a card is drawn from a pack of cards C Obtaining 2 Heads or obtaining 2 Tails when a

coin is tossed Obtaining 2 Heads or obtaining 1 Head when a coin is tossed twice E Obtaining a 3 or obtaining a 6 when a die is rolled

✔ D

Chapter 12 Probability

425

statistics AND probability • Chance 4 The Australian cricket team has won 12 of the

last 15 Test matches. What is the experimental probability of Australia losing its next Test match? A C

4 5 1 4

✔ B

D

13 The Venn diagram below shows the results of a

survey completed by a Chinese restaurateur to find out the food preferences of his regular customers.

1 5 3 4

x

Fried rice

 he team’s win or loss depends upon how other 11 a Whether it rains or not on Thursday is not determined by what happened b T players bat and bowl or how the other team plays. on Monday, Tuesday or Wednesday. It can still rain on Thursday. c There is an equal chance of having a boy or a girl.

5 A card is drawn from a well-shuffled pack of

10

52 cards. What is the theoretical probability of drawing: a an ace 131 b a spade 14 2 c a queen or a king 13 d not a heart? 43 6 The odds for a horse to win a race are 4–3. a What is the probability that this horse will win

7

8

9

10

11

12

5

7

E 1

3

12 5

Chicken wings

8 Dim sims

Determine the number of customers: surveyed 50 the race? 37 showing a preference for fried rice only 7 b What is the probability that this horse will not showing a preference for fried rice 25 win the race? 47 s howing a preference for chicken wings and c Charlie bets $12 that this horse will win. If the dim sims. 8 horse wins, what is Charlie’s payout? $28 b A customer from this group won the draw for A die is rolled five times. a lucky door prize. Determine the probability 1 that this customer: a What is the probability of rolling five 6s? 7776 7775    i likes fried rice 12 b What is the probability of not rolling five 6s? 7776  ii likes all three — fried rice, chicken wings Alan and Mary own 3 of the 8 dogs in a race. What and dim sims 503 is the probability that: 6 iii prefers chicken wings only. 25 a one of Alan’s or Mary’s dogs will win? 83 c A similar survey was conducted a month b none of Alan’s or Mary’s dogs will win? 85 later with another group of 50 customers. A die is rolled. Event A is obtaining an even This survey yielded the following results: number. Event B is obtaining a 3. 2 customers liked all three foods; 6 preferred a Are events A and B mutually exclusive? Yes fried rice and chicken wings; 7 preferred 1 1 b Calculate P(A) and P(B). P(A) = 2 and P(B) = 6 chicken wings and dim sims; 8 preferred fried 2 c Calculate P(A ß B). 3 rice and dim sims; 22 preferred fried rice; A card is drawn from a shuffled pack of 52 playing 23 preferred chicken wings; and 24 preferred cards. Event A is drawing a club and event B is dim sims. drawing an ace.   i Display this information on a Venn diagram.  ii What is the probability of selecting a a Are events A and B mutually exclusive? No customer who prefers all three foods, if a b Calculate P(A), P(B) and P(A ¶ B). 1 4 1 1 1 c Calculate P(A ß B). P(A) = 4 , P(B) = 13, P(A ¶ B) = 52 random selection is made? 25 13 14 A pair of dice is rolled and the sum of the numbers Discuss the accuracy of the following statements. shown is noted. a It did not rain on Monday, Tuesday or a Show the sample space in a two-way table. Wednesday, so it will not rain on Thursday. b In how many different ways can the sum of b A cricket team lost because two of its batsmen 7 be obtained? 6 scored ducks. c Are all outcomes equally likely? No. Frequency of c The Rams family had a boy, then a girl and numbers is different. d Complete the given table. then another boy. They must have a girl next. Sum 2 3 4 5 6 7 8 9 11 12 Comment on the contradictions involved in these If you were defeated, the opponent was the winner. statements. Frequency a I was defeated by a loser. b The slowest motocross racer in the competition e What are the relative frequencies of the won the race. following sums? 1 i 2 36  ii 7 16 iii 11 181 c The most popular person did not get elected.

The person elected was the most popular choice for the position. 426 Maths Quest 10 for the Australian Curriculum

a

  i  ii iii iv

Sum 2 Frequency 1

3 2

4 3

5 4

6 5

7 6

8 5

9 4

10 11 12 3 2 1

stAtistiCs AND probAbility • ChANCe f What is the probability of obtaining the

19 Determine the probability of drawing 2 aces from a following sums? pack of cards if: 1 1 1 i 2 36 ii 7 6 iii 11 18 a the first card is replaced before the second one 1 g If a pair of dice is rolled 300 times, how many is drawn 169 1 times do you expect to obtain the sum of 7? 50 b the first card drawn is not replaced. 221 15 A tetrahedral die is numbered 0, 1, 2 and 3. Two of 20 On grandparents day at a school a group of these dice are rolled and the sum of the numbers grandparents was asked where they most like (the number on the face that the die sits on) is to take their grandchildren — the beach (B) or taken. shopping (S). The results are illustrated in the a Show the possible outcomes in a two-way Venn diagram below. Use the Venn diagram given table. to calculate the following probabilities relating b Are all the outcomes equally likely? No to the place grandparents most like to take their c Which total has the least chance of being grandchildren. rolled? 0 and 6 d Which total has the best chance of being ξ B S rolled? 3 e Which sums have the same chance of being 5 8 2 rolled? 0 and 6, 1 and 5, 2 and 4 16 An eight-sided die is rolled three times to see 10 whether 5 occurs. a Draw a tree diagram to show the sample space. a What is the probability that a randomly b Calculate: selected grandparent preferred to take their 1 i P(three 5s) 512 grandchildren to the beach or shopping? 15 = 3 343 ii P(no 5s) 25 5 512 21 b What is the probability that a randomly iii P(two 5s) 512 selected grandparent preferred to take their 11 iv P(at least two 5s). 256 grandchildren to the beach, given that they 17 A tetrahedral die (four faces labelled 1, 2, 3 and 4) preferred to take their grandchildren shopping? is rolled and a coin is tossed simultaneously. 8 4 21 Two marbles are chosen, without replacement, =5 a Show all the outcomes on a two-way table. 10 from a jar containing only red and green marbles. b Draw a tree diagram and list all outcomes and The probability of selecting a green marble and their respective probabilities. then a red marble is 0.72. The probability of c Calculate the probability of getting a Head on selecting a green marble on the first draw is 0.85. the coin and an even number on the die. 14 What is the probability of selecting a red marble 18 A bag contains 20 pears, of which 5 are bad. Cathy on the second draw if the first marble drawn was picks 2 pears (without replacement) from the bag. green? 0.847 What is the probability that: a both pears are bad? 191 eBook plus 21 17 a Die outcomes b both pears are good? 38 15 c one of the two pears is good? 38 Interactivities 1 2 3 4

15 a

Die 1 outcomes

0

2

1

0

Coin outcomes

Die 2 outcomes 3

(0, 0) (0, 1) (0, 2) (0, 3)

3

16 a

7 – 8

Test yourself Chapter 12 int-2858

T

(T, 1) (T, 2) (T, 3) (T, 4)

Word search Chapter 12 int-2856

1– 4

(2, 0) (2, 1) (2, 2) (2, 3) H

(3, 0) (3, 1) (3, 2) (3, 3) 1

1 – 8

(H, 1) (H, 2) (H, 3) (H, 4)

b

1 (1, 0) (1, 1) (1, 2) (1, 3) 2

H

f

fÅ

2 1 – 8

f

7 – 8

fÅ

1 – 8

f

7 – 8

fÅ

f = outcome of 5

1– 8 7– 8 7– 8 7– 8 7– 8

1– 8 1– 8 1– 8

3 f fÅ

1– 2

Outcomes Probability 1 —– fff 512 7 —– fffÅ 512

f fÅ

ffÅf ffÅfÅ

f fÅ

fÅff fÅffÅ

f fÅ

fÅfÅf fÅfÅfÅ

7 —– 512 49 —– 512 7 —– 512 49 —– 512 49 —– 512 343 —– 512

1– 4 1– 4

1– 4

1– 4

1– 2

T 1– 4

1– 4 1– 4

Outcomes Probability 1– 1– 1– 1 H1 2 ì 4 = 8 H2

1– 2

H3

1– 2

4

H4

1– 2

1

T1

2 3

2 3

4

ì

1– 4

=

1– 8

ì

1– 4

=

1– 8

ì

1– 4

=

1– 8

ì ì

1– 4 1– 4

=

T2

1– 2 1– 2

=

1– 8 1– 8

T3

1– 2

ì

1– 4

=

1– 8

T4

1– 2

ì

1– 4

=

Crossword Chapter 12 int-2857

1–

8 — 1

Chapter 12 probability

427

eBook plus

ACtivities

Chapter opener Digital doc

• Hungry brain activity (doc-5285): Chapter 12 (page 379) Are you ready? Digital docs

(page 380)

• SkillSHEET 12.1 (doc-5286): Set notation • SkillSHEET 12.2 (doc-5287): Simplifying fractions • SkillSHEET 12.3 (doc-5288): Determining complementary events • SkillSHEET 12.4 (doc-5289): Addition and subtraction of fractions • SkillSHEET 12.5 (doc-5290): Multiplying fractions for calculating probabilities 12A Review of probability Interactivity

• Random number generator (int-0089) (page 381) Digital docs

• Activity 12-A-1 (doc-5110): Review of probability (page 392) • Activity 12-A-2 (doc-5111): General probability problems (page 392) • Activity 12-A-3 (doc-5112): Tricky probability problems (page 392) • SkillSHEET 12.1 (doc-5286): Set notation (page 384) • SkillSHEET 12.2 (doc-5287): Simplifying fractions (page 392) • SkillSHEET 12.6 (doc-5291): Working with Venn diagrams (page 392) • SkillSHEET 12.7 (doc-5292): Writing odds as probabilities (page 395) • SkillSHEET 12.8 (doc-5293): Writing probabilities as odds (page 395) 12B Complementary and mutually exclusive events Digital docs

• Activity 12-B-1 (doc-5113): Complementary and mutually exclusive events (page 401) • Activity 12-B-2 (doc-5114): Harder complementary and mutually exclusive events (page 401) • Activity 12-B-3 (doc-5115): Tricky complementary and mutually exclusive events (page 401) • SkillSHEET 12.3 (doc-5288): Determining complementary events (page 401) • SkillSHEET 12.4 (doc-5289): Addition and subtraction of fractions (page 402) • SkillSHEET 12.9 (doc-5294): Distinguishing between complementary and mutually exclusive events (page 403) • WorkSHEET 12.1 (doc-5295): Introducing probability (page 403) 12C Two-way tables and tree diagrams elesson

• Games at Wimbledon (eles-1032) (page 405) 428

Maths Quest 10 for the Australian Curriculum

Digital docs

• Activity 12-C-1 (doc-5116): Review of two-way tables and tree diagrams (page 410) • Activity 12-C-2 (doc-5117): Practice with two-way tables and tree diagrams (page 410) • Activity 12-C-3 (doc-5118): Tricky two-way tables and tree diagrams (page 410) • SkillSHEET 12.5 (doc-5290): Multiplying fractions for calculating probabilities (page 410) • WorkSHEET 12.2 (doc-5296): Tree diagrams (page 412) 12D Independent and dependent events Interactivities

• A pack of cards (int-2787) (page 413) • Random numbers (int-0085) (page 415) Digital docs

(page 415)

• Activity 12-D-1 (doc-5119): Simple independent and dependent events • Activity 12-D-2 (doc-5120): Independent and dependent events • Activity 12-D-3 (doc-5121): Tricky independent and dependent events 12E Conditional probability Digital docs

(page 419)

• Activity 12-E-1 (doc-5122): Introducing conditional probability • Activity 12-E-2 (doc-5123): Practice with conditional probability • Activity 12-E-3 (doc-5124): Tricky conditional probability problems 12F Subjective probability Digital docs

• Activity 12-F-1 (doc-5125): Subjective probability (page 421) • Activity 12-F-2 (doc-5126): Harder subjective probability (page 421) • Activity 12-F-3 (doc-5127): In-depth subjective probability (page 421) • WorkSHEET 12.3 (doc-5297): Subjective probability (page 422) Chapter review Interactivities

(page 427)

• Test yourself Chapter 12 (int-2858): Take the end-ofchapter test to test your progress • Word search Chapter 12 (int-2856): an interactive word search involving words associated with this chapter • Crossword Chapter 12 (int-2857): an interactive crossword using the definitions associated with the chapter To access eBookPLUS activities, log on to www.jacplus.com.au

statistics aND probability • Data represeNtatioN aND iNterpretatioN

13

13A 13B 13C 13D 13E 13F

Measures of central tendency Measures of spread Box-and-whisker plots The standard deviation [suitable for 10A] Comparing data sets Skewness

What Do yoU kNoW ?

Univariate data

1 List what you know about data. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of data. eBook plus

Digital doc

Hungry brain activity Chapter 13 doc-5298

opeNiNG QUestioN

A new drug for the relief of cold symptoms has been developed. To test the drug, 40 people were exposed to a cold virus. Twenty patients were then given a dose of the drug while another 20 patients were given a placebo. (In medical tests a control group is often given a placebo drug. The subjects in this group believe that they have been given the real drug but in fact their dose contains no drug at all.) All participants were then asked to indicate the time when they first felt relief of symptoms. The number of hours from the time the dose was administered to the time when the patients first felt relief of symptoms are detailed below. Group A (drug) 25 29 32 45 18 21 37 42 62 13 42 38 44 42 35 47 62 17 34 32 Group B (placebo) 25 17 35 42 35 28 20 32 38 35 34 32 25 18 22 28 21 24 32 36 Does the drug work? How do drug companies analyse these results?

statistics aND probability • Data represeNtatioN aND iNterpretatioN

are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■click■on■the■SkillSHEET■icon■next■to■the■question■ on■the■Maths■Quest■eBookPLUS■or■ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 13.1 doc-5299

eBook plus

Digital doc

SkillSHEET 13.2 doc-5300

eBook plus

Digital doc

SkillSHEET 13.3 doc-5301

eBook plus

Digital doc

Finding the mean of a small data set 1 Find■the■mean■of■the■following■data■sets. a 4,■6,■2,■6,■4,■3,■7,■3,■2,■9 ■ 4.6 10.3375 b 10.4,■10.5,■10.7,■10.4,■10.1,■10.2,■10.4,■10.0 c 164,■136,■171,■144,■128,■130,■165,■170,■120,■124,■124,■143 143.25 Finding the median of a small data set 2 Find■the■median■of■each■of■the■following■data■sets. a 2,■6,■8,■4,■5,■6,■7 ■ 6 12.5 b 13,■10,■15,■12,■18,■17,■12,■12 c 52,■45,■23,■83,■9,■45,■71,■84,■90,■183 61.5 d 9.2,■9.3,■9.4,■9.3,■9.4,■9.5,■9.9,■9.4,■9.7,■9.8,■9.0,■10.0,■9.4,■9.2,■9.9 9.4 Finding the mode of a small data set 3 Find■the■mode■of■each■of■the■following■data■sets. a 2,■4,■3,■1,■6,■7,■3,■4,■3,■2,■7 ■ 3 b 34,■82,■94,■81,■70,■45,■32,■46,■48 No■mode 2■and■3 c 1,■2,■4,■3,■5,■2,■6,■7,■3,■3,■2 Finding the mean, median and mode from a stem-and-leaf plot 4 Find■the■mean,■median■and■mode■of■the■data■presented■in■

SkillSHEET 13.4 doc-5302

eBook plus

Digital doc

Mean■=■37,■median■=■39,■mode■=■43

Presenting data in a frequency distribution table 5 Place■the■following■set■of■scores■into■a■frequency■distribution■table■including■a■column■

for■f■■ì■x■and■cumulative■frequency. 8■ 7■ 4■ 9■ 6■ 7■ 9■ 6■ 5■ 4 7■ 9■ 8■ 6■ 5■ 8■ 9■ 4■ 5■ 8

SkillSHEET 13.5 doc-5303

Digital doc

SkillSHEET 13.6 doc-5304

Drawing statistical graphs 6 Use■the■data■from■question■5■to■draw■a■combined■frequency■histogram■and■polygon. Frequency

eBook plus

5 4 3 2 1 4

430

Key:■ 3■■|■■4■=■34 Stem Leaf 2 1■ 4■ 5■ 3 0■ 6 4 2■ 3■ 3■ 7 5 9

the■stem-and-leaf■plot■shown■at■right.

5

6 7 Score

8

9

Maths Quest 10 for the australian curriculum

Score (x)

Frequency (f)

fìx

Cumulative frequency

4

3

12

■3

5

3

15

■6

6

3

18

■9

7

3

21

12

8

4

32

16

9

4

36

20

statistics AND probability • data representation and interpretation

13A

Measures of central tendency ■■

Measures of central tendency are summary statistics that measure the middle (or centre) of the data. These are known as the mean, median and mode. •• The mean is the average of all observations in a set of data. •• The median is the middle observation in an ordered set of data. •• The mode is the most frequent observation in a data set.

Ungrouped data Mean To obtain the mean of a set of ungrouped data, all numbers (scores) in the set are added together and then the total is divided by the number of scores in that set.

■■

Mean = ■■

Symbolically this is written x =

sum of all scores number of scores

∑x . n

Median ■■ The median is the middle value of any set of data arranged in numerical order. In the set of n numbers, the median is located at the n + 1 th score. The median is: 2 •• the middle score for an odd number of scores arranged in numerical order •• the average of the two middle scores for an even number of scores arranged in numerical order. Mode ■■ The mode is the score that occurs most often in a set of data. Sets of data may contain: 1. no mode; that is, each score occurs once only 2. one mode 3. more than one mode. Worked Example 1

For the data set 6, 2, 4, 3, 4, 5, 4, 5, find the: a  mean b  median c  mode. Think a

1

Write

Calculate the sum of the scores; that is, S x.

a S x = 6 + 2 + 4 + 3 + 4 + 5 + 4 + 5

= 33

2

Count the number of scores; that is, n.

n=8

3

Write the rule for the mean.

x=

4

Substitute the known values into the rule.

=

5

Evaluate.

= 4.125

6

Answer the question.

∑x n 33 8

The mean is 4.125. Chapter 13 Univariate data

431

statistics AND probability • data representation and interpretation b

b 23444556

1

Check that scores are arranged in numerical order.

2

Locate the position of the median using n +1 the rule , where n = 8. This places 2 the median as the 4.5th score; that is, between the 4th and 5th score.

Median =

Obtain the average of the two middle scores.

Median =

3

=

n+1 th score 2 8+1 2

th score

= 4.5th score 23444556

=

4+4 2 8 2

=4

c

4

Answer the question.

1

Systematically work through the set and make note of any repeated values (scores).

2

Answer the question.

The median is 4. éé c 23444556 å å å The mode is 4.

Calculating mean, median and mode from a frequency distribution table ■■ If data are presented in a frequency distribution table, the formula used to calculate the mean ∑( f × x ) . is x = n ■■ Here, each value (score) in the table is multiplied by its corresponding frequency; then all the f ì x products are added together and the total sum is divided by the number of observations in the set. ■■ To find the median we find the position of each score from the cumulative frequency column. ■■ The mode is the score with the highest frequency. Worked Example 2

For the table at right find the:    a  mean b  median c  mode.

Score (x) 4 5 6 7 8 Total

Think

432

1

Rule up a table with four columns titled Score (x), Frequency (  f  ), Frequency ì score (  f ì x) and Cumulative frequency (cf  ).

2

Enter the data and complete both the■ f ì x and cumulative frequency columns.

Maths Quest 10 for the Australian Curriculum

Frequency (  f )  1  2  5  4  3 15

Write

Frequency Cumulative Score Frequency ì score frequency ( x) (  f ) (  f ì x) (cf ) 4 1  4 1 5 2 10 1+2=3 6 5 30 3+5=8 7 4 28 8 + 4 = 12 8 3 24 12 + 3 = 15 n = 15 S(  f ì x) = 96

statistics AND probability • data representation and interpretation

a

b

c

x=

∑( f × x ) n

1

Write the rule for the mean.

2

Substitute the known values into the rule and evaluate.

x=

3

Answer the question.

The mean of the data set is 6.4.

1

Locate the position of the median n +1 using the rule , where n = 15. 2 This places the median as the 8th score.

2

Use the cumulative frequency column to find the 8th score and answer the question.

1

The mode is the score with the highest frequency.

2

Answer the question.

a

96 15 = 6.4

b The median is the

15 + 1 th or 8th score. 2

The median of the data set is 6.

c

The score with the highest frequency is 6. The mode of the data set is 6.

Grouped data Mean When the data are grouped into class intervals, the actual values (or data) are lost. In such cases we have to approximate the real values with the midpoints of the intervals into which these values fall. For example, when measuring heights of students in a class, if we found that 4 students had a height between 180 and 185 cm, we have to assume that each of those 4 students is 182.5 cm tall. The formula used for calculating the mean is the same as for data presented in a frequency table: ∑( f × x ) x= n Here x represents the midpoint (or class centre) of each class interval, f is the corresponding frequency and n is the total number of observations in a set.

■■

Median The median is found by drawing a cumulative frequency polygon (ogive) of the data and estimating the median from the 50th percentile.

■■

Modal class We do not find a mode because exact scores are lost. We can, however, find a modal class. This is the class interval that has the highest frequency.

■■

Worked Example 3

For the given data:          a  estimate the mean b  estimate the median c  find the modal class.

Class interval   60 – < 70   70 – < 80   80 – < 90   90 – <100 100 – <110 110 – <120 Total

Frequency  5  7 10 12  8  3 45 Chapter 13 Univariate data

433

statistics AND probability • data representation and interpretation

Think

Write

1

Draw up a table with 5 columns headed Class interval, Class centre (x), Frequency (  f  ), Frequency ì class centre (  f ì x) and Cumulative frequency (cf  ).

2

Complete the x, f ì x and cf columns.

Class FreFrequency Cumulative Class centre quency ì class centre frequency interval (x) (  f  ) ( f ì x) (cf  )   65  5   325  5   60– <70   75  7   525 12   70– <80   85 10   850 22   80– <90   95 12 1140 34   90– <100  8   840 42 100– <110 105  3   345 45 110– <120 115 n = 45 S(f ì x) = 4025

c

434

∑( f × x ) n

Write the rule for the mean.

2

Substitute the known values into the rule and evaluate.

x=

3

Answer the question.

The mean for the given data is approximately 89.4.

1

Draw a combined cumulative frequency histogram and ogive, labelling class centres on the horizontal axis and cumulative frequency on the vertical axis. Join the end-points of each class interval with a straight line to form the ogive.

2

Locate the middle of the cumulative frequency axis, which is 22.5, and label it.

3

Draw a horizontal line from this point to the ogive and then vertically to the horizontal axis.

4

Read off the value of the median from the x-axis and answer the question.

1

The modal class is the class interval with the highest frequency.

2

Answer the question.

a

4025 45 ≈ 89.4

b Cumulative frequency

b

x=

1

Cumulative frequency

a

45 40 35 30 25 20 15 10 5 0

45 40 35 30 25 20 15 10 5 0

65 75 85 95105115 Data

65 75 85 95105115 Data

The median for the given data is approximately 90.

c

Maths Quest 10 for the Australian Curriculum

The modal class is the 90 – <100 class interval.

statistics aND probability • Data represeNtatioN aND iNterpretatioN

reMeMber

For■ungrouped■data■the■following■measures■of■central■tendency■are■used. 1.■ The■mean■is■the■sum■of■scores■in■a■given■set■of■data■divided■by■the■number■of■scores■in■ the■set. ∑x x= ■ ■ is■used■when■a■list■of■scores■is■given. n ∑( f × x ) x= ■ ■ is■used■when■a■frequency■distribution■table■is■given. n 2.■ The■median■is: (a)■the■middle■score■for■an■odd■number■of■scores■arranged■in■numerical■order (b)■the■average■of■the■two■middle■scores■for■an■even■number■of■scores■arranged■in■ numerical■order. n +1 Its■location■is■determined■by■fi■nding■the■score■in■the■ th■position. 2 3.■ The■mode■is■the■score■that■occurs■most■often■in■a■set■of■data. For■grouped■data■the■following■measures■of■central■tendency■are■used. ∑( f × x ) ,■where■x■represents■the■midpoint■of■a■class■interval. 4.■ The■mean■is■ x = n 5.■ The■median■can■be■estimated■from■the■ogive■by■fi■nding■the■50th■percentile. 6.■ The■modal■class■is■given■by■the■class■interval■with■the■highest■frequency. exercise

13a iNDiviDUal pathWays eBook plus

Activity 13-A-1

Mean, median and mode doc-5128 Activity 13-A-2

Practice with mean, median and mode doc-5129 Activity 13-A-3

Mean, median and mode in depth doc-5130

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Digital doc

SkillSHEET 13.1 doc-5299

Measures of central tendency flUeNcy 1 We1 ■For■each■of■the■following■sets■of■data■fi■nd■the: i mean■ ii■ median■ a 3,■5,■6,■8,■8,■9,■10 1 a■ i■ 7 b 4,■6,■7,■4,■8,■9,■7,■10 b i■ 6.875 c 17,■15,■48,■23,■41,■56,■61,■52 c i■ 39.125 d 4.5,■4.7,■4.8,■4.8,■4.9,■5.0,■5.3 d i■ 4.857 1 1 1 1 1 e i■ 12 e 7 2 ,■10 4 ,■12,■12 4 ,■13,■13 2 ,■13 2 ,■14

iii mode. ii 8 iii 8 ii 7 iii 4,■7 ii 44.5 iii No■mode ii 4.8 iii 4.8 ii 12.625 iii 13.5

2 The■back-to-back■stem-and-leaf■plot■at■right■shows■

the■test■results■of■25■Year■10■students■in■Mathematics■ and■Science.■Find■the■mean,■median■and■mode■for■ each■of■the■two■subjects. Key:■ 3■■|■■2■=■32 Leaf Stem Science 8■7■3 3 9■6■2■2■1 4 8■7■6■1■1■0 5 9■7■4■3■2 6 8■5■1■0 7 7■3 8 9

Leaf Mathematics 2■9 0■6■8 1■3■5 2■6■7■9 3■6■7■8 0■4■4■6■8■9 2■5■8

Science:■mean■=■57.6,■median■=■57,■mode■=■42,■51■ Mathematics:■mean■=■69.12,■median■=■73,■mode■=■84 chapter 13 Univariate data

435

statistics aND probability • Data represeNtatioN aND iNterpretatioN 3 a■ i■ 5.83 ii 6 iii 6 b i■ 14.425 ii 15 iii 15 eBook plus

Digital doc

SkillSHEET 13.2 doc-5300

eBook plus

Digital doc

SkillSHEET 13.3 doc-5301

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Digital doc

SkillSHEET 13.4 doc-5302

3 We2 ■For■the■data■shown■in■each■of■the■following■frequency■distribution■tables,■fi■nd■the: i mean■ ii■ median■ iii mode. a

Score (x) 4 5 6 7 8 Total

Frequency ( f ) ■3 ■6 ■9 ■4 ■2 24

b

Score (x) 12 13 14 15 16 Total

Frequency ( f ) ■4 ■5 10 12 ■9 40

4 The■following■data■show■the■number■of■

bedrooms■in■each■of■the■10■houses■in■a■ particular■neighbourhood:■2,■1,■3,■4,■2,■3,■2,■ 2,■3,■3. a Calculate■the■mean■and■median■number■ of■bedrooms. Mean■=■2.5,■median■=■2.5 b A■local■motel■contains■20■rooms.■Add■ this■observation■to■the■set■of■data■and■ recalculate■the■values■of■the■mean■and■ median. Mean■=■4.09,■median■=■3 c Compare■the■answers■obtained■ in■parts■a■and■b■and■complete■the■ following■statement:■When■the■ set■of■data■contains■an■unusually■ large■value(s),■called■an■outlier,■the■ Median ■(mean/median)■is■the■ better■measure■of■central■tendency,■as■it■ is■less■affected■by■this■extreme■value. 5 We3 ■For■the■given■data: a estimate■the■mean ■ 72 23 b estimate■the■median 73 Class interval 40■–■<50 50■–■<60 60■–■<70 70■–■<■80 80■–■<■90 90■–■<100 Total

c fi■nd■the■modal■class. 70■–■<80

Frequency ■2 ■4 ■6 ■9 ■5 ■4 30

6 Calculate■the■mean■of■the■grouped■data■shown■in■the■table■below. 124.83

Class interval 100■–■109 110■–■119 120■–■129 130■–■139 140■–■149 Total 436

Maths Quest 10 for the australian curriculum

Frequency ■3 ■7 10 ■6 ■4 30

statistics aND probability • Data represeNtatioN aND iNterpretatioN 7 Find■the■modal■class■of■the■data■shown■in■the■table■below. 65■–■<70■ 9 b Class Cumulative interval Frequency frequency ■ 0–9 ■5 ■5 10–19 ■5 10 20–29 ■5 15 30–39 ■3 18 40–49 ■5 23 50–59 ■3 26 60–69 ■3 29 70–79 ■1 30 Total 30 Mean■=■$32.50

Class interval 50–■<55 55■–■<60 60–■<65 65■–■<70 70–■<75 75■–■<80 Total

Frequency ■1 ■3 ■4 ■5 ■3 ■2 18

8 Mc ■The■number■of■textbooks■sold■by■various■bookshops■during■the■second■week■of■

December■was■recorded.■The■results■are■summarised■in■the■table■below.

Cumulative frequency

c

30 25 20 15 10 5 0

10 20 30 40 50 60 70 80

Amount spent ($)

Median■=■$30

Number of books sold

Frequency

220■–■229

■2

230■–■239

■2

240■–■249

■3

250■–■259

■5

260■–■269

■4

270■–■279

■4 Total

20

a The■modal■class■of■the■data■is■given■by■the■class■interval(s): A 220■–■229■and■230■–■239 ✔ B 250■–■259 C 260■–■269■and■270■–■279 D of■both■A■and■C b The■class■centre■of■the■fi■rst■class■interval■is: A 224 ✔ B 224.5 c The■median■of■the■data■is■in■the■interval: A 230■–■239 B 240■–■249 d The■estimated■mean■of■the■data■is: A 251 B 252

C 224.75 ✔ C

D 225

250■–■259

C 253

D 260■–■269 ✔ D

254

UNDerstaNDiNG eBook plus

9 A■random■sample■was■taken,■composed■of■30■people■shopping■at■a■Coles■supermarket■on■a■

Tuesday■night.■The■amount■of■money■(to■the■nearest■dollar)■spent■by■each■person■was■recorded■ as■follows: SkillSHEET 13.5 6,■32,■66,■17,■45,■1,■19,■52,■36,■23,■28,■20,■7,■47,■39■ doc-5303 6,■68,■28,■54,■9,■10,■58,■40,■12,■25,■49,■74,■63,■41,■13 a Find■the■mean■and■median■amount■of■money■spent■at■the■checkout■by■the■people■in■this■ sample. Mean■=■$32.93,■median■=■$30 b Group■the■data■into■class■intervals■of■10■and■complete■the■frequency■distribution■table.■ Use■this■table■to■estimate■the■mean■amount■of■money■spent. T he■mean■is■slightly■ c Add■the■cumulative■frequency■column■to■your■table■and■fi■ll■it■in.■Hence,■construct■the■ underestimated;■the■ median■is■exact.■The■ ogive.■Use■the■ogive■to■estimate■the■median. estimate■is■good■ d Compare■the■mean■and■the■median■of■the■original■data■from■part■a■with■the■mean■and■the■ enough■as■it■provides■ median■obtained■for■grouped■data■in■parts■b■and■c.■Were■the■estimates■obtained■in■parts■b■ a■guide■only■to■the■ and■c■good■enough?■Explain■your■answer. amount■that■may■ Digital doc

be■spent■by■future■ customers.

chapter 13 Univariate data

437

statistics aND probability • Data represeNtatioN aND iNterpretatioN

eBook plus

Digital doc

SkillSHEET 13.6 doc-5304

10 a■ ■Add■one■more■number■to■the■set■of■data■3,■4,■4,■6■so■that■the■mean■of■a■new■set■is■equal■to■ its■median. ■ 3 4 ,■5,■5,■5,■6■ b Design■a■set■of■fi■ve■numbers■so■that■mean■=■median■=■mode■=■5. (one■possible■solution) c In■the■set■of■numbers■2,■5,■8,■10,■15,■change■one■number■so■that■the■median■remains■ One■possible■solution■is■to■exchange■15■with■20. unchanged■while■the■mean■increases■by■1. 11 Thirty■men■were■asked■to■reveal■the■number■of■hours■they■spent■doing■housework■each■week.■

Frequency

15

Age of emergency ward patients

10 5 0

7.5 22.5 37.5 52.5 67.5 82.5

Age

The■results■are■detailed■below. ■ ■Frequency■ 1 5 2 12 2 6 2 8 14 18 column:■16,■ 0 1 1 8 20 25 3 0 1 2 6,■4,■2,■1,■1 7 10 12 1 5 1 18 0 2 2 a Present■the■data■in■a■frequency■distribution■table.■(Use■class■intervals■of■0–4,■5–9■etc.) b Use■your■table■to■estimate■the■mean■number■of■hours■that■the■men■spent■doing■housework. 0■–4■hours c Find■the■median■class■for■hours■spent■by■the■men■at■housework. 6.8 d Find■the■modal■class■for■hours■spent■by■the■men■at■housework. 0■–4■hours reasoNiNG

12 The■data■at■right■give■the■age■of■25■patients■

438

Cumulative frequency (%)

Cumulative frequency

18 16 ■6 75 24 admitted■to■the■emergency■ward■of■a■hospital. 23 82 75 25 21 ■ ■Frequency■column:■ a Present■the■data■in■a■frequency■distribution■ 1,■13,■2,■0,■1,■8 43 19 84 76 31 table.■(Use■class■intervals■of■0–<15, 78 24 20 63 79 15–<30■and■so■on.) 80 20 23 17 19 b Draw■a■histogram■of■the■data. A symmetrical■or■ c What■word■could■you■use■to■describe■the■ bimodal■(as■if■the■ pattern■of■the■data■in■this■distribution? data■come■from■two■ d Use■your■table■to■estimate■the■mean■age■of■ separate■graphs) 44.1 patients■admitted. e Find■the■median■class■for■age■of■patients■ admitted. 15–<30 f Find■the■modal■class■for■age■of■patients■ admitted. 15–<30 g Draw■an■ogive■of■the■data. h Use■the■ogive■to■determine■the■median■age. 28 26 100% i Do■any■of■your■statistics■(mean,■median■or■ 24 22 20 mode)■give■a■clear■representation■of■the■ 18 16 No typical■age■of■an■emergency■ward■patient? 14 50% 12 10 j Give■some■reasons■which■could■explain■the■ 8 6 4 pattern■of■the■distribution■of■data■in■this■ 2 question. Class■discussion 0 15 30 45 60 75 90 Age 13 The■batting■scores■for■two■cricket■players■over■6■innings■are■as■follows: Player■A■ 31,■34,■42,■28,■30,■41 Player■B■ 0,■0,■1,■0,■250,■0 a Find■the■mean■score■for■each■player. Player■A■mean■=■34.33,■Player■B■mean■=■41.83 b Which■player■appears■to■be■better,■based■upon■mean■result? Player■B c Find■the■median■score■for■each■player. Player■A■median■=■32.5,■Player■B■median■=■0 P layer■A■is■more■ d Which■player■appears■to■be■better■when■the■decision■is■based■on■the■median■result? Player■A consistent.■One■ e Which■player■do■you■think■would■be■the■most■useful■to■have■in■a■cricket■team■and■why?■ large■score■can■ How■can■the■mean■result■sometimes■lead■to■a■misleading■conclusion? distort■the■mean. 14 The■resting■pulse■rate■of■20■female■athletes■was■measured.■The■results■are■detailed■below. 50■ 52■ 48■ 52■ 71■ 61■ 30■ 45■ 42■ 48 43■ 47■ 51■ 62■ 34■ 61■ 44■ 54■ 38■ 40 a Construct■a■frequency■distribution■table.■(Use■class■sizes■of■1–<10,■10–<20■etc.) b Use■your■table■to■estimate■the■mean■of■the■data. 50.5 ■Frequency■column:■3,■8,■5,■3,■1 Maths Quest 10 for the australian curriculum

Cumulative frequency

Ogive of pulse rate of female athletes

Cumulative frequency (%)

statistics AND probability • data representation and interpretation

100%

20 15

50%

10 5 30 50 70 Beats per minute

Find the median class of the data. 40 –<50 Find the modal class of the data. 40 –<50 Draw an ogive of the data. (You may like to use a graphics calculator for this.) Approximately 48 beats/min Use the ogive to determine the median pulse rate. 15   MC  In a set of data there is one score that is extremely small when compared to all the others. This outlying value is most likely to: ✔ A have greatest effect upon the mean of the data B have greatest effect upon the median of the data C have greatest effect upon the mode of the data D have very little effect on any of the statistics as we are told that the number is extremely small. 16 The following frequency table gives the number of employees in different salary brackets for a small manufacturing plant. Check with your teacher. c d e f

Position

Number of Salary ($) employees

Machine operator

18  000

50

Machine mechanic

20  000

15

Floor steward

24  000

10

Manager

62  000

 4

Chief executive officer

80  000

 1

a Workers are arguing for a pay rise but the management of the factory claims that

workers are well paid because the mean salary of the factory is $22  100. Are they being honest? b Suppose that you were representing the factory workers and had to write a short submission in support of the pay rise. How could you explain the management’s claim? Quote some other statistics in favour of your case. reflection    17 Design a set of five numbers with: Under what circumstances might a mean = median = mode the median be a more reliable b mean > median > mode measure of centre than the mean? c mean < median = mode.

Answers will very. Examples given. a 3, 4, 5, 5, 8 b 4, 4, 5, 10 c 2, 3, 6, 6, 12

13B

Measures of spread ■■

■■

A music store proprietor has stores in Newcastle and Wollongong. The number of CDs sold each day over one week is recorded below. Newcastle: 45, 60, 50, 55, 48, 40, 52 Wollongong:  20, 85, 50, 15, 30, 60, 90 In each of these data sets consider the measures of central tendency. Newcastle:  Mean = 50 Wollongong:  Mean = 50 Median = 50 Median = 50 No mode No mode With these measures being the same for both data sets we could come to the conclusion that both data sets are very similar; however, if we look at the data sets, they are very different. We can see that the data for Newcastle are very clustered around the mean while the Wollongong data spread out more. As well as using measures of central tendency to analyse a data set, we use measures of spread to look at how spread out a data set is. Chapter 13 Univariate data

439

statistics AND probability • data representation and interpretation

Range ■■

The most basic measure of spread is the range. It is defined as the difference between the highest and the lowest values in the set of data.

Range = highest score - lowest score or Range = Xmax - Xmin

Worked Example 4

Find the range of the given data set: 2.1, 3.5, 3.9, 4.0, 4.7, 4.8, 5.2. Think

Write

1

Identify the lowest score of the data set.

Lowest score = 2.1

2

Identify the highest score of the data set.

Highest score = 5.2

3

Write the rule for the range.

Range = highest score - lowest score

4

Substitute the known values into the rule.

= 5.2 - 2.1

5

Evaluate.

= 3.1

Interquartile range ■■

Now let us consider another two sets of scores. Which of these two data sets has the greater spread of scores?     Set A: 40, 42, 46, 48, 50, 54, 100     Set B: 32, 34, 45, 52, 66, 75, 89

■■ ■■ ■■ ■■ ■■

I n Set A the range equals 60, while in Set B the range equals 57. However, when the scores are examined closely, there is only one score in Set A (100) that makes the range so large. All other scores are relatively close together. It takes only one large or small score to increase the range of a whole data set. For this reason, the interquartile range is considered a more reliable measure of spread than the range. Another way of measuring the difference in spread is by dividing the data set into quarters. The number that marks the end of the first quarter of an ordered data set is called the lower quartile and is denoted by Q1 (or the 25th percentile). The number that marks the end of the third quarter is called the upper quartile and is denoted by Q3 (or the 75th percentile). The difference between the upper and lower quartiles is called the interquartile range (IQR). It considers the middle 50% of the data. IQR = Q3 - Q1

■■

440

The lower quartile, upper quartile and the interquartile range of a set of data may be calculated using the following steps. 1. Order the set of data. 2. Locate the median that divides the set of data into two halves. (a) For an odd number of scores, the median will be one of the original scores. It should not be included in either the lower or upper half of the scores. (b) For an even number of scores the median will lie halfway between two scores. It will divide the data into two equal sets. 3. Locate and calculate Q1, the median of the lower half of the data. 4. Locate and calculate Q3, the median of the upper half of the data. 5. Obtain the interquartile range by calculating the difference between the upper and lower quartiles; that is, IQR = Q3 - Q1.

Maths Quest 10 for the Australian Curriculum

statistics AND probability • data representation and interpretation

Worked Example 5

Calculate the interquartile range (IQR) of the following set of data: 3, 2, 8, 6, 1, 5, 3, 7, 6. Think

Write

1

Arrange the scores in order.

123356678

2

Locate the median and use it to divide the data into two halves. Note: The median is the 5th score in this data set and should not be included in either half of the data.

1 2 3 3    5   6 6 7 8

3

Find Q1, the median of the lower half of the data.

Q1 =

4

Find Q3, the median of the upper half of the data.

5

Calculate the interquartile range.

■■

2+3 2 5 = 2 = 2.5 6+7 Q3 = 2 13 = 2 = 6.5 IQR = Q3 - Q1 = 6.5 - 2.5 =4

When data are presented in a frequency distribution table, either ungrouped or grouped, the best way to find the interquartile range is to use the ogive as done earlier when finding the median. In this case, however, we find Q1 and Q3 by dividing the data into quarters as shown in the worked example that follows.

Worked Example 6

The following frequency distribution table gives the number of customers who order different volumes of concrete from a readymix concrete company during the course of a day. Find the interquartile range of the data. Volume (m3) 0.0 –<0.5 0.5 –<1.0 1.0 –<1.5

Frequency 15 12 10

Think 1

To find the 25th and 75th percentiles from the ogive, first add a class centre column and a cumulative frequency column to the frequency distribution table and fill them in.

Volume (m3) 1.5 –<2.0 2.0 –<2.5 2.5 –<3.0

Frequency 8 2 4

Write

Volume 0.0 –<0.5 0.5 –<1.0 1.0 –<1.5 1.5 –<2.0 2.0 –<2.5 2.5 –<3.0

Class centre 0.25 0.75 1.25 1.75 2.25 2.75

f 15 12 10  8  2  4

cf 15 27 37 45 47 51

Chapter 13 Univariate data

441

Draw■the■ogive.■A■percentage■axis■will■be■ useful.

3

Find■the■upper■quartile■(75th■percentile)■and■ lower■quartile■(25th■percentile)■from■the■ogive.

4

The■interquartile■range■is■the■difference■ between■the■upper■and■lower■quartiles.

100%

50 40 30 20 10

75% 50% 25% 0 5 75 25 5 5 75 0.2 0. 1. 1.7 2.2 2. Volume (m3)

Cumulative frequency (%)

2

Cumulative frequency

statistics aND probability • Data represeNtatioN aND iNterpretatioN

Q3■=■1.6■m3 Q1■=■0.4■m3 IQR■=■Q3■-■Q1 =■1.6■-■0.4 =■1.2■m3

reMeMber

1.■ Range■=■highest■score■-■lowest■score or Range■=■Xmax■-■Xmin 2.■ The■difference■between■the■upper■and■lower■quartiles■is■called■the■interquartile■range,■ IQR.■IQR■=■Q3■-■Q1.■The■IQR■considers■the■middle■50%■of■the■data. 3.■ For■continuous■data,■the■quartiles■can■be■estimated■from■the■ogive. exercise

13b iNDiviDUal pathWays eBook plus

Activity 13-B-1

Range and quartiles doc-5131 Activity 13-B-2

Practice with range and quartiles doc-5132 Activity 13-B-3

Range and quartiles in depth doc-5133

442

Measures of spread flUeNcy 1 We4 ■Find■the■range■for■each■of■the■following■sets■of■data. ■ 15 a 4,■3,■9,■12,■8,■17,■2,■16 b 49.5,■13.7,■12.3,■36.5,■89.4,■27.8,■53.4,■66.8 77.1 1

3

1

2

1

3

c 7 2 ,■12 4 ,■5 4 ,■8 3 ,■9 6 ,■3 4 9 2 We5 ■Calculate■the■interquartile■range■(IQR)■for■the■following■sets■of■data. ■ 7 a 3,■5,■8,■9,■12,■14 7 b 7,■10,■11,■14,■17,■23 8.5 c 66,■68,■68,■70,■71,■74,■79,■80 39 d 19,■25,■72,■44,■68,■24,■51,■59,■36 3 The■following■stem-and-leaf■plot■shows■the■mass■of■newborn■babies■(rounded■to■the■nearest■

100■g).■Find■the: a range■of■the■data ■ 3.3■kg b IQR■of■the■data. 1.5■kg

Maths Quest 10 for the australian curriculum

Key:■ 1*  |  9■=■1.9■kg Stem Leaf 1* 9 2 2■4 2* 6■7■8■9 3 0■0■1■2■3■4 3* 5■5■6■7■8■8■8■9 4 0■1■3■4■4 4* 5■6■6■8■9 5 0■1■2■2

Cumulative frequency

4 Use the ogive at right to determine the interquartile range of the data. 22 cm

Cumulative frequency (%)

statistics AND probability • data representation and interpretation

100%

50 40 30

50%

20 10

100 120 140 160 180 Height (cm)

5   WE 6  The following frequency distribution table gives the amount of time spent by 50 people on shopping for Christmas presents. Estimate the IQR of the data. 0.8

Time (h)

0 –<0.5 0.5 –<1

Frequency

1

1–<1.5

1.5 –<2

2–<2.5

7

15

13

2

2.5 –<3 3 –<3.5 3.5 –<4 8

2

2

6   MC  Calculate the interquartile range of the following data: A 8

B 18

17, 18, 18, 19, 20, 21, 21, 23, 25 ✔ C 4

D 20

Understanding 7 The following frequency distribution table shows the life expectancy of 40 household batteries.

50 – <55

55 – <60

60 – <65

65 – <70

70 – <75

75 – <80

Frequency

4

10

12

8

5

1

Draw an ogive curve which represents the data in the table above. Use the ogive to answer the following questions.    i What is the median score?   62.5   Q1 = 58, Q3 = 67  ii What are the upper and lower quartiles?  9 iii What is the interquartile range?   14 iv How many batteries lasted less than 60 hours?  6  v How many batteries lasted 70 hours or more? 8 Calculate the IQR for the following data. a b

Cumulative frequency



55 50 45 40 35 30 25 20 15 10 5 0

120 130 140 150 160 170 180 190 200

Class interval

IQR = 24

Class interval

Frequency

120 – <130

 2

130 – <140

 3

140 – <150

 9

150 – <160

14

160 – <170

10

170 – <180

 8

180 – <190

 6

190 – <200

 3

Cumulative frequency

Life (hours)

40 35 30 25 20 15 10 5 0

50 55 60 65 70 75 80

Battery life (h)

9 For each of the following sets of data, state: i the range and ii the IQR of each set. a 6, 9, 12, 13, 20, 22, 26, 29 a i  Range = 23 ii IQR = 13.5 b 7, 15, 2, 26, 47, 19, 9, 33, 38 b i  Range = 45 ii IQR = 27.5 c 120, 99, 101, 136, 119, 87, 123, 115, 107, 100 c i  Range = 49 ii IQR = 20 Chapter 13 Univariate data

443

statistics aND probability • Data represeNtatioN aND iNterpretatioN reasoNiNG 10 As■newly■appointed■coach■of■Terrorolo’s■Meteors■netball■team,■Kate■decided■to■record■each■

player’s■statistics■for■the■previous■season.■The■number■of■goals■scored■by■the■leading■goal■ shooter■was: 1,■3,■8,■18,■19,■23,■25,■25,■25,■26,■27,■28,■ 28,■28,■28,■29,■29,■30,■30,■33,■35,■36,■37,■40. a Find■the■mean■of■the■data.■ b■ Find■the■median■of■the■data. 28 ■ 25.5 c Find■the■range■of■the■data.■ 39 d■ Find■the■interquartile■range■of■the■data. 6 e There■are■three■scores■that■are■much■lower■than■most.■Explain■the■effect■these■scores■have■ on■the■summary■statistics. T he■three■lower■scores■affect■the■mean■but■not■the■median■or■mode. 11 The■following■back-to-back■stem-and-leaf■plot■shows■the■ages■of■30■pairs■of■men■and■women■

when■entering■their■fi■rst■marriage. ■ ■Men:■ ■mean■=■32.3;■median■=■32.5;■range■=■38;■IQR■=■14 Key:■ 1■■|■■6■=■16■years■old ■ ■ ■ Women:■ ■mean■=■29.13;■median■=■27.5;■range■=■36;■IQR■=■13 Leaf Stem Leaf  Men Women 9■9■8 1 6■7■7■8■9 9■9■8■8■7■6■4■4■3■2■0 2 0■0■1■2■3■4■5■6■7■7■8■9 reflectioN    9■8■8■8■6■5■5■4■3■2 3 0■1■2■2■3■4■7■9 What do measures of spread 6■3■0■0 4 1■2■4■8 tell us about a set of data? 6■0 5 2 a Find■the■mean,■median,■range■and■interquartile■range■of■each■set. b Write■a■short■paragraph■comparing■the■two■distributions. T ypically,■women■marry■

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WorkSHEET 13.1 doc-5311

13c

younger■than■men,■although■ the■spread■of■ages■is■similar.

box-and-whisker plots five-point summary ■■ ■■

A■fi■ve-point■summary■is■a■list■consisting■of■the■lowest■score,■lower■quartile,■median,■upper■ quartile■and■greatest■score■of■a■set■of■data. A■fi■ve-point■summary■gives■information■about■the■spread■of■a■set■of■data,■as■shown■in■the■ example■below. Xmin Q1 Median■(Q2) Q3 Xmax 4

15

21

23

28

WorkeD exaMple 7

From the following five-point summary: 29 37 39 44 48 find: a the median b the interquartile range thiNk

Write

The■fi■gures■are■presented■in■the■order■of■lowest■score,■ lower■quartile,■median,■upper■quartile,■greatest■score.

Xmin■=■29,■Q1■=■37,■median■=■39,■Q3■=■44,■ Xmax■=■48

a The■median■is■39.

a Median■=■39

b The■interquartile■range■is■the■difference■between■

b IQR■=■Q3■-■Q1

the■upper■and■lower■quartiles. c The■range■is■the■difference■between■the■greatest■

score■and■the■lowest■score.

444

c the range.

Maths Quest 10 for the australian curriculum

=■44■-■37 =■7

c

Range■=■Xmax■-■Xmin =■48■-■29 =■19

statistics AND probability • data representation and interpretation

Box-and-whisker plots ■■ ■■ ■■ ■■ ■■

A box-and-whisker plot (or boxplot) is a graph of the five-point summary. It is a powerful way to show the spread of data. Box-and-whisker plots consist of a central divided box with attached whiskers. The box spans the interquartile range. The median is marked by a vertical line inside the box. The whiskers indicate the range of scores:

Indicates the lowest score Xmin ■■ ■■

Indicates the lower quartile Q1

Indicates the median Med.

Indicates the upper quartile Q3

Indicates the greatest score Xmax

Box-and-whisker plots are always drawn to scale. 4 15 21 23 28 They are presented either with the five-point summary figures attached as labels (diagram at right) or with a scale presented alongside the box-and-whisker plot like the diagram below. They can also be drawn vertically. 0

5

10

15

20

25

30

Scale

Identification of extreme values ■■ ■■

Extreme values (outliers) often make the whiskers appear longer than they should and hence give the appearance that the data are spread over a much greater range than they really are. If an extreme value or outlier occurs in a set of data it can be denoted by a small cross on the box-and-whisker plot. The whisker is then shortened to the next largest (or smallest) figure. The box-and-whisker plot below shows that the lowest score was 5. This was an extreme value as the rest of the scores were located within the range 15 to 42. 0

5

10

15

20

25

30

35

40

45 Scale

ì Worked Example 8

The following stem-and-leaf plot gives the speed of 25 cars caught by a roadside speed camera. Key:  8  |  2 = 82 km/h, 8*  |  6 = 86 km/h Stem Leaf  8 2  2  4  4  4  4   8* 5  5  6  6  7  9  9  9  9 0  1  1  2  4   9* 5  6  9 10 0  2 10* 11 4 a Prepare a five-point summary of the data. b Draw a box-and-whisker plot of the data. (Identify any extreme values.) c Describe the distribution of the data. Chapter 13 Univariate data

445

statistics AND probability • data representation and interpretation

Think 1

2

25+1 th 2

First identify the positions of the median and upper and lower quartiles. There are 25 pieces of data. The median is the n +1 th score. The lower quartile is the 2 median of the lower half of the data. The upper quartile is the median of the upper half of the data (each half contains 12 scores).

The median is the

Mark the position of the median and upper and lower quartiles on the stemand-leaf plot.

Key:  8  |  2 = 82 km/h 8*  |  6 = 86 km/h

score — that is, the13th score.

The Q1 is the 12 +1 th score in the lower half — that is, 2

the 6.5th score. That is, halfway between the 6th and 7th scores. The Q3 is halfway between the 6th and 7th scores in the upper half of the data.

Stem  8   8*  9   9* 10 10* 11

Q1 Leaf Median 2  2  4  4  4  4| 5  5  6  6  7  9  9  9 0  1  1  2  4| 5  6  9 0  2 Q3 4

a

Write the five-point summary: The lowest score is 82. The lower quartile is between 84 and 85; that is, 84.5. The median is 89. The upper quartile is between 94 and 95; that is, 94.5. The greatest score is 114.

a Five-point summary: 82, 84.5, 89, 94.5, 114

b

Start by ruling a suitable scale. Remember to include the units of measurement. The box represents the interquartile range and thus runs from 84.5 to 94.5. The median is a vertical line in the box at 89. The whiskers should extend to the lowest score (82) and the highest score (114). But the score 114 is a great deal higher than any of the others in the set and might be regarded as an extreme value. It should be indicated by a cross and the whisker will extend only as far as 102 (the second largest number in the set).

b

Even when the extreme value is excluded the data appear to be skewed with high values being spread over a much greater range.

c

c

446

Write

Maths Quest 10 for the Australian Curriculum

80

90

100

110

km/h ì

The data are skewed (positively) and include one extremely high value.

statistics aND probability • Data represeNtatioN aND iNterpretatioN

reMeMber

1.■ A■fi ve-point summary■is■a■list■consisting■of■the■lowest■score,■lower■quartile,■median,■ upper■quartile■and■greatest■score■of■a■set■of■data. 2.■ A■box-and-whisker plot■is■a■graphical■representation■of■a■fi■ve-point■summary■and■is■a■ powerful■tool■to■show■the■spread■of■data. 3.■ The■box■spans■the■interquartile■range;■the■median■is■marked■by■a■vertical■line■inside■the■ box■and■the■whiskers■extend■to■the■lowest■and■greatest■scores. 4.■ Box-and-whisker■plots■are■always■drawn to scale. 5.■ If■an■extreme■value■(outlier)■occurs■in■a■set■of■data,■it■can■be■denoted■by■a■small■cross;■ the■whisker■is■then■shortened■to■the■next■largest■(or■smallest)■value. exercise

13c iNDiviDUal pathWays

box-and-whisker plots flUeNcy 1 We7 ■From■the■following■fi■ve-point■summary■fi■nd:

6,■11,■13,■16,■32

eBook plus

Activity 13-C-1

Constructing boxplots doc-5134 Activity 13-C-2

Boxplots and outliers doc-5135 Activity 13-C-3

Boxplots with decimals doc-5136

a the■median ■ 13 b the■interquartile■range 5 c the■range. 26 2 From■the■following■fi■ve-point■summary■fi■nd:

101,■119,■122,■125,■128 a the■median ■ 122 6 b the■interquartile■range c the■range. 27 3 From■the■following■fi■ve-point■summary■fi■nd:

39.2,■46.5,■49.0,■52.3,■57.8 ■ 49.0 a the■median b the■interquartile■range 5.8 18.6 c the■range. 4 The■box-and-whisker■plot■

50

70

90

110

130

150 Points

at■right■shows■the■distribution■ of■fi■nal■points■scored■by■a■ football■team■over■a■season’s■ roster. a What■was■the■team’s■greatest■points■score? ■ 140 56 b What■was■the■team’s■least■points■score? 90 c What■was■the■team’s■median■points■score? 84 d What■was■the■range■of■points■scored? 26 e What■was■the■interquartile■range■of■points■scored? 5 The■box-and-whisker■plot■at■right■shows■ Number of 30 35 40 45 50 55 60 honey bears the■distribution■of■data■formed■by■counting■ the■number■of■honey■bears■in■each■of■a■ large■sample■of■packs. a What■was■the■largest■number■of■honey■ bears■in■any■pack? ■ 58 b What■was■the■smallest■number■of■honey■bears■in■any■pack? 31 c What■was■the■median■number■of■honey■bears■in■any■pack? 43 d What■was■the■range■of■numbers■of■honey■bears■per■pack? 27 e What■was■the■interquartile■range■of■honey■bears■per■pack? 7 chapter 13 Univariate data

447

statistics AND probability • data representation and interpretation

Questions 6 to 8 refer to the following box-and-whisker plot. 5

10

15

20

25

30 Score

ì 6   MC  The median of the data is: A 20 ✔ B 23

C 25

7   MC  The interquartile range of the data is: A 23 B 26

✔ C

D 31

5

D 20 to 25

8   MC  Which of the following is not true of the data represented by the box-and-whisker

plot? One-quarter of the scores are between 5 and 20. Half of the scores are between 20 and 25. The lowest quarter of the data is spread over a wide range. Most of the data are contained between the scores of 5 and 20.

a b c ✔ d



Understanding 9 The number of sales made each day by a salesperson is recorded over a 2-week period:

25, 31, 28, 43, 37, 43, 22, 45, 48, 33 a Prepare a five-point summary of the data. (There is no need to draw a stem-and-leaf plot of the data. Just arrange them in order of size.) (22, 28, 35, 43, 48) 20 30 40 50 Sales b Draw a box-and-whisker plot of the data. 10 The data below show monthly rainfall in millimetres.



0 10 20 30 40 50 Rainfall (mm)

J

F

M

A

M

J

J

A

S

O

N

D

10

12

21

23

39

22

15

11

22

37

45

30

a Prepare a five-point summary of the data. (10, 13.5, 22, 33.5, 45) b Draw a box-and-whisker plot of the data.

Key:  1  |  8 = 18 years Stem Leaf 1 8  8  9  9  9 2 0  0  0  1  1  3  4  6  9 3 0  1  2  7 4 2  5 5 3  6  8 The distribution is positively skewed, with most of the offenders 6 6 being young drivers. 7 4

11   WE 8  The stem-and-leaf plot at right details the age 10

30

50

70 Age

of 25 offenders who were caught during ■ (18, 20, 26, 43.5, 74) random breath testing. a Prepare a five-point summary of the data. b Draw a box-and-whisker plot of the data. c Describe the distribution of the data.

12 The following stem-and-leaf plot details the price at which 30 blocks of land in a particular

suburb sold for.

Key:  12  |  4 = $124  000 Stem Leaf 12 4  7  9 13 0  0  2  5  5 14 0  0  2  3  5  5  7  9  9 15 0  0  2  3  7  7  8 16 0  2  2  5  8 17 5

a Prepare a five-point summary of the data. (124  000, 135  000, 148  000, 157  000, 175  000) b Draw a box-and-whisker plot of the data. (You might like to use a calculator for this question.) 120 140 160 180 ($ì1000)

448

Maths Quest 10 for the Australian Curriculum

statistics AND probability • data representation and interpretation Key:  12  |  1 = 121 Reasoning Stem Leaf 13 The following data detail the number of hamburgers 12 1  5  6  9 sold by a fast food outlet every day over a 4-week ■ 13 1  2  4 14 3  4  8  8 period. 15 0  2  2  2  5  7 M T W T F S S 16 3  5 17 2  9 125 144 132 148 187 172 181 18 1  1  1  2  3  7  8

134

157

152

126

155

183

188

131

121

165

129

143

182

181

152

163

150

148

152

179

181

B f

D f

Size

13d

T  he distribution is positively skewed, with first-time mothers being under the age of 30. There is one outlier (48) in this group.

Size

Size ✔ C f

Age

A f

ì

22 21 18 33 17 23 22 24 24 20 25 29 32 18 19 22 23 24 28 20 31 22 19 17 23 48 25 18 23 20 a Prepare a stem-and-leaf plot of the data. (Use a class size of 5.) b Draw a box-and-whisker plot of the data. Indicate any extreme values appropriately. c Describe the distribution in words. What does the distribution say about the age that mothers have their first baby? (You might like to use a calculator for this question.) 15   MC  Match the box-and-whisker plot at right with its most likely histogram.

15 25 35 45

Key:  1*  |  7 = 17 years Stem Leaf 1* 7  7  8  8  8  9  9 2 0  0  0  1  2  2  2  2  3  3  3  3  4  4  4 2* 5  5  8  9 3 1  2  3 3* 4 4* 8

a Prepare a stem-and-leaf plot of the data. On most days the (Use a class size of 10.) 120 140 160 180 Number sold hamburger sales are b Draw a box-and-whisker plot of the data. less than 160. Over the (You might like to use a calculator for this question.) weekend the sales figures c What do these graphs tell you about hamburger sales? spike beyond this. 14 The following data show the ages of 30 mothers upon the birth of their first baby.

reflection 

Size

 

What advantages and disadvantages do box-and-whisker plots have as a visual form of representing data?

The standard deviation ■■ ■■ ■■ ■■ ■■

The standard deviation is the most useful measure of the spread of a data set. This is because every score in the data set is used to calculate the standard deviation. The standard deviation shows how much dispersion there is from the mean. A low standard deviation indicates that the data values tend to be close to the mean. A high standard deviation indicates that the data values tend to be spread out over a large range. How the standard deviation is calculated is beyond this course; however, for our purposes we can obtain the result using either a scientific or graphics calculator. Chapter 13 Univariate data

449

statistics AND probability • data representation and interpretation

■■

The standard deviation (s ) can be found using the statistics function of your calculator. Enter the scores into the calculator using the statistics function. The standard deviation can be obtained using the sx function. This will vary between different models of calculator and you may need to check with your teacher to find out how to retrieve the standard deviation on your calculator.

Worked Example 9

For each of the following data sets find the standard deviation. a  7, 5, 6, 3, 4, 8, 2, 3, 7, 8 b 

Score

Frequency

1

 2

2

13

3

14

4

20

5

 1

Think a

b

Write

1

Enter the scores into the statistics mode or spreadsheet page of your scientific or graphics calculator.

2

Write the answer from your calculator.

1

Enter the scores into the statistics mode or spreadsheet page of your scientific or graphics calculator.

2

Write the answer from your calculator.

■■ ■■

a

s = 2.1 b

s = 0.94

Standard deviation can be used to measure consistency. When the standard deviation is low we are able to say that the scores in the data set are more consistent with each other.

Worked Example 10

The price of a fertiliser spray, in cents per litre, is recorded at 8 independent outlets in two different suburbs.   Suburb A: 93.9   97.9   92.4   93.9   98.5   92.3   97.9   99.9   Suburb B: 95.9   94.9   96.9   93.9   97.9   93.5   96.9   98.9 In which suburb is the price of the fertiliser spray more consistent? Think

450

Write

1

Enter the statistics for Suburb A into your calculator and retrieve the standard deviation.

s = 3.02

2

Enter the statistics for Suburb B into your calculator and retrieve the standard deviation.

s = 1.91

3

Make a conclusion based on which Suburb has the lower standard deviation.

The price of the fertiliser spray is more consistent in Suburb B.

Maths Quest 10 for the Australian Curriculum

statistics aND probability • Data represeNtatioN aND iNterpretatioN

reMeMber

1.■ The■standard■deviation■is■the■most■useful■measure■of■spread. 2.■ The■standard■deviation■is■found■by■entering■the■data■set■into■the■statistics■function■of■ your■scientifi■c■or■graphics■calculator. 3.■ A■lower■standard■deviation■indicates■that■the■data■are■more■bunched■or■clustered■while■ a■higher■standard■deviation■indicates■that■the■data■are■more■spread■out. 4.■ Standard■deviation■can■be■used■to■measure■consistency■(a■low■standard■deviation■ indicating■greater■consistency.) exercise

13D iNDiviDUal pathWays eBook plus

Activity 13-D-1

Standard deviation doc-5137

the standard deviation flUeNcy 1 We9a ■Find■the■standard■deviation■of■each■of■the■following■data■sets. a 3,■5,■8,■2,■7,■1,■6,■5 b 11,■8,■7,■12,■10,■11,■14 ■ 2.29 2.19 c 25,■15,■78,■35,■56,■41,■17,■24 d 5.2,■4.7,■5.1,■12.6,■4.8 20.17 3.07 2 We9b ■Find■the■standard■deviation■of■each■of■the■following■data■sets. a

b

Score

Frequency

Score

Frequency

Practice with standard deviation doc-5138

1

1

16

15

2

5

17

24

Activity 13-D-3

3

9

18

26

4

7

19

28

5

3

20

27

Score

Frequency

Score

Frequency

■8

15

65

15

10

19

66

15

12

18

67

16

14

■7

68

17

16

■6

69

16

18

■2

70

15

71

15

72

12

Activity 13-D-2

Standard deviation in depth doc-5139

c

d

2 a■ 1.03 b 1.33 c 2.67 d 2.22

3 Complete■the■following■frequency■distribution■table■and■use■the■table■to■fi■nd■the■standard■ deviation■of■the■data■set. 10.82

Class

Class centre

Frequency

■ 1–10

■6

11–20

15

21–30

25

31–40

■8

41–50

■6 chapter 13 Univariate data

451

statistics AND probability • data representation and interpretation 4 First-quarter profit increases for 8 leading companies are given below as percentages.

2.3  0.8  1.6  2.1  1.7  1.3  1.4  1.9 Calculate the standard deviation for this set of data and express your answer correct to 2 decimal places. 0.45% 5 The heights in metres of a group of army recruits are given below. 1.8    1.95    1.87    1.77    1.75    1.79    1.81    1.83    1.76    1.80    1.92    1.87    1.85    1.83 Calculate the standard deviation for this set of data and express your answer correct to 0.06 m 2 decimal places. 6 Times (to the nearest tenth of a second) for the heats in Key:  11  |  0 = 11.0 s the open 100 m sprint at the school sports are given at right. Stem Leaf Calculate the standard deviation for this set of data and ■ 11 0 express your answer correct to 2 decimal places. 0.49 s 11 2  3 11 4  4  5 11 6  6 11 8  8  9 12 0  1 12 2  2  3 12 4  4 12 6 12 9

7 The number of outgoing phone calls from an office each day over a 4-week period is shown on

the stem plot below.

Key:  1  |  3 = 13 calls Stem Leaf 0 8  9 1 3  4  7  9 2 0  1  3  7  7 3 3  4 4 1  5  6  7  8 5 3  8

Calculate the standard deviation for this set of data and express your answer correct to 2 decimal places. 15.10 calls 8   MC  A new legal aid service has been operational for only 5 weeks. The number of people who have made use of the service each day during this period is set out below. Key:  1  |  6 = 16 people Stem Leaf 0 2  4 0 7  7  9 1 0  1  4  4  4  4 1 5  6  6  7  8  8  9 2 1  2  2  3  3  3 2 7 The standard deviation (to 2 decimal places) of these data is: ✔ B 6.34 C 6.47

A 6.00 452

Maths Quest 10 for the Australian Curriculum

D 15.44

statistics aND probability • Data represeNtatioN aND iNterpretatioN UNDerstaNDiNG 9 We10 ■The■following■data■represent■the■scores■of■2■golfers■on■10■rounds■of■golf■at■the■same■

course. Greg:■ 65■ 74■ 76■ 68■ 72■ 77■ 66■ 69■ 74■ 70 Adam■is■more■consistent■because■he■ Adam:■ 72■ 71■ 72■ 73■ 74■ 70■ 71■ 68■ 69■ 71 has■the■lower■standard■deviation. Which■golfer■is■the■more■consistent■player? (1.7■compared■with■3.9) 10 Mc ■The■data■given■below■represent■Anna’s■marks■in■each■of■fi■ve■tests■in■four■different■ subjects.■In■which■subject■does■Anna■achieve■the■most■consistent■results? A English:■■ 60■ 82■ 75■ ■ 47■ 90 B Maths:■ 72■ 74■ 35■ 100■ 89 ✔ C Science:■ 50■ 57■ 65■ ■ 46■ 50 D Geography:■ 60■ 70■ 65■ ■ 85■ 79 reasoNiNG 11 Two■classes■of■students,■each■with■30■students,■sit■for■a■short■quiz.■The■results■of■the■quiz■are■

shown■in■the■tables■below.■Each■mark■is■out■of■10. Class■A

Class■B

Mark

Frequency

Mark

Frequency

■5

■4

■5

8

■6

■9

■6

7

■7

12

■7

6

■8

■3

■8

3

■9

■1

■9

1

C lass■A■is■more■consistent■ because■the■standard■ 10 ■1 10 5 deviant■is■lower. a For■each■class■fi■nd■the■standard■deviation■of■the■marks. ■Class■A:■1.13;■Class■B:■1.74 b Which■class■had■the■more■consistent■results?■Justify■your■answer. 12 A■company■that■makes■batteries■needs■to■test■a■batch■of■batteries■to■determine■if■they■are■of■

satisfactory■quality.■The■results■of■the■testing■are■shown■below.

The■batch■is■ unsatisfactory.■ Although■the■mean■is■ greater■than■13■hours,■ the■batch■fails■as■the■ standard■deviation■ is■greater■than■the■ required■6■hours.

eBook plus

Digital doc

WorkSHEET 13.2 doc-5318

Life of battery (hours)

Class centre

Frequency

0–<5

■ 2.5

■6

5–<10

■ 7.5

16

10–<15

12.5

18

15–<20

17.5

15

20–<25

22.5

■5

25–<30

27.5

■5

a Complete■the■class■centre■column■of■the■frequency■distribution■table. b Find■the■mean■and■standard■deviation■of■the■data. Mean■=■13.4,■standard■deviation■=■6.73 c For■the■batch■to■be■considered■satisfactory,■the■ mean■of■the■distribution■must■be■greater■than■ reflectioN   

13■hours■and■the■standard■deviation■must■be■less■ than■6■hours.■Determine■if■this■is■a■satisfactory■ batch■and■explain■your■answer.

What does the standard deviation tell us about a set of data?

chapter 13 Univariate data

453

statistics aND probability • Data represeNtatioN aND iNterpretatioN

13e eBook plus

Interactivity Parallel boxplots

comparing data sets When■multiple■data■displays■are■used■to■display■similar■sets■of■data,■comparisons■and■ conclusions■can■then■be■drawn■about■the■data. ■■ We■can■use■back-to-back stem-and-leaf plots■and■multiple■or■parallel■box-and-whisker■plots■ to■help■compare■statistics■such■as■the■median,■range■and■interquartile■range. ■■

int-2788

WorkeD exaMple 11

A shop compares the number of customers on weekdays and weekends. Twenty weekdays and twenty weekend days are chosen for the sample. The back-to-back stem-and-leaf plot at right shows the result. a Find the median number of customers on weekdays and weekends. b Calculate the range of customer numbers on weekdays and weekends. c What conclusions can be made from the display about the average number of customers on weekdays and weekends?

1 | 2 = 12 customers Leaf  Stem Weekday 7 0 86311 1 9666554331 2 952 3 5 4 5

Key:

thiNk

7

Write

a There■are■20■scores■in■each■set■and■so■the■median■

24 + 25 2 =■24.5■ 16 + 16 Weekends:■Median■=■ 2 =■16

a Weekdays:■Median■=■

will■be■the■average■of■the■10th■and■11th■scores.

b For■each■data■set,■subtract■the■lowest■score■from■

b Weekdays:■Range■=■45■-■7

=■38■ Weekends:■Range■=■57■-■7 =■50

the■highest■score.

c Write■your■conclusion■from■observing■that■there■

are■generally■fewer■customers■on■weekends■and■the■ results■are■more■consistent■except■for■one■outlier.

■■ ■■ ■■ ■■

454

Leaf  Weekend 788 1124456667 2558 16

c

There■are■generally■fewer■customers■ on■weekends.■There■is■one■outlier■in■ the■weekend■scores,■causing■the■range■ to■be■larger.■However,■apart■from■this■ outlier,■the■weekend■scores■are■less■ spread■out.

In■the■above■worked■example■we■could■have■found■the■interquartile■range■as■further■evidence■ that■the■scores■on■weekends■are■generally■less■spread■out. The■most■common■method■for■comparing■data■sets■is■to■compare■the■summary■statistics■from■ the■data■sets. The■measures■of■centre■such■as■mean■and■median■are■used■to■compare■the■typical■score■in■a■ data■set. Measures■of■spread■such■as■range,■interquartile■range■and■standard■deviation■are■used■to■make■ assessments■about■the■consistency■of■scores■in■the■data■set.

Maths Quest 10 for the australian curriculum

statistics aND probability • Data represeNtatioN aND iNterpretatioN

WorkeD exaMple 12

Below are the scores for two students in eight mathematics tests throughout the year. John: 45, 62, 64, 55, 58, 51, 59, 62 Penny: 84, 37, 45, 80, 74, 44, 46, 50 a Use the statistics function on a calculator to find the mean and standard deviation for each student. b Which student had the better overall performance on the eight tests? c Which student was more consistent over the eight tests? thiNk

Write

a Enter■the■statistics■into■a■calculator■and■use■the■x■

a John:■

x■=■57,■s ■=■6 Penny:■ x■=■57.5,■s ■=■17.4

function■for■the■mean■and■the■s■function■for■the■ standard■deviation. b Compare■the■mean■for■each■student.■The■student■

b Penny■performed■slightly■better■overall■as■

with■the■higher■mean■performed■better■overall.

her■mean■mark■was■higher■than■John’s.

c Compare■the■standard■deviation■for■each■student.■

The■student■with■the■lower■standard■deviation■ performed■more■consistently.

c

John■was■the■more■consistent■student■ because■his■standard■deviation■was■much■ lower■than■Penny’s.

reMeMber

1.■ When■multiple■displays■are■used■for■two■or■more■sets■of■data,■we■can■compare■and■ contrast■the■data■sets■and■determine■whether■any■relationship■exists■between■them.■ 2.■ A■multiple■stem-and-leaf■plot■allows■for■a■quick■comparison■of■the■data■from■which■we■ can■easily■compare■medians,■ranges■and■interquartile■ranges. 3.■ The■summary■statistics■from■two■data■sets■can■be■compared■quickly■on■a■box-andwhisker■plot. 4.■ The■most■commonly■used■comparisons■are■summary■statistics■to■compare■what■is■a■ typical■score■and■what■the■spread■of■the■data■is. exercise

13e iNDiviDUal pathWays eBook plus

Activity 13-E-1

Comparing data 1 doc-5140 Activity 13-E-2

Comparing data 2 doc-5141 Activity 13-E-3

Comparing data 3 doc-5142

comparing data sets UNDerstaNDiNG 1 We11 ■The■back-to-back■stem-and-leaf■plot■drawn■below■shows■the■number■of■days■that■both■a■

group■of■boys■and■girls■were■absent■from■school■over■a■two-year■period. Key:■ 2■■|■■1■=■21■days Leaf  Stem Boys 0 7■4■1■0 1 9■9■7■6■6■5■3■1■1■0 2 8■7■7■5■2 3 2 4 5

Leaf  Girls 1■7■ 2■4■7■9■9■ 1■3■3■4■6■6 4■4■4■8 3■6■ 4 Boys:■median■=■26;■girls:■median■=■23.5

oth■sets■have■similar■ a Calculate■the■median■number■of■days■absent■for■both■boys■and■girls. B medians■but■the■girls■ b Calculate■the■range■for■both■boys■and■girls. Boys:■range■=■32;■girls:■range■=■53 have■a■far■greater■ c Comment■on■the■distribution■of■days■absent■for■each■group. range■of■absenteeism■ than■boys. chapter 13 Univariate data

455

statistics AND probability • data representation and interpretation 2 A bank surveys the average morning and afternoon waiting times for customers. The figures

were taken each Monday to Friday in the morning and afternoon for one month. The stem-andleaf plot below shows the results. a Morning: median = 2.45; afternoon: median = 1.6 b M  orning: range = 3.8; afternoon: range = 5 c T  he waiting time is generally shorter in the afternoon. One outlier in the afternoon data causes the range to be larger. Otherwise the afternoon data are far less spread out.

Key:  1  |  2 = 1.2 minutes Leaf  Stem Morning 7 0 86311 1 9666554331 2 952 3 5 4 5

Leaf   Afternoon 788 1124456667 2558 16 7

a Find the median morning waiting time and the median afternoon waiting time. b Calculate the range for morning waiting times and the range for afternoon waiting times. c What conclusions can be made from the display about the average waiting time at the

bank in the morning compared with the afternoon? 3 In a class of 30 students there are 15 boys and 15 girls. Their heights are measured (in metres) and are listed below. Key:  16  |  1 = 1.61 m Boys: 1.65, 1.71, 1.59, 1.74, 1.66, 1.69, 1.72, 1.66, ■ 1.65, 1.64, 1.68, 1.74, 1.57, 1.59, 1.60 Girls:  1.66, 1.69, 1.58, 1.55, 1.51, 1.56, 1.64, 1.69, ■ 1.70, 1.57, 1.52, 1.58, 1.64, 1.68, 1.67

Leaf  Boys 997 98665540 4421

Stem 15 16 17

Leaf   Girls 1256788 4467899 0

Display this information in a back-to-back stem-and-leaf plot. Key:  1  |  5 = 15 vehicles number of vehicles sold by the Ford and Holden Leaf  Stem Leaf   dealerships in a Sydney suburb each week for a threeFord Holden month period. 74 0 39 a State the median of both distributions. 952210 1 111668 b Calculate the range of both distributions. 8544 2 2279 c Calculate the interquartile range of both distributions. 0 3 5 d Show both distributions on a box-and-whisker plot.

4 The stem-and-leaf plot at right is used to display the

a Ford: median = 15; Holden: median = 16 b F  ord: range = 26; Holden: range = 32 c F  ord: IQR = 14; Holden: IQR = 13.5 d Ford Holden 0 5 10 15 20 2530 35 40

456

Maths Quest 10 for the Australian Curriculum

statistics AND probability • data representation and interpretation 5 The box-and-whisker plot drawn below displays statistical data for two AFL teams over a

season. a Brisbane Lions b B  risbane Lions: range = 63; Sydney Swans: range = 55 c B  risbane Lions: IQR = 40; Sydney Swans: IQR = 35

Sydney Swans Brisbane Lions 50 60 70 80 90 100 110 120 130 140 150 Points

a Which team had the higher median score? b What was the range of scores for each team? c For each team calculate the interquartile range. 6 Tanya measures the heights (in m) of a group of Year 10 boys and girls and produces the

following five-point summaries for each data set.  lthough boys and girls A have the same median height, the spread of heights is greater among boys as shown by the greater range and interquartile range. 7

There are generally more cold drinks sold in summer as shown by the higher median. The spread of data is similar as shown by the IQR although the range in winter is greater. 8

✔

9

Girls Boys

Boys:  1.45, 1.56, 1.62, 1.70, 1.81 Girls:  1.50, 1.55, 1.62, 1.66, 1.73

1.4 1.5 1.6 1.7 1.8 1.9 Height

Draw a box-and-whisker plot for both sets of data and display them on the same scale. What is the median of each distribution? Boys: median = 1.62; girls: median = 1.62 What is the range of each distribution? Boys: range = 0.36; girls: range = 0.23 What is the interquartile range for each distribution? Boys: IQR = 0.14; girls: IQR = 0.11 Comment on the spread of the heights among the boys and the girls. The box-and-whisker plots at right Summer show the average daily sales of cold drinks at the school canteen in summer Winter and winter. a Calculate the range of sales in 0 5 10 15 20 25 30 35 40 Daily sales of cold both summer and winter. drinks Summer: range = 23; b Calculate the interquartile range winter: range = 31 of the sales in both summer and winter. Summer: IQR = 14; winter: IQR = 11 c Comment on the relationship between the two data sets, both in terms of measures of centre and measures of spread.   MC  Andrea surveys the age of people at Movie A two movies being shown at a local cinema. The box-and-whisker plot at right shows the Movie B results. 0 10 20 30 40 50 60 70 80 Age Which of the following conclusions could be drawn based on the above information? A Movie A attracts an older audience than Movie B. B Movie B attracts an older audience than Movie A. C Movie A appeals to a wider age group than Movie B. D Movie B appeals to a wider age group than Movie A.   MC  Note: There may be more than one correct answer. The figures below show the age of the first 10 men and women to finish a marathon. a b c d e

Men: 28, 34, 25, 36, 25, 35, 22, 23, 40, 24 Women:  19, 27, 20, 26, 30, 18, 28, 25, 28, 22 Which of the following statements is correct? The mean age of the men is greater than the mean age of the women. The range is greater among the men than among the women. C The interquartile range is greater among the men than among the women. D The standard deviation is greater among the men than among the women.

✔ A ✔ B ✔ ✔

Chapter 13 Univariate data

457

statistics aND probability • Data represeNtatioN aND iNterpretatioN reasoNiNG 10 We12 ■Cory■recorded■his■marks■for■each■test■that■he■did■in■English■and■Science■throughout■the■

year. English:■ 55,■64,■59,■56,■62,■54,■65,■50 Science:■ 35,■75,■81,■32,■37,■62,■77,■75

10 a■ ■Cory■achieved■a■better■average■mark■in■Science■ (59.25)■than■he■did■in■English■(58.125). b Cory■was■more■consistent■in■English■(s ■=■4.9)■ than■he■was■in■Science■(s ■=■19.7)

a In■which■subject■did■Cory■achieve■the■better■average■mark? b In■which■subject■was■Cory■more■consistent?■Explain■your■answer. 11 b The■drivers■are■ generally■driving■ faster■on■the■back■ street. c The■spread■of■ speeds■is■greater■ on■the■main■road■ as■indicated■by■the■ higher■standard■ deviation. 12

The■police■set■up■two■radar■speed■checks■on■a■back■street■of■Sydney■and■on■a■main■road.■In■ both■places■the■speed■limit■is■60■km/h.■The■results■of■the■fi■rst■10■cars■that■have■their■speed■ checked■are■given■below. ■Back■street:■x■=■61,■s ■=■4.3;■ Back■street:■ 60,■62,■58,■55,■59,■56,■65,■70,■61,■64 main■road:■x■=■58.8,■s ■=■12.1 Main■road:■ 55,■58,■59,■50,■40,■90,■54,■62,■60,■60 a Calculate■the■mean■and■standard■deviation■of■the■readings■taken■at■each■point. b On■which■road■are■drivers■generally■driving■faster? c On■which■road■is■the■spread■of■the■reading■taken■greater?■Justify■your■answer.

Nathan■and■Timana■are■wingers■in■their■local■rugby■league■team.■The■number■of■tries■they■have■ scored■in■each■season■are■listed■below. Nathan:■ 25,■23,■13,■36,■1,■8,■0,■9,■16,■20 Timana’s■lower■ Timana:■ 5,■10,■12,■14,■18,■11,■8,■14,■12,■19 Nathan:■mean■=■15.1;■Timana:■mean■=■12.3

range■and■IQR■ a Calculate■the■mean■number■of■tries■scored■by■each■player. shows■that■he■ is■the■more■ b What■is■the■range■of■tries■scored■by■each■player? Nathan:■range■=■36;■Timana:■range■=■14 consistent■player. c What■is■the■interquartile■range■of■tries■scored■by■each■player? Nathan:■IQR■=■15;■Timana:■IQR■=■4 d Which■player■would■you■consider■to■be■the■more■consistent■player?■Justify■your■answer.

13 In■boxes■of■Smarties■it■is■advertised■that■there■are■50■Smarties■in■each■box.■Two■machines■

are■used■to■distribute■the■Smarties■into■the■boxes.■The■results■from■a■sample■taken■from■each■ machine■are■shown■in■the■stem-and-leaf■plot■below.

a■

Key:■ 5■■|■■1■=■51■ ■ 5*■■|■■6■=■56 Leaf  Stem Machine A 4 4 9■9■8■7■7■6■6■5 4* 4■3■2■2■2■1■1■1■0■0■0■0■0■0 5 5■5 5*

Leaf  Machine B 5■7■8■9■9■9■9■9■9■9■9 0■0■0■0■0■1■1■1■1■1■2■2■3■ 9

Machine A Machine B 40 42 44 46 47 48 50 52 54 56 58 60 Number of Smarties in a box

b Machine■A:■mean■=■49.88,■ standard■deviation■=■2.88;■ Machine■B:■mean■=■50.12,■ standard■deviation■=■2.44 c Machine■B■is■more■reliable,■as■ shown■by■the■lower■standard■ deviation■and■IQR.■The■range■is■ greater■on■machine■B■only■because■ of■a■single■outlier. a Display■the■data■from■both■machines■on■parallel■box-and-whisker■plots. b Calculate■the■mean■and■standard■deviation■of■the■number■of■matches■distributed■from■

both■machines. c Which■machine■is■the■more■dependable?■Justify■your■answer. 458

Maths Quest 10 for the australian curriculum

statistics AND probability • data representation and interpretation 14 a Go to the internet or some other information source and find the maximum daily

temperature in two cities each day over a period of one month.

Student’s own work. b Enter the results into a spreadsheet. c Use the AVERAGE function to find the mean daily temperature in each location. d Use the STDEV function to find the standard deviation of the daily temperature in each

location. e Draw a line graph to show the maximum daily

temperature in each location. f Describe the similarities and differences of the temperature pattern in each location.

 

Skewness ■■ ■■

When comparing data sets, it is advisable to consider the ■ distribution of the data within the sets. Consider the graph that has been drawn at right. This graph is symmetrical, and we can see that the mean, median and mode are all equal to 3. The majority of scores are clustered around the mean. This is an example of a normal distribution.

Frequency

13f

reflection 

What statistical values should be used when comparing data sets?

10 9 8 7 6 5 4 3 2 1

■■

The second graph shows a data set in which the scores are not clustered and there are two modes at either end of the distribution. In this example, although it is still symmetrical there are two modes, 1 and 5, while the mean and median are still 3. This graph can be described as bi-modal.

Frequency

1 2 3 4 5 Score 10 9 8 7 6 5 4 3 2 1

1 2 3 4 5 Score

The figure at right shows the distribution of a set of scores on an exam. a Is the graph symmetrical? b What is the mode? c Can the mean and median be seen from the graph and, if so, what are their values?

Frequency

Worked Example 13 10 9 8 7 6 5 4 3 2 1

1617 18 19 20 Score Chapter 13 Univariate data

459

statistics AND probability • data representation and interpretation

Think

Write

a Consider the columns either side of the middle. If

a The graph is symmetrical.

they are equal, then the graph is symmetrical. b Look for the score or scores that occur the most

often. c Since the graph is symmetrical, the middle score

will be the mean and the median.

b Mode = 17 and 19 c

Mean = 18 Median = 18

When a graph is not symmetrical, the mean and median cannot be easily seen from the graph. ■■ Consider the graph at right where the scores are gathered to the lower end of the distribution. ■■ The way in which the data are gathered to one end of the distribution is called the skewness. ■■ When a greater number of scores is distributed at the lower end of the distribution, the data are said to be positively skewed.

Frequency

■■

10 9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 Score

Similarly, when most of the scores are distributed at the upper end, the data are said to be negatively skewed, as ■ shown at right. Frequency

■■

10 9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 Score

20 16 12 8 4 51–60 61–70 71–80 81–90 91–100

The distribution at right shows the results of a Mathematics exam at a certain school. a What is the modal class? b Describe the skewness of the data set.

Frequency

Worked Example 14

Exam score Think a Locate the modal class by considering which class

occurs most often. Which column is the largest? b Observe where the majority of data are located. In

this case it is at the upper end of the distribution.

460

Maths Quest 10 for the Australian Curriculum

Write a Modal class = 81 - 90 b The data are negatively skewed.

statistics aND probability • Data represeNtatioN aND iNterpretatioN

■■

There■are■many■reasons■why■a■data■set■may■be■skewed.■In■the■case■of■an■exam,■an■easier■exam■ may■lead■to■negatively■skewed■data■with■more■students■obtaining■a■higher■mark,■while■a■more■ diffi■cult■exam■may■lead■to■more■students■at■the■lower■end■of■the■distribution■and■hence■the■ data■will■be■positively■skewed. reMeMber

1.■ A■distribution■is■symmetrical■when■the■data■are■equally■distributed■around■the■mean. 2.■ When■the■data■are■symmetrical,■the■median■and■mean■will■both■be■the■middle■score. 3.■ When■the■data■are■clustered■around■the■mean,■the■standard■deviation■is■smaller. 4.■ When■the■majority■of■scores■are■at■the■lower■end■of■a■distribution,■it■is■said■to■be■ positively■skewed. 5.■ When■the■majority■of■scores■are■at■the■upper■end■of■the■distribution,■it■is■said■to■be■ negatively■skewed. exercise

skewness

13f

eBook plus

Activity 13-F-1

Skewed data doc-5143

1 We13 ■The■fi■gure■at■right■shows■the■distribution■of■a■set■of■scores. a Is■the■graph■symmetrical? Yes b What■is■the■mode? 8 c Can■the■mean■and■median■be■seen■from■the■graph■and,■ if■so,■what■are■their■values? Both■equal■8.

Frequency

flUeNcy

iNDiviDUal pathWays

5 4 3 2 1

Activity 13-F-2

6 7 8 9 10 Score 2 Consider■the■distribution■shown■at■right. a Are■the■data■symmetrical? No b What■is■the■modal■class? 31–40 c Can■the■mean■and■median■be■seen■from■the■graph■and,■ if■so,■what■are■their■values? No.■They■can,■however,■be■calculated.

Activity 13-F-3

30 25 20 15 10 5

1–10 11–20 21–30 31–40 41–50

Tricky skewed data doc-5145

Frequency

More skewed data doc-5144

Score

Frequency

3 The■table■below■shows■the■number■of■goals■scored■by■a■soccer■team■throughout■a■season.

7 6 5 4 3 2 1 0

1 2 3 4 5 Number of goals

a b c d

Number of goals

Frequency

0

5

1

7

2

7

3

7

4

7

5

5

Show■this■information■in■a■frequency■histogram. Are■the■data■symmetrical? Yes Yes.■Both■equal■2.5. What■is■the■mode? 1,■2,■3■and■4 Can■the■mean■and■median■be■seen■for■this■distribution■and,■if■so,■what■are■their■values?■ chapter 13 Univariate data

461

statistics AND probability • data representation and interpretation Understanding

Frequency

4   WE 14  Consider the distribution shown at right. a What is the modal class? 4 b Describe the skewness of the distribution. Negatively skewed

70 60 50 40 30 20 10

1 2 3 4 5 Score

35 30 25 20 15 10 5 1–10 11–20 21–30 31–40 41–50

Frequency

5 The table below shows the number of goals scored by a netball team throughout a season.

Number of goals

Number of goals

Frequency

  1-10

13

11–20

16

21–30

27

31–40

33

41–50

31

6   MC  The distribution represented by the graph at right is: A positively skewed ✔ B negatively skewed C symmetrical D normally distributed

Frequency

a Draw a frequency histogram of the data. b Describe the data set in terms of its skewness. Negatively skewed 10 9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 Score Reasoning 7 A movie is shown at 30 cinemas across Sydney on its opening day. The number of people 10 9 8 7 6 5 4 3 2 1 1–50 51–100 101–150 151–200 201–250

Frequency

attending at each cinema is shown in the table below.

Number of people

a b c d 462

Number of people

Frequency

   1–50

 3

  51–100

 2

101–150

 6

151–200

 9

201–250

10 The distribution is negatively

Present the data in a frequency histogram. skewed. Reasons could include Are the data symmetrical? No the size of cinemas or the target audience of the movie. What is the modal class? 201–250 Describe the skewness of the distribution and explain possible reasons for the skewness.

Maths Quest 10 for the Australian Curriculum

statistics aND probability • Data represeNtatioN aND iNterpretatioN

shown■in■the■table■below.

Hours

WorkSHEET 13.3 doc-5325

Number of students Maths

51–60

■7

■6

61–70

10

■7

71–80

■8

12

81–90

■8

■9

91–100

■2

■6

Maths■has■a■greater■ standard■deviation■ (12.6)■compared■to■ Science■(11.9).

10 A■new■drug■for■the■relief■of■cold■symptoms■has■been■developed.■To■test■the■drug,■40■people■

■■

10 20 30 40 50 60 70 Nouns

Group A

Group B

Digital doc

Number of students Science

9 Draw■an■example■of■a■graph■that■is: Answers■will■vary.■Check■with■your■teacher. a symmetrical b positively■skewed■with■one■mode c negatively■skewed■with■two■modes.

10 a■ Key:■ 2■■|■■3■=■2.3■hours Leaf Stem Group■A 8■ 7■ 3 1 9■ 5■ 1 2 8■ 7■ 5■ 4■ 2■ 2 3 7■ 5■ 4■ 2■ 2■ 2 4 5 2■ 2 6 eBook plus

Mark

a Is■either■distribution■symmetrical? No b If■either■distribution■is■not■symmetrical,■state■whether■it■is■positively■or■negatively■ skewed. Science:■positively■skewed,■Maths:■negatively■skewed c Discuss■the■possible■reasons■for■any■skewness. The■science■test■may■have■been■more■diffi■cult. d State■the■modal■class■of■each■distribution. Science:■61–70,■Maths:■71–80 e In■which■subject■is■the■standard■deviation■greater?■Explain■your■answer.

■ Leaf ■ Group■B 7■ 8 0■ 1■ 2■ 4■ 5■ 5■ 8■ 8 2■ 2■ 2■ 4■ 5■ 5■ 5■ 6■ 8 2 ■

b Five-point■summary ■ ■ Group■A:■ 13■ 27■ 36■ 43■ 62 ■ ■ Group■B:■ 17■ 23■ 30■ 35■ 42

Five-point■summary x Range IQR s

c Student■comparison Statistics

Group A

13■ 27■ 36■ 43■ 62 35.85■hours 49■hours 16■hours 13■hours

Group B

17■ 23■ 30■ 35■ 42 28.95■hours 25■hours 12■hours 7■hours

8 Year■10■students■at■Merrigong■High■School■sit■exams■in■Science■and■Maths.■The■results■are■

were■exposed■to■a■cold■virus.■Twenty■patients■were■then■given■a■dose■of■the■drug■while■another■ 20■patients■were■given■a■placebo.■(In■ medical■tests■a■control■group■is■often■ given■a■placebo■drug.■The■subjects■in■ this■group■believe■that■they■have■been■ given■the■real■drug■but■in■fact■their■dose■ contains■no■drug■at■all.)■All■participants■ were■then■asked■to■indicate■the■time■ when■they■fi■rst■felt■relief■of■symptoms.■ The■number■of■hours■from■the■time■the■ dose■was■administered■to■the■time■when■ the■patients■fi■rst■felt■relief■of■symptoms■ are■detailed■below. Group A (drug) 25 29 32 45 42 38 44 42 Group B (placebo) 25 17 35 42 34 32 25 18

18 35

21 47

37 62

42 17

62 34

13 32

35 22

28 28

20 21

32 24

38 32

35 36

a Detail■the■data■on■a■back-to-back■stem-and-leaf■plot. b Display■the■data■for■both■groups■on■a■box-and-whisker■

plot. c Make■comparisons■of■the■data.■Use■statistics■in■your■ Student■decision,■justifying■answer answer. d Does■the■drug■work?■Justify■your■answer. e What■other■considerations■should■be■taken■into■account■

when■trying■to■draw■conclusions■from■an■experiment■of■ this■type? Class■discussion

reflectioN 

 

What is a quick way to remember the shape of positively and negatively skewed graphs?

chapter 13 Univariate data

463

statistics AND probability • data representation and interpretation

Summary Measures of central tendency

For ungrouped data the following measures of central tendency are used. The mean is the sum of scores in a given set of data divided by the number of scores in the set. ∑x x=    is used when a list of scores is given. n ∑( f × x ) x=    is used when a frequency distribution table is given. n ■■ The median is: (a) the middle score for an odd number of scores arranged in numerical order (b) the average of the two middle scores for an even number of scores arranged in numerical order. n +1 th position. Its location is determined by finding the score in the 2 ■■ The mode is the score that occurs most often in a set of data. ■■

For grouped data the following measures of central tendency are used. ∑( f × x ) ■■ The mean is x = , where x represents the midpoint of a class interval. n ■■ The median can be estimated from the ogive by finding the 50th percentile. ■■ The modal class is given by the class interval with the highest frequency. Measures of spread ■■ ■■ ■■

Range = highest score - lowest score  or  range = Xmax - Xmin The difference between the upper and lower quartiles is called the interquartile range, IQR. IQR = Q3 - Q1. The IQR considers the middle 50% of the data. For continuous data, the quartiles can be estimated from the ogive. Box-and-whisker plots

■■ ■■ ■■ ■■ ■■

A five-point summary is a list consisting of the lowest score, lower quartile, median, upper quartile and greatest score of a set of data. A box-and-whisker plot is a graphical representation of a five-point summary and is a powerful tool to show the spread of data. The box spans the interquartile range; the median is marked by a vertical line inside the box and the whiskers extend to the lowest and greatest scores. Box-and-whisker plots are always drawn to scale. If an extreme value (outlier) occurs in a set of data, it can be denoted by a small cross; the whisker is then shortened to the next largest (or smallest) value. The standard deviation

■■ ■■ ■■ ■■

The standard deviation is the most useful measure of spread. The standard deviation is found by entering the data set into the statistics function of your scientific or graphics calculator. A lower standard deviation indicates that the data are more bunched or clustered while a higher standard deviation indicates that the data are more spread out. Standard deviation can be used to measure consistency (a low standard deviation indicating greater consistency.) Comparing data sets

■■

464

When multiple displays are used for two or more sets of data, we can compare and contrast the data sets and determine whether any relationship exists between them.

Maths Quest 10 for the Australian Curriculum

statistics aND probability • Data represeNtatioN aND iNterpretatioN

■■ ■■ ■■

A■multiple■stem-and-leaf■plot■allows■for■a■quick■comparison■of■the■data■from■which■we■can■ easily■compare■medians,■ranges■and■interquartile■ranges. The■summary■statistics■from■two■data■sets■can■be■compared■quickly■on■a■box-and-whisker■ plot. The■most■commonly■used■comparisons■are■summary■statistics■to■compare■what■is■a■typical■ score■and■what■the■spread■of■the■data■is. Skewness

■■ ■■ ■■ ■■ ■■

A■distribution■is■symmetrical■when■the■data■are■equally■distributed■around■the■mean. When■the■data■are■symmetrical,■the■median■and■mean■will■both■be■the■middle■score. When■the■data■are■clustered■around■the■mean,■the■standard■deviation■is■smaller. When■the■majority■of■scores■are■at■the■lower■end■of■a■distribution,■it■is■said■to■be■positively■ skewed. When■the■majority■of■scores■are■at■the■upper■end■of■the■distribution,■it■is■said■to■be■negatively■ skewed.

MAPPING YOUR UNDERSTANDING

Homework Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■429. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

chapter 13 Univariate data

465

statistics AND probability • data representation and interpretation

Chapter review

4 a

Year 10 Year 8 0

10 20 30 40 50 60 70

b Y  ear 8: mean = 26.83, median = 27, range = 39, IQR = 19, sd = 11.45 Year 10: mean = 40.7, median = 39.5, range = 46, IQR = 20, sd = 12.98

Key:  2  |  6 = 26 wpm Leaf  Stem Leaf   1 Find the mean, median and mode for each of the Year 8 Year 10 following sets of data: 99 0 a 7, 15, 8, 8, 20, 14, 8, 10, 12, 6, 19 9865420 1 79 b Key:  1  |  2 = 12 988642100 2 23689 Stem Leaf 9776410 3 02455788 1 26 86520 4 1258899 a Mean = 11.55; median = 10; mode = 8 2 178 5 03578 3 0 3 3 4 6 8 b Mean = 36; median = 36; mode = 33, 41 6 003 c Mean = 72.18; median = 72; mode = 72 4 01159 5 136 a Using a calculator or otherwise, construct Fluency

Score (x)

Frequency (  f  )

70

2

71

6

72

9

73

7

74

4

a pair of parallel box-and-whisker plots to represent the two sets of data. b Find the mean, median, range, interquartile range and standard deviation of each set. c Compare the two distributions, using your answers to parts a and b. 5 Consider the box-and-whisker plot drawn

below.

2 For each of the following data sets, find the range. a 4, 3, 6, 7, 2, 5, 8, 4, 3 6 6 b

x

13

14

15

16

17

18

19

f

 3

 6

 7

12

 6

 7

 8

0

c Key:  1  |  8 = 18 20

Stem 1 2 3

Leaf 7889 12445777899 0001347

5

10 15 20

a Find the median. 20 b Find the range. 24 c Find the interquartile range.

8

(in kg) obtained from each of 20 lambs.

interquartile range. a 18, 14, 15, 19, 20, 11, 16, 19, 18, 19 4 b Key:  9  |  8 = 9.8 8.5 Stem Leaf 8 7889 (3.9, 4.4, 4.9, 5.85, 6.8) 9 0 2 4 4 5 7 7 7 8 9 9 3.5 4.5 5.5 6.5 kg 10 0 1 1 1 3 7

4 The following back-to-back stem-and-leaf plot

shows the typing speed in words per minute (wpm) of 30 Year 8 and Year 10 students.

4.5 6.2 5.8 4.7 4.0 3.9 6.2 6.8 5.5 6.1 5.9 5.8 5.0 4.3 4.0 4.6 4.8 5.3 4.2 4.8 a Detail the data on a stem-and-leaf plot. (Use a

class size of 0.5 kg.) b Prepare a five-point summary of the data. c Draw a box-and-whisker plot of the data.

Find the standard deviation of each of the following data sets. a 58, 12, 98, 45, 60, 34, 42, 71, 90, 66 24.4 b

1.1

x

1

2

 3

4

5

f

2

6

12

8

5

c Key:  1  |  4 = 14

Stem 0 1 2

Maths Quest 10 for the Australian Curriculum

40 Score

6 The following data give the amount of cut meat

3 For each of the following data sets, find the

466

25 30 35

The typing speed of Year 10 students is about 13 to 14 wpm faster than that of Year 8 students. The spread of data in Year 8 is slightly less than in Year 10.

c

7.3

Leaf 1344578 00012245789 022357

Key:  3*  |  9 = 3.9 kg Stem Leaf 3* 9 4 0  0  2  3 4* 5  6  7  8  8 5 0  3 5* 5  8  8  9 6 1  2  2 6* 8

statistics AND probability • data representation and interpretation 8   MC  The Millers obtained a number of quotes on

the price of having their home painted. The quotes, to the nearest hundred dollars, were: 4200  5100  4700  4600  ■ 4800  5000  4700  4900

  The standard deviation for this set of data, to the nearest whole number is: A 12 ✔ B 14 C 17 D 35 10 Each week, varying amounts of a chemical are added to a filtering system. The amounts required (in mL) over the past 20 weeks are shown in the stem-and-leaf plot below. Key:  3  |  8 represents 0.38 mL Stem Leaf Class 2 1 interval Frequency 2 2  2   0–9  2 2 4  4  4  5 10–19  7 2 6  6 20–29  6 2 8  8  9  9 30–39  6 3 0 3 2  2 40–49  3 3 4  5 50–59  3 3 6 60–69  3 3 8 Total 30   Calculate to 2 decimal places the standard deviation of the amounts used. 0.05 mL problem solving 11 A sample of 30 people was selected at random

from those attending a local swimming pool. ■ Their ages (in years) were recorded as ■ follows:

✔

The standard deviation for this set of data, to the nearest whole dollar, is: A 260 B 278 C 324 D 325

9   MC  The number of Year 12 students who, during

semester 2, spent all their spare periods studying in the resource centre is shown on the stem-and-leaf plot below.

Cumulative frequency

Key:  2  |  5 = 25 students Stem Leaf 0 8 1 2 5  6  6  7 3 0  2  3  6  9 4 7  9 5 6 6 1

19   7  40  37  40  36  20  29 

c

e f

10 20 30 40 50 60 70

Age

41  29  59  24 

17  21  65  10 

23  62  55 18  16  10 68  15   9 30

a Find the mean and the median age of the people in this sample. Mean = 32.03; median = 29.5 b Group the data into class intervals of

d 30 25 20 15 10 5 0

58  32  33  38 

g

10 (0–9 etc) and complete the frequency distribution table. Use the frequency distribution table to estimate the mean age. Mean = 31.83 Calculate the cumulative frequency and, hence, Median = 30 plot the ogive. Estimate the median age from the ogive. Compare the mean and median of the original data in part a with the estimates of the mean and the median obtained for the grouped data in parts c and e. Were the estimates good enough? Explain your answer. Yes, they were fairly close to the

Estimates from parts c and e were fairly accurate.

mean and median of the raw data.

Chapter 13 Univariate data

467

statistics AND probability • data representation and interpretation 12 Consider the box-and-whisker plot below which

shows the number of weekly sales of houses by two real estate agencies.

15 The table below shows the number of cars that

are garaged at each house in a certain street each night.

 ane and Roarne had a higher median H and a lower spread and so they appear to have performed better.

HJ Looker Hane & Roarne 9 10 Number of weekly sales HJ Looker: median = 5; Hane and Roarne: median = 6

0 1 2

3 4

5 6

7

8

a What is the median number of weekly sales for

each real estate agency? b Which agency had the greatest range of sales? HJ Looker c Which agency had the greatest interquartile range of sales? HJ Looker d Which agency performed better? Explain your

Number of cars

Frequency

1

9

2

6

3

2

4

1

5

1

answer. 13 Kloe compares her English and Maths marks. The results of eight tests in each subject are shown below. English: 76, 64, 90, 67, 83, 60, 85, 37 Maths: 80, 56, 92, 84, 65, 58, 55, 62 a Calculate Kloe’s mean mark in each subject. b Calculate the range of marks in each subject. c Calculate the standard deviation of marks in

each subject. d Based on the above data, in which subject

Frequency

10 9 8 7 6 5 4 3 2 1 0

9 8 7 6 5 4 3 2 1 12 34 5 Number of cars

1 2 3 4 5 Score

a Are the data symmetrical? Yes b Can the mean and median of the data be seen? If so, what are their values? Yes. Both are 3. c What is the mode of the data? 3

a Show these data in a frequency histogram. b Are the data positively or negatively skewed?

Justify your answer. 16 Find the mean, median and mode of this data set:

2, 5, 6, 2, 5, 7, 8. Comment on the shape of the distribution. Mean = 5, median = 5, mode = 2 and 5. The distribution is positively skewed and bimodal.

468

Maths Quest 10 for the Australian Curriculum

Positively skewed — a greater number of scores is distributed at the lower end of the distribution.

a English: mean = 70.25; Maths: mean = 69 b E  nglish: range = 53; Maths: range = 37 c English: s  = 16.1; Maths: s  = 13.4 d K  loe has performed more consistently in Maths as the range and standard deviation are both lower. Frequency

would you say that Kloe has performed more consistently? 14 Consider the data set represented by the frequency histogram below.

statistics aND probability • Data represeNtatioN aND iNterpretatioN 17 Mc ■A■data■set■has■a■mean■of■75■and■a■standard■

✔

deviation■of■5.■Another■score■of■50■is■added■to■the■ data■set.■Which■of■the■following■will■occur? A The■mean■will■increase■and■the■standard■ deviation■will■increase. B The■mean■will■increase■and■the■standard■ deviation■will■decrease. C The■mean■will■decrease■and■the■standard■ deviation■will■increase. D The■mean■will■decrease■and■the■standard■ deviation■will■decrease.

18 Mc ■Note:■There■may■be■more■than■one■correct■

✔ ✔ ✔

answer. A■data■set■has■a■mean■of■60■and■a■standard■ deviation■of■10.■A■score■of■100■is■added■to■the■ data■set.■This■score■becomes■the■highest■score■in■ the■data■set.■Which■of■the■following■will■increase? A Mean B Standard■deviation■ C Range D Interquartile■range eBook plus

Interactivities

Test yourself Chapter 13 int-2861 Word search Chapter 13 int-2859 Crossword Chapter 13 int-2860

chapter 13 Univariate data

469

eBook plus

activities

Chapter opener

(page 429) •■ Hungry■brain■activity■(doc-5298):■Chapter■13 Digital doc

Are you ready?

(page 430) •■ SkillSHEET■13.1■(doc-5299):■Finding■the■mean■of■a■ small■data■set •■ SkillSHEET■13.2■(doc-5300):■Finding■the■median■of■ a■small■data■set •■ SkillSHEET■13.3■(doc-5301):■Finding■the■mode■of■a■ small■data■set •■ SkillSHEET■13.4■(doc-5302):■Finding■the■mean,■ median■and■mode■from■a■stem-and-leaf■plot •■ SkillSHEET■13.5■(doc-5303):■Presenting■data■in■a■ frequency■distribution■table •■ SkillSHEET■13.6■(doc-5304):■Drawing■statistical■ graphs Digital docs

13A Measures of central tendency Digital docs

•■ Activity■13-A-1■(doc-5128):■Mean,■median■and■ mode■(page 435) •■ Activity■13-A-2■(doc-5129):■Practice■with■mean,■ median■and■mode■(page 435) •■ Activity■13-A-3■(doc-5130):■Mean,■median■and■ mode■in■depth■(page 435) •■ SkillSHEET■13.1■(doc-5299):■Finding■the■mean■of■a■ small■data■set■(page 435) •■ SkillSHEET■13.2■(doc-5300):■Finding■the■median■of■ a■small■data■set■(page 436) •■ SkillSHEET■13.3■(doc-5301):■Finding■the■mode■of■a■ small■data■set■(page 436) •■ SkillSHEET■13.4■(doc-5302):■Finding■the■mean,■ median■and■mode■from■a■stem-and-leaf■plot■ (page 436) •■ SkillSHEET■13.5■(doc-5303):■Presenting■data■in■a■ frequency■distribution■table■(page 437) •■ SkillSHEET■13.6■(doc-5304):■Drawing■statistical■ graphs■(page 438) 13B Measures of spread Digital docs

•■ Activity■13-B-1■(doc-5131):■Range■and■quartiles■ (page 442) •■ Activity■13-B-2■(doc-5132):■Practice■with■range■and■ quartiles■(page 442) •■ Activity■13-B-3■(doc-5133):■Range■and■quartiles■in■ depth■(page 442) •■ WorkSHEET■13.1■(doc-5311):■Univariate■data■I■ (page 444)

470

Maths Quest 10 for the australian curriculum

13C Box-and-whisker plots

(page 447) •■ Activity■13-C-1■(doc-5134):■Constructing■ boxplots •■ Activity■13-C-2■(doc-5135):■Boxplots■and■outliers •■ Activity■13-C-3■(doc-5136):■Boxplots■with■ decimals Digital docs

13D The standard deviation Digital docs

•■ Activity■13-D-1■(doc-5137):■Standard■deviation■ (page 451) •■ Activity■13-D-2■(doc-5138):■Practice■with■standard■ deviation■(page 451) •■ Activity■13-D-3■(doc-5139):■Standard■deviation■in■ depth■(page 451) •■ WorkSHEET■13.2■(doc-5318):■Univariate■data■II■ (page 453) 13E Comparing data sets Interactivity (page 454) •■ Parallel■boxplots■(int-2788) Digital docs (page 455) •■ Activity■13-E-1■(doc-5140):■Comparing■data■1 •■ Activity■13-E-2■(doc-5141):■Comparing■data■2 •■ Activity■13-E-3■(doc-5142):■Comparing■data■3

13F Skewness Digital docs

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stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

14 bivariate data

14a Identifying related pairs of variables 14b Graphing bivariate data 14c Scatterplots WhAt Do yoU kNoW ? 1 List what you know about data that shows relationships between two variables (such as height and mass data for a group of people). Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of data of this type. eBook plus

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opeNiNg QUestioN

If you graphed the height against the mass for all the people in this group, would the result be a straight line?

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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Distinguishing qualitative from quantitative data 1 State■whether■the■following■pieces■of■data■are■numerical■or■non-numerical. a The■monthly■temperatures ■ Numerical Non-numerical b The■brands■of■soft■drinks Non-numerical c The■types■of■dogs Numerical d The■balance■in■a■bank■account Distinguishing discrete from continuous data 2 State■whether■the■following■pieces■of■data■are■discrete■or■continuous. a The■daily■temperature ■ Continuous b Your■height Continuous ■m Continuous c The■time■taken■to■swim■100■ Discrete d The■number■of■children■in■your■family ■ Number■of■kilograms:■independent;■ total■cost:■dependent Determining independent and dependent variables 3 For■each■of■the■following,■state■the■independent■and■the■dependent■variables. a The■number■of■kilograms■of■potatoes■purchased■and■the■total■cost. b The■number■of■swimmers■in■a■public■swimming■pool■and■the■temperature■for■that■day. Age:■independent;■height:■dependent c The■height■and■age■of■a■student. Reading scales (how much is each interval worth?) 4 For■the■scales■shown,■how■much■is■each■interval■worth? a 20 30 a■ 1■unit b 0.1■of■a■unit b 5 6

Temperature:■independent;■ number■of■swimmers:■ dependent

Reading a column graph Cost of grapes (per kg) between January and May

Digital doc

SkillSHEET 14.5 doc-5331

4.40 4.20 Price ($)

4.00 3.80 3.60 3.40 3.20

The■cost■of■grapes■was■ recorded■over■5■months.

3.00

January February March Month

April

May

5 Use■the■above■column■graph■to■answer■the■following■questions. a What■was■the■price■of■grapes■in■March? ■ $3.20■in■March b In■which■month■did■grapes■cost■the■least? In■February■grapes■cost■$3.05. c During■how■many■months■was■the■cost■of■grapes■recorded? d In■which■month■was■the■cost■of■grapes■the■highest,■and■what■was■this■cost? 472

Grapes■cost■the■most■in■May■when■they■were■$4.33■per■kilogram. Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

Digital doc

SkillSHEET 14.6 doc-5332

Reading line graphs 6 The■line■graph■at■right■shows■the■height■

of■a■child■(Timmy)■over■5■years. a How■tall■was■Timmy■at■the■start■of■the■ measurement■period? ■ 145■■cm b How■much■did■Timmy■grow■in■the■ fi■rst■year? 5■■cm c How■much■did■Timmy■grow■over■the■ fi■ve■years? 20■■cm d How■many■years■did■it■take■for■Timmy■ to■grow■10■■cm? 2■years

170

Increase in Timmy’s height between 2005 and 2010

160 Height (cm)

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150 140 130 120 110 100 2005 2006 2007 2008 2009 2010 Years

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Plotting coordinate points 7 State■the■size■and■the■direction■of■the■moves■from■the■origin■of■a■Cartesian■plane■needed■to■

locate■each■of■the■following■points. a A(3,■2) 3■right,■2■up b B(5,■0) 5■right c C(0,■7) 7■up d D(2.5,■1.5) 2.5■right,■1.5■up

Chapter 14 bivariate data

473

statistics AND probability • Data representation and interpretation

14A

Identifying related pairs of variables Univariate and bivariate data ■■

■■

Univariate data have only one variable for each piece of data. This is the type of data we have considered so far. For example, we have looked at data like the heights of a group of students and the number of children per family in a group of families. Bivariate data have pairs of variables for each piece of data; for example, a person’s height and weight.

Variables ■■

■■ ■■

Variables (like height and weight) are shown as symbols, that can take the place of a range of numbers or pieces of data. For example, we might let the height of a tree be represented by the variable h, or the length of the shadow of the tree by the variable l. By using variables and their symbols in this way, we are able to write relationships between variables in terms of an equation. Data can be qualitative (nominal, ordinal) or quantitative (discrete, continuous). In this chapter we deal mainly with quantitative data.

Worked Example 1

Classify the following types of data. a  Time      b  Shoe size      c  Eye colour Think

Write

a Time can be measured in various units, such as years,

a Time measurements give quantitative,

days, minutes and seconds. It is recorded in numbers; that is, how many there are of these units (and any of their fractions, so it is continuous). b Shoes come only in set sizes. You cannot get sizes in

continuous data.

b Shoe sizes are quantitative, discrete

data.

between these. c

Eye colour does not measure how much or how many. It represents a description of the colour of the eye.

c

Eye colour is qualitative, nominal data.

Dependent and independent variables ■■ ■■ ■■

■■ ■■

Bivariate data consist of two variables. One variable is generally the dependent variable, and the other the independent variable. The dependent variable, as the name suggests, is the one whose value depends on the other variable. The independent variable takes on values that do not depend on the value of the other variable. When data are expressed in the form of a table, generally the independent variable is written in the first row or the first column. The independent variable is placed on the x-axis and the dependent variable on the y-axis.

Worked Example 2

State the dependent and the independent variable in each of the following pairs of data. a Age and height of a child b Cost of bus fare and distance travelled in the bus c Number of people at a football match and the number of drinks sold there. 474

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Think

Write

a Generally, a child grows taller with increasing

a Height is the dependent variable, and age the

age (height depends on age). b The fare is more when you travel further (fare

independent variable. b Cost is the dependent variable, and distance the

depends on distance). c

More drinks will be sold if more people are there (number of drinks sold depends on number of people).

independent variable. c

Number of drinks sold is the dependent variable, while number of people at the match is the independent variable.

Identifying a relationship ■■

■■

The options we have when considering two variables are as follows. •• There is a relationship between them. •• There is no relationship between them. If there is a relationship, it can be classified as being strong, moderate or weak. The relationship can also be regarded as positive or negative. •• If one variable increases as the other increases, the relationship is positive. •• If one variable increases as the other decreases, the relationship is negative.

Worked Example 3

For each of the following pairs of variables, indicate:  i whether you think there is a relationship between them, and if so   ii whether the relationship is positive or negative, and iii the strength of the relationship. a A person’s height (h) and mass (m) b The length of a song (l ) and its position on a CD ( p) c The speed of travel (s) and the time taken to reach a destination (t). Think a

b

c

Write

i

Generally as a person grows taller, their mass changes.

ii

With increasing height, generally mass also increases.

The relationship is positive.

iii

This change tends to be quite consistent.

The relationship is quite strong.

Short songs are not put at the beginning, and long songs at the end. There is no particular order to the songs.

a There is a relationship between h and m.

b There is no relationship between the l and p.

i

As speed changes, the time taken to reach a destination also changes.

c

ii

With a faster speed it takes less time to reach a destination.

The relationship is negative.

iii

The change tends to be consistent since travelling at twice the speed will halve the time.

There is a strong negative relationship.

There is a relationship between s and t.

Chapter 14 Bivariate data

475

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

reMeMber

1.■ Data■can■be■qualitative■(nominal,■ordinal)■or■quantitative■(discrete,■continuous). 2.■ The■dependent■variable■is■the■one■whose■value■depends■on■the■other■variable.■The■ independent■variable■takes■on■values■which■do■not■depend■on■the■value■of■the■other■ variable.■The■independent■variable■is■recorded■in■the■fi■rst■row■or■column■of■a■table,■and■ placed■on■the■x-axis■of■a■graph. 3.■ If■there■is■a■relationship■between■two■variables: •■ it■can■be■positive■or■negative •■ it■can■be■strong,■moderate■or■weak.

exerCise

14A iNDiViDUAl pAthWAys eBook plus

Activity 14-A-1

Pairs of variables doc-5146 Activity 14-A-2

More pairs of variables doc-5147 Activity 14-A-3

a■ Dependent:■time■spent■travelling■to■school;■ independent:■distance■to■school b Dependent:■heart■rate■of■a■runner;■independent:■running■speed c Dependent:■value■of■CD■collection;■independent:■number■of■ CDs■in■collection d Dependent:■amount■of■computer■memory■used■by■fi■le;■ independent:■length■of■fi■le e Dependent:■cost■of■second-hand■car;■independent:■age■of■car

Advanced pairs of variables doc-5148

476

Identifying related pairs of variables flUeNCy 1 We1 ■Classify■the■following■types■of■data,■using■two■words■from■the■following:■qualitative,■

quantitative,■nominal,■ordinal,■discrete■and■continuous. a The■number■of■children■in■your■school ■ Quantitative,■discrete b The■types■of■bicycles■students■ride■to■school Qualitative,■nominal Quantitative,■continuous c The■heights■of■students■in■your■class d The■mass■of■your■textbooks Quantitative,■continuous Qualitative,■nominal e The■languages■you■speak■at■home f The■time■it■takes■you■to■travel■to■school Quantitative,■continuous Quantitative,■discrete g The■number■of■cousins■each■person■in■your■class■has ■m. Quantitative,■continuous h The■time■it■takes■you■to■run■100■ Discrete■data■can■be■counted■in■ 2 Explain■the■difference■between■discrete■and■continuous■data. exact■values;■continuous■data■can■ be■measured■in■a■continuous■scale. 3 Explain■whether■data■can■be: a nominal■and■discrete ■ If■data■is■nominal,■it■is■qualitative■in■nature,■so■it■cannot■also■be■discrete. If■data■is■ordinal,■this■implies■an■order,■which■is■a■qualitative■ b ordinal■and■continuous. classifi■cation.■This■means■that■it■cannot■also■be■continuous. 4 We2 ■State■the■dependent■and■the■independent■variable■in■each■of■the■following■pairs■of■

data. a Time■spent■travelling■to■school■and■distance■to■school b The■heart■rate■of■a■runner■and■the■running■speed c The■value■of■a■CD■collection■and■the■number■of■CDs■in■the■collection d The■amount■of■computer■memory■used■by■a■fi■le■and■the■length■of■the■fi■ le e The■cost■of■a■second-hand■car■and■the■age■of■the■car.

b i There■is■a■ relationship■ between■c■and■s. ii Positive iii Strong c i There■is■a■ relationship■ between■l■and■t. ii Positive iii Strong d i There■is■a■ relationship■ between■p■and■a. ii Positive iii Moderate e i There■is■a■ relationship■ between■h■and■a. ii Positive iii Strong

5 We3 ■For■each■of■the■following■pairs■of■variables,■indicate: i whether■you■think■there■is■a■relationship■between■them■and,■if■so, ii whether■the■relationship■is■positive■or■negative,■and iii the■strength■of■the■relationship. a Length■of■your■foot■(f )■and■length■of■your■hair■(h) ■ No■relationship b Cost■of■a■TV■set■(c)■and■the■size■of■its■screen■(s) c Length■of■a■race■(l)■and■the■time■taken■(t) d Size■of■an■animal’s■paw■(p)■and■the■size■of■the■animal■(a) e Height■of■a■person■(h)■and■their■arm■span■(a) f Time■it■takes■a■train■to■travel■from■A■to■B■(t)■and■the■time■it■takes■a■bus■to■travel■from■ A■to■B■(b). No■relationship

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation Understanding 6 It was a hot day at the beach. The lifesavers

noticed there was an increase of 20% in the number of people seeking help with sunburn. On the same day, the café also recorded an increase of 20% in their ice-cream sales. Since both of these variables increased, explain whether there is a relationship between the two, and if so, the strength of the relationship. There is no relationship between sunburn and ice-cream sales. The increase in both is influenced by the weather.

14B

reflection   How can we know if there is a relationship between two variables?

Graphing bivariate data ■■ ■■ ■■

When a set of bivariate data is collected, it is generally first organised into a table. A graph is then drawn to display the trends in the data. This visual representation shows at a glance whether there is a relationship between the two variables and, if so, the strength of the relationship.

Drawing column graphs from data tables ■■ ■■ ■■

When we dealt previously with column graphs for univariate data, we used qualitative data with a frequency distribution table. When we draw column graphs from bivariate data, both variables are quantitative data. As mentioned previously, when drawing a graph we must make sure we place the independent variable on the x-axis and the dependent variable on the y-axis.

Worked Example 4 1

This table shows data collected by measuring the length of the shadow of a stick every 2 hour from noon until 4.00 pm. Time (t)

Length of shadow (l) (cm)

12.00 pm

  20

12.30 pm

  32

  1.00 pm

  48

  1.30 pm

  60

  2.00 pm

  76

  2.30 pm

  88

  3.00 pm

100

  3.30 pm

112

  4.00 pm

128

Draw a column graph to display the data. Chapter 14 Bivariate data

477

statistics AND probability • Data representation and interpretation

Think

Write/DRAW

The length of the shadow depends on the time of day, so time is the independent variable, so it must be placed on the x-axis. Choose a suitable scale.

2

Draw columns for each reading.

Draw up axes with time on the x-axis and length of shadow on the y-axis.

Length of shadow (cm)

1

Shadow length over time

130 120 110 100 90 80 70 60 50 40 30 20 10 0

12.00 12.30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 Noon pm pm pm pm pm pm pm pm Time

Using a column graph to create a scatterplot ■■ ■■

Once a column graph has been drawn, the height of each column can be used to plot points of the bivariate data. These types of graphs are called scatterplots.

Worked Example 5 a Use the column graph showing length of a shadow over time from Worked example 4 to create a

scatterplot of the bivariate data.

b Comment on the trend shown by the scatterplot. Think 1

Place dots at the tops of the columns to represent the length of the shadow at that time.

a

Length of shadow (cm)

a

Write/DRAW Shadow length over time

130 120 110 100 90 80 70 60 50 40 30 20 10 0

478

Maths Quest 10 for the Australian Curriculum

12.00 12.30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 Noon pm pm pm pm pm pm pm pm Time

statistics AND probability • Data representation and interpretation

Erase the columns, leaving the dots on the graph. Length of shadow (cm)

2

Shadow length over time

130 120 110 100 90 80 70 60 50 40 30 20 10 0

b Examine the points on the

graph, looking for trends. As the time increases, the length of the shadow increases.

■■

■■

12.00 12.30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 Noon pm pm pm pm pm pm pm pm Time

b The scatterplot shows a strong, positive relationship between the

time of the reading and the length of the shadow.

Note that this trend applies only to the data shown. It is obvious that the trend in this case ■ will not continue much longer, as the sun will soon set and there will be no shadow. Would ■ the trend be the same if the experiment was conducted in the morning? It is important to understand that the trends of data collected from an experiment may change under different circumstances. Even when we do not know what the particular variables represent, we can still determine the type of relationship between them.

Worked Example 6

m

This column graph shows the relationship between two variables, p and m. a Produce a scatterplot from the column graph. b Comment on the trend shown by the scatterplot.



0

p Chapter 14 Bivariate data

479

statistics AND probability • Data representation and interpretation

Think

Place dots at the tops of the columns then erase the columns, leaving the dots on the graph.

a

m

a

Write/DRAW

0 b

Examine the points on the graph, looking for trends. As the value of p increases, the value of m generally decreases.

■■

p

b The scatterplot shows a negative relationship. As the value of p

increases, the value of m generally decreases. The points do not all lie on a straight line. The negative relationship is fairly strong.

When working with bivariate data, it is a good idea to examine the data and consider the questions that could be posed about the data.

Consider this scatterplot, which shows the number of sick days taken in a year by 10 employees, and relates this to the number of children they have.   Write two questions which an employer may be hoping to answer from this data. Think

Number of sick days

Worked Example 7

12 8 4 0

0 2 4 6 Number of children

Write

Think of questions that an employer would like to know about the number of sick days taken by her staff.

Do those employees with more children have more sick days?   If the staff allocation is 5 sick days a year, which staff members are taking more than their allocation?   Which employees need assistance with regard to the number of sick days they take?

remember

1. Scatterplots can be drawn from column graphs. 2. The position of the points on a scatterplot indicates the relationship between the two variables. 3. Examine collected bivariate data and consider the questions that could be answered. 480

Maths Quest 10 for the Australian Curriculum

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

exerCise

14b

graphing bivariate data flUeNCy

iNDiViDUAl pAthWAys

1 We 4 ■This■table■shows■the■number■of■goals■scored■in■a■ball■game.

eBook plus

Activity 14-B-1

Scatterplots doc-5149

Time after start (min)

0

10

20

30

40

50

60

70

80

90

100

Number of goals

0

■5

12

17

19

25

28

30

35

38

■ 42

Draw■a■column■graph■to■display■the■data. 2 We 5 ■a■ Use■the■column■graph■you■created■in■Question■1■to■create■a■scatterplot. b Comment■on■the■trend■shown■by■the■scatterplot. 3 We 6 ■This■column■graph■shows■the■relationship■between■two■variables■r■and■b.

Activity 14-B-2

More scatterplots doc-5150 Activity 14-B-3

Advanced scatterplots doc-5151

The■trend■is■positive■and■ strong,■indicating■that■ as■the■match■progressed■ goals■were■scored■at■a■ steady■rate.

70 60 100

b

30 20 10

90 20

b

40

10

20

30

40

50

0

10

0

1

2

3

4

5

6

r

7

8

Relationship between r and b

1 2 3 4 5 6 7 8 9 10 11 12 r

30 40 50 60 70 80 Time after start of match (min)

0

Goals scored in a match

70 60 50 40 30 20 10

50

9 10 11 12

a Produce■a■scatterplot■from■the■column■graph. The■trend■is■positive,■but■only■moderate. b Comment■on■the■trend■shown■by■the■scatterplot.

a■

Total number of goals

4 We 7 ■Write■one■question■that■could■be■asked■for■each■of■the■following■sets■of■bivariate■

data■collected. Do■people■with■more■ formal■education■earn■ more■money?

a Length■of■newborn■baby■and■height■of■father ■ Are■long■babies■born■to■tall■fathers? b Length■of■formal■education■and■income■in■a■job c Age■and■exercise■performed. Does■the■amount■of■exercise■performed■decrease■with■age? UNDerstANDiNg

(in■weeks)■spent■by■a■person■on■a■healthy■diet■and■the■ corresponding■mass■lost■(in■kg). Study■the■scatterplot■and■state■whether■each■of■the■ following■is■true■or■false. a The■number■of■weeks■that■the■person■stays■on■a■diet■is■ the■independent■variable. ■ True b The■y-coordinates■of■the■points■represent■the■time■ spent■by■a■person■on■a■diet. False c There■is■evidence■to■suggest■that■the■longer■the■person■ stays■on■a■diet,■the■greater■the■loss■in■mass. True d The■time■spent■on■a■diet■is■the■only■factor■that■ contributes■to■the■loss■in■mass. False

Loss in mass (kg)

5 Each■point■on■the■scatterplot■at■right■shows■the■time■

Number of weeks

Chapter 14 bivariate data

481

statistics AND probability • Data representation and interpretation e The relationship between

a

Temperature (èC)

the number of weeks on ■ the diet and the number ■ of kilograms lost is ■ positive. True 6   MC  The scatterplot that best represents the relationship between the amount of water consumed daily by a certain household for a number of days in summer and the daily temperature is:

✔ b

Water usage (L)

Water usage (L)

Temperature (èC)

e

Temperature (èC)

d

Water usage (L)

c

Water usage (L)

482

of sides and the sum of interior angles for a number of polygons.■   Which of the following statements is not true? a The relationship between the number of sides and the angle sum of the polygon is perfectly linear. b The increase in the number of sides causes the increase in the size of the angle sum. c The number of sides depends on the sum of the angles. d The angle sum is the dependent variable. e The relationship between the two variables is positive.

Maths Quest 10 for the Australian Curriculum

Sum of angles (è)

7   MC  The scatterplot at right shows the number

✔

Water usage (L)

Temperature (èC)

Temperature (èC)

1300 1200 1100 1000 900 800 700 600 500 400 300 200 3

4

5 6 7 8 9 10 Number of sides

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN 8 MC ■After■studying■a■non-linear■scatterplot,■it■was■concluded■that■there■was■evidence■that■the■

greater■the■level■of■one■variable,■the■smaller■the■level■of■the■other■variable.■The■scatterplot■must■ have■shown■a: a strong,■positive■relationship b strong,■negative■relationship c moderate,■positive■relationship ✔ d moderate,■negative■relationship e weak,■positive■relationship. 9 MC ■In■which■of■the■following■is■no■relationship■evident■between■the■variables? ✔ a y

b y

c y

x x d y

e y

x

eBook plus

Digital doc

x

x

10 Give■an■example■of■a■situation■where■the■scatterplot■may■look■like■the■ones■below. a y b y

WorkSHEET 14.1 doc-5334

x

0

10 a■ Number■of■tickets■sold■ and■the■total■money■raised■ for■a■number■of■different■ charity■concerts. b Number■of■items■sold■and■ the■price■of■the■item.

14C eBook plus

Interactivity Scatterplots

int-2789

x

0

refleCtioN Why is a scatterplot an important tool for analysing data?

scatterplots ■■ ■■ ■■

An■ideal■way■to■determine■whether■there■is■a■relationship■between■two■variables■is■through■ the■use■of■scatterplots.■ As■we■saw,■the■position■of■the■points■shows■the■direction■and■strength■of■the■relationship.■ The■direction■shows■whether■the■relationship■is■positive■or■negative,■while■the■strength■ indicates■whether■the■relationship■is■strong,■medium■or■weak.■ Consider■two■variables,■x■and■y. y

x Perfectly linear positive relationship

y

y

x Strong positive relationship

x Moderate positive relationship Chapter 14 bivariate data

483

statistics AND probability • Data representation and interpretation y

y

Weak positive relationship

x

x No relationship

  The scattering of the points can also slope in a downwards direction, indicating a negative relationship.   The first graph shows a perfectly linear negative relationship, while the second one shows a moderate negative relationship.

y

y

x

x

Worked Example 8

Data were collected to investigate whether the outside temperature is related to the number of people preferring to spend time in a recreation room at an island resort. Temperature (èC) 25 33 30 22 15 18 27 22 28 20 People in room

26 18 19 31 46 40 20 36 31 42

a Propose a question that this set of data might be able

to answer.

b Draw a scatterplot of the data. c Describe the trend shown by the scatterplot. d What advice would you give the management regarding the use of the room? Think a Look at the data and consider a

Write/DRAW a Does the outside temperature have an influence on the number

question it could answer. b The temperature outside is the

of people spending time in the recreation room? b

independent variable, so it should be placed on the x-axis. Choose a suitable scale for both axes and plot the points.

Outside temperature and people in room

People in room

50 40 30 20 10 0

c

Look at the direction and strength of the plotted points.

d Provide useful tips for the

management.

484

c

10 20 30 40 50 Temperature outside (èC)

The general trend is a negative one. As the temperature increases, fewer people spend time in the recreation room. The relationship between the two variables is moderate to strong.

d It seems that as the temperature increases, more people prefer to

spend their time outside. A suggestion could then be to provide more activities outside for the guests to enjoy at these times.

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Making predictions ■■ ■■ ■■ ■■

■■ ■■ ■■ ■■

■■

After a scatterplot has been drawn, it is possible to draw a straight line which is a reasonable estimation of the trend of the data. Unless we have a perfectly linear relationship, there is no single straight line that would go through all the points. A line can be drawm which is as close as possible to all the data points. This line is called the line of best fit and is positioned by eye so that there is an equal number of points above and below the line. Alternative methods for fitting this line will be explained in chapter 23. Using this line, we are then able to make predictions about values of data within the data set. It is important to remember that predictions made outside the data set are not considered reliable, as the trend may not continue in the same manner. The data graphed in the previous worked example were quantitative and discrete on the y-axis, while the x-axis data were quantitative and continuous. Since both axes represent quantitative data, we can use the line of best fit to make predictions about the temperature of the room with a particular number of people present, or the number of people present when the temperature was at a particular value. It must be remembered that values predicted in this way are simply estimations, unless the relationship is perfectly linear.

Worked Example 9 Outside temperature and people in room

50 People in room

Consider the scatterplot from Worked example 8. a Draw the line of best fit. b Use the line of best fit to predict:    i the temperature outside when 22 people are in the room   ii the number of people in the room when the outside temperature is 17  èC iii the number of people in the room when the outside temperature is 50  èC. c Comment on your answers from part b.

40 30 20 10 0

Think

Align a ruler to draw a line with roughly the same number of points above and below the line. The line does not necessarily have to pass through any of the points.

Write/DRAW a

Outside temperature and people in room

50 People in room

a

10 20 30 40 50 Temperature outside (èC)

40 Line of best fit

30 20 10 0

10 20 30 40 50 Temperature outside (èC) Chapter 14 Bivariate data

485

statistics AND probability • Data representation and interpretation

c

Note that these answers are only estimates.      i Rule a horizontal line from 22 on the y-axis to meet the line of best fit. From that point, draw a line vertically to the x-axis and read this value.    ii Rule a vertical line from 17  èC on the x-axis to meet the line of best fit. From that point, draw a line horizontally to the y-axis and read this value. Note that because this variable represents discrete data, the answer must be rounded to the nearest whole number, if necessary. iii It seems not possible to draw a line from 50  èC to the line of best fit (even if the line was extended).

b

Examine the answers for appropriateness.

c

■■

Outside temperature and people in room

50 People in room

b

42 40 Line of best fit

30 (i) 22

20

(ii) 17èC

10

31èC 0

10 20 30 40 50 Temperature outside (èC)

     i When 22 people are in the room, the outside temperature is about 31  èC.    ii When the temperature is 17  èC outside, about 42 people are in the room. iii A temperature of 50  èC is probably not possible. Even if a line could be drawn from 50  èC to the line of best fit, the answer would not be appropriate.

The answers to parts i and ii are estimates only. Part iii sends a warning to always check for the appropriateness of an answer.

One of our previous worked examples looked at the length of the shadow of a stick during part of the day. Both of these data values are quantitative and continuous. In this case we can therefore use the line of best fit to make predictions from one variable to the other for a continuous set of values. After all, the shadow length doesn’t just jump from one recorded value to the next; it grows continuously in length, even while recording is not taking place.

This scatterplot from Worked example 5 shows the recorded shadow length of a stick over a time period from noon to 4 pm. a Draw the line of best fit. b Use this line to predict:    i the length of the shadow at 2.15 pm.   ii the time when the shadow length would be 50  cm. iii  the time when there is no shadow. c Comment on your answers to part b.

Length of shadow (cm)

Worked Example 10 Shadow length over time

130 120 110 100 90 80 70 60 50 40 30 20 10 0

486

Maths Quest 10 for the Australian Curriculum

Noon 12.30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 pm pm pm pm pm pm pm pm Time

statistics AND probability • Data representation and interpretation

Think a Draw a line that is as close as

Write/Draw a Length of shadow (cm)

possible to all the points.

Shadow length over time

130 120 110 100 90 80 70 60 50 40 30 20 10

Line of best fit

0

b      i Locate the position of

b Length of shadow (cm)

2.15 pm on the x-axis. (It is half way between 2 pm and 2.30 pm.) From here, draw a vertical line until it meets the line of best fit. From this point, draw a horizontal line to the y-axis. Read the shadow length on the y-axis. (Note that since length is a continuous data variable, the answer does not have to be a whole number.)

Shadow length over time

130 120 110 100 90 80 70 60 50 40 30 20 10

(ii) 50 cm (i) 2.15 pm 1.06 pm Noon 12.30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 pm pm pm pm pm pm pm pm Time

     i The shadow is about 81  cm long at 2.15 pm.    ii The shadow length would be 50  cm at about 1.06 pm.

   ii Locate 50  cm on the y-axis. From here, draw a horizontal line until it meets the line of best fit. From this point, draw a vertical line to the x-axis. Read the time on the x-axis. (Note that since time is a continuous data variable, the answer does not have to be a whole number.) iii Look at the graph to see when the shadow length is 0. Examine the answers to part b and make a comment.

Line of best fit

81 cm

0

c

Noon 12.30 1.00 1.30 2.00 2.30 3.00 3.30 4.00 pm pm pm pm pm pm pm pm Time

iii At no time within the range of the data is there no shadow. c

The answers to parts i and ii are estimated answers. You would think that there would be no shadow at noon. However, this is only the case when the sun is directly overhead. This depends on the season and on latitude.

Chapter 14 Bivariate data

487

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

■■

In■providing■answers■for■questions■posed■about■bivariate■data,■it■is■important■to■be■able■to■ justify■why■a■particular■conclusion■is■reached.■It■is■not■advisable■to■simply■state■an■answer■ without■also■providing■a■reason. reMeMber

1.■ Scatterplots■can■be■used■to■observe■the■direction■and■strength■of■a■relationship. 2.■ A■line■of■best■fi■t■can■be■drawn■by■eye,■with■approximately■an■equal■number■of■points■ above■and■below■the■line. 3.■ Predictions■made■from■the■line■of■best■fi■t■within■the■data■set■are■considered■reliable.■ Those■made■beyond■the■data■set■are■not■reliable. exerCise

14C iNDiViDUAl pAthWAys

The■direction■of■the■relationship■is■positive;■the■greater■the■English■ mark,■the■greater■the■history■mark,■generally.■However,■as■the■ points■on■the■scatterplot■vary■quite■a■bit■from■a■straight■line,■the■ relationship■is■only■moderate.

scatterplots flUeNCy

1 We8 ■The■table■below■shows■the■marks■obtained■by■nine■students■in■English■and■History.

eBook plus

Activity 14-C-2

More relationships doc-5153 Activity 14-C-3

b The■direction■of■the■relationship■is■negative;■the■greater■the■ temperature,■the■fewer■the■pies■sold.■The■points■on■the■scatterplot■lie■ close■to■a■straight■line,■so■the■relationship■is■strong. c The■school■canteen■should■stock■more■pies■during■cooler■weather,■ and■fewer■pies■during■hot■weather. Note:■Best■fi■t■lines■are■indicated■as■a■guide■only.

Advanced relationships doc-5154

488

55

20

27

33

73

18

37

51

79

History

72

37

53

74

73

44

59

55

84

80 History

Types of relationships doc-5152

English

40 0

a Draw■a■scatterplot■of■the■data. b Describe■the■trend■shown■by■the■scatterplot.

0

40 80 English

2 The■table■below■shows■the■daily■temperature■and■the■number■of■hot■pies■sold■at■the■school■

canteen. Temperature (èC)

24

32

28

23

16

■ 14

26

20

29

21

No. of pies sold

56

20

24

60

84

120

70

95

36

63

a Draw■a■scatterplot■of■the■data. b Describe■the■trend■shown■by■the■scatterplot. c What■advice■could■you■give■the■managers■of■the■school■canteen?■ 3 Draw■a■line■of■best■fi■t■for■the■following■scatterplots,■which■show■relationships■between■the■

variables■x■and■y.■Remember■to■try■to■place■an■approximately■equal■number■of■points■above■ and■below■the■line. a

b

y

c

y

■

y

■

x d

x e

y

y

x

Number of pies sold

Activity 14-C-1

80 0

x

Maths Quest 10 for the Australian Curriculum

x

40 0 20 Temperature (èC)

statistics AND probability • Data representation and interpretation y 70 60 50 40 30 20 10 0

4   WE9  Use the given scatterplot and line of best

fit to predict: a the value of y when x = 45 38 b the value of x when y = 15. 18

5 Analyse the graph at right and use the line of

best fit to predict: a the value of y when the value of x is: 36 130 i 7 460 ii 22 290 iii b the value of x when the value of y is: i 120 37 480 6 ii 260 24 iii

Earnings ($)

6 6 a  and  b Note: Answers may vary depending on the line of best fit drawn. 130 120 110 100 90 80 70 60 50 40 30 20 10 0

2



4

y 600 500 400 300 200 100 0

10 20 30 40 50 60 70 80 x

10 15 20 25 30 35 40 45 x

5

  WE  10  A random sample of ten Year 10 students who have part-time jobs was selected. Each student was asked to state his/her average number of hours worked per week and average weekly earnings (to the nearest dollar). The results are summarised in the table below.

Hours worked Weekly earnings ($)

 4 23

 8 47

15 93

  18 122

10 56

 5 33

12 74

  16 110

14 78

 6 35

23  3

56  8

95 14

a Draw a scatterplot of the data. b Draw the line of best fit. c Use your line of best fit to determine: i how many hours per week would $100 Approximately 16 hours return 6 8 10 12 14 16 18 ii how much would be earned by Hours worked week About $55 working 9 hours per About $6.25 iii the average rate of pay for the students. 7 The data in this table show the distance

travelled by 10 cars, and the amount of fuel (to the nearest litre) used for their journeys. Distance travelled (km) Petrol used (L)

a  and  b

Petrol used (L)

Note: Answers may vary depending on the line of ■ best fit drawn. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

52  7

36  5

83  9

12  2

44  7

67  9

74 12

a Construct a scatterplot of the data. b Draw the line of best fit. c Use your line of best fit to determine: i how much petrol would be used on a 60-km journey About 8 litres ii how far a car could travel on 10 L of petrol About 70 km About 7 km/L iii the average petrol consumption in km/L.



Understanding 8 As a part of her project Rachel is growing a crystal. Every day she measures the crystal’s mass 10 20 30 40 50 60 70 80 90 100

Distance travelled (km)

using special laboratory scales and notes it in her book. The table below shows the results of her experiment. Day number Mass (g)

1 2.5

2 3.7

3 4.2

4 5.0

5 6.1

8 8.4

9 8.9

10 11 12 15 16 11.2 11.6 12.8 16.1 17.3 Chapter 14 Bivariate data

489

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

Measurements■on■days■6,■7,■13■ and■14■are■missing,■since■these■ were■2■consecutive■weekends■ and,■hence,■Rachel■did■not■have■a■ chance■to■measure■her■crystal,■ which■is■kept■in■the■school■ laboratory. a Construct■a■scatterplot■of■the■ data,■and■draw■in■the■line■of■ best■fi■t. b For■her■report,■Rachel■would■ like■to■fi■ll■in■the■missing■ 7.1■■g,■8.1■■g,■ 13.9■■g,■14.9■■g measurements■for■days■ 6,■7,■13■and■14.■Give■ an■estimate■of■these■ measurements. c Rachel■fell■sick■and■couldn’t■ record■the■mass■of■the■crystal■ for■the■last■two■days■of■the■ experiment■(days■17■and■18).■ What■would■you■predict■these■ 17.8■■g,■18.8■■g masses■to■be? d What■is■the■average■daily■ increase■in■mass■of■the■ About■1■■g crystal?

Mass (g)

reAsoNiNg 9 Consider■this■scatterplot■and■line■of■best■fi■t.

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

y

10 20 30 40 50 60 70 x



1 2 3 4 5 6 7 8 9 10111213141516

Day

eBook plus

Digital doc

Prediction■of■y-values■when■x■=■15■and■x■=■60■ would■be■considered■unreliable■as■these■x-values■ are■beyond■the■range■of■the■data.■For■an■x-value■ of■40,■the■predicted■y-value■would■be■considered■ reliable■as■this■is■within■the■range■of■the■data.

The■line■of■best■fi■t■is■used■to■predict■values■of■y■when■x■=■15,■x■=■40■and■when■x■=■60.■ Comment■on■these■predictions. 10 This■scatterplot■is■used■to■predict■the■value■of■y■when■x■=■300. y

WorkSHEET 14.2 doc-5335

500 400 This■prediction■would■be■ 300 considered■unreliable,■ 200 as■the■scattering■of■the■ 100 points■indicates■that■ 0 there■is■no■relationship■ between■x■and■y.

refleCtioN 100 200 300 400 500 600 700 x

Comment■on■the■prediction

490

Maths Quest 10 for the Australian Curriculum

Since lines of best fit are not really accurate, why should we use them at all?

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

summary Identifying related pairs of variables ■■ ■■

■■

Data■can■be■qualitative■(nominal,■ordinal)■or■quantitative■(discrete,■continuous). The■dependent■variable■is■the■one■whose■value■depends■on■the■other■variable.■The■ independent■variable■takes■on■values■which■do■not■depend■on■the■value■of■the■other■variable.■ The■independent■variable■is■recorded■in■the■fi■rst■row■or■column■of■a■table,■and■placed■on■the■ x-axis■of■a■graph. If■there■is■a■relationship■between■two■variables: •■ it■can■be■positive■or■negative •■ it■can■be■strong,■moderate■or■weak. Graphing bivariate data

■■ ■■ ■■

Scatterplots■can■be■drawn■from■column■graphs. The■position■of■the■points■on■a■scatterplot■indicates■the■relationship■between■the■two■ variables. Examine■collected■bivariate■data■and■consider■the■questions■that■could■be■answered. Scatterplots

■■ ■■ ■■

Scatterplots■can■be■used■to■observe■the■direction■and■strength■of■a■relationship. A■line■of■best■fi■t■can■be■drawn■by■eye,■with■approximately■an■equal■number■of■points■above■ and■below■the■line. Predictions■made■from■the■line■of■best■fi■t■within■the■data■set■are■considered■reliable.■Those■ made■beyond■the■data■set■are■not■reliable.

MaPPING YOUR UNdeRSTaNdING

Homework book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■471. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

Chapter 14 bivariate data

491

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

Chapter review 5 This■table■shows■the■maximum■and■minimum■daily■

flUeNCy

temperatures■in■a■city■over■a■one-week■period.

1 MC ■The■classifi■cation■of■data■describing■the■

number■of■iPods■sold■during■the■year■is: quantitative,■discrete. b qualitative,■nominal. c qualitative,■ordinal. d quantitative,■continuous. e none■of■these. 2 MC ■The■classifi■cation■of■data■which■describe■a■ person’s■mass■is: a quantitative,■discrete. b qualitative,■nominal. c qualitative,■ordinal. ✔ d quantitative,■continuous. e none■of■these. ✔ a

■1 ■2 ■3 ■4

5

■6

7

Maximum (èC)

12 13 10 11

9

10

8

Minimum (èC)

■3 ■3 ■2 ■1

0

■4

2

Use■the■table■to■answer■the■following■questions. a What■was■the■maximum■temperature■on■day■3? b Which■day■had■the■lowest■minimum■ ■ 10■■èC Day■5 temperature? c Which■day■was■the■coldest? Day■7 d Which■day■had■the■warmest■overnight■ temperature? Day■6 e What■was■the■temperature■range■(variation)■on■ day■2? 10■■èC f Which■day■had■the■smallest■range■of■ temperatures? Days■6■and■7

3 MC ■The■data■which■describe■sandwich■types■at■a■

takeaway■outlet■are: qualitative,■nominal. b quantitative,■discrete. c qualitative,■ordinal. d quantitative,■continuous. e none■of■these.

Day

✔ a

6 Consider■this■table■showing■the■age■and■height■of■a■

child■over■6■years. Age (years) Height (cm) 1 2 3 4 5 6

The■relationship■is■ positive■and■strong.■ As■the■child■grows■ older■his/her■height■ also■increases. 4 For■each■of■the■following■pairs,■decide■which■of■the■

492

Maths Quest 10 for the Australian Curriculum

■ Height

a Which■variable■is■the■dependent■variable? b Which■variable■should■be■graphed■on■the■ x-axis? Age c Draw■a■column■graph■to■display■the■data d Describe■the■relationship■between■the■

■

two■variables. 7 Look■at■the■following■graph,■showing■the■change■in■

size■of■an■iceblock■over■time. Size of iceblock over a period of 11 minutes 1000 Size (mm3)

variables■is■independent■and■which■is■dependent. a Number■of■hours■spent■studying■for■a■ Mathematics■test■and■the■score■on■that■test. b Daily■amount■of■rainfall■(in■mm)■and■daily■ attendance■at■the■Botanical■Gar■dens. c Number■of■hours■per■week■spent■in■a■gym■and■ the■annual■number■of■visits■to■the■doctor.■ d Amount■of■computer■memory■taken■by■an■ essay■and■the■length■of■the■essay■(in■words). e The■cost■of■care■in■a■childcare■centre■and■ attendance■in■the■childcare■centre. f The■cost■of■the■property■(real■estate)■and■the■ age■of■the■property. g The■cut-off■ENTER■score■for■a■certain■course■ and■the■number■of■applications■for■that■course. h The■heart■rate■of■a■runner■and■the■running■ speed.

■ 80 ■ 85 ■ 88 ■ 93 ■ 99 104

800 600 400 200 0

1

2

3

4

5 6 7 Time (min)

8

9

10 11

No■relationship Positive,■strong Positive,■moderate Negative,■moderate

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN 9 MC ■Which■of■the■following■graphs■best■

depicts■a■strong■negative■relationship■between■ the■two■variables? b y

b■ d■ f■ h■

a y

Positive,■perfectly■linear■ Negative,■moderate■ No■relationship■ Negative,■perfectly■linear■

x

x

c y

✔ d y

x

x

11

a■ c e g

e y

The■relationship■is■negative;■as■time■increases■the■size■of■the■ ice■block■decreases.■It■is■moderately■strong,■but■not■linear.



x a What■are■the■units■on■the■x-axis? ■ Minutes 10 MC ■What■type■of■relationship■is■shown■by■the■ b What■are■the■units■on■the■y-axis? mm3 graph■on■the■right? y c Which■variable■is■the■independent■ a Strong■positive■relationship variable? Time b Moderate■positive■relationship d What■is■the■scale■on■the■y-axis? 1■unit■=■100■mm3 ✔ c Moderate■negative■relationship e What■type■of■data■is■shown■on: d Strong■negative■relationship i the■x-axis ■ Quantitative,■continuous e None■of■the■above x Quantitative,■continuous ii the■y-axis? 11 State■the■type■of■relationship■between■x■and■y■for■ f How■long■did■it■take■for■half■the■iceblock■ each■of■the■following■scatterplots.■ 3■minutes to■melt? a y b y g Describe■the■relationship■between■

the■two■variables. 8 MC ■A■researcher■administers■different■amounts■

✔

of■fertiliser■to■a■number■of■trial■plots■of■potato■ crop.■She■then■measures■the■total■mass■of■potatoes■ harvested■from■each■plot.■When■drawing■the■ scatterplot,■the■researcher■should■graph: a mass■of■harvest■on■the■x-axis■because■it■is■the■ independent■variable,■and■amount■of■fertiliser■ on■the■y-axis■because■it■is■the■dependent■ variable. b mass■of■harvest■on■the■y-axis■because■it■is■the■ independent■variable,■and■amount■of■fertiliser■ on■the■x-axis■because■it■is■the■dependent■ variable. c mass■of■harvest■on■the■x-axis■because■it■is■the■ dependent■variable,■and■amount■of■ferti■liser■on■ the■y-axis■because■it■is■the■independent■variable. d mass■of■harvest■on■the■y-axis■because■it■is■the■ dependent■variable,■and■amount■of■fertiliser■ on■the■x-axis■because■it■is■the■independent■ variable. e none■of■the■above.

x c

y

x d

y

x e

y

g

y

x f

y

h

y

x

x

x

x Chapter 14 bivariate data

493

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

i

Since■the■cheaper■bags■sell■better,■have■a■greater■stock■of■them■than■the■more-expensive■bags. c Draw■a■scatterplot■of■the■data. j y

y

d What■type■of■relationship■exists■between■the■

variables? e What■advice■would■you■give■Eugene■after■ x k

x

y

l

y

examining■his■records■of■sales? 2 The■relationship■between■two■variables■x■and y■is■

shown■in■this■scatterplot. y 50

x

45

x

m y

n

40

y

35 30 25

As■the■price■increases,■ the■number■of■bags■ sold■decreases.■ This■means■that■the■ relationship■is■negative.■ The■points■vary■quite■ a■bit■from■a■straight■ line,■so■this■indicates■ that■the■relationship■is■ moderately■strong.

20 o

x x

y

x

i Negative,■weak j Positive,■moderate k Positive,■moderate l Negative,■moderate m Negative,■strong n Positive,■weak o Positive,■moderate

probleM solViNg 1 Eugene■is■selling■handbags■at■the■local■market.■

During■the■day■he■keeps■records■of■his■sales.■The■ table■below■shows■the■number■of■bags■sold■over■ one■weekend■and■their■corresponding■prices■(to■the■ nearest■dollar).

15 10 5 5 10 15 20 25 30 35 40 x

■

■ a Use■the■line■of■best■fi■t■to■predict■the■value■of■y■

Birth mass (kg)

■

a What■type■of■data■is■this? Birth■mass b Which■is■the■dependent■variable? c Suggest■a■question■that■could■be■answered■by■

■ Price ($) of a bag

this■data. 30 35 40 45 50 55 60 65 70 75 80

Number of 10 12 ■ 8 ■ 6 ■ 4 ■ 3 ■ 4 ■ 2 ■ 2 ■ 1 ■ 1 bags sold a Which■is■the■dependent■variable? ■ Number■of■bags■sold b What■would■be■a■relevant■question■that■could■

be■answered■upon■examination■of■the■data? 494

1.1 1.5 1.8 2.1 2.2 2.5 2.8 3.1 3.1 3.4

Does■the■number■of■bags■sold■depend■on■the■price? Maths Quest 10 for the Australian Curriculum

d Construct■a■scatterplot■of■the■data■and■draw■in■

the■line■of■best■fi■t. e Full■term■of■gestation■is■considered■to■be■

40■weeks,■although■some■pregnancies■last■ longer.■Use■your■line■of■best■fi■t■to■predict■the■ mass■of■a■baby■born■after: i 41■weeks 3.7■kg ii 42■weeks. 4■kg

The■two■sets■of■data■are■ What■infl■uence■on■the■birth■mass■ quantitative■and■continuous. does■the■gestation■period■have?

Gestation time 31 32 33 34 35 36 37 38 39 40 (weeks)



when■the■value■of■x■is: i 10 12.5 ii 35. 49 b Use■the■line■of■best■fi■t■to■predict■the■value■of■x■ when■the■value■of■y■is: i 15 12 ii 30. 22.5 3 This■table■shows■the■gestation■time■and■the■birth■ mass■of■10■babies.

During■weeks■36■to■40■of■the■gestation■period,■the■ birth■mass■increased■about■0.3■■kg■per■week.■This■is■ supported■by■readings■from■the■line■of■best■fi■t.

f Many■babies■are■born■prematurely.■Use■your■

5 For■his■birthday,■Ari■was■given■a■small■white■rabbit.■

line■of■best■fi■t■to■predict■the■mass■of■a■baby■ born■at■30■weeks. 1■kg g What■was■the■gestation■time■(to■the■nearest■ week)■of■a■baby■born■with■a■birth■mass■of■ 2.4■kg? 36■weeks h Consider■your■question■from■part■c.■Supply■ an■answer,■indicating■how■you■came■to■this■ conclusion. 4 As■preparation■for■a■Mathematics■test,■a■group■of■ 22■students■was■given■a■revision■sheet■containing■ 60■questions.■The■table■below■shows■the■number■ of■questions■from■the■revision■sheet■successfully■ completed■by■each■student■and■the■mark,■out■of■ 100,■of■that■student■on■the■test.

To■monitor■the■rabbit’s■development,■Ari■decided■to■ measure■it■once■a■week.■The■table■below■shows■the■ length■of■the■rabbit■for■various■weeks.

Number of questions

■ 9 12 37 60 ■ 55 40 10 25 50 48 60

Test result

18 21 52 95 100 67 15 50 97 85 89

Week number

1

Length (cm)

20 21 23 24 25 30 32 35 36 37 39

2

3

4

6

8 10 13 14 17 20

a Construct■a■scatterplot■of■the■data,■and■draw■

the■line■of■best■fi■t. b As■can■be■seen■from■the■table,■Ari■did■not■

measure■his■rabbit■on■weeks■5,■7,■9,■11,■12,■ 15,■16,■18■and■19.■Use■your■line■of■best■fi■t■to■ predict■the■length■of■the■rabbit■for■those■weeks. c Predict■the■length■of■the■rabbit■for■the■next■ 3■weeks. 42■■cm,■43■■cm,■44■■cm d Comment■on■your■predictions■for■parts■b■and■c.

25■■cm,■27■■cm,■29■■cm,■31■■cm,■33■■cm,■36■■cm,■37■■cm,■39■■cm,■40■■cm

stAtistiCs AND probAbility • DAtA represeNtAtioN AND iNterpretAtioN

The■predictions■for■part■ b■are■quite■reliable,■as■ 50 48 35 29 ■ 19 44 49 20 16 58 52 they■have■been■made■ within■the■limits■of■the■ data.■The■predictions■in■ Test result 97 85 62 54 ■ 30 70 82 37 28 99 80 part■c■for■the■3■weeks■ beyond■the■upper■limit■ a Which■of■these■variables■is■the■dependent■ of■the■data■would■not■ variable■and■which■is■the■independent■variable? be■considered■reliable. b Suggest■a■question■that■could■be■answered■by■

Number of questions

this■set■of■data. c Draw■a■scatterplot■of■the■data. d What■type■of■relationship■exists■between■the■

two■variables? e What■answer■would■you■give■to■the■question■ you■suggested■in■part■b?■Provide■evidence■to■

support■your■answer. 4 a■ The■test■result■is■the■dependent■variable,■ while■the■number■of■questions■is■the■ independent■variable. b Does■completing■more■of■these■revision■ questions■contribute■to■a■higher■test■mark? c Length (cm)

Test result

100 90 80 70 60 50 40 30 20 10 0

L

5 10 15 20 25 30 35 40 45 50 55 60

Number of questions

39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20



eBook plus

Interactivities

Test yourself Chapter 14 int-2864 Word search Chapter 14 int-2862 Crossword Chapter 14 int-2863

0 1 2 3 4 5 6 7 8 9 1011121314151617181920 n

Week

d The■relationship■is■positive,■and■ moderately■strong. e There■is■evidence■to■suggest■that■ completing■more■revision■questions■is■ benefi■cial,■since■the■relationship■is■positive■ and■moderately■strong. Chapter 14 bivariate data

495

eBook plus

ACtiVities

chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■14■(doc-5326)■ (page 471) are you ready? Digital docs (pages 472–3) •■ SkillSHEET■14.1■(doc-5327):■Distinguishing■ qualitative■from■quantitative■data •■ SkillSHEET■14.2■(doc-5328):■Distinguishing■ discrete■from■continuous■data •■ SkillSHEET■14.3■(doc-5329):■Determining■ independent■and■dependent■variables •■ SkillSHEET■14.4■(doc-5330):■Reading■scales■(how■ much■is■each■interval■worth?) •■ SkillSHEET■14.5■(doc-5331):■Reading■a■column■ graph •■ SkillSHEET■14.6■(doc-5332):■Reading■line■graphs •■ SkillSHEET■14.7■(doc-5333):■Plotting■coordinate■ points

14a Identifying related pairs of variables Digital docs (page 476) •■ Activity■14-A-1■(doc-5146):■Pairs■of■variables •■ Activity■14-A-2■(doc-5147):■More■pairs■of■variables •■ Activity■14-A-3■(doc-5148):■Advanced■pairs■of■ variables

14b Graphing bivariate data Digital docs

•■ Activity■14-B-1■(doc-5149):■Scatterplots■(page 481) •■ Activity■14-B-2■(doc-5150):■More■scatterplots■ (page 481)

496

Maths Quest 10 for the Australian Curriculum

•■ Activity■14-B-3■(doc-5151):■Advanced■scatterplots■ (page 481) •■ WorkSHEET■14.1■(doc-5334):■Bivariate■data■I■ (page 483) 14c Scatterplots Digital docs

•■ Activity■14-C-1■(doc-5152):■Types■of■relationships■ (page 488) •■ Activity■14-C-2■(doc-5153):■More■relationships■ (page 488) •■ Activity■14-C-3■(doc-5154):■Advanced■relationships■ (page 488) •■ WorkSHEET■14.2■(doc-5335):■Bivariate■data■II■ (page 490) Interactivity

•■ Scatterplots■(int-2789)■(page 483) chapter review Interactivities (page 495) •■ Test■yourself■Chapter■14■(int-2864):■Take■the■end-ofchapter■test■to■test■your■progress. •■ Word■search■Chapter■14■(int-2862):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■14■(int-2863):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

statistics aND probability • Data represeNtatioN aND iNterpretatioN

15

15A Populations and samples 15B Primary and secondary data 15C Evaluating inquiry methods and statistical reports 15D Statistical investigations What Do you kNoW ?

statistics in the media

1 List what you know about how the media reports data. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of how the media reports data. eBook plus

Digital doc

Hungry brain activity Chapter 15 doc-5336

opeNiNg QuestioN

When scientists conduct experiments and tests in laboratories, how are their reports released to the public?

statistics aND probability • Data represeNtatioN aND iNterpretatioN

are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

Digital doc

SkillSHEET 15.1 doc-5337

eBook plus

Digital doc

SkillSHEET 15.2 doc-5338

Determining suitability of questions for a survey 1 Lauren■was■preparing■a■questionnaire■for■a■survey■on■graphics■calculator■usage■in■the■

classroom.■Would■the■following■be■suitable■questions? a Do■you■own■(or■have■access■to)■a■graphics■calculator? Suitable b Do■you■agree■that■graphics■calculators■are■too■expensive? Not■suitable■(irrelevant) c How■frequently■(on■average)■would■you■use■a■graphics■calculator■in■a■maths■lesson? Suitable Finding proportions 2 A■school■has■430■students■in■the■junior■school,■260■in■the■middle■school■and■170■in■the■senior■ 1 school.■Determine■the■proportion■of■students■in■each■of■the■three■sections. Junior■school:■ 2

13

Middle■school:■ 43 17

Senior■school:■ 86

Digital doc

SkillSHEET 15.3 doc-5339

eBook plus

Digital doc

Distinguishing between types of data 3 Decide■whether■the■following■data■are■categorical■or■numerical.■For■categorical■data,■state■

whether■they■are■ordinal■or■nominal.■For■numerical■data,■state■whether■they■are■continuous■or■ discrete. a Height■of■students■in■Year■10 Numerical,■continuous b Pets■owned■by■students Categorical,■nominal c Position■in■the■under-15■cross-country■race Categorical,■ordinal Reading bar graphs 4 The■graph■at■right■represents■the■

favourite■television■shows■of■ 500■teenagers. a What■are■the■most■popular■and■ least■popular■television■shows? Most■popular:■cartoons b How■many■teenagers■prefer■ least■popular:■ watching■comedy■television■ documentaries■and■ lifestyle■programs shows? 50 c How■many■more■teenagers■ prefer■soaps■to■thriller■ television■shows? 40 SkillSHEET 15.4 doc-5340

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Digital doc

SkillSHEET 15.5 doc-5341

Favourite television shows Television shows

eBook plus

Comedy Soaps Police Drama News Documentaries Cartoons Science Fiction Lifestyle Thriller

0

5%

10% 15% 20% Percentage favouring

25%

Determining independent and dependent variables 5 For■each■of■the■following,■state■the■independent■and■the■dependent■variables. a The■number■of■kilograms■of■potatoes■purchased■and■the■total■cost. b The■number■of■swimmers■in■a■public■swimming■pool■and■the■day■temperature. c The■height■and■age■of■a■student. a■ Number■of■kilograms:■independent total■cost:■dependent b Temperature:■independent number■of■swimmers:■dependent c Age:■independent height:■dependent

498

maths Quest 10 for the australian curriculum

statistics AND probability • Data representation and interpretation

15A

Populations and samples Populations ■■ ■■ ■■ ■■

The term population refers to a complete set of individuals, objects or events belonging to some category. When data are collected from a whole population, the process is known as a census. It is often not possible, nor cost-effective, to conduct a census. For this reason, samples have to be selected carefully from the population. A sample is a subset of its population.

Worked Example 1

List some of the problems you might encounter in trying to collect data on the following populations. a The life of a mobile phone battery b The number of possums in a local area c The number of males in Australia d The average cost of a loaf of white bread Think

write

For each of these scenarios, consider how the data might be collected, and the problems in obtaining this data. a The life of a mobile phone battery

a The life of a mobile phone battery can not be measured

until it is dead. The battery life also depends on how the phone is used, and how many times it has been recharged. b The number of possums in a local area

b It would be almost impossible to find all the possums

in a local area in order to count them. The possums also may stray into other areas. c

The number of males in Australia

d The average cost of a loaf of white

bread

c

The number of males in Australia is constantly changing. There are births and deaths every second.

d The price of one particular loaf of white bread varies

widely from one location to another. Sometimes the bread is on ‘Special’ and this would affect the calculations.

■■ ■■

Since collecting data from a population is sometimes not an option, sampling provides a viable alternative. Many methods for obtaining appropriate samples have been discussed previously.

Samples ■■ ■■

■■

Surveys are conducted on samples. Ideally the sample should reveal generalisations about the population. A random sample is generally accepted as being an ideal representation of the population from which it was drawn. However, it must be remembered that different random samples from the same population can produce different results. This means that we must be cautious about making predictions about a population from surveys conducted on samples. A sample size must be sufficiently large. As a general rule, the sample size should be about N , where N is the size of the population. It is a misconception that a larger sample will produce a more reliable prediction of the characteristics of its population. Chapter 15 Statistics in the media

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statistics AND probability • Data representation and interpretation

Worked Example 2

A die was rolled 50 times and the following results were obtained. 6  5  3  1  6  2  3  6  2  5  3  4  1  3  2  6  4  5  5  4  3  1  2  1  6  4  5  2  3  6  1  5  3  3  2  4  1  4  2  3  2  6  3  4  6  2  1  2  4  2. a Determine the mean of the population (to 1 decimal place). b A suitable sample size for this population would be 7( 50 ö 7.1).   i  Select a random sample of 7 scores, and determine the mean of these scores.    ii  Select a second random sample of 7 scores, and determine the mean of these. iii  Select a third random sample of 20 scores, and determine the mean of these. c Comment on your answers to parts a and b. Think

write

a Calculate the mean by first finding the

a Population mean

sum of all the scores, then dividing by the number of scores (50).

b

i

Use a calculator to randomly generate 7 scores from 1 to 50 Relate these numbers back to the scores, then calculate the mean.

∑x n 169 = 50 = 3.4 =

b The 7 scores randomly selected are numbers 17, 50,

11, 40, 48, 12, 19 in the set of 50 scores. These correspond to the scores 4, 2, 3, 3, 2, 4, 5. 23 The mean of these scores = 7 = 3.3

ii Repeat bi to obtain a second set of

Ignore the second and third attempts to select 7 random numbers because of repeated numbers. The second set of 7 scores randomly selected is numbers 16, 49, 2, 42, 31, 11, 50 of the set of 50. These correspond to the scores 6, 4, 5, 6, 1, 3, 2. 27 The mean of these scores = 7 = 3.9

7 randomly selected scores. This second set of random numbers produced the number 1 twice. Try again. Another attempt produced the number 14 twice. Try again. A third attempt produced 7 different numbers. This set of 7 random numbers will then be used to, again, calculate the mean of the scores. iii Repeat for a randomly selected

The set of 20 randomly selected numbers produced a total of 68. 68 Mean of 20 random scores = 20 = 3.4

20 scores. c

Comment on the results.

c

The population mean is 3.4. The means of the two samples of 7 are 3.3 and 3.9. This shows that, even though the samples are randomly selected, their calculated means may be different. The mean of the sample of 20 scores is 3.4. This indicates that by using a bigger sample the result is more accurate than those obtained with the smaller samples.

Note: This example deals with quite small numbers. It does, however, illustrate the fact that randomly selected samples provide an estimate of population statistics, but different random samples often produce different results. 500

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

To sample or to conduct a census? ■■

■■

The particular circumstances determine whether data are collected from a population, or from a sample of the population. For example, suppose you collected data on the height of every Year 10 student in your class. If your class was the only Year 10 class in the school, your class would be the population. If, however, there were several Year 10 classes in your school, your class would be a sample of the Year 10 population. In the previous worked example, it was seen that different random samples can produce different results. For this reason, it is important to acknowledge that there could be some uncertainty when using sample results to make predictions about the population.

Worked Example 3

For each of the following situations, state whether the information was obtained by census or survey. Justify why that particular method was used. a A roll call is conducted each morning at school to determine which students are absent. b TV ratings are collected from a selection of viewers to discover the popular TV shows. c Every hundredth light bulb off an assembly production line is tested to determine the life of that type of light bulb. d A teacher records the examination results of her class. Think

write

a Every student is recorded as

a This is a census. If the roll call only applied to a sample of the

being present or absent at the roll call. b Only a selection of the TV

students, there would not be an accurate record of attendance at school. A census is essential in this case. b This is a survey. To collect data from the whole viewer

audience contributed to this data. c

Only 1 bulb in every 100 is tested.

d Every student’s result is

recorded.

population would be time-consuming and expensive. For this reason, it is appropriate to select a sample to conduct the survey. c

This is a survey. Light bulbs are tested to destruction (burn-out) to determine their life. If every bulb was tested in this way, there would be none left to sell! A survey on a sample is essential.

d This is a census. It is essential to record the result of every

student.

remember

1. The term population refers to a complete set of individuals, objects or events belonging to some category. 2. When data are collected from a whole population, the process is known as a census. 3. Surveys are conducted on samples. Ideally the sample should reveal generalisations about the population. 4. Different random samples from the same population can produce different results. 5. As a general rule, the sample size should be about N , where N is the size of the population. 6. It is a misconception that a larger sample will produce a more reliable prediction of the characteristics of its population. 7. The particular circumstances determine whether data are being collected from the population, or from a sample of the population. 8. It is important to acknowledge that there could be some uncertainty when using sample results to make predictions about the population. Chapter 15 Statistics in the media

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statistics aND probability • Data represeNtatioN aND iNterpretatioN

exercise

15a iNDiviDual pathWays eBook plus

Activity 15-A-1

Populations and samples doc-5155 Activity 15-A-2

More populations and samples doc-5156

populations and samples

When■was■it■fi■rst■put■into■the■machine?■How■old■ was■the■battery■before■being■purchased?■How■ frequently■has■the■computer■been■used■on■battery?

flueNcy 1 We1 ■List■some■of■the■problems■you■might■encounter■in■trying■to■collect■data■from■the■ following■populations. Can’t■always■see■if■a■residence■has■a■dog;■ A■census■is■very■time-consuming;■Perhaps■ a The■life■of■a■laptop■computer■battery. could■approach■council■for■dog■registrations. b The■number■of■dogs■in■your■neighbourhood This■number■is■never■constant■ c The■number■of■fi■sh■for■sale■at■the■fi■sh■markets with■ongoing■purchases,■and■ d The■average■number■of■pieces■of■popcorn■in■a■bag■of■popcorn continuously■replenishing■stock. 2 We2 ■The■data■below■show■the■results■of■the■rolled■die■from■Worked■example■2.

6■ 5■ 3■ 1■ 6■ 2■ 3■ 6■ 2■ 5■ 3■ 4■ 1■ 3■ 2■ 6■ 4■ 5■ 5■ 4■ 3■ 1■ 2■ 1■ 6■ 4■ 5■ These■answers■will■vary■with■ 2■ 3■ 6■ 1■ 5■ 3■ 3■ 2■ 4■ 1■ 4■ 2■ 3■ 2■ 6■ 3■ 4■ 6■ 2■ 1■ 2■ 4■ 2. the■samples■chosen.

Activity 15-A-3

The■mean■of■the■population■is■3.4.■Select■your■own■samples■for■the■following■questions. Select■a■random■sample■of■7■scores,■and■determine■the■mean■of■these■scores. ■Census.■The■ Select■a■second■random■sample■of■7■scores,■and■determine■the■mean■of■these. airline■must■have■ a■record■of■every■ Select■a■third■random■sample■of■20■scores,■and■determine■the■mean■of■these. passenger■on■ Would■have■to■sample■ Comment■on■your■answers■to■parts■a,■b■and■c. every■fl■ight. in■this■case■as■a■census■ 3 We3 ■In■each■of■the■following■scenarios,■state■whether■the■information■was■obtained■by■ would■involve■opening■ census■or■survey.■Justify■why■that■particular■method■was■used. every■packet. a Seating■for■all■passengers■is■recorded■for■each■aeroplane■fl■ight. b Movie■ratings■are■collected■from■a■selection■of■viewers■to■discover■the■best■movies■for■ the■week. Survey.■It■would■be■impossible■to■interview■everyone. c Every■hundredth■soft■drink■bottle■off■an■assembly■production■line■is■measured■to■ Census.■The■instructor■ determine■the■volume■of■its■contents. Survey.■A■census■would■involve■opening■every■bottle. must■have■an■accurate■ d A■car■driving■instructor■records■the■number■of■hours■each■learner■driver■has■spent■driving. record■of■each■learner■ driver’s■progress. 4 For■each■of■the■following,■state■whether■a■census■or■a■survey■has■been■used. a Two■hundred■people■in■a■shopping■centre■are■asked■to■nominate■the■supermarket■where■ they■do■most■of■their■grocery■shopping. Survey b To■fi■nd■the■most■popular■new■car■on■the■road,■500■new■car■buyers■are■asked■what■make■ and■model■they■purchased. Survey c To■fi■nd■the■most■popular■new■car■on■the■road,■data■are■obtained■from■the■transport■ department. Census d Your■Year■10■Maths■class■completed■a■series■of■questions■on■the■amount■of■maths■ homework■for■Year■10■students. Survey Drawing■numbers■from■a■hat,■ In depth populations and samples doc-5157

a b c d

using■a■calculator,■…..

uNDerstaNDiNg

To■conduct■a■statistical■investigation,■Gloria■needs■to■obtain■information■from■630■students. a What■size■sample■would■be■appropriate? About■25 b Describe■a■method■of■generating■a■set■of■random■numbers■for■this■sample. A■local■council■wants■the■opinions■of■its■residents■regarding■its■endeavours■to■establish■a■new■ sporting■facility■for■the■community.■It■has■specifi■cally■requested■all■residents■over■10■years■of■ age■to■respond■to■a■set■of■on-line■questions. a Is■this■a■census■or■a■survey? b What■problems■could■you■encounter■collecting■data■this■way? 7 A■poll■was■conducted■at■a■school■a■few■days■before■the■election■for■Head■Boy■and■Head■Girl.■ Residents■may■not■ After■the■election,■it■was■discovered■that■the■polls■were■completely■misleading.■Explain■how■ all■have■internet■ access.■Only■ this■could■have■happened. The■sample■could■have■been■biased.■The■questionnaire■may■have■been■unclear. those■who■are■ 8 A■sampling■error■is■said■to■occur■when■results■of■a■sample■are■different■from■those■of■the■ highly■motivated■ population■from■which■the■sample■was■drawn.■Discuss■some■factors■which■could■introduce■ are■likely■to■ respond. sampling■errors. Sample■size,■randomness■of■sample

The■council■is■ 5 probably■hoping■ it■is■a■census,■but■ it■will■probably■be■ a■survey■because■ 6 not■all■those■over■ 10■will■respond.

502

maths Quest 10 for the australian curriculum

statistics aND probability • Data represeNtatioN aND iNterpretatioN Populations■growing■ very■rapidly,■large■ number■of■expatriate■ 9 Since■1961,■a■census■has■been■conducted■in■Australia■every■5■years.■Some■people■object■to■the■ census■on■the■basis■that■their■privacy■is■being■invaded.■Others■say■that■the■expense■involved■ workers■in■China■ have■a■different■ could■be■directed■to■a■better■cause.■Others■say■that■a■sample■could■obtain■statistics■which■are■ background■and■forms■ just■as■accurate.■What■are■your■views■on■this?■Justify■your■statements. need■to■be■modifi■ed■ Answers■will■vary.■Check■with■your■teacher.■ for■them,■people■ reasoNiNg from■Hong■Kong■ working■on■mainland■ 10 Australia■has■a■very■small■population■compared■with■other■countries■like■China■and■India.■ China,■large■migrant■ These■are■the■world’s■most■populous■nations,■so■the■problems■we■encounter■in■conducting■a■ population■in■New■ census■in■Australia■would■be■insignifi■cant■compared■with■those■encountered■in■those■countries.■ Delhi,■often■migrants■ don’t■have■residency■ What■different■problems■would■authorities■come■across■when■conducting■a■census■there? permits■(so■the■truth■ 11 The■game■of■Lotto■involves■picking■the■same■6■numbers■in■the■range■1■to■45■as■has■been■ of■their■answers■ randomly■selected■by■a■machine■containing■45■numbered■balls.■The■balls■are■mixed■ is■questionable),■ many■people■live■in■ thoroughly,■then■8■balls■are■selected■representing■the■6■main■numbers,■plus■2■extra■numbers,■ inaccessible■areas,■ called■supplementary■numbers. some■families■in■ Here■is■a■list■of■the■number■of■times■each■number■had■been■drawn■over■a■period■of■time,■and■ China■have■more■than■ 1■child■and■do■not■ also■the■number■of■weeks■since■each■particular■number■has■been■drawn. disclose■this.

NUMBER OF WEEKS SINCE EACH NUMBER DRAWN

NUMBER OF TIMES EACH NUMBER DRAWN SINCE DRAW 413

1

2

3

4

5

6

7

8

1

5

2

1

1

7

-

4

9

10

11

12

13

14

15

16

3

3

1

5

5

7

-

4

17

18

19

20

21

22

23

24

9

-

9

2

2

12

10

8

25

26

27

28

29

30

31

32

5

11

17

2

3

3

-

22

33

34

35

36

37

38

39

40

4

3

-

1

12

-

6

-

41

42

43

44

45

6

1

7

-

31

2

Digital doc

9

10

4

5

6

7

8

11

12

13

14

15

16

17

18

19

20

21

22

23

24

217 233 240 226 238 240 253 228 25

26

27

28

29

30

31

32

252 239 198 229 227 204 230 226 33

34

35

36

37

38

39

40

246 233 232 251 222 221 219 259 41

42

43

44

45

245 242 237 221 224

There■is■quite■a■variation■in■the■frequency■of■particular■ numbers■drawn.■For■example,■the■number■45■has■not■ been■drawn■for■31■weeks,■while■most■have■been■drawn■ within■the■last■10■weeks.■In■the■long■term,■one■should■ fi■nd■the■frequency■of■drawing■each■number■is■roughly■ the■same.■It■may■take■a■long■time■for■this■to■happen,■as■ only■8■numbers■are■drawn■each■week.

WorkSHEET 15.1 doc-5342

3

228 213 250 233 224 221 240 223

If■these■numbers■are■randomly■chosen,■explain■the■ differences■shown■in■the■tables.

eBook plus

15b

1

246 238 244 227 249 241 253 266

reflectioN 

 

A well-known saying about statistics is: Statistics means never having to say you’re certain. What does this saying mean?

primary and secondary data primary data ■■ ■■ ■■

Primary data■collection■involves■collecting■data■yourself. This■means■that■you■have■ownership■of■the■data,■and■no■one■else■has■access■to■the■data■until■it■ is■released■or■published. A■variety■of■methods■of■collecting■primary■data■is■available.■These■include■observation,■ measurement,■survey,■experiment■or■simulation. chapter 15 statistics in the media

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statistics AND probability • Data representation and interpretation

Observation ■■ ■■

■■ ■■

This involves recording the behavioural patterns of people, objects and events in a systematic manner. The data can be collected as a disguised observation (respondents are unaware they are being observed) or undisguised observation (the respondent is aware). CCT cameras are an example of people knowing that their movements are being recorded, but are not always aware of where the recording takes place. Observations can be in a natural environment (for example, in a food hall), or a contrived environment (a food-tasting session for a food company). Mechanical devices (video cameras, closed circuit television, counting devices across a road) can also be used.

Measurement ■■ ■■

Measurement involves using some measuring device to collect data. This generally involves conducting an experiment of some type. •• The height of everyone in your class can be measured. •• The mass of all new-born babies can be collected. •• A pedometer measures the number of paces the wearer takes.

Surveys ■■ ■■

■■

Surveys involve designing a questionnaire to interview people. Often the questionnaire requires many rewrites to obtain one which is clear and unbiased. The interview can be in person — face to face or by telephone. The advantage of this method is that you are able to see the reactions of those you are interviewing, and explain particular questions, if necessary. Most frequently these days, email is used; however, there are advantages and disadvantages to using this type of survey. •• Advantages –– Can cover a large number of people or organisations –– A wide geographic coverage is possible –– It avoids embarrassment on the part of the respondent –– There is no interviewer bias –– The respondent has time to consider responses –– It’s relatively cheap •• Disadvantages –– The questions have to be relatively simple –– The response rate is often quite low (inducements often given as an incentive to return) –– The reliability of the answers is questionable –– No control over who actually completes the questionnaire –– Problems with incomplete questionnaires

Experiment ■■ ■■ ■■ ■■ ■■

Generally, when conducting an experiment the data collected are quantitative. Particular care should be taken to ensure that the experiment is conducted in a manner which would produce similar results if repeated. Care must be taken with the recording of results. The results must be in a form which can readily be analysed. All results need to be recorded, including the weird or unexpected outcomes

Simulation ■■ ■■ ■■

504

Experiments such as rolling a die, tossing a coin or drawing a card from a deck may be conducted to model some real-life situation. Simulations occur in areas such as business, engineering, medical and scientific research. They are often used to imitate real-life situations which may be dangerous, impractical or too expensive to explore by other means.

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Worked Example 4

It is widely believed that there is equal chance of having a boy or girl with each birth. Genetics and the history of births in a family sometimes have a great influence on the sex of the child as well. Ignore those factors in this question. a Design an experiment to simulate the chance of giving birth to a boy or a girl. b Describe how your experiment could be conducted to determine the number of children a couple should have, on average, to ensure they have offspring of both sexes. Think

write

a Use a device that can simulate two

outcomes which are equally likely. This could be a random number generator to generate two integers, say a 0 (representing a boy) and 1 (representing a girl). A fair coin could be tossed, such that, a Head represents a boy, and a Tail represents a girl. b

1

2

Describe how the experiment will be conducted.

Display the table of results.

a A fair coin will be tossed with a Head representing ■

a boy (B), and a Tail representing a girl (G).

b The experiment will be conducted 50 times, and a

record kept of each experiment. For each experiment, the coin will be tossed until both sexes result. This may mean that there could be 7 trials in an experiment (GGGGGGB) before both sexes are represented. The table below shows the results of the 50 experiments. Exp. No. of no. Results trials  1 BG 2  2 GGB 3  3 BG 2  4 GGGGB 5   5 BBBBBBG 7  6 GGGB 4  7 BBG 3  8 BBG 3  9 BBBBG 5 10 GB 2 11 BG 2 12 GGGB 4 13 BBG 3 14 BBG 3 15 GB 2 16 BG 2 17 GGB 3 18 GB 2 19 GGB 3 20 BBBG 4 21 BG 2 22 GB 2 23 GGGGB 5 24 BG 2 25 GGGGB 5

Exp. no. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

No. of Results trials GGGB    4 GGGGB    5 GGGB    4 BG    2 BBBG    4 BG    2 GB    2 GGGB    4 BG    2 GGGGGGB    7 BBBBBBG    7 GB    2 BG    2 GGB    3 GGGGB    5 BBG    3 BBBBBG    6 GGB    3 GGB    3 BBBG    4 BBG    3 GGGGGGB    7 BG    2 BBG    3 GGGGGB    6 Total 175 This table shows that 175 trials were undertaken in 50 experiments where each experiment resulted in both sexes. Chapter 15 Statistics in the media

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statistics AND probability • Data representation and interpretation

175

3

Determine the average number of children required to produce offspring of both sexes.

Average number of children =

4

Write a conclusion.

The average number of children a couple should have to reach the goal of having both sexes is 4.

■■ ■■ ■■ ■■

50

= 3.5

Before collecting any primary data, it must be clear what data are to be collected. A decision must be made as to the method of collection. The advantages and disadvantages of the collection method must be acknowledged. The reason for the data collection should be clear from the outset.

Worked Example 5

You have been asked to obtain primary data to determine the methods of transport used to travel to school by the students at your school. The data collected are to provide support for the Student Council’s proposal for a school bus. a What data should be collected? b Outline possible methods which could be used to collect this data. c Decide which method you consider to be the best option, and discuss its advantages and disadvantages. Think

write

a Outline the various forms of transport

a The modes of transport available to students at the

available to the students. b Consider all the alternatives for collecting

school are: car, bus, train, bicycle and walking b Several methods could be used to collect the data.

the data.

c

•• C  ould stand at the school gate one morning and ask students as they arrive •• A questionnaire could be designed •• Students could be asked to write their mode of transport on a piece of paper and place in a collection tin.

1

Decide on best option.

2

Discuss advantages and disadvantages.

c

The first option of standing at the school gate is very time-consuming, and students could arrive at the back gate. The third option does not seem reliable, as some students may not comply, and other students may place multiple pieces of paper in the collection tin. The second option seems the best of the three. The advantages of a questionnaire include: •• There is a permanent record on paper. •• It is not time-consuming to distribute or collect. •• Students can complete it at their leisure. Disadvantages include: •• Students may not return it. •• Expense involved in producing copies.

Note: This example does not represent the views of all those collecting such data. It merely serves to challenge students to explore and discuss available options. 506

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

■■ ■■

Sometimes the primary data required are not obvious at the outset of the investigation. For example, you are asked to investigate the claim: Most students do not eat a proper breakfast before school. What questions would you ask to prove or refute this claim?

Secondary data ■■ ■■

■■ ■■ ■■ ■■ ■■ ■■

Secondary data are data which have already been collected by someone else. The data can come from a variety of sources: •• Paper — books, journals, magazines, company reports •• Electronic — online databases, internet, broadcasts, DVDs •• Government sources — ABS provides a wealth of statistical data •• General business sources — academic institutions, stockbroking firms, sporting clubs •• Media — newspapers, TV reports. Secondary data sources often provide data which would not be possible for an individual to collect. The data can be qualitative or quantitative. The accuracy and reliability of the data sometimes needs to be questioned, depending on its source. The age of the data should always be considered. Often the data which surrounds us passes by unnoticed. It is important to learn the skills to be able to critically analyse secondary data.

Worked Example 6

Subway advertises the energy and fat content of some of their subs on their napkins. a What information can you gain from this data? b Subway advertise that they have a range of subs with less than 6 grams of fat. Comment on this claim. c This could be the starting point of a statistical investigation. How could you proceed from here? d Investigations are not conducted simply for the sake of investigating. Suggest some aims for investigating further.

Think a Look at the data on the napkin to gain as

much information as possible.

write a The napkin reveals the following information:

•• A higher energy content of a sub does not necessarily mean that its fat content is higher. •• As the fat content of a sub increases, generally the saturated fat content also increases. •• The addition of some types of protein (ham, turkey, beef, chicken) increases the energy content of the sub. •• These data are only for those subs on white or wheat bread with salads and meat. •• The addition of condiments (sauces) or cheese will alter these figures. •• An apple slice has much less energy and fat ■ than a sub. Chapter 15 Statistics in the media

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statistics AND probability • Data representation and interpretation b Examine the data to discover if there is

b All the subs displayed have less than 6 grams of

evidence to support the claim. Make further comment.

c

What would be the next step in the investigation?

d What are some interesting facts which could

be revealed through a deeper investigation?

fat, so Subway’s claim is true. It must be remembered that the addition of cheese and sauce to these subs would increase their fat content. Also, if the sub was on any bread other than white or wheat, the fat content could go beyond 6 grams. c

The napkin displays a toll-free phone number for further information. Their web site also contains additional detailed information.

d Suggested aims for investigating further could be:

•• How much extra fat is added to a sub by the addition of cheese and/or sauce? •• What difference does a different type of bread make to the fat content of the sub? •• Which sub contains the highest fat content? •• What is the sugar content of the subs?

remember

1. Primary data collection •• This involves collecting data yourself. •• You have ownership of the data, and no one else has access to the data until it is released or published. •• A variety of methods of collecting the data is available including observation, measurement, survey, experiment or simulation. 2. Secondary data collection •• This is data which has already been collected by someone else. •• The data can come from a variety of sources including: paper, electronic, government sources, general business sources and the media. •• Secondary data sources often provide data which would not be possible for an individual to collect. •• The data can be qualitative or quantitative. •• The accuracy and reliability of the data sometimes needs to be questioned, depending on its source. •• The age of the data should always be considered. •• It is important to learn the skills to be able to critically analyse secondary data. Exercise

15b

Primary and secondary data

fluency These are simply examples of 1   WE 4  Devise an experiment to simulate each of the following situations and specify the device simulations which used to represent the outcomes. Coin could be flipped (Heads represents True, while Tails represents False) could be conducted. a A true/false test in which answers are randomly distributed. b A casino game with outcomes grouped in colours of either red or black. Coin could be flipped (Heads represents ‘red’, while Tails represents ‘black’) 508 Maths Quest 10 for the Australian Curriculum

statistics aND probability • Data represeNtatioN aND iNterpretatioN Spinner■with■4■equal■sectors■(each■sector■representing■a■different■toy)

iNDiviDual pathWays eBook plus

Activity 15-B-1

Data collection doc-5158 Activity 15-B-2

Further data collection doc-5159 Activity 15-B-3

Advanced data collection doc-5160

Breakfast■cereal■boxes■containing■4■different■types■of■plastic■toys. In■a■group■of■six■people,■one■person■is■to■be■chosen■as■the■leader. A■choice■of■three■main■meals■on■a■restaurant’s■menu,■all■of■which■are■equally■popular. Five■possible■holiday■destinations■offered■by■a■travel■agent;■such■that■all■destinations■are■ equally■available■and■equally■priced. g Five■types■of■takeaway■fast■foods■available■in■one■area,■where■one■pizza■is■twice■ as■popular■as■each■of■the■others■types■of■takeaway■food■(the■other■4■are■equally■ popular). 2 We5 ■You■have■been■asked■to■obtain■primary■data■from■students■at■your■school■to■determine■ internet■access■students■have■at■home.■The■data■collected■are■to■provide■support■for■opening■the■ computer■room■for■student■use■at■night. a What■data■should■be■collected? b Outline■possible■methods■which■could■be■used■to■collect■this■data. c Decide■which■method■you■consider■to■be■the■best■option,■and■discuss■its■advantages■and■ disadvantages. Answers■will■vary.■Check■with■your■teacher. 3 We6 ■This■label■shows■the■nutritional■information■of■Brand■X■rolled■oats. c d e f

d R oll■a■die■(each■face■represents■ a■particular■person) e Spinner■with■3■equal■sectors■ (each■one■representing■a■ particular■meal) f Spinner■with■5■equal■sectors■ (each■one■representing■a■ particular■destination) g Spinner■with■5■sectors,■one■ which■will■have■an■angle■size■ of■120è,■while■the■other■4■each■ have■an■angle■size■of■60è■(each■ one■representing■a■particular■ fast■food)

Nutrition Information Servings Per Package: 25 Serving Size 30g Per 100g Per Serving 30g %Dl* Per Serving Energy 486kJ 6% 1620kJ Protein 4.3g 9% 14.3g Fat - Total 2.8g 4% 9.3g - Saturated 0.5g 2% 1.7g - Trans Less than 0.1g - Less than 0.1g - Polyunsaturated 1.0g 3.2g - Monounsaturated 1.3g 4.4g Carbohydrate 16.8g 5% 56g - Sugars 0.9g 1% 3.0g Dietary Fibre 3.1g 10% 10.4g Sodium 0.7mg 0.1% 2mg

Answers■will■vary,■ however■some■possible■ suggestions■include: Which■students■have■ internet■access■at■ home?■ Do■the■students■need■ access■at■night?■ What■hours■would■be■ suitable?■ How■many■would■ make■use■of■this■ facility?

* % DI = Percentage daily intake

a What■information■can■you■gain■from■this■data? b This■could■be■the■starting■point■of■a■statistical■investigation.■How■could■you■proceed■from■

here? c Suggest■some■aims■for■investigating■further.





Answers■will■vary,■ however■some■ possible■suggestions■ include: a Census,■survey,■ questionnaire,■ interview,■ observation,■ experiment,■online■response,■… b i■ Measurement ii Observation iii Newspaper■ recordings iv Survey

4 a■ b

Provide■a■list■of■methods■you■could■use■to■collect■primary■data. Describe■which■method■you■would■use■to■collect■the■following■primary■data. i Heights■of■trees■along■the■footpaths■of■a■tree-lined■street ii Number■of■buses■that■transport■students■to■your■school■in■the■morning iii Sunrise■times■during■summer iv Student■opinion■regarding■length■of■lessons

uNDerstaNDiNg

For■questions■5■and■6,■design■an■experiment■to■simulate■the■situation,■carry■out■the■experiment■ and■give■the■results■of■the■experiment. 5 A■mouse■in■a■maze■can■make■left■or■right■turns■at■each■junction.■Assuming■each■turn■is■equally■ likely,■how■many■junctions■on■average■must■the■mouse■go■through■before■each■type■of■turn■will■ have■been■made? Student’s■own■response 6 A■restaurant■menu■features■4■desserts■which■are■assumed■to■be■equally■popular.■How■many■ dessert■orders■must■be■fi■lled■(on■average)■before■the■owner■can■be■sure■all■types■will■have■been■ ordered? Student’s■own■response chapter 15 statistics in the media

509

statistics AND probability • Data representation and interpretation 7 This label shows the nutritional information of Brand Y rolled oats. nutrition information Servings per package: 30 Serving size: 30g Avg. Quantity Avg. Quantity Per serving 30g Per 100g Energy 480kJ (115Cal) 1600kJ (383Cal) Protein 3.2g 10.5g Fat, total 2.4g 8.0g - saturated LESS THAN 1g 1.5g Carbohydrate 18.3g 61.0g - sugars 0.0g 0.0g Dietary Fibre, total 3.3g 10.0g Sodium LESS THAN 5mg LESS THAN 5mg ingredients Oats (100%) attention

THIS PRODUCT CONTAINS GLUTEN. storage Store in a cool, dry place. Student’s own response

Compare the nutritional information with that on the Brand X label on page 509. 8 Comment on this claim.

The claim is false. It is not a logical deduction.

We did a survey on 100 people regarding eating chocolate. 60 of these people said they regularly ate chocolate. We then measured the heights of all 100 people. **** The result **** Eating chocolate makes you taller!!

510

ì 100%). Sleepmaker Umbria 42% off

1999

10 The following claim has been made regarding secondary data.

There is at least 40% off these beds.

( 1800 4299

3899

Sleepmaker Casablanca 40% off ( 800 ì 100%),

Sealy Posturepedic 41% off ( 1600 ì 100%),

11 Sealy Posturepremier 40% off ( 1000 ì 100%), 2499

9 Russel operates a computer software sales outlet. He keeps a log of all complaints from customers. Suggest how he could organise his log. Student’s own response

There’s a lot more secondary data than primary data, it’s a lot cheaper and it’s easier to acquire. Comment on this statement. Student’s own response Reasoning 11 The local Bed Barn was having a sale on selected beds by Sealy

and Sleepmaker. Four of the beds on sale were: Sealy Posturepremier on sale for $1499 a saving of $1000 Sealy Posturepedic on sale for $2299 a saving of $1600 Sleepmaker Casablanca on sale for $1199 a saving of $800 Sleepmaker Umbria on sale for $2499 a saving of $1800 The store claimed that all these beds had been discounted by 40%. Comment on whether this statement is true, supporting your comments with sound mathematical reasoning.

Maths Quest 10 for the Australian Curriculum

reflection 

 

When using secondary data from other countries, what different unit classifications could you encounter?

statistics aND probability • Data represeNtatioN aND iNterpretatioN

15c eBook plus

evaluating inquiry methods and statistical reports ■■

Interactivity Compare statistical reports

■■

int-2790

■■

Data■investigations■involve■collecting■data,■recording■the■data,■analysing■the■data■then■ reporting■the■data.

Data collection methods

■■ ■■ ■■

Collection■methods■involve■gathering■primary■data,■or■using■secondary■data■from■stored■records. Primary■data■can■be■collected■by■observation,■measurement,■survey,■experiment■or■simulation■ as■discussed■earlier. Secondary■data■can■be■collected■electronically■or■via■a■hard■copy. It■is■important■to■be■able■to■justify■the■particular■method■chosen■for■each■of■these■processes. Sometimes■alternative■methods■are■just■as■appropriate.

WorkeD example 7

You have been given an assignment to investigate which year level uses the school library, after school, the most. a Explain whether it is more appropriate to use primary or secondary data in this case. Justify your choice. b Describe how the data could be collected. Discuss any problems which might be encountered. c Explain whether an alternative method would be just as appropriate. thiNk a No■records■have■

Write a Since■records■are■not■kept■on■the■library■use,■secondary■data■is■not■an■option.

been■kept■on■library■ use.

b The■data■can■be■

Primary■data■collection■could■be■either■sampling■or■census.■A■suffi■ciently■ large■sample■size■could■be■chosen;■this■would■take■less■time■than■conducting■ a■census,■although■it■would■not■be■as■accurate. Sampling■would■be■considered■appropriate■in■this■case. b A■questionnaire■could■be■designed■and■distributed■to■a■randomly-chosen■

collected■via■a■ questionnaire■or■in■ person. c

A■census■is■the■other■ option.

sample.■The■problem■here■would■be■the■non-return■of■the■forms. Observation■could■be■used■to■personally■interviewed■students■as■they■entered■ the■library.■This■would■take■more■time,■but■random■interview■times■could■be■ selected. c

A■census■could■be■conducted,■either■by■questionnaire■or■observation.■This■ should■yield■a■more■accurate■outcome.

WorkeD example 8

Which method would be the most appropriate to collect the following data? Suggest an alternative method in each case. a The number of cars parked in the staff car park each day. b The mass of books students carry to school each day. c The length a spring stretches when weights are added to it. d The cost of mobile phone plans with various network providers. thiNk a Observation

Write a The■best■way■would■probably■be■observation■by■visiting■the■staff■car■park■to■

count■the■number■of■cars■there. An■alternative■method■would■be■to■conduct■a■census■of■all■workers■to■ask■if■ they■parked■in■the■staff■car■park.■This■is■probably■not■as■good. chapter 15 statistics in the media

511

statistics AND probability • Data representation and interpretation b Measurement

b The mass of the books could be measured by weighing each student’s pack

on scales. A random sample would probably yield a reasonably accurate result. c

c

Experiment

d Internet search

Conduct an experiment and measure the extension of the spring with various weights. There is probably no alternative to this method.

d An internet search would enable data to be collected.

Alternatively, a visit to mobile phone outlets would yield similar results.

Analysing the data ■■ ■■ ■■ ■■

Once the data have been collected and collated, a decision must be made with regard to the best methods for analysing the data. A measure of central tendency should be chosen — mean, median or mode A measure of spread (range, interquartile range) indicates how the data is distributed. An appropriate graph gives a visual representation of the data.

Graphing statistical data ■■ ■■ ■■ ■■

Statistical data can be graphed in a variety of ways — line graphs, bar graphs, histograms, stem plots, box plots, etc. These have all been discussed in detail previously. In media reports it is common to see line and bar graphs. Because graphs give a quick visual impression, the temptation is to not look at them in great detail. Often these graphs can be quite misleading. It is easy to manipulate a graph to give an impression which is supported by the creator of the graph. This is achieved by careful choice of scale on the horizontal and vertical axes. •• Shortening the horizontal axis tends to highlight the increasing/decreasing nature of the trend of the graph. Lengthening the vertical axis tends to have the same effect. •• Lengthening the horizontal and shortening the vertical axes tends to level out the trends.

Worked Example 9

This report shows the annual change in median house prices in the local government areas (LGA) of Queensland from 2008–09 to 2009–10. a Draw a bar graph which would give the impression that the percentage annual change was much the same throughout the whole state. b Construct a bar graph to give the impression that the percentage annual change in Brisbane was far greater than that in the other local government areas.

512

Maths Quest 10 for the Australian Curriculum

HOUSES

Suburb/locality

Median house price 2008–09 2009–10

Annual change

Brisbane (LGA) Ipswich City (LGA) Redland City (LGA) Logan City (LGA) Moreton Bay (LGA) Gold Coast City (LGA) Toowoomba (LGA) Sunshine Coast (LGA) Fraser Coast (LGA) Bundaberg (LGA) Gladstone (LGA) Rockhampton (LGA) Mackay (LGA) Townsville City (LGA) Cairns (LGA)

$530,000 $323,000 $467,500 $360,000 $399,000 $505,000 $289,500 $470,000 $307,400 $282,000 $370,000 $315,250 $398,000 $375,000 $365,000

11.6% 4.2% 7.5% 5.9% 7.3% 8.6% 7.6% 5.6% 3.2% 2.5% 0.0% 5.1% 3.9% 4.5% 2.8%

$475,000 $310,000 $435,000 $340,000 $372,000 $465,000 $269,000 $445,000 $297,750 $275,000 $370,000 $300,000 $383,000 $359,000 $355,000

statistics AND probability • Data representation and interpretation

Think a

Annual % change

a To flatten out trends,

Write/draw

lengthen the horizontal axis and shorten the vertical axis.

% house price changes in QLD 2008–9 to 2009–10 10 5

Cairns

Townsville

Mackay

Rockhampton

Gladstone

Bundaberg

Fraser Coast

Sunshine Coast

Toowoomba

Gold Coast

Moreton Bay

Logan

Redland

Ipswich

Brisbane

0

Area

% house price changes in QLD 2008–9 to 2009–10

12 11 10 9 8 7 6 5 4 3 2 0

Brisbane Ipswich Redland Logan Moreton Bay Gold Coast Toowoomba Sunshine Coast Fraser Coast Bundaberg Gladstone Rockhampton Mackay Townsville Cairns

shorten the horizontal axis and lengthen the vertical axis.

b

Annual % change

b To accentuate trends,

Area Worked Example 10

Consider the data displayed in the table of Worked example 9. Use the data collected for the median house prices in 2009–10. a  Explain whether this data would be classed as primary or secondary data. b  Why does this data show median house prices rather than the mean or modal house price? c  Calculate a measure of central tendency for the data. Explain the reason for this choice. d  Give a measure of spread of the data, giving a reason for the particular choice. e  Display the data in a graphical form, explaining why this particular form was chosen. Think a This is data which has been collected

by someone else. b Median is the middle price, mean is the

average price, and mode is the most frequently-occurring price.

Write a This is secondary data because it has been collected by

someone else. b The median price is the middle one. It is not affected by

outliers as the mean is. The modal house price may only occur for two house sales with the same value. On the other hand, there may not be any mode. The median price is the most appropriate in this case. Chapter 15 Statistics in the media

513

statistics AND probability • Data representation and interpretation c

Which measure of central tendency is the most appropriate one?

d Consider the range and the interquartile

range as measures of spread.

e

Consider the graphing options.

c

The measures of central tendency are the mean, median and mode. The mean is affected by high values (i.e. $530  000) and low values (i.e. $282  000). These are not typical values, so the mean would not be appropriate. There is no modal value, as all the house prices are different. The median house price is the most suitable measure of central tendency to represent the house prices in the Queensland local government areas. The median value is $370 000.

d The five-number summary values are:

Lowest score = $282  000 Lower quartile = $315  250 Median = $370  000 Upper quartile = $467  500 Highest score = $530  000 Range = $530  000 - $282  000 = $248  000 Interquartile range = $467  500 − $315  250 = $152  250 The interquartile range is a better measure for the range as the house prices form a cluster in this region. e Of all the graphing options, the box plot seems the most appropriate as it shows the spread of the prices as well as how they are grouped around the median price.

280000

■■

340000 400000 460000 520000 Median house price 2009-10 ($)

Secondary sources of data provide a great starting point for investigations.

Worked Example 11

The Australian women’s national basketball team, the Opals, competed at the 2008 Olympic Games in Beijing, winning a silver medal. These are the heights (in metres) of the 12 team members: 1.73, 1.65, 1.8, 1.83, 1.96, 1.88, 1.63, 1.88, 1.83, 1.88, 1.8, 1.96 Provide calculations and explanations as evidence to verify or refute the following statements. a  The mean height of the team is greater than their median height. b  The range of the heights of the 12 players is almost 3 times their interquartile range. c Only 5 players are on the court at any one time. A team of 5 players can be chosen such that their mean, median and modal heights are all the same. 514

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Think a

b

c

Write a Mean =

∑ x 21.83 = = 1.82 m n 12

1

Calculate the mean height of the 12 players.

2

Order the heights to determine the median.

3

Comment on the statement.

1

Determine the range and the interquartile range of the ■ 12 heights.

2

Compare the two values.

Range = 0.33 m Interquartile range = 0.115 m Range 0.33 = = 2.9 Interquartile range 0.115

3

Comment on the statement.

Range = 2.9 ì interquartile range This is almost 3 times, so the statement is true.

1

Choose 5 players whose mean, median and modal heights are all equal. Trial and error is appropriate here. There may be more than one answer.

2

Comment on the statement.

The heights of the players, in order, is: 1.63, 1.65, 1.73, 1.8, 1.8, 1.83, 1.83, 1.88, ■ 1.88, 1.88, 1.96, 1.96 There are 12 scores, so the median is the average of the 6th and 7th scores. 1.83 + 1.83 Median = = 1.83 m 2 The mean is 1.82 m, while the median is 1.83 m. This means that the mean is less than the median, so the statement is not true. b Range = 1.96 - 1.63 = 0.33 m Lower quartile is the average of 3rd and 4th scores. 1.73 + 1.8 Lower quartile = = 1.765 m 2 Upper quartile is average of 3rd and 4th scores from the end. 1.88 + 1.88 Upper quartile = = 1.88 m 2 Interquartile range = 1.88 - 1.765 = 0.115 m

c

Three players have a height of 1.88 m. If a player shorter and one taller are chosen both the same measurement from 1.88 m, this would make the mean, median and mode all the same. Choose players with heights: 1.8, 1.88, 1.88, 1.88, 1.96 9.4 Mean = = 1.88 m 5 Median = 3rd score = 1.88 m Mode = Most frequent score = 1.88 m The 5 players with heights 1.8 m, 1.88 m, 1.88 m, ■ 1.88 m, 1.96 m have a mean, median and modal height of 1.88 m. It is true that a team of 5 such players can be chosen.

Statistical reports ■■

Reported data must not be simply taken at face value; all reports should be examined with a critical eye. Chapter 15 Statistics in the media

515

statistics AND probability • Data representation and interpretation

Worked Example 12

This is an excerpt from an article which appeared in a newspaper on Father’s Day. It was reported to be a national survey findings of a Gallup Poll of data from 1255 fathers of children aged 17 and under.

The great Aussie dads survey Thinking about all aspects of your life, how happy would you say you are? % I am very happy.................................................26 I am fairly happy................................................49 Totally happy.....................................................75 Some days I’m happy and some days I’m not...............................................................21 I am fairly unhappy.............................................3 I am very unhappy...............................................1 Total unhappy.....................................................4 How often, if ever, do you regret having children? Every day.............................................................1 Most days............................................................2 Some days.........................................................18 Never.................................................................79 Which one of these best describes the impact of having children on your relationship with your partner? We’re closer than ever.......................................29 We don’t spend as much time together as we should . ...................................................40 We’re more like friends now than lovers..........21 We have drifted apart..........................................6 None of the above...............................................4 Which one of these best describes the allocation of cooking and cleaning duties in your household? My partner does nothing/I do everything...........1 I do most of it....................................................11 We share the cooking and cleaning..................42 My partner does most of it................................41 I do nothing/my partner does everything............4 None of the above...............................................1 Which of these aspects of your children’s future do you have concerns about? Their safety........................................................70 Being exposed to drugs.....................................67 Their health.......................................................54

% Bullying or cyber-bullying.................................50 Teenage violence...............................................50 Their ability to afford a home...........................50 Alcohol consumption and binge drinking........47 Achieving academic success..............................47 Feeling pressured into sex.................................41 Being able to afford the lifestyle they expect to have....................................................38 Climate change..................................................23 Having them living with you in their mid 20s..............................................................14 None of the above...............................................3 What is the best thing about being a dad? The simple pleasures of family life....................61 Enjoying the successes of your kids.................24 The unpredictability it brings.............................9 The comfort of knowing that you will be looked after in later life.......................................3 None of the above...............................................3

Key findings

75% 79% 67% 57%

of Aussie dads are totally happy have never regretted having children are worried about their children being exposed to drugs would like more intimacy with their partner

“Work-life balance is definitely an issue for dads in 2010.” David Briggs Galaxy principal

Source: The Sunday Mail, 5 Sept. 2010, pp. 14–15.

a  Comment on the sample chosen. b  Discuss the percentages displayed. c  Comment on the claim that 57% of dads would like more intimacy with their partner. 516

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Think

Write

a How is the sample chosen?

a The results of a national survey such as this should reveal the

Is it truly representative of the population of Australian dads? b Look at the percentages in

outlook of the whole nation’s dads. There is no indication of how the sample was chosen, so without further knowledge we tend to accept that it is representative of the population. A sample of 1255 is probably large enough. b For the first question regarding happiness, the percentages total

each of the categories.

c

more than 100%. It seems logical that, in a question such as this, the respondents would tick only one box, but obviously this has not been the case. In the question regarding aspects of concern of ‘your children’s future’, these percentages also total more than 100%. It seems appropriate here that dads would have more than one concerning area, so it is possible for the percentages to total more than 100%. In each of the other three questions, the percentages total 100%, which is appropriate.

Look at the tables to try to find the source of this figure.

■■

c

Examining the reported percentages in the question regarding ‘relationship with your partner’, there is no indication how a figure of 57% was determined.

Frequently media reports make claims where the reader has no hope of confirming their truth.

Worked Example 13

This article appeared in a newspaper. Read the article, then answer the following questions.

Sponges are toxic Washing dishes can pose a serious health risk, with more than half of all kitchen sponges containing high levels of dangerous bacteria, research shows. A new survey dishing the dirt on washing up shows more than 50 per cent of kitchen sponges have high levels of E coli, which can cause severe cramps and diarrhoea, and staphylococcus aureus, which releases toxins that can lead to food poisoning or toxic shock syndrome.

Microbiologist Craig Andrew-Kabilafkas of Australian Food Microbiology said the Westinghouse study of more than 1000 households revealed germs can spread easily to freshly washed dishes. The only way to safeguard homes from sickness was to wash utensils at very high temperatures in a dishwasher.

Source: The Sunday Mail, 5 Sept. 2010, p. 36.

a  Comment on the sample used in this survey. b  Comment on the claims of the survey. c  Is the heading of the article appropriate? Think a Look at sample size and

selection of sample.

Write a The report claims that the sample size was more than 1000.

There is no indication how the sample was selected. The point to keep in mind is whether this sample is truly representative of the population consisting of all households. We have no way of knowing. Chapter 15 Statistics in the media

517

statistics aND probability • Data represeNtatioN aND iNterpretatioN b What■are■the■results■of■the■

b The■survey■claims■that■50%■of■kitchen■sponges■have■high■levels■

survey?

c

of■E. coli■which■can■cause■severe■medical■problems. The■study■was■conducted■by■Westinghouse,■so■it■is■not■surprising■ they■recommend■using■a■dishwasher.

Examine■the■heading■in■the■ light■of■the■contents■the■article.

c

The■heading■is■sensational,■designed■to■catch■the■attention■of■ readers.

remember

1.■ Collecting■data •■ Primary■data■can■be■gathered■in■a■variety■of■ways. •■ The■particular■method■chosen■for■a■statistical■investigation■should■be■justifi■ed. •■ Secondary■data■are■gathered■from■stored■records. 2.■ Analysing■data •■ A■measure■of■central■tendency■should■be■chosen■—■mean,■median■or■mode •■ A■measure■of■spread■(range,■interquartile■range)■indicates■how■the■data■is■distributed. •■ An■appropriate■graph■gives■a■visual■representation■of■the■data. 3.■ Evaluating■reported■data •■ The■data■should■be■examined■with■a■critical■eye. •■ Often■graphs■can■be■misleading. –■ ■Shortening■the■horizontal■axis■and■lengthening■the■vertical■axis■tends■to■highlight■ the■increasing/decreasing■nature■of■the■trend■of■the■graph.■ –■ ■Lengthening■the■horizontal■and■shortening■the■vertical■axes■tends■to■level■out■the■ trends. exercise

15c iNDiviDual pathWays eBook plus

Activity 15-C-1

Collecting and analysing data doc-5161 Activity 15-C-2

More collecting and analysing data doc-5162 Activity 15-C-3

Detailed collecting and analysing data doc-5163

evaluating inquiry methods and statistical reports flueNcy 1 We7,8 ■You■have■been■given■an■assignment■to■investigate■which■Year■level■has■the■greatest■

number■of■students■who■are■driven■to■school■each■day■by■car. a Explain■whether■it■is■more■appropriate■to■use■primary■or■secondary■data■in■this■case.■ Justify■your■choice. ■Primary.■There■is■probably■no■secondary■data■available. b Describe■how■the■data■could■be■collected.■Discuss■any■problems■which■might■be■ encountered. Answers■will■vary.■Check■with■your■teacher.■ Answers■will■vary.■Check■ c Explain■whether■an■alternative■method■would■be■just■as■appropriate. with■your■teacher.■ 2 We9 ■You■run■a■small■company■that■is■listed■on■the■Australian■Stock■Exchange■(ASX).■During■

the■past■year■you■have■given■substantial■rises■in■salary■to■all■your■staff.■However,■profi■ts■have■ not■been■as■spectacular■as■in■the■year■before.■This■table■gives■the■fi■gures■for■the■salary■and■ profi■ts■for■each■quarter. 1st quarter

2nd quarter

3rd quarter

4th quarter

Profi■ts $’000■■000

6

5.9

6

6.5

Salaries $’000■■000

4

5

6

7

Draw■two■graphs,■one■showing■profi■ts,■the■other■showing■salaries,■which■will■show■you■in■ the■best■possible■light■to■your■shareholders. 518

maths Quest 10 for the australian curriculum

statistics AND probability • Data representation and interpretation 3   WE 10  The data below were collected from a real estate agent and show the sale prices of ten

blocks of land in a new estate. $150 000, $190 000, $175 000, $150 000, $650 000, $150 000, $165 000, ■ $180 000, $160 000, $180 000 a Calculate a measure of central tendency for the data. Explain the reason for this

choice. b Give a measure of spread of the data, giving a reason for the particular choice. c Display the data in a graphical form, explaining why this particular form was chosen. d The real estate agent advertises the new estate land as:

Own one of these amazing blocks of land for only $150 000 (average)! Comment on the agent’s claims. 4   WE 11  Use the data for the heights of the Opal players in Worked example 11 (page 514) to answer the following question. Provide calculations and explanations as evidence to verify or refute the following statements. a The mean height of the team is closer to the lower quartile than it is to the median. b Half the players have a height within the interquartile range. c Which 5 players could be chosen to have the minimum range in heights? 5 This table below shows the number of shoes of each size that were sold over a week at a shoe store. a Mean = $215  000, median = $170  000, mode = $150  000. The median best represents these land prices. The mean is inflated by one large score, and the mode is the lowest price. b Range = $500  000, interquartile range = $30  000. The interquartile range is the better measure of spread. c 150000

300000 450000 Price

600000

This dot plot shows how 9 of the scores are grouped close together, while the score of $650 000 is an outlier. d The agent is quoting the modal price, which is the lowest price. This is not a true reflection of the average price of these blocks of land.

Size

Number sold

 4

 5

 5

 7

 6

19

 7

24

 8

16

 9

 8

10

 7

a True. Mean = 1.82 m, lower quartile = 1.765 m, median = 1.83 m b True. This is the definition of interquartile range. c Players with heights 1.83 m, 1.83 m, 1.88 m, 1.88 m, ■ 1.88 m

Calculate the mean shoe size sold. 7.1 Determine the median shoe size sold. 7 Determine the modal shoe size sold. 7 Explain which measure of central tendency has the most meaning to the proprietor. The mode has the most meaning as this size sells the most. 6 The resting pulse of 20 female athletes was measured and is shown below. 50  62  48  52  71  61  30  45  42  48  43  47  51  52  34  61  44  54  38  40 a b c d

a Represent the data in a distribution table using appropriate groupings. b Find the mean, median and mode of the data. c Comment on the similarities and differences between the three values.

store Check with your teacher. Answers depend on groupings used.

Understanding 7 The batting scores for two cricket players over six innings were recorded as follows.

Player A  31, 34, 42, 28, 30, 41 Player B  0, 0, 1, 0, 250, 0 Player B was hailed as a hero for his score of 250. Comment on the performance of the two players.

Player B appears to be the better player if the mean result is used. However, Player A is the moreconsistent player.

Chapter 15 Statistics in the media

519

statistics AND probability • Data representation and interpretation 8 A small manufacturing plant employs 80 workers. This table below shows the structure of

the plant. Position

The median and modal salary is $18 000 and a only 15 out of 80 (less than 20%) earn more than the mean. b

Salary ($)

Number of employees

Machine operator

18  000

50

Machine mechanic

20  000

15

Floor steward

24  000

10

Manager

62  000

 4

Chief Executive Officer

80  000

 1

Workers are arguing for a pay rise, but the management of the factory claims that workers are well paid because the mean salary of the factory is $22  100. Explain whether this is a sound argument. The statement is true, but misleading as most of the employees earn $18 000. Suppose that you were representing the factory workers and had to write a short submission in support of the pay rise. How could you explain the management’s claim? Provide some other statistics to support your case.

■■ ■■

Points which could be mentioned. 10.1% is only just ‘double digit’ growth. 2006–08 showed mid to low 20% growth. Growth has been declining since 2008. ■■ Share price has rebounded, but not to its previous high. ■■ Share price scale is not consistent. Most increments are 30c, except for $27.70 to $28.10 (40c increment). Note also the figure of 20.80 — probably a typo instead of 26.80.

9   WE 12,13  This report from Woolworths appeared in a newspaper.

It’s a record ■■ ■■ ■■ ■■

Woolworths posted 10.1% gain in annual profit to $2.02b 11th consecutive year of double-digit growth Flags 8% to 11% growth in the current financial year Sales rose 4.8% to $51.2b

SHARES REBOUND $ 28.40 28.10 27.70 27.40 27.10 20.80 26.50 26.20 25.90 25.60

2.4% Yesterday

■■ ■■

Wants to increase its share of the fresh food market Announced $700m off-market share buyback Final fully franked dividend 62¢ a share

NET PROFIT +$2.02b

$b 2

+25.7%

1.5 1

+12.8%

+10.1%

+27.5% +24.3%

0.5 May 26

Aug 26

Source: IRESS Source: The Courier Mail, 27 Aug. 2010, pp. 40–1.

Comment on the report. 520

■■

Maths Quest 10 for the Australian Curriculum

0

2006

2007

2008

2009

2010

statistics AND probability • Data representation and interpretation Reasoning

during the period 13 July to 13 September 2010. The higher the Australian dollar, the cheaper it is for Australian companies to import goods from overseas, and the cheaper they should be able to sell their goods to the Australian public. The manager of Company XYZ produced a graph to support his claim that, because there hasn’t been much change in the Aussie dollar over that period, there hasn’t been any change in the price he sells his imported goods to the Australian public. Draw a graph which would support his claim. Explain how you were able to achieve this effect.

13 July

Time

Aussie dollar

10 Shorten the y-axis and expand the x-axis. US c

13 September

10 This graph at right shows the fluctuation in the Australian dollar in terms of the US dollar

reflection 

0 90 c

US¢ 92.8

US 93.29¢

90.9 88.8 86.8 84.8 82.8 80.8

 

Jul 13

Sep 13 Source: IRESS

What is the point of drawing a misleading graph in a report?

80 c

15D

AUSSIE

Source: The Courier Mail, 14 Sept. 2010, p. 25.

Statistical investigations Using primary data ■■ ■■

■■

This section deals with the steps involved in carrying out a statistical investigation with primary data collection. For this exercise we will assume you have been given this task. Which pizza on the market is the best value for money? This is a very broad investigation, and each stage of the investigation must be carefully planned. •• Collecting the data •• Organising the data •• Performing calculations •• Analysing the data •• Reporting the results

Collecting the data ■■

At this initial stage, questions should be posed with regard to the data. •• What data should be collected? –– Best value for money involves the price and size of the pizza. Data on both of these need to be collected. –– Stores have different prices for different sizes. –– Would size best be measured as area or mass? –– Not all pizzas are round; some are rectangular. –– What about the variety of toppings? A standard one should be chosen. –– Should frozen pizzas be included? •• How should the data be collected? –– It is not possible to buy every pizza on the market, so what alternatives are there? –– A store is probably not willing to allow their pizzas to be weighed, so mass is most likely out of the question. –– Will the store allow their pizzas to be measured? •• What problems are likely to be encountered? –– How many different companies market pizza? Chapter 15 Statistics in the media

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statistics AND probability • Data representation and interpretation

Organising the data ■■

■■

■■

The data should be organised into some sort of table format. •• What format is appropriate for this investigation? •• A table with column headings Price and Measurements will organise this data. Take time to design the table so figures required for calculation are readily visible. •• What calculations are required at this stage? •• Measurements are required to calculate the area of each pizza. Think forward and add extra columns for future calculations. •• What further calculations are needed? •• Area and Value for money need to be calculated. Provide two extra columns for these.

Performing calculations ■■ ■■

What calculations need to be performed? The Area and Value for money are required in this case. How should this be calculated? Dividing price by area gives ($/cm2), while dividing area by price gives (cm2/$).

Analysing the data ■■ ■■ ■■

■■

Are there any anomalies, or obvious calculation errors? Do the calculated results ‘make sense’? In this case, if Value for money is calculated in units of $/cm2, the pizza with the smallest of these values is the best value for money. Using units of cm2/$, the pizza with the highest of these values gives the best value for money. Would the inclusion of graphs be appropriate?

Reporting the results ■■ ■■ ■■

The results should be reported in a clear, concise manner. Justify any conclusions. Are there any anomalies or exceptions to mention?

Using secondary data ■■

The procedure for undertaking a statistical investigation using secondary data is similar to that for primary data, the difference being that you sometimes have to search for data in several areas before you find the appropriate source. Suppose you were given this assignment. There have been __________ Prime Ministers of Australia since 1901 until this day. There have been _________ elections. __________ Prime Ministers have been defeated at a general election. There have been ________ changes of Prime Minister without an election. The average length these Prime Ministers served in office is ____________ . Undertake a statistical investigation to complete the details.

■■

■■

■■

522

Collecting the data •• What data should be collected? •• Where can this data be found? The internet is probably a good starting point, but not all sites are reliable. •• If there are multiple sources for the data, are they all in agreement? •• How many of these statements require calculations? Organising the data •• Design a table to record all the data. •• Consider how many columns are necessary. •• Leave columns for calculations. Performing calculations •• There is at least one calculation here — to determine the average length of time served in office. Are there any more?

Maths Quest 10 for the Australian Curriculum

statistics AND probability • Data representation and interpretation

■■

■■

Analysing the data •• Do all the calculated values make sense? •• Would a graph be appropriate? Reporting the results •• Complete the details. •• Acknowledge the source of the secondary data.

Investigating media reports ■■

Frequently reports in the media provide a good starting point for an interesting investigation. Here are a few suggestions. Media report 1 Here is an article on an analysis of the speech habits of two high-profile parliamentarians — Julia Gillard and Tony Abbott. It was written by Roly Sussex, a professor of English.

Roly Sussex

Word Limit

[email protected] We all have individual features in the way we speak. Our tone of voice, an intonation, a rhythm, a favourite word or phrase — the things that make us quickly recognised even on bad telephone connections. But if you are a public figure, and especially a political leader, your speech habits will be picked up, criticised, satirised and caricatured. As I discussed in previous weeks, our current political leaders show a wide variety of habits and idiosyncrasies. Our new PM, Julia Gillard is a very consecutive speaker. Like Kevin Rudd before her, she speaks in long complete sentences. But unlike Rudd, her sentences contain a fair number of pre-programmed mantras and phrases: “happy to be judged”, “enhanced the capacity”, “regional neighbours”. Her pronunciation is also distinctive. She grew up in South Australia, and so says W instead of L at the end of a syllable. Her “milk” is MIWK, and her “football” is FOOTBAW. And she is our first Prime Minister to have high rising tone, the rising intonation at the end of a clause (rising pitch). Her favourite word is “obviously” and she has also quickly assumed the leader’s “I”. The closer you get to the top job, the more the ego asserts itself in grammar. Especially in answer to a probing question: “I’m not going to be .  .  .”. Former PM John Howard perfected this technique, and it is piously observed by our current leaders. Treasurer Wayne Swan is acquiring it. Aha.

Compared to the PM, Opposition Leader Tony Abbott is a less fluent speaker. He is an “um”, “look” and “ah” man. His sentences contain pauses, sometimes for reflection, sometimes for emphasis, sometimes both. He is somewhat less given to mantra, and greetings-wise he is more a man of the people: he says “G’day”. But Gillard and Abbott share three features which are now so ingrained under the fingernails of our pollies that they won’t scrub off. One feature is repetition. “As I said in my speech .  .  .” says the PM, “.  .  . as I said in my speech”. Well, yes, we know that. We heard the speech. Abbott, on the other hand, repeats repeated negatives about the Government: “Spin .  .  . contradiction .  .  . incompetent .  .  . disarray .  .  .”. It’s like swearing — the more you use these words, the less meaning they convey. The second shared feature is the preprogrammed response. A trigger in the question presses a specific answer-button. “Asylum seekers”, “deficit”, “mining super-tax”, “health” and similar key issues prompt the automatic rehearsed rejoinder. You know it’s pre-programmed because you’ll always hear the same words, whenever the trigger is pressed. The third thing they share is that they won’t say “yes” or “no”. Both respond to a “can you tell us, yes or no?” with streams of verbal flimflam. Interviewers should give up trying to prise a clear yea/nay out of either of them. But the public does have a right to know, yes or no, where they stand on issues, and we aren’t getting what we crave.

Source: The Courier Mail, 14–15 Aug. 2010, p. 25.

There is no doubt that these comments are true. How could you find evidence of this? Chapter 15 Statistics in the media

523

statistics AND probability • Data representation and interpretation

Media report 2

Single Women Earn More WASHINGTON: The income of one group of US women is catching up to and even overtaking men, a study shows. They are single women in their 20s without children, who live in large cities and work full-time, according to a study of census data by Reach Advisors, a New York–based strategy and research firm focused on emerging shifts in the consumer landscape. These young women earn on average 8 per cent more than men in their age group, but in some cities, such as Atlanta in Georgia and Memphis, Tennessee, women earn about one-fifth more than men. On average, American women who

work full-time earn about 80 per cent of what men earn. The report says that one reason for the finding is that girls are “going to college in droves”. Nearly three-quarters of girls who complete high school go on to university, compared with two-thirds of boys. 1 Women are 1 2 times more likely than men to graduate from university and to obtain a masters degree or higher. Census data released in April showed that 58 per cent of all US masters degrees or PhDs were awarded to women. As women go further in their education, they are also delaying getting married and starting a family.

Source: The Weekend Australian, 4–5 Sept. 2010, p. 20.

Is this report really true? Is it perhaps only true in America? (The heading seems to suggest that it is universally true.) What is the status of women in other parts of the world? Further investigation could reveal interesting comparisons. Media report 3

Egg shortage is no yolk Producers lay plan to meet need Peddy Hintz Blame MasterChef or the Heart Foundation, but it’s getting harder to find the right kind of eggs at the supermarket — and it’s likely to stay that way until Christmas. Queensland egg producers are struggling to keep up with demand but the boom in sales has also been matched by the recent interest in keeping backyard chickens. The winter shortage of eggs on Australian supermarket shelves will mean that instead of sitting in a coolroom for a week, eggs are being transferred to shelves almost straight from the supplier. “It’s currently taking only about 48 hours from being laid to getting onto the shelves so the eggs that people do buy will be a really good, fresh product,” chief executive of Sunny Queen, one of the country’s biggest suppliers, John O’Hara said. The Australian Egg Corporation has put the increased demand for eggs down Source: The Courier Mail, 28–9 Aug. 2010, p. 13.

524

Maths Quest 10 for the Australian Curriculum

to revised Heart Foundation guidelines raising the number of eggs recommended for a healthy diet from two a week to six. But, Mr O’Hara said, cooking shows such as MasterChef had also led to a rise in demand. Current estimates have Australians eating 203–205 eggs per person a year, compared with 195 last year, 156 the year before and a low of 132 10 years ago. UNSCRAMBLING EGGS Annual egg consumption (per person) Australia: 205 Japan: 320 US: 230 UK: 230 NZ: 214 Egg sales 65% caged 24–25% free range 10% cage free Fresh is best The most effective way to test if an egg is fresh is to put it in water. The more it sinks, the fresher the egg. If it floats, it’s nearly off.

statistics aND probability • Data represeNtatioN aND iNterpretatioN

Note■the■catchy■heading■on■this■article.■Does■the■advice■from■the■Heart■Foundation■or■ cooking■shows■like■Master■Chef■really■have■that■much■effect■on■egg■sales?■How■does■egg■ consumption■in■Australia■compare■with■that■in■the■other■countries■mentioned?■This■is■worthy■ of■further■investigation.

remember

1.■ The■following■steps■are■involved■in■a■statistical■investigation. •■ Collecting■the■data •■ Organising■the■data •■ Performing■calculations •■ Analysing■the■data •■ Reporting■the■results 2.■ Media■reports■provide■a■starting■point■for■further■investigations.

exercise

15D iNDiviDual pathWays eBook plus

Activity 15-D-1

Analysing reports doc-5164 Activity 15-D-2

Analysing reports in depth doc-5165 Activity 15-D-3

Analysing reports in greater depth doc-5166

statistical investigations uNDerstaNDiNg

This■section■has■guided■you■through■undertaking■a■statistical■investigation.■Apply■this■ knowledge■in■answering■these■questions. The■questions■in■this■exercise■relate■to■student■investigations,■ so■there■will■be■a■variety■of■answers. 1 a■ ■Write■a■plan■detailing■how■you■would■collect■primary■data■to■undertake■an■investigation■to■

determine■which■pizza■on■the■market■is■the■best■value■for■money. b Undertake■your■investigation. c Report■on■the■results■of■your■fi■ndings. 2 Undertake■the■investigation■on■the■history■of■Prime■Ministers■in■Australia.■Report■your■

fi■ndings. 3 Find■evidence■from■speeches■of■Julia■Gillard■and■Tony■Abbott■to■support■Roly■Sussex’s■report■

on■the■speech■habits■of■these■two■politicians. 4 Do■single■women■really■earn■more?■Investigate. 5 What’s■the■story■on■egg■consumption■in■Australia? 6 Search■for■a■media■article■you■would■like■to■investigate.■Provide■a■full■report■on■your■

fi■ndings.

There■have■been■ 27■Prime■Ministers■ reasoNiNg of■Australia■since■ 1901■until■this■day. There■have■been■42■ 7 Below■are■a■few■statistics■on■Facebook■users.■These■fi■gures■are■those■reported■in■the■ Year■2010. elections. ■■ There■are■more■than■400■million■active■users. 10■Prime■Ministers■ have■been■defeated■ ■■ 70%■of■Facebook■users■are■outside■the■US. at■a■general■ ■■ 50%■of■active■users■log■on■to■Facebook■in■any■given■day. election. ■■ More■than■60■million■updates■are■posted■each■day. There■have■been■21■ ■■ More■than■3■billion■photos■are■uploaded■to■the■site■each■month. changes■of■Prime■ Minister■without■an■ ■■ The■average■user■has■130■friends■on■the■site. election. ■■ The■average■user■spends■more■than■55■minutes■per■day■on■Facebook. The■average■ The■top■10■countries■on■Facebook■represent■just■a■little■over■half■of■the■Facebook■users.■ length■these■Prime■ Ministers■served■ China■(population■1.3■billion)■and■India■(1.2■billion)■do■not■appear■in■the■top■10■list.■Write■ in■offi■ce■is■(This■ a■report■summarising■the■usage■of■Facebook■throughout■the■world.■(See■overleaf■for■some■ changes■daily).

statistics.)

chapter 15 statistics in the media

525

statistics aND probability • Data represeNtatioN aND iNterpretatioN

These■are■the■top■10■countries■on■Facebook. Country

Population (millions)

% of world population

Users (millions)

1

USA

310.3

4.5

111.2

2

UK

62

0.9

23.5

3

Indonesia

237.6

3.5

19.5

4

Turkey

72.6

1.1

18.7

5

France

65.4

1.0

15.9

6

Italy

60.4

0.9

14.9

7

Canada

34.3

0.5

13.4

8

Philippines

94

1.4

10.6

Digital doc

9

Spain

46.1

0.7

8.9

WorkSHEET 15.2 doc-5343

10

108.4

1.6

8.2

eBook plus

Mexico

reflectioN 

 

What would you consider to be the most important factor in reporting the results of a statistical investigation?

526

maths Quest 10 for the australian curriculum

statistics AND probability • Data representation and interpretation

Summary Populations and samples ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

The term population refers to a complete set of individuals, objects or events belonging to some category. When data are collected from a whole population, the process is known as a census. Surveys are conducted on samples. Ideally the sample should reveal generalisations about the population. Different random samples from the same population can produce different results. As a general rule, the sample size should be about N , where N is the size of the population. It is a misconception that a larger sample will produce a more reliable prediction of the characteristics of its population. The particular circumstances determine whether data are being collected from the population, or from a sample of the population. It is important to acknowledge that there could be some uncertainty when using sample results to make predictions about the population. Primary and secondary data

■■

■■

Primary data collection •• This involves collecting data yourself. •• You have ownership of the data, and no one else has access to the data until it is released or published. •• A variety of methods of collecting the data is available including observation, measurement, survey, experiment or simulation. Secondary data collection •• This is data which has already been collected by someone else. •• The data can come from a variety of sources including: paper, electronic, government sources, general business sources and the media. •• Secondary data sources often provide data which would not be possible for an individual to collect. •• The data can be qualitative or quantitative. •• The accuracy and reliability of the data sometimes needs to be questioned, depending on its source. •• The age of the data should always be considered. •• It is important to learn the skills to be able to critically analyse secondary data. Evaluating inquiry methods and statistical reports

■■

■■

■■

Collecting data •• Primary data can be gathered in a variety of ways. •• The particular method chosen for a statistical investigation should be justified. •• Secondary data are gathered from stored records. Analysing data •• A measure of central tendency should be chosen — mean, median or mode •• A measure of spread (range, interquartile range) indicates how the data is distributed. •• An appropriate graph gives a visual representation of the data. Evaluating reported data •• The data should be examined with a critical eye. •• Often graphs can be misleading. – Shortening the horizontal axis and lengthening the vertical axis tends to highlight the increasing/decreasing nature of the trend of the graph. –  Lengthening the horizontal and shortening the vertical axes tends to level out the trends. Chapter 15 Statistics in the media

527

statistics aND probability • Data represeNtatioN aND iNterpretatioN Statistical investigations ■■

■■

The■following■steps■are■involved■in■a■statistical■investigation. •■ Collecting■the■data •■ Organising■the■data •■ Performing■calculations •■ Analysing■the■data •■ Reporting■the■results Media■reports■provide■a■starting■point■for■further■investigations.

MAPPING YOUR UNDERSTANDING

Homework Book

528

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■497. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

maths Quest 10 for the australian curriculum

statistics AND probability • Data representation and interpretation

Chapter review

1 a You would need to open every can to determine this. b F  ish are continuously dying, being born, being caught. c Approaching work places and public transport offices

Fluency

6 The table below shows the number of students in

each year level from Years 7 to 12. 6 Check with your teacher. a This graph should look relatively flat, with little decline in the Years 11 and 12 region. b T  his graph should show a sharp decline in the Years 11 and 12 region.

1 List some problems you might encounter in trying

to collect data from the following populations. a The average number of mL in a can of soft Year Number of students drink. b The number of fish in a dam.  7 230 c The number of workers who catch public transport to work each weekday morning.  8 200 2 a Calculate the mean of the integers 1 to 100. 50.5  9 189 b      i Randomly select 10 numbers in the range 1 to 100. 10 175 ii Calculate the mean of these numbers. c      i Randomly select 20 numbers in the range 11 133 1 to 100. Answers will vary. ii Calculate the mean of these numbers. 12 124 d Comment on the similarities/differences between your means calculated in parts a, b and c. Draw two separate graphs to illustrate the 3 For each of the following investigations, state following. whether a census or a survey has been used. a The principal of the school claims a high a The average price of petrol in Canberra was retention rate in Years 11 and 12 (that is, most estimated by averaging the price at 30 petrol of the students from Year 10 continue on to stations in the area. Survey complete Years 11 and 12). b The performance of a cricketer is measured by b The parents claim that the retention rate of looking at his performance in every match he students in Years 11 and 12 is low (that is, a has played. Census large number of students leave at the end of c Public opinion on an issue is sought by a Year 10). telephone poll of 2000 homes. Survey 7 Records from a school were examined to determine 4 Traffic lights (red, amber, green) are set so that the number of absent days of both boys and girls each colour shows for a set amount of time. over the two years of Year 9 and Year 10. The result Describe how you could use a spinner to simulate is shown in this stem-and-leaf plot. the situation so that you could determine (on average) how many sets of lights you must Key: 2 | 1 = 21 days encounter in order to get two green lights in Leaf■ Stem Leaf■ succession. Boys Girls 5   MC  John and Bill play squash each week. In any 0 17 given game they are evenly matched. A device 7410 1 24799 which could not be used to represent the outcomes 9976653110 2 133466 of the situation is: 87752 3 4448 A a die 2 4 36 B a coin 5 4 Boys: range = 32; girls: range = 53 Boys: median = 26; girls: median = 23.5 C a circular spinner divided into 2 equal a Calculate the median number of days absent sectors for both boys and girls. ✔ D a circular spinner divided into 5 equal b Calculate the range for both boys and girls. sectors c Comment on the distribution of days absent for E a circular spinner divided into 4 equal each group. sectors Both sets have similar medians, Use a spinner of 3 equal sectors, each sector having an angle size of 120° and representing a particular colour. Twirl the spinner until a green/green combination has been obtained. This is defined as one experiment. Count the number of trials required for this experiment. Repeat this procedure a number of times and determine an average.

but the girls have a greater range of absenteeism than the boys.

Chapter 15 Statistics in the media

529

statistics AND probability • Data representation and interpretation 8 15 boys and 15 girls were randomly chosen from

a group of 900 students. Their heights (in metres) were measured as shown below. Boys: 1.65, 1.71, 1.59, 1.74, 1.66, 1.69, 1.72, 1.66, 1.65, 1.64, 1.68, 1.74, 1.57, 1.59, 1.60 Girls: 1.66, 1.69, 1.58, 1.55, 1.51, 1.56, 1.64, 1.69, 1.70, 1.57, 1.52, 1.58, 1.64, 1.68, 1.67 a Ford: median = 15; Holden: median = 16 b Ford: range = 26; Holden: range = 32 c Ford: IQR = 14; Holden: IQR = 13.5 a State the median of both distributions. b Calculate the range of both distributions. c Calculate the interquartile range of both

distributions. 10 The box plots drawn below display statistical data

for two AFL teams over a season. Sydney Swans Brisbane Lions

a Which team had the higher median score? b What was the range of scores for each team? c For each team calculate the interquartile range. 11 Tanya measures the heights (in m) of a group of

Year 10 boys and girls and produces the following five-point summaries for each data set. Girls Boys:  1.45, 1.56, 1.62, 1.70, 1.81 Girls:  1.50, 1.55, 1.62, 1.66, 1.73

Boys 1.4 1.5 1.6 1.7 1.8 1.9 Height (m)

display them on the same scale. The sample is an appropriate size as

900 = 30.

a Comment on the size of the sample. b Display the data as a back-to-back stem plot. c Compare the heights of the boys and girls.

of vehicles sold by the Ford and Holden dealerships in a Sydney suburb each week for a three-month period. 9 d Ford Holden 0 5 10 15 20 2530 35 40 Number of vehicles sold

b What is the median of each distribution? c What is the range of each distribution? d What is the interquartile range for each

distribution? e Comment on the spread of the heights among

the boys and the girls. 12 The box plots below show the average daily sales

of cold drinks at the school canteen in summer and winter.

Key: 1 | 5 = 15 vehicles Summer Leaf■ Stem Leaf■ Ford Holden 74 0 39 Winter 952210 1 111668 0 5 10 15 20 25 30 35 40 Daily sales 8544 2 2279 of cold 0 3 5 drinks Although boys and girls have the same median height,

Maths Quest 10 for the Australian Curriculum

the spread of heights is greater among boys as shown by the greater range and interquartile range.

11 b Boys: median = 1.62 m; girls: median = 1.62 m c Boys: range = 0.36 m; girls: range = 0.23 m d Boys: IQR = 0.14 m; girls: IQR = 0.11 m

a Draw a box plot for both sets of data and

9 The stem plot below is used to display the number

530

50 60 70 80 90 100 110 120 130 140 150 Points

Brisbane Lions

b Brisbane Lions: range = 65; Sydney Swans: range = 55 c Brisbane Lions: IQR = 40; Sydney Swans: IQR = 35

 he boys are generally better than the girls, with the mean of the boys being T 1.66 m and that of the girls being 1.62 m. The five-number summaries are: Boys:  1.57 m, 1.6 m, 1.66 m, 1.71 m, 1.74 m Girls:  1.51 m, 1.56 m, 1.64 m, 1.68 m, 1.7 m

d Show both distributions on a parallel box plot.

Year 8: mean = 26.83, median = 27, range = 39, IQR = 19    Year 10: mean = 40.7, median = 39.5, range = 46, IQR = 20 The typing speed of Year 10 students is about 13 to 14 wpm faster than that of Year 8 students. The spread of data in Year 8 is slightly less than the spread in Year 10.

 here are generally more cold drinks sold in summer as T shown by the higher median. The spread of data is similar as shown by the IQR although the range in winter is greater.

statistics AND probability • Data representation and interpretation Summer: IQR = 13; winter: IQR = 11 a Calculate the range of sales in both summer and winter. Summer: range = 23; winter: range = 32 b Calculate the interquartile range of the sales in

both summer and winter. c Comment on the relationship between the two

data sets, both in terms of measures of location and measures of spread. 13 A movie theatre has taken a survey of the ages of

people at a showing of two of their movies. The results are shown in these box plots. Movie A Movie B 0

10 20 30 40

50 60 70

80 Age

a Find the mean and the median age of the people in this sample. Mean = 32.03; median = 29.5 b Group the data into class intervals of

c d e f

✔

Which of the following conclusions could be drawn based on the above information? A Movie A attracts an older audience than Movie B. B Movie B attracts an older audience than Movie A. C Movie A appeals to a wider age group than Movie B. D Movie B appeals to a wider age group than Movie A. E More people went to Movie A.

g

10 (0–9 etc) and complete the frequency distribution table. Use the frequency distribution table to estimate the mean age. Mean = 31.83 Calculate the cumulative frequency and, hence, Median = 30 plot the ogive. Estimate the median age from the ogive. Compare the mean and median of the original data in part a with the estimates of the mean and the median obtained for the grouped data in parts c and e. Were the estimates good enough? Explain your Estimates answer.

2 The typing speed (words per minute) was

recorded for a group of Year 8 and Year 10 students. The results are displayed in this ■ back-to-back stem plot.

from parts c and e were fairly accurate.

Yes, they were fairly close to the mean and median of the raw data.

problem solving 1 A sample of 30 people was selected at random

from those attending a local swimming pool. ■ Their ages (in years) were recorded as follows: 19, 7, 58, 41, 17, 23, 62, 55, 40, 37, 32, 29, 21, 18, 16, 10, 40, 36, 33, 59, 65, 68, 15, 9, 20, 29, 38, 24, 10, 30.

Key: 2 | 6 = 26 wpm Leaf■ Stem Leaf■ Year 8 Year 10 99 0 9865420 1 79 988642100 2 23689 9776410 3 02455788 86520 4 1258899 5 03578 003 6

Write a report comparing the typing speeds of the two groups. Chapter 15 Statistics in the media

531

Hane■and■Roarne■had■a■higher■median■and■a■lower■ spread,■so■they■appear■to■have■performed■better.

statistics aND probability • Data represeNtatioN aND iNterpretatioN 3 These■parallel■box■plots■show■the■number■of■

5 There■has■been■a■rise■in■supermarket-own■brands■

weekly■house■sales■by■two■real■estate■agencies■over■ a■3-month■period. HJ Looker Hane & Roarne 0 1 2

3 4

5 6

7

8

9 10 Number of weekly sales

Prepare■a■report■to■compare■the■performance■of■ the■two■agencies. 4 This■taste■test■on■corn■chips■appeared■in■a■newspaper.

in■Australia.■These■are■commonly■available■in■ supermarkets■like■Woolworths,■Coles■and■Aldi.■It■ has■been■said■that■these■brands■account■for■almost■ one-quarter■of■all■grocery■sales.■It■has■also■been■ claimed■that■the■quality■of■supermarket-own■brands■ is■comparable■with■the■equivalent■market-leading■ brand,■at■a■much■reduced■cost. Assume■you■are■planning■undertake■a■study■ of■a■particular■grocery■line■(e.g.,■baked■beans,■ breakfast■cereal,■……).■Write■a■plan■of■how■you■ would■undertake■this■study. Student’s■plan■for■an■investigation.

TasTe TesT: corn chips

Interactivities

Test yourself Chapter 15 int-2867

Byron Bay Chilli Co. Corn chips

Word search Chapter 15 int-2865

230g $3.40 ($1.48 per 100g) • Made in Australia • Fat 24.9g/100g • Saturated fat 11.8g/100g • Sodium 44mg/100g Verdict: Crisp, thick chips with fresh corn flavour and low sodium content. No preservatives, no GM corn. Put Byron Bay in the title and things cost more!

Crossword Chapter 15 int-2866

CCs Corn Chips

200g $2.69 ($1.34 per 100g) • Made in Australia • Fat 24.4g/100g • Saturated fat 10.6g/100g • Sodium 550 mg/100g Verdict: Chips are quite thin and very salty to taste. Very high sodium content.

IGA Black and Gold Plain Corn chips

230g $1.99 (87 cents per 100g) • Made in Australia • Fat 24.9g/100g • Saturated fat 11.8g/100g • Sodium 415mg/100g Verdict: Cheap in comparison and quite nice. A good budget option as they are as good as the popular branded ones.

Doritos Corn Chips

200g $2.99 ($1.50 per 100g) • Made in Australia • Fat 23.2g/100g • Saturated Fat 11.2g/100g • Sodium 435mg/100g Verdict: Very crisp and fresh, but really no better than the home brand which is $1 cheaper. Source: The Sunday Mail, 4 Apr. 2010, p. 26.

Comment■on■the■information■displayed.

532

maths Quest 10 for the australian curriculum

eBook plus

They■are■all■made■in■Australia■and■ have■comparable■fat■and■saturated■fat■ contents.■The■Byron■Bay■Chilli■corn■ chips■have■a■much■lower■salt■content■ than■the■other■three■varieties.■The■ verdict■comments■require■a■mention.

eBook plus

activities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■15■(doc-5336)■ (page 497) Are you ready? Digital docs (page 498) •■ SkillSHEET■15.1■(doc-5337):■Determining■ suitability■of■questions■for■a■survey •■ SkillSHEET■15.2■(doc-5338):■Finding■proportions •■ SkillSHEET■15.3■(doc-5339):■Distinguishing■ between■types■of■data •■ SkillSHEET■15.4■(doc-5340):■Reading■bar■ graphs •■ SkillSHEET■15.5■(doc-5341):■Determining■ independent■and■dependent■variables

15A Populations and samples Digital docs

Digital docs (page 518) •■ Activity■15-C-1■(doc-5161):■Collecting■and■analysing■ data •■ Activity■15-C-2■(doc-5162):■More■collecting■and■ analysing■data •■ Activity■15-C-3■(doc-5163):■Detailed■collecting■and■ analysing■data

15D Statistical investigations Digital docs

•■ Activity■15-D-1■(doc-5164):■Analysing■reports■ (page 525) •■ Activity■15-D-2■(doc-5165):■Analysing■reports■in■ depth■(page 525) •■ Activity■15-D-3■(doc-5166):■Analysing■reports■in■ greater■depth■(page 525) •■ WorkSHEET■15.2■(doc-5343):■Statistics■in■the■media■ (page 526) Chapter review

•■ Activity■15-A-1■(doc-5155):■Populations■and■samples■ Interactivities (page 532) (page 502) •■ Test■yourself■Chapter■15■(int-2867):■Take■the■end-of•■ Activity■15-A-2■(doc-5156):■More■populations■and■ chapter■test■to■test■your■progress■ samples■(page 502) •■ Word■search■Chapter■15■(int-2865):■an■interactive■ •■ Activity■15-A-3■(doc-5157):■In■depth■populations■ word■search■involving■words■associated■with■this■ and■samples■(page 502) chapter •■ WorkSHEET■15.1■(doc-5342):■Populations■and■ •■ Crossword■Chapter■15■(int-2866):■an■interactive■ samples■(page 503) crossword■using■the■defi■nitions■associated■with■the■ 15B Primary and secondary data chapter Digital docs (page 509) To access eBookPLUS activities, log on to •■ Activity■15-B-1■(doc-5158):■Data■collection■ •■ Activity■15-B-2■(doc-5159):■Further■data■collection www.jacplus.com.au •■ Activity■15-B-3■(doc-5160):■Advanced■collection 15C Evaluating inquiry methods and statistical reports Interactivity

•■ Compare■statistical■reports■(int-2790)■(page 511)

chapter 15 statistics in the media

533

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ict activity

climate change SEARCHLIGHT ID: PRO-0100

Scenario Climate■change■is■upon■us■and■has■become■one■of■ the■great■challenges■facing■humanity.■Our■fossil■ fuel■driven■economies■are■producing■large■volumes■ of■greenhouse■gases■(water■vapour,■carbon■dioxide,■ methane■and■ozone)■that■are■warming■the■planet.■ As■our■planet■heats■up,■the■ice■sheets■at■the■poles■ slowly■melt,■causing■sea■levels■to■rise.■Islands■in■ the■Pacifi■c■Ocean■are■already■being■overcome■by■ water,■leading■the■inhabitants■of■islands,■such■as■

534

maths Quest 10 for the australian curriculum

Tuvalu■Island■in■the■Pacifi■c■Ocean,■to■announce■ that■they■are■abandoning■their■homeland■due■to■ rising■sea■levels.■In■Sydney,■many■well■known■ suburbs■could■be■threatened■by■rising■sea■levels■in■ the■future,■including■Caringbah,■Kurnell,■Cromer,■ Manly■Vale,■Newcastle,■the■central■coast,■Homebush■ Bay,■Newington■Silverwater,■Arncliffe,■Marrickville■ and■Sydney■Airport.■Climate■change■could■cause■ the■extinction■of■many■species■as■ecosystems■are■ damaged■by■rising■temperatures.■ In■order■to■address■these■apocalyptic■issues,■we■ need■to■understand■the■role■of■human■activity■in■ climate■change.■It■will■be■your■job■to■investigate■and■ understand■the■relationships■that■underpin■global■ warming.■You■can■then■make■recommendations■to■our■ political■leaders■and■take■action■yourself■to■help■save■ our■planet.

Task

You■must■have■a■gmail■ SUGGESTED account■and■internet■ You■will■need■to■analyse■real data sets■to■develop■a■ SOFTWARE access■to■use■the■ mathematical■understanding■of■climate■change■issues.■ • Microsoft Excel Google■data■tools. The■analysis■will■involve■the■use■of■scatter■plots,■box• Geogebra and-whisker■plots■and■fi■ve■number■summaries.■Scatter■ ■■ You■will■need■ • Internet connection Microsoft■Excel■and■ plots■will■be■used■to■investigate■and■comment■on■ • Internet browser GeoGebra■installed■ with Adobe Flash relationships■between■two■climate■change■variables.■ player installed. on■your■computer. Data■sets■will■be■compared■using■box-and-whisker■ • Use the World Bank Go■to■projectsPLUS■ plots,■dot■plots■and■histograms.■Environmental■data■ weblink in your on■your■eBook,■set■ will■be■graphed,■such■that■the■independent■variable■ eBookPLUS to locate up■a■group■and■then■ is■time.■At■the■end■of■your■project,■your■improved■ banks of data in open■the■Media Centre■ mathematical■understanding■of■climate■change■will■ Excel form. to■locate■everything■ allow■you■to■make■key■recommendations■on■how■we■ you■need. can■meet■the■environmental■challenges■of■the■future.■ ■■ Open■the■Word■documents■titled■Lesson■1,■ Process Lesson■2■etc.■Follow■the■instructions■in■each■ You■will■use■Microsoft Excel,■Google Fusion Tables■ document■to■complete■your■project. ■■ At■various■stages■of■your■project,■you■will■need■to■ and■Google Public Data Explorer■to■investigate■ global■environmental■data. access■data■sets■in■Microsoft■Excel■fi■les. ■■

Ict activity — projectsplus

535

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

16

16A 16B 16C 16D 16E 16F

Purchasing goods Buying on terms Successive discounts Compound interest Depreciation Loan repayments

WhAt Do you knoW ?

financial maths

1 List what you know about financial maths. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shos your class’s knowledge of financial maths. eBook plus

Digital doc

Hungry brain activity Chapter 16 doc-5344

oPening Question

Which is the better interest rate — 6.25% p.a. simple interest, or 6% p.a. compound interest?

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

Are you ready?

Try■the■questions■below.■If■you■have■diffi■culty■with■any■of■them,■extra■help■can■be■obtained■by■ completing■the■matching■SkillSHEET.■Either■search■for■the■SkillSHEET■in■your■eBookPLUS■or■ ask■your■teacher■for■a■copy. eBook plus

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SkillSHEET 16.1 doc-5345

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SkillSHEET 16.2 doc-5346

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Digital doc

SkillSHEET 16.3 doc-5347

Converting a percentage to a decimal 1 Convert■each■of■the■following■percentages■to■a■decimal. a 24% 0.24 b 17.5% 0.175 c 3% 0.03 3 d 9 4 % 0.0975 Finding simple interest 2 Find■the■simple■interest■earned■on■an■investment■of: a $6000■at■7%■p.a.■for■5■years $2100 b $14■■000■at■9.5%■p.a.■for■8■years $10■640 c $100■■000■at■7.6%■p.a.■for■3■years $22■800 d $45■■000■at■3.5%■p.a.■for■15■months. $1968.75 Finding a percentage of a quantity (money) 3 Find■each■of■the■following. a 15%■of■$200 $30 b 8%■of■$540 $43.20 c 2.5%■of■$44 $1.10 1 d 6 %■of■$1250 $81.25 2

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Digital doc

SkillSHEET 16.4 doc-5348

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Digital doc

SkillSHEET 16.5 doc-5349

538

Finding percentage discount 4 Find■the■percentage■discount■given■on■each■of■the■following. a Normal■price■$90,■sale■price■$72 20% b Normal■price■$450,■sale■price■$427.50 5% c Normal■price■$1750,■sale■price■$1400 20% d Normal■price■$5920,■sale■price■$4000 32.4% Decreasing a quantity by a percentage 5 Find■the■purchase■price■of■each■item■after■a■discount■is■allowed■on■the■marked■price. a $2300■with■a■10%■discount $2070 b $590.00■with■a■25%■discount $442.50 c $896.00■with■a■12.5%■discount $784 1 d $5800■with■a■6 4 %■discount■ $5437.50

maths Quest 10 for the Australian curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

16A

Purchasing goods ■■

■■ ■■ ■■

There■are■many■different■payment■options■when■purchasing■major■goods,■such■as■fl■at■screen■ televisions■and■computers.■Payment■options■include: •■ cash •■ credit■card •■ lay-by •■ deferred■payment •■ buying■on■terms •■ loan. The■cost■of■purchasing■an■item■can■vary■depending■on■the■method■of■payment■used. Some■methods■of■payment■involve■borrowing■money■and,■as■such,■mean■that■interest■is■ charged■on■the■money■borrowed. The■simple■interest■formula■can■be■used■to■calculate■the■interest■charged■on■borrowed■money, P ×r ×T I= 100 where:■ I■is■the■simple■interest■($) P■is■the■principal■or■amount■borrowed■or■invested■($) r■is■the■rate■of■interest■per■time■period T■is■the■time■for■which■the■money■is■invested■or■borrowed. ■ If■T■is■in■years,■then■r■is■the■rate■of■interest■per■annum■(%■p.a.).

WorkeD exAmPle 1

Find the simple interest on $4000 invested at 4.75% p.a. for 4 years. think

Write

1

Write■the■formula■and■the■known■values■of■the■ variables.

I=

P ×r ×T ,■where■P■=■$4000,■r■=■4.75%,■T■=■4 100

2

Substitute■known■values■to■fi■nd■I.

I=

$4000 × 4.75 × 4 100

3

Calculate■the■value■of■I.

■=■$760

■■

What■are■the■ways■of■purchasing■the■item■shown■in■the■advertisement■below?

120 c m HD T V 5 year ty warran ■ ■ ■ ■

High definition HDMI ports 16 : 9 aspect ratio 1080i

$1200 chapter 16 financial maths

539

number AND algebra • Money and financial mathematics

Payment options Cash ■■ ■■

With cash, the marked price is paid on the day of purchase with nothing more to pay. A cash-paying customer can often negotiate, with the retailer, to obtain a lower price for the item.

Lay-by ■■ ■■

With lay-by, the item is held by the retailer while the customer makes regular payments towards paying off the marked price. In some cases a small administration fee may be charged.

Credit cards ■■ ■■ ■■

With a credit card, the retailer is paid instantly from the credit card provider, generally a financial lender. The customer takes immediate possession of the goods. The financial lender later bills the customer — collating all purchases over a monthly period and billing the customer accordingly. The entire balance shown on the bill can often be paid with no extra charge, but if the balance is not paid in full, interest is charged on the outstanding amount, generally at a very high rate.

Worked Example 2

The ticketed price of a mobile phone is $600. Andrew decides to purchase the phone using his credit card. After 1 month the credit card company charges interest at a rate of 15% p.a. Calculate the amount of interest that Andrew must pay on his credit card after 1 month. Think

Write

1

Write the formula and the known values of the 1 variables. Remember that 1 month = 12 year.

P ×r ×T 100 1 P = $600, r = 15%, T = 12

2

Substitute known values to find I.

I=

3

Calculate the value of I.

I=

600 × 15 × 1 100 × 12

= $7.50

remember

1. There are alternatives to consider when deciding on how to pay for a major purchase. P ×r ×T 2. The simple interest formula is I = , where P = principal, r = interest rate and 100 T = time. 3. Credit card companies calculate interest on a monthly basis. Exercise

16a

Purchasing goods FLUENCY 1   WE 1  Find the simple interest payable on a loan of $8000 at 6% p.a. for 5 years.

$2400

2 Find the simple interest on each of the following loans. a $5000 at 9% p.a. for 4 years $1800 b $4000 at 7.5% p.a. for 3 years c $12  000 at 6.4% p.a. for 540

2 12

Maths Quest 10 for the Australian Curriculum

years $1920

d $6000 at 8% p.a. for

1 12

$900

years $720

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

inDiViDuAl PAthWAys eBook plus

Activity 16-A-1

Simple interest doc-5167 Activity 16-A-2

Harder simple interest doc-5168 Activity 16-A-3

Tricky simple interest doc-5169

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Digital docs

SkillSHEET 16.1 doc-5345

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SkillSHEET 16.2 doc-5346

3 Find■the■simple■interest■on■each■of■the■following■investments. a $50■■000■at■6%■p.a.■for■6■months $1500 b $12■■500■at■12%■p.a.■for■1■month $125 c $7500■at■15%■p.a.■for■3■months $281.25 d $4000■at■18%■p.a.■for■18■months $1080 4 Calculate■the■monthly■interest■charged■on■each■of■the■following■outstanding■credit■card■

balances. a $1500■at■15%■p.a. $18.75 c $2750■at■18%■p.a. $41.25 e $5690■at■21%■p.a. $99.58

b $4000■at■16.5%■p.a. $55.00 d $8594■at■17.5%■p.a. $125.33

unDerstAnDing 5 We2 ■The■ticketed■price■of■a■mobile■phone■is■$800.■Elena■decides■to■purchase■the■phone■using■

her■credit■card.■After■1■month■the■credit■card■company■charges■interest■at■a■rate■of■15%■p.a.■ Calculate■the■amount■of■interest■that■Elena■must■pay■on■her■credit■card■after■1■month. $10 6 Reece■decides■to■purchase■a■new■stereo■system■using■her■credit■card.■The■ticketed■price■of■the■ stereo■system■is■$900.■When■Reece’s■credit■card■statement■arrives,■it■shows■that■she■will■pay■ no■interest■if■she■pays■the■full■amount■by■the■due■date. a If■Reece■pays■$200■by■the■due■date,■what■is■the■balance■owing?■ $700 b If■the■interest■rate■on■the■credit■card■is■18%■p.a.,■how■much■interest■will■Reece■be■ charged■in■the■month? $10.50 c What■will■be■the■balance■that■Reece■owes■at■the■end■of■the■month? $710.50 d At■this■time■Reece■pays■another■$500■off■her■credit■card.■How■much■interest■is■Reece■ then■charged■for■the■next■month? $3.16 e Reece■then■pays■off■the■entire■remaining■balance■of■her■card.■What■was■the■true■cost■of■ the■stereo■including■all■the■interest■payments? $913.66 7 Carly■has■an■outstanding■balance■of■$3000■on■her■credit■card■for■June■and■is■charged■interest■at■ a■rate■of■21%■p.a. a Calculate■the■amount■of■interest■that■Carly■is■charged■for■June. $52.50 b Carly■makes■the■minimum■repayment■of■$150■and■makes■no■other■purchases■using■the■ credit■card■in■the■next■month.■Calculate■the■amount■of■interest■that■Carly■will■be■charged■ for■July. $50.79 c If■Carly■had■made■a■repayment■of■$1000■at■the■end■of■June,■calculate■the■amount■of■ interest■that■Carly■would■then■have■been■charged■for■July. $35.92 d How■much■would■Carly■save■in■July■had■she■made■the■higher■repayment■at■the■end■of■ June? $14.87 8 Shane■buys■a■new■home■theatre■system■using■his■credit■card.■The■ticketed■price■of■the■bundle■is■ $7500.■The■interest■rate■that■Shane■is■charged■on■his■credit■card■is■18%■p.a.■Shane■pays■off■the■ credit■card■at■a■rate■of■$1000■each■month. a Complete■the■table■below. Month January February March April May June July August

Balance owing $7500.00 $6612.50

Interest $112.50 $99.19

$5711.69

$85.68

$4797.37

$71.96

$3869.33

$58.04

$2927.37

$43.91

$1971.28

$29.57

$1000.85

$15.01

Payment $1000.00 $1000.00 $1000.00 $1000.00 $1000.00 $1000.00 $1000.00 $1015.86

Closing balance $6612.50 $5711.69 $4797.37 $3869.33 $2927.37 $1971.28 $1000.85

$0

b What■is■the■total■amount■of■interest■that■Shane■pays? $515.86 c What■is■the■total■cost■of■purchasing■the■home■theatre■system■using■his■credit■card?

$8015.86

chapter 16 financial maths

541

number AND algebra • Money and financial mathematics Reasoning 9 Design a table that compares the features of each method of payment: cash, lay-by and

credit card. 10 Choose the most appropriate method of payment for each of the described scenarios below. Explain your choice. Scenario 1: Andy has no savings and will not be paid for another two weeks. Andy would like to purchase an HD television and watch tomorrow’s football final. reflection    Scenario 2: In September Lena spots on special a home What can you do to remember theatre system which she would like to the simple interest formula? purchase for her family for Christmas.

16b

S1: Credit card — payment is delayed, but possession is immediate S2: Lay-by, or cash if she has savings, would like to negotiate a lower price and has somewhere to store it.

Buying on terms ■■

When buying an item on terms: •• a deposit is paid •• the balance is paid off over an agreed period of time with set payments •• the set payments may be calculated as a stated arbitrary amount or interest rate •• total monies paid will exceed the initial cash price.

Worked Example 3

The cash price of a computer is $2400. It can also be purchased on the following terms: 25% deposit and payments of $16.73 per week for 3 years. Calculate the total cost of the computer purchased on terms as described. Think

Write

1

Calculate the deposit.

Deposit = 25% of $2400 = 0.25 ì $2400 = $600

2

Calculate the total of the weekly repayments.

Total repayment = $16.73 ì 52 ì 3 = $2609.88

3

Add these two amounts together to find the total cost.

Total cost = $600 + $2609.88 = $3209.88

■■

In some examples we need to be able to calculate the amount of each regular repayment using the terms of the purchase.

Worked Example 4

A diamond engagement ring has a purchase price of $2500. Michael buys the ring on the following terms: 10% deposit with the balance plus simple interest paid monthly at 12% p.a. over 3 years. a Calculate the amount of the deposit. b What is the balance owing after the initial deposit? c Calculate the interest payable. d What is the total amount to be repaid? e Find the amount of each monthly repayment. Think a Calculate the deposit by finding 10% of $2500.

542

Maths Quest 10 for the Australian Curriculum

Write a

Deposit = 10% of $2500 = 0.1 ì $2500 = $250

number AND algebra • Money and financial mathematics b Find the balance owing by subtracting the

b

Balance = $2500 - $250 = $2250

c

I=

deposit from the purchase price. c

Find the simple interest on $2250 at 12% p.a. for 3 years.

d Find the total repayment by adding the balance

owing with the interest payable. e

Find the monthly repayment by dividing the total repayment by the number of months over which the ring is to be repaid.

P ×r ×T , where P = $2250, r = 12%, T = 3 100 = $2250 ì 0.12 ì 3 = $810

d Total repayment = $2250 + $810

= $3060

e

Monthly repayment = $3060 ó 36 = $85

Loans ■■ ■■ ■■ ■■ ■■

Money can be borrowed from a bank or other financial institution, in order to pay cash for an item. Interest is charged on the amount of money borrowed. Both the money borrowed and the interest charged must be paid back. The interest rate on a loan is generally lower than the interest rate offered on a credit card or when buying on terms. The calculation of loan payments is done in the same way as for buying on terms; that is, calculate the interest and add it to the principal before dividing into equal monthly repayments.

remember

1. When buying an item on terms we usually pay a deposit with the balance plus interest paid in weekly or monthly instalments over an agreed period of time. 2. To calculate the total cost of a purchase, add the deposit to the total of the regular repayments. 3. The amount of each repayment is found by following these steps: (a) Calculate the deposit. (b) Find the balance owing by subtracting the deposit from the cash price. (c) Find the total repayments by adding the interest to the balance owing. (d) Divide the total amount to be repaid by the number of regular repayments that must be made. 4. Loan repayments may be calculated in the same way; however, there is no deposit made.

Exercise

16b

Buying on terms Fluency

1 The best deal is the one with the lowest cost — 20% deposit and weekly payments of $20 over ■ 3 years.

Calculate the total cost of a $3000 purchase given the terms described below. a   i 12% deposit and monthly payments of $60 over 5 years $3960 ii 20% deposit and weekly payments of $20 over 3 years $3720 iii 15% deposit and annual payments of $700 over 5 years $3950 b Which of these options is the best deal for a purchaser? Chapter 16 Financial maths

543

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

inDiViDuAl PAthWAys eBook plus

Activity 16-B-1

Buying on terms doc-5170 Activity 16-B-2

Buying on difficult terms doc-5171 Activity 16-B-3

Buying on tricky terms doc-5172

2 Calculate■the■amount■of■each■repayment■for■a■$5000■purchase■given■the■terms■described■

below. a 10%■deposit■with■the■balance■plus■simple■interest■paid■monthly■at■15%■p.a.■over■ 5■years■ $131.25 b 10%■deposit■with■the■balance■plus■simple■interest■paid■fortnightly■at■12%■over■ 5■years■ $55.38 c 20%■deposit■with■the■balance■plus■simple■interest■paid■monthly■at■10%■over■ 3■years■ $144.44 3 Calculate■the■total■repayment■and■the■amount■of■each■monthly■repayment■for■each■of■the■ following■loans. a $10■■000■at■9%■p.a.■repaid■over■4■years $13■■600,■$283.33 b $25■■000■at■12%■p.a.■repaid■over■5■years $40■■000,■$666.67 c $4500■at■7.5%■p.a.■repaid■over■18■months $5006.25,■$278.13 d $50■■000■at■6%■p.a.■repaid■over■10■years $80■■000,■$666.67 e $200■■000■at■7.2%■p.a.■repaid■over■20■years $488■■000,■$2033.33 unDerstAnDing 4 We3 ■The■cash■price■of■a■bedroom■suite■is■$4200.■The■bedroom■suite■can■be■purchased■on■the■

5

eBook plus

6

Digital doc

SkillSHEET 16.3 doc-5347

7

8

9

544

following■terms:■20%■deposit■and■weekly■repayments■of■$43.94■for■2■years.■Calculate■the■total■ cost■of■the■bedroom■suite■if■you■bought■it■on■terms. $5409.76 Guy■purchases■a■computer■that■has■a■cash■price■of■$3750■on■the■following■terms:■$500■deposit■ with■the■balance■plus■interest■paid■over■2■years■at■$167.92■per■month.■What■is■the■total■amount■ that■Guy■pays■for■the■computer? $4530.08 Robert■wants■to■buy■a■used■car■with■a■cash■price■of■$12■■600.■The■dealer■offers■terms■of■10%■ deposit■and■monthly■repayments■of■$812.70■for■2■years. a Calculate■the■amount■of■the■deposit. $1260 b Calculate■the■total■amount■to■be■paid■in■monthly■repayments. $19■■504.80 c What■is■the■total■amount■Robert■pays■for■the■car? $20■■764.80 d How■much■more■than■the■cash■price■of■the■car■does■Robert■pay?■(This■is■the■interest■ charged■by■the■dealer.) $8164.80 Kylie■wants■to■purchase■an■entertainment■system■that■has■a■cash■price■of■$5800.■She■purchases■ the■entertainment■system■on■terms■of■no■deposit■and■monthly■repayments■of■$233.61■for■ 3■years. a Calculate■the■total■amount■that■Kylie■pays■for■the■entertainment■system. $8409.96 b Calculate■the■amount■that■Kylie■pays■in■interest. $2609.96 c Calculate■the■amount■of■interest■that■Kylie■pays■each■year. $869.99 d Calculate■this■amount■as■a■percentage■of■the■cash■price■of■the■entertainment■system. 15% We4 ■A■used■car■has■a■purchase■price■of■$9500.■Dayna■buys■the■car■on■the■following■terms:■ 25%■deposit■with■balance■plus■interest■paid■at■12%■p.a.■interest■over■3■years. a Calculate■the■amount■of■the■deposit. $2375 b What■is■the■balance■owing? $7125 c Calculate■the■interest■payable. $2565 d What■is■the■total■amount■to■be■repaid? $9690 e Find■the■amount■of■each■monthly■repayment. $269.17 A■department■store■offers■the■following■terms:■one-third■deposit■with■the■balance■plus■interest■ paid■in■equal,■monthly■instalments■over■18■months.■The■interest■rate■charged■is■9%■p.a.■Ming■ buys■a■lounge■suite■with■a■ticketed■price■of■$6000. a Calculate■the■amount■of■the■deposit. $2000 b What■is■the■balance■owing? $4000 c Calculate■the■interest■payable. $540 d What■is■the■total■amount■to■be■repaid? $4540 e Find■the■amount■of■each■monthly■repayment. $252.22

maths Quest 10 for the Australian curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics 10 Calculate■the■monthly■payment■on■each■of■the■following■items■bought■on■terms.■ (Hint:■Use■the■steps■shown■in■question■8.) a Dining■suite:■cash■price■$2700,■deposit■10%,■interest■rate■ 12%■p.a.,■term■1■year $226.80

b Video■camera:■cash■price■$990,■

deposit■20%,■interest■rate■15%■p.a.,■ term■6■months $141.90

c Car:■cash■price■$16■■500,■deposit■25%,■ interest■rate■15%■p.a.,■term■5■years $360.94

d Mountain■bike:■cash■price■$3200,■

one-third■deposit,■interest■rate■9%■p.a.,■ 1 term■2 2 ■years $87.11

e Watch:■cash■price■$675,■no■deposit,■interest■ rate■18%■p.a.,■term■9■months $85.13

11 Fred■wants■to■purchase■his■fi■rst■car.■He■has■saved■$1000■as■a■deposit■but■the■cost■of■the■car■is■

$5000.■Fred■takes■out■a■loan■from■the■bank■to■cover■the■balance■of■the■car■plus■$600■worth■of■ on-road■costs. a How■much■will■Fred■need■to■borrow■from■the■bank? $4600 b Fred■takes■the■loan■out■over■4■years■at■9%■p.a.■interest.■How■much■interest■will■Fred■need■ to■pay? $1656 c What■will■be■the■amount■of■each■monthly■payment■that■Fred■makes? $130.33 d What■is■the■total■cost■of■the■car■after■paying■off■the■loan,■including■the■on-road■costs?■ Give■your■answer■to■the■nearest■$. $7256 12 mc ■Kelly■wants■to■borrow■$12■■000■for■some■home■improvements.■Which■of■the■following■ loans■will■lead■to■Kelly■making■the■lowest■total■repayment? A Interest■rate■6%■p.a.■over■4■years B Interest■rate■7%■p.a.■over■3■years ✔ C

1

Interest■rate■5.5%■p.a.■over■3 2 ■years

D Interest■rate■6.5%■p.a.■over■5■years

E Interest■rate■7.5%■p.a.■over■3■years reAsoning

eBook plus

Digital doc

WorkSHEET 16.1 doc-5350

13 mc ■Without■completing■any■calculations■explain■which■of■the■following■loans■will■be■the■ This■option■has■the■lowest■interest■rate■and■ best■value■for■the■borrower. A Interest■rate■8.2%■p.a.■over■5■years time■frame■when■compared■to■all■others. B Interest■rate■8.2%■over■4■years C Interest■rate■8%■over■6■years D Interest■rate■8%■over■5■years ✔ E Interest■rate■8%■over■4■years reflection    14 Explain■how,■when■purchasing■an■item,■making■

a■deposit■using■existing■savings■and■taking■out■ a■loan■for■the■balance■can■be■an■advantage.

When buying on terms, what arrangements are the most beneficial to the buyer?

The■larger■the■deposit■the■smaller■the■loan■and■hence■the■interest■ charged.■Loans■generally■offer■a■lower■rate■than■buying■on■terms.

chapter 16 financial maths

545

number AND algebra • Money and financial mathematics

16c

Successive discounts ■■



■■ ■■

Consider the case of Tony who is a mechanic. Tony purchases his hardware from Tradeways hardware store, which is having a 10%-off sale. Tradeways also offers a 5% discount to tradespeople. Tony purchases hardware that has a total value of $800. What price does Tony pay for these supplies? After the 10% discount, the price of the supplies is 90% of $800 = 0.90 ì $800 = $720 The 5% trades discount is then applied. 95% of $720 = 0.95 ì $720 = $684 So the price Tony pays is $684. Now let us consider what single discount Tony has actually received. Amount of discount = $800 - $684 = $116 $116 Percentage discount = ì 100% $800 = 14.50% So we can conclude that the successive discounts of 10% followed by a further 5% is equivalent to receiving a single discount of 14.50%. When two discounts are applied one after the other, the total discount is not the same as a single discount found by adding the two percentages together. The order of calculating successive discounts does not affect the final answer.

Worked Example 5

A furniture store offers a discount of 15% during a sale. A further 5% discount is then offered to customers who pay cash. a Find the price paid by Lily, who pays cash for a bedroom suite priced at $2500. b What single percentage discount does Lily receive on the price of the bedroom suite? Think a

b

1

Subtract 15% from 100% to find the percentage paid.

2

Calculate 85% of the price.

3

Subtract 5% from 100% to find the next percentage paid.

4

Calculate 95% of $2125.

1

Calculate the amount of discount received.

2

546

Write

Express the discount as a percentage of the original marked price.

Maths Quest 10 for the Australian Curriculum

a

100% - 15% = 85% 85% of $2500 = 0.85 ì $2500 = $2125 100% - 5% = 95% 95% of $2125 = 0.95 ì $2125 = $2018.75

b Discount = $2500 – $2018.75

= $481.25

$481.25 ì 100% $2500 = 19.25%

Percentage discount =

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

■■

The■single■discount■that■is■equivalent■to■successive■discounts■can■also■be■worked■out■by■ working■out■a■percentage■of■a■percentage,■as■shown■in■the■worked■example■below.■

WorkeD exAmPle 6

Find the single percentage discount that is equivalent to successive discounts of 15% and 5%. think

Write

1

Subtract■15%■from■100%■to■fi■nd■the■percentage■paid■ after■the■fi■rst■discount.

100%■-■15%■=■85%

2

Subtract■5%■from■100%■to■fi■nd■the■percentage■paid■ after■the■second■discount.

100%■-■5%■=■95%

3

Find■95%■of■85%.■This■is■actually■the■percentage■of■ the■marked■price■that■the■customer■pays.

95%■of■85%■=■0.95■ì■0.85 =■0.8075 =■80.75%

4

Subtract■the■percentage■from■100%■to■fi■nd■the■single■ percentage■discount.■This■answer■should■be■less■than■ 15%■+■5%.

Discount■=■100%■-■80.75% =■19.25%

Note:■The■single■percentage■discount■for■successive■discounts■is■always■less■than■the■sum■of■the■ individual■discounts. remember

1.■ When■two■separate■percentage■discounts■are■given,■they■must■be■calculated■one■after■ the■other.■Their■order■does■not■affect■the■fi■nal■answer. 2.■ The■single■discount■received■is■not■the■total■of■the■two■percentage■discounts;■rather,■it■ will■always■be■slightly■less. exercise

16c 13A inDiViDuAl PAthWAys eBook plus

Activity 16-C-1

Successive discounts doc-5173 Activity 16-C-2

Difficult successive discounts doc-5174 Activity 16-C-3

Tricky successive discounts doc-5175

successive discounts fluency 1 In■each■of■the■following,■an■item■is■reduced■in■price.■Calculate■the■percentage■discount,■correct■

to■1■decimal■place. a A■jumper,■usually■$29.95,■is■reduced■to■$24.95. 16.7% b A■video■game,■usually■$60,■is■reduced■to■$53.90. 10.2% c A■child’s■bike,■usually■$158,■is■reduced■to■$89. 43.7% d A■new■car,■usually■$29■500,■is■reduced■to■$24■950. 15.4% e A■plot■of■land,■priced■at■$192■■000,■is■reduced■to■ $177■■500■for■a■quick■sale. 7.6% 2 We 6 ■Calculate■the■single■percentage■discount■that■is■equivalent■to■ successive■discounts■of■15%■and■10%. 23.5% 3 mc ■The■single■percentage■discount■that■is■equivalent■to■ successive■discounts■of■10%■and■20%■is: A 10% B 18% ✔ C 28% D 30% E 35% chapter 16 financial maths

547

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

eBook plus

Digital doc

SkillSHEET 16.4 doc-5348

4 Find■the■single■percentage■discount■that■is■equivalent■to■each■of■the■following■successive■discounts. a 15%■and■20% 32% b 12%■and■8% 19.04% c 10%■and■7.5% 16.75% d 50%■and■15% 57.5% 5 Calculate■the■single■percentage■discount■that■is■equivalent■to■two■successive■10%■discounts. 19% unDerstAnDing 6 We 5 ■A■supplier■of■electrical■parts■offers■tradespeople■a■20%■trade■discount.■If■accounts■are■

settled■within■7■days,■a■further■5%■discount■is■given. a Calculate■the■price■paid■by■an■electrician■for■parts■to■the■value■of■$4000■if■the■account■is■ settled■within■7■days. $3040 b What■single■percentage■discount■does■the■electrician■receive■on■the■price■of■the■electrical■ parts? 24%

eBook plus

Digital docs

SkillSHEET 16.5 doc-5349 SkillSHEET 16.6 doc-5351

7 At■a■confectionary■wholesaler,■customers■have■their■accounts■reduced■by■10%■if■they■are■paid■

within■7■days. a Jacinta■pays■her■$100■account■within■7■days.■How■much■does■she■actually■pay? $90 b If■customers■pay■cash,■they■receive■a■further■5%■discount.■How■much■would■Jacinta■pay■ if■she■pays■cash? $85.50 c By■how■much■in■total■has■her■account■been■reduced? $14.50 d What■is■the■single■percentage■discount■equivalent■to■these■successive■discounts? 14.5% 8 A■fabric■supplier■offers■discounts■to■fashion■stores■and■a■further■discount■if■the■store’s■account■ is■paid■with■14■days.■‘David’s■Fashion■Stores’■have■ordered■fabric■to■the■value■of■$2000■from■ the■fabric■supplier. a If■fashion■stores■receive■a■reduction■of■8%,■how■much■does■‘David’s■Fashion■Stores’■owe■ on■its■account? $1840 b This■amount■is■reduced■by■a■further■5%■for■payment■within■14■days.■How■much■needs■to■ be■paid■now? $1748 c What■has■been■the■total■reduction■in■the■cost? $252 d What■do■the■successive■discounts■of■8%■and■5%■equal■as■a■single■percentage■discount? 12.6% 9 Tony■is■a■mechanic■who■wants■to■buy■equipment■worth■$250■at■a■hardware■store.■Tony■receives■ 15%■off■the■marked■price■of■all■items■and■then■a■further■5%■trade■discount. a Calculate■the■amount■that■is■due■after■Tony■is■given■the■fi■rst■15%■discount. $212.50 b From■this■amount,■apply■the■trade■discount■of■5%■to■fi■nd■the■amount■due. $201.88 c How■much■is■the■cash■discount■that■Tony■receives? $48.12 d Calculate■the■amount■that■would■have■been■due■had■Tony■received■a■single■discount■of■ 20%.■Is■this■the■same■answer? $200,■no e Calculate■the■amount■of■cash■discount■that■Tony■receives■as■a■percentage■of■the■original■bill. 19.25% f Would■the■discount■have■been■the■same■had■the■5%■discount■been■applied■before■the■15%■ discount? Yes g Calculate■the■single■percentage■discount■that■is■equivalent■to■successive■discounts■of■10%■ and■20%. 28% 10 A■car■has■a■marked■price■of■$25■■000. a Find■the■price■paid■for■the■car■after■successive■discounts■of■15%,■10%■and■5%. $18■■168.75 b What■single■percentage■discount■is■equivalent■to■successive■discounts■of■15%,■10%■and■5%? 27.325% reAsoning 11 Is■a■12.5% discount followed by a 2.5% discount,■the■same■single■discount■as■a■2.5% discount 

Single■discount■=■ 1■-■(1■-■a)■ì■(1■-■b),■ 12 where■a■and■b■are■ successive■discounts■ (as■decimals).

548

followed by a 12.5% discount?■Investigate■and■explain■your■answer■giving■mathematical■ evidence. Yes.■Both■lead■to■a■single■discount■of■14.69% Derive■a■mathematical■formula■to■calculate■the■ reflection    single■discount■(expressed■as■a■decimal)■ In what situations might a successive generated■by■two■successive■discounts,■a■and■b  discount be applied? (expressed■as■decimals).

maths Quest 10 for the Australian curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

16D eBook plus

Interactivity Compound interest

compound interest ■■ ■■

int-2791 ■■ ■■

Interest■on■the■principal■in■a■savings■account,■or■short■or■long■term■deposit,■is■generally■ calculated■using■compound■interest■rather■than■simple■interest. When■interest■is■added■to■the■principal■at■regular■intervals,■increasing■the■balance■of■the■ account,■and■each■successive■interest■payment■is■calculated■on■the■new■balance,■it■is■called■ compound interest. Compound■interest■can■be■calculated■by■calculating■simple■interest■one■period■at■a■time. The■amount■to■which■the■initial■investment■grows■is■called■the■compounded value.

WorkeD exAmPle 7

Kyna invests $8000 at 8% p.a. for 3 years with interest paid at the end of each year. Find the compounded value of the investment by calculating the simple interest on each year separately. think

Write

1

Write■the■initial■(fi■rst■year)■principal.

Initial■principal■=■$8000

2

Calculate■the■interest■for■the■fi■rst■year.

Interest■for■year■1■=■8%■of■$8000 =■$640

3

Calculate■the■principal■for■the■second■year■by■adding■ the■fi■rst■year’s■interest■to■the■initial■principal.

Principal■for■year■2■=■$8000■+■$640■ =■$8640

4

Calculate■the■interest■for■the■second■year.

Interest■for■year■2■=■8%■of■$8640 =■$691.20

5

Calculate■the■principal■for■the■third■year■by■adding■ the■second■year’s■interest■to■the■second■year’s■ principal.

Principal■for■year■3■=■$8640■+■$691.20 ■=■$9331.20

6

Calculate■the■interest■for■the■third■year.

Interest■for■year■3■=■8%■of■$9331.20 =■$746.50

7

Calculate■the■future■value■of■the■investment■by■ adding■the■third■year’s■interest■to■the■third■year’s■ principal.

Compounded■value■after■3■years =■$9331.20■+■$746.50 =■$10■■077.70■

To■calculate■the■actual■amount■of■interest■received,■we■subtract■the■initial■principal■from■the■ future■value. ■■ In■the■example■above,■compound■interest■=■$10■■077.70■-■$8000 ■ ■ ■ =■$2077.70 ■ We■can■compare■this■with■the■simple■interest■earned■at■the■same■rate. ■■

■ ■ ■ ■ ■ ■ ■■

P ×r ×T 100 8000 × 8 × 3 = 100 =■$1920

I=

The■table■below■shows■a■comparison■between■the■interest■earned■on■an■investment■of■ $8000■earning■8%■p.a.■at■both■simple■interest■(I )■and■compound■interest■(CI )■over■an■ eight■year■period. Year 1 2 3 4 5 6 7 8 Total (I ) $640.00 $1280.00 $1920.00 $2560.00 $3200.00 $3840.00 $4480.00 $5120.00 Total (CI ) $640.00 $1331.20 $2077.70 $2883.91 $3754.62 $4694.99 $5710.59 $6807.44 chapter 16 financial maths

549

number AND algebra • Money and financial mathematics

■■

We can develop a formula for the future value of an investment rather than do each example by repeated use of simple interest. Consider Worked example 7. Let the compounded value after each year, n, be An.

  After 1 year, A1 = 8000 ì 1.08 (increasing $8000 by 8%)   After 2 years, A2 = A1 ì (1.08) = 8000 ì 1.08 ì 1.08 (substituting the value of A1) = 8000 ì 1.082   After 3 years, A3 = A2 ì 1.08 = 8000 ì 1.082 ì 1.08 (substituting the value of A2) = 8000 ì 1.083   The pattern then continues such that the value of the investment after n years equals: $8000 ì 1.08n. ■■

■■

This can be generalised for any investment: A = P(1 + R)n where A = amount (or future value) of the investment P = principal (or present value) R = interest rate per compounding period expressed as a decimal n = number of compounding periods. To calculate the amount of compound interest (CI ) we then use the formula CI = A - P

Worked Example 8

William has $14  000 to invest. He invests the money at 9% p.a. for 5 years with interest compounded annually. a Use the formula A = P(1 + R)n to calculate the amount to which this investment will grow. b Calculate the compound interest earned on the investment. Think a

Write

Write the compound interest formula.

2

Write down the values of P, R and n.

P = $14  000, R = 0.09, n = 5

3

Substitute the values into the formula.

A = $14  000 ì 1.095

4

Calculate.

= $21  540.74 The investment will grow to $21  540.74.

b Calculate the compound interest earned.

■■ ■■ ■■

550

a A = P(1 + R)n

1

b CI = A - P

= $21  540.74 - $14  000 = $7540.74 The compound interest earned is $7540.74.

In the above example, interest is paid annually. Interest can be paid more regularly — it may be paid six-monthly (twice a year), ■ quarterly (4 times a year), monthly or even daily. This is called the compounding period. The time and interest rate on an investment must reflect the compounding period. For example, an investment over 5 years at 6% p.a. compounding quarterly will have: n = 20 (5 ì 4) and R = 0.015 (6% ó 4).

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

■■

To■fi■nd■n:■

■■

To■fi■nd■R:■

n■=■number■of■years■ì■compounding■periods■per■year R■=■interest■rate■per■annum■ó■compounding■periods■per■year WorkeD exAmPle 9

Calculate the future value of an investment of $4000 at 6% p.a. for 2 years with interest compounded quarterly. think

Write

1

Write■the■compound■interest■formula.

A■=■P(1■+■R)n

2

Write■the■values■of■P,■R■and■n.

P■=■$4000,■R■=■0.015,■n■=■8

3

Substitute■the■values■into■the■formula.

A■=■$4000■ì■1.0158

4

Calculate.

=■$4505.97 The■future■value■of■the■investment■is■$4505.97.

remember

1.■ The■future■value■of■an■investment■under■compound■interest■can■be■found■by■calculating■ the■simple■interest■for■each■year■separately. 2.■ The■compound■interest■formula■is■A■=■P(1■+■R)n,■where■A■is■the■amount■to■which■the■ investment■grows■and■P■is■the■principal■or■initial■amount■invested.■The■compound■ interest■earned■is■then■calculated■using■the■formula■CI■=■A■-■P. 3.■ In■the■formula,■n■is■the■number■of■compounding■periods■over■the■term■of■the■ investment:■ n■=■number■of■years■ì■compounding■periods■per■year. 4.■ In■the■formula,■R■is■the■interest■rate■(as■a■decimal)■per■compounding■period:■ R■=■interest■rate■per■annum■ó■compounding■periods■per■year. exercise

16D inDiViDuAl PAthWAys eBook plus

Activity 16-D-1

Compound interest puzzle 1 doc-5176 Activity 16-D-2

Compound interest puzzle 2 doc-5177 Activity 16-D-3

Compound interest puzzle 3 doc-5178

compound interest fluency 1 Use■the■formula■A■=■P(1■+■R)n■to■calculate■the■amount■to■which■each■of■the■following■

investments■will■grow■with■interest■compounded■annually. a $3000■at■4%■p.a.■for■2■years $3244.80 b $9000■at■5%■p.a.■for■4■years $10■■939.56 c $16■■000■at■9%■p.a.■for■5■years $24■■617.98 d $12■■500■at■5.5%■p.a.■for■3■years $14■■678.02 e $9750■at■7.25%■p.a.■for■6■years $14■■838.45 f $100■■000■at■3.75%■p.a.■for■7■years $129■■394.77 2 Calculate■the■compounded■value■of■each■of■the■following■investments. a $870■for■2■years■at■3.50%■p.a.■with■interest■compounded■six-monthly $932.52 1 2

b $9500■for■2 ■years■at■4.6%■p.a.■with■interest■compounded■quarterly c

$10■■650.81

1 $148■■000■for■3 2■years■at■9.2%■p.a.■with■interest■compounded■six-monthly

$202■■760.57

d $16■■000■for■6■years■at■8%■p.a.■with■interest■compounded■monthly $25■■816.04 e $130■■000■for■25■years■at■12.95%■p.a.■with■interest■compounded■quarterly $3■■145■■511.41 chapter 16 financial maths

551

number AND algebra • Money and financial mathematics UNDERSTANDING 3   WE 7  Danielle invests $6000 at 10% p.a. for 4 years with interest paid at the end of each year.

4 5

6 7 8 9

10

11

12

✔

13

14

Find the compounded value of the investment by calculating the simple interest on each year separately. $8784.60 Ben is to invest $13  000 for 3 years at 8% p.a. with interest paid annually. Find the amount of interest earned by calculating the simple interest for each year separately. $3376.26   WE 8  Simon has $2000 to invest. He invests the money at 6% p.a. for 6 years with interest compounded annually. $2837.04 a Use the formula A = P(1 + R)n to calculate the amount to which this investment will grow. b Calculate the compound interest earned on the investment. $837.04   WE 9  Calculate the future value of an investment of $14  000 at 7% p.a. for 3 years with interest compounded quarterly. $17  240.15 A passbook savings account pays interest of 0.3% p.a. Jill has $600 in such an account. Calculate the amount in Jill’s account after 3 years, if interest is compounded quarterly. $605.42 Damien is to invest $35  000 at 7.2% p.a. for 6 years with interest compounded six-monthly. Calculate the compound interest earned on the investment. $18  503.86 Sam invests $40  000 in a one-year fixed deposit at an interest rate of 7% p.a. with interest compounding monthly. a Convert the interest rate of 7% p.a. to a rate per month. 0.5833% b Calculate the value of the investment upon maturity. $42  891.60   MC  A sum of $7000 is invested for 3 years at the rate of 5.75% p.a., compounded quarterly. The interest paid on this investment, to the nearest dollar, is: A $1208 C $8208 D $8308 E $8508 ✔ B $1308   MC  After selling their house and paying off their mortgage, Mr and Mrs Fernhill have $73  600. They plan to invest it at 7% p.a. with interest compounded annually. The value of their investment will first exceed $110 000 after: A 5 years C 8 years D 10 years E 15 years ✔ B 6 years   MC  Maureen wishes to invest $15 000 for a period of 7 years. The following investment alternatives are suggested to her. The best investment would be: A simple interest at 8% p.a. B compound interest at 6.7% p.a. with interest compounded annually C compound interest at 6.6% p.a. with interest compounded six-monthly D compound interest at 6.5% p.a. with interest compounded quarterly E compound interest at 6.4% p.a. with interest compounded monthly   MC  An amount is to be invested for 5 years and compounded semi-annually at 7% p.a. Which of the following investments will have a future value closest to $10  000? A $700 B $6500 D $9000 e $9900 ✔ C $7400 Jake invests $120 000 at 9% p.a. for a 1-year term. For such large investments interest is compounded daily. a Calculate the daily percentage interest rate, correct to 4 decimal places. Use 1 year = 365 days. 0.0247% b Calculate the compounded value of Jake’s investment on maturity. $131  319.80 c Calculate the amount of interest paid on this investment. $11  319.80 d Calculate the extra amount of interest earned compared with the case where the interest is calculated only at the end of the year. $519.80

REASONING a   i  $17  745.95   ii  $17  786.61   iii  $17  807.67 iv  $17  821.99 552

15 Daniel has $15 500 to invest. An investment over a 2-year term will pay interest of 7% p.a. a Calculate the compounded value of Daniel’s investment if the compounding period is: i  1 year ii  6 months iii  3 months iv  monthly. b Explain why it is advantageous to have interest compounded on a more frequent basis.

Maths Quest 10 for the Australian Curriculum



T  he interest added to the principal also earns interest.

number AnD AlgebrA • money AnD finAnciAl mAthemAtics 16 Jasmine■invests■$6000■for■4■years■at■8%■p.a.■simple■interest.■David■also■invests■$6000■for■

4■years,■but■his■interest■rate■is■7.6%■p.a.■with■interest■compounded■quarterly.■ a Calculate■the■value■of■Jasmine’s■investment■on■maturity. $7920 Because■David’s■ b Show■that■the■compounded■value■of■David’s■investment■is■greater■than■Jasmine’s■ interest■is■compounded,■ investment. David’s■investment■=■$8108.46 the■interest■is■added■ c Explain■why■David’s■investment■is■worth■more■than■Jasmine’s■investment■despite■ to■the■principal■each■ receiving■a■lower■rate■of■interest. quarter■and■earns■ interest■itself. 17 Quan■has■$20■■000■to■invest■over■the■next■3■years.■He■has■the■choice■of■investing■his■money■at■ 6.25%■p.a.■simple■interest■or■6%■p.a.■compound■interest. a Calculate■the■amount■of■interest■that■Quan■will■earn■if■he■selects■the■simple■interest■ option. $3750■interest b Calculate■the■amount■of■interest■that■Quan■will■earn■if■the■interest■is■compounded: i annually $3820.32■interest ii six■monthly $3881.05 iii quarterly. $3912.36 Compound■quarterly■ c Clearly■Quan’s■decision■will■depend■on■the■compounding■period.■Under■what■conditions■ gives■the■best■return. should■Quan■accept■the■lower■interest■rate■on■the■compound■interest■investment? d Consider■an■investment■of■$10■■000■at■8%■p.a.■simple■interest■over■5■years.■Use■a■trialeBook plus and-error■method■to■fi■nd■an■equivalent■rate■of■ Digital doc compound■interest■over■the■same■period. reflection    WorkSHEET 16.2 e Will■this■equivalent■rate■be■the■same■if■we■change:■ doc-5352 How is compound interest i the■amount■of■the■investment? Yes calculated differently to simple ii the■period■of■the■investment? No interest? I f■we■assume■that■interest■is■compounded■annually,■ an■equivalent■return■of■R■=■7%■would■be■achieved.

16e eBook plus

eLesson What is depreciation?

eles-0182

Depreciation

■■

■■ ■■

Depreciation■is■the■reduction■in■the■value■of■an■item■as■it■ages■over■a■period■of■time.■For■ example,■a■car■that■is■purchased■new■for■$45■■000■will■be■worth■less■than■that■amount■1■year■ later■and■less■again■each■year. Depreciation■is■usually■calculated■as■a■percentage■of■the■yearly■value■of■the■item. To■calculate■the■depreciated■value■of■an■item■use■the■formula■ A■=■P(1■-■R)n

■■

■ here■A■is■the■depreciated■value■of■the■item,■P■is■the■initial■value■of■the■item,■R■is■the■ w percentage■that■the■item■depreciates■each■year■expressed■as■a■decimal■and■n■is■the■number■of■ years■that■the■item■has■been■depreciating■for. This■formula■is■almost■the■same■as■the■compound■interest■formula■except■that■it■subtracts■a■ percentage■of■the■value■each■year■instead■of■adding.

WorkeD exAmPle 10

A farmer purchases a tractor for $115 000. The value of the tractor depreciates by 12% p.a. Find the value of the tractor after 5 years. think

Write

1

Write■the■depreciation■formula.

A■=■P(1■-■R)n

2

Write■the■values■of■P,■R■and■n.

P■=■$115■■000,■R■=■0.12,■n■=■5

3

Substitute■the■values■into■the■formula.

A■=■$115■■000■ì■(0.88)5

4

Calculate.

■=■$60■■689.17 The■value■of■the■tractor■after■5■years■ is■$60■■689.17.

chapter 16 financial maths

553

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

eBook plus

Interactivity Different rates of depreciation

int-1155

■■ ■■ ■■

In■many■cases,■depreciation■can■be■a■tax■deduction. When■the■value■of■an■item■falls■below■a■certain■value■it■is■said■to■be■written off.■That■is■to■say■ that,■for■tax■purposes,■the■item■is■considered■to■be■worthless.■ Trial-and-error■methods■can■be■used■to■calculate■the■length■of■time■that■the■item■will■take■to■ reduce■to■this■value.

WorkeD exAmPle 11

A truck driver buys a new prime mover for $500■■000. The prime mover depreciates at the rate of 15% p.a. and is written off when its value falls below $100■■000. How long will it take for the prime mover to be written off? think

Write

1

Make■an■estimate■of,■say,■n■=■5.■Use■the■depreciation■ formula■to■fi■nd■the■value■of■the■prime■mover■after■ 5■years.

Consider■n■=■5. A■=■P(1■-■R)n ■=■$500■■000■ì■(0.85)5 ■=■$221■■852.66

2

Because■the■value■will■still■be■greater■than■$100■■000,■try■a■ larger■estimate,■say,■n■=■10.

Consider■n■=■10.■ A■=■P(1■-■R)n =■$500■■000■ì■(0.85)10 =■$98■■437.20

3

As■the■value■is■below■$100■■000,■check■n■=■9.

Consider■n■=■9. A■=■P(1■-■R)n =■$500■■000■ì■(0.85)9 =■$115■■808.47

4

Because■n■=■10■is■the■fi■rst■time■that■the■value■falls■below■ $100■■000,■conclude■that■it■takes■10■years■to■be■written■off.

The■prime■mover■will■be■written■off■in■ 10■years.

remember

1.■ Depreciation■is■the■reducing■value■of■a■major■asset■over■time. 2.■ Depreciation■is■usually■calculated■as■a■percentage■of■the■yearly■value■of■the■item. 3.■ The■depreciation■formula■is■A■=■P(1■-■R)n,■where■A■is■the■depreciated■value■of■the■ item, P is■the■initial■value,■R■is■the■percentage■depreciation■per■annum■expressed■as■a■ decimal■and■n■is■the■number■of■years■that■the■item■has■been■depreciating■for. exercise

16e inDiViDuAl PAthWAys eBook plus

Activity 16-E-1

Depreciation doc-5179 Activity 16-E-2

Harder depreciation doc-5180

Depreciation fluency 1 Calculate■the■depreciated■value■of■an■item■for■the■initial■value,■depreciation■rate■and■time,■in■

years,■given■below. a Initial■value■of■$30■■000■depreciating■at■16%■p.a.■over■4■years■ $14■936.14 b Initial■value■of■$5000■depreciating■at■10.5%■p.a.■over■3■years■ $3584.59 c Initial■value■of■$12■■500■depreciating■at■12%■p.a.■over■5■years■ $6596.65 unDerstAnDing 2 We10 ■A■laundromat■installs■washing■machines■and■clothes■dryers■to■the■value■of■

$54■■000.■If■the■value■of■the■equipment■depreciates■at■a■rate■of■20%■p.a.,■fi■nd■the■value■of■the■ equipment■after■5■years. $17■694.72 554

maths Quest 10 for the Australian curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics

inDiViDuAl PAthWAys eBook plus

Activity 16-E-3

Tricky depreciation doc-5181

3 A■drycleaner■purchases■a■new■machine■for■$38■■400.■The■machine■depreciates■at■16%■p.a. a Calculate■the■value■of■the■machine■after■4■years. $19■■118.26 b Find■the■amount■by■which■the■machine■has■depreciated■over■this■period■of■time. $19■■281.74 4 A■tradesman■values■his■new■tools■at■$10■■200.■For■tax■purposes,■their■value■depreciates■at■a■rate■

of■15%■p.a.■ a Calculate■the■value■of■the■tools■after■6■years. $3846.93 b Find■the■amount■by■which■the■value■of■the■tools■has■depreciated■over■these■6■years. $6353.07 c Calculate■the■percentage■of■the■initial■value■that■the■tools■are■worth■after■6■years. 38% 5 A■taxi■is■purchased■for■$52■■500■with■its■value■depreciating■at■18%■p.a. a Find■the■value■of■the■taxi■after■10■years. $7216.02 b Calculate■the■accumulated■depreciation■over■this■period. $45■■283.98 6 A■printer■depreciates■the■value■of■its■printing■presses■by■25%■p.a.■Printing■presses■are■ purchased■new■for■$2.4■million.■What■is■the■value■of■the■printing■presses■after:

a 1■year

$1.8■million

b 5■years

$569■■531.25

c 10■years?

$135■■152.44

7 mc ■A■new■computer■workstation■costs■$5490.■With■26%■p.a.■reducing-value■depreciation,■

the■workstation’s■value■at■the■end■of■the■third■year■will■be■close■to: ✔ B $2225 C $2811 D $3082 E $3213 mc ■The■value■of■a■new■photocopier■is■$8894.■Its■value■depreciates■by■26%■in■the■fi■rst■year,■ 21%■in■the■second■year■and■16%■reducing■balance■in■the■remaining■7■years.■The■value■of■the■ photocopier■after■this■time,■to■the■nearest■dollar,■is: A $1534 B $1851 C $2624 D $3000 E $3504 mc ■A■company■was■purchased■8■years■ago■for■$2.6■million.■With■a■depreciation■rate■of■ 12%■p.a.,■the■total■amount■by■which■the■company■has■depreciated■is■closest■to: A $0.6■million B $1.0■million ✔ C $1.7■million D $2.0■million E $2.3■million mc ■Equipment■is■purchased■by■a■company■and■is■depreciated■at■the■rate■of■14%■p.a.■The■ number■of■years■that■it■will■take■for■the■equipment■to■reduce■to■half■of■its■initial■value■is: ✔ B 5■years A 4■years C 6■years D 7■years E 8■years mc ■An■asset,■bought■for■$12■■300,■has■a■value■of■$6920■after■5■years.■The■depreciation■rate■is■ close■to: A 10.87% B 16.76% C 18.67% D 21.33% E 27.34% We11 ■A■farmer■buys■a■light■aeroplane■for■crop-dusting.■The■aeroplane■costs■$900■■000.■The■ aeroplane■depreciates■at■the■rate■of■18%■p.a.■and■is■written■off■when■its■value■falls■below■ $150■■000.■How■long■will■it■take■for■the■aeroplane■to■be■written■off? 10■years A $1684

8

✔

9

10

11 ✔

12

chapter 16 financial maths

555

number AND algebra • Money and financial mathematics 13 A commercial airline buys a jumbo jet for $750 million. The value of this aircraft depreciates

at a rate of 12.5% p.a. a Find the value of the plane after 5 years, correct to the nearest million dollars. $385 million b How many years will it take for the value of the jumbo jet to fall below $100 million? 16 years

reasoning

n

A = P (1 − R) A = (1 − R) n P A = (1 − R) P R = 1− n

n

14 A machine purchased for $48  000 will have a value of $3000 in 9 years. a Use a trial-and-error method to find the rate at which the machine is depreciating per annum. 27% n b Consider the equation x = a n , a = x . Verify your answer to part a using this relationship. 15 Camera equipment purchased for $150  000 will

have a value of $9000 in 5 years. Find the rate of annual depreciation using trial ■ and error first and then algebraically with the n relationship, ‘if x = a then a = n x ’. Compare and contrast each method.

A P

16f

Loan repayments ■■

reflection 

 

How and why is the formula for depreciation different to compound interest?

a Approx 43% b T  rial and error: can be time consuming, answer is often an estimate; algebraic solution: correct answer calculated immediately from equation

The simple interest formula is used to calculate the interest on a flat rate loan.

Worked Example 12

Calculate the interest payable on a loan of $5000 to be repaid at 12% p.a. flat interest over 4 years. Think

Write

Write the simple interest formula.

I=

2

List the known values.

P = $5000, r = 12%, T = 4

3

Substitute the values into the formula.

I=

4

Calculate the interest.

= $2400 The interest payable is $2400.

■■

556

P ×r ×T 100

1

5000 × 12 × 4 100

The total amount that would have to be repaid under the loan in Worked example 12 is $7400, and this could be made in 4 equal payments of $1850. With a flat-rate loan, the interest is calculated on the initial amount borrowed regardless of the amount of any repayments made.

Maths Quest 10 for the Australian Curriculum

number AND algebra • Money and financial mathematics

In contrast, taking a reducible-interest-rate loan means that each annual amount of interest is based on the amount owing at the time. ■■ Consider the same loan of $5000, this time at 12% p.a. reducible interest and an agreed annual repayment of $1850. At the end of each year, the outstanding balance is found by adding the amount of interest payable and then subtracting the amount of each repayment.

■■

Interest for year 1 = 12% of $5000 = 0.12 ì $5000 = $600 Balance for year 2 = $5000 + $600 - $1850 = $3750 Interest for year 2 = 12% ì $3750 = 0.12 ì $3750 = $450 Balance for year 3 = $3750 + $450 - $1850 = $2350 Interest for year 3 = 12% of $2350 = 0.12 ì $2350 = $282 Balance for year 4 = $2350 + $282 - $1850 = $782 Interest for year 4 = 12% of $782 = 0.12 ì $782 = $93.84 I n the fourth year, a payment of only $875.84 is required to fully repay the loan. The total amount of interest charged on this loan is $1425.84, which is $974.16 less than the same loan calculated using flat-rate interest.

Worked Example 13

Calculate the amount of interest paid on a loan of $10 000 that is charged at 9% p.a. reducible interest over 3 years. The loan is repaid in two annual instalments of $4200 and the balance at the end of the third year. Think

Write

1

Calculate the interest for the first year.

Interest for year 1 = 9% of $10  000 = 0.09 ì $10  000 = $900

2

Calculate the balance at the start of the second year.

Balance for year 2 = $10  000 + $900 - $4200 = $6700

3

Calculate the interest for the second year.

Interest for year 2 = 9% of $6700 = 0.09 ì $6700 = $603

4

Calculate the balance at the start of the third year.

Balance for year 3 = $6700 + $603 - $4200 = $3103

5

Calculate the interest for the third year.

Interest for year 3 = 9% of $3103 = 0.09 ì $3103 = $279.27 Chapter 16 Financial maths

557

number AnD AlgebrA • money AnD finAnciAl mAthemAtics 6

Calculate■the■amount■of■the■fi■nal■repayment■and■ ensure■that■the■loan■is■fully■repaid.

7

Find■the■total■amount■of■interest■paid■by■adding■each■ year’s■amount.

Balance■remaining■at■end■of■year■3 =■$3103■+■$279.27 =■$3382.27 Interest■charged■=■$900■+■$603■+■$279.27 =■$1782.27

remember

1.■ Loans■can■be■charged■by■calculating■either■fl■at■(simple)■interest■or■by■reducible■ interest. 2.■ To■calculate■the■cost■of■a■fl■at-rate■interest■loan,■use■the■simple■interest■formula. 3.■ To■calculate■the■cost■of■a■loan■at■a■reducible■interest■rate,■calculate■the■amount■of■ interest■payable■each■year■and■then■recalculate■the■outstanding■balance■of■the■loan■after■ each■payment■is■made■before■calculating■the■next■year’s■interest. exercise

16f inDiViDuAl PAthWAys eBook plus

Activity 16-F-1

Repaying a loan doc-5182 Activity 16-F-2

Harder loan repayments doc-5183 Activity 16-F-3

Difficult loan repayments doc-5184

loan repayments fluency 1 We12 ■Calculate■the■interest■payable■on■a■loan■of■$10■■000■to■be■repaid■at■15%■p.a.■fl■at-rate■ interest■over■3■years. $4500 2 Calculate■the■interest■payable■on■each■of■the■following■loans. a $20■■000■at■8%■p.a.■fl■at-rate■interest■over■5■years $8000 b $15■■000■at■11%■p.a.■fl■at-rate■interest■over■3■years $4950 c $7500■at■12.5%■p.a.■fl■at-rate■interest■over■2■years $1875 d $6000■at■9.6%■p.a.■fl■at-rate■interest■over■18■months $864 e $4000■at■21%■p.a.■fl■at-rate■interest■over■6■months $420 unDerstAnDing 3 Larry■borrows■$12■■000■to■be■repaid■at■12%■p.a.■fl■at■rate■of■interest■over■4■years. a Calculate■the■interest■that■Larry■must■pay. $5760 b What■is■the■total■amount■that■Larry■must■repay? $17■■760 c If■Larry■repays■the■loan■in■equal■annual■instalments,■calculate■the■amount■of■each■ repayment. $4440 4 We 13 ■Calculate■the■amount■of■interest■paid■on■a■loan■of■$12■■000■that■is■charged■at■10%■p.a.■

reducible■interest■over■3■years.■The■loan■is■repaid■in■two■annual■instalments■of■$5000■and■the■ balance■at■the■end■of■the■third■year. $2422 5 Calculate■the■total■amount■that■is■to■be■repaid■on■a■loan■of■$7500■at■12%■p.a.■reducible■interest■ over■3■years■with■two■annual■repayments■of■$3400■and■the■balance■repaid■at■the■end■of■the■ third■year. $9264 6 Brian■needs■to■borrow■$20■■000.■He■fi■nds■a■loan■that■charges■15%■p.a.■fl■at-rate■interest■over■ 4■years. a Calculate■the■amount■of■interest■that■Brian■must■pay■on■this■loan. $12■■000 b Calculate■the■total■amount■that■Brian■must■repay■on■this■loan. $32■■000 c Brian■repays■the■loan■in■4■equal■annual■instalments.■Calculate■the■amount■of■each■ instalment. $8000 d Brian■can■borrow■the■$20■■000■at■15%■p.a.■reducible■interest■instead■of■fl■at-rate■interest.■If■ Brian■makes■the■same■annual■repayment■at■the■end■of■the■fi■rst■three■years■and■the■balance■ in■the■fourth,■calculate■the■amount■of■money■that■Brian■will■save. $4966.87 558

maths Quest 10 for the Australian curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics 7 Georgia■borrows■$12■■000■at■10%■p.a.■reducible■interest■over■3■years.■Georgia■repays■the■loan■

in■two■equal■annual■payments■of■$4900■and■the■balance■at■the■end■of■the■third■year. a Calculate■the■amount■of■interest■that■Georgia■must■pay■on■this■loan. $2453 b Georgia■fi■nds■that■she■can■afford■to■repay■$5200■each■year.■How■much■does■Georgia■save■ by■making■this■higher■repayment? $93 8 Frank■borrows■$25■■000■at■12%■p.a.■reducible■interest■over■3■years■with■two■annual■repayments■ of■$11■■000■and■the■balance■repaid■at■the■end■of■the■third■year. a Find■the■total■amount■of■interest■that■Frank■pays■on■this■loan. $6004.80 b What■is■the■average■amount■of■interest■charged■on■this■loan■per■year? $2001.60 c By■writing■your■answer■to■part■b■as■a■percentage■of■the■initial■amount■borrowed,■fi■nd■the■ equivalent■fl■at■rate■of■interest■on■the■loan. 8% 9 Felicity■borrows■$8000■at■8%■p.a.■reducible■interest■over■3■years,■repaying■the■loan■in■ two■annual■payments■of■$3200■and■the■balance■repaid■at■the■end■of■the■third■year. a Using■the■method■described■in■question■8,■fi■nd■the■equivalent■fl■at■rate■of■interest. 5.4% b Find■the■equivalent■fl■at■rate■of■interest■charged■if■Felicity■increases■the■amount■of■each■ annual■repayment■to■$4000. 4.6% reAsoning 10 Natalie■has■the■choice■of■two■loans■of■$15■■000.■Each■loan■is■to■be■taken■over■a■three-year■term■

eBook plus

Digital doc

WorkSHEET 16.3 doc-5353

with■annual■repayments■of■$6350.■Loan■A■is■charged■at■9%■fl■at-rate■interest;■Loan■B■is■charged■ at■10%■reducible■interest.■As■Natalie’s■fi■nancial■planner,■construct■a■detailed■report■to■advise■ Natalie■which■loan■would■be■better■for■her■to■take.■ Loan■B■better■(total■savings■$1053.50) 11 Chris■borrows■$13■■500■at■10%■p.a.■reducible■interest■ over■2■years,■making■an■annual■repayment■of■$7800■and■ reflection    the■balance■repaid■at■the■end■of■the■second■year.■Show■ How does a loan at reducible that■if■interest■is■added■every■six■months,■at■which■ interest compare with the same time■a■repayment■of■$3900■is■made,■a■saving■of■ loan at flat-rate interest? approximately■$350■is■made. Actual■savings■$355.15■

chapter 16 financial maths

559

number AND algebra • Money and financial mathematics

Summary Purchasing goods ■■

There are alternatives to consider when deciding on how to pay for a major purchase.

■■

The simple interest formula is I = T = time.

■■

Credit card companies calculate interest on a monthly basis.

P ×r ×T , where P = principal, r = interest rate and 100

Buying on terms ■■

When buying an item on terms we usually pay a deposit with the balance plus interest paid in weekly or monthly instalments over an agreed period of time.

■■

To calculate the total cost of a purchase, add the deposit to the total of the regular repayments.

■■

The amount of each repayment is found by following these steps: (a) Calculate the deposit. (b) Find the balance owing by subtracting the deposit from the cash price. (c) Find the total repayments by adding the interest to the balance owing. (d) Divide the total amount to be repaid by the number of regular repayments that must be made.

■■

Loan repayments may be calculated in the same way, however there is no deposit made. Successive discounts

■■

When two separate percentage discounts are given, they must be calculated one after the other. Their order does not affect the final answer.

■■

The single discount received is not the total of the two percentage discounts; rather, it will always be slightly less. Compound interest

■■

The future value of an investment under compound interest can be found by calculating the simple interest for each year separately.

■■

The compound interest formula is A = P(1 + R)n, where A is the amount to which the investment grows and P is the principal or initial amount invested. The interest earned is then calculated using the formula CI = A - P.

■■

In the formula, n is the number of compounding periods over the term of the investment: n = number of years ì compounding periods per year.

■■

In the formula, R is the interest rate (as a decimal) per compounding period: R = interest rate per annum ó compounding periods per year. Depreciation

560

■■

Depreciation is the reducing value of a major asset over time.

■■

Depreciation is usually calculated as a percentage of the yearly value of the item.

■■

The depreciation formula is A = P(1 - R)n, where A is the depreciated value of the item, P is the initial value, R is the percentage depreciation per annum expressed as a decimal and n is the number of years that the item has been depreciating for.

Maths Quest 10 for the Australian Curriculum

number AnD AlgebrA • money AnD finAnciAl mAthemAtics Loan repayments ■■

Loans■can■be■charged■by■calculating■either■fl■at■(simple)■interest■or■by■reducible■interest.

■■

To■calculate■the■cost■of■a■fl■at-rate■interest■loan,■use■the■simple■interest■formula.

■■

To■calculate■the■cost■of■a■loan■at■a■reducible■interest■rate,■calculate■the■amount■of■interest■ payable■each■year■and■then■recalculate■the■outstanding■balance■of■the■loan■after■each■payment■ is■made■before■calculating■the■next■year’s■interest. MAPPING YOUR UNDERSTANDING

Homework Book

Using■terms■from■the■summary,■and■other■terms■if■you■wish,■construct■a■concept■map■that■ illustrates■your■understanding■of■the■key■concepts■covered■in■this■chapter.■Compare■your■ concept■map■with■the■one■that■you■created■in■What do you know?■on■page■537. Have■you■completed■the■two■Homework sheets,■the■Rich task■and■two■Code puzzles■in■ your■Maths Quest 10 Homework Book?

chapter 16 financial maths

561

number AND algebra • Money and financial mathematics

Chapter review Fluency 1 Calculate the simple interest that is earned on $5000 at 5% p.a. for 4 years. $1000 2 Jim invests a sum of money at 9% p.a. Which one

✔

of the following statements is true? A Simple interest will earn Jim more money than if compound interest is paid annually. B Jim will earn more money if interest is compounded annually rather than monthly. C Jim will earn more money if interest is compounded quarterly rather than six-monthly. D Jim will earn more money if interest is compounded annually rather than six-monthly. E It does not matter whether simple interest or compound interest is used to calculate the growth of Jim’s investment.

3 Find the single discount that is equivalent to successive discounts of 12.5% and 5%. 16.875% 4 Which one of the following statements is correct? ✔ A Successive discounts of 10% and 15% are less

than a single discount of 25%. B Successive discounts of 10% and 15% are

equal to a single discount of 25%. C Successive discounts of 10% and 15% are

greater than a single discount of 25%. D Successive discounts of 10% and 15% are

equal to successive discounts of 12% and 13%. E Successive discounts of 10% and 15% are

equal to successive discounts of 13% and 12%. 5 Brendan has a credit card with an outstanding

balance of $3600. The interest rate charged on the loan is 18% p.a. Calculate the amount of interest that Brendan will be charged on the credit card for the next month. $54 6 An LCD television has a cash price of $5750. It

can be purchased on terms of 20% deposit plus weekly repayments of $42.75 for 3 years. Calculate the total cost of the television if it is purchased on terms. $7819 7 Erin purchases a new entertainment unit that has

a cash price of $6400. Erin buys the unit on the following terms: 10% deposit with the balance plus interest to be repaid in equal monthly repayments over 4 years. The simple interest rate charged is 12% p.a. a Calculate the amount of the deposit. $640 b Calculate the balance owing after the deposit has been paid. $5760 562

Maths Quest 10 for the Australian Curriculum

c Calculate the interest that will be charged. $2764.80 d What is the total amount that Erin has to repay? e Calculate the amount of each monthly $8524.80 repayment. $177.60 8 A new car has a marked price of $40  000. The car

can be purchased on terms of 10% deposit and monthly repayments of $1050 for 5 years. a Find the total cost of the car if it is purchased on terms. $67  000 b Calculate the amount of interest paid. $27  000 c Calculate the amount of interest paid per year. $5400 d Calculate the interest rate charged. 15% p.a. 9 The single discount that is equivalent to successive discounts of 15% and 20% is: A 10% B 18% C 28% D 30% ✔ E 32% 10 A car dealership offers a 10% discount on the price

of service of a car purchased at the dealership. a Calculate the price paid for a service valued at $290.00 by a person who purchased their car at the dealership. $261 b During November, the dealership offers an extra 15% discount on all services and mechanical repairs. Calculate the price Callum, who purchased his car at the dealership, pays for a service in November. $221.85 c What is the total discount given on this service? $68.15 d Determine the single percentage discount that would be equivalent to the successive discounts of 10% and 15% that Callum receives. $23.5% 11 Ryan invests $12  500 for 3 years at 8% p.a. with interest paid annually. By calculating the amount of simple interest earned each year separately, determine the amount to which the investment will grow. $15  746.40 12 Calculate the compound interest earned on $45  000 at 12% p.a. over 4 years if interest is compounded: a annually $25  808.37 b six-monthly $26  723.16 c quarterly $27  211.79 d monthly. $27  550.17 13 A new computer server costs $7290. With 22% p.a. reducing-value depreciation, the server’s value at the end of the third year will be close to: A $1486 B $2257 C $2721 D $3023 ✔ E $3460

number AnD AlgebrA • money AnD finAnciAl mAthemAtics 14 An■asset,■bought■for■$34■■100,■has■a■value■of■$13■■430■

2 Gavin■borrows■$18■■000■over■5■years■from■the■

after■5■years.■The■depreciation■rate■is■close■to: A 11% ✔ B 17% C 18% D 21% E 22% 15 The■value■of■a■new■car■depreciates■by■15%■p.a.■ Find■the■value■of■the■car■after■5■years■if■it■was■ purchased■for■$55■■000. $24■■403.80

bank.■The■loan■is■charged■at■8.4%■p.a.■fl■at-rate■ interest.■The■loan■is■to■be■repaid■in■equal■monthly■ instalments.■Calculate■the■amount■of■each■monthly■ repayment. $426 3 A■building■society■advertises■investment■accounts■ at■the■following■rates: a 3.875%■p.a.■compounding■daily 3.95%■p.a.■fl■at■rate b 3.895%■p.a.■compounding■monthly 3.97%■p.a.■fl■at■rate c 3.9%■p.a.■compounding■quarterly. 3.96%■p.a.■fl■at■rate Peter■thinks■the■fi■rst■account■is■the■best■one■ Problem solVing because■the■interest■is■calculated■more■frequently.■ 1 The■value■of■a■new■tractor■is■$175■■000.■The■value■ Paul■thinks■the■last■account■is■the■best■one■ of■the■tractor■depreciates■by■22.5%■p.a. because■it■has■the■highest■interest■rate.■Explain■ a Find■the■value■of■the■tractor■after■8■years. $22■■774.65 whether■either■is■correct. b What■percentage■of■its■initial■value■is■the■ Neither■is■correct.■The■best■ tractor■worth■after■8■years? 13% option■is■to■choose■3.895%■p.a.■ compounding■monthly.

eBook plus

Interactivities

Test yourself Chapter 16 int-2870 Word search Chapter 16 int-2868 Crossword Chapter 16 int-2869

chapter 16 financial maths

563

eBook plus

ActiVities

Chapter opener Digital doc

•■ Hungry■brain■activity■Chapter■16■(doc-5344)■ (page 537) Are you ready? Digital docs (page 538) •■ SkillSHEET■16.1■(doc-5345):■Converting■a■ percentage■to■a■decimal •■ SkillSHEET■16.2■(doc-5346):■Finding■simple■ interest •■ SkillSHEET■16.3■(doc-5347):■Finding■a■percentage■ of■a■quantity■(money) •■ SkillSHEET■16.4■(doc-5348):■Finding■percentage■ discount •■ SkillSHEET■16.5■(doc-5349):■Decreasing■a■quantity■ by■a■percentage

16A Purchasing goods

(page 541) Activity■16-A-1■(doc-5167):■Simple■interest Activity■16-A-2■(doc-5168):■Harder■simple■interest Activity■16-A-3■(doc-5169):■Tricky■simple■interest SkillSHEET■16.1■(doc-5345):■Converting■a■ percentage■to■a■decimal •■ SkillSHEET■16.2■(doc-5346):■Finding■simple■interest Digital docs

•■ •■ •■ •■

16B Buying on terms Digital docs

•■ Activity■16-B-1■(doc-5170):■Buying■on■terms■ (page 544) •■ Activity■16-B-2■(doc-5171):■Buying■on■diffi■cult■terms■ (page 544) •■ Activity■16-B-3■(doc-5172):■Buying■on■tricky■terms■ (page 544) •■ SkillSHEET■16.3■(doc-5347):■Finding■a■percentage■ of■a■quantity■(money)■(page 544) •■ WorkSHEET■16.1■(doc-5350):■Buying■on■terms■ (page 545) 16C Successive discounts Digital docs

•■ Activity■16-C-1■(doc-5173):■Successive■discounts■ (page 547) •■ Activity■16-C-2■(doc-5174):■Diffi■cult■successive■ discounts■(page 547) •■ Activity■16-C-3■(doc-5175):■Tricky■successive■ discounts■(page 547) •■ SkillSHEET■16.4■(doc-5348):■Finding■percentage■ discount■(page 548)\ •■ SkillSHEET■16.5■(doc-5349):■Decreasing■a■quantity■ by■a■percentage■(page 548) •■ SkillSHEET■16.6■(doc-5351):■Expressing■one■ quantity■as■a■percentage■of■another■(page 548)

564

maths Quest 10 for the Australian curriculum

16D Compound interest Interactivity

•■ Compound■interest■(int-2791)■(page 549) Digital docs

•■ Activity■16-D-1■(doc-5176):■Compound■interest■ puzzle■1■(page 551) •■ Activity■16-D-2■(doc-5177):■Compound■interest■ puzzle■2■(page 551) •■ Activity■16-D-3■(doc-5178):■Compound■interest■ puzzle■3■(page 551) •■ WorkSHEET■16.2■(doc-5352):■Compound■interest■ (page 553) 16E Depreciation Interactivity

•■ Different■rates■of■depreciation■(int-1155)■(page 554) eLesson

•■ What■is■depreciation?■(eles-0182)■(page 553) Digital docs (pages 554–5) •■ Activity■16-E-1■(doc-5179):■Depreciation •■ Activity■16-E-2■(doc-5180):■Harder■depreciation •■ Activity■16-E-3■(doc-5181):■Tricky■depreciation 16F Loan repayments Digital docs

•■ Activity■16-F-1■(doc-5182):■Repaying■a■loan■ (page 558) •■ Activity■16-F-2■(doc-5183):■Harder■loan■repayments■ (page 558) •■ Activity■16-F-3■(doc-5184):■Diffi■cult■loan■ repayments■(page 558) •■ WorkSHEET■16.3■(doc-5353):■Loan■repayments■ (page 559) Chapter review

(page 563) •■ Test■yourself■Chapter■16■(int-2870):■Take■the■end-ofchapter■test■to■test■your■progress •■ Word■search■Chapter■16■(int-2868):■an■interactive■ word■search■involving■words■associated■with■this■ chapter •■ Crossword■Chapter■16■(int-2869):■an■interactive■ crossword■using■the■defi■nitions■associated■with■the■ chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

probleM solving

17 problem solving II

opening QUesTion

There are many strategies that can be used when solving problems. List 6 strategies.

problem solving

2

1  −   1 2   1 What is the value of    4   ? 4   2 Rachel is flying from Sydney to Perth. She left Sydney at 11:00 pm on January 6 and arrived in Perth five hours later. 1 am, 7 January (Perth a What is the time in Sydney when Rachel’s plane landed in Perth? 4 am, 7 January is 3 hours behind Sydney during daylight b What is the time in Perth when Rachel’s plane landed? saving time.) 3 The numbers 1 through 10 were written on pieces of paper and placed into a hat. Greg chose one piece of paper from the hat, and without replacing that number, then chose a second piece of paper from the hat. a Is Greg’s first choice dependent upon his second choice? Explain. b Is Greg second choice dependent upon his first choice? Explain. 4 Asuka sells musical instruments at discount prices. She had a drum kit on sale for 15% off the retail price of $5000. After two months the drum kit did not sell, and Asuka decided to apply an extra 10% discount to the existing sale price. a What is the total amount saved by the customer? $1175 b What is the final price of the drum kit? $3825 c Explain how a 25% discount on the retail price would compare with the successive 25% discount gives a final price of $3750. The discounts? a No. Greg’s first choice of

customer would be $75 better off.

1 0

1

0.5

y

2

y = 4-x

2

3 x

a number is independent of his second choice of a number. The ten numbers will always be in the hat on the first draw, thus he has an equal chance to pick any of them. b G  reg’s second choice is dependent upon his first choice. When Greg chooses a number in the first draw and does not replace that number, then he changes the sample space and the probability for the second draw.

5 I have a combination of $10 notes and $5 notes in my wallet. If I have 27 notes altogether and their total value is $190, how many of each type of note do I have? 16 ì $5 notes and 11 ì $10 notes 6 a Graph the equation y = 4-x using the following table:

x y becomes smaller and approaches 0, but never actually reaches 0.

566

−2.5

−2.0

−1.5

−1.0

−0.5

0

0.5

1.0

1.5

2.0

2.5

y b Describe what happens to y as x becomes larger. c Describe what happens to y as x becomes smaller.

Maths Quest 10 for the Australian Curriculum

y approaches infinity as x becomes smaller.

problem solving 7 Mariah the Mathematics teacher wanted to give her students a chance to win a reward at the

8 a Since the interest rate is lower for Loan 2 than for Loan 1, Thomas should choose Loan 2 if he decides to pay the loan off at the end of the first, second or third year. b L  oan 1 at term amounts to $9444.63. Loan 2 at the end of 4 years amounts to $9523.42. Thomas should choose Loan 1. c T  homas should choose Loan 1. At the end of its term (3 years), it amounts to less than Loan 2 at 4 years, 1 year before its term is finished. d T  homas may not have the money to pay off Loan 1 in 3 years. He may need the extra 2 years to accumulate his funds.

9 10 11

12 13

14





a Future population in n years = 350(1 + 0.1)n. b 12 years c 19 years d L  ance has assumed that every 19 years there will be approximately 2140 additional people. e L  ance has assumed 15 that the growth is linear, whereas it is actually exponential. Drawing a graph would help him see the growth.

end of the term. She placed 20 cards into a box, and wrote the word ON on 16 cards, and OFF on 4 cards. After a student chooses a card, that card is replaced into the box for the next student to draw. If a student chooses an OFF card, then they do not have to attend school on a specified day. If they choose an ON card, then they do not receive a day off. a Mick, a student, chose a random card from the box. What is the probability he received a day off? 15 b Juanita, a student, chose a random card from the box after Mick. What is the probability 4 that she did not receive a day off? 5 1 c What is the probability that Mick and Juanita both received a day off? 25 Thomas went to an electronics store to buy a flat screen HD TV together with some accessories. The store offered him two different loans to buy the television and equipment.   The following agreement was struck with the store. •• Thomas will not be penalised for paying off the loans early. •• Thomas does not have to pay the principal and interest until the end of the loan period. Loan 1  $7000 for 3 years at 10.5% p.a. compounding yearly Loan 2  $7000 for 5 years at 8% p.a. compounding yearly a Explain which loan Thomas should choose if he decides to pay off the loan at the end of the first, second or third year. b Explain which loan Thomas should choose for these two options. Paying off Loan 1 at term Paying off Loan 2 at the end of four years c Thomas considers the option to pay off the loans at the end of their terms. Explain how you can determine the better option without further calculations. d Why would Thomas decide to choose Loan 2 instead of Loan 1 (paying over its full term), even if it cost him more money? 77 Express 4.27 as an improper fraction. 18 What is the probability of choosing a red card or the Ace of Clubs from a standard pack of cards? 27 52 n The mean mass of a group of n people is m kg. If another people whose mean mass is 2 5m kg join the group, the mean mass changes to (m + 6) kg. Evaluate m. 72 4 What is the cost of buying 2000 shares in the mining company DIGGER at $10.47 each if there is a transaction fee of 0.1% OR $20, whichever is the larger? $20  960.94 A and B are complementary events and P(A) = a. Determine: a P(B) 1 - a b P(A ß B) 1 c P(A ¶ B) 0 Lance lives in a small town with a population of 350 people. The town was predicted to grow at rate of 10% per year. a Write an equation to model the growth of this town over n years. b How many years will it be until the population is over 1000 people? (Whole years only) c How many years will it be until the population is over 2000 people? (Whole years only) d Lance, using the data from part c, assumed that the town should have approximately 10  000 people 95 years from now. Explain why his reasoning is incorrect. e Explain how you can help Lance see how the population will change over the next 100 years. A TV cost $700 cash. I buy it on terms, that are $200 deposit plus $30 per month for 20 months. a How much more do I actually pay? $100 b Express this amount as a percentage of the cash price. 14.29% Chapter 17 Problem solving II

567

problem solving 16 Write the following expression in its simplest form.

18 x (3 x − 12)2

×

15 x + 30 15 ÷ x ( x + 2) ( x − 4)2

2

17 A paint shop produces a unique type of paint by mixing two kinds of paint together — indoor

paint and outdoor paint. The unique paint is subject to the following conditions:

If y = 10, the first i At least 20 litres of paint (indoor and outdoor combined) must be mixed. equation becomes ii Indoor paint requires 3 units of dye, while outdoor paint requires 8 units of dye and x í 10 whereas the second equation there is a maximum of 110 units of dye available. becomes x Ç 10. a If x litres of indoor paint and y litres of outdoor paint are mixed, write two inequations This means that the for the conditions given. x + y í 20 and 3x + 8y Ç 110 only possible value for x here is equal b If 10 litres of outdoor paint is required, how much dye can be used for the indoor to 10. paint? No more than 30 units of dye

c Discuss the effect that the information in part b has on the two inequations from part a. 18 What is the probability of rolling two even numbers in one throw of two unbiased, 6-sided 1

19

20 Since the depreciation of 40% is on a lower value each year, the amount Jan can deduct from her taxable income decreases every year.

21

Teacher to check.

22 23 p = 250n + 15, n is independent variable and p dependent variable — number of people depend on number of screens.

t=

568

p 100

dice? 4 A calculator company took a poll of 80 Year 10 students to find out what types of calculators students had used during their academic careers: Scientific calculators: 70 Graphing calculators: 50 Graphing and Scientific calculators: 40 What is the probability that a student had used a graphing calculator given that they had 4 also used a scientific calculator? 7 Jan bought a computer for her business at a cost of $2500. She elected to use the diminishing value method (compound depreciation), instead of the straight-line method of depreciation. Her accountant told her that she was entitled to depreciate the cost of the computer over 5 years at 40% per year. a How much was the computer worth at the end of the first year? $1500 b By how much could Jan reduce her taxable income at the end of the first year? (The amount Jan can reduce her taxable income is equal to how much value the asset lost from one year to the next.) $1000 c Explain whether the amount she can deduct from her taxable income will increase or decrease at the end of the second year. A Shend is a type of tropical pumpkin grown by the people of Outer Thrashia. The diameter (D m) of a Shend increases over a number of months (m) according to the rule D = 0.25 ì (10)0.01m. a Determine the diameter of the pumpkin after 4 months. 0.27 m b If the Shend is not harvested it will explode when it reaches a critical diameter of 0.5 metres. Show that it takes approximately 30 months for an un-harvested Shend to explode? There are 6 black discs and 9 red discs in a jar. Two discs are withdrawn simultaneously. 17 What is the probability that the discs are the same colour? 35 Natalina is going to build a movie theatre with n screens. At each screen, there will be 250 seats for the audience to watch that movie. In addition to audience members, there are 15 employees on the premises at any given time (selling tickets and popcorn and so on). According to building regulations, she must have one toilet for every 100 people in the building. a Write an equation relating the number of screens (n) to the total number of people who can possibly be in the building (p) any one time. Which variable is dependent? Which variable is independent? Explain. b Write an equation relating the total number of people who can possibly be in the building (p) to the number of toilets (t). c What kind of numbers do p, t and n have to be? Integers

Maths Quest 10 for the Australian Curriculum

problem solving d If Natalina builds 5 theatre screens, what is the minimum number of toilets she must also build? At least 13 e If Natalina can only supply eight toilets, what is the maximum number of screens she can build? 3 24 Jason and Paul go to the shopping car park on weekends to see if they can find any loose

∑ x 1146 = = 38.2 n 30 Elena: Mean =

26

Midpoint Frequency x f   4.5 0 14.5 4 24.5 7 34.5 5 44.5 4 54.5 4 64.5 6

fx    0   58 171.5 172.5 178 218 387

Anthea: Mean =

∑ fx 1185 = = 39.5 30 ∑f

The median is the 50th percentile which corresponds to the 55% to 64% interval.

Cumulative frequency

Stephanie can only work out which class interval her test result is in — the 65% to 25 74% interval.

change that people have dropped. Over the past year they have kept track of how much money they have found. They found twelve $2 coins, three fifty-cent pieces, thirty-nine 20c coins, thirty-eight 10c coins, and one hundred and fifty-two 5c coins. What is the probability that the next coin they find will be worth more than ten cents? State your answer as a percent to the nearest percent. 22% A set of examination results is displayed in the following cumulative frequency histogram and 100 ogive. 80 a Stephanie knows that her result is in the 60 85th percentile. Explain whether Stephanie can work out her exact result. 40 b What is the class median? 20 c Stephanie’s twin, Betty, knows that she got 70% for her exam. Can she compare her 0 30 40 50 60 70 80 90 exam result to Stephanie’s result? No. Examination mark d Approximately in which percentile would Betty’s result lie? 85th — the same as Stephanie. Dimitri constructed a back-to-back stem plot to compare the ages of the students in his dance class. Class Class   Ages of students attending at the Ballroom Dancing Studio interval mid-point Frequency Females

Males

9

1

123

7

2

022245

5

3

0017

52

4

67

320

5

2

4421

6

44

0–9 10–19 20–29 30–39 40–49 50–59 60–69

  4.5 14.5 24.5 34.5 44.5 54.5 64.5

0 4 7 5 4 4 6

Range of females = 45 years; range of males = 53 years

a Compare the range of distribution of the ages of males and females from this set of data. Females: 64 years; males: 22 years b What is the mode for this set of data? c One of Dimitri’s students, Anthea, used a grouped frequency distribution table of all the

Class interval 0–9 10–19 20–29 30–39 40–49 50–59 60–69

students to calculate the mean age of the students at Dimitri’s Ballroom Dancing Studio. Create the grouped frequency distribution table using class intervals of 0–9, 10–19 etc. d Anthea correctly calculated the mean from the grouped frequency distribution table to be 39.5. Elena, another student, correctly used the original data in the back-to-back stem plot and calculated the mean to be 38.2. Show how they both got their answers. e What is the reason for the difference in the two answers? Prove P(AÅ) ì P(B) = P(AÅ ¶ B) 27 Events AÅ and BÅ are independent. Prove that events AÅ and B are independent. 28 Tylar is offered a new Beta Brava, a sleek 4 cylinder sports car for $56  000. He borrows the A  nthea used the money and repays it at $12  500 per year for 5 years. midpoints of the class All new cars depreciate in value; the Beta Brava is no different. It loses 10% in the first intervals, whereas Elena year and then 5% of the previous year’s value each year thereafter. used the exact values. a What is the car worth after 5 years ? $41  051 b Taking the depreciation into account and considering how much he paid for his car, how much has Tylar lost, on average, each year over the 5 years? Approx $4300 Chapter 17 Problem solving II

569



c

i The two equations are the same, so the graphs lie on top of each other. ii The gradient is doubled, the y-intercept is unchanged, the x-intercept is halved. iii The gradient is halved, the y-intercept is halved, the x-intercept is unchanged. iv The gradient is unchanged, the y-intercept is doubled, the x-intercept is doubled. These are parallel lines.

problem solving 29 The equation 2x + 3y = 6 is changed according to the following rules. i The whole equation is doubled. ii Only the x-value is doubled. a i  4x + 6y = 12 ii  4x + 3y = 6 iii  2x + 6y = 6 iv  2x + 3y = 12 iii Only the y-value is doubled. iv The y-intercept is doubled. a Write the new equation in each case. b Draw the graph of each new equation on the same axes as the original equation. c Describe the effect of each transformation on the graph of the original equation. 30 Andrew does not know the answer to two questions on a multiple choice exam. The first

question has four choices and the second question he does not know has five choices. 3 a What is the probability that he will get both questions wrong? 5 b If he is certain that one of the choices cannot be the answer in the first question, how 8 will this change the probability that he will get both questions wrong? 15 31 Lara is employed as a salesperson. She is offered two methods of calculating her income. Method 1: Commission only of 13% on all sales Method 2: $350 per week plus a commission of 4.5% on all sales

Lara’s research shows that the average sales total per employee per month is $14 382. a If Lara were to choose her method based on the average employee sales total, from

which method of payment would she receive the most income, in dollars per annum? b Compare the difference, in dollars, between the two methods of payment based on the

average sales total. Justify your answer with calculations. 32 Annie’s and Barbara’s ages add to 25, and have a difference of 11. If Annie is the older of the two, how old are both Annie and Barbara? Annie is 18 and Barbara is 7. 33 Rosetta and Theo have been married for 16 years but have kept separate investment accounts during that time. The graph below shows the value of their investments over time where D  ifference of $3530 in interest has been paid annually. favour of Method 2.

Value of investment ($)

Value of investments

R  = 10.5%, This is the percentage interest rate per annum. So, R = 10.41% p.a.

Theo’s account Rosetta’s account

A 25000 20000 15000 10000 5000 0

570

Method 2

2

4

6 8 10 12 14 16 Time (years)

n

a What were the values of Rosetta’s and Theo’s initial investments? $5000 b How much did Rosetta’s investment change yearly over time? What does this represent $1000, interest earned per annum. in real terms? A = 5000 + 1000n c Give the equation to describe the value of Rosetta’s investment over time (A). n

R   compounded  100  annually. Determine the rate at which Theo’s investment is growing? e After initially investing the same amount of money, the graph shows that after approximately 13 years their investments had grown to the same amount. Use your equations to show that this is so. C  heck with your teacher. R  osetta had a higher return in the f Compare the two investments over the 16 year period. d Theo’s investment grew according to the general formula A = P  1 +

Maths Quest 10 for the Australian Curriculum

first 13 years then Theo had the higher return after that.

35 36

37

80 60 50 10 Test B

50 20 Test A

40

70

 he interquartile range is the same for both tests (IQR = 30). Xmax b T This indicates that the spread of the results across the middle 80 group of the class for both tests is the same.  ased on the median score, the students appear to have done 90 c B better in Test B. The middle groups stayed the same, the strong maths students did better, but the struggling students did worse. Median QL

QU

summary for each. b Which statistic appears to be the

38

a

Xmin

34 The marks for two Mathematics tests, A and B, for a class are presented in the box plots below. a Compare the five point

50%



Mean = 5.4; median = 5.5; mode = 6 The median is best because it allows for the range of values and is also between the mean and mode.

39



40

Mathematics test A

Students may have found topic B easier to understand.

problem solving

Mathematics test B same for both tests? What does this indicate? 0 10 20 30 40 50 60 70 80 90 100 c In which test did the students Test B may have been an easier test than Test A. appear to perform better? d What can the teacher deduce about the tests from these results? e What can the teacher deduce about her students’ learning from these results? Three contestants A, B and C enter a race. If A is three times as likely to win as B and four times as likely to win as C, find the probability that either A or B wins. 16 19 My weight is k kg. I put on a kg per month for m months after which my weight increases at the compound rate of b% per month for c months. I start a diet and my weight drops at the compound rate of d% per month for e months. What is my final weight? Your big brother has buried your mp3 player in the backyard to tease you. He’s given you the following Cartesian co-ordinate clues to help you find it, where x and y are the horizontal and c e vertical distances from the back door in metres. y b   d   Shrub 8 (k + am)  1 + 1−    The back door is at (0, 0)  100   100  6 The tree is at (0, 3) 4 Stump Tree 2 The shrub is at (8, 7) 0 The stump is at (8, 3) 2 4 6 8 10 12 14 x The player is on the line y = − 45 x + 10 which bisects the line connecting the tree and shrub and it is also 3 metres from the stump.   Draw a diagram to represent this information and mark the possible locations of your player on the Cartesian plane. In the scientific area of Genetics, probability is used to assist in determining the likelihood of inherited characteristics.   For example; a widow’s peak hairline is dominant; a straight hairline is recessive. Consider a mother who is heterozygous dominant (Ww) for the widow’s peak and a father who is homozygous recessive (ww). W w a Complete this table. Ww ww w b Use your table to determine the probability Ww ww w that their offspring will have a widow’s peak. Palmo is in his third year as an apprentice cabinet maker. He earns $855 gross salary per week. a Calculate his income for the year, if he receives a 17.5% four week holiday loading. $45 058.50 b Palmo purchased cutting tools for $5000. In the first three years they depreciated at a constant rate to $3635. How much did the tools depreciate each year? $455 per year Penny is a softball player. The number of runs she scored in her first eight games were 2, 3, 4, 5, 6, 6, 8, 9. a Calculate the mean, median and mode for the number of runs to 1 decimal place. Discuss which statistic is the best indicator of the measure of centre in this case. b Based on the above figures:   0.5 i what is the probability that in future games, Penny will score at least six runs? ii in how many of her next five games would you expect her to score at least six runs?   2 or 3 c Penny’s coach claims that it is possible for her to achieve a median of 7 if she keeps up her practice.   4 more games scoring at least 8 runs in each i What is the minimum number of games she must play to have a median run score of 7? At least how many runs will she need in each game? ii Give a possible set of scores for these games that would allow Penny to achieve this. d Given your answer to part b above, how likely is this? Not very likely

  2  3  4  5  6  6  8  8  8  8  8  9 Chapter 17 Problem solving II

571

$15 Loss

$20 Loss 0.462

$20

$0 0.0231

Profit/ Loss $8 loss

41 This board game consists of 25 squares, each with side

42 43

N  o this is not a realistic model as is it does not take into account changes to climate, rain, runoff from mountains, glaciers etc. 44

45

lengths of 12 cm. The triangles are equilateral in shape, with side lengths of 8 cm. A coin is tossed and lands on the game board. a What is the probability it will land on a square that contains a triangle? 0.24 b What is the probability it will land on a triangle? 0.0462 c What is the probability it will land on a green or blue 0.0231 triangle? d It costs $1 per throw and pays $2.50 for landing on a square that contains a triangle, $5.00 for landing on a triangle and $15 for landing on a green or blue triangle.   If you played 20 games what could be your potential return per game on each? Would you make a profit or loss on your outlay? e If you play 100 games, would you made a profit? Show your reasoning. f What entices people to play these games? The difference between two numbers is 3. If six times the larger number minus twice the smaller number is 46, determine the two numbers. The numbers are 7 and 10. The surface area of a lake is evaporating at a rate of 5% per year due to climate change. To model this situation, a relationship between the surface area of the lake (S km2) over time is given by S = 20  000 ì 0.95x, where x is the time in years. Yes, because the relationship involves a variable as an exponent. a Explain whether this is an exponential relationship. 20  000 km2 b What is the surface area of the lake initially? c What will the surface area be in 10 years’ time? S = 11  975 km2 d Plot a graph for this relationship. e What will the surface area be in 100 years’ time? In 100 years, S = 118 km2 f Explain whether this is a realistic model. A bakery employs experienced bakers at $25 per hour and apprentices at $16 per hour. The manager has a budget which will allow her to employ four experienced bakers to work a 40 hour week. a How many apprentices who will only work for 30 hours per week can be employed for the same amount of money? About 8 apprentices can work for the same amount of money. b If the manager want to employ a combination of experienced bakers and apprentices, how many of each could she employ for $4000 per week? Bailey has built a wooden ramp to practise his skateboard tricks. A

Experienced bakers Apprentices Total $1000 each 30 h ì $16 = $480 cost 4 ($4000)

0

B

$4000

3 ($3000)

2 ($480) = $960 $3960

2 ($2000)

4 ($480) = $1920 $3920

1 ($1000)

6 ($480) = $2880 $3880

0

8 ($480) = $3840 $3840

E

D

2.2 m

3.6 m F

12 m

C

a Find the distance travelled up the ramp (i.e. distance FB). 12.2 m b What angle does the ramp make with the ground? 10.4è c After construction, Bailey checks that the ramp is ‘square’ by measuring the diagonals. Find the length of AF. 12.72 m d Two such ramps are made so that they can be placed 10 m apart. If Bailey can cover the

distance from the top of one ramp, along the ground, then to the top of the other, on his skateboard, in 12 seconds, how fast is he going? (Answer in km/h). 10.3 km/h 572

Maths Quest 10 for the Australian Curriculum

e 1 00 games — potentially 30 wins so get $85, but paid $100 to play, so still losing. f P  eople look at short term gains, but fail to consider the long term maths calculations.

Potential Probability wins Outlay Winnings 0.24 0.24 ì 20 $20 2.50 ì 4.8 = 4.8 = $12 0.0462 0.924 ö 1 $20 $5

problem solving

problem solving 46 Briana’s normal rate of pay is $15.25 per hour. Last week she was paid for 11 hours, at time-

47 149 cm and 171 cm. The average height of the two students must equal 160 cm. 48

Toyota (22)

Nissan (18)

2

10 4

and-a-half. a If Briana was paid at double the hourly rate, how many hours would she need to work the next week, to earn the same amount of money? 8.25 hours b Briana’s boss is offering her the choice for two shifts, the rate paid depending on the work days offered. Shift 1:  15 hours at the normal rate plus 5 hours at time-and-a-half. Or Shift 2:  10 hours at the normal rate plus 8 hours at double time. Which shift offers a better wage and by how much? Shift 2 offers $53.38 more. The height of each student in a Year 10 class was measured and it was found that the mean height was 160 cm. Two students were absent. When their heights were included in the data for the class, the mean height did not change.   Suggest a pair of heights that are possible for the two absent students. What reasoning could be used to find a pair of possible heights? The IT department for a school can buy mini-laptops from an educational supplier and receive a discount. If the IT manager orders 150 mini-laptops (for all the Year 10 students), the cost is $30 000 and if she orders 80 mini-laptops the cost is $16  000. If she only had $20  000 to spend how many mini-laptops could she buy? 100 mini-laptops

6

4 6

1 2 Subaru

(17)

49 At the entrance to a car rally, 35 people were surveyed and asked which of three models of

4WD rally cars they preferred — Toyota, Nissan or Subaru. Six of the group liked all three types of 4WD vehicles. Eight of the group liked Toyota and Nissan, 10 liked Toyota and Subaru and 12 liked Nissan and Subaru. Also 22 of the group liked Toyota, 18 liked Nissan and 17 liked Subaru. Two people didn’t like any of the models of 4WD rally cars. a Display this information on a Venn diagram. b Determine the probability of selecting a person who: 2 i liked Toyota only  7 18 ii does not like the Subaru.   35 11 c Find the probability that a person likes Subaru or Nissan but not Toyota. 35 Chapter 17 Problem solving II

573

problem solving 50 Bronwyn decides to buy a new laptop. The ticketed price is $1200. When Bronwyn’s credit

Class interval Frequency

10 20 30 40 50 60 70 80 Age

x

Males Females

b

Mean Range IQR

Males Females 28.2 31.1 70 57 18 22

card statement arrives, it shows that she will not pay any interest if she pays the full amount by the due date. 0−4  7 a If Bronwyn pays $300 by the due date, what is the balance owing? $900 5−9  8 b If the interest rate on the credit card is 22% p.a., how much interest will Bronwyn be charged on the balance owing in the next month? $16.50 10−14  5 c How much will Bronwyn owe at the end of that month? $916.50 15−19  4 d Bronwyn now pays $600 off her credit card. How much interest is she charged the 20–24  1 $322.30 $5.80 following month? Total 25 e Bronwyn then pays off the remaining balance on her credit card. How much does she pay? f How much has the laptop cost her, including all the interest payments? $1222.30 51 The data below shows the number of times 25 Year 10 students have used a computer in the last week. 10, 19, 7, 0, 1, 6, 22, 3, 9, 15, 3, 6, 13, 2, 16, 8, 5, 4, 11, 10, 16, 4, 8, 5, 13 a Group the data into a frequency table in class intervals of size 5. b Represent the grouped data as a histogram. 5 0th percentile is about 8.5 computers used, 30th percentile is 5 computers used. This c Add a frequency polygon to the histogram. means that 50% of the data lies below 8.5 and d Construct a cumulative frequency polygon. 30% of the data lies below 5. e From the cumulative frequency polygon, estimate how many students used a computer 10 students more than 10 times in the week. f Add a percentage cumulative frequency axis to the graph. g Estimate the 50th percentile and the 30th percentile. Interpret these results. 52 The initial alcohol content in a glass of red wine is about 12%. If the initial concentration is given by C0 and t represents the number of days after the wine has been opened, then the reduction in concentration of alcohol is given by: C = C0 ì 0.12kt a Show that the value of k is 0.24, if the initial concentration is 100 mg/L and 60 mg/L after 1 day. Teacher to check. b State the exponential equation. C = 100 ì 0.120.24t c Draw the graph of this relationship. d Find the concentration remaining after 5 days. 7.85 mg/L 53 The following data show the ages of a Male Female group of 30 males and 30 females as they 98 0 5 enter hospital for the first time. 9 9 8 8 8 6 3 21 1 77899 a Construct a pair of parallel box plots to 87764320 2 0012455679 represent the two sets of data, showing 86310 3 013358 working out for the median and 1st and 3rd quartiles. 752 4 2368 T  ypically males b Calculate the mean, range and IQR for 53 5 134 seem to enter both sets of data. T  here is one outlier hospital for the first 6 2 c Determine any outliers if they exist. — a male aged 78. time at a younger 8 7 age than females. d Write a short paragraph comparing the data.

2 5

Red

3 5 2 5

2 5

Red

Black 3 5

3 5

Red 2 5

Black 3 5

Black Red

Black Red 2 5

2 5

Black Red

Black Red

2 3 5 5

3 5

Black 3 5

receiving a higher rate of interest. 55 A pencil case contains 3 black pens and 2 red pens. The pencil case is shaken, one falls out and is put back in the case. This is repeated twice more. Each pen has an equal chance of falling out. a Represent this information on a tree diagram. b Find the probability of getting three black pens. 0.216 0.352 c Find the probability of getting at least two red pens.

Maths Quest 10 for the Australian Curriculum

B  ecause Nathan’s interest is compounded, the interest is added to the principal each quarter and earns interest itself.

a

0



574

54 Fiona invests $8000 for 4 years at 6% p.a. simple interest. Nathan also invests $8000 for  athan will have 4 years; however, his interest rate is 5.6% p.a. compounded quarterly. N $9993.03 at maturity. a Calculate the value of Fiona’s investment at maturity. $9920 b Show that the compounded value of Nathan’s investment is greater than that of Fiona’s. c Explain why Nathan’s investment after 4 years is greater than Fiona’s even though she is

problem solving A Investment value ($)

12000 10000

56 Catherine invests $2000 in a term deposit account which pays interest at a rate of 4.5% per annum

8000 6000 4000 2000 0

10 20 30 40 x Time (years)

57

58

59

60

61

62

on the balance at the beginning of each year. 104.5% a After one year what is her investment worth? $2090 b What percentage of her original investment is her bank balance at the end of the year? $2184.05 c What would Catherine’s investment be worth after two years? d Plot the graph of this exponential function. e Use the graph to predict when her investment will be worth $10  0 00. Approximately 36.5 years A well-known problem in mathematics is called the ‘handshake’ problem. a 3 people are in a room. How many handshakes are required so that everyone shakes hands with everyone else once? 3 6 b Repeat the above exercise for 4 people. n( n − 1) c Develop a general rule for the number of handshakes for n people. 2 In preparation for the Christmas office party, Fred is put in charge of providing coffee. He determines that each cup of coffee requires 3.2 grams of coffee ($19 per kg), 6.4 grams of sugar ($0.98 per kg) and 10.5 grams of milk ($1.40 per kg) plus a plastic cup which costs $2.00 for 24 cups. a Determine the cost of providing a single cup of coffee. $0.165 b If Fred has $25 to spend, how many cups of coffee can he make, assuming that he charges 10 cents a cup, and half the people pay? 217 cups The cost of a mobile phone from Company A is $40 per month plus $0.25 per SMS, while Company B offers a plan for $30 a month but $0.30 per SMS. How many SMSs would make the plans the same monthly cost? 200 Virgin Blue buys a new plane so that extra flights can be arranged between Sydney, Australia and Wellington, New Zealand. The plane costs $1  200  000. It depreciates at a rate of 16.5% p.a. and is written off when its value falls below $150  000. How long can Virgin Blue use this plane before it is written off? 12 years At a baby shower, we started discussing baby statistics. One of the women told us she had heard a report that for every 100 babies born, there were 6 more boys than girls. If we were to 0.47 randomly pick a child from a representative group, what is the probability of picking a girl? The numbat is an Australian animal which is in danger of becoming extinct because of habitat loss and foxes. Since it only eats white ants, its source of food is also diminishing. The number of numbats, t weeks after their habitat has been lost is modelled by the function 500 N = 60 + numbats per hectare. t+2 a How many numbats were there before their habitat was lost? 310 101 b How many numbats are there 10 weeks after their habitat loss? 23 weeks c How long after habitat loss are there only 80 numbats per hectare? d According to this model, will the numbats die out completely? No

Chapter 17 Problem solving II

575

T  he median relies on the middle value of the data and won’t change much if an extra value is added. The mean however has increased because this large value will change the average of the numbers. The mean is used as a measure of central tendency if there are no outliers or if the data are symmetrical. The median is used as a measure of central tendency if there are outliers or the data are skewed.

problem solving

2, 3, 3, 2, 2, 3, 2, 4, 3, 1, 1, 0 a Calculate the mean and median number of pets. Mean = 2.17, median = 2 b The empty block of land at the end of the street was bought by a Cattery and now houses 20 cats. Recalculate the mean and median. Mean = 3.54, median = 2 c Explain why the answers are so different, and which measure of central tendency is best 64

65

66

n

P

 1

120  000

 2

60  000

 3

40  000

 4

30  000

 5

24  000

 6

20  000

 8

15  000 67

10

12  000

12

10  000

T  he differences in this case were minimal; however, the grouped data mean is not based on the actual data but on the frequency in each interval and the interval midpoint. It is unlikely to yield an identical value to the actual mean. The spread of the scores within the class interval has a great effect on the grouped data mean. 576

63 The following data show the number of pets in each of the 12 houses in Coral Avenue, Rosebud.

68

69

70

used for certain data. Erin borrows $12  000 for a new car at 9% p.a. over 4 years. a Calculate the total amount to be repaid if the interest is compounded monthly. $17 176.86 $5176.86 b How much will be paid in interest for this loan? c How much would each repayment be in order to repay the loan in equal monthly $357.85 instalments? Based on her performance throughout the year, Mary was given a probability of 0.7 of winning her first tennis tournament. If the probability of winning both her first and second tennis tournaments is 0.56, what is her probability of winning the second tennis tournament? 0.8 A syndicate won first prize in Tattslotto but isn’t sure how many people participated. The total amount won is $120  000. The amount won by each person is a whole number of dollars. a Make a table of values showing the amount, P, won by each person when the prize money is shared among n people from 1 to 12 inclusively. 120 000 P= b Sketch the function. n c Write a rule that relates the number of people to the amount of prize won. k. Inverse variation; k = 120  000 d What type of variation (function) is this? State the value of e If 20 people had been in the syndicate, how much would each have received? $6000 f If each person only received $1500, how many people would have been in the 80 people syndicate? The amount of money in an investment plan (V ) grows exponentially with compound interest according to the rule V = PAn , where n = the time (in years), P = the initial investment and A is the compound interest term.   After 1 year (n = 1), the amount of money in the investment is $2662.50, while after 2 years the amount is $2835.56. a Determine the values of P and A. P = 2500, A = 1.065 b Determine the annual rate of interest as a percentage. 6.5% In the game of draw poker, a player is dealt 5 cards from a deck of 52. To obtain a flush, all 5 cards must be of the same suit. a Determine the probability of getting a diamond flush. 0.000 495 b Determine the probability of getting any flush. 0.001 981 The number of Year 10 students in all the 40 schools in the Northern District of the Education Department was recorded as follows: 56, 134, 93, 67, 123, 107, 167, 124, 108, 78, 89, 99, 103, 107, 110, 45, 112, 127, 106, 111, 127, 145, 87, 75, 90, 123, 100, 87, 116, 128, 131, 106, 123, 87, 105, 112, 145, 115, 126, 92 a Using a interval of 10, produce a table showing the frequency for each interval. 106.75 b Use the table to estimate the mean. 107.15 c Calculate the mean of the ungrouped data. d Compare the results from parts b and c and explain any differences. An electronics store is having trouble selling the latest mp3 player. The original price was $99 but on October 1 it was reduced 10%. On October 8 it was reduced a further 10%. On October 12 the regional manager decided to increase all prices by 5%. On October 15 the local manager decided to reduce the price by another 10% anyway. a Calculate the prices on all 4 dates after the discounts/increases have been applied. b What is the ‘final’ percentage discount after Oct 15? 23.5%

Maths Quest 10 for the Australian Curriculum

October 1: $89.10, October 8: $80.19, October 12: $84.20, October 15: $75.78

It is 3 times more likely that the spinner will land on A. Therefore I would not play this game as I should be winning $9 not $8.

problem solving 71 You use this spinner to play a game. Explain whether it is a fair

game and whether you would play if the scoring was: a if the pointer lands in A, you win $8, if the pointer lands in B, you win $3 b if the pointer lands in A, you win $18, if the pointer lands in B, you win $5. I would play this game as a fair game would only

B A

give me $15. To get $18 would be in my favour.

72 Moore’s Law states that the capacity of computer hard drives doubles every 18 months.

Assume that on 1 January 2010 the capacity of a hard drive was 512 GB (1 GB = 220 bytes). a Sketch a graph showing hard drive capacity going back at least 10 years. b What was the capacity on 1 January 2000? Approx. 5 GB c Predict the capacity on 1 January 2016. 8192 GB 73 Complete this table regarding compound interest. Annual interest rate

Compounding period

Length of investment

$1000

6.4%

Monthly

11 months

$2000

7.82%

Daily

77 days

$2003.30

$4000

8.08%

Quarterly

8 years

$7585.60

Principal

Value of investment $1060.26

74 In the game of blackjack, players are initially given 2 cards from a deck of 52. Face cards

(jack, queen, king) are worth 10 points, an ace is either 1 or 11 (player’s choice) and other cards (2–10) are worth their numerical value. The value of the 2 cards dealt is added together; for example; 2 + king = 2 + 10 = 12. a What is the probability of getting a total of 10 with the first 2 cards? 0.053 b What is the probability of getting a total of 20 with the first 2 cards? 0.103 75 A gold and copper bracelet weighs 238 grams. The volume of the bracelet is 15 cm3. Gold weighs 19.3 grams per cm3, and copper weighs 9 grams per cm3. What percentage of gold (by volume and by mass) is in the bracelet? 66.7% by volume and 81.1% by mass 76 A small car yard has 60 cars, half of which are Toyotas. Since running a car on LPG rather than petrol has become one of the most popular features, 25 cars are LPG based, including 20 of the Toyotas. There are some 4WD vehicles in the car yard, 5 of which are Toyota and 6 of which have LPG systems. Unfortunately only one of the Toyota 4WDs is also LPG based. 18 of the vehicles have none of the characteristics already mentioned. Draw a Venn diagram and then answer the following questions. If a customer looks at a vehicle, find the probability that the vehicle is: Tovota LPG System 17 a a 4WD vehicle 60 1 19 b an LPG Toyota 3 6 0 7 c not LPG based 12 1 1 4 5 d not a Toyota 2 7 10 e either a Toyota or a 4WD 7 2 5 f neither LPG based nor 4WD 1 4 WD 18 g an LPG system but is not a Toyota 12 59 h not an LPG based Toyota 4WD. 60 77 Patrick and Trisha were told that they needed a deposit of $34 000 before they could borrow enough money to buy their new home. At that time Trisha had saved $6200 and Patrick had saved $7400. They both worked and although they paid rent of $1280 per month they are able to save $260 per week. a When they made enquiries about the loan, how much less than the deposit did they have? $20  400 b At their current rate of savings, how long will it take them to save the rest of the 1 deposit? 79 weeks (1 2 years) Chapter 17 Problem solving II

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78

79

80 T  his game has a total 2 expected loss of $ 36 , so the game is not fair and is biased against the player.

2

3

4

5

6

7

8

9

10

11

12

W1

L2

W3

L4

W5

L5

W4

L3

W2

L1

W1

Where, for example, W5 means you win $5, L2 means you lose $2 and so on. Determine the expected value and comment on your result. 81 A survey recorded the number of years of education of the parents of a class of Year 10 students. The results were as follows

Frequency  2  4  8 17  6  5  4  3  1 50

Cumulative frequency  2  6 14 31 37 42 46 49 50

If the deposit is 10% of the cost of the home, what was the cost of the house. $340  000 $306  000 How much money will they have to borrow once they get their deposit? How much interest is charged for the first month, if the interest rate is 9% p.a. reducible. $2295 If Trisha and Patrick use their savings every week and the amount they would have spent $2406.67 on rent to repay the loan, how much can they afford to pay every month? $305  888.33 g How much would they owe at the beginning of the second month? h What will happen if interest rates rise to 9.5% p.a. at the beginning of the second month? i What would you advise Trisha and Patrick? Payments will be more than they are able to afford. 100 people seated at different tables in a Mexican restaurant were asked if their party had ordered any of the following items: Burritos, Fajitas, or Tacos. The following information was found. 23 people had ordered none of these items. 11 people had ordered all three of these items. 29 people had ordered fajitas or tacos but did not order burritos. Burritos Fajitas 41 people had ordered tacos. 15 9 12 46 people had ordered at least two of these items. 13 people had ordered burritos and tacos but had not ordered fajitas. 11 13 7 26 people had ordered burritos and fajitas. 10 a Place this information in a Venn diagram. b How many people ordered Burritos only? 9 Tacos 23 c How many people ordered Fajitas? 45 Radioactive decay for a certain substance can be modelled using the relationship M M = Mo ì (1.0122)-t, where Mo is its original mass and M is its mass after t days. Prove M = 0.5 0 a Prove that the substance has a half-life of 57.3 days. = (1.0122)-57.3 21.54% b Calculate the percentage of mass lost by the substance after 20 days. To determine whether a game is fair the expected value is calculated. In its simplest form: expected value = P(win) ì (prize) - P(loss) ì (cost of game) If there is more than one prize, with an associated probability, then add up the various P(win) ì prize terms. $0 a A game is fair if expected value = ______ b Consider the following game based on the total of 2 dice. c d e f



F  ind a less-expensive house, save more money then pay more off the principal each month, find a loan with a lower interest rate, save for a larger deposit.

problem solving

Years of education  9 10 11 12 13 14 15 16 17 Total

12, 15, 12, 11, 13, 17, 10, 12, 14, 16, 12, 13, 11, 9, 11, 12, 12, 16, 12, 14, 12, 13, 11, 11, 14, 12, 11, 10, 15, 12, 13, 12, 12, 11, 10, 10, 12, 12, 14, 13, 11, 9, 12, 15, 16, 13, 12, 14, 15, 12

Parents of year 10 students

8

10 12 14 16 Years of education

18

a Produce a table showing frequency and cumulative frequency for each number of

years. b Produce a cumulative frequency polygon. c From the results of part a and part b, produce a box-plot. 82 The annual rate of inflation is very similar to the annual interest rate on a compound interest

investment. Determine the annual rate of inflation in the following cases. Give your answers as a percentage rounded to 2 decimal places. a In 2005 the price of a family-sized pizza was $11. The exact same pizza in 2010 is $17. 578

9.10%

Maths Quest 10 for the Australian Curriculum

x

A  t the top end of the heights, there is a possible outlier of 189 cm. Removal of this value results in a mean of 171 cm and a median of 171 cm. The mean is reduced slightly, while the median is unchanged. Option 2 is less expensive, despite having a higher   At the lower end of the heights, there are possible outlier s of 159 cm annual interest rate, because it is paid off in only and 160 cm. Removal of these values results in a mean of 172.5 cm and a 10 months instead of 2 years. The TV would cost median of 171.5 cm. The mean and median are increased slightly. $217 with option 1 and $212 with option 2.   Removal of both the upper and lower outliers results in a mean of 171.9 cm and a median of 171 cm. The mean is increased slightly, while the median is unchanged.

problem solving b In 2000 the price of a litre of petrol was $0.98. In 2009 the price was $1.45. 4.45% 5.76% c In 1972 the price of a can of soft drink was $0.25 while in 2010 the price is $2.10. 83 Ethan wants to borrow $30  000 to buy a new Toyota corolla. He finds a bank that will give

him the loan at 12% flat-rate interest over 5 years. a Calculate the amount of interest that Ethan will pay on this loan. $18  000 b Calculate the total amount that Ethan must repay on his loan. $48  000 c Ethan decides to repay the loan in 5 equal yearly instalments. Calculate the amount of each instalment. $9600 $8117.09 d Ethan could have borrowed $30  000 at 12% reducible interest instead of flat-rate interest.

If he makes the same repayments for 4 years, how much money will Ethan have saved? 84 a A Year 10 boy is talking with a Year 10 girl and asks her if she has any brothers or sisters. 1 She says, ‘Yes, I have one’. What is the probability that she has a sister? 3 b A Year 10 boy is talking with a Year 10 girl and asks her if she has any brothers or sisters. She says, ‘Yes, I have an older one’. What is the probability that she has a sister? 85 A recent survey of the heights (in cm) of a group of Year 10 boys resulted in the following data.

1 2

172, 178, 159, 168, 167, 172, 177, 171, 169, 172, 170, 189, 173, 177, 169, 168, 171, 180, 174, 160, 175, 171, 173, 168, 170, 171, 172, 174, 168, 170 a Calculate the mean. 171.6 cm 171 cm b Calculate the median. c Examine the data and identify any possible outliers. If they exist, discuss the effect of

the outliers on the mean and the median. 86 An electronics store offers its customers 2 choices.

Option 1: 20% down then make equal payments (based on the money owing at the beginning of the loan) every 6 months for 2 years. The interest rate (simple) is 8% (annual) of the amount owing at the time of the payment. Option 2: No money down, with 10 equal payments (based on the money owing at the beginning of the loan), 1 per month. The interest rate (simple) is 12% (annual) of the amount owing at the time of the payment. In both cases, the balance owing at the end of the time period is to be paid in full. Calculate the better deal on a $200 TV. Justify your conclusions with appropriate calculations. 87 The times, in seconds, of the duration of 20 TV advertisements shown in the 6–8 pm time slot are recorded below. 16  60  35  23  45  15  25  55  33  20  22  30  28  38  40  18  29  19  35  75 a From the data, determine the: i mode 35 s 29.5 s ii median iii mean, write your answer correct to 2 decimal places 33.05 s iv range 60 s 21 29.5 39 21 s v lower quartile 39 s vi upper quartile 15 20 25 30 35 40 45 50 55 60 65 70 range. 18 s vii interquartile b Using your results from part a, construct a box plot for the time, in seconds, for the

75

t

20 TV advertisements in the 6–8 pm time slot. c From your box plot, determine: i the percentage of advertisements that are more than 39 seconds in length 25% ii the percentage of advertisements which last between 21 and 39 seconds     50% iii the percentage of advertisements which are more than 21 seconds in length 75%

The types of TV advertisements during the 6−8 pm time slot were categorised as Fast Food, Supermarkets, Program information, Retail (clothing, sporting goods, furniture). Chapter 17 Problem solving II

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problem solving

A frequency table for the frequency of these advertisements being shown during this time slot is shown below. Type

Pictogram, pie chart or bar chart.

Frequency

Fast food

7

Supermarkets

5

Program information

3

Retail

5

Categorical d What type of data has been collected in the table? e What percentage of advertisements are advertisements for fast food outlets? 35% f What would be good options for a graphical representation of this type of data? 88 In the game of major league baseball, a very rare event is what is called a ‘perfect game’. This

Number of bacteria

occurs when the pitcher gets all men out. Since there are 9 innings and 3 outs/inning, this means that the pitcher must get 27 men in a row out.   The most recent perfect game was by Mark Buehrle, July 23, 2009. a Given that the probability a batter who faces Mark Buehrle does not get out is 0.313, determine the probability that he throws a perfect game. 0.000  039  6 b Given that there are about 2000 games per year, about how often is a perfect game About once in 12.6 years thrown. 89 Two experiments are performed to record the growth of different bacteria. Bacteria A molecules grow according to the formula: A = 500(1.031)t, where t = time in days. Bacteria B molecules grow according to the formula: B = 200(1.086)t, where t = time in days. a On the same set of axes, sketch graphs of these 2 models. b Estimate, graphically when the number of molecules of Bacteria A is the same as the number of molecules of Bacteria B. Approx 17.5 days c Using a calculator, spreadsheet or another method, calculate the time the two bacteria are equal in number. Give your answer to 2 decimal places. 17.63 days 90 When all of Saphron’s team players turn up for their twice weekly netball training the chance that they then win their Saturday game is 0.65. If not all players are at the training session then the chance of winning their Saturday game is 0.40. Over a four week period, Saphron’s players all turn up for training three times. 800 600 400 0.75

200 0

5

10 15 20 Time

y = 200 (1.086)t y = 500 (1.031)t

0.25

0.65

W

0.35

WÅ

0.40

W

0.60

WÅ

T

TÅ

a Using a tree diagram, with T to represent all players training and W to represent a win,

represent the winning chance of Saphron’s netball team. b Using the tree diagram constructed in part a, determine the probability of Saphron’s team winning their Saturday game. Write your answer correct to 4 decimal places. 0.5875 c Determine the exact probability that Saphron’s team did not train given that they won 8 their Saturday game. 47 580

Maths Quest 10 for the Australian Curriculum

problem solving 91 The speeds, in km/h, of 55 cars travelling along a major road are recorded below.

Speed

Frequency

60–64

 1

65–69

 1

70–74

10

75–79

13

80–84

 9

85–89

 8

90–94

 6

95–99

 3

100–104

 2

105–109

 1

110–114

 1

Total

55

a By finding the midpoint for each class interval, determine the mean speed, in km/h,

of the cars travelling along the road. Write your answer correct to two decimal places. 82.73 km/h b The speed limit along the road is 75 km/h. A speed camera is set to photograph the license plates of cars travelling 7% more than the speed limit. A speeding fine is automatically sent to the owners of the cars photographed. Based on the 55 cars 30 cars recorded, how many speeding fines were issued? c Drivers of cars travelling 5 km/h up to 15 km/h over the speed limit are fined $135. Drivers of cars travelling more than 15 km/h and up to 25 km/h over the speed limit are fined $165 and drivers of cars recorded travelling more than 25 km/h and up to 35 km/h are fined $250. Drivers travelling more than 35 km/h pay a $250 fine in addition to having their driver’s license suspended. If it is assumed that this data is representative of the speeding habits of drivers along a major road and there are 30  000 cars travelling along this road on any given month. Determine: i The amount, in dollars, collected in fines throughout the month. Write your answer correct to the nearest cent.   $2 607 272.73 ii How many drivers would expect to have their licenses suspended throughout the month?   About 545 92 Calculate the total interest paid on a 6 month loan of $14  000 with 10% p.a. reducible interest. Each month $2000 is paid with the balance at the end of the 6 months. $461.96 93 Consider the case of a ‘random’ 2-digit number (00, 01  .  .  .  98, 99) and determining the probability that the 2 digits are different.   The 1st number can be anything (10 digits) while the 2nd number can be one of 9 that is different from the 1st. Therefore the probability that the 2 digits are different 10 × 9 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = = 0.9 10 × 10 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10   An Australian bank note consists of 8 digits. a Write the formula for the probability that all 8 digits are different. b Calculate this probability exactly. 0.018  144 (about 1.8%) Chapter 17 Problem solving II

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problem solving 94 Year 10 student Ruby is a babysitter. Her babysitting fees are $12.50 per hour for up to three

Hours 1 2 3 4 5 6

Fees OLD 12.50 25.00 37.50 51.50 65.50 79.50

Fees NEW 12.50 25.00 37.50 70.00 70.00 70.00

Difference ($)    0    0    0 +18.50   +4.50   -9.50

hours. After three hours, her babysitting fees are at the half-hourly rate of $7.00. a Write an expression that can be used to determine Ruby’s babysitting fees, R, for the following: i for any time, t, up to three hours R = 12.5t, t Ç 3 ii for any time, t, after three hours. R = 14t - 4.50, t > 3 b Use an appropriate expression from part a. If Ruby receives $72.50 for babysitting, how 5.5 hours many hours did she spend babysitting? c Ruby decides to restructure her babysitting fees. For the first 3 hours her fee is $12.50 per hour. For between four to six hours she charges a flat fee of $70. By comparing her two fee structures, determine the expected amount, in dollars, Ruby would receive using the new fee structure for each hour up to and including 6 hours. 95 Sixty Year 10 students were surveyed about driving. 80% of the students said that they currently had their learner’s permit. Of the students with a learner’s permit, 40% of the students said that they averaged more than 5 hours each week in driving practice and of these students, 15% had also driven in wet conditions. 25% of the students with a learner’s permit said that they had driven in wet conditions. a Determine the probability that a student with a learner’s permit, selected at random had: i practised driving for more than 5 hours each week and in dry conditions 0.25 ii not practised driving for more than 5 hours but had driven in wet conditions 0.10 0.375 iii practised for more than 5 hours given that they had driven in wet conditions b A different group of 150 Year 10 students were surveyed. What would be the predicted 30 students number of students who would have a learner’s permit and had driven in wet conditions? 96 The test scores, out of a total score of 50, for two classes A and B are shown in the stem plot below.



Reducing value allows you to claim $300, $210, $147 over the 3 years for a total of $657. Straight line allows you to claim $200 each year over 5 years for a total of $1000. Although the reducing value depreciation is greater in the years 1 and 2, the sum over the life of the item is greater in the straight line case.

Class A 5 9753 97754 886551 320 0

0 1 2 3 4 5

Class B 124 145 005 155 157789 00

Based on the comparison between Class A’s IQR (16.5) and Class B’s IQR (32.5), Ms Vinculum was correct in her statement.

the test showed that the students’ ability was more closely matched than the students’ ability in Class B’. By finding the measure of centre, first and third quartiles, and the measure of spread for the test scores for each class, explain if Ms. Vinculum’s statement was correct. b Would it be correct to say that Class A performed better on the test than Class B? Justify your answer by comparing the quartiles and median for each class. No 97 There are two major methods of calculating depreciation; the usual reducing value depreciation and what is called straight-line depreciation. In the latter you are allowed to claim the same, fixed amount each year until there is no value left. For example, over 5 years 1 one can claim 5 of the item’s initial value each year.   Compare the two methods of depreciation in the following cases. a Original price = $1000 Reducing value depreciation of 30% for 3 years Straight-line depreciation of 20% of original value each year for 5 years b Original price = $5000 Reducing value depreciation of 20% for 5 years Straight-line depreciation of 16.7% of original value each year for 6 years

Maths Quest 10 for the Australian Curriculum



582

Reducing value: $1000, $800, $640, $512, $409.60 for a total of $3361.60 Straight line allows $835 per year over 6 years for a total of $5000 (actually $5010). In this case the reducing value method is only better in the 1st year.

a Ms. Vinculum teaches both classes and made the statement that ‘Class A’s performance on

problem solving 98 In a mixture of dried fruit and nuts, the nuts 1

make up 6 of the total mass. An extra 4 kg of nuts is added to the mixture, changing the nut composition of the mixture to 1 of its mass. What 4

is the total mass of the fruit and nut mixture now? 40 kg

99 You often see side-by-side escalators in department stores, one travelling up and the other

travelling down at the same speed. Consider this scenario. •• Ann walks down the down escalator, stepping on every step, and taking 1 second per step. She takes 24 steps to reach the bottom. •• Zoe walks down the up escalator, stepping on every step, and taking 1 second per step. She takes 120 steps to reach the bottom. The escalators are moving at the same speed, but in opposite directions. If they had not been moving, how many steps would be visible on each escalator? (They both have the same number of steps.) 40 steps 100 Juanita receives an annual salary of $48  500. She receives a pay increase of 3.5% each year for three years. At the end of the three years, her pay is increased by an additional x% for t years.   Ida receives an annual salary of $52  750. She also receives a pay increase at the same time at Juanita. Ida’s pay is increased by 2% each year for n years. Student’s own work a Show that if Juanita and Ida both receive the same annual salary at the end of n years, t x   then 1.041 × 1.02t =  1 +  100  b If t = 5 years, show that x = 2.82% 101 Helena is the owner of the web store ‘Warm as Toast’ which makes and sells sheepskin slippers. It costs Helena $1000 for machinery and materials plus a cost of $15 for each slipper C = 1000 + 15s made. a Write an equation that determines the cost, C in dollars, of each slipper, s, made. b Helena is able to sell each pair of slippers for $25. Write an equation that determines the revenue, R, in dollars, Helena makes on each slipper sold, s. R = 25s c Using your equations from parts a and b, how many slippers does Helena need to sell to meet her costs? 100 d Helena makes a profit of $2500. By writing an equation that determines the profit, P, in dollars, Helena makes on each pair of slippers sold, determine the number of slippers Helena sold to make this profit. 350 slippers 102 When walking home from school during the summer months, Harold buys either an ice cream or a drink from the corner shop. If Harold bought an ice cream the previous day then there is a 30% chance that he will buy a drink the next day. If he bought a drink the previous day then there is a 40% that he will buy an ice cream the next day. On Monday, Harold bought an ice cream. Determine the probability that he buys an ice cream on Wednesday. 0.61 Chapter 17 Problem solving II

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problem solving 103 Fred purchases a plasma TV on a payment plan. He pays 15% deposit and then equal monthly

1

A and B are not independent because P(A ¶ B) = 12

104

105





and P(A) ì P(B) =

5 . 108

106

107

108 109

110

111

584

repayments of $105.50 for 24 months. The TV is priced at $2650. a Determine the amount Fred pays in deposit. Write your answer to the nearest cent. $397.50 b After the deposit is paid, determine the amount owing on the TV. $2252.50 $2929.50 c How much does Fred pay for the TV after he has made 24 equal repayments? d Determine the amount Fred pays extra for the TV. $279.50 e Using your value from part d determine the percentage interest Fred pays on the money he borrowed for the TV. Base your calculations on simple interest. Write your answer 6.2% p.a. correct to 1 decimal place. A sheet of paper is torn in half, and the pieces placed on top of one another. These pieces together are torn in half, and the new pieces are placed on top of one another. This process continues through 20 such tears, with the pieces being placed on top of each other each time. If the original piece of paper was 0.01 mm thick, what is the height of the final pile of pieces of paper? Give your answer in metres, correct to 1 decimal place. 10.5 m The amount of energy the sun delivers to the surface of the Earth each second is known as the Solar Constant. It is measured to be 1.35 ì 106 ergs/cm2 each second. If 1 watt is equal to 1 ì 107 ergs/s, what is the Solar Constant each second, expressed in watts/m 2. 1350 watts/m2 Two standard dice are rolled. Event A is the subset of that at least one of the numbers of the upper face is a multiple of 3 Event B is the subset of the sum of the numbers of the upper face is greater than 10. Mutually exclusive events, A and B, follow the condition of P(A ¶ B) = 0. a Using the condition for mutually exclusivity, show that events A and B are not mutually 1 exclusive. P(A ¶ B) = 12 ò 0 so A and B are not mutually exclusive. If A and B are independent events, then P(A ¶ B) = P(A) ì P(B) b Using the condition for independent events, can events A and B be described as independent? Justify your answer using calculations. Theresa’s home office equipment is depreciated based on reducing balance method at a rate of x% per year until the scrap value is less than $300. The office equipment had an initial value of $3700. After 7 years the office equipment will, for the first time, have a value below the scrap value of $300. Determine the rate of depreciation, x, correct to 2 decimal places. 30.16% A container holds 100 mL of a salt solution. How many mL of water must be added to the solution to decrease the saltiness by 20%? 25 mL An amoeba is a microscopic organism. It is able to reproduce itself in 3 minutes. One amoeba is placed in a jar. In 3 hours the jar is completely filled with amoeba. How long did it take for the jar to be one-quarter filled with amoeba? 2 h 54 min Gloria is investing $25 000 in a bank account. She is offered two different interest options: Option A: 4.5% p.a. compounded monthly Option B: 4.75% p.a. compounded half yearly Gloria decides to invest using Option B for a period of five years. Show, by finding the amount Gloria’s investment has grown after 5 years, that she receives $319.08 more using Option B than Option A. Student’s own work Bianca and Hannah modify the game of darts for a fund 6 1 raising school fete. The dart board Bianca and Hannah use is shown in the diagram at right   The dart board has a 40 cm diameter. The bull’s eye 12 7 (red centre) has a diameter of 5 cm. The inner section, (i.e. 5 11 8 2 numbers 7 to 12 circle) has a radius of 10 cm. 10 9 The rules of their game are as follows: To score, the player has to hit either the same number 4 three times, or the same number twice and the bull’s eye. 3 It costs $2 to enter and players have three shots at the dartboard.

Maths Quest 10 for the Australian Curriculum

problem solving

P($5) = 0.12503, P($10) = 0.039  063, P($15) = 0.12502 ì 0.0156, P($20) = 0.039  062 ì 0.0156. The probabilities of obtaining any of the required scores to receive a payout are very low. Bianca and Hannah would be unlikely to have to pay out any money, so would collect $300 for their fundraising.

  The table below shows how much players would receive for certain scores. Score

112

113 114

115

Payout ($)

The same three numbers in the range 1 to 6

  $5

The same three numbers in the range 7 to 12

$10

Bull’s eye with any same two numbers in the range 1 to 6

$15

Bull’s eye with any same two numbers in the range 7 to 12

$20

If 150 players enter Bianca and Hannah’s dart game, what would be the expected amount, in dollars, they would pay out? Justify your answer by determining the probabilities of obtaining each of the required scores to receive a payout. Tracy places her money in an investment account that earns x% interest per year for five years. At the end of each year the amount of interest for the 12 months is added to the sum in the account. At the end of the first year her $2500, with interest added, has grown to $2575. At the end of the second year the amount of money in her account was $2652.25. a If the rate of interest Tracy’s money earned over the first two years was constant, determine the value of x. 3 b At the end of the second year, the rate of interest was increased. This rate did not change for the next two years. At the end of the fourth year, the amount of money in Tracy’s account was $2896.32. Determine the new rate of interest Tracy’s account earned during 4.5% the third and fourth years. Write your answer correct to 1 decimal place. c After interest had been paid into Tracy’s account at the end of the fourth year, she was able to deposit another $1000 into the account. The interest rate earned on the account was increased by 1.5%. Determine the amount of money Tracy will have in her account $4130.10 at the end of the fifth year. Write your answer to the nearest cent. Andy and Tom have regular games of chess. So far, Tom has won 25% of the time and Andy has won 18 games. How many games has Tom won? 6 games Andy, Bill and Cam each have a mobile phone with a different payment plan. Andy’s plan 40c flag fall plus 45c for every 30 sec or part thereof of a call Bill’s plan 35c flag fall plus 92c per minute or part thereof of a call Cam’s plan 36c flag fall plus 46c per 30 second or part thereof of a call The boys call each other regularly, with calls lasting up to 3 min. Give an analysis of call costs for each of them for calls lasting up to 3 mins. This Venn diagram shows the relationship between four sets A, B, C and D. ξ

A

B

1

2

3

16

4

5

6

7

8

9

10

11

12

13

14

15

C

D

What is the sum of the elements contained within the region (A ¶ B) ß (C ¶ D)? 58 Chapter 17 Problem solving II

585

problem solving 116 The tennis coach at school is retiring, and eight members of his squad decide to buy him a

117

118

119

121

Round 1

Round 4

X

Round 2

Round 3

X

X Bye

Round 3

X

X

X

X

Bye

X

X

X

X X

X

Bye

X X

X

X

X

X

X

X

X X

X

Champion

X

X Round 2

X

X

X

Round 1 X

The draw for 9 players could look like this.

Champion

120

farewell gift. They each contribute the same amount of money. When another two members of the squad found out about the plan, they said they would join in, paying their share of the cost of the gift. The original eight contributors then each received $3 back. What was the cost of the gift? $120 In financial circles, there is a formula that is commonly used to determine the length of time it takes to double a sum of money invested at a compound interest rate of R% p.a. n Number of years to double money = . R Using an investment of $1000 and an interest rate of 10% p.a., determine a value for n. 70 There are 40 children in a room and they are either right-handed or left-handed. There are 17 right-handed boys, and 9 left-handed children. Of the 40 children, 21 are girls. How many left-handed girls are in the room? 7 Suppose you buy a rare stamp for $15, sell it for $20, buy it back for $22, and finally sell it for $30. How much money did you make, or lose, in buying and selling this stamp? $13 profit Note: The answer is not $15. In any leap year, the calendar for January is the same as the calendar for another month in the year. Which month is this? July A draw for a tennis tournament looks like a tree diagram. This could be the draw for eight players.

X

X

X

In this type of draw, a player is out of the tournament after one loss. With an odd number of players, one of the competitors is given a ‘bye’, which means the player does not compete in that round. No competitor can be given more than one bye in a tournament.   Consider a tournament of this type with nine tennis players. a Draw a diagram showing how this draw could be played. 4 b What is the maximum number of games the champion will have to play? 3 c What would be the minimum number of games the champion could play? 122 Sandy’s savings account pays a simple interest rate of 3.5% on daily balances (the interest is calculated daily on the balance in the account). Her statement for July looks like this. Date 1/7 3/7 7/7 21/7 28/7 31/7 586

Maths Quest 10 for the Australian Curriculum

Deposit

Withdrawal

$100 $500 $725 $85

Balance $4200

problem solving a Complete the Balance column for the month of July. b In order to calculate the interest she has earned for the month, complete the following table. Total interest earned is $13.24.

Date 4 4300 × 3.5 × 365 100

6/7

4

$4300

$1.65

7 365

20/7

14

$4800

$6.44

27/7

7

$4075

$2.74

4 365

31/7

4

$4160

$1.60

Balance $4200 $4300 $4800 $4075 $4160 $4160

123

  $85

124

Date 1/7 3/7 7/7 21/7 28/7 31/7

$100 $500

$725

Withdrawal

2 365

$4200

100

Deposit

4200 × 3.5 ×

Interest earned

2

100 4160 × 3.5 ×

Interest calculation

2/7

14 4800 × 3.5 × 365 100 4075 × 3.5 ×

Number of days Balance

125

126

$0.81

100

What is Sandy’s total interest for the month? In a school election for House Captain there are six candidates — let’s call them A, B, C, D, E and F. After the election, the following facts were released. •• A won with 50 votes. •• B was second. •• F was last with 5 votes. •• No two candidates received the same number of votes. •• 100 students voted. What is the greatest number of votes B could have received? 24 votes I have 6 bags of marbles, each containing fewer than 20 marbles. There is a total of 90 marbles in five of the bags. The sixth bag contains 5 fewer marbles than the average number in all six bags. How many marbles are in the sixth bag? 12 In my drawer I have socks of 5 different colours, and 10 of each colour. They are not in pairs, and randomly distributed in the drawer. In the dark I go to my drawer to get 4 socks of the same colour. How many would I need to select to be certain of getting 4 of the same colour? 16 A die in the shape of a dodecahedron has regular pentagonal faces numbered 1 to 12. The sum of the numbers on each pair of opposite faces equals 13.   With its face numbered 1 uppermost, it is adjacent to faces numbered 2, 4, 6, 5 and 3, reading clockwise. The faces adjacent to face number 6 would be 1, 4, 10, 11 and 5.

3 5

2 1 6

1

4

5

4

6 11

10

If the 6 was sitting uppermost, draw a diagram to show the numbers adjacent to its face. 127 Guests at a luxury resort are told that the entry code to the sauna is a 3-digit odd number with

no repeated digits. (It can not start with 0.) They are given this code written on a piece of paper.   Suppose you lose this piece of paper, and wish to have a sauna. What is the maximum number of numbers you will have to try to gain entry to the sauna? 320 128 A long rectangular table has boys and girls seated on either side, with no-one seated at the ends. An equal number of children sit on both sides. On one side of the table there are 11 boys, while on the other side, there are 5 girls. How many more boys than girls are seated at the table? 12 Chapter 17 Problem solving II

587

problem solving 129 The local take-away cafe is trying to promote its business by giving away drink vouchers.

Here’s the deal. •• Every disposable drink cup comes with a voucher attached to the cup. •• Every 4 vouchers collected can be traded for a free drink, served in a new cup, with a voucher attached. Last month I spent $56 on $2 take-away drinks at the cafe. If I used all my vouchers when I accumulated them, how many drinks did I have? 37 130 An interesting biological fact has been discovered about bees. •• A male bee has only one parent — a mother. •• A female bee has two parents — a mother and a father. If we look back 5 generations into the ancestry of a male bee, draw a diagram to show how many ancestors we would find.

Looking back 5 generations, there would be 8 ancestors for the male bee.

5th generation 4th generation

M

F

F

F

M

3rd generation

F

2nd generation

M

1st generation

M

F F F

Maths Quest 10 for the Australian Curriculum

F F

F F

Male bee

588

M

M

F

M

number AND algebra • REAL NUMBERS

18

18A 18B 18C 18D 18E 18F 18G 18H

Number classification review Surds Operations with surds Fractional indices Negative indices Logarithms Logarithm laws Solving equations

What do you know ?

Real numbers

1 List what you know about real numbers. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of real numbers.

opening question

Manning’s formula is a formula used to estimate the flow of water down a river in a flood event, measured in metres per 2

1

R3 S 2 second. The formula is v = , where n R is the hydraulic radius, S is the slope of the river and n is the roughness coefficient. What will be the flow of water in the river if R = 8, S = 0.0025 and n = 0.625?

number and algebra • real numbers

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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590

Identifying surds a, b and d 1 Which of the following are surds? a c

10 d 4 2

7 49

b

Simplifying surds 2 Simplify each of the following. a 48

98 7 2 d 3 72 18 2 b

4 3

c 5 12 10 3

Adding and subtracting surds 3 Simplify each of the following. a 2 6 − 4 3 + 7 3 − 5 6

3 3−3 6

b 2 32 − 5 45 − 4 180 + 10 8 28 2 − 39 5

Multiplying and dividing surds 4 Simplify each of the following. a c

7 × 10 6 2



70

3

Evaluating numbers in index form 5 Evaluate each of the following. a 72 49 6 244.140 625 c (2.5)

Using the index laws 6 Simplify each of the following. a x3 ì x7 x10 4a2 c 24a3b ó 6ab5 b4

maths quest 10 + 10a for the australian Curriculum

b 2 3 × 4 6 24 2 d

5 6 2 10 3 2

b 34 81 d (0.3)4 0.0081

b 4y3 ì 5y8 20y11 d (2m4)2 4m8

number and algebra • real numbers

18a eBook plus

Interactivity Classifying numbers

number classification review ■

■

int-2792

■

The number systems used today evolved from a basic and practical need of primitive people to count and measure magnitudes and quantities such as livestock, people, possessions, time and so on. As societies grew and architecture and engineering developed, number systems became more sophisticated. Number use developed from solely whole numbers to fractions, decimals and irrational numbers.

The Real Number System contains the set of rational and irrational numbers. It is denoted by the symbol R. The set of real numbers contains a number of subsets which can be classified as shown in the chart below. Real numbers R Irrational numbers I (surds, non-terminating and non-recurring decimals, π ,e)

Negative Z–

Rational numbers Q Integers Z

Zero (neither positive nor negative)

Non-integer rationals (terminating and recurring decimals) Positive Z+ (Natural numbers N)

rational numbers (q ) ■

A rational number (ratio-nal) is a number that can be expressed as a ratio of two whole a numbers in the form , where b ò 0. b Rational numbers are given the symbol Q. Examples are: 1 2 3 9 , , , , 5 7 10 4

i

7, -6, 0.35, 1. 4 Chapter 18 real numbers

591

number AND algebra • REAL NUMBERS

Integers (Z ) ■■

Rational numbers may be expressed as integers. Examples are: 5 1

= 5,

−4 1

= -4,

27 1

15

= 27, -  1 = -15

The set of integers consists of positive and negative whole numbers and 0 (which is neither positive nor negative). They are denoted by the letter Z and can be further divided into subsets. That is: Z = {.  .  ., -3, -2, -1, 0, 1, 2, 3, .  .  .} Z + = {1, 2, 3, 4, 5, 6, .  .  .} Z - = {-1, -2, -3, -4, -5, -6, .  .  .} ■■ Positive integers are also known as natural numbers (or counting numbers) and are denoted by the letter N. That is: N = {1, 2, 3, 4, 5, 6, .  .  .} ■■ Integers may be represented on the number line as illustrated below. ■■

-3 -2 -1 0 1 2 3 Z The set of integers

1 2 3 4 5 6 N The set of positive integers or natural numbers

Z- -6 -5 - 4 -3 -2 -1 The set of negative integers

Note: Integers on the number line are marked with a solid dot to indicate that they are the only points in which we are interested.

Non-integer rationals ■■

Rational numbers may be expressed as terminating decimals. Examples are: 7 10

■■ ■■

1

5

9

= 0.7, 4 = 0.25, 8 = 0.625, 5 = 1.8

These decimal numbers terminate after a specific number of digits. Rational numbers may be expressed as recurring decimals (non-terminating or periodic decimals). For example:



1 3



9 11



5 6



3 13

= 0.333  333 .  .  . or 0.3   (or 0.81) = 0.818  181 .  .  . or 0.81 = 0.833  333 .  .  . or 0.83   (or 0.230769) = 0.230  769  230  769 .  .  . or 0.230769

These decimals do not terminate, and the specific -3.743 3 1– digit (or number of digits) is repeated in a pattern. 2 1.63 -2–4 3.6 Recurring decimals are represented by placing a dot or line above the repeating digit or pattern. - 4 -3 -2 -1 0 1 2 3 4 Q ■■ Rational numbers are defined in set notation as: Q = rational numbers a  Q =  , a, b ∈ Z , b ≠ 0  where Æ means ‘an element of’. b   ■■

Irrational numbers (I ) ■■

■■

An irrational number (ir-ratio-nal) is a number that cannot be expressed as a ratio of two a whole numbers in the form , where b ò 0. b Irrational numbers are given the symbol I. Examples are: 7, 13, 5 21,

592

Maths Quest 10 + 10A for the Australian Curriculum

7 , p, e 9

number AND algebra • REAL NUMBERS

■■

Irrational numbers may be expressed as decimals. For example:

5 = 2.236  067  977  5 .  .  . 0.03 = 0.173  205  080  757 .  .  . 18 = 4.242  640  687  12 .  .  . 2 7 = 5.291  502  622  13 .  .  . p  = 3.141  592  653  59 .  .  . e = 2.718  281  828  46 .  .  . ■■ These decimals do not terminate, and the digits do not repeat themselves in any particular pattern or order (that is, they are non-terminating and non-recurring). ■■ Rational and irrational numbers belong to the set of π– - 1–2 - 12 - 5 2 4 π real numbers (denoted by the symbol R). They can be positive, negative or 0. The real numbers may be represented on a number line as shown at right (irrational - 4 -3 -2 -1 0 1 2 3 4 R numbers above the line; rational numbers below it). ■■ To classify a number as either rational or irrational: 1. Determine whether it can be expressed as a whole number, a fraction or a terminating or recurring decimal. 2. If the answer is yes, the number is rational; if the answer is no, the number is irrational.

p (pi) ■■ ■■

■■

The symbol p (pi) is used for a particular number; that is, the circumference of a circle whose diameter length is 1 unit. It can be approximated as a decimal that is non-terminating and non-recurring. Therefore, p  is classified as an irrational number. (It is also called a transcendental number and cannot be expressed as a surd.) In decimal form, p  = 3.141  592  653  589  793  23  .  .  . It has been calculated to 29  000  000 (29 million) decimal places with the aid of a computer.

Worked Example 1

Specify whether the following numbers are rational or irrational. 1 5

a     b 

25   c  13   d  3p    e  0.54   f  3 64    g  3 32    h 

Think a b

c

d

e f

1 5

1

Evaluate 25.

2

The answer is an integer, so classify 25.

1

Evaluate 13.

2

The answer is a non-terminating and non-recurring decimal; classify 13.

1

Use your calculator to find the value of 3p.

2

The answer is a non-terminating and non-recurring decimal; classify 3p.

0.54 is a terminating decimal; classify it accordingly. 2

1 27

Write

is already a rational number.

1

3

Evaluate 3 64 . The answer is a whole number, so classify 3 64 .

a

1 5

is rational. 25 = 5

b

25 is rational. 13 = 3.605  551  275  46 .  .  .

c

13 is irrational. d 3p  = 9.424  777  960  77 .  .  .

3p  is irrational. e

0.54 is rational.

f

3

64 = 4

3

64 is rational.

Chapter 18 Real numbers

593

number AND algebra • REAL NUMBERS

g

h

1

Evaluate 3 32 .

2

The result is a non-terminating and non-recurring decimal; classify 3 32 .

1

Evaluate

2

The result is a number in a rational form.

3

g

1 . 27

h

3

32 = 3.174  802  103  94 .  .  .

3

32 is irrational.

3

1 1 = 27 3

3

1 is rational. 27

remember

a 1. Rational numbers (Q) can be expressed in the form , where a and b are whole b numbers and b ò 0. They include whole numbers, fractions and terminating and recurring decimals. a 2. Irrational numbers (I) cannot be expressed in the form , where a and b are whole b numbers and b ò 0. They include surds, non-terminating and non-recurring decimals, and numbers such as p and e. 3. Rational and irrational numbers together constitute the set of real numbers (R). Exercise

18A

Number classification review fluency 1   WE 1  Specify whether the following numbers are rational (Q) or irrational (I ). a 4

Q

0.04 Q

f

k -2.4

Q

25 Q 9

p

b

4 Q 5

c

7 Q 9

d

2 I

e

7 I

9 Q 4

j

0.15 Q

g 2

1 2

Q

h

5 I

i

l

100 Q

m

14.4 I

n

1.44 Q

o p

s

1000 I

t 7.216  349  157 .  .  .

q 7.32

Q

r − 21 I

I

1 y 3 0.0001 I Q 16 2 Specify whether the following numbers are rational (Q), irrational (I ) or neither. 1 11 0 1 a b 625 Q c d e -6 Q Q Q Q 7 8 4 8 u − 81 Q

594

v 3p

w

I

f

3

81 I

g − 11 I

k

3

21 I

l

p

64 16

u

22π 7

π 7

2 25

q

Q I

v

3

h m

I

I

3

r

3

62 I

x

1.44 Q 4

i

(−5)2 I

n - 

6 2

s

I

−1.728 Q w 6 4 Q

Maths Quest 10 + 10A for the Australian Curriculum

3

π

I

j

3 11

Q

o

27 Q

x 4 6

I

t y

8 Undefined 0 1 Q 100 1

Q

4

( 2)

4

Q

I

number AND algebra • REAL NUMBERS 3   MC  Which of the following best represents a rational number? A p

4 9

✔ B

9 12

C

d

3

3

e

5

e

12

4   MC  Which of the following best represents an irrational number? A − 81

B

6 5

C

3

343

✔ d

22

5   MC  Which of the following statements regarding the numbers -0.69, A

π is the only rational number. 3

π , 49 is correct? 3

7 and 49 are both irrational numbers.

B ✔ C

7,

-0.69 and 49 are the only rational numbers.

D -0.69 is the only rational number.

7 is the only rational number. 1 11 6   MC  Which of the following statements regarding the numbers 2 , -  , 624, 3 99 is correct? 2 3 E

11 3

A -  and B

624 are both irrational numbers.

624 is an irrational number and 3 99 is a rational number. 624 and 3 99 are both irrational numbers.

✔ C

1 2

11 3

D 2 is a rational number and -  is an irrational

number.

E

18b

3

99 is the only rational number.

reflection 

 

Why is it important to understand the real number system?

Surds ■■

A surd is an irrational number that is represented by a root sign or a radical sign, for example: ,3 ,4 Examples of surds include: 7, 5, 3 11 , 4 15 Examples that are not surds include: 9, 16 , 3 125 , 4 81

■■

Numbers that are not surds can be simplified to rational numbers, that is: 9 = 3, 16 = 4 , 3 125 = 5, 4 81 = 3

Worked Example 2

Which of the following numbers are surds? 1 a  16    b  13   c     d  3 17    e  4 63    e  3 1728 16 Think a

Write

1

Evaluate 16 .

2

The answer is rational (since it is a whole number), so state your conclusion.

a

16 = 4 16 is not a surd.

Chapter 18 Real numbers

595

number AND algebra • REAL NUMBERS

b

c

d

e

f

1

Evaluate 13.

2

The answer is irrational (since it is a nonrecurring and non-terminating decimal), so state your conclusion.

1

Evaluate

2

The answer is rational (a fraction); state your conclusion.

1

Evaluate 3 17 .

2

The answer is irrational (a nonterminating and non-recurring decimal), so state your conclusion.

1

Evaluate 4 63 .

2

The answer is irrational, so classify 4 63 accordingly.

1

Evaluate 3 1728 .

2

The answer is rational; state your conclusion.

1 . 16

13 = 3.605  551  275  46 .  .  .

b

13 is a surd.

1 1 = 16 4

c

1 is not a surd. 16 d

e

f

3

17 = 2.571  281  590  66 .  . .

3

17 is a surd.

4

63 = 2.817  313  247  26 .  .  .

4

63 is a surd.

3

1728 = 12

3

1728 is not a surd. So b, d and e are surds.

Proof that a number is irrational ■■ ■■

■■

In Mathematics you are required to study a variety of types of proofs. One such method is called proof by contradiction. This method is so named because the logical argument of the proof is based on an assumption that leads to contradiction within the proof. Therefore the original assumption must be false. a An irrational number is one that cannot be expressed in the form (where a and b are b integers). The next worked example sets out to prove that 2 is irrational.

Worked Example 3

Prove that 2 is irrational. Think

596

Write

a Let 2 = , where b ò 0 b

1

Assume that 2 is rational; that is, it a can be written as in simplest form. b We need to show that a and b have no common factors.

2

Square both sides of the equation.

2=

3

Rearrange the equation to make a2 the subject of the formula.

a2 = 2b2

Maths Quest 10 + 10A for the Australian Curriculum

a2 b2 [1]

number and algebra • real numbers 4

If x is an even number, then x = 2n.

\ a2 is an even number and a must also be even; that is, a has a factor of 2.

5

Since a is even it can be written as a = 2r.

\ a = 2r

6

Square both sides.

a2 = 4r2 But a2 = 2b2

[2] from [1]

7

Equate [1] and [2].

\ 2b2 = 4r2 4r 2  b 2 = 2 = 2r2 \ b2 is an even number and b must also be even; that is, b has a factor of 2.

8

Repeat the steps for b as previously done for a.

Both a and b have a common factor of 2. This contradicts the original assumption that a 2 = , where a and b have no common factor. b \ 2 is not rational. \ It must be irrational.

■

■

The dialogue included in the worked example should be present in all proofs and is an essential part of the communication that is needed in all your solutions. Note: An irrational number written in surd form gives an exact value of the number; whereas the same number written in decimal form (for example, to 4 decimal places) gives an approximate value.

remember

A number is a surd if: 1. it is an irrational number (equals a non-terminating, non-recurring decimal) 2. it can be written with a radical sign (or square root sign) in its exact form.

exerCise

18b

surds fluenCy

eBook plus

Digital doc

SkillSHEET 18.1 doc-5354

1 We2 Which of the numbers below are surds? b d f g h i l m o q r s t w z a

81

b

g

3 4

h

3

48

c

16

d

1.6

e

0.16

f

3 27

i

1000

j

1.44

k 4 100

l

m

3

32

n

361

o

3

100

p

3

125

q

s

3

169

t

7 8

u

4

16

2 v ( 7 )

w

y

5

32

z

80

3

11 2 + 10

6+ 6

r 2p

33

x

0.0001

Chapter 18 real numbers

597

number AND algebra • REAL NUMBERS

 6 ,  9

2   MC  The correct statement regarding the set of numbers  ✔ A

3

27 and

9 are the only rational numbers of the set.

B

6 is the only surd of the set. 9

C

6 and 9

D

20 and

E

9 and

20 are the only surds of the set. 54 are the only surds of the set. 20 are the only surds of the set.  1 ,  4

3   MC  Which of the numbers of the set 

A

✔ d

3

 1 , 21, 3 8  are surds? 8 

1 , 27

21 only

B

1 only 8

1 and 21 only 8

e

1 and 21 only 4  

12 is a surd.

c p is irrational but not a surd.

1 and 3 8 8

C

4   MC  Which statement regarding the set of numbers π , A

 20 , 54 , 3 27 , 9  is: 

 1 , 12 , 16 , 3 + 1 is not true? 49 

12 and 16 are surds.

✔ B

12 and 3 + 1 are not rational.

d

E p is not a surd.



5   MC  Which statement regarding the set of numbers 6 7 ,



not true? A

✔ c

E

144 when simplified is an integer. 16 7 6 is smaller than 9 2 .

B

 144 , 7 6 , 9 2 , 18 , 25  is 16 

144 and 25 are not surds. 16

d 9 2 is smaller than 6 7.

18 is a surd.

understanding 6 Complete the following statement by selecting appropriate words, suggested in brackets: Any perfect square

6

a is definitely not a surd, if a is .  .  . (any multiple of 4; a perfect square and cube).

m = 4 7 Find the smallest value of m, where m is a positive integer, so that 3 16m is not a surd. Reasoning 8   WE 3  Prove that the following numbers are irrational, using a proof by contradiction: Check with your teacher. a

3

b

5

c

7

598

Maths Quest 10 + 10A for the Australian Curriculum

reflection 

 

How can you be certain that

a is a surd?

number AND algebra • REAL NUMBERS

18c

Operations with surds Simplifying surds ■■ ■■ ■■

To simplify a surd means to make a number (or an expression) under the radical sign ( ) as small as possible. To simplify a surd (if it is possible), it should be rewritten as a product of two factors, one of which is a perfect square, that is, 4, 9, 16, 25, 36, 49, 64, 81, 100 and so on. We must always aim to obtain the largest perfect square when simplifying surds so that there are fewer steps involved in obtaining the answer. For example, 32 could be written as 4 × 8 = 2 8 ; however, 8 can be further simplified to 2 2 , so 32 = 2 × 2 2 ; that is 32 = 4 2 . If, however, the largest perfect square had been selected and 32 had been written as 16 × 2 = 16 × 2 = 4 2 , the same answer would be obtained in fewer steps.

Worked Example 4

Simplify the following surds. Assume that x and y are positive real numbers. a 

1 384       b  3 405        c  -  175        d  5 180 x 3 y5 8

Think a

b

c

Write

384 = 64 × 6

a

1

Express 384 as a product of two factors where one factor is the largest possible perfect square.

2

Express 64 × 6 as the product of two surds.

= 64 × 6

3

Simplify the square root from the perfect square (that is, 64 = 8).

=8 6

1

Express 405 as a product of two factors, one of which is the largest possible perfect square.

2

Express 81 × 5 as a product of two surds.

= 3 81 × 5

3

Simplify 81.

= 3× 9 5

4

Multiply together the whole numbers outside the square root sign (3 and 9).

= 27 5

1

Express 175 as a product of two factors in which one factor is the largest possible perfect square.

2

Express 25 × 7 as a product of 2 surds.

1 = − × 25 × 7 8

3

Simplify 25.

1 =− ×5 7 8

4

Multiply together the numbers outside the square root sign.

b 3 405 = 3 81 × 5

c

−

1 1 175 = − 25 × 7 8 8

=−

5 7 8 Chapter 18 Real numbers

599

number AND algebra • REAL NUMBERS

d

5 180 x 3 y 5 = 5 36 × 5 × x 2 × x × y 4 × y

1

Express each of 180, x3 and y5 as a product of two factors where one factor is the largest possible perfect square.

2

Separate all perfect squares into one surd and all other factors into the other surd.

= 5 × 36 x 2 y 4 × 5 xy

3

Simplify 36 x 2 y 4 .

= 5 × 6 × x × y 2 × 5 xy

4

Multiply together the numbers and the pronumerals outside the square root sign.

= 30 xy 2 5 xy

d

Addition and subtraction of surds ■■

■■

Surds may be added or subtracted only if they are alike.   Examples of like surds include 7, 3 7 and −5 7. Examples of unlike surds include 11, 5 , 2 13 and −2 3. In some cases surds will need to be simplified before you decide whether they are like or unlike, and then addition and subtraction can take place. The concept of adding and subtracting surds is similar to adding and subtracting like terms in algebra.

Worked Example 5

Simplify each of the following expressions containing surds. Assume that a and b are positive real numbers. a  3 6 + 17 6 − 2 6    b  5 3 + 2 12 − 5 2 + 3 8    c  Think a

b

Write

All 3 terms are alike because they contain the same surd ( 6 ). Simplify. 1

1 100 a 3 b2 + ab 36 a − 5 4 a2 b 2

Simplify surds where possible.

a 3 6 + 17 6 − 2 6 = (3 + 17 − 2) 6

= 18 6 b 5 3 + 2 12 − 5 2 + 3 8

= 5 3+2 4×3−5 2 +3 4×2 = 5 3 + 2× 2 3 − 5 2 +3× 2 2 = 5 3+4 3−5 2 +6 2

c

2

Add like terms to obtain the simplified answer.

1

Simplify surds where possible.

=9 3+ 2 c

1 100 a3b 2 + ab 36a − 5 4 a 2 b 2 1 = × 10 a 2 × a × b 2 + ab × 6 a − 5 × 2 × a b 2 1 = × 10 × a × b a + ab × 6 a − 5 × 2 × a b 2 = 5ab a + 6ab a − 10 a b

2

600

Add like terms to obtain the simplified answer.

Maths Quest 10 + 10A for the Australian Curriculum

= 11ab a − 10 a b

number AND algebra • REAL NUMBERS

Multiplication and division of surds Multiplying surds ■■

■■

To multiply surds, multiply together the expressions under the radical signs. For example, a × b = ab , where a and b are positive real numbers. When multiplying surds it is best to first simplify them (if possible). Once this has been done and a mixed surd has been obtained, the coefficients are multiplied with each other and then the surds are multiplied together. For example, m a × n b = mn ab

Worked Example 6

Multiply the following surds, expressing answers in the simplest form. Assume that x and y are positive real numbers. a  11 × 7     b  5 3 × 8 5     c  6 12 × 2 6     d  15 x 5 y2 × 12 x 2 y Think a

b

Write

Multiply the surds together, using

a

11 × 7 = 11 × 7

a × b = ab (that is, multiply expressions under the square root sign). Note: This expression cannot be simplified any further.

= 77

Multiply the coefficients together and then multiply the surds together.

b 5 3 ×8 5 = 5×8× 3 × 5

= 40 × 3 × 5 = 40 15

c

1

Simplify 12 .

c

6 12 × 2 6 = 6 4 × 3 × 2 6 = 6×2 3 ×2 6 = 12 3 × 2 6

2

Multiply the coefficients together and multiply the surds together.

= 24 18

3

Simplify the surd.

= 24 9 × 2 = 24 × 3 2 = 72 2

d

1

Simplify each of the surds.

d

15 x 5 y 2 × 12 x 2 y = 15 × x 4 × x × y 2 × 4 × 3 × x 2 × y = x 2 × y × 15 × x × 2 × x × 3 × y = x 2 y 15 x × 2 x 3 y

2

3

Multiply the coefficients together and the surds together.

= x 2 y × 2 x 15 x × 3 y

Simplify the surd.

= 2 x 3 y 9 × 5 xy

= 2 x 3 y 45 xy

= 2 x 3 y × 3 5 xy = 6 x 3 y 5 xy

Chapter 18 Real numbers

601

number AND algebra • REAL NUMBERS

■■

When working with surds, we sometimes need to multiply surds by themselves; that is, square them. Consider the following examples: ( 2 )2 = 2 × 2 = 4 = 2 ( 5 )2 = 5 × 5 = 25 = 5

■■

■■

Observe that squaring a surd produces the number under the radical sign. This is not surprising, because squaring and taking the square root are inverse operations and, when applied together, leave the original unchanged. When a surd is squared, the result is the number (or expression) under the radical sign; that is, ( a )2 = a, where a is a positive real number.

Worked Example 7

Simplify each of the following. a  ( 6 )2 b  ( 3 5 )2 Think

Write

Use ( a )2 = a, where a = 6.

a b

2 a ( 6) = 6

Square 3 and use ( a )2 = a to square 5 .

1 2

b (3 5 )2 = 32 × ( 5 )2

=9ì5

= 45

Simplify.

Dividing surds ■■ ■■

a a To divide surds, divide the expressions under the radical signs; that is, = , where a and b b b are whole numbers. When dividing surds it is best to simplify them (if possible) first. Once this has been done, the coefficients are divided next and then the surds are divided.

Worked Example 8

Divide the following surds, expressing answers in the simplest form. Assume that x and y are positive real numbers. a 

c 

55 5

b 



9 88 6 99

d 



48 3 36 xy 25 x 9 y11

Think a

602

1

Write

Rewrite the fraction, using

a b

=

a . b

Maths Quest 10 + 10A for the Australian Curriculum

a

55 5

=

55 5

number AND algebra • REAL NUMBERS

b

c

d

2

Divide the numerator by the denominator (that is, 55 by 5).

3

Check if the surd can be simplified any further.

1

Rewrite the fraction, using

2

Divide 48 by 3.

= 16

3

Evaluate 16 .

=4

1

Rewrite surds, using

2

a

a b

=

= 11

a . b

b

3

=

9 88 6 99

Simplify the fraction under the radical by dividing both numerator and denominator by 11.

=

9 8 6 9

3

Simplify surds.

=

9×2 2 6×3

4

Multiply the whole numbers in the numerator together and those in the denominator together.

=

18 2 18

5

Cancel the common factor of 18.

= 2

1

Simplify each surd.

b

c

d

9 88

48 3

=

=

a . b

48

6 99

36 xy 9 11

25 x y

= =

2

=

Cancel any common factors — in this case xy.

6 xy 8

5 x × x × y10 × y 6 xy 5 x 4 y 5 xy 6 5x 4 y 5

Rationalising denominators ■■ ■■

If the denominator of a fraction is a surd, it can be changed into a rational number. In other words, it can be rationalised. As discussed earlier in this chapter, squaring a simple surd (that is, multiplying it by itself) results in a rational number. This fact can be used to rationalise denominators as follows. a b

■■

×

b b

=

ab , where b

b b

=1

If both numerator and denominator of a fraction are multiplied by the surd contained in the denominator, the denominator becomes a rational number. The fraction takes on a different appearance, but its numerical value is unchanged, because multiplying the numerator and denominator by the same number is equivalent to multiplying by 1.

Chapter 18 Real numbers

603

number AND algebra • REAL NUMBERS

Worked Example 9

Express the following in their simplest form with a rational denominator. a 

6 13

       b 

2 12 3 54

       c 

17 − 3 14 7

Think a

b

Write

1

Write the fraction.

2

Multiply both the numerator and denominator by the surd contained in the denominator (in this case 13). This has the 13 same effect as multiplying the fraction by 1, because = 1. 13

1

Write the fraction.

2

Simplify the surds. (This avoids dealing with large numbers.)

a

6 13 = =

b

6 13

2 12 3 54 2 12 3 54

=

= =

Multiply both the numerator and denominator by 6. (This has the same effect as multiplying the fraction by 1, because

6

= 1.)

6 Note: We need to multiply only by the surd part of the denominator (that is, by 6 rather than by 9 6).

4

c

604

Simplify 18.

5

Divide both the numerator and denominator by 6 (cancel down).

1

Write the fraction.

2

Multiply both the numerator and denominator by 7. Use grouping symbols (brackets) to make it clear that the whole numerator must be multiplied by 7.

Maths Quest 10 + 10A for the Australian Curriculum

c

13

78 13

=

3

13

×

2 4×3 3 9×6 2×2 3 3×3 6 4 3 9 6 4 3 9 6

×

=

4 18 9×6

=

4 9×2 9×6

=

4×3 2 54

=

12 2 54

=

2 2 9

6 6

17 − 3 14 7 =

( 17 − 3 14 ) 7

×

7 7

number AND algebra • REAL NUMBERS

3

=

Apply the Distributive Law in the numerator. a(b + c) = ab + ac

= 4

17 × 7 − 3 14 × 7 7× 7 119 − 3 98 7

119 − 3 49 × 2 7 119 − 3 × 7 2 = 7 119 − 21 2 = 7 =

Simplify 98.

Rationalising denominators using conjugate surds ■■

■■

The product of pairs of conjugate surds results in a rational number. (Examples of pairs of conjugate surds include 6 + 11 and 6 − 11, a + b and a − b, 2 5 − 7 and 2 5 + 7 .)   This fact is used to rationalise denominators containing a sum or a difference of surds. To rationalise the denominator that contains a sum or a difference of surds, multiply both numerator and denominator by the conjugate of the denominator. Two examples are given below: 1. To rationalise the denominator of the fraction

1 a+ b 1

2. To rationalise the denominator of the fraction

, multiply it by , multiply it by

a− b a− b

.

a+ b

. a− b a+ b   A quick way to simplify the denominator is to use the difference of two squares identity:



( a − b )( a + b) = ( a )2 − ( b )2 =a-b

Worked Example 10

Rationalise the denominator and simplify the following. a  b 

1 4− 3 6+3 2 3+ 3

Think a

Write

1

Write down the fraction.

2

Multiply the numerator and denominator by the conjugate of the denominator. (4 + 3 ) = 1.) (Note that (4 + 3 )

a

1 4− 3 =  

1 (4 − 3 )

×

(4 + 3 ) (4 + 3 )

Chapter 18 Real numbers

605

number AND algebra • REAL NUMBERS

3

4

=

Apply the Distributive Law in the numerator and the difference of two squares identity in the denominator.

 

Simplify.

   

b

1

Write down the fraction.

2

Multiply the numerator and denominator by the conjugate of the denominator. (3 − 3 ) = 1.) (Note that (3 − 3 )

3

4

Multiply the expressions in grouping symbols in the numerator, and apply the difference of two squares identity in the denominator. Simplify.

4+ 3 (4) 2 − ( 3 ) 2

=

4+ 3 16 − 3

=

4+ 3 13

6 +3 2

b

3+ 3 =  

=

( 6 + 3 2) (3 + 3 )

×

(3 − 3 ) (3 − 3 )

6 ×3+ 6 × − 3 +3 2 ×3+3 2 × − 3 (3)2 − ( 3 )2

 

         

=

3 6 − 18 + 9 2 − 3 6 9−3

=

− 18 + 9 2 6

=

− 9×2 +9 2 6

=

−3 2 + 9 2 6

=

6 2 6

 = 2

remember

1. To simplify a surd means to make a number (or an expression) under the radical sign as small as possible. 2. To simplify a surd, write it as a product of two factors, one of which is the largest possible perfect square. 3. Only like surds may be added and subtracted. 4. Surds may need to be simplified before adding and subtracting. 5. When multiplying surds, simplify the surd if possible and then apply the following rules: (a) a × b = ab (b) m a × n b = mn ab , where a and b are positive real numbers. 606

Maths Quest 10 + 10A for the Australian Curriculum

number and algebra • real numbers

6. When a surd is squared, the result is the number (or the expression) under the radical sign: ( a )2 = a, where a is a positive real number. 7. When dividing surds, simplify the surd if possible and then apply the following rule: a÷ b=

a

=

a b

b where a and b are whole numbers, and b ò 0. 8. To rationalise a surd denominator, multiply the numerator and denominator by the surd contained in the denominator. This has the effect of multiplying the fraction by 1, and thus the numerical value of the fraction remains unchanged, while the denominator becomes rational: a a b ab = × = b b b b where a and b are whole numbers and b ò 0. 9. To rationalise the denominator containing a sum or a difference of surds, multiply both the numerator and denominator of the fraction by the conjugate of the denominator. This eliminates the middle terms and leaves a rational number. exerCise

18C

operations with surds fluenCy

eBook plus

Digital doc

SkillSHEET 18.2 doc-5355

1 We4a Simplify the following surds. a 12 2 3

b

24 2 6

c

27 3 3

d

125 5 5

e 54 3 6

f

112 4 7

g

68 2 17

h

180 6 5

j

162 9 2 2 We4b, c Simplify the following surds.

k

245 7 5

l

448 8 7

b 8 90 24 10

c 9 80 36 5

d 7 54 21 6

f −7 80 −28 5

g 16 48 64 3

h

i

88 2 22

a 2 8

4 2

e −6 75 −30 3 i

1 162 9

j

2

1 192 2 3 4

k

1 1 15 135 3 9

1 392 2 2 7 3 3 7 l 175 2 10

3 We4d Simplify the following surds. Assume that a, b, c, d, e, f, x and y are positive real

numbers. a

a 2 4a 16

b

72a 2 6a 2

13ab 2ab

e

338a3b3

f

68a3b 5 2ab 2 17ab g

54 c3d 2 2cd

i

6 162c 7 d 5

j 2 405c 7 d 9 18c3d 4 5cd

c

k

90 a 2 b 3a 10 b

d

a 4 13a 2 2 338

125 x 6 y 4 5x 3 y 2 5 h 5 80 x 3 y 2 20 xy 5x 1 88ef 2

22ef

l

1 7e 5 f 5 2ef 392e11 f 11 2

4 We5a Simplify the following expressions containing surds. Assume that x and y are positive

real numbers. a 3 5 + 4 5 7 5

b 2 3 + 5 3 + 3 8 3

c 8 5 + 3 3 + 7 5 + 2 3 15 5 + 5 3

4 11 d 6 11 − 2 11

e 7 2 + 9 2 − 3 2 13 2

f 9 6 + 12 6 − 17 6 − 7 6 −3 6

g 12 3 − 8 7 + 5 3 − 10 7 17 3 − 18 7

h 2 x + 5 y + 6 x − 2 y 8 x + 3 y Chapter 18 real numbers

607

number and algebra • real numbers

eBook plus

5 We5b Simplify the following expressions containing surds. Assume that a and b are positive

real numbers.

Digital doc

SkillSHEET 18.3 doc-5356

15 10 − 10 15 + 10 12 30 − 16 15

a

10( 2 − 3 ) 200 − 300

b

125 − 150 + 600 5( 5 + 6 )

c

27 − 3 + 75 7 3

d 2 20 − 3 5 + 45 4 5

e 6 12 + 3 27 − 7 3 + 18 14 3 + 3 2

f

150 + 24 − 96 + 108 3 6 + 6 3

g 3 90 − 5 60 + 3 40 + 100

h 5 11 + 7 44 − 9 99 + 2 121 −8 11 + 22

i

2 30 + 5 120 + 60 − 6 135

j 6 ab − 12ab + 2 9ab + 3 27ab 12 ab + 7 3ab

k

1 1 1 7 2+2 3 48 + 12 98 + 2 2 3 3

l

1 7 18 + 3 72 15 2 32 − 8 6

6 We5c Simplify the following expressions containing surds. Assume that a and b are positive

real numbers. a 7 a − 8a + 8 9a − 32a

31 a − 6 2a

c

150 ab + 96ab − 54 ab 6 6ab

d 16 4 a 2 − 24 a + 4 8a 2 + 96a

e

a 3 a 2a 8a3 + 72a3 − 98

f

g

9a3 + 3a 5 3a a + a 2 3a

2 h 6 a 5b + a 3b − 5 a 5b (a + a) ab

4 ab ab + 3a 2 b b

i

−6ab 2a + 4 a 2 b3 3a

k

eBook plus

Digital doc

SkillSHEET 18.4 doc-5357

32a + 2 6a + 8a 2 b 10 a − 15 27a + 8 12a + 14 9a 52 a − 29 3a

ab ab + 3ab a 2 b + 9a3b3 32a3b 2 − 5ab 8a + 48a 5b6

a + 2 2a

j

a3b + 5 ab − 2 ab + 5 a3b 3 ab (2a + 1)

l

4 a 2 b + 5 a 2 b − 3 9a 2 b −2a b

7 We6 Multiply the following surds, expressing answers in the simplest form. Assume that a,

b, x and y are positive real numbers. a

2× 7

14

b

6× 7

d

10 × 10 10

e

g 5 3 × 2 11

10 33

j 10 6 × 3 8 120 3 m 6a 5b 2 2b

1 1 1 36a + 128a − 144 a 2 4 6

8× 6

21 × 3 3 7

f

27 × 3 3 27

h 10 15 × 6 3 180 5

i

4 20 × 3 5 120

l

1 48 × 2 3 2 2 9 3

k

1 1 60 × 40 2 6 10 5 5 12a 7 b × 6a3b 4

p

c

1 48 × 2 2 4

2 6A

n

xy × x 3 y 2 x 2 y y

o

q

15 x 3 y 2 × 6 x 2 y 3

r

8 We7 Simplify each of the following. a ( 2 )2

42





4 3

3a 4 b 2 × 6a 5b 3 3a 4 b 2 2ab 1 15a3b3 × 3 3a 2 b6 2

3 x 2 y 2 10 xy

b ( 5 )2 5

c ( 12 )2 12

d ( 15 )2 15

e (3 2 ) 2 18

f (4 5 )2 80

g (2 7 )2 28

h (5 8 )2 200

2

9 2 4 a b 5ab 2

9 We8 Simplify the following surds, expressing answers in the simplest form. Assume that a,

b, x and y are positive real numbers. a e 608

15 3

5

b

18

3 4

f



4 6

8 2

2

65 2 13



c 5 2

maths quest 10 + 10a for the australian Curriculum

g

60 10



6

96 2 3 8

d

h

128 8

4

7 44 14 11

1

number and algebra • real numbers 11 a 2 + 2 b

3 10 − 2 33 6

i

12 5 − 5 6 10

c d

a

SkillSHEET 18.7 doc-5360

3 10 + 6 14 4

f

5 6 3

f

k

3 22 − 4 10 g 6 eBook plus

xy x5y7

12 x8 y12

×

2040

j

x 2 y3

30

a e

eBook plus

i

Digital doc

2 xy 3 y

n



x2y5

2 2a 2 b 4 5a3b6

16 xy

x y

l

×

10 a 9 b3 4 a 3 3 a7b

8x 7 y 9

5



2 15 6

5 2 2

b

10 2

g

5 7 14

l



5 14 7 8



7 3

7 3 3



16 3 6 5

h

8 15 15



4

c

2 3 2 15 5 5

6 + 12

11 3 7

m

5



4 11 11

d



3 35 5

i

8 21 49

n

8 3





7 7

15 − 22

b

3 3 5+6 7

f

8 7 12 − 5 6

j

6 3

c

6 4 2 +3 8

g

2 3 6 2− 5

k

4 8

1 5+2



5−2

b

5 3 15 15 − 20 6 −6 + 6 2 + 10 − 2 5 d 13 2 3 5+4 2 f

eBook plus

Digital doc

WorkSHEET 18.1 doc-5363

8 4 6 3 6 5 2

5 6 6



2 3

8 60 28



j

4 3

4 15 15

7

3 5

6 2 − 15

d

10 3 11 − 4 5

h

18 6 3−5 5

l

7 20

5 2 7−2 5 12 3 5+7 3 5 24

h

12 − 7 19 − 4 21 5 12 + 7

e g

3 6 − 15

i

6+2 3

1 8− 5



2 2+ 5 3

c

8−3

12 2 − 17



8+3 3 −1

31

15 − 3 − 5 + 1 4

refleCtion 

5 +1

 

Under what circumstance might you need to rationalise the denominator of a fraction

5− 3 4 2− 3

fractional indices

4 8 11 + 4 13

2 11 − 13



4 10 + 15 − 4 6 − 3 29

1

■

12 − 10 j 16

■

2 2 From this we can conclude that (a 2 ) = ( a ) and further conclude that a 2 = a .

■

We can similarly show that a 3 = 3 a .

■

This pattern can be continued and generalised to produce a n = n a .

30 + 7 2 20

10 3

2 18 + 3 2

(a 2 )2 = a (using the Fourth Index Law, (am)n = am ì n) 2 Now, from our work on surds we know that ( a ) = a.

l



2 35 8 105 o 7 3 14

14 − 5 2 i 6

6 15 − 25 70

x y

2 21 7

Consider the expression a 2 . Now consider what happens if we square that expression.

k

3 4

12

e

■

11 h

2



12 We10 Rationalise the denominator and simplify. a

21 − 15 3

x 4 y3

k

11 We9c Express the following in their simplest form with a rational denominator.

SkillSHEET 18.8 doc-5361

18d

2 17

understanding

Digital doc

SkillSHEET 18.9 doc-5362

15 7

4 5

10 We9a, b Express the following in their simplest form with a rational denominator.

Digital doc

e

1

m

9 10 5 eBook plus

9 63

1

1

1

1 1

Chapter 18 real numbers

609

number AND algebra • REAL NUMBERS

Worked Example 11

Evaluate each of the following without using a calculator. 1

1

a  9 2              b  64 3 Think

Write 1

a

b

1

1

Write 9 2 as 9.

2

Evaluate. 1 64 3

1

Write

2

Evaluate.

a 92 =

9

=3 as 3 64 .

1 64 3

b

= 3 64 =4

Worked Example 12

Use a calculator to find the value of the following, correct to 1 decimal place. 1

1

a  10 4              b  200 5 Think

Write 1

a 10 4 = 1.778  279  41

a Use a calculator to produce the answer.

ö 1.8

b Use a calculator to produce the answer.

1 200 5

b

= 2.885  399  812 ö 2.9

1 ■■

Consider the expression (a m ) n . Using our work so far on fractional indices, we can say 1

(a m ) n = n a m . ■■

1

m

We can also say (a m ) n = a n using the index laws. m

■■ ■■

We can therefore conclude that a n = n a m . Such expressions can be evaluated on a calculator either by using the index function, which is usually either ^ or xy and entering the fractional index, or by separating the two functions for power and root.

Worked Example 13 2

Evaluate 3 7 , correct to 1 decimal place. Think

Write 2

2

3 7 ≈ 1.4

Use a calculator to evaluate 3 7 . 1 ■■

610

We can also use the index law a 2 = a to convert between expressions that involve fractional indices and surds.

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • REAL NUMBERS

Worked Example 14

Write each of the following expressions in simplest surd form. 1

3

a  10 2              b  5 2 Think

Write

a Since an index of

1 2

is equivalent to taking the square root, this term can be written as the square root of 10.

b

1

a 10 2 = 10 3

1

A power of 32 means square root of the number cubed.

2

Evaluate 53.

= 125

3

Simplify 125.

=5 5

■■

53

b 52 =

In Year 9 you would have studied the index laws and all of these laws are valid for fractional indices.

Worked Example 15

Simplify each of the following.

1

 2 2 1 1 2  x3  a  m 5 × m 5          b  ( a2 b3 ) 6          c   3   y 4  Think

Write 1

a

2

a m5 × m5

1

Write the expression.

2

Multiply numbers with the same base by adding the indices.

1

Write the expression.

2

Multiply each index inside the grouping symbols (brackets) by the index on the outside.

= a6b6

3

Simplify the fractions.

= a3b2

3

= m5 1

b

b (a 2 b 3 ) 6 2 3

1 1

1

c

1

Write the expression.

 2 2 x3  c   3  y 4  1

2

Multiply the index in both the numerator and denominator by the index outside the grouping symbols.

=

x3 3 8 y

Chapter 18 Real numbers

611

number AND algebra • REAL NUMBERS

remember

1. Fractional indices are those that are expressed as fractions. 2. Numbers with fractional indices can be written as surds, using the following identities: m

1 an

n m m n = a         a n = a = ( a ) 3. All index laws are applicable to fractional indices.

Exercise

18D

n

Fractional indices Fluency 1   WE 11  Evaluate each of the following without using a calculator if necessary. 1

1

1

b 25 2 5

a 16 2 4 1

c 812 9

1

d 8 3 2

1

e 27 3 3

f 125 3 5

2   WE 12  Use a calculator to evaluate each of the following, correct to 1 decimal place if necessary. 1

1

a 814 3 d

1 5 2

c 3 3 1.4

1 75

f 8 9 1.3

e

2.2

1

b 16 4 2

1

1.5

3   WE 13  Use a calculator to find the value of each of the following, correct to 1 decimal place. 5

3

d

2

b 100 9 12.9

a 12 8 2.5

c 50 3 13.6

3

4 (0 .6) 5

2

 3 4 e   0.8  4

0.7

 4 3 f   0.9  5

4   WE 14  Write each of the following expressions in simplest surd form. 1

a 7 2

7

1

1

b 12 2 2 3

c 72 2 6 2

5

3

d 2 2 4 2

5

e 3 2 3 3

f 10 2 100 10

5 Write each of the following expressions with a fractional index. 1

a

5 52 3

1

1

10 10 2

c

e 2 t 2t 2

f

b

3

1

1

3

6 63 6   WE 15a  Simplify each of the following. Leave your answer in index form. d

m m 2

x x2

3

1

1

4

a 4 5 × 4 5 4 5 3

2

3

1

1

1

23

d x 4 × x 5 x 20

1

3

8

e 5m 3 × 2m 5 10 m15

2

3

20

g −4 y 2 × y 9 −4 y 9

h

1

5

c a 2 × a 3 a 6

b 2 8 × 2 8 2 2

5 1 7 b × 4 b 7 2b 7 2

i

5 x 3 × x 2 5x 2

1

3

9 2 8 a × 0.05a 4 0.02a 8 5

2

f

7

7 Simplify each of the following. 2 3

1 3

3

a a 3 b 4 × a 3 b 4 ab 2 3

d 6m 7 × 612

1

3 2 1 1

2

19 2 1 4 5 m n 2m 28 n 5 3

1 1

4

1

5

b x 5 y 9 × x 5 y 3 x 5 y 9 1 1 1

19 5 5

e x 3 y 2 z 3 × x 6 y 3 z 2 x 6 y 6 z 6

Maths Quest 10 + 10A for the Australian Curriculum

3 4

8 17

c 2ab 3 × 3a 5 b 5 6a 5 b 15 2 3 1

3 3

2 9

f 2a 5 b 8 c 4 × 4 b 4 c 4 8a 5 b 8 c

number AND algebra • REAL NUMBERS 8 Simplify each of the following. 1

1

2

1

a 3 2 ÷ 3 3 3 6 d

6 a7

3 ÷ a 7



1

3

5

3 a7

e

3 x2

÷

1 x4

4

5 x4



1

c 122 ÷ 12 2 12 2

b 5 3 ÷ 5 4 512

f

11

m5

m 45 5

m9 3

1 g 3 x 20 2

h

4 3

d

2 1 ÷ 5 x 3 y 4

4

5 2

5 7

a x 3 y 2 ÷ x 3 y 5 x 3 y 5 4 10 x 5 y

1 n3 3

21n 3

4x 5 9 Simplify each of the following.

3

2

7 n2

3

2x 4

i

2 2

3 3



1

4

20 b 4 7

11

4

3

3 11

1 8 56 m n 3

c m 8 n 7 ÷ 3n 8 7 1

7

5a 4 b 5 1 20 20 a b e 1 1

7

b 20

3 4

4

b a 9 b 3 ÷ a 5 b 5 a 45 b 15

2 3 2 x 15 y 4

25b 5 5

5

f

20 a 5 b 4

1

p 8 q 4 1 24 12 p q 7

2 1

7 p3q6

10 Simplify each of the following. a

( )

d

1 (a 3 )10

3 3 5 2 4

9 2 20



( )

14 3 15 p7

g 4

b

3 a 10

e

2 4 p5

h

( ) (m ) (x )

1 2 4 53

i

(7 ) ( 2b ) (3m )

b a c b

3c m c

a 3b 4

c

( )

x5y4

)

 3  3 a2  a4  2 f   b  b 3



1 56

3 4 8 9

1 m6

n m p n

x

c f

m p

1 5

6

6

75

1 1 3 2

1

1

23 b6 b

a

Understanding 11   WE 15b, c  Simplify each of the following. a

( )

d

(

1 1 1 2 a 2 b 3

1 1

a4b6

)

1 1 3 3 3 3a 3 b 5 c 4



b (a

1 1 1 1 33 a 9 b 5 c 4

e

(

4

3 4 b)

3

1 1 2 2 2 x2y3z5

3 7 x5y8

2

2

1 1 1 x4 y3z 5

6 7

1

2

2

2  33 5 5 b b   h  4  8  9  c 27 c

 4 8  m 5  m 5 g  7  7  8  n 4 n

1

1

7

 4x7  2 22 x 2 i  3 3  2 y 4  y8

12   MC  Note: There may be more than one correct answer.

If ( )

m 3 n a4

is equal to

1 a4,

then m and n could not be:

A 1 and 3 ✔ C 3 and 8 13 Simplify each of the following.

B 2 and 6

✔ D

a

a 8 a4

b

3

b9

d

x 4 4x2 16

e

3

27m 9 n15 3m3n5

h

5

g

3

4 and 9 c

4

m16 m4

8 y 9 2y3

f

4

16 x8 y12 2x2y3

32 p5q10 2pq2

i

3

216a6 b18 6a2b6

b3

Chapter 18 Real numbers

613

number and algebra • real numbers reasoning 14 At the start of this chapter we looked at Manning’s formula, which is used to calculate the flow 2

1

R3S 2 of water in a river during a flood situation. Manning’s formula is v = , where R is the n hydraulic radius, S is the slope of the river and n is the roughness coefficient. This formula is used by meteorologists and civil engineers to analyse potential flood situations. We were asked to find the flow of water in metres per second in the river if R = 8, S = 0.0025 and n = 0.625. a Use Manning’s formula to find the flow of water in the river. 0.32 m/s the 16 640 L/s b To find the volume of water flowing through the river, we multiply the flow rate by 59 904 000 L/hr average cross-sectional area of the river. If the average cross-sectional area is 52 m2, find the That is 16 640 ì 60 ì 60. volume of water (in L) flowing through the river each second. (Remember 1 m3 = 1000 L.) c If water continues to flow at this rate, what will be the total amount of water to flow through in refleCtion    one hour? Justify your answer. How will you remember the rule d Use the Internet to find the meaning of the terms for fractional indices? hydraulic radius and roughness coefficient.

The hydraulic radius is the measure of a channel flow efficiency. The roughness coefficient is the resistance of the bed of a channel to the flow of water in it.

eBook plus

Digital doc

WorkSHEET 18.2 doc-5364

18e

negative indices ■

Consider the following division 23

23 2

4

= 2−1 (using the Second Index Law).

8 1 Alternatively, 4 = = . 16 2 2

■

1 We can conclude that 2−1 = . 2 In general form: 1 1 −n and a = n . a a When using a calculator to evaluate expressions that involve negative indices, we need to familiarise ourselves with the keys needed. a −1 =

■

614

maths quest 10 + 10a for the australian Curriculum

number AND algebra • REAL NUMBERS

Worked Example 16

Evaluate each of the following using a calculator. a  4-1       b  2-4 Think

Write

a Use a calculator to evaluate 4-1.

a 4-1 = 0.25

b Use a calculator to evaluate 2-4.

b 2-4 = 0.0625

■■

1 Consider the index law a −1 = . Now let us look at the case in which a is fractional. a −1

 a Consider the expression   .  b  a   b

−1

=

1 a b

= 1×

b a

b a We can therefore consider an index of -1 to be a reciprocal function. =

■■

Worked Example 17

Write down the value of each of the following without the use of a calculator.  2

−1

 1

−1

a                 b    3 5

 1 4

−1

c   1  Think

Write

 2 a To evaluate    3 b

c

−1

2 take the reciprocal of . 3 −1

1

 1 To evaluate    5

2

Write

1

1 Write 1 as an improper fraction. 4

2

Take the reciprocal of

1 take the reciprocal of . 5

−1

a

 2   3  1   5

−1

b

5 as a whole number. 1

5 . 4

=

3 2

=

5 1

=5 c

 1  1  4

−1

 5 =   4 =

−1

4 5

Chapter 18 Real numbers

615

number AND algebra • REAL NUMBERS

remember

1

or the x-1 function. xy 2. An index of -1 can be considered as a reciprocal function and applying this to fractions −1 b  a gives us the rule   = .  b a

1. To evaluate an expression that involves negative indices, use the

Exercise

18E

Negative indices Fluency

1 a

1 5

= 0.2

b

1 3

= 0.3

c

1 8

= 0.125

1

d 10 = 0.1 e

1 8

= 0.125

f

1 9

= 0.1

1   WE 16  Evaluate each of the following using a calculator. a 5-1 b 3-1 -1 d 10 e 2-3 -2 g 5 h 10-4

c 8-1 f 3-2

2 Find the value of each of the following, correct to 3 significant figures. a 6 b 7-1 0.143 c 6-2 0.0278 -1 0.167 -3 -3 d 9 0.001  37 e 6 0.004  63 f 15-2 0.004  44 -2 -4 g 16 h 5 0.001  60 0.003  91

3 Find the value of each of the following, correct to 2 significant figures. a (2.5) b (0.4)-1 2.5 c (1.5)-2 0.44 -1 0.40 g = 0.04 -2 -3 d (0.5) e (2.1) 0.11 f (10.6)-4 0.000  079 4.0 1 -3 -4 g (0.45) h (0.125) 4100 11 h 10 000 = 0.0001 4 Find the value of each of the following, correct to 2 significant figures. -1 -0.33 a (-3) b (-5)-1 -0.20 c (-2)-2 0.25 -2 -1 0.063 d (-4) e (-1.5) -0.67 f (-2.2)-1 -0.45 -1 -2 g (-0.6) h (-0.85) 1.4 -1.7 1 25

5   WE 17  Write down the value of each of the following without the use of a calculator.

  a    5

−1

 1

−1

4



5 4

1

or 1 4

 1

e   2  2 −1

10 3

1

or 3 3

−1

f   4  4

 7

−1

 1

−1



c    8

8 7

1

or 1 7

−1

 13   20 

d  

20 13

7

or 113

−1

 1  10  10 

g   8  8

−1

h  

−1

−1

4  1  1  1 2 10 k  1  11 l  5  11 9    2 4 10 6 Find the value of each of the following, leaving your answer in fraction form if necessary.

i

 1   1  2

−1

 3  10 

b  

2 3

j  2   

−2

−2

 1 a   4  2

−3

 2 b   6 4  5

−2

 2 c   3 8  3

1

−2

16  1 4  1 f  2  81 9  4 2 7 Find the value of each of the following.

e  1    −1

  a  −  - 2   2 3

−2

 2 e  −    3

616

3

9 4

 3 5

−1 5

b  −  - 3   −2

 1 f  −  25   5

Maths Quest 10 + 10A for the Australian Curriculum

 1 3

3

−3

g  1   

 1 4

−1

c  −   



1 2

−2

 1 d   16  4  1 5

−3

27 64

h  2   

-4

d  − 

−1

2 g  −1  - 3  



125 1331

−1

1  -10  10  −2

 3 h  −2    4

16 121

number and algebra • real numbers

eBook plus

Digital doc

reasoning 8 Consider the expression 2-n. Explain what happens to

refleCtion 

the value of this expression as n increases.

WorkSHEET 18.3 doc-5365

As the value of n increases, the value of 2-n gets closer to 0.

 

How can division to used to explain negative indices?

logarithms

18f

■

■

The index, power or exponent in the statement y = ax is also known as a logarithm (or log for short). Logarithm (or index or power or exponent) x y=a Base This statement y = ax can be written in an alternative form as loga y = x, which is read as ‘the logarithm of y to the base a is equal to x’. These two statements are equivalent. ax = y « Index form

■

loga y = x Logarithmic form

For example, 32 = 9 can be written as log3 9 = 2. The log form would be read as ‘the logarithm of 9, to the base of 3, is 2’. In both forms, the base is 3 and the logarithm is 2.

Worked example 18

Write the following in logarithmic form. a 104 = 10 000 b 6x = 216 think a

b

Write a 104 = 10 000

1

Write the given statement.

2

Identify the base (10) and the logarithm (4) and write the equivalent statement in logarithmic form. (Use ax = y « loga y = x, where the base is a and the log is x.)

1

Write the given statement.

2

Identify the base (6) and the logarithm (x) and write the equivalent statement in logarithmic form.

log10 10 000 = 4

b 6x = 216

log6 216 = x

Worked example 19

Write the following in index form. a log2 8 = 3

b log25 5 =

1 2

think a

Write

1

Write the statement.

2

Identify the base (2) and the log (3) and write the equivalent statement in index form. Remember that the log is the same as the index.

a log2 8 = 3

23 = 8

Chapter 18 real numbers

617

number AND algebra • REAL NUMBERS

b

1 2

b log25 5 =

Write the statement.

1

 1 Identify the base (25) and the log   2 and write the equivalent statement in index form.

■■

■■

1 2

25 2 = 5

In the previous examples, we found that: log2 8 = 3 « 23 = 8 and log10 10  000 = 4 « 104 = 10  000.   We could also write log2 8 = 3 as log2 23 = 3 and log10 10  000 = 4 as log10 104 = 4. Can this pattern be used to work out the value of log3 81? We need to find the power when the base of 3 is raised to that power to give 81.

Worked Example 20

Evaluate log3 81. Think

Write

1

Write the log expression.

log3 81

2

Express 81 in index form with a base of 3.

= log3 34

3

Write the value of the logarithm.

=4

remember

1. Logarithm is another name for an index, power or exponent.   For example, in the statement 23 = 8, the logarithm is 3. 2. The logarithm of a number to any positive base is the index when the number is expressed as a power of the base.   That is, ax = y « loga y = x, where a > 0, y > 0. 3. One way of evaluating the logarithm of a number is to write the number in index form to the given base.   That is, loga ax = x.   For example, log3 81 = log3 34 = 4. Exercise

18F

Logarithms Fluency

log10 1000 = 3

1   WE 18  Write the following in logarithmic form. a 42 = 16 log4 16 = 2 b 25 = 32 log2 32 = 5 c 34 = 81 log3 81 = 4 d 62 = 36 log6 36 = 2 3 2 3 e 1000 = 10 f 25 = 5 log5 25 = 2 g 4 = x log4 x = 3 h 5x = 125 log5 125 = x 1

i

7x = 49 log7 49 = x j p4 = 16 logp 16 = 4 k 9 2 = 3 log9 3 = 12 1

1 1 log2 2 = -1 o a0 = 1 loga 1 = 0 2 2   MC  The statement w = ht is equivalent to: A w = logt h B h = logt w C t = logw h ✔ D t = logh w 1 m 2 = 8 3 log8 2 = 3

618

n 2−1 =

Maths Quest 10 + 10A for the Australian Curriculum

l

0.1 = 10-1 log10 0.1 = -1 3

p 4 2 = 8 log4 8 =

3 2

number AND algebra • REAL NUMBERS 106 = 1  000  000

3   WE 19  Write the following in index form. a log2 16 = 4 24 = 16 b log3 27 = 3 33 = 27 d log5 125 = 3 53 = 125

e log16 4 =

1

1 2

f log4 64 = x 4x = 64 1

1 = log 49 7 49 2 = 7 2

j

10-2 = 0.01 k log8 8 = 1 81 = 8 log10 0.01 = -2

i

c log10 1  000  000 = 6

=4

g

1 log81 9 = 812 = 9 2 1 1 l log64 4 = 64 3 = 4 3

h log3 x = 5 35 = x

4   MC  The statement q = logr p is equivalent to: ✔ A q = r p C r = pq 5   WE 20  Evaluate the following logarithms. a log2 16 4 c log11 121 2 5 e log3 243 g log5 1 0

 1 log3   -1  3  1   100  -2

k log10 

6 Write the value of each of the following. a log 10 1 0 c log10 100 2 e log10 10  000 4 a log10 g = k implies that g = 10k so g2 = (10k)2. That is, g2 = 102k; therefore, log10 g2 = 2k.

1 16 2

i

B p = rq D r = qp b d f h

log4 16 2 log10 100  000 5 log2 128 7 log9 3 12

j

log6 6 1

l

log125 5

1 3

b log10 10 1 3 d log10 1000 f log10 100  000 5

Understanding

7 Use your results to question 6 to answer the following. a Between which two whole numbers would log10 7 lie? 0 and 1 b Between which two whole numbers would log10 4600 lie? 3 and 4 b logx y = 2 implies 1 and 2 c Between which two whole numbers would log10 85 lie? that y = x2, so d Between which two whole numbers would log 12   7 50 lie? 4 and 5 10 1 2 2 and 3 e Between which two whole numbers would log 110 lie? x = y and 10 1 therefore logy x = . f Between which two whole numbers would log10 81  000 lie? 4 and 5 2 Reasoning 8 a If log10 g = k, find the value of log10 g2. Justify your answer. b If logx y = 2, find the value of logy x. Justify your answer. c By referring to the equivalent index statement, explain

reflection 

why x must be a positive number given log4 x = y, for all  he equivalent exponential statement is x = 4y, values of y. T

 

How are indices and logarithms related?

and we know that 4y is greater than zero for all values of y. Therefore, x is a positive number.

18G

Logarithm laws ■■

■■

From previous work, you will be familiar with the index laws. am 1. am ì an = am + n 2. n = a m − n 3. (am)n = amn a 1 4. a0 = 1 5. a1 = a 6. a−1 = a We can use these index laws to produce equivalent logarithm laws. Chapter 18 Real numbers

619

number AND algebra • REAL NUMBERS

Law 1 ■■

If x = am and y = an, then loga x = m and loga y = n (equivalent log form). Now xy = am ì an (First Index Law). or xy = am + n (equivalent log form) So loga (xy) = m + n (substituting for m and n). or loga (xy) = loga x + loga y log a x + log a y = log a (xy)

■■

This means that the sum of two logarithms with the same base is equal to the logarithm of the product of the numbers.

Worked Example 21

Evaluate log10 20 + log10 5. Think

Write

1

Since the same base of 10 is used in each log term, use loga x + loga y = loga (xy) and simplify.

2

Evaluate. (Remember that 100 = 102.)

log10 20 + log10 5 = log10 (20 ì 5) = log10 100 =2

Law 2 ■■

If x = am and y = an, then loga x = m and loga y = n (equivalent log form). x am = y an x = am −n y

Now or

(Second Index Law).

So

 x log a   = m − n  y

(equivalent log form)

or

 x log a   = log a x − log a y  y

(substituting for m and n).

 x log a x - log a y = log a    y ■■

This means that the difference of two logarithms with the same base is equal to the logarithm of the quotient of the numbers.

Worked Example 22

Evaluate log4 20 - log4 5. Think

620

1

Since the same base of 4 is used in each  x log term, use log a x − log a y = log a   and  y simplify.

2

Evaluate. (Remember that 4 = 41.)

Maths Quest 10 + 10A for the Australian Curriculum

Write

 20  log4 20 - log4 5 = log4   5 = log4 4 =1

number AND algebra • REAL NUMBERS

Worked Example 23

Evaluate log5 35 + log5 15 - log5 21. Think

Write

1

Since the first two log terms are being added, use loga x + loga y = loga (xy) and simplify.

2

To find the difference between the two remaining log terms, use  x log a x − log a y = log a   and simplify.  y

3

Evaluate. (Remember that 25 = 52.)

■■

log5 35 + log5 15 - log5 21   = log5 (35 ì 15) - log5 21   = log5 525 - log5 21  525  = log 5   21      = log5 25   =2

Once you have gained confidence in using the first two laws, you can reduce the number of steps of working by combining the application of the laws. In Worked example 23, we could write:  35 × 15  log 5 35 + log 5 15 − log 5 21 = log 5   21  = log5 25 =2



Law 3 ■■

If x = am, then loga x = m (equivalent log form). Now or So or or

xn = (am)n xn = amn loga xn = mn loga xn = (loga x) ì n loga xn = n loga x.

(Third Index Law). (equivalent log form) (substituting for m)

log a xn = n log a x ■■

This means that the logarithm of a number raised to a power is equal to the product of the power and the logarithm of the number.

Worked Example 24

Evaluate 2 log6 3 + log6 4. Think

Write

2 log6 3 + log6 4 = log6 32 + log6 4 = log6 9 + log6 4

1

The first log term is not in the required form to use the log law relating to sums. Use loga xn = n loga x to rewrite the first term in preparation for applying the first log law.

2

Use loga x + loga y = loga (xy) to simplify the two log terms to one.

= log6 (9 ì 4) = log6 36

3

Evaluate. (Remember that 36 = 62.)

=2

Chapter 18 Real numbers

621

number AND algebra • REAL NUMBERS

Law 4 ■■

As

a0 = 1 loga 1 = 0

(Fourth Index Law), (equivalent log form).

log a 1 = 0 ■■

This means that the logarithm of 1 with any base is equal to 0.

Law 5 ■■

As

a1 = a loga a = 1

(Fifth Index Law), (equivalent log form).

log a a = 1 ■■

This means that the logarithm of any number a with base a is equal to 1.

Law 6 ■■

Now

 1 log a   = log a x −1  x

(Sixth Index Law)

or

 1 log a   = −1 × log a x  x

(using the fourth log law)

or

 1 log a   = − log a x.  x  1 log a   = − log a x  x

Law 7 ■■

Now or or

loga ax = x loga a loga ax = x ì 1 loga ax = x.

(using the third log law) (using the fifth log law)

log a ax = x remember

The index laws can be used to produce the following logarithm laws. 1. loga x + loga y = loga (xy) 3. loga xn = n loga x 5. loga a = 1

 x 2. log a x − log a y = log a    y 4. loga 1 = 0  1 6. log a   = − log a x  x

7. loga ax = x Exercise

18G

Logarithm laws Fluency 1 Use a calculator to evaluate the following, correct to 5 decimal places. a log10 50 1.698  97 b log10 25 1.397  94 c log10 5 0.698  97

d log10 2 0.301  03

2 Use your answers to question 1 to show that each of the following statements is true. b log10 50 - log10 2 = log10 25 Teacher to check. a log10 25 + log10 2 = log10 50 c log10 25 = 2 log10 5 d log10 50 - log10 25 - log10 2 = log10 1 622

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • REAL NUMBERS 3   WE 21  Evaluate the following. a log6 3 + log 6 2 1 c log10 25 + log10 4 2 e log6 108 + log6 12 4

b log4 8 + log4 8 3 d log8 32 + log8 16 3 f log14 2 + log14 7 1

4   WE 22  Evaluate the following. a log2 20 - log 25 2 c log4 24 - log4 6 1 e log6 648 - log6 3 3

b log3 54 - log3 2 3 d log10 30  000 - log 10 3 4 f log2 224 - log2 7 5

5   WE 23  Evaluate the following. 36 2 a log3 27 + log3 2 - log c log6 78 - log6 13 + log6 1 1

b log4 24 - log4 2 - log4 6 2 d log2 120 - log2 3 - log2 5 3

6 Evaluate 2 log4 8. 3 7   WE 24  Evaluate the following. a 2 log10 5 + log 10 4 2

b log3 648 - 3 log3 2 4

1

c 4 log5 10 - log5 80 3

1 log2 16 - 2 log2 5 3 2

d log2 50 +

8 Evaluate the following.

 1

a log 88 1

b log5 1 0

c log2   -1 2

d log 4 4 5 5

e log6 6-2 -2

f log20 20 1

g log 21 0

h log3   -2 9

j

 1

log 5 5

 1  1  - 2 3

k log 3  

1 2

i

 1 log4   - 12 2

l

log 2 8 2

7 2

Understanding 9 Use the logarithm laws to simplify each of the following. a loga 5 + log a 8 loga 40 b loga 12 + loga 3 - loga 2 loga 18 c 4 logx 2 + logx 3 logx 48 d logx 100 - 2 logx 5 logx 4 e 3 loga x - log a x2 loga x f 5 loga a - loga a4 1 g logx 6 - logx 6x -1 h loga a7 + loga 1 7 i

log p p

1 2

 1 a

k 6 log a   -6  

3 2

j

log k k k

l

 1  log a  3  - 1 3  a

10   MC  Note: There may be more than one correct answer. a The equation y = 10x is equivalent to: ✔ B x = log10 y A x = 10y C x = logx 10 D x = logy 10 b The equation y = 104x is equivalent to: A x = log10 4 y 1

C x = 10 4

y

✔ B

x = log10 4 y

✔ D

x=

1 log10 y 4

c The equation y = 103x is equivalent to: ✔

1 A x = log10 y 3 C x = log10 y – 3

✔ B

x = log10

1 y3

D x = 10y – 3 Chapter 18 Real numbers

623

number AND algebra • REAL NUMBERS d The equation y = manx is equivalent to: A x = ✔ C

x=

 m  y 

1 my a n

n

B x = log a 

1 (log a y − log a m) n

✔ D

x=

1  y log   n a  m

11 Simplify, and evaluate where possible, each of the following without a calculator. a log2 8 + log2 10 b log3 7 + log3 15 c log10 20 + log10 5 d log6 8 + log6 7 e log2 20 - log2 5 f log3 36 - log3 12

a log2 80 b log3 105 c log10 100 = 2 d log6 56 e log2 4 = 2 f log3 3 = 1 g log5 12.5 h log2 3 i log4 5 12 1 j log10. 4 k log3 4 l log2 3 m log3 20 n log4 2 =

1 2

1 1 i log4 25 + log4 + log2 9 3 5 4 1 j log10 5 - log10 20 k log3 - log3 l log2 9 + log2 4 - log2 12 5 5 m log3 8 - log3 2 + log3 5 n log4 24 - log4 2 - log4 6   MC  a The expression log10 xy is equal to: A log10 x ì log10 y B log10 x - log10 y ✔ C log10 x + log10 y D y log10 x b The expression log10 xy is equal to: A x log10 y ✔ B y log10 x C 10 logx y D log10 x + log10 y 1 c The expression log2 64 + log2 10 is equal to: 3 B log2 80 ✔ A log2 40 64 C log2 D 1 10 g log5 100 - log5 8

h log2

Reasoning 7 (Let y = 5log 7 13 For each of the following, write the possible strategy you intend to use. and write an a Evaluate (log3 81)(log3 27). 12 (Evaluate each logarithm separately and then find the product.) equivalent statement in log a 81 4 (First simplify the numerator by reflection    . logarithmic form.) b Evaluate expressing 81 as a power of 3.) 5

log a 3

c Evaluate 5

log 5 7

. In each case, explain how you obtained your final answer.

18H

Solving equations ■■

The equation loga y = x is an example of a general logarithmic equation. Laws of logarithms and indices are used to solve these equations.

Worked Example 25

Solve for x in the following equations. a  log2 x = 3 c  log3 x4 = -16

b  log6 x = -2 d  log5 (x - 1) = 2

Think a

624

Write

1

Write the equation.

2

Rewrite using = y « loga y = x. Rearrange and simplify.

3

What technique will you use to remember the log laws?

ax

Maths Quest 10 + 10A for the Australian Curriculum

a

log2 x = 3 23 = x x=8

number AND algebra • REAL NUMBERS b

c

d

b

1

Write the equation.

2

Rewrite using ax = y « loga y = x.

3

Rearrange and simplify.

1

Write the equation.

2

Rewrite using loga xn = n loga x.

3

Divide both sides by 4. ax

Rewrite using

5

Rearrange and simplify.

1

Write the equation.

2

Rewrite using

3

Solve for x.

6-2 = x 1 x= 2 6 1 = 36 c

log3 x4 = -16 4 log3 x = -16 log3 x = -4 3-4 = x 1 x= 4 3 1 = 81

= y « loga y = x.

4

ax

log6 x = -2

d log5 (x - 1) = 2

= y « loga y = x.

52 = x - 1 x - 1 = 25 x = 26

Worked Example 26

Solve for x in log x 25 = 2, given that x > 0. Think

Write

logx 25 = 2

1

Write the equation.

2

Rewrite using ax = y « loga y = x.

3

Solve for x. Note: x = -5 is rejected as a solution because x > 0.

x2 = 25 x = 5 (because x > 0)

Worked Example 27

Solve for x in the following.  1 3

a  log2 16 = x      b  log3   = x      c  log9 3 = x   Think a

1

Write

Write the equation. ax

= y « loga y = x.

2

Rewrite using

3

Write 16 with base 2.

4

Equate the indices.

a log2 16 = x

2x = 16 = 24 x=4 Chapter 18 Real numbers

625

number AND algebra • REAL NUMBERS

b

1

Write the equation.

2

Rewrite using ax = y « loga y = x.

 1

b log 3   = x  3

3x = =

4

1 with base 3. 3 Equate the indices.

1

Write the equation.

3

c

31

3x = 3-1

Write

ax

1 3 1

x = -1 c

log9 3 = x

= y « loga y = x.

9x = 3

2

Rewrite using

3

Write 9 with base 3.

4

Remove the grouping symbols.

32x = 31

5

Equate the indices.

2x = 1

6

Solve for x.

(32)x = 3

x=

1 2

Worked Example 28

Solve for x in the equation log2 4 + log2 x - log2 8 = 3. Think

Write

1

Write the equation.

2

Simplify the left-hand side. Use loga x + loga y = loga (xy) and

log2 4 + log2 x - log2 8 = 3  4 × x =3 log 2   8 

 x log a x − log a y = log a   .  y Simplify.

4

Rewrite using ax = y « loga y = x.

5

Solve for x.

■■

■■

626

 x log 2   = 3  2

3

23 =

x 2

x = 2 ì 23 =2ì8 = 16

When solving an equation like log2 8 = x, we could rewrite it in index form as 2x = 8. This can be written with the same base of 2 to produce 2x = 23. Equating the indices gives us a solution of x = 3. Can we do this to solve the equation 2x = 7? Consider the method shown in the next worked example. It involves the use of logarithms and the log10 function on a calculator.

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • REAL NUMBERS

Worked Example 29

Solve for x, correct to 3 decimal places, if a  2x = 7              b  3-x = 0.4 Think a

b

Write

2x = 7

a

1

Write the equation.

2

Take log10 of both sides.

3

Use the logarithm-of-a-power law to bring the power, x, to the front of the logarithmic equation.

4

Divide both sides by log10 2 to get x by itself.

5

Use a calculator to evaluate the logarithms and write the answer correct to 3 decimal places.

1

Write the equation.

2

Take log10 of both sides.

3

Use the logarithm of a power law to bring the power, x, to the front of the logarithmic equation.

4

Divide both sides by log10 3 to get the -x by itself.

−x =

5

Use a calculator to evaluate the logarithms and write the answer correct to 3 decimal places.

-x = -0.834

6

Divide both sides by -1 to get x by itself.

■■

log10 2x = log10 7 x log10 2 = log10 7

Therefore, x =

log10 7 log10 2

= 2.807

b

3-x = 0.4 log10 3-x = log10 0.4 -x log10 3 = log10 0.4

log10 0.4 log10 3

x = 0.834

Therefore, we can state the following rule: If ax = b, then x =

log10 b . log10 a

  This rule applies to any base, but since your calculator has base 10, this is the most commonly used for this solution technique.

remember

1. In a logarithmic equation the unknown, x, can be: (a) the number, log2 x = 5 (b) the base, logx 8 = 3 (c) the logarithm, log2 4 = x. 2. The laws of logarithms and indices can be used to solve these equations. log10 b 3. If ax = b, then x = . log10 a Chapter 18 Real numbers

627

number AND algebra • REAL NUMBERS

Exercise

18H

Solving equations Fluency 1   WE 25  Solve for x in the following. a log5 x = 2 25 b log3 x = 4 81 d log4 x = -2 161 e log10 x2 = 4 100, -100 g log3 (x + 1) = 3 26 h log5 (x - 2) = 3 127 1 j log10 (2x + 1) = 0 0 k log2 (-x) = -5 - 32 m log5 (1 - x) = 4 -624 n log10 (5 - 2x) = 1 -2.5

c f i l

log2 x = -3 18 log2 x3 = 12 16 log4 (2x - 3) = 0 2 log3 (-x) = -2 - 19

2   WE 26  Solve for x in the following, given that x > 0. a logx 9 = 2 3

 1 8 2 3 g logx 6 = 2 6 h logx 4 = 3 4 3   WE 27  Solve for x in the following. a log2 8 = x 3 d logx 125 =

2 3

b logx 16 = 4 2

3 625 4

c logx 25 = 125

 1 = -2 8  64 

2 e logx   = -3  

 1 5 1 e log4 2 = x 2 g log6 1 = x 0 i log 1 2 = x -1

f logx 

b log3 9 = x 2

 1 = x -2  16  2 f log8 2 = x 5 h log8 1 = x 0 j log 1 9 = x -2

c log5   = x -1  

d log4 

2

3

4   WE 28  Solve for x in the following. a log2 x + log2 4 = log2 20 5 c log3 x - log3 2 = log3 5 10 e log4 8 - log4 x = log4 2 4 g log6 4 + log6 x = 2 9 i 3 - log10 x = log10 2 500 k log2 x + log2 6 - log2 3 = log2 10 5 m log3 5 - log3 x + log3 2 = log3 10 1

b d f h j l n

log5 3 + log5 x = log5 18 6 log10 x - log10 4 = log10 2 8 log3 10 - log3 x = log 3 5 2 log2 x + log2 5 = 1 25 5 - log4 8 = log4 x 128 log2 x + log2 5 - log2 10 = log2 3 6 log5 4 - log5 x + log5 3 = log5 6 2

5   MC  a The solution to the equation log 7  343 = x is: A x = 2 C x = 1 ✔ B x = 3 b If log8 x = 4, then x is equal to: ✔ A 4096 B 512 C 64

1 c Given that logx 3 = , x must be equal to: 2 A 3 B 6 d If log a  x = 0.7, then log a  x2 is equal to: ✔ B 1.4 A 0.49 6 Solve for x in the following equations. a 2x = 128 7

b 3x = 9 2

e 5x = 625 4

f 64x = 8

i

3x =

1 3

1

- 2

m 3 x + 1 = 27 3 628

j 5

2

4x = 8

n 2 x − 1 =

D 2

C 81

✔ D

C 0.35 c 7 x = g 6 x =

1 2

1 32 2

9

- 2

9

D 0.837

1 -2 49 6

k 9 x = 3 3

3 2

Maths Quest 10 + 10A for the Australian Curriculum

D x = 0

o 4 x + 1 =

d 9x = 1 0

1 2

h 2 x = 2 2

3 4

1 8 2

l 11

- 4

2x =

1 4 2

3 2 5

- 2

number and algebra • real numbers understanding 7 We 29 Solve the following equations, correct to 3 decimal places. a 2x = 11 3.459 b 2x = 0.6 -0.737 c x d 3 = 1.7 0.483 e 5x = 8 1.292 f g 0.4x = 5 -1.756 h 3x + 2 = 12 0.262 i j 8-x = 0.3 0.579 k 10-2x = 7 -0.423 l

3x = 20 2.727 0.7x = 3 -3.080 7-x = 0.2 0.827 82 - x = 0.75 2.138

8 The decibel (dB) scale for measuring loudness, d,

is given by the formula d = 10 log10 (I ì 1012), where I is the intensity of sound in watts per square metre. a Find the number of decibels of sound if the intensity is 1. 120 b Find the number of decibels of sound produced by a jet engine at a distance of 50 metres if the intensity is 10 watts per square metre. 130 c Find the intensity of sound if the sound level of a pneumatic drill 10 metres away is 90 decibels. 0.001 d Find how the value of d changes if the intensity is doubled. Give your answer to the 3 dB are added. nearest decibel. e Find how the value of d changes if the intensity is 10 times as great. 10 dB are added. f By what factor does the intensity of sound have to be multiplied in order to add 20 decibels to the sound level? 100

reasoning 9 The Richter scale is used to describe the energy of earthquakes. A formula for the Richter

2 log10 K – 0.9, where R is the Richter scale value for an earthquake that 3 releases K kilojoules (kJ) of energy. a Find the Richter scale value for an earthquake that releases the following amounts of energy: N o; see answers to 9a i & ii above. i 1000 kJ 1.1 ii 2000 iii 3000 kJi 1.3 kJ 1.418 kJ 1.77 kJ 3.1 iv 10 000 v 100 000 kJi 2.43 vi 1 000 000 b Does doubling the energy released double the Richter scale value? Justify your answer. c Find the energy released by an earthquake of: i magnitude 4 on the Richter scale 22 387 211 kJ T he energy is increased ii magnitude 5 on the Richter scale 707 945 784 kJ by a factor of 31.62. 22 387 211 385 kJ. iii magnitude 6 on the Richter scale. d What is the effect (on the amount of energy released) of increasing the Richter scale value by 1? e Why is an earthquake measuring 8 on the Richter scale so much more devastating than one that measures 5? scale is: R =

It releases 31.623 times more energy. eBook plus

Digital doc

WorkSHEET 18.4 doc-6754

refleCtion 

 

Tables of logarithms were used in classrooms before calculators were used there. Would using logarithms have any effect on the accuracy of calculations?

Chapter 18 real numbers

629

number AND algebra • REAL NUMBERS

Summary Number classification review ■■

■■

■■

a Rational numbers (Q) can be expressed in the form , where a and b are whole numbers b and b ò 0. They include whole numbers, fractions and terminating and recurring decimals. a Irrational numbers (I) cannot be expressed in the form , where a and b are whole b numbers and b ò 0. They include surds, non-terminating and non-recurring decimals, and numbers such as p and e. Rational and irrational numbers together constitute the set of real numbers (R).

Surds

A number is a surd if: it is an irrational number (equals a non-terminating, non-recurring decimal) ■■ it can be written with a radical sign (or square root sign) in its exact form. ■■

Operations with surds ■■ ■■ ■■ ■■ ■■

To simplify a surd means to make a number (or an expression) under the radical sign as small as possible. To simplify a surd, write it as a product of two factors, one of which is the largest possible perfect square. Only like surds may be added and subtracted. Surds may need to be simplified before adding and subtracting. When multiplying surds, simplify the surd if possible and then apply the following rules: (a)  a × b = ab

■■ ■■

(b)  m a × n b = mn ab , where a and b are positive real numbers. When a surd is squared, the result is the number (or the expression) under the radical sign: ( a )2 = a, where a is a positive real number. When dividing surds, simplify the surd if possible and then apply the following rule: a÷ b=

■■

b

=

a b

where a and b are whole numbers, and b ò 0. To rationalise a surd denominator, multiply the numerator and denominator by the surd contained in the denominator. This has the effect of multiplying the fraction by 1, and thus the numerical value of the fraction remains unchanged, while the denominator becomes rational: a b

■■

a

=

a b

×

b b

=

ab b

where a and b are whole numbers and b ò 0. To rationalise the denominator containing a sum or a difference of surds, multiply both the numerator and denominator of the fraction by the conjugate of the denominator. This eliminates the middle terms and leaves a rational number. Fractional indices

■■

Fractional indices are those that are expressed as fractions. Numbers with fractional indices can be written as surds, using the following identities:

■■

n m m n a n = n a         a n = a = ( a ) All index laws are applicable to fractional indices.

■■

1

630

Maths Quest 10 + 10A for the Australian Curriculum

m

number AND algebra • REAL NUMBERS Negative indices ■■ ■■

1

or the x-1 function. xy An index of -1 can be considered as a reciprocal function and applying this to fractions −1 b  a gives us the rule   = .  b a To evaluate an expression that involves negative indices, use the

Logarithms ■■ ■■

■■

Logarithm is another name for an index, power or exponent. For example, in the statement 23 = 8, the logarithm is 3. The logarithm of a number to any positive base is the index when the number is expressed as a power of the base. That is, ax = y « loga y = x, where a > 0, y > 0. One way of evaluating the logarithm of a number is to write the number in index form to the given base. That is, loga ax = x. For example, log3 81 = log3 34 = 4. Logarithm laws

■■

The index laws can be used to produce the following logarithm laws. 1. loga x + loga y = loga (xy)  x 2. log a x − log a y = log a    y 3. loga xn = n loga x 4. loga 1 = 0 5. loga a = 1  1 6. log a   = − log a x  x 7. loga ax = x Solving equations

■■

■■ ■■

In a logarithmic equation the unknown, x, can be: (a) the number, log2 x = 5 (b) the base, logx 8 = 3 (c) the logarithm, log2 4 = x. The laws of logarithms and indices can be used to solve these equations. log10 b If ax = b, then x = . log10 a Mapping your understanding

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 589.

Chapter 18 Real numbers

631

number AND algebra • REAL NUMBERS 2 a Irrational, since equal to non-recurring and non-terminating decimal b Rational, since can be expressed as a whole number c Rational, since given in a rational form

Chapter review Fluency

8 Simplify the following, giving answers in the

simplest form.

6 , 0.81, 5, 1   MC  Which of the given numbers, 12 3 π -3.26, 0.5 , , are rational? 5 12 ✔ A

0.81, 5, -3.26, 0.5 and

B

6 π and 12 5

C

6 3 , 0.81 and 12 12

d 5, -3.26 and

6 12

a 7 12 + 8 147 − 15 27

1 3 1 100 a 5b 5 3ab ab 64 a3b3 − ab 16ab + 2 4 5ab 9 Simplify each of the following. b

3 12

a 3 × 5 15 b 2 6 × 3 7 6 42 d Rational, since it is a recurring decimal c 3 10 × 5 6 30 15 d ( 5 )2 5 e I rrational, since 10 Simplify the following, giving answers in the equal to nonrecurring and nonsimplest form. terminating decimal 1 a 675 × 27 27

5

b 10 24 × 6 12 720 2

2 For each of the following, state whether the number

11 Simplify the following.

is rational or irrational and give the reason for your answer: 2 a 12 b 121 c 9 d 0.6 e 3 0.08

a

2 , 5 7 , 9 4 , 6 10 , 7 12 , 12 64 }, are surds?

A 9 4 , 12 64

a

20 3 , m , 3 8m m

B 3 2 and 7 12 only C 3 2 , 5 7 and 6 10 only ✔ D

2m ,

25m ,

3 2 , 5 7 , 6 10 and 7 12

4 Which of





2m , 25m ,

surds a if m = 4?

m , 16

m , 16

20 m

50 5 2

c 2 32 8 2

20 3 , m , 3 8m are m

6

180 6 5

632

2 5+4

30 12

d

b d



2 6

3 −1

4

2− 3

1

b 20 2 4.5

1

1

c 10 3 2.2

d 50 4 2.7

place.

B 2 x 4 y 3 14 y

2

3

a 20 3 7.4

b 2 4 1.7

c

D 14 x 4 y 3 2

2 25 5 11 x y 5 64 − 1 x 2 y 5 xy

b - 

4

Maths Quest 10 + 10A for the Australian Curriculum

3 (0.7) 5

2

0.8

 2 3 d   0.8  3

15 Write each of the following in simplest surd form.

the simplest form.

72 x 3 y 4 2 xy

6

14 Evaluate each of the following, correct to 1 decimal

d 5 80 20 5

7 Simplify the following surds. Give the answers in a 4 648 x 7 y 9

3 5

1

simplified to: 14 x 4 y 3 2 y

or

3

a 64 3 4

6   MC  The expression 392 x y may be

✔ c



10



2

8 7

a 196 x 4 y 3 2 y

6 45

5−2 3 +1 13 Evaluate each of the following, correct to 1 decimal place if necessary.

b if m = 8? b

3 20

b

3



10

c

5 Simplify each of the following. a

30

( 7 )2 1 2 12 6 4 3 14 12 Rationalise the denominator of each of the following. 2 6 3 2 c

3   MC  Which of the numbers of the given set,

{3

25 3

1

a 2 2

1

2

3

c 5 2 5 5

b 18 2 3 2 4

d 8 3 16

number and algebra • real numbers 16 Evaluate each of the following, without using a

26 Use the logarithm laws to simplify each of the

calculator. Show all working. a

3 4 16

1 × 814

1 6 × 16 2

b

1

(

)

1 2 2 2 125 3 − 27 3

following. a loga 16 + loga 3 - loga 2 b log x x x

4

c 4 loga x - log a x2 loga x2 or 2 loga x

17 Evaluate each of the following, giving your answer

as a fraction. 1 1 1 a 4-1 4 b 9-1 9 c 4-2 16 d 10-3 18 Find the value of each of the following, correct to 3 significant figures. a 12-1 0.0833 b 7-2 0.0204 -1 c (1.25) 0.800 d (0.2)-4 625 19 Write down the value of each of the following.  2

−1

 1

−1

1 1000

−1

b  

12

−1

 1 4 13 4 250 may be simplified to: d  3   

c   5  5 20 mC The expression A 25 10

✔ B

C 10 5

c 2 x+1 = 8 2

3 decimal places. a 2x = 25 4.644 c 9-x = 0.84 0.079

D 5 50

2 98 − 3 72 is equal to: −4 2

C −2 4

B -4

1 Answer the following. Explain how you reached

D 4 2

your answer. 3 a What is the hundred’s digit in 3 ? 9 b What is the one’s digit in 6 704? 6 ? 0 c What is the thousand’s digit in 91000 x 2 a Plot a graph of y = 4 by first producing a table of values. Label the y-intercept and the equation of any asymptotes. b Draw the line y = x on the same set of axes. c Use the property of inverse graphs to draw the graph of y = log4 x. Label any intercepts and the equation of any asymptotes. d Use a graphics calculator or graphing software to check your graphs. 3

22 mC When expressed in its simplest form,

equal to: x x 2

B

8x is 32

x3 4

x3 x x D 2 4 23 Find the value of the following, giving your answer in fraction form. C

 2 a    5

−1

1

22

−2

 2 1 b   2 4  3

a, b, c

24 Find the value of each of the following, leaving

your answer in fraction form. 1 a 2-1 2 b 3-2 c 4-3

1 64

25 Evaluate the following. a log12 18 + log12 8 2 b log4 60 - log4 15 1 c log9 98 8 d 2 log3 6 - log3 4 2

b 0.6x = 7 -3.809

problem solVing

3

✔ A

5 2

30 Solve for x in the following equations, correct to

5 10

21 mC When expressed in its simplest form, ✔ A

 1 x 27 Solve for x in the following, given that x > 0. 1 a log2 x = 9 512 b log5 x = -2 25 c logx 25 = 2 5 d logx 26 = 6 2 e log3 729 = x 6 f log7 1 = x 0 28 Solve for x in the following. a log5 4 + log5 x = log5 24 6 b log3 x - log3 5 = log3 7 35 29 Solve for x in the following equations. 1 1 1 x -2 -2 a 6 x = b 7 = 36 7 d 5 log x   -5  

 7 3  17 10 

1

a    3

loga 24

3 2

 1

1 9

−1

d   2  2

y=0

y = 4x

y

y = log4x

1 0

y=x

1

x=0

x

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Test Yourself Chapter 18 int-2873 Word search Chapter 18 int-2871 Crossword Chapter 18 int-2872

Chapter 18 real numbers

633

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(page 590) SkillSHEET 18.1 (doc-5354): Identifying surds SkillSHEET 18.2 (doc-5355): Simplifying surds SkillSHEET 18.3 (doc-5356): Adding and subtracting surds SkillSHEET 18.4 (doc-5357): Multiplying and dividing surds SkillSHEET 18.5 (doc-5358): Evaluating numbers in index form SkillSHEET 18.6 (doc-5359): Using the index laws

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• • • • • •

18A Number classification review Interactivity

• Classifying numbers (int-2792) (page 591) 18B Surds Digital doc

• SkillSHEET 18.1 (doc-5354): Identifying surds (page 597) 18C Oprations with surds

(pages 607–9) • SkillSHEET 18.2 (doc-5355): Simplifying surds • SkillSHEET 18.3 (doc-5356): Adding and subtracting surds • SkillSHEET 18.4 (doc-5357): Multiplying and dividing surds • SkillSHEET 18.7 (doc-5360): Rationalising denominators • SkillSHEET 18.8 (doc-5361): Conjugate pairs • SkillSHEET 18.9 (doc-5362): Applying the difference of two squares rule to surds • WorkSHEET 18.1 (doc-5363): Real numbers I Digital docs

634

maths quest 10 + 10a for the australian Curriculum

18D Fractional indices Digital doc

• WorkSHEET 18.2 (doc-5364): Real numbers II (page 614) 18E Negative indices Digital doc

• WorkSHEET 18.3 (doc-5365): Real numbers III (page 617) 18H Solving equations Digital doc

• WorkSHEET 18.4 (doc-6754): Real numbers IV (page 629) Chapter review Interactivities (page 633) • Test yourself Chapter 18 (int-2873): Take the end-ofchapter test to test your progress • Word search Chapter 18 (int-2871): an interactive word search involving words associated with this chapter • Crossword Chapter 18 (int-2872): an interactive crossword using the definitions associated with the chapter

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number AND algebra • Patterns and algebra

19

19A Polynomials 19B Adding, subtracting and multiplying polynomials 19C Long division of polynomials 19D Polynomial values 19E The remainder and factor theorems 19F Factorising polynomials 19G Solving polynomial equations What do you know ?

Polynomials

1 List what you know about polynomials. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of polynomials.

opening question

What does the graph of y = (x - 2)(x + 1)(x - 3) look like?   How can we solve cubic equations such as x3 + 3x2 - 17x - 2 = 0?

number and algebra • patterns and algebra

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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SkillSHEET 19.1 doc-5366

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636

Expanding the product of two linear factors 1 Expand each of the following. a (x + 1)(x - 3) x2 - 2x - 3 b (x + 6)2 x2 + 12x + 36 c (3x - 4)(2x + 5) 6x2 + 7x - 20 Substitution into quadratic equations 2 Find the value of y for these parabolas when x = 2. a y = x2 - 4 0 b y = x2 + 9x - 4 18 c y = -3x2 + 5x + 1 -1 Factorising quadratic trinomials 3 Factorise each of the following. a x2 + x - 6 (x - 2)(x + 3) b x2 - 5x - 6 (x + 1)(x - 6) c 2x2 - 5x - 3 (2x + 1)(x - 3) Factorising difference of two squares expressions 4 Factorise each of the following. a x2 - 4 (x + 2)(x - 2) b 25 - x2 (5 + x)(5 - x) c 3x2 - 147 3(x + 7)(x - 7) Solving quadratic equations 5 Solve each of the following. a (x + 1)(x - 3) = 0 x = -1 or 3 b x2 - 2x - 15 = 0 x = -3 or 5 3 c 2x2 - 7x + 6 = 0 x = 2 or 2

maths quest 10 + 10a for the australian Curriculum

number and algebra • patterns and algebra

19a

polynomials ■

■

■

■

A polynomial in x, sometimes denoted P(x), is an expression containing only non-negative integer powers of x. The degree of a polynomial in x is the highest power of x in the expression. For example: 3x + 1 is a polynomial of degree 1, or linear polynomial. is a polynomial of degree 2, or quadratic polynomial. x2 + 4x - 7 x -5x3 + is a polynomial of degree 3, or cubic polynomial. 2 10 is a polynomial of degree 0 (think of 10 as 10x0). Expressions containing a term similar to any of the following terms are not polynomials: 1 x , 2x, sin x, etc. , x-2, x For example, the following are not polynomials. 2 x2 + sin x + 1 -5x4 + x3 - 2 x 3x2 - 4x + x In the expression P(x) = 6x3 + 13x2 - x + 1 x is the variable. 6 is the coefficient of x3. 13 is the coefficient of x2. -1 is the coefficient of x. 6x 3, 13x 2, -x and +1 are all terms. The constant term is +1. The degree of the polynomial is 3.

■ ■ ■

The leading term is 6x3 because it is the term that contains the highest power of x. The leading coefficient is 6. Any polynomial with a leading coefficient of 1 is called monic. An example of where polynomials are useful is shown below.

The surface area, S, of a plant hothouse of length L and height x can be approximated by the polynomial S(x) = p x2 + Lp x – 4.

Chapter 19 polynomials

637

number AND algebra • Patterns and algebra

remember

1. A polynomial in x, sometimes denoted P(x), is an expression containing only nonnegative integer powers of x. 2. The degree of a polynomial in x is the highest power of x in the expression. Exercise

19A

Polynomials Fluency 1 State the degree of each of the following polynomials. b 65 + 2x7 7 a x3 - 9x2 + 19x + 7 3 d x6 - 3x5 + 2x4 + 6x + 1

2 a x b x c x d x e y f u g e h g i f

g 18 −

e5 6

6

e y8 + 7y3 - 5

c 3x2 - 8 + 2x

8

h 2g - 3 1

5 

2

4

f

1 5 u u − + 2u − 6 2 3

i

1.5f  6 - 800f

6

2 State the variable for each polynomial in question 1. 3 Which polynomials in question 1 are: a linear? Polynomial 1h b quadratic? Polynomial 1c c cubic? Polynomial 1a d monic? Polynomials 1a, 1d and 1e 4 State whether each of the following is a polynomial (P) or not (N). a 7 x + 6 x 2 +

5 x

b 33 - 4p

N

d 3 x 4 − 2 x 3 − 3 x − 4

N

e k-2 + k - 3k3 + 7

4 c6 − 3c3 + 1 h 2x - 8x + 1 N P 2 5 Consider the polynomial P(x) = -2x3 + 4x2 + 3x + 5. a What is the degree of the polynomial? 3 b What is the variable? x c What is the coefficient of x2? 4 d What is the value of the constant term? 5 e Which term has a coefficient of 3? 3x f Which is the leading term? -2x3 6 Consider the polynomial P(w) = 6w7 + 7w6 - 9. a What is the degree of the polynomial? 7 b What is the variable? w c What is the coefficient of w6? 7 d What is the coefficient of w? 0 e What is the value of the constant term? -9 f Which term has a coefficient of 6? 6w7 7 Consider the polynomial f (x) = 4 - x2 + x4. a What is the degree of the polynomial? 4 b What is the coefficient of x4? 1 c What is the leading term? x4 d What is the leading coefficient? 1 g

638

Maths Quest 10 + 10A for the Australian Curriculum

x2 +x 9

P

f 5r − r 9 +

1 3

c

P N

i

sin x + x2

P N

5

number AND algebra • Patterns and algebra Understanding 8 A sports scientist determines the following equation for the velocity of a breaststroke swimmer

during one complete stroke: v(t) = 63.876t6 – 247.65t 5 + 360.39t 4 – 219.41t 3 + 53.816t 2 + 0.4746t a What is the degree of the polynomial? 6 b What is the variable? t c How many terms are there? 6 d Use a graphics calculator or graphing software to draw the Check with your teacher. graph of this polynomial. e Match what happens during one complete stroke with points on the graph. reflection 

 

How can you tell what the degree of a polynomial is?

19B

Adding, subtracting and multiplying polynomials ■■

To add or subtract polynomials, we simply add or subtract any like terms in the expressions.

Worked Example 1

Simplify each of the following. a (5x3 + 3x2 - 2x - 1) + (x4 + 5x2 - 4) b (5x3 + 3x2 - 2x - 1) - (x4 + 5x2 - 4) Think a

b

Write a (5x3 + 3x2 - 2x - 1) + (x4 + 5x2 - 4)

1

Write the expression.

2

Remove any grouping symbols, watching any signs.

= 5x3 + 3x2 - 2x - 1 + x4 + 5x2 - 4

3

Identify any like terms and change the order.

= x4 + 5x3 + 3x2 + 5x2 - 2x - 1 - 4

4

Simplify by collecting like terms.

= x4 + 5x3 + 8x2 - 2x - 5

1

Write the expression.

2

Remove any grouping symbols, watching any signs.

= 5x3 + 3x2 - 2x - 1 - x4 - 5x2 + 4

3

Identify any like terms and change the order.

= -x4 + 5x3 + 3x2 - 5x2 - 2x - 1 + 4

4

Simplify by collecting like terms.

= -x4 + 5x3 - 2x2 - 2x + 3

■■

b (5x3 + 3x2 - 2x - 1) - (x4 + 5x2 - 4)

If we expand linear factors, for example, (x + 1)(x + 2)(x - 7), we may also get a polynomial as the following worked example shows. Chapter 19 Polynomials

639

number and algebra • patterns and algebra

Worked example 2

Expand and simplify: a x(x + 2)(x - 3)

b (x - 1)(x + 5)(x + 2)

think a

b

Write a x(x + 2)(x - 3)

1

Write the expression.

2

Expand the last two linear factors.

= x(x2 - 3x + 2x - 6) = x(x2 - x - 6)

3

Multiply the expression in the grouping symbols by x.

= x3 - x2 - 6x

1

Write the expression.

2

Expand the last two linear factors.

= (x - 1)(x2 + 2x + 5x + 10)

3

Multiply the expression in the second pair of grouping symbols by x and then by -1.

= (x - 1)(x2 + 7x + 10) = x3 + 7x2 + 10x - x2 - 7x - 10

4

Collect like terms.

= x3 + 6x2 + 3x - 10

b (x - 1)(x + 5)(x + 2)

remember

1. To add or subtract polynomials, add or subtract any like terms in the expression. 2. When expanding linear factors: (a) expand two factors first, and then multiply by the remaining linear factors, one at a time (b) collect like terms at each stage (c) (x + 2)3 may be written as (x + 2)(x + 2)(x + 2). exerCise

19b

adding, subtracting and multiplying polynomials FluenCy 1 We1a Simplify each of the following. a (x4 + x3 - x2 + 4) + (x3 - 14) x4 + 2x3 - x2 - 10 b (x6 + x4 - 3x3 + 6x2) + (x4 + 3x2 + 5) x6 + 2x4 - 3x3 + 9x2 + 5 c (x3 + x2 + 2x - 4) + (4x3 - 6x2 + 5x - 9) 5x3 - 5x2 + 7x - 13 d (2x4 - 3x3 + 7x2 + 9) + (6x3 + 5x2 - 4x + 5) 2x4 + 3x3 + 12x2 - 4x + 14 e (15x4 - 3x2 + 4x - 7) + (x5 - 2x4 + 3x2 - 4x - 3) x5 + 13x4 - 10

eBook plus

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SkillSHEET 19.1 doc-5366

640

2 We1b Simplify each of the following. a (x4 + x3 + 4x2 + 5x + 5) - (x3 + 2x2 + 3x + 1) x4 + 2x2 + 2x + 4 b (x6 + x3 + 1) - (x5 - x2 - 1) x6 - x5 + x3 + x2 + 2 c (5x7 + 6x5 - 4x3 + 8x2 + 5x - 3) - (6x5 + 8x2 - 3) 5x7 - 4x3 + 5x d (10x4 - 5x2 + 16x + 11) - (2x2 - 4x + 6) 10x4 - 7x2 + 20x + 5 e (6x3 + 5x2 - 7x + 12) - (4x3 - x2 + 3x - 3) 2x3 + 6x2 - 10x + 15 3 We2a Expand and simplify each of the following. a x(x + 6)(x + 1) x3 + 7x2 + 6x b x(x - 9)(x + 2) x3 - 7x2 - 18x 3 2 c x(x - 3)(x + 11) x + 8x - 33x d 2x(x + 2)(x + 3) 2x3 + 10x2 + 12x 3 e -3x(x - 4)(x + 4) 48x - 3x f 5x(x + 8)(x + 2) 5x3 + 50x2 + 80x

maths quest 10 + 10a for the australian Curriculum

a x3 + 13x2 + 26x - 112 c 4x4 + 3x3 - 37x2 - 27x + 9 e -6x3 - 71x2 - 198x + 35 g 54x3 + 117x2 - 72x i 20x4 – 39x3 - 50x2 + 123x - 54

b 3x3 + 26x2 + 51x - 20 d 10x3 - 49x2 + 27x + 36 f 21x4 - 54x3 – 144x2 + 96x h 24x3 - 148x2 + 154x + 245 j 4x3 + 42x2 + 146x + 168

number and algebra • patterns and algebra



19C

eBook plus

Interactivity Long division of polynomials

int-2793

g x2(x + 4) x3 + 4x2 i (5x)(-6x)(x + 9) -30x3 - 270x2

h -2x2(7 - x) j -7x(x + 4)2

2x3 - 14x2 -7x3 - 56x2 - 112x

4 We2b Expand and simplify each of the following. a (x + 7)(x + 2)(x + 3) x3 + 12x2 + 41x + 42 b c (x - 1)(x - 4)(x + 8) x3 + 3x2 - 36x + 32 d e (x + 6)(x - 1)(x + 1) x3 + 6x2 - x - 6 f g (x + 11)(x + 5)(x - 12) x3 + 4x2 - 137x - 660 h i (x + 2)(x - 7)2 x3 - 12x2 + 21x + 98 j

(x - 2)(x + 4)(x - 5) x3 - 3x2 - 18x + 40 (x - 1)(x - 2)(x - 3) x3 - 6x2 + 11x - 6 (x - 7)(x + 7)(x + 5) x3 + 5x2 - 49x - 245 (x + 5)(x - 1)2 x3 + 3x2 - 9x + 5 (x + 1)(x - 1)(x + 1) x3 + x2 - x - 1

5 Expand and simplify each of the following. a (x - 2)(x + 7)(x + 8) c (4x - 1)(x + 3)(x - 3)(x + 1) e (1 - 6x)(x + 7)(x + 5) g -9x(1 - 2x)(3x + 8) i (3 - 4x)(2 - x)(5x + 9)(x – 1)

(x + 5)(3x - 1)(x + 4) (5x + 3)(2x - 3)(x - 4) 3x(7x - 4)(x - 4)(x + 2) (6x + 5)(2x - 7)2 2(7 + 2x)(x + 3)(x + 4)

6 Expand and simplify each of the following. a (x + 2)3 b (x + 5)3 3 c (x - 1) d (x - 3)4 3 e (2x - 6) f (3x + 4)4

b d f h j

reFleCtion 

 

How do you add or subtract polynomials?

long division of polynomials ■

■

The reverse of expanding is factorising (expressing a polynomial as a product of its linear factors). Before learning how to factorise, you must be familiar with long division of polynomials. You may remember in earlier levels doing long division questions. 3 2 a x + 6x + 12x + 8

Consider 745 ó 3, or 3) 745 . The process used is as follows. b x3 + 15x2 + 75x + 125

3 into 7 goes 2 times. Write 2 at the top.

2 3) 745

2ì3=6 Write the 6.

2 3) 745 6

Subtract to get 1.

c x3 - 3x2 + 3x - 1 d x4 – 12x3 + 54x2 - 108x + 81 e 8x3 - 72x2 + 216x - 216 f 81x4 + 432x3 + 864x2 + 768x + 256

2 3) 745 6 1

Bring down the 4 to form 14.

2 3) 745 6 14

3 into 14 goes 4. Write 4 at the top.

24 3) 745 6 14

4 ì 3 = 12

24 3) 745 6 14 12

Write the 12.



Chapter 19 polynomials

641

number AND algebra • Patterns and algebra

24 3) 745 6 14 12  2

Subtract to get 2.

24 3) 745 6 14 12   25

Bring down the 5 to form 25. 3 into 25 goes 8. Write 8 at the top.

248 3) 745 6 14 12   25

8 ì 3 = 24

248 3) 745 6 14 12   25   24

Write the 24. Divisor

Subtract to get 1. Answer: 248 remainder 1 ■■

Quotient Dividend

248 3) 745 6 14 12   25   24    1

Remainder

The same process can be used to divide polynomials by polynomial factors. Consider (x3 + 2x2 - 13x + 10) ó (x - 3)    or

x - 3) x3 + 2x2 - 13x + 10

x into x3 goes x2 times (consider only the leading terms). Write x2 at the top.

x - 3)

x3

x2 + 2x2 - 13x + 10

x2 ì (x - 3) = x3 - 3x2 Write the x3 - 3x2.

x2 x - 3) x3 + 2x2 - 13x + 10 x3 - 3x2

Subtract. (x3 - x3 = 0, 2x2 - -3x2 = 5x2)

x2 x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 x3

Note: Subtracting a negative is the same as changing the sign and adding. 642

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • Patterns and algebra

Bring down the -13x.

x into

5x2

goes 5x. Write +5x at the top.

x2 x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x x3

x2 +   5x x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x x3

Write the 5x2 - 15x.

x2 +   5x x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x 5x2 - 15x

Subtract. Note: 5x2 - 5x2 = 0, -13x - -15x = +2x

x2 +   5x x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x 5x2 - 15x 2x

5x ì (x - 3) =

5x2

- 15x

x3

x3

Bring down the 10.

x into 2x goes 2. Write +2 at the top.

2 ì (x - 3) = 2x - 6

Write the 2x - 6. Subtract to get 16.

Answer: x2 + 5x + 2 remainder 16

x2 +   5x x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x 5x2 - 15x 2x + 10 x3

x2 +   5x +   2 x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x 5x2 - 15x 2x + 10 x3

x2 +   5x +   2 x - 3) + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x 5x2 - 15x 2x + 10 2x - 6 x3

x2 +   5x +   2 x - 3) x3 + 2x2 - 13x + 10 x3 - 3x2 5x2 - 13x 5x2 - 15x 2x + 10 2x - 6 16

Quotient

Remainder

Chapter 19 Polynomials

643

number AND algebra • Patterns and algebra

Worked Example 3

Perform the following long divisions and state the quotient and remainder. a  (x3 + 3x2 + x + 9) ó (x + 2)   b  (x3 - 4x2 - 7x - 5) ó (x - 1)   c  (2x3 + 6x2 - 3x + 2) ó (x - 6) Think a

b

1

Write the question in long division format.

2

Perform the long division process.

3

Write the quotient and remainder.

1

Write the question in long division format.

2

Perform the long division process.

3

c

Write

Quotient is

Write the question in long division format.

2

Perform the long division process.

Write the quotient and remainder.

Q

x2 +

R

x - 1; remainder is 11.

x2 - 3x - 10

Q

x3 - x2 -3x2 - 7x -3x2 + 3x -10x - 5 -10x + 10 -15

R

b x - 1) x3 - 4x2 - 7x - 5

Quotient is x2 - 3x - 10; remainder is –15.

Write the quotient and remainder.

1

3

x2 + x - 1 a x + 2) + 3x2 + x + 9 x3 + 2x2   x2 + x   x2 + 2x -x + 9 -x - 2 11 x3

c

2x2 +    18x + 105

Q

2x3 - 12x2 18x2 -    3x 18x2 - 108x 105x +    2 105x - 630 632

R

x - 6) 2x3 + 6x2 -     3x +    2

Quotient is 2x2 + 18x + 105; remainder is 632.

Worked Example 4

State the quotient and remainder for (x3 - 7x + 1) ó (x + 5). Think 1

Write the question in long division format. Note that there is no x2 term in this equation. Include 0x2 as a ‘placeholder’.

2

Perform the long division process.

3

644

Write the quotient and remainder.

Maths Quest 10 + 10A for the Australian Curriculum

Write

x2 - 5x + 18 x + 5) + 0x2 - 7x + 1 x3 + 5x2 -5x2 - 7x -5x2 - 25x 18x + 1 18x + 90 -89

Q

x3

R

Quotient is  x2 - 5x + 18; remainder is –89.

number AND algebra • Patterns and algebra

Worked Example 5

Find the quotient and the remainder when x4 - 3x3 + 2x2 - 8 is divided by the linear expression x + 2. Think 1

2

3

4

Write

Set out the long division with each polynomial in descending powers of x. If one of the powers of x is missing, include it with 0 as the coefficient. Divide x into

x4

and write the result above.

Multiply the result result underneath.

x3

by x + 2 and write the

Subtract and bring down the remaining terms to complete the expression.

-5x3

and write the result above.

x + 2) x4 - 3x3 + 2x2 + 0x - 8

x + 2)

x4

x3 - 3x3 + 2x2 + 0x - 8

x3 x + 2) - 3x3 + 2x2 + 0x - 8 x4 + 2x3 x4

x3 - 5x2 x + 2) - 3x3 + 2x2 + 0x - 8 -(x4 + 2x3) -5x3 + 2x2 + 0x - 8 x4

x3 - 5x2 + 12x - 24

x + 2) x4 - 3x3 + 2x2 + 0x - 8

5

Divide x into

6

Continue this process to complete the long division.

7

The polynomial x3 - 5x2 + 12x - 24, at the top, is the quotient.

The quotient is x3 - 5x2 + 12x - 24.

8

The result of the final subtraction, 40, is the remainder.

The remainder is 40.

-(x4 + 2x3) -5x3 + 2x2 + 0x - 8 -(-5x3 - 10x2) 12x2 + 0x - 8 -(12x2 + 24x) -24x - 8 -(-24x - 48) 40

remember

Long division of polynomials is similar to long division with numbers. The highest power term is the main one considered at each stage. The key steps are: 1. determine how many times the first term of the divisor goes into the first term of dividend 2. multiply and write the result underneath 3. subtract. (If necessary, change the sign and add.) 4. bring down the next term 5. repeat the process until no pronumerals remain to be divided 6. state the quotient and remainder. Chapter 19 Polynomials

645

number AND algebra • Patterns and algebra

Exercise

19C

Long division of polynomials fluency 1   WE 3a  Perform the following long divisions and state the quotient and remainder. a (x3 + 4x2 + 4x + 9) ó (x + 2) x2 + 2x, 9 b (x3 + 2x2 + 4x + 1) ó (x + 1) x2 + x + 3, -2 c (x3 + 6x2 + 3x + 1) ó (x + 3) x2 + 3x - 6, 19 d (x3 + 3x2 + x + 3) ó (x + 4) x2 - x + 5, -17 e (x3 + 4x2 + 3x + 4) ó (x + 2) x2 + 2x - 1, 6 f (x3 + 6x2 + 2x + 2) ó (x + 2) x2 + 4x - 6, 14 g (x3 + x2 + x + 3) ó (x + 1) x2 + 1, 2 h (x3 + 8x2 + 5x + 4) ó (x + 8) x2 + 5, -36 i (x3 + x2 + 4x + 1) ó (x + 2) x2 - x + 6, -11 j (x3 + 9x2 + 3x + 2) ó (x + 5) x2 + 4x - 17, 87 2   WE 3b  State the quotient and remainder for each of the following. a (x3 + 2x2 - 5x - 9) ó (x - 2) x2 + 4x + 3, -3 b (x3 + x2 + x + 9) ó (x - 3) x2 + 4x + 13, 48 c (x3 + x2 - 9x - 5) ó (x - 2) x2 + 3x - 3, -11 d (x3 - 4x2 + 10x - 2) ó (x - 1) x2 - 3x + 7, 5 e (x3 - 5x2 + 3x - 8) ó (x - 3) x2 - 2x - 3, -17 f (x3 - 7x2 + 9x - 7) ó (x - 1) x2 - 6x + 3, -4 g (x3 + 9x2 + 2x - 1) ó (x - 5) x2 + 14x + 72, 359 h (x3 + 4x2 - 5x - 4) ó (x - 4) x2 + 8x + 27, 104 3   WE 3c  Divide the first polynomial by the second and state the quotient and remainder. a 3x3 - x2 + 6x + 5, x + 2 3x2 - 7x + 20, -35 b 4x3 - 4x2 + 10x - 4, x + 1 4x2 - 8x + 18, -22 c 2x3 - 7x2 + 9x + 1, x - 2 2x2 - 3x + 3, 7 d 2x3 + 8x2 - 9x - 1, x + 4 2x2 - 9, 35 3 2 2 e 4x - 10x - 9x + 8, x - 3 4x + 2x - 3, -1 f 3x3 + 16x2 + 4x - 7, x + 5 3x2 + x - 1, -2 4 Divide the first polynomial by the second and state the quotient and remainder. a 6x3 - 7x2 + 4x + 4, 2x - 1 3x2 - 2x + 1, 5 b 6x3 + 23x2 + 2x - 31, 3x + 4 2x2 + 5x - 6, -7 3 2 2 c 8x + 6x - 39x - 13, 2x + 5 4x - 7x - 2, -3 d 2x3 - 15x2 + 34x - 13, 2x - 7 x2 - 4x + 3, 8 e 3x3 + 5x2 - 16x - 23, 3x + 2 x2 + x - 6, -11 f 9x3 - 6x2 - 5x + 9, 3x - 4 3x2 + 2x + 1, 13 5 State the quotient and remainder for each of the following. a

− x 3 − 6 x 2 − 7 x − 16 x +1

c

−2 x 3 + 9 x 2 + 17 x + 15 2x + 1

-x2 - 5x - 2, -14

b

−3 x 3 + 7 x 2 + 10 x − 15 x−3

-x2 + 5x + 6, 9

d

4 x 3 − 20 x 2 + 23 x − 2 −2 x + 3

-3x2 - 2x + 4, -3

-2x2 + 7x - 1, 1

6   WE 4  State the quotient and remainder for each of the following. a (x3 - 3x + 1) ó (x + 1) x2 - x - 2, 3 b (x3 + 2x2 - 7) ó (x + 2) x2, -7 3 2 2 c (x - 5x + 2x) ó (x - 4) x - x - 2, -8 d (-x3 - 7x + 8) ó (x - 1) -x2 - x - 8, 0 2 e (5x + 13x + 1) ó (x + 3) 5x - 2, 7 f (2x3 + 8x2 - 4) ó (x + 5) 2x2 - 2x + 10, -54 3 2 g (-2x - x + 2) ó (x - 2) -2x - 4x - 9, -16 h (-4x3 + 6x2 + 2x) ó (2x + 1) -2x2 + 4x - 1, 1 7   WE 5  Find the quotient and the remainder when each polynomial is divided by the linear

7

7

20

20

x3 - 3 x2 + 9 x + 3 27 , –3 27

646

expression given. x4 + x3 + 3x2 - 7x, x - 1 x3 + 2x2 + 5x - 2, - 2 x4 - 13x2 + 36, x - 2 x3 + 2x2 - 9x - 18, 0 x5 - 3x3 + 4x + 3, x + 3 x4 - 3x3 + 6x2 - 18x + 58, -171 2x6 - x4 + x3 + 6x2 - 5x, x + 2 6x4 - x3 + 2x2 - 4x, x - 3 6x3 + 17x2 + 53x + 155, 465 3x4 - 6x3 + 12x, 3x + 1

a b c d e f

Maths Quest 10 + 10A for the Australian Curriculum

reflection 

 

Can you think of an alternative way to divide polynomials?

2x5 - 4x4 + 7x3 - 13x2 + 32x - 69, 138

number AND algebra • Patterns and algebra

19D

Polynomial values ■■ ■■

Consider the polynomial P(x) = x3 - 5x2 + x + 1. The value of the polynomial when x = 3 is denoted by P(3) and is found by substituting x = 3 into the equation in place of x. That is: P(3) = (3)3 - 5(3)2 + (3) + 1 P(3) = 27 - 5(9) + 3 + 1 P(3) = 27 - 45 + 4 P(3) = -14

Worked Example 6

If P(x) = 2x3 + x2 - 3x - 4, find: a  P(1)     b  P(-2)     c  P(a)     d  P(2b)     e  P(x + 1). Think a

b

c

d

e

Write

1

Write the expression.

2

Replace each occurrence of x with 1.

3

Simplify.

1

Write the expression.

2

Replace each occurrence of x with -2.

3

Simplify.

1

Write the expression.

2

Replace each occurrence of x with a.

3

No further simplification is possible, so stop here.

1

Write the expression.

2

Replace each occurrence of x with 2b.

3

Simplify.

1

Write the expression.

2

Replace each occurrence of x with (x + 1).

3

Expand the right-hand side and collect like terms.

a

P(x) = 2x3 + x2 - 3x - 4 P(1) = 2(1)3 + (1)2 - 3(1) - 4 =2+1-3-4 = -4

b

P(x) = 2x3 + x2 - 3x - 4 P(-2) = 2(-2)3 + (-2)2 - 3(-2) - 4 = 2(-8) + (4) + 6 - 4 = -16 + 4 + 6 - 4 = -10

c

P(x) = 2x3 + x2 - 3x - 4 P(a) = 2a3 + a2 - 3a - 4

d

P(x) = 2x3 + x2 - 3x - 4 P(2b) = 2(2b)3 + (2b)2 - 3(2b) - 4 = 2(8b3) + 4b2 – 6b + 4 = 16b3 + 4b2 – 6b + 4

e

P(x) = 2x3 + x2 - 3x - 4 P(x + 1) = 2(x + 1)3 + (x + 1)2 - 3(x + 1) - 4 = 2(x + 1)(x + 1)(x + 1) + (x + 1)(x + 1) - 3(x + 1) - 4 = 2(x + 1)(x2 + 2x + 1) + x2 + 2x + 1 - 3x - 3 - 4 = 2(x3 + 2x2 + x + x2 + 2x + 1) + x2 - x - 6 = 2(x3 + 3x2 + 3x + 1) + x2 - x - 6 = 2x3 + 6x2 + 6x + 2 + x2 - x - 6 = 2x3 + 7x2 + 5x - 4

Chapter 19 Polynomials

647

number and algebra • patterns and algebra

remember

P(a) means the value of P(x) when x is replaced by a and the polynomial is evaluated. exerCise

19d

polynomial values FluenCy

eBook plus

Digital doc

SkillSHEET 19.2 doc-5367

1 We6 If P(x) = 2x3 - 3x2 + 2x + 10, find the following. a P(0) 10 b P(1) 11 c P(2) 18 d P(3) 43 e P(-1) 3 f P(-2) -22 g P(-3) -77 h P(a) 2a3 - 3a2 + 2a + 10 3 2 i P(2b) 16b - 12b + 4b + 10 j P(x + 2) 2x3 + 9x2 + 14x + 18 3 2 k P(x - 3) 2x - 21x + 74x - 77 l P(-4y) –128y3 - 48y2 - 8y + 10 2 Copy the following table.

Column Column Column Column Column Column Column Column Column 1 2 3 4 5 6 7 8 9 Rem Rem Rem Rem when when when when divided divided divided divided by by by by (x - 1) (x - 2) (x + 1) (x + 2)

P(x)

P(1)

P(2)

P(-1)

P(-2)

a

4

15

0

-5

4

15

0

-5

b

10

28

-2

-8

10

28

-2

-8

c

3

11

-7

-21

3

11

-7

-21

d

-7

-19

5

-7

-7

-19

5

-7

Complete columns 2 to 5 of the table for each of the following polynomials. a P(x) = x3 + x2 + x + 1 b P(x) = x3 + 2x2 + 5x + 2 c P(x) = x3 - x2 + 4x - 1 d P(x) = x3 - 4x2 - 7x + 3 understanding 3 Find the remainder when each polynomial in question 2 is divided by (x - 1) and complete

column 6 of the table. 4 Find the remainder when each polynomial in question 2 is divided by (x - 2) and complete column 7 of the table. 5 Find the remainder when each polynomial in question 2 is divided by (x + 1) and complete column 8 of the table. 6 Find the remainder when each polynomial in question 2 is divided by (x + 2) and complete column 9 of the table. 648

maths quest 10 + 10a for the australian Curriculum

number and algebra • patterns and algebra 7 Copy and complete: a A quick way of finding the remainder when P(x) is divided by (x + 8) is to calculate P(-8) __________. b A quick way of finding the remainder

when P(x) is divided by (x - 7) is to P(7) calculate __________. c A quick way of finding the remainder when P(x) is divided by (x - a) is to P(a) calculate __________.

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Digital doc

WorkSHEET 19.1 doc-5375

19e

reFleCtion 

 

Is there a quick way to find a remainder when dividing polynomials?

the remainder and factor theorems the remainder theorem ■

■

In the previous exercise, you may have noticed that: the remainder when P(x) is divided by (x - a) is equal to P(a). That is, R = P(a). This is called the remainder theorem. We could have derived this result as follows. If 13 is divided by 4, the quotient is 3, and the remainder is 1. That is: 13 ó 4 = 3 +

■

■

■

1 4

and

13 = 4 ì 3 + 1 Similarly, if P(x) = x3 + x2 + x + 1 is divided by (x - 2), the quotient is x2 + 3x + 7 and the remainder is 15. That is: 15 (x3 + x2 + x + 1) ó (x - 2) = x2 + 3x + 7 + and x−2 (x3 + x2 + x + 1) = (x2 + 3x + 7)(x - 2) + 15 In general, if P(x) is divided by (x - a), the quotient is Q(x) and the remainder is R, we can write: R P (x) ó (x - a) = Q (x) + and ( x − a) P (x) = (x - a)Q (x) + R Substituting x = a into this last expression yields P(a) = (a - a)Q(x) + R = 0 ì Q(x) + R =R (as before).

the factor theorem ■ ■

■

■

The remainder when 12 is divided by 4 is zero, since 4 is a factor of 12. Similarly, if the remainder (R) when P(x) is divided by (x - a) is zero, then (x - a) must be a factor of P(x). Since R = P(a), all we need to do is to find a value of a that makes P(a) = 0, and we can say that (x - a) is a factor. If P(a) = 0, then (x - a) is a factor of P(x). This is called the factor theorem. Imagine P(x) could be factorised as follows: P(x) = (x - a)Q(x), where Q(x) is ‘the other’ factor of P(x). Then we have: P(a) = (a - a)Q(a) = 0 ì Q(a) =0 So if P(a) = 0, (x - a) is a factor. Chapter 19 polynomials

649

number AND algebra • Patterns and algebra

Worked Example 7

Without actually dividing, find the remainder when x3 - 7x2 - 2 x + 4 is divided by: a  x - 3           b  x + 6. Think a

b

Write

1

Name the polynomial.

2

The remainder when P(x) is divided by (x - 3) is equal to P(3).

The remainder when P(x) is divided by (x + 6) is equal to P(-6).

a Let P(x) = x3 - 7x2 - 2x + 4

R = P(3) = 33 - 7(3)2 - 2(3) + 4 = 27 - 7(9) - 6 + 4 = 27 - 63 - 6 - 4 = -46 b

R = P(-6) = (-6)3 - 7(-6)2 - 2(-6) + 4 = -216 - 7(36) + 12 + 4 = -216 - 252 + 12 + 4 = -452

Worked Example 8

The remainder when x3 + kx2 + x - 2 is divided by (x - 2) is equal to 20. Find the value of k. Think

Write

1

Name the polynomial.

Let P(x) = x3 + kx2 + x - 2.

2

The remainder when P(x) is divided by (x - 2) is equal to P(2).

R = P(2) = 23 + k(2)2 + 2 - 2 = 8 + 4k

3

We are given R = 20. Put 8 + 4k = 20.

Since R = 20, 8 + 4k = 20

4

Solve for k.

4k = 12 k=3

remember

1. The remainder when P(x) is divided by (x – a) is equal to P(a). That is: R = P(a). This is known as the remainder theorem. 2. If P(a) = 0, then (x – a) is a factor of P(x). This is known as the factor theorem. Exercise

19e

The remainder and factor theorems fluency

650

1   WE 7  Without actually dividing, find the remainder when x3 + 3x2 - 10x - 24 is divided by: a x - 1 -30 b x + 2 0 c x - 3 0 d x + 5 -24 e x - 0 -24 f x - k g x + n h x + 3c. -27c3 + 27c2 + 30c - 24 k3 + 3k2 - 10k - 24 -n3 + 3n2 + 10n - 24 Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • Patterns and algebra 2 Find the remainder when the first polynomial is divided by the second without performing long

division. a x3 + 2x2 + 3x + 4, x - 3 58 c x3 + 3x2 - 3x + 1, x + 2 11 e 2x3 + 3x2 + 6x + 3, x + 5 -202 g x3 + x2 + 8, x - 5 158 i -x3 + 8, x + 3 35

b d f h j

x3 - 4x2 + 2x - 1, x + 1 -8 x3 - x2 - 4x - 5, x - 1 -9 -3x3 - 2x2 + x + 6, x + 1 6 x3 - 3x2 - 2, x - 2 -6 x3 + 2x2, x - 7 441

Understanding 3   WE 8  a The remainder when x3 + k x + 1 is divided by (x + 2) is -19. Find the value of k. 6 b The remainder when x3 + 2x2 + m x + 5 is divided by (x - 2) is 27. Find the value of m. 3 c The remainder when x3 - 3x2 + 2x + n is divided by (x - 1) is 1. Find the value of n. 1 d The remainder when ax3 + 4x2 - 2x + 1 is divided by (x - 3) is -23. Find the value of a. -2 e The remainder when x3 - bx2 - 2x + 1 is divided by (x + 1) is 0. Find the value of b. 2 f The remainder when -4x2 + 2x + 7 is divided by (x - c) is -5. Find a possible whole number value of c. 2 g The remainder when x2 - 3x + 1 is divided by (x + d ) is 11. Find the possible values of d. -5, 2 h The remainder when x3 + ax2 + bx + 1 is divided by (x - 5) is -14. When the cubic polynomial is divided by (x + 1), the remainder is -2. Find a and b. a = -5, b = -3 4   MC  Note: There may be more than one correct answer. a When x3 + 2x2 - 5x - 5 is divided by (x + 2), the remainder is: A -5 B -2 C 2 b Which of the following is a factor of 2x3 + 15x2 + 22x - 15? ✔ C (x + 3) A (x - 1) B (x - 2) c When x3 - 13x2 + 48x - 36 is divided by (x - 1), the remainder is: A -3 B -2 C -1 d Which of the following is a factor of x3 - 5x2 - 22x + 56? ✔ A (x - 2) B (x + 2) ✔ C (x - 7)

✔ D

5

✔ D

(x + 5)

✔ D

0

✔ D

(x + 4)

5 Find one factor of each of the following cubic polynomials. a x3 - 3x2 + 3x - 1 (x - 1) b x3 - 7x2 + 16x - 12 (x - 3) or (x - 2) c x3 + x2 - 8x - 12 (x - 3) or (x + 2) d x3 + 3x2 - 34x - 120 (x - 6) or (x + 4) or (x + 5) Reasoning 6 Prove that each of the following is a linear factor of x3 + 4x2 - 11x - 30 by substituting values into the cubic function: (x + 2), (x - 3), (x + 5). Show P(-2) = 0, P(3) = 0 and P(-5) = 0. 7 Avoid division and show that the first polynomial is exactly divisible by the second (that is, the

second polynomial is a factor of the first). a x3 + 5x2 + 2x - 8, x - 1 Show P(1) = 0 b x3 - 7x2 - x + 7, x - 7 Show P(7) = 0 c x3 - 7x2 + 4x + 12, x - 2 Show P(2) = 0 d x3 + 2x2 - 9x - 18, x + 2 Show P(–2) = 0 e x3 + 3x2 - 9x - 27, x + 3 Show P(–3) = 0 f -x3 + x2 + 9x - 9, x - 1 Show P(1) = 0 g -2x3 + 9x2 - x - 12, x - 4 Show P(4) = 0 h 3x3 + 22x2 + 37x + 10, x + 5 Show P(–5) = 0

19f

reflection 

 

How are the remainder and factor theorems related?

Factorising polynomials Using long division ■■

Once one factor of a polynomial has been found (using the factor theorem as in the previous section), long division may be used to find other factors. In the case of a cubic polynomial, one — possibly two — other factors may be found. Chapter 19 Polynomials

651

number AND algebra • Patterns and algebra

Worked Example 9

Use long division to factorise the following. a  x3 - 5x2 - 2 x + 24     b  x3 - 19x + 30     c  -2 x3 - 8x2 + 6x + 4 Think a

Write

1

Name the polynomial.

2

Look for a value of x such that P ( x) = 0. For cubics containing a single x3, try a factor of the constant term (24 in this case). Try P (1).

a P(x) = x3 - 5x2 - 2x + 24

P(1) = 13 - 5 ì 12 - 2 ì 1 + 24 = 1 - 5 - 2 + 24 = 18 ò0 P(2) = 23 - 5 ì 22 - 2 ì 2 + 24 = 8 - 20 - 4 + 24 ò0 P(-2) = (-2)3 - 5 ì (-2)2 - 2 ì (-2) + 24 = -8 - 20 + 4 + 24 = -28 + 28 =0 (x + 2) is a factor.

P(1) ò 0, so (x - 1) is not a factor. Try P(2). P(2) ò 0, so (x - 2) is not a factor. Try P(-2). P(-2) does equal 0, so (x + 2) is a factor.

b

652

x2 -   7x + 12 x + 2) - 5x2 -   2x + 24 x3 + 2x2 -7x2 -   2x -7x2 - 14x 12x + 24 12x + 24 0 2 P(x) = (x + 2)(x - 7x + 12)

3

Divide (x + 2) into P(x) using long division to find a quadratic factor.

4

Write P(x) as a product of the two factors found so far.

5

Factorise the quadratic factor if possible.

1

Name the polynomial. Note: There is no x2 term, so include 0x2.

2

Look at the last term in P(x), which is 30. This suggests it is worth trying P(5) or P(-5). Try P(-5). P(-5) = 0 so (x + 5) is a factor.

3

Divide (x + 5) into P(x) using long division to find a quadratic factor.

4

Write P(x) as a product of the two factors found so far.

P(x) = (x + 5)(x2 - 5x + 6)

5

Factorise the quadratic factor if possible.

P(x) = (x + 5)(x -2)(x - 3)

Maths Quest 10 + 10A for the Australian Curriculum

x3

P(x) = (x + 2)(x - 3)(x - 4) b

P(x) = x3 - 19x + 30 P(x) = x3 + 0x2 - 19x + 30 P(-5) = (-5)3 - 19 ì (-5) + 30 = -125 + 95 + 30 =0 So (x + 5) is a factor. x2 -   5x +   6 3 x + 5) x + 0x2 - 19x + 30 x3 + 5x2 -5x2 - 19x -5x2 - 25x   6x + 30   6x + 30 0

number AND algebra • Patterns and algebra c

c

Let P(x) = -2x3 - 8x2 + 6x + 4

1

Write the given polynomial.

2

Take out a common factor of -2. (We could take out +2 as the common factor, but taking out -2 results in a positive leading term in the part still to be factorised.)

3

Let Q(x) = (x3 + 4x2 - 3x - 2). (We have already used P earlier.)

Let Q(x) = (x3 + 4x2 - 3x - 2).

4

Evaluate Q(1). Q(1) = 0, so (x - 1) is a factor.

Q(1) = 1 + 4 - 3 - 2 =0 So (x - 1) is a factor.

5

Divide (x - 1) into Q(x) using long division to find a quadratic factor.

6

Write the original polynomial P(x) as a product of the factors found so far.

7

In this case, it is not possible to further factorise P(x).

■■

= -2(x3 + 4x2 - 3x - 2)

x2 + 5x + 2 x - 1) + 4x2 - 3x - 2 x3 -   x2 5x2 - 3x 5x2 - 5x 2x - 2 2x - 2 0 x3

P(x) = -2(x - 1)(x2 + 5x + 2)

Note: In these examples, P(x) may have been factorised without long division by finding all three values of x that make P(x) = 0 (and hence three factors) and then checking that the three factors multiply to give P(x).

Using short division ■■ ■■

The process of long division can be quite time (and space) consuming. An alternative is short division, which may take a little longer to understand, but is quicker once mastered. Consider P(x) = x3 + 2x2 - 13x + 10. Using the factor theorem, we can find that (x - 1) is a factor of P(x). So, P(x) = (x - 1)(  ?  ). Actually, we know more than this: as P ( x) begins with x3 and ends with +10, we could write P(x) = (x - 1)(x2 + ? - 10)

■■

The x2 in the second pair of grouping symbols produces the desired x3 (the leading term in P(x)) when the expressions are multiplied. The -10 in the second pair of grouping symbols produces +10 (the last term in P(x)) when the expressions are multiplied. Imagine expanding this version of P(x). Multiplying x in the first pair of grouping symbols by x2 in the second would produce x3, which is what we want, but multiplying -1 in the first pair of grouping symbols by x2 in the second gives -1x2. Since P(x) = x3 + 2x2 - 13x + 10, we really need +2x2, not -1x2. That is, we need +3x2 more. To get this, the ? must be 3x, because when x in the first pair of grouping symbols is multiplied by 3x in the second pair, +3x2 results. That is, we have deduced P(x) = (x - 1) (x2 + 3x - 10). Factorising the expression in the second pair of grouping symbols gives P(x) = (x - 1)(x + 5)(x - 2) Chapter 19 Polynomials

653

number AND algebra • Patterns and algebra

■■ ■■

This procedure, which we will call short division, can be confusing at first, but with persistence it can be a quick and easy method for factorising polynomials. The following worked example is a repeat of a previous one, but explains the use of short, rather than long, division.

Worked Example 10

Use short division to factorise x3 - 5x2 - 2 x + 24. Think

Write

1

Name the polynomial.

Let P(x) = x3 - 5x2 - 2x + 24.

2

Look for a value of x such that P(x) = 0. Try P(-2).

P(-2) = (-2)3 - 5 ì (-2)2 - 2 ì (-2) + 24 = -8 - 20 + 4 + 24 = -28 + 28 =0 So (x + 2) is a factor.

P(-2) does equal 0, so (x + 2) is a factor. 3

Look again at the original P(x) = x3 - 5x2 - 2x + 24. The first term in the grouping symbols must be x2, and the last term must be 12. P(x) = (x + 2)(x2   + 12)

4

Imagine the expansion of the expression in step 3. We have x3 and 2x2, but require -5x2. We need an extra -7x2. We get this by inserting a -7x term in the second pair of grouping symbols.

P(x) = (x + 2)(x2 - 7x + 12)

5

Factorise the expression in the second pair of grouping symbols if possible.

P(x) = (x + 2)(x - 3)(x - 4)

remember

To factorise a polynomial: 1. let P(x) = the given polynomial 2. use the factor theorem to find a linear factor (try factors of the constant term) 3. use long or short division to find the remaining factor 4. factorise the remaining factor if possible. Exercise

19F

Factorising polynomials fluency

1   WE 9a  Use long division to factorise each dividend. (x + 1)(x + 3)(x + 6) (x + 1)(x + 2)(x + 9) (x + 3)(x + 4)(x + 7)

654

a x + 1) x3 + 10x2 + 27x + 18

b x + 2) x3 + 8x2 + 17x + 10

(x + 1)(x + 2)(x + 5)

c x + 9) x3 + 12x2 + 29x + 18

d x + 1) x3 + 8x2 + 19x + 12

(x + 1)(x + 3)(x + 4)

e x + 3) x3 + 14x2 + 61x + 84

f x + 7) x3 + 12x2 + 41x + 42

(x + 2)(x + 3)(x + 7)

Maths Quest 10 + 10A for the Australian Curriculum

number and algebra • patterns and algebra g x + 2) x3 + 4x2 + 5x + 2 i

x + 5) x3 + 14x2 + 65x + 100

k x) x3 + 7x2 + 12x m x + 1) x3 + 6x2 + 5x eBook plus

Digital doc

SkillSHEET 19.3 doc-5368

(x + 1)2(x + 2)

h x + 3) x3 + 7x2 + 16x + 12

(x + 4)(x + 5)2 j

x(x + 3)(x + 4) x(x + 1)(x + 5)

l

x(x + 5)(x + 8)

x + 5) x3 + 10x2 + 25x

n x + 6) x3 + 6x2

x(x + 5)2

x2(x + 6)

2 We9, 10 Factorise the following as fully as possible. a x3 + x2 - x - 1 (x - 1)(x + 1)2 b x3 - 2x2 - x + 2 (x - 2)(x - 1)(x + 1) 3 2 2(x + 5) (x + 1) c x + 7x + 11x + 5 d x3 + x2 - 8x - 12 (x - 3)(x + 2)2 3 2 2 e x + 9x + 24x + 16 (x + 1)(x + 4) f x3 - 5x2 - 4x + 20 (x - 5)(x - 2)(x + 2) 3 2 g x + 2x - x - 2 (x - 1)(x + 1)(x + 2) h x3 - 7x - 6 (x - 3)(x + 1)(x + 2) 3 2 2 i x + 3x - 4 (x - 1)(x + 2) j x3 + x2 + x + 6 (x + 2)(x2 - x + 3) 3 2 (x + 1)(x + 2)(x + 5) k x + 8x + 17x + 10 l x3 + x2 - 9x - 9 (x - 3)(x + 1)(x + 3) m x3 - x2 - 8x + 12 (x - 2)2(x + 3) n x3 + 9x2 - 12x - 160 (x - 4)(x + 5)(x + 8) 3 Factorise as fully as possible. a 2x3 + 5x2 - x - 6 (2x + 3)(x - 1)(x + 2) b c 3x3 + 2x2 - 12x - 8 (3x + 2)(x - 2)(x + 2) d e 5x3 + 9x2 + 3x - 1 (5x - 1)(x + 1)2 f g 4x3 + 16x2 + 21x + 9 (x + 1)(2x + 3)2 h i 10x3 + 19x2 - 94x - 40 (x + 4)(2x - 5)(5x + 2) j 4 Factorise as fully as possible. a 3x3 - x2 - 10x x(x - 2)(3x + 5) b 4x3 + 2x2 - 2x 2x(x + 1)(2x - 1) c 3x3 - 6x2 - 24x 3x(x - 4)(x + 2) d -2x3 - 12x2 - 18x -2x(x + 3)2 e 6x3 - 6x2 6x2(x - 1) f -x3 - 7x2 - 12x -x(x + 4)(x + 3) g -x3 - 3x2 + x + 3 -(x - 1)(x + 1)(x + 3) h -2x3 + 10x2 - 12x -2x(x - 3)(x - 2) i -6x3 - 5x2 + 12x - 4 -(x + 2)(2x - 1)(3x - 2) j -5x3 + 24x2 - 36x + 16 -(x - 2)2(5x - 4) k -x5 - x4 + 21x3 + 49x2 - 8x - 60

19g

x) x3 + 13x2 + 40x

(x + 2)2(x + 3)

3x3 + 14x2 + 7x - 4 (3x - 1)(x + 1)(x + 4) 4x3 + 35x2 + 84x + 45 (4x + 3)(x + 3)(x + 5) x3 + x2 + x + 1 (x + 1)(x2 + 1) 6x3 - 23x2 + 26x - 8 (x - 2)(2x - 1)(3x - 4) 7x3 + 12x2 - 60x + 16 (7x - 2)(x - 2)(x + 4)

reFleCtion 

 

Explain the steps in factorising polynomials

-(x - 1)(x + 3)(x - 5)(x + 2)2

solving polynomial equations ■

■ ■

■

A polynomial equation of the form P(x) = 0 may be solved by factorising P(x) and applying the Null Factor Law. The Null Factor Law applies to polynomial equations just as it does for quadratics. If P(x) = (x - a)(x - b)(x - c) = 0, then the solutions can be found as follows. Let each factor = 0: x-a=0 x-b=0 x-c=0 Solving each of these equations produces the solutions x=a x=b x = c. If P(x) = k(lx - a)(mx - b)(nx - c) = 0, then the solutions can be found as follows. Let each factor = 0: lx - a = 0 mx - b = 0 nx - c = 0 Solving each of these equations produces the solutions a b c x= x= x= . l m n Note: The coefficient k used in this example does not produce a solution in x. Chapter 19 polynomials

655

number AND algebra • Patterns and algebra

Worked Example 11

Solve: a  x3 = 9x b  -2 x3 + 4x2 + 70x = 0 c  2 x3 - 11x2 + 18x - 9 = 0. Think a

b

c

656

Write a x3 = 9x

1

Write the equation.

2

Rearrange so all terms are on the left.

x3 - 9x = 0

3

Take out a common factor of x.

x(x2 - 9) = 0

4

Factorise the expression in the grouping symbols using the difference of squares rule.

x(x + 3)(x - 3) = 0

5

Use the Null Factor Law to solve.

x = 0, x + 3 = 0 or x - 3 = 0 x = 0, x = -3 or x = 3

1

Write the equation.

2

Take out a common factor of -2x.

-2x(x2 - 2x - 35) = 0

3

Factorise the expression in the grouping symbols.

-2x(x - 7)(x + 5) = 0

4

Use the Null Factor Law to solve.

-2x = 0, x - 7 = 0 or x + 5 = 0 x = 0, x = 7 or x = -5

1

Name the polynomial.

2

Use the factor theorem to find a factor (search for a value a such that P(a) = 0). Consider factors of the constant term (that is, factors of 9 such as 1, 3). The simplest value to try is 1.

b -2x3 + 4x2 + 70x = 0

c

Let P(x) = 2x3 - 11x2 + 18x - 9.

P(1) = 2 - 11 + 18 - 9 =0 So (x - 1) is a factor. 2x2 -   9x + 9 x - 1) - 11x2 + 18x - 9 3 2x -   2x2 -9x2 + 18x -9x2 +   9x 9x - 9 9x - 9 0 P(x) = (x - 1)(2x2 - 9x + 9) 2x3

3

Use long or short division to find another factor of P(x).

4

Factorise the quadratic factor.

P(x) = (x - 1)(2x - 3)(x - 3)

5

Consider the factorised equation to solve.

For (x - 1)(2x - 3)(x - 3) = 0

6

Use the Null Factor Law to solve.

x - 1 = 0, 2x - 3 = 0 or x - 3 = 0 3 x = 1, x = 2 or x = 3

Maths Quest 10 + 10A for the Australian Curriculum

number and algebra • patterns and algebra

remember

To solve a polynomial equation: 1. let P(x) = . . . 2. use the factor theorem (try P(1), etc.) to find a factor of the form (x - a) 3. use long or short division to find the quotient 4. factorise the quotient if possible 5. let each linear factor equal zero and solve for x in each case. 6. If P(x) = (x – a)(x – b)(x – c) = 0, then the solutions are: x=a x=b x = c. 7. If P(x) = k(lx – a)(mx – b)(nx – c) = 0, then the solutions are: a b c x= x= x= . l m n exerCise

19g

solving polynomial equations FluenCy

eBook plus

Digital doc

SkillSHEET 19.4 doc-5369

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SkillSHEET 19.5 doc-5370

1 We11a, b Solve the following. a x3 - 4x = 0 -2, 0, 2 c 2x3 - 50x = 0 -5, 0, 5 e x3 + 5x2 = 0 -5, 0 g -4x3 + 8x = 0 1 0, 5 i 4x2 - 20x3 = 0 3 2 k x - 8x + 16x = 0 0, 4 m 9x2 = 20x + x3 0, 4, 5

x3 - 16x = 0 -4, 0, 4 -3x3 + 81 = 0 3 x3 - 2x2 = 0 0, 2 1 -4 , 0 − 2 , 0, 2 12x3 + 3x2 = 0 x3 - 5x2 + 6x = 0 0, 2, 3 x3 + 6x2 = 7x -7, 0, 1 x3 + 6x = 4x2 0 2 We11c Use the factor theorem to solve the following. a x3 - x2 - 16x + 16 = 0 -4, 1, 4 b x3 - 6x2 - x + 30 = 0 -2, 3, 5 c x3 - x2 - 25x + 25 = 0 -5, 1, 5 d x3 + 4x2 - 4x - 16 = 0 -4, -2, 2 e x3 - 4x2 + x + 6 = 0 -1, 2, 3 f x3 - 4x2 - 7x + 10 = 0 -2, 1, 5 g x3 + 6x2 + 11x + 6 = 0 -3, -2, -1 h x3 - 6x2 - 15x + 100 = 0 -4, 5 i x3 - 3x2 - 6x + 8 = 0 -2, 1, 4 j x3 + 2x2 - 29x + 42 = 0 -7, 2, 3 1 -6, - 2 , -1 k 2x3 + 15x2 + 19x + 6 = 0 1 3 l -4x3 + 16x2 - 9x - 9 = 0 - 2 , 2 , 3 1 3 2 m -2x - 9x - 7x + 6 = 0 -3, -2, 2 3 2 -2, -1, 1 n 2x + 4x - 2x - 4 = 0 3 mC Note: There may be more than one correct answer. Which of the following is a solution to x3 - 7x2 + 2x + 40 = 0? B -4 ✔ A 5 ✔ C -2 D 1 3 2 4 mC A solution of x - 9x + 15x + 25 = 0 is x = 5. How many other (distinct) solutions are there? ✔ B 1 A 0 C 2 D 3 b d f h j l n

Chapter 19 polynomials

657

number and algebra • patterns and algebra 5 Solve P(x) = 0. a P(x) = x3 + 4x2 - 3x - 18 -3, 2 1 b P(x) = 3x3 - 13x2 - 32x + 12 -2, 3, 6 3 -4, 2 c P(x) = -x + 12x - 16 d P(x) = 8x3 + 10x2 - 38x + 20 1 e P(x) = x4 + 2x3 - 13x2 - 14x + 24 -4, -2, 1, 3 f P(x) = -72 - 42x + 19x2 + 7x3 - 2x4 -2, - 32 , 3, 4 g P(x) = x4 + 2x3 - 7x2 - 8x + 12 -3, -2, 1, 2 h P(x) = 4x4 + 12x3 - 24x2 - 32x -4, -1, 0, 2

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WorkSHEET 19.2 doc-5376

658

6 Solve each of the following equations. -2, 1, 4 a x3 - 3x2 - 6x + 8 = 0 3 2 -3, -1, 3 b x + x - 9x - 9 = 0 c 3x3 + 3x2 - 18x = 0 -3, 0, 2 d 2x4 + 10x3 - 4x2 - 48x = 0 -4, -3, 0, 2 3 e 2x4 + x3 - 14x2 - 4x + 24 = 0 -2, 2 , 2 4 2 f x - 2x + 1 = 0 -1, 1

maths quest 10 + 10a for the australian Curriculum

reFleCtion 

 

Can you predict the number of solutions a polynomial might have?

number AND algebra • Patterns and algebra

Summary Polynomials ■■

■■

A polynomial in x, sometimes denoted P(x), is an expression containing only non-negative integer powers of x. The degree of a polynomial in x is the highest power of x in the expression. Adding, subtracting and multiplying polynomials

■■ ■■

To add or subtract polynomials, add or subtract any like terms in the expression. When expanding linear factors: (a) expand two factors first, and then multiply by the remaining linear factors, one at a time (b) collect like terms at each stage (c) (x + 2)3 may be written as (x + 2)(x + 2)(x + 2). Long division of polynomials

Long division of polynomials is similar to long division with numbers. The highest power term is the main one considered at each stage. The key steps are: ■■ determine how many times the first term of the divisor goes into the first term of dividend ■■ multiply and write the result underneath ■■ subtract. (If necessary, change the sign and add.) ■■ bring down the next term ■■ repeat the process until no pronumerals remain to be divided ■■ state the quotient and remainder. Polynomial values ■■

P(a) means the value of P(x) when x is replaced by a and the polynomial is evaluated. The remainder and factor theorems

■■

■■

The remainder when P(x) is divided by (x – a) is equal to P(a). That is: R = P(a). This is known as the remainder theorem. If P(a) = 0, then (x – a) is a factor of P(x). This is known as the factor theorem. Factorising polynomials

To factorise a polynomial: let P(x) = the given polynomial ■■ use the factor theorem to find a linear factor (try factors of the constant term) ■■ use long or short division to find the remaining factor ■■ factorise the remaining factor if possible. ■■

Solving polynomial equations

To solve a polynomial equation: let P(x) = .  .  . ■■ use the factor theorem (try P(1), etc.) to find a factor of the form (x - a) ■■ use long or short division to find the quotient ■■ factorise the quotient if possible ■■

Chapter 19 Polynomials

659

number AND algebra • Patterns and algebra

■■ ■■

■■

let each linear factor equal zero and solve for x in each case. If P(x) = (x – a)(x – b)(x – c) = 0, then the solutions are: x = a    x = b    x = c. If P(x) = k(lx – a)(mx – b)(nx – c) = 0, then the solutions are: a b c x =      x =     x = . l m n

Mapping your understanding

Using terms from the summary, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 635.

660

Maths Quest 10 + 10A for the Australian Curriculum

number and algebra • patterns and algebra

Chapter review FluenCy 1 mC Which of the following is not a polynomial? A x 3 −

x2 + 7x − 1 3

2

x + 3x + 2 2 Consider the polynomial ✔ C

B a4 + 4a3 + 2a + 2 D 5

1

f ( x ) = − 7 x 4 + x 5 + 3. What is the degree of f (x)? 5 1 What is the coefficient of x4? - 7 What is the constant term? 3 What is the leading term? x5 3 mC The expansion of (x + 5)(x + 1)(x - 6) is: A x3 - 30 B x3 + 12x2 - 31x + 30 ✔ C x3 - 31x - 30 D x3 + 5x2 - 36x - 30 a b c d

4 mC x3 + 5x2 + 3x - 9 is the expansion of: A (x + 3)3 B x(x + 3)(x - 3) ✔ C (x - 1)(x + 3)2 D (x - 1)(x + 1)(x + 3) 5 Expand: a (x - 2)2(x + 10) x3 + 6x2 - 36x + 40 b (x + 6)(x - 1)(x + 5) x3 + 10x2 + 19x - 30 c (x - 7)3 x3 - 21x2 + 147x - 343 d (5 - 2x)(1 + x)(x + 2). -2x3 - x2 + 11x + 10 6 mC Consider the following long division.

x2 + x + 2 3 x + 4) x + 5x2 + 6x - 1 x3 + 4x2 x2 + 6x x2 + 4x 2x - 1 2x + 8 -9 a The quotient is: A -9 B 4 C x + 4 ✔ D x2 + x + 2

b The remainder is: ✔ A -9 C 4

B 2 D 2x - 1

7 Find the quotient and remainder when the first

polynomial is divided by the second in each case. a x3 + 2x2 - 16x - 3, x + 2 x2 - 16, 29 b x3 + 3x2 - 13x - 7, x - 3 x2 + 6x + 5, 8 c -x3 + x2 + 4x - 7, x + 1 -x2 + 2x + 2, -9 8 mC If P(x) = x3 - 3x2 + 7x + 1, then P(-2) equals: ✔ B -33 A -34 C -9 D 7 9 If P(x) = -3x3 + 2x2 + x - 4, find: a P(1) -4 b P(-4) c P(2a). -24a3 + 8a2 + 2a - 4

216

10 Without dividing, find the remainder when x3 + 3x2 - 16x + 5 is divided by x - 1. -7 11 Show that x + 3 is a factor of x3 - 2x2 - 29x - 42. 12 Factorise x3 + 4x2 - 100x - 400.

(x - 10)(x + 4)(x + 10)

13 Solve: Show P(-3) = 0. 1 a (2x + 1)(x - 3)2 = 0 - 2 , 3 b x3 - 9x2 + 26x - 24 = 0 2, 3, 4 c x4 - 4x3 - x2 + 16x - 12 = 0 –2, 1, 2, 3 problem solVing 1 Let P(x) = an xn + an - 1 xn - 1 + . . . + a1 x + a0 be

a ploynomial where the coefficients are integers. Also let P(w) = 0 where w is an integer. Show that w is a factor of a0

Teacher to check. For example, given P(x) = x3 - x2 - 34x - 56 and P(7) = 0 À (x - 7) is a factor and 7 is a factor of 56.

eBook plus

Interactivities

Test yourself Chapter 19 int-2876 Word search Chapter 19 int-2874 Crossword Chapter 19 int-2875

Chapter 19 polynomials

661

eBook plus

aCtiVities

Are you ready?

(page 636) SkillSHEET 19.1 (doc-5366): Expanding the product of two linear factors SkillSHEET 19.2 (doc-5367): Substitution into quadratic equations SkillSHEET 19.3 (doc-5368): Factorising quadratic trinomials SkillSHEET 19.4 (doc-5369): Factorising difference of two squares expressions SkillSHEET 19.5 (doc-5370): Solving quadratic equations

Digital docs

• • • • •

19B Adding, subtracting and multiplying polynomials Digital doc

• SkillSHEET 19.1 (doc-5366): Expanding the product of two linear factors (pages 640) 19C Long division of polynomials Interactivity

• Long division of polynomials (int-2793) (page 641) 19D Polynomial values

19F Factorising polynomials Digital doc

• SkillSHEET 19.3 (doc-5368): Factorising quadratic trinomials (page 655) 19G Solving polynomial equations Digital docs

• SkillSHEET 19.4 (doc-5369): Factorising difference of two squares expressions (page 657) • SkillSHEET 19.5 (doc-5370): Solving quadratic equations (page 657) • WorkSHEET 19.2 (doc-5376): Polynomials II (page 658) Chapter review

(page 661) • Test yourself Chapter 19 (int-2876): Take the end-ofchapter test to test your progress • Word search Chapter 19 (int-2874): an interactive word search involving words associated with this chapter • Crossword Chapter 19 (int-2875): an interactive crossword using the definitions associated with the chapter Interactivities

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• SkillSHEET 19.2 (doc-5367): Substitution into quadratic equations (page 648) • WorkSHEET 19.1 (doc-5375): Polynomials I (page 649)

662

maths quest 10 + 10a for the australian Curriculum

To access eBookPLUS activities, log on to www.jacplus.com.au

number AND algebra • Linear and non-linear relationships

20

20A 20B 20C 20D 20E

Functions and relations Exponential functions Cubic functions Quartic functions Transformations

What do you know ?

Functions and relations

1 List what you know about functions and relations. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of functions and relations.

opening question

The sails of the opera house are formed from sections of what shape?

number and algebra • linear and non-linear relationships

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

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SkillSHEET 20.1 doc-5378

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SkillSHEET 20.2 doc-5379

Finding the gradient and y-intercept 1 Find the gradient and y-intercept of the following straight lines. a y = 3x + 4 Gradient = 3, y-intercept = 4 b 2x + y = 3 Gradient = -2, y-intercept = 3 c 5x - 2y - 8 = 0 Gradient = 5 , y-intercept = -4 2 a

Sketching straight lines 2 Sketch the graph of each straight line. a y = 2x + 1 c y = 2

1

b y = -4x + 2

1 -— 2

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SkillSHEET 20.3 doc-5380

c

Sketching parabolas 3 Sketch the graph of each parabola. a y = x2 c y = -2x2

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SkillSHEET 20.4 doc-5381

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SkillSHEET 20.5 doc-5382

b y = - 4 d y = (x - 2)2

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SkillSHEET 20.7 doc-5384

f y = 2x + 3 Neither

Substitution into index expressions 7 Substitute the following values for x into y = 3x and evaluate. a x = 4 81 b x = 1 3 1 c x = 0 1 d x = -3 27 y

y = x2 (1,1)

664

y=2

Finding points of intersection 6 Find any points of intersection of the lines y = 2x + 3 and y = 4x - 5. (4, 11)

b

y

x

0

2 x

Identifying equations of straight lines and parabolas 5 Identify each of the following as a straight line, a parabola or neither. a y = 2x2 + 7 Parabola b x2 + y2 = 25 Neither c 2x + y - 3 = 0 Straight line d y = (x - 2) (x - 4) Parabola

SkillSHEET 20.6 doc-5383

eBook plus

2

Completing the square 4 Complete the square to change each of these quadratic equations into turning point form. a y = x2 + 6x + 11 y = (x + 3)2 + 2 b y = x2 - 4x - 1 y = (x - 2)2 - 5 c y = x2 + x - 2 y = (x + 1 )2 - 9 2 4

1 2

eBook plus

x

0

y

y

x2

e y =  x + 6 Straight line

0

y = 2x + 1

0

eBook plus

3 a

b

y

y = x2 - 4

c

y 0

-2 0 -4

2

y = -2x 2 x (1, -2)

x

maths quest 10 + 10a for the australian Curriculum

d

y 4 y = (x - 2)2 0

2

x

y = -4x + 2

1 — 2

x

number AND algebra • Linear and non-linear relationships

Functions and relations

20A

Relations ■■

A relation is a set of ordered pairs of values such as all the points on the circle x2 + y2 = 4 or all the points on the exponential y = 2x. Relations can be grouped into the following four categories. y

One-to-one relations ■■

A one-to-one relation exists if for any x-value there is only one corresponding y-value and vice versa. For example:

0

One-to-many relations ■■

y

A one-to-many relation exists if for any x-value there is more than one y-value, but for any y-value there is only one x-value. For example:

x

0

Many-to-one relations ■■

y

A many-to-one relation exists if there is more than one x-value for any y-value but for any x-value there is only one y-value. For example: 0

Many-to-many relations ■■

x

y

A many-to-many relation exists if there is more than one x-value for any y-value and vice versa. For example:

x

y

x

0

0

x

Worked Example 1

What type of relation does each graph represent? y y a        b 

0

x

0

Think a

1

      c 

y 0

x

x

Write

For some x-values there is more than one y-value. A line through some x‑values shows that 2 y-values are available:

a One-to-many relation

y x = -1 0

x

Chapter 20 Functions and relations

665

number AND algebra • Linear and non-linear relationships

For any y-value there is only one x‑value. A line through any y-value shows that only one x-value is available:

2

y y=1 x

0

b

c

1

For any x-value there is only one y‑value.

2

For any y-value there is only one x‑value.

1

For any x-value there is only one y‑value.

2

For some y-values there is more than one x‑value.

b One-to-one relation

c

Many-to-one relation

Functions ■■

Relations that are one-to-one or many-to-one are called functions. That is, a function is a relation where for any x-value there is at most one y-value. For example: y y 1.  2. 

0

x

0

x

Vertical line test ■■

A function is determined from a graph if a vertical line, drawn anywhere on the graph cannot intersect with the curve more than once.

Worked Example 2

State whether or not each of the following relations are functions. y a           b  y

0

x 0

Think a It is possible for a vertical line to intersect with

x Write a Not a function

the curve more than once. b It is not possible for any vertical line to intersect

with the curve more than once.

666

Maths Quest 10 + 10A for the Australian Curriculum

b Function

number AND algebra • Linear and non-linear relationships

Function notation ■■

Consider the relation y = 2x, which is a function. The y-values are determined from the x-values, so we say ‘y is a func­tion of x’, which is abbreviated to y = f (x). So, the rule y = 2x can also be written as f (x) = 2x. If x = 1, then y = f (1) =2ì1 =2 If x = 2, then y = f (2) =2ì2 = 4, and so on.

Evaluating functions ■■

For a given function y = f (x), the value of y when x = 1 is written as f (1), the value of y when x = 5 is written as f (5), the value of y when x = a as f (a), etc.

Worked Example 3

If f (x) = x2 - 3, find: a  f (1) b  f (-2) c  f (a) d  f (2a). Think a

b

c

d

Write

1

Write the rule.

2

Substitute x = 1 into the rule.

3

Simplify.

1

Write the rule.

2

Substitute x = -2 into the rule.

3

Simplify.

1

Write the rule.

2

Substitute x = a into the rule.

1

Write the rule.

2

Substitute x = 2a into the rule.

3

Simplify the expression if possible.

a

f (x) = x2 - 3 f (1) = 12 - 3 =1-3 = -2

b

f (x) = x2 - 3 f (-2) = (-2)2 - 3 =4-3 =1

c

f (x) = x2 - 3 f (a) = a2 - 3

d

f (x) = x2 - 3 f (2a) = (2a)2 - 3 = 22a2 - 3 = 4a2 - 3

Identifying features of functions ■■

We can identify features of certain functions by observing what happens to the function value (y value) when x approaches a very small value such as 0 (x ç 0) or a very large value such as Ñ (x ç Ñ). Chapter 20 Functions and relations

667

number AND algebra • Linear and non-linear relationships

Worked Example 4

Describe what happens to these functions as the value of x increases, that is, as x ç Ñ. 1 a  f (x) = x2      b  f (x) = 2-x      c  f (x) = + 1 x Think a

b

c

Write a f (x) = x2

1

Write the function.

2

Substitute large x values into the function, such as x = 10  000 and x = 1  000  000.

f (10  000) = 100  000  000 f (1  000  000) = 1 ì 1012

3

Write a conclusion.

As x ç Ñ, f (x) also increases; that is, f (x) ç Ñ.

1

Write the function.

2

Substitute large x values into the function, such as x = 10  000 and x = 1  000  000.

3

Write a conclusion.

1

Write the function.

2

Substitute large x values into the function, such as x = 10  000 and x = 1  000  000.

f (10  000) = 1.0001 f (1  000  000) = 1.000  001

3

Write a conclusion.

As x ç Ñ, f (x) ç 1.

b f (x) = 2-x

f (10  000) ö 0 f (1  000  000) ö 0 As x ç Ñ, f (x) ç 0. 1 c f (x) = + 1 x

Points of intersection ■■

If two functions are drawn on the one set of axes, there may be a point or points where the curves intersect. The function equations can be solved simultaneously to find the coordinates of these points of intersection.

Worked Example 5

1 Find any points of intersection between f (x) = 2x + 1 and g(x) = . x Think 1

Write the two equations.

f (x) = 2x + 1 1 g( x ) = x

2

Points of intersection are common values between the two curves. To solve the equations simultaneously, equate both functions.

For points of intersection: 1 2x + 1 = x

3

Rearrange the resulting equation and solve for x.

2x2 + x = 1 2 2x + x - 1 = 0 (2x - 1)(x + 1) = 0 1 x = 2 or -1

4

Substitute the x values into either function to find the y values.



f ( 1 ) = 2 ì 12 + 1 = 2



f (-1) = 2 ì -1 + 1 = -1

Write the coordinates of the two points of intersection.

Points of intersection are ( 1 , 2) and (-1, -1).

5

668

Write

Maths Quest 10 + 10A for the Australian Curriculum

2

2

number AND algebra • Linear and non-linear relationships

remember

1. A function is a relation so that for any x-value there is at most one y-value (one-to-one or many-to-one relations). 2. Vertical line test: The graph of a function cannot be crossed more than once by any vertical line. y

y

0

x

x

0

Function Not a function 3. f (x) = .  .  . is used to describe ‘a function of x’. To evaluate the function, for example when x = 2, find f (2) by replacing each occurrence of x on the RHS with 2. 4. Substitute appropriate x values to describe what happens to functions as x ç Ñ (x approaches infinity) or x ç 0 (x approaches zero). 5. To find points of intersection, solve function equations simultaneously to find both x and y values. Exercise

20A

Functions and relations Fluency 1   WE 1  What type of relation does each graph represent?



a One-to-many b Many-to-one c Many-to-one d One-to-one e One-to-one f Many-to-one g Many-to-many h Many-to-one i One-to-one j Many-to-one k Many-to-one l Many-to-one

y

a

0

d

h

x

y

0

k

x

x

0

y

x

y

x

0 l

x

x

0

i

y

x

0

f

y

0

y

c

y

0

x

y

y

0

e

0

j

x

y

0

g

b

x

y

0

x

Chapter 20 Functions and relations

669

number AND algebra • Linear and non-linear relationships b, c, d, e, f, h, i, j, k, l 2   WE 2  Use the vertical line test to determine which of the relations in question 1 are functions. 3   WE 3  a If f ( x) = 3x + 1, find i f (0),   1 ii f (2), 7 b If g ( x) = x + 4 , find i g (0), 2 ii g (-3),

iii f (-2) and 1

–5

iii g (5) and 3

iv f (5) respectively. 16 iv g (-4) respectively. 0

1 c If g ( x) = 4 - , find x  1 , 2

i g (1),   3 d If f ( x) = (x + i f (0),   9 e If h ( x) =

ii g 

3)2,

2

 1  2

iii g  −

 1  5

and 6

iv g  −

respectively. 9

find ii f (-2), 1

iii f (1) and 16

iv f (a) respectively.

24 , find x

a2 + 6a + 9

i h (2),   12

ii h (4), 6

iii h (-6) and -4

iv h (12) respectively. 2

understanding 4   MC  Note: There may be more than one correct answer.

Which of the following relations is a function? y B x2 + y2 = 9 ✔ C y = 8x - 3 ✔ a a 3 b 3 5 c − 2 x x 10 d 2 − x 2 x 10 − x − 3 5 e x+3 10 − x +1 f x −1

0

x

y

0

Which of the following relations are functions? a y = 2x + 1 c y = 2x e x2 + 4x + y2 + 6y = 14

6 Given that f ( x ) =

✔ d

a, b, c, f b y = x2 + 2 d x2 + y2 = 25 f y = −4x

10 − x find: x

a f (2) d f (x2)

b f (-5) e f (x + 3)

c f (2x) f f (x - 1)

7 Find the value (or values) of x for which each function has the value given. a f (x) = 3x - 4, f (x) = 5 3 b g(x) = x2 - 2, g(x) = 7 -3 or 3 c f (x) =

1 , f ( x) = 3 x

1 3

e g(x) = x2 + 3x, g(x) = 4 -4 or 1 Reasoning 8   WE 4  Describe what happens to: a f (x) = x2 + 3 as x ç Ñ f (x) ç Ñ b f (x) = 2x as x ç - Ñ f (x) ç 0

1 as x ç Ñ f (x) ç 0 x d f (x) = x3 as x ç - Ñ f (x) ç -Ñ c f (x) =

e f (x) = -5x as x ç - Ñ. f (x) ç 0 670

Maths Quest 10 + 10A for the Australian Curriculum

d h(x) = x2 - 5x + 6, h(x) = 0 2 or 3 f f (x) = 8 − x , f (x) = 3 -1

x

number and algebra • linear and non-linear relationships 9 We5 Find any points of intersection between the following curves. a f (x) = 2x - 4 and g(x) = x2 - 4 (0, -4), (2, 0)

2 2 (1, -2), (- 3 , 3) x c f (x) = x2 - 4 and g(x) = 4 - x2 (2, 0), (-2, 0) b f (x) = -3x + 1 and g(x) = -

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eLesson Exponential growth

d f (x) =

3 x 4

-

1 64

and

x2

+

y2

reFleCtion 

 

How do you determine the difference between functions and relations?

= 25 (3, -4)

exponential functions ■

■

eles-0176

Exponential functions can be used to model many real situations involving natural growth and decay. Exponential growth is when a quantity grows by a constant percentage in each fixed period of time. Examples of exponential growth include growth of investment at a certain rate of compound interest and growth in the number of cells in a bacterial colony.

Exponential growth y 10 8 6 4 2 -4

■

■

-2

0

y = ax

2

4

Exponential decay Exponential decay is when a quantity decreases by a y constant percentage in each fixed period of time. 10 Examples of exponential decay include yearly loss of -x y=a 8 value of an item (called depreciation) and radioactive or 1 6 y = decay. ax 4 Both exponential growth and decay can be modelled by 2 exponential functions of the type y = kax (y = k ì a x). -4 -2 0 2 4 The difference is in the value of the base a. When a > 1, there is exponential growth and when 0 < a < 1 there is exponential decay. The value of k corresponds to the initial quantity that is growing or decaying.

x

x

Worked example 6

The number of bacteria, N, in a Petri dish after x hours is given by the equation N = 50 ì 2 x.

a b c d

Determine the initial number of bacteria in the Petri dish. Determine the number of bacteria in the Petri dish after 3 hours. Draw the graph of the function of N against x. Use the graph to estimate the length of time it will take for the initial number of bacteria to treble. Chapter 20 Functions and relations

671

number AND algebra • Linear and non-linear relationships

Think a

b

c

Write/draw a N = 50 ì 2x

1

Write the equation.

2

Substitute x = 0 into the given formula and evaluate. (Notice that this is the value of k for equations of the form y = k ì a x.)

When x = 0, N = 50 ì 20 = 50 ì 1 = 50

3

Write the answer in a sentence.

The initial number of bacteria in the Petri dish is 50.

1

Substitute x = 3 into the formula and evaluate.

2

Write the answer in a sentence.

1

Draw a set of axes, labelling the horizontal axis as x and the vertical axis as N.

2

Plot the points generated by the answers to parts a and b.

3

Calculate the value of N when x = 1 and x = 2 and plot the points generated.

4

Join the points plotted with a smooth curve.

b When x = 3, N = 50 ì 23

= 50 ì 8 = 400



After 3 hours there are 400 bacteria in the Petri dish. c

At x = 1, N = 50 ì 21   At x = 2, N = 50 ì 22 = 50 ì 2 = 50 ì 4 = 100 = 200 N 500

N = 50 ì 2x

400 300 200 100 0

d

5

Label the graph.

1

Determine the number of bacteria required.

1

Draw a horizontal line from N = 150 to the curve and from this point draw a vertical line to the x-axis.

3

x

d Number of bacteria = 3 ì 50

= 150



2

2

N 500

N = 50 ì 2x

400 300 200 100 0

672

3

The point on the x-axis will be the estimate of the time taken for the number of bacteria to treble.

4

Write the answer in a sentence.

Maths Quest 10 + 10A for the Australian Curriculum

1

2

3

x

The time taken will be approximately 1.6 hours.

number AND algebra • Linear and non-linear relationships

Worked Example 7

A new computer costs $3000. It is estimated that each year it will be losing 12% of the previous year’s value. a Determine the value, $V, of the computer after the first year. b Determine the value of the computer after the second year. c Determine the equation which relates the value of the computer to the number of years, n, it has been used. d Use your equation to determine the value of the computer in 10 years’ time. Think a

b

c

d

Write a V0 = 3000

1

State the original value of the computer.

2

Since 12% of the value is being lost each year, the value of the computer will be 88% or (100 - 12)% of the previous year’s value. Therefore, the value after the first year (V1) is 88% of the original cost.

V1 = 88% of 3000 = 0.88 ì 3000 = 2640

3

Write the answer in a sentence.

The value of the computer after 1 year is $2640.

1

The value of the computer after the second year, V2, is 88% of the value after the first year.

b V2 = 88% of 2640



= 0.88 ì 2640 = 2323.2

2

Write the answer in a sentence.

1

The original value is V0.

2

The value after the first year, V1, is obtained by multiplying the original value by 0.88.

V1 = 3000 ì 0.88

3

The value after the second year, V2, is obtained by multiplying V1 by 0.88, or by multiplying the original value, V0, by (0.88)2.

V2 = (3000 ì 0.88) ì 0.88 = 3000 ì (0.88)2

4

The value after the third year, V3, is obtained by multiplying V2 by 0.88, or V0 by (0.88)3.

V3 = 3000 ì (0.88)2 ì 0.88 = 3000 ì (0.88)3

5

By observing the pattern we can generalise as follows: the value after the nth year, Vn, can be obtained by multiplying the original value, V0, by 0.88 n times; that is, by (0.88)n.

Vn = 3000 ì (0.88)n

1

Substitute n = 10 into the equation obtained in part c to find the value of the computer after 10 years.

2

Write the answer in a sentence.

■■

The value of the computer after the second year is $2323.20. c

V0 = 3000

d When n = 10,

V10 = 3000 ì (0.88)10 = 835.50

The value of the computer after 10 years is $835.50.

Sometimes the relationship between the two variables closely resembles an exponential pattern, but cannot be described exactly by an exponential function. In such cases, part of the data are used to model the relationship with exponential growth or the decay function. Chapter 20 Functions and relations

673

number AND algebra • Linear and non-linear relationships

Worked Example 8

The population of a certain city is shown in the table below. Year

1985

1990

1995

2000

2005

2010

Population (ì 1000)

128

170

232

316

412

549

Assume that the relationship between the population, P, and the year, x, can be modelled by the function P = ka x, where x is the number of years after 1985. The value of P must be multiplied by 1000 in order to find the actual population. a State the value of k, which is the population, in thousands, at the start of the period. b Use a middle point in the data set to find the value of a, correct to 2 decimal places. Hence, write the formula, connecting the population, P, with the number of years, x, since 1985. c For the years given, find the size of the population using the formula obtained in part b. Compare it with the actual size of the population in those years. d Predict the population of the city in the years 2015 and 2020. Think

Write/display

a From the given table, state the value of k

that corresponds to the population of the city in the year 1985. b

a k = 128

b P = kax

1

Write the given formula for the population of the city.

2

Replace the value of k with the value found in a.

P = 128 ì ax

3

Using a middle point of the data, replace x with the number of years since 1985 and P with the corresponding value.

Middle point is (1995, 232). When x = 10, P = 232, so 232 = 128 ì a10

4

Solve the equation for a.

a10 =

232 128 = 1.8125



a = 10 1.8125 a = 1.0613.  .  .

c

d

674

5

Round the answer to 2 decimal places.

6

Rewrite the formula with this value of a.

1

Draw a table of values and enter the given years, the number of years since 1985, x, and the population for each year, P. Round values of P to the nearest whole number.

2

Comment on the closeness of the fit.

1

Find the value of x, the number of years after 1985.

2

Substitute this value of x into the formula and evaluate.

a ö 1.06 So P = 128 ì (1.06)x c

Year

1985

1990

1995

2000

2005

2010

x

0

5

10

15

20

25

P

128

171

229

307

411

549

The values for the population obtained using the formula closely resemble the actual data. d For the year 2015, x = 30.

Maths Quest 10 + 10A for the Australian Curriculum

P = 128 ì (1.06)30 = 735.166 87. . .

number AND algebra • Linear and non-linear relationships

3

Round to the nearest whole number.

P ö 735

4

Answer the question in a sentence.

The predicted population for 2015 is 735  000.

5

Repeat for the year 2020.

For the year 2020, x = 35. P = 128 ì (1.06)35 = 983.819  .  .  . P ö 984 The predicted population for 2020 is 984  000.

remember

In the function y = kax: 1. k represents the initial amount or quantity 2. a is the base. If a > 1, the function represents exponential growth. If 0 < a < 1, it represents exponential decay. 3. To find the value of a: (a) in the case of exponential growth, add the % increase to 100% and change the resulting percentage into a decimal (b) in the case of exponential decay, subtract the % decrease from 100% and change the resulting percentage into a decimal. Exercise

Exponential functions Fluency

5 4 3 0

n 0

2

4

6

8

10

N

12 000 10 000 8 000 6 000 4 000 2 000

A = 5000 ì (1.075)n A

14 000 12 000 10 000 8 000 6 000 4 000 2 000

1

2

N = 2000 ì 3x

x

20b

1   WE 6  The number of micro-organisms, N, in a culture dish after x hours is given by the

equation N = 2000 ì 3x. a Determine the initial number of micro-organisms in the dish. 2000 b Determine the number of micro-organisms in a dish after 5 hours. 486  000 c Draw the graph of N against x. d Use the graph to estimate the number of hours needed for the initial number of microorganisms to quadruple. 1.26  h 2 The value of an investment (in dollars) after n years is given by A = 5000 ì (1.075)n. a Determine the size of the initial investment. $5000 b Determine the value of the investment (to the nearest dollar) after 6 years. $7717 c Draw the graph of A against n. d Use the graph to estimate the number of years needed for the initial investment to double. 10 years 3   MC  a The function P = 300 ì (0.89)n represents an: A exponential growth with the initial amount of 300 B exponential growth with the initial amount of 0.89 ✔ C exponential decay with the initial amount of 300 D exponential decay with the initial amount of 0.89 E exponential decay with the initial amount of 300 ì 0.89 b The relationship between two variables, A and t, is described by the function A = 45 ì (1.095)t, where t is the time, in months, and A is the amount, in dollars. This function indicates: A a monthly growth of $45 B a monthly growth of 9.5 cents ✔ D a monthly growth of 9.5% C a monthly growth of 1.095% E a yearly growth of 9.5% Chapter 20 Functions and relations

675

number and algebra • linear and non-linear relationships 4 mC The graph of y = 2x + 1 - 1 is best represented by: y ✔ A B

y

1 0 -1

x

0

y

C

y

1 1

x

x

-1

D

0

1

-1

0

x

-1

y

E

2

-1

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0

x

5 mC The graph of y = 3x - 2 + 2 has an asymptote and y-intercept respectively at: 1 9 1 29

A y = 0, 2 ✔ B

y = 2,

C y = 2, 2 D y = 2, 1

8 9

E y = 0, 2 understanding eBook plus

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SkillSHEET 20.9 doc-5387

676

6 We7 A new washing machine costs $950. It is estimated that each year it will be losing 7% of

the previous year’s value. a Calculate the value of the machine after the first year. $883.50 b Calculate the value of the machine after the second year. $821.66 c Determine the equation that relates the value of the machine, $V, to the number of years, V = 950 ì (0.93)n n, that it has been used. d Use your equation to find the value of the machine in 12 years’ time. $397.67 7 A certain radioactive element decays in such a way that every 50 years the amount present decreases by 15%. In 1900, 120 mg of the element was present. 102 mg a Calculate the amount present in 1950. b Calculate the amount present in the year 2000. 86.7 mg

maths quest 10 + 10a for the australian Curriculum

number AND algebra • Linear and non-linear relationships A 140 120 100 80 60 40 20

A = 120 ì (0.85)t

0 50 200

C 100 80 60 40 20 0

c Determine the rule that connects the amount of the element present, A, with the number 83.927 mg of 50-year intervals, t, since 1900. A = 120 ì (0.85)t

t

d Calculate the amount present in the year 2010. Round your answer to 3 decimal places. e Graph the function of A against t. f Use the graph to estimate the half-life of this element (that is, the number of years needed for half the initial amount to decay). Approximately 210 years

8 When a shirt made of a certain fabric is washed, it loses 2% of its colour. a Determine the percentage of colour that remains after: i two washes   96.04% C = 100 ì (0.98)w ii five washes. 90.39% C = 100(0.98)w b Write a function for the percentage of colour, C, remaining after w washings. c Draw the graph of C against w. d Use the graph to estimate the number of washes after which there is only 85% of the original colour left. 8 washings 5 10 15 20 w 9   WE 8  The population of a certain country is shown in the table below.

Year 1990 1995 2000 2005 2010 Population 118 130 144 159 175 Calculated population is less accurate after 10 years.

Year

Population (in millions)

1990

118

1995

130

2000

144

2005

160

2010

178

Assume that the relationship between the population, P, and the year, n, can be modelled by the formula P = kan, where n is the number of years since 1990. a = 1.02; P = 118 ì (1.02)n a State the value of k. 118 (million) b Use the middle point of the data set to find the value of a rounded to 2 decimal places. Hence, write the formula that connects the two variables, P and n. c For the years given in the table, find the size of the population, using your formula. Compare the numbers obtained with the actual size of the population. 288 (million) d Predict the population of the country in the year 2035. 10 The temperature in a room (in degrees Celsius), recorded at 10-minute intervals after the air conditioner was turned on, is shown in the table below. Time (min)

0

10

20

30

40

Temperature (èC)

32

26

21

18

17

Assume that the relationship between the temperature, T, and the time, t, can be modelled by the formula T = cat, where t is the time, in minutes, since the air conditioner was turned on. a State the value of c. 32 b Use the middle point in the data set to find the value of a to 2 decimal places. 0.98 c Write the rule connecting T and t. T = 32 ì (0.98)t 2 6.1, 21.4, 17.5, 14.3; d Using the rule, find the temperature in the room 10, 20, 30 and 40 minutes after the air values are close except conditioner was turned on and compare your numbers with the recorded temperature. for t = 40. Comment on your findings. (Give answers correct to 1 decimal place.) 11 The population of a species of dogs (D) increases exponentially and is described by the equation D = 60(1 - 0.6t ) + 3, where t represents the time in years. a Calculate the initial number of dogs. 3 dogs b Calculate the number of dogs after 1 year. 27 dogs c Determine the time taken for the population to reach 50 dogs. 3 years Chapter 20 Functions and relations

677

number and algebra • linear and non-linear relationships 12 Carbon-14 decomposes in such a way that the amount present can be calculated using the

equation, Q = Q0(1 - 0.038)t, where Q is measured in milligrams and t in centuries. a If there is 40 mg present initially, how much is present in 10 years’ time and 2000 years’ time? b How many years will it take for there to be less than 10 mg? More than 35.78 centuries

a i 39.85 mg ii 18.43 mg

reasoning 13 Fiona is investing $20 000 in a fixed term deposit earning 6% p.a. interest. When Fiona has A = 20 000 ì 1.06x $30 000 she intends to put a deposit on a house. a Determine an exponential function that will model the growth of Fiona’s investment. b Graph this function. c Determine the length of time (correct to the nearest year) that it will take for Fiona’s 7 years investment to grow to $30 000. d Suppose Fiona had been able to invest at 8% p.a. How much quicker would Fiona’s investment have grown to the $30 000 she needs? 6 years — 1 year quicker e Alvin has $15 000 to invest. Find the interest rate at which Alvin must invest his money, if his investment is to grow to $30 000 in less than 8 years. 9.05% p.a. 14 A Petri dish containing a bacteria colony was exposed to an antiseptic. The number of bacteria

Investment ($)

within the colony, B, over time, t, in hours is shown in the diagram below. 30000 25000 20000 15000 10000 5000 0 1 2 3 4 5 6 7 8 Years

(‘000) B 120 100 80 60 40 20 0

(1, 84) (2, 58.8) (3, 41.16) (4, 28.81) 1 2 3 4 Hours

t Approximately 20 200

a Using the graph above, predict the number of bacteria in the Petri dish after 5 hours. b Using the points from the graph, show that if B can be modelled by the function B (in thousands) = abt, then a = 120 and b = 0.7. Teacher to check. c After 8 hours, another type of antiseptic was added to the Petri dish. Within three hours,

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678

the number of bacteria in the Petri dish had decreased to 50. If the number of bacteria decreased at a constant rate, show that the total of number of bacteria that had decreased within two hours was approximately 6700. Teacher to check. 15 One hundred people were watching a fireworks display at a local park. As the fireworks were set off, more people started to arrive to see the show. The number of people, P, at time, t minutes, after the start of the fireworks display, can be modelled by the function, P = abt. a If after 5 minutes there were approximately 249 people, show that the number of people arriving at the park to watch the fireworks increased by 20% each minute. The fireworks display lasted for 40 minutes. After 40 minutes, people started to leave the park. The number of people leaving the park could be modelled by an exponential function. 15 minutes after the fireworks ceased there were only 700 people in the park. a = 100, b = 1.20, increase = 20%/min reFleCtion    b Derive an exponential function that What are the main differences between can determine the number of people, a graph modelling exponential growth N, remaining in the park after the compared with one showing decay? fireworks had finished at any time, m, in minutes. N = 146 977 ì 0.70m

maths quest 10 + 10a for the australian Curriculum

number AND algebra • Linear and non-linear relationships

20c

Cubic functions ■■

■■

Cubic functions are those where the highest power of x is 3. These include functions such as y = x3 or y = (x + 1)(x − 2)(x + 3). These can be compared with quadratic functions such as y = x2 or y = (x + 1)(x − 2). The graphs of cubic functions have predictable properties, just as the graphs of quadratic functions do.

Worked Example 9

Plot the graph of y = x3 - 1 by completing a table of values. Think 1

2

Prepare a table of values, taking x values from -3 to 3. Fill in the table by substituting each x value into the given equation to find the corresponding y value. Draw a set of axes and plot the points from the table. Join them with a smooth curve.

Write

x

  -3

-2

-1

0

1

2

 3

y

-28

-9

-2

-1

0

7

26

y 25 20 15 10 5 -3 -2 -1 0 -1 -5 -10 -15 -20 -25

1 2 3

x

y = x3 - 1

Worked Example 10

Plot the curve of y = x(x - 2)(x + 2) by completing a table of values. Think 1

2

Prepare a table of values, taking x values from -3 to 3. Fill in the table by substituting each x value into the given equation. Draw a set of axes and plot the points from the table. Join them with a smooth curve.

Write

x

  -3

-2

-1

0

 1

2

 3

y

-15

 0

 3

0

-3

0

15

y 15

-3 -2 -1 0 1 -15

2 3

x

y = x (x - 2)(x + 2)

Chapter 20 Functions and relations

679

number AND algebra • Linear and non-linear relationships

■■

A good sketch of a cubic function shows: 1. x- and y-intercepts 2. the behaviour of the function at extreme values of x, that is, as x approaches infinity (x ç + Ñ) and as x approaches negative infinity (x ç - Ñ) 3. the general location of turning points. Note that for cubic functions, ‘humps’ are not symmetrical as they are for parabolas, but are skewed to one side. The graphs below show the two main types of cubic graph. y

y

Point of Inflection

Turning points

x x

  Consider the general factorised cubic y = (x - a)(x - b)(x - c).   The x-intercepts occur when y = 0, that is, when x = a or x = b or x = c. The y-intercept occurs when x = 0, that is, the y-intercept is

y = (0 - a)(0 - b)(0 - c) = -abc y

c

b

a

x

-abc

Worked Example 11

Sketch the following, showing all intercepts: a  y = (x - 2)(x - 3)(x + 5) b  y = (x - 6)2(4 - x) c  y = (x - 2)3. Think a

680

Write

1

Write the equation.

2

The y-intercept occurs where x = 0. Substitute x = 0 into the equation.

Maths Quest 10 + 10A for the Australian Curriculum

a y = (x - 2)(x - 3)(x + 5)

y-intercept: if x = 0, y = (-2)(-3)(5) = 30 Point: (0, 30)

number AND algebra • Linear and non-linear relationships

3

Solve y = 0 to find the x-intercepts.

4

Combine the above steps to sketch.

x-intercepts: if y = 0, x - 2 = 0, x - 3 = 0 or x + 5 = 0 x = 2, x = 3 or x = -5 Points: (2, 0), (3, 0), (-5, 0) y

30 -5 b

2

x

b y = (x - 6)2(4 - x)

1

Write the equation.

2

Substitute x = 0 to find the y-intercept.

y-intercept: if x = 0, y = (-6)2(4) = 144 Point: (0, 144)

3

Solve y = 0 to find the x-intercepts.

x-intercepts: if y = 0, x - 6 = 0  or  4 - x = 0 x = 6  or x=4 Points: (6, 0), (4, 0)

4

Combine all information and sketch the graph. Note: The curve just touches the x-axis at x = 6. This occurs with a double factor such as (x - 6)2.

y 144

4 c

3

c

6

x

y = (x - 2)3

1

Write the equation.

2

Substitute x = 0 to find the y-intercept.

y-intercept: if x = 0, y = (-2)3 = -8

3

Solve y = 0 to find the x-intercepts.

x-intercept: if y = 0, x-2=0 x=2

4

Combine all information and sketch the graph. Note: The point of inflection is at x = 2. This occurs with a triple factor such as (x - 2)3.

y

2

x

-8

Chapter 20 Functions and relations

681

number AND algebra • Linear and non-linear relationships

remember 1 a

y

0

x

1 2 3

-6

To sketch a cubic function: 1. find the y-intercept (let x = 0) 2. find the x-intercepts (let y = 0) 3. use all available information to sketch the graph.

y 20

y = (x +1)(2x - 5)(x - 4)

-1

Exercise

Cubic functions

20c b

y 30

-2 0

5 x

3

c

y 7 x

-1 0 -42

-6

x

4

5– 2

y 30

2a

1   WE 9, 10, 11  Sketch the following, showing all intercepts. a y = (x - 1)(x - 2)(x - 3) b y = (x - 3)(x - 5)(x + 2) c y = (x + 6)(x + 1)(x - 7) d y = (x + 4)(x + 9)(x + 3) e y = (x + 8)(x - 11)(x + 1) f y = (2x - 6)(x - 2)(x + 1) g y = (2x - 5)(x + 4)(x - 3) h y = (3x + 7)(x - 5)(x + 6) i y = (4x - 3)(2x + 1)(x - 4) j y = (2x + 1)(2x - 1)(x + 2) b k y = (x - 3)2(x - 6) l y = (x + 2)(x + 5)2 2 Sketch the following (a mixture of positive and negative cubics). a y = (2 - x)(x + 5)(x + 3) b y = (1 - x)(x + 7)(x - 2) c y = (x + 8)(x - 8)(2x + 3) d y = (x - 2)(2 - x)(x + 6) e y = x(x + 1)(x - 2) f y = -2(x + 3)(x - 1)(x + 2) g y = 3(x + 1)(x + 10)(x + 5) h y = -3x(x - 4)2 2 i y = 4x (x + 8) j y = (5 - 3x)(x - 1)(2x + 9) c k y = (6x - 1)2(x + 7) l y = -2x2(7x + 3)

-5

y

A

y

2

x

-14

0 1 2 x

y

-8

y

B

0 y

-7

3   MC  Which of the following is a reasonable sketch of y = (x + 2)(x - 3)(2x + 1)?

d

-3

- 3– 0 2

8

x

2

x

-192

108

-9

-4 -3

e -8

-88

f

11

0

x

y -2 - 1– 2

2

3

-2 -

2 x D 3 x

4   MC  The graph shown could be that of: A y = x2(x + 2) B y = (x + 2)3 ✔ C y = (x - 2)(x + 2)2 D y = (x - 2)2(x + 2)

y 12

-1 0

682

✔ C

y -1

-3

x

0

1– 2

x

Maths Quest 10 + 10A for the Australian Curriculum

d

3 x

1– 2

y -6

0 -24

y 1– 2

2

3 x

y

-2

2

-8

x

number AND algebra • Linear and non-linear relationships y

5   MC  The graph at right has the equation: A y = (x + 1)(x + 2)(x + 3) ✔ B y = (x + 1)(x - 2)(x + 3) C y = (x - 1)(x + 2)(x + 3) D y = (x - 1)(x + 2)(x - 3)

-3

-1

x

2

-6 6   MC  If a, b and c are positive numbers, the equation of the graph shown below could be: y A y = (x - a)(x - b)(x - c) B y = (x + a)(x - b)(x + c) C y = (x + a)(x + b)(x - c) ✔ D y = (x - a)(x + b)(x - c) -b reflection 

c

a

x

 

Is it possible to get symmetrical ‘humps’ for the graphs of cubic functions?

20d

Quartic functions ■■

■■

Quartic functions are those where the highest power of x is 4. These include functions such as y = x4 or y = (x + 1)(x − 2)(x + 3)(x − 5). These can be compared with cubic functions such as y = x3 or y = (x + 1)(x − 2)(x + 3). It is necessary, when sketching the graphs of quartic functions, to find all the intercepts on both the x- and y-axes. This is best achieved by factorising the expression.

Basic shapes of quartic graphs ■■

The direction of a quartic graph is determined by the coefficient of the x4 term. This is similar to the effect the coefficient of x2 has on the shape of a parabola. Consider the coefficient of x4 to be a. When a is positive (a > 0) 1. y = ax4

2. y = ax4 + cx2, c í 0

y

3. y = ax2(x - b)(x - c)

y

y

b

x

0

4. y = a(x - b)2(x - c)2

5. y = a(x - b)(x - c)3

6. y = a(x - b)(x - c)(x - d )(x - e)

y

b 0

x

0

y

b

c x

0

c

x

0

y

c

x b

c

0

d

ex

Chapter 20 Functions and relations

683

number AND algebra • Linear and non-linear relationships y

When a is negative (a < 0) If a was negative in each of the previous graphs, they would be reflected in the x-axis.   The inverted form of y = x4 is shown below — y = -x4. ■■

x

0

To find the x-intercepts of a quartic function, let y = 0, and solve the equation for x.

y = -x 4

Worked Example 12

Sketch the graph of y = x4 - 2x3 - 7x2 + 8x + 12, showing all intercepts. Think

Write

1

Find the y-intercept.

When x = 0, y = 12. The y-intercept is 12.

2

Let P(x) = y.

Let P(x) = x4 - 2x3 - 7x2 + 8x + 12.

3

Find two linear factors of the quartic expressions, if possible, using the factor theorem.

P(1) = (1)4 - 2(1)3 - 7(1)2 + 8(1) + 12 = 12 ò0 P(–1) = (-1)4 - 2(-1)3 - 7(-1)2 + 8(-1) + 12 =0 (x + 1) is a factor. P(2) = (2)4 - 2(2)3 - 7(2)2 + 8(2) + 12 =0 (x - 2) is a factor.

4

Find the product of the two linear factors.

(x + 1)(x - 2) = x2 - x - 2

5

Use long division to divide the quartic by the quadratic factor x2 - x - 2.

x2

x2 -   x -   6 - x - 2) - 7x2 + 8x + 12 4 3 x -   x - 2x2 -x3 - 5x2 + 8x -x3 +   x2 + 2x -6x2 + 6x + 12 -6x2 + 6x + 12 0 x4

2x3

6

Express the quartic in factorised form.

y = (x + 1)(x - 2)(x2 - x - 6) = (x + 1)(x - 2)(x - 3)(x + 2)

7

To find the x-intercepts, solve y = 0.

If 0 = (x + 1)(x - 2)(x - 3)(x + 2) x = -1, 2, 3, -2.

8

State the x-intercepts.

The x-intercepts are -2, -1, 2, 3.

9

Sketch the graph of the quartic.

y 12

-2 -10

684

Maths Quest 10 + 10A for the Australian Curriculum

2 3

x

number AND algebra • Linear and non-linear relationships

remember

1 a

y 24

-3 -1 0

b

y

Quartic graphs 1. General equation is y = ax4 + bx3 + cx2 + dx + e. 2. Basic shape of quartic graphs: (a) If a > 0: y = a(x - b)(x - c)(x - d)(x - e) 2

4

y

x

y

y

x

0

10

b

Exercise

b 0

c x

c

x

y = a(x - b)(x - c)3

(b) If a < 0, then the reflection in the x-axis of the types of graph in the figures above is obtained.

x

5

y

y = ax2(x - b)(x - c)

y = ax4 + cx2, c ≥ 0

-2-10 1

0

ex

b c 0d

Quartic functions

20d

Understanding c

1   WE 12  Sketch the graph of each of the following showing all intercepts. You may like to verify

y 32 x

-2 0 1 2

-4

d

y

-2 0

-5

e

x

3

y - 3

-3

-1 0

3

x

the shape of the graph using a graphics calculator. y = (x - 2)(x + 3)(x - 4)(x + 1) y = (x2 - 1)(x + 2)(x - 5) y = 2x4 + 6x3 - 16x2 - 24x + 32 y = x4 + 4x3 - 11x2 - 30x y y = x4 + 4x3 - 12x - 9 36 y = x4 - 4x2 + 4 y = 30x - 37x2 + 15x3 - 2x4 x - 3–2 0 2–3 -3 2 y = 6x4 + 11x3 - 37x2 - 36x + 36 2   MC  Consider the function f (x) = x4 - 8x2 + 16. a When factorised, f (x) is equal to: A (x + 2)(x - 2)(x - 1)(x + 4) B (x + 3)(x - 2)(x - 1)(x + 1) ✔ D (x - 2)2(x + 2)2 C (x - 2)3(x + 2) b The graph of f  ( x) is best represented by: a b c d e f g h

-2

-9

f

y

A

y 16

✔ B

0

2

x

y 4

- 2

-16

0

x

2

-2

y 16

C

x

2

y

D

g

4

y

0

0

2

5– 2

3

x

-2

0

2

x

-2

0

2

x

Chapter 20 Functions and relations

685

number and algebra • linear and non-linear relationships 3 a

reasoning

y

3 Sketch the graph of each of the following functions. a y = x(x - 1)3 3 c b y = (2 - x)(x2 - 4)(x + 3)

d

y y = x4 - x2 x 0 4 3 2 1 y = 9x - 30x + 13x + 20x + 4 b y = -(x - 2)2(x + 1)2 y y = x4 - 6x2 - 27 0 -1 3(x - 3) x 0 y = (x + 2) -3 -2 2 y = 4x2 - x4 Verify your answers using a graphics calculator. -24 (3, -30) 4 The function f (x) = x4 + ax3 - 4x2 + bx + 6 has x-intercepts (2, 0) and (-3, 0). Find the values eBook plus a = 4, b = -19 of a and b. 5 The functions y = (a - 2b)x4 - 3x - 2 and Digital doc WorkSHEET 20.3 y = x4 - x3 + (a + 5b)x2 - 5x + 7 both have an x-intercept doc-5389 of 1. Find the value of a and b. a = 3, b = -1

c d e f g h

y 400 300 200

1

x

(-1, 36)

100

-2 -1

reFleCtion 

0 1

eBook plus

■

Interactivity Polynomial transformations

int-2794

■ x

2

Once the basic shape of the graph of a particular function is known, it is not difficult to predict the shape of a related function, which is a transformation of the basic function. Transformations of parabolas have been dealt with previously, but for the sake of comparison with other functions, it will be included in this chapter. Other functions considered are circles, hyperbolas, exponential functions, cubic and quartic functions. This is essentially a summary of transformations of functions discussed previously.

The basic quadratic function is y = x2. The shape of its graph is:

(-2, -16) −16

y = x2

y

0

-3

x

3

(0, 0)

-27

g (-3, 6)

■ -2

3

x

Adding or subtracting a constant to y = x2 moves the curve up or down the y-axis. y

-24

h

y = x2 + 2

y -2

(-3, -45)

0

y

y = x2 y = x2 - 3

y = x2 2

x

x

(0, 2) x

686

x

Vertical translation

y

0

x

What are the basic differences between cubic and quartic functions?

y

f

3

 

quadratic functions

y −1 0

2

transformations

20e

3 e

(-2, 400)

maths quest 10 + 10a for the australian Curriculum

(0, -3)

number AND algebra • Linear and non-linear relationships

Horizontal translation ■■

If the graph of y = x2 is translated b units horizontally, the equation becomes y = (x - b)2. y

y

y = x2

(0, 1)

y = (x - 2)2

(0, 4)

x

(2, 0)

y = (x + 1)2 y = x2

x

(-1, 0)

Dilation ■■

If the graph of y = x2 is dilated by factor a, the graph becomes narrower if a >1 and wider if 0 < a < 1. y

y

y = 2x2 y = x2

(0, 0)

x

y = 1–4 x2 y = x2

x

(0, 0)

Reflection ■■

If the x2 term is positive, the graph is concave up, while if there is a negative sign in front of the x2 term, the graph is concave down.

y

y = x2

x

(0, 0)

y = -x2

Circles ■■

y

The equation of a circle with centre (0, 0) and radius r is x2 + y2 = r2.

P(x, y) y

r

x

x

Translation ■■

If the circle is translated b units to the right, parallel to the x-axis, and k units upwards, parallel to the y-axis, the equation of the circle, centre (h, k) becomes (x - h)2 + (y - k)2 = r2.

y y k

P(x, y) (y - k) (x - h) h

x x

Chapter 20 Functions and relations

687

number AND algebra • Linear and non-linear relationships

Hyperbolas ■■ ■■

k The hyperbola is a function of the form xy = k or y = . x 1 The graph of y = has the shape x y 2 1

-3 -2 -1

y = —1x

0

x

1 2 3 -1 -2

Dilation

k 1 Graphs of the form y = are the same basic shape as y = , with y-values dilated by a factor x x of k. y

-2 -1

8 4

y = —4x

0

x

1 2 -4 -8

Negative values of k ■■

Negative values of k cause the graph to be reflected across the y-axis. y 6 3 -3 -2 -1 0 -3

-

y = —x3 1 2 3 x

-6

Exponential functions ■■

These functions are of the form y = ax, where a ò 1. The basic shape has a y-intercept of 1. y

y = 2x

18 16 14 12 10 8 6 4 2 -4 -3 -2 -1 0

688

Maths Quest 10 + 10A for the Australian Curriculum

1 2 3 4

x

number AND algebra • Linear and non-linear relationships

Functions for the form y = k ì ax ■■

Multiplying by a factor of k causes the y-intercept to move to the point (0, k). 24 22 20 18 16 14 12 10 8 6 4 2

y y = 3 ì 2x

y=0

-3 -2 -1 0

x

1 2 3

Functions with a negative exponent ■■

This causes the graph to be reflected in the y-axis. y y = 3-x

28 26 24 22 20 18 16 14 12 10 8 6 4 2

-3 -2 -1 0

y=1

y=0 x

1 2 3

Cubic functions ■■

The basic form of a cubic function is y = x3. This can also be expressed in the form y = a(x − b)3 + c, where a = 1, b = 0 and c = 0. y

y = x3

x

Translation ■■

If a ò 1, b ò 0 and c ò 0, the graph is translated +b units in the x direction, +c units in the y direction, and dilated by a factor of a in the y direction.

y

y = a(x - b)3 + c

(b, c) x

Chapter 20 Functions and relations

689

number AND algebra • Linear and non-linear relationships

Reflection ■■

The cubic function can be expressed in factor form as y = a(x - b)(x - c)(x - d), where b, c and d are the x-intercepts. If the value of a is negative, this causes the curve to be reflected in the x-axis. y

y = a(x - b)(x - c)(x - d) where a > 0

y = -(x + 2)(x - 1)(x - 3) y

-2 b

c d

3 x

1

x

Quartic functions ■■

The basic form of the quartic function y = ax4, when a is positive, has the following shape.

y y = ax4 where a > 0 x

0

Reflection ■■

y

Negative values of a cause the graph to be reflected in the x-axis.

x

0

y = -x 4

Transformation in general polynomials ■■

With knowledge of the transformations which occur in the functions just discussed, it is possible to generate many other graphs without knowing the equation of the original function. Consider a basic polynomial y = P(x) and what happens to the shape of the curve as the function is changed.

Worked Example 13

Use the sketch of y = P(x) shown at right to sketch: a y = P(x) + 1 b y = P(x) - 1 c y = -P(x)

y

0

x

y = P(x) 690

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • Linear and non-linear relationships

Think a

Write/draw

1

Sketch the original y = P(x).

2

Consider the x-values. They remain unchanged — there is no horizontal translation.

a

y

x

0

y = P(x) 3

Consider the y-values. They are increased by 1 — the curve is shifted up 1 unit.

y

1 x

0

y = P(x) + 1

b

4

Sketch the graph of y = P(x) + 1 using a similar scale to the original.

1

Sketch the original y = P(x).

2

Consider the x-values. They remain unchanged — there is no horizontal translation.

b

y

x

0

y = P(x) 3

Consider the y-values. They are decreased by 1 — the curve is shifted down 1 unit.

y

0

-1

x

y = P(x) - 1 4

Sketch the graph of y = P(x) - 1 using a similar scale to the original. Chapter 20 Functions and relations

691

number AND algebra • Linear and non-linear relationships c

1

Sketch the original y = P(x).

2

Consider the x-values. They remain unchanged — there is no horizontal translation.

c

y

x

0

y = P(x) 3

Consider the y-values. They will all change sign — the curve will be reflected in the x-axis. That is, negative becomes positive and positive becomes negative.

y

y = -P(x)

x

0

4

Sketch the graph of y = -P(x) using a similar scale to the original.

remember

To sketch general polynomials: 1. sketch the original curve 2. examine the changes in the x-values 3. examine the changes in the y-values 4. draw the final sketch using a similar scale. Exercise

Transformations

20e

Understanding y

y = -P(x)

x 1 -1

y y = P(x + 2)

692

2 Consider the sketch of y = P(x). Sketch: a y = P(x) + 1 b y = -P(x) c y = P(x + 2)

0

y = P(x) + 1 y = P(x)

1   WE 13  Use the sketch of y = P(x) shown at right to sketch: a y = P(x) + 1 b y = P(x) - 2 c y = -P(x) d y = 2P(x)

y

0

x

0

y = P(x)

1

y = P(x)

y

x

0

y = 2P(x) y = P(x) y = P(x) + 1 y = P(x) - 2 x

y = -P(x)

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • Linear and non-linear relationships Reasoning 3 Draw any polynomial y = P(x). Discuss the similarities and differences between the graphs of They have the same x-intercepts, but y = -P(x) is a reflection of y = P(x) in the x-axis. y = P(x) and y = -P(x). 4 Draw any polynomial y = P(x). Discuss the similarities and differences between the graphs of They have the same x-intercepts, but the y-values in y = 2P(x) are all twice as large. y = P(x) and y = 2P(x). 5 Draw any polynomial y = P(x). Discuss the similarities and differences between the graphs of y = P(x) and y = P(x) - 2. The entire graph is moved down 2 units. The shape is identical. 6 Consider the sketch of y = P(x).

 ive a possible equation for each of the following in terms of G P(x).

y

a y = -P(x) b y = P(x) - 3 c y = 2P(x)

a

0 -1

y

b

c

y

x y = P(x)

y

1 0

x 0

x

0

x

-2

-3 -4

reflection 

 

Why is it important to understand how transformations can affect the shape of a graph?

Chapter 20 Functions and relations

693

number AND algebra • Linear and non-linear relationships

Summary Functions and relations ■■ ■■

■■

■■ ■■

A function is a relation so that for any x-value there is at most one y-value (one-to-one or many-to-one relations). y y Vertical line test: The graph of a function cannot be crossed more than once by any vertical line. x 0 f (x) = .  .  . is used to describe ‘a function of x x’. To evaluate the function, for example 0 when x = 2, find f (2) by replacing each occurrence of x on the RHS with 2. Function Not a function Substitute appropriate x values to describe what happens to functions as x ç Ñ (x approaches infinity) or x ç 0 (x approaches zero). To find points of intersection, solve function equations simultaneously to find both x and y values. Exponential functions

In the function y = kax: ■■ k represents the initial amount or quantity ■■ a is the base. If a > 1, the function represents exponential growth. If 0 < a < 1, it represents exponential decay. ■■ To find the value of a: (a) in the case of exponential growth, add the % increase to 100% and change the resulting percentage into a decimal (b) in the case of exponential decay, subtract the % decrease from 100% and change the resulting percentage into a decimal. Cubic functions y

To sketch a cubic function: find the y-intercept (let x = 0) ■■ find the x-intercepts (let y = 0) ■■ use all available information to sketch the graph. ■■

20

-1

y = (x +1)(2x - 5)(x - 4)

5– 2

x

4

Quartic functions

Quartic graphs ■■ General equation is y = ax4 + bx3 + cx2 + dx + e. ■■ Basic shape of quartic graphs: (a) If a > 0: y = a(x - b)(x - c)(x - d)(x - e) y

y

x

0

b

0

y

b c 0d

ex

y

c x

b 0

c

x

y = ax2(x - b)(x - c) y = a(x - b)(x - c)3 y= + c≥0 (b) If a < 0, then the reflection in the x-axis of the types of graph in the figures above is obtained. ax4

694

cx2,

Maths Quest 10 + 10A for the Australian Curriculum

number AND algebra • Linear and non-linear relationships Transformations

To sketch general polynomials: sketch the original curve ■■ examine the changes in the x-values ■■ examine the changes in the y-values ■■ draw the final sketch using a similar scale. ■■

Mapping your understanding

Using terms from the summary above, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 663.

Chapter 20 Functions and relations

695

number AND algebra • Linear and non-linear relationships

Chapter review Fluency

8   MC  Which of the following shows the graph of

y = -2(x + 5)3 - 12?

1 Which of the following are functions? a y y a b

y

✔ A

y

B x

x

0

x

0

x

d y = 3 If a b c

y

C

2 Which of the following are functions? a, c, d a y = 2x - 7 b x2 + y2 = c y = 2x

(5, -12)

1 x +1

1

y

9 Sketch: a y = x(x - 2)(x + 11) b y = x3 + 6x2 - 15x + 8 c y = -2x3 + x2

8 x

-11

x

x

2

10   MC  The rule for the graph shown below could be:

4 Sketch each of the following curves, showing all

intercepts. a y = (x - 1)(x + 2)(x - 3) b y = (2x + 1)(x + 5)2

(5, 12)

x

y

-8

y

D

30

f ( x ) = 4 − x 2 , find: f (0) 2 f (1) 3 f (2) 0

(-5, 12)

(-5, -12)

y

y

f(x)

y

25 y = (2x + 1)(x + 5)2

5 Give an example of the equation of a cubic that

1 0 -5 - — 2

would just touch the x-axis and cross it at another point. 6 Match each equation with its type of curve. a y = x2 + 2 D A  circle b x2 + y2 = 9 A B  cubic

2 C  exponential x+2 d g(x) = 6-x C D  parabola e h(x) = (x + 1)(x - 3)(x + 5) B E  hyperbola

x

0

x

2

A f (x) = x(x + 2)3 C f (x) = x2(x - 2)2

1– 2

B f (x) = -x(x - 2)2 ✔ D

f (x) = x(x - 2)3

11   MC  The graph of y = (x + 3)2(x - 1)(x - 3) is best

c f ( x ) = E

represented by: ✔ A

b

y

y

7   MC  The equation for this graph could be: y

y

-3

6

-3 -2 0 1 3

c -3

696

1

3

1

3

1

3

x

x

x

y = (x - 1)(x + 2)(x - 3)

A B C ✔ D

0

0

-1

5

d

y

y

x

y = (x - 5)(x + 1)(x + 3) y = (x - 3)(x - 1)(x + 5) y = (x - 3)(x + 1)(x + 5) y = (5 - x)(1 + x)(3 + x)

Maths Quest 10 + 10A for the Australian Curriculum

-3

0

1

3

x

-3

0

x

x

number and algebra • linear and non-linear relationships 12 Sketch the graph of y = x4 - 7x3 + 12x2 + 4x - 16,

showing all intercepts. 13 Consider the sketch of y = P(x) shown at right. y Sketch y = -P(x).

6 The number of hyenas, H, in the zoo is given by

y

y

d Determine the rule for this increase in temperature. T = 50 ì 2x e If the stove is left on, what would the predicted temperature be in 6 hours? 3200 èC

H = 20(100.1t ), where t is the number of years since 0 counting started. At the same time, the number of -1 2 4 dingoes, D, is given by D = 25(100.05t ). x 0 1 1 0 1x a Calculate the number of: -1 -16 i hyenas 20 ii dingoes 25 H yenas after 3 years; dingoes after 4 years when counting began. 14 Draw any polynomial y = P(x). Discuss the b Calculate the numbers of each after: similarities and differences between the graphs of i 1 year H = 25; D = 28 y = P(x) and y = P(x) + 3. The entire graph is moved up ii 18 months. H = 28; D = 30 3 units. The shape is identical. c Which of the animals is the first to reach a problem solVing population of 40 and by how long? 1 Find the value(s) of x for which: d After how many months are the populations a f (x) = x2 + 7 and f (x) = 16 x = ê 3 equal and what is this population? 1 1 7 The temperature in a green house is monitored b g( x ) = and g(x) = 3 x = 2 3 x −2 when the door is left open. The following c h( x ) = 8 + x and h(x) = 6 x = 28 measurements are taken. 2 Describe what happens to f (x) = -2x as x ç Ñ and Time (min) 0 5 10 15 20 x ç - Ñ. Temperature (èC) 45 35 27 23 16 3 Find any points of intersection between f (x) = x2 - 4 and g(x) = x3 + x2 - 12 (2, 0) a Determine an exponential equation to fit the 45 èC collected data. T = 45 ì 0.95t 4 The concentration of alcohol (mg/L) in a bottle of kt b State the initial temperature of the green house. champagne is modelled by C = C0 ì 0.33 where c What will the temperature be after 30 minutes? t represents the time in days after the bottle is It is discovered that one of the temperature opened. If the initial concentration is 80 mg/L and 10 èC readings is incorrect. the concentration after 1 day is 70 mg/L, find the d Recalculate all the temperatures using the concentration remaining after: exponential rule found in part a. 52.67 mg/L a 3 days e If the original incorrect temperature was b 1 week 31.524 mg/L omitted from the data, does this change c 18 hours. 72.4 mg/L the rule? No No. The line T = 0 is an asymptote. 5 A hot plate used as a camping stove is cooling f Will the temperature ever reach 0  èC? Explain. down. The formula which describes this cooling pattern is T = 500 ì 0.5t where T is the temperature Time (min) 0 5 10 15 20 in degrees Celsius and t is the time in hours. Temperature (èC) 45 35 27 21 16 a What is the initial temperature of the stove? 500 èC b What is the temperature of the stove after eBook plus 2 hours? 125 èC c Decide when the stove will be cool enough to Interactivities touch and give reasons. Test yourself Chapter 20 A second camp stove is set up and turned on. Its int-2879 Word search Chapter 20 heating temperature is measured and listed below. x

As x ç Ñ, f (x) ç -Ñ As x ç -Ñ, f (x) ç 0

After about 23 months; 31 animals

Time (hours)

0

Temperature (èC)

50

1

2

3

4

100 200 400 800

int-2877 Crossword Chapter 20 int-2878

B etween 5 and 6 hours once it has cooled to below 15 èC

Chapter 20 Functions and relations

697

eBook plus

aCtiVities

Are you ready? Digital docs (page 664) • SkillSHEET 20.1 (doc-5378): Finding the gradient and y-intercept • SkillSHEET 20.2 (doc-5379): Sketching straight lines • SkillSHEET 20.3 (doc-5380): Sketching parabolas • SkillSHEET 20.4 (doc-5381): Completing the square • SkillSHEET 20.5 (doc-5382): Identifying equations of straight lines and parabolas • SkillSHEET 20.6 (doc-5383): Finding points of intersection • SkillSHEET 20.7 (doc-5384): Substitution into index expressions

20A Functions and relations Digital doc

• WorkSHEET 20.1 (doc-5385): Functions and relations (page 671) 20B Exponential functions eLesson

• Exponential growth (eles-0176) (page 671) Digital docs

• SkillSHEET 20.8 (doc-5386): Converting a percentage to a decimal (page 676)

698

maths quest 10 + 10a for the australian Curriculum

• SkillSHEET 20.9 (doc-5387): Decreasing a quantity by a percentage (page 676) • WorkSHEET 20.2 (doc-5388): Exponential growth and decay (page 678) 20D Quartic functions Digital doc

• WorkSHEET 20.3 (doc-5389): Cubic and quartic functions (page 686) 20E Transformations Interactivity

• Polynomial transformations (int-2794) (page 686) Chapter review Interactivities (page 697) • Test yourself Chapter 20 (int-2879): Take the end-ofchapter test to test your progress • Word search Chapter 20 (int-2877): an interactive word search involving words associated with this chapter • Crossword Chapter 20 (int-2878): an interactive crossword using the definitions associated with the chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

measurement AND geometry • Geometric reasoning

21

21A Angles in a circle 21B Intersecting chords, secants and tangents 21C Cyclic quadrilaterals 21D Tangents, secants and chords What do you know ?

Circle geometry

1 List what you know about circle geometry. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of circle geometry.

opening question

How does an understanding of shapes and angles help us design the most effective sails for yachts?

measurement and geometry • geometric reasoning

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

Digital doc

SkillSHEET 21.1 doc-5390

Using 1 a b c

tests to prove congruent triangles

True or false? Congruent triangles are identical triangles in every way. True State four tests that can be used to prove congruent triangles. Prove that DABC ô DADC, giving reasons.

A

S SS (all corresponding sides equal in length), SAS (two corresponding sides equal in length, included angle equal), ASA (two angles equal, one pair of corresponding sides equal B D in length), RHS (right-angled triangles with the hypotenuses C and one other pair of corresponding sides equal in length) Corresponding sides and angles of congruent triangles 2 The two triangles below are congruent. ±QPR is common. A Q ±PQR = ±PST (corresponding angles are equal as QR || ST) P ±PRQ = ±PTS (corresponding angles are equal as QR || ST) B DPQR ~ DPST (equiangular) C R Triangle 1 Triangle 2

AC is common. ±BAC = ±DAC (given) AB = AD (given) DABC ô DADC (SAS) eBook plus

Digital doc

SkillSHEET 21.2 doc-5391

a b c d eBook plus

Digital doc

SkillSHEET 21.3 doc-5392

Using 3 a b c

Which side in triangle 2 corresponds to side AB in triangle 1? RP Which side in triangle 1 corresponds to side PQ in triangle 2? BC 1. ±RQP Name the angle in triangle 2 that corresponds to ±ACB in triangle Name the angle in triangle 1 that corresponds to ±PRQ in triangle 2. ±BAC False. Sides may be different.

tests to prove similar triangles

P

True or false? Similar triangles have equal angles and sides. State four tests than can be used to prove similar triangles. Prove that DPQR ~ DPST, giving reasons.

Q

A AA or equiangular (all corresponding angles equal), SSS (all corresponding sides in same ratio), SAS (two pairs of corresponding sides in same ratio and included angle equal), RHS (both right-angled triangles with the hypotenuses and one other pair of corresponding sides in same ratio) Angles in a triangle eBook plus 4 Find the value of the pronumeral in each of the following triangles. Digital doc a c b a SkillSHEET 21.4 b

S

55è

doc-5393

43è

Digital doc

SkillSHEET 21.5 doc-5394

700

T

a a = 84è b b = 88è c c = 75è

53è c

46è eBook plus

R

More angle relations 5 Find the value of the pronumeral in each of the following. a b 77è b x a 28è

maths quest 10 + 10a for the australian curriculum

130è

a x = 62è b a = 77è, b = 103è c y = 45è c 31è 140è y 49è 95è

measurement and geometry • geometric reasoning

21a eBook plus

Interactivity Angles in a circle

angles in a circle introduction ■

■

int-2795

eBook plus

Digital docs

SkillSHEET 21.1 doc-5390 SkillSHEET 21.2 doc-5391 SkillSHEET 21.3 doc-5392

■

■

■ ■

In circle geometry, there are many theorems that can be used to solve problems. It is important that we are also able to prove these theorems. To prove a theorem is true: 1. state the aim of the proof 2. use given information and previously established theorems to establish the result 3. give a reason for each step of the proof 4. state a clear conclusion. In proving circle geometry theorems, the most useful tool is to be able to prove both congruent and similar triangles. The SkillSHEETs may assist you in revising these concepts. You may also need to construct new lines on the diagram. By convention, we will use the letter O to represent the centre of the circles. In this chapter we will prove and then use 14 different theorems. First, let’s recall the definitions for various parts of a circle.

parts of a circle Part (name) Centre

Circumference

Radius

Diameter

Chord

Description

Diagram

The middle point, equidistant from all points on the circumference. It is usually shown by a dot and labelled O.

The outside length or the boundary forming the circle. It is the circle’s perimeter.

A straight line from the centre to any point on the circumference.

A straight line from one point on the circumference to another, passing through the centre.

A straight line from one point on the circumference to another.

O

O

O

O

O

(continued) chapter 21 circle geometry

701

measurement AND geometry • geometric reasoning

Part (name) Segment

Description The area of the circle between a chord and the circumference. The smaller segment is called the minor segment and the larger segment is the major segment.

Diagram

O

O

Sector

An area of a circle enclosed by 2 radii and the circumference.

O

O

Arc

A portion of the circumference. O

O

Tangent

A straight line that touches the circumference at one point only. O

Secant

702

A chord extended beyond the circumference on one side.

Maths Quest 10 + 10A for the Australian Curriculum

O

measurement AND geometry • geometric reasoning

Angles in a circle

C

■■

In the diagram at right, chords AC and BC form the angle ACB. We say that arc AB has subtended angle ACB.

■■

Theorem 1  Code The angle subtended at the centre of a circle is twice the angle subtended at the circumference, standing on the same arc.

A

B

R

Proof: Let ±PRO = x and ±QRO = y O RO = PO = QO (radii of the same Q circle are equal) P ±RPO = x and ±RQO = y R ±POM = 2x (exterior angle of triangle) xy and ±QOM = 2y (exterior angle of triangle) ±POQ = 2x + 2y Q O = 2(x + y) P M which is twice the size of ±PRQ = x + y.   The angle subtended at the centre of a circle is twice the angle subtended at the circumference, standing on the same arc. ■■

Theorem 2  Code R All angles that have their vertex on the circumference S and are subtended by the same arc are equal. O Proof: Q Join P and Q to O, the centre of the circle. P P Q Let ±PSQ = x ±POQ = 2x (angle at the centre is twice the angle at the circumference) ±PRQ = x (angle at the circumference is half the angle of the centre) ±PSQ = ±PRQ. Angles at the circumference subtended by the same arc are equal.   The application of the first two circle geometry theorems can be seen in the following worked example.

Worked Example 1

Find the values of the pronumerals in the diagram at right, giving reasons for your answers.

Think

46è

O y

x

Write

1

Angles x and 46è are angles subtended by the same arc and both have their vertex on the circumference.

x = 46è 

2

Angles y and 46è stand on the same arc. The 46è angle has its vertex on the circumference and y has its vertex at the centre. The angle at the centre is twice the angle at the circumference.

y = 2 ì 46è  = 92è

Chapter 21 Circle geometry

703

measurement AND geometry • geometric reasoning

■■

Theorem 3  Code Angles subtended by the diameter, that is, angles in a semicircle, are right angles.   In the diagram at right, PQ is the diameter. Angles a, b c b and c are right angles. This theorem is in fact a special Q case of Theorem 1. O

P

Proof: a ±POQ = 180è (straight line)   Let S refer to the angle at the circumference subtended by the diameter. In the figure, S could be at the points where a, b and c are represented on the diagram. ±PSQ = 90è (angle at the circumference is half the angle at the centre) Angles subtended by a diameter are right angles.

Constructing a tangent There are a number of ways to construct a tangent to a circle. One of the techniques is outlined below. 1. Draw a circle of radius 5 cm and centre O. 2. Draw a radius. 3. Call the point of intersection of the radius and the circumference, P. 4. Extend this radius through P to the point Q, 5 cm outside the circle. 5. Using O and Q as centres, draw intersecting arcs above and below the line OQ. 6. Draw a straight line joining the points of intersection. This line is the tangent. 7. What do you notice about the angle between OQ and the tangent? 8. Investigate another technique for constructing a tangent to a circle. 9. Write a set of instructions for this method of constructing a tangent.

O

■■

P

Q

Theorem 4  Code If a radius is drawn to any point on the circumference and a tangent is drawn at the same point, then the radius will be perpendicular to the tangent.   In the diagram at right, the radius is drawn to a point, P, on the circumference. The tangent to the circle is also drawn at P. The radius and the tangent meet at right angles, that is, the angle at P equals 90è.

O

Worked Example 2

Find the values of the pronumerals in the diagram at right, giving a reason for your answer.

z s O

Think

704

Write

1

Angle z is subtended by the diameter. Use an appropriate theorem to state the value of z.

z = 90è 

2

Angle s is formed by a tangent and a radius, drawn to the point of contact. Apply the corresponding theorem to find the value of s.

s = 90è 

Maths Quest 10 + 10A for the Australian Curriculum

P

measurement AND geometry • geometric reasoning

■■

Theorem 5  Code The angle formed by two tangents meeting at an external point is bisected by a straight line joining the centre of the circle to that external point. Proof: R O

S T

Consider DSOR and DSOT. OR = OT (radii of the same circle are equal) OS is common. ±ORS = ±OTS = 90è (angle between a tangent and radii is 90è) \ DSOR @ DSOT (RHS) So ±ROS = ±TOS and ±OSR = ±OST ( corresponding angles in congruent triangles are equal) The angle formed by two tangents meeting at an external point is bisected by a straight line joining the centre of the circle to the external point.

Worked Example 3

Given that BA and BC are tangents to the circle, find the values of the pronumerals in the diagram at right. Give reasons for your answers.

A r O 68è q

t u

B

s C Think

Write

1

Angles r and s are angles formed by the tangent and the radius, drawn to the same point on the circle. State their size.

s = r = 90è 

2

In the triangle ABO, two angles are already known and so angle t can be found using our knowledge of the sum of angles in a triangle.

DABO: t + 90è + 68è = 180è  t + 158è = 180è t = 22è

3

±ABC is formed by the two tangents, so the line BO, joining the vertex B with the centre of the circle, bisects this angle. This means that angles t and u are equal.

±ABO = ±CBO  ±ABO = t = 22è, ±CBO = u u = 22è

4

AOB and COB are similar triangles.

In AOB and COB r + t + 68è = 180è  s + u + q = 180è  r = s = 90è (proved previously) t = u = 22è (proved previously) \ q = 68è

Chapter 21 Circle geometry

705

measurement and geometry • geometric reasoning

remember

1. An angle with its vertex at the centre of the circle is twice the size of an angle subtended by the same arc, but with the vertex at the circumference. 2. Angles with their vertices on the circumference, subtended by the same arc, are equal. 3. Angles subtended by the diameter are right angles. 4. A tangent and a radius, drawn to the same point on a circle, meet at a 90è angle. 5. An angle formed by two tangents is bisected by the line joining the vertex of that angle to the centre of the circle. exercise

21a

angles in a circle Note: It is acceptable to use a code as a reason for a statement in geometric proofs. Fluency 1 We1 Find the values of the pronumerals in each of the following, giving reasons for your



answers.

Digital doc SkillSHEET 21.4 doc-5393

a

b

30è

P

c

Q

x

x 25è y A

e

x

f

g

B

B

h

i O•

50è

O x•

42è

A 80è O x

A

A

x

30è •O x

y

40è

a x = 30è (theorem 2) b x = 25è, y = 25è (theorem 2 for both angles) c x = 32è (theorem 2) d x = 40è, y = 40è (theorem 2 for both angles) e x = 60è (theorem 1) f x = 40è (theorem 1) g x = 84è (theorem 1) h x = 50è (theorem 2); y = 100è (theorem 1) i x = 56è (theorem 1)

R

•

Digital doc SkillSHEET 21.5 doc-5394

32è

B

d

eBook plus

S

x

x

A

y

B

B 28è •

O

a s = 90è, r = 90è (theorem 3 for both angles) b u = 90è (theorem 4); t = 90è (theorem 3) c m = 90è, n = 90è (theorem 3 for both angles) d x = 52è (theorem 3 and angle sum in a triangle = 180è) e x = 90è (theorem 4) f x = 90è (theorem 4); y = 15è (angle sum in a triangle = 180è)

eBook plus



2 We2 Find the values of the pronumerals in each of the following figures, giving reasons for

your answers. a

b •

t

u

s

•

c

m •

•

r

f

e

d 38è

•

x

O •

x 75è • O

x y

706

n

maths quest 10 + 10a for the australian curriculum

3   WE 3  Given that AB and DB are tangents, find the value of the pronumerals in each of the

following, giving reasons for your answers. a

b

A x y w

O • 70è

B

A r B

t O

40è

•

z D c

s D d

A y O• z

20è x

O A s 70è x

B

y B

D rz D e

f

D

D

20è y

z x yO

15è

B

z B

x •

A

A

•



understanding

•

a x = z = 90è (theorem 4); y = w = 20è (theorem 5 and angle sum in a triangle = 180è) b s = r = 90è (theorem 4); t = 140è (angle sum in a quadrilateral = 360è) c x = 20è (theorem 5); y = z = 70è (theorem 4 and angle sum in a triangle = 180è) d s = y = 90è (theorem 4); x = 70è (theorem 5); r = z = 20è (angle sum in a triangle = 180è) e x = 70è (theorem 4 and angle sum in a triangle = 180è); y = z = 20è (angle sum in a triangle = 180è) f x = y = 75è (theorem 4 and angle sum in a triangle = 180è); z = 75è (theorem 1)

measurement AND geometry • geometric reasoning

O

4   MC  Note: There may be more than one correct answer.

In the diagram at right, which angle is subtended by the same arc as ±APB? A ±APC b ±BPC c ±ABP ✔ d ±ADB 5   MC  Note: There may be more than one correct answer. Referring to the diagram at right, which of the statements is true? A 2±AOD = ±ABD ✔ b ±AOD = 2±ACD c ±ABF = ±ABD ✔ d ±ABD = ±ACD

D P A C

B B

C

F O A

D

Reasoning 6 Values are suggested for the pronumerals in the diagram

below. AB is a tangent to a circle and O is the centre. In each case give reasons to justify suggested values. B a s = t = 45è Base angles of a right-angled isosceles triangle b r = 45è r + s = 90è, s = 45è À r = 45è c u = 65è u is the third angle in DABD, which is right-angled. d m = 25è m is the third angle in DOCD, which is right-angled.  AOC and ±AFC stand on the same arc with ±AOC e n = 45è ±

A rs 25è C

t

m

O u n D

F

at the centre and ±AFC at the circumference.

Chapter 21 Circle geometry

707

OR = OP (radii of the circle) ±OPR = x (equal angles lie opposite equal sides) ±SOP = 2x (exterior angle equals the sum of the two interior opposite angles) OR = OQ (radii of the circle) ±OQR = y (equal angles lie opposite equal sides) ±SOQ = 2y (exterior angle equals the sum of the two interior opposite angles) Now ±PRQ = x + y and ±POQ = 2x + 2y = 2(x + y). Therefore ±POQ = 2 ì ±PRQ.

measurement and geometry • geometric reasoning 7 Set out below is the proof of this result: The angle at the centre of a circle is twice the angle at

the circumference standing on the same arc. R a O b P

Q

Copy and complete the following to show that ±POQ = 2 ì ±PRQ. Construct a diameter through R. Let the opposite end of the diameter be S. Let ±ORP = x and ±ORQ = y. OR = OP (____________________) ±OPR = x (____________________) ±SOP = 2x (exterior angle equals ____________________) OR = OQ (____________________) ±OQR = __________ (____________________) ±SOQ = __________ (____________________) Now ±PRQ = __________ and ±POQ = __________. Therefore ±POQ = 2 ì ±PRQ.

R y

x

O P

S

Q

8 Prove that the segments formed by drawing tangents from an external point to a circle are Check with your teacher. equal in length. 9 Prove that an angle formed by two tangents is bisected by the line joining the vertex of that Check with your teacher. angle to the centre of the circle. 10 Use the figure drawn below to prove that angles subtended by the same arc are equal.

eBook plus

R

Digital doc WorkSHEET 21.1 doc-5395

reFlection 

O P

21b

Check with your teacher.

S  

What are the common steps in proving a theorem?

Q

intersecting chords, secants and tangents intersecting chords In the diagram below, chords PQ and RS intersect at X. P S X

R

Q

1. Measure lengths PX, QX, RX and SX and complete the table below. Line segment

PX

QX

RX

SX

Length 2. Calculate the following: PX ì QX and RX ì SX 3. What do you notice about the results in step 2 above? 4. Draw another circle and a pair of intersecting chords and repeat steps 1 to 3 above. 708

maths quest 10 + 10a for the australian curriculum

measurement AND geometry • geometric reasoning

The results of this activity can be generalised for any circle as follows. ■■

Theorem 6  Code If the two chords intersect inside a circle, then the point of intersection divides each chord into two segments so that the product of the lengths of the segments for both chords is the same. PX ì QX = RX ì SX or a ì b = c ì d R Proof: Join PR and SQ. Consider DPRX and DSQX. ±PXR = ±SXQ (vertically opposite angles are equal) ±RSQ = ±RPQ (angles at the circumference standing on the same arc are equal) ±PRS = ±PQS (angles at the circumference standing on the same arc are equal) DPRX ~ DSQX (equiangular) PX RX (ratio of sides in similar triangles is equal) = SX QX or PX ì QX = RX ì SX

P a

d

S

c X b Q

Worked Example 4

Find the value of the pronumeral. A 4 6 C

5 D X m B

Think

Write

1

Chords AB and CD intersect at X. Point X divides each chord into two parts so that the products of the lengths of these parts are equal. Write this as a mathematical statement.

AX ì BX = CX ì DX 

2

Identify the lengths of the line segments.

AX = 4, BX = m, CX = 6, DX = 5

3

Substitute the given lengths into the formula and solve for m.

4m = 6 ì 5 30 m= 4 = 7.5

Intersecting secants In the diagram below, chords CD and AB are extended to form secants CX and AX respectively. They intersect at X. Measure lengths AX, XB, XC and DX and calculate the products AX ì XB and XC ì DX. What do you notice? C D X

B

A

Chapter 21 Circle geometry

709

measurement AND geometry • geometric reasoning

Your observation from the activity above can be generalised as follows. ■■

Theorem 7  Code If two secants intersect outside the circle as shown, then the following relationship is always true: AX ì XB = XC ì DX or a ì b = c ì d.

X

d b

D B

c

a

C

A

Proof: Join D and A to O, the centre of the circle. C Let ±DCA = x. D ±DOA = 2x (angle at the centre is twice the angle O at the circumference standing on the X A same arc) B Reflex ±DOA = 360è - 2x ( angles in a revolution add to 360è) ±DBA = 180è - x (angle at the centre is twice the angle at the circumference standing on the same arc) ±DBX = x (angle sum of a straight line is 180è) ±DCA = ±DBX Consider DBXD and DCXA. ±BXD is common. ±DCA = ±DBX (shown previously) ±XAC = ±XDB (angle sum of a triangle is 180è) DAXC ~ DDXB (equiangular) AX XC = DX XB or AX ì XB = XC ì DX

Worked Example 5

Find the value of the pronumeral. C y D 6 X

7

Think

710

B

5

A

Write

1

Secants XC and AX intersect outside the circle at X. Write the rule connecting the lengths of XC, DX, AX and XB.

XC ì DX = AX ì XB 

2

State the length of the required line segments.

XC = y + 6 AX = 7 + 5 = 12

3

Substitute the length of the line segments and solve the equation for y.

(y + 6) ì 6 = 12 ì 7 6y + 36 = 84 6y = 48 y=8

Maths Quest 10 + 10A for the Australian Curriculum

DX = 6 XB = 7

measurement AND geometry • geometric reasoning

Intersecting tangents ■■

In the diagram below right, tangents AC and BC intersect at C and AC = BC.

■■

Theorem 8  Code If two tangents meet outside a circle, then the lengths from the external point to where they meet the circle are equal. Proof: Join A and B to O, the centre of the circle. Consider DOCA and DOCB. OC is common. OA = OB (radii of the same circle are equal) ±OAC = ±OBC (radius is perpendicular to tangent through the point of contact) DOCA @ DOCB (RHS) AC = BC (corresponding sides of congruent triangles are equal). If two tangents meet outside a circle, the lengths from the external point to the point of contact are equal.

A C B A C

O B

Worked Example 6

Find the value of the pronumeral. B 3 C m A Think

Write

1

BC and AC are tangents intersecting at C. State the rule that connects the lengths BC and AC.

AC = BC 

2

State the lengths of BC and AC.

AC = m, BC = 3

3

Substitute the required lengths into the equation to find the value of m.

m=3

Chords and radii ■■

In the diagram below, the chord AB and the radius OC intersect at X at 90è; that is, ±OXB = 90è. OC bisects the chord AB; that is, AX = XB. O A

X C

B

Chapter 21 Circle geometry

711

measurement AND geometry • geometric reasoning

■■

Theorem 9  Code If a radius and a chord intersect at right angles, then the radius bisects the chord. Proof: O A

■■

■■

B

X C

Join OA and OB. Consider DOAX and DOBX. OA = OB (radii of the same circle are equal) ±OXB = ±OXA (given) OX is common. DOAX @ DOBX (RHS) AX = BX (corresponding sides in congruent triangles are equal) If a radius and a chord intersect at right angles, then the radius bisects the chord. The converse is also true: If a radius bisects a chord, the radius and the chord meet at right angles. Theorem 10  Chords equal in length are equidistant from the centre. This theorem states that if the chords MN and PR are of equal length, then OD = OC. Proof:

A

A

P B C

D

O

R

N

Construct OA ^ MN and OB ^ PR. Then OA bisects MN and OB bisects PR (Theorem 9) Because MN = PR, MD = DN = PC = CR. Construct OM and OP, and consider DODM and DOCP. MD = PC (shown above) OM = OP (radii of the same circle are equal) ±ODM = ±OCP = 90è (by construction) DODM @ DOCP (RHS) So OD = OC (corresponding sides in congruent triangles are equal) Chords equal in length are equidistant from the centre.

Worked Example 7

Find the values of the pronumerals, given that AB = CD.

A

C Maths Quest 10 + 10A for the Australian Curriculum

G E m 3 n O

B

F

D

2.5

H

P B C

D N

M

712

M

O

R

measurement AND geometry • geometric reasoning

Think

Write

1

Since the radius OG is perpendicular to the chord AB, the radius bisects the chord.

AE = EB 

2

State the lengths of AE and EB.

AE = m, EB = 3

3

Substitute the lengths into the equation to find the value of m.

m=3

4

AB and CD are chords of equal length and OE and OF are perpendicular to these chords. This implies that OE and OF are equal in length.

OE = OF 

5

State the lengths of OE and OF.

OE = n, OF = 2.5

6

Substitute the lengths into the equation to find the value of n.

n = 2.5

remember

1. Code

2. Code C

P D

S

X

X

R

Q

A

B

AX ì XB = XC ì DX

PX ì QX = RX ì SX 3. Code

4. Code

If OC ^ AB, AX = XB.

A O

C A B

AC = BC 5. Code M A

P B

X C

B

If MN = PR, then OD = OC.

C D

O

R

N

Exercise

21b

Intersecting chords, secants and tangents Fluency

a m = 3 b m = 3 c m = 6

1   WE 4  Find the value of the pronumeral in each of the following. c a b D C A m 4 6 A 2 6 X 9 X B m 8 C D B

C A m 4 X m D

9 B

Chapter 21 Circle geometry

713

measurement AND geometry • geometric reasoning

5

c

6

d

8

a n = 1 b m = 7.6 c n = 13 d m = 4



2   WE 5  Find the value of the pronumeral in each of the following. a b 4 2 4.5 m 3 n

5

4

6

3

n

7 m

3   WE 6  Find the value of the pronumerals in each of the following. a b c 5 7

a x = 5 b m = 7 c x = 2.5, y = 3.1

x

x

3.1

2.5

y

m 4   WE 7  Find the value of the pronumeral in each of the following. a b x 3.3 2.8 O O x

c



a x = 2.8 b x = 3.3 c x = 5.6 d m = 90è

d m

5.6 2.5 2.5 x O

O

understanding 5   MC  Note: There may be more than one correct answer.

In which of the following figures is it possible to find the value of m through solving a linear equation? a

✔ b

7

7

2 5

2 m

m

✔ c

3

✔ d

m m

4 4

2 3

714

Maths Quest 10 + 10A for the Australian Curriculum

1

2

measurement AND geometry • geometric reasoning 6 Find the length, ST, in the diagram below. ST = 3 cm Q

P

5 cm 9 cm

R 4 cm T

S

Reasoning 7 Prove the result: If a radius bisects a chord, then the radius meets the chord at right angles. Remember to provide reasons for your statements. Check with your teacher. 8 Prove the result: Chords that are an equal distance from the centre are equal in length. Provide Check with your teacher. reasons for your statements. reflection    9 Prove that the line joining the centres of two intersecting

circles bisects their common chord at right angles. Provide reasons for your statements.

What techniques will you use to prove circle theorems?

Check with your teacher.

21c

Cyclic quadrilaterals Quadrilaterals in circles

1. Construct a circle of radius 10 cm. 2. Mark points A, B, C and D on different points on the circumference. 3. Join points A and B, B and C, C and D and A and D by straight lines to B A construct a quadrilateral. 4. Accurately measure the interior angles at points A, B, C and D. C 5. Repeat steps 1 to  4 for another circle. 6. What is the relationship between D (a)  ±ABC and ±ADC (b)  ±BAD and ±BCD? 7. In each circle, extend AD to form an exterior angle at A and measure the exterior angle. What is the relationship between this exterior angle at A and ±BCD? ■■ A cyclic quadrilateral has all four vertices on the circumference of a B A circle; that is, the quadrilateral is inscribed in the circle.   In the diagram at right, points A, B, C and D lie on the circumference; C hence, ABCD is a cyclic quadrilateral.   It can also be said that points A, B, C and D are concyclic; that is, the D circle passes through all the points. ■■

■■

Theorem 11  Code The opposite angles of a cyclic quadrilateral are supplementary (add to 180è). Proof: B Join A and C to O, the centre of the circle. A Let ±ABC = x. C Reflex ±AOC = 2x (a ngle at the centre is twice the angle at the O circumference standing on the same arc) D Reflex ±AOC = 360è - 2x (angles in a revolution add to 360è) ±ADC = 180è - x (angle at the centre is twice the angle at the circumference standing on the same arc) ±ABC + ±ADC = 180è Similarly, ±DAB + ±DCB = 180è. Opposite angles in a cyclic quadrilateral are supplementary. The converse is also true: If opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Chapter 21 Circle geometry

715

measurement AND geometry • geometric reasoning

Worked Example 8

Find the values of the pronumerals in the diagram at right. Give reasons for your answers.

P 120è

Q 75è y R

x S Think

Write

1

PQRS is a cyclic quadrilateral, so its opposite angles are supplementary. First find the value of x by considering a pair of opposite angles ±PQR and ±RSP and forming an equation to solve.

±PQR + ±RSP = 180è (the opposite angles of a cyclic quadrilateral are supplementary.) ±PQR = 75è, ±RSP = x x + 75è = 180è x = 105è

2

Find the value of y by considering the other pair of opposite angles (±SPQ and ±QRS).

±SPQ + ±QRS = 180è  ±SPQ = 120è, ±QRS = y y + 120è = 180è y = 60è

■■

Theorem 12  Code The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Proof: Q T

P b a

a R

S

±QPS + ±QRS = 180è (opposite angles of a cyclic quadrilateral) ±QPS + ±SPT = 180è (adjacent angles on a straight line) Therefore ±SPT = ±QRS. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Worked Example 9

Find the value of the pronumerals in the diagram at right.

A 50è D y

B C x

Think

716

Write

1

ABCD is a cyclic quadrilateral. The exterior angle, x, is equal to its interior opposite angle, ±DAB.

x = ±DAB, ±DAB = 50è  So x = 50è.

2

The exterior angle, 100è, is equal to its interior opposite angle, ±ADC.

±ADC = 100è, ±ADC = y  So y = 100è.

Maths Quest 10 + 10A for the Australian Curriculum

100è

measurement and geometry • geometric reasoning

remember

1. A cyclic quadrilateral has all four of its vertices on the circumference of a circle. 2. Opposite angles of a cyclic quadrilateral are supplementary. 3. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

exercise

21c

cyclic quadrilaterals Fluency

eBook plus

Digital doc SkillSHEET 21.6 doc-5396

1 We8 Find the values of the pronumerals in each of the following. a c b 65è m 92è 95è y

x



a x = 115è, y = 88è b m = 85è c n = 25è d x = 130è e x = y = 90è f x = 45è, y = 95è

d

n

155è

e

x 50è

f O

x

y O

y

85è

x

2 We9 Find the values of the pronumerals in each of the following. a c b y a x = 85è, y = 80è x 80è b x = 110è, y = 115è x y c x = 85è 85è d x = 150è 115è e x = 90è, y = 120è f m = 120è, n = 130è 110è e

d 150è

x

x y

135è

95è x

f 120è

130è

120è n m

3 mc Note: There may be more than one correct answer.

✔

Which of the following correctly states the relationship between x, y and z in the diagram shown? A x = y and x = 2z B x = 2y and y + z = 180è C z = 2x and y = 2z D x + y = 180è and z = 2x

x O z y

chapter 21 circle geometry

717

measurement and geometry • geometric reasoning understanding 4 The steps below show you how to set out the proof that the opposite

A

angles of a cyclic quadrilateral are equal. a Find the size of ±DOB. 2x b Find the size of the reflex angle DOB. 360è - 2x 180è - x c Find the size of ±BCD. d Find ±DAB + ±BCD. 180è 5 mc Note: There may be more than one correct answer.

B

x O

C D q

r

t p

s

a Which of the following statements is always true for the diagram shown? ✔ A r = t B r = p C r = q D r = s b Which of the following statements is correct for the diagram shown? ✔ A r + p = 180è ✔ D t = r ✔ B q + s = 180è ✔ C t + p = 180è reasoning eBook plus

Digital doc WorkSHEET 21.2 doc-5397

21d

6 Prove that the exterior angle of a cyclic quadrilateral is

equal to the interior opposite angle. Check with your teacher.

reFlection 

 

What is a cyclic quadrilateral?

tangents, secants and chords the alternate segment theorem ■

■

■

■

Consider the figure shown. Line BC is a tangent to the circle at the point A. A line is drawn from A to anywhere on the circumference, point D. The angle ±BAD defines a segment (the shaded area). The unshaded part of the circle is called the alternate segment to ±BAD. Now consider angles subtended by the chord AD in the alternate segment, such as the angles marked in red and blue. The alternate segment theorem states that these are equal to the angle that made the segment, namely: ±BAD = ±AED and ±BAD = ±AFD

Theorem 13 Code The angle between a tangent and a chord is equal to the angle in the alternate segment. Proof: We are required to prove that ±BAD = ±AFD. Construct the diameter from A through O, meeting the circle at G. Join G to the points D and F. ±BAG = ±CAG = 90è (radii ^ tangent at point of contact) ±GFA = 90è (angle in a semicircle is 90è) ±GDA = 90è (angle in a semicircle is 90è)

718

maths quest 10 + 10a for the australian curriculum

D

O

B

A

C

E

D

O F

B

A

C

G

D

O F

B

A

C

measurement AND geometry • geometric reasoning

Consider DGDA. We know that ±GDA = 90è. ±GDA + ±DAG + ±AGD = 180è 90è + ±DAG + ±AGD = 180è ±DAG + ±AGD = 90è ±BAG is also a right angle. ±BAG = ±BAD + ±DAG = 90è Equate the two results. ±DAG + ±AGD = ±BAD + ±DAG Cancel the equal angles on both sides. ±AGD = ±BAD Now consider the fact that both triangles DAG and DAF are subtended from the same chord (DA). ±AGD = ±AFD (angles in the same segment standing on the same arc are (equal) Equate the two equations. ±AFD = ±BAD

Worked Example 10

Find the value of x and y, giving reasons. A x B D

y

62è

C

T Think

Write

1

Use the alternate segment theorem to find x.

x = 62è (angle between a tangent and a chord is equal to the angle in the alternate segment)

2

The value of y is the same as x because x and y are subtended by the same chord BT.

y = 62è (angles in the same segment standing on the same arc are equal)

Tangents and secants ■■

Theorem 14  Code If a tangent and a secant intersect as shown, the following relationship is always true: XA ì XB = (XT)2 or a ì b = c2.

A a b

Proof: Join BT and AT. Consider DTXB and DAXT. ±TXB is common. ±XTB = ±XAT (angle between a tangent and a chord is equal to the angle in the alternate segment) ±XBT = ±XTA (angle sum of a triangle is 180è) DTXB ~ DAXT (equiangular) XB XT So = XT XA or XA ì XB = (XT)2.

X

B c

T

A B X

T

Chapter 21 Circle geometry

719

measurement AND geometry • geometric reasoning

Worked Example 11

Find the value of the pronumeral. A

m 5

B

X 8

T Think

Write

1

Secant XA and tangent XT intersect at X. Write the rule connecting the lengths of XA, XB and XT.

XA ì XB = (XT)2 

2

State the values of XA, XB and XT.

XA = m + 5, XB = 5, XT = 8

3

Substitute the values of XA, XB and XT into the equation and solve for m.

(m + 5) ì 5 = 82 5m + 25 = 64 5m = 39 m = 7.8

remember

1. The angle formed by a tangent and a chord is equal to the angle in the alternate segment.  B

A

C

D

2. If a tangent and a secant intersect as shown, then XA ì XB = (XT)2.  A B X T

Exercise

21d

Tangents, secants and chords Fluency 1   WE 10  Find the value of the pronumerals in the following. a b x 70è

Maths Quest 10 + 10A for the Australian Curriculum

59è

47è y

720

a x = 70è b x = 47è, y = 59è

x

measurement AND geometry • geometric reasoning 2   WE 11  Find the value of the pronumerals in the following. a b 5 4 12 p

a p = 6 b q = 8

q 4

3 Line AB is a tangent to the circle as shown in the figure on

B y

the right. Find the values of the angles labelled x and y.

x

A

O 21è

Questions 4 to 6 refer to the figure on the right. The line MN is a tangent to the circle, and EA is a straight line. The circles have the same radius. x = 42è, y = 132è

D

M

F O

E

G C

A

B

4 Find 6 different right angles. MAC, NAC, FDA, FBA, EDG, EBG

N

5   MC  If ±DAC = 20è, then ±CFD and ±FDG are respectively: ✔ B 70è and 40è A 70è and 50è C 40è and 70è D 70è and 70è 6   MC  A triangle similar to FDA is: A FDG C EDA

B FGB ✔ D

GDE A y

x = 42è, y = 62è 7 Find the values of the angles x and y in the figure at right.

O B

x

42è 62è

understanding 8 Show that if the sum of the two given angles in question 7 is 90è, then the line AB must be a Answers will vary. diameter. 9 Find the value of x in the figure at right, given that the line underneath the circle is a tangent. 60è

x 100è

O

20è 10 In the figure at right, express x in terms of a and b. This is the same drawing as in question 9. x = 180è - a - b

x a

O

b Chapter 21 Circle geometry

721

measurement AND geometry • geometric reasoning 11 Two tangent lines to a circle meet at an angle y, as shown in

10è

the figure at right. Find the values of the angles x, y and z. x = 80è, y = 20è, z = 80è

z O x

12 Solve question 11 in the general case (see the figure at right)

y

a

and show that y = 2a. This result is important for space navigation (imagine the circle to be the earth) in that an object at y can be seen by people at x and z at the same time.

z O

Answers will vary.

y

x 13 In the figure at right, find the values of the angles x, y and z.

z

y

75è

x = 85è, y = 20è, z = 85è

x

O 20è 14   MC  Examine the figure at right. The angles x and y

(in degrees) are respectively: A 51 and 99 B 51 and 129 ✔ D 51 and 122 C 39 and 122

19 x

51 y

O

Questions 15 to 17 refer to the figure at right. The line BA is a tangent to the circle at point B. Line AC is a chord that meets the tangent at A.

C x

y D

O z 50è B

15 Find the values of the angles x and y. x = 50è, y = 95è

45è A

16   MC  The triangle which is similar to triangle BAD is: B BCD ✔ A CAB C BDC D AOB 17   MC  The value of the angle z is: A 50è ✔ C 95è

B 85è D 100è

Reasoning 18 Find the values of the angles x, y and z in the figure at right. The line AB is tangent to the circle at B. x = 33è, y = 55è, z = 22è

O D

A

722

Maths Quest 10 + 10A for the Australian Curriculum

z

33è y 92è x B

C

measurement AND geometry • geometric reasoning 19 Find the values of the angles x, y and z in the figure below. The line AB is tangent to the circle at B. The line CD is a diameter. x = 25è, y = 65è, z = 40è C

x

O

y D 25è

z

A

B 20 Solve question 19 in the general case; that is, express angles x, y and z in terms of a (see the figure below). x = a, y = 90è - a, z = 90è - 2a C x

O B

y D a

z

A

21 Prove that, when two circles touch, their centres and the point of contact are collinear. Check with your teacher. reflection 

 

Describe the alternate segment of a circle.

Chapter 21 Circle geometry

723

measurement AND geometry • geometric reasoning

Summary Angles in a circle ■■ ■■ ■■ ■■ ■■

An angle with its vertex at the centre of the circle is twice the size of an angle subtended by the same arc, but with the vertex at the circumference.  Angles with their vertices on the circumference, subtended by the same arc, are equal.  Angles subtended by the diameter are right angles.  A tangent and a radius, drawn to the same point on a circle, meet at a 90è angle.  An angle formed by two tangents is bisected by the line joining the vertex of that angle to the centre of the circle.  Intersecting chords, secants and tangents

■■

Code P S X

R

Q

PX ì QX = RX ì SX ■■

Code C D X

A

B

AX ì XB = XC ì DX ■■

Code A C B

AC = BC ■■

Code O A

B

X C

If OC ^ AB, AX = XB. ■■

Code M A

P B C

D

O

R

N

If MN = PR, then OD = OC. 724

Maths Quest 10 + 10A for the Australian Curriculum

measurement AND geometry • geometric reasoning Cyclic quadrilaterals ■■ ■■ ■■

A cyclic quadrilateral has all four of its vertices on the circumference of a circle. Opposite angles of a cyclic quadrilateral are supplementary.  The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.  Tangents, secants and chords

■■

The angle formed by a tangent and a chord is equal to the angle in the alternate segment.  B

A

C

D ■■

If a tangent and a secant intersect as shown, then XA ì XB = (XT)2.  A B X T

Mapping your understanding

Using terms from the summary above, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 699.

Chapter 21 Circle geometry

725

measurement AND geometry • geometric reasoning a x = 50è d x = 90è g x = 55è j x = 100è

Chapter review Fluency

Note: All questions in this exercise can be done without a calculator. 1 Determine the values of the pronumerals in each of

the following. a



b x = 48è, y = 25è e y = 90è h x = 125è k m = 40è



c f i l

x = y = 28è, z = 56è x = 140è x = 70è x = 90è, y = 60è, z = 40è

2 Find the value of the pronumeral in each case. a a x = 90è b x = 20è x c x = 55è d x = 125è O

b 50è

x

48è

25è

b x y

x

O 70è x

c

d

28è

x

y

O z

O

c x

e

f

y

O x

x

d x

70è

x

h

O

O

110è

250è

i

O

70è

O

g

110è

j

O

3 Find the value of m in each of the following. a a m = 3 6 b m = 12 m 10

x

x

110è

5

100è b k

m

70è

l

8 50è y

O

z 30è

x

726

Maths Quest 10 + 10A for the Australian Curriculum

m 4

6

measurement AND geometry • geometric reasoning

c

3 c m = 9 d m = 11.7

5

5   MC  Note: There may be more than one correct

answer. Which of the following statements is true for the diagram shown?

4 3

m

A

C

O

d 10

8

B

7.5

✔ ✔

4   MC  Note: There may be more than one correct

answer. In which of the following figures is it possible to get a reasonable value for the pronumeral? ✔ a

AO = BO B AC = BC C ±OAC = ±OBC D ±AOC = 90è

✔ A

m

6 Two chords, AB and CD, intersect at E as shown. If AE = CE, prove that EB = ED. CE ì ED = AE ì EB AE = CE (given) \ ED = EB C B A

4

6

E

5 m

D 7 Two circles intersect at X and Y. Two lines, AXB

✔ b

±AYC = ±AXC

6 3 2

±BXD = ±BYD But ±AXC = ±BXD À ±AYC = ±BYD

m

c

and CXD, intersect one circle at A and C, and the other at B and D, as shown. Prove that ±AYC = ±BYD. C

B

X

D

A

2

8

Y

5 m

8 Name at least five pairs of equal angles in the

following diagram. ±PQT & ±PST, ±PTS & ±RQS, ±TPQ & ±QSR, ±QPS & ±QTS, ±TPS & TQS, ±PQS & ±PTS, ±PUT & ±QUS, ±PUQ & TUS

✔ d

R Q

7

m

S

U O

4

3

P T Chapter 21 Circle geometry

727

measurement AND geometry • geometric reasoning 9 Find the values of the pronumerals in the following

figures.



a 85è

a x = 95è, y = 80è b x = 99è c x = 78è, y = 92è d x = 97è, y = 92è

problem solving 1 Find the values of the pronumerals in the following

figures.

y

100è

a x = 42è b y = 62è c p = 65è

a x

42è x

b

b

y 56è

81è

x

c O 130è

c y

78è

p

x

92è

2 Find the value of the pronumerals in the following. a x

4 6

d 88è y

97è

x b k 4

10   MC  Which of the following statements is not

always true for the diagram below? a

b

8 c m

d

c

e 5

A B C ✔ D 728

±a + ±c = 180è ±b + ±d = 180è ±e + ±c = 180è ±a + ±e = 180è

Maths Quest 10 + 10A for the Australian Curriculum

4

n

a x = 5 b k = 12 c m = 6, n = 6

measurement and geometry • geometric reasoning

f

d 7

2 x 2 d x = 7 e b = 4, a = 2 f w = 3, x = 5

x w 3

e

a 6

1 b

eBook plus

5.5 8 11

Interactivities

Test yourself Chapter 21 int-2882 Word search Chapter 21 int-2880 Crossword Chapter 21 int-2881

chapter 21 circle geometry

729

eBook plus

activities

Are you ready? Digital docs (page 700) • SkillSHEET 21.1 (doc-5390): Using tests to prove congruent triangles • SkillSHEET 21.2 (doc-5391): Corresponding sides and angles of congruent triangles • SkillSHEET 21.3 (doc-5392): Using tests to prove similar triangles • SkillSHEET 21.4 (doc-5393): Angles in a triangle • SkillSHEET 21.5 (doc-5394): More angle relations

21A Angles in a circle Interactivity

• Angles in a circle (int-2795) (page 701) Digital docs

• SkillSHEET 21.1 (doc-5390): Using tests to prove congruent triangles (page 701) • SkillSHEET 21.2 (doc-5391): Corresponding sides and angles of congruent triangles (page 701) • SkillSHEET 21.3 (doc-5392): Using tests to prove similar triangles (page 701) • SkillSHEET 21.4 (doc-5393): Angles in a triangle (page 706) • SkillSHEET 21.5 (doc-5394): More angle relations (page 706)

730

maths quest 10 + 10a for the australian curriculum

• WorkSHEET 21.1 (doc-5395): Circle geometry I (page 708) 21C Cyclic quadrilaterals Digital docs

• SkillSHEET 21.6 (doc-5396): Angles in a quadrilateral (page 717) • WorkSHEET 21.2 (doc-5397): Circle geometry II (page 718) Chapter review Interactivities (page 729) • Test yourself Chapter 21 (int-2882): Take the end-of chapter test to test your progress • Word search Chapter 21 (int-2880): an interactive word search involving words associated with this chapter • Crossword Chapter 21 (int-2881): an interactive crossword using the definitions associated with the chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

measurement AND geometry • Pythagoras & trigonometry

22

22A 22B 22C 22D 22E 22F

The sine rule The cosine rule Area of triangles The unit circle Trigonometric functions Solving trigonometric equations

What do you know ?

Trigonometry II

1 List what you know about trigonometry. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of trigonometry.

opening question

How would you find the height of this water tower if you could not climb up to the top?

measurement and geometry • Pythagoras & trigonometry

are you ready?

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. eBook plus

Digital doc

SkillSHEET 22.1 doc-5398

Labelling right-angled triangles 1 Label the sides of these triangles as opposite, adjacent and hypotenuse with respect to the angle

marked q. a

b Hypotenuse

Adjacent

Opposite

q

Hypotenuse Opposite

q Adjacent

eBook plus

Digital doc

SkillSHEET 22.2 doc-5399

eBook plus

Digital doc

SkillSHEET 22.3 doc-5400

Calculating sin, cos or tan of an angle 2 Evaluate each of the following, correct to 2 decimal places. a sin 23è 0.39 b cos 47è28Å 0.68

Finding side lengths in right-angled triangles 3 Find the value of x, correct to 1 decimal place. a 3.4 cm

b

6 cm

x

14 cm

Digital doc

SkillSHEET 22.4 doc-5401

eBook plus

Digital doc

SkillSHEET 22.5 doc-5402

20è

Calculating the angle from a sin, cos or tan ratio 4 Find the value of q, correct to the nearest degree. a cos q =

1 2

60è

b sin q = 0.866

c tan q = 1 45è

60è

Finding angles in right-angled triangles 5 Find the value of q, correct to the nearest minute. a b 12 cm 36è52

58è13Å

20 cm 15 cm 17 cm

732

38.5 cm

x

35è

eBook plus

c tan 19è39Å 0.36

maths quest 10 + 10a for the australian Curriculum

measurement AND geometry • Pythagoras & trigonometry

22A

The sine rule Exact values ■■ ■■

Most of the trigonometric values that we will deal with in this chapter are only approximations. However, angles of 30è, 45è and 60è have exact values of sine, cosine and tangent. B

2

30è

2

60è A ■■

■■

opp 1 À sin 30è = hyp 2 adj 3 cos B = À cos 30è = hyp 2 opp 1 3 tan B = À tan 30è = or adj 3 3

■■

C

Consider an equilateral triangle, ABC, of side length 2 units.   If the triangle is perpendicularly bisected, then two congruent triangles, ABD and CBD, are obtained. From triangle ABD it can be seen that altitude BD creates a right-angled triangle with angles of 60è and 30è and base length (AD) of 1 unit. The altitude BD is obtained using Pythagoras’ theorem. B (AB)2 = (AD)2 + (BD)2 22 = 12 + (BD)2 30è 4 = 1 + (BD)2 2 2 4 - 1 = (BD) 3 (BD)2 = 3 60è BD = 3 A D Using triangle ABD and the three trigonometric ratios the following exact 1 values are obtained: sin B =

■■

D 2

opp À sin 60è = hyp adj cos A = À cos 60è = hyp opp tan A = À tan 60è = adj sin A =

3 2 1 2 3 or 3 1

Consider a right-angled isosceles triangle EFG whose equal sides are of 1 unit. The hypotenuse EG is obtained by using Pythagoras’ theorem. (EG)2 = (EF)2 + (FG)2 = 12 + 12 =2 EG = 2 Using triangle EFG and the three trigonometric ratios, the following exact values are obtained: opp 1 2 sin E = À sin 45è = or hyp 2 2 adj 1 cos E = À cos 45è = or 2 hyp 2 2 opp 1 tan E = À tan 45è = or 1 adj 1

G

E

2

1

45è 1

F

Chapter 22 Trigonometry II

733

measurement AND geometry • Pythagoras & trigonometry

When working with non–right-angled triangles, it is usual to B label the angles A, B and C, and the sides a, b and c, so that side a c a is the side opposite angle A, side b is the side opposite angle B and side c is the side opposite angle C. A C b ■■ In a non–right-angled triangle, a perpendicular line, h, can be drawn from the angle B to side b. This divides the triangle into two right-angled triangles, ABD B and CBD. ■■ Using triangle ABD and the sine trigonometric ratio for c a h h right-angled triangles, we obtain sin A = . Using triangle CBD c A C D b and the sine trigonometric ratio for right-angled triangles, we h– h h c = sin A and –a = sin C obtain sin C = . a ■■ Transposing each equation to make h the subject, we obtain: h = c sin A and h = a sin C. Since h is common to both triangles the two equations may be equated and we get c sin A = a sin C. Dividing both sides of the equation by sin A gives: a sin C c= sin A Dividing both sides of the equation by sin C gives: c a = sin C sin A ■■

■■

■■

In a similar way, if a perpendicular line is drawn from angle A to side a, the two right-angled triangles would give h = c sin B and h = b sin C. b c This would give: = sin B sin C From this, the sine rule can be stated. In any triangle ABC:

B c h A

a b c = = sin A sin B sin C

■■

Notes c 1. When using this rule, depending on the values given, any combination of the two equalities may be used to solve a A particular triangle. 2. To solve a triangle means to find all unknown side lengths and angles. The sine rule can be used to solve non–right-angled triangles if we are given: 1. two angles and one side length 2. two side lengths and an angle opposite one of these side lengths.

Worked Example 1

In the triangle ABC, a = 4 m, b = 7 m and B = 80è. Find A, C and c. Think 1

Write/draw

Draw a labelled diagram of the triangle ABC and fill in the given information.

B c A

734

b h = c sin B and h = b sin C

Maths Quest 10 + 10A for the Australian Curriculum

b=7

80è a = 4 C

C

B a b

C

measurement AND geometry • Pythagoras & trigonometry

2

Check that one of the criteria for the sine rule has been satisfied.

The sine rule can be used since two side lengths and an angle opposite one of these side lengths have been given.

3

Write down the sine rule to find A.

To find angle A: a b = sin A sin B

4

Substitute the known values into the rule.

5

Transpose the equation to make sin A the subject.

6

Evaluate.

7

Round off the answer to degrees and minutes.

8

Determine the value of angle C using the fact that the angle sum of any triangle is 180è.

C ö 180è - (80è + 34è15Å) = 65è45Å

9

Write down the sine rule to find c.

To find side length c: c b = sin C sin B

10

Substitute the known values into the rule.

11

Transpose the equation to make c the subject.

12

Evaluate. Round off the answer to 2 decimal places and include the appropriate unit.

4 7 = sin A sin 80 ° 4 sin 80è = 7 sin A 4 sin 80 ° sin A = 7  4 sin 80 °  A = Sin −1    7 ö 34.246  004  71è ö 34è15Å

c b = sin 65 ° 45′ sin 80 ° c=

7 sin 65 ° 45′ sin 80 °

ö 6.48 m

The ambiguous case ■■

When using the sine rule there is one important issue to consider. If we are given two side lengths and an angle opposite one of these side lengths, then two different triangles may be drawn. For example, if a = 10, c = 6 and C = 30è, two possible triangles could be created. B

B c=6 A

■■

a = 10 30è

a = 10

c=6 C

A

30è

C

I n the first case (above left), angle A is an acute angle, while in the second case (above right), angle A is an obtuse angle. When using the sine rule to find an angle, we have to use the inverse sine function. If we are finding an angle, given the sine value, it is important to remember that an angle between 0è and 90è has the same sine value as its supplement. For example, sin 40è = 0.6427, and sin 140è = 0.6427. Chapter 22 Trigonometry II

735

measurement AND geometry • Pythagoras & trigonometry

Worked Example 2

In the triangle ABC, a = 10 m, c = 6 m and C = 30è. Find two possible values of A, and hence two possible values of B and b. Case 1 Think 1

Draw a labelled diagram of the triangle ABC and fill in the given information.

Write/draw B a = 10

c=6

30è

A

C

2

Check that one of the criteria for the sine rule has been satisfied.

The sine rule can be used since two side lengths and an angle opposite one of these side lengths have been given.

3

Write down the sine rule to find A.

To find angle A: a c = sin A sin C

4

Substitute the known values into the rule.

5

Transpose the equation to make sin A the subject.

6

Evaluate angle A.

7

Round off the answer to degrees and minutes.

8

Determine the value of angle B, using the fact that the angle sum of any triangle is 180è.

9

Write down the sine rule to find b.

10

Substitute the known values into the rule.

11

Transpose the equation to make b the subject. Evaluate. Round off the answer to 2 decimal places and include the appropriate unit.

12

10 6 = sin A sin 30 ° 10 sin 30è = 6 sin A sin A =

10 sin 30 ° 6

 10 sin 30 °  A = sin −1    6 ö 56.442  690  24è A ö 56è27Å B ö 180è - (30è + 56è27Å) = 93è33Å To find side length b: b c = sin B sin C b 6 = sin 93 ° 33′ sin 30 ° b=

6 sin 93 ° 33′ sin 30 °

ö 11.98 m

The values we have just obtained are only one set of possible answers for the given dimensions of the triangle ABC.   We are told that a = 10 m, c = 6 m and C = 30è. Since side a is larger than side c, it follows that angle A will be larger than angle C. Angle A must be larger than 30è; therefore it may be an acute angle or an obtuse angle. 736

Maths Quest 10 + 10A for the Australian Curriculum

measurement AND geometry • Pythagoras & trigonometry

Case 2 Think 1

Write/draw B

Draw a labelled diagram of the triangle ABC and fill in the given information.

a = 10

c=6

30è

A

C

2

Write down the alternative value for angle A. Simply subtract the value obtained for A in Case 1 from 180è.

To find the alternative angle A: If sin A = 0.8333, then A could also be: A ö 180è - 56è27Å = 123è33Å

3

Determine the alternative value of angle B, using the fact that the angle sum of any triangle is 180è.

B ö 180è - (30è + 123è33Å) = 26è27Å

4

Write down the sine rule to find the alternative b.

To find side length b: b c = sin B sin C

5

Substitute the known values into the rule.

6

Transpose the equation to make b the subject.

7

Evaluate. Round off the answer to 2 decimal places and include the appropriate unit.

b 6 = sin 26 ° 27′ sin 30 ° b=

6 sin 26 ° 27′ sin 30 °

ö 5.34 m

Hence, for this example there were two possible solutions as shown by the diagrams below. B

B a = 10

c=6 A ■■

■■ ■■ ■■

C

a = 10 30è

A

C

The ambiguous case does not work for each example. Consider Worked example 1, in which we were required to solve the triangle ABC given a = 4 m, b = 7 m and B = 80è. For angle A, we obtained A = 34è15Å. However, angle A could also have been A = 145è45Å (since there are two possible values of A between 0è and 180è whose sine is the same; that is, sin 34è15Å = 0.5628 and sin 145è45Å = 0.5628).   We will now see whether or not A = 145è45Å is a possible solution.   To obtain C subtract angles A and B from 180è.

■■

30è

c=6

C = 180è - (80è + 145è45Å) = 180è - 225è45Å = -45è45Å (not possible)

Hence, for Worked example 1 only one possible solution exists. It would be useful to know, before commencing a question, whether or not the ambiguous case exists and, if so, to then find both sets of solutions. The ambiguous case exists if C is an acute angle and a > c > a sin C, or any equivalent statement; for example, if B is an acute angle and a > b > a sin B, and so. In Worked example 2, where a = 10 m, c = 6 m and C = 30è, there were two possible solutions because C was an acute angle and a > c > a sin C, since 10 > 6 > 10 ì 0.5. In Worked example 1, where a = 4 m, b = 7 m and B = 80è, there was only one possible solution because even though B was an acute angle, the condition a > b > a sin B could not be satisfied. Chapter 22 Trigonometry II

737

measurement AND geometry • Pythagoras & trigonometry

Worked Example 3

To calculate the height of a building, Kevin measures the angle of elevation to the top as 52è. He then walks 20 m closer to the building and measures the angle of elevation as 60è. How high is the building? Think 1

Write/draw

Draw a labelled diagram of the situation and fill in the given information.

C

h 52è

120è

A

60è

B 20

x

D x – 20

2

Check that one of the criteria for the sine rule has been satisfied for triangle ABC.

The sine rule can be used for triangle ABC since two angles and one side length have been given.

3

Determine the value of angle ACB, using the fact that the angle sum of any triangle is 180è.

±ACB = 180è - (52è + 120è) = 8è

4

Write down the sine rule to find b (or AC).

To find side length b of triangle ABC: b c = sin B sin C

5

Substitute the known values into the rule.

6

Transpose the equation to make b the subject.

7

Evaluate. Round off the answer to 2 decimal places and include the appropriate unit.

8

Draw a diagram of the situation, that is, triangle ADC, labelling the required information. Note: There is no need to solve the rest of the triangle in this case as the values will not assist in finding the height of the building.

b 20 = sin 120 ° sin 8 ° b=

20 × sin 120 ° sin 8 °

ö 124.45 m C 124.45 m

h

52è A

738

D

9

Write down what is given for the triangle.

Have: angle and hypotenuse

10

Write down what is needed for the triangle.

Need: opposite side

11

Determine which of the trigonometric ratios is required (SOH–CAH–TOA).

sin θ =

12

Substitute the given values into the appropriate ratio.

sin 52 ° =

13

Transpose the equation and solve for h.

124.45 sin 52è = h h = 124.45 sin 52è

14

Round off the answer to 2 decimal places.



15

Answer the question.

The height of the building is 98.07 m.

Maths Quest 10 + 10A for the Australian Curriculum

O H h 124 . 45

ö 98.07

measurement and geometry • Pythagoras & trigonometry

remember

1. Angles of 30è, 45è and 60è have exact values of sine, cosine and tangent. q

30è

sin q

1 2

1

cos q

3 2

1

tan q

1 3

=

45è

2

2 3 3

60è

=

2 2

3 2

=

2 2

1 2

1

3

2. The sine rule states that for any triangle ABC: a b c = = sin A sin B sin C 3. When using this rule it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. 4. The sine rule can be used to solve non–right-angled triangles if we are given: (a) two angles and one side length (b) two side lengths and an angle opposite one of these side lengths. 5. The ambiguous case exists if C is an acute angle and a > c > a sin C. exerCise

22a

the sine rule FluenCy 1 We 1 In the triangle ABC, a = 10, b = 12 and B = 58è. Find A, C and c.

44è58Å, 77è2Å, 13.79

2 In the triangle ABC, c = 17.35, a = 26.82 and A = 101è47Å. Find C, B and b.

39è18Å, 38è55Å, 17.21

3 In the triangle ABC, a = 5, A = 30è and B = 80è. Find C, b and c. 70è, 9.85, 9.4 C = 51è, b = 54.66, 4 In the triangle ABC, c = 27, C = 42è and A = 105è. Find B, a and b. 33è, 38.98, 21.98 c = 44.66 5 In the triangle ABC, a = 7, c = 5 and A = 68è. Find the perimeter of the triangle. 19.12 A = 60è, b = 117.11, c = 31.38

6 Find all unknown sides and angles for the triangle ABC, given A = 57è, B = 72è and a = 48.2. 7 Find all unknown sides and angles for the triangle ABC, given a = 105, B = 105è and C = 15è. 8 Find all unknown sides and angles for the triangle ABC, given a = 32, b = 51 and A = 28è. 9 Find the perimeter of the triangle ABC if a = 7.8, b = 6.2 and A = 50è.

24.17

10 mC Note: There may be more than one correct answer.

In a triangle ABC, B = 40è, b = 2.6 and c = 3. The value of C is approximately: ✔ B 48è B = 48è26Å, C = 103è34Å, c = 66.26; or B = 131è34Å, C = 20è26Å, c = 23.8 D 133è We 2 In the triangle ABC, a = 10, c = 8 and C = 50è. Find two possible values of A, and hence two possible values of b. A = 73è15Å, b = 8.73; or A = 106è45Å, b = 4.12 In the triangle ABC, a = 20, b = 12 and B = 35è. Find two possible values for the perimeter of the triangle. 51.9 or 44.86 Find all unknown sides and angles for the triangle ABC, given A = 27è, B = 43è and c = 6.4. Find all unknown sides and angles for the triangle ABC, given A = 100è, b = 2.1 and C = 42è. Find all unknown sides and angles for the triangle ABC, given A = 25è, b = 17 and a = 13.

A 47è ✔ C 132è 11 C = 110è, a = 3.09, 12 b = 4.64 13 B = 38è, 14 a = 3.36, c = 2.28 15

B = 33è33Å, C = 121è27Å, c = 26.24; or B = 146è27Å, C = 8è33Å, c = 4.57

Chapter 22 trigonometry II

739

measurement and geometry • Pythagoras & trigonometry understanding 16 We 3 To calculate the height of a building, Kevin measures the angle of elevation to the top

as 48è. He then walks 18 m closer to the building and measures the angle of elevation as 64è. How high is the building? 43.62 m reasoning 17 A river has parallel banks which run directly east–west. Kylie takes a bearing to a tree on the

18

19

20

21

✔

22

23

opposite side. The bearing is 047è T. She then walks 10 m due east, and takes a second bearing to the tree. This is 305è T. Find: a her distance from the second measuring point to the tree 6.97 m b the width of the river, to the nearest metre. 4 m A ship sails on a bearing of S20èW for 14 km; then it changes direction and sails for 20 km and drops anchor. Its bearing from the starting point is now N65èW. a How far is it from the starting point? 13.11 km b On what bearing did it sail the 20 km leg? N20è47ÅW A cross-country runner runs at 8 km/h on a bearing of 150è T for 45 mins; then he changes direction to a bearing of 053è T and runs for 80 mins at a different speed until he is due east of the starting point. a How far was the second part of the run? 8.63 km b What was his speed for this section? 6.48 km/h c How far does he need to run to get back to the starting point? 9.90 km From a fire tower, A, a fire is spotted on a bearing of N42èE. From a second tower, B, the fire is on a bearing of N12èW. The two fire towers are 23 km apart, and A is N63èW of B. How far is the fire from each tower? 22.09 km from A and 27.46 km from B mC A boat sails on a bearing of N15èE for 10 km and then on a bearing of S85èE until it is due east of the starting point. The distance from the starting point to the nearest kilometre is: A 10 km B 38 km C 113 km D 114 km mC A hill slopes at an angle of 30è to the horizontal. A tree which is 8 m tall is growing at an angle of 10è to the vertical and is part-way up the slope. The vertical height of the top of the tree above the slope is: A 7.37 m ✔ B 8.68 m C 10.84 m D 15.04 m A cliff is 37 m high. The rock slopes outward at an angle of 50è to the horizontal and then cuts back at an angle of 25è to the vertical, meeting the ground directly below the top of the cliff. Carol wishes to abseil from the top of the cliff to the ground as shown in the diagram. Her climbing rope is 45 m long, and she needs 2 m to secure it to a tree at the top of the cliff. Will the rope be long enough to allow her to reach the ground? Yes, she needs 43 m altogether.

50è 25è Rope

eBook plus

Digital doc WorkSHEET 22.1 doc-5403

740

reFleCtion 

 

In what situations can the sine rule be used?

maths quest 10 + 10a for the australian Curriculum

Rock 37 m

measurement AND geometry • Pythagoras & trigonometry

22b

The cosine rule ■■

In any non–right-angled triangle ABC, a perpendicular line can be drawn from angle B to side b. Let D be the point where the perpendicular line meets side b, and let the length of the perpendicular line be h. Let the length AD = x units. The perpendicular line creates two rightangled triangles, ADB and CDB. B c A

■■

■■ ■■

x

h

a

D b

b–x

C

Using triangle ADB and Pythagoras’ theorem, we obtain: c2 = h2 + x2 [1] Using triangle CDB and Pythagoras’ theorem, we obtain: a2 = h2 + (b - x)2 [2] Expanding the brackets in equation [2]: a2 = h2 + b2 - 2bx + x2 Rearranging equation [2] and using c2 = h2 + x2 from equation [1]: a2 = h2 + x2 + b2 - 2bx = c2 + b2 - 2bx = b2 + c2 - 2bx From triangle ABD, x = c cos A; therefore a2 = b2 + c2 - 2bx becomes a2 = b2 + c2 - 2bc cos A This is called the cosine rule and is a generalisation of Pythagoras’ theorem. In a similar way, if the perpendicular line was drawn from angle A to side a or from angle C to side c, the two right-angled triangles would give c2 = a2 + b2 - 2ab cos C and b2 = a2 + c2 - 2ac cos B respectively. From this, the cosine rule can be stated: In any triangle ABC a2 = b2 + c2 - 2bc cos A b2 = a2 + c2 - 2ac cos B c2 = a2 + b2 - 2ab cos C B

c A ■■

a C

b

The cosine rule can be used to solve non–right-angled triangles if we are given: 1. three sides of the triangle 2. two sides of the triangle and the included angle (the angle between the given sides).

Worked Example 4

Find the third side of triangle ABC given a = 6, c = 10 and B = 76è. Think 1

Write/draw

Draw a labelled diagram of the triangle ABC and fill in the given information.

B c = 10 A

76è b

a=6 C Chapter 22 Trigonometry II

741

measurement AND geometry • Pythagoras & trigonometry

2

Check that one of the criteria for the cosine rule has been satisfied.

Yes, the cosine rule can be used since two side lengths and the included angle have been given.

3

Write down the appropriate cosine rule to find side b.

To find side b: b2 = a2 + c2 - 2ac cos B

4

Substitute the given values into the rule.

5

Evaluate.

6

Round off the answer to 2 decimal places. ■■ ■■

= 62 + 102 - 2 ì 6 ì 10 ì cos 76è ö 106.969 372 5 b ≈ 106.969 372 5 ö 10.34

Note: Once the third side has been found, the sine rule could be used to find other angles if necessary. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject. b2 + c2 − a2 a2 = b2 + c2 - 2bc cos A   À cos A = 2bc 2 a + c2 − b2 b2 = a2 + c2 - 2ac cos B   À cos B = 2ac 2 a b2 − c2 + c2 = a2 + b2 - 2ab cos C   À cos C = 2ab

Worked Example 5

Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm. Think 1

Draw a labelled diagram of the triangle, call it ABC and fill in the given information. Note: The smallest angle will correspond to the smallest side.

Write C b=7 A

a=4 c=9

B

Let a = 4 b=7 c=9

742

2

Check that one of the criteria for the cosine rule has been satisfied.

The cosine rule can be used since three side lengths have been given.

3

Write down the appropriate cosine rule to find angle A.

cos A =

b2 + c2 − a2 2bc

4

Substitute the given values into the rearranged rule.

=

72 + 92 − 4 2 2×7×9

5

Evaluate.

=

114 126

6

Transpose the equation to make A the subject by taking the inverse cos of both sides.

7

Round off the answer to degrees and minutes.

Maths Quest 10 + 10A for the Australian Curriculum

 114  A = cos-1   126  ö 25.208  765  3è ö 25è13Å

measurement AND geometry • Pythagoras & trigonometry

Worked Example 6

Two rowers, Harriet and Kate, set out from the same point. Harriet rows N70èE for 2000 m and Kate rows S15èW for 1800 m. How far apart are the two rowers? Think 1

Write

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

N

2000 m 70° C

A Harriet

15° 1800 m B Kate 2

Check that one of the criteria for the cosine rule has been satisfied.

The cosine rule can be used since two side lengths and the included angle have been given.

3

Write down the appropriate cosine rule to find side c.

To find side c: c2 = a2 + b2 - 2ab cos C

4

Substitute the given values into the rule.

5

Evaluate.

6

Round off the answer to 2 decimal places.

7

Answer the question.

= 20002 + 18002 - 2 ì 2000 ì 1800 cos 125è ö 11  369  750.342 c ≈ 11369750.342 ö 3371.91 The rowers are 3391.91 m apart.

remember

1. In any triangle ABC:

a2 = b2 + c2 - 2bc cos A b2 = a2 + c2 - 2ac cos B c2 = a2 + b2 - 2ab cos C

2. The cosine rule can be used to solve non–right-angled triangles if we are given: (a) three sides of the triangle (b) two sides of the triangle and the included angle (that is, the angle between the two given sides). 3. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject. b2 + c2 − a2 a2 = b2 + c2 - 2bc cos A   À cos A = 2bc b2 = a2 + c2 - 2ac cos B   À cos B =

a2 + c2 − b2 2ac

c2 = a2 + b2 - 2ab cos C   À cos C =

a2 + b2 − c2 2ab

Chapter 22 Trigonometry II

743

measurement and geometry • Pythagoras & trigonometry

exerCise

22b

the cosine rule FluenCy 1 We 4 Find the third side of triangle ABC given a = 3.4, b = 7.8 and C = 80è. 2 In triangle ABC, b = 64.5, c = 38.1 and A = 58è34Å. Find a.

7.95

55.22

3 In triangle ABC, a = 17, c = 10 and B = 115è. Find b, and hence find A and C.

23.08, 41è53Å, 23è7Å

4 We 5 Find the smallest angle in the triangle with sides 6 cm, 4 cm and 8 cm. 28è57Å 5 6 7 8

(Hint: The smallest angle is opposite the smallest side.) A = 61è15Å, In triangle ABC, a = 356, b = 207 and c = 296. Find the largest angle. 88è15Å B = 40è, In triangle ABC, a = 23.6, b = 17.3 and c = 26.4. Find the size of all the angles. C = 78è45Å We 6 Two rowers set out from the same point. One rows N30èE for 1500 m and the other rows S40èE for 1200 m. How far apart are the two rowers? 2218 m Maria cycles 12 km in a direction N68èW and then 7 km in a direction of N34èE. a How far is she from her starting point? 12.57 km b What is the bearing of the starting point from her finishing point? S35è1ÅE

understanding 9 A garden bed is in the shape of a triangle, with sides of length 3 m, 4.5 m and 5.2 m. a Calculate the smallest angle. 35è6Å b Hence, find the area of the garden. (Hint: Draw a diagram, with the longest length as the base of the triangle.) 6.73 m2 10 A hockey goal is 3 m wide. When Sophie is 7 m from one post and 5.2 m from the other, she shoots for goal. Within what angle, to the nearest degree, must the shot be made if it is to score a goal? 23è 11 An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly? 89.12 m 12 A plane flies in a direction of N70èE for 80 km and then on a bearing of S10èW for 150 km. a How far is the plane from its starting point? 130 km b What direction is the plane from its starting point? S22è12ÅE 13 Ship A is 16.2 km from port on a bearing of 053è T and ship B is 31.6 km from the same port on a bearing of 117è T. Calculate the distance between the two ships. 28.5 km 14 A plane takes off at 10.00 am from an airfield and flies at 120 km/h on a bearing of N35èW. A

second plane takes off at 10.05 am from the same airfield and flies on a bearing of S80èE at a speed of 90 km/h. How far apart are the planes at 10.25 am? 74.3 km reasoning 15 Three circles of radii 5 cm, 6 cm and 8 cm are positioned so that they

just touch one another. Their centres form the vertices of a triangle. Find the largest angle in the triangle. 70è49Å 16 For the given shape at near right,

8

5 cm

6 cm

determine: x 150è 8 cm a the length of the diagonal 8.89 m B 7 b the magnitude (size) of angle B 77è 60è c the length of x. x = 10.07 m 10 m 17 From the top of a vertical cliff 68 m high, an observer reFleCtion    notices a yacht at sea. The angle of depression to the In what situations would you yacht is 47è. The yacht sails directly away from the cliff, use the sine rule rather than and after 10 minutes the angle of depression is 15è. How the cosine rule? fast does the yacht sail? 1.14 km/h 744

maths quest 10 + 10a for the australian Curriculum

measurement AND geometry • Pythagoras & trigonometry

22c

Area of triangles ■■

1

The area of any triangle is given by the rule area = 2 bh, where b is the base and h is the perpendicular height of the triangle.

h b ■■

■■

■■

B

However, often the perpendicular height is not given directly and needs to be calculated first. In the triangle ABC, b is the base and h is the perpendicular height of the triangle. Using the trigonometric ratio for sine: h sin A = c Transposing the equation to make h the subject, we obtain: h = c sin A Therefore, the area of triangle ABC becomes:

c A

h

a

b

C

1

area = 2 bc sin A ■■

Depending on how the triangle is labelled, the formula could read: 1

1

1

area = 2 ab sin C    area = 2 ac sin B    area = 2 bc sin A ■■

The area formula may be used on any triangle provided that two sides of the triangle and the included angle (that is, the angle between the two given sides) are known.

Worked Example 7

Find the area of the triangle shown.

7 cm

120è

Think 1

9 cm

Write

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

B c = 7 cm 120è

a = 9 cm

A

C

Let a = 9 cm, c = 7 cm, B = 120è 2

Check that the criterion for the area rule has been satisfied.

The area rule can be used since two side lengths and the included angle have been given.

3

Write down the appropriate rule for the area.

Area = 2 ac sin B

4

Substitute the known values into the rule.

= 2 ì 9 ì 7 ì sin 120è

5

Evaluate. Round off the answer to 2 decimal places and include the appropriate unit.

ö 27.28 cm2

■■

1

1

Note: If you are not given the included angle, you will need to find it in order to calculate the area. This may involve using either the sine or cosine rule. Chapter 22 Trigonometry II

745

measurement AND geometry • Pythagoras & trigonometry

Worked Example 8

A triangle has known dimensions of a = 5 cm, b = 7 cm and B = 52è. Find A and C and hence the area. Think 1

Write/draw

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

B 52è A

a=5

b=7

C

Let a = 5, b = 7, B = 52è 2

Check whether the criterion for the area rule has been satisfied.

The area rule cannot be used since the included angle has not been given.

3

Write down the sine rule to find A.

To find angle A: a b = sin A sin B

4

Substitute the known values into the rule.

5

Transpose the equation to make sin A the subject.

6

Evaluate.

7

Round off the answer to degrees and minutes.

8

Determine the value of the included angle, C, using the fact that the angle sum of any triangle is 180è.

9

Write down the appropriate rule for the area.

10

Substitute the known values into the rule.

ö 2 ì 5 ì 7 ì sin 93è45Å

11

Evaluate. Round off the answer to 2 decimal places and include the appropriate unit.

ö 17.46 cm2

5 7 = sin A sin 52 ° 5 sin 52è = 7 sin A 5 sin 52 ° sin A = 7  5 sin 52 °  A = sin −1    7 ö 34.254  151  87è ö 34è15Å C ö 180è - (52è + 34è15Å) = 93è45Å 1

Area = 2 ab sin C 1

Heron’s formula ■■

■■

If we know the lengths of all the sides of the triangle but none of the angles, we could use the cosine rule to find an angle and then use 1 bc sin A to find the area. Alternatively, we could 2 use Heron’s formula to find the area. Heron’s formula states that the area of a triangle is: Area =

s(s − a)(s − b)(s − c)

where s is the semi-perimeter of the triangle; that is, s = 12 (a + b + c) Note: The proof of this formula is beyond the scope of this course. 746

Maths Quest 10 + 10A for the Australian Curriculum

measurement and geometry • Pythagoras & trigonometry

Worked examPle 9

Find the area of the triangle with sides of 4 cm, 6 cm and 8 cm. think 1

Write/draW

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

C 4 cm B

6 cm A

8 cm

Let a = 4, b = 6, c = 8 2

Determine which area rule will be used.

Since three side lengths have been given, use Heron’s formula.

3

Write down the rule for Heron’s formula.

Area = s(s − a)(s − b)(s − c)

4

Write down the rule for s, the semi-perimeter of the triangle.

5

Substitute the given values into the rule for the semi-perimeter.

s = 12 (a + b + c) = 12 (4 + 6 + 8) =9 Area = 9(9 − 4)(9 − 6)(9 − 8)

6

Substitute all of the known values into Heron’s formula.

7

Evaluate.

= 9 × 5× 3×1 = 135 ö 11.618 950 04

8

Round off the answer to 2 decimal places and include the appropriate unit.

ö 11.62 cm2

remember

1. If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, the following rules may be used to determine the area of that triangle. 1 1 1 Area = 2 ab sin C Area = 2 ac sin B Area = 2 bc sin A 2. Alternatively, if the lengths of three sides of a triangle are known, Heron’s formula may be used to find the area of the triangle: Area = s(s − a)(s − b)(s − c) where s is the semi-perimeter of the triangle; that is, s = 12 (a + b + c) exerCise

22C

area of triangles FluenCy 1 We 7 Find the area of the triangle ABC with a = 7, b = 4 and C = 68è. 12.98 2 Find the area of the triangle ABC with a = 7.3, c = 10.8 and B = 104è40Å.

38.14

3 Find the area of the triangle ABC with b = 23.1, c = 18.6 and A = 82è17Å. 212.88 4 A triangle has a = 10 cm, c = 14 cm and C = 48è. Find A and B and hence the area. A = 32è4Å, B = 99è56Å, area = 68.95 cm2

Chapter 22 trigonometry II

747

measurement and geometry • Pythagoras & trigonometry A = 125è14Å, C = 16è46Å, area = 196.03 mm2

A = 39è50Å, B = 84è10Å, area = 186.03 m2 5 We 8 A triangle has a = 17 m, c = 22 m and C = 56è. Find A and B and hence the area.

6 A triangle has b = 32 mm, c = 15 mm and B = 38è. Find A and C and hence the area. 7 mC In a triangle, a = 15 m, b = 20 m and B = 50è. The area of the triangle is: ✔ C 149.4 m2 A 86.2 m2 B 114.9 m2 D 172.4 m2 8 We 9 Find the area of the triangle with sides of 5 cm, 6 cm and 8 cm. 14.98 cm2 9 Find the area of the triangle with sides of 40 mm, 30 mm and 5.7 cm. 570.03 mm2 10 Find the area of the triangle with sides of 16 mm, 3 cm and 2.7 cm. 2.15 cm2 11 mC A triangle has sides of length 10 cm, 14 cm and 20 cm. The area of the triangle is: ✔ B 65 cm2 A 41 cm2 C 106 cm2 D 137 cm2 12 A piece of metal is in the shape of a triangle with sides of length 114 mm, 72 mm and 87 mm. Find its area using Heron’s formula. 3131.41 mm2 13 A triangle has the largest angle of 115è. The longest side is 62 cm and another side is 35 cm. Find the area of the triangle. 610.38 cm2 14 A triangle has two sides of 25 cm and 30 cm. The angle between the two sides is 30è. Find: 2 a its area 187.5 cm b the length of its third side 15.03 cm c its area using Heron’s formula. 187.47 cm2 understanding 15 The surface of a fish pond has the shape shown in the diagram at right.

How many goldfish can the pond support if each fish requires 0.3 m2 surface area of water? 17 goldfish

1m 2m

5m 4m

16 Find the area of this quadrilateral.

22.02 m2

3.5 m 8m

4m

60è 5m reasoning 17 A parallelogram has diagonals of length 10 cm and 17 cm. An angle between them is

125è. Find: a the area of the parallelogram Area = 69.63 cm2 b the dimensions of the parallelogram. Dimensions are 12.08 cm and 6.96 cm. 18 A lawn is to be made in the shape of a triangle, with sides of length 11 m, 15 m and 17.2 m. How much grass seed, to the nearest kilogram, is needed if it is sown at the rate of 1 kg per 5 m2? 17 kg 748

maths quest 10 + 10a for the australian Curriculum

measurement and geometry • Pythagoras & trigonometry 19 A bushfire burns out an area of level grassland shown in the diagram. (Note: This is a

sketch of the area and is not drawn to scale.) What is the area, in hectares, of the land that is burned? 52.2 hectares

400 m

20 An earth embankment is 27 m long and has a vertical cross-section

shown in the diagram. Find the volume of earth needed to build the embankment. 175 m3

km

2 km River

1.8

200 m Road

2m

130è 100è 80è 50è 5m

21 mC A parallelogram has sides of 14 cm and 18 cm and an angle between them of 72è. The

area of the parallelogram is: B 172.4 cm2 D 252 cm2 ✔ C 239.7 cm2 22 mC An advertising hoarding is in the shape of an isosceles triangle, with sides of length 15 m, 15 m and 18 m. It is to be painted with two coats of purple paint. If the paint covers 12 m2 per litre, the amount of paint needed, to the nearest litre, would be: A 9 L ✔ B 18 L reFleCtion    C 24 L List three formulas for finding the area of a triangle. D 36 L A 118.4 cm2

eBook plus

Digital doc WorkSHEET 22.2 doc-5404

22d

the unit circle ■ ■ ■

A unit circle is a circle with a radius of 1 unit. The unit circle can be divided into 4 quadrants. As you can see from the diagram, all angles in quadrant 1 are between 0è and 90è. All angles in quadrant 2 are between 90è and 180è, in quadrant 3 between 180è and 270è, and in quadrant 4 between 270è and 360è.

90è y

180è

2nd 1st quadrant quadrant 3rd 4th quadrant quadrant

0è x 360è

270è Chapter 22 trigonometry II

749

measurement AND geometry • Pythagoras & trigonometry

Worked Example 10

State the quadrant of the unit circle in which each of the following angles is found. a  145è                    b  282è Think

Write

a The given angle is between 90è and 180è.

a 145è is in quadrant 2.

b The given angle is between 270è and 360è.

b 282è is in quadrant 4.

State the appropriate quadrant. State the appropriate quadrant.

■■ ■■

■■

■■

So far we have looked at triangles constructed in quadrant 1 of the unit circle, with the angle q  being less than 90è. However, triangles can be drawn in other parts of the circle and we need to know what happens when angles become greater than 90è. We can certainly use a calculator to find sine, cosine and tangent values for angles greater than 90è, but it is important to understand where these values have come from. If a right-angled triangle containing angle q is 90è constructed in quadrant 1 of the unit circle, then y the value of sin q can be found by measuring the 1 P length of the opposite side and the value of cos q by measuring the length of the adjacent side. 0è q The point of intersection of the radius (which is one 180è x –1 1 of the arms of angle q ) with the unit circle is P. From cos q 360è the diagram at right observe that cos q represents the x‑coordinate of point P and sin q represents its –1 y‑coordinate. This observation provides us with the 270è technique for finding sine and cosine of any angle in the unit circle, as shown at right. To find the value of sine and/or cosine of any angle q  from the unit circle, follow these steps: 1. Draw a unit circle. 2. Construct the required angle so that its vertex is at the origin and the angle itself is measured from 0è (as marked on the x-axis) in an anticlockwise direction. Label the point of intersection of the radius and the unit circle, P. 3. Use a ruler to find the coordinates of point P. 4. Interpret the results: x = cos q  and y = sin q, where x and y are coordinates of P. sin q

■■

The 4 quadrants of the unit circle For the following activity and exercise, we will need to be able to read values for sine and cosine from a unit circle. Constructing your own unit circle Step 1. Using graph paper, carefully draw a circle centred at the origin and with a radius of 5 cm. Label the x- and y-axes. Step 2. On your graph, mark in 0è, 90è, 180è, 270è and 360è. Step 3. Since we need a unit circle, 5 cm will represent 1 unit; that is, 5 cm = 1 unit. Carefully mark a scale on each axis, where each centimetre represents 0.2 units. (Draw as carefully as possible, since you will need to read values from your axes in exercise 22D.) Use a unit circle to investigate the following. 1. If P has coordinates (x, y) and is located on the unit circle, what is the highest value that the x-coordinate can take? Hence, what is the largest value that the cosine of an angle can take? 2. What is the lowest value that the x-coordinate can take? Hence, what is the smallest value that the cosine of an angle can take? 750

Maths Quest 10 + 10A for the Australian Curriculum

measurement AND geometry • Pythagoras & trigonometry

3. What is the highest value that the y-coordinate can take? Hence, what is the largest value that the sine of an angle can take? 4. What is the lowest value that the y-coordinate can take? Hence, what is the smallest value that the sine of an angle can take? 5. Note that P could be in any of the four quadrants (depending on the size of the angle). Hence, its coordinates could take either positive or negative values, or zero.   Copy and complete the table below to summarise whether sine and cosine are positive or negative for angles in each of the four quadrants. 1st quadrant sin q

positive (+)

cos q

positive (+)

2nd quadrant

3rd quadrant

4th quadrant

negative (-) negative (-)

6. Copy and complete the following sentences. Sine is positive in the ______ and _______ quadrants and is negative in the _____ and _______ quadrants. Cosine is positive in the ______ and ______ quadrants and is negative in the ____ and _______ quadrants. sin θ 7. Use the identity tan q = to work out whether the tangent of an angle in each quadrant cos θ is positive or negative. Copy and complete the following sentence: Tangent is positive in the _____ and ____ quadrants and is negative in the _______ and _______ quadrants. Worked Example 11

Find the value of each of the following using the unit circle. a  sin 200è            b  cos 200è Think

Write/draw

Draw a unit circle and construct an angle of 200è. Label the point corresponding to the angle of 200è on the circle P. Highlight the lengths, representing the x- and y-coordinates of point P.

90è y 1

q = 200è 180è

-1

x

y

1

0è x 360è

-1 270è a The sine of the angle is given by the

a sin 200è = -0.3

b Cosine of the angle is given by the

b cos 200è = -0.9

y-coordinate of P. Find the y-coordinate of P by measuring the distance along the y-axis. State the value of sin 200è. (Note: The sine value will be negative as the y-coordinate is negative.) x-coordinate of P. Find the x-coordinate of P by measuring the distance along the x-axis. State the value of cos 200è. (Note: Cosine is also negative in quadrant 3, as the x-coordinate is negative.)

Chapter 22 Trigonometry II

751

measurement AND geometry • Pythagoras & trigonometry

■■

The results obtained in worked example 11 can be verified with the aid of a calculator: sin 200è = -0.342  020  143 and cos 200è = -0.939  692  62.   Rounding these values to 1 decimal place would give -0.3 and -0.9 respectively, which match the values obtained from the unit circle. remember

1. The unit circle is divided into four quadrants, as shown.

180è

2nd 1st quadrant quadrant 3rd 4th quadrant quadrant

90è y 1

0è x 360è

180è

q cos q

–1

P sin q

90è y

1

0è x 360è

–1 270è

270è

2. Sine and cosine of any angle, q, are given as follows: x = cos q and y = sin q, where x and y are coordinates of point P on the unit circle, corresponding to the given angle. 3. -1 Ç sin q Ç 1 and -1 Ç cos q Ç 1. 4. Sine is positive in quadrants 1 and 2 and negative in quadrants 3 and 4. 5. Cosine is positive in quadrants 1 and 4 and negative in quadrants 2 and 3. Exercise

22d

The unit circle Fluency 1   WE 10  State which quadrant of the unit circle each of the following angles is in. a 60è 1st b 130è 2nd c 310è 4th d 260è 3rd e 100è 2nd f 185è 3rd g 275è 4th h 295è 4th 2   MC  If q = 43è, the triangle drawn to show this would be in: B quadrant 2 ✔ A quadrant 1 C quadrant 3 D quadrant 4 3   MC  If q = 295è, the triangle drawn to show this would be in: A quadrant 1 B quadrant 2 ✔ D quadrant 4 C quadrant 3 4   WE 11  Find the value of each of the following using the unit circle. a sin 20è 0.35 b cos 20è 0.95 c cos 100è -0.17 d sin 100è 0.99 e sin 320è -0.64 f cos 320è 0.77 g sin 215è -0.57 h cos 215è -0.82 5 Use the unit circle to find each of the following. a sin 90è 1 c sin 180è 0 e sin 270è -1 g sin 360è 0

752

Maths Quest 10 + 10A for the Australian Curriculum

b d f h

cos 90è 0 cos 180è -1 cos 270è 0 cos 360è 1

measurement AND geometry • Pythagoras & trigonometry understanding 6 On the unit circle, use a protractor to measure an angle of

9

10

tan 20è

8

sin 20è

7

y

30è from the positive x‑axis. Mark the point P on the circle. Use this point to construct a triangle in quadrant 1 as shown. P a Find cos 30è. (Remember that the length of the adjacent sin 30è 30è side of the triangle is cos 30è.) 0.87 x O b Find sin 30è. (This is the length of the opposite side of cos 30è the triangle.) 0.50 c Check your answers in a and b by finding these values with a calculator. Using your graph of the unit circle, measure 150è with y a protractor and mark the point P on the circle. Use this point to draw a triangle in quadrant 2 as shown. P a What angle does the radius OP make with the 150è sin 150è negative x-axis? 30è x O b Remembering that x = cos q, use your circle to find cos 150è the value of cos 150è. -0.87 c How does cos 150è compare to cos 30è? cos 150è = -cos 30è d Remembering that y = sin q, use your circle to find the value of sin 150è. 0.5 e How does sin 150è compare with sin 30è? sin 150è = sin 30è y On the unit circle, measure 210è with a protractor and mark the point P on the circle. Use this point to draw a triangle in quadrant 3 as shown. 210è cos 210è a What angle does the radius OP make with the x negative x-axis? 30è O sin 210è b Use your circle to find the value of cos 210è. -0.87 P c How does cos 210è compare to cos 30è? cos 210è = -cos 30è d Use your circle to find the value of sin 210è. -0.50 e How does sin 210è compare with sin 30è? sin 210è = -sin 30è On the unit circle, measure 330è with a protractor and y mark the point P on the circle. Use this point to draw a triangle in quadrant 4 as shown. a What angle does the radius OP make with the positive 330è cos 330è x-axis? 30è b Use your circle to find the value of cos 330è. 0.87 x O sin 330è c How does cos 330è compare to cos 30è? cos 330è = cos 30è P d Use your circle to find the value of sin 330è. -0.50 e How does sin 330è compare with sin 30è? sin 330è = -sin 30è On the unit circle, draw an appropriate triangle for the angle of 20è in quadrant 1. y a Find sin 20è. 0.34 b Find cos 20è. 0.94 c Draw a tangent line and extend the hypotenuse of the triangle to meet the tangent as shown. 20è x   Accurately measure the length of the tangent between cos 20è the x-axis and the point where it meets the hypotenuse and, hence, state the value of tan 20è. 0.36 sin 20è d What is the value of ? 0.36 cos 20è sin 20è e How does tan 20è compare with ? They are equal. cos 20è Chapter 22 Trigonometry II

753

measurement AND geometry • Pythagoras & trigonometry 11 On the unit circle, draw an appropriate triangle for the angle of 135è in quadrant 2.

tan 135è

y 135è x

a Find sin 135è. 0.71 b Find cos 135è. -0.71 c Draw a tangent line and extend the hypotenuse of the triangle to meet the tangent as

tan 300è

tan 220è

shown.   Accurately measure the length of the tangent to where it meets the hypotenuse to find the value of tan 135è. -1 sin 135è d What is the value of   ? -1 cos 135è sin 135è e How does tan 135è compare with   ? They are equal. cos 135è f How does tan 135è compare with tan 45è? tan 135è = -tan 45è 12 On the unit circle, draw an appropriate triangle for the angle y of 220è in quadrant 3. a Find sin 220è. -0.64 b Find cos 220è. -0.77 220è c Draw a tangent line and extend the hypotenuse of the x triangle to meet the tangent as shown.   Find tan 220è by accurately measuring the length of the tangent to where it meets the hypotenuse. 0.84 sin 220è d What is the value of   ? 0.83 cos 220è sin 220è e How does tan 220è compare with   ? They are approx. equal. cos 220è f How does tan 220è compare with tan 40è? (Use a calculator.) tan 220è = tan 40è 13 On the unit circle, draw an appropriate triangle for the angle y of 300è in quadrant 4. a Find sin 300è. -0.87 b Find cos 300è. 0.5 300è c Draw a tangent line and extend the hypotenuse of the x triangle to meet the tangent as shown.   Find tan 300è by accurately measuring the length of the tangent to where it meets the hypotenuse. -1.73 sin 300è d What is the value of   ? -1.74 cos 300è sin 300è e How does tan 300è compare with   ? They are approx. equal. cos 300è f How does tan 300è compare with tan 60è? (Use a calculator.) tan 300è = -tan 60è reflection    14   MC  In a unit circle, the length of the radius is equal to: What is the length of the A sin q B cos q diameter of the unit circle? ✔ D 1 C tan q 754

Maths Quest 10 + 10A for the Australian Curriculum

measurement and geometry • Pythagoras & trigonometry

trigonometric functions

22e

sine and cosine graphs ■

eBook plus

The graphs of y = sin x and y = cos x are shown below, sketched from 0è to 360è y

y

y = sin x

1

Interactivity Trigonometric functions

y = cos x

1

int-2796

90è

180è

270è

360è

x

90è

-1 ■

■

■

180è

270è

360è

x

-1

Trigonometric graphs repeat themselves continuously in cycles, and hence they are called periodic functions. The period of the graph is the distance between repeating peaks or troughs. In the example above, the period between the repeating peaks is 360è The amplitude of the graph is half the distance between the maximum and minimum values of the function. In the example above, the distance is half of two units (the distance between -1 and 1), hence the amplitude is 1 unit. Amplitude can also be described as the amount by which the graph goes above and below its mean value. In the above example, the mean value lies along the x axis.

Worked examPle 12

Sketch the graphs of a y = 2 sin x and b y = cos 2x for 0è Ç x Ç 360è think a

1 2

b

Write/draW

The graph must be drawn from 0è to 360è.

Label the graph y = 2 sin x.

1

The graph must be drawn from 0è to 360è.

3

y

y = 2 sin x

2

Compared to the graph of y = sin x each value of sin x has been multiplied by 2, therefore the amplitude of the graph must be 2.

3

2

a

180è

■ ■

x

-2 b

y

y = cos 2x

1

Compared to the graph of y = cos x, each value of x has been multiplied by 2, therefore the period of the graph must become 180è. Label the graph y = cos 2x.

360è

90è

180è

270è

360è

x

-1

For the graph of y = asin nx, or y = acos nx, the amplitude is a and the period becomes

360° . n

If the graph has a negative value of a, the amplitude is the positive value of a. i.e. The amplitude is always | a | Chapter 22 trigonometry II

755

measurement AND geometry • Pythagoras & trigonometry

Worked Example 13

For each of the following graphs, state: i  the amplitude ii  the period. a y = 2 sin 3x x b y = cos 3 c

y 3 2 1 -1

180è

360è

540è

720è

x

-2 -3

Think

Write

a The value of a is 2.

a

b The value of a is 1.

ii Period =

b

and minimum values is 2.5. The graph goes 2.5 units above and below the mean value. The distance between repetitions is 720è

i Amplitude = 1

360 1 3 = 1080è

360° Period is . n

c Half the distance between the maximum

i Amplitude = 2

360 3 = 120è

360° Period is . n

ii Period =

c

i Amplitude = 2.5 ii Period = 720è

remember

1. Trigonometric graphs repeat themselves continuously in cycles and hence they are called periodic functions. 2. The period of the graph is the distance between repetitions. For y = sin x and y = cos x, this is 360è. 3. The amplitude of the graph is the amount the graph goes above and below its mean value. It is half the distance between the maximum and minimum values. For y = sin x and y = cos x the mean value is along the x-axis, hence the amplitude is 1 unit. 360° 4. For the graph of y = asin nx, or y = acos nx the amplitude is a and the period is n

756

Maths Quest 10 + 10A for the Australian Curriculum

measurement and geometry • Pythagoras & trigonometry

exerCise

22e

trigonometric functions FluenCy 1 Using your calculator (or the unit circle if you wish), complete the following table.

1 0 –1

y = sin x

90è 180è 270è 360è 450è 540è 630è 720è

y

x

x

0è

30è

60è

sin x

0

0.5

0.87

x sin x

90è 120è 150è 180è 210è 240è 270è 300è 330è 360è 1

0.87

0.5

0

-0.5

-0.87

-1

-0.87 -0.5

0

390è 420è 450è 480è 510è 540è 570è 600è 630è 660è 690è 720è 0.5

0.87

1

0.87

0.5

0

-0.5 -0.87

-1

-0.87

-0.5

0

a 64è, 116è, 424è, 476è c 44è, 136è, 404è, 496è e 233è, 307è, 593è, 667è

b 244è, 296è, 604è, 656è d 210è, 330è, 570è, 690è f 24è, 156è, 384è, 516è

2 On graph paper, rule x- and y-axes and carefully mark a scale along each axis. Use 1 cm = 30è

1 0 –1

y = cos x

5

6

x

0è

30è

60è

cos x

1

0.87

0.5

x cos x

90è 120è 150è 180è 210è 240è 270è 300è 330è 360è 0

-0.5 -0.87

-1

-0.87 -0.5

0

0.5

0.87

1

390è 420è 450è 480è 510è 540è 570è 600è 630è 660è 690è 720è 0.87

0.5

0

-0.5 -0.87

-1

-0.87 -0.5

0

0.5

0.87

1

7 On graph paper, rule x- and y-axes and carefully mark a scale along each axis. Use 1 cm = 30è

90è 180è 270è 360è 450è 540è 630è 720è

y

3 4

on the x-axis to show x-values from 0è to 720è. Use 2 cm = 1 unit along the y-axis to show y-values from -1 to 1. Carefully plot the graph of y = sin x using the values from the table in question 1. How long does it take for the graph of y = sin x to complete one full cycle? 360è From your graph of y = sin x, find the value of y for each of the following. a x = 42è 0.7 b x = 130è 0.8 c x = 160è 0.35 d x = 200è -0.35 e x = 180è 0 f x = 70è 0.9 g x = 350è -0.2 h x = 290è -0.9 From your graph of y = sin x, find the value of x for each of the following. a y = 0.9 b y = -0.9 c y = 0.7 d y = -0.5 e y = -0.8 f y = 0.4 Using your calculator (or the unit circle if you wish), complete the following table.

x

8

9 It is a very similar graph with the same 10 shape; however, the sine graph starts at (0, 0), whereas the cosine graph starts at (0, 1). 11 120è, 240è, 480è, 600è 127è, 233è, 487è, 593è

12

on the x-axis to show x-values from 0è to 720è. Use 2 cm = 1 unit along the y-axis to show y-values from -1 to 1. Carefully plot the graph of y = cos x using the values from the table in The graph would continue with the cycle. question 6. If you were to continue the graph of y = cos x, what shape would you expect it to take? Is the graph of y = cos x the same as the graph of y = sin x? How does it differ? What features are the same? Using the graph of y = cos x, find a value of y for each of the following. a 48è 0.7 b 170è -0.98 c 180è -1 d 340è 0.9 -0.5 e 240è f 140è -0.8 g 40è 0.8 h 165è -0.96 Using the graph of y = cos x, find a value of x for each of the following. a y = -0.5 b y = 0.8 37è, 323è, 397è, 683è c y = 0.7 46è, 314è, 406è, 674è d y = -0.6 e y = 0.9 26è, 334è, 386è, 694è f y = -0.9 154è, 206è, 514è, 566è Using your calculator (or the unit circle if you wish), complete the following table. x

0è

30è

60è

tan x

0

0.58

1.73 undef. -1.73 -0.58

90è 120è 150è 180è 210è 240è 270è 300è 330è 360è 0

0.58

1.73 undef. -1.73 -0.58

x

390è 420è 450è 480è 510è 540è 570è 600è 630è 660è 690è 720è

tan x

0.58

1.73 undef. -1.73 -0.58

0

0.58

1.73 undef. -1.73 -0.58

0

0

Chapter 22 trigonometry II

757

measurement and geometry • Pythagoras & trigonometry

14

90è

15

0º

16

180è 360è 540è 720è

17

-1

1 x

20 For each of the following, state: i the period

180è

y = cos x

b y = sin x, for x Æ [0è, 720è] d y = 2 cos x, for x Æ [-360è, 0è]

19 We 13 For each of the graphs in question 18, state i the period ii the amplitude.

90è

a y = 3cos 2x i 180è ii 3

-1

d y = -180è -90è

1

y

on the x-axis to show x-values from 0è to 720è. Use 2 cm = 1 unit along the y-axis to show y-values from -2 to 2. Carefully plot the graph of y = tan x using the values from the table in The graph would continue repeating every 180è as above. question 12. If you were to continue the graph of y = tan x, what shape would you expect it to take? Is the graph of y = tan x the same as the graphs of y = sin x and y = cos x? How does it differ? What features are the same? Using the graph of y = tan x, find a value of y for each of the following. a 60è 1.7 b 135è –1 c 310è –1.2 d 220è 0.8 –0.8 e 500è f 590è 1.2 g 710è –0.2 h 585è 1 Using the graph of y = tan x, find a value of x for each of the following. 158è, 338è, 518è, 698è a y = 1 45è, 225è, 405è, 585è b y = 1.5 56è, 236è, 416è, 596è c y = –0.4 11è, 191è, 371è, 551è f y = –1 135è, 315è, 495è, 675è d y = –2 117è, 297è, 477è, 657è e y = 0.2

18 We 12 Sketch the following graphs a y = cos x, for x Æ [-180è, 180è] c y = sin 2x, for x Æ [0è, 360è]

b

y

y = sin x

x

y

y = tan x

180è

270è

360è

450è

540è

630è

720è

x

13 On graph paper, rule x- and y-axes and carefully mark a scale along each axis. Use 1 cm = 30è

1 x sin i 1440è ii 2 4

ii the amplitude. b y = 4sin 3x

1 2

a i b i c i d i

c y = 2cos

i 120è ii 4

e y = -sin x i 360è ii 1

ii ii ii ii

360è 360è 180è 360è

x 2

1 1 1 2

i 720è ii 2

f y = -cos 2x

i 180è

ii 1

18 c

21 mC Parts a to c refer to the graph below.

y

y

y = sin 2x

1

18 a

3 90è

2

x

180è 270è 360è

-1

1 90è

-1

180è

x d

y

y = 2 cos x

2

-2 -3

-360è -270è -180è -90è

x

Period = 1080è Amplitude = 2

-2

a The amplitude of the graph is: A 180o B 90è ✔ C 3 b The period of the graph is: B 360è C 90è ✔ A 180o c The equation of the graph could be:

758

✔ D

y = 3cos 2x

E 6

D 3

E -3

B y = sin x

C y = 3cos

E y = 3sin 2x

540è

understanding 22 Sketch each of the following graphs, stating the period and amplitude of each. y Period = 180è x y = -3 sin 2x a y = 2cos , for x Æ [0è, 360è] 3 Amplitude = 3 -2

2

y

y = 2 cos –3x

1080è

x

A y = cos x

D -3

3

b y = -3sin 2x, for x Æ [0è, 360è]

maths quest 10 + 10a for the australian Curriculum

90è -3

180è 270è 360è x

x 3

measurement and geometry • Pythagoras & trigonometry

x 180è 270è 360è -1 -2

90è

y = cos x + 1

360è

x

y

2 1

M ax value of sin x = 1, hence Sketch the graph of y = cos 2x for x Æ [0è, 360è] max value of y = 2 ì 1 + 3 = 5 ii What is the minimum value of y for this graph? –1 Min value of sin x = -1, hence ii What is the maximum value of y for this graph? 1 min value of y = 2 ì -1 + 3 = 1 Using the answers obtained in part a write down the maximum and minimum values of y = cos 2x + 2. i 3 ii 1 c What would be the maximum and minimum values of the graph of y = 2sin x + 3. Explain how you obtained these values. 25 a Complete the table below by filling in the exact values of y = tan x 24 a b

1

-1

180è

y = cos 2x y

Period = 720è

y

reasoning

y y = tan x

90è



Amplitude = 3 3 x c y = 3sin , for x Æ [-180è, 180è] y = 3 sin –2x 2 x 0 d y = -cos 3x, for x Æ [0è, 360è] -180è -90è 90è 180è e y = 5cos 2x, for x Æ [0è, 180è] -3 f y = -sin 4x, for x Æ [0è, 180è] 23 Use technology to sketch the graphs of each of the following for 0è Ç x Ç 360è a y = cos x + 1 b y = sin 2x - 2 c y = cos (x - 60è) d y = 2sin 4x + 3

180è

x

x

0è

30è

y = tan x

0

3 3

60è 3

90è undef

120è

150è

180è

3 3

0

− 3

−

b Sketch the graph of y = tan x for [0è,180è] At x = 90è, y is undefined. c What happens at x = 90è? d For the graph of y = tan x, x = 90è is called an asymptote. Write down when the next asymptote would occur. T he period = 180è, x = 270è amplitude is e State the period and amplitude of y = tan x reFleCtion    undefined. 26 a Sketch the graph of y = tan 2x, for [0è,180è] For the graph of y = a tan nx, what b When do the asymptotes occur? would be the period and amplitude? c State the period and amplitude of y = tan 2x Period = 90è and amplitude is undefined.

22F eBook plus

Interactivity Trigonometry

int-2797

solving trigonometric equations solving trigonometric equations graphically ■

■

■

Because of the periodic nature of circular functions, there are infinitely many solutions to trigonometric equations. Equations are usually solved within a particular domain (x values), to restrict the number of solutions. The sine graph below shows the solutions between 0è and 360è for the equation sin xè = 0.6

23 b

c

y

-1 -2 -3 -4

90è

c = 45è and x = 135è

180è 270è 360è

x

1

■

■ ■

-1

1 0.6

y = cos (x - 60è) 120è 240è 360è

y = sin 2x - 2

23 d

y

y

x

-1

180è

360è

x

y 5 4 3 2 1

y = 2sin 4x + 3

90è

180è 270è 360è

x

In the example above, it can clearly be seen that there are two solutions to this equation, which are approximately x = 37è and x = 143è The smaller the period, the greater the number of solutions within a particular domain. It is difficult to obtain accurate answers from a graph. More accurate answers can be obtained using technology. Chapter 22 trigonometry II

759

measurement AND geometry • Pythagoras & trigonometry

Solving trigonometric equations algebraically Exact answers can be found for some trigonometric equations using the table at the top of page 739. Worked Example 14

Solve the following equations a  sin x =

3 , x Æ [0è, 360è] 2

b  cos 2 x = - 

1 2

, x Æ [0, 360è]

Think a

b

Write

 3  2 

a x = sin-1 

1

The inverse operation of sine is sin-1.

2

The first solution in the given domain from the table at the top of page 739 is x = 60è.

3

Since sine is positive in the first and second quadrants, another solution must be x = 180è - 60è = 120è.

1

The inverse operation of cosine is cos-1.

2

From the table of values, cos-1

3

Cosine is negative in the second and third quadrants, which give the first two solutions to the equation as: 180è - 45è and 180è + 45è.

2x = 135è, 225è

4

Solve for x by dividing by 2.

x = 67.5è, 112.5è

5

Since the domain in this case is [0è, 360è], and the period has been halved, there must be 4 solutions altogether. The other 2 solutions can be found by adding the period on to each solution.

1 2

There are two solutions in the given domain, x = 60è and x = 120è. b 2x = cos-1  −1 

  2

= 45è.

360° = 180è 2 x = 67.5è + 180è, 112.5è + 180è x = 67.5è, 112.5è, 247.5è, 292.5è The period =

remember

1. Because of the periodic nature of circular functions, there are infinitely many solutions to trigonometric equations. 2. Equations are usually solved within a particular domain (x values), to restrict the number of solutions. 3. The smaller the period, the greater the number of solutions within a particular domain. 4. Solutions can be found for some trigonometric equations algebraically, using the exact values table.

760

Maths Quest 10 + 10A for the Australian Curriculum

measurement and geometry • Pythagoras & trigonometry

exerCise

22F

solving trigonometric equations FluenCy 1 Use the graph at right to find approximate

y

answers to the following equations for the domain 0 Ç x Ç 360è. Check your answers using a calculator. Calculator answers iii cos x = 0.9 25.84è, 334.16è iii cos x = 0.3 72.54è, 287.46è iii cos x = -0.2 101.54è, 258.46è iv cos x = -0.6 126.87è, 233.13è 2 Solve the following equations for the domain 0è Ç x Ç 360è



1

180è

360è

-1

120è, 240è

1 3 1 a sin x = d cos x = 30è, 150è b sin x = 60è, 120è c cos x = 2 2 2 1 e sin x = 1 90è f cos x = -1 g sin x = h sin x = 180è 2 210è, 330è 3 3 i cos x = j cos x = k sin x = 1 90è l cos x = 0 2 2

30è, 330è



x

1

135è, 225è

2 1

225è, 315è

2

90è, 270è

150è, 210è

understanding 3 We 14 Solve the following equations for the given values of x. 75è, 105è, 255è, 285è 30è, 60è, 210è, 240è 3 3 a sin 2x = b cos 2x = , x Æ [0è, 360è] , x Æ [0è, 360è] 2 2 -165è, -135è, -45è, -15è, 75è, 105è

1 5è, 75è, 195è, 255è, 375è, 435è, 555è, 615è 52.5è, 82.5è, 142.5è, 172.5è

1 2

c sin 2x = , x Æ [0è, 720è]

d sin 3x =

−1

, x Æ [-180è, 180è] 2 -165è, -135è, -45è, -15è, 75è, 105è 1 f sin 3x = , x Æ [-180è, 180è] 2 3 0è, 90è, 150è, h cos 3x = 0, x Æ [0è, 360è]

1 2 g cos 4x = -1, x Æ [0è, 90è] 45è e sin 4x = - , x Æ [0è, 180è]

210è, 270è, 330è

reasoning 4 Solve the following equations for x Æ [0è, 360è] a 2sin x - 1 = 0 c

b 2cos x =

30è, 150è

2cos x - 1 = 0 45è, 315è

3 30è, 330è

2sin x + 1 = 0 225è, 315è

d

5 Sam measured the depth of water at the end of the Intergate jetty at various times on Friday

13 August 2010. The table below provides her results.

1 0.00 am, 10.30 pm, 11.00 am, 11.30 pm, noon

Time

6 am

Depth

1.5

7

8

9

10 11 12 pm

1.8 2.3 2.6 2.5 2.2

1.8

1

2

3

4

5

6

7

8

9

1.2 0.8 0.5 0.6 1.0 1.3 1.8 2.2 2.5

Plot the data. Determine: 1 ii the period 12 2 h ii the amplitude. 1.05 m Sam fishes from the jetty when the depth is a maximum. Specify these times for the next 3 days. d Sam’s mother can moor her yacht when the depth is above 1.5 m. During what periods can she moor the yacht on Sunday 16 January? a b c

U ntil 2.15 am; from 8.15 am to 2.45 pm; after 8.45 pm

reFleCtion 

 

Explain why sine and cosine functions can be used to model situations which occur in nature such as tide heights and sound waves.

Chapter 22 trigonometry II

761

measurement AND geometry • Pythagoras & trigonometry

Summary The sine rule ■■

Angles of 30è, 45è and 60è have exact values of sine, cosine and tangent. q 

30è

sin q

1 2

1

cos q

3 2

1

tan q ■■

■■ ■■

■■

1 3

=

45è

2

2 3 3

60è

=

2 2

3 2

=

2 2

1 2

1

3

The sine rule states that for any triangle ABC: a b c = = sin A sin B sin C When using this rule it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. The sine rule can be used to solve non–right-angled triangles if we are given: (a) two angles and one side length (b) two side lengths and an angle opposite one of these side lengths. The ambiguous case exists if C is an acute angle and a > c > a sin C. The cosine rule

■■

■■

■■

In any triangle ABC:

a2 = b2 + c2 - 2bc cos A b2 = a2 + c2 - 2ac cos B c2 = a2 + b2 - 2ab cos C The cosine rule can be used to solve non–right-angled triangles if we are given: (a) three sides of the triangle (b) two sides of the triangle and the included angle (that is, the angle between the two given sides). If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject.



a2 = b2 + c2 - 2bc cos A À cos A =

b2 + c2 − a2 2bc



b2 = a2 + c2 - 2ac cos B À cos B =

a2 + c2 − b2 2ac



c2 = a2 + b2 - 2ab cos C À cos C =

a2 + b2 − c2 2ab

Area of triangles ■■

762

If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, the following rules may be used to determine the area of that triangle. 1 1 1 Area = 2 ab sin C    Area = 2 ac sin B    Area = 2 bc sin A

Maths Quest 10 + 10A for the Australian Curriculum

measurement AND geometry • Pythagoras & trigonometry

■■

Alternatively, if the lengths of three sides of a triangle are known, Heron’s formula may be used to find the area of the triangle: Area =

s(s − a)(s − b)(s − c)

where s is the semi-perimeter of the triangle; that is, s = 12 (a + b + c) The unit circle

The unit circle is divided into four quadrants, as shown. 90è y

180è

2nd 1st quadrant quadrant 3rd 4th quadrant quadrant

90è y 1

0è x 360è

270è ■■ ■■ ■■ ■■

180è

–1

q cos q

P sin q

■■

1

0è x 360è

–1 270è

Sine and cosine of any angle, q, are given as follows: x = cos q and y = sin q, where x and y are coordinates of point P on the unit circle, corresponding to the given angle. -1 Ç sin q Ç 1 and -1 Ç cos q Ç 1. Sine is positive in quadrants 1 and 2 and negative in quadrants 3 and 4. Cosine is positive in quadrants 1 and 4 and negative in quadrants 2 and 3. Trigonometric functions

■■ ■■ ■■

■■

Trigonometric graphs repeat themselves continuously in cycles and hence they are called periodic functions. The period of the graph is the distance between repetitions. For y = sin x and y = cos x, this is 360è. The amplitude of the graph is the amount the graph goes above and below its mean value. It is half the distance between the maximum and minimum values. For y = sin x and y = cos x the mean value is along the x-axis, hence the amplitude is 1 unit. 360° For the graph of y = asin nx, or y = acos nx the amplitude is a and the period is n Solving trigonometric equations

■■ ■■ ■■ ■■

Because of the periodic nature of circular functions, there are infinitely many solutions to trigonometric equations. Equations are usually solved within a particular domain (x values), to restrict the number of solutions. The smaller the period, the greater the number of solutions within a particular domain. Solutions can be found for some trigonometric equations algebraically, using the exact values table. Mapping your understanding

Using terms from the summary above, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 731.

Chapter 22 Trigonometry II

763

measurement and geometry • Pythagoras & trigonometry

Chapter review

y = tan x y

FluenCy

10 mC The value of sin 53è is equal to: A cos 53è ✔ B cos 37è C sin 37è D tan 53è

1 Find the value of x, correct to 1 decimal place. y

450è

–1

11 Simplify 360è

0

x

270è

55è

y = sinx

90è

1

180è

14.2 cm

x

x

sin 53è . tan 53è sin 37è

90è

270è

13 Draw a sketch of y = cos x from 0è Ç x Ç 360è. 14 Draw a sketch of y = tan x from 0è Ç x Ç 360è.

Non-calculator questions

12 cm

15 Label this triangle so that

2 Find the value of q, correct to the nearest minute.

y

450è

360è

B

270è

–1

90è

0

10.2 m b = 22.11 m, c = 5.01 m, C = 10è

x y = . sin 46è sin 68è

y = cosx

180è

1

105è 3.7 m

given a = 25 m, A = 120è and B = 50è. 4 Find the value of x, correct to 1 decimal place. 4.8 cm

3.6 cm

y x

A

A

68è

C

C

16 State the period and amplitude of each of the

following graphs. a y = 2sin 3x c

b y = -3cos 2x

y 1

5.6 cm 90è

5 Find the value of q, correct to the nearest degree. 34è 6 cm 6 cm 10 cm 6 A triangle has sides of length 12 m, 15 m and 20 m. Find the magnitude (size) of the largest angle. 94è56Å

46è

180è

360è

-1 17 Sketch the following graphs. a y = 2sin x, x Æ [0è, 360è] b y = cos 2x, x Æ [-180è, 180è]

x

a Period = 120è, amplitude = 2 b Period = 180è, amplitude = 3 c Period = 180è, amplitude = 0.5

40è

B

x

3 Find all unknown sides and angles of triangle ABC,

x

360è

12 Draw a sketch of y = sin x from 0è Ç x Ç 360è.

75è

20è31Å

180è

0è

18 Use technology to write down the solutions to the

maths quest 10 + 10a for the australian Curriculum

a x = 191.54, 348.46 b x = 22.79, 157.21, 202, 79, 337.21 c x = 88.09, 271.91 d x = 7.24, 172.76, 187.24, 352.76

764



following equations for the domain 0è Ç x Ç 360è to 2 decimal places. 7 A triangle has two sides of 18 cm and 25 cm. The a sin x = -0.2 b cos 2x = 0.7 angle between the two sides is 45è. Find: c 3cos x = 0.1 d 2sin 2x = 0.5 a its area 159.10 cm2 19 Solve each of the following equations. b the length of its third side 17.68 cm 1 a sin x = - , x Æ [0è, 360è] 210è, 330è c its area using Heron’s formula. 159.09 cm2 2 8 If q = 290è, the triangle to show this would be 3 30è, 330è b cos x = , x Æ [0è, 360è] drawn in which quadrant? 4th quadrant 2 9 On the unit circle, draw an appropriate triangle for 1 45è, 315è c cos x = , x Æ [0è, 360è] the angle 110è in quadrant 2. 2 a Find sin 110è and cos 110è, correct to 2 1 0.94, -0.34 decimal places. 45è, 135è d sin x = , x Æ [0è, 360è] b Find tan 110è, correct to 2 decimal places. -2.75 2

measurement and geometry • Pythagoras & trigonometry 20 mC The equation which represents the graph

below could be:

1 A satellite dish is placed on top of an apartment

60è

120è

-1 -2 -3 -4 -5

y

-2

21

22

23

25

B y = 2cos 3x D y = 2sin 2x

2 A yacht sets sail from a mariner and sails on a

bearing of 065èT for 3.5 km. It then turns and sails on a bearing of 127èT for another 5 km. a How far is the yacht from the mariner? 7.3 km a Use technology to help sketch the graph of b On what bearing to the nearest minute should y = 2sin 2x - 3 the yacht travel if it was to sail directly back to b Write down the period and the amplitude of the the mariner? 281è57ÅT a . Period = 180è, amplitude = 2 graph in part 3 Australian power points supply voltage, V, in volts, Sketch the graphs of each of the following, stating where V = 240 sin 18 000t and t is measured in ii the period a i Period = 180è 22 seconds. ii Amplitude = 2 ii the amplitude. a Complete the table below and sketch the graph, b i Period = 90è a y = 2cos 2x, x Æ [0è, 360è] showing the fluctuations in voltage over time. ii Amplitude = 3 b y = 3sin 4x, x Æ [0è, 180è] c i Period = 120è ii Amplitude = 2 t V c y = -2cos 3x, x Æ [-60è, 60è] d i Period = 180è d y = 4sin 2x, x Æ [-90è, 90è] 0 0.000 ii Amplitude = 4 Solve each of the following equations for the given 0.005 240 values of x. 0.010 0 3 a cos 2x = , x Æ [0è, 360è] 15è, 165è, 195è, 345è 0.015 -240 2 0.020 0 1 -70è, 10è, 50è b sin 3x = , x Æ [-90è, 90è] 2 0.025 240 1 0.030 0 1 12.5è, 157.5è, c sin 2x = , x Æ [0è, 360è] 2 292.5è, 337.5è 0.035 -240 1 0.040 0 15è, 105è, 135è d cos 3x = , x Æ [0è, 180è] 2 b Find the times at which the maximum voltage 0, 45è, 90è, 135è, 180è e sin 4x = 0, x Æ [0è, 180è] output occurs. f cos 4x = -1, x Æ [0è, 360è] 45è, 135è, 225è, 315è c How many seconds are there between times of Solve the following for x Æ [0è, 360è] maximum voltage output? 0.02 s a 2cos x - 1 = 0 60, 300 d How many periods eBook plus b 2sin x = - 3 240, 300 (or cycles) are there Interactivities per second? c - 2cos x + 1 = 0 45, 315 Test yourself Chapter 22 50 cycles per second int-2885 d 2sin x + 1 = 0 225, 315 Word search Chapter 22 y Sketch the graph of y = tan 2x, x Æ [0è, 180è]. y = tan 2x Period = 90, int-2883 amplitude is Write down the period, amplitude and the equations Crossword Chapter 22 undefined. int-2884 of any asymptotes. x 90è

180è

Maximum voltage occurs at t = 0.005 s, 0.025 s

24

450 m

t (second)

-3

A y = 3sin 2x C y = 3cos 2x ✔ E y = 2sin 3x

48.3è

.035

-1

0.22è x

.010 .020 .030 .040

1

.005 .015

y = 2sin 2x - 3

2

240

-240

360è

x

3

V (Volts)

building as shown in the diagram below. Find the height of the satellite dish. 3.9 m

y

180è

Problem solVing

Asymptotes are at x = 45è and x = 135è.

Chapter 22 trigonometry II

765

eBook plus

aCtiVities

Are you ready?

(page 732) SkillSHEET 22.1 (doc-5398): Labelling right-angled triangles SkillSHEET 22.2 (doc-5399): Calculating sin, cos or tan of an angle SkillSHEET 22.3 (doc-5400): Finding side lengths in right-angled triangles SkillSHEET 22.4 (doc-5401): Calculating the angle from a sin, cos or tan ratio SkillSHEET 22.5 (doc-5402): Finding angles in right-angled triangles

Digital docs

• • • • •

22A The sine rule Digital doc

• WorkSHEET 22.1 (doc-5403): The sine rule (page 740) 22C Area of triangles Digital doc

• WorkSHEET 22.2 (doc-5404): Cosine rule and area of triangles (page 749)

766

maths quest 10 + 10a for the australian Curriculum

22E Trigonometric functions Interactivity

• Trigonometric functions (int-2796) (page 755) 22F Solving trigonometric equations Interactivity

• Trigonometry (int-2797) (page 759) Chapter review

(page 765) • Test yourself Chapter 22 (int-2885): Take the end-of-chapter test to test your progress • Word search Chapter 22 (int-2883): an interactive word search involving words associated with this chapter • Crossword Chapter 22 (int-2884): an interactive crossword using the definitions associated with the chapter Interactivities

To access eBookPLUS activities, log on to www.jacplus.com.au

statistics AND probability • Data representation and interpretation

23 Interpreting data

23A Bivariate data 23B Lines of best fit 23C Time series What do you know ? 1 List what you know about interpreting bivariate data. Create a concept map to show your list. 2 Share what you know with a partner and then with a small group. 3 As a class, create a large concept map that shows your class’s knowledge of interpreting bivariate data.

opening question

Does an athlete’s performance continue to improve with increasing age?

statistics and probability • data representation and interpretation

are you ready? eBook plus

Digital doc

SkillSHEET 23.1 doc-5405

eBook plus

Digital doc

SkillSHEET 23.2 doc-5406

eBook plus

Digital doc

SkillSHEET 23.3 doc-5407

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SkillSHEET 23.4 doc-5408

Try the questions below. If you have difficulty with any of them, extra help can be obtained by completing the matching SkillSHEET. Either search for the SkillSHEET in your eBookPLUS or ask your teacher for a copy. Substitution into a linear rule 1 Substitute -1 for x in each of the following equations to calculate the value of y. a y = 4x - 2 -6 b y = 3 - x 4 c y = -2 - 5x 3

Solving linear equations that arise when finding x- and y-intercepts a i y = 2 b i y = -3 2 For each of the following equations, substitute: −3 c i y = i x = 0 to find the corresponding value of y 2 ii y = 0 to find the corresponding value of x a 2x + 3y = 6 b x - 3y = 9 c 4y = 3x - 6

Transposing linear equations to standard form 3 Write the following equations in the form y = mx + c. a 2y + 4x = 8 y = -2x + 4 b 8x - 2y = 10 y = 4x - 5

Measuring the rise and the run 4 Find the gradient for each of the following straight lines. a b y y 10 20 5 10 5 10 x

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eBook plus

Digital doc

SkillSHEET 23.6 doc-5410

c 2x + 3y + 5 = 0

a 1

b 2

c

c -1

y 10 5

-10 -5 0 5 10 x -10 -20

-10

SkillSHEET 23.5 doc-5409

ii x = 2

2 5 y=− x− 3 3

-10 -5 0 -5

eBook plus

ii x = 3 ii x = 9

-10 -5 0 -5

5 10 x

-10

Finding the gradient given two points 5 Find the gradient of the line passing through each of the following pairs of points. a (1, 2) and (3, 7) 5 b (-1, -4) and (2, 3) 7 c (6, -1) and (-2, 1) 2

3

−1 4

Graphing linear equations using the x- and y-intercept method 6 Graph each line with the following equations using the x- and y-intercept method. a 5y - 4x = 20 b 4y - 2x = 5 c 3y + 4x = -12 a

y 5y - 4x = 20 -5

4

b

y 4y - 2x = 5

c

y

1 1–4 0

x

-2 1–2

0

x

-3

0

x

-4 3y + 4x = -12

768

maths quest 10 + 10a for the australian curriculum

statistics AND probability • Data representation and interpretation

23A

Bivariate data ■■ ■■ ■■

Bivariate data is really ‘two-variable data’. The list of bivariate data can be considered as numerical pairs of the type: (x1, y1), (x2, y2),  .  .  .  ,  (xn, yn) The easiest way to visualise bivariate data is by constructing a scatterplot.

Scatterplots ■■ ■■

Each piece of data on a scatterplot is shown by a point on a Cartesian plane. The x-coordinate of the point is the value of the independent variable, while the y-coordinate is the corresponding value of the dependent variable.

Worked Example 1

The following table shows the total revenue from selling tickets for a number of different chamber music concerts. Represent these data on a scatterplot. Number of tickets sold Total revenue ($)

400

200

450

350

8  000

3  600

8  500

7  700

300

500

400

5  800 6  000 11  000 7  500

350

250

6  600

5  600

Write

1

Determine the nature of the variables with reasoning.

2

Rule up a set of axes on graph paper. Title the graph. Label the horizontal axis ‘Number of tickets sold’ and the vertical axis ‘Total revenue ($)’.

3

Scale the horizontal and vertical axes.

4

Plot the points on the scatterplot. In each pair of values, treat the number of tickets as the horizontal coordinate and the corresponding total revenue as the vertical coordinate. For example, the first pair of values in the table is represented by the point with coordinates (400, 8000).

The total revenue depends on the number of tickets being sold, so the number of tickets is the independent variable and the total revenue is the dependent variable.

11 000

Revenue obtained from selling music concert tickets

10 000 Total revenue ($)

Think

250

9000 8000 7000 6000 5000 4000 3000 0

200 250 300 350 400 450 500 Number of tickets sold

Correlation ■■ ■■

It is useful to determine whether any relationship exists between the two variables, and if it does, what type of relationship it is. The relationship between the variable is called correlation, and can be classified according to three properties. •• Form — whether it is linear or non-linear •• Direction — whether it is positive or negative •• Strength — whether it is strong, moderate or weak. These classifications are qualitative rather than quantitative. Chapter 23 Interpreting data

769

statistics AND probability • Data representation and interpretation

Linear and non-linear relationships ■■

If the scatterplot is in the shape of a straight, narrow band and a straight line seems to have a reasonable fit, the relationship between the two variables can be called linear. y

y

x Linear relationships ■■

x

If this is not the case, the relationship is non-linear. y

y

x Non-linear relationships ■■ ■■

x

Non-linear relationships can be classified further as being quadratic, exponential and so on. No correlation exists between the two variables if the points on the scatterplot appear to be randomly spread over the set of axes. y

x No correlation

Positive and negative correlation ■■

Positive correlation occurs if one variable increases as the other variable increases. The data points on a scatterplot appear to form a path, directed from the bottom left to the top right corner. y

x Positive correlation ■■

Negative correlation occurs if one variable tends to decrease with the increase of the other. The points on the scatterplot form a path directed from the top left to the bottom right corner. y

x Negative correlation 770

Maths Quest 10 + 10A for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Strength of the correlation ■■

The narrower the path, the stronger the correlation between the two variables. y

y

x Strong correlation ■■

y

x Moderate correlation

x Weak correlation

A perfectly linear correlation exists if the points on the scatterplot form a straight line. y

y

x Perfectly linear relationships

x

Worked Example 2

State the type of the relationship between the variables x and y, suggested by the scatterplot at right.

y

x Think

Write

Carefully analyse the scatterplot and comment on its form, direction and strength.

The points on the scatterplot form a narrow path that resembles a straight ‘corridor’ (that is, it would be reasonable to fit a straight line to it). Therefore the relationship is linear.   The path is directed from the bottom left corner to the top right corner and the value of y increases as x increases. Therefore the correlation is positive.   Furthermore the points are quite tight; that is, they form a thin corridor. So the correlation can be classified as being strong.   There is a strong, positive, linear relationship between x and y.

Correlation and causation ■■

■■

Even a strong correlation does not necessarily mean that the increase or decrease in the level of one variable causes an increase or decrease in the level of the other. It is best to avoid statements such as: ‘An increase in rainfall causes an increase in the wheat growth.’ The following guidelines should be closely followed in order to draw a conclusion about the relationship between the two variables based on the scatterplot. •• If the correlation between x and y is weak, we can conclude that there is little evidence to show that the larger x is, the larger (positive correlation) or smaller (negative correlation) y is. Chapter 23 Interpreting data

771

statistics AND probability • Data representation and interpretation

•• If the correlation between x and y is moderate, we can conclude that there is evidence to show that the larger x is, the larger (positive correlation) or smaller (negative correlation) y is. •• If the correlation between x and y is strong, we can conclude that the larger x is, the larger (positive correlation) or smaller (negative correlation) y is. Worked Example 3

Mary sells business shirts in a department store. She always records the number of different styles of shirt sold during the day. The table below shows her sales over one week. Price ($)

14

18

20

21

24

25

28

30

32

35

Number of shirts sold

21

22

18

19

17

17

15

16

14

11

a Construct a scatterplot of the data. b State the type of correlation between the two variables and, hence, draw a corresponding conclusion. Think

write

(independent variable) on the horizontal axis and ‘Number of shirts sold’ (dependent variable) on the vertical axis.

a

Number of shirts sold

a Draw the scatterplot showing ‘Price ($)’

28 26 24 22 20 18 16 14 12 10 5 10 15 20 25 30 35 Price ($)

b

1

2

Carefully analyse the scatterplot and comment on its form, direction and strength.

Draw a conclusion corresponding to the analysis of the scatterplot.

b The points on the plot form a path that resembles a

straight, narrow band, directed from the top left corner to the bottom right corner. The points are close to forming a straight line. There is a strong, negative, linear correlation between the two variables. The price of the shirt appears to affect the number sold; that is, the more expensive the shirt the fewer sold.

remember

1. Bivariate data involve two sets of related variables for each piece of data. 2. Bivariate data are best represented on a scatterplot. On a scatterplot each piece of data is shown by a single point whose x-coordinate is the value of the independent variable, and whose y-coordinate is the value of the dependent variable. 3. The relationship between two variables is called correlation. Correlation can be classified as linear, non-linear, positive, negative, weak, moderate or strong. 4. If the points appear to be scattered about the scatterplot in no particular order, then no correlation between the two variables exists. If the points form a straight line, then the relationship between the variables is perfectly linear. 5. When drawing conclusions based on the scatterplot, it is important to distinguish between the correlation and the cause. Strong correlation between the variables does not necessarily mean that an increase in one variable causes an increase or decrease in the other. 772

Maths Quest 10 + 10A for the Australian Curriculum

statistics and probability • data representation and interpretation

exercise

23a

bivariate data fluency

eBook plus

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1 For each of the following pairs, decide which of the variables is independent and which is

dependent. a Number of hours spent studying for a Mathematics test and the score on that test. b Daily amount of rainfall (in mm) and daily attendance at the Botanical Gardens. c Number of hours per week spent in a gym and the annual number of visits to the doctor. d Amount of computer memory taken by an essay and the length of the essay (in words). e The cost of care in a childcare centre and attendance at the childcare centre. f The cost of the property (real estate) and the age of the property. g The entry requirements for a certain tertiary course and the number of applications for that course. h The heart rate of a runner and the running speed. 2 We1 The following table shows the cost of a wedding reception at 10 different venues. Represent the data on a scatterplot. No of guests

30

40

50

60

70

80

90

100

110

120

Total cost (ì $1000)

1.5

1.8

2.4

2.3

2.9

4

4.3

4.5

4.6

4.6

3 We2 State the type of relationship between x and y for each of the following scatterplots. a y b y c y

x

x e y

d y

a Perfectly linear, positive b No correlation c Non-linear, negative, moderate d Strong, positive, linear e No correlation f Non-linear, positive, strong g Perfectly linear, negative h Moderate negative, linear

j

m y

f y

h y

g y

x y

x i

y

x

x k y

x

l

y

x n y

x

x

x

x



i Weak, negative, linear j Non-linear, moderate, positive k Positive, moderate, linear l Non-linear, strong, negative m Strong, negative, linear n Weak, positive, linear o Non-linear, moderate, positive



Independent a Number of hours b Rainfall c Hours in gym d Lengths of essay e Cost of care f Age of property g Number of applicants h Running speed

Dependent Test results Attendance Visits to the doctor Memory taken Attendance Cost of property Cut-off OP score Heart rate

SkillSHEET 23.7 doc-5411

x o y

x

x chapter 23 Interpreting data

773

statistics and probability • data representation and interpretation 4 We3 Eugene is selling leather bags at the local market. During the day he keeps records

eBook plus

of his sales. The table below shows the number of bags sold over one weekend and their corresponding prices (to the nearest dollar).

Digital doc

Number of bags sold

SkillSHEET 23.8 doc-5413

12 11 10 9 8 7 6 5 4 3 2 1

Price ($) of a bag

30

35

40

45

50

55

60

65

70

75

80

Number of bags sold

10

12

8

6

4

3

4

2

2

1

1

a Construct a scatterplot of the data. b State the type of correlation between the two variables and,

hence, draw a corresponding conclusion.

N egative, linear, moderate. The price of the bag appeared to affect the numbers sold; that is, the more expensive the bag, the fewer sold.

30 35 40 45 50 55 60 65 70 75 80 Cost ($)

understanding

Price ($1000)

5 The table below shows the number of bedrooms and the price of each of 30 houses. 420 400 380 360 340 320 300 280 260 240 220 200 180 160 140 2

3

4

5

6

Number of bedrooms

7

8 7 6 5 4

Price (ì $1000)

Number of bedrooms

Price (ì $1000)

2

180

3

279

3

243

2

160

2

195

3

198

3

240

6

408

3

237

2

200

4

362

2

226

2

155

2

205

4

359

4

306

7

420

4

316

3

297

5

369

2

200

5

383

1

195

2

158

2

212

3

265

1

149

4

349

2

174

3

286

a Construct a scatterplot of the data. b State the type of correlation between the number of bedrooms and the price of the house

and, hence, draw a corresponding conclusion.

Number of questions completed

3

Number of bedrooms

c Suggest other factors that could contribute to the price of the house.

6 The table below shows the number of questions solved by each student on a test, and the

corresponding total score on that test. Number of questions

2

4

7

5

2

6

3

9

4

8

3

6

Total score (%)

22

39

69 100 56

18

60

36

87

45

84

32

63

10

100 90 80 70 60 50 40 30 20 10 0

1

2

Price (ì $1000)

V arious answers; location, age, number of people interested in the house, and so on.

M oderate positive linear correlation. There is evidence to show that the larger the number of bedrooms, the higher the price of the house. 9 10

1

Number of bedrooms

Total score (%)

774

a Construct a scatterplot of the data. b What type of correlation does the scatterplot suggest? Strong, positive, linear correlation c Give a possible explanation as to why the scatterplot is not perfectly linear.

V arious answers — some students are of different ability levels and maths quest 10 + 10a for the australian curriculum they may have attempted the questions but had incorrect answers.

statistics AND probability • Data representation and interpretation 7 A sample of 25 drivers who had obtained a full licence within the last month was asked to

Number of lessons

Number of accidents

Number of lessons

Number of accidents

 5

6

 5

5

20

2

20

3

15

3

40

0

25

3

25

4

10

4

30

1

35

0

15

4

 5

5

35

1

15

1

 5

4

10

3

30

0

20

1

15

2

40

2

20

3

25

2

10

4

10

5

a Represent these data on a scatterplot. b Specify the relationship suggested by the scatterplot. Weak, negative, linear relation c Suggest some reasons why this scatterplot is not perfectly linear. Reasoning 8 Each point on the scatterplot below shows the time (in weeks) spent by a person on a healthy

diet and the corresponding mass lost (in kg).

Loss in mass

6 5 4 3 2 1 0

Number of lessons

5 10 15 20 25 30 35 40

recall the approximate number of driving lessons they had taken (to the nearest 5), and the number of accidents they had had while being on P plates. The results are summarised in the table which follows.

V  arious answers, such as some drivers are better than others, live in lower traffic areas, traffic conditions etc.

Number of weeks

Study the scatterplot and state whether each of the following statements is true or false. a The number of weeks that the person stays on a diet is the independent variable. T b The y-coordinates of the points represent the time spent by a person on a diet. F c There is evidence to suggest that the longer the person stays on a diet, the greater the loss in mass. T d The time spent on a diet is the only factor that contributes to the loss in mass. F e The correlation between the number of weeks on a diet and the number of kilograms lost is positive. T Chapter 23 Interpreting data

775

Number of accidents

statistics and probability • data representation and interpretation 9 mc The scatterplot that best represents the relationship between the amount of water

✔ B

Water usage (L)

A

Temperature (èC)

consumed daily by a certain household for a number of days in summer and the daily temperature is:

Water usage (L) Temperature (èC)

d

Water usage (L)

C

Temperature (èC)

Water usage (L)

Temperature (èC)

✔

sides and the sum of interior angles for a number of polygons. Which of the following statements is not true? A The correlation between the number of sides and the angle sum of the polygon is perfectly linear. B The increase in the number of sides causes the increase in the size of the angle sum. C The number of sides depends on the sum of the angles. d The correlation between the two variables is positive.

Sum of angles (è)

10 mc The scatterplot below shows the number of

1300 1200 1100 1000 900 800 700 600 500 400 300 200 3

4

5 6 7 8 9 10 Number of sides

11 mc After studying a scatterplot, it was concluded that

✔

23b eBook plus

Interactivity Applying lines of best fit

reflection How could you determine whether the change in one variable causes the change in another variable?

lines of best fit ■

■

int-2798 ■

776

there was evidence that the greater the level of one variable, the smaller the level of the other variable. The scatterplot must have shown a: A strong, positive correlation B strong, negative correlation C moderate, positive correlation d moderate, negative correlation

Although a relationship between two variables can be linear, quadratic, exponential and so on, only linear relationships will be considered here. The method used is to fit a straight line to the scatterplot. This is positioned by eye, so that there are approximately an equal number of points on either side of the line, with the points being as close to the line as possible. Such a line is called a line of best fit. It should be noted that the position of this line is rather subjective. Once a line of best fit has been placed on the scatterplot, an equation for this line can be established, using the coordinates of any two points on the line. These two points do not necessarily have to be actual data points, but if any data points do lie on the line, these are chosen as their values are known accurately.

maths quest 10 + 10a for the australian curriculum

statistics AND probability • Data representation and interpretation

■■

The equation for the line passing through these two selected points can then be calculated.   The equation through the two points (x1, y1) and (y2, y2) is given by: y −y y = mx + c, where m = 2 1 . x2 − x1

Worked Example 4

The data in the table below show the cost of using the internet at a number of different internet cafes based on hours used per month. Hours used per month

10

12

20

18

10

13

15

17

14

11

Total monthly cost ($)

15

18

30

32

18

20

22

23

22

18

a Construct a scatterplot of the data. b Draw in the line of best fit. c Find the equation of the line of best fit in terms of the variables n (number of hours) and

C (monthly cost).

Think

Write/draw

a Draw the scatterplot placing the

a Total monthly cost ($)

independent variable (hours used per month) on the horizontal axis and the dependent variable (total monthly cost) on the vertical axis.

32 30 28 26 24 22 20 18 16 14 10 11 12 13 14 15 16 17 18 19 20 Hours used per month

1

Carefully analyse the scatterplot.

2

Position the line of best fit so there is approximately an equal number of data points on either side of the line and so that all points are close to the line. Note: With the line of best fit, there is no single definite solution.

b Total monthly cost ($)

b

32 30 28 26 24 22 20 18 16 14

(20, 30)

(13, 20)

10 11 12 13 14 15 16 17 18 19 20 Hours used per month c

1

Select two points on the line which are not too close to each other.

2

Calculate the gradient of the line.

3

Write the rule for the equation of a straight line.

c

Let (x1, y1) = (13, 20) and (x2, y2) = (20, 30). y2 − y1 x2 − x1 30 − 20 m= 20 − 13 10 = 7 y = mx + c m=

Chapter 23 Interpreting data

777

statistics AND probability • Data representation and interpretation

10 x+c 7 10 20 =  (13) + c 7 130 c = 20 7 140 − 130 = 7 10 = 7 y=

4

Substitute the known values into the equation.

5

Substitute one pair of coordinates (say, 13, 20) into the equation to calculate c.

6

Write the equation.

y=

10 10 x+ 7 7

7

Replace x with n (number of hours used) and y with C (the total monthly cost) as required.

C=

10 10 n+ 7 7

■■

The line of best fit can be used to graphically predict the value of one variable from that of another. Because of the subjective nature of the line, it should be noted that predictions are not accurate values, but rough estimates. Although this is the case, predictions using this method are considered valuable when no other methods are available.

Worked Example 5

Use the given scatterplot and line of best fit to predict: a the value of y when x = 10 b the value of x when y = 10.

Think a

1

2

778

y 45 40 35 30 25 20 15 10 5 0

5

10 15 20 25 30 35 40 x

write

Locate 10 on the x-axis and draw a vertical line until it meets with the line of best fit. From that point, draw a horizontal line to the y-axis. Read the value of y indicated by the horizontal line.

Write your answer.

a

y 45 40 35 30 25 20 15 10 5 0

5

10 15 20 25 30 35 40 x

When x = 10, y = 35

Maths Quest 10 + 10A for the Australian Curriculum

statistics AND probability • Data representation and interpretation b

1

2

b

Locate 10 on the y-axis and draw a horizontal line until it meets with the line of best fit. From that point draw a vertical line to the x-axis. Read the value of x indicated by the vertical line.

45 40 35 30 25 20 15 10 5 0

5

10 15 20 25 30 35 40 x

When y = 10, x = 27

Write your answer.

■■

y

If the equation of the line of best fit is known, or can be calculated, predictions can be made algebraically by substituting known values into the equation.

Worked Example 6

The table below shows the number of boxes of tissues purchased by hayfever sufferers and the number of days affected by hay fever during the blooming season in spring. Number of days affected by hayfever (d)

3

12

14

7

9

5

6

4

10

8

Total number of boxes of tissues purchased (T )

1

4

5

2

3

2

2

2

3

3

a Construct a scatterplot of the data and draw a line of best fit. b Determine the equation of the line of best fit. c Interpret the meaning of the gradient. d Use the equation of the line of best fit to predict the number of boxes of tissues purchased by people

suffering from hayfever over a period of: i  11 days ii  15 days. Think a

1

write

Draw the scatterplot showing the independent variable (number of days affected by hayfever) on the horizontal axis and the dependent variable (total number of boxes of tissues purchased) on the vertical axis.

a

T 5 4 3 2 1 0

3 4 5 6 7 8 9 10 11 12 13 14 d Chapter 23 Interpreting data

779

statistics AND probability • Data representation and interpretation

2

T

Position the line of best fit on the scatterplot so there is approximately an equal number of data points on either side of the line.

5

(14, 5)

4 3 2 1

(3, 1)

0 b

1

Select two points on the line which are not too close to each other.

2

Calculate the gradient of the line.

3

Write the rule for the equation of a straight line.

4

Substitute the known values into the equation, substituting a known coordinate pair (say, 3, 1) to calculate c.

5

Replace x with d (number of days with hay fever) and y with T (total number of boxes of tissues used) as required.

c

1

Using the line of best fit, interpret the meaning of the gradient.

d

i

1

Substitute the value d = 11 into the equation and evaluate.

3 4 5 6 7 8 9 10 11 12 13 14 d

b Let (x1, y1) = (3, 1) and (x2, y2) = (14, 5).

m=

y2 − y1 x2 − x1

m=

5 −1 4 = 14 − 3 11

y = mx + c 4 x+c 11 4 1 = (3) + c 11 12 c=111 −1 = 11 4 1 y= x11 11 y=

T=

c

d

4 1 d11 11

The gradient indicates an increase in consumption of tissues as the length of the illness continues. A 4 hayfever sufferer is using on average 11 (or about 0.36) of a box of tissues per day. i When d = 11,

4 1 × 11 − 11 11 1 =4− 11

T=

10

= 3 11 2

780

Interpret and write your answer.

Maths Quest 10 + 10A for the Australian Curriculum

In 11 days the hayfever sufferer will need about 4 boxes of tissues.

statistics and probability • data representation and interpretation ii

1

Substitute the value d = 15 into the equation and evaluate.

ii When d = 15,

4 1 × 15 − 11 11 60 1 = − 11 11

T=

4

= 5 11 2

eBook plus

Interactivity Extrapolation

int-1154

Interpret and write your answer.

In 15 days the hayfever sufferer will need 6 boxes of tissues.

interpolation and extrapolation ■

■

■

■

Interpolation is the term used for predicting a value of a variable from within the range of the given data. Extrapolation occurs when the value of the variable being predicted is outside the range of the given data. T In the previous worked example, the values of the given set of data ranged from 3 days to 14 days. 5 (14, 5) Extrapolation This means that the predicted value for 11 days (outside the 4 was an example of interpolation, whereas the given range) predicted value for 15 days was an example of 3 Interpolation extrapolation. (inside the given range) Predictions involving interpolation are considered 2 to be quite reliable. Those involving extrapolation 1 can be viewed with caution, as they rely on the (3, 1) trend of the line remaining unchanged beyond the 0 range of the data. 3 4 5 6 7 8 9 10 11 12 13 14 d

reliability of predictions ■

■

When predictions of any type are made, it is always good to know whether they are reliable or not. If the line of best fit is used to make predictions, they can be considered to be reliable if each of the following is observed. • The number of observations (that is, points constituting the scatterplot) is reasonably large • the scatterplot indicates reasonably strong correlation between the variable • the predictions are made using interpolation.

least squares lines ■

■

y

Least squares regression involves an exact mathematical approach to fitting a line of best fit to bivariate data which shows a strong linear correlation. Consider the regression line shown at right. The vertical lines give an indication of how well the line best ‘fits’ the data. The line of best fit is placed so that these ‘error’ lines are minimised, by balancing the errors above and below the line.

x chapter 23 Interpreting data

781

statistics AND probability • Data representation and interpretation

■■

Least-squares regression takes these error lines, forms squares, and minimises the sum of the squares. y

■■

The actual calculation of the equation of a least squares regression line is complicated; however, a calculator can generate the equation with ease.

x

Correlation coefficient ■■

■■

■■

Once a relationship between two variables has been established, it is helpful to develop a quantitative value to indicate the strength of the relationship. One method is to calculate a correlation coefficient (r). This is easily done using a calculator, but a manual method is shown below. The formula for the correlation coefficient r is: ∑ ( x − x )( y − y ) r= ∑ ( x − x )2 ∑ ( y − y )2 where x and y are the two sets of scores x and y are the means of those scores the symbol “ representing the sum of the expressions indicated. The correlation coefficient is a value in the range −1 to +1. The value of −1 indicates a perfect negative relationship between the two variables, while the value of +1 indicates a perfect positive relationship. For values within this range, a variety of descriptors are used, typically as described in the following table, for positive values of r. Value of r Description 1 Perfect 0.9 – <1.0 Very strong/very high 0.7 – <0.9 Strong/high 0.4 – <0.7 Moderate/reasonable 0.2 – <0.4 Weak/slight/low 0.0 – <0.2 Negligible/very weak 0 Nil/no Similar descriptors can also be used for negative values of r.

Worked Example 7

The percentages from two tests (an English and Maths) for a group of 5 students are as shown: a Calculate the correlation coefficient between the two sets of results. b Based on this value, describe the relationship between the English and Maths results for this group of students. 782

Maths Quest 10 + 10A for the Australian Curriculum

Student 1 2 3 4 5

English (%) 95 85 80 70 60

Maths (%) 85 95 70 65 70

statistics AND probability • Data representation and interpretation

Think a

1

write

Draw up a table to calculate all the necessary data: x , y , (x - x ), (y, - y ), (x - x )2, (y - y )2

a

x

“

y

(x - x ) (x - x )2 (y - y ) (y - y )2 (x - x )(y - y )

  95   85

  17

289

   8

  64

136

  85   95

   7

  49

  18

324

126

  80   70

   2

   4

  -7

  49

-14

  70   65

  -8

  64

-12

144

  96

  60   70

-18

324

  -7

  49

126

630

470

390 385

730

Mean   78   77 2

Substitute into the formula to calculate the correlation coefficient r.

b Describe the relationship.

r= =

∑ ( x − x )( y − y ) ∑ ( x − x )2 ∑ ( y − y )2 470

730 × 630 = 0.69 b A correlation coefficient of 0.69 indicates that the relationship

between English and Maths marks for this group of students is only moderate. This seems to indicate that students who are good at English are not necessarily good at Maths, and vice versa.

remember

1. If the scatterplot indicates a linear relationship between two variables, the linear model of the relationship can be established as follows: (a) position a line of best fit into the scatterplot (b) select any two points on the line and determine the equation of the line. The equation of the line passing through two points (x1, y1) and (x2, y2) is given by: y − y1 y = mx + c   where   m = 2 . x2 − x1 2. The line of best fit can be used for predicting the value of one variable when given the value of the other. This can be done graphically, or if the equation of the line is known, algebraically (by substituting known values into the equation of the line of best fit). 3. When the value that is being predicted using the line of best fit is within the given range, the process is called interpolation. When the value that is being predicted using the line of best fit is outside the given range, the process is called extrapolation. 4. Only predictions made using interpolation can be considered reliable. 5. Least squares regression involves a mathematical approach to fitting a line of best fit to bivariate data which shows a strong linear correlation. It takes error lines, forms squares, and minimises the sum of the squares. A calculator is best used for the calculations. 6. The correlation coefficient r is a quantitative measure of the correlation between two variables. The value of r lies in the range -1 to +1. The closer the value of r lies to zero, the weaker the correlation between the two variables. Chapter 23 Interpreting data

783

statistics and probability • data representation and interpretation

exercise

23b eBook plus

Digital doc

U sing (23, 3) and (56, 8), the 5 16 equation is P = d − . 33 33

SkillSHEET 23.5 doc-5409

lines of best fit understanding 1 We4 The data in the table below show the distances travelled by 10 cars and the amount of petrol used for their journeys (to the nearest litre). Note: Answers may vary depending on the line of best fit drawn. a Construct a scatterplot of the data. b Draw in the line of best fit. c Determine the equation of the line of best fit in terms of the variables d (distance

travelled) and P (petrol used). Distance travelled (km) d

52

36

83

12

44

67

74

23

56

95

Petrol used (L) P

7

5

9

2

7

9

12

3

8

14

2 A random sample of ten Year 10 students who have part-time jobs was selected. Each student

was asked to state the average number of hours they work per week and their average weekly earnings (to the nearest dollar). The results are summarised in the table below. Hours worked h

4

8

15

18

10

5

12

16

14

6

Weekly earnings ($) E

23

47

93

122

56

33

74

110

78

35

E 130 120 110 100 90 80 70 60 50 40 30 20 10 0

3 We5 Use the given scatterplot and line of best fit to predict:

2

4

6

8 10 12 14 16 18 h

Hours worked

y 70 60 50 40 30 20 10 0

Petrol used (L)

Earnings ($)

a Construct a scatterplot of the data using technology. b Draw in the line of best fit using technology. c Write the equation of the line of best fit, in terms of variables h (hours worked) and E (weekly earnings). U sing (8, 47) and (12, 74), the equation is E = 6.75h - 7. d Interpret the meaning of the gradient. On average, students were paid $6.75 per hour.

10 20 30 40 50 60 70 80 x

P 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 100 d

Distance travelled (km)

a the value of y when x = 45 38 b the value of x when y = 15. 18

4 Analyse the graph below and use the line of best fit to predict: y y-values: 600 i 466.60 ii 290.95 500 x-values: 400 i 36.60 ii 24.64 300 200 100 0 5 10 15 20 25 30 35 40 45 x

iii 127.01 iii 5.86

a the value of y when the value of x is: i 7 460 ii 22 290 b the value of x when the value of y is:

iii 36

130

i 120 39 ii 260 24 iii 480. 6 c Determine the equation of the line of best fit, if it is known that it passes through the points (5, 490) and (40, 80). y = -11.71x + 548.57 d Use the equation of the line to verify the values obtained from the graph in parts a and b. 784

maths quest 10 + 10a for the australian curriculum

Cost of food ($)

statistics AND probability • Data representation and interpretation C 165 160 155 150 145 140 135 130 125 120 115 110 105 100 95 90 85 80 75 70

5   WE6  The table below shows the average weekly expenditure on food for households of

various sizes. Number of people in a household

1

2

4

7

5

4

3

5

Cost of food ($ per week)

70

100

150

165

150

140

120

155

Number of people in a household

2

4

6

5

3

1

4

Cost of food ($ per week)

90

160

160

160

125

75

135

a Construct a scatterplot of the data and draw in the line of 1

2

3

4

5

6

Number of people

7

n

best fit. b Determine the equation of the line of best fit. Write it

in terms of variables n (for the number of people in a household) and C (weekly cost of food). c Interpret the meaning of the gradient. d Use the equation of the line of best fit to predict the weekly food expenditure for a family of: O  n average, weekly cost of i 8 $206.25 ii 9 $225.00 iii 10. $243.75 food increases by $18.75 for every 6 The following table shows the gestation time and the birth mass of 10 babies.  sing (1, 75) and (5, U 150), the equation is C = 18.75n + 56.25.

Mass (kg)

Gestation time (weeks)

31

32

33

34

35

36

37

38

39

40

Birth mass (kg) 1.080 1.470 1.820 2.060 2.230 2.540 2.750 3.110 3.080 3.370

M 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 31 32 33 34 35 36 37 38 39 40

Weeks



T  here is a strong positive relationship between the number of hours spent studying and the marks obtained. This seems to indicate that, greater dedication to studying will produce better results.

Positive, strong, linear correlation

Use technology to answer the following questions. a Construct a scatterplot of the data. What type of correlation does the scatterplot suggest? b Draw in the line of best fit and determine its equation. Write it in terms of the variables M = 0.247t - 6.408 t (gestation time) and M (birth mass). c What does the value of the gradient represent? d Although full term of gestation is considered to be 40 weeks, some pregnancies last longer. Use the equation obtained in part b to predict the birth mass of babies born after t 3.719 kg; 3.966 kg 41 and 42 weeks of gestation. e Many babies are born prematurely. Using the equation obtained in part b, predict the 1.002 kg birth mass of a baby whose gestation time was 30 weeks. f If the birth mass of the baby was 2.390  kg, what was his or her gestation time (to the 36 weeks nearest week)? 7   WE7  The number of hours spent studying, and the marks obtained by a group of students on a test are shown in this table. Hours spent studying

45

30

90

60

105

65

90

80

55

75

Marks obtained

40

35

75

65

90

50

90

80

45

65

 W  ith every week of gestation the mass of the baby increases by 247 g.

extra person.

a Calculate the correlation coefficient between the two sets of data. r = 0.9 b Based on this value, describe the relationship between the number of hours spent

studying, and the mark obtained. Reasoning 8 As a part of her project, Rachel is growing a crystal. Every day she measures the crystal’s

mass using special laboratory scales and records it. The table below shows the results of her experiment. Day number Mass (g)

1

2

3

4

5

8

9

2.5

3.7

4.2

5.0

6.1

8.4

9.9

10

11

12

15

16

11.2 11.6 12.8 16.1 17.3 Chapter 23 Interpreting data

785

Mass (g)

statistics and probability • data representation and interpretation

M 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Measurements on days 6, 7, 13 and 14 are missing, since these were 2 consecutive weekends and, hence, Rachel did not have a chance to measure her crystal, which is kept in the school laboratory. a Construct the scatterplot of the data and draw in the line of best fit. b Determine the equation of the line of best fit. Write the equation, using variables d (day of the experiment) and M (mass of the crystal). M = 0.973d + 1.285 c Interpret the meaning of the gradient. Each day Rachel’s crystal gains 0.973 g in mass. d For her report, Rachel would like to fill in the missing measurements (that is, the mass of the crystal on days 6, 7, 13 and 14). Use the equation of the line of best fit to help Rachel find these measurements. Is this an example of interpolation or extrapolation? Explain your answer. 7 .123 g; 8.096 g; 13.934 g; 14.907 g; interpolation (within the given range of 1–16) e Rachel needed to continue her experiment for 2 more days, but she fell ill and had to miss school. Help Rachel to predict the mass of the crystal on those two days (that is, day 17 and 18), using the equation of the line of best fit. Are these predictions reliable? Explain 1 2 3 4 5 6 7 8 9 10111213141516 d your answer. 1 7.826 g; 18.799 g; predictions are not reliable, since they were obtained using extrapolation. Day 9 mc Consider the figure at right. y The line of best fit on the scatterplot at right is used to predict the values of y when x = 15, x = 40 and x = 60. a Interpolation would be used to predict the value of y when the value of x is: A 15 and 40 B 15 and 60 C 15 only ✔ d 40 only e 60 only 10 20 30 40 50 60 70 x b The prediction of the y-value(s) can be considered reliable when: A x = 15 and x = 40 B x = 15, x = 40 and x = 60 d x = 40 and x = 60 ✔ C x = 40 e x = 60 y 10 mc The scatterplot at right is used to predict the value of y when x = 300. 500 This prediction is: 400 A reliable, because it is obtained using 300 interpolation 200 B not reliable, because it is obtained using 100 extrapolation 0 C not reliable, because only x-values can be 100 200 300 400 500 600 700 x predicted with confi dence eBook plus d reliable because the scatterplot contains large Digital doc number of points reflection WorkSHEET 23.1 ✔ e not reliable, because there is no correlation doc-5412 Why is extrapolation considered between x and y to be not reliable?

23c eBook plus

eLesson Fluctuations and cycles

eles-0181

786

time series ■

■

A time series is a sequence of measurements taken at regular intervals (that is, daily, weekly, monthly and so on) over a certain period of time. Time series plots are similar to scatterplots. However, they are usually drawn as a series of points with straight lines joining adjacent points in time. Time is plotted on the x-axis, with the other relevant data plotted on the y-axis. Plots of this type are frequently seen in newspapers and magazines. They include daily temperatures, monthly employment rates and daily share prices.

maths quest 10 + 10a for the australian curriculum

statistics AND probability • Data representation and interpretation

■■

■■

The purpose of these plots is to analyse general trends, and to make predictions for the future. The value of the variable may go up and down in an erratic pattern. These are called fluctuations. However, over a long period of time, the time series will usually suggest a certain trend, called a long-term trend. Trends can be classified as being: •• linear or non-linear Data

Data

Linear trend

t

Non-linear trend

t

Upward trend

t

•• downward or upward Data

Data

Downward trend

t

•• stationary in the mean (that is, no trend). Data

Stationary in the mean (no trend)

t

Worked Example 8

Classify the trend suggested by the time series graph below as being linear or non-linear, and upward, downward or stationary in the mean (no trend). Data

t Chapter 23 Interpreting data

787

statistics AND probability • Data representation and interpretation

Think

Write

Carefully analyse the given graph and comment on whether the graph resembles a straight line or not and whether the values of y increase or decrease over time.

The time series graph does not resemble a straight line and overall the level of the variable, y, decreases over time. The time series graph suggests a non-linear downward trend.

Worked Example 9

The data below show the average daily mass of a person (to the nearest 100 g), recorded over a period of 4 weeks. 63.6, 63.8, 63.5, 63.7, 63.2, 63.0, 62.8, 63.3, 63.1, 62.7, 62.6, 62.5, 62.9, 63.0, 63.1, 62.9, 62.6, 62.8, 63.0, 62.6, 62.5, 62.1, 61.8, 62.2, 62.0, 61.7, 61.5, 61.2 a  Plot these masses as a time series graph. b  Comment on the trend. Think 1

Draw the points on a scatterplot with day on the horizontal axis and mass on the vertical axis.

a

Mass (kg)

a

Write/draw 64.0 63.8 63.6 63.4 63.2 63.0 62.8 62.6 62.4 62.2 62.0 61.8 61.6 61.4 61.2 61.0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 Day

b

2

Join the points with straight line segments.

1

Carefully analyse the given graph and comment on whether the graph resembles a straight line or not and whether the values of y (in this case, mass) increase or decrease over time.

b The graph resembles a straight line that slopes

downwards from left to right (that is, mass decreases with increase in time). Although a person’s mass fluctuates daily, the time series graph suggests a downward trend. That is, overall, the person’s mass has decreased over the 28-day period.

Trend lines ■■ ■■

788

Trend lines can be compared with lines of best fit. They indicate the general trend of the data. The main use of trend lines is in forecasting, or making predictions about the future. This obviously involves extrapolation, which has limited reliability. In this case, no future information is available, so the predictions are based on the assumption that the current trend will continue into the future.

Maths Quest 10 + 10A for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Worked Example 10

The graph at right shows the average cost of renting a one-bedroom flat, as recorded over a 10-year period. a If appropriate, draw in a line of best fit and comment on the type of the trend. b Assuming that the current trend will continue, use the line of best fit to predict the cost of rent in 5 years’ time.

300

Cost of rent ($)

280 260 240 220 200 180 160 140 1

Think 1

10 Year

15

Write/draw

Analyse the given graph and observe what occurs over a period of time. Decide whether a straight line can fit the data. Draw a line of best fit if appropriate.

a

300 280 Cost of rent ($)

a

5

260 240 220 200 180 160 140 1

Comment on the type of trend observed.

1

Extend the line of best fit drawn in part a. The last entry corresponds to the 10th year and we need to predict the cost of rent in 5 years’ time; that is, in the 15th year.

2

Locate 15th year on the time axis and draw a vertical line until it meets with the line of best fit. From the trend line (line of best fit) draw a horizontal line to the cost axis.

10 Year

15

The graph illustrates that the cost of rent increases steadily over the years. The plot indicates a linear trend. The time series graph indicates an upward linear trend. b

300 280 Cost of rent ($)

b

2

5

260 240 220 200 180 160 140 1

5

10 Year

15

Chapter 23 Interpreting data

789

statistics AND probability • Data representation and interpretation

3

Read the cost from the vertical axis.

4

Write your answer.

Assuming that the cost of rent will continue to increase at the present rate, in 5 years we can expect the cost of rent to reach $260 per week.

remember

1. Time series graphs are line graphs with the time plotted on the horizontal axis. 2. Time series are used for analysing general trends and for making predictions for the future. 3. Predictions involving time series graphs are always based on the assumption that the current trend will continue in the future. Exercise

23c

Time series fluency 1   WE 8  Classify the trend suggested by each time series graph as being linear or non-linear, and Data

c

Data

b

Data

d

Data

t

t



a Linear, downward b Non-linear, upward c Non-linear, stationary in the mean d Linear, upward e Non-linear, downward f Non-linear, stationary in the mean g Non-linear, stationary in the mean h Linear, upward

upward, downward or stationary in the mean (no trend). a

t e

Data

g

Data

t f

Data

h

Data

t

t 790

Maths Quest 10 + 10A for the Australian Curriculum

t

t

statistics AND probability • Data representation and interpretation 2   WE 9  The data below show the average daily

Temperature (èC)

Understanding 18.0 17.8 17.6 17.4 17.2 17.0 16.8 16.6 16.4 16.2 16.0 15.8 15.6 15.4 15.2 15.0 14.8 14.6 14.4 14.2 14.0

May temperature

3 The data below show the quarterly sales (in thousands of dollars) recorded by the owner of a

sheepskin product store over a period of 4 years. Quarter

2006

2007

2008

2009

1

  57

  59

  50

  52

2

100

102

  98

100

3

125

127

120

124

4

  74

  70

  72

  73

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Year

a Plot the time series. b The time series plot displays seasonal fluctuations of period 4 (since there are four

quarters). Explain in your own words what this means. Also write one or two possible reasons for the occurrence of these fluctuations. No trend c Overall, does the time series plot indicate upward, downward or no trend? 4 The table below shows the total monthly revenue (in thousands of dollars) obtained by the owners of a large reception hall. The revenue comes from rent and catering for various functions over a period of 3 years.

100 95 90 85 80 75 70 65 60 55 50 45 40

2007

60

Feb. Mar. Apr. May June July Aug. Sept. Oct. 65

40

45

40

50

45

Nov. Dec.

50

55

50

55

  70

2008

70

65

60

65

55

60

60

65

70

75

80

  85

2009

80

70

65

70

60

65

70

75

80

85

90

100

Revenue ($1000)

Construct a time series plot for these data. Describe the graph (peaks and troughs, long-term trend, any other patterns). Try to give possible reasons for monthly fluctuations. Does the graph show seasonal fluctuations of period 12? Are there any patterns that Yes. Peaks in December, troughs in April. repeat from year to year? The owner of a motel and caravan park 90 in a small town keeps records of the total 80 number of rooms and total number of 70 camp sites occupied per month. The time series plots based on his records are shown 60 at right. 50 Peaks around a Describe each graph, discussing general 40 Christmas holidays trend, peaks and troughs and so on. 30 and a minor peak at Explain particular features of the Easter. No camping 20 graphs and give possible reasons. in colder months. 10 b Compare the two graphs and write a short paragraph commenting on any Jan April Aug Dec similarities and differences between Month Check with your teacher. them. Camp sites Motel rooms a b c d

Number of rooms/sites occupied

P  eaks around Christmas where people have lots of parties, troughs around April where weather gets colder 5 and people less inclined to go out.



Jan.

Chapter 23 Interpreting data

G  eneral upward trend with peaks around December and troughs around April.

2007

2008

2009

1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 Month

Day

b S  heepskin products more popular in the third quarter (presumably winter) — discount sales, increase in sales, and so on.

temperatures recorded in May. 17.6, 17.4, 18.0, 17.2, 17.5, 16.9, 16.3, 17.1, 16.9, 16.2, 16.0, 16.6, 16.1, 15.4, 15.1 15.5, 16.0, 16.0, 15.4, 15.2, 15.0, 15.5, 15.1, 14.8, 15.3, 14.9, 14.6, 14.4, 15.0, 14.2 a Plot these temperatures as a time series graph. Linear downward trend b Comment on the trend.

791

statistics AND probability • Data representation and interpretation 6   WE 10  The graph below shows enrolments in the Health and Nutrition course at a local

college over a 10-year period. 120 100 90

120 110 100 90 80 70 60 50 40 30 20 10 0

70 60 50 40

Enrolment



Enrolment

80

30 20 10 1

2

3

Upward linear

4

5

6

7

8

9 10 11 12 13 14 15

Year

0

1

2

3

4

5 6 Year

7

8

9 10

a If appropriate, draw in a line of best fit and comment on the type of the trend. b Assuming that the trend will continue, use the line of best fit to predict the enrolment for In 15th year the expected amount = 122 the course in 5 years’ time; that is, in the 15th year. Reasoning 7 In June a new childcare centre was opened. The number of children attending full time

(according to the enrolment at the beginning of each month) during the first year of operation is shown in the table below.

y=

4 45 x+ 7 7

June

July

Aug.

Sept.

Oct.

Nov.

Dec.

Jan.

Feb.

Mar.

Apr.

May

6

8

7

9

10

9

12

10

11

13

12

14

a Plot this time series. Yes, the graph shows an upward trend. b Is the child care business going well? Justify your answer. c Draw a line of best fit and find its equation, using coordinates of any two points on the

line. (Let June = 1, July = 2 and so on.)

d Use your equation of the line of best fit to predict the enrolment in the centre during the

second year of operation at the beginning of: i August 15 ii January 18 (The assumption made was that business will continue on a linear upward trend.) What assumptions have you made? 8 The graph below shows the monthly sales of a certain book since its publication. Explain in your own words why linear trend forecasting of the future sales of this book is not appropriate. Sales Number of children

The trend is non-linear, therefore unable to forecast future sales.

14 13 12 11 10 9 8 (1, 7) 7 6

(8, 11)

June July Aug Sep Oct Nov Dec Jan Feb Mar Apr May

Month

Time 792

Maths Quest 10 + 10A for the Australian Curriculum

statistics and probability • data representation and interpretation 9 a Choose an object or subject that is of interest to you and which can be observed and Answers will vary.

measured during one day. For example, you might decide to measure your own pulse rate. b Prepare a table where you will record your results every hour within the school day. For

example, for the pulse rate the table might look like this. Time

8 am

9 am

10 am

11 am

12 pm

1 pm

2 pm

3 pm

Pulse rate c Take your measurements at the regular time intervals you have decided on and record

them in the table.

eBook plus

Digital doc WorkSHEET 23.2 doc-5415

d Plot the time series obtained as a result of your experiment. e Describe the graph and comment on the trend. f If appropriate, draw in a line of best fit and predict the values (that is, your pulse rate) for

the next 2–3 hours. g Take the actual measurements during the hours you have made predictions for. Compare

your predictions with the actual measurements. Were your predictions accurate? Why or why not? reflection Why are predictions in the future appropriate for time series even though they involve extrapolation?

chapter 23 Interpreting data

793

statistics AND probability • Data representation and interpretation

Summary Bivariate data ■■ ■■

■■ ■■

■■

Bivariate data involve two sets of related variables for each piece of data. Bivariate data are best represented on a scatterplot. On a scatterplot each piece of data is shown by a single point whose x-coordinate is the value of the independent variable, and whose y-coordinate is the value of the dependent variable. The relationship between two variables is called correlation. Correlation can be classified as linear, non-linear, positive, negative, weak, moderate or strong. If the points appear to be scattered about the scatterplot in no particular order, then no correlation between the two variables exists. If the points form a straight line, then the relationship between the variables is perfectly linear. When drawing conclusions based on the scatterplot, it is important to distinguish between the correlation and the cause. Strong correlation between the variables does not necessarily mean that an increase in one variable causes an increase or decrease in the other. Lines of best fit

■■

■■

■■

■■ ■■

■■

If the scatterplot indicates a linear relationship between two variables, the linear model of the relationship can be established as follows: (a) position a line of best fit into the scatterplot (b) select any two points on the line and determine the equation of the line. The equation of the line passing through two points (x1, y1) and (x2, y2) is given by: y − y1 y = mx + c   where   m = 2 . x2 − x1 The line of best fit can be used for predicting the value of one variable when given the value of the other. This can be done graphically, or if the equation of the line is known, algebraically (by substituting known values into the equation of the line of best fit). When the value that is being predicted using the line of best fit is within the given range, the process is called interpolation. When the value that is being predicted using the line of best fit is outside the given range, the process is called extrapolation. Only predictions made using interpolation can be considered reliable. Least squares regression involves a mathematical approach to fitting a line of best fit to bivariate data which shows a strong linear correlation. It takes error lines, forms squares, and minimises the sum of the squares. A calculator is best used for the calculations. The correlation coefficient r is a quantitative measure of the correlation between two variables. The value of r lies in the range -1 to +1. The closer the value of r lies to zero, the weaker the correlation between the two variables. Time series

■■ ■■ ■■

Time series graphs are line graphs with the time plotted on the horizontal axis. Time series are used for analysing general trends and for making predictions for the future. Predictions involving time series graphs are always based on the assumption that the current trend will continue in the future. Mapping your understanding

Using terms from the summary above, and other terms if you wish, construct a concept map that illustrates your understanding of the key concepts covered in this chapter. Compare your concept map with the one that you created in What do you know? on page 767.

794

Maths Quest 10 + 10A for the Australian Curriculum

statistics AND probability • Data representation and interpretation

Chapter review

S  trong, positive, linear correlation; the larger the number of completed revision questions, the higher the mark on the test.

3 The graph shows the number of occupants of a

large nursing home over the last 14 years.

1 As preparation for a Mathematics test, a group of

20 students was given a revision sheet containing 60 questions. The table below shows the number of questions from the revision sheet successfully completed by each student and the mark, out of 100, of that student on the test.

130 Number of occupants

Number of 9 12 37 60 55 40 10 25 50 48 60 questions Test result 18 21 52 95 100 67 15 50 97 85 89 Number of 50 48 35 29 19 44 49 20 16 58 52 questions

120 110 100 90 80 70 60 50

Test result 97 85 62 54 30 70 82 37 28 99 80 a State which of the variables is dependent and

19 9 19 6 9 19 7 9 19 8 9 20 9 0 20 0 0 20 1 0 20 2 0 20 3 0 20 4 0 20 5 0 20 6 0 20 7 0 20 8 09

Number of questions — independent; mark on a test — dependent

Fluency

which is independent.

Year

b Construct a scatterplot of the data. c State the type of correlation between the two

Linear downwards a Comment on the type of trend displayed. b Explain why it is appropriate to draw in a line of best fit. The trend is linear. c Draw a line of best fit and use it to predict the

variables suggested by the scatterplot and draw a corresponding conclusion. d Suggest why the relationship is not perfectly linear. Different abilities of the students 2 a Use the line of best fit shown on the graph below

to predict the value of y, when the value of x is: i 10 12.5 ii 35. 49 b Use the line of best fit to predict the value of x,

when the value of y is:

number of occupants in the nursing home in Assumes that 3 years’ time. About 65 occupants d What assumption has been made when the current trend will predicting figures for part c ? continue.

4 The table below shows the advertised sale price

i 15 12 ii 30. 22.5 c Find the equation of a line of best fit if it is

known that it passes through the points (5, 5) 22 7 and (20, 27). y= x− 15 3 d Use the equation of the line to algebraically verify the values obtained from the graph in parts a and b. i  12.33  ii  49  and  i  11.82  ii  22.05

(‘000s dollars) and the land size (m2) for ten vacant blocks of land. Land size (m2)

Sale price (ì $1000)

  632

36

1560

58

  800

40

1190

44

  770

41

1250

52

1090

43

25

1780

75

20

1740

72

  920

43

y 50 45 40 35 30

15 10 5

P = 31.82a + 13  070.4, where P is the sale price a Construct a scatterplot and determine the and a is the land area. equation of the line of best fit. x b What does the gradient represent? 5 10 15 20 25 30 35 40 The price of land is approximately $31.82 per square metre. Chapter 23 Interpreting data

795

 = 0.91t + 2.95, where P is the number of P pirouettes and t is the number of hours of training.

5 a C = 0.15p + 11.09, where C is the money spent at the canteen and p represents the pocket money received. b S  tudents spend 15 cents at the canteen per dollar received for pocket money.

statistics AND probability • Data representation and interpretation

796

c Using the line of best fit, predict the

c Using your line of best fit, predict the number

approximate sale price, to the nearest 000’ dollars for a block of land with an area of 1600 m2. $64  000 d Using the line of best fit, predict the approximate land size, to the nearest 10 square metres, you could purchase with $50  000. 1160 m2

of pirouettes that could be complete if a student undertakes 14 hours of training. d Professional ballet dancers may undertake up to 30 hours of training a week. Using your line best fit, predict the number of pirouettes they should be able to do in a row. Comment on Approximately 15 pirouettes your findings.

5 The table below shows, for fifteen students, the

amount of pocket money they receive and spend at the school canteen in an average week. Pocket money ($) 30 40 15 25 40 15 30 30 25 15 50 20 35 20 10

Canteen spending ($) 16 17 12 14 16 14 16 17 15 13 19 14 17 15 13

a Construct a scatterplot and determine the

equation of the line of best fit. b What does the gradient represent? c Using your line of best fit, predict the amount

of money spent at the canteen for a student receiving $45 pocket money a week. $18 d Using your line best fit, predict the amount of money spent at the canteen by a student who receives $100 pocket money each week? Does this seem reasonable? Explain. 6 The table below shows, for 10 ballet students, the

number of hours a week spent training and the number of pirouettes in a row they can complete.

7 The table below shows the heights of 10 students

and the distances along the ground between their feet as they attempt to do the splits. Height (cm)

Distance stretched (cm)

134.5

150

156

160

133.5

147

145

160

160

162

135

149

163

163

138

149

152

158

159

160

Using the data, estimate the distance a person 1.8-m tall can achieve when attempting the splits. Write a detailed analysis of your result. Include: –  an explanation of the method(s) used –  any plots or formula generated –  comments on validity of the estimate – any ways the validity of the estimate could be improved. problem solving 1 For his birthday, Ari was given a small white rabbit.

To monitor the rabbit’s growth, Ari decided to measure it once a week. The table below shows the length of the rabbit for various weeks. Week number n

1 2 3 4 6 8 10 13 14 17 20

Length (cm) l 20 21 23 24 25 30 32 35 36 37 39 Training (h) 11 11 2 Number of pirouettes

8

4 16 11 16 5

3

15 13 3 12 7 17 13 16 8

5

a Construct a scatterplot of the data. L = 1.062n + 19.814 b Draw a line of best fit and determine its equation. c As can be seen from the table, Ari did not

measure his rabbit on weeks 5, 7, 9, 11, 12, 15,

16, 18 and 19. Use the equation of the line of a Construct a scatterplot and determine the Ballet students can do best fit to predict the length of the rabbit for equation of the line of best fit. approximately 0.91 those weeks. b What does the gradient represent? pirouettes for each 25.124 cm; 27.248 cm; 29.372 cm; 31.496 cm; 32.558 cm; hour of training. 35.774 cm; 36.806 cm; 38.93 cm; 39.992 cm Maths Quest 10 + 10A for the Australian Curriculum

statistics and probability • data representation and interpretation Interpolation (within the given range of 1–20) Not reliable, because extrapolation has been used.

d Were the predictions made in part c an example

of interpolation or extrapolation? Explain. e Predict the length of the rabbit in the next three weeks (that is, weeks 21–23), using the line of best fit from part c. 42.116 cm; 43.178 cm; 44.24 cm f Are the predictions that have been made in part e reliable? Explain. g Check your results using technology. 2 Laurie is training for the long jump, hoping to make the Australian Olympic team. His best jump each year is shown in the table below.

a b c d

Age (a)

Best jump (B) (metres)

8

4.31

9

4.85

10

5.29

11

5.74

12

6.05

13

6.21

14

—

15

6.88

16

7.24

17

7.35

18

7.57

Best jump (metres)

B

8 7 6 5 4 3 2 1

0

1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Age

a

1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Age

int-2887

c A small range may not give a fair indication if a data set shows a strong linear correlation. Try to increase the range of the data set by taking more measurements or undertaking more research. d A small number of data points 0 1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 may not be able to establish with Age confidence the existence of a strong linear correlation. Try to increase d Yes. Using points (9, 4.85) and (16, 7.24), the number of data points by taking B = 0.341a + 1.781; estimated best jump more measurements or undertaking = 8.6 m. more research. e No, trends work well over the short term but long term are affected by other variables. f 24 years old: 9.97 m; 28 years old: 11.33 m. It is unrealistic to expect his jumping distance to increase indefinitely. g Equal first chapter 23 Interpreting data 797 Best jump (metres)

Best jump (metres)

c

b

8 7 6 5 4 3 2 1 0

improvement in this situation? What problems are there with using a line of best fit? f There will also be Olympic Games held when Laurie is 24 years old and 28 years old. Using extrapolation, what length would you predict Laurie could jump at these two ages? Is this realistic? g When Laurie was 14, he twisted a knee in training and did not compete for the whole season. In that year, a national junior championship was held. The winner of that championship jumped 6.5 metres. Use your line of best fit to predict whether Laurie would have won that championship. 3 The existence of the following situations is often considered an obstacle to making estimates from data. a Outlier b Extrapolation c Small range of data d small Number of data points Explain why each of these situations is considered an obstacle to making estimates of data and how each might be overcome.

a Outliers can unfairly skew data and as such dramatically alter the line of best fit. Identify and remove Plot the points generated by the table on a any outliers from the data before determining the line of best fit. scatterplot. b Extrapolation involves making Join the points generated with straight line eBook plus estimates outside the data range segments. and this is considered unreliable. Interactivities Draw a line of best fit and determine its When extrapolation is required, consider the data and the likelihood Test yourself Chapter 23 equation. int-2888 that the data would remain linear The next Olympic Games will occur when Word search Chapter 23 if extended. When giving results, Laurie is 20 years old. Use the equation of the int-2886 make comment on the validity of Crossword Chapter 23 line of best fit to estimate Laura’s best jump the estimation.

that year and whether it will pass the qualifying mark of 8.1 metres. a

e Is a line of best fit a good way to predict future

8 7 6 5 4 3 2 1

eBook plus

activities

Are you ready?

(page 768) SkillSHEET 23.1 (doc-5405): Substitution into a linear rule SkillSHEET 23.2 (doc-5406): Solving linear equations that arise when finding x- and y-intercepts SkillSHEET 23.3 (doc-5407): Transposing linear equations to standard form SkillSHEET 23.4 (doc-5408): Measuring the rise and the run SkillSHEET 23.5 (doc-5409): Finding the gradient given two points SkillSHEET 23.6 (doc-5410): Graphing linear equations using the x- and y-intercept method

Digital docs

• • • • • •

23A Bivariate data Digital docs

• SkillSHEET 23.7 (doc-5411): Determining independent and dependent variables (page 773) • SkillSHEET 23.8 (doc-5413): Determining the type of correlation (page 774) 23B Lines of best fit Interactivities

• Applying lines of best fit (int-2798) (page 776) • Extrapolation (int-1154) (page 781)

798

maths quest 10 + 10a for the australian curriculum

Digital docs

• SkillSHEET 23.5 (doc-5409): Finding the gradient given two points (page 784) • WorkSHEET 23.1 (doc-5412): Lines of best fit (page 786) 23C Time series eLesson

• Fluctuations and cycles (eles-0181) (page 786) Digital doc

• WorkSHEET 23.2 (doc-5415): Time series (page 793) Chapter review Interactivities (page 797) • Test yourself Chapter 23 (int-2888): Take the end-ofchapter test to test your progress • Word search Chapter 23 (int-2886): an interactive word search involving words associated with this chapter • Crossword Chapter 23 (int-2887): an interactive crossword using the definitions associated with the chapter

To access eBookPLUS activities, log on to www.jacplus.com.au

Answers CHAPTER 1

Indices Are you ready? 1 a Base is 3, power is 4 b Base is 2, power is 5 c Base is 15, power is 7 2 a 16 b 125 3 a 3 b 11 4 a 8 b 10 5 a 2 b 5 6 a 4 b 6 7 a 5 b 10 d 4 e 2 8 a 4.7958 b 10.0995 d 3.9149 e 2.2240

e i

3 3  x y 4

f

4m5

j

7b3

c c c c c c f c f

4096 17 5 9 1 6 5 6.3246 4.9324

d a4b7 h 6a2b l 4x8y6 4

d 3  a4

g

m3n

h

k

5 2 2  m p 4

l

1 2  y 2 1  xy2 2

b 1 e 4 h -7

c 1 f -3 i 4

4 a a6

b 16a20

c

d 4  n8

e a6b3

f 9a6b4

g 16m12n20

h

j

12

625m

k

8

n m -243 5 a D 6 a C 7 a 64 d 48

b D b E

343 x

i

3

l

8 y15 n 49

b 72 e 1600

a4 b6 81a 4

625b12 o -32

c B

d D c 625 f 27 125

h 1 b ab

i 4 c manb

e n3 - pm2 - q

f amp + np

9 1 ê 3 10 1 ô 1

2 ô 10

3 ô 11

4 ô 100

5 ô 101

6 ô 110

7 ô 111

8 ô 1000

9 ô 1001

10 ô 1010

Exercise 1B — Negative indices 1 1 1 a 5 b 4 y x

d

4 5a

3

e

3x 2 y

3

c f

e 2 y 3x

a9 1 4 m3 n 4

f 1

i

3m3 n3

m

27q 9

j n

8 p6

7b 3

l

2a 4

6

c

x6 y 5y

g

6x3 1 15

32a m

k

20

b6

o

4 a8

d 3

h

m 2 n2 4q

8

l

14

p

4 a2b5 4 y12 x5 3 a8b12

1 8a6 b6

h

32 27

k 125

l

3 4

1 36

c

e

1 16

f

5 36

i

27 25

j 4

3 n8

g 48

b

25

3b 4 n5

8 9

1 8

=12

2a 4 3 2 m3 a 2

d

3 a

1 81

4 a 0.001  371  742 b 0.000  048  225 c 0.000  059  499 d 256 e 7.491  540  923 f 5  419  228.099 5 a D b C c B 6 a B b D c C d E 25 m2 n2 c 7 6 7 a b 8 a b m n 8 a r6 - s6 b m10 + 2m5n5 + n10 c 1 d p2 2r 4 9 2 10 63m 11 x = 3 12 a The power is reduced by 1. b Each answer is divided by 2 to get the next answer. 1 c If the pattern continues we will get 2-1 = , 21 1 1 2-2 = = 2 , etc. which illustrates a-n = n . 4 2 a Exercise 1C — Fractional indices 1 a 4 b 5 d 2 e 4 g 2 h 125 j 10  000  000 k 8 2 a 1.44 b 2.24 d 1.26 e 2.54 g 0.54 h 0.81 4

1

c f i l c f i

3 a 4 5

b 2 2

c a 6

23

8

d x 20

e 10 m15

g

2

b

a 2b3

i

20 −4 y 9

h

9 0.02a 8

9 3 216 9 1.48 0.66 0.86 5

f 2b

5 7 7

i 5 x 2

3

4 5

8 17

4 a ab 2

9 b x 5 y

c 6a 5 b 15

19

2

d 2m 28 n 5

19 5 5

e x 6 y 6 z 6

2

Answers Answers 7A 1A ➜ 1C 7E

g 20 8 a x 3yz a2 x d 3 x b

27 6 3  m n 64

1 8 m 81

k

1

2 a

3 a 1 d 3 g 3

9

bc

h a6

5

j 2ab2

Exercise 1A — Review of index laws 1 a a7 b a6 c b8 5 13 5 7 3 e m n f a b c g m6n4p5 4 9 8 7 i 10a b j 36m n k 12x6y6 5 2 a a b a c b3 3b4

6a 3

g

9

f 8a 5 b 8 c Answers

799

1

5

1

5 a 3 6

b 512

c 12 2

3 a

3

5

11

d a 7

e x 4

f m 45

1 g 2

3 x 20

h

7 a d g

7

c

e

9 2 20

b

3 a 10

e

2 4 p5

8 a a b

4

1 a 4

c

11 7 20 b 20

f

1 56 1 m6 m p

h x 1 6

b

6 7 x5y4 1 4

d 1 3

e 5 x y z

1 5

7 5 20 b

4

3 11

b a 45 b 15

2 3 2 x 15 y 4

1 4

i

3

5 7

6 a x 3 y 5 d

2 1 3 n

d

f

1 8 56 m n 3 5 1 p 24 q 12

1 7

c

6 75

f

1 1 23 b 6

i

b 3c

1 9

1 5

a b c

a mc

8

2

b5

h

7

n4

i

8

e 36a20b10

f

15 15

j 2 a d

8a 7 64 y 36 x 24

Answers

4 81x 2 y14

b

x 4 y6

e 24a24b7 h

625 81b 20 c 28

c f

17

75q 5

4 a 3b 3 15

f

48x11y6

7

i x 10 y 10

2 p11

b

x4

c

n9

d

4 m9

4 m5 9 n15

g

3 p4

h

5q 9

2b 12 17

3a 24

4 x 12 21

3 y 20 5

b

13

2a

25

e

128 x 23 y 4

56a11b6 81 4 y 36

c

1024 b 2 81a

f 6m19n19

27 x16 11

4b 2

h

1

7

3 2 c 30

125 8

b 1 b y = 4

2 n13 m9 15b 2 c 26 6 7 p12

11 a m 6 n c

7 − 3 3

−

×

7 6

27 128m29 n26 27h12 8g 6 5 1 3

i x 3 y 8 z 2

or

6

3

m

b g −6 h3 n 2

n7

7 − 5 6

e a6b-8 or

d 2-2 or

1 4

14

a6

15 14 f d 15 or d

b8 12 a P0 = 20, k = 0.3 b 79 koalas c During the 6th year. 13 a 79% b 56% c 31% Chapter review Fluency 1 D 2 C 5 A 6 A 9 B

10 a 9x10y10

35 1

g p 3 q 2 800

i

7 5 8 p 45 q 18

5

5a b

1

c m4 f 2x2 y3 i 6a2 b6 c 4.98 swings

c

h 8m 4 n 4

4 7

f

m 2 n4 3 y2

9 E 10 A

c B

b 48a5b16

7 11

2

7 1 8 a 5y - 1

Exercise 1D — Combining index laws

g 12 x 8 y 15

i

6 a

3 y8

d 500p8q18

3a 4

36 x 6 y

11

1 7 22x2

1 a 54a10b9

h

16m12 n g 3

c 27

9 a E b C 10 a E b B 11 a a4 b b3 d 4x2 e 2y3 g 3m3 n5 h 2pq2 12 a 2.007 s b 20.07 s 13 Check with your teacher.

b7

2

m5

e

8

c

1

d

1 a2

3y

b 8n2

1

5 a

1 4

b3

g

4 a

e

3 a 3b 4 1 33

g

3a 2 2 4 x5

d

3 C 7 B b

16 p28

13ab3c 2 6

81q12 11 a 16

12 a

8 11 2

a b

13 a 8

4 C 8 C c 1000m15n6

3

b - 2 b b

y2 17

5x 3 2

c

m12

16 n8 c 0

1

41 33

14 a 30a 20 b 20

b

4

c

2

1

x 20 y 9

15 a 1 3

16 a −2a + 2a13 17 a 5b 2 18 a 46

d 43 — in this case, division is not closed on natural numbers. e -2 — in this case, subtraction is not closed on natural numbers. f 4 — in this case, division is closed on natural numbers. 11 a (a + 2b) + 4c = a + (2b + 4c) b (x ì 3y) ì 5c = x ì (3y ì 5c) c 2p ó q ≠ q ó 2p d 5d + q = q + 5d e 3z + 0 = 0 + 3z = 3z 1 1 f 2 x × = × 2x = 1 2x 2x g (4x ó 3y) ó 5z ò 4x ó (3y ó 5z) h 3d - 4y ≠ 4y - 3d

2a 6 3

b2

b 4

1 2a 2 b 2

b 6xy2 b

9y

4

4

c 2 3 m

15

32 x

1

b - 18

Problem solving 1 1 36

2 a 8 b 6 c 2

Exercise 2B — Adding and subtracting algebraic fractions

Chapter 2

Linear algebra Are you ready? b x2y and 14 yx 2 1 a abc and 3acb c -2q2p and 2pq2 2 a -3x + 2 b -5a - 9 c -2p - 2q + 8 3 a 6 b 3ab c -4pq b 11 4 a 1 5 c 5 12

5 a 6 a

1 4 5 6

b b

24 1 9 1 19

c

12 2 23 2 3

c f i l c f i

0 3 -12 -5 1 36 15

c

1 12

c

Exercise 2A — Substitution 1 a 5 b 2 d 6 e -17 g 30 h 12 j 27 k 30 2 a -11 b -1 d 30 e -24 g -125 h 1

3 a

b - 121

7 12

d 1 1

e

3

4 a 17 f 68 5 a D

b 30 g 46

1 576

c 8 h 113.1 b C

49 72

17 99

e

1 35

g

15 x − 4 27

h

2 a

5y 12

d

14 x 9

e

3w 28

f -

g

89 y 35

h

32 x 15

i

7 x + 17 10

j

7 x + 30 12

k

2 x − 11 30

l

19 x + 7 6

c

38 21x

f

9 20x

26 21

d

5 21

5 8x 8 d 3x

3 a

e 1.5 j 624.6

f 6 − 5 x 30 15 − 2 x i 3x 13 x c 12

15 − 16 x 40 3y b 40

5 12x 7 e 24x

b

51 10x

y 5

1 6x

4 a

3 x 2 + 14 x − 4 ( x + 4)( x − 2)

b

2 x 2 + 3 x + 25 ( x + 5)( x − 1)

c

2 x 2 + 6 x − 10 (2 x + 1)( x − 2)

d

4 x 2 − 17 x − 3 ( x + 1)(2 x − 7)

e

7x2 + x ( x + 7)( x − 5)

f

2x2 + 6x + 7 ( x + 1)( x + 4)

g

− x 2 + 7 x + 15 ( x + 1)( x + 2)

h

x−7 ( x + 3)( x − 2)

i

x 2 + 3x + 9 ( x + 2)(3 x − 1)

j

5 − 5x 5 = ( x − 1)(1 − x ) x − 1

k

3x + 7 ( x + 1)2

h

i −

l

3x − 4

Answers 1D ➜ 2B

37 100x

c B

6 3.9  cm 7 65.45  cm3 8 361  m 9 a -1 — in this case, addition is closed on integers. b -1 — in this case, subtraction is closed on integers. c 2 — in this case, multiplication is closed on integers. d -1 — in this case, division is closed on integers. e -2 — in this case, subtraction is closed on integers. f - 12 — in this case, division is not closed on integers. 10 a 10 — in this case, addition is closed on natural numbers. b -4 — in this case, subtraction is not closed on natural numbers. c 12 — in this case, multiplication is closed on natural numbers.

c 1

g

f 48 d 4 i 5

(1 )

b

1 a

( x − 1)2

Answers

801

Exercise 2C — Multiplying and dividing algebraic fractions 4x 3x 4y 9x 1 a b c d y y x 4y

e

−5 x 4y

f

3w 2x

g

−3 x j 245 2y 2 5 2 a b 3x − 2 x−3 2x x +1 e f ( x + 1)2 2(2 x − 3) 9 i 2 32 x ( x − 2)

i

3 a

3 5

e

1 25

8y i 9

2

b

2 9

f

35 6

or

32 xy j 15

12z x 9 c 2( x − 6)

k

g

a 10(a + 3)

j

3x 10( x − 1)

c

1 3

k

2z 7x −x l 6w 1 d x+3

h

h

35d 8(d − 3)

d 3

4 y2 g 7

55 6

9 4 a (3 x − 7)( x + 3)

c

6z 7x

2y2 h 25

2 3

l

y2

d

13 9( x − 4)( x + 1)

i y = 21 12

3 a t = 100

b y = ê17

c q = 6.25

d f = ê1.2

e h = 16 49

f p = ê 83

h j =

i a =

4 a a = 4 d f = 9 g s =

802

4 65

c i = 3 f r = 5 25

h t =

i a =

5 a f = 40 d m = 18 6 a x = 1 13

b g = 30 e n = 28 b y = 9

d k = 1 12

e n = 5 23

7 a k = 25

b m = 16

d u = -418 8 a B 9 a x = -5 d x = -11 g v = -20 10 a x = -1 d g = -2

e b b e h b e

Answers

ê1 23

b b = 6 e q = 118 9 45

8 x = 11 E d = -1 h = -2 r = -3 v=1 t=3

-7 12

c r = -10 f p = 62.4 c m = 4 25 f c = 1 13

c p = -11 37 f c c f i c f

-2 75

d u = g d = -6 14 a A

b e h b e h

v=3 C p=7 t=5 g = -0.8 l=2 e = -23 13

i f = -12 14

b=5 t=3 h = -2 15 c=2 y = -118 m = 15

c f i c f i

w=2 r = 2 13 a=0 r = 2 23 g=7 p = 1 23

c t = 21

12 12

f r = 7 12 i x = 1 c B

e f = h h = -12 b D

Exercise 2E — Solving equations with algebraic fractions and multiple brackets 1 a x = 20 b x = 3 5 c x = 29 31

8

36

-2 8 11

10 43

d x = -7

e x =

g x = -5

h x = -2

i x = 5 3

j x = 2 11

k x = -2

l x = -6

12

h x =

5 7

i x =

3 a x =

5 17

b x = 15

4 a x =

5 19

3 2

g x = 3

c x = -6 2 9

j x = 3 b x =

2 13

d x =

13 20

2

i x =

4

f x =

e x = -1 1 f x = -192 31 4

f x =

c x = -172

b x = 18 3

h m = 16 85

g g =

b y = -4 45

x=2 f=7 g = -1 13 x = -1 k=1 w=1

11

g a = 0.425

ê14 31

13 a x = -15

11 a d g 12 a d g

e x = - 3 or x = -3 2

Exercise 2D — Solving linear equations 1 a a = 24 b k = 121 c g = 2.9 d r = 3 e h = 0.26 f i = -2 h q = 16 g t = 5 i x = 0 2 a f = 12 b i = -60 c z = -7 e w = -5 13 d v = 7 f k = 10

225 484

h k = -36

2 a x = 4

1 b ( x + 2)( x − 9)

21( x − 3) x+5

g j = -3 83

g x =

4 7

h x = 12

k x = 52

1 31 58

c x =

10

d x = -19 l x = 1 5 8

4 11 14

d x = -315 17

e x = 5 20

f x = -110

g x = 1 2

h x = -4 9

i x = 1.5

j x = -4 1

k x = 3

l x = 1

3 C

4 B

c -5d - 5c

d 7y2 - 5y

43

13 3

Chapter review Fluency 1 D 2 B 5 D 6 a 7c - 13 b -7k + 3m

61

26

7 35 8 a (a + 3b) + 6c = a + (3b + 6c) b 12a - 3b ≠ 3b - 12a 1 1 = × 7p = 1 c 7 p × 7p 7p d (x ì 5y) ì 7z = x ì (5y ì 7z) e 12p + 0 = 0 + 12p = 12p f (3p ó 5q) ó 7r ≠ 3p ó (5q ó 7r) g 9d + 11e = 11e + 9d h 4a ó b ≠ b ó 4a 9 a 96 — in this case, multiplication is closed on natural numbers. b 1 — in this case, division is not closed on natural 3 numbers. c -4 — in this case, subtraction is not closed on natural numbers.

10 a

7y 6

b

22 15x 8y 11 a x

c

d b

25z 4x

7 x + 18 10

4 a -2 5 a y = 4 6 a 10 m

3 x 2 + 2 x − 17 ( x + 3)( x + 2) 5 c x+3

e

f

12 a p = 88

y2 50 b s = 3.01

d r = -35

e x = 144

f x = -132

g y = 60 13 a b = 4

h a = ê6 b t = 2

i k = 12 c p = -2

1

c x = - 14

d

5 6

14 a x =

1 2

2

f x = 1 6

1

c x = 2

e x = 12 9

6 7

b x = 22 2

b

f x = -16 21

e x = 3 8

Problem solving 1 a $3 per adult ticket; $5 per child’s ticket b 240 c 60 d P = 3a + 5c, where a = number of adults and c = number of children e $1380 2 a C = 250 + 40h b 18 hours 45 minutes c 18750 d Printing is the cheaper option by $1375.

c

CHAPTER 3

Coordinate geometry Are you ready? 1 a Rise = 6, run = 2 2 a Positive 3 a y = x + 3

b Rise = -2, run = 5 b Negative

d

y 6

x

-2

-1

0

1

2

4

y

 1

 2

3

4

5

2 -4

-2

b y = x - 2

0 -2

y 4

-2

-1

 0

 1

2

y

-4

-3

-2

-1

0

-4

-2

0 -2

y=x+3

2

4x

4x

-4

c y = 2x

y 4

x

-2

-1

0

1

2

y

-4

-2

0

2

4

y = 2x

2 -4

-2

0 -2 -4

y -17 -12 -7 -2 3 8

x -6 -4 -2 0 2 4

y 13 12 11 10 9 8

x

y

0

-240

1

-140

2

-40

3

60

4

160

5

260

x

y

-3

18

y 20 15 y = -5x + 3 10

-2

13

5

-1

8

0

3

1

-2

2

-7

y=x-2

2

2

4x

e

y

x -1  0  1  2  3  4

y = 5x - 12

10 5 -2 -1-5 -10 -15

1

x

2 3 4 5

-20

y 14 y = -0.5x + 10 12 10 8 6 4 2 -6 -4 -2 0

2 4 6

x

y 300 y = 100x - 240 250 200 150 100 50 0 -50 -100 -150 -200 -250

12

-3 -2 -1 0 1 -5 -10

3 4 5

2

x

x

Answers

Answers 2C ➜ 3A

x

2

x

1 2

-15 -20 -25

1

3

d x = 5

Exercise 3A — Sketching linear graphs y 1 a x y 35 -5 -25 y = 10x + 25 30 25 -4 -15 20 -3 -5 15 10 -2 5 5 -1 15 -5 -4 -2 -1-5 0 25 -3 -10 1 35

3

b x = 6 5

d x = 1 15 a x =

2x ( x − 1)(9 x + 1) c b = 16

b 2 b x = 3 b 5 cm

803

f

2 a

b

c

3 a

x

y

-3

19

-2

15

-1

11

0

7

1

3

2

-1

5

20

-4

14

-2

8

0

2

2

-4

4

-10

6

-16

6

-2

5

-1

4

0

3

1

2

2

1

3

0

x

y

-6

15

-4

11

-2

7

0

3

2

-1

4

-5

6

-9

l

y 10 5

y

1 2 3

y 20 15 y = -2x + 3 10 5 0 -10 -5 -5

5 10

y = -5x + 20

0

0

x

x

4



y

y = - 1–2x - 4 -8

x

-4

4 a

y

y = 4x + 1

b

y

y = 3x - 7

(1, 5)

5

01

b

y

1

4

01

-2 0 -2

d

2

-5x + 3y = 10

10 x

y 20

x

5

x

o

2

4x

7

c

3

2

1 2

-4 -5x - 3y = 10

4x

d

y

y 4

-4 -2 0 -2

x (1, -4)

4

x

4x

5x + 3y = 10

2

n

5

6x - 4y = -24

-10 -5 0 -5

-10

4x

-4 -2 0 -2

10 x

y = 2x - 10

10 20 x

y 10 5

0

y = -x + 3

y 10

-10

-10

y 4

Answers

-10 10x + 30y = -150

m

y 5 -2x + 8y = -20 -10 -5 0 5 10 x -5

5 5x + 30y = -150 -30 -20 -10 0 -5

10 x

5

-10 -5 0 5 -5 -9x + 4y = 36

-3 -2 -1 0

-4

804

x

50 x

10 x

5

j

-15 -10 -5 0 -5

2 2

10

y 5

k

6 5 4 3 2 1

2

c

5 10

i

h

y 20

-100 -50 0 -10

y

y 5x - 3y = 10 4

-2 0 -2

g

y 20 15 y = -3x + 2 10 5 -10 -5 0 -5 -10 -15 -20

-5 0 -5

x

3

-x + 6y = 120

-6

-3

2

y 10 4x + 4y = 40 5

-10 -5 0 5 10 x -5 2x - 8y = 20

-3 -2 -1 0 1 -5

y

y

f

5

y = 7 - 4x 10

x

x

y

e

y 20 15

y 0 1

(1, 1) 01

x

y = -2x + 3

-4

-9

x

y = -5x - 4

(1, -9)

e

f

y

e

y

y = 1–2 x - 2

y = - 2–7 x + 3 3

0 -1 -2

g

2

(2, -1)

x

7

h

b

y y = 2x

y

y = 5x

5 x

d

c (-5, 0), (0, 25)

y

y = 1–2 x

1– 2

x

0 1

0

-3

x

1

y = -3x y

f

y = 2–3 x

2

3

y

x

- –25

b

-10 -5 0 -5

5 10 x

d

y 10

-5 0 -5 -10

x

y = - –25 x

y 5 -10 -5 0 -5

5 10 x

-10

y = -10

y 10 5

5 5 10 x x = 10

e y = 12 x + 3

g y = 7x - 5

h y = -3x - 15

-10 -5 0 -5 -10 x = -10

5x

1 2

d y = x +

f y = 4 x - 4

c y = 1 x 2

b y = 2x - 1 1 2

e y = -2x - 2

4 a y = 3x + 3 c y = -4x + 2 e y = -x - 4 g y = 5x + 2.5 i y = -2.5x + 1.5 5 a y = 5x - 19 c y = -4x - 1 e y = 3x - 35 g y = -2x + 30 i y = 0.5x - 19

b d f h j b d f h j

d y = − 43 x c y = − 12 x + f y = -x - 8

7 2

y = -3x + 4 y = 4x + 2 y = 0.5x - 4 y = -6x + 3 y = 3.5x + 6.5 y = -5x + 31 y = 4x - 34 y = -3x + 6 y = 2x - 4.5 y = -0.5x + 5.5

Exercise 3C — The distance between two points on a straight line 1 AB = 5, CD = 2 10 or 6.32, EF = 3 2 or 4.24,

GH = 2 5 or 4.47, IJ = 5, KL = 26 or 5.10,



MN = 4 2 or 5.66, OP = 10 or 3.16

Answers

Answers 3B ➜ 3C

5

1

1

d y = 2x - 8

3 a y = x + 3

y = 10

10

Exercise 3B — Determining linear equations 1 a y = 2x + 4 b y = -3x + 12 c y = -x + 5

2 a y = 2x b y = -3x

y 0

0

y = -12

8 a (2, 0), (0, -8) b (- 12, 0), (0, 3)

x

0 1

y

x

7 a x-intercept: -0.5; y-intercept: 0.4 b x-intercept: 0.5; y-intercept: -0.4 c x-intercept: 0; y-intercept: 0 d x-intercept: -3; y-intercept: 12 e x-intercept: -4; y-intercept: -4 f x-intercept: -1; y-intercept: -0.5 g x-intercept: 2.75; y-intercept: 2.2 h x-intercept: 7; y-intercept: 3.5 i x-intercept: 9.75; y-intercept: -3.9 23 j x-intercept: 13 ö 1.77; y-intercept: 4.6

(1, -6)

0 1

c

y

-12

2

50 x

-100 -50 0 -5

0

x

1

-6 -7

5 x

-10

x

1

y=0 5 10 x

x = -100 y 10

-10

y=x-7

0

5

i 0

6 a

x=0

x

5

y

10 x

5

y 5 -10 -5 0 -5

h

-5 0 -5

y = 8x

i

y 10 5

(5, 3.5)

0

e

g

(1, 8)

8

1.5

c

-10 -5 0 -50

y

y = 0.6x + 0.5

5 a

x

f

y = 100

50

(7, 1)

1 0

y 3.5

y 100

805

b d f h

2 a 5 c 10 e 6.71 g 13

13 7.07 14.42 13

a2 + 4b2 i j 3 a 2 + b 2 3, 4 and 5 Answers will vary. 6 a AB = 4.47, BC = 2.24, CD = 4.47, DA = 2.24 b AC = 5, BD = 5 c Rectangle 7 B 8 D 9 a 12 b 5 c 13 d -2.2 10 Answers will vary. Exercise 3D — The midpoint of a line segment

1 a (-3, -3 12 )

b

c (-1, 1)

d (0, 1 12 )

e (2a, 12b) 2 (-3, -10) 3 a (3, 1) c 6.32 4 D 5 C 6 a    i  (-1, 4)

f (a + b, 12a)

1 (7 2 ,

0)

b 4.47

  ii 

1 (1 2,

806

Answers

8

25 a m = − 5 b m = 185 26 E 27 B 28 a 5.10 km b (6.5, 5.5) d y = 2x - 18 e (10, 2) 29, 30  Answers will vary. 31 Answers will vary. Chapter review Fluency 1 A 4 C 7 A 10

2 D 5 A 8 C

-10 -8

-6

-4

-2

0

2

4

65

45

35

25

15

5

-5 -15 -25 -35

55

-10 -8 -6 -4 -2 0 -20 -40 -60 -80

11 a

c (1, 4)

8

10

( 2– , 0) 3

2 4 6 8 10 x

b

y

y y = -5x + 15 (0, 15) 10 (1, 10) 15

y = 3x - 2 (1, -1)

1

0 1 -2 (0, -2)

x (3, 0) 0 1

y

1 0 -1

d

(0, 1) y

y=x+3

6

y 80 60 40 20

c No f No

b x + 2y = 0 b 5x + 3y - 8 = 0 b y = -7 c D d B b 1 d Isosceles triangle

3 B 6 C 9 C

y

c

b y = -4x + 9 d 5y + 2x + 13 = 0 f x - 3y + 17 = 0

c 2 f 7.071 km

x

and perpendicular lines Yes c No Yes f No

Yes Yes

21 E b 3y + 2x + 1 = 0

1)

iii  3.9 iv  7.8 b Answers will vary.   ii  (1, -0.5) 7 a    i  (1, -0.5) b Answers will vary. 8 a    i  (-2, 2)   ii  8.94 iii  9.55 iv  9.55 b Isosceles. PC could be the perpendicular height of the triangle. 9 y = -3x - 2 10 3y - 2x + 14 = 0 Exercise 3E — Parallel 1 a No b d No e 2 b, f ; c , e 3 Answers will vary. 4 Answers will vary. 5 Answers will vary. 6 a Yes b d Yes e 7 y = 2x - 9 8 3x + 2y - 8 = 0 9 a y = 3x + 2 c 3x - 2y - 8 = 0 e x + 5y + 5 = 0 g x - 3y - 14 = 0 10 a 2x - y + 5 = 0 11 a 3x - 5y + 2 = 0 12 a x = 1 13 a B b C 14 a (2, 5) c Answers will vary. 15 y = -x - 3 16 4x - 6y + 23 = 0 17 a y = -x + 5 b 18 Answers will vary. 19 Answers will vary.

20 B 22 a y = -2x + 1 23 a, e; b, f ; c , h; d, g 1 24 y = − 2 x + 32

=–2–3 x

+1 ( 3– , 0)

y 4 (2 1– , 0)

2

(3, -1)

3

7

x

0

5 y = 7–5 x - 3 -3 (0, -3)

6

12 a x-intercept = 7 , y-intercept c = 6 b x-intercept = c x-intercept =

40 3 21 16

1

(133), y-intercept c = -5 3

5

(116 ), y-intercept c = − 4

d x-intercept = -5.6, y-intercept c = 2.8 y 13 a b y 2x - 3y = 6

0 -2

3

x

3

x -1 0

x y = -3x

x

d

5x + y = -3 x

0

- –53

0

-3

b

y

0

-2

Cost ($)

y 7 y=7

x

0

x

0

e

Cost ($)

b y = -x - 4

1 y = − 3 x + 2 3 y = - 4

b y = -2x - 5 b y = -3x + 4

1

d y = 35 x − 18 15

c y = 2 x + 6 61 Answers will vary. Answers will vary. (0, -18) Answers will vary. Answers will vary. x + 2y - 2 = 0 2x + 3y - 9 = 0 3x + 2y - 21 = 0 3x - 2y + 16 = 0 4

ii

5 4

•

0

2

4 6 8 10 12 Time (hours)

CHAPTER 4 ii

1 (− 2 , 1)

Simultaneous linear equations and inequations

1 1 (4 2 , 1 2 )

iii b Answers will vary. Problem solving 1 a Number of hours

•

200

c Since the gradient of the path AB is 45, which is the same as the gradient of the known path of travel from the common point A, the direction of travel is toward B. d dAB = 0.8 metres. Yes, guard ball A will collide with guard ball B as it will not be deviated from its linear path under 1 metre of travel.

iv 5x - 4y - 25 = 0

1 10

Pay ($)

b C = 22.50h + 160 c Approx $436

•

0

2

4

6

0

27

54

81

b Pay = $13.50 ì (number of hours worked)

8

10

108 135

Are you ready? 1 a -6 2 a i y = 2 b i y = -3     c i y = - 3 2

3 a y = -2x + 4

b 4

ii x = 3 ii x = 9   ii x = 2

b y = 4x - 5

c 3

c y = − 23 x −

Answers 3D ➜ 3E

iii 4x + 5y - 61 = 0 v (9, 5) b Square

10

41 b y = 45 x + 20 or 25x - 20y + 41 = 0

29 a i − 5

30 a i

4 6 8 Number of rides

4 a 7x - 3y - 1 = 0 b 3x + 7y - 49 = 0 c -7 5 a a = -7 b B(-7, 50), C(-4, 90) c 40.1 metres 6 a Since the gradient of SA = the gradient of SO = -0.8, the points S, A and O are collinear. Player Y will displace guard ball A.

d y = 6

18 a y = 7x - 13

10

300

0

f x = 5

c y = 12 x + 5

8

100

d y = 4x

17 a y = 3x - 4

19 20 21 22 23 24 25 26 27 28

2

400

16 a y = 2x - 2

6

12.50 17.50 22.50 27.50 32.50 37.50

500

3(y - 5) = 6(x + 1)

c

4

d $30 3 a x

0

2

40 30 20 10

0

y 7 (0, 7) - –27

0

b Cost = $2.50 ì number of rides + $12.50 c

d

y

1 2 3 4 Hours worked (h)

d $91.13 2 a Number of rides

x 0 1 y = -4x -4

x

1 y = 1–2 x

x = -2

15

y

1– 2)

50 40 30 20 10 0

Total cost ($)

(1,

1– 2

c

x

-3

-3

14 a

c

y x+y+3=0

Pay ($)

y

c

5 3

Answers

807

4 a 1

b 2

c -1

5 2

7 3

1 -4

5 a

b

6 a

d

y 5y - 4x = 20

0

15  − 2 



3

1

j (1, 0.3)

x

x

g (1 12, 3 12 ) c

4 a (2, 1) d (1, 3)

y

g (4, 2) 0

-3

x

j 5 a d g j 6 a

-4 3y + 4x = -12

7 a False

b False

c True

Exercise 4A — Graphical solution of simultaneous linear equations 1 a (2, 1) b (1, 1) c (0, 4) d (2, -1) e (-2, -4) f (-0.5, 1.5) 2 a No b Yes c Yes d No e Yes f No g No h Yes i No j Yes 3 a (3, 2) b (4, 3) c (-3, 4) d (-2, 2) e (2, 0) f (3, 0) g (-2, 4) h (3, 8)

j (2, 5)

4 a (3, 5) b (-2, 4) e (5, 1) f (6, -2) 5 a No solution c No solution e (3, 1) g No solution 6 y = 4x - 16 7 a Northern beach C = 20 + 12t Southern beach D = 8 + 18t b Northern beaches in red, southern beaches in blue c Time > 2 hours d Time = 2 hours, cost = $44

k (5, 3) c g b d f h

l

3

(5, 7) d (-2, -5) (-4, 7) h (3, 4) (2, -1) (1, 9) No solution (2, 1)

D C

80 60 40

C = 20 + 12t D = 8 + 18t

20 0

1

2 3 4 Time (hours)

Exercise 4B — Solving simultaneous linear equations using substitution 1 a (2, 3) b (2, -1) c (3, -2) d (7, 6) e (3, 6) f (2, 1) g (-1, -2) h (-4, 0) i (-1, -2)

j (6, -2) 808

Answers

k

e (1, -7)

f (- 2 , -4)

1

h

 1 4 ,  − 5 5 

i

(-3, -1.5)

k

 4 4  − ,  5 5

l

(1, -1)

1 (3, 1 2 )

l

(-3, -5)

5

4

h (2, 15 )

i

b (3, 5) e (2, 4)

c (3, 3) f (5, 2)

h (-3, 4)

i

(-3, -1 2 )

k b e h k b

l c f i l c

(2, 1.8) (-2, 6) (3, 1) (1, 3) (-3, 5) (-3, 5)

f

1  , 3

(-3, 5) (3, 3) (7, 0) (2, -2) (-8, 18) (4, 0)

e (8, 5)

(1, 1)

1

1

− 3 

7 Ann 61 kg, Beth 58 kg, Celine 54 kg

(2, 2 )

100

(-6, -5) (5, 2) (5, -1) (6, 3) (-1.5, -3) (1, 3)

d (4, 3)

120

Cost

i (- 12 , 112)

c (2, -6)

Exercise 4C — Solving simultaneous linear equations using elimination 1 a (3, 1) b (-2, 3) c (-2, 6) 2 a (5, -1) b (2, 3) c (-3, 1) 3 a (6, 3) b (-3, -7) c (2, -5) d (-3, 5) e (-5, -8) f (2, -2)

1 1–4 0

b (5, 23)

3 26 chickens

y 4y - 2x = 5 -2 1–2

3 ,  2

g  − 2, − 2 

4

-5

b

c

2 a (-6, -23)

Exercise 4D — Problem solving using simultaneous linear equations 1 Maths mark = 97, English mark = 66 2 18 nuts, 12 bolts 3 8 and 3 4 9 and 7 5 6 and 5 6 Length = 12 m and width = 8 m 7 Lemons cost 55 cents and oranges cost 25 cents. 8 Length 60 m and width 20 m 9 Eight 20-cent coins and three 50-cent coins 10 Twelve $1 coins and nine $2 coins 11 Paddlepops cost $1.20 and a Magnum costs $2.10. 12 Cost of the Golden rough = 35 cents and cost of the Redskin = 25 cents 13 Fixed costs = $87, cost per person = $23.50 14 PE mark is 83 and Science mark is 71. 15 Mozzarella costs $6.20, Swiss cheese costs $5.80. 16 x = 3 and y = 4 17 Fixed costs = $60, cost per person = $25 18 $4 each for DVDs and $24 each for zip disks Exercise 4E — Solving linear inequations 1 a x > 2 b a > -1 c y í 7 d m í 4 e p < 1 f x < 7 g m Ç 9 h a Ç 7 i x > 3 j m í 2 k q í -4 l a > -8 2 a m > 3 b p Ç 2 c a < 4 d x í 5 e p > -5 f x Ç -7 g m í -0.5 h b > -0.5 i m > 18 j x < 8 k a Ç -14 l m í 25 3 a m < 4.5 b x í 3 c p > 4 d n Ç 2 e b < 5 f y > 2

g m Ç -1

h a í -5

i

b < -4 1

j c Ç -1

k p > -2

l

a í -7

3

4 a d g j

b e h k

m > 3 a Ç -3 b < 4 m Ç 3

c f i l

a í 2 x > 6 a > 5 b Ç -16

11

g

a < -1 x Ç 2 m < 2 m í 1

b x í -18 e x í 5

c x < -10 f x < -1 4

6 a m < -2 d p í -5 g p < 0

b p í -3 e y Ç -3 h a í 1

c a Ç 5 f x > 7 i x > -3

k b Ç 3 n x > -18 1

l x < -3 o a Ç 40

7 B 8 a x < -1 d a >  5

b m Ç -3 e m í 1 1

c x > 17 f m í -12

9 a k > 2

b a > -5

c m Ç 1 2

d x > 5

e y í 7

f d < -2

g p í −6 7

h x í -5

i

m < -2

j a < 9

k p í 3

l

x > -4 1

5

j a Ç -11 m k > 8

2

8

3

i

0

l

y

y

3 a B 4 a y = 12 x + 3

(0, 0) 0

b D

c A

30 15

y

12

s

30

Exercise 4G — Solving simultaneous linear inequations 1 a True b False c False d True e True f False g True h False i False j False 2 Note: the shaded region is the region required. y a x + y > 3 6

2

f

y

y

-6

(0, 4)

(0, -2)

x

0

-2

0

2

4

6

x

-2 -4

(2, 0) 0

-4

x

b x + 2y Ç 6

y 3 2 1 -4 -2-10 -2

x 2

4

6

Answers

Answers 4A ➜ 4G

x

0

d 15 large and 15 small dogs 6 a y > x + 2 b Answers will vary.

4

e

x

b y í 12 x + 3, x > 2, y Ç 7 5 a l + s Ç 30 b At least 12 small dogs l c

x

0 (3, 0)

(6, 0) x

0

(0, 3) x

x

(1, 3)

(-2, 0) 0 (0, -2)

(-7, 0) 0

x

k

0

d

y

(0, -2)

(6, 0)

y

j

y

(0, 7)

(0, -6)

c

x

(0, -7)

2

x

0

x

5

3

(0, 1) 0

(7, 0) 0

10 a 5x > 10 b x - 3 Ç 5 c 7 + 3x < 42 11 a –6.5 < x < –2 −d − b −c − b b < x < a a 12 a S > 47 b No c Answers will vary. 13 a CA = 700 + 20x b CB = 1200 + 15x c 700 + 20x < 1200 + 15x, x < 100 d x > 100 e x = 100 Exercise 4F — Sketching linear inequations y y 1 a b

(-1, 0 )

y

(-5, 0)

5 a x Ç 7 d x > 10 1

2

h

y

809

c 3x - 2y > 12 

4 Note: The shaded region is the region required.

y 2 1 -4 -2 -10 -2 -3 -4 -5 -6 -7

d 4x + y í -8 

a 2

2 0

–1

3

2

1

–2

x

5

4

x+y<3

–4 –6

2 1 -2 -10 -2 -3 -4 -5 -6 -7 -8 -9

e y í x + 4 

2x - y í 4

4

y

-4

y

6

x

4

x

2

b

y

6

3x + 2y > 12

4 2

x

0

-2

-2

y

-4

6

-6

6

4

2

10

8

x + 5y Ç 10

4

c

2 -6

-2

-4

0

2

x

4

2y > x - 2

2

-2

1

-4 -1

f y < 3 - 3x 

y

3

y

0

3

2

1

-1 -2

3

x

5

4

y<3-x

-3

2

d

1 -1

0

1

2

x

3

-1 -3

3 a A b C c B d E 810

Answers

-2

0

x

3

2

1

-2

y < 4 - 2x

-6

e

y

y - 2x Ç 5

6 4

x+y>4

2

2 x

1

-6

-2

-4

y

0

-1

-4

0

x

6

4

2

-2 -4

10 9 8 7 6 5 4 3 2 1 -1

y > 2x + 4

2

y 10 9 8 7 6 5 4 3 2 1 -3 -2 -1 0

h 2x + y í 8

y

4

-2

g y - 3x < 9 

6

-6

f

1

2

3

4

x

y 20 18 16 14 12 10 8 6 4 2 0

-2 -4

3x + y > 17

y<8 x 1

2

3

4

5

5 2–3

6

g

m

y

15

y 10 8 6 4 2

10 5

x + 2y í 10

0

5

-2 -20 -4 -6 -8 -10

x

10

-5 -10

3x + y > 15

-15

h

y > 2x - 3

0

3

2

1

x

5

4

x<5

i 18 16 14 12 10 8 6 4 2

y

-4

k

3y - 2x < 6

0

-2 -1 -2 -3 -4 -5 -6

6 5 4 3 2 1

y

b 6

4

2

x

x yÇx+2

4

y í 4 - 2x

0

2

x



y yÇx+2

4 2

x

2x + 3y Ç 6

-2 0

2

x y > 4 - 2x

6 Note: The shaded region is the region required. y

y < 2x

y + 2x > 3

4

x - 2y > 0 x

1 2

7 a r + x Ç 2000 b r Ç 600 c r í 0, x í 0. Amount of money cannot be negative. x d e Answers will vary.

y - 2x í 9

2

-1 0

x

3

2

-2 -20 -4 -6 -8 –10

4

y

x

y

-4

3

y

1 11– 2

10 8 6 4 2

2

2000

6

Note: The shaded region is the region required.

x

x+yÇ4 0

600

Answers 4G ➜ 4G

-1 -1 -2 -3 -4 -5 -6

l

ii 

2 2x + y < 0 1

0

-2

1

2

-6 -3-20 2 4 6 8 10 12 14 16 -4 -6

-6

y>4

–2 0

6 5 4 3 2 1

x

8

5 Note: The shaded region is the region required. a   i  y

y í 2x - 2

y-x>4

6

y

6 5 4 3 2 1

-1 -10 -2 -3 -4 -5 -6

-2

-1 -2 -3 -4 -5

4

2x - 3y í 18

y í 2x

7 6 5 4 3 2 1

-6

2

n

y

j

x+y>7

2000 r

Answers

811

10

8 a 100a + 75b í 450 b 50a + 75b í 300 b c

d Answers will vary.

6

2

0

1

2

3

4

5

6

a

Mass of chocolate chips in grams (c)

700

Note: The shaded region is the region required.

600 500

-4

-2

0

2

4

y

65

45

35

25

15

5

-5 -15 -25 -35

 

-10 -8 -6 -4 -2 0 -20 -40 -60 -80

11 a

6

  b

y

300

y = -5x + 15 (1, 10)

10

0 1 –2

100 p + c Ç 400

x 0 1

0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

c

 d

y

1 0 -1

600 500

(3, -1)

y

x

3

x

4

y =–2–3 x + 1

700

0

400 200

5 y=

-3

7– x 5

-3

0.3p + 0.6c í 180

100 0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

700

6

12 a x-intercept = 7 , y-intercept c = 6 b x-intercept =

40 3 21 16

1

(133), y-intercept c = -5

5 (116 ),

3 c x-intercept = y-intercept c = - 4 d x-intercept = -5.6, y-intercept c = 2.8

13 a

b

y

y

2x - 3y = 6

600 500

0 -2

400

3

x

3

-1 0

300

x y = -3x

200 100 0 100 200 300 400 500 600700 Mass of peanuts in grams (p)

Chapter review Fluency 1 A 2 D 3 B 4 A 5 C 6 E 7 C 8 D 9 A

Answers

10

y 15

(1, 1)

1

200

300

8

2 4 6 8 10

400

c

d

y

x

0

- –53

y x+y+3=0

5x + y = -3

f Answers will vary.

812

55

y = 3x - 2

c 0.3p + 0.6c í 180 d Mass of chocolate chips in grams (c)

-6

80 60 40 20

9 a p + c Ç 400 b

Mass of chocolate chips in grams (c)

-10 -8

Note: The shaded region is the region required.

4

e

x

-3

-3

-3

14 a

0

b

y

1– 2

(1, 1–2 ) 0

1 y = 1–2 x

x

y

x 0 1 y = -4x -4

x

x

c

y

x = -2

24 a (2, 7)

7 y=7 x

0

-2

15

y

d

d x

0

25 a (5, 2) d (1, 3) 26 a (0, 3) 27 a

y 7 (0, 7) - –27

 7 7  − ,  3 3

16 a y = 2x - 2 3

c (-3, -1) f (4, 2) c (2, 1) Note: The shaded region is the region required.

1

2y - 3x í 12

18

3

b y = - 32 x + 12

-6

20 Note: The shaded region is the region required. y a b y

c

-5 0

d

y

c

5x + y < 10

10

y í 2x + 10

x

-1 0

y 5

y = 5x

0 1

x y < 5x

x

x

-12 y

f

x=7

y 1

7

xí7

x

-2

h

y 9 2x + y í 9

21 a (3, 1) 22 a No 23 a (-2, 1) b (0, -2) c (5, 2)

0

y

x + 2y < 11

0

2

4

6

0

27

54

81

8

10

108 135

50 40 30 20 10

1 2 3 4 Hours worked (h)

d $91.13 2 a Number of rides Cost ($)

y > -12

x

b Pay = $13.50 ì (number of hours worked) c

0 • 12 x

–16• 4x - 3y í 48

0 -12

x

x

0

2

4

6

8

10

12.50 17.50 22.50 27.50 32.50 37.50

b Cost = $2.50 ì number of rides + $12.50 c

x

b (2, 3) b Yes

40 30 (12.50) 20 10 0

2

(10, 37.50)

4 6 8 10 Number of rides

Answers 4G ➜ 4G

i

0

y 5

x

–9 2

y Ç –21 x + 1

6

2 4 6 8 10 12 14 16

Pay ($)

Pay ($)

0

4

y

Problem solving 1 a Number of hours

Cost ($)

4

2

y + 3x > 0

10 9 8 7 6 5 4 3 2 1

0 -8 -6 -4 -2 -1 -2

y > 3x - 12 0

x

6

7 6 5 4 3 2 1

-2 -10 -2 -3 -4

-4

27 5

1 yÇx+1

4

y

d y = 5 x - 5

19 a y = -x + 8 +

b

b y = -3x + 4

c y = 2 x + 6

2

-2

-6

d y = 6

18 a y = 7x - 13

0

0

-2

-4

b y = -2x - 5

1 y = 2 x + 5

2 y = 5 x

-4

yÇx+4

f x = 5

17 a y = 3x - 4

g

5

4

d y = 4x

e y = - 4

e

f ( 2, -7)

y

6

-6

b y = -x - 4

c y = − 13 x + 2

c

e (-14, -53)

2

3(y - 5) = 6(x + 1)

c

c (-2, 2)

b (-2, 3) e (2, -2) b (-3, -3) yí3

x

0

b (-5, -3)

d $30 Answers

813

5 a

500

m

25è

•

A

100 0

0

2

4

6 8 10 12 Time

b Cost = 22.5 ì time + 160 c $435.63 4 a Numbers are 9 and 14. b Length = 11 metres, width = 6 metres c Chupa-chups cost 45 cents and Whizz fizzes cost 55 cents. 5 Milk $1.75, bread $2.35 6 13 kangaroos and 8 cockatoos 7 Rollercoaster ride $6, Ferris wheel ride $4, Gravitron ride $8 8 a d = b + 10 b 7000 = 70b + 40d c b = 60 and d = 70 d Number of seats in ‘Bleachers’ is 4200; the number of seats in the ‘Dress circle’ is 2800. e $644  000 9 a CG = 114 + 0.20k b CS = 90 + 0.32k c 200 km d 114 + 0.20k < 90 + 0.32k \ k > 200 e k < 200 10 a 5400 + 260d = CH b 61 days 11 a n < 16 800 km b Mick travelled less than 16  800 km for the year and his costs stayed below $16  000. CHAPTER 5

Trigonometry I Are you ready? 1 a 0.685 c 0.749 2 a i  15è33Å b i  63è16Å c i  27è10Å 3 a H O

A c

b 1.400 ii  15è32Å41ë ii  63è15Å32ë ii  27è10Å16ë b A q q

O H A

q

4 a x = 30 ì tan (15è) b x =

4.2 tan(28°)

c x = 5.3 ì tan (64è) 814

Answers

H O

S

B 120è

5k

Cost

N

300 •

20è

km

400

200

b

N

•

180

3 a

7.

5

70 km

N

km 60è

C

Exercise 5A — Pythagoras’ theorem 1 a 7.86 b 33.27 c 980.95 d 12.68 e 2.85 f 175.14 2 a 36.36 b 1.62 c 15.37 d 0.61 e 2133.19 f 453.90 3 23.04  cm 4 12.65  cm 5 a 14.14  cm b 24.04  cm c 4.53  cm 6 a 97.47  cm b 334.94  cm c 6822.90  cm2 7 a 6.06 b 4.24 c 4.74 8 18.03  cm 9 17.32  cm 10 19.23  cm 11 65.82  cm; 2501.16  cm2 12 39  m 13 Yes 14 4.34  km 15 38.2  m 16 20.61  m 17 130  mm 18 a 386.13  mm b 62.09  cm c 2.33  km d 16.15  cm e 541.70  cm f 2615.61  m g 478.97  mm h 369.87  km 19 54.67  mm 20 a 28  cm b 588  cm2 21 36.37  cm 22 552.86  cm2 23 21.46 diagonals, so would need to complete 22 24 1600  mm 25 5889.82  m 26 7.07  cm 27 $81.60 28 185  cm 29 Students own working. Exercise 5B — Pythagoras’ theorem in three dimensions 1 a 17.32 b 12.25 c 15.12 2 12.21, 12.85 3 4.84  m, 1.77  m 4 11.31, 5.66 5 31.62  cm 6 6  cm 7 12.65  cm 8 23  mm 9 No: maximum stick can be only 115  cm long. 10 3.41  cm 11 a i  283.02  m ii 240.21  m iii 150.33  m b 141.86  m

12 14.72  cm 13 13.38  cm 14 42.27  cm 15 1.3  m, 5.98  m2 16 Students’ own working Exercise 5C — Trigonometric ratios 1 a 0.5000 b 0.7071 c 0.4663 d 0.8387 e 8.1443 f 0.7193 2 a 0.6944 b 0.5885 c 0.5220 d -1.5013 e 0.9990 f 0.6709 g 0.8120 h 0.5253 i -0.8031



25 30

c tan (q ) = 45

b cos (q ) =

2.7 17 e sin (35è) = p t 7 20 g sin (15è) = h tan (q ) = 31 x 9 a a

d tan (q ) =

H

41è A

Exercise 5F — Angles of elevation and depression 1 8.74  m 2 687.7  m 3 a 176.42  m b 152.42  m 4 65è46Å 5 16.04  m 6 a h = x tan (47è12Å)  m; h = (x + 38) tan (32è50Å)  m b x = 76.69  m c 84.62  m 7 a h = x tan (43è35Å)  m; h = (x + 75) tan (32è18Å)  m b 148.40  m c 141.24  m 8 0.033  km or 33  m 9 21è 10 a 8.43  m b 56.54  m

11 44.88  m 12 a

.3 f sin (a ) = 14 17.5

O

b O = 34  mm, A = 39  mm, H = 51  mm c   i sin (41è) = 0.67 ii cos (41è) = 0.76 iii tan (41è) = 0.87 d a = 49è e   i sin (49è) = 0.76 ii cos (49è) = 0.67 iii tan (49è) = 1.15 f They are equal. g They are equal. h The sin of an angle is equal to the cos of its complement angle.

b 15.27  m 13 66  m 14 a 54è 15 a 2.16 m/s, 7.77 km/h

b 0.75  m b 54.5è

Exercise 5G — Bearings and compass directions 1 a 020èT b 340èT c 215èT d 152èT e 034èT f 222èT 2 a N49èE b S48èE c S87èW d N30èW e N86èE f S54èW 3 a 3  km 325èT b 2.5  km 112èT c 8  km 235èT d 4  km 090èT, then 2.5  km 035èT e 12  km 115èT, then 7  km 050èT f 300  m 310èT, then 500  m 220èT

4 a

b

N

135è

km

40è

N

m

0k

14

c

1.3

km

N

d

30è

50è

7k

32è

m 8k

N N 260è 120è 0.8 km N

240è

N 40è

m

Answers 5A ➜ 5G

100è 30 km

N

N

0

6 1.05  m 7 a x = 30.91 cm, y = 29.86 cm, z = 39.30 cm b 2941.54 cm2

15 m

23

Exercise 5D — Using trigonometry to calculate side lengths 1 a 8.660 b 7.250 c 8.412 2 a 0.79 b 4.72 c 101.38 3 a 33.45  m b 74.89  m c 44.82  m d 7.76  mm e 80.82  km f 9.04  cm 4 a x = 31.58  cm b y = 17.67  m c z = 14.87  m d p = 67.00  m e p = 21.38  km, q = 42.29  km f a = 0.70  km, b = 0.21  km 5 a 6.0  m b 6.7  m

1.76 m

3.1

i cos (a ) = 9.8

x

42è

m



5k

8 a sin (q ) = 12 15



km

e f i b i  sin (a) = g l c i  sin (b ) = k n d i  sin (g  ) = m b e i  sin (b ) = c v f i  sin (g  ) = u

7 a i  sin (q) =

Exercise 5E — Using trigonometry to calculate angle size 1 a 67è b 47è c 69è 2 a 54è47Å b 33è45Å c 33è33Å 3 a 75è31Å21ë b 36è52Å12ë c 37è38Å51ë 4 a 41è b 30è c 49è d 65è e 48è f 37è 5 a a = 25è47Å, b = 64è13Å b d = 25è23Å, e = 64è37Å c x = 66è12Å, y = 23è48Å 6 a r = 57.58, l = 34.87, h = 28.56 b 714 cm2 c  29.7è 7 a   i  29.0è ii  41.4è iii 51.3è b   i  124.42 km/h ii 136.57 km/h iii  146.27 km/h

km

-0.9613 0.1320 53è 41è 0è52Å 26è45Å 64è46Å59ë 88è41Å27ë 20.361 1.192 4909.913 14.814 e iii tan (q) = d i iii tan (a) = h l iii tan (b ) = j n iii tan (g  ) = o b iii tan (b ) = a v iii tan (g  ) = t

l o c f c f c f c f i l

2.1

0.9880 -0.5736 24è 86è 6è19Å 44è48Å 64è1Å25ë 36è52Å12ë 71.014 226.735 32.259 0.904 d ii cos (q) = f h ii cos (a) = g j ii cos (b ) = k o ii cos (g  ) = m a ii cos (b ) = c t ii cos (g  ) = u

k n b e b e b e b e h k

40

j 0.4063 m 1.7321 3 a 50è d 71è 4 a 54è29Å d 72è47Å 5 a 26è33Å54ë d 48è5Å22ë 6 a 2.824 d 2.828 g 7.232 j 0.063

S

Answers

815

N m 0k 22

70è180 km

30è

N

km

20è

320

S

5 a i  13.38  km ii 14.86  km b i  N B

N

130è 80

20 km

42è A

iii 222èT ii 51.42  km iii 61.28  km iv 310èT

ii 38.97  km iii 22.5  km iv 030èT

N B 130è

N

A

20

km

42è

80

km

N

210è

45

km

C

D

6 215èT 7 1.732  km 8 a 9.135  km b 2.305  km c 104è10Å T 9 684.86  km 10 a 60è43Å T b 69è27Å T c 204è27Å T 11 a q = 60è, a = 40è b 2.5 km c 2.198 km d 1.22 km Exercise 5H — Applications 1 a 36è52Å b 53è8Å c 2.4  m 2 a 14è29Å b 31  cm 3 6.09  m 4 19è28Å 5 62è33Å 6 a 11è32Å b 4è25Å 7 a   i  35.36  cm ii 51.48  cm iii 51.48  cm iv  57.23  cm v 29è3Å vi 25è54Å b   i  25.74  cm ii 12.5  cm iii 25è54Å iv  28.61  cm 8 a 77è b 71è56Å c 27.35  cm 9 a 7.05  cm b 60è15Å c 8.12  cm 10 a 28.74  cm b 40.64  cm c 66è37Å 11 a 26.88  cm b 11.07  cm d tan θ 12 a 90 m ì tan q2 b h = tan θ1 + tan θ 2 c 250 m Chapter review Fluency 1 E 2 D 3 E 7 E 8 B 9 B 11 a x = 113.06 cm 12 9.48  cm 14 17.6  m

816

Answers

Chapter 6

Surface area and volume

km

C

c i 

16 67.98  km 17 4.16  km 18 40è32Å Problem solving 1 a h = tan (47è48Å)x  m h = tan (36è24Å) (x + 64)  m b 129.10  m c 144.32  m 2 a 11.04  cm b 15.6  cm c 59è2Å 3 a 27.42 km b N43èW or 227èT 4 a 1280.6 m b 12:02:16.3 pm 5 33.29 m, 21.27 m

4 E 5 D 6 B 10 A b x = 83.46 mm 13 8.25  mm 15 26.86  m

Are you ready? 1 a 3.6 ì 106  mm2 b 2 ì 10-6  km2 4 2 c 5.2 ì 10 m 2 a 24  m2 b 30  cm2 c 4.9  cm2 2 2 3 a 150  cm b 232  cm c 1.22  m2 4 a 3.4 ì 106  cm3 b 2.5 ì 10-4  m3 c 6.5 ì 103  mm3 5 a 125  cm3 b 160  cm3 c 0.03  m3 Exercise 6A — Area 1 a 16  cm2 b 48  cm2 c 75  cm2 2 2 d 120  cm e 706.86  cm f 73.5  mm2 g 254.47  cm2 h 21  m2 i 75  cm2 2 Part e = 225p  cm2; part g = 81p  cm2 3 a 20.7  cm2 b 7.64  cm2 4 a 113.1  mm2 b 188.5  mm2 5 a i 12p  cm2 ii 37.70  cm2 b i 69π   mm2 ii 108.38  mm2 2 c i 261p  cm2 ii 819.96  cm2 6 E 7 D 8 a 123.29  cm2 b 1427.88  m c 52  cm2 d 30.4  m2 e 78  cm2 f 2015.5  cm2 9 a 125.66  cm2 b 102.87  cm2 c 13.73  m2 d 153.59  m2 e 13.86  m2 f 37.5  m2 10 11  707.93  cm2 11 21  m2 12 60 13 $840 14 a 260.87  m2 b 195.71  m2 c 130.43  m2 d 97.85  m2 e 37.5% f 18.75  m 15 a 50 = x + y b y = 50 – x c Area = 50x − x2 d x

0

Area (m2)

0 225 400 525 600 625 600 525 400 225 0

5

10

15

20

25

30

e No, impossible to  make a rectangle. f 600 Area

e

500 400 300 200 100 0 10 20 30 40 50 60 70 x

35

40

45 50

g x = 25 h y = 25 i Square j 625  m2 k r = 15.915  m l 795.77  m2 2 m 170.77  m 16 a Students’ work b 2020.83  m; horizontal 17 a Circular area, 1790.49 m2; rectangular area, 1406.25 m2  1 2 n m2; rectangular (square) b Circular area,  4π   1 2 4 area,  n  m2. Circular area is always or 1.27  16  π 1  1  4π ÷ 16  times larger. Exercise 6B — Total surface area 1 a 600  cm2 b 384  cm2 c 2 d 27  m 2 a 113.1  m2 b 6729.3  cm2 c 2 d 452.4  cm 3 a 1495.4  cm2 b 502.7  cm2 4 a 506.0  cm2 b 9.4  m2 c d 224.1  cm2 5 a 13.5  m2 b 90  m2 c d 9852.0  mm2 e 125.6  cm2 f 6 a 880  cm2 b 3072.8  cm2 c d 70.4  cm2 e 193.5  cm2 f 7 B 8 63 9 11  216  cm2 10 a 70.0  m2 b $455 11 a 3063.1  cm2 b $168.47

1440  cm2 8.2  m2 340.4  cm2 11  309.7  cm2 1531.4  cm2 75  cm2 1547.2  cm2

12 a q = 120 è

b x = 1; y = 3 c 3 3  cm2

d 6 3  cm2

e 32

13 a

8p  m2

7   m b 2

c 2 7π   m2 d 7 × 7 × 1 14 Calculation is correct. 15 a 6.6 m2 c Cheapest: 30 cm by 30 cm, $269.50; 20 cm by 20 cm (individually) $270; 20 cm by 20 cm (boxed) $276.50 16 r =

3 3a 2

Chapter review Fluency 1 D 4 A 5 a 84 cm2 d 56.52 cm2 6 a 300 cm2 7 a 499.86 cm2 8 a 18  692.48 cm2 b 1495.40 cm2 d 642 cm2 9 a 343 cm3 d 1.45 m3 g 297 cm3

2 C

3 E

b 100 cm2 e 60 cm2 b 224.52 cm2 b 44.59 cm2



c 6.5 cm2 f 244.35 cm2 c 160 cm2 c 128.76 cm2

c 804.25 cm2 e 873.36 mm2 b 672 cm3 e 1800 cm3 h 8400 cm3



f 760 cm2 c 153  938.04 cm3 f 1256.64 cm3 i 7238.23 mm3

Problem solving 1 a 62  m2 b $7290 3 2 2 V = π r h, the volume will be 1.5 times as large as the 2 original volume. 3 V = 3lwh, the volume will be 3 times as large as (or triple) the original volume. 4 a 3605.55  cm2 b $180.33 c 18062.1  cm3 d 9155.65  cm3 5 a 1.33  m b 910.81  m2 c $655.85 d 303.48  m3 e 11 trucks f 12 minutes

Chapter 7

Quadratic expressions Are you ready? 1 a 12x + 20 b 10x2 - 15x c -12x + 8x2 2 2 2 a x - 4 b 4x - 12x + 9 c 6x2 - 11x - 10 3 a 4x(x + 2) b -3x(5x + 3) c x(6x - 1) 4 a (x + 2)(3x + 4) b (x - 1)(4x - 1) c -(x + 3)(2x + 1)

Answers

Answers 5H ➜ 6C

Exercise 6C — Volume 1 a 27  cm3 b 74.088  m3 c 3600  cm3 d 94.5  cm3 2 a 450  mm3 b 360  cm2 3 a 6333.5  cm3 b 19.1  m3 c 280  cm3 d 288  mm3 4 a 7.2  m3 b 14 137.2  cm3 c 1436.8  mm3 d 523 598.8  cm3 5 a 377.0  cm3 b 2303.8  mm3 6 a 400  cm3 b 10  080  cm3 c 576  cm3 7 a 1400  cm3 b 10  379.20  cm3 c 41.31  cm3 d 48.17  cm3 e 218.08  cm3 f 3691.37  cm3 8 a Vnew = 27l3, the volume will be 27 times as large as the original volume. b Vnew = 18l2, the volume will be 18 of the original volume.

c Vnew = 2p r2h, the volume will be twice as large as the original volume. d Vnew = p r2h, the volume will remain the same. e Vnew = 3lwh, the volume will be 3 times as large as the original value. 9 E 10 7438.34  cm3 11 4417.9 L 12 10 215.05  cm3 13 a H = 6  m b 112  m3 c 19 bins d 112  000 L e 1.95  m from floor 14 a i 4.57  cm ii 262.5  cm2 b i 14.15  cm ii 323.27  cm2 c i 33.3  cm ii 434.28  cm2 d Sphere. Costs less for a smaller surface area. 15 Required volume = 1570.79 cm3; tin volume =1500 cm3; muffin tray volume = 2814.72 cm3. Marion should use the tin with approximately 70 cm3 mixture left over. 16 Increase radius of hemispherical section to 1.92 m. 17 Cut squares of side length, s = 0.3 m or 0.368 m from the corners. 18 Volume of water needed; 30.9 m3.

817

1 ( x − 3)( x + 7)

5 a x + 3

b

6 a 2 6

b 6 3

c

x+2 2( x + 3)

c 36 3

Exercise 7A — Expanding algebraic expressions 1 a 2x + 6 b 4x - 20 c 21 - 3x d -x - 3 e x2 + 2x f 2x2 - 8x g 15x2 - 6x h 10x - 15x2 i 8x2 + 2x j 4x3 - 6x2 k 6x3 - 3x2 l 15x3 + 20x2 2 2 2 a x - x - 12 b x - 2x - 3 c x2 - 5x - 14 d x2 - 6x + 5 e -x2 - x + 6 f x2 - 6x + 8 g 2x2 - 17x + 21 h 3x2 - x - 2 2 i 6x - 17x + 5 j 21 - 17x + 2x2 2 k 15 + 14x - 8x l 110 + 47x - 21x2 2 3 a 2x - 4x - 6 b 8x2 - 28x - 16 c -2x2 + 12x + 14 d 2x3 - 2x e 3x3 - 75x f 6x3 - 54x 3 2 g 2x - 12x + 18x h 5x3 - 30x2 + 40x i -6x3 - 6x2 + 120x 4 a x3 + 2x2 - x - 2 b x3 - 2x2 - 5x + 6 c x3 - 5x2 - x + 5 d x3 - 6x2 + 11x - 6 3 2 e 2x - 7x - 5x + 4 f 6x3 - 7x2 + 1 2 5 a x - x - 2 b -2x2 + 4x + 10 c 5x2 - 6x - 5 d 19x - 23 e -5x - 1 f -2x + 6

g x2 - 2x - 3 + 3x h 6 + 2 2 x − 3 3 x − 6 x 2 − 5 x 6 a A b C 7 B 8 a x2 - 2x + 1 b x2 + 4x + 4 c x2 + 10x + 25 d 16 + 8x + x2 e 49 - 14x + x2 f 144 - 24x + x2 2 g 9x - 6x + 1 h 144x2 - 72x + 9 2 i 25x + 20x + 4 j 4 - 12x + 9x2 k 25 - 40x + 16x2 l 1 - 10x + 25x2 9 a 2x2 - 12x + 18 b 4x2 - 56x + 196 2 c 3x + 6x + 3 d -4x2 - 12x - 9 e -49x2 + 14x - 1 f 8x2 - 24x + 18 g -12 + 108x - 243x2 h -45 + 330x - 605x2 2 i -16x - 16x - 4 10 a x2 - 49 b x2 - 81 c x2 - 25 2 2 d x - 1 e 4x - 9 f 9x2 - 1 g 49 - x2 h 64 - x2 i 9 - 4x2 11 a (x + 1)(x - 3) b x2 - 2x - 3 c 6  cm, 2  cm, 12  cm2 b 12 a (x + 1) m xm

(x + 2) m

c (x + 1)(x + 2) d x2 + 3x + 2 e 4  m2, 12  m2 13 a (x + 2)2 b 5(x + 2)2 c 5x2 + 20x + 20 3 2 d 500  cm e 100  cm , 100 tiles x 14 a x2 + x b 5x2 + 21x + 20 c i ii 1.50 m 2 818

Answers

15 a = 4, b = 4, c = -24, d = 0, e = 3 16 a (x + 2)(x – 1)(x – 3) b 56  cm3 c 0 d No; you can’t have a negative volume. e x > 3 f 18  cm3 g x = 6 h x3 - 2x2 – 5x + 6 Exercise 7B — Factorising expressions with three terms 1 a (x + 2)(x + 1) b (x + 3)(x + 1) c (x + 8)(x + 2) d (x + 4)2 e (x - 3)(x + 1) f (x - 4)(x + 1) g (x - 12)(x + 1) h (x - 6)(x + 2) i (x + 4)(x - 1) j (x + 5)(x - 1) k (x + 7)(x - 1) l (x + 5)(x - 2) m (x - 3)(x - 1) n (x - 4)(x - 5) o (x + 14)(x - 5) 2 a -2(x + 9)(x + 1) b -3(x + 2)(x + 1) c -(x + 2)(x + 1) d -(x + 10)(x + 1) e -(x + 2)(x + 5) f -(x + 12)(x + 1) g -(x + 3)(x + 4) h -(x + 2)(x + 6) i 2(x + 2)(x + 5) j 3(x + 1)(x + 10) k 5(x + 20)(x + 1) l 5(x + 4)(x + 5) 3 a (a - 7)(a + 1) b (t - 4)(t - 2) c (b + 4)(b + 1) d (m + 5)(m - 3) e (p - 16)(p + 3) f (c + 16)(c - 3) g (k + 19)(k + 3) h (s - 19)(s + 3) i (g + 8)(g - 9) j (v - 25)(v - 3) k (x + 16)(x - 2) l (x - 15)(x - 4) 4 a C b B

5 C 6   i d ii b iii b iv a v c vi d 7 a (2x + 1)(x + 2) b (2x - 1)(x - 1) c (4x + 3)(x - 5) d (2x - 1)(2x + 3) e (x - 7)(2x + 5) f (3x + 1)(x + 3) g (3x - 7)(2x - 1) h (4x - 7)(3x + 2) i (5x + 3)(2x - 3) j (4x - 1)(5x + 2) k (3x + 2)(4x - 1) l (3x - 1)(5x + 2) 8 a 2(x - 1)(2x + 3) b 3(3x + 1)(x - 7) c 12(2x + 1)(3x - 1) d -3(3x + 1)(2x - 1) e -30(2x + 1)(x - 3) f 3a(4x - 7)(2x + 5) g -2(4x - 3)(x - 2) h -(2x - 7)(5x + 2) i -(8x - 1)(3x - 4) j -2(3x - y)(2x + y) k -5(2x - 7y)(3x + 2y) l -12(5x + 3y)(10x + 7y) 9 a w2 + 5w - 6 b (w + 6)(w - 1) c (x + 5)(x - 2) 10 a x(x + 5) b x(x + 5) c (x - 1)2 d (x + 9)(x + 5) e (x - 15)(x - 6) f (x - 10)(x - 3) 11 (x - 0.5)(x + 1.5) 12 a (x - 5)(x + 1) b (x - 5) cm c x = 15 cm d 160  cm2 e 3000(x - 5)(x + 1)  cm2 or (3000x2 - 12 000x - 15 000)  cm2 13 a (2x + 3)(3x + 1) b P = 10x + 8 c x = 8 metres

14 a c e 15 a c

SA = 3x2 + 16x (3x + 4)(x + 4) 275  m2 Yellow = 3 cm ì 3 cm Black = 3 cm ì 6 cm White = 6 cm ì 6 cm

b Total area = 3x2 + 6x + 16 d l = 21  m; w = 7  m; d = 2  m f 294  m3 b Yellow = 0.36 m2 Black = 0.72 m2 White = 0.36 m2

Exercise 7C — Factorising expressions with two or four terms 1 a x(x + 3) b x(x - 4) c 3x(x - 2) d 4x(x + 4) e 3x(3x - 1) f 8x(1 - x) g 3x(4 - x) h 4x(2 - 3x) i x(8x - 11) 2 a (x - 2)(3x + 2) b (x + 3)(5 - 2x) c (x - 1)(x + 5) d (x + 1)(x - 1) e (x + 4)(x - 2) f (x - 3)(4 - x) 3 a (x + 1)(x - 1) b (x + 3)(x - 3) c (x + 5)(x - 5) d (x + 10)(x - 10) e ( y + k)( y - k) f (2x + 3y)(2x - 3y) g (4a + 7)(4a - 7) h (5p + 6q)(5p - 6q) i (1 + 10d)(1 - 10d) 4 a 4(x + 1)(x - 1) b 5(x + 4)(x - 4) c a(x + 3)(x - 3) d 2(b + 2d )(b - 2d ) e 100(x + 4)(x - 4) f 3a(x + 7)(x - 7) g 4p(x + 8)(x - 8) h 4(3x + 2)(3x - 2) i 3(6 + x)(6 - x) 5 a C b B c B

e (5p - 4t + 3t)(5p - 4t - 3t) f (6t − 1 + 5v )(6t − 1 − 5v ) 13 a E b A c D 14 B 15 a (x - 5)(x + 5) b (x - 5)  cm, (x + 5)  cm c 2  cm, 12  cm d 24  cm2 e 120  cm2 or 6 times bigger 16 a r metres b (r + 1) m c A1 = pr2  m2 d A2 = p(r + 1)2  m2 e A = p(r + 1)2 - pr2 = p(2r + 1) m2 f 34.56  m2 17 a Annie = (x + 3)(x + 2) m2 Bronwyn = 5(x + 2) m2 b (x + 3)(x + 2) – 5(x + 2) c (x + 2)(x - 2) = x2 – 4 d Width = 5  m e Annie has 30  cm2 and Bronwyn has 25  cm2.

6 C 7 a (x + 11)(x - 11)

b (x + 7)(x - 7)

Exercise 7D — Factorising by completing the square 1 a (x + 5)2 - 25 b (x + 3)2 - 9 c (x - 2)2 - 4 d (x + 8)2 - 64 e (x - 10)2 - 100 f (x + 4)2 - 16 g (x - 7)2 - 49 h (x + 25)2 - 625 i (x - 1)2 - 1

c (x + 15)(x - 15)

d (2x + 13)(2x - 13)

2 a (x - 2 + 11)(x - 2 - 11)

e (3x + 19)(3x - 19)

f 3(x + 22 )(x - 22)

g 5(x + 3)(x - 3)

h 2(x + 2)(x - 2) b (x - 4)(x + 6) d (x - 1)(x + 7) f (10 - x)(x + 2) h (7 - x)(5x + 1) b (x + y)(2 + a) d (x + y)(4 + z) f (n - 7)(m + 1) h 7(m - 3)(n + 5) j a(3 - b)(a + c) l m(m + n)(2 - n) b (m + 2)(n - 3) d (s + 3)(s - 4t) f (1 + 5z)(xy - z) b ( p - q)( p + q - 3) d (x + y)(7 + x - y) f (7g + 6h)(7g - 6h - 4)

c (x - 5 + 13)(x - 5 - 13) d (x + 3 + 19)(x + 3 - 19) e (x + 8 + 65)(x + 8 - 65) f (x - 7 + 6)(x - 7 - 6) g (x + 4 + 7)(x + 4 - 7) h (x - 2 + 17 )(x - 2 - 17) i (x - 6 + 11)(x - 6 - 11) 3 a (x - 1 +

5 )(x 2

b (x - 3 +

21 )(x 2

-3-

21 ) 2

c (x + 1 +

21 )(x 2

+1-

21 ) 2

d (x + 3 +

13 )(x 2

+3-

13 ) 2

e (x + 5 +

17 )(x 2

+5-

17 ) 2

f (x + 5 +

33 )(x 2

+5-

33 ) 2

g (x - 7 +

53 )(x 2

-7-

53 ) 2

h (x - 9 +

29 )(x 2

-9-

29 ) 2

i (x - 1 +

13 )(x 2

-1-

13 ) 2

2

2

2

2

2 2

2 2

2

-12

2

2

2

2

2

2

2

2

5 ) 2

Answers 7A ➜ 7D

i 12(x + 3)(x - 3) 8 a (x - 3)(x + 1) c (x - 5)(x + 1) e (6 - x)(x + 8) g 8(x - 3) i (x - 22)(9x + 2) 9 a (x - 2y)(1 + a) c (x - y)(a + b) e (f - 2)(e + 3) g 3(2r - s)(t + u) i 2(8 - j)(4 + k) k x(5 + y)(x + 2) 10 a (y + 7)(x - 2) c (q + 5)( p - 3) e (b + d)(a2 - c) 11 a (a - b)(a + b + 4) c (m + n)(m - n + l) e (1 - 2q)(5p + 1 + 2q) 12 a (x + 7 + y)(x + 7 - y) b (x + 10 + y)(x + 10 - y) c (a - 11 + b)(a - 11 - b) d (3a + 2 + b)(3a + 2 - b)

b (x + 1 + 3)(x + 1 - 3)

Answers

819

4 a 2(x + 1 + 3)(x + 1 - 3) b 4(x - 1 + 6)(x - 1 - 6) c 5(x + 3 + 2 2)(x + 3 - 2 2) d 3(x - 2 + 17)(x - 2 - 17 ) e 5(x - 3 + 7)(x - 3 - 7) f 6(x + 2 + 5)(x + 2 - 5) g 3(x + 5 + 2 3)(x + 5 - 2 3) h 2(x - 2 + 11)(x - 2 - 11) i 6(x + 3 + 14)(x + 3 - 14) 5    i  d ii b   iv  a v c vii  d viii e 6 a B b E

iii c vi d

7 E 8 a = 0.55; b = 5.45 Exercise 7E — Mixed factorisation 1 3(x + 3) 2 (x + 2 + 3y)(x + 2 - 3y) 3 (x + 6)(x - 6) 4 (x + 7)(x - 7) 5 (5x + 1)(x - 2) 6 5(3x - 4y) 7 (c + e)(5 + d) 8 5(x + 4)(x - 4) 9 -(x + 5)(x + 1) 10 (x + 4)(x - 3)

11 (m + 1)(n + 1) 13 4x(4x - 1) 15 3(3 - y)(x + 2) 17 4(x2 + 2) 19 (x + 5)(x - 5) 21 (x + 5)(x + 1) 23 (x + 2)(x - 2) 25 (y + 1)( x - 1) 27 7(x + 2)(x - 2) 29 (2 + r)( p - s)

12 (x + 7)(x - 7) 14 5(x + 10)(x + 2) 16 (x - 4 + y)(x - 4 - y) 18 (g + h)(f + 2) 20 5(n + 1)(2m - 1) 22 (x + 1)(x - 11) 24 (a + b)(c - 5) 26 (3x + 2)(x + 1) 28 -4(x + 6)(x + 1) 30 3(x + 3)(x - 3)

31 (u + v)(t - 3) 32 (x + 11 )(x - 11) 33 (4x - 1)(3x - 1) 34 (x + 1)(x - 3) 35 (x + 6)(x - 2) 36 4(x - 1)(x + 4) 37 3(x + 2)(x + 8) 38 (3 + x)(7 - x) 39 4(3 - x + 2y)(3 - x - 2y) 40 3(y + x) (y - x) 41 4(x + 2) 42 (3x - 4y)(x - 2y) 43 (x + 7)(x + 4) 44 (x + 2)(x - 5) 45 2(2x + 3)(x + 3) ( x + 5)( x − 2) ( x + 2)( x + 2) × 46 a ( x + 2)( x − 2) ( x − 4)( x + 2) b c 47 a d g j 820

( x + 5) ( x − 2) ( x + 2) ( x − 2)

×

( x + 2) ( x + 2) ( x − 4 ) ( x + 2)

x+5 x−4 x −1 x +1 b x−6 2x + 3 2x − 1 x+2 e x+4 x+5 4(b + 2) p( p + 7) h 5 ( p + 3)( p − 2) 5(d − 3 + 5e) 4(4 d + 3)

Answers

18 x ( x − 5) x−6 f x+3 5(m + 2 + n) i 2(2m − 5)

c

Chapter review Multiple choice 1 E 2 D 3 E 4 C 5 C 6 A 7 E 8 C 9 a 3x2 - 12x b -21x2 - 7x c x2 - 6x - 7 d 2x2 - 11x + 15 e 12x2 - 23x + 5 f 6x2 - 3x - 84 g 2x3 + 15x2 - 8x - 105 h 3x2 - 5x + 65 i 5x2 + 12x - 3 10 a x2 - 14x + 49 b 4 - 4x + x2 c 9x2 + 6x + 1 d -18x2 + 24x - 8 e -28x2 - 140x - 175 f -160x2 + 400x - 250 2 g x - 81 h 9x2 - 1 2 i 25 - 4x 11 a 2x(x - 4) b -4x(x - 3) c ax(3 - 2x) d (x + 1)(x + 2) e 2(2x - 5)(4 - x) f (x - 4)(x + 1) 12 a (x + 4)(x - 4) b (x + 5)(x - 5) c 2(x + 6)(x - 6) d 3(x + 3y)(x - 3y) e 4a(x + 2y)(x - 2y) f (x - 1)(x - 7) 13 a (x - y)(a + b) b (x + y)(7 + a) c (x + 2)( y + 5) d (1 + 2q)(mn - q) e (5r + 1)( pq - r) f (v - 1)(u + 9) g (a - b)(a + b + 5) h (d - 2c)(d + 2c - 3) i (1 + m)(3 - m) 14 a (2x + 3 + y)(2x + 3 - y) b (7a - 2 + 2b)(7a - 2 - 2b) c (8s - 1 + 3t )(8s - 1 - 3t ) 15 a (x + 9)(x + 1) b (x - 9)(x - 2) c (x - 7)(x + 3) d (x + 7)(x - 4) e -(x - 3)2 f 3(x + 13)(x - 2) g -2(x - 5)(x + 1) h -3(x - 6)(x - 2) i (4x - 1)(2x + 1) j (3x - 1)(2x + 1) k 4(2x + 3)(x - 1) l 5(7x - 3)(3x + 1) m -2(3x - 5)(2x - 7) n -3(3x - 1)(5x + 2) o -30(2x + 3)(x + 3)

16 a (x + 3 + 2 2)(x + 3 - 2 2) b (x - 5 + 2 7)(x - 5 - 2 7) c (x + 2 + 6)(x + 2 - 6) d (x - 25 +

17 )(x 2

- 25 -

17 ) 2

e (x + 72 +

53 )(x 2

+ 72 -

53 ) 2

f 2(x + 92 + 85 )(x + 92 - 85 ) 2 2 17 a 3x(x - 4) b (x + 3 + 7 )(x + 3 - 7 ) c (2x + 5)(2x - 5) d (2x + 5)(x + 2) e (a + 2)(2x + 3) f -3(x - 2)(x + 3) ( x − 2)( x − 1) 18 a 2( x + 4) b 7 c 8 x ( x − 4) 5( x + 1) Problem solving 1 a (x + 2)2 c 32x2 + 128x + 128 2 a 4r c 4pr    2 e 4p  (2r + 1)

b 32(x + 2)2 d 32  768  cm3 b 2r + 2 d (4pr    2 + 8r + 4)p f 28p m2

3 a (x – 7)(x + 2) c 35

b x – 7 cm d 1036 cm2

g - 32 +

4 Division by zero in Step 5

i

CHAPTER 8

Quadratic equations Are you ready? 1 a x(x - 3) b 4x(x + 3) c 12x(3x - 1) 2 a -2 and 3 b -2 and -3 c 2 and -3 3 a 2 6 b 10 2 c 36 2 4 a 0 b -16 c -38 5 a x = -2 b x = 3 c x = 1.5 Exercise 8A — Solving quadratic equations 1 a -7, 9 b -2, 3 c 2, 3 d 0, 3 e 0, 1 f -5, 0 g 0, 3 h -2, 0 i - 12 , 12

j -1.2, -0.5 2 a d g

l - 2 , 3

k 0.1, 0.75

1 , 1 2 - 67 , 1 12 0, 12 , 3

b -2, e h

- 23

3 2 , 5 3 0, 12 , - 25

c

1 , 4

f

- 85, 23

i 0, -3, 25

d - 23 , 0

e 0, 1 12

f 0, 13

h -  33 , 0

i 0, 1 14

4 a -2, 2

b -5, 5

c -2, 2

d -7, 7

e -1 13 , 1 13

f -2 12 , 2 12

g - 23 , 23

h - 12 , 12

i - 15 , 15

j -4, 4

k − 5, 5

l - 

5 a -2, 3 d 3, 5 g 5 j -3, 7 6 B 7 C 8 a - 12 , 3

b e h k

b 23, -1

c -2, 15

d 13, 1 12

e - 143 , 1

f 14 , 13

g -1 13 , 2 12

h -1 43 , -113

i - 25 ,

j 1 12, 2 23

k - 25 , 16

l 3, 4

-4, -2 1 -2, 5 -5, 6

c f i l

11 , 3

11 3

-1, 7 -1, 4 2, 6 3, 4

1 2

d 4 + 2 3 , 4 − 2 3

e 5 + 2 6 , 5 − 2 6

f 1 + 3 , 1 − 3

g −1 + 6 , −1 − 6

h −2 + 10 , − 2 − 10

i −2 + 15 , − 2 − 15 5 3 , 2 2

c

7 2

+

33 7 , 2 2

e 11 + 2

-

5 2

33 2

-

117 11 , 2 2

b - 25 +



-



117 2

d

1 2

+

f - 12 +

29 , 2 21 1 , 2 2 5 , 2

- 25 -

29 2

21 2

- 12 -

+

37 5 ,2 2

-

37 2

65 2

11 a -3, 1 b -4.24, 0.24 c -1, 3 d -0.73, 2.73 e 0.38, 2.62 f -0.30, 3.30 g -1.19, 4.19 h -2.30, 1.30 i -2.22, 0.22 12 No real solutions — when we complete the square we get the sum of two squares, not the difference of two squares and we cannot factorise the expression. 13 8 and 9 or -8 and -9 14 6 and 8, -6 and -8 15 9 or -10 16 2 or -2 23 17 8 or -10 12

18 6 seconds 19 a l = 2x b m 45

c

x cm

c + (2x)2 = 452, 5x2 = 2025 d Length 40 cm, width 20 cm 20 8 m, 6 m 21 a - 73 b x = 0 22 a

c x = ê  11 3

5 2

b (2 + x) m, (4 + x) m c (2 + x)(4 + x) = 24 d x = 2, 4 m wide, 6 m long 23 a (l - 4) cm b l - 8, l - 4 c (l - 8)(l - 4) = 620 d 31 cm e 836 cm2 24 a CAnnabel(28) = $364 800, CBetty(28) = $422 400 b 10 knots c Speed can only be a positive quantity, so the negative solution is not valid. Exercise 8B — The quadratic formula 1 a a = 3, b = -4, c = 1 b a = 7, b = -12, c = 2 c a = 8, b = -1, c = -3 d a = 1, b = -5, c = 7 e a = 5, b = -5, c = -1 f a = 4, b = -9, c = -3 g a = 12, b = -29, c = 103 h a = 43, b = -81, c = -24 i a = 6, b = -15, c = 1 −3 ± 13 5 ± 17 −5 ± 21 2 a b c 2 2 2 7 ± 45 d 2 ± 13 e −1 ± 2 3 f 2 9 ± 73 g h 3 ± 2 3 i −4 ± 31 2 1 ± 21 5 ± 33 j k l −1 ± 2 2 2 2 3 a -0.54, 1.87 b -1.20, 1.45 c -4.11, 0.61 d -0.61, 0.47 e 0.14, 1.46 f 0.16, 6.34 g -1.23, 1.90 h -1.00, 1.14 i -0.83, 0.91 j -0.64, 1.31 k -0.35, 0.26 l -1.45, 1.20 m 0.08, 5.92 n -0.68, 0.88 o -0.33, 2.00 4 C Answers

Answers 7E ➜ 8B

c −3 + 10 , − 3 − 10

+

5 2

4m

b −1 + 3 , −1 − 3

3 2

-

h

2m

9 a 2 + 2 , 2 − 2

10 a

65 9 ,2 2

37 2

2x cm

c 0, 7



3

- 2 -

x2

b -5, 0

7 2

+

7

3 a 0, 2 g 0,

9 2

37 , 2

821

6 B 5 C 7 C 8 a 0.5, 3 b 0, 5 c -1, 3 d 0.382, 2.618 e 0.298, 6.702 f 2, 4 g No real solution h -1, 8 i -4.162, 2.162 j -2, 1 k -7, 1.5 l No real solution m 2, 7 n - 12 , 13 o No real solution 9 a 2p r2 + 14p r - 231 = 0 b 3.5 cm c 154 cm2 10 a x(x + 30) b x(x + 30) = 1500 c 265 mm 11 a Pool A: 3 23 m by 6 23 m; Pool B: 3 13 m by 7 13 m b The area of each is 24 49 m2. 12 25 m, 60 m Exercise 8C — Solving quadratic equations by inspecting graphs 1 a x = -2, x = 3 b x = 1, x = 10 c x = -5, x = 5 d x = 2 e x = -1, x = 4 f x ö -1.4, x ö 4.4 g x = -25, x = 10 h x = 0 i x ö -2.3, x ö 1.3 j x ö -1.5, x = 1 2 a–j Confirm by substitution of above values into quadratic equations. 3 150 m 4 7 m 5 b x = -0.72, 1.39 c The answer for part b are the x-coordinates of the intersection of the quadratic in part a. Exercise 8D — Finding solutions to quadratic equations by interpolation and using the discriminant 1 a -4.5, 1.5 b -0.87, 1.5 c -4.6, 1.1 2 a -11 b 0 c 169 d 0 e 37 f 0 g 52 h -7 i -4 j 109 k 129 l 1 3 a No real solutions b 1 rational solution c 2 rational solutions d 1 rational solution e 2 irrational solutions f 1 rational solution g 2 irrational solutions h No real solutions i No real solutions j 2 irrational solutions k 2 irrational solutions l 2 rational solutions

4 a No real solutions

b 2 12

c -11, 2

d - 23

−3 ± 37 ö -4.541, 1.541 e 2 1 ± 13 1 ö -0.869, 1.535 f 5 g 3 h No real solutions i No real solutions

j k 822

−5 ± 109 ö -2.573, 0.907 6 −7 ± 129 ö -4.589, 1.089      l   5, 6 4

Answers

5 a b c 6 a

a = 3, b = 2, c = 7 -80 No real solutions a = -6, b = 1, c = 3

b 73 1 ± 73 d 12 8 C 10 k = -1

c 2 real solutions

7 A 9 C 11 m = 1, 8 12 n > - 49 13 p2 can only give a positive number which, when added to 24, is always a positive solution. 14 a 0.4 m b 0.28 m c 2.20 m d 2.5 m e i  Yes   ii  No f 1.25 m Exercise 8E — Solving a quadratic equation and a linear equation simultaneously 1 (-4, 1) and (1, 6) 2 a (-4, 12) and (-3, 10) b (-2, -5) and (6, 35) c (3, -2) and (5, 0) 3 (2, 4) 4 D = -8 5 a (-2, 4) and (5, 18) b (-2, -9) and (-1, -8) c (4, 10) d (-7, 18) and (-1, 6) e (1, 1) and (3, 9) f (1, 4) and (10, 22) 6 (-3, 1) and (-2, 1) 7 a (1, -5) b No, but the straight line is vertical and intersects at one point only. 8 (-2, 0) and (2, 0) y 9 a (3, 32)

(2, 21)

5 -3 -2

-1

x

1 — 11

b

y (2, 8)

-6

(1, 0) -6

c

-8

y

14 -7

-2 (-3, -4)

5 x

x

d

y

j

y

10

(-2, 28)

6

2

3– 2

5

x

10 6

k

6 — 11

2

e

y

x

5

21

y

(-2, 9) 3

-1 -6

-3

x

6

l

f

(5, 16)

11

y

y

7 x 54 — 13

x

6 8 (3, -15)

(3, 70) -48

-54

(-2, -80)

40

m

28

y 16

(-4, 0) -7

(-1, 7)

x

-4

g

-2 — - 16 9

y

(5, 14)

-9

4

6

(-4, -20)

n

x

-36

y (1, 20)

(-8, 20)

12

-61

h

y

-17–4

x

-4 -3

o

-2

-17

x

y

8 x -16

9

(1, -21) 9– 4

y

-5 (9, 39)

p

1

Answers 8C ➜ 8E

i

x

y

(6, 16) -4 (-3, -9)

- 3–4

3 6

x 4

-24

2 4 (2, 0)

x

Answers

823

10 1.322 km and 2.553 km x2 11 y = , (2, 2) and (-2, 2) 2

Exercise 9A — Plotting parabolas

1

4

2 B 4 D

g 2, 1

b e h b

-6, -1 5, -2 5, 6 -2, -1

e

1 - 2 , 4 5 5 , 3 2

h

2

c f i c

-8, -3 4, -7 7, -5 1 , -3 2

f

2 - 3,

2

i -7,

1 2 1 4

8 a -4 ê 17 b -1 ê 6 c -1, 9 4 10 a -0.651, 1.151 b -0.760, 0.188 c 0.441, -0.566 11 a -0.571, 0.682 b -0.216, 3.836 c -0.632, 0.632 12 -3, 7 13 -3, 1 14 a 2 irrational solutions b 2 rational solutions c No real solutions 15 a (-8, 22) and (2, 2) b (5, 10) c No solution Problem solving 1 -8 and 7 2 Length = 6 m, width = 3 m 3 a 2p r(r + 10) = 245 b 3.0 cm c 188 cm2 4 - 25 8 5 k > 9 and k < 1 6 a 6 m b 6 m 7 24 8 a y = 2x2 - 5x - 2 b No parabola is possible. The points are on the same straight line. 9 12( 5 + 2) cm

Chapter 9

2 a

824

2

Answers

2 3

y

1

y = –4 x2

1

-3 -2-1 0 1 2 3 x





-3 -2-1 0 1 2 3 x

x = 0, (0, 0) x = 0, (0, 0) 3 Placing a number greater than 1 in front of x2 makes the graph thinner. Placing a number greater than 0 but less than 1 in front of x2 makes the graph wider.

4 a

b

y 10 8 6

4 2 (0, 3)

x

-3-2-10 1 2 3

x = 0, (0, 1), 1 y

y = x2 - 3



-3-2-10 1 2 3

y y = x2 - 1 8

6

6

4

4

(0, -1)

x

-3-2-10 1 2 3 -2

x

x = 0, (0, 3), 3

d

2



y = x2 + 3

6

2

c

y

8

4



12 10

y = x2 + 1

2

-3-2-1 1 2 3 x -2

x = 0, (0, -3), -3 x = 0, (0, -1), -1 5 Adding a number raises the graph of y = x2 vertically that number of units. Subtracting a number lowers the graph of y = x2 vertically that number of units. b

y 16

+ 1)2

12

c -38 c x = 32 or x = 1.5

3 7  (x + + 1 b  x −  + c 2(x - 1)2 + 4  2 4 2± 2 −1 ± 5 b 2 2 1 ± 7 −1 ∓ 7 = −3 3 x = -2 or x = -3 b x = 1 or x = -2 x = 2 or x = -2 1 1 b x = 2 or x = - 3 x = - 2 or x = -2

b

2

(-5, 16) y = (x

b -16 b x = 3

x

y = 3x2

20

1)2

c x = 32 or x =

y

6 a

Are you ready? 1 a 0 2 a x = -2

5 a c 6 a

(0, 0)

-4 -3–2-1 1 2 3 -2

30 25 20 15 10 5

Functions

c

x = 0, (0, 0)

6

d 2, -7

4 a

y = x2

8

Chapter review Fluency 1 B 3 A 5 (3x + 4) m 6 a -5, -3 d 2, -6 g 3, 1 7 a -2, -6

3 a

y 10

8 4

c

(1, 4)

-6-5-4-3-2-1 01 2 x



x = -1, (-1, 0), 1 y 2 10 y = (x - 2) 8 6

0 12345 x

x = 2, (2, 0), 4

-6 -4 -2 0

2 x

x = -2, (-2, 0), 4

d

4 2



y y = (x + 2)2 16 12 8 4



y 2 10 y = (x - 1) 8 6 4 2 0 12345 x

x = 1, (1, 0), 1

7 Adding a number moves the graph of y = x2 horizontally to the left by that number of units. Subtracting a number moves the graph of y = x2 horizontally to the right by that number of units. 8 a

b

y 1



-3-2-1 01 2 3 4 x -2 -3 -4 -5 -6 -7 -8 y = -x2 + 1



x = 0, (0, 1), 1

c

d

y

y

-8

y = -(x + 2)2



x = -2, (-2, 0), -4

y y = (x - 5)2 + 1

y = -3(x - 1)2 + 2

x = -2, (-2, -9), min, -5

y y = x2 + 4x - 5 10 5 x

-6 -4 -2 0 -5 -10

f

x = -1, (-1, 16), max, 15

y 20 15 10

-10

5

-12 y = -x2 - 3

-6 -4 -2 0 -5

x = 0, (0, -3), -3

g

4 x y = -x2 - 2x + 15

2

-10



x = -1, (-1, 27), max, 24

y = -3x2 - 6x + 24 y

25

20 15 10 5 -6 -4 -2 -5 0 2 x -10 -15 -20 -25

h

26

x

4

-8

9 The negative sign inverts the graph of y = x2. The graphs with the same turning points are: y = x2 + 1 and y = -x2 + 1; y = (x - 1)2 and y = -(x - 1)2; y = (x + 2) and y = -(x + 2)2; y = x2 - 3 and y = -x2 - 3. They differ in that the first graph is upright while the second graph is inverted. 10 a

-25

2

-6

-6



e

x = 1, (1, 0), -1

-4

-4

-10 -15 -20

y = -(x - 1)2

0 12345 x -2

-6 -4 -2 0 1 x -2

-2 0 -5

y 0 -2-1 1 2 3 4 5 x -2 -3 -4 -5 -6 -7 -8 -9

x = 1, (1, 2), max, -1

y 5

d

x = 2, (2, 1) min, 5

y

20 y = (x - 2)2 + 1 16 12

b

1 0 12 3 4 5 6

x = 5, (5, 1), min, 26 y = 2(x + 2)2 - 3

y



x

x = -2, (-2, -3), min, 5 y 4 3 2 1 0 123456 -2 -3 -4 -5

4

x

14 C

15 A h 18 16 14 12 10 8 6 4 2

x

x = 3, (3, 4), max, -5

2

11 a I f the x2 term is positive, the parabola has a minimum turning point. If the x2 term is negative, the parabola has a maximum turning point. b If the equation is of the form y = a(x - b)2 + c, the turning point has coordinates (b, c). c The equation of the axis of symmetry can be found from the x-coordinate of the turning point. That is, x = b. 12 C 13 B 16 a

y = -(x - 3)2 + 4

-2 0

b

h = -(t - 4)2 + 16

0 1234567 8

i  16  m

t

Answers 9A ➜ 9A

c

-8 -6 -4 -2 0 -4

4



16 12 8 4



8

x

 ii   8  s Answers

825

17 a



c It is possible to have 0, 1 or 2 points of intersection. y

h 18 16 14 12 10 8 6 4 2

0

t

1 2 3

x

0

b   i  18  m

 ii  Yes, by 3  m

iii  1.5  s iv  3  s 18 a



y

y

x

0 0

x



y

y

x

0 0



x

y

0

x

Exercise 9B — Sketching parabolas using the basic graph of y = x 2 1 a Narrower, TP (0, 0) b Wider, (0, 0) c Narrower, TP (0, 0) d Narrower, TP (0, 0) e Wider, TP (0, 0) f Wider, TP (0, 0) g Narrower, TP (0, 0) h Narrower, TP (0, 0) 2 a Vertical 3 up, TP (0, 3) b Vertical 1 down, TP (0, -1) c Vertical 7 down, TP (0, -7) 1

1

d Vertical 4 up, TP (0, 4 ) 1



e Vertical 2 down, TP (0, - 12 ) f Vertical 0.14 down, TP (0, -0.14) g Vertical 2.37 up, TP (0, 2.37)

y

0

x

h Vertical 3 up, TP (0, 3) 3 a Horizontal 1 right, (1, 0) b Horizontal 2 right, (2, 0) c Horizontal 10 left, (-10, 0) d Horizontal 4 left, (-4, 0) 1

1

1

1

e Horizontal 2 right, ( 2 , 0) b An infinite number of points of intersection occur when the two equations represent the same parabola, with the effect that the two parabolas superimpose. For example y = x2 + 4x + 3 and 2y = 2x2 + 8x + 6. 826

Answers

f Horizontal 5 left, (- 5 , 0) g Horizontal 0.25 left, (-0.25, 0) h Horizontal 3 left, (− 3, 0) 4 a (0, 1), max b (0, -3), min c (-2, 0), max d (0, 0), min

e (0, 4), max f (0, 0), max g (5, 0), min h (0, 1) min 5 a Narrower, min b Narrower, max c Wider, min d Wider, max e Narrower, max f Wider, min g Narrower, min h Wider, max 6 a   i  Horizontal translation 1 left   ii  (-1, 0) iii  y y = (x + 1)2

f   i  Horizontal translation 4 right   ii  (4, 0) iii  y 2 y = (x - 4)

y = x2

0



x

(4, 0)

g   i  Reflected, wider (dilation)   ii  (0, 0) iii y 2

y = x2

y=x



x

(-1, 0) 0

b   i  Reflected, narrower (dilation)   ii  (0, 0) iii y

x y = - 2– x2

(0, 0)

5

h   i  Narrower (dilation)   ii  (0, 0) iii y y = 5x2

2

y=x

y = x2

x

0 y = -3x2

x

(0, 0)





c   i  Vertical translation 1 up   ii  (0, 1) iii  y y = x2 + 1

i   i  Reflected, vertical translation 2 up   ii  (0, 2) iii y y = x2 (0, 2)

y = x2

x

0

(0, 1) x

0



d   i  Wider (dilation)   ii  (0, 0) iii  y y = x2

y = -x + 2 j   i  Reflected, horizontal translation 6 right   ii  (6, 0) iii y y = x2 2

(6, 0) 0

(0, 0)



x

e   i  Vertical translation 3 down   ii  (0, -3) iii  y y = x2

y = -(x - 6) k   i  Reflected, vertical translation 4 down   ii  (0, -4) iii y y = x2 2

0

y = x2 - 3

x y = -x2 - 4

x

0



x

Answers 9B ➜ 9B

y = 1–3 x2

(0, -3)

Answers

827

l   i  Reflected, horizontal translation 1 left   ii  (-1, 0) iii y y = x2

p   i Narrower (dilation), reflected, horizontal translation 1 right, vertical translation 3 down 2   ii  (1, - 32 ) iii

(-1, 0)

y = x2

y

x

0

y = -(x + 1)2



0 (1, - 3 ) 2

m   i Narrower (dilation), horizontal translation 1 left, vertical translation 4 down   ii  (-1, -4) iii y

x y = - 74 (x - 1)2 - 32

7 a 10  cm b 5  cm c 5  cm d y = (x - 5)2

y = x2

x

0

Exercise 9C — Sketching turning point form 1 a (1, 2), min c (-1, 1), min e (5, 3), max

2

y = 2(x + 1) - 4



(-1, -4)

1

n   i Wider (dilation), horizontal translation 3 right, vertical translation 2 up   ii  (3, 2) iii y

(3, 2) 1

y = –2 (x - 3)2 + 2

y = x2

x

0

o   i Wider (dilation), reflected, horizontal translation 1 2 left, vertical translation 4 up   ii  (-2, 14 ) iii

y

y = x2

3

g (- 2 , - 4 ), min i (-0.3, -0.4), min 2 a i  (-3, -5) ii Min b i  (1, 1) ii Max c i  (-2, -4) ii Max d i  (3, 2) ii Min e i  (-1, 7) ii Max 1 1 f i  (- 5, - 2 ) ii Min 3   i b  y = -(x - 2)2 + 3 iii f  y = (x + 1)2 - 3 v c  y = x2 - 1 4 a A b C d C e B 5 a i  -3 b i  12 c i  -18 d i  -5 e i  4 f i  4 6 a   i  (4, 2) iv  18 vi  y

parabolas in



b (-2, -1), min d (2, 3), max f (-2, -6), min 1 2 h ( 3 , 3 ), min iii Narrower iii Same iii Narrower iii Wider iii Wider iii Wider

ii e  y = -x2 + 1 iv d  y = -(x + 2)2 + 3 vi a  y = (x - 1)2 - 3 c B

ii -3, 1 ii 2 ii No x-intercepts ii -1, 5 ii No x-intercepts ii -3 - 5, -3 + 5 (approx. -5.24, -0.76) ii Min iii Same width v No x-intercepts

y = (x - 4)2 + 2

18 (-2, 14 ) 0

x (4, 2)

  828

Answers

y = – 13 (x + 2)2 + 4

0 12 34

b   i  (3, -4) iv  5

ii Min v 1, 5

x

iii Same width

vi 

y

h   i  (1, 3)   ii  Min iii  Narrower iv  5   v  No x-intercepts

y = (x - 3)2 - 4

5 x

0 1 2 3 45 -4

(-1, 2)

i   i  (-2, 1) ii Max iii Narrower iv  -11 1 1   v  -2 (approx. -2.58, -1.42) , -2 + 3 3 vi  y x 0 1 -2 + — 3

1 -2 - — 3

x

iii Same width

  v  -5 - 3, -5 + 3 (approx. -6.73, -3.27) vi  y = (x + 5)2 - 3 y 22

 

-11 y = -3(x + 2)2 + 1

7 a 2(x - 43 )2 b x = 43 ê

73 8

=0

73 4

c ( 43 , - 73 ), minimum 8

-5 + 3

8 a y = -23(x + 4)2 + 6 b (-7, 0)

0x

(-5, -3)

ii Max

iii Same width

  v  1 - 2, 1 + 2 (approx. -0.41, 2.41) vi  y 1+ 2 x

1 2

y = -(x - 1) + 2   f   i  (-2, -3) ii Max iii Same width iv  -7 v No x-intercepts vi  y -2 (-2, -3)

9 a

b c d e

p ($) 1.9 1.4 1.0

0

(1, 2)

-1 0

x

0

(-2, 1)

0

1- 2 2 1

y = 2(x - 1)2 + 3

(1, 3)

 

3 2 1

-1   d   i  (-5, -3) ii Min iv  22

  -5 - 3 e   i  (1, 2) iv  1

y 5

(3, -4)

  c   i  (-1, 2) ii Min iii Same width iv  3 v No x-intercepts vi  y = (x + 1)2 + 2 y

0

x

-3

t (Hours

Exercise 9D — Sketching parabolas of the form y = ax 2 + bx + c 1 a y = (x + 2)2 - 6, (-2, -6) b y = (x + 6)2 - 40, (-6, -40) c y = (x - 4)2 - 10, (4, -10) d y = (x - 1)2 + 11, (1, 11) 5 5 3 3 e y = (x + 2 )2 - 4 , (- 2 , - 4 ) 1

9

x

7

g y = (x + 2 )2 -

41 , 4

7

41

(- 2 , -  4 )

h y = 2(x + 1)2 + 6, (-1, 6) i y = 3(x - 2)2 - 6, (2, -6) 2 a y = (x + 1)2 - 6, x-intercepts are -1 ê 6 (ö -3.4, 1.4)

y = x2 + 2x - 5 y

-1 - 6 -11

y = -(x + 3)2 - 2

-1 0

x -1 + 6

-5 (-1, -6) -6

Answers

Answers 9C ➜ 9D

2

(-3, -2)

5

after 12 pm.) 10 a 0.5 m b (15 + 4 15) m c Maximum height is 8 metres when horizontal distance is 15 metres.

9

  y = -(x + 2) - 3 g   i  (-3, -2) ii Max iii Same width iv  -11 v No x-intercepts vi  y -3 -2 -1 0 -2

3

$1.90 $1 3  pm $1.40

f y = (x + 12 )2 - 4 , (- 2 , - 4 )

-7

 

vi 

829

b y = (x - 2)2 + 3, no x-intercepts

h y = -5(x - 1)2 - 30, no x-intercepts y 0

y y = x2 - 4x + 7

7 x

0



(1, -30)

-30 -35

(2, 3)

y = -5x2 + 10x - 35



c y = (x + - 12, x-intercepts are −6 ± 48 = -3 ê 2 3 (ö -6.5, 0.5) 2 3)2

x

1

1

3

i y = -7(x + 2 )2 + 50 4 , x-intercepts are −1 ± 29 (ö -3.2, 2.2) 2

y

y (-1– , 50 3–) 49 2

4

-3 + 2√3 0

-3 - 2√3

x

–3

-1 - √29 2

(–3, –12)

1 5 d y = (x - 2 )2 - 5 4 , x-intercepts

y

5 ± 21 are (ö 0.2, 4.8) 2 1 0

y = x2 -5x + 1 5 - √21 2

(2 1–2, -5 1–4)

1

5

e y = -(x + 2 )2 + 7 4 , x-intercepts are −5 ± 29 (ö -5.2, 0.2) 2 -5 - √29 2

(-21–2 , 71–4 )

2

0

-4

(- 1–, -12 1–)



2

c

3

y

y

y = x2 - 12x + 32

32

x

0

-12

4



4

y = x2 - 8x - 9

-1 0

d

x

y (-3, 1) x

-4 -2 0

x

9

8 (6, -4)

(0, -9) -5 + √29 2 x

1

1 0

-8

(4, -25)



e

y = -x2 - 6x - 8

f

(-3, 36) y

y = -x2 -5x + 1 1

b

y = x2 + x - 12 y

5 + √21 2 x

71–4

x

0 y = -7x2 - 7x + 49

y

-2 –2



1

2 –2

3 a

-1 + √29 2

– 1–

y (1, 36) 35

27 3

f y = -(x - 2 )2 - 2 4 , no x-intercepts y 0

x

1– 2



( 1–2, -2 3–4)

-3

g

y = x2 + 4x - 5

3

1

g y = 3(x + 2 )2 - 12 4 , x-intercepts are −1 ± 17 (ö -2.6, 1.6) 2 y y = 3x2 + 3x - 12 - 1–

830

Answers

0



0 -5 y = -x2 + 2x + 35

1

x

4 a

(-2, -9) -1–

b

y

2

x

-9

0

y = 2x2 - 17x - 9 9 x

y 14

y = 3x2 - 23x + 14

02 – 3

-12 (- 1–, -12 3–) 2

4

7 x

-5

2

-3 - √17 2



-1 + √17 2

x

y 0

-5

y = -x2 + x - 3



0 3 -9 y = -x2 - 6x + 27



(4 1–4, -451–8 )



7 x 1) (3 5–6, -30— 12

c

y = 5x2 + 27x + 10

d

y

y = 6x2 + 7x - 3

y 14

10 a

y

12

10 x - 2–5 0

-5

e

2

9) 7 , -26— (-2 — 20 10

y

0 1 –

- 3–

f

(1 3– , 10 1–8 ) 4

10

x

3

-3

8

1) 7 , -5 — (-— 24

12

6

y

(2 3– , 36 1–8 ) 4

4 2

21 4 - 1– 0

y = -2x2 + 7x + 4

g

y

h

5 — 1 (— 12, 724)

- 3– 0 7 x 2 y = -2x2 + 11x + 21 y

31 25 — (1— 36, 4872)

6

0

- 2– 3

i

x

3 – 2

y = -6x2 + 5x + 6

y



x 0 2– 7– 2 -14 9 2 y = -18x + 67x - 14

y = 2x2 - 7x + 8

5x

0

-5

x

4

2

-2 y = x2 + 2x + 5 y=

x2 +

4x + 5

y = x2 - 2x + 5 y = x2 - 4x + 5

b If p < 0, the turning point is on the right side of the y-axis. If p > 0 the turning point is on the left side of the y-axis. As the magnitude of p increases the turning point moves away from the y-axis. All graphs have the same y-intercept (0, 5). 11 a b h = 0 h (25, 2500) c 2500  m 2500 d 25  s after launching e 50  s

8 50 t

0

(13–4 , 17–8 )



x

0

f1(x) = x2 + 6·x + 5

200

0

13 a

h 17



-1.72 0

h = -4.9t2 + 1.5t + 17

b 2  s c 0.15  s d 17.11  m

(10, 200)

20 x

t

2.02

14 a A = 2x(150 - x)  m2 b A 11 250

f2(x) = 2·x + 1

b 2x + y = 40  m d A = 2x(20 - x)  m2 f y

Answers 9C ➜ 9D

5 a B b C 6 a iv b vii c vi d iii e i f viii g ii h v 7 a A: y = 2(x - 3)2 + 4, y = 2x2 - 12x + 22; B: y = -3(x + 1)2 - 1, y = -3x2 - 6x - 4 b Translated 4 units to the left and 5 units down, reflected in the x-axis, and dilated by 32 in the y-direction. 8 (-0.32, 3.18) and (-4.68, -1.18) 9 a (-2, -3) y b

12 a A = xy  m2 c y = (40 - 2x)  m e (10, 200) g Maximum area is 200  m2, paddock is 10  m wide and 20  m long.

(75, 11 250)

1 0

1

(-2, -3)

x

0

150 x

c 11  250  m2, 75  m and 150  m −5 45 15 a a = 64 , b = 16 , c = −85 b (18, 20) 16 −5 c y = 64 (x - 18)2 + 20

Answers

831

y

6

Exercise 9E — Exponential functions and their graphs y 1 a

8

y = 10x

100

y = 3 ì 2x y ì 2x

10 6 4

y = 1–5 ì 2x x 1 2 3

2 -3 -2 -1 0

10

(0, 1)

2 x

-1

-4

-3

-2

-1

0

1

1 1 10 000 10 00

1 10 0

1 10

1

10

x y

0

-2

1

2

3

4



100 1000 10 000

b Providing a realistic scale is difficult. y y = 4x 2 a 100 80 60 40 (0, 1)

20

(1, 4)

-4 -3 -2 -1 0 1 2 3 4 x

y

b

x

2x

3 ì 2x

1 5

-3

0.125

0.375

0.025

-2

0.25

0.75

0.05

-1

0.5

1.5

0.1

 0

1

 3

0.2

 1

2

 6

0.4

 2

4

12

0.8

 3

8

24

1.6

7 The coefficient, k, affects the steepness of the graph: the larger the value of k, the steeper the graph. 8 y -x

y = 5x

100 80

y=2

60

10 8

40 (0, 1)

20

6

(1, 5)

4

-4 -3 -2 -1 0 1 2 3 4 x

c

y

ì 2x

(0, 1)

2 -3 -2 -1 0

y = 6x

x

1 2 3

100 80 60

9

40 (0, 1) 20

y-intercept at (0, 1) Equation of horizontal asymptote is y = 0.

(1, 6)

y

y = 3-x

y = 3x

10

-4 -3 -2 -1 0 1 2 3 4 x

8 6

3

y = 4-x

y = 3-x

y = 4x

y 100

4

y = 3x

(0, 1)

2

80

x

0

60

y = 2-x

40 20 -4 -3 -2 -1 0

y = 2x

(0, 1) 1

2

3

4 x

10 The negative index reflects the graph in the y-axis. 11 a y 1 x y = ( –2 )

10 8

4 Increasing the value of a increases the steepness of the graph where x is positive and flattens the graph where x is negative. y 5 a x 60

6 4 2 -3 -2 -1 0

y=2ì3

x

1 2 3

50 40 30 20 (0, 2)

10

832

Answers

-3

-2

-1

0

1

2

3

y

8

4

2

1

0.5

0.25

0.125

x

-4 -3 -2 -1 0 1 2 3 4 x

b 2

x

c y = 0

 1 b   = (2−1 ) x = 2− x  2

12

y y = (1.8)x

y 10

b

y = (1.5)x

y = 2x

8 y = (1.2)x

6

1

13 a

4

x

0

y

y = 2x - 1

(0, 1) y = 10 ì (1.3)x

-6

-4

2 (0, 0) 2

0

-2

4

6

8

10 x

-2 10

c

y

(0, 16)

15 x

0

y = 2x + 4

b 10 c y = 0 14 a, b, c 

10

y = 2x

5

(0, 1)

-10 -5 0 -5 y

(0, 1) 2

6

1 -6 -5 -4 -3 -2 -1

2

-12 -10 -8 -6 -4 -2 0 -2 y = 3x - 3 -4

2

4

x

6

b

y = 4x - 3

n V

4

b

2

-8 -6 -4 -2 0 -2

2

4

6

8

A

2

3

x

A = 1000 ì (1.1)n

1

4

5

6 n

x

40 000

V = 40 000 ì (0.85)n 1

2

3

4

5 n

c As n increases, the value of the car decreases. d $17 748 25 a 190 s (bacteria A); 110 s (bacteria B) b Bacteria A starts at 20 000; bacteria B starts at 260 000. c Bacteria A d 240 s 26 a 65 536 b 2.3 ì 1018 x c 2048 ì ( 2 ) d   i  18 days   ii  25 days   iii  38 days Answers

Answers 9E ➜ 9E

1 2 3 4 5x

3

V

0

9 8 7 6 5 4 3 2 1

2

0 1 2 3 4 5 40 000 34 000 28 900 24 565 20 880 17 748

15 000

21 Moves the graph horizontally 22 a y

-2 -3

1

c $1331 24 a

y = 4x

y = 2x 0 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1

(0, 0.5)

0 1 2 3 4 5 6 1000 1100 1210 1331 1464.10 1610.51 1771.56

0

8

y = 2x + 4

-1

y = 2x - 1

1000

10

y = 4x + 1

y = 2x

23 a n A

15 a y = 0 b y = 2 c y = -3 16 Moves the graph vertically 17 a ç iv b ç i c ç ii d ç iii 18 B 19 B 20 a, b, c  y

6

y

8 4

y = 3x

x

10

d

10

y = 3x + 2

5

833

Exercise 9F — The hyperbola 1 See the table at the bottom of the page*. y

1 2 3

y = -10 —– x

1 2 3 2 3

-5 -4 -3 -2 -1 0

x

0

y 10

— y = 10

10 -3 -2 -1

5

x

-10

x

-10

6 2 a   i 

y (1, 6)

y

6 y=— x

5 y=—

5

0

x

0

(1, -6) — y = -6

x

1

x

x

7 The negative reflects the curve y =

ii 

8

y

x — y = 20

20 0

x

y

-2

-1

0

1

-0.25 -0.33 -0.5 -1 Undefined



x

1

-3

k in the x-axis. x 2 1

3

4

0.5 0.33

y 1

y = —— x-1 1

-1

iii 

0 -1

y —– y = 100

100 0

x

x=1 x

1

x

1 2

Equation of vertical asymptote is x = 1. 9 a y 1

y = —— x-2

b   i  x = 0, y = 0    ii  x = 0, y = 0 iii  x = 0, y = 0 y 3

1 0 1 -— 2

x

2 3

x=2 (1, 4) (1, 3) (1, 2) 0

b

4 y=— x 3 y=— x 2 y=— x x

y 1

y = —— x-3 1 1 -— 3

0

x

34

x=3

4 It increases the y-values by a factor of k and hence dilates the curve by a factor of k.

1 *

834

x

-5

-4

-3

-2

-1

0

1

2

3

4

5

y

-2

-2.5

-3.3

-5

-10

Undefined

10

5

3.3

2.5

2

Answers

y

c

y 5

e 1

y = —— x+1 -2 -1

1 -1

10 The a translate the graph left or right, and x = a becomes the vertical asymtote. 11 a y



y 8

d

e

x

4

0 1 -2

-1 -3

x

2

4

12 x

8

-13

f



y -2

(2, 2)

y -4 -5

1 -3 -6

2

4 x

3

7

y = —— x-1

-2 6 -3

-8

-9

c

y

y 3

6 x

-3

-10

b

b



y 7

-4

-4

x

Centre (0, 0), radius 103

5 (1, 2)

y = —— x+1

3

1 3

-3 –

Centre (0, 0), radius 5

2 a

1 3– x

1 3

-3 –

-5



-1 0 -4

1 3

5 x

x

1

y

f

3–

-5

0

x = -1

(-2, 4)



y 10

2 x

10

-5

15 x

5

-10

-5

3 a (x + 2)2 + (y + 4)2 = 22 b (x - 5)2 + (y - 1)2 = 42 y y c

y 1

2— 2 -2

0

-3 1

-4

5 y = —— x+2

12 Check with your teacher.

1 1    b  y = x−3 x + 10

b

Centre (0, 0), radius 7

c



y 6 -6

6 x -6



4 x

Centre (0, 0), radius 6

Centre (0, 0), radius 4

d

y 9 -9

9 x -9

Centre (0, 0), radius 9

-2

12

4

x

e x2 + (y - 9)2 = 102 f (x - 1)2 + (y + 2)2 = 32 y y 1 -2

9

-4

-7

-10

6

-10 -1

4 10

x

-2

1

4 x

Answers 9F ➜ 9G

-4

7 x

14

x

19

y 4

7



x

14

7

-3

-7

9

c (x - 7)2 + (y + 3)2 = 72 d (x + 4)2 + (y - 6)2 = 82 y y 4

Exercise 9G — The circle 1 a y

5

-6

x

(-3, -5)

Possible answers: a  y =

5 1

x

-4 -2 -2

-5

4 D 5 B 6 (x - 5)2 + (y - 3)2 = 16 Chapter review Fluency 1 D 2 A 3 D 4 A 5 B

Answers

835

6 a (4, -15) b (-2, 9) 7 x -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 y 16 7 0 -5 -8 -9 -8 -5 0 7 16 y 16 14 12 10 8 6 4 2



y

y = 10 ì 3x

450 400 350 300 250 200 150 (0, 10) 100 50 -3 -2 -1 0

12

y = 10-x

x

0 -8 -6-4 -2 -2 2 4 6 8 -4 -6 -8 (-4, -9) -10

1

(0, 1) 1

13 a

y

3 x

2

y 140 120 100 80 60 40 20

-3 -2 -1 0

TP (-4, -9); x-intercepts: -7 and -1 8 a TP (3, 1); no x-intercepts; y-intercept: (0, 10) 10

11

2

y 10

y = (x - 3)2 + 1

8 6

(3, 1)

y = (1.2)x

2

x

-3 -2 -1 0

5 5 , −1 + ; b TP (-1, -5); x-intercepts: −1 − 2 2 y-intercept: (0, -5) y

5– 2

-1 -



(-1, -5)

9 TP (-1, 16); x-intercepts: -5 and 3; y-intercept: (0, 15) y (-1, 16)

15

y = -x2 - 2x + 15

0

-5

y = 5 ì 3x y = 2 ì 3x y = 1–2 ì 3x

-3 -2 -1 0

1

x

-3

y

0.008

b

-2 0.04

-1 0.2

y

y = 5x

160 140 120 100 80 60 40 (0, 1) 20 -4 -3 -2 -1 0

Answers

0

1

2

3

1

5

25

125

1

2

3 4x

1

2

3

x

b Changing the sign of the index reflects the graph in the y-axis. 16 a y (1, 4)

0 (1, 5)

x

3

y = (2.5)

45 40 35 30 25 20 15 10 5

-3 -2 -1 0

10 a

2

b Increasing the value of k makes the graph steeper. 15 a y -x x y = (2.5)

x

3

3 x

2

36 32 28 24 20 16 12 8 4

5– 2

x

0 -3

1

b Increasing the value of a makes the graph steeper for positive x-values and flatter for negative x-values. 14 a y

y = 2(x + 1)2 - 5 -1 +

836

y = (1.5)x

4

(0, 1)

0



x

3

4 y=— x

x

y

b

2 a x

0

x (1, -2)

-3 y = —— x-2

0

0

b 4  m c 2 s d 4 s 3 a

y 3 — 2

(2, 4)

h = 4t - t2

— y = -2

17

h

4 t

h 21 h = -x + 4x + 21

(2, 25)

-3 0

7

2

x

2

x

(3, -3)

1 . 18 Check with your teacher. Possible answer is y = x+3 19 a y 2 2 4 x + y = 16

-4

4 a

4 x

0

b 25  m c 2  m d 7  m h

(2, 20) h = -5t2 + 20t

-4

b 

20 a

y

(x - 5)2 + (y + 3)2 = 64

0

8 (5, -3)

x

x 2 + 4x + y 2 - 2y = 4

(-2, 1) 3

b

x

y 8 0 (-4, -4)

x

x 2 + 8x + y 2 + 8y = 32

4 2 -2

0

2

4x

-2 -4 y = -x2 + 4x - 1 y = x2 - 4x + 7

5 a [0, 12] b 32 m c 11:41 am to 6:19 pm 6 a P < x2 - 5x b 6.25 m c i Check with your teacher. ii Dilation by a factor of 0.48 d 28.6% 7 a Check with your teacher. b When x = 0.3, b = 10.7. Therefore if p is greater than 10.7 cm the platform would hit the bridge. Chapter 10

Deductive geometry Are you ready? 1 a DABC 2 a CA c ±BCA

b ±ACB c AC b PQ d ±QPR AD AE DE 3 DADE ~ DABC, = = AB AC BC 4 a Parallelogram b Trapezium c Kite Exercise 10A — Congruence review 1 a I and III, SAS b I and II, AAS c II and III, RHS d I and II, SSS

Answers

Answers 10A ➜ 10A

21 x2 + y2 = 36 Problem solving 1 a y = -(x - 2)2 + 3 = -x2 + 4x - 1 b y 6

4 t

b 4 s c 2 s d The ball is never above a height of 20  m.

y

0

0

837

2 a x = 3 cm b x = 85è c x = 80è, y = 30è, z = 70è d x = 30è, y = 7 cm e x = 40è, y = 50è, z = 50è, m = 90è, n = 90è 3 a Use SAS b Use SAS. c Use ASA. d Use ASA. e Use SSS. 4 C, D 5 a x = 110è, y = 110è, z = 4 cm, w = 7 cm b x = 70è c x = 30è, y = 65è 6 The third sides are not necessarily the same. 7 Corresponding sides are not the same. 8 Use SSS. Exercise 10B — Similarity review 1 a i and iii, RHS b i and ii, SAS c i and iii, SSS d i and iii, AAA e i and ii, SSS 2 a Triangles PQR and ABC b Triangles ADB and ADC c Triangles PQR and TSR d Triangles ABC and DEC e Triangles ABC and DEC 3 Check with your teacher. AB BC AC 4 a = = AD DE AE b f = 9, g = 8 5 x = 4 1 6 x = 20è, y = 2 4 7 x = 3, y = 4 Exercise 10C — Congruence and proof 1 Use AAS. 2 Check with your teacher. 3 Use SAS; then corresponding sides in congruent triangles are equal. NO = OP. 4 Check with your teacher. 5 Use SAS; then alternate angles in congruent triangles are equal. Hence AB || CD. 6 Use AAS. 7 Use RHS. 8 Use AAS. 9 Use RHS or AAS; then corresponding sides and angles in congruent triangles are equal. 10 Use RHS. 11 a B

y x x A

y D

C

b ±ADB = 90è given ±ABC = 90è given ±BAD = xè given ±BAC = xè given À ±ABD = 90è - x and ±ACB = 90è - x. À DBAD ~ DCAB. 838

Answers

AD AB = AB AC \AB2 = AD ì AC d ±BDC = 90è given ±ABC = 90è given ±ACB = 90è - x ±DCB = 90è - xè À ±DBC = xè ±BAC = xè given \ DBCD ~ DACB CD BC e = BC AC \ BC2 = CD.AC f AB2 + BC2 = AD ì AC + CD ì AC = AC (AD + CD) = AC ì AC \ AB2 + BC2 = AC2. g Students to do.

c

Exercise 10D — Quadrilaterals: definitions and properties 1 a True b True c True d False e True f False g False h False 2 None are true. 3 a, c, f 4 a, b, c, f, g, h 5 a, c, d, e, f 6 a, b, c, d, e, f, g, h 7 Rhombus, square 8 Rectangle, square 9 Parallelogram, rhombus, rectangle, square 10 Square 11 a

b 6 sides c 7 sides d Table size

Number of sides hit

5 cm ì 3 cm

6

7 cm ì 2 cm

7

4 cm ì 3 cm

5

4 cm ì 2 cm

1

6 cm ì 3 cm

1

9 cm ì 3 cm

2

12 cm ì 4 cm

2

e If the ratio of the sides is written in simplest form then the pattern is m + n - 2.

f There are two routes for the ball when hit from B. Either 2 or 3 sides are hit. The ball does not end up in the same hole each time.   A suitable justification would be a diagram — student to draw. g Isosceles triangles and parallelograms. The triangles are congruent. h The shapes formed are parallelograms. There is only one possible path although the ball could be hit in either of two directions initially. i Given m : n is the ration length to width in simplest form. When m is even and n is odd the destination pocket will be the upper left. When m and n are both odd, the destination pocket will be the upper right. When m is odd and n is even the destination pocket will be the lower right. j Students to investigate.

4 a Similar, scale factor = 1.5 b Not similar c Similar, scale factor = 2 5 a x = 48è, y = 4.5 cm b x = 86è, y = 50è, z = 12 cm c x = 60è, y = 15 cm, z = 12 cm 6 Use equiangular test. 7 Use equiangular test. 8 A

Exercise 10E — Quadrilaterals and proof 1 Use AAS to show DZWX @ DZYX. 2 Use AAS to show DAED @ DCEB and hence AE = EC and DE = EB. 3 a Use SAS. b ±AED = ±CED (corresponding angles in congruent triangles equal) and ±AED + ±CED = 180è (angle sum of straight lines is 180è) \ ±AED = ±CED = 90è c Corresponding angles in congruent triangles are equal. 4 Use SAS to show DDAE @ DBAE. Hence, DE = EB. (See previous question.) 5 Use co-interior angles and parallel lines. 6 Use SAS. AC = BD (corresponding sides in congruent triangles are equal). 7 AX || DY because ABCD is a parallelogram AX = DY (given) \ AXYD is a parallelogram since opposite sides are equal and parallel. 8 a Use SAS. b Use SAS. c Opposite sides are equal. 9 AC = DB (diameters of the same circle are equal) AO = OC and OD = OB (radii of the same circle are equal) \ ABCD is a rectangle. (Diagonals are equal and bisect each other.) 10 Check with your teacher. 11 PS = QR (corresponding sides in congruent triangles are equal) PS || QR (alternate angles are equal) \ PQRS is a parallelogram since one pair of opposite sides are parallel and equal. 12 MP = MQ (radii of same circle) PN = QN (radii of same circle) and circles have equal radii. \ All sides are equal. \ PNQM is a rhombus.



C

D

Bisect ±BAC AB = AC (given) ±BAD = ±DAC AD is common. \ DABD @ DACD (SAS) \ ±ABD = ±ACD (corresponding sides in congruent triangles are equal) 9 A, B, C, D 10 a False b True c True 11 a Use SAS. b Use SAS. c Use SAS. d They are all the same length. e B and C 12 Use SAS. PQ = PS (corresponding sides in congruent triangles are equal) 13 Rhombus, square 14 A quadrilateral is a rhombus if: a all sides are equal b the diagonals bisect each other at right angles c the diagonals bisect the angles they pass through. 15 WZ || XY (co-interior angles are supplementary) and WZ = XY (given) \ WXYZ is a parallelogram since one pair of sides is parallel and equal.

16

A

D

B

C

±ABD = ±ADB (angles opposite the equal sides in an isosceles triangle are equal) ±ABD = ±BDC (alternate angles equal as AB || DC) \ ±ADB = ±BDC \ Diagonals bisect the angles they pass through. 17 Corresponding sides are not the same. 18 ±FEO = ±OGH (alternate angles equal as EF || HG) ±EFO = ±OHG (alternate angles equal as EF || HG) ±EOF = ±HOG (vertically opposite angles equal) \ DEFO ~ DGHO (equiangular) 19 A rhombus is a parallelogram with two adjacent sides equal in length. 20 Rectangle, square

Answers

Answers 10B ➜ 10E

Chapter review Fluency 1 a I and III, ASA or SAS b I and II, RHS 2 a x = 8 cm b x = 70è c x = 30è, y = 60è, z = 90è 3 a Use SAS. b Use ASA.

B

839

CHAPTER 11

Problem solving I 1 23.83 cm 2 81x4 - 216x3y + 216x2y2 - 96xy3 + 16y4 y 3 3 + ø 9 - 8g g

4

3 - ø 9 - 8g 4



2

( 43 , g - 98 )

x-intercepts: x =

1 e (y-intercept - 45 ) and h (y-intercept = 14 ), gradient = - 20 ;

19 a A = 80p cm2 b If radius and height are both halved, the surface area is one-quarter its original value.

x

0

16 400 students 17 11.75 cm 18 a (y-intercept -12) and f (y-intercept = 2), gradient = –2; b (y-intercept 5) and d (y-intercept = 6), gradient = -1; c (y-intercept - 13 ) and g (y-intercept =  72 ), gradient = 3;

20 y =

3 ± 9 − 8g ; 4

− 4k  h  x − 2  + k h2  2

9 3 y-intercept: y = g; turning point:  , g −  4 8 4 1231.5 cm3 5 54.28è 6 8.3 cm by 1.7 cm f + 3be  f + 3be  ; y = a 7 x =  − 3b ae + d  ae + d 

8 a x = 8 b x = -4 c x = 12 9 No, Mary will need 64.5 cm of ribbon. 10 17.05è 11 -q Ç x Ç p 2 3  L − l 4π  L − l  12 π  l+ cm 3     2  3  2  13 a This is a quadratic equation, which means that there is a possibility of two different answers. Marlon has one of the two parts of the answer correct. b No. x(x - 3) = 10 x2 - 3x = 10 x2 - 3x - 10 = 0 (x - 5)(x - 2) = 0 (x - 5) = 0   or   (x - 2) = 0 x = 5     x = 2 y 14

2

e f 21   ≤ A ≤    4  4 7 18 22 y = -  x + 5 5 23 229.1 m P P 24 by 4 4 25 Any false statement that occurs during the solving of simultaneous equations indicates the lines are parallel, and have no points of intersection. 26 47 cm for the circle and 53 cm for the square 27 a y = x2 + 2x b y = -4x2 + 14x - 15 c y = x2 - 4x + 4 28 a 1072 cm3 b 9.4 cm 29 a 28.3 m b The image width doubles. 30 a Total length = 4l - 8, where l is the length of the lawn. b Cost = 23(4l - 8) + 100 c $1296 31 a a = 2 b b = 4 c c = -3 d The equation of the quadratic is y = 2x2 + 4x - 3. y

25 20 15

0

1

10

x

5 -3

-2

-1

-6

32 a L =

840

1





= 2 ì 2b ì 4 2b





= 4 2b2 cm2.

Answers

1

-5 y = -8

15 a 4 2b b 19.5è, 70.5è, 90è. 1 c Area = 2 base ì height

0

130 − x 2 2x

130 x − x 3 2 c x = 6.6 cm, L = 6.55 cm 33 Check with your teacher.

b V =

1

34 V = 3 p r2 s 2 − r 4

2

3x

35 a b2 + 8b + 28a + 4c - 4ac - 12 < 0 b b2 + 8b + 28a + 4c - 4ac - 12 = 0 c b2 + 8b + 28a + 4c - 4ac - 12 > 0 36 a 55 bottles b 20r cm c 17.32r cm d 3nr cm 37 132 passengers — 72 Virgin Green passengers, 60 Qintas passengers 38 a 75è58Å b 73 cm 39 a 3rd is (x + 2) cm; 4th is (x + 3) cm; 5th is (x + 4) cm b 5 cm 100 % larger than the c Circumference of 4th circle is x +3 3rd circle’s circumference. 19 40 a 24 b

5 24

c 32 3 cm 1 ± 10 41 x = 3 42 a 7 3 metres b 35 15 m2 7 3 metres 6 ii  18 43 a A = nx + 96.25n b $3500 44 a B (12.5, 0) and C (37.5, 0) 22 2 94 x - x + 90 = 0 b i  375 15 ii  (17.1, 18) c i  Translated 25 units to the right

c i 

ii  ( 75 , 125 ) 2 6 45 a x 2 = b

10

100 3

π c The cylinder has larger surface area by 57.25 cm2. 46 a 97.2 km/h b 140.4 km/h in 2 seconds 47 a r

b 43 r

163è20Å 16è40Å b km

c Check with your teacher. 48 a At t = 0, h = h0 = the starting height. b h 30

25 km N 121è40Å a km d km

19 km

10 1

2

3 t

c 22.25 m d 0.375 seconds e 1.55 seconds (2 x − 1)(3 x + 1) 1 49 a = 6(2 x − 1)(3 x + 1) 6

b 23.3 km east and 33.9 km south of its starting point c 145è30ÅT 62 a (x - 5)(x + 2) b (x - 5) is the shorter length. c x = 17 d 228 cm2 63 The rug is 80 cm wide and 400 cm long. 64 a 4r cm b (2r + 20) cm c 4p r2 cm2 Answers

Answers 11 ➜ 11

c km 58è20Å

20

0

b Solving 36x2 - 6x - 6 = 97 026 using any method gives x = -51.833 metres and x = 52. Ignore the negative solution because x > 0 for measurement units. Possible dimensions could be: 3(2x - 1) by 2(3x + 1) 2(2x - 1) by 3(3x + 1) 6(2x - 1) by 1(3x + 1) (2x - 1) by 6(3x + 1). Or any possible combination for numbers whose product is 6, such as 1.5 and 4 50 a i  Width: 2.95 m to 3.05 m, length: 4.45 m to 4.55 m ii  0.67% b 49.23 m2 c $760 51 a p = 72è, s = 108è b ABCD is trapezoidal with AD||BC. ±BAD = ±CDA = 72è ±ABC = ±BCD = 108è 52 The square numbers are 1, 4, 9, 16, 25, 36, ………… The difference between these numbers is 3, 5, 7, 9, 11... If this continues to 75, it is the 37th number, so 382 – 372 = 75. So, the two natural numbers are 37 and 38. 53 a Mr Barnes has (x + 2)(x + 5) m2; Mr Snowdon has 4(x + 2) m2. b (x + 2)(x + 5) - 4(x + 2) c (x + 2)(x + 1) d The carpet has a width of 3 m. e Mr Barnes bought 18 m2 and Mr Snowdon bought 12 m2. 54 The side of the square is 4 5 cm. 55 Check with your teacher. 56 Check with your teacher. 57 Use similar triangles. 58 240 students 59 a V = x(x + 10)(x + 7) b 44 400 cm3 c 40 cm 60 a k = 12 k = 12 makes the two equations represent the same line, giving an infinite number of solutions. All other values of k generate two parallel lines. b k ≠ 12 N 61 a

841

d (4p r2 + 80p r + 400p ) cm2 e 80p (r + 5) cm2 f 1.131 m2 g 30 cm 65 The factors of 24 are: 1 and 24; 2 and 12, 3 and 8 and 4 and 6. To make the first bracket equal 1, then x must be 7 and to make the second bracket equal 24, then y must be 28. This pattern continues until all possibilities are found. They are: Factors  1  2  3  4  6  8 12 24

x  7  8  9 10 12 14 18 30

24 12  8  6  4  3  2  1

y 28 16 12 10  8  7  6  5

69 a r = -4 b s < 70 Dan is 25 years old. 71 19.85 m xm 1m 72 a

c t =

c (2 73 + 30 ) m of pipe is required. 75 Approximately 98.3 cm 76 a b = -8, c = 12 b y = x2 - 8x + 12 c (4, -4) d y 15 10 5 0

2

4

6

10 x

8

-5

77 a 8 m b 57è c 7.6 m 78 a First ripple’s radius is 3 cm, second ripple’s radius is 15 cm. y b

66 5704 mm 67 360 adults and 190 children 68 a The cloth is x cm wide and 4x cm long. b P = 10x, A = 4x2 c Length = 120 cm, width = 30 cm. d Perimeter = (10x + 48) cm, area = (4x2 + 60x + 144) cm2 e The area has increased by 1944 cm2. 25 8

73 m long. 2 b The chain length is (2 3 − 1) m, so (8 3 − 4) m of chain is required.

74 a The struts each need to be

15 10 5

-15 -10

0

-5

9 - 32

5

10

-10 -15

c 2.4 cm/s d 1 minute 23 seconds after it is dropped 79 The centre of the racket travels 5.24 m. 80 a i  6 cm

xm

3m 10 cm

Temperature degrees celsius

b (x + 1)(x + 3) d 5.24 m by 3.24 m T 73 a

c 5 m by 3 m

8 cm

10 cm

12 cm



25

ii  10 cm

20

8 cm

15 10 cm

10

2

4 6 Hours

b 21èC c Decreasing d Increasing e 5 èC after 4 hours f 21èC Answers

8 cm 12 cm

5 0

842

15 x

-5

8h

iii 

6 cm

10 cm

8 cm

12 cm

10 cm

6 cm

b Area of rectangle = 12 ì 8 = 96 cm2. Area of parallelogram = 12 ì 8 = 96 cm2, 12 × 16 = 96 cm2, Area of triangle = 2 1 Area of trapezium = 2 (18 + 6) ì 8 = 96 cm2.

90 a y = -2x2 - 12x - 14 b a = -2, b = -12, c = -14 c y (-3, 4)

4 2 (-1.6, 0) (-4.4, 0) x 0 -5 -4 -3 -2 -1 -2 1 2 3

c Perimeter of rectangle = 40 cm. Perimeter of parallelogram = 44 cm, Perimeter of triangle = 48 cm, Perimeter of trapezium = 44 cm. The triangle has the largest perimeter, while the rectangle has the smallest. 81 a 0 b (6, 6) h c

-4 -6 -8 y = -2x2 - 12x - 14 -10 -12 (0, -14) -14 -16

Height (metres)

8 6 4

2 0

2

12 d

4 6 8 10 Distance (metres)

d 12 m e 6 m f 6 m 82 a 60è b 3.98 km c 71è d 1.34 km 83 True: the tip travels 30.2 m. 84 a 56 941 cm3 b 11 938 cm2 y 85 a 12

Height (metres)

10 8 6 4 2 0

1 2 3 4 5 6 Horizontal distance (metres)

x

b 8 m c 11 m above the water d 5.83 m 86 a 24.5 m3 b The dimensions of the smaller skip are half those of the larger one. 87 7.6è 88 x = 13.75, y = 11.4 89 a Centre is (1, 2). b Radius is 2. c x-intercept 1; y-intercepts (− 3 + 2) and ( 3 + 2) d y



4

(1, 2)



1 0 -1 -2

1

2

3

4

5x

1 12 1 12 1 12 1 12 1 12 1 12 1 12 1 12

1 1 = 13 + 156 1 = 141 + 84 1 = 15 +

= = = = =

1 60 1 1 + 48 16 1 1 + 36 18 1 1 + 30 20 1 1 + 28 21 1 1 + 24 24

104 The factors of a number are generally written in pairs, producing an even number of factors. With a perfect square, one of these factors will be paired with itself, producing an odd number of factors. This occurs for all perfect squares. 105 15.9 cm 106 The total number of tiles needed for r rows is r2. Answers

Answers 11 ➜ 11



2

-1



5

3

-2

91 180è 92 The area of material required is 1.04 m2. If Tina is careful in placing the pattern pieces, she may be able to cover the footstool. 93 a Prove the equation is y = 15 - 0.15x2. b The height at the edge of the road is 5.4 m. c The road needs to be 15.5 m wide. 94 - 49 95 a 50 minutes b 8.03 am or 8.07 am c 8.11 am d 8.05 am e 25 minutes f Between 8.03 am and 8.07 am 96 For example: take the two numbers 48 and 60. Their HCF is 12. Difference = 60 - 48 = 12 60 ó 12 = 5 and 48 ó 12 = 4 The two numbers are exactly divisible by 12. This theorem says, then, that the HCF of 48 and 60 is 12, which is the case. 97 Matt is travelling at 45 km/h and Steve is travelling at 60 km/h. 98 The perimeter of the octagon is 61 cm. 99 Check with your teacher. 100 17.4 circuits 101 Approximately 40 400 years from now 102 Check with your teacher. 103 The 8 different ways are:

843

 et the cost of an apple be a cents, and the cost of a 107 a L banana be b cents. 6a + 4b = 700 [1] 1a + 9b = 450 [2] Subtract [2] from [1]. 5a - 5b = 250 Divide through by 5. a - b = 50 b An apple costs 50c more than a banana. 108 a 1.25 mL b 1.47 mL c Answers may vary. The two answers are only slightly different. Because of an inability to measure to that degree of accuracy in the home, they both provide a good guide to a safe amount to administer. d Approx 1 and 10 years 109 y = 2x2 - 5x + 1 110 11

12

1 2

10 9

3

8

4 7

6

5

The total of the numbers here in each region is 26. 111 Call the people A, B, C and D who take times of 1, 2, 5 and 10 minutes respectively. A and B go over 2 min B returns 2 min C and D go over 10 min A returns 1 min A and B go over 2 min Total time = 2 + 2 + 10 + 1 + 2 = 17 min 112 35 tennis balls 113 58 railings 114 If the cut was vertical, a rectangle (or straight line if the cut was on the very edge) would result. A horizontal cut would result in a circle. A cut which goes through the sides at an angle would produce an ellipse or a parabola. An ellipse results if the cut is simply through the sides. A parabola results if the cut is through the side and the base (or top).

115 Answers may vary. The value of n must be greater than 1 because the second number would be 0 if n were 1. 116 The deck should be 5 m wide and 10 m long on each side. 117 The original number could have been 187, 781, 286, 682, 385, 583 or 484. 118 There is no solution to the equation. It is not possible for the square root of a number to be negative. Squaring both sides of the equation has produced an invalid solution. 844

Answers

119 a There are 5 different shapes.

b They all have a perimeter of 10 units, except for the last one which has a perimeter of 8 units. 120 112 km 121 Check with your teacher. 122 505 1 1 1 123 = + n n + 1 n( n + 1) 124 Fold the left vertical line forward, so that page 3 sits on top of page 4. Fold the entire bottom half backwards, so that page 5 sits behind page 4. Fold forward the left vertical fold, so that page 3 sits on top of page 2. Finally, fold forward the left vertical fold, so that page 6 sits on page 7. The pages are now in order from page 1 to page 8. 125 9 trains 126 The claim appears to be true. Further investigation would need to be conducted to determine if it worked in all cases — for example, if the lines were parallel, at right angles, vertical or horizontal. 127 6450 multi-packs 128 Take the total, subtract 16, then divide by 4 to get the first number in the square. The other numbers are 1, 7 and 8 larger. 129 11 times 130 a x2 but no y2 gives vertical parabola, y2 but no x2 gives a horizontal parabola. b The value of a represents the horizontal stretching factor. Positive a, open to right, negative a open to left.

c

x + 4 = (y + 1)2

5

y

8 6

2

9 a

1 -4

-3

-2

10 a i  20

ii 

1

ii 

-1

b i  16 11 A

-2

12 a

0

-1

(-4, -1)

1

x

2

(0, -3)

-3

d 13 a

-4

c

Probability b 4

1

1 4

2 3 1 12

c 13

5

c 8

1

2

5

1 4

or 0.25

3

6

3 10

or 0.3

4

3

3 20

or 0.15

5

4

1 5

ii

1

6 1 2

12 3 iv  20 = 5 4 a A ¶ B c AÅ ¶ BÅ

b   i 

5 a

1 6

=

8

10

ii 

4 20

=

1 5

v 

8 20

=

2 5

b

4 17

9 10

1

b 10

17 6

Tennis



iii

Volleyball

x

Walking 10

15

38

8 2

17 6

b 96

20

c i 

35 96

14

d i 

63 96

18

8

23

21

= 32

ii 96

16 a iii 

2 20

=

x = 30

Volleyball

1 10

c

Soccer 1

7 2

b XÅ ¶ Y d A ¶ C ¶ BÅ 1 8

1

ii 96 = 12

4

5

4

7 1

1

b i  2 ii 6

6 A 7 a

12

38

1

c i  2

Tennis 1

iii  30

2

iv  5

7

v  15

Answers 12A ➜ 12A

19

2 10 20

15

4 16

10

Tennis

x

x

Walking

2

3

1 9

6

8

1.00

B 5 11

17

15

c 10

A

38

Volleyball

or 0.2

“ f = 20

b 10

10 2



or 0.1

x

Walking

Tennis

1 10

17

7 20

8

7

2

7 13

1

Volleyball 15

1

3

f 2

1

d 0

c 2

133

b 312

3 a

1

iii  16

e 0 b

b Drawing a red card

b 156

1 5

iv  40

c 2

1 5 11 20

Exercise 12A — Review of probability 1 x Relative frequency f

2 a

7

iii  80

1 4

b

c 23

b 9

3 a Not drawing an ace c Obtaining a 4 or a 5 5 a

9

1 5 17 80

1 13 12 13

15 a i 

Are you ready? 1 a Set A: 3, Set B: 4, Set C: 4

4 a

c 5

14 a C b D c E

Chapter 12

2 a

b 10 3

(0, 1)

(-3, 0)

4

1

17 30

8

ii 15 Answers

845

17 a

7 C

x = 35

Calculator

1 10

b 10

9

c 50

10 a

9 50

b 50

41

c 25

11 a

15 16

b 9

8 a

Graph book

47

9 D 7

18

5 5

b i  25 d i  18 a

12 No, getting 1 Tail is possible too. 1 13 a 8

ii 23

18 35 1 5

ii 7

5

iii 

6 7

iii 

ii

3 10

12

1 7 1 7

iv  35

b 15 a

c 9–4

16 a

19 4–3 1

A

Overlaying AÅ and BÅ shows AÅ ¶ BÅ as the area surrounding A and B

b 47

1 a P(A) 2 a M

d 1

N

b P(M ) + P(N ) - P(M ¶ N ) c i  False ii  True 3 a

1 13

4 20% or

b 1 5

1

5 a  i  2 16

6 25 846

iii Cannot be determined

12 13

Answers

1

ii 2

b Yes

18 a

5 9

19 a

8 14

20 a

1 13

b

The union of A and B is shown in brown, leaving the surrounding area as (A ß B)Å

c 0.35

b 25

19

c 25

9

2

c 13

3

b 13

4

b 7 or

4 7

2

10

1

b 14 or 7

c 14 or

1

5 7

4

b 4

c 13

21 a Yes

B

Exercise 12B — Complementary and mutually exclusive events

9 25 7 26

17 C

20 a No. P(Azi rolls a 5) = 8 and P(Robyn rolls a 5) = 16 1 b Yes. P(Azi wins) = 2 and P(Robyn wins) = 12 21 Yes. Both have a probability of 12. 22 a The person with the 6-sided die has less chance of winning. 1 For the 8-, 12- & 16-sided dice: P(mult 4) = 4 ; for 1 6-sided die, P(mult 4) = 6 b Answers may vary, check with your teacher.   i Example — rolling an even number ii Example — rolling a 3 iii Example — rolling a number greater than 3 23 P(Alex wins) = 15 ; P(Rene wins) = 25 24 Yes, equivalent fractions; 166 = 83 25 B

7 8

2

14 3

b $50

A

d 25

7

1 2

22 Answers may vary, check with your teacher. a i No. There are many other foods one could have. ii Having Weet Bix and not having Weet Bix b i No. There are other means of transport; for example, catching a bus. ii Walking to a friend’s place and not walking to a friend’s place c i No. There are other possible leisure activities. ii Watching TV and not watching TV d i No. The number 5 can be rolled too. ii Rolling a number less than 5 and rolling a number 5 or greater e Yes. There are only two possible outcomes; passing or failing. 23 No. The number 2 is common to both events. 24 a i T

ii F

v F

vi F

b i

3 16

ii 1

c i

5 16

ii

iii T

iv F

5

iv 16

8

iii 8

3 16

iii

3

3 4

Exercise 12C — Two-way tables and tree diagrams 1 a 0.2 b 0.1 c 0.2 d 0.5 e 0.4 f 0.8 Card outcomes 2 i Coin outcomes

c i 

13

12

ii a  18

b  0.5 c  0.5

Club,

Spade,

Diamond,

Heart,

H

H,

H,

H,

H,

T

T,

T,

T,

T,

3 a 

Die 1 outcomes

1

3

4

5

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

1– 2

1– 2

B

1– 2

R

1– 2

B

1– 2

R

1– 2

B

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

a

1 8

b 8

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

d

1 8

e 8

7 a

1 12

Coin outcomes

2

3

4

5

6

7

R

9

10

1– 3

H (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (H, 7) (H, 8) (H, 9) (H, 10)

c

3

4

1 4

5

6

7

8

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8)

Yellow octahedron outcomes

2

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8)

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8)

7

(7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (7, 7) (7, 8)

8

(8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6) (8, 7) (8, 8)

Green octahedron outcomes 3

4

5

6

7

8

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8)

2

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8)

3

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8)

4

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (4, 7) (4, 8)

5

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (5, 7) (5, 8)

1– 2

1– 2

(7, 1) (7, 2) (7, 3) (7, 4) (7, 5) (7, 6) (7, 7) (7, 8)

8

(8, 1) (8, 2) (8, 3) (8, 4) (8, 5) (8, 6) (8, 7) (8, 8)

R B

RBR RBB

R B

BRR BRB

R B

BBR BBB

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

3

c 8

3

7

f

1– 6

1– 2

1– 6

1– 2 1– 3

1– 6

1 2

2 Outcomes Probability 1– RR 9

R

G

RG

B

RB

R

GR

G

GG

B

GB

R

BR

G

BG

B

BB

1 — 18 1– 6 1 — 18 1 — 36 1 — 12 1 — 6 1 — 12 1– 4

— 1

  b {(R, R), (R, G), (R, B)} 1 c 3 d 8 a

7 18

1

2

B

1– 2

1– 2

G

1– 2

B

1– 2

G

1– 2

B

1– 2

1– 2 1– 2

1– 2

1– 4 1– 2

1– 4

2 1– 4

1– 4

1– 2

1– 4

3 1– 4

  b No 1

1

1– 2

1– 2

1– 4

1

1– 4

1– 2

1– 2

G

1– 2

1– 2

1– 2

  b 3 3 c 8 d They are equally likely. 7 e 8 9 a 1

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (6, 7) (6, 8)

7

1 8

1– 2

Outcomes Probability 1– RRR 8 1– RRB 8

3 B G

Outcomes Probability 1– BBB 8 1– BBG 8

B G

BGB BGG

B G

GBB GBG

B G

GGB GGG

1– 8 1– 8 1– 8 1– 8 1– 8 1– 8

— 1

2 Outcomes Probability 1– 11 4

1 2

12

3

13

1

21

2

22

3

23

1

31

2

32

3

33

3

1– 8 1– 8 1– 8 1 — 16 1 — 16 1– 8 1 — 16 1 — 16

— 1

Answers 12C ➜ 12C

1

6

1– 2

1– 2

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (1, 7) (1, 8)

2

1– 2

B

1

1

G

1– 2

Green octahedron outcomes 2

1– 2

1– 3

1– 6

T (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) (T, 7) (T, 8) (T, 9) (T, 10)

1 5

1– 2

1– 3

8

3 R B

1– 2

1

Die outcomes

b

Yellow octahedron outcomes

R

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

1

c

2

4

5 a 

 

1

6

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

1

b

2

1

  b 4 a

6

Die 2 outcomes

9

c i  4     ii  2     iii  8     iv  16 Answers

847

10

1 1– 6

t

5– 6

Outcomes Probability 1 — tt 36

2 1– 6 5– 6

1– 6

t' 5– 6

t t'

tt'

5 — 36

t

t't

5 — 36

t'

14 a

1

b c d

B Outcomes Probability 1 — SS 16 S

2

1– 4 3– 4

S

SS'

3 — 16

S

S'S

3 — 16

S'

S'S'

S'

3– 4

1– 4

S' 3– 4

9 —

16 — 1

S = outcome of spade Sample space = {SS, SSÅ, SÅS, SÅSÅ}

b c d

1 16 9 16 3 8

12

1 1– 3

X

1– 4 1– 4

Y

1– 4

1– 3

1– 3

1– 3

1– 3

W 1– 4

Z

1– 3

1– 3

1– 3

  a 13 a

1 4

2 Outcomes Probability 1 — Y XY 12 1– 1 — 3W XW 12 1 — XZ Z 12 1 — X YX 12 1– 1 — 3W YW 12 1 — YZ Z 12 1 — X WX 12 1– 1 3Y — WY 12 1 — WZ Z 12 1 X — ZX 12 1– 1 3 — ZY Y 12 1 — ZW W 12 — 1

1

b 4

4 — 10

6 — 10

B

G

3– 9

B

BB

2 — 15

6– 9

G

BG

4 — 15

d 848

2 15 8 15

Answers

1– 3

1– 3

1– 3

1– 3

E

 

ADF

1– 3

H 1– 3

AEF AEG AEH BCF BCG BCH

F G

BDF BDG BDH BEF

H

1– 3 1– 3

1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18 1 — 18

ADG ADH

F G

H F –1 3 G

C

D

F G

1– 3

1 — 18 1 — 18 1 — 18

ACG ACH

H

1– 3

1– 3

F G H

BEG BEH

— 1

Sample space = {ACF, ACG, ..., BEG, BEH} 1 1 b 2 c 6 15 There is a 50% chance that a third room will be needed. 16 Susan would have 16% chance of passing the exam if the last three questions had the standard 4 choices. This chance is reduced to 48% with the inclusion of two questions offering 6 possible answers. 17 i a Outcomes Probability

R

1– 2

1– 2

G

1– 2

R

RR

1– 4

1– 2

G

RG

1– 4

1– 2

R

GR

1– 4

1– 2

G

GG

1–

4 — 1

  1

1

b 2 c 2 ii As the first counter is not replaced, the probability of drawing the second counter is altered. This is reflected in the probabilities along the branches of the tree 1 diagram; P(2 counters of the same colour) = 3; 2

P(2 counters of different colours) = 3 .

c 1 2– 4

2– 4

R

G

Outcomes Probability 1– RR 6

1– 3

R

2– 3

G

RG

1– 3

2– 3

R

GR

1– 3

1– 3

G

GG

1–

6 —

c 3

1   Exercise 12D — Independent and dependent events 1 a 0.28 b 0.12 c 0.42 d 0.18 1 1 1 2 a Yes b i  2 ii 6 c 12

e No

3 40

4– 9

B

GB

5– 9

G

GG

4 — 15 1–

3 — 1

  b

1– 3

2 Outcomes Probability

1

1– 3

1– 3

D

ACF

F G H

1– 3

E

1– 3

1– 3

1– 3

1– 3

1– 2

1– 4

 

1– 3

1– 2

1

1– 3

A

36 — 1

1 36 5 18 25 36 11 36

11 a

C

1– 3

t = outcome of 3

a

3 Outcomes Probability 1– 3

25 —

t't'

2

1

1

5

4 36

7 a

12 a

c 25

48

c 77

b 125 b D

3 77 1 37

b 77

1 5 1 17 26 145

b

1

8 a 9 0.9 1 10 14 11 a

64

16

5 a 25 6 a C

b 1369

b

1 5 1 221 136 435

1

8

4

d 25 d

18 77

73

c 1369

1

c 10

1

d 3

25

c 102 221

13 a b c 435 14 No. Coin tosses are independent events. No one toss affects the outcome of the next. The probability of a Head or Tail on a fair coin is always 0.5. Greg has a 50% chance of tossing a Head on the next coin toss as was the chance in each of the previous 9 tosses. 15 No. As events are illustrated on a tree diagram, the individual probability of each outcome is recorded. The probability of a dependent event is calculated (altered according to the previous event) and can be considered as if it was an independent event. As such, the multiplication law of probability can be applied along the branches to calculate the probability of successive events. Exercise 12E — Conditional probability 41

1 a P(J) = 90

12

b P(H  |  J) = 41 2 a P(S) = 13 30 b P(S  |  (C ß S)) = 13 28 3 a 0.3 b

3 7

4 a

9 13

b

3 5 15

5 0.58 or 26 6 0.22 or

5 23

6



ii  P(B  |  A) = 16

iii  P(C  |  A) =

1 6

iv  P(C  |  B) = 0 10 A 11 Conditional probability is when the probability of one event depends on the outcome of another event. 12 a 0.0875 b 0.065 13 a 0.585 b 0.1525 or 15.25%.

Chapter review Fluency 1 A 2 B 5 B 6 B 7 a A

3 B

4 D B

x

Answers

Answers 12D ➜ 12F

7 0.9 8 0.8375 9 a D b i  P(A  |  B) = 1

Exercise 12F — Subjective probability 1 a The outcome depends upon whether it is a Test match or a one-day game and how effective the bowlers and batsmen are; not forgetting the pitch usually favours spin bowling. b The outcome depends on which team is better on the day and which team can adjust to the conditions. c No. The third one has an equal chance of being a girl or a boy. d This is not necessarily true. Current position and form of both teams should be used as a gauge. e It does not mean it will rain again on Friday. f There is no certainty about that. It depends upon the condition and location of your house. g Cricket games are not won or lost by the attractiveness of the uniform. h It is possible to get 6 Heads in a row on a normal coin. i They will have a good chance but there is no certainty. The country with the best competitors on the day of each event will win. j This is dependent on the person’s own interests. 2 a You still have a chance. b No horse is certain to win. Lots of problems can occur on the track. c This is not true. Even though Heads and Tails have equal chances, it does not mean half the results will show Heads. d Favourites do not always win. e Sometimes outsiders pay well, if you back the right one! You can lose more money than you win. 3 Answers will vary. Class discussion required as there are many factors to consider. 4 a There is a contradiction. The job was never hers. She had to do well to win the position. b The team may have had a lead but a match is only won when finished. c No horse is certain to win. 5 Experimental probability is based on data collected from trials. The more trials undertaken, the closer the experimental probability will reflect theoretical probability.   Theoretical probability is based on mathematical models. A theoretical probability does not guarantee a particular outcome in real life situations.   Subjective probability is based on judgements and opinions and hence may be biased. Subjective probability may approach theoretical probability if the assigned probability is based on real experiences and judgements made from an objective and educated position. 6 Answers will vary. Class discussion may be required. Example only: medical — our town is so far away from any major airports that it is unlikely our residents will need immunisation from swine flu.

849

b

A

B

14 a

x

Die 2 outcomes 2

1

x

A

B

8 a 107 9 a 4–7 b 1–6 c 8–25

b 27

c 175

3 7

b 47

c $28

f i  g 50 15 a

6

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

1 36 1 36

ii ii

1 52

11 a W  hether it rains or not on Thursday is not determined by what happened on Monday, Tuesday or Wednesday. It can still rain on Thursday. b The team’s win or loss depends upon how other players bat and bowl or how the other team plays. c There is an equal chance of having a boy or a girl. 12 a If you were defeated, the opponent was the winner. b The slowest motocross rider could not win the race if he/she crossed the finish line first. c The person elected was the most popular choice for the position. 13 a i  50 ii 7 iii 25 iv 8 3 6 b i  12 ii 50 iii 25 c i 

n(x ) = 50 Fried rice

Chicken wings

4

10 6

2 11

0

Die 1 outcomes

and P(B) =

1 6

1 ii 25

b c d e 16 a

Dim sims

6

7

8

9

10 11 12

4

5

6

5

4

3



iii 181 iii 181

1 6 1 6



2

2

1

3

(0, 0) (0, 1) (0, 2) (0, 3)

1 (1, 0) (1, 1) (1, 2) (1, 3) 2

(2, 0) (2, 1) (2, 2) (2, 3)

3

(3, 0) (3, 1) (3, 2) (3, 3)

No 0 and 6 3 0 and 6, 1 and 5, 2 and 4 1

1– 8

7– 8

f

fÅ

2 1– 8

f

7– 8

fÅ

1– 8

f

7– 8

fÅ

3 f fÅ

1– 8 7– 8 7– 8 7– 8

1– 8 1– 8 1– 8

7– 8

ffÅf ffÅfÅ

f fÅ

fÅff fÅffÅ

f fÅ

fÅfÅf fÅfÅfÅ

ii 343 512

21 512

11 iv 256

iii  17 a

Outcomes Probability 1 —– fff 512 7 —– fffÅ 512

f fÅ

f = outcome of 5 1 b   i  512

Die outcomes

12 5

5

Die 2 outcomes

1 Coin outcomes

1 2

4 13

Answers

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

0

c 23 10 a No b P(A) = 14 , P(B) = 131 , P(A ¶ B) =

850

5

3

d 43

c

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

2

c 132

b 85 9 a Yes b P(A) =

4

Frequency 1

b 14

8 a

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

4

5 a  131

7775 7776 3 8

3

3

e i 

b

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

2

4 B

7 a

2

Sum

3 D

1 7776

6

b 6 c No. Frequency of numbers is different. d

Problem solving 1 C 2 D

6 a

5

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

 

C

4

1 Die 1 outcomes

c

3

2

3

4

H

(H, 1) (H, 2) (H, 3) (H, 4)

T

(T, 1) (T, 2) (T, 3) (T, 4)

7 —– 512 49 —– 512 7 —– 512 49 —– 512 49 —– 512 343 —– 512

1

b 1– 4

H 1– 2

1– 4 1– 4

1– 4

1– 4

1– 2

T

1– 4 1– 4

1– 4

b

1 4 1 19 1 169 1 221

20 a

15 25

=

3 5

b

8 10

=

4 5

c 18 a 19 a

4 a Mean = 2.5, median = 2.5 b Mean = 4.09, median = 3 c Median 2 5 a 72 3 b 73 c 70 – <80 6 124.83 7 65 – <70 8 a B b B c C 9 a Mean = $32.93, median = $30 b

Outcomes Probability 1– 1– 1– 1 H1 2 ì 4 = 8 H2

1– 2

ì

1– 4

=

1– 8

3

H3

1– 2

ì

1– 4

=

1– 8

4

H4

1– 2

ì

1– 4

=

1– 8

1

T1

ì

T2

ì

1– 4 1– 4

=

2

1– 2 1– 2

=

1– 8 1– 8

3

T3

1– 2

ì

1– 4

=

1– 8

4

T4

1– 2

ì

1– 4

=

2

21 b 38

1–

8 — 1

Class interval

Frequency

Cumulative frequency

  0–9

 5

 5

10–19

 5

10

20–29

 5

15

30–39

 3

18

40–49

 5

23

50–59

 3

26

60–69

 3

29

70–79

 1

30

Total

30

c 15 38

21 0.847 Chapter 13

Univariate data

4

3

12

 3

5

3

15

 6

6

3

18

 9

7

3

21

12

8

4

32

16

9

4

36

20

4

5

6 7 Score

8

9

Exercise 13A — Measures of central tendency 1 a i  7 ii 8 iii 8 b i  6.875 ii 7 iii 4, 7 c i  39.125 ii 44.5 iii No mode d i  4.857 ii 4.8 iii 4.8 e i  12 ii 12.625 iii 13.5 2 Science: mean = 57.6, median = 57, mode = 42, 51 Maths: mean = 69.12, median = 73, mode = 84 3 a i  5.83 ii 6 iii 6 b i  14.425 ii 15 iii 15

Cumulative frequency

   Median = $30

30 25 20 15 10 5 0

10 20 30 40 50 60 70 80

Amount spent ($)

d The mean is slightly underestimated; the median is exact. The estimate is good enough as it provides a guide only to the amount that may be spent by future customers. 10 a 3 b 4, 5, 5, 5, 6 (one possible solution) c One possible solution is to exchange 15 with 20. 11 a Frequency column: 16, 6, 4, 2, 1, 1 b 6.8 c 0 –4 hours d 0 –4 hours 12 a Frequency column: 1, 13, 2, 0, 1, 8 Age of emergency b 15 ward patients

10 5 0

7.5 22.5 37.5 52.5 67.5 82.5

Age

c Asymmetrical or bimodal (as if the data come from two separate graphs). d 44.1 e 15–<30 f 15–<30 Answers

Answers 13A ➜ 13A

5 4 3 2 1

Mean = $32.50 c

Frequency

6

Frequency

Are you ready? 1 a 4.6 b 10.3375 c 143.25 2 a 6 b 12.5 c 61.5 d 9.4 3 a 3 b No mode c 2 and 3 4 Mean = 37, median = 39, mode = 43 5 Score Frequency Cumulative (x) (f) fìx frequency

d D

851

26 24 22 20 18 16 14 12 10 8 6 4 2

8

100%

0 15 30 45 60 75 90 Age

Cumulative frequency

50%

Cumulative frequency (%)

Cumulative frequency

g

Cumulative frequency

of female athletes

Cumulative frequency (%)

h 28 i No j Class discussion 13 a Player A mean = 34.33, Player B mean = 41.83 b Player B c Player A median = 32.5, Player B median = 0 d Player A e Player A is more consistent. One large score can distort the mean. 14 a Frequency column: 3, 8, 5, 3, 1 b 50.5 c 40 –<50 d 40 –<50 e Ogive of pulse rate

15 50%

10 5 30 50 70 Beats per minute

f Approximately 48 beats/min 15 A 16 Check with your teacher. 17 Answers will very. Examples given. a 3, 4, 5, 5, 8 b 4, 4, 5, 10 c 2, 3, 6, 6, 12

Cumulative frequency

Exercise 13B — Measures of spread 1 a 15 b 77.1 c 9 2 a 7 b 7 c 8.5 3 a 3.3 kg b 1.5 kg 4 22 cm 5 0.8 6 C 7 a 40 35 30 25 20 15 10 5 0

50 55 60 65 70 75 80

Battery life (h)

b i  62.5 ii  Q1 = 58, Q3 = 67 iii  9 iv  14  v  6 852

Answers

   IQR = 24

120 130 140 150 160 170 180 190 200

Class interval

9 a i  Range = 23 ii IQR = 13.5 b i  Range = 45 ii IQR = 27.5 c i  Range = 49 ii IQR = 20 10 a 25.5 b 28 c 39 d 6 e The three lower scores affect the mean but not the median or mode. 11 a Men: mean = 32.3; median = 32.5; range = 38; IQR = 14 Women: mean = 29.13; median = 27.5; range = 36; IQR = 13 b Typically, women marry younger than men, although the spread of ages is similar. Exercise 13C — Box-and-whisker 1 a 13 b 5 2 a 122 b 6 3 a 49.0 b 5.8 4 a 140 b 56 d 84 e 26 5 a 58 b 31 d 27 e 7 6 B 7 C 9 a (22, 28, 35, 43, 48) b 20 30 40 50 Sales

100%

20

55 50 45 40 35 30 25 20 15 10 5 0

plots c 26 c 27 c 18.6 c 90

c 43 8 D

10 a (10, 13.5, 22, 33.5, 45) b 0 10 20 30 40 50 Rainfall (mm) 11 a (18, 20, 26, 43.5, 74) b 10 30 50 70 Age d 39

c The distribution is positively skewed, with most of the offenders being young drivers. 12 a (124  000, 135  000, 148  000, 157  000, 175  000) b 120 140 160 180 ($ì1000) 13 a Key:  12  |  1 = 121 Stem Leaf 12 1  5  6  9 13 1  2  4 14 3  4  8  8 15 0  2  2  2  5  7 16 3  5 17 2  9 18 1  1  1  2  3  7  8 b 120 140 160 180 Number sold c On most days the hamburger sales are less than 160. Over the weekend the sales figures spike beyond this.

14 a Key:  1*  |  7 = 17 years Stem Leaf 1* 7  7  8  8  8  9  9 2 0  0  0  1  2  2  2  2  3  3  3  3  4  4  4 2* 5  5  8  9 3 1  2  3 3* 4 4* 8 b 15 25 35 45 Age ì

c The distribution is positively skewed, with first-time mothers being under the age of 30. There is one outlier (48) in this group. 15 C Exercise 13D — The standard deviation 1 a 2.29 b 2.19 c 20.17 d 3.07 2 a 1.03 b 1.33 c 2.67 d 2.22 3 10.82 4 0.45% 5 0.06 m 6 0.49 s 7 15.10 calls 8 B 9 Adam is more consistent because he has the lower standard deviation. (1.7 compared with 3.9) 10 C 11 a Class A: 1.13; Class B: 1.74 b Class A is more consistent because the standard deviant is lower. 12 a Life of battery (hours) Class centre Frequency   0–<5

  2.5

 6

  5–<10

  7.5

16

10–<15

12.5

18

15–<20

17.5

15

20–<25

22.5

 5

25–<30

27.5

 5

b Mean = 13.4, standard deviation = 6.73 c The batch is unsatisfactory. Although the mean is greater than 13 hours, the batch fails as the standard deviation is greater than the required 6 hours.

Holden 0 5 10 15 20 2530 35 40

5 a Brisbane Lions b Brisbane Lions: range = 63; Sydney Swans: range = 55 c Brisbane Lions: IQR = 40; Sydney Swans: IQR = 35 6 a Girls Boys 1.4 1.5 1.6 1.7 1.8 1.9 Height

b Boys: median = 1.62; girls: median = 1.62 c Boys: range = 0.36; girls: range = 0.23 d Boys: IQR = 0.14; girls: IQR = 0.11 e Although boys and girls have the same median height, the spread of heights is greater among boys as shown by the greater range and interquartile range. 7 a Summer: range = 23; winter: range = 31 b Summer: IQR = 14; winter: IQR = 11 c There are generally more cold drinks sold in summer as shown by the higher median. The spread of data is similar as shown by the IQR although the range in winter is greater. 8 A 9 A, B, C, D 10 a Cory achieved a better average mark in Science (59.25) than he did in English (58.125). b Cory was more consistent in English (s  = 4.9) than he was in Science (s  = 19.7) 11 a Back street: x = 61, s  = 4.3; main road: x = 58.8, s  = 12.1 b The drivers are generally driving faster on the back street. c The spread of speeds is greater on the main road as indicated by the higher standard deviation. 12 a Nathan: mean = 15.1; Timana: mean = 12.3 b Nathan: range = 36; Timana: range = 14 c Nathan: IQR = 15; Timana: IQR = 4 d Timana’s lower range and IQR shows that he is the more consistent player. 13 a Machine A Machine B 40 42 44 46 47 48 50 52 54 56 58 60 Number of Smarties in a box

b Machine A: mean = 49.88, standard deviation = 2.88; Machine B: mean = 50.12, standard deviation = 2.44 c Machine B is more reliable, as shown by the lower standard deviation and IQR. The range is greater on machine B only because of a single outlier. 14 Students’ own work. Exercise 13F — Skewness 1 a Yes b 8 c Both equal 8. 2 a No b 31–40 c No. They can, however, be calculated. Answers

Answers 13B ➜ 13F

Exercise 13E — Comparing data sets 1 a Boys: median = 26; girls: median = 23.5 b Boys: range = 32; girls: range = 53 c Both sets have similar medians but the girls have a far greater range of absenteeism than boys. 2 a Morning: median = 2.45; afternoon: median = 1.6 b Morning: range = 3.8; afternoon: range = 5 c The waiting time is generally shorter in the afternoon. One outlier in the afternoon data causes the range to be larger. Otherwise the afternoon data are far less spread out. 3 Key:  16  |  1 = 1.61 m Leaf  Stem Leaf   Boys Girls 997 15 1256788 98665540 16 4467899 4421 17 0

4 a Ford: median = 15; Holden: median = 16 b Ford: range = 26; Holden: range = 32 c Ford: IQR = 14; Holden: IQR = 13.5 d Ford

853

Frequency

3 a

c Student comparison Statistics

7 6 5 4 3 2 1 0

1 2 3 4 5 Number of goals

b Yes c 1, 2, 3 and 4 d Yes. Both equal 2.5. 4 a 4 5 a

b Negatively skewed

35 30 25 20 15 10 5

1–10 11–20 21–30 31–40 41–50

Frequency

Five-point summary x Range IQR s

Number of goals

1–50 51–100 101–150 151–200 201–250

Frequency

Number of people

b No c 201–250 d The distribution is negatively skewed. Reasons could include the size of cinemas or the target audience of the movie. 8 a No b Science: positively skewed, Maths: negatively skewed c The science test may have been more difficult. d Science: 61–70, Maths: 71–80 e Maths has a greater standard deviation (12.6) compared to Science (11.9). 9 Answers will vary. Check with your teacher. 10 a Key:  2  |  3 = 2.3 hours Leaf Stem Leaf Group A Group B 8  7  3 1 7  8 9  5  1 2 0  1  2  4  5  5  8  8 8  7  5  4  2  2 3 2  2  2  4  5  5  5  6  8 7  5  4  2  2  2 4 2 5 2  2 6 b Five-point summary Group A:  13  27  36  43  62 Group B:  17  23  30  35  42 Group B



854



Answers

d Student decision, justifying answer e Class discussion Chapter review Fluency 1 a Mean = 11.55; median = 10; mode = 8 b Mean = 36; median = 36; mode = 33, 41 c Mean = 72.18; median = 72; mode = 72 2 a 6 b 6 c 20 3 a 4 b 8.5 4 a Year 10

0

10 9 8 7 6 5 4 3 2 1

10 20 30 40 50 60 70 Nouns

Group B 17  23  30  35  42 28.95 hours 25 hours 12 hours 7 hours

Year 8

b Negatively skewed 6 B 7 a

Group A

Group A 13  27  36  43  62 35.85 hours 49 hours 16 hours 13 hours

Hours

10 20 30 40 50 60 70

b Year 8: mean = 26.83, median = 27, range = 39, IQR = 19, sd = 11.45 Year 10: mean = 40.7, median = 39.5, range = 46, IQR = 20, sd = 12.98 c The typing speed of Year 10 students is about 13 to 14 wpm faster than that of Year 8 students. The spread of data in Year 8 is slightly less than in Year 10. 5 a 20 b 24 c 8 6 a Key:  3*  |  9 = 3.9 kg Stem Leaf 3* 9 4 0  0  2  3 4* 5  6  7  8  8 5 0  3 5* 5  8  8  9 6 1  2  2 6* 8 b (3.9, 4.4, 4.9, 5.85, 6.8) c 3.5 4.5 5.5 6.5 kg 7 a 24.4 8 A

b 1.1 9 B

c 7.3 10 0.05 mL

Problem solving 11 a Mean = 32.03; median = 29.5 b Class interval Frequency   0–9

 2

10–19

 7

20–29

 6

30–39

 6

40–49

 3

50–59

 3

60–69

 3 Total

30

30 25 20 15 10 5 0

10 20 30 40 50 60 70

Age

Frequency

e Median = 30 f Estimates from parts c and e were fairly accurate. g Yes, they were fairly close to the mean and median of the raw data. 12 a HJ Looker: median = 5; Hane and Roarne: median = 6 b HJ Looker c HJ Looker d Hane and Roarne had a higher median and a lower spread and so they appear to have performed better. 13 a English: mean = 70.25; Maths: mean = 69 b English: range = 53; Maths: range = 37 c English: s  = 16.1; Maths: s  = 13.4 d Kloe has performed more consistently in Maths as the range and standard deviation are both lower. 14 a Yes b Yes. Both are 3. c 3 15 a 9 8 7 6 5 4 3 2 1

12 34 5 Number of cars

b Positively skewed — a greater number of scores is distributed at the lower end of the distribution. 16 Mean = 5, median = 5, mode = 2 and 5. The distribution is positively skewed and bimodal. 17 C 18 A, B and C Chapter 14

Bivariate data

4 5

you ready? a Numerical b Non-numerical c Non-numerical d Numerical a Continuous b Continuous c Continuous d Discrete a Number of kilograms: independent; total cost: dependent b Temperature: independent; number of swimmers: dependent c Age: independent; height: dependent a 1 unit b 0.1 of a unit a $3.20 in March b In February grapes cost $3.05. c The cost of grapes was recorded over 5 months. d Grapes cost the most in May when they were $4.33 per kilogram.

a c a c

145  cm 20  cm 3 right, 2 up 7 up

b d b d

5  cm 2 years 5 right 2.5 right, 1.5 up

Exercise 14A — Identifying related pairs of variables 1 a Quantitative, discrete b Qualitative, nominal c Quantitative, continuous d Quantitative, continuous e Qualitative, nominal f Quantitative, continuous g Quantitative, discrete h Quantitative, continuous 2 Discrete data can be counted in exact values; continuous data can be measured in a continuous scale. 3 a If data is nominal, it is qualitative in nature, so it cannot also be discrete. b If data is ordinal, this implies an order, which is a qualitative classification. This means that it cannot also be continuous. 4 a Dependent: time spent travelling to school; independent: distance to school b Dependent: heart rate of a runner; independent: running speed c Dependent: value of CD collection; independent: number of CDs in collection d Dependent: amount of computer memory used by file; independent: length of file e Dependent: cost of second-hand car; independent: age of car 5 a No relationship b   i There is a relationship between c and s. ii Positive iii Strong c   i There is a relationship between l and t. ii Positive iii Strong d   i There is a relationship between p and a. ii Positive iii Moderate e   i There is a relationship between h and a. ii Positive iii Strong f No relationship 6 There is no relationship between sunburn and ice-cream sales. The increase in both is influenced by the weather. Exercise 14B — Graphing bivariate data 1

Answers 14A ➜ 14B

Are 1 2 3

6 7

Goals scored in a match Total number of goals

Cumulative frequency

c Mean = 31.83 d

50 40 30 20 10 0

10

20

30 40 50 60 70 80 Time after start of match (min)

90

100

Answers

855

c The school canteen should stock more pies during cooler weather, and fewer pies during hot weather. Note: Best fit lines are indicated as a guide only. 3 a y b y

2 a Goals scored in a match Total number of goals



50 40 30 20 10

x

x

c

0

10

20

30 40 50 60 70 80 Time after start of match (min)

90

d

y

y

100

b The trend is positive and strong, indicating that as the match progressed goals were scored at a steady rate. 3 a

x

e

70 60 50 40 30 20 10 0

x 1 2 3 4 5 6 7 8 9 10 11 12 r

b The trend is positive, but only moderate. 4 a Are long babies born to tall fathers? b Do people with more formal education earn more money? c Does the amount of exercise performed decrease with age? 5 a True b False c True d False e True 6 B 7 C 8 D 9 A 10 a Number of tickets sold and the total money raised for a number of different charity concerts. b Number of items sold and the price of the item.

4 5 6

a 38 b  18 a i  460 ii  290 iii  130 b i  37 ii  24 iii  6 a  and  b Note: Answers may vary depending on the line of best fit drawn.

Exercise 14C — Scatterplots 1 a 40

0

40 80 English

Number of pies sold

b The direction of the relationship is positive; the greater the English mark, the greater the history mark, generally. However, as the points on the scatterplot vary quite a bit from a straight line, the relationship is only moderate. 2 a 80 0

40 0 20 Temperature (èC)

b The direction of the relationship is negative; the greater the temperature, the fewer the pies sold. The points on the scatterplot lie close to a straight line, so the relationship is strong. 856

Answers

130 120 110 100 90 80 70 60 50 40 30 20 10 0

2

4

6

8 10 12 14 16 18

Hours worked

c   i Approximately 16 hours ii About $55 iii About $6.25 7 a  and  b Note: Answers may vary depending on the line of best fit drawn.

Petrol used (L)

History

80

0

x

y

Earnings ($)

b

Relationship between r and b

14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 100

Distance travelled (km)

c   i About 8 litres ii About 70 km iii About 7 km/L

1 2 3 4 5 6 7 8 9 10111213141516

Day

9

b 7.1  g, 8.1  g, 13.9  g, 14.9  g c 17.8  g, 18.8  g d About 1  g Prediction of y-values when x = 15 and x = 60 would be considered unreliable as these x-values are beyond the range of the data. For an x-value of 40, the predicted y-value would be considered reliable as this is within the range of the data. 10 This prediction would be considered unreliable, as the scattering of the points indicates that there is no relationship between x and y. Chapter review Fluency 1 A 2 D 3 A 4 Independent Dependent a Number of hours Test results b Rainfall Attendance c Hours in gym Visits in the doctor d Lengths of essay Memory taken e Attendance Cost of care f Age of property Cost of property g Number of applicants Cut-off ENTER score h Running speed Heart rate 5 a 10  èC b Day 5 c Day 7 d Day 6 e 10  èC f Days 6 and 7 6 a Height b Age c Age and height of child 120

80 60

30 35 40 45 50 55 60 65 70 75 80 Cost ($)

d As the price increases, the number of bags sold decreases. This means that the relationship is negative. The points vary quite a bit from a straight line, so this indicates that the relationship is moderately strong. e Since the cheaper bags sell better, have a greater stock of them than the more-expensive bags. 2 a i  12.5 ii  49 b i  12 ii  22.5 3 a The two sets of data are quantitative and continuous. b Birth mass c What influence on the birth mass does the gestation period have? d 3.6

40 20 0

1

2

3 4 Age

5

6

d The relationship is positive and strong. As the child grows older his/her height also increases. 7 a Minutes b mm3 c Time d 1 unit = 100 mm3

12 11 10 9 8 7 6 5 4 3 2 1

3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0

Answers 14C ➜ 14C

Height (cm)

100

i  Quantitative, continuous ii  Quantitative, continuous 3 minutes The relationship is negative; as time increases the size of the ice block decreases. It is moderately strong, but not linear. 8 D 9  D 10  C 11 a Positive, perfectly linear b No relationship c Negative, moderate d Positive, strong e No relationship f Positive, moderate g Negative, perfectly linear h Negative, moderate i Negative, weak j Positive, moderate k Positive, moderate l Negative, moderate m Negative, strong n Positive, weak o Positive, moderate Problem solving 1 a Number of bags sold b Does the number of bags sold depend on the price? c Number of bags sold

e f g

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Mass (kg)

Mass (g)

8 a

31 32 33 34 35 36 37 38 39 40

Weeks

Answers

857

e f g h

i  3.7 kg ii  4 kg 1 kg 36 weeks During weeks 36 to 40 of the gestation period, the birth mass increased about 0.3  kg per week. This is supported by readings from the line of best fit. 4 a The test result is the dependent variable, while the number of questions is the independent variable. b Does completing more of these revision questions contribute to a higher test mark? c

Test result

100 90 80 70 60 50 40 30 20 10 0

5 10 15 20 25 30 35 40 45 50 55 60

Number of questions

Length (cm)

d The relationship is positive, and moderately strong. e There is evidence to suggest that completing more revision questions is beneficial, since the relationship is positive and moderately strong. L 5 a 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20

0 1 2 3 4 5 6 7 8 9 1011121314151617181920 n

Week

b 25  cm, 27  cm, 29  cm, 31  cm, 33  cm, 36  cm, 37  cm, 39  cm, 40  cm c 42  cm, 43  cm, 44  cm d The predictions for part b are quite reliable, as they have been made within the limits of the data. The predictions in part c for the 3 weeks beyond the upper limit of the data would not be considered reliable. Chapter 15

Statistics in the media Are you ready? 1 a Suitable b Not suitable (irrelevant) c Suitable

2 Junior school:

1 2

Middle school:

13 43

17

Senior school: 86

858

Answers

3 a Numerical, continuous b Categorical, nominal c Categorical, ordinal 4 a Most popular: cartoons least popular: documentaries and lifestyle programs b 50 c 40 5 a Number of kilograms: independent total cost: dependent b Temperature: independent number of swimmers: dependent c Age: independent height: dependent Exercise 15A — Populations and samples 1 a When was it first put into the machine? How old was the battery before being purchased? How frequently has the computer been used on battery? b Can’t always see if a residence has a dog; A census is very time-consuming; Perhaps could approach council for dog registrations. c This number is never constant with ongoing purchases, and continuously replenishing stock. d Would have to sample in this case as a census would involve opening every packet. 2 These answers will vary with the samples chosen. 3 a Census. The airline must have a record of every passenger on every flight. b Survey. It would be impossible to interview everyone. c Survey. A census would involve opening every bottle. d Census. The instructor must have an accurate record of each learner driver’s progress. 4 a Survey b Survey c Census d Survey 5 a About 25 b Drawing numbers from a hat, using a calculator, ….. 6 a The council is probably hoping it is a census, but it will probably be a survey because not all those over 10 will respond. b Residents may not all have internet access. Only those who are highly motivated are likely to respond. 7 The sample could have been biased. The questionnaire may have been unclear. 8 Sample size, randomness of sample 9 Answers will vary. Check with your teacher. 10 Populations growing very rapidly, large number of expatriate workers in China have a different background and forms need to be modified for them, people from Hong Kong working on mainland China, large migrant population in New Delhi, often migrants don’t have residency permits (so the truth of their answers is questionable), many people live in inaccessible areas, some families in China have more than 1 child and do not disclose this. 11 There is quite a variation in the frequency of particular numbers drawn. For example, the number 45 has not been drawn for 31 weeks, while most have been drawn within the last 10 weeks. In the long term, one should find the frequency of drawing each number is roughly the same. It may take a long time for this to happen, as only 8 numbers are drawn each week.

Exercise 15B — Primary and secondary data 1 These are simply examples of simulations which could be conducted. a Coin could be flipped (Heads represents ‘True’, while Tails represents ‘False’) b Coin could be flipped (Heads represents ‘red’, while Tails represents ‘black’) c Spinner with 4 equal sectors (each sector representing a different toy) d Roll a die (each face represents a particular person) e Spinner with 3 equal sectors (each one representing a particular meal) f Spinner with 5 equal sectors (each one representing a particular destination) g Spinner with 5 sectors, one which will have an angle size of 120è, while the other 4 each have an angle size of 60è (each one representing a particular fast food) 2 Answers will vary, however some possible suggestions include: Which students have internet access at home? Do the students need access at night? What hours would be suitable? How many would make use of this facility? 3 Answers will vary. Check with your teacher. 4 Answers will vary, however some possible suggestions include: a Census, survey, questionnaire, interview, observation, experiment, on-line response,  .  .  . b   i Measurement ii Observation iii Newspaper recordings iv Survey    5–7 Student’s own response 8 The claim is false. It is not a logical deduction. 9 Student’s own response 10 Student’s own response

11 Sealy Posturepremier 40% off ( 1000 ì 100%), 2499 1600 Sealy Posturepedic 41% off ( 3899 ì 100%), 800 Sleepmaker Casablanca 40% off (1999 ì 100%), 1800 Sleepmaker Umbria 42% off ( 4299 ì 100%).

There is at least 40% off these beds.

Mean salaries

3 a Mean = $215  000, median = $170  000, mode = $150  000. The median best represents these land prices. The mean is inflated by one large score, and the mode is the lowest price. b Range = $500  000, interquartile range = $30  000. The interquartile range is the better measure of spread. c 150000

300000 450000 Price

600000

This dot plot shows how 9 of the scores are grouped close together, while the score of $650 000 is an outlier. d The agent is quoting the modal price, which is the lowest price. This is not a true reflection of the average price of these blocks of land. 4 a True. Mean = 1.82 m, lower quartile = 1.765 m, median = 1.83 m b True. This is the definition of interquartile range. c Players with heights 1.83 m, 1.83 m, 1.88 m, 1.88 m, 1.88 m 5 a 7.1 b 7 c 7 d The mode has the most meaning as this size sells the most. 6 Check with your teacher. Answers depend on groupings used. 7 Player B appears to be the better player if the mean result is used. However, Player A is the moreconsistent player. 8 a The statement is true, but misleading as most of the employees earn $18 000. b The median and modal salary is $18 000 and only 15 out of 80 (less than 20%) earn more than the mean. 9 Points which could be mentioned. ■■ 10.1% is only just ‘double digit’ growth. ■■ 2006–08 showed mid to low 20% growth. Growth has been declining since 2008. ■■ Share price has rebounded, but not to its previous high. ■■ Share price scale is not consistent. Most increments are 30c, except for $27.70 to $28.10 (40c increment). Note also the figure of 20.80 — probably a typo instead of 26.80. 10 Shorten the y-axis and expand the x-axis. US c Aussie dollar

90 c 80 c 0

13 July

Answers 15A ➜ 15C

Exercise 15C — Evaluating inquiry methods and statistical reports 1 a Primary. There is probably no secondary data available. b, c  Answers will vary. Check with your teacher. 2 Company profits



13 September Time

Answers

859

Chapter review Fluency 1 a You would need to open every can to determine this. b Fish are continuously dying, being born, being caught. c Approaching work places and public transport offices 2 a 50.5 b, c, d   Answers will vary. 3 a Survey b Census c Survey 4 Use a spinner of 3 equal sectors, each sector having an angle size of 120° and representing a particular colour. Twirl the spinner until a green/green combination has been obtained. This is defined as one experiment. Count the number of trials required for this experiment. Repeat this procedure a number of times and determine an average. 5 D 6 Check with your teacher. a This graph should look relatively flat, with little decline in the Years 11 and 12 region. b This graph should show a sharp decline in the Years 11 and 12 region. 7 a Boys: median = 26; girls: median = 23.5 b Boys: range = 32; girls: range = 53 c Both sets have similar medians, but the girls have a greater range of absenteeism than the boys. 8 a The sample is an appropriate size as 900 = 30. b Key: 16 | 1 = 1.61 Leaf Stem Leaf Girls Boys 1256788 9 9 7 15 4467899 9 8 6 6 5 5 4 0 16 0 4 4 2 1 17 c The boys are generally better than the girls, with the mean of the boys being 1.66 m and that of the girls being 1.62 m. The five-number summaries are: Boys:  1.57 m, 1.6 m, 1.66 m, 1.71 m, 1.74 m Girls:  1.51 m, 1.56 m, 1.64 m, 1.68 m, 1.7 m 9 a Ford: median = 15; Holden: median = 16 b Ford: range = 26; Holden: range = 32 c Ford: IQR = 14; Holden: IQR = 13.5 d Ford Holden 0 5 10 15 20 2530 35 40 Number of vehicles sold

10 a Brisbane Lions b Brisbane Lions: range = 65; Sydney Swans: range = 55 c Brisbane Lions: IQR = 40; Sydney Swans: IQR = 35 860

Answers

11 a

Girls Boys 1.4 1.5 1.6 1.7 1.8 1.9 Height (m)

b Boys: median = 1.62 m; girls: median = 1.62 m c Boys: range = 0.36 m; girls: range = 0.23 m d Boys: IQR = 0.14 m; girls: IQR = 0.11 m e Although boys and girls have the same median height, the spread of heights is greater among boys as shown by the greater range and interquartile range. 12 a Summer: range = 23; winter: range = 32 b Summer: IQR = 13; winter: IQR = 11 c There are generally more cold drinks sold in summer as shown by the higher median. The spread of data is similar as shown by the IQR although the range in winter is greater. 13 A Problem solving 1 a Mean = 32.03; median = 29.5 b Class interval Frequency 0–9

 2

10–19

 7

20–29

 6

30–39

 6

40–49

 3

50–59

 3

60–69

 3

Total

30

c Mean = 31.83 d Cumulative frequency

Exercise 15D — Statistical investigations The questions in this exercise relate to student investigations, so there will be a variety of answers. 2 There have been 27 Prime Ministers of Australia since 1901 until this day. There have been 42 elections. 10 Prime Ministers have been defeated at a general election. There have been 21 changes of Prime Minister without an election. The average length these Prime Ministers served in office is (This changes daily).

30 25 20 15 10 5 0

10 20 30 40 50 60 70

Age

e Median = 30 f Estimates from parts c and e were fairly accurate. g Yes, they were fairly close to the mean and median of the raw data. 2 Year 8: mean = 26.83, median = 27, range = 39, IQR = 19 Year 10: mean = 40.7, median = 39.5, range = 46, IQR = 20 The typing speed of Year 10 students is about 13 to 14 wpm faster than that of Year 8 students. The spread of data in Year 8 is slightly less than the spread in Year 10. 3 Hane and Roarne had a higher median and a lower spread, so they appear to have performed better. 4 They are all made in Australia and have comparable fat and saturated fat contents. The Byron Bay Chilli corn chips have a much lower salt content than the other three varieties. The verdict comments require a mention. 5 Student’s plan for an investigation.

Chapter 16

Financial maths Are you ready? 1 a 0.24 b 2 a $2100 b 3 a $30 b 4 a 20% b 5 a $2070 b

c c c c c

0.175 $10 640 $43.20 5% $442.50

Exercise 16A — Purchasing goods 1 $2400 2 a $1800 b $900 d $720 3 a $1500 b $125 d $1080 4 a $18.75 b $55.00 d $125.33 e $99.58 5 $10 6 a $700 b $10.50 d $3.16 e $913.66 7 a $52.50 b $50.79 d $14.87 8 a Month

Balance owing

January

$7500.00

February

d d d d d

0.03 $22 800 $1.10 20% $784

0.0975 $1968.75 $81.25 32.4% $5437.50

c $1920 c $281.25 c $41.25 c $710.50 c $35.92

Payment

Closing balance

$112.50

$1000.00

$6612.50

$6612.50

$99.19

$1000.00

$5711.69

March

$5711.69

$85.68

$1000.00

$4797.37

April

$4797.37

$71.96

$1000.00

$3869.33

May

$3869.33

$58.04

$1000.00

$2927.37

June

$2927.37

$43.91

$1000.00

$1971.28

July

$1971.28

$29.57

$1000.00

$1000.85

August

$1000.85

$15.01

$1015.86

$0

Interest

b $515.86 c $8015.86 9 Payment Immediate Immediate option payment possession Cash

✓

Possible Possible extra price cost negotiation

✓

✓

Lay-by

✓

Payment option

✓

✓ Extra cost

Payment

Possession

Cash

immediate

immediate

nil

Price

Lay-by

intervals

delayed

limited

–

Credit card

delayed

immediate

possible

–

negotiable

10 S1: Credit card — payment is delayed, but possession is immediate S2: Lay-by, or cash if she has savings, would like to negotiate a lower price and has somewhere to store it.

Exercise 16C — Successive discounts 1 a 16.7% b 10.2% c 43.7% d 15.4% e 7.6% 2 23.5% 3 C 4 a 32% b 19.04% c 16.75% d 57.5% 5 19% 6 a $3040 b 24% 7 a $90 b $85.50 c $14.50 d 14.5% 8 a $1840 b $1748 c $252 d 12.6% 9 a $212.50 b $201.88 c $48.12 d $200, no e 19.25% f Yes g 28% 10 a $18  168.75 b 27.325% 11 Yes. Both lead to a single discount of 14.69% 12 Single discount = 1 - (1 - a) ì (1 - b), where a and b are successive discounts (as decimals). Exercise 16D — Compound interest 1 a $3244.80 b $10  939.56 c $24  617.98 d $14  678.02 e $14  838.45 f $129  394.77 2 a $932.52 b $10  650.81 c $202  760.57 d $25  816.04 e $3  145  511.41 3 $8784.60 4 $3376.26 5 a $2837.04 b $837.04 6 $17  240.15 7 $605.42 8 $18  503.86 9 a 0.5833% b $42  891.60 10 B 11 B 12 C 13 C 14 a 0.0247% b $131  319.80 c $11  319.80 d $519.80

Answers

Answers 15D ➜ 16D

Credit card

Exercise 16B — Buying on terms 1 a   i $3960 ii $3720 iii $3950 b The best deal is the one with the lowest cost — 20% deposit and weekly payments of $20 over 3 years. 2 a $131.25 b $55.38 c $144.44 3 a $13  600, $283.33 b $40  000, $666.67 c $5006.25, $278.13 d $80  000, $666.67 e $488  000, $2033.33 4 $5409.76 5 $4530.08 6 a $1260 b $19  504.80 c $20  764.80 d $8164.80 7 a $8409.96 b $2609.96 c $869.99 d 15% 8 a $2375 b $7125 c $2565 d $9690 e $269.17 9 a $2000 b $4000 c $540 d $4540 e $252.22 10 a $226.80 b $141.90 c $360.94 d $87.11 e $85.13 11 a $4600 b $1656 c $130.33 d $7256 12 C 13 E. This option has the lowest interest rate and time frame when compared to all others. 14 The larger the deposit the smaller the loan and hence the interest charged. Loans generally offer a lower rate than buying on terms.

861

  ii  $17  786.61 15 a   i  $17  745.95   iii  $17  807.67 iv  $17  821.99 b The interest added to the principal also earns interest. 16 a $7920 b David’s investment = $8108.46 c Because David’s interest is compounded, the interest is added to the principal each quarter and earns interest itself. 17 a $3750 interest b i  $3820.32 interest ii  $3881.05 iii  $3912.36 c Compound quarterly gives the best return. d If we assume that interest is compounded annually, an equivalent return of R = 7% would be achieved. e i  Yes ii  No Exercise 16E — Depreciation 1 a $14 936.14 b $3584.59 2 $17 694.72 3 a $19  118.26 b $19  281.74 4 a $3846.93 b $6353.07 5 a $7216.02 b $45 283.98 6 a $1.8 million b $569  531.25 7 B 8 A 9 C 10 B 11 A 12 10 years 13 a $385 million b 16 years 14 a 27% n b A = P (1 − R) A = (1 − R) n P A n = (1 − R) P

c $6596.65 c 38% c $135  152.44

A P 15 a Approx 43% b Trial and error: can be time consuming, answer is often an estimate; algebraic solution: correct answer calculated immediately from equation R = 1− n

Exercise 16F — Loan repayments 1 $4500 2 a $8000 b $4950 c d $864 e $420 3 a $5760 b $17  760 c 4 $2422 5 $9264 6 a $12  000 b $32  000 c $8000 d $4966.87 7 a $2453 b $93 8 a $6004.80 b $2001.60 c 9 a 5.4% b 4.6% 10 Loan B better (total savings $1053.50) 11 Actual savings $355.15 Chapter review Fluency 1 $1000 2 C 3 16.875% 5 $54 6 $7819 7 a $640 b $5760 c $2764.80 e $177.60

862

Answers

8 a $67  000 b $27  000 c $5400 d 15% p.a. 9 E 10 a $261 b $221.85 c $68.15 d $23.5% 11 $15  746.40 12 a $25  808.37 b $26  723.16 c $27  211.79 d $27  550.17 13 E 14 B 15 $24  403.80 Problem solving 1 a $22  774.65 b 13% 2 $426 3 a 3.95% p.a. flat rate b 3.97% p.a. flat rate c 3.96% p.a. flat rate Neither is correct. The best option is to choose 3.895% p.a. compounding monthly.

CHAPTER 17

Problem solving II 1 4 2 a 4 am, 7 January b 1 am, 7 January (Perth is 3 hours behind Sydney during daylight saving time.) 3 a No. Greg’s first choice of a number is independent of his second choice of a number. The ten numbers will always be in the hat on the first draw, thus he has an equal chance to pick any of them. b Greg’s second choice is dependent upon his first choice. When Greg chooses a number in the first draw and does not replace that number, then he changes the sample space and the probability for the second draw. 4 a $1175 b $3825 c 25% discount gives a final price of $3750. The customer would be $75 better off. 5 16 ì $5 notes and 11 ì $10 notes y 6 a 2 1 0.5 0

$1875 $4440

4 A d $8524.80

1

2

3 x

b y becomes smaller and approaches 0, but never actually reaches 0. c y approaches infinity as x becomes smaller. 7 a

8%

y = 4-x

1 5

b

4 5

c

1 25

8 a Since the interest rate is lower for Loan 2 than for Loan 1, Thomas should choose Loan 2 if he decides to pay the loan off at the end of the first, second or third year. b Loan 1 at term amounts to $9444.63. Loan 2 at the end of 4 years amounts to $9523.42. Thomas should choose Loan 1. c Thomas should choose Loan 1. At the end of its term (3 years), it amounts to less than Loan 2 at 4 years, 1 year before its term is finished. d Thomas may not have the money to pay off Loan 1 in 3 years. He may need the extra 2 years to accumulate his funds.

d

77

9 18 10

27 52

11 72 12 $20  960.94 13 a 1 - a b 1 c 0 14 a Future population in n years = 350(1 + 0.1)n. b 12 years c 19 years d Lance has assumed that every 19 years there will be approximately 2140 additional people. e Lance has assumed that the growth is linear, whereas it is actually exponential. Drawing a graph would help him see the growth. 15 a $100 b 14.29% 16 2 17 a x + y í 20 and 3x + 8y Ç 110 b No more than 30 units of dye c If y = 10, the first equation becomes x í 10 whereas the second equation becomes x Ç 10. This means that the only possible value for x here is equal to 10. 1

18 4 4

19 7

Midpoint x

Frequency f

0–9

  4.5

0

fx    0

10–19

14.5

4

  58

20–29

24.5

7

171.5

30–39

34.5

5

172.5

40–49

44.5

4

178

50–59

54.5

4

218

60–69

64.5

6

387

Anthea: Mean =

∑ fx 1185 = = 39.5 ∑f 30

∑ x 1146 = = 38.2 n 30 e Anthea used the midpoints of the class intervals, whereas Elena used the exact values. 27 Prove P(AÅ) ì P(B) = P(AÅ ¶ B) 28 a $41  051 b Approx $4300 29 a i  4x + 6y = 12 ii  4x + 3y = 6 iii  2x + 6y = 6 iv  2x + 3y = 12 Elena: Mean =

y

b i 

20 a $1500 b $1000 c Since the depreciation of 40% is on a lower value each year, the amount Jan can deduct from her taxable income decreases every year. 21 a 0.27 m b Teacher to check. 22

Class interval

4

2x +3 4x 2 y= +6 6 y= 12 0 -2 2

-4

17 35

4 x

-2

20–29

24.5

7

30–39

34.5

5

40–49

44.5

4

50–59

54.5

4

60–69

64.5

6

2

-4

0

-2

2x

+6

y=

6

4 x

2

-2 -4

iii 

2x

2x +

-4

Answers 16E ➜ 16F

4

4

6

0

14.5

y

ii 

y=

  4.5



+3

0–9 10–19

-4

4x

23 a p = 250n + 15, n is independent variable and p dependent variable — number of people depend on number of screens. p b t = 100 c Integers d At least 13 e 3 24 22% 25 a Stephanie can only work out which class interval her test result is in — the 65% to 74% interval. b The median is the 50th percentile which corresponds to the 55% to 64% interval. c No. d 85th — the same as Stephanie. 26 a Range of females = 45 years; range of males = 53 years b Females: 64 years; males: 22 years c Class Class interval mid-point Frequency

y +3

y=

6y =

6

-2

4 6

2

0

2

4 x

-2 -4

Answers

863

2x

38 a

y +3

y=

4

2x

6

+3

2

-4

0

-2

y=

4 x

2

-2 -4

c i The two equations are the same, so the graphs lie on top of each other. ii The gradient is doubled, the y-intercept is unchanged, the x-intercept is halved. iii The gradient is halved, the y-intercept is halved, the x-intercept is unchanged. iv The gradient is unchanged, the y-intercept is doubled, the x-intercept is doubled. These are parallel lines. 30 a

3 5

b

8 15

31 a Method 2 b Difference of $3530 in favour of Method 2. 32 Annie is 18 and Barbara is 7. 33 a $5000 b $1000, interest earned per annum. c A = 5000 + 1000n d R = 10.5%, This is the percentage interest rate per annum. So, R = 10.41% p.a. e Check with your teacher. f Rosetta had a higher return in the first 13 years then Theo had the higher return after that. 34 a Xmin QL Median QU Xmax Test A

20

40

50

70

80

Test B

10

50

60

80

90

b The interquartile range is the same for both tests (IQR = 30). This indicates that the spread of the results across the middle group of the class for both tests is the same. c Based on the median score, the students appear to have done better in Test B. The middle groups stayed the same, the strong maths students did better, but the struggling students did worse. d Test B may have been an easier test than Test A. e Students may have found topic B easier to understand. 35

16 19 c

b   d   36 (k + am)  1 + 1−  100   100 

37

e

y Shrub

8 6 4 Tree 2 0

864

Stump

2

Answers

4

6

8

10 12 14 x

W

12

w

w

Ww

ww

w

Ww

ww

b 50% 39 a $45 058.50 b $455 per year 40 a Mean = 5.4; median = 5.5; mode = 6 The median is best because it allows for the range of values and is also between the mean and mode. b i  0.5 ii  2 or 3 c i  4 more games scoring at least 8 runs in each ii  2  3  4  5  6  6  8  8  8  8  8  9 d Not very likely 41 a 0.24 b 0.0462 c 0.0231 d Potential Profit/ Probability wins Outlay Winnings Loss 0.24

0.24 ì 20 $20 = 4.8

2.50 ì 4.8 $8 loss = $12

0.0462

0.924 ö 1 $20

$5

$15 Loss

0.0231

0.462

$0

$20 Loss

$20

e 100 games — potentially 30 wins so get $85, but paid $100 to play, so still losing. f People look at short term gains, but fail to consider the long term maths calculations. 42 The numbers are 7 and 10. 43 a Yes, because the relationship involves a variable as an exponent. b 20  000 km2 c S = 11  975 km2 y d 20000

Surface area (km2)

iv 

15000 10000 5000

0

20

40

60 Years

80

100 x

e In 100 years, S = 118 km2 f No this is not a realistic model as is it does not take into account changes to climate, rain, runoff from mountains, glaciers etc. 44 a About 8 apprentices can work for the same amount of money. b Experienced bakers $1000 each

Apprentices 30 h ì $16 = $480

Total cost

4 ($4000)

0

$4000

3 ($3000)

2 ($480) = $960

$3960

2 ($2000)

4 ($480) = $1920

$3920

1 ($1000)

6 ($480) = $2880

$3880

0

8 ($480) = $3840

$3840

(22)

(18)

2

10

6

4

b i 

6

(17)

2 7 18

ii  35 c

11 35

50 a $900 b $16.50 c $916.50 d $5.80 e $322.30 f $1222.30 51 a Class interval

b, c

5−9

 8

10−14

 5

15−19

 4

20–24

 1 25

y

Frequency

8 6 4 2

5 10 15 20 25 x Number of computers used

100 80 60 40 20 1

2 3 4 Time (days)

Males Females

5 10 15 20 25 x Number of computers used

0

b

Males

Females

Mean

28.2

31.1

Range

70

57

IQR

18

22

c There is one outlier — a male aged 78. d Typically males seem to enter hospital for the first time at a younger age than females. 54 a $9920 b Nathan will have $9993.03 at maturity. c Because Nathan’s interest is compounded, the interest is added to the principal each quarter and earns interest itself. 3 55 a

Cumulative frequency

3 5

25 15

2 5

5 5 10 15 20 25 x Number of computers used

e 10 students

3 5

Black

2 5

Red

3 5

Black

2 5

Red

5

2 3 5 5

Black

20 10

x

10 20 30 40 50 60 70 80 Age

y

0

5 x

Red

2 5

3 5

2 5

3 5 2 5

Black Red

Black Red Black Red Black Red

Answers 17 ➜ 17

d

20

Frequency  7

0

40

d 7.85 mg/L 53 a

0−4

0

60

0

Total



80

0

Concentration mg/L

Subaru

y 100

g 50th percentile is about 8.5 computers used, 30th percentile is 5 computers used. This means that 50% of the data lies below 8.5 and 30% of the data lies below 5. 52 a Teacher to check. b C = 100 ì 0.120.24t y c

4

1 2

f

Percentage cumulative frequency

45 a 12.2 m b 10.4è c 12.72 m d 10.3 km/h 46 a 8.25 hours b Shift 2 offers $53.38 more. 47 149 cm and 171 cm. The average height of the two students must equal 160 cm. 48 100 mini-laptops 49 a Toyota Nissan

b 0.216 c 0.352 Answers

865

67 a P = 2500, A = 1.065 b 6.5% 68 a 0.000 495 b 0.001 981 69 a Frequency Interval (  f ) Midpoint ì (  f )

56 a $2090 b 104.5% c $2184.05 A d Investment value ($)

12000

40–49

 1

44.5 ì 1 = 44.5

8000

50–59

 1

54.5 ì 1 = 54.5

6000

60–69

 1

64.5 ì 1 = 64.5

4000

70–79

 2

74.5 ì 2 = 149

2000

80–89

 4

84.5 ì 4 = 338

90–99

 4

94.5 ì 4 = 378

100–109

 8

104.5 ì 8 = 836

110–119

 6

114.5 ì 6 = 687

120–129

 8

124.5 ì 8 = 996

130–139

 2

134.5 ì 2 = 269

140–149

 2

144.5 ì 2 = 289

10000

10 20 30 40 x Time (years)

0

e Approximately 36.5 years 57 a 3 b 6 n( n −1) c 2 58 a $0.165 b 217 cups 59 200 60 12 years 61 0.47 62 a 310 b 101 c 23 weeks d No 63 a Mean = 2.17, median = 2 b Mean = 3.54, median = 2 c The median relies on the middle value of the data and won’t change much if an extra value is added. The mean however has increased because this large value will change the average of the numbers. The mean is used as a measure of central tendency if there are no outliers or if the data are symmetrical. The median is used as a measure of central tendency if there are outliers or the data are skewed. 64 a $17 176.86 b $5176.86 c $357.85 65 0.8 66 a See the table at the bottom of the page*. b

150–159

 0

154.5 ì 0 = 0

160–169

 1

164.5 ì 1 = 164.5

TOTAL

40

4270

b 106.75 c 107.15 d The differences in this case were minimal; however, the grouped data mean is not based on the actual data but on the frequency in each interval and the interval midpoint. It is unlikely to yield an identical value to the actual mean. The spread of the scores within the class interval has a great effect on the grouped data mean. 70 a October 1: $89.10, October 8: $80.19, October 12: $84.20, October 15: $75.78 b 23.5% 71 a It is 3 times more likely that the spinner will land on A. Therefore I would not play this game as I should be winning $9 not $8. b I would play this game as a fair game would only give me $15. To get $18 would be in my favour. y 72 a 500

100000

400

80000

Capacity (GB)

Prize money ($)

P 120000

60000 40000 20000 0

2

4 6 8 10 12 14 Number of people

n

300 200 100

120 000 n d Inverse variation; k = 120  000 e $6000 f 80 people

c P =

0 1/1/95

1/1/01

1/1/07

1/1/10 x

Date

b Approx. 5 GB c 8192 GB

66 a *

866

n

1

2

3

4

5

6

8

10

12

P

120  000

60  000

40  000

30  000

24  000

20  000

15  000

12  000

10  000

Answers

19

6 4

b

0

1

5

7

17 60

b

1 3

c

7 12

d

1 2

e

7 10

f

2 5

g

1 12

h

59 60

77 a $20  400 1 b 79 weeks (1 2 years) c $340  000 d $306  000 e $2295 f $2406.67 g $305  888.33 h Payments will be more than they are able to afford. i Find a less-expensive house, save more money then pay more off the principal each month, find a loan with a lower interest rate, save for a larger deposit. 78 a Burritos Fajitas 15 11

12 7

10 Tacos

b 9

30 20

18

a

13

40

10 4 WD

9

50

Cumulative frequency

73 $1060.26, $2003.30, $7585.60 74 a 0.053 b 0.103 75 66.7% by volume and 81.1% by mass 76 Tovota LPG System

23

c

8

10 12 14 16 Years of education

18

Parents of year 10 students

8

10 12 14 16 Years of education

18

x

2 a 8 c 83 a c

9.10% 5.76% $18  000 $9600

b 4.45%

84 a

1 3

b

b $48  000 d $8117.09 1 2

85 a 171.6 cm b 171 cm c At the top end of the heights, there is a possible outlier of 189 cm. Removal of this value results in a mean of 171 cm and a median of 171 cm. The mean is reduced slightly, while the median is unchanged.   At the lower end of the heights, there are possible outliers of 159 cm and 160 cm. Removal of these values results in a mean of 172.5 cm and a median of 171.5 cm. The mean and median are increased slightly.   Removal of both the upper and lower outliers results in a mean of 171.9 cm and a median of 171 cm. The mean is increased slightly, while the median is unchanged. 86 Option 2 is less expensive, despite having a higher annual interest rate, because it is paid off in only 10 months instead of 2 years. The TV would cost $217 with option 1 and $212 with option 2. 87 a i 35 s ii 29.5 s iii 33.05 s iv 60 s v 21 s vi 39 s vii 18 s b 21 29.5 39 15 20 25 30 35 40 45 50 55 60 65 70 75

t

c i 25% ii 50% iii 75% d Categorical e 35% f Pictogram, pie chart or bar chart. 88 a 0.000  039  6 b About once in 12.6 years Answers

Answers 17 ➜ 17

c 45 M 79 a Prove = 0.5 = (1.0122)-57.3 M0 b 21.54% 80 a $0 2 b This game has a total expected loss of $ 36 , so the game is not fair and is biased against the player. 81 a Years of Cumulative education Frequency frequency  9  2  2 10  4  6 11  8 14 12 17 31 13  6 37 14  5 42 15  4 46 16  3 49 17  1 50 Total 50

0

867

Number of bacteria

89 a

98 99 100 101 102 103 104 105 106

40 kg 40 steps Student’s own work a C = 1000 + 15s b R = 25s c 100 d 350 slippers 0.61 a $397.50 b $2252.50 c $2929.50 d $279.50 e 6.2% p.a. 10.5 m 1350 watts/m2 1 a P(A ¶ B) = 12 ò 0 so A and B are not mutually exclusive. 1 b A and B are not independent because P(A ¶ B) = 12

800 600 400 200 0

5

10 15 20 Time

y = 200 (1.086)t y = 500 (1.031)t

b Approx 17.5 days c 17.63 days 90 a 0.65 0.75

0.25

W

T 0.35

WÅ

0.40

W

0.60

WÅ

TÅ



b 0.5875 c

8 47

91 a 82.73 km/h b 30 cars c  i  $2 607 272.73 ii  About 545 92 $461.96 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 93 a 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 b 0.018  144 (about 1.8%) 94 a i  R = 12.5t, t Ç 3 ii  R = 14t - 4.50, t > 3 b 5.5 hours c Fees Fees Difference Hours OLD NEW ($)

5

and P(A) ì P(B) = 108.

07 30.16% 1 108 25 mL 109 2 h 54 min 110 Student’s own work 111 P($5) = 0.12503, P($10) = 0.039  063, P($15) = 0.12502 ì 0.0156, P($20) = 0.039  062 ì 0.0156. The probabilities of obtaining any of the required scores to receive a payout are very low. Bianca and Hannah would be unlikely to have to pay out any money, so would collect $300 for their fundraising. 112 a 3 b 4.5% c $4130.10 113 6 games 114 For each 30-sec block, the lowest-cost calls are shown in blue, while the highest-cost calls are shown in red.

1

12.50

12.50

   0

2

25.00

25.00

   0

3

37.50

37.50

   0

Call length

4

51.50

70.00

+18.50

30 sec

5

65.50

70.00

  +4.50

40 + 45 = 85

35 + 92 = 127

36 + 46 = 82

6

79.50

70.00

  -9.50

1 min

40 + 45 ì 2 = 130

35 + 92 = 127

36 + 46 ì 2 = 128

1 min 30 sec

40 + 45 ì 3 = 175

35 + 92 ì 2 = 219

36 + 46 ì 3 = 174

2 min

40 + 45 ì 4 = 220

35 + 92 ì 2 = 219

36 + 46 ì 4 = 220

2 min 30 sec

40 + 45 ì 5 = 265

35 + 92 ì 3 = 311

36 + 46 ì 5 = 266

3 min

40 + 45 ì 6 = 310

35 + 92 ì 3 = 311

36 + 46 ì 6 = 312

95 a i  0.25 ii  0.10 iii  0.375 b 30 students 96 a Based on the comparison between Class A’s IQR (16.5) and Class B’s IQR (32.5), Ms Vinculum was correct in her statement. b No 97 a Reducing value allows you to claim $300, $210, $147 over the 3 years for a total of $657. Straight line allows you to claim $200 each year over 5 years for a total of $1000. Although the reducing value depreciation is greater in the years 1 and 2, the sum over the life of the item is greater in the straight line case. b Reducing value: $1000, $800, $640, $512, $409.60 for a total of $3361.60 Straight line allows $835 per year over 6 years for a total of $5000 (actually $5010). In this case the reducing value method is only better in the 1st year. 868



Answers

15 1 116 117 118 119 120

Andy (cents)

Bill (cents)

Cam (cents)

If calls last a full number of minutes, as opposed to half-minute calls, Bill’s plan seems to be a good one. The 30-sec plans seem to be best for calls lasting just over the minute. 58 $120 70 7 $13 profit July

121 a The draw for 9 players could look like this. Round 1

Round 2

X

Round 3

27 320 1 128 12 129 37 130 Looking back 5 generations, there would be 8 ancestors for the male bee.

Round 4

X

X

X

X

Bye

X

5th generation

X

X X

X

Bye

X

4th generation

Champion X X

X

X Bye

Deposit

Withdrawal

Balance $4200

F

2nd generation

M

M

M

F F

F F

F

Chapter 18

Real numbers

$4300

Are you ready? 1 a, b and d

7/7

$500

$4800

2 a 4 3

$4075

3 a 3 3 − 3 6

b 28 2 − 39 5

4 a 70

c

$725   $85

$4160 $4160

b Total interest earned is $13.24. Number of days 2

4

14

7

4

Interest calculation

Balance

4200 × 3.5 ×

$4200

2 365

$0.81

4 365

$1.65

14 365

$6.44

7 365

$2.74

4 365

$1.60

100 4300 × 3.5 ×

$4300

100 4800 × 3.5 ×

$4800

100 4075 × 3.5 ×

$4075

100 4160 × 3.5 ×

$4160

100

Interest earned

23 24 votes 1 124 12 125 16 126 The faces adjacent to face number 6 would be 1, 4, 10, 11 and 5. 1

5

4

6 11

10

b 7 2

b 24 2

c 10 3

3

5 a 49 c 244.140  625

b 81 d 0.0081

6 a x10

c

b 20y11

d 18 2

d

2 2

4a2

d 4m8 b4 18A — Number classification review b Q c Q d I e I g Q h I i Q j Q l Q m I n Q o I q Q r I s I t I v I w I x Q y I b Q c Q d Q e Q g I h Q i I j Undefined l I m I n Q o Q q I r I s Q t Q v Q w Q x I y Q

Exercise 1 a Q f Q k Q p Q u Q 2 a Q f I k I p Q u I 3 B 4 D 5 C 6 C Exercise 18B — Surds 1 b d f g h i l m o q r s t w z 2 A 3 D 4 B 5 C 6 Any perfect square 7 m = 4 8 Check with your teacher.

Exercise 18C — Operations with surds

1 a 2 3

b 2 6

c 3 3

d 5 5

e 3 6

f 4 7

g 2 17

h 6 5

i 2 22

j 9 2

k 7 5

l 8 7 Answers

Answers 18A ➜ 18C

31/7

3rd generation

F

M

$100

31/7

27/7

M

F F

3/7

28/7

20/7

M

Male bee

21/7

6/7

F

1st generation

1/7

2/7

F

X

b 4 c 3 122 a Date

Date

F

X

X X

M

869

2 a 4 2

b 24 10

c 36 5

d 21 6

e −30 3

f −28 5

i

j 2 3

g 64 3 1 k 15 3

h 2 2 3 l 7 2

2

3 a 4a

b 6a 2

d 13a

2

e 13ab 2ab

2

3 2

g 5 x y

h 20 xy 5 x

5

j 18c d

3 4

k

5cd

2

f 2ab 17ab i 54 c3d 2 2cd 5 5

l 7e f

22ef

b 8 3

c 15 5 + 5 3

d 4 11

e 13 2

f −3 6

g 17 3 − 18 7

h 8 x + 3 y

5 a 10( 2 − 3 )

b 5( 5 + 6 )

c 7 3

d 4 5

e 14 3 + 3 2

f 3 6 + 6 3

g 15 10 − 10 15 + 10

h −8 11 + 22

i 12 30 − 16 15 7 2+2 3 k 2

j 12 ab + 7 3ab

6 a 31 a − 6 2a

b 52 a − 29 3a

c 6 6ab

d 32a + 2 6a + 8a 2

g 3a a + a

i 4 ab ab + 3a 2 b b 2 3

k −6ab 2a + 4 a b 7 a 14

b

e 3 7 i 120 m

f 27

g 10 33

j 120 3

k 2 6

h 180 5 2 l 2 3

n x 2 y y

o 3a 4 b 2 2ab

8 a 2 e 18

b 5 f 80

p 6a 5b 2 2b 9 r a 2 b 4 5ab 2 c 12 d 15 g 28 h 200

9 a 5

b 2

c

2 2

q 3 x y

10 xy

e

3 4

f

i 1

4 5

j 2 17

m 2 xy 3 y

n

5 2

6

g 2 3 k

8 21 49

n

8 105 7

o

10 3

11 a 2 + 2

x y

b

3 10 − 2 33 6

c

12 5 − 5 6 10

d

9 10 5

e

3 10 + 6 14 4

f

5 6 3

g

3 22 − 4 10 6

h

i

14 − 5 2 6

j

k

6 15 − 25 70

l

8 11 + 4 13 31

e 12 2 − 17 15 − 3 − 5 + 1 4

8 15 15

l

21 − 15 3 12 − 10 16 30 + 7 2 20

b

2 2+ 5 3

d

15 15 − 20 6 13

f

19 − 4 21 5

h

−6 + 6 2 + 10 − 2 5 2

4 10 + 15 − 4 6 − 3 29

Exercise 18D — Fractional indices 1 a 4 b 5 d 2 e 3 2 a 3 b 2 d 2.2 e 1.5 3 a 2.5 b 12.9 d 0.7 e 0.8

c f c f c f

4 a 7

b 2 3

c 6 2

d 4 2

e 3 3

f 100 10

5 a

1 52

b

1 10 2

9 5 1.4 1.3 13.6 0.9

1

c x 2

d 4

3

1

d m 2

e 2t 2

h 1

4

1

5

6 a 4 5

b 2 2

c a 6

l

2 3 4

x y

4 a 3

5 2 2

b

7 3 3

c

4 11 11

d

4 6 3

e

2 21 7

f

10 2

g

2 15 5

h

3 35 5

1

f 6 3

23

8

5

d x 20

e 10 m15

f 2b 7

20

9

7

g −4 y 9

h 0.02a 8

i 5 x 2

3

10 a

Answers

m

i

l −2a b d 10

2 6 5

5 7 14

g

c 4 3

42

k

a + 2 2a

j 3 ab (2a + 1) 3a

4 15 15

c

2 h (a + a) ab

3a

j

12 a 5 − 2

l 15 2

f 2

5 6 6

2ef

4 a 7 5

e a 2a

870

c 3a 10 b

i

7 a ab 2 8 17

c 6a 5 b 15 19 5 5

e x 6 y 6 z 6

4

5

b x 5 y 9 19

2

d 2m 28 n 5 2 9

f 8a 5 b 8 c

1

5

1

8 a 3 6

b 512

c 12 2

d

e

f

g

3 a7 3

2

1 20 x 2

1 3 n 3

h

9 a

5 7 x3y5

d

2 3 2 x 15 y 4

10 a

9 2 20

d

3 a 10

g

2 4 p5

11 a

1 1 a4b6

d

1 1 1 1 33 a 9 b 5 c 4

g

5 x4

i

7 4 a 45 b 15

b

11

c 7

c

e

1 m6

f 2 3 b 6

b e

1 1 1 x4 y3z 5

1

f

i

8 c 27

1

6 7 x5y4

d

= 0.1

e

1 8

g

1 25

= 0.04

h

1 10 000

2 a 0.167 d 0.001  37 g 0.003  91 3 a 0.40 d 4.0 g 11 4 a -0.33 d 0.063 g -1.7

b e h b e h b e h

0.143 0.004  63 0.001  60 2.5 0.11 4100 -0.20 -0.67 1.4

5 a

5 4

e 2 i

2 3

1

or 1 4

b

10 3

f 4 j

4 9

1

or 3 3

= 0.125

2 b3

8 7

3

1 8

= 0.125

f

1 9

= 0.1

g 8 k

10 11

h 10 l

2 11

d -10 h

16 121

log2 32 = 5 log6 36 = 2 log5 25 = 2 log5 125 = x logp 16 = 4

l log10 0.1 = -1

m log8 2 = 3

1

n log2 2 = -1

o loga 1 = 0

p log4 8 =

1 49 2

e 5

1

b 33 = 27 e

1 16 2

3 2

c 106 = 1  000  000

=4

f 4x = 64 1

=7

i 812 = 9

h 35 = x

f 7

l 64 3 = 4 d 5

g 0

h

1 2 1

i -1 j 1 k -2 l 3 6 a 0 b 1 c 2 d 3 e 4 f 5 7 a 0 and 1 b 3 and 4 c 1 and 2 d 4 and 5 e 2 and 3 f 4 and 5 8 a log10 g = k implies that g = 10k so g2 = (10k)2. That is, g2 = 102k; therefore, log10 g2 = 2k. 1 b logx y = 2 implies that y = x2, so x = y 2 and therefore 1 logy x = . 2 c The equivalent exponential statement is x = 4y, and we know that 4y is greater than zero for all values of y. Therefore, x is a positive number.

c 0.25 f -0.45 20 13

125 1331

1

c

d

g

j 10-2 = 0.01 k 81 = 8 4 B 5 a 4 b 2 c 2

c 0.44 f 0.000  079

1

f 25

2 - 3

h

1 2

g

c 0.0278 f 0.004  44

or 1 7

c -4

d 53 = 125

1 7 22 x 2

= 0.0001

c

5

d 16

7

or 113

Exercise 18G — Logarithm laws 1 a 1.698  97 b 1.397  94 c 0.698  97 d 0.301  03 2 Teacher to check. 3 a 1 b 3 d 3 e 4 4 a 2 b 3 d 4 e 3

5 a 2 6 3 7 a 2

c f c f

2 1 1 5

1 2

c 1

d 3

b 4

c 3

d 3

b

Answers

Answers 18D ➜ 18G

3

1 10

g

b - 3

2 D 3 a 24 = 16

a2

12 a C, D 13 a a4 b b3 c m4 d 4x2 e 2y3 f 2x2y3 g 3m3n5 h 2pq2 i 6a2b6 14 a 0.32 m/s b 16  640 L/s c 59  904  000 L/hr That is 16  640 ì 60 ì 60. d The hydraulic radius is the measure of a channel flow efficiency. The roughness coefficient is the resistance of the bed of a channel to the flow of water in it.

5

9 4

k log9 3 =

b a 3c m c

y8

Exercise 18E — Negative indices 1 1 1 a = 0.2 b = 0.3

27 64

Exercise 18F — Logarithms 1 a log4 16 = 2 b c log3 81 = 4 d e log10 1000 = 3 f g log4 x = 3 h i log7 49 = x j

1

2 b5

h

7 n4

c

16 81

f

3

8 As the value of n increases, the value of 2-n gets closer to 0.

6 75 1

3 a 3b 4

4 9

e

3

b

i

c 3 8

3

3 11 1 8 56 m n

1 56

h x

1

7 a - 2

7

5 20 b 4

1 f p 24 q 12 7

m p

b 6 4

e

5

1 e a 20 b 20 4

8 m5

11 m 45

6 a 4

871

8 a 1 d 5

b 0 e -2

c -1 f 1

g 0

h -2

i - 2

1 2

k - 2

j

1

9 a loga 40 e loga x i

1

b loga 18 f 1

1 2

j

3 2

l c logx 48 g -1

d logx 4 h 7

k -6

l - 3

1

10 a B b B, D c A, B 11 a log2 80 b log3 105 d log6 56 e log2 4 = 2 g log5 12.5 h log2 3 1

7 2

j log10. 4

k log3 4

m log3 20

n log4 2 =

d C, D c log10 100 = 2 f log3 3 = 1 i log4 5 l log2 3

1 2

12 a C b B c A 13 a 12 (Evaluate each logarithm separately and then find the product.) b 4 (First simplify the numerator by expressing 81 as a power of 3.) c 7 (Let y = 5log 7 and write an equivalent statement in logarithmic form.) Exercise 18H — Solving equations 5

1 8

1 a 25

b 81

c

e 100, -100

f 16

g 26

h 127

1 - 32

j 0

m -624 2 a 3 e 2 3 a 3

n b f b

-2.5 2 8 2

c 125 g 6 c -1

d 625 h 4 d -2

f

2 5

g 0

h 0

1 2

i -1 4 a 5

b 6

j -2 c 10

f 2

g 9

h

k 5 5 a B 6 a 7

l 6 b A b 2

m 1 c D c -2

f

1 2

g

1 2

k

3 4

l - 2

h 5

2 5

Answers

2m ,

c 8 2

b 3ab ab

9 a 15

b 6 42

10 a 27

b 720 2

c 30 15

11 a 3 c

10 4 3

20 m

d 20 5

8 a 25 3

d 5

b 6 or

30 12

d

1 2

6 3 13 a 4 14 a 7.4

b

2 4 b 4.5 b 1.7

c 2 5 + 4

d 2 − 3

c 2.2 c 0.8

d 2.7 d 0.8

15 a 2 16 a 1

b 3 2 b 4

c 5 5

d 16

12 a

1 4

18 a 0.0833

n - 2

m , 16

1 b − x 2 y 5 xy 4

j 128

5

25m ,

7 a 72 x 3 y 4 2 xy

i 500

i - 2

b

b 6 5

17 a

n 2 d B d 0

20 3 , m , 3 8m m

5 a 5 2 6 C

e 4

3 2

m 2

l

4 a

d 8

19 a

b

1 12

1 9

c

b 0.0204 b

3 17

1 16

d

1 1000

c 0.800

d 625

c 5

d

4 13

20 B 21 A 22 A

e 4

1

j

9

o - 4

23 a 2 2

d 0.483 h 0.262 l 2.138 0.001

24 a

7 a 3.459 b -0.737 c 2.727 e 1.292 f -3.080 g -1.756 i 0.827 j 0.579 k -0.423 8 a 120 b 130 c d 3 dB are added. e 10 dB are added. f 100 9 a i  1.1 iii  1.3 iii  1.418 iv  1.77 iv  2.43 vi  3.1 b No; see answers to 9a i & ii above. c i  22  387  211 kJ ii  707  945  784 kJ iii  22  387  211  385 kJ. 872

1 - 9

i 2

e

k

d

1 16

d The energy is increased by a factor of 31.62. e It releases 31.623 times more energy. Chapter review Fluency 1 A 2 a Irrational, since equal to non-recurring and non-terminating decimal b Rational, since can be expressed as a whole number c Rational, since given in a rational form d Rational, since it is a recurring decimal e Irrational, since equal to non-recurring and non-terminating decimal 3 D

3 2 11

1

1 2

25 a 2

1

b 2 4 b

1 9

c

b 1

b

d 2

3 2

d -5

c loga x2 or 2 loga x 1 25

27 a 512

b

d 2 28 a 6

e 6 b 35

29 a -2

b - 2

Problem solving 1 a 9 b 6 c 0

d 2

c 8

26 a loga 24

30 a 4.644

1 64

1

b -3.809

c 5 f 0 c

5 2

c 0.079

2 a, b, c 

y = 4x

y

y=0

y=x y = log4x

1 0

x

1

x=0 Chapter 19

Polynomials Are you ready? 1 a x2 - 2x - 3 b x2 + 12x + 36 c 6x2 + 7x - 20 2 a 0 b 18 3 a (x - 2)(x + 3) b (x + 1)(x - 6) c (2x + 1)(x - 3) 4 a (x + 2)(x - 2) b (5 + x)(5 - x) c 3(x + 7)(x - 7) 5 a x = -1 or 3 b x = -3 or 5 3 c x = 2 or 2 Exercise 19A — Polynomials 1 a 3 b 7 c 2 f 5 g 5 h 1 2 a x b x c x f u g e h g 3 a Polynomial 1h b Polynomial 1c c Polynomial 1a d Polynomials 1a, 1d and 1e 4 a N b P c P f P g P h N 5 a 3 b x c 4 f -2x3 6 a 7 b w c 7 f 6w7 7 a 4 b 1 c x4 8 a 6 b t c 6 d, e  Check with your teacher.

c -1

d i d i

6 6 x f

e 8

d N i N d 5

e N

d 0

e -9

d 1

e 3x

Exercise 19C — Long division of polynomials 1 a x2 + 2x, 9 b x2 + x + 3, -2 c x2 + 3x - 6, 19 d x2 - x + 5, -17 e x2 + 2x - 1, 6 f x2 + 4x - 6, 14 g x2 + 1, 2 h x2 + 5, -36 i x2 - x + 6, -11 j x2 + 4x - 17, 87 2 a x2 + 4x + 3, -3 b x2 + 4x + 13, 48 c x2 + 3x - 3, -11 d x2 - 3x + 7, 5 e x2 - 2x - 3, -17 f x2 - 6x + 3, -4 g x2 + 14x + 72, 359 h x2 + 8x + 27, 104 2 3 a 3x - 7x + 20, -35 b 4x2 - 8x + 18, -22 c 2x2 - 3x + 3, 7 d 2x2 - 9, 35 2 e 4x + 2x - 3, -1 f 3x2 + x - 1, -2 4 a 3x2 - 2x + 1, 5 b 2x2 + 5x - 6, -7 c 4x2 - 7x - 2, -3 d x2 - 4x + 3, 8 e x2 + x - 6, -11 f 3x2 + 2x + 1, 13 5 a -x2 - 5x - 2, -14 b -3x2 - 2x + 4, -3 c -x2 + 5x + 6, 9 d -2x2 + 7x - 1, 1 6 a x2 - x - 2, 3 b x2, -7 2 c x - x - 2, -8 d -x2 - x - 8, 0 e 5x - 2, 7 f 2x2 - 2x + 10, -54 2 g -2x - 4x - 9, -16 h -2x2 + 4x - 1, 1 7 a x3 + 2x2 + 5x - 2, - 2 b x3 + 2x2 - 9x - 18, 0 c x4 - 3x3 + 6x2 - 18x + 58, -171 d 2x5 - 4x4 + 7x3 - 13x2 + 32x - 69, 138 e 6x3 + 17x2 + 53x + 155, 465 7 7 20 20 f x3 - 3 x2 + 9 x + 3 27 , –3 27 Exercise 19D — Polynomial values 1 a 10 b 11 c 18 d 43 e 3 f -22 g -77 h 2a3 - 3a2 + 2a + 10 i 16b3 - 12b2 + 4b + 10 j 2x3 + 9x2 + 14x + 18 k 2x3 - 21x2 + 74x - 77 l –128y3 - 48y2 - 8y + 10

Answers

Answers 18H ➜ 19D

Exercise 19B — Adding, subtracting and multiplying polynomials 1 a x4 + 2x3 - x2 - 10 b x6 + 2x4 - 3x3 + 9x2 + 5 c 5x3 - 5x2 + 7x - 13 d 2x4 + 3x3 + 12x2 - 4x + 14 e x5 + 13x4 - 10 2 a x4 + 2x2 + 2x + 4 b x6 - x5 + x3 + x2 + 2 c 5x7 - 4x3 + 5x d 10x4 - 7x2 + 20x + 5 e 2x3 + 6x2 - 10x + 15 3 a x3 + 7x2 + 6x b x3 - 7x2 - 18x c x3 + 8x2 - 33x

e y

d 2x3 + 10x2 + 12x e 48x - 3x3 f 5x3 + 50x2 + 80x g x3 + 4x2 h 2x3 - 14x2 i -30x3 - 270x2 j -7x3 - 56x2 - 112x 4 a x3 + 12x2 + 41x + 42 b x3 - 3x2 - 18x + 40 c x3 + 3x2 - 36x + 32 d x3 - 6x2 + 11x - 6 e x3 + 6x2 - x - 6 f x3 + 5x2 - 49x - 245 3 2 g x + 4x - 137x - 660 h x3 + 3x2 - 9x + 5 i x3 - 12x2 + 21x + 98 j x3 + x2 - x - 1 3 2 5 a x + 13x + 26x - 112 b 3x3 + 26x2 + 51x - 20 c 4x4 + 3x3 - 37x2 - 27x + 9 d 10x3 - 49x2 + 27x + 36 e -6x3 - 71x2 - 198x + 35 f 21x4 - 54x3 – 144x2 + 96x g 54x3 + 117x2 - 72x h 24x3 - 148x2 + 154x + 245 i 20x4 – 39x3 - 50x2 + 123x - 54 j 4x3 + 42x2 + 146x + 168 6 a x3 + 6x2 + 12x + 8 b x3 + 15x2 + 75x + 125 c x3 - 3x2 + 3x - 1 d x4 – 12x3 + 54x2 - 108x + 81 e 8x3 - 72x2 + 216x - 216 f 81x4 + 432x3 + 864x2 + 768x + 256

873

2 to 6 Column 1

Column 2

Column 3

Column 4

Column 5

P(x) a b c d

P(1)    4   10    3   -7

P(2)   15   28   11 -19

P(-1)  0 -2 -7  5

P(-2)   -5   -8 -21   -7

Column 7 Rem when divided by (x - 2)   15   28   11 -19

Column 8 Rem when divided by (x + 1)  0 -2 -7  5

1

Column 9 Rem when divided by (x + 2)   -5   -8 -21   -7 1 5

g − 2 , 0, 2

h - 4 , 0

i 0,

Exercise 19E — The remainder and factor theorems 1 a -30 b 0 c 0 d -24 e -24 f k3 + 3k2 - 10k - 24 g -n3 + 3n2 + 10n - 24 h -27c3 + 27c2 + 30c - 24 2 a 58 b -8 c 11 d -9 e -202 f 6 g 158 h -6 i 35 j 441 3 a 6 b 3 c 1 d -2 e 2 f 2 g -5, 2 h a = -5, b = -3 4 a D b C, D c D d A, C, D 5 a (x - 1) b (x - 3) or (x - 2) c (x - 3) or (x + 2) d (x - 6) or (x + 4) or (x + 5) 6 Show P(-2) = 0, P(3) = 0 and P(-5) = 0. 7 a Show P(1) = 0 b Show P(7) = 0 c Show P(2) = 0 d Show P(–2) = 0 e Show P(–3) = 0 f Show P(1) = 0 g Show P(4) = 0 h Show P(–5) = 0

j 0, 2, 3 m 0, 4, 5 2 a -4, 1, 4 d -4, -2, 2 g -3, -2, -1 j -7, 2, 3

k n b e h k

l -7, 0, 1

Exercise 19F — Factorising polynomials 1 a (x + 1)(x + 3)(x + 6) b (x + 1)(x + 2)(x + 5) c (x + 1)(x + 2)(x + 9) d (x + 1)(x + 3)(x + 4) e (x + 3)(x + 4)(x + 7) f (x + 2)(x + 3)(x + 7) g (x + 1)2(x + 2) h (x + 2)2(x + 3) i (x + 4)(x + 5)2 j x(x + 5)(x + 8) k x(x + 3)(x + 4) l x(x + 5)2 m x(x + 1)(x + 5) n x2(x + 6) 2 a (x - 1)(x + 1)2 b (x - 2)(x - 1)(x + 1) c (x + 1)2(x + 5) d (x - 3)(x + 2)2 2 e (x + 1)(x + 4) f (x - 5)(x - 2)(x + 2) g (x - 1)(x + 1)(x + 2) h (x - 3)(x + 1)(x + 2) i (x - 1)(x + 2)2 j (x + 2)(x2 - x + 3) k (x + 1)(x + 2)(x + 5) l (x - 3)(x + 1)(x + 3) m (x - 2)2(x + 3) n (x - 4)(x + 5)(x + 8) 3 a (2x + 3)(x - 1)(x + 2) b (3x - 1)(x + 1)(x + 4) c (3x + 2)(x - 2)(x + 2) d (4x + 3)(x + 3)(x + 5) e (5x - 1)(x + 1)2 f (x + 1)(x2 + 1) g (x + 1)(2x + 3)2 h (x - 2)(2x - 1)(3x - 4) i (x + 4)(2x - 5)(5x + 2) j (7x - 2)(x - 2)(x + 4) 4 a x(x - 2)(3x + 5) b 2x(x + 1)(2x - 1) c 3x(x - 4)(x + 2) d -2x(x + 3)2 e 6x2(x - 1) f -x(x + 4)(x + 3) g -(x - 1)(x + 1)(x + 3) h -2x(x - 3)(x - 2) i -(x + 2)(2x - 1)(3x - 2) j -(x - 2)2(5x - 4) k -(x - 1)(x + 3)(x - 5)(x + 2)2

7 a P(-8)

b P(7)

c P(a)

Exercise 19G — Solving polynomial equations 1 a -2, 0, 2 b -4, 0, 4 c -5, 0, 5 d 3 e -5, 0 f 0, 2

874

Column 6 Rem when divided by (x - 1)  4 10  3 -7

Answers

m -3, -2,

1 2

0, 4 0 -2, 3, 5 -1, 2, 3 -4, 5 1 -6, - 2 , -1

c f i l

-5, 1, 5 -2, 1, 5 -2, 1, 4 1 3 - 2 , 2 , 3

n -2, -1, 1

3 A, C 4 B 5 a -3, 2

b -2, 3, 6

c -4, 2 e -4, -2, 1, 3

d 1 3 f -2, - 2 , 3, 4

g -3, -2, 1, 2 6 a -2, 1, 4 c -3, 0, 2 3 e -2, 2 , 2

h -4, -1, 0, 2 b -3, -1, 3 d -4, -3, 0, 2 f -1, 1

Chapter review Fluency 1 C 2 a 5

b - 7

1

1

c 3 d x5 3 C 4 C 5 a x3 + 6x2 - 36x + 40 b x3 + 10x2 + 19x - 30 c x3 - 21x2 + 147x - 343 d -2x3 - x2 + 11x + 10 6 a D b A 7 a x2 - 16, 29 b x2 + 6x + 5, 8 c -x2 + 2x + 2, -9 8 B 9 a -4 b 216 c -24a3 + 8a2 + 2a - 4 10 -7 11 Show P(-3) = 0. 12 (x - 10)(x + 4)(x + 10) 1 13 a - 2 , 3 b 2, 3, 4 c –2, 1, 2, 3 Problem solving 1 Teacher to check. For example, given P(x) = x3 - x2 - 34x - 56 and P(7) = 0 À (x - 7) is a factor and 7 is a factor of 56.

Chapter 20

6 a 3

Functions and relations

d

Are you ready? 1 a Gradient = 3, y-intercept = 4 b Gradient = -2, y-intercept = 3 c Gradient = 25 , y-intercept = -4 y

2 a

1 1 -— 2

c



b

y

y = 2x + 1

2

x

0

y = -4x + 2

0

x

1 — 2

y=2

y



b

y

y = x2

0

-2x 2



d

10 − x +1 x −1 1 3

e -4 or 1 b f (x) ç 0

f -1 c f (x) ç 0

d f (x) ç -Ñ

e f (x) ç 0 2

9 a (0, -4), (2, 0)

b (1, -2), (- 3 , 3)

c (2, 0), (-2, 0)

d (3, -4)

N = 2000 ì 3x

12 000 10 000 8 000 6 000 4 000 2 000

2

x

y = (x - 2)2 2

x

3 a 4 A 5 B 6 a c 7 a b c d e

c 1

d

Exercise 20A — Functions 1 a One-to-many c Many-to-one e One-to-one g Many-to-many i One-to-one k Many-to-one 2 b, c, d, e, f, h, i, j, k, l 3 a i  1 ii 7

and relations b Many-to-one d One-to-one f Many-to-one h Many-to-one j Many-to-one l Many-to-one

5

4

x

b Neither d Parabola f Neither

1 27

2

4

6

$883.50 V = 950 ì (0.93)n 102 mg 86.7 mg A = 120 ì (0.85)t 83.927 mg

b $821.66 d $397.67

A 140 120 100 80 60 40 20

A = 120 ì (0.85)t

t

f Approximately 210 years 8 a i  96.04% b C = 100(0.98)w c C

iii –5

iv 16

ii 1

iii 3

iv 0

100

c i  3

ii 2

iii 6

iv 9

80

ii 1

iii 16

iv

e i  12

ii 6

iii -4

iv 2

+ 6a + 9

n

b D

b i  2 d i  9

10

C

0 50 200

a2

8

ii 90.39%

C = 100 ì (0.98)w

60 40 20 0

5

10

15

Answers 19E ➜ 20B

b 3

3

A = 5000 ì (1.075)n

0

4 a y = (x + 3)2 + 2 b y = (x - 2)2 - 5 1 9 c y = (x + 2 )2 - 4 5 a Parabola c Straight line e Straight line 6 (4, 11) 7 a 81

2

b $7717 d 10 years

14 000 12 000 10 000 8 000 6 000 4 000 2 000

4

x (1, -2)

1

2 a $5000 c A

y

0

4 A, C, D 5 a, b, c, f

f

d 2 or 3 8 a f (x) ç Ñ

y = x2 - 4

-2 0 -4

x

y=

10 − x−3 x+3

c

0

(1,1)

y

e

b -3 or 3

x

0

c

− x2

x2

5 − 2x x

Exercise 20B — Exponential functions 1 a 2000 b 486  000 c d 1.26  h N

y

0

10

c

7 a 3

2

3 a

b 3

20 w

d 8 washings Answers

875

0 1 2 3 4 5 6 7 8 Years

y

k

0

y 30

2

y 30

d

y

- 3– 0

8

2

12

2

x

-3 y

g

11

-4 -3

x

0

-10

y 12

f x

-5

-1

0

-1 0

-6

i

y 12 - 1– 0 2

0

3

x

y

j

0

1

x

-45

x

y

k

4

y -7

x

- 1– 0 2

-2

Answers

x

0

5– 3

y

l

x

-2

876

5

-8 0

- 7– 3

j 3– 4

x

x

-210

5– 2

-4

3

4

y

h

60

2

1 x

0

0

y

i

-2 y

h

- 9– 2

y

x

y

f

y

150

-9

g

2

-24

x

y

7 x

0

0

5 x

3

d

y -1

0 1 2 x

-6

e

108

-88

-14

x

y

c

-1 0

-1 0 -42

-8

0

x

y -7

x

1 2 3

y

e

b

0

-2

-192

-2 0

-6

-3

-8

-6

c

-5

2 a

14 a Approximately 20  200 b, c Teacher to check. 15 a a = 100, b = 1.20, increase = 20%/min b N = 146 977 ì 0.70m Exercise 20C — Cubic functions y 1 a b

50

6 x

3

-54

-5

y

l

0

Investment ($)

9 a 118 (million) b a = 1.02; P = 118 ì (1.02)n c Year 1990 1995 2000 2005 2010 Population 118 130 144 159 175 Calculated population is less accurate after 10 years. d 288 (million) 10 a 32 b 0.98 c T = 32 ì (0.98)t d 26.1, 21.4, 17.5, 14.3; values are close except for t = 40. 11 a 3 dogs b 27 dogs c 3 years 12 a i 39.85  mg ii 18.43  mg b More than 35.78 centuries 13 a A = 20  000 ì 1.06x b 30000 c 7 years 25000 20000 d 6 years — 1 year 15000 quicker 10000 5000 e 9.05% p.a.

1– 2

x



3 C 4 C 5 B 6 D

7 0 1– 6

- 3– 7

x

0

x

Exercise 20D — Quartic functions y 1 a b

Exercise 20E — Transformations y = 2P(x) 1 y

y

24

y = P(x) y = P(x) + 1 y = P(x) - 2

10 -3 -1 0

2

4

x

-2-10 1

x

5

x

0 y

c

y

d

y = -P(x)

32

-4

- 3 -3

-1 0

3

5– 2

3

2

36

- 3–2 0

2– 3

0

-3 -2

x

2

-24

x

d

x

2

y

b

y

(-2, 400)

(3, -30)

Chapter review Fluency 1 a 2 a, c, d 3 a 2 b 3 c 0 y 4 a 6

y 400 300

-2 0 1 3

200

-1

0

1

x (-1, 36)

0 1

y



(-3, 6) -2 -24

4 a = 4, b = -19 5 a = 3, b = -1

0

x

3

0

3

x

h

y -2

(-3, -45)

0

y 25 y = (2x + 1)(x + 5)2 1 0 -5 - — 2

-27

(-2, -16) −16

y = (x - 1)(x + 2)(x - 3)

x

b

x

2

-3

g

3

2

x

Answers 20C ➜ 20E

−1 0

2

y

f

y

x

100

-2 -1

e

x

1

3 They have the same x-intercepts, but y = -P(x) is a reflection of y = P(x) in the x-axis. 4 They have the same x-intercepts, but the y-values in y = 2P(x) are all twice as large. 5 The entire graph is moved down 2 units. The shape is identical. 6 a y = -P(x) b y = P(x) - 3 c y = 2P(x)

y

-3

1

x

x

y

0

0

y = -P(x) 0

h 2

y = P(x) + 1 y = P(x)

x - 2

0

c

-1

4

y

2 a D b B 3 a

y

2

y

f

-9

g

x

3

y = P(x + 2)

y

e

-2 0

-5

x

-2 0 1 2

x

5 Check with your teacher. Possible answer is y = (x - 1)(x - 2)2. 6 a D b A c E d C e B 7 D 8 A Answers

877

d Time (min)

y

9 a

Temperature (èC)

0

5

10

15

20

45

35

27

21

16

e No f No. The line T = 0 is an asymptote. -11

2

x

Chapter 21

y

b

Circle geometry Are you ready? 1 a True b SSS (all corresponding sides equal in length), SAS (two corresponding sides equal in length, included angle equal), ASA (two angles equal, one pair of corresponding sides equal in length), RHS (rightangled triangles with the hypotenuses and one other pair of corresponding sides equal in length) c AC is common. ±BAC = ±DAC (given) AB = AD (given) DABC ô DADC (SAS) 2 a RP b BC c ±RQP d ±BAC 3 a False. Sides may be different. b AAA or equiangular (all corresponding angles equal), SSS (all corresponding sides in same ratio), SAS (two pairs of corresponding sides in same ratio and included angle equal), RHS (both right-angled triangles with the hypotenuses and one other pair of corresponding sides in same ratio) c ±QPR is common. ±PQR = ±PST (corresponding angles are equal as QR || ST) ±PRQ = ±PTS (corresponding angles are equal as QR || ST) DPQR ~ DPST (equiangular) 4 a a = 84è b b = 88è c c = 75è 5 a x = 62è b a = 77è, b = 103è c y = 45è

8 -8

c

1

x

1– 2

x

y

10 D 11 A 12

y -1

0

2

4

x

-16

y

13

-1

0

1

x

14 The entire graph is moved up 3 units. The shape is identical. Problem solving

1 a x = ê 3

1

b x = 23

c x = 28

2 As x ç Ñ, f (x) ç -Ñ As x ç -Ñ, f (x) ç 0 3 (2, 0) 4 a 52.67  mg/L b 31.524  mg/L c 72.4  mg/L 5 a 500  èC b 125  èC c Between 5 and 6 hours once it has cooled to below 15  èC d T = 50 ì 2x e 3200  èC 6 a i 20 ii 25 b i H = 25; D = 28 ii H = 28; D = 30 c Hyenas after 3 years; dingoes after 4 years d After about 23 months; 31 animals 7 a T = 45 ì 0.95t b 45  èC c 10  èC 878

Answers

Exercise 21A — Angles in a circle 1 a x = 30è (theorem 2) b x = 25è, y = 25è (theorem 2 for both angles) c x = 32è (theorem 2) d x = 40è, y = 40è (theorem 2 for both angles) e x = 60è (theorem 1) f x = 40è (theorem 1) g x = 84è (theorem 1) h x = 50è (theorem 2); y = 100è (theorem 1) i x = 56è (theorem 1) 2 a s = 90è, r = 90è (theorem 3 for both angles) b u = 90è (theorem 4); t = 90è (theorem 3) c m = 90è, n = 90è (theorem 3 for both angles) d x = 52è (theorem 3 and angle sum in a triangle = 180è) e x = 90è (theorem 4) f x = 90è (theorem 4); y = 15è (angle sum in a triangle = 180è) 3 a x = z = 90è (theorem 4); y = w = 20è (theorem 5 and angle sum in a triangle = 180è) b s = r = 90è (theorem 4); t = 140è (angle sum in a quadrilateral = 360è) c x = 20è (theorem 5); y = z = 70è (theorem 4 and angle sum in a triangle = 180è)

d s = y = 90è (theorem 4); x = 70è (theorem 5); r = z = 20è (angle sum in a triangle = 180è) e x = 70è (theorem 4 and angle sum in a triangle = 180è); y = z = 20è (angle sum in a triangle = 180è) f x = y = 75è (theorem 4 and angle sum in a triangle = 180è); z = 75è (theorem 1) 4 D 5 B, D 6 a Base angles of a right-angled isosceles triangle b r + s = 90è, s = 45è À r = 45è c u is the third angle in DABD, which is right-angled. d m is the third angle in DOCD, which is right-angled. e ±AOC and ±AFC stand on the same arc with ±AOC at the centre and ±AFC at the circumference. 7 OR = OP (radii of the circle) ±OPR = x (equal angles lie opposite equal sides) ±SOP = 2x (exterior angle equals the sum of the two interior opposite angles) OR = OQ (radii of the circle) ±OQR = y (equal angles lie opposite equal sides) ±SOQ = 2y (exterior angle equals the sum of the two interior opposite angles) Now ±PRQ = x + y and ±POQ = 2x + 2y = 2(x + y). Therefore ±POQ = 2 ì ±PRQ. 8 Check with your teacher. 9 Check with your teacher. 10 Check with your teacher. Exercise 21B — Intersecting chords, secants and tangents 1 a m = 3 b m = 3 c m = 6 2 a n = 1 b m = 7.6 c n = 13 d m = 4 3 a x = 5 b m = 7 c x = 2.5, y = 3.1 4 a x = 2.8 b x = 3.3 c x = 5.6 d m = 90è 5 B, C, D 6 ST = 3 cm 7 Check with your teacher. 8 Check with your teacher. 9 Check with your teacher.

Chapter review Fluency 1 a x = 50è c x = y = 28è, z = 56è e y = 90è g x = 55è i x = 70è k m = 40è

2 a x = 90è c x = 55è 3 a m = 3 c m = 9 4 A, B, D 6 CE ì ED = AE ì EB AE = CE (given) \ ED = EB 7 ±AYC = ±AXC

b x = 48è, y = 25è d x = 90è f x = 140è h x = 125è j x = 100è l x = 90è, y = 60è, z = 40è b x = 20è d x = 125è b m = 12 d m = 11.7 5 A, B, C

±BXD = ±BYD But ±AXC = ±BXD À ±AYC = ±BYD 8 ±PQT & ±PST, ±PTS & ±RQS, ±TPQ & ±QSR, ±QPS & ±QTS, ±TPS & TQS, ±PQS & ±PTS, ±PUT & ±QUS, ±PUQ & TUS 9 a x = 95è, y = 80è b x = 99è c x = 78è, y = 92è d x = 97è, y = 92è 10 D Problem solving 1 a x = 42è b y = 62è c p = 65è 2 a x = 5 b k = 12 c m = 6, n = 6 d x = 7 e b = 4, a = 2 f w = 3, x = 5

Answers

Answers 21A ➜ 21D

Exercise 21C — Cyclic quadrilaterals 1 a x = 115è, y = 88è b m = 85è c n = 25è d x = 130è e x = y = 90è f x = 45è, y = 95è 2 a x = 85è, y = 80è b x = 110è, y = 115è c x = 85è d x = 150è e x = 90è, y = 120è f m = 120è, n = 130è 3 D 4 a 2x b 360è - 2x c 180è - x d 180è 5 a A b A, B, C, D 6 Check with your teacher.

Exercise 21D — Tangents, secants and chords 1 a x = 70è b x = 47è, y = 59è 2 a p = 6 b q = 8 3 x = 42è, y = 132è 4 MAC, NAC, FDA, FBA, EDG, EBG 5 B 6 D 7 x = 42è, y = 62è 8 Answers will vary. 9 60è 10 x = 180è - a - b 11 x = 80è, y = 20è, z = 80è 12 Answers will vary. 13 x = 85è, y = 20è, z = 85è 14 D 15 x = 50è, y = 95è 16 A 17 C 18 x = 33è, y = 55è, z = 22è 19 x = 25è, y = 65è, z = 40è 20 x = a, y = 90è - a, z = 90è - 2a 21 Check with your teacher.

879

Chapter 22

Trigonometry II Are you ready? 1 a Hypotenuse

Opposite

q Adjacent

b Adjacent

q

Hypotenuse Opposite



2 a 0.39 3 a 3.4 cm 4 a 60è 5 a 36è52

b b b b

0.68 38.5 cm 60è 58è13Å

c 0.36 c 45è

Exercise 22A — The sine rule 1 44è58Å, 77è2Å, 13.79 2 39è18Å, 38è55Å, 17.21 3 70è, 9.85, 9.4 4 33è, 38.98, 21.98 5 19.12 6 C = 51è, b = 54.66, c = 44.66 7 A = 60è, b = 117.11, c = 31.38 8 B = 48è26Å, C = 103è34Å, c = 66.26; or B = 131è34Å, C = 20è26Å, c = 23.8 9 24.17 10 B, C 11 A = 73è15Å, b = 8.73; or A = 106è45Å, b = 4.12 12 51.9 or 44.86 13 C = 110è, a = 3.09, b = 4.64 14 B = 38è, a = 3.36, c = 2.28 15 B = 33è33Å, C = 121è27Å, c = 26.24; or B = 146è27Å, C = 8è33Å, c = 4.57 16 43.62 m 17 a 6.97 m b 4 m 18 a 13.11 km b N20è47ÅW 19 a 8.63 km b 6.48 km/h c 9.90 km 20 22.09 km from A and 27.46 km from B 21 C 22 B 23 Yes, she needs 43 m altogether. Exercise 22B — The cosine rule 1 7.95 2 55.22 3 23.08, 41è53Å, 23è7Å 4 28è57Å 5 88è15Å 6 A = 61è15Å, B = 40è, C = 78è45Å 7 2218 m 8 a 12.57 km b S35è1ÅE 9 a 35è6Å b 6.73 m2 10 23è 11 89.12 m 12 a 130 km b S22è12ÅE 13 28.5 km 14 74.3 km 15 70è49Å 16 a 8.89 m b 77è c x = 10.07 m 17 1.14 km/h

880

Answers

Exercise 22C — Area of triangles 1 12.98 2 38.14 3 212.88 4 A = 32è4Å, B = 99è56Å, area = 68.95 cm2 5 A = 39è50Å, B = 84è10Å, area = 186.03 m2 6 A = 125è14Å, C = 16è46Å, area = 196.03 mm2 7 C 8 14.98 cm2 9 570.03 mm2 10 2.15 cm2 11 B 12 3131.41 mm2 13 610.38 cm2 14 a 187.5 cm2 b 15.03 cm c 187.47 cm2 15 17 goldfish 16 22.02 m2 17 a Area = 69.63 cm2 b Dimensions are 12.08 cm and 6.96 cm. 18 17 kg 19 52.2 hectares 20 175 m3 21 C 22 B Exercise 22D — The unit circle 1 a 1st b 2nd c 4th d 3rd e 2nd f 3rd g 4th h 4th 2 A 3 D 4 a 0.35 b 0.95 c -0.17 d 0.99 e -0.64 f 0.77 g -0.57 h -0.82 5 a 1 b 0 c 0 d -1 e -1 f 0 g 0 h 1 6 a 0.87 b 0.50 7 a 30è b -0.87 c cos 150è = -cos 30è d 0.5 e sin 150è = sin 30è 8 a 30è b -0.87 c cos 210è = -cos 30è d -0.50 e sin 210è = -sin 30è 9 a 30è b 0.87 c cos 330è = cos 30è d -0.50 e sin 330è = -sin 30è 10 a 0.34 b 0.94 c 0.36 d 0.36 e They are equal. 11 a 0.71 b -0.71 c -1 d -1 e They are equal. f tan 135è = -tan 45è 12 a -0.64 b -0.77 c 0.84 d 0.83 e They are approx. equal. f tan 220è = tan 40è 13 a -0.87 b 0.5 c -1.73 d -1.74 e They are approx. equal. f tan 300è = -tan 60è 14 D

Exercise 22E — Trigonometric functions 1 x 0è 30è 60è 90è sin x 0 0.5 0.87 1 x 390è 420è 450è 480è sin x 0.5 0.87 1 0.87 y

2

90è 180è 270è 360è 450è 540è 630è 720è

–1

150è 0.5 540è 0

y = sin x

1 0

120è 0.87 510è 0.5

x

3 360è 4 a 0.7 b 0.8 c 0.35 d -0.35 e 0 f 0.9 g -0.2 h -0.9 5 a 64è, 116è, 424è, 476è b 244è, 296è, 604è, 656è c 44è, 136è, 404è, 496è d 210è, 330è, 570è, 690è e 233è, 307è, 593è, 667è f 24è, 156è, 384è, 516è 6 See the table at the bottom of the page*. 7 y 1

180è 0 570è -0.5

210è -0.5 600è -0.87

240è -0.87 630è -1

330è -0.5 720è 0

360è 0

1

y = cos x

90è 180è 270è 360è 450è 540è 630è 720è

0

300è -0.87 690è -0.5

14 The graph would continue repeating every 180è as above. 15 Quite different. y = tan x has undefined values (asymptotes) and repeats every 180è rather than 360è. It also gives all y values, rather than just values between –1 and 1. 16 a 1.7 b –1 c –1.2 d 0.8 e –0.8 f 1.2 g –0.2 h 1 17 a 45è, 225è, 405è, 585è b 56è, 236è, 416è, 596è c 158è, 338è, 518è, 698è d 117è, 297è, 477è, 657è e 11è, 191è, 371è, 551è f 135è, 315è, 495è, 675è y 18 a y = cos x

-180è -90è –1

270è -1 660è -0.87

b

540è

x

y = sin x

180è 360è 540è 720è

x

-1

c

y

y = sin 2x

1

90è

180è 270è 360è

x

-1

720è

0º

y 1

d

y 360è

180è

-1

x

8 The graph would continue with the cycle. 9 It is a very similar graph with the same shape; however, the sine graph starts at (0, 0), whereas the cosine graph starts at (0, 1). 10 a 0.7 b -0.98 c -1 d 0.9 e -0.5 f -0.8 g 0.8 h -0.96 11 a 120è, 240è, 480è, 600è b 37è, 323è, 397è, 683è c 46è, 314è, 406è, 674è d 127è, 233è, 487è, 593è e 26è, 334è, 386è, 694è f 154è, 206è, 514è, 566è 12 See the table at the bottom of the page**. 13 y = tan x

180è

90è

y

y = 2 cos x

2

-360è -270è -180è -90è

x

x -2

270è

450è

630è

6 * x

cos x x

cos x 12 ** x tan x x tan x

0è 1 390è 0.87 0è 0 390è 0.58

30è 0.87 420è 0.5 30è 0.58 420è 1.73

60è 0.5 450è 0 60è 1.73 450è undef.

90è 0 480è -0.5 90è undef. 480è -1.73

120è -0.5 510è -0.87 120è -1.73 510è -0.58

150è -0.87 540è -1 150è -0.58 540è 0

180è -1 570è -0.87 180è

0 570è 0.58

210è -0.87 600è -0.5 210è 0.58 600è 1.73

240è -0.5 630è 0

270è 0 660è 0.5

300è 0.5 690è 0.87

330è 0.87 720è 1

360è 1

240è 1.73 630è undef.

270è undef. 660è -1.73

300è -1.73 690è -0.58

330è -0.58 720è 0

360è 0

Answers

Answers 22A ➜ 22E

90è

881

19 a b c d 20 a b c d

i i i i i i i i

ii ii ii ii ii ii ii ii

360è 360è 180è 360è 180è 120è 720è 1440è

e i 360è f i 180è 21 a C b A c D

1 1 1 2 3 4 2 1 2

ii 1 ii 1

22 a y Period = 1080è 2 Amplitude = 2 y = 2 cos –x 3

540è

1080è

23 a

y 2 1 -1 -2

b

c

90è

180è 270è 360è

90è

180è 270è 360è

y 1

y = cos (x - 60è) 120è 240è 360è

d

-3

c Period = 720è y Amplitude = 3 3 y = 3 sin –2x

0

-180è -90è

90è

180è

x

y 5 4 3 2 1

y = 2sin 4x + 3

90è

24 a

y

180è 270è 360è

180è

-1

e

y 5



y = 5cos 2x

180è

Period = 180è Amplitude = 5

y

b

y f Period = 90è y = -sin 4x 1 Amplitude = 1 90è -1

882

Answers

180è

x

x

i –1 ii 1 b i 3 ii 1 c Max value of sin x = 1, hence max value of y=2ì1+3=5 Min value of sin x = -1, hence min value of y = 2 ì -1 + 3 = 1 25 a x 0 30è 60è 90è 120è 150è 180è

x

-5

360è

-1

d y Period = 120è y = -cos 3x 1 Amplitude = 1 x

x

y = cos 2x

1

-3

120è 240è 360è

x

-1

b y Period = 180è y = -3 sin 2x 3 Amplitude = 3 90è

x

y = sin 2x - 2

-2

180è 270è 360è x

x

y -1 -2 -3 -4

x

y = cos x + 1

0

3 3

3

undef − 3 −

3 3

y y = tan x

90è

180è

x

c At x = 90è, y is undefined. d x = 270è e The period = 180è, amplitude is undefined.

0

4 3.6 cm 5 34è 6 94è56Å 7 a 159.10 cm2 8 4th quadrant 9 a 0.94, -0.34 10 B 11 tan 53è 12 y

13

Depth (m)

2.5

•

2.0 1.5

•

• • 1.05 m

0.5

•

• •

•

1.55 m

• 1

1.0

•

ö 12–2 hours

•

1.05 m

• 0.5 m

• •

450è

x

y y = cosx

1 0 –1

x

14 y = tan x y 180è

0è

360è

x 90è

15

270è

B

3.0

2.6 m

360è

0 –1

450è

Exercise 22F — Solving trigonometric equations 1 Calculator answers iii 25.84è, 334.16è iii 72.54è, 287.46è iii 101.54è, 258.46è iv 126.87è, 233.13è 2 a 30è, 150è b 60è, 120è c 120è, 240è d 135è, 225è e 90è f 180è g 210è, 330è h 225è, 315è i 30è, 330è j 150è, 210è k 90è l 90è, 270è 3 a 30è, 60è, 210è, 240è b 75è, 105è, 255è, 285è c 15è, 75è, 195è, 255è, 375è, 435è, 555è, 615è d -165è, -135è, -45è, -15è, 75è, 105è e 52.5è, 82.5è, 142.5è, 172.5è f -165è, -135è, -45è, -15è, 75è, 105è g 45è h 30è, 90è, 150è, 210è, 270è, 330è 4 a 30è, 150è b 30è, 330è c 45è, 315è d 225è, 315è 5 a

b -2.75

y = sinx

1

270è

b c = 45è and x = 135è c Period = 90è and amplitude is undefined.

c 159.09 cm2

b 17.68 cm

360è

x

270è

180è

90è

90è

180è

y = tan 2x

90è

y

180è

26 a

y x A

46è 68è C

16 a Period = 120è, amplitude = 2 b Period = 180è, amplitude = 3 c Period = 180è, amplitude = 0.5 17 a y

• •

2 1

180è

360è

x

Answers 22F ➜ 22F

-1 -2

5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 am pm

y = 2sin x

Time (hours) 1

b i 12 2 h ii 1.05 m c 10.00 am, 10.30 pm, 11.00 am, 11.30 pm, noon d Until 2.15 am; from 8.15 am to 2.45 pm; after 8.45 pm Chapter review Fluency 1 14.2 cm 2 20è31Å 3 b = 22.11 m, c = 5.01 m, C = 10è

y

b

1

-180è

y = cos 2x

180è

x

-1

18 a x = 191.54, 348.46 b x = 22.79, 157.21, 202, 79, 337.21 c x = 88.09, 271.91 d x = 7.24, 172.76, 187.24, 352.76 Answers

883

19 a 210è, 330è b 30è, 330è c 45è, 315è d 45è, 135è 20 E 21 a y

Problem solving 1 3.9 m 2 a 7.3 km 3 a

180è

-1 -2 -3 -4 -5

360è

x

y = 2sin 2x - 3

b Period = 180è, amplitude = 2 y 22 a    i Period = 180è y = 2cos 2x ii Amplitude = 2 2

180è

360è

x

b 281è57ÅT t

V

0.000

    0

0.005

  240

0.010

    0

0.015

-240

0.020

    0

0.025

  240

0.030

    0

0.035

-240

0.040

    0

-2 V (Volts)

b    i Period = 90è y = 3sin 4x ii Amplitude = 3 3 y

240 .010 .020 .030 .040 .005 .015

90è

180è

c

y 2

-60è

y = -2cos 3x

60è



   i Period = 120è ii Amplitude = 2

y 4

-90è

y = 4sin 2x

90è

Interpreting data    i Period = 180è ii Amplitude = 4

x

-4

23 a 15è, 165è, 195è, 345è b -70è, 10è, 50è c 112.5è, 157.5è, 292.5è, 337.5è d 15è, 105è, 135è e 0, 45è, 90è, 135è, 180è f 45è, 135è, 225è, 315è 24 a 60, 300 b 240, 300 c 45, 315 d 225, 315 25 y y = tan 2x

90è

180è

b Maximum voltage occurs at t = 0.005 s, 0.025 s c 0.02 s d 50 cycles per second Chapter 23

x

-2

d

x

Are you ready? 1 a -6 b 4 2 a i y = 2 b i y = -3 −3 c i y = 2 3 a y = -2x + 4 b y = 4x - 5 2 5 c y = − x − 3 3 4 a 1 b 2

5 a

5 2

b

6 a 5y - 4x = 20

884

Period = 90, amplitude is undefined. Asymptotes are at x = 45è and x = 135è.

Answers

4 0

-5

b

7 3

y

x

y 4y - 2x = 5 1 1–4 -2 1–2



t (second)

-240

x

-3

.035

0

x

ii x = 3 ii x = 9

c 3

ii x = 2

c -1 c

−1 4

5 a

y

0

x Price ($1000)

-3

-4 3y + 4x = -12

Dependent Test results Attendance Visits to the doctor Memory taken Attendance Cost of property Cut-off OP score Heart rate

30 40 50 60 70 80 90 100 110 120 Number of guests

Number of bags sold

12 11 10 9 8 7 6 5 4 3 2 1 30 35 40 45 50 55 60 65 70 75 80 Cost ($)

b Negative, linear, moderate. The price of the bag appeared to affect the numbers sold; that is, the more expensive the bag, the fewer sold.

3

4

5

6

7

100 90 80 70 60 50 40 30 20 10 0

1

2

3

4

5

6

7

8

9 10

Number of questions completed

b Strong, positive, linear correlation c Various answers — some students are of different ability levels and they may have attempted the questions but had incorrect answers. 7 a 6 5 4 3 2 1 0

5 10 15 20 25 30 35 40

Number of lessons

b Weak, negative, linear relation c Various answers, such as some drivers are better than others, live in lower traffic areas, traffic conditions etc. 8 a T b F c T d F e T 9 B 10 C 11 D Exercise 23B — Lines of best fit Note: Answers may vary depending on the line of best fit drawn. P 1 a, b  14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Answers 23A ➜ 23B

3 a Perfectly linear, positive b No correlation c Non-linear, negative, moderate d Strong, positive, linear e No correlation f Non-linear, positive, strong g Perfectly linear, negative h Moderate negative, linear i Weak, negative, linear j Non-linear, moderate, positive k Positive, moderate, linear l Non-linear, strong, negative m Strong, negative, linear n Weak, positive, linear o Non-linear, moderate, positive 4 a

2

Number of bedrooms

b Moderate positive linear correlation. There is evidence to show that the larger the number of bedrooms, the higher the price of the house. c Various answers; location, age, number of people interested in the house, and so on. 6 a Total score (%)

4.6 4.4 4.2 4.0 3.8 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4

1

Number of accidents

Cost ($1000)

Exercise 23A — Bivariate data 1 Independent a Number of hours b Rainfall c Hours in gym d Lengths of essay e Cost of care f Age of property g Number of applicants h Running speed 2

420 400 380 360 340 320 300 280 260 240 220 200 180 160 140

Petrol used (L)

c

10 20 30 40 50 60 70 80 90 100 d

Distance travelled (km)

Answers

885

130 120 110 100 90 80 70 60 50 40 30 20 10 0

2

4

6

8 10 12 14 16 18 h

Hours worked

Cost of food ($)

c Using (8, 47) and (12, 74), the equation is E = 6.75h - 7. d On average, students were paid $6.75 per hour. 3 a 38 b 18 4 a i 460 ii 290 iii 130 b i 39 ii 24 iii 6 c y = -11.71x + 548.57 d y-values: i 466.60 ii 290.95 iii 127.01 x-values: i 36.60 ii 24.64 iii 5.86 C 5 a 165 160 155 150 145 140 135 130 125 120 115 110 105 100 95 90 85 80 75 70

1

2

3

4

5

6

Number of people

7

n

b Using (1, 75) and (5, 150), the equation is C = 18.75n + 56.25. c On average, weekly cost of food increases by $18.75 for every extra person. d i $206.25 ii $225.00 iii $243.75 M 6 a 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0

17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10111213141516 d

Day

b M = 0.973d + 1.285 c Each day Rachel’s crystal gains 0.973 g in mass. d 7.123 g; 8.096 g; 13.934 g; 14.907 g; interpolation (within the given range of 1–16) e 17.826 g; 18.799 g; predictions are not reliable, since they were obtained using extrapolation. 9 a D b C 10 E Exercise 23C — Time series 1 a Linear, downward b Non-linear, upward c Non-linear, stationary in the mean d Linear, upward e Non-linear, downward f Non-linear, stationary in the mean g Non-linear, stationary in the mean h Linear, upward 2 a May temperature 18.0

Temperature (èC)

Mass (kg)

b M = 0.247t - 6.408 c With every week of gestation the mass of the baby increases by 247 g. d 3.719 kg; 3.966 kg e 1.002 kg f 36 weeks 7 a r = 0.9 b There is a strong positive relationship between the number of hours spent studying and the marks obtained. This seems to indicate that, greater dedication to studying will produce better results. M 8 a 18

Mass (g)

Earnings ($)

c Using (23, 3) and (56, 8), the equation is 5 16 P = d − . 33 33 E 2 a, b 

17.8 17.6 17.4 17.2 17.0 16.8 16.6 16.4 16.2 16.0 15.8 15.6 15.4 15.2 15.0 14.8 14.6 14.4 14.2 14.0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Day

31 32 33 34 35 36 37 38 39 40 t

Weeks



886

Positive, strong, linear correlation

Answers

b Linear downward trend

b Yes, the graph shows an upward trend. 4 45 c y = x + 7 7 d i 15 ii 18 (The assumption made was that business will continue on a linear upward trend.) 8 The trend is non-linear, therefore unable to forecast future sales. 9 Answers will vary.

130 125 120 115 110 105 100 95 90 85 80 75 70 65 60 55 50 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Quarter

2006

2007

2008

2009 Year

b Sheepskin products more popular in the third quarter (presumably winter) — discount sales, increase in sales, and so on. c No trend 4 a 100 Revenue ($1000)

95 90 85 80 75 70 65 60 55 50 45 40 2007

2008

2009

Year

b General upward trend with peaks around December and troughs around April. c Peaks around Christmas where people have lots of parties, troughs around April where weather gets colder and people less inclined to go out. d Yes. Peaks in December, troughs in April. 5 a Peaks around Christmas holidays and a minor peak at Easter. No camping in colder months. b Check with your teacher. 6 a 120

Enrolment

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15

5 10 15 20 25 30 35 40 45 50 55 60

13 12 11 10 9 8 (1, 7) 7 6

(8, 11)

June July Aug Sep Oct Nov Dec Jan Feb Mar Apr May

Month

c Strong, positive, linear correlation; the larger the number of completed revision questions, the higher the mark on the test. d Different abilities of the students 2 a i 12.5 ii 49 b i 12 ii 22.5 22 7 c y = x − 15 3 d i 12.33 ii 49 and i 11.82 ii 22.05 3 a Linear downwards b The trend is linear. c About 65 occupants d Assumes that the current trend will continue. 4 a P = 31.82a + 13  070.4, where P is the sale price and a is the land area. b The price of land is approximately $31.82 per square metre. c $64  000 d 1160 m2 5 a C = 0.15p + 11.09, where C is the money spent at the canteen and p represents the pocket money received. b Students spend 15 cents at the canteen per dollar received for pocket money. c $18 d $26. This involves extrapolation which is considered unreliable. It does not seem reasonable that, if a student receives more money, they will eat more or have to purchase more than any other student. 6 a P = 0.91t + 2.95, where P is the number of pirouettes and t is the number of hours of training. b Ballet students can do approximately 0.91 pirouettes for each hour of training. c Approximately 15 pirouettes

Answers

Answers 23C ➜ 23C

Year

Upward linear b In 15th year the expected amount = 122 7 a 14 Number of children

100 90 80 70 60 50 40 30 20 10 0

Number of questions

1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 Month

110 100 90 80 70 60 50 40 30 20 10 0

Chapter review Fluency 1 a Number of questions — independent; mark on a test — dependent b

Test result

Sales (ì $1000)

3 a

887

Length (cm)

Problem solving 1 a L

0 1 2 3 4 5 6 7 8 9 1011121314151617181920 n

Week

Best jump (metres)

b L = 1.062n + 19.814 c 25.124 cm; 27.248 cm; 29.372 cm; 31.496 cm; 32.558 cm; 35.774 cm; 36.806 cm; 38.93 cm; 39.992 cm d Interpolation (within the given range of 1–20) e 42.116 cm; 43.178 cm; 44.24 cm f Not reliable, because extrapolation has been used. 2 a

888

8 7 6 5 4 3 2 1

Answers

Best jump (metres)

8 7 6 5 4 3 2 1 0

c

1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Age

1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Age

B

8 7 6 5 4 3 2 1

0

39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20

0

b

Best jump (metres)

d Approximately 30 pirouettes. This estimate is based on extrapolation which is considered unreliable. To model this data linearly as the number of hours of training becomes large is unrealistic. 7 About 170 cm; Data was first plotted as a scatter plot. (145, 160) was identified as an outlier and removed from the data set. A line of best fit was then fitted to the remaining data and its equation determined as d = 0.5h + 80, where d is the distance stretched and h is the height. Substitution was used to obtain the estimate.   The estimation requires extrapolation and cannot be considered reliable. The presence of the outlier may indicate variation in flexibility rather than a strong linear correlation between the data. Estimate is based on a small set of data. More data should be collected in order to determine the suitability of least squares regression.

1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Age

a

d Yes. Using points (9, 4.85) and (16, 7.24), B = 0.341a + 1.781; estimated best jump = 8.6 m. e No, trends work well over the short term but long term are affected by other variables. f 24 years old: 9.97 m; 28 years old: 11.33 m. It is unrealistic to expect his jumping distance to increase indefinitely. g Equal first 3 a Outliers can unfairly skew data and as such dramatically alter the line of best fit. Identify and remove any outliers from the data before determining the line of best fit. b Extrapolation involves making estimates outside the data range and this is considered unreliable. When extrapolation is required, consider the data and the likelihood that the data would remain linear if extended. When giving results, make comment on the validity of the estimation. c A small range may not give a fair indication if a data set shows a strong linear correlation. Try to increase the range of the data set by taking more measurements or undertaking more research. d A small number of data points may not be able to establish with confidence the existence of a strong linear correlation. Try to increase the number of data points by taking more measurements or undertaking more research.

Glossary 2-dimensional: a description of a plane shape. The dimensions are given in two directions, such as length and width or length and height. 3-dimensional: a shape that occupies space (a solid). That is, one that has dimensions in three directions — length, width and height. Addition law of probability: if A and B are mutually exclusive events, then P(A or B) = P(A) + P(B) or P(A ß B) = P(A) + P(B) Adjacent angles: angles at a point that share a common ray and a common vertex A

Area: the amount of flat surface enclosed by the shape. It is measured in square units, such as square metres, m2, or square kilometres, km2. Area of triangle (using sin): if the perpendicular height of a triangle is not known, but two sides and the included angle are known, the area of any triangle ABC can be calculated using the rule: 1 1 1 Area = ab sin C,  Area = ac sin B  or  Area = bc sin A 2 2 2 B

c a

C

A b

B O Algebraic expression: an expression formed by numbers and algebraic symbols using arithmetic operations. For example, 4x + 3y - 2 is an algebraic expression. Algebraic fractions: fractions that contain pronumerals (letters) Algebraic term: an algebraic expression that forms a ‘separable’ part of some other algebraic expression. For example, in the expression 4x + 2y - 3, 4x and 2y are algebraic terms, while -3 is a constant term. Alternate angles: angles on alternate sides of a transversal. On parallel lines, alternate angles are equal.

C Associative Law: a method of combining two numbers or algebraic expressions is associative if the result of the combination of these objects does not depend on the way in which the objects are grouped. Addition and multiplication obey the Associative Law, but subtraction and division are not associative. Asymptote: a line that a graph approaches but never meets Average: see Mean Average speed: The total distance travelled during a journey divided by the total time taken. It is given by the formula: total distance travelled . average speed = total time taken Back-to-back stem-and-leaf plot: a method for comparing two data distributions by attaching two sets of ‘leaves’ to the same ‘stem’ in a stem-and-leaf plot; for example, comparing the pulse rate before and after exercise

Amplitude: half the distance between the maximum and minimum values of a function Angle of depression: the angle measured down from the horizontal line (through the observation point) to the line of vision Horizontal

Pulse rate Before After 9888 6 8664110 7 8862 8 6788 60 9 02245899 4 10 0 4 4 0 11 8 12 4 4 13 14 6 Bar graph: a graph drawn in a similar way to a column graph, with horizontal bars instead of vertical columns. Categories are graphed on the vertical axis and the frequencies (numbers) on the horizontal axis. Base: the digit at the bottom of numbers written in index form. For example, in 64, the base is 6. This tells us that 6 is multiplied by itself four times. Bi-modal: describes data whose distribution has two modes Bisect: cut into two equal parts Bivariate data: sets of data where each piece is represented by two variables

Angle of depression Line of sight Object Angle of elevation: the angle measured up from the horizontal line (through the observation point) to the line of vision Object Line of sight Angle of elevation Horizontal Arc (of a circle): a portion of the circumference of a circle

Glossary

889

Boxplots (box-and-whisker plots): a graphical representation of the 5-number summary; that is, the lowest score, lower quartile, median, upper quartile and highest score, for a particular set of data

65 70 75 80 85 90 95 100 105 110 Pulse rate Brackets: also called grouping symbols Capacity: the maximum amount of fluid that can be contained in an object. It is usually applied to the measurement of liquids and is measured in units such as millilitres (mL), litres (L) and kilolitres (kL). Cartesian coordinate system: the position of any point in the Cartesian plane can be represented by an ordered pair of numbers (x, y). These are called the coordinates of the point. y 5 4 3 2 1 0 -5 -4 -3 -2 -1 -1

x-coordinate (4, 2) 1 2 3 4 5x

Cartesian plane: the area formed by a horizontal line with a scale (x-axis) joined to a vertical line with a scale (y-axis). The point of intersection of the lines is called the origin. Categorical (data): data that cannot be measured or counted but can be categorised; for example, eye colour or television programs Census: collection of data from a population (e.g. all Year 10 students) rather than a sample Centre (of circle): middle point of a circle, equidistant (equal in distance) from all points on its circumference Chord: straight line from one point on the circumference of a circle to another point on the circumference

P

O

Q Circle (equation): the general equation of a circle, with centre (h, k) and radius r is: (x - h)2 + ( y - k)2 = r 2.

Glossary

G b C A

y-coordinate

-2 -3 -4 -5

890

Circumference: distance around the outside of a circle. It is given by the rule 2p r or p D, where r is the radius and D is the diameter of the circle. Class interval: a subdivision of a set of data. For example, students’ heights may be grouped into class intervals of 150 cm - 154 cm, 155 cm - 159 cm. Closure Law: when an operation is performed on an element (or elements) of a set, the result produced must also be an element of that set. Coefficient: the number part of a term, generally written in front of the pronumeral Co-interior angles: angles that lie on the same side of a transversal that cuts across a pair of lines. For a pair of parallel lines, co-interior angles are supplementary (add to 180è). Q

D B

a F

P Collinear points: points that all lie on the same straight line Column graph: a graph in which equal width columns are used to represent the frequencies (numbers) of different categories Common factor (common divisor): a factor that is common to each element of the set; for example, 3x is a common factor of the elements 9x2 and 12x Commutative Law: a method of combining two numbers or algebraic expressions is commutative if the result of the combination does not depend on the order in which the objects are given. For example, the addition of 2 and 3 is commutative, since 2 + 3 = 3 + 2. However, subtraction is not commutative, since 2 - 3 ò 3 - 2. Compass (conventional) bearings: directions measured in degrees from the north–south line in either a clockwise or anticlockwise direction. To write the compass bearing we need to state whether the angle is measured from the north or south, the size of the angle and whether the angle is measured in the direction of east or west; for example, N27èW, S32èE. Complement (of a set): the complement of a set, A, written AÅ, is the set of elements that are in x but not in A Complementary angles: two angles that add to 90è; for example, 24è and 66è are complementary angles Complementary events: events that have no common elements and together make up the sample space. If A and AÅ are complementary events, then P(A) + P(AÅ) = 1. Completing the square: a procedure used to transform an algebraic expression into a perfect square Composite number: a number that has more than two factors. For example, 6 is a composite number because it has factors 1, 2, 3 and 6. Composite figure: a figure made up of more than one basic shape

Compound graphs: column and bar graphs that display two or more sets of data simultaneously. They are drawn with each column or bar representing combined sets of data. Individual columns or bars are multicoloured, one colour for each set. Compound interest: the interest earned by investing a sum of money (the principal) when each successive interest payment is added to the principal for the purpose of calculating the next interest payment. The formula used for compound interest is: A = P(1 + R)n, where A is the amount to which the investment grows, P is the principal or initial amount invested, R is the interest rate per compounding period (as a decimal) and n is the number of compounding periods. The compound interest is calculated by subtracting the principal from the amount: CI = A - P. Concentration: a measure of the strength of a solution. The measured units can be, for example, g/mL. Concyclic (points): points that lie on the circumference of a circle Conditional probability: where the probability of an event is conditional (depends) on another event occurring first. For two events A and B, the conditional probability of event B, given that event A occurs, is denoted by P(B | A) and can be calculated using the formula: P ( A ∩ B) , P(A) ò 0. P(B  |  A) = P( A) Cone: A solid formed by taking a circular base and a point not in the plane of the circle, called the vertex, which lies above or below the circle, and joining the vertex to each point on the circumference of the circle.

Constant of proportionality (or variation) k: used to prove that a proportionality relationship (direct or inverse) exists between 2 or more variables (or quantities) Continuous (data): numerical data that can take any value within a certain range. They are generally associated with measuring; for example, the heights of students. Coordinates (x, y): two numbers that give the position of a point on the Cartesian plane. The first number is the x-coordinate and the second number is the y-coordinate. Correlation: a measure of the relationship between two variables. Correlation can be classified as linear, nonlinear, positive, negative, weak, moderate or strong. Correlation coefficient r: the value of r indicates the strength of the relationship between two variables. Its range is -1 Ç r Ç + 1, -1 being a strong negative relationship and +1 being a strong positive relationship. The closer the value of r is to 0, the less strong the relationship between the variables. Corresponding angles: angles that are in corresponding positions with respect to a transversal. On parallel lines, corresponding angles are equal. Q

G C

D B

F

A P

Radius r

Cosine (cos) ratio: the ratio of the adjacent side to the hypotenuse in a right-angled triangle. adjacent  . So, cos q  = hypotenuse

Height h

Slant height l

Hypotenuse

Opposite

q Vertex Congruent figures: figures that are identical; that is, they have exactly the same shape and size D DÅ AÅ A

Adjacent Cosine rule: in any triangle ABC, c2 = a2 + b2 - 2ab cos C.

B c a A

B C CÅ BÅ Congruent triangles: there are four standard congruence tests for triangles: SSS (side, side, side), SAS (side, included angle, side), ASA (two angles and one side) and RHS (right angle, hypotenuse, side) Conjugate surds: surds that, when multiplied together, result in a rational number. For example, ( a + b ) and ( a − b ) are conjugate surds, because ( a + b ) × ( a − b ) = a − b. Constant: a term or expression whose value does not vary

b C Counting numbers: the non-negative integers; that is, one of the numbers 0, 1, 2, 3, .  .  . Cross-section: the shape (plane section) produced when a solid is cut through by a plane parallel to the base. For example, the cross-section of a cone is a circle. Cube: a polyhedron with 6 faces. All faces are squares of the same size. Cubic functions: the basic form of a cubic function is y = ax3. These functions can have 1, 2 or 3 roots. Glossary

891

Cumulative frequency: the total of all frequencies up to and including the frequency for a particular score in a frequency distribution Cumulative frequency polygon: a line graph that is formed when the cumulative frequencies of a set of data are plotted against the end points of their respective class intervals and then joined up by straight-line segments. It is also called an ogive. Cyclic quadrilateral: a quadrilateral that has all four vertices on the circumference of a circle. That is, the quadrilateral is inscribed in the circle. Cylinder: a solid that has parallel circular discs of equal radius at the ends. The centres of the discs form the axis of the cylinder.

Axis

Cross-section is a circle

Cylinder Data: various forms of information Decimal number system: the base 10, place-value system most commonly used for representing real numbers Degree (angle): a unit used to measure the size of an angle Degree (of a polynomial): the degree of a polynomial in x is the highest power of x in the expression. Denominator: the lower number of a fraction that represents the number of equal fractional parts a whole has been divided into Density: the ratio of mass to volume of a substance. It is given by the formula: mass density = . volume It is measured in units such as g/cm3. Dependent events: successive events in which one event affects the occurrence of the next Dependent variable: this variable is graphed on the y-axis. Depreciation: the reduction in the value of an item as it ages over a period of time. The formula used is A = P(1 - R)n, where A is the depreciated value of the item, P is its initial value, R is the percentage the item depreciates each year (expressed as a decimal) and n is the number of years the item has depreciated. Diameter: the straight line from one point on the circumference of a circle to another on the circumference, passing through the centre. Dilated (Quadratics): occurs when graphs are made thinner or wider Dilation (Geometry): occurs when figures are made larger (enlarged) or smaller (reduced) in proportion Direct variation: describes a particular relationship between two variables (or quantities); that is, as one variable increases so does the other variable. The graph of the relationship is a straight line, passing through the origin and the rule used to relate the two variables is y = kx. 892

Glossary

Discrete data: numerical data in which the information can take only certain exact values, usually whole numbers. They are associated with counting. Discriminant: referring to the quadratic equation ax2 + bx + c = 0, the discriminant is given by D = b2 - 4ac. It is the expression under the square-root sign in the quadratic formula and can be used to determine the number and type of solutions of a quadratic equation. Disjoint sets: these sets have no elements in common with each other. Distance formula: the distance between two points A(x1, y1) and B(x2, y2) is given by the formula ( x2 − x1 )2 + ( y2 − y1 )2 . Distributive Law: the product of one number with the sum of two others equals the sum of the products of the first number with each of the others; for example 4(6 + 2) = 4 ì 6 + 4 ì 2. It is also applicable to algebra; for example, 3x(x + 4) = 3x2 + 12x. Dividing by a fraction: when dividing by a fraction, multiply by the reciprocal of the fraction, then simplify 3 2 the expression. For example, 6 ó = 6 ì = 4. 2 3 Dodecahedron: a regular polyhedron (platonic solid) with 12 faces, all of which are regular pentagons. Dot plot: this graphical representation uses one dot to represent a single observation. Dots are placed in columns or rows, so that each column or row corresponds to a single category or observation.

0

1

2

3 4 5 6 7 Passengers Edges: straight lines where pairs of faces of a polyhedron meet Eighth Index Law: terms with fractional indices can be 1

m

written as surds. For example, a n = n a and a n = n a m . Element: an element of a set is a member of that set; for example; 5 is an element of the set of counting numbers. Elimination method: a method used to solve simultaneous equations. This method combines the two equations into a third equation involving only one of the variables. Ellipse: a plane figure in the shape of an oval Enlargement (dilation): a scaled-up (or down) version of a figure in which the transformed figure is in proportion to the original figure; that is, the two figures are similar Equally-likely outcomes: outcomes in a probability experiment that have the same chance of occurring Equating: the process of writing one expression as equal to another Equation: a statement that asserts that two expressions are equal in value. An equation must have an equal sign. For example, x + 4 = 12. Equilateral triangle: a triangle with all sides equal in length, and all angles equal to 60è

Fourth Index Law: to remove brackets, multiply the indices inside the brackets by the index outside the brackets. Where no index is shown, assume that it is 1. So, (am)n = amn. a Fraction: numbers represented in the form , where a and b b are whole numbers and b is not equal to zero. Frequency: the number of times a particular score appears Frequency polygon: a special type of line graph, which uses the same scaled axes as the histogram. The midpoints of the tops of the histogram columns are joined by straight line intervals. The polygon is closed by drawing lines at each end down to the score- or x-axis. Frequency table: a means of organising a large set of data. It shows the number of scores (frequencies) that belong to each group or class interval. Function: a process that takes a set of x-values and produces a related set of y-values. For each distinct x-value, there is only one related y-value. They are usually defined by a formula for f (x) in terms of x; for example, f (x) = x2. Gradient (slope) m: this is a measure of the steepness of a line or plane. The gradient of a line is given by rise y2 − y1 m= = and is constant anywhere along that line. run x2 − x1

y B(x2, y2) y2 - y1 (Rise) A(x1, y1)

x2 - x1

x (Run) Heron’s formula: this formula is used to find the area of a triangle when all three sides are known. The formula is A = s(s − a)(s − b)(s − c) , where a, b and c are the lengths of the sides and s is the semi-perimeter or a+b+c . s= 2 Histogram: a special type of column graph, in which no gaps are left between columns and each column straddles an x-axis score. The x-axis scale is continuous and usually a half-interval is left before the first column and after the last column. y 10 0

Frequency

Equivalent fractions: fractions that can be reduced to the same basic fraction; that is, fractions that have the same 1 2 3 4 value, for example, = = = 3 6 9 12 Estimate: information about a population extrapolated from a sample of the population Euler’s rule: a rule that links the number of faces, F, the number of vertices, V, and the number of edges, E, of a polyhedron. Euler’s formula: F + V − E = 2. Evaluate: determine a value for an expression Event: a set of favourable outcomes in each trial of a probability experiment Expanding (algebra): this is the process of multiplying everything inside the brackets by what is directly outside the brackets. Expanding is the opposite of factorising. Expected frequency: the number of times a particular event is expected to occur when a chance experiment is repeated a number of times Exponent: see Index Exponential decay: a quantity that decreases by a constant percentage in each fixed period of time. This growth can be modelled by exponential functions of the type y = kax, where 0 < a < 1. Exponential functions: relationships of the form y = ax, where a ≠ 1, are called exponential functions with base a. Exponential growth: a quantity that grows by a constant percentage in each fixed period of time. This growth can be modelled by exponential functions of the type y = kax, where a > 1. Expression: this is a collection of two or more numbers or variables, connected by operations. For example, 12 - 2, 2a + 3b. Expressions do not contain an equal sign. Extrapolation: the process of predicting a value of a variable outside the range of the data Faces: 2-dimensional, closed shapes that form the surfaces of a polyhedron Factor: a factor of a given number is a whole number that divides it exactly. For example, 1, 2, 3, 4, 6 and 12 are the factors of 12. Factor theorem: if P(x) is a polynomial, and P(a) = 0 for some number a, then P(x) is divisible by (x - a). Factorising: breaking down a number or expression into smaller factors that can be numeric or algebraic. The process of factorising an algebraic expression involves changing it from a sum (or difference) into a product of factors. Fifth Index Law: to remove brackets containing a product, raise every part of the product to the index outside the brackets. So, (ab)m = ambm. Finite: a fixed number or amount. For example, the decimal 0.25 has a fixed number of decimal places. First Index Law: when terms with the same base are multiplied, the indices are added. So, am ì an = am + n. Five-number-summary: a method for summarising a data set using five statistics: the minimum value, the lower quartile, the median, the upper quartile and the maximum value FOIL: a diagrammatic method of expanding a pair of brackets. The letters in FOIL represent the order of the expansion: First, Outer, Inner and Last.

8 6 4 2

0 155 160 165 170 175 180 185 190 195 200 x Height Horizontal: a line is said to be horizontal if it is parallel to the horizon of the Earth. Horizontal lines have a gradient of zero and are parallel to the x-axis. Glossary

893

Hypotenuse: the longest side of a right-angled triangle. It is the side opposite the right angle. Icosahedron: a regular polyhedron (platonic solid) with 20 faces, all of which are equilateral triangles Identity Law: when 0 is added to an expression or the expression is multiplied by 1, the value of the variable does not change. For example, x + 0 = x and x ì 1 = x. Image (similar figures): the enlarged (or reduced) figure produced Improper fraction: a fraction in which the numerator is greater than the denominator Independent events: successive events that have no effect on each other Independent variable: this is the x-axis (or horizontal) variable Index (power or exponent): the number expressing the power to which a number or pronumeral is raised. For example, in the expression 32, the index is 2. Plural: indices. Inequality signs: signs used in inequations. They are < (less than), > (greater than), Ç (less than or equal to) and í (greater than or equal to). Inequations: similar to equations, but contain an inequality sign instead of an equal sign. For example, x = 3 is an equation, but x < 3 is an inequation. Infinite: never-ending; for example, the decimal 0.3 is nonterminating and therefore its number of decimal places cannot be counted. Integers (Z): These include the positive and negative whole numbers, as well as zero; that is, .  .  ., -3, -2, -1, 0, 1, 2, .  .  . Interpolation: the process of predicting a value of a variable from within the range of the data Interquartile range: the difference between the upper (or third) quartile, QU (or Q3), and the lower (or first) quartile, QL (or Q1); that is, IQR = QU - QL = Q3 - Q1. It is the range of approximately the middle half of the data. Intersection (of sets): region that represents the common elements of two or more sets. A ¶ B denotes the intersection of sets A and B. Inverse variation: describes a particular relationship between two variables (or quantities); that is, as one variable increases, the other decreases. The rule used to k relate the two variables is y = . x Irrational numbers (I): numbers that cannot be written as fractions. Examples of irrational numbers include surds, p and non-terminating, non-recurring decimals. Isometric drawing: a 2-dimensional representation of a 3-dimensional shape in which vertical lines remain vertical, horizontal lines are drawn at an angle and parallel edges remain parallel. Inverse Law: when the additive inverse of a number or pronumeral is added to itself, the sum is equal to 0. When the multiplicative inverse of a number or pronumeral is multiplied by itself, the product is equal to 1. So, 1 x + (-x) = 0 and x ì = 1. x Iterations: repeated calculations

894

Glossary

Inverse operation: the operation that reverses the effect of the original operation. Addition and subtraction are inverse operations; multiplication and division are inverse operations. Isosceles triangle: a triangle with two sides equal in length Intercepts: points where a curve crosses the x- or y-axis Kite: a quadrilateral with two pairs of adjacent sides equal. A kite may be convex or non-convex.

Like terms: terms that contain exactly the same pronumeral (letter) part; for example, 3ab and 7ab are like terms but 5a is not. Line of best fit: a straight line that best fits the data points of a scatterplot that appear to follow a linear trend. It is positioned on the scatterplot so that there is approximately an equal number of data points on either side of the line, and so that all the points are as close to the line as possible. Line of vision: the straight line from an observation point to the object being viewed Line symmetry: a figure has line symmetry if one or more lines (‘line of symmetry’ or ‘axis of symmetry’) can be drawn that divide the figure into two mirror images. Linear equation: an equation involving pronumerals of degree 1. The general form of a linear equation in one variable is ax + b = 0. Linear graphs: consist of an infinite number of points that can be joined to form a straight line Linear modelling: applies the principle of linear equations to represent practical situations Logarithm: the power to which a given positive number b, called the base, must be raised in order to produce the number x. The logarithm of x, to the base b, is denoted by logb x. Algebraically: logb x = y ä by = x; for example, log10 100 = 2 because 102 = 100. Logarithm Laws: Law 1: loga x + loga y = loga (xy) x Law 2: loga x - loga y = loga    y  Law 3: loga x n = nloga x Law 4: loga 1 = 0 Law 5: loga a = 1 1 Law 6: loga   = -loga x x Law 7: loga ax = x Lower (or first) quartile: the score that marks the end of the first quarter in an ordered set of data. It is denoted by QL or Q1. It is calculated by finding the median of the lower half of the data. Lowest common denominator (LCD): the lowest number that denominators of all fractions considered in a problem will divide equally into

Lowest common multiple (LCM): the lowest multiple that two or more numbers have in common Many-to-one correspondence: a function or mapping that takes the same value for at least two different elements of its domain Maximum turning point: the highest point of a parabola that is inverted Mean: one measure of the centre of a set of data. It is given ∑x sum of all scores . When data are or x = by mean = number of scores n ∑( f × x ) presented in a frequency distribution table, x = . n Measures of central tendency: mean, median and mode Measures of spread: range, interquartile range, standard deviation Median: one measure of the centre of a set of data. It is the middle score for an odd number of scores arranged in numerical order. If there is an even number of scores, the median is the mean of the two middle scores when they are ordered. Its location is determined by the rule

n +1

. 2 For example, the median value of the set 1 3 3 4 5 6 8 9 9 is 5, while the median value for the set 1 3 3 4 5 6 8 9 9 10 is the mean of 5 and 6 (5.5). Midpoint: the midpoint of a line segment is the point that divides the segment into two equal parts. The coordinates of the midpoint M between the two points P(x1, y1) and  x + x2 y1 + y2  Q(x2, y2) is given by the formula  1 , .  2 2  y Q(x2, y2) y 2

y y1

0

M(x, y) T P(x1, y1) x1

S x

Multiple: a number that is the product of a given number and any whole number greater than zero. For example, the multiples of 3 are 3, 6, 9, 12, 15, .  .  . Multiplication law of probability: if events A and B are independent, then: P(A and B) = P(A) ì P(B) or P(A ¶ B) = P(A) ì P(B). Mutually exclusive events: events that cannot occur together. On a Venn diagram, two mutually exclusive events will appear as disjoint sets. Natural numbers: the set of positive integers, or counting numbers; that is, the set 1, 2, 3, .  .  . Net: a 2-dimensional plan of a solid that can be cut out and folded to form that solid. Below is the net of a cube.

x2 x

Mixed number (numeral): a number that consists of a whole number part and a fractional part, 1 for example, 2 3 Minimum turning point: the lowest point of a parabola which is upright Minutes (angle): units of angular measurement, where 1 degree (1è) = 60 minutes (60Å) Modal class: the term used when analysing grouped data. It is the class interval with the highest frequency. Mode: one measure of the centre of a set of data. It is the score that occurs most often. There may be no mode (all scores occur once), one mode or more than one mode (two or more scores occur equally frequently). Monic: a monic pronumeral or monic expression is one in which the coefficient of the leading term is 1. For example, x2 + 4x - 3 is monic, while 6x2 + 4x - 3 is not.

Nominal (data): a type of categorical data in which the information is divided into subgroups; for example, eye colour (hazel, blue, green) Non-recurring decimals: decimals that have no repeating digits or pattern; for example 5.482  786  2.  .  . Non-terminating decimals: decimals that have an infinite number of decimal places Null Factor Law: if a ì b = 0, then either a = 0 or b = 0 or both a = 0 and b = 0; used when solving quadratic equations Numerator: the upper number of a fraction that represents the number of equal fractional parts Numerical (data): data that can be measured or counted Object (similar figures): the original figure is classed as the object Octahedron: a regular polyhedron (platonic solid) with 8 faces, all of which are equilateral triangles Odds: relates to probabilities in gambling. They are given as 5 ratios, such as 5-1, 1 or 5 : 1. Ogive (cumulative frequency polygon): a graph formed by joining the top right-hand corners of the columns of a cumulative frequency histogram One-to-one correspondence: refers to the relationship between two sets such that every element of the first set corresponds to one and only one element of the second set Ordinal (data): a type of categorical data in which the information is in some type of ranked order; for example, first, second, third, .  .  . Order of rotational symmetry: the number of times a figure coincides with its original position in turning through one full rotation. For example, an equilateral triangle has rotational symmetry of order three and a square has rotational symmetry of order four. Orthogonal drawing: a drawing that consists of the front view, the top view and the side views of an object Outcome: the result obtained when a probability experiment is conducted

Glossary

895

Outlier: a piece of data that is considerably different from the rest of the values in a set of data; for example, 24 is the outlier in the set of ages {12, 12, 13, 13, 13, 13, 13, 14, 14, 24}.

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Age Palindromic numbers: numbers that are the same if read forwards or backwards, for example 33, 16 561 Parabola: the graph of a quadratic function has the shape of a parabola. For example, the typical shape is that of the graph of y = x2. y 9 8 7 6 5 4 3 2 1

0 -4 -3 -2 -1 -1

1 2 3 4x

Most often this simply means adding a vertical percentage axis on the right-hand side of a cumulative frequency graph. Percentile: the value below which a given percentage of all scores lie. For example, the 20th percentile is the value below which 20% of the scores in the set of data lie. Perimeter: the distance around the boundary of a 2-dimensional shape Periodic functions: functions that have graphs that repeat themselves continuously in cycles, for example, graphs of y = sin x and y = cos x. The period of the graph is the distance between repeating peaks or troughs. Perpendicular: perpendicular lines are at right angles to each other. The product of the gradients of two perpendicular lines is -1. Pi (π): the Greek letter p represents the ratio of the circumference of any circle to its diameter. The number 22 p is irrational, with an approximate value of , and a 7 decimal value of p = 3.141 59  .  .  .. Pie chart (graph): see Sector graph Platonic solids: five regular polyhedra, that is, five polyhedra whose faces are regular congruent polygons: tetrahedron (4 faces); cube (6 faces); octahedron (8 faces); dodecahedron (12 faces); icosahedron (20 faces) Plotting: placing points on a Cartesian plane, using their coordinates Polygon: a plane figure bounded by line segments

-2

Before

Time

After

Parallel: parallel lines in a plane never meet, no matter how far they are extended. Parallel lines have the same gradient. Parallel boxplots: two or more boxplots drawn on the same scale to visually compare the five-number summaries of the data sets. These boxplots compare the pulse rates of the same group of people before and after exercise.

70 80 90 100 110 120 130 140 150 Pulse rate Parallelogram: a quadrilateral with both pairs of opposite sides parallel A

D

B C Payout: winnings paid, for example, on races. A payout is made on the ratio given by the odds, with the initial investment being returned plus the winning amount. Percentage: a fraction whose denominator is 100; for 28 example, 28% = 100 Percentage cumulative frequency polygon: a cumulative frequency polygon expressed as a percentage of the total. 896

Glossary

Polyhedron: a solid in which each face is a polygon; plural: polyhedra Population: the whole group from which a sample is drawn Power: see Index Primary data: data collected by the user Prime factor: a prime number that divides a given number exactly; for example, the prime factors of 42 are 2, 3 and 7 Prime number: a number that has only two different factors — itself and one. For example, 3 is a prime number because its only factors are 1 and 3. Note that 1 is not a prime number because its two factors 1 ì 1 are the same. Prism: A solid comprising two congruent parallel faces (bases) and the (lateral) faces that connect them. The lateral faces are parallelograms. If they are all right-angled (i.e. rectangles) the prism is a right prism; if they are not all right-angled, then the prism is an oblique prism.

Right rectangular prism

Oblique rectangular prism

Right triangular prism

Probability: the likelihood or chance of a particular event (result) occurring. number of favourable outcomes P(event) = . number of possible outcomes The probability of an event occurring ranges from 0 (impossible — will not occur) to 1 (certainty — will definitely occur). Product: the result of a multiplication Pronumerals: letters used in place of numbers Proportion: corresponding elements are in proportion if there is a constant ratio; for example, circumference = p diameter for all circles Pyramid: a group of solids with any polygon as the base. Its other faces are triangles that meet at a common vertex. Pyramids are named according to their base. For example, a pyramid with a square base is a square pyramid.

Square-based Triangular-based Hexagonal-based pyramid pyramid pyramid Pythagoras’ theorem: in any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. This is often expressed as c2 = a2 + b2. A

c

b

a B C Quadrant: a sector with an arc equal to a quarter of a circle (and therefore centre angle of 90è) Quadrant 1: the quarter of the unit circle where the value of the angle being considered is between 0è and 90è. That is, the x- and y-coordinates are both positive. Quadrants of a Cartesian plane: four regions of the Cartesian plane produced by the intersection of the x- and y-axes Quadratic equation: the general form of the quadratic equation is ax2 + bx + c = 0. Quadratic formula: gives the roots of the quadratic equation ax2 + bx + c = 0. It is expressed as: −b ± b 2 − 4 ac . x= 2a Quadratic trinomial: an algebraic expression that contains three terms, in which the highest power of the pronumeral is a squared term; for example, 4x2 - 3x + 7 Quantiles: percentiles expressed as decimals. For example, the 95th percentile is the same as the 0.95 quantile. Quantitative data: data that can be counted (discrete data) or measured (continuous data), for example, the number of students enrolled in a school (discrete), the heights in centimetres of the students in a class (continuous) Quartic functions: the basic form of a quartic function is y = ax4. If the value of a is positive, the curve is upright,

while a negative value of a results in an inverted graph. A maximum of 4 roots can result. Quartiles: Values that divide an ordered set into four (approximately) equal parts. There are three quartiles — the first (or lower) quartile Q1, the second quartile (or median) Q2 and the third (or upper) quartile Q3. Quotient: the result of dividing one number or algebraic expression by another Radius: the straight line from a circle’s centre to any point on its circumference Random number: a number whose value is governed by chance, and cannot be predicted in advance Range: the difference between the highest and lowest scores in a set of data; that is, range = highest score − lowest score Rate: a particular kind of ratio where the two quantities are measured in different units; for example, km/h, $/g Ratio: the comparison of two or more quantities of the same kind. A ratio has no units. Rational numbers (Q): numbers that can be written as fractions, where the denominator is not zero Rationalising the denominator: a method used to express the denominator as a rational number. Both the numerator and denominator of a fraction are multiplied by the surd (or conjugate surd) contained in the denominator. Real numbers (R): the set of all rational and irrational numbers 1 Rectangular hyperbola: the graph of y = is called a x rectangular hyperbola. The x- and y-axes are asymptotes.

y y=

0

1 x

x

Recurring decimals: These decimals have one or more digits repeated continuously; for example, 0.999  .  .  .  . They can be expressed exactly by placing a dot or horizontal line over the repeating digits; for example, ••

8.343  434 = 8.34 or 8.34. Reflected (Quadratics): one parabola is a mirror image of the other. Reflection (Geometry): the image is a mirror image of the object. Relative frequency: represents the frequency of a particular score divided by the total sum of the frequencies. It is given by the rule: frequency of the score relative frequency of a score = . total sum of frequencies Remainder theorem: if a polynomial P(x) is divided by x - a, where a is any real number, the remainder is P(a).

Glossary

897

Revolution (angle): the size of a revolution is 360è.

360è

Rhombus: a parallelogram with all sides equal

Right angle: the size of a right angle is 90è.

X

A O B Rotation: turning a figure about a fixed point, called the centre of rotation Sample: part of a population chosen so as to give information about the population as a whole Sample space: a list of all the possible outcomes obtained from a probability experiment. It is written as x or S, and the list is enclosed in a pair of curled brackets {}. Sampling: obtaining data from a small group of subjects (often people) within a larger population. This smaller group should be representative of the larger population. Scale factor: the ratio of the corresponding sides in similar figures, where the enlarged (or reduced) figure is referred to as the image and the original figure is called the object. image length scale factor = object length Scalene triangle: a triangle with no two sides equal in length Scatterplot: a graphical representation of bivariate data that displays the degree of correlation between two variables. Each piece of data on a scatterplot is shown by a point. The x-coordinate of this point is the value of the independent variable and the y-coordinate is the corresponding value of the dependent variable. Scientific notation: a method of expressing a number as the product of a power of 10, and a decimal that has just one digit to the left of the decimal point; for example 54 267 would be written as 5.4267 ì 104 and 0.005  426 7 as 5.4267 ì 10-3. Secant: a chord of a circle that is extended beyond the circumference on one side Second Index Law: when terms with the same base are divided, the indices are subtracted. So, am ÷ an = am – n. Secondary data: data collected by others Sector: part of a circle bounded by two radii and an arc

898

Glossary

Sector graph: a type of graph mostly used to represent categorical data. A circle is used to represent all the data, with each category being represented by a sector of the circle, whose size is proportional to the size of that category compared to the total. Segment: a region of a circle between a chord and the circumference. The smaller segment is called the minor segment and the larger one is called the major segment. Semicircle: part (half) of a circle bounded by a diameter and an arc joining the ends of the diameter Set: a collection of similar elements Seventh Index Law: a term with a negative index can be 1 expressed with a positive index using this law. So, a-n = n a 1 and −n = an. a Similar figures: figures that have identical shape but different size. The corresponding angles in similar figures are equal in size, and the corresponding sides are in the same ratio, called a scale factor. Similar triangles: triangles that have similar shape but different size. There are four standard tests to determine whether two triangles are similar: AAA (angle, angle, angle), SAS (side, angle, side), SSS (side, side, side) and RHS (right angle, hypotenuse side). Simple interest: the interest accumulated when the interest payment in each period is a fixed fraction of the principal. P ×r ×T The formula used is I = , where I is the interest 100 earned (in $) when a principal of $P is invested at an interest rate of r% p.a. for a period of T years. Simple random sampling: a survey that ensures all subjects have an equal chance of inclusion in the sample Simplify: to write an expression in its simplest form by the use of algebraic or arithmetical techniques Simultaneous equations (linear): two (or more) linear graphs that have the same solution Sine (sin) ratio: the ratio of the opposite side to the hypotenuse in a right-angled triangle. So, opposite sin q  = . hypotenuse a b c Sine rule: in any triangle ABC, = = sin A sin B sin C B

c a A b C Sixth Index Law: to remove brackets containing a fraction, multiply the indices of both the numerator and denominator  a m a m by the index outside the brackets. So,   = m . b b Sketch: the drawing of a graph highlighting its special features; for example, the y-intercept and gradient or the x- and y-intercepts of a straight line. Sketches are not drawn on graph paper and scales are not shown along the axes.

Skewed: if a distribution’s shape is not symmetric, it can be described as being positively skewed (tailing off to the upper end of the distribution) or negatively skewed (tailing off to the lower end of the distribution).

Shape

Negative skew Positive skew Symmetric 70

90

110 Values

130

150

Slant height (of cone): the distance from any point on the circumference of the circular base of the cone to the vertex of the cone Solid: a 3-dimensional object Solve: to find a solution to a problem or an equation Speed: the rate that describes how quickly distance changes over a period of time. It is given by the formula distance speed = . time Standard deviation: a measure of the variability of spread of a data set. It gives an indication of the degree to which the individual data values are spread around the mean. Stem-and-leaf plot: a display that provides simultaneously a rank order of individual scores and the shape of the distribution. The stem is used to group the scores and the leaves indicate the individual scores within each group. The stem-and-leaf plot for the pulse rates of a group of students is shown. Pulse rate 6 8889 7 0114668 8 2688 9 06 10 4 11 0 A back-to-back stem-and-leaf plot has two sets of data displayed — one on either side of the common stem. Stratified random sampling: sampling of a population that consists of identifiable groups or strata, so that each group receives fair representation. Each group should make up the same proportion of the sample as it does of the full population. Subjective probability: probability that is based on one or more of the following: judgements, opinions, assessments, estimations and conjectures by individuals. It may also involve beliefs, emotions and bias. Subset: a smaller set within another set. It is denoted by the symbol ´. Substitution: the replacement of a variable by a number. For example, substituting x = 2 in the expression 5x gives 5 ì 2 = 10. Substitution method: a method used to solve simultaneous equations. It is useful when one (or both) of the equations has one of the variables as the subject.

Subtended (angle): an angle standing on an arc of a circle, with its vertex on the circumference or at the centre of the circle Supplementary (angles): angles that add to 180è Summary statistics: measures such as mean, mode, median and range, used in analysing a set of data Surds: roots of numbers that do not have an exact answer, so they are irrational numbers. Surds themselves are exact numbers; for example, 6 or 3 5 . Symmetrical: the identical size, shape and arrangement of parts of an object on opposite sides of a line or plane Tangent (to a circle): a straight line that touches the circumference of a circle or a curve at one point only

T

Tangent (tan) ratio: the ratio of the opposite side to the adjacent side in a right-angled triangle. So, opposite tan q  = . adjacent Terminating decimals: decimals that have a fixed number of places; for example, 0.6 and 2.54 Tetrahedron: a regular polyhedron (platonic solid) with 4 faces, all of which are equilateral triangles Theorems: rules or laws Theoretical probability: given by the rule number of favourable outcomes P(event = or number of possible outcomes n( E ) , where n(E) = number of times or ways an        P ( E ) = n(S ) event, E, can occur and n(S) = number of elements in the sample space or number of ways all outcomes can occur, given all the outcomes are equally likely Third Index Law: any term (excluding 0) with an index of 0 is equal to 1. So, a0 = 1. Time series: a sequence of measurements taken at regular intervals (daily, weekly, monthly and so on) over a certain period of time. They are used for analysing general trends and making predictions for the future. Total surface area (TSA): the area of the outside surface of a 3-dimensional figure Transformations (Quadratics): changes that occur to the basic parabola y = x2 in order to obtain another graph. Examples of transformations are translations, reflections or dilations. Translated (Quadratics): moving a parabola horizontally (left/right) or vertically (up/down) Translation (Geometry): occurs when an object moves up, down, left or right without flipping, turning or changing size Transposing: changing the order of terms in an equation

Glossary

899

Transversal: a line that meets two or more other lines in a plane

Trapezium: a quadrilateral with at least one pair of opposite sides parallel

Travel graphs: graphs that represent the relationship between distance and time. The distance covered is shown on the vertical axis and the time taken on the horizontal axis. Tree diagrams: branching diagrams that list all the possible outcomes of a probability experiment. This diagram shows the outcomes when a coin is tossed twice. Head, Head Head

Head

Tail

Tail

Head

A

Head, Tail

Tail, Tail Trend line: the line of best fit that is drawn on a time series graph, which is used to forecast future values Trial: the number of times a probability experiment is conducted Triangle: a 3-sided polygon Trigonometric ratios: three different ratios of one side of a triangle to another. The three ratios are the sine, cosine and tangent ratios. Trinomial: an expression consisting of three terms; for example, x2 + 3x - 5 True bearings: directions that are written as the number of degrees (3 digits) from north in a clockwise direction, followed by the word true or T; for example, due east would be 090è true or 090èT Truncated cone: a cone with its top cut off Turning point: the point at which the graph of a quadratic function (parabola) changes direction (either up or down) Two-way tables: a table that lists all the possible outcomes of a probability experiment in a logical manner

900

Glossary

B

Tail, Head

Tail

Hair colour Red Brown Blonde Black Total

Uniform cross-section: a solid has a uniform cross-section if cross-sections taken parallel to its base are always the same size and shape. Cross-sections parallel to the base of prisms are uniform, whereas cross-sections parallel to the base of pyramids are not. Union (of sets): represents the combination of elements of two or more sets. A ß B denotes the union of sets A and B. Unit circle: a circle with its centre at the origin and having a radius of 1 unit Unit fraction: a fraction that has a numerator of one; for 1 1 1 example, , , 2 5 10 Univariate data: data relating to a single variable Universal set (x ): the largest set that contains all possible elements of the data considered Upper quartile: the score that marks the end of the third quarter in an ordered set of data. It is denoted by QU or Q3. It is calculated by finding the median of the upper half of the data. Venn diagrams: a series of circles, representing sets, within a rectangle, which represents the universal set. They show the relationships between the sets.

Hair type 1 8 1 7 17

1 4 3 2 10

Total 2 12 4 9 27

Vertex: plural: vertices; a point where two or more sides of a polygon or edges of a solid meet. For example, a square has 4 vertices and a cube has 8 vertices. Vertical: vertical lines are parallel to the y-axis and have an undefined (infinite) gradient. Vertical line test (function): the graph of a function cannot be crossed more than once by any vertical line. Vertically opposite angles: when two lines intersect, four angles are formed at the point of intersection, and two pairs of vertically opposite angles result. Vertically opposite angles are equal.

Y

B O

A

X

Volume: the amount of space a 3-dimensional object occupies. The units used are cubic units, such as cubic centimetres (cm3) and cubic metres (m3). Wedge: a piece of wood, metal etc., thick at one end, tapering to a thin edge. x-intercept: the point where a graph intersects the x-axis y-intercept: the point where a graph intersects the y-axis. In the equation of a straight line, y = mx + c, the constant term, c, represents the y-intercept of that line.

Index Addition Law of probability  398–400 algebra see linear algebra algebraic expressions, expanding  221–7, 242 binomial expansion  221–2 difference of two squares rule  223–4 FOIL method  222–3 perfect squares  223 algebraic fractions adding and subtracting  33–6, 50 exercises  36–7, 39–40 multiplying and dividing  37–9, 50 simplifying  220 alternate segment theorem  718–19 amplitude of graphs  755 angles, naming  326 angles of depression  161–4, 178 angles of elevation  161–4, 178 arcs  702 area  185 common shapes  184 composite figures  187–8 exercises  189–93 Heron’s formula  746–7 of triangles  745–9, 762–3 using formulas to find  184, 185–7 see also total surface area area units, conversion of  184 Associative Law  30 bar graphs, reading  498 bearings  165 and compass directions  165–72, 178 exercises  169–72 true bearings  165, 166 bi-modal graphs  459 binomial expansion  221–2 bivariate data  474, 769, 794 correlation  769–72 drawing column graphs from data tables  477–8 eBookplus activities  496 exercises  476–7, 481–3, 773–6, 784–6 graphing  477–83, 491 identifying related pairs of variables  474–7, 491 identifying a relationship  475–6 lines of best fit  776–86 review exercises  492–5 scatterplots  483–90, 769 variables  474–5 box-and-whisker plots  464 exercises  447–9 five-point summary  444–5 identification of extreme values  445–7 multiple or parallel  454 buying on terms  542–5, 560 capacity, volume  207–8 Cartesian plane  57 cash payments  540 causation, and correlation  771–2

census  499, 501 central tendency, measures of  431–9, 464 exercises  435–9 grouped data  433–5 mean  432, 433 median  431, 433 modal class  433–5 mode  431 ungrouped data  431–3 chords intersecting  708–9, 724 as parts of a circle  701, 725 and radii  711–13 circle geometry angles in a circle  701–8, 724 constructing a tangent  704–6 cyclic quadrilaterals  715–18, 725 eBookplus activities  730 exercises  706–8, 713–15, 720–3, 726–9 intersecting chords, secants and tangents  708–15, 724 parts of a circle  701–2 tangents, secants and chords  718–23, 725 theorem 1  703 theorem 2  703 theorem 3  704 theorem 4  704 theorem 5  705 theorem 6  709 theorem 7  710 theorem 8  711 theorem 9  712 theorem 10  712 theorem 11  715 theorem 12  716 theorem 13  718–19 theorem 14  719 circles  315, 320 angles in  703–4, 724 area formula  185 centre of  701 equation of  315–18 exercises  317–18 transformation  686 translation  687 see also unit circles circumference  701 climate change project  534–5 Closure Law  30 column graphs drawing from data tables  477–8 reading  472 using to create a scatterplot  478–80 Commutative Law  29–30 compass directions and bearings  165–72, 178 exercises  169–72 true bearings  165, 166 complementary events  382, 396–7, 423 determining  380 exercises  400–3

Index

901

composite figures area  187–8 volume  206–7 composite solids, total surface area  196–9 compound interest  549–53, 560 conditional probability  417–20, 424 cones, total surface area  195 congruence exercises  329–32, 338–9 and proof  336–9, 347 review  327–32, 347 congruent triangles corresponding sides and angles  326, 700 tests to prove  327–9, 700 continuous data  472 coordinate geography determining linear equations  64–8 distance between two points  68–71 eBookplus activities  88 exercises  85–7 midpoint on line segment  71–4 parallel and perpendicular lines  74–82 sketching linear graphs  57–63 coordinate points, plotting  473 correlation  769 and causation  771–2 linear and non-linear relationships  770 positive and negative correlation  770 strength of  771 correlation coefficient  782–3 cosine, calculating  732 cosine graphs  755–9 cosine ratio (CAH)  147–9, 177 calculating the angle from  732 cosine rule  741–4, 762 credit card payments  540 cube roots calculating  2 estimating  2 linking with cubes  2 using a calculator to evaluate  2 cubes linking with cube roots  2 total surface area  184, 193 volume  184, 203 cubic functions  679–83, 694 reflection  690 transformation  689–90 translation  689 cuboid  193 cyclic quadrilaterals  715–18, 725 cylinders total surface area  194 volume  203 data see bivariate data; univariate data data analysis  512 graphing statistical data  512–15 in statistical investigations  522 data collection evaluating methods of  511–12 in statistical investigations  521

902

Index

data interpretation see interpreting data data organisation, in statistical investigations  522 data sets comparing  454–9, 464–5 exercises  455–9 data types distinguishing between  498 exercises  508–10 qualitative data  472 quantitative data  472 see also primary data; secondary data deductive geometry congruence and proof  336–9, 347 congruence review  327–32, 347 eBookplus activities  351 eBookplus ICT activity  352–3 exercises  349–50 quadrilaterals: definitions and properties  340–3, 347 quadrilaterals and proof  344–6, 347–8 similarity review  332–6, 347 dependent events  413–17, 423 dependent variables  472, 474–5, 498 depreciation  553–6, 560 depression, angles of  161–4, 178 diagrams, drawing from given directions  132 diameter  701 discounts percentage discounts  538 successive  546–8, 560 discrete data  472 disjoint sets  386 distance between two points  68–71 elevation, angles of  161–4, 178 ellipse, area formula  185 equation of a straight line  64–6 equations  40 solving using laws of logarithms and indices  624–9, 631 expanding brackets  220 expanding a pair of brackets  220 experimental probability  381–2 experiments, primary data  504 exponential functions  306, 694 exercises  309–12, 675–8 exponential decay  671 exponential growth  671 and their graphs  306–12, 319 transformation  688–9 extrapolation  781 extreme values, identification of  445 factor theorem  649–51, 659 factorising by completing the square  236–40, 242–3 by taking out a common binomial factor  220 by taking out the highest common factor  220, 248 difference of two squares expressions  636 finding a factor pair that add to a given number  248 mixed factorisation  240–1 polynominals  651–5 quadratic trinomials  636 factorising expressions with two terms  231–2, 242 a2 - b2  232 exercises  234–6

factorising expressions with three terms  227–31, 242 ax 2 + bx + c when a = 1  227–8 ax 2 + bx + c when a ò 1  228–9 exercises  229–31 factorising expressions with four terms  231, 232–6, 242 figures, naming  326 financial maths buying on terms  542–5, 560 compound interest  549–53, 560 depreciation  553–6, 560 eBookplus activities  564 exercises  540–2, 543–5, 547–8, 551–3, 554–6, 558–9, 562–3 loan repayments  556–9, 560 payment options  540 purchasing goods  539–42, 560 successive discounts  546–8, 560 five-point summary, box-and-whisker plots  444–5 flood in backyard project  352–3 FOIL method, expanding algebraic expressions  222–3 formulas, rearranging  132 fractional indices  12–16, 23, 609–14, 630 fractions addition  28, 380 division  28 multiplication  28 multiplying for calculating probabilities  380 simplifying  380 subtraction  28, 380 frequency distribution table calculation of mean, median and mode from  432–3 presentation of data  430 function notation  667 functions  666–9 circles  315–18, 320 cubic functions  679–83, 694 eBookplus activities  324 evaluating  667 exercises  321–3 exponential functions  671–8, 694 exponential functions and their graphs  306–12, 319 function notation  667 hyperbola  312–15, 320 identifying features of  667–8 plotting parabolas  281–6, 319 points of intersection  668–9 quartic functions  683–6, 694 sketching parabolas of the form y = ax 2 + bx + c  298–306, 319 sketching parabolas in turning point form  292–8, 319 sketching parabolas using basic graph of y = x 2  287–92 transformations  686–93, 695 vertical line test  666 functions and relations  665–71, 694 eBookplus activities  698 exercises  696–7 relations  665–6 gradient finding  664 finding given two points  90, 768 linear graphs  57 lines  56

Heron’s formula  746–7 highest common factor, finding  28 horizontal lines  77–9 hyperbolas  312–15, 320 dilation  688 negative values of k  688 transformation  688 hypotenuse, finding  134 Identity Law  30 independent events  413–17, 423 independent variables  472, 474–5, 498 index expressions, substitution into  664 index form  2 evaluating numbers in  590 using calculator to evaluate numbers in  2 index laws  3–7, 23 First Index Law  3 Second Index Law  3 Third Index Law  3 Fourth Index Law  4 Fifth Index Law  4 Sixth Index Law  4 Seventh Index Law  8 Eighth Index Law  12–13 combining  17–20, 23 exercises  5–7, 20–22, 24–5 using  590 indices combining index laws  17–23 eBookplus activities  26 exercises  24–5 fractional indices  12–16 negative indices  7–12 review of index laws  3–7 inequations  106 inquiry methods, evaluating  511–15, 518–21, 527 integers  592 interest compound interest  549–53, 560 simple interest  538 interpolation  781 interpreting data bivariate data  769–76, 794 eBookplus activities  798 exercises  773–6, 790–3, 795–7 lines of best fit  776–86, 794 time series  786–93, 794 interquartile range  440–2 Inverse Law  30 irrational numbers  592–3 lay-by payments  540 least squares regression  781–2, 794 like terms, collecting  28 line graphs, reading  473 line segments, midpoint of  71–4 line of vision  161 linear algebra adding and subtracting algebraic fractions  33–7 eBookplus activities  54 exercises  52–3 multiplying and dividing algebraic fractions  37–40

Index

903

linear algebra (continued) solving equations with algebraic fractions and multiple brackets  45–9 solving linear equations  40–4 substitution  29–33 linear equations horizontal and vertical lines  77–9 parallel lines  74–5, 79–82 perpendicular lines  75–6, 79–82 transposing to standard form  90, 768 linear equations, determining  64–6 equation of a straight line  64–6 exercises  67–8 point–gradient method  66 straight line parallel or perpendicular to another straight line  76–7 linear equations, solving  40–7, 50 exercises  43–4, 48–9 involving algebraic fractions  45–9, 50 with multiple brackets  45, 50 that arise when finding x- and y- intercepts  56, 90, 768 where pronumeral appears on both sides  41–3 linear graphs  57 Cartesian plane  57 gradient/slope  57 quadrants  57 linear graphs, sketching  57–62 exercises  62–3 gradient–intercept method  59–60 x- and y-intercept method  58–9, 90 y = c and x = a form  61–2 y = mx form  60 linear inequations exercises  108–10, 113–16, 118–22 multiplying or dividing both sides by negative numbers  107–8 sketching  110–16, 124 solving  106–10, 123 solving simultaneous  116–22, 124 linear relationships  770 lines describing gradient  56 identifying equations of straight lines  664 naming  326 plotting using table of values  56 sketching straight lines  664 lines of best fit  485, 776–86, 794 correlation coefficient  782–3 exercises  784–6 interpolation and extrapolation  781 least squares regression  781–2 reliability of predictions  781 loan repayments  556–9, 560 loans  543 logarithm laws  619–24, 631 exercises  622–4 Law 1  620 Law 2  620–1 Law 3  621 Law 4  622 Law 5  622 Law 6  622 Law 7  622 logarithms  617–19, 631

904

Index

Manning’s formula  589 many-to-many relations  665 many-to-one relations  665 mean calculation from frequency distribution table  432–3 grouped data  433–5 of small data set  430 of stem-and-leaf plot  430 ungrouped data  431 measurement, primary data  504 measures of central tendency see central tendency, measures of measures of spread see spread, measures of media reports, investigating  523–5 median calculation from frequency distribution table  432–3 grouped data  433–5 small data set  430 of stem-and-leaf plot  430 ungrouped data  431 midpoint formula  71 mixed factorisation  240–1 modal class, grouped data  433–5 mode calculation from frequency distribution table  432–3 small data set  430 of stem-and-leaf plot  430 ungrouped data  431 Multiplication Law of probability  413, 418 mutually exclusive events  397–8, 400–3, 423 negative correlation  770 negative indices  7–12, 23, 614–17, 631 non-integer rationals  592 non-linear relationships  770 normal distribution  459 Null Factor Law  249 number classification exercises  594–5 integers  592 irrational numbers  592–3 non-integer rationals  592 p (pi)  593 rational numbers  591 review  591–4, 630 number laws  29–30 Associative Law  30 Closure Law  30 Commutative Law  29–30 Identity Law  30 Inverse Law  30 observation, primary data  504 odds  390–2 for and against  390 payouts  390 one-to-many relations  665 one-to-one relations  665 outliers  445 parabolas  287 identifying equations of  664 parabolas, plotting  281, 281–6, 319 dilation  281, 287 exercises  284–6

finding turning point when equation is not in turning point form  298–302 horizontal translation  288–9 maximum turning point  281 minimum turning point  281 reflection  281, 289–90 transformation  281 translation  281, 287–9 turning points  281, 298–302 vertical translation  287–8 parabolas, sketching  664 exercises  291–2, 296–8, 302–6 form y = ax2 + bx + c  298–306, 319 in turning point form  292–8, 306 using basic graph of y = x2  287–92, 319 parallel lines  74–5 exercises  79–82 and simultaneous linear equations  93–4 parallelograms area formula  185 definition  340 properties  340 test  344 percentage discounts  538 percentages converting to a decimal  538 decreasing a quantity by  538 of a quantity of money  528 periodic functions  755 perpendicular lines  75–6, 79–82 p (pi)  593 points, distance between two  68–71 points of intersection  664, 668–9 polynomial equations, solving  655–8, 659 polynomial values  647–9, 659 polynominals  637–9, 659 adding  639–41, 659 degree of  637 eBookplus activities  662 exercises  661 factor theorem  649–51, 659 factorising  651–5, 659 factorising using long division  651–3 factorising using short division  653–4 leading coefficient  637 leading term  637 long division of  641–6, 659 multiplying  639–41, 659 polynomial values  647–9, 659 remainder theorem  649, 650–1, 659 subtracting  639–41, 659 populations  499 exercises  502–3 and samples  499–503, 527 positive correlations  770 predictions from scatterplots  485–8 reliability of  781 primary data  503–7, 527 exercises  508–10 experiments  504 measurement  504 observation  504 simulations  504

in statistical investigations  521 surveys  504 probability Addition Law of  398–400 complementary events  382, 396–7, 400–3, 423 conditional probability  417–20, 423 eBookplus activities  428 equally-likely outcomes  381 events  381 exercises  392–6, 400–3, 410–12, 415–17, 419–20, 425–7 experimental probability  381–2 frequency  381 independent and dependent events  413–17, 423 Multiplication Law of  413, 418 mutually exclusive events  397–8, 400–3, 423 odds  390–2 outcome  381 relative frequency  382–3 review  381–92, 423 sample space S  381 subjective probability  420–2, 423 theoretical probability  383–4 tree diagrams  405–12, 423 trial  381 two-way tables  403–5, 423 Venn diagrams  384–90 problem solving exercises  355–78, 565–88 proof of congruence  336–9, 347 quadrilaterals  344–6, 347 proportions, finding  498 purchasing goods  539–42, 560 exercises  540–2 payment options  540 pyramids, volume  205–6 Pythagoras’ theorem  68, 134, 177 exercises  137–40, 143–5 finding the hypotenuse  134 finding a shorter side  135–7 similar right-angled triangles  133–4 in three dimensions  140–5, 177 using  56 quadrants  57 quadratic equations eBook activities  278 exercises  276–7 solving  636 substitution into  248, 280, 636 quadratic equations, solving  249–55, 274 by completing the square  250–2 by inspecting graphs  258–63, 274 by interpolation  263, 274 by using the discriminant  264–8 confirming solutions  260–1 exercises  253–5, 261–3, 267–8, 272–3 and linear equations simultaneously  269–73, 274–5 Null Factor Law  249 type ax2 + bx + c = 0 where a = 1  280 type ax2 + bx + c = 0 where a ò 1  281 using quadratic formula  255–8, 274, 280 quadratic expressions eBookplus activities  246 exercises  244–5

Index

905

quadratic expressions (continued) expanding algebraic expressions  221–7 factorising by completing the square  236–40 factorising expressions with three terms  227–31 factorising expressions with two or four terms  231–6 mixed factorisation  240–1 quadratic formula  255–8, 274, 280 quadratic functions dilation  687 horizontal translation  687 reflection  687 transformation  686 vertical translation  686 quadratic trinomials  227, 636 quadrilaterals in circles  715–18, 725 definitions  340, 347 exercises  341–3, 345–6 identifying  326 and proof  344–6, 347–8 properties  340, 347 tests  344, 347–8 qualitative data  472 quantitative data  472 quartic functions  683–6, 694 basic shapes of quartic graphs  683–5 exercises  685–6 reflection  690 transformation  690 quartic graphs, basic shapes  683–5 radii and chords  711–13 as parts of a circle  701 random samples  499 range  440 rational numbers  591 rationalising denominators surds  603–5 using conjugate surds  605–7 real numbers eBookplus activities  634 exercises  632–3 fractional indices  609–14, 630 logarithm laws  619–24, 630 logarithms  617–19, 631 negative indices  614–17, 632 number classification review  591–5, 630 operations with surds  599–600, 630 solving equations  624–9, 631 surds  595–9, 630 rectangles area formula  185 definition  340 properties  340 test  344 rectangular prisms total surface area  184, 193 volume  184, 203 relations  665–6 many-to-many relations  665 many-to-one relations  665 one-to-many relations  665 one-to-one relations  665

906

Index

relative frequency  382–3 remainder theorem  649, 650–1, 659 rhombus area formula  185 definition  340 properties  340 test  344 right-angled triangles finding angle in  732 finding side lengths  732 labelling the sides  132, 732 rise, measuring  56, 90, 768 rounding size of angle to nearest minute and second  132 to a given number of decimal places  132 run, measuring  56, 90, 768 samples  499–500 exercises  502–3 and populations  499–503, 527 scales, reading  472 scatterplots  478, 483–90, 491, 769 line of best fit  485 making predictions  485–8 using column graphs to create  478–80 secants intersecting  709–10, 724 as parts of circles  702 and tangents  719–20, 725 secondary data  507–8, 527 exercises  508–10 in statistical investigations  522–3 sectors area formula  185 as parts of circles  702 segments alternate segment theorem  718–19 as parts of circles  702 set notation  380 sets, Venn diagrams  384 similar triangles, tests to prove  333, 700 similarity exercises  335–6 review  332–6, 347 testing triangles for  333 similarity statements, writing  326 simple interest, finding  538 simulations, primary data  504 simultaneous linear equations  91 eBookplus activities  130 elimination method of solving  99–101, 123 exercises  94–6, 98–9, 101–3, 105–6, 125–9 graphical solution  91–3, 94–6, 123 parallel lines  93–4 problem solving using  103–6 solving  116–22, 124 substitution method of solving  96–9, 123 sine, calculating  732 sine graphs  755–9 sine ratio (SOH)  147–9, 177 calculating the angle from  732 sine rule  769 ambiguous case  735–9 exact values  733–5

skewness  459–63, 465 exercises  461–3 negatively skewed data  460 positively skewed data  460 small data sets finding the mean  430 finding the median  430 finding the mode  430 spheres total surface area  194 volume  205 spread, measures of  439–44, 464 exercises  442–4 interquartile range  440–2 range  440 square roots calculating  2 estimating  2 linking with squares  2 using a calculator to evaluate  2 squares completing  280, 664 linking with square roots  2 squares (quadrilateral shape) area formula  185 definition  340 properties  340 standard deviation  449–53, 464 statistical data, graphing  512–15 statistical graphs, drawing  430 statistical investigations data analysis  522 data collection  521 exercises  525–6 investigating media reports  523–5 organising the data  522 performing calculations  522 reporting results  522 steps in  528 using primary data  521 using secondary data  522–3 statistical reports, evaluating  515–21, 527 statistics in the media eBook plus ICT activity  534–5 eBookplus activities  533 evaluating inquiry methods and statistical reports  511–20 exercises  529–32 populations and samples  499–503 primary and secondary data  503–10 statistical investigations  521–6 stem-and-leaf plots back-to-back  454 mean, median and mode  430 straight lines see lines subjective probability  420–2, 424 subsets  386 substitution  29–31, 50 exercises  31–3 into index expressions  664 into a linear rule  90, 768 into quadratic equations  248, 280, 636 number laws  29–30 to check if given point makes inequation a true statement  90 to solve simultaneous linear equations  96–9

successive discounts  546–8, 560 surds  595–8, 630 adding  590, 600 dividing  590, 602–3 exercises  597–9, 607–9 identifying  590 multiplying  590, 601–2 operations with  599–609, 630 proof by contradiction  596 proof a number is irrational  596–7 rationalising denominators  603–5 rationalising denominators using conjugate surds  605–7 simplifying  220, 248, 590, 599–600 subtracting  590, 600 surface area see total surface area survey questions, determining suitability  498 surveys  501, 504 symmetrical graphs  459 tangent, calculating  732 tangent ratio (TOA)  147–9, 177 calculating the angle from  732 tangents intersecting  711, 724 as parts of circles  702 and secants  719–20, 725 theoretical probability  383–4 time series  786–93, 794 exercises  790–3 trend lines  788–90 total surface area  213 of composite solids  196–9 cones  195 of cubes  184, 193 cylinders  194 exercises  199–202, 214–15, 216 other solids  195–6 rectangular prisms  184, 193 spheres  194 transformation  686–93, 695 circles  686 cubic functions  689–90 exercises  692–3 exponential functions  688–9 in general polynominals  690–2 hyperbolas  688 parabolas  281 quadratic functions  686 quartic functions  690 trapezium area formula  185 definition  340 properties  340 tree diagrams  403–5, 423 trend lines  788–90 triangles angle relations  700 angles in  700 area formula  185, 186 area of  745–9, 762–3 Heron’s formula  746–7 testing for similarity  333, 700 see also congruent triangles; Pythagoras’ theorem; right-angled triangles

Index

907

triangular prisms, volume  203 trigonometric equations exercises  761 solving  759–61, 763 solving algebraically  760 solving graphically  759 trigonometric functions  763 amplitude of graphs  755 exercises  757–9 periodic functions  755 sine and cosine graphs  755–6 trigonometric ratios  177 angles and the calculator  145–6 cosine ratio  147–9, 177 exercises  149–51 sine ratio  147–9, 177 tangent ratio  147–9, 177 trigonometry angles of elevation and depression  161–4, 178 applications  172–6, 178 area of triangles  745–9, 762–3 calculating angle size  156–60, 177 calculating side lengths  151–5, 177 compass directions and bearings  165–72, 178 cosine rule  741–4, 762 eBookplus activities  182, 764 exercises  154–5, 158–60, 173–6, 179–81, 739–40, 764–5 sine rule  733–40, 762 unit circles  749–54 see also Pythagoras’ theorem; trigonometric functions; trigonometric ratios trinomials  227 TSA see total surface area two squares expressions, factorising difference of  636 two-way tables  403–5, 423 unit circles  749–54, 763 constructing  750–2 exercises  752–4 four quadrants of  750–2

908

Index

univariate data  474 box-and-whisper plots  444–9 comparing data sets  454–9 eBookplus activities  470 exercises  466–9 measures of central tendency  431–9 measures of spread  439–44 skewness  459–63 standard deviation  449–53 universal set, Venn diagrams  384 variables  474 dependent and independent  474–5 identifying related pairs  474–7, 491 Venn diagrams  384–90 complement of a set  385 disjoint sets  386 intersection of sets  384 sets  384 subsets  386 union of the sets  385 universal set  384 vertical line test  666 vertical lines  77–9 equation of  248, 280 volume  203, 204–5, 213 capacity  207–8 composite figures  206–7 cubes  184, 203 cylinders  203 exercises  208–12 pyramids  205–6 rectangular prisms  184, 203 review exercises  216, 217 spheres  205 triangular prisms  203 volume units, conversion of  184 y-intercept finding  664 stating from graph  56