MathPower 10 Practise Masters

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MATHPOWER 10ONTARIO EDITION

Practice Masters

™

Shirley BarrettRichmond Hill, Ontario

Contributing WriterJanice NixonToronto, Ontario

http://www.mcgrawhill.ca/

McGraw-Hill Ryerson Limited

MATHPOWER™ 10, Ontario EditionPractice Masters

Copyright © 2001, McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rightsreserved. This publication may be reproduced for classroom purposes without the prior written permission ofMcGraw-Hill Ryerson Limited. McGraw-Hill Ryerson Limited shall not be held responsible for content if anyrevisions, additions, or deletions are made to any material provided in editable format on the enclosed CD-ROM.

This package contains the MATHPOWER™ 10, Ontario Edition, Practice Masters and one CD-ROM.

ISBN 0-07-560802-2

http://www.mcgrawhill.ca

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Printed and bound in Canada

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The Geometer’s Sketchpad® is a registered trademark of Key Curriculum Press.CBL™ and CBR™ are trademarks of Texas Instruments Incorporated.Adobe, Acrobat, and the Acrobat logo are registered trademarks of Adobe Systems Incorporated. This product isnot endorsed or sponsored by Adobe Systems Incorporated, publisher of Adobe® Acrobat® Reader 4.0 orAdobe® Acrobat® 4.0.

Canadian Cataloguing in Publication Data

Barrett, Shirley, date-Mathpower 10, Ontario ed. Practice Masters

ISBN 0-07-560802-2

Mathematics – Study and teaching (Secondary). 2. Mathematics – Problems,exercises, etc. I. Title. II. Title. Mathpower ten, Ontario edition.

QA107.M37648 2000 Suppl. 2 510 C00-931907-7

PUBLISHER: Diane WymanEDITORIAL CONSULTING: Michael J. Webb Consulting Inc.ASSOCIATE EDITOR: Mary Agnes ChallonerSENIOR SUPERVISING EDITOR: Carol AltiliaCOPY EDITOR AND PROOFREADER: Julia KeelerPERMISSIONS EDITOR: Ann LudbrookEDITORIAL ASSISTANTS: Joanne Murray, Erin PartonPRODUCTION SUPERVISOR: Yolanda PigdenPRODUCTION COORDINATOR: Jennifer VassiliouINTERIOR DESIGN: The ArtPlus GroupELECTRONIC PAGE MAKE-UP: First Folio Resource Group, Inc.COVER DESIGN: Dianna LittleCOVER ILLUSTRATIONS: Citrus MediaCOVER IMAGE: Peter Pearson/Stone



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To the Teacher v

Tips for Learning Math

Getting Started 1Know Your Textbook and How to Use It 2Classroom Learning 3Homework 4Seeking Help 5Preparing for Tests and Exams 6Writing Tests and Exams 7Problem Solving Skills 8Solving Word Problems 9Goals 10

CHAPTER 1 Linear Systems

Practice1.1 Investigation: Ordered Pairs and

Solutions 111.2 Solving Linear Systems Graphically 121.3 Solving Linear Systems by Substitution 131.4 Investigation: Equivalent Equations 141.5 Solving Linear Systems by Elimination 151.6 Investigation: Translating Words Into

Equations 161.7 Solving Problems Using Linear Systems 17Answers CHAPTER 1 Linear Systems 19

CHAPTER 2 Analytic Geometry

Practice2.1 Length of a Line Segment 212.2 Investigation: Midpoints of Horizontal

and Vertical Line Segments 222.3 Midpoint of a Line Segment 232.4 Verifying Properties of Geometric

Figures 242.5 Distance From a Point to a Line 25Answers CHAPTER 2 Analytic Geometry 27

CHAPTER 3 Polynomials

Practice3.1 Polynomials 293.2 Multiplying Binomials 303.3 Special Products 313.4 Common Factors 323.5 Factoring x2 + bx + c 333.6 Factoring ax2 + bx + c, a ≠ 1 343.7 Factoring Special Quadratics 35Answers CHAPTER 3 Polynomials 37

CHAPTER 4 Quadratic Functions

Practice4.1 Functions 394.2 Graphing y = x2 + k, y = ax2,

and y = ax2 + k 404.3 Graphing y = a(x – h)2 + k 414.4 Graphing y = ax2 + bx + c by

Completing the Square 424.5 Investigation: Sketching Parabolas in

the Form y = ax(x – s) + t 434.6 Investigation: Finite Differences 444.7 Technology: Equations of Parabolas of

Best Fit 454.8 Technology: Collecting Distance and Time

Data Using CBRTM or CBLTM 46Answers CHAPTER 4 Quadratic Functions 47

CHAPTER 5 Quadratic Equations

Practice5.1 Solving Quadratic Equations by

Graphing 515.2 Solving Quadratic Equations by

Factoring 525.3 Investigation: Graphing Quadratic

Functions by Factoring 535.4 The Quadratic Formula 54Answers CHAPTER 5 Quadratic Equations 55

CONTENTS

CHAPTER 6 Trigonometry

Practice6.1 Technology: Investigating Similar Triangles

Using The Geometer’s Sketchpad® 576.2 Similar Triangles 586.3 The Tangent Ratio 596.4 The Sine Ratio 606.5 The Cosine Ratio 616.6 Solving Right Triangles 626.7 Problems Involving Two Right Triangles 636.8 Technology: Relationships Between Angles

and Sides in Acute Triangles 646.9 The Sine Law 656.10 The Cosine Law 66Answers CHAPTER 6 Trigonometry 67

APPENDIX A:

Review of Prerequisite Skills

PracticeAdding Polynomials 71

Angle Properties I(Interior and Exterior Angles of Trianglesand Quadrilaterals) 72

Angle Properties II(Angles and Parallel Lines) 73

Common Factoring 74

Congruent Triangles 75

Evaluating Expressions I(Variables in Expressions) 76

Evaluating Expressions II(Expressions With Integers) 77

Evaluating Expressions III(Applying Formulas) 78

Evaluating Expressions IV(Non-Linear Relations) 79

Evaluating Radicals(The Pythagorean Theorem) 80

Expanding and Simplifying Expressions I(The Distributive Property) 81

Expanding and Simplifying Expressions II(Multiplying a Polynomial by a Monomial) 82

Exponent Rules I(Powers With Whole Number Bases) 83

Exponent Rules II

(Powers With Integral Bases) 84

Exponent Rules III(Multiplying Monomials by Monomials) 85

Exponent Rules IV(Powers of Monomials) 86

Exponent Rules V(Dividing Monomials by Monomials) 87

Graphing Equations I(Graphing Linear Equations) 88

Graphing Equations II(Methods for Graphing Linear Equations) 89

Graphing Equations III(Intersecting Lines) 90

Greatest Common Factors 91

Like Terms 92

Polynomials 93

Slope I(Using Points) 94

Slope II(Linear Equations: Slope and y-Intercept Form) 95

Slope III(Parallel and Perpendicular Lines) 96

Solving Equations I (Using Addition and Subtraction) 97

Solving Equations II (Using Division and Multiplication) 98

Solving Equations III (Multi-Step Equations) 99

Solving Equations IV (With the Variable on Both Sides) 100

Solving Equations V (With Brackets) 101

Solving Equations VI (With Fractions and Decimals) 102

Solving Proportions 103

Subtracting Polynomials 104

Transformations I(Translations) 105

Transformations II(Reflections) 106

Transformations III(Dilatations) 107

Answers APPENDIX AReview of Prerequisite Skills 109

To the Teacher

Tips for Learning MathThese first ten masters were designed to help students start the term with some insights intohow to succeed at Math.

Emphasis is on • examining attitudes toward Math • identifying and using available resources • being prepared for class • using class time wisely • making time for and establishing a place to do homework • doing homework daily • staying on top of new learning • seeking help• studying productively• applying problem solving skills • setting a goal for Math

You may wish to assign these masters to • all students at the beginning of the term, to be completed the first week and showing a

completed Goals master• students who display a need for some assistance in any of the areas covered

While the masters are intended to be completed by individual students, you may wish tosuggest that students work as partners to discuss the ideas on the masters.

Encourage students to keep the Tips for Learning Math masters to help them reach the goalsthey set.

PracticeIn each chapter, there is a one-page Practice master for each numbered section.

Each master • provides additional practice with the skills and concepts new to the section• can be assigned as needed • includes a variety of practice in four categories: knowledge/understanding, problem solving,

communication, and application

Review of Prerequisite SkillsIn Appendix A, after the Practice masters for Chapters 1 to 6, there are 37 masters for Review ofPrerequisite Skills.

These masters• provide additional practice with the prerequisite skills identified in the student text at the

beginning of each chapter, and in Appendix A at the back of the text• can be assigned as needed

Copyright © 2001 McGraw-Hill Ryerson Limited To the Teacher v

Answers

Answers are provided for all the masters •at the end of each chapter for the Practice masters•at the end of Appendix A for the Review of Prerequisite Skills masters

The answers include —• numerical solutions• word solutions• diagram solutions• graphical solutions

Permission to reproduce these pages is provided as you may wish to post answers for studentsto check their work.

Installing and Using MATHPOWER™ 10, Ontario Edition, Practice Masters CD-ROM

To use the MATHPOWER™ 10, Ontario Edition, Practice Masters CD-ROM requires Adobe® Acrobat®Reader 4.0, which is included on the CD. If Acrobat Reader is not already installed on your computer,insert the MATHPOWER™ 10, Ontario Edition, Practice Masters CD in the CD-ROM drive, then followthese steps:

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Copyright © 2001 McGraw-Hill Ryerson Limitedvi To the Teacher

Getting StartedMATHPOWER™ 10, Ontario Edition

Here you are starting into a new Math course. This page and the other Tips for Learning Mathpages are just for you.

You are an important resource in your success in Math. Take a look at your strengths and weaknesses when it comes to Math, and how you think andfeel about Math.

Complete each statement or circle the letter of the response that best suits you.

1. When I hear the word Math, the first few things that come into my mind are

2. What I like the least about Math is

3. What I like the most about Math is

4. In my last Math course, my mark was

a) better than b) worse than c) about the same as the marks I received in most of my other subjects.

5. In the last few years, my Math marks have been

a) staying about the same b) getting worse c) getting better

6. I plan to take Math

a) never after this course b) to the end of high school c) at university or college

7. My attitude toward Math is

a) more positive than b) less positive than c) about the same asthe attitude I had when I was in grade 7.

8. Something that I would like to change about my attitude to Math is

But you are not alone. You have other resources.

9. Circle the letter of each resource that you feel you have available, and specify when appropriate.

a) my textbook, MATHPOWER™ 10, Ontario Edition

b) my Math teacher,

c) my classmates,

d) friends taking the same Math course, but in a different class,

e) friends or family who have already taken this level of Math,

f) computer and Internet access,

Copyright © 2001 McGraw-Hill Ryerson Limited Tips for Learning Math 1

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Know Your Textbook and How to Use ItMATHPOWER™ 10, Ontario Edition

MATHPOWER™ 10, Ontario Edition was written for students your age, and, throughout the chapters, uses real information that will, hopefully, interest you.

Find five introductions or problems with real information that interests you.

1. Page Topic

2. Page Topic

3. Page Topic

4. Page Topic

5. Page Topic

Read pages xii to xxi in your textbook. Then, identify each statement as True or False.

6. Examples with fully worked solutions are provided. ________________

7. Many of the Applications and Problem Solving questions are connected to other subjectsand other Math topics. ________________

8. Communicate Your Understanding means that I have an opportunity to demonstrate myunderstanding of a topic ________________

9. Core sections are numbered and usually include an Investigation to actively involve me inmy own learning. ________________

10. There are Cumulative Review sections, at the end of Chapters 2, 4, and 6 that I can usewhen studying. ________________

11. Detailed instructions involving Technology are provided in two appendixes at the back ofthe text to help me recall essential skills. ________________

12. A Glossary is found near the end of the book to help me understand Math terms.________________

13. I can use the Review of Key Concepts sections and Chapter Test sections to test myself.________________

14. Circle the letter of each thing you can learn about when using MATHPOWER™ 10, Ontario Edition.a) how Math is connected to other disciplinesb) how people use Math in their careersc) how to use problem solving strategies to tackle problemsd) how to use technology, such as calculators and computer spreadsheetse) how different topics in Math are interconnected

15. Circle the letter of each worthwhile use of the Answers that appear near the end of thebook.a) verify whether my answers are right b) if an answer is not right, work backward from the given answer to understand where I mademy mistake, and then correct it

Copyright © 2001 McGraw-Hill Ryerson Limited2 Tips for Learning Math

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Classroom LearningMATHPOWER™ 10, Ontario Edition

Whether your Math class is 30 min or 70 min, you will gain the most from that time with yourMath teacher and your classmates, if you• come to class prepared• listen actively• take clear notes• use your time productively

1. Circle the letter of each thing that you bring to Math class.a) textbook, MATHPOWER™ 10, Ontario Edition b) notebookc) pens and/or pencils d) erasere) ruler f) calculator g) diskette h) other tools _____________________i) completed and checked homework j) a positive attitudek) questions to ask about concepts that confuse me l) enthusiasm

2. Circle the letter of each way that you actively listen in Math class. Place an X beside the letterof each way you would like to try.a) anticipating what is coming nextb) trying to connect new concepts with familiar conceptsc) sketching diagrams to illustrate my understandingd) asking questions to clarify my understandinge) identifying when the teacher says something particularly important

3. Circle the letter of each way your Math teacher points out important information.a) says it directlyb) implies with tone of voicec) repeatsd) writes it on the boarde) highlights it on the board, in a special place or in a special wayf) other ways ____________________________________________________________________________

4. Circle the letter of each technique that you use to make your Math notes clear. Place an Xbeside the letter of each technique you would like to try.a) record dateb) record textbook page referencesc) start a new paged) write legiblye) make listsf) draw diagramsg) highlight key pointsh) other techniques _______________________________________________________________________

5. Circle the letter of each good reason for using your Math class time productively.a) The teacher is available to clarify and provide direction.b) Classmates are present to work with.c) Ideas are fresh.d) I will have less homework.


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HomeworkMATHPOWER™ 10, Ontario Edition

In a subject like Math, where the concepts build day after day and year after year, it is vital thatyou keep up daily. Always do your homework.

You must do your homework daily, and to get it done you need to • make time for it• have a work space• be free of distractions to be able to concentrate

1. Circle the letter of each thing that competes with Math homework for your time.a) other homeworkb) special assignments or projects c) sports practices/gamesd) music lessons/practicee) watching television or videosf) listening to the radiog) playing video games or playing cardsh) visiting with friendsi) talking on the telephonej) household choresk) part-time jobl) shoppingm) others _____________________________________________________________________

2. Circle the letter of each work space that you use for doing your Math homework. Place an Xbeside the letter of a space you would like to try.a) own room b) a shared roomc) the kitchen d) the dining roome) another room at home f) the libraryg) another space ________________________________________________

Whenever and wherever you do your Math homework, you must have all the tools you need,and you must have minimal distractions.

Complete each statement.

3. The best time of day for me to be alert and have enough time to complete my Math

homework is

4. My distractions include

5. I can avoid

this distraction by doing this


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Seeking HelpMATHPOWER™ 10, Ontario Edition

In a subject like Math, where the concepts build day after day and year after year, it is vital thatyou keep up daily. When you do not understand something, seek help immediately.

Use your textbook, MATHPOWER™ 10, Ontario Edition. The Examples in the numbered sectionsare typical questions that use the concepts and skills of the section. Detailed step-by-step solutions are provided for each.

Ask your Math teacher for help. When asking for help, prepare to use your time and yourteacher’s time productively. Identify the point at which you first became confused. Have specificquestions about what you don’t understand.

1. Find out when your Math teacher is available, and circle the letter of each time that applies.a) during Math classb) before Math classc) after Math classd) before schoole) after schoolf) during lunchg) other times ______________________________________________________________________

A different explanation from the textbook’s or from your teacher’s can sometimes help.

2. Circle the letter of each way that you currently work with others. Place an X beside the letterof each way you would like to try.a) working in a group or with a partner in Math classb) discussing problems with friends taking the same Math course, but in a different classc) discussing problems with friends or family who have already taken this level of Mathd) working in a study group or with a study partner outside of Math classe) other ways _______________________________________________________________


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Preparing for Tests and ExamsMATHPOWER™ 10, Ontario Edition

You will be prepared for writing tests and exams, if you• develop good work habits in class and use class time productively• do your homework every day• seek help when you first become confused

The most important step in studying for a test or an exam is not one you can do the night before.It is staying on top of things right from the start.

When you are informed of a test or exam, you should find out all the specifics.

1. Circle the letter of each thing that you find out about a test or exam when you are informedof it. Place an X beside the letter of each thing you are going to start to find out about.a) exact date, time, and locationb) exact content to be coveredc) time allowed for itd) format — true/false, multiple choice, fill in the blanks, full solutions, combination e) resources that I should have with me and know how to use, such as a calculatorf) value of it, relative to all tests, assignments, projects, and examsg) resources available to help me study

2. Circle the letter of each resource that you have used in the past to prepare for Math tests andexams. Place an X beside the letter of each resource you would like to try in the future.a) my notesb) Examples and Solutions in my textbookc) Achievement Check Review, and Chapter Test sections in my textbookd) Cumulative Review sections in my textbooke) MATHPOWER™ 10, Ontario Edition, Self-Check Assessment Mastersf) teacher-provided old or sample tests or examsg) a study group or partnerh) others ___________________________________________________________________________

The amount of time you need to study depends upon • how well you have already prepared by keeping on top of things, that is, your existing

knowledge base• the amount of material to be covered

Focus your time on the important concepts and skills, and on what you are less confident about.

Studying is a cyclical process. Review the material that is to be tested. Then, test yourself usingthe Self-Check Assessment Master that you can get from your teacher for the appropriate chap-ter. Next, check your results using the Answers to the Self-Check, which you can also get fromyour teacher. Then, for any question you got wrong, use the table on the second page of the Self-Check to identify the textbook section and the specific worked example. Review the example andtry the question again.


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Writing Tests and ExamsMATHPOWER™ 10, Ontario Edition

When you write a test or exam, your mind should be in top shape.

1. Circle the letter of each thing that you do before taking a test or exam to keep your mindsharp. Place an X beside the letter of each thing you would like to start doing.a) getting a good sleep the night beforeb) eating a nutritious breakfastc) planning ahead and being prepared in order to be relaxed when I begind) other things ______________________________________________________________________________

When writing a test or an exam, some people feel anxious.

2. Circle the letter of each technique that you have used to avoid text or exam anxiety in thepast. Place an X beside the letter of each technique you would like to try.a) being prepared b) avoiding thoughts about past less-than-successful experiencesc) exercisingd) eating nutritious meals and snacks e) learning deep breathing and other relaxation techniques f) getting adequate, but not too much, sleepg) other techniques _________________________________________________________________________

Just as in Math class, when you write a test or exam, you want to be prepared and use your timewisely.

3. Circle the letter of each strategy that you have used in the past to help you use the time for atest or exam wisely. Place an X beside the letter of each strategy you would like to try.a) having everything I need, such as pencils, eraser, ruler, and calculator, ready for useb) reading the entire test quickly to get an overview and opening my mind to ideas that couldhelp with later questionsc) using the marking scheme to help me determine the relative importance of questionsd) pacing myself, not rushing, but getting down to worke) reading the instructions, questions, and diagrams or graphs carefully to make sure I answerwhat is askedf) identifying the questions that I am most confident about, and doing them firstg) if I do questions out of order, checking them off and numbering them carefully to avoid confusing myself and my teacherh) predicting or estimating answers, and recording them to check against my final answersi) checking the reasonableness of my answersj) showing all my work, step by stepk) writing legibly l) marking questions that I want to return tom) in multiple-choice questions, ruling out obviously wrong choicesn) if my answer is not the same as one of the choices in a multiple-choice question, workingbackward from choices that I have not ruled out


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Problem Solving SkillsMATHPOWER™ 10, Ontario Edition

ProblemAt a handball tournament, 63 games were played. Each competitor played each of the other competitors three times. How many competitors were there?

Answer the following questions in your notebook.Understand the Problem1. How many games would have been played if each competitor played eachof the other competitors only once?2. Explain how answering question 1 allows you to proceed with solving theproblem.

Think of a PlanConsider just one game between competitors.Try a simpler problem, draw diagrams, and look for a pattern.3. Explain what each tree diagram shows.

Carry Out the Plan4. Draw the next diagram.5. Think about how the number of competitors is related to the number ofgames. 2 to 1 3 to 3 4 to 6 5 to 10Ask yourself:a) Is some value being added to, subtracted from, multiplied by, or dividedinto the number of competitors to get the number of games?b) Have I seen this relationship before?c) If I consider the number of competitors to the number of games to be 2 to 2 3 to 6 4 to 12 5 to 20I could return to the actual second terms by dividing by 2 or multiplying by 1–

2.

How does this help?6. What is the relationship?7. Using the relationship, determine how many competitors play 21 gameswith each competitor playing each of the others once (or 63 games with eachcompetitor playing each of the others three times).8. Write a final statement to answer the original question.

Look Back9. Solve the problem another way.


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Diagram 1 A B

A B C

A B

A B

C DC

C DD E

BC

B

D

BCDE

CD

C

E

DE

Diagram 2

Diagram 3

Diagram 4

Understand the Problem

Thinkof a Plan

Carry Outthe Plan

LookBack

Solving Word ProblemsMATHPOWER™ 10, Ontario Edition

Almost every time you learn a new Math skill, you have opportunities to use it in applied situations.

When you learned to solve equations like the following, you were also presented with problemsthat required creating and solving an equation from given information.

3x = 8(x – 5)

3x = 8x – 40

–5x = –40

x = 8

ProblemPlane 1 flew from Calgary to Montreal at 750 km/h. Plane 2 flew the same route at 600 km/hand took an hour longer. What is the flying distance from Calgary to Montreal?

This problem is presented to you after you have learned to solve rational equations. It is reason-able to expect that you should create and solve an equation from the information given in orderto solve the problem.

In this problem, you will need to use the relationship: distance = speed × time. This table show-ing the known values can help you get organized.

Plane Distance (km) Speed (km/h) Time (h)

1 750

2 600

1. How are the distances for the two planes related? ________________________________________________

2. How are the times for the two planes related? ____________________________________________________

3. Let x be the time for Plane 1 to make the flight. Complete the table above.

4. Write the equation using the relationship between the distances. _________________________

5. Solve the equation.

6. Substitute the value for x into each expression for distance.

7. Write a final statement to answer the problem, and check your answer.


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Copyright © 2001 McGraw-Hill Ryerson Limited

GoalsMATHPOWER™ 10, Ontario Edition

You can put what you have read in these Tips for Learning Math pages into practice. Set a goalfor yourself in Math this term.

A goal must be measurable.I am going to improve my Math mark by 10% this term.I am going to get at least a B on every Math test this term.

1. Write your goal.

A goal must be supported by actions.I will always be prepared for class.I will do my homework daily.I will join a study group.When I miss a class, I will catch up within one day.

2. Write your actions.

A goal will have obstacles. Identify them up front.I am on the soccer team. Two close friends are not. They will want me to do things with them, when Ishould be doing homework and getting caught up after missed classes.

3. Write your obstacles.

Obstacles need to be met head on.I am going to set aside time right after dinner to do homework, and I am going to tell my friends that Iwon’t be available until later.I will ask my two friends who are not on the soccer team to help me get caught up after missed classes.

4. Write how you are going to confront your obstacles.


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1. Which of the four ordered pairs satisfy theequation?a) 3x + y = –4 (0, –4), (–1, 1),

(1, –7)

b) (2, 21), (0, 19), (–4, 17), (6, 22)

c) 2x – y + 17 = 0 (0, 17), (4, –25), (–8.5, 0), (–5, –7)

d) 2y = 3x (0, 0),

(–6, –9)

2. Complete each ordered pair, so that itsatisfies the equation.a) x = –y + 6 (0, �), (6, �), (�, 9), (�, –2)

b) y = – 1 (5, �), (–5, �), (�, 0)

c) y + 3x = 13 (�, 4), (0, �), (�, 0), (–1, �)

d) 7x = 2y (�, 0), (2, �), (–1, �)

3. Complete each ordered pair so that it meetsthe stated condition.a) x + 2y = 4

y = 2x – 3 (�, 0) satisfies the first equation but not thesecond.b) 3x + 1 = y

2y – 6x = 2(�, �) satisfies neither equation.

c) x + y = –10.5y = 2x – 3

satisfies both equations.

d)

(8, �) satisfies the second equation but not thefirst.

4. Problem Solving To run a 30-s ad during aweekday morning, one radio station charges afixed cost of $400, plus $150 for each day the adis run. A second radio station charges a fixedcost of $300, plus $200/day. The costs can bemodelled by the following equations, where C isthe total cost and n is the number of days forwhich the ad is run.station 1: C = 150n + 400station 2: C = 200n + 300a) Find the missing element in the ordered pair (�, 700) that satisfies both equations and is inthe form (n, C).

b) Communication Explain what your orderedpair from part a) means for both radio stations.

c) Which station has the lower cost for runninga 30-s ad for 5 weekday mornings?Hint: Find (5, �) for each equation.

y x

y x

= −

= −

14

4 3 16

−

52

, �

�, 12

,

13

, , �

35

x

12

12

, , , , 13

13

y x= +12

19

−

43

0, ,

1.1 Investigation: Ordered Pairs and SolutionsMATHPOWERTM 10, Ontario Edition, pp. 4–5

• To verify that an ordered pair satisfies an equation in two variables,a) substitute the values for x and y in the original equationb) evaluate the left side of the equation and the right side of the equationc) check that L.S. = R.S.

• To verify that an ordered pair satisfies a system of equations,a) substitute the values for x and y in each equationb) evaluate L.S. and R.S. in each equationc) check that L.S. = R.S. in each equation

Copyright © 2001 McGraw-Hill Ryerson Limited Chapter 1 11

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1. Solve by graphing. Check each solution.a) y = x – 3 b) y = 2x – 1

y = 3 – x y = –x + 5

c) x + 3y = 2 d) 3x + 2y = 6 x – y = 6 y = 4 – x

e) 2x – 5y = 10 f)x + 3y = –6

g) 2x + 3y = –1 h) x + y = 8 3x + 5y = –3 x – y = 3

i) 3x – 4y = 12 j) 6x + 3y = 11 2x + 5y = 4 0.2x + y = –0.6

k) 3.74x – y = 2 x = 5

2. Communication Without graphing,determine whether each system has onesolution, no solution, or infinitely manysolutions. Explain your thinking.a) y = 2x – 5

4x – 2y = 10

b) 3x – y = 176x + 2y = –8

c) 12x + 8y + 4 = 015x + 10y = 5

3. Application The arms of an angle lie on thelines x + 4y = 9 and 3x – 2y = 13. What are thecoordinates of the vertex of the angle?

4. Problem Solving One equation of a systemis 2x – y = 5. Write a second equation so that thesystem hasa) no solution

b) infinitely many solutions

c) one solution, (3, 1)

y x

x y

= −

− + =

12

1

3 6 6 0

1.2 Solving Linear Systems GraphicallyMATHPOWERTM 10, Ontario Edition, pp. 6–15

• To solve a system of linear equations graphically,a) graph the equations using a graphing calculator, graphing software, or paper and pencilb) determine the coordinates of the point of intersectionc) check the solution by substituting it in each of the original equations

• The number of solutions to a linear system isa) exactly one, if the lines intersectb) none, if the lines are parallel and distinctc) infinitely many, if the lines coincide

Copyright © 2001 McGraw-Hill Ryerson Limited12 Chapter 1

Name

1. Solve each equation for the specified variable.a) x + 2y = 7; x b) 3x – y = 5; y

c) 4x + y – 11 = 0; y d) x + 6y + 2 = 0; x

2. Solve each system of equations bysubstitution. If there is exactly one solution,check the solution.a) x – y = 7 b) 4x + 3y = 7

2x + 5y = 21 3x + y = –1

c) p + 6q = 11 d) 5m – 4n = 11 4p – q = –6 3m = n + 15

e) 7s + 5t = 13 f) 11x – y = 21 t – s = –1 9x + 2y = –11

g) x – 2y = 5 h) 6x + 2y = 5 4x = 8y + 20 y = 11 – 3x

3. Communication Solve the system bysubstitution. Describe the difficulty youencounter.3x + 5y = 17x + 9y = 5

4. Application Simplify each system, and solveby substitution. Check each solution.a) 3(x + 5) – 2(y – 1) = 6

x + 4(y + 3) = 13

b) 2(x – 1) – (y – 7) = 134(x + 2) + 3(y + 1) = 37

5. Problem Solving The perimeter of anenvelope is about 68 cm. The length is equalto 2.4 times the width. Use the equations2l + 2w = 68 and l = 2.4w to find the dimensionsof the envelope.

1.3 Solving Linear Systems by SubstitutionMATHPOWERTM 10, Ontario Edition, pp. 16–23

• To solve a system of two linear equations in two variables by substitution,a) solve one equation for one of its variablesb) substitute the expression from part a) in the other equation and solve for the variablec) substitute the value of the variable found in part b) in one of the original equations to find the

value of the other variabled) check the solution in each of the original equations

• If the statement that results from a solution isa) not true for any value of a variable, the lines are parallel and distinct, and there is no solutionb) true for all values of a variable, the lines coincide and there are infinitely many solutions


Name

1. Communication Are the equationsequivalent? Explain.a) 2x – y = 3

y = 2x – 3

b)

c) y = 5x – 210x – 2y – 4 = 0

d)

2. Write three equivalent equations for eachgiven equation.a) x + y = 0.4

b) 4x – 3y = 7

c)

d) 0.5y = 2x – 1.3

3. Application Describe an easy way to writean equivalent equation for each given equation.

a)

b) 0.2y – 3.1x = 5.1

c) 22x – 33y = 11

d)

4. Here are two systems of equations.System J System Ky = 2x – 4 3y = 6x – 12 2x + y = –4 –2x – y = 4a) Find the intersection point of system J.

b) Communication Will the intersection pointof system K be the same or different? Explain.

c) Write the missing numbers in system L thatmake it an equivalent system. System Lx = _____y = _____

5. Add the equations in system K in question 4.Use your result to write two other equivalentsystems.

6. Write three different systems of equationsthat are equivalent to system A. Write eachequation in two variables.System Ax = –2

y = 32

23

12

1y x− = −

13

23

7x y− =

13

4 8x y− =

x y

x y

− =

− + =

23

32

y x

y x

= − +

= +

213

3 6 1

1.4 Investigation: Equivalent EquationsMATHPOWERTM 10, Ontario Edition, pp. 24–25

• To write an equation equivalent to a given equation, multiply each term in the given equation by thesame number.

• To write a system of equations equivalent to a given system, a) write an equation equivalent to each of the given equations (1) and (2)

ORb) add the given equations (1) and (2), and then write the result (3) with either of the original

equations: (1) and (3) or (2) and (3)


Name

1. Solve each system of equations byelimination. Check the solution if there is exactlyone solution.a) 3x + y = 17 b) 5x – 3y = 19

2x – y = –2 5x + 4y = 33

c) 6x + 7y = 23 d) 4x + 9y = 2 2x + 7y = 31 4x – 3y = 10

e) x + 2y = –4 f) 8x + 5y = 12 3x – 4y = 18 2x + 3y = 10

g) 2x – 5y = 29 h) 3x – 4y = 5 x – 2.5y = 14.5 5x + 3y = –11

i) 7a + 3b = 47 j) 3x – 8y = 2 2a + 5b = 30 5x + 3y = 1

k) l) 0.2x + 0.3y = 2 0.5x + 0.4y = 2.9

m) 4(x + 3) – 3(y – 2) = 4 3(x + 4) + 2(y + 4) = 1

2. Communication Solve by any method.Explain why you chose that method.a) 5x – 4y = 11 b) 14x + 33y = 37

y = 3x – 15 17x – 22y = 107

c) 7x – 5y = 6 d) 2x – 5y = 03x + 4y = 19 y = 1.4x + 0.8

3. Application Find the coordinates of thevertices of a triangle whose sides lie on thefollowing three lines.4x – 3y = –183(x + 4) + 2(y + 3) = 30y = 7x – 11

x y

x y6 4

6

56 3

11

+ =

− =

1.5 Solving Linear Systems by EliminationMATHPOWERTM 10, Ontario Edition, pp. 26–33

• To solve a linear system in two variables by elimination,a) clear decimals and fractions, if necessaryb) rewrite the equations with like terms in the same column, if necessaryc) multiply one or both equations by numbers to obtain two equations in which the coefficients of one

variable are the same or oppositesd) add or subtract the equations to eliminate a variable, and solve the resulting equation for the

remaining variablee) substitute the value from d) into one of the original equations to find the value of the other variablef) check the solution in each of the original equations


Name

1. Mitch invested $m at 9%/a and $n at 7.5%/a.Write an algebraic expression for each situation.a) the total amount of money Mitch invested

b) the interest Mitch earned at 7.5% in one year

c) the total interest Mitch earned in one year

2. Let x represent the larger of two numbers,and y, the smaller. Write an equation for eachsituation.a) The sum of the two numbers is negative five.

b) The difference between the larger numberand the smaller is twelve.

c) Three times the smaller number subtractedfrom twice the larger gives a result of nine.

d) The sum of half the larger number andone-third the smaller number equals zero.

3. Write a system of equations for each pair ofrelations.a) x y x y b) x y x y

2 –5 6 1 3 9 3 10 –3 3 –2 1 3

2–2 –1 0 –5 0 0

1 0–3 0 –1 –6 –3 –9

0

4. Communication Describe in words each ofthe four relations in question 3.

5. A dry cleaner charges $5 to alter a pair ofpants and $7 to alter a suit jacket. Last month,45 jackets and pants were altered for $275.a) Write an equation that relates the number ofpants, p, and the number of jackets, j, to the totalnumber.

b) Write an equation that relates p and j to theamount earned.

6. Application Introduce variables and writeeach of the following as a system of equationsin two variables.a) Two angles are complementary. The measureof one angle is 10° more than twice the measureof the other angle.

b) The length of a rectangle is 8 cm less thanthree times the width. Half the perimeter is60 cm.

c) A square and an equilateral triangle havelengths of sides such that the sum of theperimeters is 105 cm. The length of a side ofthe equilateral triangle is half the length of aside of the square.

− 12

12

1.6 Investigation: Translating Words Into EquationsMATHPOWERTM 10, Ontario Edition, pp. 36–37

• To express a situation described in words as an equation in two variables, a) identify the unknowns, and assign variables to the unknownsb) determine how the unknowns are relatedc) write an equation that shows the relationship between the unknowns


Name

1. Communication Chris has $3.85 in dimesand quarters. There are 25 coins in all. Howmany of each type of coin does he have?a) If d represents the number of dimes and qrepresents the number of quarters, what twoequations describe the problem?

b) Is the correct solution for the problem thenumbers 9 and 16? Explain.

Problem Solving

2. A supermarket sells 2-kg and 4-kg bags ofsugar. A shipment of 1100 bags of sugar has atotal mass of 2900 kg. How many 2-kg bags and4-kg bags are in the shipment?

3. The school car wash charged $5 for a car and$6 for a van. A total of 86 cars and vans werewashed on Saturday, and the amount earnedwas $475. How many vans were washed onSaturday?

4. In hockey, a team receives 2 points for a winand 1 point for a tie. During a hockey season of60 games, the Rockets lost 28 games but earned51 points. How many games did the team win?

5. A lab technician needs to combine some30% alcohol solution and 35% alcohol solutionto make 5 L of 33% alcohol solution. How many

litres of the 30% and of the 35% solution will beused?

6. A plane makes a trip of 5040 km in 7 h, flyingwith the wind. Returning against the wind, theplane makes the trip in 9 h. What is the speed ofthe wind?

7. To attend a wedding, Barbara starts drivingwest from Woodstock at 80 km/h. One hourlater, Barbara’s parents leave Woodstock anddrive along the same road at 100 km/h. At whatdistance from Woodstock will Barbara’s parentspass her?

8. The manager of a bulk food store mixed somejellybeans that cost $1.99/kg with gumdropsthat cost $2.99/kg to form 50 kg of a mixturethat cost $2.23/kg. How many kilograms of eachtype of candy were in the mixture?

9. What are the values of x and y?

(4x + y)°70°

(4x – 3y)°

1.7 Solving Problems Using Linear Systems MATHPOWERTM 10, Ontario Edition, pp. 38–47

• To solve a problem using a linear system,a) read the problem carefully, identify the unknowns, and assign variables to the unknownsb) determine how the unknowns are relatedc) write a system of equations that shows relationships between the unknownsd) solve the system of equationse) check the solution, using the facts given in the problem


Name

1.1 Investigation: Ordered Pairs andSolutions

1. a) (0, –4), (1, –7)

b) (0, 19), (–4, 17), (6, 22)

c) (0, 17), (–8.5, 0) d) (0, 0), (–6, –9)

2. a) 6, 0, –3, 8 b) 2, –4,

c) 3, 13, 16 d) 0, 7,

3. a) 4 b) Answers may vary. (0, 0) c) –8 d) 2

4. a) 2 b) The total cost of running a 30-s ad for 2 days

is $700, which is the same for both radio stations.

c) station 1: 150(5) + 400 = 1150,

station 2: 200(5) + 300 = 1300; station 1

1.2 Solving Linear Systems Graphically

1. a) (3, 0) b) (2, 3) c) (5, –1) d) (–2, 6) e) (0, –2)

f) no solution; parallel lines g) (4, –3)

h) i) j)

k) (5, 16.7)

2. a) The equations are equivalent, so there are

infinitely many solutions.

b) The slopes are 3 and –3, so the lines do not

coincide and there is one solution.

c) Both lines have a slope of and the y-intercepts

are different, so there is no solution.

3. (5, 1)

4. Answers may vary.

a) y = 2x + 5 b) y = 2x – 5 c) x + y = 4

1.3 Solving Linear Systems bySubstitution

1. a) x = 7 – 2y b) y = 3x – 5 c) y = 11 – 4x

d) x = –6y – 2

2. a) (8, 1) b) (–2, 5) c) (–1, 2) d) (7, 6)

e) f) (1, –10)

g) infinitely many solutions h) no solution

3. (2, –1); any substitution you try has fractions

4. a) (–3, 1) b) (5, 2)

5. width is 10 cm and length is 24 cm

1.4 Investigation: Equivalent Equations

1. a) yes; write equation (1) in y = mx + b form

b) no; the numerical coefficient of x in (2) should be

–6, not 6

c) yes; write equation (1) in standard form and then

multiply each term by 2

d) no; multiply equation (1) by –1 to get


a) 2x + 2y = 0.8, 3x + 3y = 1.2, 4x + 4y = 1.6

b) 8x – 6y = 14, 12x – 9y = 21, 0.4x – 0.3y = 0.7

c) x – 12y = 24, 2x – 24y = 48, 3x – 36y = 72

d) 5y = 20x – 13, y = 4x – 13––5

, 2y = 8x – 5.2


a) multiply by 3 to get x – 2y = 21

b) multiply by 10 to get 2y – 31x = 51

c) divide by 11 to get 2x – 3y = 1

d) multiply by 6 to get 4y – 3x = –6

4. a) (0, –4) b) same, because the equations are

equivalent: equation (1) in system J is multiplied by 3

and equation (2) is multiplied by –1 to give the

corresponding equations in system K

c) 0, –4

5. Order of equations may vary.

System M System N–8x + 2y = –8 –8x + 2y = –8–2x – y = 4 3y = 6x – 12


System B System C System D

2x + 2y = –1

x = –2 x = –2y = 32

x y+ = − 12

x y+ = − 12

− + = −x y23

32

, 12

− 32

6427

, 2927

−

7623

, 1223

−

112

, 52

− 72

17

,133

,

− 45

, 53

13

, , 12

−

43

, , 0

Answers CHAPTER 1 Linear Systems


1.5 Solving Linear Systems byElimination

1. a) (3, 8) b) (5, 2) c) (–2, 5) d)

e) (2, –3) f) (–1, 4) g) no solution h) (–1, –2)

i) (5, 4) j) k) (18, 12) l) (1, 6)

m) (–5, –2)

2. a) (7, 6); Use graphing or substitution because the

coefficient of y in the second equation is 1.

b) (5, –1); Use graphing or elimination. (Solve for y,

then substitute this value to find x.)

c) Use graphing or elimination. (Solve

for one variable; then, instead of substituting this

value, solve for the other variable.)

d) Use graphing or substitution or

elimination. (Multiply the second equation by 5.)

3. (0, 6), (3, 10), (2, 3)

1.6 Investigation: Translating WordsInto Equations

1. a) m + n b) 0.075n c) 0.09m + 0.075n

2. a) x + y = –5 b) x – y = 12

c) 2x – 3y = 9 d)

3. a) x + y = –3 and x – y = 5

b) 3x = y and x – 1 = 2y

4. a) The sum of two numbers is negative three. A

larger first number minus a smaller second number is

five.

b) Three times the first number equals the second

number. One subtracted from a first number equals

twice the second number.

5. a) p + j = 45 b) 5p + 7j = 275

6. a) a° + b° = 90° and a° = 2b° + 10°

b) l = 3w – 8 and l + w = 60

c) 4s + 3e = 105 and 2e = s or

1.7 Solving Problems Using LinearSystems

1. a) The equation for the number of coins is

d + q = 25. The equation for the value of the coins is

0.10d + 0.25q = 3.85.

b) You need to include the units with the numbers.

The correct solution is 9 quarters and 16 dimes.

2–9. The variables used may vary.

2. s + l = 1100 and 2s + 4l = 2900; 750 2-kg bags and

350 4-kg bags

3. 5c + 6v = 475 and c + v = 86; 45 vans

4. w + t = 60 – 28 and 2w + t = 51; 19 games won

5. 0.3a + 0.35b = 1.65 and a + b = 5; 2 L of 30% alcohol

solution and 3 L of 35% alcohol solution

6. and 80 km/h

7. b + 80 = p and 400 km

8. 50 – j = g and 1.99j + 2.99g = 2.23 × 50;

38 kg of jellybeans and 12 kg of gumdrops

9. (4x – 3y)° = 70° and (4x + y)° + 70°= 180° or

(4x + y)° + (4x – 3y)° = 180°; x = 25, y = 10

b p80 100

= ;

p w− = 50409

;p w+ = 50407

e s= 12

12

13

0x y+ =

− −

45

, ; 825

11943

, ; 11543

27

, 17

−

223

, −


1. Determine the length of the line segmentjoining each pair of points. Express each lengthas an exact solution and as an approximatesolution, to the nearest tenth.a) (3, 7) and (–1, –5)

b) (0, 5) and (6, 10)

2. Determine the radius of the circle with centre(–5, 6) and point (2, –7) on its circumference.Round the radius to the nearest tenth, ifnecessary.

3. Classify each triangle as equilateral, isosceles,or scalene. Then, find each perimeter, to thenearest tenth.a) W(2, 3), X(–1, –2), Y(5, –2)

b) B(–4, –1), C(2, –1)

4. Find the perimeter of parallelogram ABCD.

5. Communication Explain why �POR is aright triangle.

Applications

6. The vertices of a right triangle are (2, 2), (5, 8),and (–2, 4). Find the area of the triangle.

7. Three points, A(–3, 1), B(2, 4), and C(7, 7), lieon a straight line. Show that B is the midpoint ofAC.

8. The coordinates of the endpoints of thediameter of a circle are (–3, –5) and (3, 3). Findthe length of the radius of the circle.

9. a) Verify that the quadrilateral with verticesD(2, 6), E(3, 3), F(–3, 1), and G(–4, 4) is arectangle.

b) Determine the length of its diagonals, to thenearest tenth.

20

2

–2

–2

x

y

P(3, 3)

O(0, 0)

R(4, –4)

20

2

–2

–2

x

y

A(–2, 2)

D(–5, –2)

B(6, 2)

C(3, –2)

A − −( )1 27 1, ,

2.1 Length of a Line SegmentMATHPOWERTM 10, Ontario Edition, pp. 66–73

• To find the length of a line segment joining (x1, y1) and (x2, y2), use the formula

• An equation of the circle with centre O(0, 0) and radius r is x2 + y2 = r2.

l x x y y= − + −( ) ( ) .2 12

2 12


Name

1. Count units to find the coordinates of themidpoint of each line segment.a) b)

c) d)

2. Given the endpoints of each line segment,find the coordinates of the midpoint withoutplotting the points.a) A(–11, 3) and B(–3, 3)

b) K(12, 0) and L(–1, 0)

c) X(2.1, –5) and W(10.7, –5)

d) and

3. CD is a line segment joining the points C(–13, –3) and D(–1, –3). Find the coordinates ofthe three points that divide CD into 4 equalparts.

4. Communication The midpoint of ahorizontal line segment is M(2, –2). Whatcoordinates are possible for the endpoints of theline segment?

5. Count units to find the coordinates of themidpoint of each line segment.a) b)

c) d)

6. Given the endpoints of each line segment,find the coordinates of the midpoint withoutplotting the points.a) E(13, 1) and F(13, 11)

b) R(0, –8) and S(0, 6)

c) K(–7, 0) and J(–7, –5)

d) and

7. HI is a line segment joining the points H(1, –11) and I(1, 5). Find the coordinates of thethree points that divide HI into 4 equal parts.

8. Communication The midpoint of a verticalline segment is M(–3, –5). What coordinates arepossible for the endpoints of the line segment?

Z 14

612

,

W

14

112

,

0–10

–10

x

y

G(–14, –4)

H(–14, –26)

20

2

–2

–2

x

y E(0.5, 2.5)

F(0.5, –2)

20

2

–2

–2

x

yC(–9, 8)

D(–9, –5)

20

2

–2

x

y

B(5, –8)

A(5, 4)

Q −

1 1

12

,T 512

112

,

100

10

–10

x

y

G(0, 4) H(24, 4)

20

2

–2

–2

x

yE(–3, 2.5)

F(2, 2.5)

20

2

–2–2

x

y

C(–8, –7) D(–1, –7)

20

2

–2

–2

x

yA(–4, 4) B(6, 4)

2.2 Investigation: Midpoints of Horizontal and Vertical Line SegmentsMATHPOWERTM 10, Ontario Edition, p. 74

• To find the midpoint, M, of a horizontal line segment joining (x1, y) and (x2, y), use the formula

• To find the midpoint, M, of a vertical line segment joining (x, y1) and (x, y2), use the formula

xy y

, .1 2

2+

x xy1 2

2+

, .


Name

1. Determine the midpoint of each line segmentwith the given endpoints.a) (–6, 2) and (4, 8)

b) (1.5, 3) and (–6, –2.5)

c) (–200, –100) and (350, 600)

d) and

e) (3a, 2b) and (–3a, 5b)

f) (–6a, 5b) and (11a, 0)

2. Find the midpoints of the sides of �DEF.

3. Communication One endpoint of a linesegment is D(5, –7). The midpoint of the linesegment is M(3.5, 1.5). Explain how to find thecoordinates of the other endpoint, E, of the linesegment.

Applications

4. The endpoints of the diameter of a circle are(–3, 11) and (2, 9). What are the coordinates ofthe centre of the circle?

5. The endpoints of line segment MN are M(–6, –10) and N(2, –2). Find the coordinates ofthe point P on the line segment MN such thatMP:PN = 3:1.

6. A square has vertices K(–4, 3), L(3, 4), M(4, –3), and N(–3, –4).a) Find the coordinates of the midpoint of eachside.

b) Find the coordinates of the point ofintersection of the diagonals.

c) Find the perimeter of the square formed byjoining the midpoints of the sides of squareKLMN.

7. Vertex V of �UVW has coordinates (4, 6). Thecoordinates of the midpoint of UV are (1, 6), andthe coordinates of the midpoint of VW are (3, 2).Find the coordinates of points U and W.

0–2

–2

x

y

F(–8, –4)

D(–2, 4)

E(5, –7)

2

2

−

52

34

, 72

14

,

2.3 Midpoint of a Line SegmentMATHPOWERTM 10, Ontario Edition, pp. 75–80

• To find the midpoint, M, of a line segment joining (x1, y1) and (x2, y2), use the midpoint formula,

.x x y y1 2 1 2

2 2+ +

,


Name

1. Communication For any three points A, B,and C, not in a line, M and N are the midpointsof AB and AC, respectively. How can you prove

that MN || BC and ?

2. �DEF has vertices D(–1, 3), E(7, 1), and F(4, 6). Classify the triangle as a) isosceles or scaleneb) right-angled or not

3. The vertices of a quadrilateral are S(1, 2),T(3, 5), U(6, 7), and V(4, 4). Verify each of thefollowing.a) STUV is a parallelogram.b) The diagonals of STUV bisect each other.c) STUV is a rhombus.d) The diagonals of STUV are perpendicular toeach other.

Applications

4. �ABC has vertices A(3, 5), B(2, 3), and C(5, 2).a) Find the equations of the three altitudes.b) Find the intersection point of any two of thealtitudes.c) Verify that this point (the orthocentre of thetriangle) is on the altitude not used in part b).

5. The sides of a triangle have the equations 2x – 3y + 13 = 0, 3x + 2y = 0, and x + 5y – 26 = 0.Verify that the triangle is an isosceles righttriangle.

6. A quadrilateral has vertices P(–3, 1), Q(3, 7),R(9, 3), and S(–1, –1).a) Verify that PQRS has no equal sides and noparallel sides.b) Find the midpoints A, B, C, and D of PQ, QR,RS, and SP.c) Verify that ABCD is a parallelogram.

MN BC= 12

2.4 Verifying Properties of Geometric FiguresMATHPOWERTM 10, Ontario Edition, pp. 88–99

• The following formulas can be used to determine characteristics of geometric figures and to verifygeometric properties.

Slope of a line segment:

Length of a line segment:

Midpoint of a line segment:

Point-slope form of the equation of a line: y – y1 = m(x – x1)Slope and y-intercept form of the equation of a line: y = mx + b

x x y y1 2 1 2

2 2+ +

,

l x x y y= − + −( ) ( )2 12

2 12

my yx x

=−−

2 2

2 1


Name

1. Communication Explain how to find theshortest distance from the point P(–5, 7) to theline x = 3.

2. In each case, write an equation for the linethat is perpendicular to the line with the givenequation, and passes through the given point.

a)

b)

c) 4x – 3y – 7 = 0; (–5, 2)

d) 2x + 5y + 3 = 0; (3, –4)

3. Find the exact value of the shortest distancefrom the given point to the given line.a) (0, 0) and y = 2x – 10

b) (0, 0) and 5x + 12y – 39 = 0

c) (3, –2) and

d) (5, –2) and 4x + 3y + 14 = 0

4. Find the shortest distance from the givenpoint to the given line. Round to the nearesttenth, if necessary.

a) (0, 0) and

b) (0, 0) and 15x – 8y – 29 = 0

c) (–4, –5) and

d) (6, 5) and 7x + y + 23 = 0

Applications

5. A line has a y-intercept of 3 and an x-interceptof 4. What is the shortest distance from theorigin to this line?

6. a) Find the exact distance from the pointA(5, 7) to the line joining B(–2, 1) and C(4, –3).

b) Find the exact length of BC.

c) Use your answers to parts a) and b) to findthe area of �ABC.

y x= −53

4

y x= − +23

5

y x= − +13

9

y x= − − − −32

2 1 7; ( , )

y x= −12

7 2 5; ( , )

2.5 Distance From a Point to a LineMATHPOWERTM 10, Ontario Edition, pp. 100–105

• To determine the distance from a given point to a line whose equation is given,a) write an equation for the perpendicular from the given point to the given lineb) find the coordinates of the point of intersection of the perpendicular and the given linec) use the distance formula


Name

2.1 Length of a Line Segment

1. a) b)

2. 14.8

3. a) isosceles; 17.7 b) equilateral; 18

4. 26

5. length of PQ = length of QR = length

of

Thus, by the Pythagorean Theorem, �PQR is a right

triangle.

6. 15 square units

7.

8. 5

9. a)

DE2 + DG2 = 10 + 40 = 50 = GE2

Thus, ∠EDG = 90°.

b) 7.1

2.2 Investigation: Midpoints ofHorizontal and Vertical Line Segments1. a) (1, 4) b) (–4.5, –7)

c) (–0.5, 2.5) d) (12, 4)

2. a) (–7, 3) b) (5.5, 0)

c) (6.4, –5) d)

3. (–4, –3), (–7, –3), and (–10, –3)

4. Answers may vary. (8, –2) and (–4, –2), or (0, –2)

and (4, –2). There are an infinite number of pairs of

coordinates on the line y = –2 that are on opposite

sides of M(2, –2).

5. a) (5, –2) b) (–9, 1.5)

c) (0.5, 0.25) d) (–14, –15)

6. a) (13, 6) b) (0, –1)

c) (–7, –2.5) d)

7. (1, 1), (1, –3), and (1, –7)

8. Answers may vary. (–3, 0) and (–3, –10), or (–3, 3)

and (–3, –13). There are an infinite number of pairs of

coordinates on the line x = –3 that are on opposite

sides of M(–3, –5).

2.3 Midpoint of a Line Segment

1. a) (–1, 5) b) (–2.25, 0.25) c) (75, 250)

d) e) (0, 3.5b) f)

2. (–5, 0), (1.5, –1.5), (–1.5, –5.5)

3. Let the coordinates of E be (x1, y1). Substitute the

values into the midpoint formula,

. Then, solve for x1

and y1 using these equations: and

4. (–0.5, 10)

5. P(0, –4)

6. a) (–0.5, 3.5), (3.5, 0.5), (0.5, –3.5), (–3.5, –0.5)b) (0, 0) c) 20

7. U(–2, 6), W(2, –2)

2.4 Verifying Properties of GeometricFigures1. Find the coordinates of M and N using the

midpoint formula. Find the slope of MN and of BC. If

the slopes are equal, the line segments are parallel.

Find the length of MN and of BC using the length

formula. The length of MN should be half the length

of BC.

2. a) isosceles b) right

3. a) slopes of ST and slopes of SV and

b) The midpoint of both SU and TV is

c) ST = TU = UV = VS = d) slope of TV = –1; slope of SU = 1

4. a) from A to BC: y = 3x – 4; from B to AC:

from C to AB:

b)

c) satisfies the equation of the other altitude177

237

,

177

237

,

y x= − +12

92

y x= +23

53

;

13

72

92

, .

TU = 23

VU = 32

;

yx y1

1 17

21 5 2 10

+ −= ( ) = ( )( )

. . , ,

x1 52

3 5+ = .

( . , . ) ,( )

3 5 1 55

27

21 1= + + −

x y

52

52

a b,

12

12

,

14

4,

214

112

,

DE GF FE GD= = = =10 40, ,

AB BC= =34 34;

PR PQ QR PR= + = + = =50 18 32 502 2 2;

32 ,18 ,

61 7 8; .160 12 6; .

Answers CHAPTER 2 Analytic Geometry


5. vertices: A(–2, 3), B(1, 5), C(–4, 6); AB = AC

slope of slope of

6. a) slope = 1;

slope = 2–5

; slope = –1

b) A(0, 4), B(6, 5), C(4, 1), D(–2, 0)

c) slopes of AB and slopes of AD and

BC = 2

2.5 Distance From a Point to a Line

1. The shortest distance is along the line through P

that is perpendicular to the vertical line x = 3. The

length of the line segment joining P(–5, 7) and Q(3, 7)

is 8.

2. a) y = –2x + 9 or 2x + y – 9 = 0

b) or 2x – 3y – 19 = 0

c) or 3x + 4y + 7 = 0

d) or 5x – 2y – 23 = 0

3. a) b) 3 c) d)

4. a) 4.2 b) 1.7 c) 2.9 d) 9.9

5.

6. a) b) c) 32523213

125

285

9020

y x= −52

232

y x= − −34

74

y x= −23

193

DC = 16

;

SP = 8 ,RS = 116 ,

QR = −5223

, ;slope =PQ = 72 ,

AC = − 32

AB = 23

,

= 13 ;



Name

1. Classify each polynomial by degree and bynumber of terms.a) 3x2 – 2x b) 4a2b3

c) 8 + 2y4 + 3y3 d) 4x5 – 2x3 + x2 + 4

2. Evaluate each expression for the givenvalue(s) of the variable(s).a) 5x2 – 4x + 9 for x = 2

b) 2x2 – 4xy – 5y2 for x = –3, y = 2

3. Write each polynomial in descending orderof x.a) 6 + 4x2 – 5x5 + 3x – 2x2

b) 3x2y4 + 4x4y2 – x3y3 + x5y – 2xy5

4. Simplify.a) (6y – 2) + (2y + 8) b) (a + 2b) + (3a – 4b)

c) (8 + 6x) – (9 + x) d) (x + y) – (x – y)

e) (3x2 + 2x – 6) + (2x2 – 4x + 7)

f) (5a2b + 2ab – 3b2) – (6a2b – 3ab + b2)

g) (3y2 – y – 6) – (2y2 + 5y – 7)

5. Simplify.a) (6x)(2x2) b) (5pq2)(–4p2q2)

c) (3ab)(–2ab2)(2a3) d) (–6x2yz)(–5y3z)

e) f)

g) h)

6. Communication Explain how to simplifyand evaluate 3x(x + 1) – 4(x2 – 3x) for x = 2.

7. Expand and simplify.a) 2z + 3(4z – 2) + 2(4 – 3z)

b) 3x(x2 + 2x – 2) + 2x(3x2 – x – 4)

c) 4m(m2 – mn – n2) – 2n(6m2 + mn + 4n2)

8. Application Write a polynomial with threeterms and degree 4.

9. Problem Solving For therectangular prism, write anexpression that representsa) the volumeb) the surface area

−328

2 4

2 3

p qp q

−−217

2 2

2

x y zxy z

243

3 2

2

a ba b−

155

6

2

xx

3.1 PolynomialsMATHPOWERTM 10, Ontario Edition, pp. 128–133

• To add polynomials, collect like terms.• To subtract a polynomial, add its opposite.• To multiply monomials, multiply the numerical coefficients. Then, multiply the variables

using the exponent rules for multiplication.• To divide monomials, divide the numerical coefficients. Then, divide the variables using

the exponent rules for division.• To multiply a polynomial by a monomial, use the distributive property to multiply each

term in the polynomial by the monomial.

4y3x

2y


Name

1. Communication Explain howthe diagram models the product.

2. Find the product.a) (a + 3)(a + 2)

b) (2 + k)(3 + k)

c) (c – 5)(c – 3)

d) (t + 5)(t – 1)

e) (3 – b)(4 + b)

f) (6v + 3)(v + 1)

g) (5 + 2x)(2 + x)

h) (y – 5)(2y – 2)

i) (m + 4)(3m – 2)

j) (4g – 3)(g + 4)

k) (2y + 3)(3y + 2)

l) (5h – 1)(2h – 3)

m) (3 – 2s)(2 – 3s)

n) (4 + 2p)(–3 – 4p)

o) (–2t – r)(–3t + r)

3. Expand and simplify.a) 2(m – 3)(m + 8)

b) 3(x + 2)(x + 3)

c) –2(y – 3)(y + 2)

d) 0.2(x + 1)(x + 2)

e) 3(6x – 2y)(2x – 3y)

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

g) (y – 4)(y – 3) – (y – 2)(y + 5)

h) (3w – 2)(w + 4) + (2w + 3)(4w – 1)

i) 6(m – 2)(m + 3) – 3(3m – 4)

j) 4(2x + 3)(2x + 3) – 10 + 3(3x – 1)(3x – 1)

Problem Solving

4. Write and simplify anexpression to represent thearea of the figure.

5. Write and simplify anexpression to represent thearea of the shaded region.

3.2 Multiplying BinomialsMATHPOWERTM 10, Ontario Edition, pp. 134–139

• To find the product of two binomials, use either of the following.a) the distributive propertyb) FOIL, which stands for the sum of the products of the First terms, Outside terms, Inside

terms, and Last terms• Verify the product of two binomials by substituting a convenient value for the variable in the

original product and in the simplified expression.

2x

3x2x2

2xy

3+

3y

x

y+

4

x – 1

x + 6

x

2y + 1

2y + x

2y – x

2y


Name

1. Expand.a) (x + 4)2 b) (y – 7)2

c) (m –2)(m + 2) d) (x – 5)(x + 5)

e) 2(6x – 3)2 f) 3(5 + 4t)2

g) (3y – 3)(3y + 3) h) (5m + 2n)(5m – 2n)

i) (3x + 4y)2 j) 2(a – 7b)2

2. Expand and simplify.a) (m – 6)2 – (m + 2)(m – 2)

b) (x + 4)(x – 3) – 3(x + 2)2

c) 3(2b – 1)2 – 2(4b – 5)2

d) (x + 5)(x – 5) + (3x – 1)(3x + 1)

e) 4x2 – (2 – 3x)2 + 6(2x – 1)(2x + 1)

f) (2a – 1)(2a + 1) – (a – 3)(a + 3)

3. Application Complete the table.

Numbers (a + b)(a – b) Product

a) 25 × 15 (20 + 5)(20 – 5)

b) (30 + 6)(30 – 6)

c) 27 × 33

d) (20 – 4)(20 + 4)

4. Expand and simplify.a) (x2 + 2)2

b) (2y2 – 3)2

c) (y2 + 3)(y2 – 3)

d) (4m2 + n2)(4m2 – n2)

e) (–3x – 5)(–3x + 5) + (x + 1)2

5. Problem Solving The length of an edge of acube is represented by the expression 3x – 2y.a) Write, expand, and simplify an expression forthe surface area of the cube.

b) If x represents 4 cm and y represents 3 cm,calculate the surface area, in square centimetres.

3.3 Special ProductsMATHPOWERTM 10, Ontario Edition, pp. 140–145

• To square a binomial, use one of the following patterns.(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

• To find the product of the sum and difference of two terms, use the following pattern.(a + b)(a – b) = a2 – b2


Name

1. Factor, if possible.a) 4x + 28 b) 3x + 17

c) 6x – 32y d) 26x2 – 13y

e) 2ax + 10ay – 8az f) 2a2 – 6a – 15

g) 8x2 + 32y3 h) 10y – 5y2 + 25y3

i) 14rst + 7rs – 6t j) 36xy – 12x2y

k) 4ab2 + 2a2c + 5b2c2

l) 3x3y2 – 12x2y3 + 18x2y + 15xy2

2. Factor, if possible. a) 3x(y – z) – 2(y – z)

b) 5y(z + 3) + x(z – 3)

c) 4t(r + 6) – (r + 6)

d) 7(a + b) – 2x(a + b)

e) 2x(3m – 5) – 3(5 – 3m)

3. Factor by grouping.a) ax – by + xb – ya

b) y2 – x + y – xy

c) ab + 9 + 3a + 3b

d) t2 – tr + 4r – 4t

e) 4x2 + 6xy + 12y + 8x

f) 3x2y – 6x2 – 2y + y2

g) 4ab2 – 12a2b – 3bc + 9ac

4. Problem Solving Write an expression for thearea of each shaded region in factored form. a)

b)

c)

3.4 Common FactorsMATHPOWERTM 10, Ontario Edition, pp. 147–152

• To factor a polynomial with a common monomial factor, remove the greatest common factorof the coefficients and the greatest common factor of the variable parts.

• To factor a polynomial with a common binomial factor, think of the binomial as one factor.• To factor a polynomial by grouping, group pairs of terms with a common factor.

3x

2y3x

4r

3y

4x2

y + 1


Name

1. Factor, if possible.a) x2 – 5x + 6

b) y2 + 2y – 3

c) m2 + 7m – 12

d) a2 + 6a + 5

e) x2 – 9x – 10

f) b2 – 7b + 10

g) y2 – 6y + 7

h) x2 + x – 20

2. Factor, if possible.a) x2 + 24x – 52

b) m2 – 18m + 45

c) x2 + 5x – 36

d) x2 – 5xy – 66y2

e) m2 + 12mn + 32n2

f) 42 + y – y2

g) 32 + 4x – x2

h) x4 + 7x2 + 12

3. Factor completely.a) 2x2 + 10x + 12

b) 3x2 + 9x – 12

c) 5x2 – 35x + 50

d) 4x2 – 16x – 48

e) 2x2 – 16x – 66

f) x3 – 13x2 + 42x

4. Application The area of a doubles tenniscourt can be represented approximately by thetrinomial x2 – x – 42.a) Factor x2 – x – 42 to find binomials thatrepresent the length and width of a doublestennis court.

b) If x represents 17.8 m, find the length andwidth of a doubles tennis court, to the nearesttenth of a metre.

5. Communication Find two values for k suchthat the trinomial can be factored over theintegers. Explain your reasoning.a) x2 – 9x + k

b) x2 – kx + 6

3.5 Factoring x2 + bx + cMATHPOWERTM 10, Ontario Edition, pp. 153–158

• To factor a trinomial in the form x2 + bx + c,a) write x as the first term in each binomial factorb) write the second terms, which are two numbers whose sum is b and whose product is c

• When factoring a trinomial, first remove any common factors.

Chapter 3 33


Name

1. Factor, if possible.a) 3y2 + y – 4 b) 3y2 + 5y + 1

c) 2a2 – 13a + 21 d) 4n2 + 7n – 5

e) 20x2 – 7x – 6 f) 18y2 + 15y – 18

g) 5x2 – 12x – 6 h) 8m2 + 6m – 20

2. Factor. a) 2x2 + 5xy – 2y2

b) 3y2 + 2yz – z2

c) 15x2 – 13xy + 2y2

d) 6m2 + 7mn + n2

e) 4a2 – 9ab – 9b2

f) 6x – 2xy – 8y2

g) 6m2 – 13mn – 5n2

h) 9x2 + 3xy – 20y2

i) 12a2 + 28ab – 24b2

3. Communication Describe how to factor18a2 – 21ab + 6b2.

4. Application The area of a rectangular lot in anew housing development can be representedapproximately by the trinomial 12x2 + 8x – 15. a) Factor the expression 12x2 + 8x – 15 to findbinomials that represent the length and width ofthe lot.

b) If x represents 21 m, what are the length andwidth of the lot, in metres?

3.6 Factoring ax2 + bx + c, a ≠ 1MATHPOWERTM 10, Ontario Edition, pp. 159–164

• To factor a trinomial in the form ax2 + bx + c, either use guess and check or break up the middle term. • To factor by guess and check, list all the possible pairs of factors and expand to see which pair gives

the correct middle term of the trinomial.• To factor by breaking up the middle term,

a) replace the middle term, bx, by two terms whose coefficients have a sum of b and a product of a × cb) group pairs of terms and remove a common factor from each pairc) remove the common binomial factor


Name

1. Factor, if possible.a) x2 – 25 b) y2 – 49

c) y4 – 1 d) z2 + 64

e) 4a2 – 9 f) 49 – 64m2

g) 169a2 – b2 h) 24 + 4x2

i) 81x2 – 121p2 j) 49 – (a – z)2

2. State whether each trinomial is a perfectsquare trinomial. If it is, factor it. a) x2 + 8x + 16 b) y2 – 14y + 49

c) z2 – 9z + 9 d) 9t2 + 6t + 1

e) 4m2 – 12m – 9 f) 4x2 – 20x + 25

g) 121 – 22m + m2 h) 16x2 + 24xy + 9y2

i) 64a2 – 30ab + 49b2

3. Factor fully, if possible.a) x2 – 196 b) 36y2 + 6y + 1

c) 16a4 + 40a + 25 d) 4x2 – 36

e) y2 + 100 f) p2 – 4pq + 4q2

g) 36x2 – 81y2 h) m3 – 25m

i) 5n3 – 30n2 + 45n j) 64x2 – 16

k) 4b2 + 121 l) x4 – 13x2 + 36

Applications

4. Evaluate each difference of squares byfactoring. a) 382 – 322

b) 552 – 452

c) 7602 – 2402

5. Determine the value(s) of k such that eachtrinomial is a perfect square.a) x2 + kx + 49 b) 9x2 + kx + 25

c) 4x2 – 12x + k d) kx2 – 40xy + 16y2

3.7 Factoring Special QuadraticsMATHPOWERTM 10, Ontario Edition, pp. 165–170

• To factor a polynomial in the form a2 – b2, use the pattern for the difference of squares. a2 – b2 = (a + b)(a – b)

• To factor a perfect square trinomial, use the patterns for squaring binomials.a2 + 2ab + b2 = (a + b)2

a2 – 2ab + b2 = (a – b)2

Chapter 3 35


3.1 Polynomials

1. a) degree 2, binomial b) degree 5, monomial

c) degree 4, trinomial

d) degree 5, polynomial of 4 terms

2. a) 21 b) 22

3. a) –5x5 + 4x3 – 2x2 + 3x + 6

b) x5y + 4x4y2 – x3y3 + 3x2y4 – 2xy5

4. a) 8y + 6 b) 4a – 2b c) 5x – 1

d) 2y e) 6x2 – 2x + 1 f) –a2b + 5ab – 4b2

g) y2 – 6y + 1

5. a) 12x3 b) –20p3q4 c) –12a5b3

d) 30x2y4z2 e) 3x4 f) –8ab

g) 3x h) –4q

6. Multiply 3x(x + 1) to get 3x2 + 3x. Then, multiply

–4(x2 – 3x) to get –4x2 + 12x. Then, collect the like

terms to get –x2 + 15x. Then, substitute 2 for each x

and evaluate to get –(2)2 + 15(2) = –4 + 30 = 26.

7. a) 8z + 2 b) 9x3 + 4x2 – 14x

c) 4m3 – 16m2n – 6mn2 – 8n3

8. Answers may vary. 2m3n + 3m – 8n

9. a) 24xy2 b) 36xy + 16y2

3.2 Multiplying Binomials

1. The length of the rectangle is 2x + 3. The width is

x + y. The area is (2x + 3)(x + y) = 2x2 + 3x + 2xy + 3y.

2. a) a2 + 5a + 6 b) 6 + 5k + k2 c) c2 – 8c + 15

d) t2 + 4t – 5 e) 12 – b – b2 f) 6v2 + 9v + 3

g) 10 + 9x + 2x2 h) 2y2 – 12y + 10 i) 3m2 + 10m – 8

j) 4g2 + 13g – 12 k) 6y2 – 13y + 6 l) 10h2 – 17h + 3

m) 6 – 13s + 6s2 n) –12 – 22p – 8p2 o) 6t2 + rt – r2

3. a) 2m2 + 10m – 48 b) 3x2 + 15x + 18

c) –2y2 + 2y + 12 d) 0.2x2 + 0.6x + 0.4

e) 36x2 – 66xy + 18y2 f) 2x2 + 10x + 10

g) –10y + 22 h) 11w2 + 20w – 11

i) 6m2 – 3m – 24 j) 43x2 + 30x + 29

4. 4x + (x + 2)(x – 1) = x2 + 5x – 2 or

x(x + 6) – (1)(x + 2) = x2 + 5x – 2

5. (2y + 1)(2y + x) – (2y – x)(2y) = 4xy + 2y + x

3.3 Special Products

1. a) x2 + 8x + 16 b) y2 – 14y + 49

c) m2 – 4 d) x2 – 25

e) 72x2 – 72x + 18 f) 75 + 120t + 48t2

g) 9y2 – 9 h) 25m2 – 4n2

i) 9x2 + 24xy + 16y2 j) 2a2 – 28ab + 98b2

2. a) –12m + 40 b) –2x2 – 11x – 24

c) –20b2 + 68b – 47 d) 10x2 – 26

e) 19x2 + 12x – 10 f) 3a2 + 8

3. Numbers (a + b)(a – b) Product

a) 25 × 15 (20 + 5)(20 – 5) 375

b) 36 × 24 (30 + 6)(30 – 6) 864

c) 27 × 33 (30 – 3)(30 + 3) 891

d) 16 × 24 (20 – 4)(20 + 4) 384

4. a) x4 + 4x2 + 4 b) 4y4 – 12y2 + 9

c) y4 – 9 d) 16m4 – n4

e) 10x2 + 2x – 24

5. a) 6(3x – 2y)2 = 54x2 – 72xy + 24y2 b) 216 cm2

3.4 Common Factors

1. a) 4(x + 7) b) does not factor

c) 2(3x – 16y) d) 13(2x2 – y)

e) 2a(x + 5y – 4z) f) does not factor

g) 8(x2 + 4y3) h) 5y(2 – y + 5y2)

i) does not factor j) 12xy(3 – x)

k) does not factor l) 3xy(x2y – 4xy2 + 6x – 5y)

2. a) (3x – 2)(y – z) b) does not factor

c) (4t – 1)(r + 6) d) (7 – 2x)(a + b)

e) (2x + 3)(3m – 5)

3. a) (x – y)(a + b) b) (y – x)(y + 1)

c) (a + 3)(b + 3) d) (t – 4)(t – r)

e) 2(x + 2)(2x + 3y) f) (3x2 + y)(y – 2)

g) (4ab – 3c)(b – 3a)

4. a) 3x(3πx – 2y) b) 2x2(5y – 1)

c) 16r2(π – 2)

Answers CHAPTER 3 Polynomials

Chapter 3 37

3.5 Factoring x2 + bx + c

1. a) (x – 3)(x – 2) b) (y + 3)(y – 1)

c) does not factor d) (a + 5)(a + 1)

e) (x – 10)(x + 1) f) (b – 5)(b – 2)

g) does not factor h) (x + 5)(x – 4)

2. a) (x + 26)(x – 2) b) (m – 15)(m – 3)

c) (x + 9)(x – 4) d) (x – 11y)(x + 6y)

e) (m + 4n)(m + 8n) f) (6 + y)(7 – y)

g) (8 – x)(4 + x) h) (x2 + 4)(x2 + 3)

3. a) 2(x + 2)(x + 3) b) 3(x + 4)(x – 1)

c) 5(x – 5)(x – 2) d) 4(x – 6)(x + 2)

e) 2(x – 11)(x + 3) f) x(x – 6)(x – 7)

4. a) (x – 7)(x + 6) b) 10.8 m by 23.8 m

5. Answers may vary. a) k = 20 because two factors

with the sum of –9 are –5 and –4.

x2 – 9x + 20 = (x – 5)(x – 4); k = 14 because two factors

with the sum of –9 are –7 and –2.

x2 – 9x + 14 = (x – 7)(x – 2)

b) k = 7 because two factors of 6 are 1 and 6 and their

sum is 7. x2 + 7x + 6 = (x + 1)(x + 6);

k = 5 because two factors of 6 are 3 and 2 and their

sum is 5. x2 + 5x + 6 = (x + 3)(x + 2)

3.6 Factoring ax2 + bx + c, a ≠ 1

1. a) (3y + 4)(y – 1) b) does not factor

c) (2a – 7)(a – 3) d) does not factor

e) (4x – 3)(5x + 2) f) 3(2y + 3)(3y – 2)

g) does not factor h) 2(m + 2)(4m – 5)

2. a) (x + 2y)(2x + y) b) (3y – z)(y + z)

c) (5x – y)(3x – 2y) d) (6m + n)(m + n)

e) (4a + 3b)(a – 3b) f) 2(x + y)(3x – 4y)

g) (2m – 5n)(3m + n) h) (3x + 5y)(3x – 4y)

i) 4(3a – 2b)(a + 3b)

3. Remove the common factor to get

3(6a2 – 7ab + 2b2). Then, factor the trinomial by guess

and test to get 3(3a – 2b)(2a – b).

4. a) (2x + 3)(6x – 5) b) 45 m by 121 m

3.7 Factoring Special Quadratics

1. a) (x + 5)(x – 5) b) (y + 7)(y – 7)

c) (y2 + 1)(y + 1)(y – 1) d) does not factor

e) (2a + 3)(2a – 3) f) (7 + 8m)(7 – 8m)

g) (13a + b)(13a – b) h) does not factor

i) (9x + 11p)(9x – 11p) j) (7 + a – z)(7 – a + z)

2. a) yes, (x + 4)2 b) yes, (y – 7)2

c) no d) yes, (3t + 1)2

e) no f) yes, (2x – 5)2

g) yes, (11 – m)2 h) yes, (4x + 3y)2

i) no

3. a) (x + 14)(x – 14) b) does not factor

c) (4a + 5)2 d) 4(x + 3)(x – 3)

e) does not factor f) (p – 2q)2

g) 9(2x + 3y)(2x – 3y) h) m(m – 5)(m + 5)

i) 5n(n – 3)2 j) 16(2x – 1)(2x + 1)

k) does not factor l) (x + 3)(x – 3)(x + 2)(x – 2)

4. a) 420 b) 1000

c) 520 000

5. a) ±14 b) ±30

c) 9 d) 25



Name

1. State whether each set of ordered pairsrepresents a function.a) (–2, 5), (–1, 10), (0, 15), (1, 20)

b) (0, 1), (1, 1), (1, 2), (2, 1), (2, 2)

c) (3, 9), (4, 6), (5, 25), (6, 30)

2. If y = –2x + 1, find the value of y for each valueof x.a) –1 b) 20 c) 0

3. If find the value of y for each value

of x.a) 4 b) –80 c) 100

4. If w = 2v2 + 1, find the value of w for eachvalue of v.a) –3 b) 1.5 c)

5. If q = –p2 + 3p + 2, find the value of q for eachvalue of p.a) 2 b) –10 c) 0.5

6. State the domain and range of each relation inparts a) and b), and state whether it is a function. a) y = –2x + 5 b) speed of 80 km/h

x y t d

2 9 0 0

1 7 0.1 8

0 5 0.25 20

–1 3 0.5 40

–2 1 1 80

c) Communication Identify the independentvariable and the dependent variable in part b).Explain your reasoning.

7. Determine the domain and range of each of the following relations.a) b)

c) d)

8. Which of the relations in question 7 arefunctions?

9. Communication a) Is the set of ordered pairs (n, t) a function, if n is a student’s name and t is the family’s home telephone number?Explain.

b) Reverse the terms of the ordered pairs to get(t, n). Is the new set of ordered pairs a function?Explain.

10. Problem Solving The volume of a sphere is

given by the function where x is the

radius of the sphere. Graph the function. Then,use your graph to find a) the volume of a sphere with radius 10 cm

b) the radius of a sphere with volume 33.5 cm3

y x= π43

3 ,

20

2

–2

–2

x

y

4

4

20

2

–2

–2

x

y

4

4 620

2

–2

–2

x

y

4

4 6

3

y x= −12

5,

4.1 FunctionsMATHPOWERTM 10, Ontario Edition, pp. 192–199

• A function is a set of ordered pairs in which, for every x, there is only one y.• If any vertical line passes through more than one point on the graph of a relation,

then the relation is not a function.• The set of the first elements in a relation is called the domain. The set of the second

elements in a relation is called the range.

120

80

120

400

40

t

d

160 20080 240 280

4.2 Graphing y = x2 + k, y = ax2, and y = ax2 + kMATHPOWERTM 10, Ontario Edition, pp. 204–216

• This table summarizes how parabolas in the form • Some geometric properties of the parabola y = ax2 + k are obtained by transforming the y = ax2 + k are summarized in this table.

function y = x2.

Sign of apositive negative


Name

1. Sketch each parabola and state the directionof the opening, the coordinates of the vertex, theequation of the axis of symmetry, the domainand range, and the maximum or minimumvalue.a) y = x2 + 4 b) y = –x2 + 2

c) y = 0.5x2 – 1 d) y = –3(x2 – 4)

2. Communication State how the graph of thesecond equation is related to the graph of thefirst equation.a) y = 2x2 + 1 and y = 2x2 – 3

b) y = –x2 and

3. Use a graphing calculator or graphingsoftware to determine any x-intercepts, to thenearest tenth.a) y = 0.25x2 – 4 b) y = –3x2 – 3

c) y = 2x2 – 9 d) y = –4x2 + 25

Applications

4. Write an equation for the parabola createdwhen each pair of transformations is applied tothe graph of y = x2.a) a reflection in the x-axis, followed by avertical translation of –3

b) a vertical translation of 2, followed by avertical stretch of scale factor 3

5. The graph of y = 4x2 + k passes through thepoint (–1, –1). Find k.

6. The graph of y = ax2 + k passes through thepoints (2, 3) and (–4, –9). Find a and k.

y x= − 12

2

20

2

–2–2 x

y

20

2

–2

–2

x

y

20

2

–2

–2

x

y

20

2

–2

–2

x

y

OperationResultingEquation Transformation

Multiply by a. y = ax2

Reflects in the x-axis, if a < 0. Stretches vertically

(narrows), if a > 1, or a < –1. Shrinks vertically

(widens), if –1 < a < 1.

Add k. y = ax2 + kShifts k units upward, if

k > 0. Shifts k unitsdownward, if k < 0.

Property(0, k) (0, k)Vertex

x = 0 x = 0Axis of Symmetry

up downDirection ofOpening

congruent congruentComparison with y = ax2


Name

1. Sketch each parabola and state the directionof the opening, the coordinates of the vertex, theequation of the axis of symmetry, the domainand range, and the maximum or minimumvalue.a) y = (x + 3)2 – 2 b) y = –(x – 4)2 – 3

c) y = 2(x – 1)2 + 1 d)

2. Without graphing, state the coordinates of thevertex and whether this vertex represents amaximum or minimum value of the function.

a) y = –3(x + 5)2 – 6 b)

c) y = 7.5(x – 2.5)2 – 9 d) y = –(x – 9)2 + 19

3. Use a graphing calculator or graphingsoftware to determine any x- or y-intercepts.Round to the nearest tenth, if necessary.a) y = (x – 1)2 – 4 b) y = –2(x + 2)2 + 8

c) y = 3(x + 1)2 + 9 d)

Applications

4. Write an equation for each parabola.a) vertex (3, –1); a = –2

b) vertex (2, 5); congruent to

c) vertex (–4, –1); y-intercept –9

d) vertex (–5, 3); through (–7, 15)

5. The vertex of a parabola is (–3, 7). They-intercept is 0. What are the x-intercepts?

y x= 12

2

y x= − − +12

4 42( )

y x= + +34

4 22( )

20

2

x

y

y x= − + +12

2 72( )

20

–2

x

y

0

2

–2

–2

x

y

4.3 Graphing y = a(x – h)2 + kMATHPOWERTM 10, Ontario Edition, pp. 217–227

• This table summarizes how parabolas in the form • Some geometric properties of the parabola y = a(x – h)2 + k are obtained by transforming the y = a(x – h)2 + k are summarized in this function y = x2. table.

Sign of apositive negativeOperation

ResultingEquation Transformation

Multiply bya.

y = ax2

Reflects in the x-axis, if a < 0. Stretches vertically

(narrows), if a > 1 or a < –1.Shrinks vertically (widens),

if –1 < a < 1.

Replace xby (x – h).

y = a(x – h)2

Shifts h units to the right, ifh > 0. Shifts h units to the

left, if h < 0.

Add k. y = a(x – h)2 + kShifts k units upward, if

k > 0. Shifts k unitsdownward, if k < 0.

Property(h, k) (h, k)Vertex

x = h x = hAxis of Symmetry

up downDirection ofOpening

congruent congruentComparison with y = ax2

20

2

–2 x

y


Name

1. Find the value of c that will make eachexpression a perfect square trinomial.a) x2 + 22x + c b) x2 – 16x + c

c) x2 – 6x + c d) x2 + 40x + c

e) cx2 + 28x + 49 f) 9x2 – 18x + c

2. Write each function in the form y = a(x – h)2 +k. Sketch the graph, showing the coordinates ofthe vertex and two other points on the graph,and the equation of the axis of symmetry.a) y = x2 – 4x – 1 b) y = –2x2 – 4x + 1

3. Without graphing, state whether eachfunction has a maximum or a minimum. Then,write each function in the form y = a(x – p)2 + qand find the minimum or maximum value andthe value of x for which it occurs.a) y = 3x2 – 18x + 1 b) y = –4x2 – 32x – 11

c) y = –7x2 + 84x + 19 d) y = 4x2 – 20x + 7

4. Communication Explain how to use algebrato find two numbers whose difference is 8 andwhose product is a minimum.

Problem Solving

5. Determine the maximum area of a triangle, insquare centimetres, if the sum of its base and itsheight is 12 cm.

6. A ball is thrown upward with an initialvelocity of 18 m/s. Its height, h metres after t seconds, is given by the equation

h = –5t2 + 18t + 1.8where 1.8 represents the height at which the ballis released by the thrower.a) What is the maximum height the ball willreach?

b) How much time elapses before the ballreaches the maximum height?

c) How long is the ball in the air, to the nearesttenth of a second?

20

2

–2 x

y

20

–2

x

y

4.4 Graphing y = ax2 + bx + c by Completing the SquareMATHPOWERTM 10, Ontario Edition, pp. 228–240

• To rewrite a quadratic function y = ax2 + bx + c in the form y = a(x – h)2 + k, use the stepsshown in the following example. y = 2x2 + 4x + 5a) Group the terms containing x: y = (2x2 + 4x) + 5b) Factor the coefficient of x2: y = 2[x2 + 2x] + 5c) Complete the square inside the brackets: y = 2[x2 + 2x + 1 – 1] + 5d) Write the perfect square trinomial as the square of a binomial: y = 2[(x + 1)2 – 1] + 5e) Expand to remove the square brackets: y = 2(x + 1)2 – 2 + 5f) Simplify: y = 2(x + 1)2 + 3

• For a quadratic function in the form y = ax2 + bx + c, find the maximum or minimum value byrewriting the function in the form y = a(x – h)2 + k to find the vertex (h, k). The maximum orminimum value of the function is k.


Name

1. Write each equation in the form y = ax(x – s) + t.a) y = x2 – 6x + 8 b) y = 2x2 + x – 5

c) y = x2 – 9x d) y = –3x2 + 12x – 0.9

2. Communication Explain how to find twopoints on the graph and deduce the coordinatesof the vertex of each function.a) y = 2x2 + 4x – 6

b) y = –x2 + 2x + 1

3. Sketch the graph of each function by writing it in the form y = ax(x – s) + t.a) y = 3x2 + 6x – 8 b) y = –x2 + 6x + 3

4. Application Verify the coordinates of thevertex of each quadratic function in questions 2

and 3, using the form st

as2 4

2

, . −

20

2

x

y

20–2

–2

x

y

4.5 Investigation: Sketching Parabolas in the Form y = ax(x – s) + tMATHPOWERTM 10, Ontario Edition, p. 241

• To sketch the graph of a quadratic function, write it in the form y = ax(x – s) + t, using the steps shown in the following example. y = 2x2 – 8x + 5a) Factor 2x from the first two terms: y = 2x(x – 4) + 5b) Substitute 0 for x in y = 2x(x – 4) + 5: y = 2(0)(0 – 4) + 5

y = 5One point on the graph is (0, 5).

c) Substitute 4 for x in y = 2x(x – 4) + 5: y = 2(4)(4 – 4) + 5y = 5

Another point on the graph is (4, 5).d) Plot the points (0, 5) and (4, 5) on a grid.e) Find the axis of symmetry that passes through the vertex.

Since (0, 5) and (4, 5) have the same y-coordinate, the points arereflection images of each other in the axis of symmetry. The equation ofthe axis of symmetry is x = 2, so the x-coordinate of the vertex is 2.

f) Substitute 2 for x in y = 2x2 – 8x + 5: y = 2(2)2 – 8(2) + 5= 8 – 16 + 5= –3

The coordinates of the vertex are (2, –3).g) Plot the vertex on the grid and draw a smooth curve through the three points.

20

2

–2

–2

x

y

(0, 5)(4, 5)

(2, –3)

y = 2x2 – 8x + 5


Name

1. Complete the following tables.a) y = 3x + 40 b) y = –5x + 8

x y 1st Difference x y 1st Difference

0 0

1 1

2 2

3 3

4 4

c) d)

x y 1st Difference x y 1st Difference

0 3 0 2

1 5 1 –4

2 7 2 –10

3 9 3 –16

4 11 4 –22

2. Communication a) Describe how the firstdifference is related to each linear equation inquestion 1 parts a) and b).

b) Where does the constant term, b, for eachequation occur in the table of values in question1 parts a) and b)?

c) Are the functions in question 1 parts c) and d)linear? Explain. Then, write an equation for eachfunction, in the form y = mx + b.

3. Application Write an equation for the linearfunction described by the ordered pairs (0, 5),(1, 4.5), (2, 4), and (3, 3.5).

4. Complete the tables for the followingquadratic functions. a) y = 2x2 + 5x + 9 b) y = x2 + 4x – 11

Difference Differencex y 1st 2nd x y 1st 2nd

0 0

1 1

2 2

3 3

4 4

5. Communication a) What is the relationshipbetween the second difference and the value of afor each function in question 4?

b) Where does the constant value, c, for eachequation occur in the table of values?

c) The first entry in the 1st Difference column isequal to a + b. Verify the b-value in eachequation.

d) Find c, a, and b for the quadratic functionshown in this table of values. Then, write anequation in the form y = ax2 + bx + c.

Differencex y 1st 2nd

0 1

1 6

2 15

3 28

4 45

4.6 Investigation: Finite DifferencesMATHPOWERTM 10, Ontario Edition, pp. 242–245

• In tables with evenly spaced x-values, first differences are calculated by subtractingconsecutive y-values, and second differences are calculated by subtracting consecutive firstdifferences.

• For a linear function, the first difference is a constant. For a quadratic function, the seconddifference is a constant.


Name

1. For each table of values, enter the data anddraw the scatter plot. Then, use trial and error tofit the equation of a quadratic function in theform y = a(x – h)2 + k, where a, h, and k areintegers, to the scatter plot.

a) x y b) x y

–1 6 –2 17

0 2 –1 12

2 0 0 9

4 6 1 8

5 12 3 12

c) x y d) x y

–3 9 –1 –10

–2 0 0 0

1 –3 1 2

2 4 2 –4

4 30 3 –18

e) x y f) x y

–5.5 0 –5 –2

–4 –2 –4 –3

–2 –3 –2 –3

0 –2 1 2

1.5 0 2 5

2. For each table of values in question 1, findthe equation of the curve of best fit usingquadratic regression. If necessary, roundcoefficients and constants to the nearesthundredth. Then, compare the equation with theequation you found in question 1.

3. Application A tennis ball is tossed upwardand its height and the elapsed time are recordedas follows.

Elapsed time (s) Height of ball (m)

0.00 0.321

0.02 0.570

0.04 0.637

0.06 0.682

0.10 0.684

0.12 0.644

0.14 0.601

a) Enter the data and draw the scatter plot ofheight versus time.

b) Communication Use trial and error to fit theequation of a quadratic function in the form y = a(x – h)2 + k to the scatter plot. Explain yourreasoning.

c) Use your equation to determine the height ofthe ball after 0.08 s.

d) Find the equation of the curve of best fitusing quadratic regression. Then, compare theequation with your result from part b).

e) Use the equation of the curve of best fit todetermine the height of the ball after 0.08 s and0.15 s.

4.7 Technology: Equations of Parabolas of Best FitMATHPOWERTM 10, Ontario Edition, pp. 246–247

• To fit an equation of a quadratic function to a scatter plot using a graphing calculator, a) use the STAT EDIT menu to enter the data as two listsb) use the STAT PLOTS menu to draw the scatter plotc) enter possible equations in the Y= editor, and use trial and error to fit an equation of a

quadratic function to the scatter plotd) compare the result from step c) with the equation of the curve of best fit found by using the

QuadReg (quadratic regression) instruction


Name

1. Communication Analyse the data in thetable above.a) At what height was the ball before it wastossed into the air?

b) What was the ball’s maximum height?

c) How high was the ball tossed in the air?

d) After how many seconds did the ball reachits maximum height?

e) Does the height at 0.06 s seem reasonable?Explain.

f) Why is it important to toss the ball straightup into the air, instead of at an angle?

g) The motion detector unit must be placedbeneath the vertical path of the ball. At whatheight could the motion detector unit have beenplaced?

h) Do you think the ball was caught before it hitthe ground?

i) What is the independent variable in thisexperiment? the dependent variable?

2. Collect your own data by doing theexperiment with a softball or a basketball. Or,use the data above. Make a scatter plot of heightversus time. Then, use the TRACE instruction tofind the approximate coordinates of the vertex ofthe scatter plot. Round coordinates to the nearesthundredth, if necessary. Use the coordinates ofthe vertex as the values of h and k in an equationof the form y = a(x – h)2 + k. By enteringequations with these values of h and k in the Y=editor, use trial and error to find a value of a foran equation that fits your scatter plot. Then, usethe quadratic regression instruction to find anequation of the curve of best fit.

3. Communication How does the equation youfound by trial and error compare with theequation of the curve of best fit produced usingquadratic regression?

4. a) Communication Use the equation topredict the height of the ball at 10 s. Do you havemuch confidence in your prediction? Explain.

b) If the ball was allowed to hit the ground, atwhat time would that have occurred? (Hint:What is the height of the ball?)

4.8 Technology: Collecting Distance and Time Data Using CBRTM

or CBLTM

MATHPOWERTM 10, Ontario Edition, pp. 248–250

Collecting Data From Tossing a BallIn this experiment, a ball is tossed vertically into the air, and the ball’s height at various times isrecorded using a CBL and a motion detector or a CBR.The following data for tossing a softball were collected, and the heights were rounded to thenearest thousandth.

Elapsed Height of Elapsed Height of Elapsed Height oftime (s) ball (m) time (s) ball (m) time (s) ball (m)

Name

0.00 1.0360.02 1.0860.04 1.1360.06 0.4570.08 1.2380.10 1.2600.12 1.2780.14 1.3020.16 1.3240.18 1.3410.20 1.356

0.22 1.3650.24 1.3710.26 1.3730.28 1.3720.30 1.3650.32 1.3580.34 1.3430.36 1.3230.38 1.3080.40 1.2810.42 1.249

0.44 1.2160.46 1.1800.48 1.1390.50 1.0940.52 1.0470.54 0.9960.56 0.9410.58 0.8860.60 0.824


4.1 Functions

1. a) function b) not a function c) function

2. a) 3 b) –39 c) 1

3. a) –3 b) –45 c) 45

4. a) 19 b) 5.5 c) 7

5. a) 4 b) –128 c) 3.25

6. a) function; domain: {2, 1, 0, –1, –2},

range: {9, 7, 5, 3, 1}

b) function; domain: {0, 0.1, 0.25, 0.5, 1},

range: {0, 8, 20, 40, 80}

c) time: independent, distance: dependent; The

distance travelled at 80 km/h depends on the time.

7. a) domain: {1, 2, 3}, range: {2, 3, 4}

b) domain: set of real numbers, range: set of real

numbers

c) domain: set of real numbers from 0 to 280, range:

set of real numbers from 0 to 120

d) domain: set of real numbers, range: set of real

numbers greater than or equal to –2

8. a) not a function b) , c) , d) functions

9. a) Yes, if every student has only 1 home phone

number. No, if a student’s family has more than 1

phone number at home, or if a family is split and

there are 2 homes where a student lives.

b) Yes, if there is only 1 student at the school from

every family. No, if there are several students at the

school from even 1 family.

10. a) 4188.8 cm3 (to nearest tenth) b) 2 cm

4.2 Graphing y = x2 + k, y = ax2, andy = ax2 + k

1. a)

b)

c)

d)

2. a) The graphs have the same shape; the graph

of y = 2x2 – 3 is the graph of y = 2x2 + 1 translated

4 units downward.

b) The graphs have the same vertex; the graph of

is flatter and is the graph of y = –x2 shrunk

vertically by a scale factor of

3. a) 4, –4 b) none c) 2.1, –2.1 d) 2.5, –2.5

4. a) y = –x2 – 3 b) y = 3(x2 + 2)

5. k = –5

6. a = –1, k = 7

4.3 Graphing y = a(x – h)2 + k

1. a)

0

2

–2

x

y

–2

(–3, –2)

y = (x + 3)2 – 2

12

.

y x= − 12

2

0

2

–2 2 x

y

y = –3(x2 – 4)

(0, 12)

20

2

–2

–2

x

y

(0, –1) y = x2 – 11–2

20

2

–2

–2

x

y

(0, 2)y = –x2 + 2

20

2

–2

–2

x

y

(0, 4)

y = x2 + 4

Answers CHAPTER 4 Quadratic Functions

up; (0, 4); x = 0;

domain: set of real numbers,

range: y ≥ 4; minimum: 4

down; (0, 2); x = 0;


range: y ≤ 2; maximum: 2

up; (0, –1); x = 0;


range: y ≥ –1; minimum: –1

down; (0, 12); x = 0;

domain: set of real

numbers, range: y ≤ 12;

maximum: 12

up; (–3, –2); x = –3;

domain: set of real

numbers, range: y ≥ –2;

minimum: –2

b)

c)

d)

2. a) (–5, –6), maximum b) (–4, 2), minimum

c) (2.5, –9), minimum d) (9, 19), maximum

3. a) x-intercepts: 3, –1; y-intercept: –3

b) x-intercepts: 0, –4; y-intercept: 0

c) no x-intercepts; y-intercept: 12

d) x-intercepts: 6.8, 1.2; y-intercept: –4

4. a) y = –2(x – 3)2 – 1 b)

c) d) y = 3(x + 5)2 + 3

5. –6, 0

4.4 Graphing y = ax2 + bx + c byCompleting the Square

1. a) 121 b) 64 c) 9

d) 400 e) 4 f) 9

2. a) y = (x – 2)2 – 5 b) y = –2(x + 1)2 + 3

3. a) min: –26 at x = 3 b) max: 53 at x = –4

c) min: 271 at x = 6 d) min: –18 at

4. Let one number be n and the other be n – 8. The

product, p, is n × (n – 8), so p = n2 – 8n and the

function is a parabola. p = (n – 4)2 – 16, so the vertex

is at (4, –16). The function reaches a minimum value

of –16 when n = 4. If n = 4, then n – 8 = 4 – 8 or –4.

The two numbers are 4 and –4.

5. 18 cm2

6. a) 18 m b) 1.8 s c) 3.7 s

4.5 Investigation: Sketching Parabolasin the Form y = ax(x – s) + t

1. a) y = x(x – 6) + 8 b)

c) y = x(x – 9) + 0 d) y = –3x(x – 4) – 0.9

2. a) Factor 2x from the first two terms to get

y = 2x(x + 2) – 6, or 2x(x – (–2)) – 6. Substitute 0 for x

to get y = –6. Substitute the value of s, –2, for x to get

y = –6. The two points (x, y) are (0, –6) and (–2, –6).

The coordinates of the vertex are (–1, –8).

b) Factor –x from the first two terms to get

y = –x(x – 2) + 1. Substitute 0 for x to get y = 1.

Substitute the value of s, 2, for x to get y = 1. The two

points (x, y) are (0, 1) and (2, 1). The coordinates of

the vertex are (1, 2).

3. a) y = 3x(x + 2) – 8 b) y = –x(x – 6) + 3

4. In question 2a), a = 2, s = –2, and t = –6, so the

coordinates of the vertex are or

(–1, –8). In question 2b), a = –1, s = 2, and t = 1, so the


(1, 2). In question 3a), a = 3, s = –2, and t = –8, so the


(–1, 11). In question 3b), a = –1, s = 6, and t = 3, so

the coordinates of the vertex are

or (3, 12).

62

31 6

4

2

,( )( )

− −

− − − −

22

83 2

4

2

,( )

22

11 2

4

2

,( )( )

− −

− − − −

22

62 2

4

2

,( )

0 x

y

2

(0, 3)

(3, 12)

(6, 3)

2

y = –x2 + 6x + 320 x

y

–2

–2

(0, –8)(–2, –8)

(–1, –11)

y = 3x2 + 6x – 8

y x x= +

−2

12

5

x = 52

0

2

x

y

(0, 1)

(–1, 3)

(–2, 1)

2–2

y = –2x2 – 4x + 1

x = –1

0

–2

x

y

(4, –1)(0, –1)

(2, –5)

2

y = x2 – 4x – 1

x = 2

y x= − + −12

4 12( )

y x= − +12

2 52( )

20

2

x

y(–2, 7)

–2

y = – (x + 2)2 + 71–2

20

2

x

y

(1, 1)

y = 2(x – 1)2 + 1

20

–2

x

y

(4, –3)

y = –(x – 4)2 – 3


down; (4, –3); x = 4;


range: y ≤ –3; maximum: –3

up; (1, 1); x = 1;

domain: set of real

numbers, range: y ≥ 1;

minimum: 1

down; (–2, 7); x = –2;


range: y ≤ 7; maximum: 7

4.6 Investigation: Finite Differences

1. a) y: 10, 13, 16, 19, 22; 1st Difference: 3

b) y: 8, 3, –2, –7, –12; 1st Difference: –5

c) 1st Difference: 2 d) 1st Difference: –6

2. a) The first difference is equal to m, the slope.

b) The y-value when x = 0 is equal to b.

c) y = 2x + 3; y = –6x + 2

3. y = –0.5x + 5

4. a) y: 9, 16, 27, 42, 61; 1st Difference: 7, 11, 15, 19;

2nd Difference: 4

b) y: –11, –6, 1, 10, 21; 1st Difference: 5, 7, 9, 11;

2nd Difference: 2

5. a) The second difference is equal to 2a.

b) The y-value when x = 0 is equal to c.

c) In part a) the second difference is 4, so a = 2. The

value of a + b is 7, so b = 5. In part b) the second

difference is 2, so a = 1. The value of a + b is 5, so b = 4.

d) c = 1, a = 2, b = 3; y = 2x2 + 3x + 1

4.7 Technology: Equations of Parabolasof Best Fit

1. Answers may vary. a) y = (x – 1.5)2

b) y = (x – 1)2 + 8 c) y = 2(x + 0.25)2 – 6

d) y = –4(x – 1)2 + 1 e) y = 0.25(x + 2)2 – 3

f) y = 0.5 (x + 3)2 – 3

2. a) y = x2 – 3x + 2 b) y = x2 – 2x + 9

c) y = 2x2 + x – 6 d) y = –4x2 + 6x

e) y = 0.24x2 + 0.98x – 2.01 f) y = 0.33x2 + 2x – 0.33

3. b) –40(x – 0.08)2 + 7; The vertex of the parabola

appears to be at (0.08, 7), so the value of h is 0.08 and

the value of k is 7. The value of a is negative because

the curve opens down. Use trial and error to guess

the value of a. c) 7 m

d) y = –49.25x2 + 8.39x + 0.37, rounded to the nearest

hundredth

e) 0.72 m, 0.52 m

4.8 Technology: Collecting Distance andTime Data Using CBR™ or CBL™

1. a) about 1 m b) 1.373 m

c) 1.373 – 1.036 = 0.337; about 34 cm

d) 0.26 s e) No, the value should be between

1.136 m and 1.238 m. It was likely due to a glitch that

caused a misreading by the motion detector unit.

f) The motion detector unit takes readings while it is

in one position. The range of the unit is limited when

measuring, so vertical is best.

g) The height could be anywhere from on the ground

(0 cm) to a flat surface less than 80 cm high; for

example, on a chair (about 45 cm) or a table (about

70 cm).

h) probably, so it wouldn’t land on the motion

detector unit

i) independent variable: Elapsed time; dependent

variable: Height

2. The parabola opens down and the vertex is about

(0.3, 1.3), so y = –a(x – 0.3)2 + 1.3. By trial and error

a = –4, so y = –4(x – 0.3)2 + 1.3. The quadratic

equation of best fit is y = –5.555x2 + 3.128x + 0.928.

3. The quadratic equations are closely related, and fit

the sample data very well except for 1 stray point.

4. a) –523.292 m; This value is not possible, because

the least possible height is 0 m. Since this value is less

than 0 m, the ball has already hit the ground.

b) When x = 0.8, y = –0.125 and when x = 0.7,

y = 0.396; try x = 0.77, y = 0.043; try x = 0.777,

y = 0.005. The ball would have hit the ground at

0.78 s.



Name

1. Solve by graphing. Round to the nearesttenth, if necessary.a) x2 – 3x – 10 = 0 b) x2 + 2x = 8

c) 0.5(x – 1)2 = 2 d) x2 = 3x + 4

e) 4x2 – 4x + 1 = 0 f) 2x2 + 7x – 4 = 0

g) 7x – 3x2 = 0 h) 3(p – 1)2 + 4 = 0

i) x2 + 5x = –6 j) 4(x + 2)2 – 9 = 0

2. Application A toy rocket was launched fromthe top of a building, 50 m above ground level.The height of the rocket above ground level, hmetres, after t seconds is given by the formulah = 50 + 45t – 5t2. How many seconds after thelaunch will the rocket hit the ground?

Problem Solving

For each problem, write a quadratic equationto find the unknown(s). Solve the equation bygraphing.

3. The length of a rectangle is 2 m more than thewidth. The area is 48 m2. Find the dimensions ofthe rectangle.

4. The sum of the squares of three consecutiveintegers is 77. Find the integers.

5. A room is in the form of a rectangular box,with length 6 m. The height is 1 m less thanthe width. If the volume is 72 m3, find thedimensions of the room.

6. The hypotenuse of a right triangle is 15 cm.The other two sides have a total length of 21 cm.Find the lengths of the two unknown sides.

5.1 Solving Quadratic Equations by GraphingMATHPOWERTM 10, Ontario Edition, pp. 270–277

• The solutions to the quadratic equation ax2 + bx + c = 0 are the x-intercepts of the quadraticfunction y = ax2 + bx + c.

• There are three possible results when solving a quadratic equation.a) two distinct real roots b) two equal real roots c) no real roots


Name

1. State the roots of each equation.a) (x – 2)(x + 7) = 0 b) (3x + 1)(2x – 3) = 0

c) 7x(x – 5) = 0 d) (2x + 5)(2x + 5) = 0

2. Write each equation in the formax2 + bx + c = 0.a) 3x2 = 5(x – 2) b) 2(x – 1)2 = 3x + 7

c) d)

3. Solve and check.a) x2 – 6x + 8 = 0 b) y2 + 10y + 24 = 0

c) q2 – 3q – 28 = 0 d) m2 + 6m – 27 = 0

e) 6t2 – 17t + 12 = 0 f) 9x2 + 30x + 25 = 0

g) 6t2 – t – 35 = 0 h) 6p2 – 16p = 0

i) (2x – 1)2 = 7x + 4 j)

k) (3x – 1)2 = 25 l)

Problem Solving

4. The hypotenuse of a right triangle is 1 mlonger than twice one of the other two sides.The third side of the triangle is 15 m. Find thelengths of the unknown side and hypotenuse.

5. The length of a flag is twice the width. Findthe width of a flag with an area of 1250 cm2.

6. Three consecutive even integers are such thatthe product of the first two is 6 less than 9 timesthe third. Find the integers.

7. A rectangle is 24 cm long and 16 cm wide.When each dimension is increased by the sameamount, the area is doubled. What are the newdimensions?

8. Communication Write a quadratic equation

whose roots are –3 and Explain your

thinking.

12

.12

1 3 52x − = .

54

515

4

2a a− =

x x2

4 313

− =45

3 22x

x= −

5.2 Solving Quadratic Equations by FactoringMATHPOWERTM 10, Ontario Edition, pp. 278–286

• To solve a quadratic equation by factoring,a) write the equation in the form ax2 + bx + c = 0b) factor ax2 + bx + cc) use the zero product propertyd) solve the two resulting equations to find the rootse) check your solutions


Name

1. Sketch the graphs of the following quadraticfunctions by locating the x-intercepts, and thenfinding the coordinates of the vertex.a) y = (x – 3)(x – 5) b) y = x2 – 1

2. Use the x-intercepts to determine thecoordinates of the points on the x-axis and thevertex for the graph of each quadratic function.a) y = (x + 5)(x – 5) b) y = x2 – 49

3. Sketch the graphs of the following quadraticfunctions by factoring to find the x-intercepts,and then deducing the coordinates of the vertex.a) y = x2 + 2x b) y = x2 – 7x + 12

c) y = x2 + 4x – 12 d) y = 6x – 0.5x2

e) y = x2 + 4x – 21 f) y = x2 – 3x – 18

g) y = –x2 – 2x + 3 h) y = x2 + 5x + 6

4. Communication Explain how to sketch agraph of each function using the interceptsand axis of symmetry.a) y = (x – 5)2

b) y = (x + 6)2

0 x

y

2

2

–2

–20 2

2

–2

–2 x

y

0

–2

2–2 x

y

0–5

–5

x

y

0 x

y

2

2

0 x

y

2–2

–2

20

2

x

y

–2

20

2

x

y

–2

–2

20

2

–2

–2

x

y

20

2

–2

x

y

5.3 Investigation: Graphing Quadratic Functions by FactoringMATHPOWERTM 10, Ontario Edition, p. 287

• To sketch the graph of a quadratic function,a) find the x-intercepts by factoring the equation in the form ax2 + bx + c = 0b) plot the points where the graph crosses the x-axis, (y = 0)c) use symmetry to find the x-coordinate of the vertexd) find the y-coordinate of the vertex by substituting the x-coordinate of the vertex into the

equation in the form y = ax2 + bx + ce) draw a smooth curve through the three points


Name

1. Solve using the quadratic formula.a) x2 – 8x + 12 = 0 b) 2y2 – 3y – 2 = 0

c) 20x2 + 27x = 14 d) 48x2 – 58x + 15 = 0

2. Solve using the quadratic formula. Expressanswers as exact roots. a) x2 – 6x + 3 = 0 b) 3x2 + 5x + 1 = 0

c) 3x2 – 6x – 8 = 0 d) 4x(x + 8) = 3

e) b2 + 3b = 1 f) 4x2 – 2x + 5 = 0

3. Solve using the quadratic formula. Round tothe nearest hundredth, if necessary.a) x2 + 13x + 9 = 0 b) 4x2 – 11x – 19 = 0

c) 1.6(y2 + 5) = 13.4y d) 18x2 + 5x + 17 = 0

e) a2 – 44 = 0 f) (x + 1)2 + (x + 3)2 = 25

4. Application Use the Pythagorean Theoremto find the value of y, to the nearest hundredth.

Problem Solving

5. The sum of the squares of three consecutiveodd integers is 875. Find the integers.

6. A plain mat is placed around a picturemeasuring 28 cm by 36 cm so that the width ofthe mat is equal on all sides. The area of the mat

is of the area of the picture. Find the width of

the mat, to the nearest millimetre.

7. A window is in the shape of a rectanglesurmounted by a semicircle. The height of therectangle is 0.4 m more than the width. The totalarea of the window is 10.4 m2. Find the widthand height of the window, to the nearesthundredth.

8. Communication Is it possible to write tworeal numbers whose sum is 4 and whoseproduct is 5? Use the quadratic formula to helpyou explain.

34

y + 2

17

y

5.4 The Quadratic FormulaMATHPOWERTM 10, Ontario Edition, pp. 288–295

• To solve a quadratic equation using the quadratic formula, write the equation in the form ax2 + bx + c = 0, a ≠ 0.

• The quadratic formula is xb b ac

a= − ± −2 4

2.


5.1 Solving Quadratic Equations byGraphing

1. a) 5, –2 b) 2, –4 c) 3, –1 d) 4, –1

e) 0.5 f) 0.5, –4 g) 0, 2.3 h) no solution

i) –2, –3 j) –0.5, –3.5

2. 10 s

3. 8 m by 6 m

4. 4, 5, 6 or –4, –5, –6

5. 6 m by 4 m by 3 m

6. 12 cm, 9 cm

5.2 Solving Quadratic Equations byFactoring

1. a) 2, –7 b) c) 0, 5 d)

2. a) 3x2 – 5x + 10 = 0 b) 2x2 – 7x – 5 = 0

c) 4x2 – 15x – 10 = 0 d) 3x2 – 4x – 4 = 0

3. a) 2, 4 b) –4, –6 c) –4, 7 d) –9, 3

e) f) g) h)

i) j) 4, –1 k) l) 3, –3

4. 8 m, 17 m

5. 25 cm

6. 10, 12, 14

7. 32 cm by 24 cm

8. x = –3 and so Expand the

equation to get Then, multiply

every term by 2 to eliminate the denominators, to get

2x2 – x + 6x – 3 = 0 or 2x2 + 5x – 3 = 0.

5.3 Investigation: Graphing QuadraticFunctions by Factoring

1. a) b)

2. a) (–5, 0), (5, 0), (0, –25)

b) (–7, 0), (7, 0), (0, –49)

3. a) b)

c) d)

e) f)

g) h)

4. a) The equation is in factored form, so there is

only one x-intercept, 5. Plot (5, 0). To find the

y-intercept, let x = 0 and find y. Plot (0, 25). A third

point is symmetrical to (0, 25) about the axis of

symmetry x = 5. Plot (10, 25). Check by substituting

(10, 25) into the equation.

0 2

–2

2

x

y

(–3, 0) (–2, 0)

y = x2 + 5x + 6

– , – 5–2

1–4

(0, 1)

(–1, 4)

(–3, 0)

y = –x2 – 2x + 3

0

2

2

–2

–2 x

y

0–2

–2

2 x

y(–3, 0) (6, 0)

y = x2 – 3x – 18

, – 203–21–4

0 2–2

–2

x

y(–7, 0) (3, 0)

(–2, –25)

y = x2 + 4x – 21

0

2

2 x

y

(0, 0) (12, 0)

(6, 18)

y = 6x – 0.5x2

0 2

–2

–2 x

y(2, 0)(–6, 0)

(–2, –16)

y = x2 + 4x – 12

0

2

x

y

(3, 0) (4, 0)

2

y = x2 – 7x + 12

, – 7–2

1–4

0

2

x

y

(–1, –1)

(–2, 0) (0, 0)2–2

y = x2 + 2x

0

2

x

y

(–1, 0)

(0, –1)

(1, 0)2–2

y = x2 – 1

0

2

x

y

(3, 0) (5, 0)

(4, –1)

2

y = (x – 3)(x – 5)

x x x2 12

332

0− + − = .

x x+( ) −

=3

12

0.x = 12

,

243

, −− 14

, 3

083

, 52

73

, −− 53

32

43

,

− 52

− 13

32

,

Answers CHAPTER 5 Quadratic Equations

b) There is one x-intercept, –6. Plot (–6, 0). To find the

y-intercept, let x = 0 and find y. Plot (0, 36). A third

point is symmetrical to (0, 36) about the axis of

symmetry x = –6. Plot (–12, 36). Check by substituting

(–12, 36) into the equation.

5.4 The Quadratic Formula

1. a) 6, 2 b) c) d)

2. a) b)

c) d)

e) f) no solution

3. a) –0.73, –12.27 b) 3.95, –1.20

c) 7.73, 0.65 d) no solution

e) ±6.63 f) –5.39, 1.39

4. y = 10.98

5. 15, 17, 19, or –19, –17, –15

6. 5.1 cm or 51 mm

7. 2.59 m wide, 2.99 m high

8. If one number is x and the other number is 4 – x,

the product of the two numbers, x(4 – x), should

equal 5. Try to solve x2 – 4x + 5 = 0. There are no real

solutions.

− ±3 132

− ±8 672

3 333

±

− ±5 136

3 6±

38

56

, − 74

25

, 212

, −


1. Use The Geometer’s Sketchpad® to • construct an obtuse triangle, such as �ABC

• construct a point, D, on side AB • create a line through D parallel to AC, andlabel the new intersection point E• hide the line, and join points D, B, and E, toform �DBEa) Measure the angles, side lengths, and areas of�ABC and �DBE.b) Communication Show that �ABC ~ �DBE.

2. CommunicationExplain why �ABC issimilar to �EDC bya) comparing themeasures of thecorresponding angles, andthe ratios of the lengths ofthe corresponding sidesb) calculating the ratio oftheir areas

3. Application Use The Geometer’s Sketchpad® toconstruct an equilateral triangle:• construct two points, A and B• construct a circle with one point as the centreand the other point as a point on the circle• construct a circle, switching which point is thecentre and which point is on the circle• construct the points C and D at theintersections of the circles• join one of the intersection points and A andB to form an equilateral triangle, such as �ABD• hide the circles and theother intersection point, C

a) Measure the angles, side lengths, and area ofthe triangle.b) Drag each vertex to enlarge the size of yourequilateral triangle. Then, measure the angles,side lengths, and area of this triangle.c) Communication Show that �ABD and theenlarged triangle are similar.

d) Communication Are all equilateral trianglessimilar? Explain and justify your reasoning.

A C

D E

B

6.1 Technology: Investigating Similar Triangles Using The Geometer’s Sketchpad®MATHPOWERTM 10, Ontario Edition, pp. 316–317

• Similar triangles have the same shape but not necessarily the same size.�PQR and �XYZ are similar. In �PQR and �XYZ,a) ∠P = ∠X, ∠Q = ∠Y, and ∠R = ∠Z. The corresponding pairs of

angles are equal.

b) and The ratios of the corresponding

sides are equal.

c) The ratio of their areas is equal to the ratio of the squares of their

corresponding sides. 14

12

2

2=

area PQRarea XYZ

�

�= =2

814

.

PRXZ

= 12

.PQXY

, QRYZ

= =12

12

,


Name

A

E

CDB

A B

C

D

P

R

X

ZYQ

1. In each diagram, the triangles are similar.

Write the ratio of the lengths of the sides

a)

b)

2. The triangles in each pair are similar. Find theunknown side lengths.a)

b)

c)

3. Communication Explain why �ABC issimilar to �AEF.

4. Find a.

5. Problem Solving Nida is 1.8 m tall and castsa shadow 1.5 m long. At the same time, amicrowave relay tower casts a shadow 32 mlong. Draw and label 2 triangles depicting theinformation. Determine the height of the tower.

6. ApplicationRanin markedout thefollowingtriangles todetermine thelength of a pond.Calculate thelength of thepond, AB, to thenearest tenth of a metre.

4

8

3

a

B

A

E

C F

10 m15 m

32 m

16 m

V

W

Z

y w

Y

X

°

°

x x√√

20 m

D

12 m

dF °°4 m

b3 mA

B C E

R

15 cm

25 cm

20 cm

Q

P

°°

xx

√

√

20 cm

T

U

S

t

s

A

B

C

D

E

F

°

°

A B

C

D E

F

° °

x

x√ √

BCEF

.

6.2 Similar TrianglesMATHPOWERTM 10, Ontario Edition, pp. 318–325

• If �ABC and �DEF are similar,a) the corresponding pairs of angles are equal

∠A = ∠D ∠B = ∠E ∠C = ∠Fb) the ratios of the corresponding sides are equal

c) the ratio of their areas is equal to the ratio of the squares of their corresponding sides

area ABCarea DEF

�

�= = =a

dbe

cf

2

2

2

2

2

2

ad

be

cf

= =

B

c b

a

A

C

D

f e

Ed

F

°°

x x


Name

3.8 m

2.8 m

13 m

E D

C

A B

1. Use a calculator to find the tangent of eachangle, to the nearest thousandth.a) 37° b) 84°

c) 15° d) 45°

e) 60° f) 72°

2. Find ∠K, to the nearest degree.a) tan K = 0.575 b) tan K = 0.243

c) tan K = 1.925 d) tan K = 2.750

e) tan K = 3.198 f) tan K = 50.375

3. Find ∠Q, to the nearest degree.

a) b)

c) d)

e) f)

4. Calculate tan D and ∠D and tan E and ∠E.Round each angle measure to the nearest degree.a) b)

5. Calculate x, to the nearest tenth of a metre.a) b)

c) d)

6. a) Find the length of PQ, to the nearest tenth of a metre.b) Classify �PQR.

7. Application Find the length of x, then thelength of y, to the nearest tenth of a metre.

8. Problem Solving The backyard of a home isin the shape of a right triangle in which one sideis twice as long as the other side. If one of thesides is the length of the house, and it is 15 mlong, find the length of the other side. Draw adiagram to show the backyard.

x39°12 m

28°y

3.7 m

45°

Q

RP

x

60°17 m

x

50°

6 m

x

28°12 m

x

3 m43°

E

8 m

9 mND

A

2 cm

4 cm E

D

tan Q = 892

tan Q = 499

tan Q = 125

tan Q = 54

tan Q = 58

tan Q = 13

6.3 The Tangent RatioMATHPOWERTM 10, Ontario Edition, pp. 326–333

• For any acute angle A in a right triangle, the tangent ratio is

or tan Aoppositeadjacent

=

tangent Alength of the side opposite A

length of the side adjacent to A=

∠∠

A

B

C

opposite

adjacent


Name

1. Use a calculator to find the sine of each angle,to the nearest thousandth.a) 62° b) 21°

c) 85° d) 45°

e) 5° f) 70°

2. Find ∠B, to the nearest degree.a) sin B = 0.990 b) sin B = 0.208

c) sin B = 0.500 d) sin B = 1.000

e) sin B = 0.345 f) sin B = 0.755

3. Find ∠G, to the nearest degree.

a) b)

c) d)

e) f)

4. Calculate sin Y. Then, find ∠Y, to the nearestdegree.a) b)

5. Calculate y, to the nearest hundredth of ametre.a) b)

c) d)

e) f)

6. Application A kite, tied to a dock, is flyingover the water. What is the height of the kiteabove the water, to the nearest tenth of a metre,if the length of the kite string is a) 60 m? b) 35 m?

7. Problem Solving �KLM is an equilateraltriangle. The length of each side of the triangle is 15 cm. Find the height of the triangle, to thenearest tenth of a centimetre.

8. Communication Explain why the sine of any acute angle in a right triangle is always less than 1.

25°

y

25°

10 my

45°17 m

y72°96 m

y

60°59 m

y

30°

15 m

y

54°28 m

Z

15 cm

X11 cm

YZ

3 cm

X

Y6 cm

sin G = 89

sin G = 111

sin G = 58

sin G = 45

sin G = 25

sin G = 12

6.4 The Sine RatioMATHPOWERTM 10, Ontario Edition, pp. 334–339

• For any acute angle A in a right triangle, the sine ratio is

or sin Aopposite

hypotenuse=

sine Alength of the side opposite A

length of the hypotenuse=

∠

A

B

C

hypotenuseopposite


Name

1. Use a calculator to find the cosine of eachangle, to the nearest thousandth.a) 23° b) 79°

c) 30° d) 50°

e) 43° f) 7°

2. Find ∠E, to the nearest degree.a) cos E = 0.982 b) cos E = 0.174

c) cos E = 0.454 d) cos E = 0.777

e) cos E = 0.999 f) cos E = 0.009

3. Find ∠V, to the nearest degree.

a) b)

c) d)

e) f)

4. Calculate cos H. Then, find ∠H, to the nearestdegree.a) b)

5. Calculate w, to the nearest tenth of acentimetre.a) b)

c) d)

6. Application Find the distance from Dani tothe clubhouse.

7. Communication How can you tell whetherthe sine or the cosine of an acute angle in a righttriangle will have the greater ratio?

8. Problem Solving A 4-m ladder leans againsta wall. The foot of the ladder makes an angle of63° with the ground. How far from the wall isthe foot of the ladder, to the nearest tenth of ametre?

clubhouse

home

Dani

d

54°1.8 km

25 cm

60°w

32 cm 70°w

27 cm

30°w

17 cm

48°w

13 m

5 m

H

4 cm

5 cm

H

cos V = 613

cos V = 1415

cos V = 111

cos V = 23

cos V = 78

cos V = 14

6.5 The Cosine RatioMATHPOWERTM 10, Ontario Edition, pp. 340–345

• For any acute angle A in a right triangle, the cosine ratio is

or cos Aadjacent

hypotenuse=

cosine Alength of the side adjacent to A

length of the hypotenuse=

∠

A

B

C

hypotenuse

adjacent


Name

1. Find all the unknown angles, to the nearestdegree, and all the unknown sides, to thenearest tenth of a unit.a) b)

c) d)

e) f)

2. Solve each triangle. Round each side lengthto the nearest tenth of a unit, and each angle, tothe nearest degree.a) b)

c) d)

3. Problem Solving A slide that is 4.2 m longmakes an angle of 35° with the ground. Howhigh is the top of the slide above the ground?

4. Problem Solving A rope is anchored to theground at its ends and is propped up in themiddle by a 1-m vertical stick. At one end, therope makes an angle of 55° with the ground.How long is the rope, to the nearest centimetre?

FE

G

45°

5 m25 mm

D

C

B

7 mm

40°

19 m

T U

S

15 cm

9 cm

Q

RP

4 m

7 m

O

M

N

5 m

7 m

L

K

JG

4 cmI

8 cm

H

3 m5 m

A C

6.6 Solving Right TrianglesMATHPOWERTM 10, Ontario Edition, pp. 346–351

• To use trigonometry to solve a right triangle, given the measure of one acute angle and the length ofone side, finda) the measure of the third angle using the angle sum in the triangleb) the measure of a second side using sine, cosine, or tangent ratiosc) the measure of the third side using a sine, cosine, or tangent ratio, or the Pythagorean Theorem

• To use trigonometry to solve a right triangle, given the lengths of two sides, finda) the measure of one angle using its sine, cosine, or tangent ratiob) the measure of the third angle using the angle sum in the trianglec) the measure of the third side using a sine, cosine, or tangent ratio, or the Pythagorean Theorem


Name

5 cm

D

E

12 cmF

24 cm

X

W

V

14 cm

1. Find BC, to the nearest centimetre.

2. Find XY, to the nearest tenth of a centimetre.

3. Find PQ, to the nearest tenth of a metre.

4. Problem Solving From a point on theground, a student sights the top and bottom of a15-m flagpole on the top of a building. The twoangles of elevation are 64.6° and 57.3°.a) Draw a diagram for the information given inthe problem.

b) How far is the student from the foot of thebuilding? Round your answer to the nearesttenth of a metre.

5. Application From two tracking stations425 km apart, a satellite is sighted at C aboveAB, making ∠CAB = 48.3° and ∠CBA = 62.6°.

Find the height of thesatellite, to the nearesttenth of a kilometre.

6. Problem Solving Two buildings are 14.7 mapart. From the top of one building, the anglesof depression of the top and bottom of thesecond building are 27.5° and 63.8°. Find theheights of the buildings, to the nearest tenth of ametre.

39.7°50 m

50.3°R

O

QS

P

40 m

65°54.5° W

V

X

Y

38.7 cm

43.4°

34°100 cm

C

D

B

A

6.7 Problems Involving Two Right TrianglesMATHPOWERTM 10, Ontario Edition, pp. 352–359

• To solve a problem involving two right triangles using trigonometry,a) draw and label a diagram showing the given information, and the length or angle measure

to be foundb) identify the two triangles that can be used to solve the problem, and plan how to use each

trianglec) solve the problem and show each step in your solutiond) write a concluding statement giving the answer


Name

48.3°

C

62.6° BA425 km

1. Communication Examine �ABC at the topof the page.

a) Explain why

b) Do you get the same results as the ratios ofthe side lengths if you use the cosine ratioinstead of the sine ratio? the tangent ratioinstead of the sine ratio?

2. a) Calculate each pairof ratios, to the nearest tenth, for �XYZ.

b) Communication Determine how manydifferent pairs of ratios are possible. Does therelationship apply to all of these pairs?

c) Communication Explain why the pairs ofratios may not be equal if you round values tothe nearest tenth.

3. Application Use The Geometer’s Sketchpad®and the instructions on Practice Master 6.1question 3 to create an equilateral triangle.a) Can you determine the value of the ratio ofany two side lengths and the ratio of the sines oftheir opposite angles without measuring?Explain.

b) Check by calculating the pair of ratios.

4. Application Use The Geometer’s Sketchpad® toconstruct an isosceles triangle:• construct two points• construct a circle with one point as the centreand the other point on the circle• construct another point on the circle• join the three points to form the triangle• hide the circle

a) Can you determine the value of the ratio ofany two side lengths and the ratio of the sines oftheir opposite angles without measuring?Explain.

b) Check by calculating the pair of ratios.

C

BA

YZXZ

and sin Xsin Y

XYXZ

and sin Zsin Y

XYYZ

and sin Zsin X

tan BACBC

≠ .

6.8 Technology: Relationships Between Angles and Sides in Acute TrianglesMATHPOWERTM 10, Ontario Edition, pp. 360–361

• In an acute triangle, such as �ABC,a) the longest side, BC, is opposite the largest angle, ∠Ab) the shortest side, AB, is opposite the smallest angle, ∠Cc) the ratio of the side lengths is equal to the ratio of the sines of their opposite angles;

and rounded to the nearest hundredthsin Bsin C

= 1 25. ,ACAB

= =14 0911 23

1 25..

.


Name

68.97°

A

B C64.84° 46.19°

11.23 cm

14.53 cm

14.09 cm

68.7°

X

45.7° 65.6°

9.6 cm

9.9 cm

7.6 cm

Y Z

1. Find the length of the indicated side, to thenearest metre.a) b)

2. Find the measure of the indicated angle, tothe nearest degree.a) b)

3. Find the indicated quantity, to the nearesttenth.a) In �KLM, ∠K = 74°, ∠L = 47.5°, and m = 37.7 cm. Find k.

b) In �ABC, ∠A = 50°, a = 9 m, and b = 8 m.Find ∠B.

4. Solve the triangle.Round each answer to thenearest whole number.

5. Application Find thearea of �ABC, to thenearest square centimetre.

6. Application Observers at points A and B,who stand on level ground on opposite sides ofa tower, measure the angle of elevation to thetop of the tower at 33° and 49°, respectively. Athird point, C, is 120 m from B. ∠ABC = 67° and∠BAC = 31°. Find the height of the tower, h, tothe nearest metre.

7. Problem Solving A rock and an oak tree areon the same side of a ravine and are 125 mapart. A birch tree is on the opposite side of theravine. The angle formed between the linejoining the rock and oak tree and the line joiningthe rock and the birch tree is 25°. The angleformed by the line joining the rock and the oaktree and the line joining the oak tree and thebirch tree is 72°.a) Draw a diagram containing the information.

b) Calculate the width of the ravine. Roundyour answer to the nearest tenth of a metre.

33° 49°67°31°

120 m

BA

C

h

47°

L

M

12 m

10 m

K

M

79°

ON34 cm

20 cm

41°

J

63°K L

27 m k

39°15 m

S

p

61°P

R

6.9 The Sine LawMATHPOWERTM 10, Ontario Edition, pp. 362–368

• There are two forms of the sine law.

• The sine law can be used to solve an acute triangle when given:a) the measures of two angles and any sideb) the measures of two sides and an angle opposite one of these sides

a b ca b csin A sin B sin C

or A sin B sin C= = = =sin


Name

CB

A

c b

a

79°

43 cm

38 m

H J

G

A

B C

84°

59°

24 cm

1. Find the missing side length, to the nearesttenth of a unit.a) b)

2. Find the measure of the indicated angle, tothe nearest degree.a) b)

3. Find the indicated quantity, to the nearesttenth.a) In �CDE, ∠E = 50°, c = 11.9 cm, andd = 13.5 cm. Find e.

b) In �KLM, k = 54.2 cm, l = 45.7 cm, andm = 36.9 cm. Find ∠K.

4. Solve each triangle. Round each calculatedvalue to the nearest whole number, if necessary.a)

b) In �NPQ, n = 8.2 cm, q = 13.7 cm, and∠P = 67°.

5. Application Find the area of �XYZ, to thenearest square metre.

6. Communication Explain whether you canuse the cosine law to find f in �DEF when givend = 19.2 cm, e = 14.7 cm, and ∠F = 39°.

7. Problem Solving Two boats left a dock atthe same time. One travelled at 7 km/h on abearing of 39°. The other travelled at 5 km/h ona bearing of 82°. How far apart were the twoboats after 3 h? Round your answer to thenearest tenth of a kilometre.

Z

17.5 m

29.1 mY

X

67°

W

YX

77 m

120 m

115 m

E

D

F

3.5 m

1.5 m

2.9 m

P

RQ

8.6 m

11.2 m

57°

6.10 The Cosine LawMATHPOWERTM 10, Ontario Edition, pp. 369–376

• There are two forms of the cosine law.

a2 = b2 + c2 – 2bc cos A or

• The cosine law can be used to solve an acute triangle when given:a) the measures of two sides and the contained angleb) the measures of three sides

cos A = + −b c abc

2 2 2

2


Name

C

hab

A D c – x Bxc

h

K

ML

7.9 cm5.1 cm80°

A

CB

19.7 cm

23.9 cm

14.1 cm

6.1 Technology: Investigating SimilarTriangles Using The Geometer’sSketchpad®

1. a) Answers will vary.

b) ∠B is common, ∠A = ∠D, and ∠E = ∠C,

or

2. a) ∠B = ∠D = 60° (equilateral triangle),

∠A = ∠E = 30° (half of an equilateral triangle),

∠C = 90° and is common;

b)

3. a) and b) Answers will vary.

c) The ratio of the two triangles’ areas is equal to the

ratio of the squares of their side lengths.

d) Yes, corresponding angles are always equal, since

they are always 60°; and the ratios of the three

corresponding side lengths are always equal, since

the side lengths in each triangle are always equal.

6.2 Similar Triangles

1. a) 2:1 b) 1:2

2. a) s = 33.3 cm, t = 26.7 cm

b) b = 5 m, d = 16 m c) y = 20 m, w = 24 m

3. ∠A is common, ∠ACB = ∠AFE (parallel lines),

∠ABC = ∠AEF (parallel lines)

4. a = 3

5.

The height is 38.4 m.

6. 17.6 m

6.3 The Tangent Ratio

1. a) 0.754 b) 9.514 c) 0.268

d) 1.000 e) 1.732 f) 3.078

2. a) 30° b) 14° c) 63°

d) 70° e) 73° f) 89°

3. a) 18° b) 32° c) 51°

d) 67° e) 80° f) 89°

4. a) tan D = 2.000; ∠D = 63°; tan E = 0.500; ∠E = 27°

b) tan D = 0.889; ∠D = 42°; tan E = 1.125; ∠E = 48°

5. a) 2.8 m b) 6.4 m c) 7.2 m d) 9.8 m

6. a) 3.7 m b) isosceles right triangle

7. x = 9.7 m and y = 18.3 m

8. 2 possible answers:

6.4 The Sine Ratio

1. a) 0.883 b) 0.358 c) 0.996

d) 0.707 e) 0.087 f) 0.940

2. a) 82° b) 12° c) 30°

d) 90° e) 20° f) 49°

3. a) 30° b) 24° c) 53°

d) 39° e) 5° f) 63°

4. a) sin Y = 0.500; ∠Y = 30°

b) sin Y = 0.733; ∠Y = 47°

5. a) 22.65 m b) 7.50 m c) 51.09 m

d) 29.66 m e) 12.02 m f) 9.06 m

6. a) 25.4 m b) 14.8 m

7. 13.0 cm

8. Since the sine has the hypotenuse as the second

term and the hypotenuse is always the longest side,

it is the ratio of a lesser number to a greater number.

6.5 The Cosine Ratio

1. a) 0.921 b) 0.191 c) 0.866

d) 0.643 e) 0.731 f) 0.993

HOUSE HOUSE15 m

30 m

15 m

7.5 m

1.8 m

1.5 m

x

32 m°

x

°

x

or 18 triangular units8 triangular units

area of ABCarea of EDC

�

�= =3

294

2

2

ABED

= =64

32

BCDC

ACEC

= =32

32

, ,

ACDE

or CBEB

2

2

2

2

area of ABCarea of DBE

ABDB

2

2

�

�=AB

DBACDE

CBEB

= = ,

Answers CHAPTER 6 Trigonometry


Name

2. a) 11° b) 80° c) 63°

d) 39° e) 3° f) 89°

3. a) 76° b) 29° c) 48°

d) 85° e) 21° f) 62°

4. a) cos H = 0.800; ∠H = 37°

b) cos H = 0.385; ∠H = 67°

5. a) 11.4 cm b) 23.4 cm

c) 10.9 cm d) 21.7 cm

6. 1.1 km

7. Since both sine and cosine have the hypotenuse as

the second term, the ratio with the greatest first term

will be greater. That is, if the opposite side is longer

than the adjacent side, the sine will be greater; if the

adjacent side is longer than the opposite side, the

cosine will be greater.

8. 1.8 m

6.6 Solving Right Triangles

1. a) AC = 4 m, ∠B = 53°, ∠C = 37°

b) DE = 13 cm, ∠D = 23°, ∠E = 67°

c) GH = 6.9 cm, ∠G = 30°, ∠I = 60°

d) LK = 4.9 m, ∠J = 44°, ∠K = 46°

e) MO = 8.1 m, ∠M = 30°, ∠O = 60°

f) QR = 12 cm, ∠Q = 37°, ∠P = 53°

2. a) ∠S = 50°, ST = 12.2 m, TU = 14.6 m

b) VX = 19.5 cm, ∠W = 54°, ∠X = 36°

c) BC = 24 mm, ∠B = 16°, ∠D = 74°

d) EF = 7.1 m, GF = 5 m, ∠E = 45°

3. 2.4 m

4. 244 cm

6.7 Problems Involving Two RightTriangles

1. 165 cm 2. 20.5 cm 3. 57.5 m

4. a) b) 27.4 m

5. 301.6 km

6. 29.9 m, 22.2 m

6.8 Technology: Relationships BetweenAngles and Sides in Acute Triangles

1. a) but does not equal the ratio of

(the adjacent side could be

either AB or BC). This is because �ABC is not a right

triangle.

b) No; the ratio of the side lengths is not equal to the

ratio of the cosines or the tangents of their opposite

angles.

2. a) 1.0; 1.3; 1.3

b) There are six possible ratios:

and the inverse of these ratios. The

relationship applies to all six ratios.

c) Rounding error can affect the calculated ratios, so

they may not be equal but are accurate to one tenth.

3. a) In an equilateral triangle, the ratio of any 2 side

lengths is 1, so the ratio of the sines of their opposite

angles should also equal 1.

b)

4. a) In �ABC where AB = AC and ∠A is less than

90°, You can’t calculate

without measuring.

b) If ∠C = ∠B = 68°, ∠A = 21° and

6.9 The Sine Law

1. a) 10 m b) 20 m

2. a) 35° b) 61°

3. a) 42.5 cm b) 42.9°

4. ∠H = 60°, ∠J = 41°, GH = 29 cm

5. 409 cm2

6. 95.8 m

sin 68sin 21

˙ .= 2 6

ACCB

or sin Bsin A

ACAB

sin Bsin C

= =1 .

sinsin

6060

1=

side 2side 3

,

side 1side 2

side 1side 3

, ,

ACBC

or ACAB

oppositeadjacent

,

tan 64 84 2 13. ˙ .=

57.3°

15 m

64.6°


7. a)

b) 50.6 m

6.10 The Cosine Law

1. a) 9.7 m b) 8.6 cm

2. a) 55° b) 55°

3. a) 10.8 cm b) 81.3°

4. a) ∠W = 74°, ∠X = 38°, ∠Y = 68°

b) p = 13 cm, ∠Q = 77°, ∠N = 36°

5. 235 m2

6. Yes, ∠F is contained between d and e. f = 12.1 cm

7. 14.3 km

25°rock oak125 m

birch

72°


Find the sum of the expressions, A and B, representedby the algebra tiles.

1.

2.

3.

Model the expressions using algebra tiles or drawingson grid paper. Then, add.

4. (x2 + 2x + 2) + (2x2 + x + 1)

5. (–2x2 + 2x) + (–x2 + x – 2)

6. (–x2 – 2x – 2) + (2x2 – x – 1)

7. 2x2 – x – 3 8. x2 – 3x + 2+ x2 – x + 1 + (–2x2) – x – 1

Simplify.

9. (3y + 4z + 6) + (2y – z – 4)

10. 2ab + 3bc + d + 2bc + 3ab – d

11. k2 – 2kj – j2 + j2 – 2kj + k2

12. s2 + 4 + t + 3 + 2t3 + 3s

Add.

13. 4a + b 14. 4m2 + 8mn + 2n2

+ 2a + 2b – 3 + m2 – 2mn + n2

15. 3r2 – 8r + 4 16. c2 + 2ac + 4+ r2 – 2r + 5 + 3c2 + 6 + a2

Simplify.

17. (4k2 + 2k – 5) + (3 – k – 2k2)

18. (x3 + 2y – 5) + (3x3 – 4y + 7)

19. (z3x + 3z – 2) + (3z3x – 4z + 6)

20. a) Write an expression in simplest form forthe perimeter of the figure.

b) If x = 8 cm, what is the perimeter?

Adding PolynomialsTo add polynomials, collect like terms and add.

Copyright © 2001 McGraw-Hill Ryerson Limited Appendix A: Review of Prerequisite Skills 71

Name

A

B

A

B

A

B

x – 2

2x + 3

Classify each triangle and find the missing anglemeasures.

1. 2.

3. 4.

Find the measures of the indicated angles.

5.

6.

7.

8.

Find the missing angle measures.

9.

10.

11.

12. If you know the measure of �DBC, how canyou find the measure of �DCA?

13. If you know the measure of �S, how canyou find the measure of �QRS?

Q

TR

S

B

C

D

A

Z

X

W

Y

c

e65°

d

vE H

G

Fr

u t

s

59°

108°

L

KN

M j

85°

69°78°

XY

Z

72°

44°

m

np

Q

R

Sfd

e

35°

M

wy

z

x

72°N

P

140°

L

K

J

50°

10°

a

D

E Fyy

A

B65°

57°

xC

Angle Properties I(Interior and Exterior Angles of Triangles and Quadrilaterals)In a triangle,• the sum of the interior angles is 180°.• the exterior angle is equal to the sum of the two interior and opposite angles.In a quadrilateral, which can be subdivided into two triangles,• the sum of the interior angles is 2 × 180°, or 360°.• the sum of the exterior angles is 360°.

Copyright © 2001 McGraw-Hill Ryerson Limited72 Appendix A: Review of Prerequisite Skills

Name

W T

U

V100°

pn

m

q

j

k

h 40°

G I

H

25° 25°z


Name

In the diagram, list the following.

1. two pairs of alternate angles

2. two pairs of corresponding angles

3. two pairs of co-interior angles

Find the missing angle measures.

4.

5.

6. Write two different methods you could use tofind the missing angle measures. What are themeasures?

Find the measures of the indicated angles.

7.

8.

9.

10.

11.

12. List all the angles in the diagram. Then,calculate the measure of each angle.

SVR

U TW

60°

A

B110° 120° C

D E

v

w x

y z

L

G

K J

H58°a

d

b

c

A

B CX

D E

Y

65° 50°d

e f

F E

C D

30°

40°

x z

y

W X

YZ

80° 50°

rs

C E

G

D F

Kr qp105°

T

R

M

Q

P S

a b

60°

x

w

L NY

s

M P

126°

X

rb

a

C F

BD

E H

GA

Angle Properties II(Angles and Parallel Lines)Parallel lines are lines in the same plane that do not intersect.A transversal is a line that crosses two or more lines, each at a different point.

Alternate Angles Corresponding Angles Co-interior Angles

Alternate angles are Corresponding angles Co-interior angles areequal, e.g., �4 = �6 are equal, e.g., �3 = �7 supplementary, e.g., �4 + �5 = 180°

1 234

5 678

1 234

5 678

1 234

5 678

Common FactoringTo factor a polynomial, determine the GCF of all the terms. Then, divide all the terms by the GCF.


Name

Determine the GCF of both terms.

1. 6x – 9

2. 9bc + 12bd

3. –2a3 – 4a2

Complete the table.

GCF of Other Polynomial Both Terms Factor

4. x2 + 2x x5. 3y 2y – 36. 8xy –3x2y2 8 – 3xy7. 2m2n – 4mn2

8. 6a3 + 9a2 – 12a

Factor each binomial.

9. 4y + 10

10. 6m2 + 9m

11. 4t3 – 6t2

12. 3p3q2r2 – 4p2q2r

Explain the error in each case and correct it.

13. 8n2 – 12n = 4(n2 – 3n)

14. 16y2z2 + 24y2z = 4y2z(4z + 6)

15. 25x6 – 5x2 = 5x2(5x3 – 1)

Factor each trinomial.

16. 5a2 + 10ab – 15b2

17. 9x3y + 3xy2 + 15xy

18. 2s3t2 – 8s2t3 + 4st

19. 6y2 – 9xy + 12x2y2

20. 12c4d + 8c3 – 16c3d

21. a) Find expressions for the length and thewidth of the rectangle if A = 3x + 5xy is theexpression for the area.

b) Write 3x + 5xy in factored form.

22. A rectangle has an area of lw, where l is thelength and w is the width.

a) Write an expression for the perimeter as asum of 4 terms.

b) Simplify the sum.

c) Factor the simplified expression.

23. a) Write an expressionin simplified form for theperimeter of the hexagon.

b) Factor the expression.

24. The surface area of a cylinderis given by the expression 2πr2 + 2πrh.

a) Factor the binomial.

b) Evaluate for r = 5 cm and h = 10 cm. Round to the nearest whole number.

25. Write a factored expression for the surfacearea of a cylinder that is open at the top.

A = 3x + 5xy

?

?

x

x

yy

yy

r

h


Name

Congruent TrianglesTwo triangles are congruent if it is possible to match up their vertices so that all pairs of corresponding angles and corresponding sides are equal. The chart gives 3 ways to state that 2 triangles are congruent.

For each pair of congruent triangles, state whichsides and which angles are equal.

1. 2.

�ABC ≅ �DEF �XYZ ≅ �STR

For each congruence relation, name the sides andangles that correspond.

3. �LMN ≅ �PQR 4. �STV ≅ �XYZ

What is the fewest number of other parts that mustbe equal, so that each pair of triangles is congruent?

5. All the angles of 1 triangle are equal to thecorresponding angles of another triangle.

6. Two sides of 1 triangle are equal to 2 sides ofanother triangle.

7. Two angles of 1 triangle are equal to 2corresponding angles of another triangle.

Find the missing measures in each pair of congruenttriangles.

8. BC = ___________EF = ____________FD = ___________

9. ∠KLM =_________∠JGH = _________∠MKL =_________

10. ∠TVS = _________∠XYZ = _________∠ZXY = _________

For each pair of triangles, identify the case thatproves that the triangles are congruent and list allthe corresponding equal parts.

11. 12.

13. 14.

P

Q

R

S

xx

H

J

K L

M

x xo o

E

F

G

D

X

Y

Z

AC

B

x x

X Y

ZR

S T

A

B

CD

E

F

SSS (side, side, side) SAS (side, angle, side) ASA (angle, side, angle)A

B C

D

E F

A

B C

D

E Foo

A

B C

D

E F

x x

o o

A

B

C D

E ?

?

F

?3 m

4 m

5 m

G

7 cm

7 cm

K M

L

?

?

?

H

J

53°

32°95°

11.4 cm

11.4 cm

8.6 cm

T

S

V

X Y

Z

65°41°

?

? ?


Name

Write an expression for each of the following.

1. the sum of 12 and 7

2. the sum of x and 10

3. the product of 9 and 12

4. the product of 5 and y

5. the product of a and b

6. the quotient m divided by v

7. 8 greater than q

8. 9 fewer than t

9. 2 times the radius, r

10. 1–2

of the distance, d

Find the values for each of the following.

11. s 3s 12. x 4x + 56 02 11 34 109 1.50 6.2

10 8.330

13. t 3t – 4 14. r y 2 – r + y3 3 42 0 15 2 52.3 1.1 3.24.1 0.8 0.97.88.3

15. Evaluate for a = 5 and b = 2.

a) a – b b) 3a + 3b

c) 2a – b d) a + 3b + 4

e) 3ab + 2 f) 22 – 2ab

16. Evaluate each expression for the given valueof the variable.

a) 5t + 3, t = 2.3 b) 3m – 2, m = 1.6

17. Evaluate for x = 1.4, y = 3.2, and z = 5.3.

a) x + y + z b) yz + xz + xy

18. The cost of a Sunday newspaper is $1.50.

a) Write an expression for the cost of n papers.

b) Calculate the cost of purchasing 8 papers.

19. The height of a jump on the moon is 6 timeshigher than a jump on Earth.

a) Write an expression for the height of a jumpon the moon.

b) Calculate the height of a jump on the moonfor a jump of 2.8 m on Earth.

20. The cost to rent a video game station for aweekend is $18 plus $4 for each game rented.

a) Write an expression for the rental cost for aweekend.

b) Calculate the cost to rent the game stationand 3 games for the weekend.

21. The width of a soccer field is 20 m greaterthan half the length of the field.

a) Write an expression for the width of the field.

b) Calculate the width and the perimeter if thelength of the field is 110 m.

Evaluating Expressions I(Variables in Expressions)Algebraic expressions are made up of numbers and letters called variables. To evaluate expressions, substitute numbers for the variables and then calculate.


Name

1. Evaluate for y = 3.

a) 3y b) 2y + 4 c) y – 5

d) –4y e) 4 – 2y f) 6 – 3y

2. Evaluate for a = –2.

a) 3a b) –4a c) 2a + 5

d) 3a – 2 e) –5 + 6a f) –0.5a – 9

3. Evaluate for c = –1 and d = 3.

a) cd b) d – c c) 2c + 2d

d) d ÷ c e) –(c – 2d) f) –2cd

4. Evaluate for r = –2.

a) r2 b) –3r2 c) –2r2 + 2

d) 2r3 e) (r + 3)2 f) 2r2 + 3r – 8

Complete each table.

5. x 2x – 2 6. m 3 + 2m2 21 10 0

–1 –1–2 –2

7. s s2 + 2s 8. a a2 – 2a –12 21 10 0

–1 –1–2 –2

9. During a recent math contest, the followinganswers were submitted as the solution to thequestion (4 – 2x)2 + 5x when x = –3.

Corrie’s answer was –11.Jillian’s answer was 35.Shara’s answer was 85.

a) Identify the correct answer.

b) Show the calculations that support youranswer in part a).

10. The CN Tower is about 547 m tall. If anobject is dropped from this height, its heightabove the ground is given by the formula

h = 547 – 5t2

where h is the height, in metres, and t is thetime, in seconds, since the object was dropped.Find the height of the object at these times afterit is dropped.

a) 5 s

b) 9 s

c) 10 s

11. The formula for the height of a model rocketis

h = –t2 + 20t

where h is the height, in metres, and t is thetime, in seconds, after liftoff.

a) What is the height of the rocket after 9 s?15 s?

b) Find the rocket’s height after 20 s. Explainyour answer.

Evaluating Expressions II(Expressions With Integers)In many formulas and expressions, variables are replaced by integers and then evaluated.


Name

1. The following formula can be used to find theapproximate mass of young adults.

M is the mass, in kilograms, and h is the height,in centimetres. Find the mass, to the nearesttenth of a kilogram, for each height.

a) 180 cm b) 210 cm

c) your height

2. To calculate the approximate number of mini-lights needed to decorate a Christmas treeto produce a full effect, the following formula isrecommended.

L is the number of lights, h is the height of thetree, in centimetres, and w is the width of thetree, in centimetres.

a) How many lights are needed for a tree thatmeasures 183 cm high and 122 cm wide?

b) How many strings of 25 mini-lights wouldbe needed for this tree?

3. If there is a discharge hole in a container, thevelocity, v, in metres per second, at which theliquid leaves the container can be calculatedusing the formula v = , where h is theheight, in metres, of the liquid above the hole.Find the velocity of discharge, to the nearesttenth of a unit.

a) height of 4 m

b) height of 0.1 m

4. Write an equation that relates equal values ofquarters and nickels, where q is the number ofquarters and n is the number of nickels.

5. An amount of money is invested in anaccount for a number of years. Compoundinterest is paid on the amount in the account.The value of the investment at any given timecan be calculated using the following formula.

V = P(1 + i)n

V is the value of the investment, P is the amountof money invested, i is the rate of interest, and nis the number of years the money has beeninvested. Find the value of each investment.

a) Amount: $200, Interest: 0.03, Years: 2

b) Amount: $650, Interest: 0.02, Years: 3

c) Amount: $1000, Interest: 0.05, Years: 5

6. Complete the columns for s and A, using theinformation below the table in parts a) and b).

a b c s A5 cm 7 cm 8 cm

11 cm 19 cm 20 cm8.4 m 3.6 m 6 m1.5 m 1.5 m 2.1 m

a) Given the side lengths a, b, and c of atriangle, calculate half the perimeter, s, using the

following formula.

b) Find the area, A, of the triangle, using the

following formula. A s s a s b s c= − − −( )( )( )

sa b c= + +

2

19 6. h

Lh w= × × 8

929

Mh= −6 90

7( )

Evaluating Expressions III(Applying Formulas)A formula uses variables to express a relationship. A formula can be used to determine the outcome of an experiment without actually carrying out the experiment and measuring the results.


Name

a) Complete the table of values for each equation.b) Use the grids to draw the graphs of the relations.The domain is R.1. y = x2 – 2 2. y = x2 + 0.5

x y x y0 01 1

–1 –12 2

–2 –23 3

–3 –3

3. The table shows therelation between the areaof a circle and itsdiameter.

a) Plot area versusdiameter, and draw asmooth curve throughthe points.

b) Use the graph tofind the area of a circlewith a diameter of3.5 cm.

c) Use the graph tofind the diameter of acircle with an area of10 cm2.

4. The cost of removing pollutants from wastewater depends on the percent of pollutantsremoved. The table shows the cost, in thousandsof dollars, for the percent of pollutants removed.

Percent of Pollutants Removed

10 20 30 40 50

Cost (thousands of dollars)

1.7 3.8 6.4 10.0 15.0

a) Plot cost versus percent of pollutants removed.

b) How much would it cost to remove 45% ofthe pollutants from the waste water?

c) It cost $14 000 to remove pollutants from thewaste water. What percent of pollutants wasremoved?

5. The equation can be used toapproximate the speed of a vehicle that hasskidded, where s is the speed, in kilometres perhour, and l is the length of the skid mark, inmetres.a) Complete the table of values, to the nearesttenth.

l 0 1 4 9 16 25 36s

b) Plot speed versusskid length, and drawa smooth curvethrough the points.

c) Use the graph to find the speed of avehicle for a skid length of 20 m.

s l= 15 9.

20

4

6

8

–2

2

x

y

20

2

4

6

–2

–2

x

y

Evaluating Expressions IV(Non-Linear Relations)A non-linear relation has a graph that is not a straight line. When the equation includes a power of x, such as y = x2 – 3, the points can be joined to form a smooth curve.

Diameter Area(cm) (cm2)

1 0.82 3.13 7.14 12.65 19.6

0

10

20

2 4 6

0

4

8

12

1618

14

10

6

2

2010 30 40 50 60 70 80

10 20 30 400

102030405060708090

100

50


Name

Find the square roots of each number.

1. 64 2. 25

3. 121 4. 0.81

5. 2.25 6. 0.04

Evaluate.

7. 8.

9. 10.

Estimate. Then, calculate, to the nearest hundredth.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

Decide whether each equation is true or false.

23.

24.

25.

26.

27.

28.

Evaluate for x = 3 and y = 2.

29. 30.

31. 32.

33. Given the area, A, of a circle and the formula

, find the radius, r, to the nearest

tenth of a unit.

a) 1256 cm2 b) 153.86 cm2

c) 4.5216 m2 d) 7.065 m2

34.

a) For a square metre, what is the length of aside, in centimetres?

b) For a square, the length of a diagonal can bedetermined using the formula d = , whered is the length of the diagonal and a is the sidelength. For a square metre, find the length of adiagonal, to the nearest tenth of a centimetre.

35. The Pythagorean relationships for righttriangles are

where c is the length of the hypotenuse, and aand b are the lengths of the other two sides.

Find the length of each unknown side.

a) b)

c) d)

b c a= −2 2a c b= −2 2c a b= +2 2

2 2a

1 m2 = 10 000 cm2

rA=

3 14.

12 2x y+xy

3 2x y+362

xy−

50 2 25 0− =

20 5 4÷ =̇

3 10 4 13 5+ =̇ .

18 5 6 5− =̇ .

4 6 24=

4 16 20+ =

0 002.0 0039.

0 68.0 5.

−7 885 66

42 000 2828

964110

− 3862

0 09.0 49.

− 10016

4 cm

3 cm

c

7 m

10 m

b

8 m

10 ma

11 cm

17 cm

b

Evaluating Radicals(The Pythagorean Theorem)Since 5 × 5 = 25 and (–5) × (–5) = 25, both 5 and –5 are square roots of 25. The radical sign, , is used only for the positive square root. 25 5=


Name

1. What is the length, width, and area of therectangle, represented by the algebra tiles?

Model the expressions using algebra tiles or drawingson grid paper. Then, expand.

2. 2(x + 2)

3. 3(x + 3)

4. 2(2x + 1)

5. 3(2x + 2)

Expand.

6. 4(x + 2)

7. 5(x – 3)

8. 0.3(x + 5)

9. 4(2x + 1)

10. 1–2

(3x – 2)

Expand. Circle the letter corresponding to the correctanswer. Rearrange the letters to find a message.

11. –3(x + 2) L –3x – 6 M 3x + 612. –(b – 4) N –b – 4 A –b + 413. 2(7 – 5f) L 14 – 10f M 9 – 7f14. –4(6n + 2) P 24n + 8 G –24n – 815. –3(4 – 2y) T 6y – 12 R –12 – 6y16. –5(2t – 2) H 10 – 10t E 10t – 1017. –4(3c + 3) S –c – 1 R –12c – 1218. 3(2 + 3k) I 6 + 9k A 9k – 6

____ ____ ____ ____ ____ ____ ____ ____

Expand and simplify.

19. 3x + 2(5x – 3)

20. 14 – 3(4n – 1–3

)

21. 3(2h – 3) + 2(h + 3)

22. –2(3y – 3) + 3(2y + 2)

23. –6 + 5(2 – k) – 4k

24. 4(3u – 1) + 2(3 – 2u)


25. 2(x2 + 2x + 1) + 3(x2 + 3)

26. 5(y – 2) – 4(2y – 1–2

)

27. 3(t2 – 2t + 1) – 4(t + 2)

28. 2(e – 4) + 4(3e + 2) – 5(2e – 4)

29. x(2x – 3) – x(4 + x)

30. a) Write the area of the large rectangle.

b) Write the area of the shaded rectangle inexpanded form.

Expanding and Simplifying Expressions I(The Distributive Property)To use the distributive property, expand the expression by removing the brackets and simplifying.

xx

+1+1+1+1+1+1

x

5

x x


Name

Expand.

1. a(a + 3) 2. s(s –5)

3. –y(y + 2) 4. b(4 – b)

5. –x(6 – x) 6. –k(k – 3)

Expand.

7. 4r(r + 3) 8. 6m(m – 2)

9. 2x(3 – x) 10. –3y(5 + y)

11. The sum of two consecutive integers can befound using the expression n + (n + 1).

a) Write an expression for the product of twoconsecutive integers.

b) Write and simplify an expression for thedifference between their product and their sum.

For each question, expand and simplify. Then, locatethe answer in the column to the right. Write theletter that follows the answer in the box below thatcorresponds to the question number. Finally, answerthe question “What shape is a square when it startsto ‘wilt’?”

12. n(n + 4) – n(n – 2) 4n2 B13. n(4 – n) + n(n – 3) 2n2 + 8n R14. n(n + 3) – (6n – 4) –n S15. n(3n – 2) + n(2n + 3) n2 – 3n + 4 H16. n(4 – n) + n(n – 6) 5n2 + n M17. n(2n + 3) – n(3 – 2n) 6n O18. n(n + 7) + n(n + 1) n U19. n(n + 1) + n(–2 – n) –2n A

16. 18. 14. 12. 15. 17. 13. 19.


20. 2a(a + 2) + 4a(a + 1)

21. 3r(r – 3) – 2r(r + 2)

22. k(4k – 2) – k(k + 3)

23. –d(3 – d) + 2d(d + 5)

24. 4x(x – 1) – x(2 – x)


25. 2(a2 + 3a – 10) – a(a + 2)

26. 3x(x2 + 2x – 8) – 2(x – 1)

27. 2(y – 1) + y(y2 – y – 2)

28. –2r(r + 5) + 3r(r – 3)

29. Write, expand, and simplify an expressionfor the area of each face of the prism and then,for the total area of all the faces.

a)

b)

3nn

2n – 3

4.2x

2x + 3

x

Expanding and Simplifying Expressions II(Multiplying a Polynomial by a Monomial)To multiply a polynomial by a monomial, use the distributive property. When more than one expansion takes place, collect like terms and simplify.


Name

Simplify.

1. 22 × 23 2. 35 × 33 3. 44 × 42

4. 103 × 10 5. 94 × 93 6. 8 × 84

7. x2 × x5 8. y3 × y3 9. z3 × z2

Find the missing exponent.

10. 32 × 3� = 34 11. 5� × 54 = 57

12. 83 × 8� = 85 13. 7� × 73 = 74

14. j5 × j� = j8 15. b� × b5 = b9

16. k × k9 = k� 17. s6 × s� = s7

Simplify.

18. 54 ÷ 52 19. 46 ÷ 43 20. 33 ÷ 32

21. 95 ÷ 92 22. 74 ÷ 73 23. 26 ÷ 24

24. m7 ÷ m5 25. p8 ÷ p6 26. a5 ÷ a4


27. 25 ÷ 2� = 23 28. 34 ÷ 3� = 32

29. 4� ÷ 42 = 44 30. 5� ÷ 53 = 5

31. n4 ÷ n� = n2 32. c� ÷ c4 = c3

33. y� ÷ y2 = y2 34. z9 ÷ z� = z

Simplify.

35. (32)3 36. (24)2 37. (73)4

38. (62)4 39. (53)2 40. (45)3

41. (x3)3 42. (s2)2 43. (r5)2


44. (33)� = 39 45. (25)� = 210

46. (5�)2 = 58 47. (4�)3 = 412

48. (g2)� = g6 49. (m3)� = m9

50. (s�)5 = s20 51. (t�)2 = t6

Find the value of each expression. Replace the blankswith the corresponding letter or symbol to decode themessage.

52. 23 × 22 E 53. 29 ÷ 22 T

54. 24 ÷ 23 P 55. (23)2 C

56. 213 ÷ 23 O 57. (26)2 K

58. 2 × 2 E 59. (24)2 *

60. 22 × 22 F 61. 22 × 2 R

62. (23)3 W 63. 212 ÷ 2 R

2 2 2 2 2 2 2 2 2 2 2 21 2 3 4 5 6 7 8 9 10 11 12

Exponent Rules I(Powers With Whole Number Bases)• To multiply powers with the same base, add the exponents.• To divide powers with the same base, subtract the exponents.• To raise a power, multiply the exponents.


Name

Complete the table.

Exponential Standard Form Base Exponent Form

1. (–2)3

2. 3 13. 54. –3 35. –2 56. –7 2

Complete the table.

Exponential Repeated Standard Form Multiplication Form

7. (–3)2 × (–3)2

8. (–4) × (–4) × (–4)

9. (–5) × (–5) × (–5)

10.

11. (+5)4 ÷ (+5)2

12.

13. (–3)5 ÷ (–3)2

14.

Write in standard form.

15. 32 × 33 16. (–2)3 × (–2)2

17. (5)4(5)3 18. (3.2)2(3.2)2

19. ((y)2)3 20. (3)4 ÷ (3)2

21. ((–2)2)5 22. (–4.5)3 ÷ (–4.5)

23. 24.

25. –(1.2)2 26. (–0.6)2

Is each statement true or false?

27. 33 = 81 28. 6(–2)3 = 48

29. y2 × y4 = y6 30. ((–2)3)3 = –512

31. (–a)4 ÷ (–a)2 = a2 32. (–5)3 ÷ (–5)2 = 5

Evaluate for s = –3 and t = 2.

33. t3 34. 6s2

35. s3 + t2 36. 2s3 ÷ 3t

37. –3st 38. –2s2 – 4t

39. The formula for the area, A, of a circle is A = πr2, where r is the radius. Complete thetable, rounding to the nearest tenth of a unit.

Radius (cm) Area (cm2)1052.51.36.2

40. If the base of a power is negative and theexponent is five, the standard form of thenumber is negative. Explain.

41. The standard forms of the following pairs ofterms are not the same. Explain.

a) (–2)4 and –24

b) ((–3)2)3 and –36

42. A manufacturing company determines itsprofit using the formula P = 120n – n2 – 220. P isthe profit, in dollars, and n is the number ofitems manufactured. How many items must thecompany produce to begin to make a profit?

( )( )−−

77

3

2

33

5

3

( ) ( )( )

− × −−

2 22

( ) ( ) ( )( ) ( )

+ × + × ++ × +

5 5 55 5

( ) ( ) ( )( )

− × − × −−

4 4 44

Exponent Rules II(Powers With Integral Bases)You can use the exponent rules when the base of a power is an integer. In general, ym × yn = ym + n ym ÷ yn = ym – n (ym)n = ym + n


Name

1. a) Tell how to multiply two monomials withthe same variable, for example, (5n)(2n2).

b) Tell how to multiply two monomialsinvolving more than one variable, for example,(2yz)(–3y3).

Multiply.

2. (2x)(x) 3. 3n × 2n

4. y × z 5. (2k)(2a)

Multiply.

6. 4v × 2w 7. 2s × 5t

8. (3a2)(4b) 9. 4f × 5g2

10. 4xy × 2z 11. (3cd)(4e)

Multiply.

12. (3a)(–2z3) 13. (–2r2)(8s)

14. –4c(5de) 15. 2xy × 3xy

16. (–3abm)(2bm) 17. –u(5ut2)

18. (2a2b3c)(–3bc2d) 19. –5r2st × 2rs2t2

20. (5x)(4y)(–3z) 21. –2d(3d)(3e)

22. (–k2mn2)(4mn)(–2kn2)

23. What is the area of the rectangle?

Write the expression for the area of each figure. Then,simplify.

24.

25.

For each of the following figures, write and simplifyan expression for

a) the area of each face

b) the total surface area

c) the volume

26.

27.

Draw and label a box with these dimensions. Then, find its volume.

28. a × 2a × 3a

29. (4c)(4c)(4c)

Exponent Rules III(Multiplying Monomials by Monomials)• To multiply two monomials with the same variable, multiply

the coefficients and add the exponents of the variable.

• To multiply two monomials involving more than one variable, multiply the coefficients and combine like variables using the exponent laws.

x

2x

x

3x

2b

4n

2n

n

1.5s

0.5s


Name

1. a) Tell how to find the power of a monomial,for example, (–a3)2.

b) Check by simplifying (–a3)(–a3).

Simplify.

2. (y)2 3. (x)4

4. (–s2)2 5. (c3)2

6. (–m5)3 7. (f 3)6

Simplify. Then, cross out the box with the answer.After you have finished, read the answer to the riddle“What number is the most restful?”

8. (st)4 9. (–x2y)3

10. (cd2)2 11. (yz)2

12. (rs)3 13. (–a2b2)3

14. (–f 3g2)2

FI FO TH ES RT LAr3s3 a2b3 c2d4 s4t4 c5d2 –x6y3

ND YW SH IN RN KSy2z2 y3z3 f 6g4 s2t2 –a6b6 x2y3

Simplify.

15. (3ty)2 16. (–2xz)3

17. (–2a2b)3 18. (3r3s)2

19. (5k3m2)2 20. (–3q2r2)3

Simplify.

21. (yz)2(y3z)

22. (–2ab)(–ab)2

23. (5s2t2)2(–st)

24. (–4k2m3)2(2km)3

25. (2r2s2t)(3rst)2

26. (4abc)2(2a2bc)(ab3c3)

27. (m2n2p2)2(mnp)(–3mn3p3)

For each cube, write and simplify an expression for

a) the area of each face

b) the volume

28.

29.

Explain the error in each case and correct it.

30. (a2)3 = a5

31. (fg3)2 = f 2g9

32. (–2x2y2)2 = –4x4y4

Exponent Rules IV(Powers of Monomials)To find the power of a monomial, find that power of the coefficient and of each variable. In general, (xmyn)a = xamyan

xy2

2a2bc3


Name

Divide.

1. 2.

3. 4.

5. 6.

Divide.

7. 14ab ÷ 7a 8. 10klm ÷ 5

9. 10.

11. –9xyz ÷ (–3xyz) 12.

Simplify.

13. 2a4b3 ÷ a2b

14.

15. 8x6y4 ÷ (–4x3y2)

16. –4w3x5 ÷ (–2w2x2)

17.

18.

Simplify.

19.

20. 4a3b2c ÷ 2bc

21. 8x5y3 ÷ 2x3y

22.

23. –9e2f 4 ÷ (–6ef 2)

24.

25. The perimeter of a square is 8s2t. Write andsimplify an expression for the length of a side.

Find the missing dimension in each rectangle.

26.

27.

28.

29. What are the dimensions of the rectangle ifthe ratio of length to width is 3 to 2?

30. What are the dimensions of each face of therectangular prism?

2012

5 3 5

2 3 4

d e fd e f

−128

7 6

2 2

s ts t

63

2 4

2

k mkm

−1218

5 5c dc

−96

3 5 2

2

f g hfg h

63

3 2

2 2

q rq r

−164

pqrpqr

84bcdbd−

124

rtrt

−−123

gg

4tt−

153

xx

−−55a

−63

s42r

Exponent Rules V(Dividing Monomials by Monomials)To divide a monomial by a monomial, divide the coefficients and combine like variables using the exponent laws.

A = 4xy

2y

A = 8r3s4 4s2

A = 16a4b3

8a2b

A = 24x2y2

6xy

A = 4xy2y

A = 8x2z2


Name

The domain of each equation is {–2, –1, 0, 1, 2}.Complete a table of values. Then, graph eachequation.

1. y = –2x + 5 2. 2x + y = 0

Graph each equation. The domain is R.

3. y = 2x + 1 4. y = –x – 3

5. x + y = –1 6. 2x – y = 3

Given the tables of values, write an equation for eachrelation.

7. x y 8. x y–2 2 –2 –6–1 3 –1 –30 4 0 01 5 1 32 6 2 6

9. Given the points on the grid, write anequation to represent the relation. Then, statethe domain.

10. The table shows the equivalent depths ofwater and heavy wet snow.

Depth ofWater (cm)

5 10 15 20 25

Depth of HeavyWet Snow (cm)

100 200 300 400 500

a) Plot the depth of heavy wet snowversus the depth of water.

b) Can you join the points to graph theequivalent depths of water and heavy wet snow? Explain.

c) Use the graph to estimate the depth of heavywet snow that is equivalent to a depth of 12 cmof water.

d) Use the graph to estimate the depth of waterthat is equivalent to a depth of 360 cm of heavywet snow.

20

2

–2

–2

x

y (4,2)

(2,1)

(0,0)

(–2,–1)

4

20

–2

2

–2

–6

–4

x

y

2

2

0–2

–2

x

y

20–2

–2

–4

x

y

20

2

–2

–2

x

y

20

2

4

–2

–2

–4

x

y

20

6

8

–2

2

4

x

y

0

100

200

300

400

500

5 10 15 20 25

Graphing Equations I(Graphing Linear Equations)To graph a linear equation, first make a table of values and substitute values of x to determine the corresponding values of y. Then, plot the coordinates. Join the points if the domain is R.


Name

Use the x- and y-intercepts to graph each line.

1. 2x + 3y = 6 2. 4x – y = 4

3. 3x + 5y – 30 = 0 4. x + y + 2 = 0

Graph each equation using the slope and they-intercept.

5. y + 1 = 2x 6. x + 4y = 8

Graph each equation using a method of your choice.Find the intercepts and slope for each line. Thedomain is R.

7. y = x + 5 8. 5x + 2y = –20

Find an equation for each line.

9. 10.

11. Write an equation of a line whose x- andy-intercepts are opposite integers, but not 0, andwhose x-intercept is positive.

12. Write an equation of a line whose y-interceptis 0 and whose slope has a negative value.

Describe the slope and the intercepts of each line.

13. y = –3 14. x = 5

15. After 10 s Beth counted 14 heartbeats, after30 s she counted 42 heartbeats, and after 35 s shecounted 49 heartbeats.

a) Plot the ordered pairs (time, heartbeats).

b) Find the slope of the line. What does theslope represent?

c) Find an equation ofthe line.

d) Use the equation to find Beth’s pulse rate inheartbeats per minute.

100

20

20 30

30

40

50

60

10

0 x

y

(0, 5)

(4, 0)

20

–4

x

y

–2

–6

–2 4 6 8–4

–8

–100

2

–2–4 x

y

4

20

–4

x

y

2

–2

–6

–2 4 6 820–2

–2

x

y

4

2

20–2

–2

x

y

–420

2

x

y4

4

20–2

–2

x

y

4

–420

2

–2

–2

x

y

40 x

y

(0, 4)

(–7, 0)

Graphing Equations II(Methods for Graphing Linear Equations)To draw the graph of a line,• use the x- and y-intercepts • use a table of values• use the slope and the y-intercept


Name

For each graph, identify the point of intersection.

1. 2.

Four pairs of lines are defined by the tables below.

a) The point of intersection of two pairs of lines canbe determined from the tables. Identify these points.

b) Graph the other two pairs of lines and determinethe points of intersection as accurately as possiblefrom the graphs.

3. x y x y 4. x y x y1 5 1 –4 0 4 0 02 3 2 –3 1 2 1 13 1 3 –2 2 0 2 24 –1 4 –1 3 –2 3 3

5. x y x y 6. x y x y–4 3 –1 –6 –1 –7 3 10 1 2 0 0 –5 3 02 0 3 2 1 –3 3 –16 –2 5 6 2 –1 3 –2

7. Complete the table of values for each relationand find the point of intersection.

a) y = x + 5 b) y = 2x – 8

x y x y0 –8

10 513 18

Make tables of values and graph each pair of lines.Find the point of intersection.

8. y = 2x + 1y = x + 3

9. 2x + y = 7x – y = 5

10. The screen of an air traffic controller showstwo planes approaching at 10 000 m. One planeis travelling in a direction described byy = 4 – 2x, the other in a direction described by x – y = –1. Determine if the planes cancontinue in the same manner. Give a reason for your answer.

420

–2

–4

x

y

2

4

20

–2

x

y

–2

2

4

20

–2

–4

–6

x

y

–2

2

20

–2

x

y

–2

2

20

–2

x

y

–4

–6

4 6

20

–2

x

y

–2

2

Graphing Equations III(Intersecting Lines)The point of intersection of two lines is a point common to both lines, that is, the point where the lines meet.


Name

Complete the table.

Number Prime Factors1. 202. 2 × 3 × 53. 184. 2 × 3 × 3 × 35. 1506. 252

Complete the table.

Expression Prime Factors7. 6x2

8. 2 × 3 × 5 × s × t × t9. 12a2bc3

10. 2 × 5 × a × a × b × b × b11. 24x2y2z

Factor fully.

12. 6m2n2

13. 51a2

14. 76r2s

Determine the GCF of each pair of numbers.

15. 12, 28

16. 15, 60

17. 24, 42

18. 54, 81

Determine the GCF of each pair of monomials.

19. 15a, 25a

20. 3x3, 12x2

21. 18xyz, 24x2y

22. 12c, 16d

23. 6st2, 8s2t

24. 4p2q2, 6p3q3

Determine the GCF of each set.

25. 6, 9, 15 26. 27, 63, 81

27. 8x, 12y, 28a 28. 15r2, 20r3, 5

29. a3, 9a2, 3a 30. s2t2, s3t2, s3t3

31. 4xyz, 12x2y2z2, 8xy2z3

32. 14c2de, 28cd2, 21ce2

33. Lyndon has a piece of cardboard 60 cm by75 cm. What is the largest size of identicalsquares he can cut from the cardboard if he usesall of it?

34. A patchwork quilt to be made of identicalsquares must measure 150 cm by 210 cm.

a) What different sizes of squares are possibleso that each size completely covers the surface?

b) What is the largest size of squares that can beused to cover the surface exactly?

35. Two rectangles share a common side. Thearea of one rectangle is 4ab and the area of thesecond rectangle is 6ab.

a) Draw a diagram showing the attachedrectangles and label their dimensions.

b) Are other dimensions possible? If so, drawand label the diagrams.

Greatest Common FactorsTo find the greatest common factor of two or more terms, first write all the factors of each term.Then, determine which factors are common to all the terms, and write the product of the common factors.


Name

1. List the like terms.

4r r3 –r 101r 5r2

–r2 (–r)2 r –2r3

2. a) State the number of terms in theexpression –x2 + 5x – 2xy + 3.

b) State the coefficients and the constant term inthe expression.

Simplify.

3. 11t – t 4. –10b2 + 3b2

5. –12y – y 6. 11m3 + 10m3

7. 5p + (–2p) 8. c2 + c2

Simplify.

9. 2x + 3x – x 10. –5y + 2y – 9y

11. 0.4d + 0.5d + 0.1d

12. –t2 – 2t2 – 3t2 + t2 + 6t2

Simplify.

13. 8y – 2z + 7y 14. –2r + 3s – 6r

15. –3a2 + 2b2 + 3a2 16. 5e3 + 2e3 – e2

Simplify, and then evaluate.

17. 4s – 2s for s = 1

18. a2 + 2a2 + a2 for a = 2

19. 2t + 3t – 3 for t = 0.5

20. –k + (–3k) – 2k + 2 for k = –3

Simplify.

21. 2c + 3 + 4d – c + d

22. m – 2 + 2n + 5 – n

23. –2 + 2z + (–3w) + 4 + z

24. 3 + 4x2 + y2 + x2 – 1

Write an expression for each perimeter in 2 differentways.

25.

26.

Using the given information, write a problem.

27. Monique travelled p kilometres the first day,(p + 6) kilometres the second day, and 2p kilometres the third day.

28. Jared read n pages each day for 3 days.

29. The number of cubes in a large box is (8c2 – 2), and in a small box is 4c2.

30. A giraffe is (z + 0.8) metres tall. An elephantis (2z – 0.1) metres tall.

31. For each problem in questions 27–30, writeand simplify the algebraic expression.

r3

2

2r

3r

1.5s

0.5s

Like TermsLike terms have the same variable raised to the same exponent. r, 4r, 101rUnlike terms have different variables or the same variable, but different exponents. 7b, –3a, x2, x


Name

Identify as a monomial, a binomial, or a trinomial.

1. 3 + 4k 2. r4 – s3 + 6

3. 4–5

x2y 4. a + 3–4

b – c

5. 0 6. 3a2t – 5a

State the degree of each monomial.

7. 2xy2z 8. 14k 9. –4ab

10. 7 11. 5xst3 12. –36wz4

State the degree of each polynomial.

13. 9t + 8s 14. 22x2 + 22y

15. n – 2p3 16. 11wxyz – 9w4

17. –2a2b3 – ab + b6 18. 3kmn + 11k2m – 10kn3

State the degree of each monomial.

19. The circumference of a circle is πd.

20. The area of a circle is πr2.

21. The volume of a cylinder is πr2h.

22. The volume of a cone is 1–3

πr2h.

Simplify each expression. Then, classify the resultingpolynomial and state the degree.

23. n + n + 1 + n + 2

24. bh – 1–2

bh

25. πr2 + πr2 + 2πrh

Arrange the terms in each polynomial in ascendingpowers of y.

26. y3 + xy2 + y + 2

27. –2x2y + 3xy3 + x3

28. 3–4

– y + y3 – y6

Arrange the terms in each polynomial in descendingpowers of x.

29. –6x + x4 + 2x3 – 10

30. 0.2mx4 – 1.3x5 + 0.4m2 + 2.1x3

31. 4b + 2–3

bx + b3x2 + x4

32. a + x

Draw and label a figure that shows the perimeter.

33. s + t + w

34. 2m + 2n

35. 4k

36. a) Write an expression for the perimeter ofthe triangle.

b) Is the degree of the polynomial for theperimeter the same as the degree of thepolynomial for each of the sides? Explain.

2x2 + 3x – 7

3x2 – 8x + 5

x2 + 5x – 3

PolynomialsA monomial is a polynomial of one term. The degree of a monomial is the sum of the exponents of its variables. For example, 4a2b3 has degree 5.

A binomial is a polynomial of two terms, and a trinomial has three terms.

The degree of a polynomial in one variable is the highest power of the variable. For example, 2x3 – 7x has degree 3.

The degree of a polynomial in two or more variables is the greatest sum of the exponents in any one term. For example, 5m3n + m2n – mn2 has degree 4.


Name

1. Calculate the slope of each line segment,where possible.

Find the slope of the line passing through each pair ofpoints.

2. (4, 5) and (0, 0)

3. (–2, –6) and (–7, –1)

4. (3, –2) and (–6, 5)

5. (3.7, 5.1) and (–1.5, 1.2)

6. (–1, 5) and (3, 5)

7. (2, –7) and (2, 4)

8. The slope of a line is . The line passes

through (4, –1). Name the coordinates of twoother points on the line.

Given a point on the line and the slope, draw thegraph of each line.

9. (–3, –1), m = –3 10. (2, –5),

11. One point of a line is in the first quadrantand another point of the line is in the secondquadrant. If the slope of the line is positive,name two points on the line.

12. The coordinates of point A are (–1, –5). If the

slope of the line is , name the coordinates of

another point on the line.

13. Given the equation of each line, find twopoints on the line and calculate the slope.a) y = 2x – 5

b) x – y = 8

14. The slope of a line is 2. The line passesthrough (–1, c), (0, –4), and (d, 4). Find the valuesof c and d.

15. A ladder is leaning against a wall. The baseof the ladder is 1.5 m from the base of the wall.The top of the ladder touches the wall at a point4 m above the ground. What is the slope of theladder?

a) Use slopes to determine whether each set of pointsis collinear or non-collinear.b) If the points are collinear, find the slope.

16. A(–2, 1), B(–1, 2), C(–4, –2)

17. M(–1, –3), N(2, 5), P(3, 4)

18. W(–3, –1), X(9, 8), Z(5, 5)

− 23

0 2

–2

x

y

–4

–6

4

20

2

–2–4

–2

x

y

–4

m = 12

12

20

2

4–2–4

–2

x

y

A

K

B

H

G

L

I

D

J

E F

4

6

C

Slope I(Using Points)To find the direction of a line and how steep it is, find the slope using thecoordinates of any two points on the line, (x1, y1) and (x2, y2).

(slope) , or myx

=∆∆

(change in y-values is the rise)(change in x-values is the run)

my yx x

=−−

2 1

2 1


Name

Find the slope and y-intercept of each line.

1. y = 2x – 5 2.

3. x + y – 1 = 0 4. x – 5 = 2y

Find the slope and y-intercept of each line.

5. 3x + y = 2 6. 4x – 3y = 9

7. 2x + y = 4 8. y = 5

Given the slope and y-intercept, write an equation ofthe line in the slope and y-intercept form. Then, writethe equation in standard form.

9. m = 5; b = 2 10. m = –1;

11. ; b = –1 12. ;

13. Find the slope and y-intercept of the linethrough the points (–2, 11) and (10, –7).

Draw the graph of each line.

14. y = 3x + 1 15.

Find the slope and y-intercept of each line. Then,write an equation of the line.

16. 17.

18. An equation of a line is y = –x + b. Find thevalue of b if the line passes through thepoint (1, –3). What is the value of m?

19. The equation 35n – t + 50 = 0 relates t, thetotal cost, in dollars, of boarding a dog in akennel, with the number of nights, n, the dog isboarded, and the cost of medical insurance.

a) Write the equation in the form y = mx + b.

b) Graph the total cost versus the number ofnights of boarding.

c) What is the total cost to board a dog at thekennel for 7 nights?

d) What is the slope of the line?

e) What does the slope represent?

f) What is the cost of medical insurance?

20

300

200

100

4 6 8

20–2

–2

x

y

–4

2

20–2

–2

x

y

–4

–4

2

20–2

–2

x

y

–420

2

–2

–2

x

y4

y x= − −12

1

b = − 13

m = − 14

m = 23

b = 12

y x= − +13

2

Slope II(Linear Equations: Slope and y-Intercept Form)When an equation is written in slope and y-intercept form, y = mx + b, m gives the direction and amount of slope of the line, and b gives the y-intercept, which is the y-coordinate of the point on the y-axis, (0, b).


Name

Given the slopes of two lines, determine whether thelines are parallel, perpendicular, or neither.

1. m1 = 2, 2. ,

3. m1 = 5, m2 = –5 4. , m2 = –1.5

Find the slope of a line perpendicular to a line withthe given slope.

5. 6. –3 7. undefined

State the slope of a linea) parallel to each lineb) perpendicular to each line

8. y = 3x – 1

9. x + 4y = 5

10. 2x – 6y – 3 = 0

11. The coordinates of six points are given.A(–2, 3), B(3, 3), C(3, –1), D(0, –1), E(–2, –1),F(–5, –1)Which points are the vertices of

a) a rectangle? b) a parallelogram?

12. The slopes of two parallel lines are –3 and

. Find the value of m.

Identify whether each pair of lines is parallel,perpendicular, or neither.

13. 3x + y – 2 = 0 and

14. 5x – y + 2 = 0 and 10x – 2y – 17 = 0

15. x + 2y – 2 = 0 and 2x – y + 3 = 0

16. Determine an equation for the line passingthrough (–3, –2) and perpendicular to 2x – 3y = 3.

17. Determine an equation for the line parallelto 3x + y – 2 = 0 and having the same x-interceptas x – 2y + 5 = 0.

Plot and join the points in order. Classify each figureas a square, a rectangle, a parallelogram, or atrapezoid.

18. C(–3, 3), D(–1, 5), 19. K(–2, 4), L(2, 3),E(3, 6), and F(–3, 0) M(–1, 1), and N(–5, 2)

20. The equation of the path of a passenger shipis given by 3x + y – 2 = 0. The equation of thepath of a cargo ship is given by 6x + 2y + 10 = 0. If the ships remain on theirgiven paths, are they likely to collide? Explain.

0

2

–2–4 x

y

4

2

0

2

–2–4 x

y

4

2

6

y x= − +13

2

m5

12

m123

=

m22 43 6

= − ..

m123

= −m236

= −

Slope III(Parallel and Perpendicular Lines)• All vertical lines never meet, so they are parallel. Also, two non-vertical lines

are parallel if they have the same slope.• A vertical line forms a right angle with a horizontal line, so they are

perpendicular. Also, two lines that are not vertical or horizontal are perpendicular if the product of their slopes is –1.


Name

To represent x as a variable, draw torepresent an x-tile and colour it green.To represent +1, draw � and colour it red.To represent –1, draw � and leave it uncoloured.On each of the following scales, draw the tiles torepresent each equation.

1. x – 2 = 6

2. x + 3 = 5

3. –4 = x – 3

4. –5 = 4 + x

What number must be added to both sides of theequation to solve it?

5. b – 11 = 25 6. r – 3 = 8

7. –5 = –2 + k 8. x – 6 = –3

9. 10. 1.3 = d – 1.4

What number must be subtracted from both sides ofthe equation to solve it?

11. c + 6 = 14 12. –3 = x + 3

13. 7 = 3 + q 14. –8 = 2 + k

15. 6 + p = –11 16. 0.6 = s + 0.4

Solve.

17. m + 3 = 9 18. x – 4 = 11

19. 4 = –3 + y 20. –14 = k – 7

21. 2 + x = –5 22. –8 = a – 6

23. 12 = 12 + e 24. –1 = –4 + r

25. 26.

27. 28.

Solve and check.

29. 2.2 + y = 6.2 30. 9.3 = a – 2.5

31. x + 4.3 = 5.8 32. –6.2 = k – 4.8

33. t – 1.4 = 4.1 34. g – 3.4 = –1.6

35. At one Winter Olympics, Canada won2 gold medals. In the same year, Canada won 4 more gold medals in the Summer games thanit did in the Winter games. Solve the equation x – 4 = 2 to determine the number of goldmedals won in the Summer games.

z − = −78

14

− + =13

0s

1112

23

= +by − =112

7

t − =12

5

=

=

=

=

Solving Equations I(Using Addition and Subtraction)An equation is like a balanced scale.By adding or subtracting the same amount from each side of a balanced scaleor an equation, the equality is maintained.To solve an equation, isolate the variable on one side of the equation by addingor subtracting the same amount from both sides of the equation.


Name


1. 2x = 4

2. 3x = –6

3.

By what number must you divide both sides tosolve each equation?

4. 6x = 60 5. 3y = –21

6. 15 = 5t 7. –36 = 9k

By what number must you multiply both sides tosolve each equation?

8. 9.

10. 11.

Find the missing number. Then, check.

12. –y = 23 13.–y × (–1) = 23 × (–1)

y = �

Solve.

14. 5t = –30 15. 3c = 24

16. –9r = 36 17. –56n = –8

18. 19.

20. 21.

22. 8 = –y 23.

Solve and check.

24. 3x = 4.8 25.

26. 27. 2.5r = –7.5

Estimate. Then, use a calculator to solve.

28. –291n = 20 661 29.

30. Circle the equation that represents thenumber of five-dollar bills in a pile that totals $60.

a) b) c) 5x = 60 d) 60x = 5x60

5=605

= x

− =c725

11

p3

2 4= .

s4

1 5= − .

− =x5

10

127

= s− =1015k

m3

3= −− =−

55

p

− =

−=

−( ) ×−

= − ×

=

m

m

m

m

48

48

44

4 8

�

38

=−y

− =84a

k7

2= −x5

5=

=− =42x

=

=

Solving Equations II(Using Division and Multiplication)To solve an equation, isolate the variable on one side of the equation bymultiplying or dividing by the same amount on both sides of the equation.


Name

Draw a flow chart to show the solution steps foreach equation.

1. 3x + 4 = 25

2. 4y – 3 = 21

3. 5 – x = 11

Solve.

4. 4t – 6 = –10 5. 5r = 9 + 6

6. 8m + 2m = 30 7. 4w + 3w = –28

8. –3 = –2 – x 9. –5 – y = 6

Solve and check.

10. 4b – 6b = 12 11. 5 + 2n = –15

12. 4 + 8 = 3m 13. 3p – 7 = 14

14. –14 = 4y + 3y 15. 8a – 3a = 15

Solve.

16. 17.

18. 19.

20. 21.

Solve and check.

22. 3x + 2x + 3 = 13

23. 6a – 3a + 5 = 14

24. 6 – 3c = 10 – 7

25. 3b + 2b – b = 15 – 7 + 4

26. 3r + r + 2r – 6 = 11 + 2 – 7

Solve.

27. 3x + 1.2 = 3.9 28. 4k – 2.5 = 1.5

29. 4 – 3.2d = 13.6 30. 0.6g – 1.6 = 0.8

31. 6.3 = 0.1 – n 32. 12 – 2.4a = 2.4

33. In 1992, the first year the Toronto BlueJays won the World Series, the Jays played162 regular-season games and12 championship games. They won 34 moregames than they lost. Solve the equationx + (x + 34) = 162 + 12 to find the number ofgames the Jays won and lost.

n2

34

14

− = −15 5

25

+ = −m

53

2+ = −kt4

3 1− =

y4

11 3= −x2

6 4= +

Solving Equations III(Multi-Step Equations)To solve equations involving more than one step, follow the process shown in the flow chart.

StopSolution

Multiply or Divide bothsides of the equation bythe same value toisolate the variable.

Add or Subtractthe same valuefrom both sidesof the equation.

Simplifyboth sides ofthe equation.

StartEquation


Name


1. 3x = x + 4

2. 2 + 3x = 6 + 4x

3. –3 + 2x = 3x – 4

Solve.

4. 5x = 4x – 4 5. 2y = 2 + 4y

6. 8r = –4r – 12 7. 5m = 10 – 5m

8. 3b = 0.64 – 2b 9. 4k = 2k + 1.38

Solve and check.

10. 3x + x = 5x – 6

11. 2t – 5t = t + 8

12. 3y + 2y = 3y + 6

13. 4k – 6k = 6 – k

14. 7m + 8m = –10 + 5m

15. 12 = 6b + 2b – 4

Solve.

16. 2a = 15.9 – 8.7 – 3a

17. 3q – 1.5q = 12 – 4.5q

18. 5k + 1 = 3k + k – 3.8

19. 12 + 7j = 14.2 + 9j + 1.8

20. Washing a car for 20 min uses x kilojoules ofenergy. Doing yard work for 20 min usesapproximately 2x kilojoules of energy. Thenumber of kilojoules used doing yard work isequal to the number used washing a car plus 65.Solve the equation x + 65 = 2x to find thenumber of kilojoules of energy used to performeach activity.

21. Seven times a number is the same as 12 more than 3 times the number.

a) Write an equation to show this relationship.

b) Find the number.

22. Six more than 5 times a number is the sameas 9 less than twice the number.

a) Write an equation to show this relationship.

b) Find the number.

=

=

=

Solving Equations IV(With the Variable on Both Sides)To solve equations with the variable on both sides, follow the process shown in the flow chart.

StopSolution



Simplifyboth sides ofthe equation.

StartEquation


Name

Solve.

1. 3(x + 1) = 24 2. 2(x – 2) = 8

3. –4(y + 3) = –16 4. 10 = 5(t + 2)

5. 2(z – 3) = –12 6. –15 = 3(k + 3)

Solve and check.

7. 4(2x – 2) = –16

8. 15 = 3(2y – 3)

9. 2(2k + 4) = 14

10. –12 = 3(2k + 2)

Solve.

11. 3(x + 2) = –9 – 2x

12. 12 + 2(k + 3) = 3k – 6

13. 8 – 3y = 2(2y – 3)

14. 3(n – 2) – 19 = 5 + 2(n + 5)

Solve and check.

15. 3(x + 2) = 9 + 2(x + 4)

16. 7 + 2(b – 3) = b + 4

17. 3(2 – 2z) = 1 – z

18. 3(4k – 1) + 2(5 – 3k) = 7k

Solve.

19. 4(x + 2) – 3(x + 1) = 2(x + 2)

20. 3(n – 5) – (2n + 2) = 2(n – 1)

21. 2(a – 4) – 3(a – 2) = 4(a + 1) + 4

22. 5(c + 4) = 4(2c – 3) – 7

23. The largest flag in the world was presentedto the city of Kao-hsiung, Taiwan, by theRepublic of China. The diagram shows thedimensions of the flag. The equation 2(2x + 70) + 2(9x + 63) = 420 represents theperimeter of the flag, in centimetres.

a) Solve the equation.

b) What are the dimensions of the flag?

24. The sides of a triangle, in centimetres, aregiven by the expressions 3(n – 2), 4(n + 3), and2(n + 4). The perimeter of the triangle is 140 cm.Find the length of each side.

9x + 63

P = 420 cm 2x + 70

Solving Equations V(With Brackets)To solve equations with brackets, follow the process shown in the flow chart.

StopSolution



Simplifyboth sides ofthe equationstarting withthe brackets.

StartEquation


Name

Solve.

1. x + 0.4 = 0.6 2. k – 1.2 = 1.8

3. 1.3 + n = 2.4 4. –1.5 = a – 1.8

5. 2.6 – z = –1.2 6. g – 3.4 = 1.65

7. 0.5x = 1.5 8. 2.4s = –4.8

9. 2 = –0.5d 10. 1.6y = –6.4

Solve and check.

11. 5x + 0.8 = 1.2

12. –1.6 = 2 – 0.4y

13. 0.5a – 1 = 2 + 0.6a

14. 1.2 + 1.4y = 1.5y + 0.63

Solve.

15. 16.

17. 18.

Solve and check.

19.

20.

21.

22.

23.

24.

25.

26.

27. The mass of a snapping turtle is given bythe expression (3x + 1.7) kg, while the mass ofthe case used to ship the turtle is given by theexpression (x + 5.2) kg. The total mass of theturtle and the case is 39.9 kg.

a) Find the mass of the turtle.

b) Find the mass of the shipping case.

15

12

1− = + −z z

− = + − +4

34

12

c c

x x+( ) − =1

6 32

34

32

−( ) =+( )k k

b b+( ) =−( )1

22

5

34

23

56

a aa− = +

23 2

1t t= −

x x4 2

1= +

− =x2

13

− =25 2

k

y5

110

= −x6

12

=

Solving Equations VI(With Fractions and Decimals)To solve equations with fractions and decimals, follow the process shown in the flow chart.

StopSolution



Multiply eachterm of anequation with a) decimals by apower of 10. b) fractions by thelowest commondenominator.

StartEquation


Name

For each ratio, write two equivalent ratios.

1. 1:4 2. 6 to 3

3. 4. 3:7

5. 1.5 to 4.5 6.

Write = or ≠ in each � to make each statement true.

7. 4–6

� 2–3

8. 3:5 � 9:15

9. 5 to 2 � 10 to 5 10. 14:10 � 5:7

Solve.

11. 12.

13. 4:a = 2:5 14. 6:2 = c:6

15. 16.

17. To make concrete, 6 bags of cement aremixed with 4 bags of sand. How many bags ofcement are needed to mix with 12 bags of sand?

18. Scientists estimate that 8 out of every 9people are right-handed. In a school of 360students, how many students would you expectto be right-handed? left-handed?

19. A basketball player makes 1 basket for every2 shots missed.

a) Write the ratio of baskets made to shotstaken.

b) How many baskets would you expect theplayer to make out of 138 shots?

20. The length of the shadow of a tree measures6.2 m, and the shadow of a fence post measures2.8 m. If the fence post is 1.4 m tall, how tall isthe tree?

21. The ratio of uncooked rice to cooked rice, byvolume, is 2 to 7. Complete the table.

Uncooked Rice Cooked Rice1 L

500 mL250 mL

280 mL1.4 L2.45 L

22. About 3 out of every 5 Canadians are far-sighted, while about 3 out of every 10 arenear-sighted, and the rest have 20/20 vision.

a) Write the ratio of Canadians who have 20/20vision to those who do not.

b) How many more times likely is a Canadianto be far-sighted than near-sighted?

c) Using the number of students in your class,how many can be expected to be far-sighted?near-sighted? to have 20/20 vision?

23. Create and solve a proportion problem usingthe following data: 8 out of 10 dentistsrecommend sugarless gum.

54 5

2.

=t

1 5 0 91 8

. ..d

=

34

75=x

y16

14

=

0 20 5..

104

Solving ProportionsRatios that make the same comparison are equivalent ratios or equal ratios. A statement that ratios are equal is a proportion.


Name

Subtract expression B from expression A, representedby the algebra tiles.

1.

2.

Write the opposite of each polynomial.

3. 5ab + 6

4. z2 – z – 4

5. –2c + 3d + e

6. –4s2 – s + 5

Model the expressions using algebra tiles or drawingson grid paper. Then, subtract.

7. (x + 4) – (x + 2)

8. (x2 + 3x + 2) – (x2 + x + 1)

9. (2x2 – 3x – 2) – (x2 – 2x – 1)

10. (–2x2 – 2x – 4) – (–x2 – x – 3)

Subtract.

11. (b + 5) – (b + 2)

12. (2c – 3) – (c – 3)

13. (–2k – 4) – (–k – 2)

14. (n2 + 3n + 2) – (n2 + 2n + 1)

15. (3w2 – w – 4) – (w2 – 3)

16. e2 + 4e + 6 17. –5f 2 – 2fg – g2

e2 – 2e – 2 f 2 + fg + g2

18. –3d2 + 4d – 2 19. 2y2 – 5y – 4–2d2 – d + 2 –y2 – 3y – 2

20. a) Write an expression for the perimeter ofthis parking lot.

b) Find the difference in length for each pair ofopposite sides.

Subtracting PolynomialsTo subtract polynomials, add the opposite of the polynomial that is being subtracted.

A

B

A

B

5m + 8

3m + 7 3m – 1

7m – 1

Draw an arrow on the grid to show each translation.

1. [4, 1] 2. (x, y) → (x − 2, y + 1)3. 3 units right 4. [0, –2]5. 5 units left 6. (x, y) → (x + 1, y)7. [2, 2] 8. (x, y) → (x − 3, y − 1)9. 2 units right, 2 units down

For questions 10–15, refer to the grid toa) describe each translation in wordsb) write the ordered pair that describes eachtranslationc) write each translation as a mapping

10. a) ____________________________________

b) _______________ c) ________________

11. a) ____________________________________

b) _______________ c) ________________

12. a) ____________________________________

b) _______________ c) ________________

13. a) ____________________________________

b) _______________ c) ________________

14. a) ____________________________________

b) _______________ c) ________________

15. a) ____________________________________

b) _______________ c) ________________

Complete the table.

16.

17.

18.

19.

Complete the table.

20.

21.

22.

23.

24. �ABC has vertices A(2, 2), B(4, 2), andC(5, 5). Draw �ABC on the grid.Draw and label each translation image.a) [–4, –3] b) (x, y) → (x, y – 3)

Transformations I(Translations)A translation, or slide, is a motion that is described by length and direction.

�XYZ has been translated 3 units right and 2 units down (3R, 2D).�X′Y′Z′ is the translation image of �XYZ. The translation can be described mathematically as the ordered pair [3, –2] or as the following mapping.(x, y) → (x + 3, y – 2)

The lengths of line segments and the sizes of angles do not change in a translation. The original figure and its image have the same sense.


Name

6

4

2

0 2 4 6 8

X

Y Z

X′

Y′ Z′

x

y

12.

13.14.

10.11.

15.

OriginalPoint

TranslationImagePoint

(2, –3) (x, y) → (x, y – 2)

(1, 0) 4 units up

(–5, 0) (x, y) → (x – 2, y + 3)

(–3, 4) [3, 0]

OriginalPoint

TranslationImagePoint

(4, 2) (–1, –3)

(5, –3) (5, –5)

(–3, –2) (–3, 1)

(0, 0) (–1, 3)

x

y

Circle the pairs of figures that are reflections. Draw the reflection line for each pair.

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

11. Draw and label the reflection image of figureABCD in each reflection line.

12. a) Draw the reflection image of each linesegment in the x-axis.

b) Use a dotted line to draw the reflectionimage of each line segment in the y-axis.

13. Write the coordinates of the image of eachpoint after a reflection in each axis.

a)

b)

c)

Draw and label the reflection images for each figurein the reflection lines l and m.14. 15.

16. a) �DEF has vertices D(–3, 1), E(0, 4), andF(–2, 3). Draw �DEF on the grid.

b) Reflect �DEF in the line l.

17. �RST is reflected in the x-axis. Thecoordinates of the image �R′S′T′ are R′(2, 2),S′(4, 3), and T′(3, 4). Write the coordinates of theoriginal figure.

18. The word MOM has a vertical reflection line.The word BED has a horizontal reflection line.

a) Write 3 other words that have a verticalreflection line.

b) Write 3 other words that have a horizontalreflection line.

x

y l

0

X Y

Z

x

y

l

m

0

y

x

A

B

G E

F

C

D

H

20

2

–2

–2–4–6 4 6

nk

lm

A

B

D

C

Transformations II(Reflections)A reflection is a transformation in which a figure is reflected or flipped over a mirror line or reflection line.�XYZ has been reflected over the mirror line l.�X′Y′Z′ is the reflection image.

The lengths of line segments and the sizes of angles do not change in a reflection. The sense of a reflection image is the reverse of the sense of the original figure.


Name

F F

FF

F F

F F

FF

F

F

FF

FF

FF

F F

Reflection Line

Point x-axis y-axis

(–2, 3)

(3, 1)

(0, 3)

T S

R

x

y

l m

m

0

y

x

Y′ Z′Z Y

X′X

H 80

6

2

4

642 10

l

1. a) Write the coordinates of each line segmentand its image in the following diagram.

AB ______________ A′B′ ___________

DE ______________ D′E′ ___________

GH ______________ G′H′ __________

JK _______________ J′K′ ___________

b) What is the scale factor for each dilatation?

AB _____ DE _____ GH _____ JK _____

2. Draw and label each dilatation image of�XYZ.

a) with dilatation centre (0, 0) and scale factor 3

b) under the mapping

3. A rectangle has vertices P(3, 2), Q(–1, 2), R(–1, –1), and S(3, –1). Write the coordinates ofthe vertices of the image of rectangle PQRSunder the mapping (x, y) → (3x, 3y).

4. Draw �ABC with vertices A(1, 2), B(–1, –1),and C(2, –2). Draw the image �A′B′C′ withvertices A′(2, 4), B′(–2, 2), and C′(4, –4).

a) What is the scale factor?

b) How do the lengths of the sides of theoriginal and the image compare?

c) How do the measures of the angles of theoriginal and the image compare?

5. Draw the image of each figure after adilatation by the given scale factor.

a) scale factor 2 b) scale factor 13

x

y

0

x

y

0

Y

X

Z

( , ) ,x y x y→

12

12

y

x

A

B

E

JG

H

K

D

D′

A′

J′

K′

B′

E′

G′

H′

20

2

–2

–2–4–6 4 6

Transformations III(Dilatations)A dilatation is a transformation that changes the size of an object.Dilatations are called enlargements or reductions, depending on theway in which the size is changed.Line m is a dilatation of line n with dilatation centre (0, 0) and a scale factor of 2. This dilatation is described mathematically as the mapping(x, y) → (2x, 2y).In a dilatation, the image and the original figure are similar. They have the same shape, but not the same size.


Name

6

4

2

0 2 4 6 8 10

n

m

(2, 6)

(1, 3)

(3, 1)

(6, 2)

x

y


Adding Polynomials

1. 5x2 + 6x + 3 2. –2x2 – x + 5 3. –x2 – 2x – 2

4. 3x2 + 3x + 3 5. –3x2 + 3x – 2 6. x2 – 3x – 3

7. 3x2 – 2x – 2 8. –x2 – 4x + 1 9. 5y + 3z + 2

10. 5ab + 5bc 11. 2k2 – 4kj

12. s2 + 3s + 2t3 + t + 7 13. 6a + 3b – 3

14. 5m2 + 6mn + 3n2 15. 4r2 – 10r + 9

16. 4c2 + 2ac + 10 + a2 17. 2k2 + k – 2

18. 4x3 – 2y + 2 19. 4z3x – z + 4

20. a) 6x + 2 b) 50 cm

Angle Properties I

1. acute scalene; x = 58° 2. right isosceles; y = 45°

3. obtuse isosceles; z = 130°

4. obtuse scalene; a = 120°

5. w = 40°, x = 108°, y = 40°, z = 68°

6. d = 35°, e = 110°, f = 145°

7. h = 40°, j = 50°, k = 50°, m = 130°, n = 80°, p = 100°,q = 80°

8. m = 72°, n = 116°, p = 64°

9. k = 128°

10. r = 90°, s = 72°, t = 121°, u = 77°, v = 103°

11. c = 60°, d = 125°, e = 50°

12.

∠ACB = ∠ABC = 180° – ∠DBC;

∠DCA = ∠BCD + ∠ACB

13.

Angle Properties II

1. ∠EDG = ∠FGD, ∠CDG = ∠HGD

2. ∠BGH = ∠GDE, ∠ADE = ∠DGH, ∠ADC = ∠DGF,∠CDG = ∠FGB

3. ∠CDG and ∠FGD, ∠EDG and ∠HGD

4. a = 126°, b = 54°, r = 54°, s = 126°

5. a = 120°, b = 60°, w = 120°, x = 120°

6. Answers will vary. r = 105°, q = 105°, p = 105°

7. s = 100°, r = 130° 8. x = 40°, y = 30°, z = 110°

9. d = 65°, e = 65°, f = 50°

10. a = 122°, b = 58°, c = 122°, d = 58°

11. v = 50°, w = 70°, x = 60°, y = 70°, z = 60°

12. ∠RST = 120°, ∠SRU = 60°, ∠RUT = 120°,∠STU = 60°, ∠RVW = ∠SVW = ∠UWV = ∠TWV= 90°

Common Factoring

1. 3 2. 3b 3. 2a2

4.

5.

6.

7.

8.

9. 2(2y + 5) 10. 3m(2m + 3) 11. 2t2(2t – 3)

12. p2q2r(3pr – 4)

13. n is also a common factor. 4n(n – 3)

14. 4 is not the greatest common factor of thecoefficients. 8y2z(2z + 3)

15. When multiplying powers, add the exponents.5x2(5x4 – 1)

16. 5(a2 + 2ab – 3b2) 17. 3xy(3x2 + y + 5)

18. 2st(s2t – 4st2 + 2) 19. 3y(2y – 3x + 4x2y)

20. 4c3(3cd + 2 – 4d)

21. a) x, 3 + 5y b) x(3 + 5y)

22. a) l + l + w + w b) 2l + 2w c) 2(l + w)

23. a) 2x + 4y b) 2(x + 2y)

24. a) 2πr(r + h) b) 471 cm2 c) πr(r + 2h)

Congruent Triangles1. AB = DE, AC = DF, BC = EF, ∠A = ∠D, ∠B = ∠E,∠C = ∠F

2. XZ = SR, XY = ST, YZ = TR, ∠X = ∠S, ∠Y = ∠T,∠Z = ∠R

∠ − − ∠QRS =

360 So o902

∠ − ∠BCD =

180 DBC;

o

2

Answers APPENDIX A: Review of Prerequisite Skills

PolynomialGCF of

Both Terms Other Factor

x2 + 2x x x + 2

6y2 – 9y 3y 2y – 3

8xy – 3x2y2 xy 8 – 3xy

2m2n – 4mn2 2mn m – 2n

6a2 + 9a2 – 12a 3a 2a2 + 3a – 4

3. LM = PQ, LN = PR, MN = QR, ∠L = ∠P, ∠M =∠Q, ∠N = ∠R

4. ST = XY, SV = XZ, TV = YZ, ∠S = ∠X, ∠T = ∠Y,∠V = ∠Z

5. 1 side

6. 1 side or the angle between the two sides

7. the side between the two angles

8. BC = 5 m, EF = 4 m, FD = 3 m

9. ∠KLM = 95°, ∠JGH = 32°, ∠MKL = 53°

10. ∠TVS = 74°, ∠XYZ = 65°, ∠ZXY = 74°

11. SAS, ∠X = ∠A, ∠Y = ∠B, ∠Z = ∠C, XY = AB,XZ = AC, YZ = BC

12. SSS, ∠D = ∠F, ∠DEG = ∠FEG, ∠DGE = ∠FGE,DE = FE, DG = FG, EG = EG

13. ASA, ∠H = ∠M, ∠J = ∠L, ∠HKJ = ∠MKL,HJ = ML, JK = LK, HK = MK

14. SAS, ∠QPR = ∠SPR, ∠Q = ∠S, ∠QRP = ∠SRP,PQ = PS, QR = SR, PR = PR

Evaluating Expressions I

1. 12 + 7 2. x + 10 3. 9 × 12 4. 5y 5. ab

6. 7. q + 8 8. t – 9 9. 2r

10. or 0.5d

11. 12.

13. 14.

15. a) 3 b) 21 c) 8 d) 15 e) 32 f) 2

16. a) 14.5 b) 2.8

17. a) 9.9 b) 28.86

18. a) 1.5n b) $12.00

19. a) 6h, where h represents the height of a jump onEarth b) 16.8 m

20. a) 18 + 4g, where g represents the number ofgames rented b) $30.00

21. a) where l represents the length of the

field in metres b) width 75 m, perimeter 370 m

Evaluating Expressions II

1. a) 9 b) 10 c) –2 d) –12 e) –2 f) –3

2. a) –6 b) 8 c) 1 d) –8 e) –17 f) –10

3. a) –3 b) 4 c) 4 d) –3 e) 7 f) 6

4. a) 4 b) –12 c) –6 d) –16 e) 1 f) –6

5. 6.

7. 8.

9. a) 85 b) (4 – 2(–3))2 + 5(–3) = (4 + 6)2 – 15 = 102 – 15= 100 – 15 = 85

10. a) 422 m b) 142 m c) 47 m

11. a) 99 m; 75 m b) 0 m; rocket hits the ground

Evaluating Expressions III

1. a) 77.1 kg b) 102.9 kg c) Answers will vary.

2. a) 192 b) 8 3. a) 8.9 m/s b) 1.4 m/s

4. 0.25q = 0.05n

5. a) $212.18 b) $689.79 c) $1276.28

12

20l + ,

12

d

mv


s 3s

6 18

2 6

1 3

4 12

9 27

0 0

10 30

30 90

t 3t – 4

3 5

2 2

5 11

2.3 2.9

4.1 8.3

7.8 19.4

8.3 20.9

x 4x + 5

0 5

1 9

3 17

10 45

1.5 11

6.2 29.8

8.3 38.2

x 2x – 2

2 2

1 0

0 –2

–1 –4

–2 –6

m 3 + 2m

2 7

1 5

0 3

–1 1

–2 1

s s2 + 2s2 8

1 3

0 0

–1 –1

–2 0

a a2 – 2a – 12 –1

1 –2

0 –1

–1 2

–2 7

r y

3 4

0 1

2 5

1.1 3.2

0.8 0.9

2 – r + y

3

3

5

4.1

2.1


6.

Evaluating Expressions IV

1. a) y = x2 – 2 2. a) y = x2 + 0.5

b) b)

3. a)

4. a)

b) $12 400 c) 48%

5. a) b)

c) about 70 km/h

Evaluating Radicals

1. 8, –8 2. 5, –5 3. 11, –11 4. 0.9, –0.9

5. 1.5, –1.5 6. 0.2, –0.2 7. 4 8. –10

9. 0.7 10. 0.3 11. 7.87 12. –6.16

13. 10.49 14. 31.05 15. 53.18 16. 204.94

17. 40.62 18. –65.67 19. 0.71 20. 0.82

21. 0.06 22. 0.04 23. F 24. T

25. F 26. T 27. F 28. F

29. –6.36 30. 5 31. 2.45 32. 10

33. a) 20 cm b) 7 cm c) 1.2 m d) 1.5 m

34. a) 100 cm b) 141.4 cm

35. a) 5 cm b) 7.14 m c) 6 m d) 13.0 cm

Expanding and Simplifying Expressions I

1. The length is (x + 3); the width is 2; and the area is2(x + 3) or 2x + 6.

2. 2x + 4 3. 3x + 9 4. 4x + 2 5. 6x + 6

6. 4x + 8 7. 5x – 15 8. 0.3x + 1.5 9. 8x + 4

10. 11. L 12. A

13. L 14. G 15. T 16. H

17. R 18. I ALL RIGHT 19. 13x – 6

20. 15 – 12n 21. 8h – 3 22. 12 23. 4 – 9k

24. 8u + 2 25. 5x2 + 4x + 11 26. –3y – 8

27. 3t2 – 10t – 5 28. 4e + 20 29. x2 – 7x

30. a) 5x b) –2x2 + 5x

23

132

1xx− − or

a b

5 cm 7 cm

11 cm 19 cm

8.4 m 3.6 m

1.5 m 1.5 m

c

8 cm

20 cm

6 m

2.1 m

s

10 cm

25 cm

9 m

2.55 m

A

17.3 cm2

102.5 cm2

9.4 m2

1.12 m2

x y

0 –2

1 –1

–1 –1

2 2

–2 2

3 7

–3 7

x y

0 0

1 15.9

4 31.8

9 47.7

16 63.6

25 79.5

36 95.4x y

0 0.5

1 1.5

–1 1.5

2 4.5

–2 4.5

3 9.5

–3 9.5

20

2

4

6

–2

–2

x

y

20

4

6

8

–2

2

x

y

0

10

20

2 4 6

Diameter (cm)

Are

a (

cm

)2

0

4

8

12

16

18

14

10

6

2

2010 30 40 50 60 70 80Percent of Pollutants Removed

Co

st (

$1

00

0s)

0

10

10 20 30 40

20

30

40

50

60

70

80

90

100

Sp

ee

d (

km

/h)

Skid Length (m)

50

b) 9.6 cm2

c) 3.6 cm

Expanding and Simplifying Expressions II

1. a2 + 3a 2. s2 – 5s 3. –y2 – 2y 4. 4b – b2

5. –6x + x2 6. –k2 + 3k 7. 4r2 + 12r 8. 6m2 – 12m

9. 6x – 2x2 10. –15y – 3y2

11. a) n(n + 1) b) n2 – n + 1

12. 6n 13. n 14. n2 – 3n + 4

15. 5n2 + n 16. –2n 17. 4n2 18. 2n2 + 8n

19. –n A RHOMBUS 20. 6a2 + 8a 21. r2 – 13r

22. 3k2 – 5k 23. 3d2 + 7d 24. 5x2 – 6x

25. a2 + 4a – 20 26. 3x3 + 6x2 – 26x + 2

27. y3 – y2 – 2 28. r2 – 19r

29. a) 3n(n) = 3n2, n(2n – 3) = 2n2 – 3n, 3n(2n – 3) = 6n2 – 9n; 2(3n2) + 2(2n2 – 3n) + 2(6n2 – 9n) = 22n2 – 24n

b) 4.2x(x) = 4.2x2, x(2x + 3) = 2x2 + 3x, 4.2x(2x + 3) = 8.4x2 + 12.6x;2(4.2x2) + 2(2x2 + 3x) + 2(8.4x2 + 12.6x) = 29.2x2 + 31.2x

Exponent Rules I

1. 25 2. 38 3. 46 4. 104 5. 97 6. 85

7. x7 8. y6 9. z5 10. 2 11. 3 12. 2

13. 1 14. 3 15. 4 16. 10 17. 1 18. 52

19. 43 20. 3 21. 93 22. 7 23. 22 24. m2

25. p2 26. a 27. 2 28. 2 29. 6 30. 4

31. 2 32. 7 33. 4 34. 8 35. 36 36. 28

37. 712 38. 68 39. 56 40. 415 41. x9 42. s4

43. r10 44. 3 45. 2 46. 4 47. 4 48. 3

49. 3 50. 4 51. 3 52–63. PERFECT*WORK

Exponent Rules II

1.

2.

3.4.5.

6.

7.

8.9.

10.

11.

12.

13.

14.

15. 243 16. –32 17. 78 125 18. 104.8576

19. y6 20. 9 21. 1024 22. 20.25

23. 9 24. –7 25. –1.44 26. 0.36

27. F 28. F 29. T 30. T

31. T 32. F 33. 8 34. 54

35. –23 36. –9 37. 18 38. –26

39.

40. Negative bases that are multiplied an oddnumber of times give a negative answer.

41. a) (–2)4 = (–2)(–2)(–2)(–2), which equals 16. –24 = –(2)(2)(2)(2), which equals –16.

b) ((–3)2)3 = (–3)2(–3)2(–3)2 or (–3)6, which equals 729.–36 = –(3)(3)(3)(3)(3)(3), which equals –729.

Exponent Rules III

1. a) Multiply the coefficients and add the exponentsof the variable, for example, (5n)(2n2) = 10n3.b) Multiply the coefficients and combine likevariables using the exponent laws, for example, (2yz)(–3y3) = –6y4z.

2. 2x2 3. 6n2 4. yz 5. 4ak

6. 8vw 7. 10st 8. 12a2b 9. 20fg2


ExponentialForm Base Exponent

StandardForm

(–2)3 –2 3 –8

31 3 1 3

5 5 1 5

(–3)3 –3 3 –27

(–2)5 –2 5 –32

72 –7 2 49

Radius (cm) Area (cm2)10 314.2

5 78.5

2.5 19.6

1.3 5.3

6.2 120.8

ExponentialForm

RepeatedMultiplication

StandardForm

(–3)2 × (–3)2 (–3) × (–3) × (–3) × (–3) 81

(–4)3 (–4) × (–4) × (–4) –64

(–5)3 (–5) × (–5) × (–5) –125

(–4)3 ÷ (–4)1 ( ) ( ) ( )( )

− × − × −−

4 4 44

16

(+5)4 ÷ (+5)2 ( ) ( ) ( ) ( )( ) ( )

+ × + × + × ++ × +

5 5 5 55 5

25

(+5)3 ÷ (+5)2 ( ) ( ) ( )( ) ( )

+ × + × ++ × +

5 5 55 5

5

(–3)5 ÷ (–3)2 ( ) ( ) ( ) ( ) ( )( ) ( )

− × − × − × − × −− × −

3 3 3 3 33 3 –27

(–2)2 ÷ (–2)1 ( ) ( )( )

− × −−

2 22

–2


10. 8xyz 11. 12cde 12. –6az3 13. –16r2s

14. –20cde 15. 6x2y2 16. –6ab2m2 17. –5u2t2

18. –6a2b4c3d 19. –10r3s3t3 20. –60xyz

21. –18d2e 22. 8k3m2n5 23. 0.75s2

24. 2x(2x) + x(4x) = 8x2

25. x(x) + x(3x + 3x + x) = 8x2

26. a) (2b)(2b) = 4b2 b) 6 × 4b2 = 24b2

c) (2b)(2b)(2b) = 8b3

27. a) 2n(n) = 2n2, 2n(4n) = 8n2, n(4n) = 4n2

b) 2(2n2) + 2(8n2) + 2(4n2) =28n2 c) n(2n)(4n) = 8n3

28. 29.

volume = 6a3 volume = 64c3

Exponent Rules IV

1. a) Find the power of the coefficient, for example,(–12), and the power of the variable, for example, (a3)2,a6. b) (–a)3(–a)3 = a6

2. y2 3. x4 4. s4 5. c6

6. –m15 7. f 18 8. s4t4 9. –x6y3

10. c2d4 11. y2z2 12. r3s3 13. –a6b6

14. f 6g4; FORTY WINKS 15. 9t2y2 16. –8x3z3

17. –8a6b3 18. 9r6s2 19. 25k6m4 20. –27q6r6

21. y5z3 22. –2a3b3 23. –25s5t5 24. 128k7m9

25. 18r4s4t3 26. 32a5b6c6 27. –3m6n8p8

28. a) (xy2)2 = x2y4 b) (xy2)3 = x3y6

29. a) (2a2bc3)2 = 4a4b2c6 b) (2a2bc3)3 = 8a6b3c9

30. The exponents are added, but should bemultiplied. a6

31. The exponent of g is squared, but should bemultiplied by 2. f 2g6

32. The square of a negative number is positive. 4x4y4

Exponent Rules V

1. 2r 2. –2s 3. a 4. 5

5. –4 6. 4 7. 2b 8. 2klm

9. 3 10. –2c 11. 3 12. –4

13. 2a2b2 14. 2q 15. –2x3y2 16. 2wx3

17. 18. 19. 2km2

20. 2a3b 21. 4x2y2 22. 23.

24. 25. 26. 2x

27. 2r3s2 28. 2a2b2

29. The length is 6xy and the width is 4xy.

30. front and back: 2x × 2y; two sides: 2x × 4xz2; topand bottom: 2y × 4xz2

Graphing Equations I

1. y = –2x + 5 2. 2x + y = 0

x y x y

–2 9 –2 4

–1 7 –1 2

0 5 0 0

1 3 1 –2

2 1 2 –4

3. y = 2x + 1 4. y = –x – 3

20–2

–2

–4

x

y

20

2

–2

–2

x

y

20

2

4

–2

–2

–4

x

y

20

6

8

–2

2

4

x

y

84

22

2s ts t=5

33d f

32

2ef− 32

5 4s t

− 23

4 5c d− 32

2 3f g h

4c4c

4c

a

3a

2a

5. x + y = –1 6. 2x – y = 3

7. y = x + 4

8. y = 3x

9. the domain is R

10. a), b)

The points can be joined because the measurementsare continuous. c) 240 cm d) 18 cm

Graphing Equations II

1. 2.

3. 4.

5. 6.

7. x-intercept –5; y-intercept 5; slope 1

8. x-intercept –4; y-intercept –10;

slope

9. 5x + 4y – 20 = 0 10. 4x – 7y + 28 = 0

11. Answers may vary. For example, y = x – 1, y = x – 2, y = x – 3

12. Answers may vary. For example, y = –x, y = –2x,

y = − 1–2

x, y = −1–3

x

13. slope 0; y-intercept –3; x-intercept none

14. slope undefined; y-intercept none; x-intercept 5

15. a)

b) slope 1.4; The slope represents the rate at whichthe heart beats. c) y = 1.4xd) 84 heartbeats per minute

100

20

20 30

30

40

50

60

10

Time (s)

Nu

mb

er

of

He

art

be

ats

− 52

20

–4

x

y

–2

–6

–2 4 6 8–4

–8

–10

5x + 2y = –20

2

0

2

–2–4 x

y

4

y = x + 5

20

2

–2

–2

x

y

4

2x + 3y = 6

0

100

200

300

400

500

5 10 15 20 25Depth of Water (cm)D

ep

th o

f H

ea

vy W

et

Sn

ow

(c

m)

y x= 12

;

20

–2

2

–2

–6

–4

x

y

2

2

0–2

–2

x

y


20–2

–2

x

y

4

–44x – y = 4

20

2

x

y4

4

3x + 5y – 15 = 020–2

–2

x

y

–4x + y + 2 = 0

20–2

–2

x

y

4

2

y + 1 = 2x

20

–4

x

y

2

–2

–6

–2 4 6 8

x + 4y = 8


Graphing Equations III

1. (1, 3) 2. (5, –4) 3. (4, –1)

4.

5. (2, 0)

6. (3, 1)

7. a) x y b) x y (13, 18)

0 5 0 –8

5 10 5 2

13 18 13 18

8. x y x y

0 1 0 3

1 3 –1 2

–1 –1 –2 1

(2, 5)

9. x y x y

0 7 0 –5

1 5 5 0

2 3

(4, –1)

10. If x = 0 for both planes as seen on the screen, thenat x = 1, both planes will be at the point (1, 2) if thespeeds allow. This assumes the plane y = 4 – 2x isgoing faster than the other. Diversionary tactics areimmediately necessary to avoid a collision.

Greatest Common Factors

Number Prime Factors1. 20 2 × 2 × 52. 30 2 × 3 × 53. 18 2 × 3 × 34. 54 2 × 3 × 3 × 35. 150 2 × 3 × 5 × 56. 252 2 × 2 × 3 × 3 × 7

Expression Prime Factors7. 6x2 2 × 3 × x × x8. 30st2 2 × 3 × 5 × s × t × t9. 12a2bc3 2 × 2 × 3 × a × a × b × c × c × c10. 10a2b3 2 × 5 × a × a × b × b × b11. 24x2y2z 2 × 2 × 2 × 3 × x × x × y × y × z

12. 2 × 3 × m × m × n × n 13. 3 × 17 × a × a

14. 2 × 2 × 19 × r × r × s 15. 4 16. 15

17. 6 18. 27 19. 5a 20. 3x2

21. 6xy 22. 4 23. 2st 24. 2p2q2

25. 3 26. 9 27. 4 28. 5

29. a 30. s2t2 31. 4xyz 32. 7c

33. 15 cm 34. a) 1 cm, 2 cm, 3 cm, 5 cm, 6 cm, 10 cm, 15 cm, 30 cm b) 30 cm

35. a) any of the diagrams shown in part b)

420

–2

–4

x

y

2

4

20

–2

x

y

–2

2

4

20

–2

x

y

–2

2

43

43

,

20

–2

x

y

–2

2

b)

Like Terms

1. 4r, –r, 101r, r; and 5r2, –r2, (–r2); and r3, –2r3

2. a) 4 b) coefficients –1, 5, –2; constant 3

3. 10t 4. –7b2 5. –13y 6. 21m3

7. 3p 8. 2c2 9. 4x 10. –12y

11. d 12. t2 13. 15y – 2z 14. –8r + 3s

15. 2b2 16. 7e3 – e2 17. 2s, 2 18. 4a2, 16

19. 5t – 3, –0.5 20. –6k + 2, 20 21. c + 5d + 3

22. m + n + 3 23. 2 + 3z – 3w 24. 2 + 5x2 + y2

25. 3r + 2r + 2r, 7r, or 2 × 2r + 3r

26. 1.5s + 0.5s + 1.5s + 0.5s, 4s

27–31. Answers will vary.

Polynomials

1. binomial 2. trinomial 3. monomial

4. trinomial 5. monomial 6. binomial

7–12. Find the sum of the exponents of its variables.

7. 4 8. 1 9. 2 10. 0 11. 5 12. 5

13–18. Find the greatest sum of the exponents in anyone term.

13. 1 14. 2 15. 3 16. 4 17. 6 18. 4

19. 1 20. 2 21. 3 22. 3

23. 3n + 3, binomial, 1 24. monomial, 2

25. 2πr2 + 2πrh, binomial, 2

26. 2 + y + xy2 + y3 27. x3 – 2x2y + 3xy3

28. 29. x4 + 2x3 – 6x – 10

30. –1.3x5 + 0.2mx4 + 2.1x3 + 0.4m2

31. 32. x + a

33–35. Answers will vary.

36. a) 6x2 – 5 b) Yes; each has a degree of 2.

Slope I

1. mCD is undefined, mEF = 0, mGH = –1,

2. 3. –1 4.

5. 6. 0 7. undefined

8. Answers may vary. For example, (0, –3), (2, –2),(6, 0), (8, 1)

9. 10.

11. Answers may vary. For example, (–1, 2) and (3, 5);(–2, 4) and (7, 11)

12. Answers may vary. For example, (–4, –3), (–7, –1),(2, –7)

13. Answers may vary. For example,

a) (0, –5) and (3, 1); slope is 2

b) (0, –8) and (8, 0); slope is 1

14. c = –6, d = 4 15.

16. non-collinear 17. non-collinear

18. collinear;

Slope II

1. m = 2, b = –5 2. b = 2

3. m = –1, b = 1 4.

5. m = –3, b = 2 6. b = –3m = 43

,

m b= = −12

52

,

m = − 13

,

m = 34

83

20

2

–2–4

–2

x

y

–4

34

− 79

54

m mIJ KL = − =53

23

,

mAB = 12

,

x b x bx b4 3 2 23

4+ + +

34

3 6− + −y y y

12

bh,

r3

2,


4

ab

6

2b

2a 3a

2ab

2 3

2ab2

3ab

2a

2b 3b

a6b4b

b4a 6a

0 2

–2

x

y

–4

–6

4


7. m = –2, b = 4 8. m = 0, b = 5

9. y = 5x + 2, 5x – y + 2 = 0

10. 2x + 2y – 1 = 0

11. 2x – 3y – 3 = 0

12. 3x + 12y + 4 = 0

13. b = 8

14.

15.

16. 3x – 4y + 2 = 0

17. x + 3y + 6 = 0

18. b = –2, m = –1

19. a) t = 35n + 50

b)

c) $295 d) 35

e) the cost per night to board the dog f) $50

Slope III

1. perpendicular 2. parallel 3. neither

4. perpendicular 5. –2 6.

7. 0 8. a) 3 b)

9. a) b) 4

10. a) b) –3

11. a) A, B, C, E b) A, B, D, F or A, B, C, E

12. m = –15 13. neither 14. parallel

15. perpendicular 16. 3x + 2y + 13 = 0

17. 3x + y + 15 = 0

18. trapezoid

19. parallelogram

20. No. The paths are parallel.

Solving Equations I

1.

2.

3.

4.

5. 11 6. 3 7. 2 8. 6 9.

10. 1.4 11. 6 12. 3 13. 3 14. 2

12

= GR RR R

= G

=G

RRR

RR

RRR

=G

RRR

RRR

0

2

–2–4 x

y

4

2

L

M

N

K

0

2

–2 x

y

4

2

6

C

F

D

E

13

− 14

− 13

13

20

300

200

100

4 6 8

Nights

Tota

l C

ost

($

)

m b y x= − = − = − −13

213

2; ; ;

m b y x= = = +34

12

34

12

; ; ;

20–2

–2

x

y

–4

y = – x – 11–2

20

2

–2

–2

x

y4

4

y = 3x + 1

m = − 32

;

y x= − −14

13

,

y x= −23

1,

y x= − + 12

,

15. 6 16. 0.4 17. 6 18. 15 19. 7

20. –7 21. –7 22. –2 23. 0 24. 3

25. 26. 27. 28. 29. 4.0

30. 11.8 31. 1.5 32. –1.4 33. 5.5 34. 1.8

35. 6

Solving Equations II

1.

2.

3.

4. 6 5. 3 6. 5 7. 9 8. 5

9. 7 10. 4 11. –8 12. –23 13. –32

14. –6 15. 8 16. –4 17. 18. 25

19. –9 20. –150 21. 84 22. –8 23. –50

24. 1.6 25. –6 26. 7.2 27. –3 28. –71

29. –7975 30. c) 5x = 60

Solving Equations III

1. START 3x + 4 Subtract 4 Divide by 3 x STOP

2. START 4y – 3 Add 3 Divide by 4 y STOP

3. START 5 – x Add xSubtract 11 x STOP, or

START 5 – x Subtract 5 Multiply or divide by –1 x STOP

4. –1 5. 3 6. 3 7. –4 8. 1

9. –11 10. –6 11. –10 12. 4 13. 7

14. –2 15. 3 16. 20 17. 32 18. 16

19. –21 20. –3 21. 1 22. 2 23. 3

24. 1 25. 3 26. 2 27. 0.9 28. 1

29. 3 30. 4 31. –6.2 32. 4

33. won 104, lost 70

Solving Equations IV1.

2.

3.

4. –4 5. –1 6. –1 7. 1 8. 0.128

9. 0.69 10. 6 11. –2 12. 3 13. –6

14. –1 15. 2 16. 1.44 17. 2 18. –4.8

19. –2 20. yard work: 130 kJ; washing car: 65 kJ21. a) 7x = 3x + 12 b) 322. a) 5x + 6 = 2x – 9 b) –5

Solving Equations V

1. 7 2. 6 3. 1 4. 0 5. –3

6. –8 7. –1 8. 4 9. 1.5 10. –3

11. –3 12. 24 13. 2 14. 40 15. 11

16. 3 17. 1 18. 7 19. 1 20. –15

21. –2 22. 13

23. a) x = 7 b) length = 126 cm, width = 84 cm

24. 36 cm, 68 cm, 36 cm

Solving Equations VI

1. 0.2 2. 3.0 3. 1.1 4. 0.3 5. 3.8

6. 5.05 7. 3 8. –2 9. –4 10. –4

11. 0.08 12. 9 13. –30 14. 5.7 15. 3

16. 17. 18. 19. –4 20. –6

21. 22. –3 23. –1 24. –11 25. 17

26. 1 27. a) 26.45 kg b) 13.45 kg

Solving Proportions

1–6. Answers may vary. 1. 2:8, 3:12

2. 2 to 1, 4 to 2 3. 4.6

143070

, 208

3012

,

− 1011

− 23

− 45

− 12

=GG

GG

G

=GG

GRR

RR

RRR

R

GGGG

=GG

G

GR RR R

17

= G

=GG

G

=GG

RR

RR

58

13

14

812



5. 1 to 3, 15 to 45 6.

7. = 8. = 9. ≠ 10. ≠ 11. y = 4

12. x = 100 13. a = 10 14. c = 18

15. d = 3.0 16. t = 1.8 17. 18

18. right-handed: 320, left-handed: 40

19. a) 1:3 b) 46 20. 3.1 m

21. Uncooked Rice Cooked Rice1 L 3.5 L

500 mL 1.75 L250 mL 875 mL80 mL 280 mL0.4 L 1.4 L0.7 L 2.45 L

22. a) 1:10 b) twice

c) Answers will vary.

23. Answers will vary.

Subtracting Polynomials

1. 2x + 1 2. –x – 7 3. –5ab – 6 4. –z2 + z + 4

5. 2c – 3d – e 6. 4s2 + s – 5

7. 2 8. 2x + 1 9. x2 – x – 1 10. –x2 – x – 1

11. 3 12. c 13. –k – 2 14. n + 1

15. 2w2 – w – 1 16. 6e + 8

17. –6f 2 – 3fg – 2g2 18. –d2 + 5d – 4

19. 3y2 – 2y – 2

20. a) 18m + 13

b) For the long sides, the difference is 2m – 9; for theshort sides, the difference is 8.

Transformations I

10. a) 3 units left, 1 unit up b) [–3, 1]

c) (x, y) → (x – 3, y + 1)

11. a) 4 units right b) [4, 0]

c) (x, y) → (x + 4, y)

12. a) 2 units right, 1 unit up b) [2, 1]

c) (x, y) → (x + 2, y + 1)

13. a) 3 units up b) [0, 3]

c) (x, y) → (x, y + 3)

14. a) 4 units right, 2 units down b) [4, –2]

c) (x, y) → (x + 4, y – 2)

15. a) 2 units left, 3 units down b) [–2, –3]

c) (x, y) → (x – 2, y – 3)

16.17.18.19.

20.

21.

22.23.

24.

Transformations II

1. 3. 5. 6.

11.

D′

B′′

C′

A′′A′′A′

B′

D′′

C′′

A

B

D

C

AIV

BIV

CIV

DIV

m

nk

lC′′′ B′′′

D′′′ A′′′

FFFFFFF F

4

2

0 2 4 6

A′ B′

x

y

–2–4–6

C′ C′′

A′′ B′′

C

A B

–2

1.

7.

5.

8.

2. 3.

4. 6. 9.

25

2050

,

OriginalPoint Translation

ImagePoint

(2, –3) (x, y) → (x, y – 2) (2, –5)

(1, 0) 4 units up (1, 4)

(–5, 0) (x, y) → (x – 2, y + 3) (–7, 3)

(–3, 4) [3, 0] (0, 4)

OriginalPoint Translation

ImagePoint

(4, 2) (x, y) → (x – 5, y – 5)(–1, –3)

(5, –3) (x, y) → (x, y – 2)(5, –5)

(–3, –2) (x, y) → (x, y + 3)(–3, 1)

(0, 0) (x, y) → (x – 1, y + 3)(–1, 3)

12.

13.

a)b)c)

14.

15.

16.

17. R(2, –2), S(4, –3), T(3, –4)

18. a) Answers may vary. MUM, TOT, TAT, TOOT,WOW

b) BOX, HIDE, CODE, BIKE, HIKE

Transformations III

1. AB: (1, 2), (3, 1); A′B′: (2, 4), (6, 2); 2

DE: (–8, 4), (–4, 4); D′E′: (–2, 1), (–1, 1);

GH: (–2, –2), (–2, –4); G′H′: (–1, –1), (–1, –2);

JK: (0, –1), (1, 0); J′K′: (0, –4), (4, 0); 4

2.

3. P′(9, 6), Q′(–3, 6), R′(–3, –3), S′(9, –3)

4.

a) 2

b) Images are twice as long as originals. c) same

5. a) b)

CB

A

A′

C′

B′

x

y

0

Z′

X′

Y′

Y′′X′′

Z′′

Y

X

Z

x

y

12

14

lEF

D D′

E′F′

x

y

H

E

F

G

A

B

C

D

B′′ F′′

E′′ G′′

H′′

C′′

D′′

A′′

0–2 2 4 6

D′

B′

A′C′

G′E′

F′

H′

x–4–6

y

2

–2


Reflection LinePoint x-axis y-axis

(–2, 3) (–2, –3) (2, 3)(3, 1) (3, –1) (–3, 1)(0, 3) (0, –3) (0, 3)

X Y

Z

X′ Y′

Z′

x

y

Y′′

Z′′

X′′

l

m

R′

x

y

S′′R′′

R

T′′

l m

T′TS′

S

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