Mathpower 10 Practise Masters 2y2648

™

MATHPOWER 10 ONTARIO EDITION Practice Masters

Shirley Barrett Richmond Hill, Ontario

Contributing Writer Janice Nixon Toronto, Ontario

Toronto Montréal Boston Burr Ridge, IL Dubuque, IA Madison, WI New York San Francisco St. Louis Bangkok Beijing Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan New Delhi Santiago Seoul Singapore Sydney Taipei

McGraw-Hill Ryerson Limited MATHPOWER™ 10, Ontario Edition Practice Masters Copyright © 2001, McGraw-Hill Ryerson Limited, a Subsidiary of The McGraw-Hill Companies. All rights reserved. This publication may be reproduced for classroom purposes without the prior written permission of McGraw-Hill Ryerson Limited. McGraw-Hill Ryerson Limited shall not be held responsible for content if any revisions, 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 1 2 3 4 5 6 7 8 9 0 0 9 8 7 6 5 4 3 2 1 0 Printed and bound in Canada Care has been taken to trace ownership of copyright material contained in this text. The publishers will gladly take any information that will enable them to rectify any reference or credit in subsequent printings. The Geometer’s Sketchpad® is a ed trademark of Key Curriculum Press. CBL™ and CBR™ are trademarks of Texas Instruments Incorporated. Adobe, Acrobat, and the Acrobat logo are ed trademarks of Adobe Systems Incorporated. This product is not endorsed or sponsored by Adobe Systems Incorporated, publisher of Adobe® Acrobat® Reader 4.0 or Adobe® Acrobat® 4.0. Canadian Cataloguing in Publication Data Barrett, Shirley, dateMathpower 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

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CONTENTS To the Teacher

v

Tips for Learning Math Getting Started Know Your Textbook and How to Use It Classroom Learning Homework Seeking Help Preparing for Tests and Exams Writing Tests and Exams Problem Solving Skills Solving Word Problems Goals

1 2 3 4 5 6 7 8 9 10

CHAPTER 3 Polynomials Practice 3.1 Polynomials 3.2 Multiplying Binomials 3.3 Special Products 3.4 Common Factors 3.5 Factoring x2 + bx + c 3.6 Factoring ax2 + bx + c, a ≠ 1 3.7 Factoring Special Quadratics Answers CHAPTER 3 Polynomials

29 30 31 32 33 34 35 37

CHAPTER 4 Quadratic Functions CHAPTER 1 Linear Systems Practice 1.1 Investigation: Ordered Pairs and Solutions 1.2 Solving Linear Systems Graphically 1.3 Solving Linear Systems by Substitution 1.4 Investigation: Equivalent Equations 1.5 Solving Linear Systems by Elimination 1.6 Investigation: Translating Words Into Equations 1.7 Solving Problems Using Linear Systems Answers CHAPTER 1 Linear Systems

11 12 13 14 15 16 17 19

Practice 4.1 Functions 4.2 Graphing y = x2 + k, y = ax2, and y = ax2 + k 4.3 Graphing y = a(x – h)2 + k 4.4 Graphing y = ax2 + bx + c by Completing the Square 4.5 Investigation: Sketching Parabolas in the Form y = ax(x – s) + t 4.6 Investigation: Finite Differences 4.7 Technology: Equations of Parabolas of Best Fit 4.8 Technology: Collecting Distance and Time Data Using CBRTM or CBLTM Answers CHAPTER 4 Quadratic Functions

39 40 41 42 43 44 45 46 47

CHAPTER 2 Analytic Geometry Practice 2.1 Length of a Line Segment 2.2 Investigation: Midpoints of Horizontal and Vertical Line Segments 2.3 Midpoint of a Line Segment 2.4 ing Properties of Geometric Figures 2.5 Distance From a Point to a Line Answers CHAPTER 2 Analytic Geometry

CHAPTER 5 Quadratic Equations 21 22 23 24 25 27

Practice 5.1 Solving Quadratic Equations by Graphing 5.2 Solving Quadratic Equations by Factoring 5.3 Investigation: Graphing Quadratic Functions by Factoring 5.4 The Quadratic Formula Answers CHAPTER 5 Quadratic Equations

51 52 53 54 55

(Powers With Integral Bases)

CHAPTER 6 Trigonometry Practice 6.1 Technology: Investigating Similar Triangles Using The Geometer’s Sketchpad® 6.2 Similar Triangles 6.3 The Tangent Ratio 6.4 The Sine Ratio 6.5 The Cosine Ratio 6.6 Solving Right Triangles 6.7 Problems Involving Two Right Triangles 6.8 Technology: Relationships Between Angles and Sides in Acute Triangles 6.9 The Sine Law 6.10 The Cosine Law Answers CHAPTER 6 Trigonometry

57 58 59 60 61 62 63 64 65 66 67

APPENDIX A: Review of Prerequisite Skills Practice Adding Polynomials

71

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

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

Angle Properties I (Interior and Exterior Angles of Triangles and Quadrilaterals)

72

Angle Properties II (Angles and Parallel Lines)

Solving Equations I (Using Addition and Subtraction)

97

73

Common Factoring

74

Solving Equations II (Using Division and Multiplication)

98

Congruent Triangles

75

Evaluating Expressions I (Variables in Expressions)

Solving Equations III (Multi-Step Equations)

99

76

Evaluating Expressions II (Expressions With Integers)

Solving Equations IV (With the Variable on Both Sides)

100

77

Evaluating Expressions III (Applying Formulas)

Solving Equations V (With Brackets)

101

78

Evaluating Expressions IV (Non-Linear Relations)

Solving Equations VI (With Fractions and Decimals)

102

79

Solving Proportions

103

Evaluating Radicals (The Pythagorean Theorem)

Subtracting Polynomials

104

80

Expanding and Simplifying Expressions I (The Distributive Property)

Transformations I (Translations)

105

81

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

Transformations II (Reflections)

106

82

Exponent Rules I (Powers With Whole Number Bases)

Transformations III (Dilatations)

107

83

Answers APPENDIX A Review of Prerequisite Skills

109

Exponent Rules II

To the Teacher Tips for Learning Math These first ten masters were designed to help students start the term with some insights into how 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 to suggest 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 goals they set.

Practice In 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 Skills In Appendix A, after the Practice masters for Chapters 1 to 6, there are 37 masters for Review of Prerequisite 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 students to 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 follow these steps: Windows: Navigate to your CD-ROM drive using Windows Explorer or My Computer; locate “ar405eng.exe” in the “WIN” directory and double-click its icon. Macintosh: If not already visible, open the CD-ROM in a window on your desktop by doubleclicking the CD-ROM’s icon. Within that window, locate “ar405eng” in the “MAC” folder and double-click to launch. After the installer launches, follow the on-screen instructions to complete the installation. When the installation is complete, restart your computer. The program is now ready to use. If you have questions regarding the use of Acrobat Reader, click the Help button for online help. To edit the MATHPOWER™ 10, Ontario Edition, Practice Masters CD-ROM requires the full Adobe® Acrobat® Writer program. Adobe® Acrobat® Reader 4.0 System Requirements Windows I486™ or Pentium® processor-based personal computer Microsoft® Windows® 95, Windows 98, or Windows NT® 4.0 with Service Pack 3 or later 10 MB of available RAM on Windows 95 and Windows 98 (16 MB recommended) 16 MB of available RAM on Windows NT (24 MB recommended) 10 MB of available hard-disk space 50 MB of additional hard-disk space for Asian fonts (required for Acrobat Reader 4.0 CD-ROM, otherwise optional) Macintosh Apple Power Macintosh or compatible computer MAC OS software version 7.1.2 or later 4.5 MB of available RAM (6.5 MB recommended) 8 MB of available hard-disk space 50 MB of additional hard-disk space for Asian fonts (required for Acrobat Reader 4.0 CD-ROM, otherwise optional)

vi

To the Teacher


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Getting Started MATHPOWER™ 10, Ontario Edition

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

6.

I plan to take Math

a)

never after this course

7.

My attitude toward Math is

about the same as

b)

getting worse

c) getting better

b)

to the end of high school

c)

at university or college

c)

about the same as

a) more positive than b) less positive than the attitude I had when I was in grade 7. 8.

c)

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,


Tips for Learning Math

1

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Know Your Textbook and How to Use It MATHPOWER™ 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 subjects and other Math topics. ________________ 8. Communicate Your Understanding means that I have an opportunity to demonstrate my understanding of a topic ________________ 9. Core sections are numbered and usually include an Investigation to actively involve me in my own learning. ________________ 10. There are Cumulative Review sections, at the end of Chapters 2, 4, and 6 that I can use when studying. ________________ 11. Detailed instructions involving Technology are provided in two appendixes at the back of the text to help me recall essential skills. ________________ 12. A Glossary is found near the end of the book to help me understand Math . ________________ 13. I can use the Review of Key Concepts sections and Chapter Test sections to test myself. ________________ 14. a) b) c) d) e)

Circle the letter of each thing you can learn about when using MATHPOWER™ 10, Ontario Edition. how Math is connected to other disciplines how people use Math in their careers how to use problem solving strategies to tackle problems how to use technology, such as calculators and computer spreadsheets how different topics in Math are interconnected

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

2



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Classroom Learning MATHPOWER™ 10, Ontario Edition

Whether your Math class is 30 min or 70 min, you will gain the most from that time with your Math teacher and your classmates, if you • come to class prepared • listen actively • take clear notes • use your time productively 1. a) c) e) g) i) k)

Circle the letter of each thing that you bring to Math class. textbook, MATHPOWER™ 10, Ontario Edition b) pens and/or pencils d) ruler f) diskette h) completed and checked homework j) questions to ask about concepts that confuse me l)

notebook eraser calculator other tools _____________________ a positive attitude enthusiasm

2. Circle the letter of each way that you actively listen in Math class. Place an X beside the letter of each way you would like to try. a) anticipating what is coming next b) trying to connect new concepts with familiar concepts c) sketching diagrams to illustrate my understanding d) asking questions to clarify my understanding e) identifying when the teacher says something particularly important 3. a) b) c) d) e) f)

Circle the letter of each way your Math teacher points out important information. says it directly implies with tone of voice repeats writes it on the board highlights it on the board, in a special place or in a special way other ways ____________________________________________________________________________

4. Circle the letter of each technique that you use to make your Math notes clear. Place an X beside the letter of each technique you would like to try. a) record date b) record textbook page references c) start a new page d) write legibly e) make lists f) draw diagrams g) highlight key points h) other techniques _______________________________________________________________________ 5. a) b) c) d)

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



3

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Homework MATHPOWER™ 10, Ontario Edition

In a subject like Math, where the concepts build day after day and year after year, it is vital that you 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. a) b) c) d) e) f) g) h) i) j) k) l) m)

Circle the letter of each thing that competes with Math homework for your time. other homework special assignments or projects sports practices/games music lessons/practice watching television or videos listening to the radio playing video games or playing cards visiting with friends talking on the telephone household chores part-time job shopping others _____________________________________________________________________

2. Circle the letter of each work space that you use for doing your Math homework. Place an X beside the letter of a space you would like to try. a) own room b) a shared room c) the kitchen d) the dining room e) another room at home f) the library g) 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

4


by doing this


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Seeking Help MATHPOWER™ 10, Ontario Edition

In a subject like Math, where the concepts build day after day and year after year, it is vital that you keep up daily. When you do not understand something, seek help immediately. Use your textbook, MATHPOWER™ 10, Ontario Edition. The Examples in the numbered sections are 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 your teacher’s time productively. Identify the point at which you first became confused. Have specific questions about what you don’t understand. 1. a) b) c) d) e) f) g)

Find out when your Math teacher is available, and circle the letter of each time that applies. during Math class before Math class after Math class before school after school during lunch 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 letter of each way you would like to try. a) working in a group or with a partner in Math class b) discussing problems with friends taking the same Math course, but in a different class c) discussing problems with friends or family who have already taken this level of Math d) working in a study group or with a study partner outside of Math class e) other ways _______________________________________________________________



5

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Preparing for Tests and Exams MATHPOWER™ 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 informed of it. Place an X beside the letter of each thing you are going to start to find out about. a) exact date, time, and location b) exact content to be covered c) time allowed for it d) 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 calculator f) value of it, relative to all tests, assignments, projects, and exams g) 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 and exams. Place an X beside the letter of each resource you would like to try in the future. a) my notes b) Examples and Solutions in my textbook c) Achievement Check Review, and Chapter Test sections in my textbook d) Cumulative Review sections in my textbook e) MATHPOWER™ 10, Ontario Edition, Self-Check Assessment Masters f) teacher-provided old or sample tests or exams g) a study group or partner h) 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 using the Self-Check Assessment Master that you can get from your teacher for the appropriate chapter. Next, check your results using the Answers to the Self-Check, which you can also get from your teacher. Then, for any question you got wrong, use the table on the second page of the SelfCheck to identify the textbook section and the specific worked example. Review the example and try the question again.

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Writing Tests and Exams MATHPOWER™ 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 mind sharp. Place an X beside the letter of each thing you would like to start doing. a) getting a good sleep the night before b) eating a nutritious breakfast c) planning ahead and being prepared in order to be relaxed when I begin d) 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 the past. 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 experiences c) exercising d) eating nutritious meals and snacks e) learning deep breathing and other relaxation techniques f) getting adequate, but not too much, sleep g) other techniques _________________________________________________________________________ Just as in Math class, when you write a test or exam, you want to be prepared and use your time wisely. 3. Circle the letter of each strategy that you have used in the past to help you use the time for a test 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 use b) reading the entire test quickly to get an overview and opening my mind to ideas that could help with later questions c) using the marking scheme to help me determine the relative importance of questions d) pacing myself, not rushing, but getting down to work e) reading the instructions, questions, and diagrams or graphs carefully to make sure I answer what is asked f) identifying the questions that I am most confident about, and doing them first g) if I do questions out of order, checking them off and numbering them carefully to avoid confusing myself and my teacher h) predicting or estimating answers, and recording them to check against my final answers i) checking the reasonableness of my answers j) showing all my work, step by step k) writing legibly l) marking questions that I want to return to m) in multiple-choice questions, ruling out obviously wrong choices n) if my answer is not the same as one of the choices in a multiple-choice question, working backward from choices that I have not ruled out



7

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Problem Solving Skills MATHPOWER™ 10, Ontario Edition

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

Understand the Problem

Answer the following questions in your notebook. Understand the Problem 1. How many games would have been played if each competitor played each of the other competitors only once? 2. Explain how answering question 1 allows you to proceed with solving the problem. Think of a Plan Consider just one game between competitors. Try a simpler problem, draw diagrams, and look for a pattern. 3. Explain what each tree diagram shows.

Think of a Plan

Diagram 1

A

B

Diagram 2

A

B B C

C

Diagram 3

A

B C B D

C D

C

D

A

B C B D E

C D E

C

D E

Diagram 4

Carry Out the Plan

Look Back

8

D

E

Carry Out the Plan 4. Draw the next diagram. 5. Think about how the number of competitors is related to the number of games. 2 to 1 3 to 3 4 to 6 5 to 10 Ask yourself: a) Is some value being added to, subtracted from, multiplied by, or divided into 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 20 I could return to the actual second 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 games with each competitor playing each of the others once (or 63 games with each competitor playing each of the others three times). 8. Write a final statement to answer the original question. Look Back 9. Solve the problem another way.



Name

Solving Word Problems MATHPOWER™ 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 problems that required creating and solving an equation from given information. 3x = 8(x – 5) 3x = 8x – 40 –5x = –40 x=8 Problem Plane 1 flew from Calgary to Montreal at 750 km/h. Plane 2 flew the same route at 600 km/h and 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 reasonable to expect that you should create and solve an equation from the information given in order to solve the problem. In this problem, you will need to use the relationship: distance = speed × time. This table showing the known values can help you get organized. Plane

Distance (km)

Speed (km/h)

1

750

2

600

Time (h)

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.



9

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Goals MATHPOWER™ 10, Ontario Edition

You can put what you have read in these Tips for Learning Math pages into practice. Set a goal for 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 ed by actions. I will always be prepared for class. I will do my homework daily. I will 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 I should 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 I won’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.

10



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1.1

Investigation: Ordered Pairs and Solutions MATHPOWERTM 10, Ontario Edition, pp. 4–5

• To that an ordered pair satisfies an equation in two variables, a) substitute the values for x and y in the original equation b) evaluate the left side of the equation and the right side of the equation c) check that L.S. = R.S. • To that an ordered pair satisfies a system of equations, a) substitute the values for x and y in each equation b) evaluate L.S. and R.S. in each equation c) check that L.S. = R.S. in each equation 1. Which of the four ordered pairs satisfy the equation? a) 3x + y = –4 (0, –4),  − 4 , 0 , (–1, 1),  3  (1, –7) b) y =

1 x + 19 2

c) 2x – y + 17 = 0

d) 2y = 3x

(2, 21), (0, 19), (–4, 17), (6, 22)

1 d) y = − x 4 4 y = 3x − 16

(0, 17), (4, –25), (–8.5, 0), (–5, –7)

(8, ) satisfies the second equation but not the first.

(0, 0),  1 , 1  ,  1 , 1  ,  2 3  3 2 (–6, –9)

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

3 x–1 5

x + y = –10.5 y = 2x – 3  5  − , satisfies both equations.  2 

c)

(5, ), (–5, ),  1 , 3

 , ( , 0)

c) y + 3x = 13

( , 4), (0, ), ( , 0), (–1, )

d) 7x = 2y

( , 0),  , 1  , (2, ), (–1, )  2

3. Complete each ordered pair so that it meets the stated condition. a) x + 2y = 4 y = 2x – 3 ( , 0) satisfies the first equation but not the second. b) 3x + 1 = y 2y – 6x = 2 ( , ) satisfies neither equation.


4. Problem Solving To run a 30-s ad during a weekday morning, one radio station charges a fixed cost of $400, plus $150 for each day the ad is run. A second radio station charges a fixed cost of $300, plus $200/day. The costs can be modelled by the following equations, where C is the total cost and n is the number of days for which the ad is run. station 1: C = 150n + 400 station 2: C = 200n + 300 a) Find the missing element in the ordered pair ( , 700) that satisfies both equations and is in the form (n, C). b) Communication Explain what your ordered pair from part a) means for both radio stations.

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

Chapter 1

11

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1.2

Solving Linear Systems Graphically MATHPOWERTM 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 pencil b) determine the coordinates of the point of intersection c) check the solution by substituting it in each of the original equations • The number of solutions to a linear system is a) exactly one, if the lines intersect b) none, if the lines are parallel and distinct c) infinitely many, if the lines coincide 1. Solve by graphing. Check each solution. a) y = x – 3 b) y = 2x – 1 y=3–x y = –x + 5

x + 3y = 2 x–y=6

d) 3x + 2y = 6 y=4–x

e)

2x – 5y = 10 x + 3y = –6

f)

g)

2x + 3y = –1 3x + 5y = –3

h) x + y = 8 x–y=3

c)

i)

3x – 4y = 12 2x + 5y = 4

j)

1 x −1 2 3x − 6 y + 6 = 0 y=

6x + 3y = 11 0.2x + y = –0.6

2. Communication Without graphing, determine whether each system has one solution, no solution, or infinitely many solutions. Explain your thinking. a) y = 2x – 5 4x – 2y = 10

b) 3x – y = 17 6x + 2y = –8

c)

12x + 8y + 4 = 0 15x + 10y = 5

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

4. Problem Solving One equation of a system is 2x – y = 5. Write a second equation so that the system has a) no solution

b) infinitely many solutions k) 3.74x – y = 2 x= 5

12

Chapter 1

c) one solution, (3, 1)


Name

1.3

Solving Linear Systems by Substitution MATHPOWERTM 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 variables b) substitute the expression from part a) in the other equation and solve for the variable c) substitute the value of the variable found in part b) in one of the original equations to find the value of the other variable d) check the solution in each of the original equations • If the statement that results from a solution is a) not true for any value of a variable, the lines are parallel and distinct, and there is no solution b) true for all values of a variable, the lines coincide and there are infinitely many solutions 1. Solve each equation for the specified variable. a) x + 2y = 7; x b) 3x – y = 5; y

c)

4x + y – 11 = 0; y

3. Communication Solve the system by substitution. Describe the difficulty you encounter. 3x + 5y = 1 7x + 9y = 5

d) x + 6y + 2 = 0; x

2. Solve each system of equations by substitution. If there is exactly one solution, check the solution. a) x – y = 7 b) 4x + 3y = 7 2x + 5y = 21 3x + y = –1

4. Application Simplify each system, and solve by substitution. Check each solution. a) 3(x + 5) – 2(y – 1) = 6 x + 4(y + 3) = 13

b) 2(x – 1) – (y – 7) = 13 4(x + 2) + 3(y + 1) = 37 c)

p + 6q = 11 4p – q = –6

d) 5m – 4n = 11 3m = n + 15

e)

7s + 5t = 13 t – s = –1

f)

11x – y = 21 9x + 2y = –11

g)

x – 2y = 5 4x = 8y + 20

h) 6x + 2y = 5 y = 11 – 3x


5. Problem Solving The perimeter of an envelope is about 68 cm. The length is equal to 2.4 times the width. Use the equations 2l + 2w = 68 and l = 2.4w to find the dimensions of the envelope.

Chapter 1

13

Name

1.4

Investigation: Equivalent Equations MATHPOWERTM 10, Ontario Edition, pp. 24–25

• To write an equation equivalent to a given equation, multiply each term in the given equation by the same 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) OR b) 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) 1. Communication Are the equations equivalent? Explain. a) 2x – y = 3 y = 2x – 3 b)

1 3 3y = 6x + 1 y = −2 x +

c)

y = 5x – 2 10x – 2y – 4 = 0

d)

x−y=

2 3

−x + y =

3 2

2. Write three equivalent equations for each given equation. a) x + y = 0.4 b) 4x – 3y = 7

c)

1 x − 4y = 8 3

d)

2 1 y − x = −1 3 2

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

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

c) Write the missing numbers in system L that make it an equivalent system. System L x = _____ y = _____ 5. Add the equations in system K in question 4. Use your result to write two other equivalent systems.

d) 0.5y = 2x – 1.3

3. Application Describe an easy way to write an equivalent equation for each given equation. a) 1 x − 2 y = 7 3 3 b) 0.2y – 3.1x = 5.1

c)

22x – 33y = 11

14

Chapter 1

6. Write three different systems of equations that are equivalent to system A. Write each equation in two variables. System A x = –2 3 y= 2


Name

1.5

Solving Linear Systems by Elimination MATHPOWERTM 10, Ontario Edition, pp. 26–33

• To solve a linear system in two variables by elimination, a) clear decimals and fractions, if necessary b) rewrite the equations with like in the same column, if necessary c) multiply one or both equations by numbers to obtain two equations in which the coefficients of one variable are the same or opposites d) add or subtract the equations to eliminate a variable, and solve the resulting equation for the remaining variable e) substitute the value from d) into one of the original equations to find the value of the other variable f) check the solution in each of the original equations 1. Solve each system of equations by elimination. Check the solution if there is exactly one solution. a) 3x + y = 17 b) 5x – 3y = 19 2x – y = –2 5x + 4y = 33

c)

6x + 7y = 23 2x + 7y = 31

d) 4x + 9y = 2 4x – 3y = 10

e)

x + 2y = –4 3x – 4y = 18

f)

8x + 5y = 12 2x + 3y = 10

g)

2x – 5y = 29 x – 2.5y = 14.5

h) 3x – 4y = 5 5x + 3y = –11

i)

7a + 3b = 47 2a + 5b = 30

j)

3x – 8y = 2 5x + 3y = 1


k)

x y + =6 6 4 5x y − = 11 6 3

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 3x + 4y = 19

d) 2x – 5y = 0 y = 1.4x + 0.8

3. Application Find the coordinates of the vertices of a triangle whose sides lie on the following three lines. 4x – 3y = –18 3(x + 4) + 2(y + 3) = 30 y = 7x – 11

Chapter 1

15

Name

1.6

Investigation: Translating Words Into Equations MATHPOWERTM 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 unknowns b) determine how the unknowns are related c) write an equation that shows the relationship between the unknowns 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 each situation. a) The sum of the two numbers is negative five. b) The difference between the larger number and the smaller is twelve. c) Three times the smaller number subtracted from twice the larger gives a result of nine.

5. A dry cleaner charges $5 to alter a pair of pants 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 of pants, p, and the number of jackets, j, to the total number.

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

6. Application Introduce variables and write each of the following as a system of equations in two variables. a) Two angles are complementary. The measure of one angle is 10° more than twice the measure of the other angle.

d) The sum of half the larger number and one-third the smaller number equals zero. 3. Write a system of equations for each pair of relations. a) x y x y b) x y x y 2 –5 6 1 3 9 3 1 0 –3 3 –2 1 3 1 2 –2 –1 0 –5 0 0 2 –3 0 –1 –6 –3 –9 1 0 1 0 − 2

4. Communication Describe in words each of the four relations in question 3.

16

Chapter 1

b) The length of a rectangle is 8 cm less than three times the width. Half the perimeter is 60 cm.

c) A square and an equilateral triangle have lengths of sides such that the sum of the perimeters is 105 cm. The length of a side of the equilateral triangle is half the length of a side of the square.


Name

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 unknowns b) determine how the unknowns are related c) write a system of equations that shows relationships between the unknowns d) solve the system of equations e) check the solution, using the facts given in the problem 1. Communication Chris has $3.85 in dimes and quarters. There are 25 coins in all. How many of each type of coin does he have? a) If d represents the number of dimes and q represents the number of quarters, what two equations describe the problem?

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

litres of the 30% and of the 35% solution will be used?

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

Problem Solving 2. A supermarket sells 2-kg and 4-kg bags of sugar. A shipment of 1100 bags of sugar has a total mass of 2900 kg. How many 2-kg bags and 4-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 were washed on Saturday, and the amount earned was $475. How many vans were washed on Saturday?

4. In hockey, a team receives 2 points for a win and 1 point for a tie. During a hockey season of 60 games, the Rockets lost 28 games but earned 51 points. How many games did the team win?

5. A lab technician needs to combine some 30% alcohol solution and 35% alcohol solution to make 5 L of 33% alcohol solution. How many


7. To attend a wedding, Barbara starts driving west from Woodstock at 80 km/h. One hour later, Barbara’s parents leave Woodstock and drive along the same road at 100 km/h. At what distance from Woodstock will Barbara’s parents her?

8. The manager of a bulk food store mixed some jellybeans that cost $1.99/kg with gumdrops that cost $2.99/kg to form 50 kg of a mixture that cost $2.23/kg. How many kilograms of each type of candy were in the mixture?

9. What are the values of x and y?

(4x + y)° (4x – 3y)° 70°

Chapter 1

17

Answers

CHAPTER 1

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

1.1 Investigation: Ordered Pairs and Solutions

4. a) (–3, 1)

1. a) (0, –4),  − 4 , 0 , (1, –7)  3 

5. width is 10 cm and length is 24 cm

b) (0, 19), (–4, 17), (6, 22) 1 d) (0, 0),  , 3 4 b) 2, –4, − , 5

c) (0, 17), (–8.5, 0) 2. a) 6, 0, –3, 8

3. a) 4

1.4 Investigation: Equivalent Equations

1 , (–6, –9) 2 5 3

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

7 1 d) 0, , 7, − 2 7

13 c) 3, 13, , 16 3

b) (5, 2)

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

b) Answers may vary. (0, 0)

c) –8

d) 2

multiply each term by 2

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

d) no; multiply equation (1) by –1 to get − x + y = −

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

2. Answers may vary.

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

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

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

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

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 h)  11 , 5  i)  76 , − 12   2 2  23 23 

g) (4, –3) j)  64 , − 29   27 27 

2 3

c) x – 12y = 24, 2x – 24y = 48, 3x – 36y = 72 13 d) 5y = 20x – 13, y = 4x – –– , 2y = 8x – 5.2 5 3. Answers may vary. a) multiply by 3 to get x – 2y = 21 b) multiply by 10 to get 2y – 31x = 51

k) (5, 16.7)

c) divide by 11 to get 2x – 3y = 1

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

d) multiply by 6 to get 4y – 3x = –6

infinitely many solutions.

4. a) (0, –4)

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

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

coincide and there is one solution. 3 c) Both lines have a slope of − and the y-intercepts 2 are different, so there is no solution.

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

3. (5, 1)

5. Order of equations may vary.

4. Answers may vary.

System M –8x + 2y = –8 –2x – y = 4

a) y = 2x + 5

b) y = 2x – 5

c) x + y = 4

1.3 Solving Linear Systems by Substitution 1. a) x = 7 – 2y

b) y = 3x – 5

c) y = 11 – 4x

d) x = –6y – 2 2. a) (8, 1) e)  3 , 1   2 2

b) (–2, 5)

c) (–1, 2)

d) (7, 6)

b) same, because the equations are

corresponding equations in system K c) 0, –4

System N –8x + 2y = –8 3y = 6x – 12

6. Answers may vary. System B 1 x+y=− 2 x = –2

System C 1 x+y=− 2 3 y= 2

System D 2x + 2y = –1 x = –2

f) (1, –10)

g) infinitely many solutions

h) no solution


Chapter 1

19

1.5 Solving Linear Systems by Elimination 1. a) (3, 8)

b) (5, 2)

e) (2, –3) i) (5, 4)

c) (–2, 5)

1.7 Solving Problems Using Linear Systems

d)  2, − 2   3

f) (–1, 4)

g) no solution

j)  2 , − 1  7 7

k) (18, 12)

h) (–1, –2)

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.

l) (1, 6)

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

m) (–5, –2)

The correct solution is 9 quarters and 16 dimes.

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

2–9. The variables used may vary.

coefficient of y in the second equation is 1.

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

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

350 4-kg bags

then substitute this value to find x.)

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

c)  119 , 115  ; Use graphing or elimination. (Solve  43 43 

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

for one variable; then, instead of substituting this value, solve for the other variable.)

5. 0.3a + 0.35b = 1.65 and a + b = 5; 2 L of 30% alcohol solution and 3 L of 35% alcohol solution

elimination. (Multiply the second equation by 5.)

5040 5040 and p − w = ; 80 km/h 7 9 p b 7. b + 80 = p and = ; 400 km 80 100

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

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

d)  − 4 , − 8  ; Use graphing or substitution or  5 25 

6. p + w =

38 kg of jellybeans and 12 kg of gumdrops

1.6 Investigation: Translating Words Into Equations 1. a) m + n

b) 0.075n

9. (4x – 3y)° = 70° and (4x + y)° + 70°= 180° or (4x + y)° + (4x – 3y)° = 180°; x = 25, y = 10

c) 0.09m + 0.075n

2. a) x + y = –5

b) x – y = 12 1 1 c) 2x – 3y = 9 d) x+ y=0 2 3 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 e =

20

Chapter 1

1 s 2


Name

2.1

Length of a Line Segment MATHPOWERTM 10, Ontario Edition, pp. 66–73

• To find the length of a line segment ing (x1, y1) and (x2, y2), use the formula l = ( x 2 − x1 ) 2 + ( y 2 − y 1 ) 2 . • An equation of the circle with centre O(0, 0) and radius r is x2 + y2 = r2. 1. Determine the length of the line segment ing each pair of points. Express each length as an exact solution and as an approximate solution, to the nearest tenth. a) (3, 7) and (–1, –5)

5. Communication Explain why POR is a right triangle. y P(3, 3) 2

O(0, 0) –2

0

x

2

–2

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

R(4, –4)

Applications 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, if necessary.

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

b) A( −1,

27 − 1), B(–4, –1), C(2, –1)

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), lie on a straight line. Show that B is the midpoint of AC.

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

9. a) that the quadrilateral with vertices D(2, 6), E(3, 3), F(–3, 1), and G(–4, 4) is a rectangle. 4. Find the perimeter of parallelogram ABCD. y A(–2, 2)

–2

B(6, 2)

2

0

2

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

x

–2

D(–5, –2)

C(3, –2)


Chapter 2

21

Name

2.2

Investigation: Midpoints of Horizontal and Vertical Line Segments MATHPOWERTM 10, Ontario Edition, p. 74

• To find the midpoint, M, of a horizontal line segment ing (x1, y) and (x2, y), use the formula  x1 + x 2  , y .  2  • To find the midpoint, M, of a vertical line segment ing (x, y1) and (x, y2), use the formula  y1 + y 2  x, .  2  5. Count units to find the coordinates of the midpoint of each line segment. y a) y b) C(–9, 8)

1. Count units to find the coordinates of the midpoint of each line segment. y y a) b) 2

A(5, 4)

B(6, 4)

A(–4, 4)

2 –2 0

2x

–2 0

2

–2

x

2

0

–2

x

2

2

–2

C(–8, –7)

D(–1, –7)

–2 0

2x

–2

c)

d)

y E(–3, 2.5) 2 F(2, 2.5) –2 0

2

D(–9, –5)

y B(5, –8)

10

G(0, 4)

H(24, 4)

c)

x

–2

0

10

y

x

E(0.5, 2.5)

d)

y

2

0 x

–10

G(–14, –4) –10

2. Given the endpoints of each line segment, find the coordinates of the midpoint without plotting 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) 1 1  1 d) T 5 , 1 and Q  −1, 1   2   2 2 3. CD is a line segment ing the points C(–13, –3) and D(–1, –3). Find the coordinates of the three points that divide CD into 4 equal parts.

–2

0

–2

2

x

–10

F(0.5, –2)

H(–14, –26)

6. Given the endpoints of each line segment, find the coordinates of the midpoint without plotting 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) 1 1 1 1 d) W  , 1  and Z  , 6  4 4 2 2 7. HI is a line segment ing the points H(1, –11) and I(1, 5). Find the coordinates of the three points that divide HI into 4 equal parts.

4. Communication The midpoint of a horizontal line segment is M(2, –2). What coordinates are possible for the endpoints of the line segment?

22

Chapter 2

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


Name

2.3

Midpoint of a Line Segment MATHPOWERTM 10, Ontario Edition, pp. 75–80

• To find the midpoint, M, of a line segment ing (x1, y1) and (x2, y2), use the midpoint formula,  x1 + x 2 y 1 + y 2  . ,  2 2  1. Determine the midpoint of each line segment with the given endpoints. a) (–6, 2) and (4, 8)

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

b) (1.5, 3) and (–6, –2.5)

c) (–200, –100) and (350, 600) d)  7 , 1  and  − 5 , 3   2 4  2 4

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

e) (3a, 2b) and (–3a, 5b)

f) (–6a, 5b) and (11a, 0) 2. Find the midpoints of the sides of DEF. D(–2, 4)

y

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 each side.

2 –2 0

2

x

–2

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

F(–8, –4) E(5, –7)

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

c) Find the perimeter of the square formed by ing the midpoints of the sides of square KLMN.

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


Chapter 2

23

Name

2.4

ing Properties of Geometric Figures MATHPOWERTM 10, Ontario Edition, pp. 88–99

• The following formulas can be used to determine characteristics of geometric figures and to geometric properties. y2 − y2 Slope of a line segment: m= x 2 − x1 Length of a line segment:

l = ( x 2 − x1 ) 2 + ( y 2 − y 1 ) 2

Midpoint of a line segment:  x1 + x2 , y1 + y 2   2 2  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 1. Communication For any three points A, B, and C, not in a line, M and N are the midpoints of AB and AC, respectively. How can you prove 1 that MN || BC and MN = BC ? 2

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

3. The vertices of a quadrilateral are S(1, 2), T(3, 5), U(6, 7), and V(4, 4). each of the following. 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 to each other.

24

Chapter 2

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 the altitudes. c) that this point (the orthocentre of the triangle) 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. that the triangle is an isosceles right triangle.

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


Name

2.5

Distance From a Point to a Line MATHPOWERTM 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 line b) find the coordinates of the point of intersection of the perpendicular and the given line c) use the distance formula 1. Communication Explain how to find the shortest distance from the point P(–5, 7) to the line x = 3.

4. Find the shortest distance from the given point to the given line. Round to the nearest tenth, if necessary. 2 a) (0, 0) and y = − x + 5 3

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

b) (0, 0) and 15x – 8y – 29 = 0

a) y =

1 x − 7 ; ( 2 , 5) 2

c) (–4, –5) and y =

5 x−4 3

3 b) y = − x − 2; ( −1, − 7 ) 2

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

c) 4x – 3y – 7 = 0; (–5, 2)

Applications

d) 2x + 5y + 3 = 0; (3, –4)

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

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

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

b) Find the exact length of BC. b) (0, 0) and 5x + 12y – 39 = 0

1 c) (3, –2) and y = − x + 9 3

c) Use your answers to parts a) and b) to find the area of ABC.

d) (5, –2) and 4x + 3y + 14 = 0


Chapter 2

25

Answers

CHAPTER 2

Analytic Geometry

2.1 Length of a Line Segment 1. a)

160 ; 12.6

b)

2.3 Midpoint of a Line Segment 1. a) (–1, 5)

61 ; 7.8

1 1 d)  ,   2 2

2. 14.8 3. a) isosceles; 17.7

b) equilateral; 18

c) (75, 250)

e) (0, 3.5b)

5 5 f)  a, b 2 2 

2. (–5, 0), (1.5, –1.5), (–1.5, –5.5)

4. 26 5. length of PQ =

b) (–2.25, 0.25)

18 , length of QR =

32 , length

3. Let the coordinates of E be (x1, y1). Substitute the values into the midpoint formula,

of PR = 50 ; PQ 2 + QR 2 = 18 + 32 = 50 = PR 2 Thus, by the Pythagorean Theorem, PQR is a right triangle.

( 3.5, 1.5) =

 x1 + 5 y1 + ( −7 )  , . Then, solve for x 1  2  2

and y1 using these equations:

6. 15 square units

x1 + 5 = 3.5 and 2

y1 + ( −7 ) = 1.5. ( x1 , y1 ) = (2, 10) 2

7. AB = 34 ; BC = 34 8. 5

4. (–0.5, 10)

9. a) DE = GF = 10 , FE = GD = 40 ,

5. P(0, –4)

DE2 + DG2 = 10 + 40 = 50 = GE2

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

Thus, ∠EDG = 90°.

7. U(–2, 6), W(2, –2)

b) 7.1

2.2 Investigation: Midpoints of Horizontal and Vertical Line Segments 1. a) (1, 4) c) (–0.5, 2.5) 2. a) (–7, 3)

b) (–4.5, –7) d) (12, 4) b) (5.5, 0)

1 1 c) (6.4, –5) d)  2 , 1   4 2 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) c) (0.5, 0.25) 6. a) (13, 6) c) (–7, –2.5)

b) (–9, 1.5) d) (–14, –15) b) (0, –1) 1 d)  , 4 4 

7. (1, 1), (1, –3), and (1, –7) 8. Answers may vary. (–3, 0) and (–3, –10), or (–3, 3)

2.4 ing Properties of Geometric Figures 1. 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 3. a) slopes of ST and VU = ; slopes of SV and 2 2 TU = 3 b) The midpoint of both SU and TV is  7 , 9  .  2 2 c) ST = TU = UV = VS = 13 d) slope of TV = –1; slope of SU = 1 4. a) from A to BC: y = 3x – 4; from B to AC: 1 9 2 5 y = x + ; from C to AB: y = − x + 2 2 3 3 b)  17 , 23   7 7

and (–3, –13). There are an infinite number of pairs of coordinates on the line x = –3 that are on opposite

c)

 17 23  , satisfies the equation of the other altitude  7 7

sides of M(–3, –5). Copyright © 2001 McGraw-Hill Ryerson Limited

Chapter 2

27

5. vertices: A(–2, 3), B(1, 5), C(–4, 6); AB = AC = 13 ; 2 3 slope of AB = , slope of AC = − 3 2 2 6. a) PQ = 72 , slope = 1; QR = 52 , slope = − ; 3 2 RS = 116 , slope = – ; SP = 8 , slope = –1 5

b) A(0, 4), B(6, 5), C(4, 1), D(–2, 0) 1 c) slopes of AB and DC = ; slopes of AD and 6 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 ing P(–5, 7) and Q(3, 7) is 8. 2. a) y = –2x + 9 or 2x + y – 9 = 0 2 19 b) y = x − or 2x – 3y – 19 = 0 3 3 3 7 x − or 3x + 4y + 7 = 0 4 4 5 23 d) y = x − or 5x – 2y – 23 = 0 2 2

c) y = −

3. a)

20

4. a) 4.2

90

b) 3

c)

b) 1.7

c) 2.9

d)

28 5

d) 9.9

5. 12 5 6. a)

28

32 13

Chapter 2

b)

52

c) 32


Name

3.1

Polynomials MATHPOWERTM 10, Ontario Edition, pp. 128–133

• To add polynomials, collect like . • 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. 1. Classify each polynomial by degree and by number of . a) 3x2 – 2x b) 4a2b3

c) 8 + 2y4 + 3y3

5. Simplify. a) (6x)(2x2)

b) (5pq2)(–4p2q2)

c) (3ab)(–2ab2)(2a3)

d) (–6x2yz)(–5y3z)

d) 4x5 – 2x3 + x2 + 4

2. Evaluate each expression for the given value(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 order of x. a) 6 + 4x2 – 5x5 + 3x – 2x2

e)

15x 6 5x 2

f)

24 a 3 b 2 −3 a 2 b

g)

−21x 2 y 2 z −7 xy 2 z

h)

−32 p 2 q 4 8p 2 q3

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

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

b) 3x2y4 + 4x4y2 – x3y3 + x5y – 2xy5 b) 3x(x2 + 2x – 2) + 2x(3x2 – x – 4) 4. Simplify. a) (6y – 2) + (2y + 8)

b) (a + 2b) + (3a – 4b)

c) (8 + 6x) – (9 + x)

d) (x + y) – (x – y)

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

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

e) (3x2 + 2x – 6) + (2x2 – 4x + 7)

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

2

2

g) (3y – y – 6) – (2y + 5y – 7)


9. Problem Solving For the rectangular prism, write an expression that represents a) the volume b) the surface area

2y

4y

3x

Chapter 3

29

Name

3.2

Multiplying Binomials MATHPOWERTM 10, Ontario Edition, pp. 134–139

• To find the product of two binomials, use either of the following. a) the distributive property b) FOIL, which stands for the sum of the products of the First , Outside , Inside , and Last • the product of two binomials by substituting a convenient value for the variable in the original product and in the simplified expression. 1. Communication Explain how the diagram models the product.

2x x + y

+3

2x 2

3x

2xy

3y

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

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

c) –2(y – 3)(y + 2) d) 0.2(x + 1)(x + 2)

b) (2 + k)(3 + k) e) 3(6x – 2y)(2x – 3y) c) (c – 5)(c – 3) f) (x + 3)(x + 2) + (x + 4)(x + 1) d) (t + 5)(t – 1) g) (y – 4)(y – 3) – (y – 2)(y + 5) 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)

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 an expression to represent the area of the figure.

4 x

x–1 x+6

l) (5h – 1)(2h – 3) m) (3 – 2s)(2 – 3s) n) (4 + 2p)(–3 – 4p)

5. Write and simplify an expression to represent the area of the shaded region.

2y + 1 2y – x 2y

2y + x

o) (–2t – r)(–3t + r)

30

Chapter 3


Name

3.3

Special Products MATHPOWERTM 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 , use the following pattern. (a + b)(a – b) = a2 – b2 1. Expand. a) (x + 4)2

3. Application Complete the table. b) (y – 7)

2

a) c) (m –2)(m + 2)

d) (x – 5)(x + 5)

e) 2(6x – 3)

(a + b)(a – b)

25 × 15

(20 + 5)(20 – 5)

Product

(30 + 6)(30 – 6)

b) c)

2

Numbers

27 × 33

2

f) 3(5 + 4t)

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)

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 a cube is represented by the expression 3x – 2y. a) Write, expand, and simplify an expression for the surface area of the cube.

f) (2a – 1)(2a + 1) – (a – 3)(a + 3) b) If x represents 4 cm and y represents 3 cm, calculate the surface area, in square centimetres.


Chapter 3

31

Name

3.4

Common Factors MATHPOWERTM 10, Ontario Edition, pp. 147–152

• To factor a polynomial with a common monomial factor, remove the greatest common factor of 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 with a common factor. 1. Factor, if possible. a) 4x + 28

b) 3x + 17

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

c) 6x – 32y

d) 26x2 – 13y

b) y2 – x + y – xy

e) 2ax + 10ay – 8az

f) 2a2 – 6a – 15

c) ab + 9 + 3a + 3b

g) 8x2 + 32y3

h) 10y – 5y2 + 25y3

d) t2 – tr + 4r – 4t

i) 14rst + 7rs – 6t

j) 36xy – 12x2y

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

k) 4ab2 + 2a2c + 5b2c2

f) 3x2y – 6x2 – 2y + y2

l) 3x3y2 – 12x2y3 + 18x2y + 15xy2

g) 4ab2 – 12a2b – 3bc + 9ac

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

4. Problem Solving Write an expression for the area of each shaded region in factored form. a) 3x

b) 5y(z + 3) + x(z – 3) 2y 3x

c) 4t(r + 6) – (r + 6) b) y+1 3y

d) 7(a + b) – 2x(a + b) 4x 2

e) 2x(3m – 5) – 3(5 – 3m)

c) 4r

32

Chapter 3


Name

3.5

Factoring x2 + bx + c MATHPOWERTM 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 factor b) write the second , which are two numbers whose sum is b and whose product is c • When factoring a trinomial, first remove any common factors. 1. Factor, if possible. a) x2 – 5x + 6

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

b) y2 + 2y – 3

b) 3x2 + 9x – 12

c) m2 + 7m – 12

c) 5x2 – 35x + 50

d) a2 + 6a + 5

d) 4x2 – 16x – 48

e) x2 – 9x – 10

e) 2x2 – 16x – 66

f) b2 – 7b + 10

f) x3 – 13x2 + 42x

g) y2 – 6y + 7

4. Application The area of a doubles tennis court can be represented approximately by the trinomial x2 – x – 42. a) Factor x2 – x – 42 to find binomials that represent the length and width of a doubles tennis court.

h) x2 + x – 20 2. Factor, if possible. a) x2 + 24x – 52 b) m2 – 18m + 45 c) x2 + 5x – 36

b) If x represents 17.8 m, find the length and width of a doubles tennis court, to the nearest tenth of a metre.

d) x2 – 5xy – 66y2 e) m2 + 12mn + 32n2

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

f) 42 + y – y2 g) 32 + 4x – x2

b) x2 – kx + 6

h) x4 + 7x2 + 12


Chapter 3

33

Name

Factoring ax2 + bx + c, a ≠ 1

3.6

MATHPOWERTM 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 whose coefficients have a sum of b and a product of a × c b) group pairs of and remove a common factor from each pair c) remove the common binomial factor 1. Factor, if possible. a) 3y2 + y – 4

f) 6x – 2xy – 8y2 2

b) 3y + 5y + 1

g) 6m2 – 13mn – 5n2 2

c) 2a – 13a + 21

2

d) 4n + 7n – 5

h) 9x2 + 3xy – 20y2 2

2

e) 20x – 7x – 6

f) 18y + 15y – 18

i) 12a2 + 28ab – 24b2 2

2

g) 5x – 12x – 6

h) 8m + 6m – 20

2. Factor. a) 2x2 + 5xy – 2y2

b) 3y2 + 2yz – z2

c) 15x2 – 13xy + 2y2

3. Communication Describe how to factor 18a2 – 21ab + 6b2.

4. Application The area of a rectangular lot in a new housing development can be represented approximately by the trinomial 12x2 + 8x – 15. a) Factor the expression 12x2 + 8x – 15 to find binomials that represent the length and width of the lot.

d) 6m2 + 7mn + n2

2

2

e) 4a – 9ab – 9b

34

Chapter 3

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


Name

3.7

Factoring Special Quadratics MATHPOWERTM 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 1. Factor, if possible. a) x2 – 25

b) y – 49

c) y4 – 1

d) z2 + 64

2

e) 4a – 9

g) 169a2 – b2

i) 81x2 – 121p2

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

2

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

2

f) 49 – 64m

h) 24 + 4x2

j) 49 – (a – z)2

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

c) z2 – 9z + 9

c) 16a4 + 40a + 25

Applications 4. Evaluate each difference of squares by factoring. a) 382 – 322

d) 9t2 + 6t + 1 b) 552 – 452

e) 4m2 – 12m – 9

f) 4x2 – 20x + 25 c) 7602 – 2402

g) 121 – 22m + m2

h) 16x2 + 24xy + 9y2

i) 64a2 – 30ab + 49b2

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

c) 4x2 – 12x + k


d) kx2 – 40xy + 16y2

Chapter 3

35

Answers CHAPTER 3 Polynomials 5. (2y + 1)(2y + x) – (2y – x)(2y) = 4xy + 2y + x

3.1 Polynomials 1. a) degree 2, binomial

b) degree 5, monomial

c) degree 4, trinomial d) degree 5, polynomial of 4 2. a) 21

b) 22 5

3

2

3. a) –5x + 4x – 2x + 3x + 6 b) x5y + 4x4y2 – x3y3 + 3x2y4 – 2xy5 4. a) 8y + 6

b) 4a – 2b

c) 5x – 1

e) 6x – 2x + 1

f) –a2b + 5ab – 4b2

5. a) 12x3

b) –20p3q4

c) –12a5b3

d) 30x2y4z2

e) 3x4

f) –8ab

g) 3x

h) –4q

d) 2y

2

2

g) y – 6y + 1

6. Multiply 3x(x + 1) to get 3x2 + 3x. Then, multiply –4(x2 – 3x) to get –4x2 + 12x. Then, collect the like to get –x2 + 15x. Then, substitute 2 for each x and evaluate to get –(2)2 + 15(2) = –4 + 30 = 26. b) 9x3 + 4x2 – 14x

7. a) 8z + 2

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

c) 4m3 – 16m2n – 6mn2 – 8n3

e) 10x2 + 2x – 24

8. Answers may vary. 2m3n + 3m – 8n

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

9. a) 24xy2

b) 36xy + 16y2

b) 216 cm2

3.4 Common Factors

3.2 Multiplying Binomials

1. a) 4(x + 7)

b) does not factor

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

c) 2(3x – 16y)

d) 13(2x2 – y)

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

e) 2a(x + 5y – 4z)

f) does not factor

2. a) a2 + 5a + 6

b) 6 + 5k + k2

c) c2 – 8c + 15

g) 8(x2 + 4y3)

h) 5y(2 – y + 5y2)

d) t2 + 4t – 5

e) 12 – b – b2

f) 6v2 + 9v + 3

i) does not factor

j) 12xy(3 – x)

g) 10 + 9x + 2x2

h) 2y2 – 12y + 10 i) 3m2 + 10m – 8

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)

j) 4g2 + 13g – 12 k) 6y2 – 13y + 6 m) 6 – 13s + 6s2

l) 10h2 – 17h + 3

n) –12 – 22p – 8p2 o) 6t2 + rt – r2

3. a) 2m2 + 10m – 48

b) 3x2 + 15x + 18

e) (2x + 3)(3m – 5)

c) –2y2 + 2y + 12

d) 0.2x2 + 0.6x + 0.4

3. a) (x – y)(a + b)

b) (y – x)(y + 1)

e) 36x2 – 66xy + 18y2

f) 2x2 + 10x + 10

c) (a + 3)(b + 3)

d) (t – 4)(t – r)

g) –10y + 22

h) 11w2 + 20w – 11

e) 2(x + 2)(2x + 3y)

f) (3x2 + y)(y – 2)

i) 6m2 – 3m – 24

j) 43x2 + 30x + 29

g) (4ab – 3c)(b – 3a)

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

4. a) 3x(3πx – 2y)

x(x + 6) – (1)(x + 2) = x2 + 5x – 2

c) 16r2(π – 2)


b) 2x2(5y – 1)

Chapter 3

37

3.5 Factoring x2 + bx + c

3.7 Factoring Special Quadratics

1. a) (x – 3)(x – 2)

b) (y + 3)(y – 1)

1. a) (x + 5)(x – 5)

b) (y + 7)(y – 7)

c) does not factor

d) (a + 5)(a + 1)

c) (y2 + 1)(y + 1)(y – 1)

d) does not factor

e) (x – 10)(x + 1)

f) (b – 5)(b – 2)

e) (2a + 3)(2a – 3)

f) (7 + 8m)(7 – 8m)

g) does not factor

h) (x + 5)(x – 4)

g) (13a + b)(13a – b)

h) does not factor

2. a) (x + 26)(x – 2)

b) (m – 15)(m – 3)

i) (9x + 11p)(9x – 11p)

j) (7 + a – z)(7 – a + z)

c) (x + 9)(x – 4)

d) (x – 11y)(x + 6y)

2. a) yes, (x + 4)2

b) yes, (y – 7)2

e) (m + 4n)(m + 8n)

f) (6 + y)(7 – y)

c) no

d) yes, (3t + 1)2

g) (8 – x)(4 + x)

h) (x2 + 4)(x2 + 3)

e) no

f) yes, (2x – 5)2

3. a) 2(x + 2)(x + 3)

b) 3(x + 4)(x – 1)

g) yes, (11 – m)2

h) yes, (4x + 3y)2

c) 5(x – 5)(x – 2)

d) 4(x – 6)(x + 2)

i) no

e) 2(x – 11)(x + 3)

f) x(x – 6)(x – 7)

3. a) (x + 14)(x – 14)

b) does not factor

4. a) (x – 7)(x + 6)

b) 10.8 m by 23.8 m

c) (4a + 5)2

d) 4(x + 3)(x – 3)

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

e) does not factor

f) (p – 2q)2

with the sum of –9 are –5 and –4.

g) 9(2x + 3y)(2x – 3y)

h) m(m – 5)(m + 5)

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

i) 5n(n – 3)2

j) 16(2x – 1)(2x + 1)

with the sum of –9 are –7 and –2.

k) does not factor

l) (x + 3)(x – 3)(x + 2)(x – 2)

x2 – 9x + 14 = (x – 7)(x – 2)

4. a) 420

b) 1000

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

c) 520 000

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

5. a) ±14

b) ±30

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

c) 9

d) 25

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)

38

Chapter 3

b) 45 m by 121 m


Name

4.1

Functions MATHPOWERTM 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 es 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. 1. State whether each set of ordered pairs represents a function. a) (–2, 5), (–1, 10), (0, 15), (1, 20)

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

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

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

4

4

2

2

0

2

4

6 x

–2

–2

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

0

4. If w = 2v2 + 1, find the value of w for each value of v. a) –3 b) 1.5 c) 3 5. If q = –p2 + 3p + 2, find the value of q for each value of p. a) 2 b) –10 c) 0.5 6. State the domain and range of each relation in parts 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 independent variable and the dependent variable in part b). Explain your reasoning.

4

x

6

–2

c)

y

d)

d

4

120 2

80

1 3. If y = x − 5, find the value of y for each value 2 of x. a) 4 b) –80 c) 100

2

40

0

–2

40

80

120

160

200

240

280 t

0

2

x

4

–2

8. Which of the relations in question 7 are functions?

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 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 4 given by the function y = πx 3 , where x is the 3 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 Copyright © 2001 McGraw-Hill Ryerson Limited

Chapter 4

39

Name

Graphing y = x2 + k, y = ax2, and y = ax2 + k

4.2


• This table summarizes how parabolas in the form y = ax2 + k are obtained by transforming the function y = x2. Resulting Equation

Operation

Reflects in the x-axis, if a < 0. Stretches vertically (narrows), if a > 1, or a < –1. Shrinks vertically (widens), if –1 < a < 1.

y = ax2

Multiply by a.

y = ax2 + k

Add k.

Transformation

Shifts k units upward, if k > 0. Shifts k units downward, if k < 0.

1. Sketch each parabola and state the direction of the opening, the coordinates of the vertex, the equation of the axis of symmetry, the domain and range, and the maximum or minimum value. a) y = x2 + 4 b) y = –x2 + 2 y

–2 2

Sign of a positive negative (0, k)

(0, k)

Axis of Symmetry

x=0

x=0

Direction of Opening

up

down

Comparison with y = ax2

congruent

congruent

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

c) y = 2x2 – 9

0

2

0

x

2

c) y = 0.5x2 – 1

d) y = –3(x2 – 4) y

y 2

0

2

d) y = –4x2 + 25

x

–2

–2

–2

Property Vertex

y 2

–2

• Some geometric properties of the parabola y = ax2 + k are summarized in this table.

Applications 4. Write an equation for the parabola created when each pair of transformations is applied to the graph of y = x2. a) a reflection in the x-axis, followed by a vertical translation of –3

x

–2 2 –2 0 –2

2 x

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

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

5. The graph of y = 4x2 + k es through the point (–1, –1). Find k.

1 b) y = –x2 and y = − x 2 2

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

40

Chapter 4


Name

Graphing y = a(x – h)2 + k

4.3


• This table summarizes how parabolas in the form y = a(x – h)2 + k are obtained by transforming the function y = x2. Resulting Equation

Operation

Multiply by a.

Transformation Reflects in the x-axis, if a < 0. Stretches vertically (narrows), if a > 1 or a < –1. Shrinks vertically (widens), if –1 < a < 1.

y = ax2

Replace x by (x – h).

y = a(x – h)2

Shifts h units to the right, if h > 0. Shifts h units to the left, if h < 0.

Add k.

y = a(x – h)2 + k

Shifts k units upward, if k > 0. Shifts k units downward, if k < 0.

1. Sketch each parabola and state the direction of the opening, the coordinates of the vertex, the equation of the axis of symmetry, the domain and range, and the maximum or minimum value. a) y = (x + 3)2 – 2 b) y = –(x – 4)2 – 3

• Some geometric properties of the parabola y = a(x – h)2 + k are summarized in this table. Property Vertex

(h, k)

(h, k)

Axis of Symmetry

x=h

x=h

Direction of Opening

up

down

Comparison with y = ax2

congruent

congruent

3. Use a graphing calculator or graphing software 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

y

y

0

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

x

2

2 –2 –2

0

Sign of a positive negative

1 d) y = − ( x − 4)2 + 4 2

x

–2

Applications

c) y = 2(x – 1)2 + 1

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

1 d) y = − ( x + 2)2 + 7 2 y

y

b) vertex (2, 5); congruent to y =

1 2 x 2

2

c) vertex (–4, –1); y-intercept –9

2 0

2

x –2

0

2

x

2. Without graphing, state the coordinates of the vertex and whether this vertex represents a maximum or minimum value of the function. 3 a) y = –3(x + 5)2 – 6 b) y = ( x + 4)2 + 2 4 c) y = 7.5(x – 2.5)2 – 9

d) vertex (–5, 3); through (–7, 15)

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

d) y = –(x – 9)2 + 19


Chapter 4

41

Name

Graphing y = ax2 + bx + c by Completing the Square

4.4


• To rewrite a quadratic function y = ax2 + bx + c in the form y = a(x – h)2 + k, use the steps shown in the following example. y = 2x2 + 4x + 5 a) Group the containing x: y = (2x2 + 4x) + 5 b) Factor the coefficient of x2: y = 2[x2 + 2x] + 5 c) Complete the square inside the brackets: y = 2[x2 + 2x + 1 – 1] + 5 d) Write the perfect square trinomial as the square of a binomial: y = 2[(x + 1)2 – 1] + 5 e) Expand to remove the square brackets: y = 2(x + 1)2 – 2 + 5 f) Simplify: y = 2(x + 1)2 + 3 2 • For a quadratic function in the form y = ax + bx + c, find the maximum or minimum value by rewriting the function in the form y = a(x – h)2 + k to find the vertex (h, k). The maximum or minimum value of the function is k. 4. Communication Explain how to use algebra to find two numbers whose difference is 8 and whose product is a minimum.

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

c) x2 – 6x + c

d) x2 + 40x + c Problem Solving

e) cx2 + 28x + 49

5. Determine the maximum area of a triangle, in square centimetres, if the sum of its base and its height is 12 cm.

f) 9x2 – 18x + c

2. Write each function in the form y = a(x – h)2 + k. Sketch the graph, showing the coordinates of the 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 y

y

0

2

x

–2

2

–2

0

2

x

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

c) y = –7x2 + 84x + 19 42

Chapter 4

d) y = 4x2 – 20x + 7

6. A ball is thrown upward with an initial velocity of 18 m/s. Its height, h metres after t seconds, is given by the equation h = –5t2 + 18t + 1.8 where 1.8 represents the height at which the ball is released by the thrower. a) What is the maximum height the ball will reach?

b) How much time elapses before the ball reaches the maximum height?

c) How long is the ball in the air, to the nearest tenth of a second? Copyright © 2001 McGraw-Hill Ryerson Limited

Name

4.5

Investigation: Sketching Parabolas in the Form y = ax(x – s) + t MATHPOWERTM 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 + 5 a) Factor 2x from the first two : y = 2x(x – 4) + 5 b) Substitute 0 for x in y = 2x(x – 4) + 5: y = 2(0)(0 – 4) + 5 y=5 One point on the graph is (0, 5). c) Substitute 4 for x in y = 2x(x – 4) + 5: y = 2(4)(4 – 4) + 5 y y=5 Another point on the graph is (4, 5). (0, 5) (4, 5) d) Plot the points (0, 5) and (4, 5) on a grid. e) Find the axis of symmetry that es through the vertex. Since (0, 5) and (4, 5) have the same y-coordinate, the points are 2 reflection images of each other in the axis of symmetry. The equation of the axis of symmetry is x = 2, so the x-coordinate of the vertex is 2. –2 0 2 x f) Substitute 2 for x in y = 2x2 – 8x + 5: y = 2(2)2 – 8(2) + 5 –2 = 8 – 16 + 5 (2, –3) = –3 y = 2x 2 – 8x + 5 The coordinates of the vertex are (2, –3). g) Plot the vertex on the grid and draw a smooth curve through the three points. 1. Write each equation in the form y = ax(x – s) + t. a) y = x2 – 6x + 8 b) y = 2x2 + x – 5

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 y

y

–2

0

2

x

–2

c) y = x2 – 9x

d) y = –3x2 + 12x – 0.9

2

0

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

x

2

4. Application the coordinates of the vertex of each quadratic function in questions 2 2 and 3, using the form  s , t − as  . 2 4 

b) y = –x2 + 2x + 1


Chapter 4

43

Name

4.6

Investigation: Finite Differences MATHPOWERTM 10, Ontario Edition, pp. 242–245

• In tables with evenly spaced x-values, first differences are calculated by subtracting consecutive y-values, and second differences are calculated by subtracting consecutive first differences. • For a linear function, the first difference is a constant. For a quadratic function, the second difference is a constant. 1. Complete the following tables. a) y = 3x + 40 b) y = –5x + 8 x

y 1st Difference

x

0

0

1

1

2

2

3

3

4

4

c)

y 1st Difference

x

d)

x

y 1st Difference

x

0

3

0

2

1

5

1

–4

2

7

2 –10

3

9

3 –16

4 11

4 –22

y 1st Difference

2. Communication a) Describe how the first difference is related to each linear equation in question 1 parts a) and b).

b) Where does the constant term, b, for each equation occur in the table of values in question 1 parts a) and b)?

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

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

44

Chapter 4

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

Difference 1st 2nd

x

0

0

1

1

2

2

3

3

4

4

y

Difference 1st 2nd

5. Communication a) What is the relationship between the second difference and the value of a for each function in question 4?

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

c) The first entry in the 1st Difference column is equal to a + b. the b-value in each equation.

d) Find c, a, and b for the quadratic function shown in this table of values. Then, write an equation in the form y = ax2 + bx + c. x

y

0

1

1

6

2

15

3

28

4

45

Difference 1st 2nd


Name

4.7

Technology: Equations of Parabolas of Best Fit MATHPOWERTM 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 lists b) use the STAT PLOTS menu to draw the scatter plot c) enter possible equations in the Y= editor, and use trial and error to fit an equation of a quadratic function to the scatter plot d) compare the result from step c) with the equation of the curve of best fit found by using the QuadReg (quadratic regression) instruction 1. For each table of values, enter the data and draw the scatter plot. Then, use trial and error to fit the equation of a quadratic function in the form y = a(x – h)2 + k, where a, h, and k are integers, to the scatter plot. a)

c)

e)

x

y

–1

Elapsed time (s)

Height of ball (m)

0.00

0.321

x

y

6

–2

17

0.02

0.570

0

2

–1

12

0.04

0.637

2

0

0

9

4

6

1

8

0.06

0.682

5

12

3

12

0.10

0.684

x

y

x

y

0.12

0.644

–3

9

–1

–10

0.14

0.601

–2

0

0

0

1

–3

1

2

2

4

2

–4

4

30

3

–18

x

y

x

y

–5.5

0

–5

–2

–4

–2

–4

–3

–2

–3

–2

–3

0

–2

1

2

0

2

5

1.5

b)

3. Application A tennis ball is tossed upward and its height and the elapsed time are recorded as follows.

d)

f)

2. For each table of values in question 1, find the equation of the curve of best fit using quadratic regression. If necessary, round coefficients and constants to the nearest hundredth. Then, compare the equation with the equation you found in question 1.


a) Enter the data and draw the scatter plot of height versus time. b) Communication Use trial and error to fit the equation of a quadratic function in the form y = a(x – h)2 + k to the scatter plot. Explain your reasoning.

c) Use your equation to determine the height of the ball after 0.08 s. d) Find the equation of the curve of best fit using quadratic regression. Then, compare the equation with your result from part b). e) Use the equation of the curve of best fit to determine the height of the ball after 0.08 s and 0.15 s.

Chapter 4

45

Name

4.8

Technology: Collecting Distance and Time Data Using CBRTM or CBLTM MATHPOWERTM 10, Ontario Edition, pp. 248–250

Collecting Data From Tossing a Ball In this experiment, a ball is tossed vertically into the air, and the ball’s height at various times is recorded 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 the nearest thousandth. Elapsed time (s) 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20

Height of ball (m) 1.036 1.086 1.136 0.457 1.238 1.260 1.278 1.302 1.324 1.341 1.356

Elapsed time (s) 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42

1. Communication Analyse the data in the table above. a) At what height was the ball before it was tossed 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 reach its maximum height? e) Does the height at 0.06 s seem reasonable? Explain. f) Why is it important to toss the ball straight up into the air, instead of at an angle? g) The motion detector unit must be placed beneath the vertical path of the ball. At what height could the motion detector unit have been placed? h) Do you think the ball was caught before it hit the ground? i) What is the independent variable in this experiment? the dependent variable? 46

Chapter 4

Height of ball (m) 1.365 1.371 1.373 1.372 1.365 1.358 1.343 1.323 1.308 1.281 1.249

Elapsed time (s) 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60

Height of ball (m) 1.216 1.180 1.139 1.094 1.047 0.996 0.941 0.886 0.824

2. Collect your own data by doing the experiment with a softball or a basketball. Or, use the data above. Make a scatter plot of height versus time. Then, use the TRACE instruction to find the approximate coordinates of the vertex of the scatter plot. Round coordinates to the nearest hundredth, if necessary. Use the coordinates of the vertex as the values of h and k in an equation of the form y = a(x – h)2 + k. By entering equations with these values of h and k in the Y= editor, use trial and error to find a value of a for an equation that fits your scatter plot. Then, use the quadratic regression instruction to find an equation of the curve of best fit. 3. Communication How does the equation you found by trial and error compare with the equation of the curve of best fit produced using quadratic regression? 4. a) Communication Use the equation to predict the height of the ball at 10 s. Do you have much confidence in your prediction? Explain. b) If the ball was allowed to hit the ground, at what time would that have occurred? (Hint: What is the height of the ball?) Copyright © 2001 McGraw-Hill Ryerson Limited

Answers CHAPTER 4 Quadratic Functions 4.1 Functions

b)

y

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

y = –x 2 + 2 (0, 2)

1. a) function

b) not a function

2. a) 3

2

c) function

b) –39

domain: set of real numbers, –2

c) 1

0

x

2

range: y ≤ 2; maximum: 2

–2

3. a) –3

b) –45

c) 45

4. a) 19

b) 5.5

c) 7

5. a) 4

b) –128

c) 3.25

c)

y

up; (0, –1); x = 0;

2

–2

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

0

(0, –1) –2

domain: set of real numbers,

x

2

1 y = – x2 – 1 2

range: y ≥ –1; minimum: –1

range: {9, 7, 5, 3, 1} b) function; domain: {0, 0.1, 0.25, 0.5, 1},

d)

y (0, 12)

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

y = –3(x 2 – 4)

domain: set of real

range: {0, 8, 20, 40, 80} c) time: independent, distance: dependent; The distance travelled at 80 km/h depends on the time.

numbers, range: y ≤ 12;

2 –2 0

maximum: 12

2 x

7. a) domain: {1, 2, 3}, range: {2, 3, 4} b) domain: set of real numbers, range: set of real

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

numbers

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

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

4 units downward.

set of real numbers from 0 to 120

numbers greater than or equal to –2

b) The graphs have the same vertex; the graph of 1 y = − x 2 is flatter and is the graph of y = –x2 shrunk 2

8. a) not a function

vertically by a scale factor of

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

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.

3. a) 4, –4

b) none 2

1 . 2

c) 2.1, –2.1

d) 2.5, –2.5

2

4. a) y = –x – 3

b) y = 3(x + 2)

5. k = –5 6. a = –1, k = 7

b) Yes, if there is only 1 student at the school from every family. No, if there are several students at the

4.3 Graphing y = a(x – h)2 + k

school from even 1 family.

1. a)

10. a) 4188.8 cm3 (to nearest tenth)

b) 2 cm

domain: set of real –2

4.2 Graphing y = x2 + k, y = ax2, and y = ax2 + k 1. a)

up; (–3, –2); x = –3;

y 2

0

x

numbers, range: y ≥ –2;

–2

(–3, –2)

y = (x + 3)2 – 2

minimum: –2

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

y

y = x2 + 4 (0, 4)

domain: set of real numbers, range: y ≥ 4; minimum: 4

2

–2

0

2

x

–2


Chapter 4

47

b)

y 0

down; (4, –3); x = 4;

x

2

–2


(4, –3)

range: y ≤ –3; maximum: –3

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

2

y = –(x – 4) – 3

c)

6. a) 18 m

y = 2(x – 1)2 + 1 2

(1, 1) 2

x

(–2, 7)

y

0

d)

2 1 y = – – (x + 2)2 + 7 2 –2

c) 3.7 s

domain: set of real

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

numbers, range: y ≥ 1;

1. a) y = x(x – 6) + 8

1 b) y = 2 x x +  − 5  2

minimum: 1

c) y = x(x – 9) + 0

d) y = –3x(x – 4) – 0.9

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

y

b) 1.8 s

down; (–2, 7); x = –2;

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


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

range: y ≤ 7; maximum: 7

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).

2 x

0

The coordinates of the vertex are (–1, –8). 2. a) (–5, –6), maximum

b) (–4, 2), minimum

b) Factor –x from the first two to get

c) (2.5, –9), minimum

d) (9, 19), maximum

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

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

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

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

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

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

the vertex are (1, 2).

d) x-intercepts: 6.8, 1.2; y-intercept: –4 1 4. a) y = –2(x – 3)2 – 1 b) y = ( x − 2)2 + 5 2 1 2 c) y = − ( x + 4) − 1 d) y = 3(x + 5)2 + 3 2

3. a) y = 3x(x + 2) – 8

b) y = –x(x – 6) + 3 y

y

(3, 12)

y = –x 2 + 6x + 3 –2

0

2

x

–2

y = 3x 2 + 6x – 8

5. –6, 0

4.4 Graphing y = ax2 + bx + c by Completing the Square 1. a) 121

b) 64

c) 9

d) 400

e) 4

f) 9

2. a) y = (x – 2)2 – 5 y

b) y = –2(x + 1)2 + 3

(0, 3)

(0, –8)

(6, 3)

2

(–1, –11)

x = –1 y

x=2

(–2, –8)

0

2

x

4. In question 2a), a = 2, s = –2, and t = –6, so the 2( −2)2   −2 coordinates of the vertex are  , − 6 −  or  2 4 

(–1, 3) 0

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

x

2

(0, –1)

(4, –1)

–2 2

y = x – 4x – 1 (2, –5)

2

(–2, 1)

(0, 1)

–2

0

2 x

y = –2x 2 – 4x + 1

3. a) min: –26 at x = 3

b) max: 53 at x = –4 5 c) min: 271 at x = 6 d) min: –18 at x = 2 4. Let one number be n and the other be n – 8. The product, p, is n × (n – 8), so p = n – 8n and the 2

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

48

Chapter 4

( −1)(2)2  2 coordinates of the vertex are  , 1 −  or 2 4 

(1, 2). In question 3a), a = 3, s = –2, and t = –8, so the 3( −2)2   −2 coordinates of the vertex are  , − 8 −  or  2 4  (–1, 11). In question 3b), a = –1, s = 6, and t = 3, so ( −1)(6)2  6 the coordinates of the vertex are  , 3 −  2 4  or (3, 12). Copyright © 2001 McGraw-Hill Ryerson Limited

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

4.8 Technology: Collecting Distance and Time Data Using CBR™ or CBL™

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

1. a) about 1 m

c) 1st Difference: 2

c) 1.373 – 1.036 = 0.337; about 34 cm

d) 1st Difference: –6

b) 1.373 m

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

d) 0.26 s

e) No, the value should be between

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

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

c) y = 2x + 3; y = –6x + 2

caused a misreading by the motion detector unit.

3. y = – 0.5x + 5

f) The motion detector unit takes readings while it is

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

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

2nd Difference: 4

measuring, so vertical is best.

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

g) The height could be anywhere from on the ground

2nd Difference: 2

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

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

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

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

70 cm).

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

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

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

detector unit

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

i) independent variable: Elapsed time; dependent

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

variable: Height 2. The parabola opens down and the vertex is about

4.7 Technology: Equations of Parabolas of 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

(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.

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


Chapter 4

49

Name

5.1

Solving Quadratic Equations by Graphing MATHPOWERTM 10, Ontario Edition, pp. 270–277

• The solutions to the quadratic equation ax2 + bx + c = 0 are the x-intercepts of the quadratic function 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 1. Solve by graphing. Round to the nearest tenth, 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 from the top of a building, 50 m above ground level. The height of the rocket above ground level, h metres, after t seconds is given by the formula h = 50 + 45t – 5t2. How many seconds after the launch will the rocket hit the ground?


Problem Solving For each problem, write a quadratic equation to find the unknown(s). Solve the equation by graphing. 3. The length of a rectangle is 2 m more than the width. The area is 48 m2. Find the dimensions of the rectangle.

4. The sum of the squares of three consecutive integers 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 than the width. If the volume is 72 m3, find the dimensions 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.

Chapter 5

51

Name

5.2

Solving Quadratic Equations by Factoring MATHPOWERTM 10, Ontario Edition, pp. 278–286

• To solve a quadratic equation by factoring, a) write the equation in the form ax2 + bx + c = 0 b) factor ax2 + bx + c c) use the zero product property d) solve the two resulting equations to find the roots e) check your solutions 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 form ax2 + bx + c = 0. a) 3x2 = 5(x – 2) b) 2(x – 1)2 = 3x + 7

c)

4x 2 = 3x − 2 5

d)

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)

5a2 15 a −5= 4 4

k) (3x – 1)2 = 25

l)

1 2 x − 1 = 3.5 2

Chapter 5

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

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

x2 x 1 − = 4 3 3

3. Solve and check. a) x2 – 6x + 8 = 0

52

Problem Solving

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

7. A rectangle is 24 cm long and 16 cm wide. When each dimension is increased by the same amount, the area is doubled. What are the new dimensions?

8. Communication Write a quadratic equation 1 whose roots are –3 and . Explain your 2 thinking.


Name

5.3

Investigation: Graphing Quadratic Functions by Factoring MATHPOWERTM 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 = 0 b) plot the points where the graph crosses the x-axis, (y = 0) c) use symmetry to find the x-coordinate of the vertex d) find the y-coordinate of the vertex by substituting the x-coordinate of the vertex into the equation in the form y = ax2 + bx + c e) draw a smooth curve through the three points 1. Sketch the graphs of the following quadratic functions by locating the x-intercepts, and then finding the coordinates of the vertex. a) y = (x – 3)(x – 5) b) y = x2 – 1 y

y

2

2

0

x

2

–2

–2

e) y = x2 + 4x – 21

f) y = x2 – 3x – 18

y –5

y –2

x

0

0

x

2

–2

–5

0

x

2

–2

2. Use the x-intercepts to determine the coordinates of the points on the x-axis and the vertex for the graph of each quadratic function. a) y = (x + 5)(x – 5) b) y = x2 – 49 g) y = –x2 – 2x + 3

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

y

2 0

x

2

–2

0

2

x

–2

c) y = x2 + 4x – 12 0

y

2

2

0 –2

2

x

–2

0

2

x

–2

d) y = 6x – 0.5x2 y

y –2

y

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

2

–2

–2

h) y = x2 + 5x + 6

2

x

–2

b) y = (x + 6)2

2

0

2


x

Chapter 5

53

Name

5.4

The Quadratic Formula MATHPOWERTM 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. −b ± b 2 − 4 ac . • The quadratic formula is x = 2a 1. Solve using the quadratic formula. a) x2 – 8x + 12 = 0 b) 2y2 – 3y – 2 = 0

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

y+2

17

c) 20x2 + 27x = 14

d) 48x2 – 58x + 15 = 0 Problem Solving

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

6. A plain mat is placed around a picture measuring 28 cm by 36 cm so that the width of the mat is equal on all sides. The area of the mat 3 is of the area of the picture. Find the width of 4 the mat, to the nearest millimetre.

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 to the nearest hundredth, if necessary. a) x2 + 13x + 9 = 0 b) 4x2 – 11x – 19 = 0

c) 1.6(y2 + 5) = 13.4y

2

e) a – 44 = 0

54

Chapter 5

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

7. A window is in the shape of a rectangle surmounted by a semicircle. The height of the rectangle is 0.4 m more than the width. The total area of the window is 10.4 m2. Find the width and height of the window, to the nearest hundredth.

d) 18x2 + 5x + 17 = 0

2

2

f) (x + 1) + (x + 3) = 25

8. Communication Is it possible to write two real numbers whose sum is 4 and whose product is 5? Use the quadratic formula to help you explain.


Answers CHAPTER 5 Quadratic Equations 5.1 Solving Quadratic Equations by Graphing 1. a) 5, –2 b) 2, –4

c) 3, –1

y

3. a)

f) 0.5, –4

g) 0, 2.3

y = x 2 – 7x + 12

y = x 2 + 2x

d) 4, –1

2

2

(–2, 0)

e) 0.5

y

b)

h) no solution

(3, 0)

(0, 0)

–2

0

x

2

0

2

(–1, –1)

i) –2, –3

(4, 0) x

7 1 –, – – 2 4

j) –0.5, –3.5 c)

2. 10 s

y

(– 6, 0) –2

3. 8 m by 6 m

d)

(2, 0) x

0

(6, 18)

y

2

–2

y = x 2 + 4x – 12

4. 4, 5, 6 or –4, –5, –6 5. 6 m by 4 m by 3 m

y = 6x – 0.5x 2

6. 12 cm, 9 cm 2

5.2 Solving Quadratic Equations by Factoring 1 3 b) − , 3 2

1. a) 2, –7 2

d) −

c) 0, 5

b) 2x – 7x – 5 = 0

c) 4x2 – 15x – 10 = 0

d) 3x2 – 4x – 4 = 0

3. a) 2, 4

b) –4, –6

c) –4, 7

3 4 , 2 3 1 i) − , 3 4

f) −

5 3

5 2

e) (–7, 0) –2

y

(3, 0)

0

2

(12, 0) x

2

f)

y

(6, 0)

(–3, 0) x

–2

–2

0

2

x

–2

y = x 2 + 4x – 21

y = x 2 – 3x – 18

d) –9, 3

5 7 , − 2 3 4 k) 2, − 3

h) 0,

g)

j) 4, –1

0

2

2. a) 3x – 5x + 10 = 0

e)

(0, 0)

(–2, –16)

8 3

l) 3, –3

4. 8 m, 17 m 3 – , – 20 1 – 2 4

5. 25 cm 6. 10, 12, 14

(–2, –25)

7. 32 cm by 24 cm

g)

1 3 equation to get x 2 − x + 3x − = 0. Then, multiply 2 2 every term by 2 to eliminate the denominators, to get

h)

y

1 1 8. x = –3 and x = , so ( x + 3) x −  = 0. Expand the  2 2

y

y = x 2 + 5x + 6

(–1, 4)

2

2

(–3, 0)

(0, 1) –2

0

2

x

y = –x 2 – 2x + 3 –2

0

(–3, 0) (–2, 0) 5 1 – – , – – –2 2 4

2 x

2x2 – x + 6x – 3 = 0 or 2x2 + 5x – 3 = 0. 4. a) The equation is in factored form, so there is

5.3 Investigation: Graphing Quadratic Functions by Factoring 1. a)

y

b)

y

y = (x – 3)(x – 5)

2

2

0

(5, 0) x

2

y-intercept, let x = 0 and find y. Plot (0, 25). A third point is symmetrical to (0, 25) about the axis of

y = x2 – 1 (3, 0)

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

(4, –1)

(–1, 0) –2

(1, 0) 0

2

x

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

(0, –1)

2. a) (–5, 0), (5, 0), (0, –25) b) (–7, 0), (7, 0), (0, –49) Copyright © 2001 McGraw-Hill Ryerson Limited

Chapter 5

55

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) 2, −

1 2

7 2 c) − , 4 5

2. a) 3 ± 6

b)

−5 ± 13 6

3 ± 33 3

d)

−8 ± 67 2

c)

d)

−3 ± 13 2 3. a) –0.73, –12.27

f) no solution

c) 7.73, 0.65

d) no solution

e) ±6.63

f) –5.39, 1.39

e)

3 5 , 8 6

b) 3.95, –1.20

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.

56

Chapter 5


Name

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)

X

P

Q

R

Y

Z

PQ 1 QR 1 PR 1 = , = , and = . The ratios of the corresponding XY 2 YZ 2 XZ 2 sides are equal.

c)

area PQR 2 1 = = . The ratio of their areas is equal to the ratio of the squares of their area XYZ 8 4 1 12 corresponding sides. = 2 4 2

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

C D

E

B

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

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

B

C

2. Communication Explain why ABC is similar to EDC by a) comparing the measures of the corresponding angles, and the ratios of the lengths of the corresponding sides b) calculating the ratio of their areas

A

E

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

d) Communication Are all equilateral triangles similar? Explain and justify your reasoning. B

D


C

Chapter 6

57

Name

6.2

Similar Triangles MATHPOWERTM 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 = ∠F b) the ratios of the corresponding sides are equal

c B

D °

A ° b

x

C

f E

e

x

F

d

a

a b c = = d e f c) the ratio of their areas is equal to the ratio of the squares of their corresponding sides area ABC a 2 b 2 c 2 = = = area DEF d 2 e 2 f 2 3. Communication Explain why ABC is similar to AEF.

1. In each diagram, the triangles are similar. BC Write the ratio of the lengths of the sides . EF a)

E B

C x

F x

√

°

A

°

√

D

B

A

4. Find a.

b)

F

F

a

8 4

3

E

°

A

D

°

B

C

2. The triangles in each pair are similar. Find the unknown side lengths. a) S P °

15 cm

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

°

20 cm 20 cm

t

x

R

x

T

√

25 cm

Q

s

b)

D

20 m A b 3m B 4 m° C

c)

C

E

X 10 m

F °

y

° √

12 m

w x

Z Y 15 m

E

d

x

32 m

V √

16 m °

√

U

6. Application A Ranin marked out the following triangles to 13 m determine the length of a pond. Calculate the length of the E pond, AB, to the nearest tenth of a metre.

B

C 2.8 m D 3.8 m

W

58

Chapter 6


Name

6.3

The Tangent Ratio MATHPOWERTM 10, Ontario Edition, pp. 326–333

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

B

length of the side opposite ∠A length of the side adjacent to ∠A

opposite C

A adjacent

opposite adjacent

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

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

c) 15°

d) 45°

e) 60°

f) 72°

28° 12 m

43° 3m

c)

50° x

2. Find ∠K, to the nearest degree. a) tan K = 0.575 b) tan K = 0.243 c) tan K = 1.925 e) tan K = 3.198

d) tan K = 2.750

d)

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

f) tan K = 50.375

x 17 m

6m

60°

P

R

45°

3.7 m Q

3. Find ∠Q, to the nearest degree. a) tan Q =

1 3

b) tan Q =

5 8

c) tan Q =

5 4

d) tan Q =

12 5

49 e) tan Q = 9

7. Application Find the length of x, then the length of y, to the nearest tenth of a metre. x 39° 12 m

89 f) tan Q = 2

4. Calculate tan D and ∠D and tan E and ∠E. Round each angle measure to the nearest degree. a) b) 4 cm A E E 2 cm D

28° y

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

8m D

9m


N

Chapter 6

59

Name

6.4

The Sine Ratio MATHPOWERTM 10, Ontario Edition, pp. 334–339

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

B

length of the side opposite ∠A length of the hypotenuse

sine A =

hypotenuse

opposite C

A

opposite or sin A = hypotenuse 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°

c)

d)

y

59 m

96 m

72°

y

60°

e)

f)

25°

y

10 m

45°

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

4 5

d) sin G =

5 8

e) sin G =

1 11

f) sin G =

8 9

y

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

3. Find ∠G, to the nearest degree. 1 2 a) sin G = b) sin G = 2 5 c) sin G =

17 m

25°

4. Calculate sin Y. Then, find ∠Y, to the nearest degree. a) b) Y Z

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

3 cm 15 cm X 6 cm Y

Z

X

11 cm

5. Calculate y, to the nearest hundredth of a metre. a) b) 30° 54° 28 m

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

y 15 m

y

60

Chapter 6


Name

6.5

The Cosine Ratio MATHPOWERTM 10, Ontario Edition, pp. 340–345

• For any acute angle A in a right triangle, the cosine ratio is length of the side adjacent to ∠A length of the hypotenuse

cosine A =

B hypotenuse A

adjacent hypotenuse

or cos A =

1. Use a calculator to find the cosine of each angle, to the nearest thousandth. a) 23° b) 79° c) 30°

d) 50°

e) 43°

f) 7°

c)

C

d) 32 cm

70°

25 cm

w w

60°

6. Application Find the distance from Dani to the clubhouse.

2. Find ∠E, to the nearest degree. a) cos E = 0.982 b) cos E = 0.174 c) cos E = 0.454

adjacent

clubhouse home d

d) cos E = 0.777

1.8 km 54°

e) cos E = 0.999

f) cos E = 0.009

Dani

3. Find ∠V, to the nearest degree. 1 7 a) cos V = b) cos V = 4 8 c) cos V =

2 3

d) cos V =

1 11

e) cos V =

14 15

f) cos V =

6 13

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

4. Calculate cos H. Then, find ∠H, to the nearest degree. a) b) 5m 5 cm

H 13 m H

4 cm

8. Problem Solving A 4-m ladder leans against a wall. The foot of the ladder makes an angle of 63° with the ground. How far from the wall is the foot of the ladder, to the nearest tenth of a metre?

5. Calculate w, to the nearest tenth of a centimetre. a) b) 27 cm 17 cm w

30°

48° w


Chapter 6

61

Name

6.6

Solving Right Triangles MATHPOWERTM 10, Ontario Edition, pp. 346–351

• To use trigonometry to solve a right triangle, given the measure of one acute angle and the length of one side, find a) the measure of the third angle using the angle sum in the triangle b) the measure of a second side using sine, cosine, or tangent ratios c) 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, find a) the measure of one angle using its sine, cosine, or tangent ratio b) the measure of the third angle using the angle sum in the triangle c) the measure of the third side using a sine, cosine, or tangent ratio, or the Pythagorean Theorem 1. Find all the unknown angles, to the nearest degree, and all the unknown sides, to the nearest tenth of a unit. E A C a) b)

2. Solve each triangle. Round each side length to the nearest tenth of a unit, and each angle, to the nearest degree. V a) S b)

5 cm 3m D

5m

19 m

F

24 cm

T

c)

d)

G

14 cm

12 cm

5m

J

W

40°

U

L

c)

d)

B

G

25 mm 8 cm

7 mm K

I

H

5m

D

7m

E

45°

F

C

4 cm

e)

4m

O

f)

Q

N

3. Problem Solving A slide that is 4.2 m long makes an angle of 35° with the ground. How high is the top of the slide above the ground?

15 cm 7m P

R 9 cm

M

62

Chapter 6

4. Problem Solving A rope is anchored to the ground at its ends and is propped up in the middle by a 1-m vertical stick. At one end, the rope makes an angle of 55° with the ground. How long is the rope, to the nearest centimetre?


X

Name

6.7

Problems Involving Two Right Triangles MATHPOWERTM 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 found b) identify the two triangles that can be used to solve the problem, and plan how to use each triangle c) solve the problem and show each step in your solution d) write a concluding statement giving the answer 1. Find BC, to the nearest centimetre. C B

A

43.4°

34° 100 cm

D

4. Problem Solving From a point on the ground, a student sights the top and bottom of a 15-m flagpole on the top of a building. The two angles of elevation are 64.6° and 57.3°. a) Draw a diagram for the information given in the problem.

2. Find XY, to the nearest tenth of a centimetre. V 54.5°

b) How far is the student from the foot of the building? Round your answer to the nearest tenth of a metre.

W

65°

38.7 cm

5. Application From two tracking stations 425 km apart, a satellite is sighted at C above AB, making ∠CAB = 48.3° and ∠CBA = 62.6°.

X

Y

C

Find the height of the satellite, to the nearest tenth of a kilometre. 3. Find PQ, to the nearest tenth of a metre. P

O

R

39.7° 50 m

Q

50.3° S 40 m


A

48.3°

62.6°

B

425 km

6. Problem Solving Two buildings are 14.7 m apart. From the top of one building, the angles of depression of the top and bottom of the second building are 27.5° and 63.8°. Find the heights of the buildings, to the nearest tenth of a metre.

Chapter 6

63

Name

6.8

Technology: Relationships Between Angles and Sides in Acute Triangles MATHPOWERTM 10, Ontario Edition, pp. 360–361

A

• In an acute triangle, such as ABC, 68.97° a) the longest side, BC, is opposite the largest angle, ∠A 11.23 cm 14.09 cm b) the shortest side, AB, is opposite the smallest angle, ∠C 64.84° 46.19° c) the ratio of the side lengths is equal to the ratio of the B C 14.53 cm sines of their opposite angles; AC 14.09 sin B = = 1.25 and = 1.25, rounded to the nearest hundredth AB 11.23 sin C 1. Communication Examine ABC at the top of the page. AC a) Explain why tan B ≠ . BC

b) Do you get the same results as the ratios of the side lengths if you use the cosine ratio instead of the sine ratio? the tangent ratio instead of the sine ratio?

3. Application Use The Geometer’s Sketchpad® and the instructions on Practice Master 6.1 question 3 to create an equilateral triangle. a) Can you determine the value of the ratio of any two side lengths and the ratio of the sines of their opposite angles without measuring? Explain.

b) Check by calculating the pair of ratios. X

2. a) Calculate each pair of ratios, to the nearest tenth, for XYZ. XY sin Z and YZ sin X

9.6 cm

Y

68.7°

45.7°

7.6 cm

65.6°

Z

9.9 cm

XY sin Z and XZ sin Y YZ sin X and XZ sin Y

4. Application Use The Geometer’s Sketchpad® to construct an isosceles triangle: • construct two points • construct a circle with one point as the centre and the other point on the circle • construct another point on the circle • the three points to form the triangle • hide the circle C

b) Communication Determine how many different pairs of ratios are possible. Does the relationship apply to all of these pairs?

c) Communication Explain why the pairs of ratios may not be equal if you round values to the nearest tenth.

A

B

a) Can you determine the value of the ratio of any two side lengths and the ratio of the sines of their opposite angles without measuring? Explain.

b) Check by calculating the pair of ratios. 64

Chapter 6


Name

6.9

The Sine Law MATHPOWERTM 10, Ontario Edition, pp. 362–368

A

• There are two forms of the sine law.

c

a b c sin A sin B sin C = = or = = sin A sin B sin C a b c B • The sine law can be used to solve an acute triangle when given: a) the measures of two angles and any side b) the measures of two sides and an angle opposite one of these sides 1. Find the length of the indicated side, to the nearest metre. a) b) S J

b

C a

5. Application Find the area of ABC, to the nearest square centimetre.

A 24 cm B

p 61°

39°

P

K

2. Find the measure of the indicated angle, to the nearest degree. a) b) M M 79° 20 cm 10 m O

K

34 cm

C

63° L

41°

15 m

N

59°

k

27 m R

84°

47° 12 m

6. Application Observers at points A and B, who stand on level ground on opposite sides of a tower, measure the angle of elevation to the top of the tower at 33° and 49°, respectively. A third point, C, is 120 m from B. ∠ABC = 67° and ∠BAC = 31°. Find the height of the tower, h, to the nearest metre. h

L A

3. Find the indicated quantity, to the nearest tenth. a) In KLM, ∠K = 74°, ∠L = 47.5°, and m = 37.7 cm. Find k.

33° 31°

49° B 67° 120 m C

7. Problem Solving A rock and an oak tree are on the same side of a ravine and are 125 m apart. A birch tree is on the opposite side of the ravine. The angle formed between the line ing the rock and oak tree and the line ing the rock and the birch tree is 25°. The angle formed by the line ing the rock and the oak tree and the line ing the oak tree and the birch tree is 72°. a) Draw a diagram containing the information.

b) In ABC, ∠A = 50°, a = 9 m, and b = 8 m. Find ∠B.

G

4. Solve the triangle. Round each answer to the nearest whole number.

79° 38 m

H

43 cm

J

b) Calculate the width of the ravine. Round your answer to the nearest tenth of a metre.


Chapter 6

65

Name

6.10 The Cosine Law MATHPOWERTM 10, Ontario Edition, pp. 369–376 C

• There are two forms of the cosine law. b2 + c 2 − a2 a2 = b2 + c2 – 2bc cos A or cos A = 2bc

b A x

• The cosine law can be used to solve an acute triangle when given: a) the measures of two sides and the contained angle b) the measures of three sides

80°

Q

5.1 cm

57°

R

11.2 m

X

19.7 cm

14.1 cm D

B

C 23.9 cm

5. Application Find the area of XYZ, to the nearest square metre.

1.5 m F

X 17.5 m

Y

3. Find the indicated quantity, to the nearest tenth. a) In CDE, ∠E = 50°, c = 11.9 cm, and d = 13.5 cm. Find e.

b) In KLM, k = 54.2 cm, l = 45.7 cm, and m = 36.9 cm. Find ∠K.

66

Chapter 6

Y

b) In NPQ, n = 8.2 cm, q = 13.7 cm, and ∠P = 67°.

2. Find the measure of the indicated angle, to the nearest degree. a) E b) A

3.5 m

B

77 m M

120 m

2.9 m

c–x c

115 m

7.9 cm

L

D

4. Solve each triangle. Round each calculated value to the nearest whole number, if necessary. a) W

1. Find the missing side length, to the nearest tenth of a unit. K a) b) P 8.6 m

a

h

67°

Z 29.1 m

6. Communication Explain whether you can use the cosine law to find f in DEF when given d = 19.2 cm, e = 14.7 cm, and ∠F = 39°.

7. Problem Solving Two boats left a dock at the same time. One travelled at 7 km/h on a bearing of 39°. The other travelled at 5 km/h on a bearing of 82°. How far apart were the two boats after 3 h? Round your answer to the nearest tenth of a kilometre.


Name

Answers CHAPTER 6 Trigonometry 6.1 Technology: Investigating Similar Triangles Using The Geometer’s Sketchpad® 1. a) Answers will vary. b) ∠B is common, ∠A = ∠D, and ∠E = ∠C, AB AC CB area of ABC AB 2 or = = , = DB DE EB area of DBE DB 2 AC 2 CB 2 or 2 DE EB 2

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°

2. a) ∠B = ∠D = 60° (equilateral triangle),

b) tan D = 0.889; ∠D = 42°; tan E = 1.125; ∠E = 48°

∠A = ∠E = 30° (half of an equilateral triangle),

5. a) 2.8 m

∠C = 90° and is common;

BC 3 AC 3 = , = , DC 2 EC 2

AB 6 3 = = ED 4 2

b) 6.4 m

c) 7.2 m

6. a) 3.7 m

d) 9.8 m

b) isosceles right triangle

7. x = 9.7 m and y = 18.3 m 8. 2 possible answers: 30 m

area of ABC 3 9  18 triangular units  = = b) area of EDC 2 2 4  or 8 triangular units 

7.5 m

2

HOUSE 15 m

HOUSE

15 m

3. a) and b) Answers will vary. c) The ratio of the two triangles’ areas is equal to the

6.4 The Sine Ratio

ratio of the squares of their side lengths.

1. a) 0.883

b) 0.358

c) 0.996

d) Yes, corresponding angles are always equal, since

d) 0.707

e) 0.087

f) 0.940

they are always 60°; and the ratios of the three

2. a) 82°

b) 12°

c) 30°

corresponding side lengths are always equal, since

d) 90°

e) 20°

f) 49°

the side lengths in each triangle are always equal.

3. a) 30°

b) 24°

c) 53°

d) 39°

e) 5°

f) 63°

6.2 Similar Triangles 1. a) 2:1

4. a) sin Y = 0.500; ∠Y = 30° 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

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

3. ∠A is common, ∠ACB = ∠AFE (parallel lines),

6. a) 25.4 m

∠ABC = ∠AEF (parallel lines)

7. 13.0 cm

4. a = 3

8. Since the sine has the hypotenuse as the second

5.

term and the hypotenuse is always the longest side,

x

x

° 1.5 m

x 1.8 m

b) 14.8 m

°

it is the ratio of a lesser number to a greater number.

32 m

6.5 The Cosine Ratio The height is 38.4 m. 6. 17.6 m Copyright © 2001 McGraw-Hill Ryerson Limited

1. a) 0.921

b) 0.191

c) 0.866

d) 0.643

e) 0.731

f) 0.993 Chapter 6

67

2. a) 11°

b) 80°

c) 63°

5. 301.6 km

d) 39°

e) 3°

f) 89°

6. 29.9 m, 22.2 m

3. a) 76°

b) 29°

c) 48°

d) 85°

e) 21°

f) 62°

4. a) cos H = 0.800; ∠H = 37°

1. a) tan 64.84 =˙ 2.13 but does not equal the ratio of

b) cos H = 0.385; ∠H = 67° 5. a) 11.4 cm

b) 23.4 cm

c) 10.9 cm

d) 21.7 cm

6.8 Technology: Relationships Between Angles and Sides in Acute Triangles opposite AC AC (the adjacent side could be , or adjacent BC AB

either AB or BC). This is because ABC is not a right

6. 1.1 km

triangle.

7. Since both sine and cosine have the hypotenuse as

b) No; the ratio of the side lengths is not equal to the

the second term, the ratio with the greatest first term

ratio of the cosines or the tangents of their opposite

will be greater. That is, if the opposite side is longer

angles.

than the adjacent side, the sine will be greater; if the

2. a) 1.0; 1.3; 1.3

adjacent side is longer than the opposite side, the

b) There are six possible ratios:

cosine will be greater.

side 1 side 1 , , side 2 side 3

side 2 , and the inverse of these ratios. The side 3

8. 1.8 m

relationship applies to all six ratios.

6.6 Solving Right Triangles

c) Rounding error can affect the calculated ratios, so

1. a) AC = 4 m, ∠B = 53°, ∠C = 37°

they may not be equal but are accurate to one tenth.

b) DE = 13 cm, ∠D = 23°, ∠E = 67°

3. a) In an equilateral triangle, the ratio of any 2 side

c) GH = 6.9 cm, ∠G = 30°, ∠I = 60°

lengths is 1, so the ratio of the sines of their opposite

d) LK = 4.9 m, ∠J = 44°, ∠K = 46°

angles should also equal 1. sin 60 b) =1 sin 60

e) MO = 8.1 m, ∠M = 30°, ∠O = 60° f) QR = 12 cm, ∠Q = 37°, ∠P = 53°

4. a) In ABC where AB = AC and ∠A is less than AC sin B =1= . You can’t calculate 90°, AB sin C AC sin B or without measuring. CB sin A

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°

b) If ∠C = ∠B = 68°, ∠A = 21° and

3. 2.4 m

sin 68 =˙ 2.6 sin 21

4. 244 cm

6.9 The Sine Law

6.7 Problems Involving Two Right Triangles

1. a) 10 m

b) 20 m

2. a) 35°

b) 61°

1. 165 cm

3. a) 42.5 cm

b) 42.9°

2. 20.5 cm b) 27.4 m

4. a) 15 m

3. 57.5 m

4. ∠H = 60°, ∠J = 41°, GH = 29 cm 5. 409 cm2 6. 95.8 m

64.6° 57.3°

68

Chapter 6


7. a)

birch

25° rock

125 m

72° oak

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


Chapter 6

69

Name

Adding Polynomials To add polynomials, collect like and add. Find the sum of the expressions, A and B, represented by the algebra tiles. 1.

Simplify. 9. (3y + 4z + 6) + (2y – z – 4)

A

10. 2ab + 3bc + d + 2bc + 3ab – d

B

11. k2 – 2kj – j2 + j2 – 2kj + k2 2.

3.

A B

12. s2 + 4 + t + 3 + 2t3 + 3s

A

Add.

B

Model the expressions using algebra tiles or drawings on grid paper. Then, add.

13.

4a + b + 2a + 2b – 3

14.

4m2 + 8mn + 2n2 + m2 – 2mn + n2

15.

3r2 – 8r + 4 + r2 – 2r + 5

16.

c2 + 2ac + 4 + 3c2 + 6 + a2

4. (x2 + 2x + 2) + (2x2 + x + 1) Simplify. 17. (4k2 + 2k – 5) + (3 – k – 2k2)

18. (x3 + 2y – 5) + (3x3 – 4y + 7)

5. (–2x2 + 2x) + (–x2 + x – 2)

19. (z3x + 3z – 2) + (3z3x – 4z + 6)

6. (–x2 – 2x – 2) + (2x2 – x – 1)

20. a) Write an expression in simplest form for the perimeter of the figure. 2x + 3

7.

2x2 – x – 3 + x2 – x + 1

8.

x2 – 3x + 2 + (–2x2) – x – 1

x–2

b) If x = 8 cm, what is the perimeter?


Appendix A: Review of Prerequisite Skills

71

Name

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°. Find the missing angle measures.

Classify each triangle and find the missing angle measures. 1.

2.

A 57°

65°

B

L 85° M

j

N 78°

x

69° K

C y

E

3.

y

4.

G

9.

D

25°

L

I

25°

F

10.

10°

z H

J a K

50°

11. M z

s G 108°

r F

W

X

Z

e

65°

Find the measures of the indicated angles. 5.

H t 59°

E v u

c d

140° w y N

Y

72° P x

6.

Q

d

35°

S f

12. If you know the measure of DBC, how can you find the measure of DCA? A B

e D

R

C m U k

7. 100° q n p

V

13. If you know the measure of S, how can you find the measure of QRS?

j 40°

h

W

8.

Q

T

72°

T

R

Z m 44°

p n Y

X S

72



Name

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

1 2 4 3 5 6 8 7

1 2 4 3 5 6 8 7

1 2 4 3 5 6 8 7

Alternate angles are equal, e.g., 4 = 6

Corresponding angles are equal, e.g., 3 = 7

In the diagram, list the following. C D

A

G

E

Find the measures of the indicated angles. 7.

F

Co-interior angles are supplementary, e.g., 4 + 5 = 180°

W

80°

X

50°

B

s

Z

H

1. two pairs of alternate angles

8.

r

Y

C

D

40° y

2. two pairs of corresponding angles 30°

3. two pairs of co-interior angles

x

F

9.

z E A

D

65°

E

50°

d

Find the missing angle measures. 4.

L

e

X

N s 126°

G

58° a

b cJ

K d

5.

11.

T

A v

Q

w 60°

R

P

a b x

M

B w 110° D

6. Write two different methods you could use to find the missing angle measures. What are the measures? C

z

E

12. List all the angles in the diagram. Then, calculate the measure of each angle. V

S

p

105° q D

x C 120°

y

R

E

r

H

P

M

S

Y

L

r

X

G

C

10.

b a

f

B

Y

K F


U

W

60° T


73

Name

Common Factoring To factor a polynomial, determine the GCF of all the . Then, divide all the by the GCF. Determine the GCF of both .

21. a) Find expressions for the length and the width of the rectangle if A = 3x + 5xy is the expression for the area.

1. 6x – 9 2. 9bc + 12bd 3

A = 3x + 5xy

3. –2a – 4a

?

b) Write 3x + 5xy in factored form.

Complete the table.

4. 5. 6. 7. 8.

?

2

GCF of Both x 3y

Polynomial x2 + 2x 2 2

8xy –3x y 2m2n – 4mn2 6a3 + 9a2 – 12a

Other Factor 2y – 3 8 – 3xy

22. A rectangle has an area of lw, where l is the length and w is the width. a) Write an expression for the perimeter as a sum of 4 .

b) Simplify the sum.

Factor each binomial. 9. 4y + 10

c) Factor the simplified expression. 10. 6m2 + 9m x

11. 4t3 – 6t2

23. a) Write an expression in simplified form for the perimeter of the hexagon.

12. 3p3q2r2 – 4p2q2r

y

y

y

y x

Explain the error in each case and correct it. 13. 8n2 – 12n = 4(n2 – 3n) 2 2

2

b) Factor the expression.

2

14. 16y z + 24y z = 4y z(4z + 6) 6

2

2

3

15. 25x – 5x = 5x (5x – 1)

Factor each trinomial.

r

24. The surface area of a cylinder is given by the expression 2πr2 + 2πrh.

h

a) Factor the binomial.

16. 5a2 + 10ab – 15b2 17. 9x3y + 3xy2 + 15xy 18. 2s3t2 – 8s2t3 + 4st 19. 6y2 – 9xy + 12x2y2 4

3

3

20. 12c d + 8c – 16c d

74


b) Evaluate for r = 5 cm and h = 10 cm. Round to the nearest whole number.

25. Write a factored expression for the surface area of a cylinder that is open at the top.


Name

Congruent Triangles Two 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. SSS (side, side, side) A

SAS (side, angle, side)

D

A

ASA (angle, side, angle) A

D

D

x B

C

E

o

B

F

C

o

E

F

x o

B

C

o

E

F

For each pair of congruent triangles, state which sides and which angles are equal.

Find the missing measures in each pair of congruent triangles.

1.

8.

A

2.

D C

X

Y

4m

A

E

ABC ≅ DEF

S

?

?

Z B

E

F

?

T

XYZ ≅ STR

9.

J 7 cm 95°

S 11.4 cm

For each congruence relation, name the sides and angles that correspond.

X

Y ?

8.6 cm

?

41° 11.4 cm

65° ? T

7 cm

?

L

10.

∠KLM =_________ ∠JGH = _________ ∠MKL =_________

32° M

K ?

53° H

G ?

3. LMN ≅ PQR

BC = ___________ EF = ____________ FD = ___________

D

5m

3m R

B

C

F

V

∠TVS = _________ ∠XYZ = _________ ∠ZXY = _________

Z

4. STV ≅ XYZ For each pair of triangles, identify the case that proves that the triangles are congruent and list all the corresponding equal parts. 11.

X x

A

C

12.

Z

What is the fewest number of other parts that must be equal, so that each pair of triangles is congruent?

E

x D

F

B

Y

G

5. All the angles of 1 triangle are equal to the corresponding angles of another triangle.

6. Two sides of 1 triangle are equal to 2 sides of another triangle.

13.

14.

H x o J

K

L

Q

o x P M

x x

R S

7. Two angles of 1 triangle are equal to 2 corresponding angles of another triangle.



75

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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. Write an expression for each of the following.

15. Evaluate for a = 5 and b = 2.

1. the sum of 12 and 7

a) a – b

b) 3a + 3b

2. the sum of x and 10

c) 2a – b

d) a + 3b + 4

3. the product of 9 and 12

e) 3ab + 2

f) 22 – 2ab

4. the product of 5 and y

16. Evaluate each expression for the given value of the variable.

5. the product of a and b

a) 5t + 3, t = 2.3

6. the quotient m divided by v

b) 3m – 2, m = 1.6

17. Evaluate for x = 1.4, y = 3.2, and z = 5.3.

7. 8 greater than q

a) x + y + z

8. 9 fewer than t

18. The cost of a Sunday newspaper is $1.50. a) Write an expression for the cost of n papers.

9. 2 times the radius, r

b) Calculate the cost of purchasing 8 papers.

1 10. – of the distance, d 2

19. The height of a jump on the moon is 6 times higher than a jump on Earth.

Find the values for each of the following. 11.

13.

76

s 6 2 1 4 9 0 10 30 t 3 2 5 2.3 4.1 7.8 8.3

3s

3t – 4

b) yz + xz + xy

12.

14.

x 0 1 3 10 1.5 6.2 8.3

r 3 0 2 1.1 0.8


4x + 5

y 2–r+y 4 1 5 3.2 0.9

a) Write an expression for the height of a jump on the moon. b) Calculate the height of a jump on the moon for a jump of 2.8 m on Earth. 20. The cost to rent a video game station for a weekend is $18 plus $4 for each game rented. a) Write an expression for the rental cost for a weekend. b) Calculate the cost to rent the game station and 3 games for the weekend. 21. The width of a soccer field is 20 m greater than half the length of the field. a) Write an expression for the width of the field. b) Calculate the width and the perimeter if the length of the field is 110 m. Copyright © 2001 McGraw-Hill Ryerson Limited

Name

Evaluating Expressions II (Expressions With Integers) In many formulas and expressions, variables are replaced by integers and then evaluated. 1. Evaluate for y = 3. a) 3y

b) 2y + 4

d) –4y

c) y – 5

e) 4 – 2y

f) 6 – 3y

Corrie’s answer was –11. Jillian’s answer was 35. Shara’s answer was 85. a) Identify the correct answer.

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

9. During a recent math contest, the following answers were submitted as the solution to the question (4 – 2x)2 + 5x when x = –3.

b) –3r2

c) –2r2 + 2

b) Show the calculations that your answer in part a).

10. The CN Tower is about 547 m tall. If an object is dropped from this height, its height above the ground is given by the formula h = 547 – 5t2 where h is the height, in metres, and t is the time, in seconds, since the object was dropped. Find the height of the object at these times after it is dropped. a) 5 s

d) 2r3

e) (r + 3)2

f) 2r2 + 3r – 8

b) 9 s c) 10 s

Complete each table. 5.

7.

x 2 1 0 –1 –2

2x – 2

s 2 1 0 –1 –2

s2 + 2s

6.

8.

m 2 1 0 –1 –2

3 + 2m

a 2 1 0 –1 –2

a2 – 2a –1

11. The formula for the height of a model rocket is h = –t2 + 20t where h is the height, in metres, and t is the time, 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. Explain your answer. Appendix A: Review of Prerequisite Skills

77

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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. 1. The following formula can be used to find the approximate mass of young adults. 6( h − 90) M= 7 M is the mass, in kilograms, and h is the height, in centimetres. Find the mass, to the nearest tenth of a kilogram, for each height. a) 180 cm

b) 210 cm

c) your height

4. Write an equation that relates equal values of quarters and nickels, where q is the number of quarters and n is the number of nickels.

5. An amount of money is invested in an for a number of years. Compound interest is paid on the amount in the . The value of the investment at any given time can be calculated using the following formula. V = P(1 + i)n

2. To calculate the approximate number of mini-lights needed to decorate a Christmas tree to produce a full effect, the following formula is recommended. h×w×8 L= 929 L is the number of lights, h is the height of the tree, in centimetres, and w is the width of the tree, in centimetres.

V is the value of the investment, P is the amount of money invested, i is the rate of interest, and n is the number of years the money has been invested. Find the value of each investment. a) Amount: $200, Interest: 0.03, Years: 2

b) Amount: $650, Interest: 0.02, Years: 3

a) How many lights are needed for a tree that measures 183 cm high and 122 cm wide?

c) Amount: $1000, Interest: 0.05, Years: 5

b) How many strings of 25 mini-lights would be needed for this tree?

6. Complete the columns for s and A, using the information below the table in parts a) and b).

3. If there is a discharge hole in a container, the velocity, v, in metres per second, at which the liquid leaves the container can be calculated using the formula v = 19.6 h , where h is the height, in metres, of the liquid above the hole. Find the velocity of discharge, to the nearest tenth of a unit. a) height of 4 m

b) height of 0.1 m 78


a 5 cm 11 cm 8.4 m 1.5 m

b 7 cm 19 cm 3.6 m 1.5 m

c 8 cm 20 cm 6m 2.1 m

s

A

a) Given the side lengths a, b, and c of a triangle, calculate half the perimeter, s, using the a+b+c following formula. s = 2 b) Find the area, A, of the triangle, using the following formula. A = s( s − a)( s − b)( s − c) Copyright © 2001 McGraw-Hill Ryerson Limited

Name

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 ed to form a smooth curve. 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 0 1 –1 2 –2 3 –3

y

x 0 1 –1 2 –2 3 –3

4. The cost of removing pollutants from waste water depends on the percent of pollutants removed. The table shows the cost, in thousands of dollars, for the percent of pollutants removed. Percent of 10 Pollutants Removed Cost 1.7 (thousands of dollars)

y

y

6

8

4

6

0

40

50

3.8

6.4 10.0 15.0

18 16 14 12 10 8 6 4 2 0

–2

30

a) Plot cost versus percent of pollutants removed.

y

2

20

10

20

30

40

50

60

70

80

4

2

2

x

–2

–2

b) Use the graph to find the area of a circle with a diameter of 3.5 cm. c) Use the graph to find the diameter of a circle with an area of 10 cm2.

0

2

Diameter (cm) 1 2 3 4 5

3. The table shows the relation between the area of a circle and its diameter. a) Plot area versus diameter, and draw a smooth curve through the points.

b) How much would it cost to remove 45% of the pollutants from the waste water? x

Area (cm2) 0.8 3.1 7.1 12.6 19.6

20

5. The equation s = 15.9 l can be used to approximate the speed of a vehicle that has skidded, where s is the speed, in kilometres per hour, and l is the length of the skid mark, in metres. a) Complete the table of values, to the nearest tenth. l s

0

1

4

b) Plot speed versus skid length, and draw a smooth curve through the points.

10

0

c) It cost $14 000 to remove pollutants from the waste water. What percent of pollutants was removed?

2


4

6

c) Use the graph to find the speed of a vehicle for a skid length of 20 m.

9

16

25

36

100 90 80 70 60 50 40 30 20 10 0

10

20

30

40


50

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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

1. 64

2. 25

3. 121

4. 0.81

33. Given the area, A, of a circle and the formula A , find the radius, r, to the nearest r= 3.14 tenth of a unit.

5. 2.25

6. 0.04

a) 1256 cm2

b) 153.86 cm2

c) 4.5216 m2

d) 7.065 m2

Find the square roots of each number.

Evaluate. 7.

16

8. − 100

9.

0.49

10.

0.09

Estimate. Then, calculate, to the nearest hundredth. 11.

62

12. − 38

13.

110

14.

964

15.

2828

16.

42 000

17. 5 66

18. −7 88

19.

0.5

20.

0.68

21.

0.0039

22.

0.002

Decide whether each equation is true or false. 23.

4 + 16 = 20

24.

4 6 = 24

25.

18 − 5 =˙ 6.5

34.

1 m2 = 10 000 cm2

a) For a square metre, what is the length of a side, in centimetres?

b) For a square, the length of a diagonal can be determined using the formula d = 2 a 2 , where d is the length of the diagonal and a is the side length. For a square metre, find the length of a diagonal, to the nearest tenth of a centimetre.

35. The Pythagorean relationships for right triangles are c=

a2 + b2

b=

c 2 − a2

Find the length of each unknown side. a)

b) c

28.

c2 − b2

where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

26. 3 10 + 4 =˙ 13.5 27.

a=

10 m

4 cm

20 ÷ 5 =˙ 4

b

7m 3 cm

50 − 2 25 = 0

Evaluate for x = 3 and y = 2. 29.

36 x − 2y

30.

3x + 2 y

c)

d)

8m

11 cm b

31.

80

xy

32.

12 x + y 2


10 m

17 cm

a


Name

Expanding and Simplifying Expressions I (The Distributive Property) To use the distributive property, expand the expression by removing the brackets and simplifying. 1. What is the length, width, and area of the rectangle, represented by the algebra tiles? x x

Expand and simplify. 19. 3x + 2(5x – 3)

+1 +1 +1 +1 +1 +1

1 20. 14 – 3(4n – – ) 3 Model the expressions using algebra tiles or drawings on grid paper. Then, expand. 2. 2(x + 2)

21. 3(2h – 3) + 2(h + 3)

22. –2(3y – 3) + 3(2y + 2)

3. 3(x + 3) 23. –6 + 5(2 – k) – 4k 4. 2(2x + 1) 24. 4(3u – 1) + 2(3 – 2u)

5. 3(2x + 2) Expand.

Expand and simplify.

6. 4(x + 2)

25. 2(x2 + 2x + 1) + 3(x2 + 3) 7. 5(x – 3) 8. 0.3(x + 5)

1 26. 5(y – 2) – 4(2y – – ) 2

9. 4(2x + 1)

27. 3(t2 – 2t + 1) – 4(t + 2)

1 10. – (3x – 2) 2

28. 2(e – 4) + 4(3e + 2) – 5(2e – 4)

Expand. Circle the letter corresponding to the correct answer. Rearrange the letters to find a message.

29. x(2x – 3) – x(4 + x)

11. 12. 13. 14. 15. 16. 17. 18.

–3(x + 2) –(b – 4) 2(7 – 5f) –4(6n + 2) –3(4 – 2y) –5(2t – 2) –4(3c + 3) 3(2 + 3k)

L N L P T H S I

–3x – 6 –b – 4 14 – 10f 24n + 8 6y – 12 10 – 10t –c – 1 6 + 9k

M A M G R E R A

3x + 6 –b + 4 9 – 7f –24n – 8 –12 – 6y 10t – 10 –12c – 12 9k – 6

30. a) Write the area of the large rectangle.

b) Write the area of the shaded rectangle in expanded form. x

x

x 5

____ ____ ____

____ ____ ____ ____ ____



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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 and simplify. Expand and simplify.

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)

20. 2a(a + 2) + 4a(a + 1)

21. 3r(r – 3) – 2r(r + 2)

22. k(4k – 2) – k(k + 3) Expand. 7. 4r(r + 3)

8. 6m(m – 2)

9. 2x(3 – x)

10. –3y(5 + y)

23. –d(3 – d) + 2d(d + 5)

24. 4x(x – 1) – x(2 – x)

11. The sum of two consecutive integers can be found using the expression n + (n + 1). a) Write an expression for the product of two consecutive integers.

Expand and simplify.

b) Write and simplify an expression for the difference between their product and their sum.

26. 3x(x2 + 2x – 8) – 2(x – 1)

25. 2(a2 + 3a – 10) – a(a + 2)

27. 2(y – 1) + y(y2 – y – 2) For each question, expand and simplify. Then, locate the answer in the column to the right. Write the letter that follows the answer in the box below that corresponds to the question number. Finally, answer the question “What shape is a square when it starts to ‘wilt’?” 12. 13. 14. 15. 16. 17. 18. 19.

n(n + 4) – n(n – 2) n(4 – n) + n(n – 3) n(n + 3) – (6n – 4) n(3n – 2) + n(2n + 3) n(4 – n) + n(n – 6) n(2n + 3) – n(3 – 2n) n(n + 7) + n(n + 1) n(n + 1) + n(–2 – n)

4n2 2n2 + 8n –n n2 – 3n + 4 5n2 + n 6n n –2n

B R S H M O U A

28. –2r(r + 5) + 3r(r – 3)

29. Write, expand, and simplify an expression for the area of each face of the prism and then, for the total area of all the faces. a)

2n – 3 3n

n

b)

x

2x + 3 4.2x

16.

82

18.

14.

12.

15.

17.


13.

19.


Name

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. Simplify.

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

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

Find the missing exponent. 10. 32 × 3 = 34

11. 5 × 54 = 57 Find the missing exponent.

12. 8 × 8 = 8

13. 7 × 7 = 7

14. j5 × j = j8

15. b × b5 = b9

16. k × k9 = k

17. s6 × s = s7

3

5

3

4

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

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

Find the missing exponent.

27. 2 ÷ 2 = 2 5

3

28. 3 ÷ 3 4

52. 23 × 22

30. 5 ÷ 5 = 5

31. n4 ÷ n = n2

32. c ÷ c4 = c3

33. y ÷ y2 = y2

34. z9 ÷ z = z

4

2

=3

29. 4 ÷ 4 = 4 2

Find the value of each expression. Replace the blanks with the corresponding letter or symbol to decode the message.

3

54. 2 ÷ 2 4

3

T

P

3 2

55. (2 )

C

O

57. (26)2

K

58. 2 × 2

E

4 2

59. (2 )

*

60. 22 × 22

F

61. 22 × 2

R

W

63. 2 ÷ 2

R

56. 213 ÷ 23

3 3

62. (2 )

21 Copyright © 2001 McGraw-Hill Ryerson Limited

53. 29 ÷ 22

E

22

23

24

25

12

26

27

28

29

210

211

212


83

Name

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 Complete the table. Exponential Form 1. (–2)3 2. 3. 5 4. 5. 6.

Is each statement true or false? Base

Exponent

3

1

–3 –2 –7

3 5 2

Standard Form

7.

Repeated Multiplication

(–4) × (–4) × (–4)

9.

(–5) × (–5) × (–5)

10.

( −4) × ( −4) × ( −4) ( −4)

( −2) × ( −2) ( −2)

Write in standard form.

4

3

31. (–a)4 ÷ (–a)2 = a2

32. (–5)3 ÷ (–5)2 = 5

33. t3

34. 6s2

35. s3 + t2

36. 2s3 ÷ 3t

37. –3st

38. –2s2 – 4t

Area (cm2)

40. If the base of a power is negative and the exponent is five, the standard form of the number is negative. Explain.

(–3)5 ÷ (–3)2

15. 32 × 33

30. ((–2)3)3 = –512

Radius (cm) 10 5 2.5 1.3 6.2

( +5) × ( +5) × ( +5) ( +5) × ( +5)

14.

29. y2 × y4 = y6

39. The formula for the area, A, of a circle is A = πr2, where r is the radius. Complete the table, rounding to the nearest tenth of a unit.

(+5)4 ÷ (+5)2

12. 13.

Standard Form

(–3)2 × (–3)2

8.

11.

28. 6(–2)3 = 48

Evaluate for s = –3 and t = 2.

Complete the table. Exponential Form

27. 33 = 81

16. (–2)3 × (–2)2 2

2

41. The standard forms of the following pairs of are not the same. Explain. a) (–2)4 and –24

17. (5) (5)

18. (3.2) (3.2)

19. ((y)2)3

20. (3)4 ÷ (3)2

b) ((–3)2)3 and –36

21. ((–2)2)5

22. (–4.5)3 ÷ (–4.5)

42. A manufacturing company determines its profit using the formula P = 120n – n2 – 220. P is the profit, in dollars, and n is the number of items manufactured. How many items must the company produce to begin to make a profit?

23.

35 33

25. –(1.2)2

84

3 24. ( −7 ) 2 ( −7 )

26. (–0.6)2



Name

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. 1. a) Tell how to multiply two monomials with the same variable, for example, (5n)(2n2).

Write the expression for the area of each figure. Then, simplify. 24.

b) Tell how to multiply two monomials involving more than one variable, for example, (2yz)(–3y3).

2x x

x

25.

Multiply.

3x

2. (2x)(x)

3. 3n × 2n

4. y × z

5. (2k)(2a)

For each of the following figures, write and simplify an expression for a) the area of each face

Multiply. 6. 4v × 2w

7. 2s × 5t

8. (3a2)(4b)

9. 4f × 5g2

10. 4xy × 2z

11. (3cd)(4e)

b) the total surface area c) the volume 26. 2b

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

4n

27. 2n

n

Draw and label a box with these dimensions. Then, find its volume. 28. a × 2a × 3a

20. (5x)(4y)(–3z)

21. –2d(3d)(3e)

22. (–k2mn2)(4mn)(–2kn2) 23. What is the area of the rectangle?

29. (4c)(4c)(4c)

0.5s 1.5s Copyright © 2001 McGraw-Hill Ryerson Limited


85

Name

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 1. a) Tell how to find the power of a monomial, for example, (–a3)2.

Simplify. 21. (yz)2(y3z) 22. (–2ab)(–ab)2

b) Check by simplifying (–a3)(–a3). 23. (5s2t2)2(–st) Simplify.

24. (–4k2m3)2(2km)3

2. (y)2

3. (x)4

4. (–s2)2

5. (c3)2

6. (–m5)3

7. (f 3)6

25. (2r2s2t)(3rst)2 26. (4abc)2(2a2bc)(ab3c3) 27. (m2n2p2)2(mnp)(–3mn3p3)

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

For each cube, write and simplify an expression for a) the area of each face b) the volume 28.

3

2 2 3

12. (rs)

13. (–a b )

xy 2

14. (–f 3g2)2 29. FI r3s3 ND y2z2

FO a2b3 YW y3z3

TH c2d4 SH f 6g4

ES s4t4 IN s2t2

RT c5d2 RN –a6b6

LA –x6y3 KS x2y3

2a 2 bc 3

Explain the error in each case and correct it. 30. (a2)3 = a5

Simplify. 15. (3ty)2

16. (–2xz)3

17. (–2a2b)3

18. (3r3s)2

19. (5k3m2)2

20. (–3q2r2)3

31. (fg3)2 = f 2g9

32. (–2x2y2)2 = –4x4y4

86



Name

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. Divide. 1.

4r 2

2.

−6 s 3

3.

−5 a −5

4.

15x 3x

4t −t

6.

5.

−12 g −3 g

25. The perimeter of a square is 8s2t. Write and simplify an expression for the length of a side.

Find the missing dimension in each rectangle. 26.

2y

Divide. 7. 14ab ÷ 7a 9.

8. 10klm ÷ 5

12rt 4rt

11. –9xyz ÷ (–3xyz)

10.

8bcd −4bd

12.

−16 pqr 4 pqr

27.

28.

Simplify. 13. 2a b ÷ a b 4 3

14.

A = 4xy

A = 8r 3 s 4

4s 2

A = 16a 4 b 3

2

8a 2 b

3 2

6q r 3q 2 r 2

15. 8x6y4 ÷ (–4x3y2) 16. –4w3x5 ÷ (–2w2x2) 17.

−9 f 3 g 5 h 2 6 fg 2 h

18.

−12c d 18c 5

29. What are the dimensions of the rectangle if the ratio of length to width is 3 to 2? A = 24x 2 y 2 6x y

5

Simplify. 6k 2 m4 19. 3km 2 20. 4a3b2c ÷ 2bc

30. What are the dimensions of each face of the rectangular prism?

21. 8x5y3 ÷ 2x3y −12 s7 t 6 22. 8s 2t 2

A = 8x 2 z 2 A = 4xy 2y

23. –9e2f 4 ÷ (–6ef 2) 24.

20d 5 e 3 f 5 12 d 2 e 3 f 4



87

Name

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. the points if the domain is R. The domain of each equation is {–2, –1, 0, 1, 2}. Complete a table of values. Then, graph each equation. 1. y = –2x + 5

Given the tables of values, write an equation for each relation. 7.

x –2 –1 0 1 2

2. 2x + y = 0

8.

x –2 –1 0 1 2

y –6 –3 0 3 6

9. Given the points on the grid, write an equation to represent the relation. Then, state the domain.

y

y

y 2 3 4 5 6

4 8

y 2

(4,2)

2

6

(2,1) –2

0

2

x

4

–2

0 (0,0)

2

4

x

(–2,–1)

–2

–2

2 –4 –2

0

2

10. The table shows the equivalent depths of water and heavy wet snow.

x

Depth of 5 Water (cm) Depth of Heavy 100 Wet Snow (cm)

Graph each equation. The domain is R. 3. y = 2x + 1

4. y = –x – 3

y

y

–2

0

2

x

2 –2 –2

0

2

x

a) Plot the depth of heavy wet snow versus the depth of water.

10

15

20

25

200

300

400

500

500 400 300

–4

b) Can you the points to graph the equivalent depths of water and heavy wet snow? Explain.

–2

5. x + y = –1

6. 2x – y = 3

y

y 2

2 –2 –2

0

2

0

x –2

–2 –4

–6

88


2

x

200 100

0

5

10

15

20

25

c) Use the graph to estimate the depth of heavy wet snow that is equivalent to a depth of 12 cm of water. d) Use the graph to estimate the depth of water that is equivalent to a depth of 360 cm of heavy wet snow.


Name

Graphing Equations II (Methods for Graphing Linear Equations) To draw the graph of a line, • use the x- and y-intercepts • use the slope and the y-intercept

• use a table of values

Use the x- and y-intercepts to graph each line.

Find an equation for each line.

1. 2x + 3y = 6

9.

2. 4x – y = 4 y

y –2

10. y

y

0

(0, 4)

(0, 5)

4 x

2

2 –2 –2

0

4 x

2

–4

x

0

(–7, 0)

–2 0

3. 3x + 5y – 30 = 0

4. x + y + 2 = 0 y

y 4 –2

0

2

x

(4, 0) x

11. Write an equation of a line whose x- and y-intercepts are opposite integers, but not 0, and whose x-intercept is positive.

2 –2 0

2

4

12. Write an equation of a line whose y-intercept is 0 and whose slope has a negative value.

x –4

Describe the slope and the intercepts of each line.

Graph each equation using the slope and the y-intercept. 5. y + 1 = 2x

13. y = –3

6. x + 4y = 8

y

y

2 2 –2

0

14. x = 5

4 x

2

–2

–2 0 –2

2

4

6

8

x

15. After 10 s Beth counted 14 heartbeats, after 30 s she counted 42 heartbeats, and after 35 s she counted 49 heartbeats. a) Plot the ordered pairs (time, heartbeats).

–4 –6

Graph each equation using a method of your choice. Find the intercepts and slope for each line. The domain is R. 7. y = x + 5

y

4

–4 –2 0 –2

60 50 40 30

c) Find an equation of the line.

8. 5x + 2y = –20 y

b) Find the slope of the line. What does the slope represent?

20 10 0

2

4

–4 –6

2

6

10

20

30

8 x

d) Use the equation to find Beth’s pulse rate in heartbeats per minute.

–8 –4

–2

0

x

–10



89

Name

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. 7. Complete the table of values for each relation and find the point of intersection.

For each graph, identify the point of intersection. 1.

2. y

a) y = x + 5

y

0

2

6 x

4

2

x 0

–2 –2

0

y

x

10

5

13

x

2

b) y = 2x – 8

–4

y –8 18

–2 –6

Four pairs of lines are defined by the tables below.

Make tables of values and graph each pair of lines. Find the point of intersection.

a) The point of intersection of two pairs of lines can be determined from the tables. Identify these points.

8. y = 2x + 1 y=x+3

y 4

b) Graph the other two pairs of lines and determine the points of intersection as accurately as possible from the graphs.

2

–2

3.

5.

x y 1 5 2 3 3 1 4 –1

x 1 2 3 4

x y –4 3 0 1 2 0 6 –2

–2

y –4 –3 –2 –1

4.

x y –1 –6 2 0 3 2 5 6

6.

x y 0 4 1 2 2 0 3 –2 x –1 0 1 2

y –7 –5 –3 –1

y

y

2

2

0 –2

2

x

–2

x 0 1 2 3

0 –2

–4

y 0 1 2 3

x y 3 1 3 0 3 –1 3 –2

2

x

0

2

x

4

x

–2

9. 2x + y = 7 x–y=5

y 4

2

0

2

–2

–4

10. The screen of an air traffic controller shows two planes approaching at 10 000 m. One plane is travelling in a direction described by y = 4 – 2x, the other in a direction described by x – y = –1. Determine if the planes can continue in the same manner. Give a reason for your answer.

–6

90



Name

Greatest Common Factors To find the greatest common factor of two or more , first write all the factors of each term. Then, determine which factors are common to all the , and write the product of the common factors. Complete the table. 1. 2. 3. 4. 5. 6.

Number 20

Determine the GCF of each set. Prime Factors 2×3×5

26. 27, 63, 81

27. 8x, 12y, 28a

28. 15r2, 20r3, 5

29. a3, 9a2, 3a

30. s2t2, s3t2, s3t3

18 2×3×3×3 150 252

31. 4xyz, 12x2y2z2, 8xy2z3

Complete the table. 7. 8. 9. 10. 11.

25. 6, 9, 15

Expression 6x2

Prime Factors

32. 14c2de, 28cd2, 21ce2

2×3×5×s×t×t 2

3

12a bc

2×5×a×a×b×b×b 2 2

24x y z

33. Lyndon has a piece of cardboard 60 cm by 75 cm. What is the largest size of identical squares he can cut from the cardboard if he uses all of it?

Factor fully. 12. 6m2n2 34. A patchwork quilt to be made of identical squares must measure 150 cm by 210 cm.

13. 51a2 14. 76r2s

a) What different sizes of squares are possible so that each size completely covers the surface?

Determine the GCF of each pair of numbers. 15. 12, 28 b) What is the largest size of squares that can be used to cover the surface exactly?

16. 15, 60 17. 24, 42 18. 54, 81 Determine the GCF of each pair of monomials. 19. 15a, 25a 3

2

20. 3x , 12x

35. Two rectangles share a common side. The area of one rectangle is 4ab and the area of the second rectangle is 6ab. a) Draw a diagram showing the attached rectangles and label their dimensions.

21. 18xyz, 24x2y 22. 12c, 16d 23. 6st2, 8s2t

b) Are other dimensions possible? If so, draw and label the diagrams.

24. 4p2q2, 6p3q3



91

Name

Like Like have the same variable raised to the same exponent. r, 4r, 101r Unlike have different variables or the same variable, but different exponents. 7b, –3a, x2, x 1. List the like . 4r –r2

3

r

(–r)2

Simplify. –r

101r

r

–2r3

2

5r r3 2

21. 2c + 3 + 4d – c + d 22. m – 2 + 2n + 5 – n 23. –2 + 2z + (–3w) + 4 + z

2. a) State the number of in the expression –x2 + 5x – 2xy + 3. b) State the coefficients and the constant term in the expression.

Write an expression for each perimeter in 2 different ways.

Simplify. 3. 11t – t

24. 3 + 4x2 + y2 + x2 – 1

25.

4. –10b2 + 3b2

2r 3

3

5. –12y – y

6. 11m + 10m

7. 5p + (–2p)

8. c2 + c2

3r

26.

0.5s

Simplify. 9. 2x + 3x – x

10. –5y + 2y – 9y

1.5s

11. 0.4d + 0.5d + 0.1d Using the given information, write a problem.

12. –t2 – 2t2 – 3t2 + t2 + 6t2

27. Monique travelled p kilometres the first day, (p + 6) kilometres the second day, and 2p kilometres the third day.

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

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 elephant is (2z – 0.1) metres tall. 31. For each problem in questions 27–30, write and simplify the algebraic expression.

20. –k + (–3k) – 2k + 2 for k = –3

92



Name

Polynomials A 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 , and a trinomial has three . 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. Identify as a monomial, a binomial, or a trinomial. 4

3

1. 3 + 4k

2. r – s + 6

4 3. – x2y 5

3 4. a + – b – c 4

Arrange the in each polynomial in ascending powers of y. 26. y3 + xy2 + y + 2 27. –2x2y + 3xy3 + x3

2

5. 0

6. 3a t – 5a

State the degree of each monomial.

3 28. – – y + y3 – y6 4

7. 2xy2z

8. 14k

9. –4ab

Arrange the in each polynomial in descending powers of x.

10. 7

11. 5xst3

12. –36wz4

29. –6x + x4 + 2x3 – 10

State the degree of each polynomial.

30. 0.2mx4 – 1.3x5 + 0.4m2 + 2.1x3

13. 9t + 8s

14. 22x2 + 22y

2 31. 4b + – bx + b3x2 + x4 3

15. n – 2p3

16. 11wxyz – 9w4

32. a + x

17. –2a2b3 – ab + b6

18. 3kmn + 11k2m – 10kn3

Draw and label a figure that shows the perimeter. 33. s + t + w

State the degree of each monomial. 19. The circumference of a circle is πd.

34. 2m + 2n

20. The area of a circle is πr . 2

21. The volume of a cylinder is πr2h. 1 22. The volume of a cone is – πr2h. 3 Simplify each expression. Then, classify the resulting polynomial and state the degree. 23. n + n + 1 + n + 2 1 24. bh – – bh 2 25. πr2 + πr2 + 2πrh


35. 4k

36. a) Write an expression for the perimeter of the triangle. 2x 2 + 3x – 7

x 2 + 5x – 3

3x 2 – 8x + 5

b) Is the degree of the polynomial for the perimeter the same as the degree of the polynomial for each of the sides? Explain.


93

Name

Slope I (Using Points) To find the direction of a line and how steep it is, find the slope using the coordinates of any two points on the line, (x1, y1) and (x2, y2). y 2 − y1 (change in y-values is the rise) ∆y (slope) m = , or m = x2 − x1 (change in x-values is the run) ∆x 1. Calculate the slope of each line segment, where possible. B

y 4

C

G A

I

2

E

F

12. The coordinates of point A are (–1, –5). If the 2 slope of the line is − , name the coordinates of 3 another point on the line.

L H

–4

–2

0

11. One point of a line is in the first quadrant and another point of the line is in the second quadrant. If the slope of the line is positive, name two points on the line.

2

6 x

4

D

–2

K

J

Find the slope of the line ing through each pair of points.

13. Given the equation of each line, find two points on the line and calculate the slope. a) y = 2x – 5

2. (4, 5) and (0, 0) b) x – y = 8

3. (–2, –6) and (–7, –1) 4. (3, –2) and (–6, 5)

14. The slope of a line is 2. The line es through (–1, c), (0, –4), and (d, 4). Find the values of c and d.

5. (3.7, 5.1) and (–1.5, 1.2) 6. (–1, 5) and (3, 5)

1 . The line es 2 through (4, –1). Name the coordinates of two other points on the line.

15. A ladder is leaning against a wall. The base of the ladder is 1.5 m from the base of the wall. The top of the ladder touches the wall at a point 4 m above the ground. What is the slope of the ladder?

Given a point on the line and the slope, draw the graph of each line. 1 9. (–3, –1), m = –3 10. (2, –5), m = 2

a) Use slopes to determine whether each set of points is collinear or non-collinear. b) If the points are collinear, find the slope.

7. (2, –7) and (2, 4) 8. The slope of a line is

y

y 2

–4

–2

0

0

2 x

–2

–2

–4

–4

–6

2

4

x

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)

94



Name

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). Find the slope and y-intercept of each line. Then, write an equation of the line.

Find the slope and y-intercept of each line. 1 2. y = − x + 2 3

1. y = 2x – 5

3. x + y – 1 = 0

16.

17. y

y

2

2

4. x – 5 = 2y –4

–2

2 x

0

–2

0

–2

–2

–4

–4

2

x

Find the slope and y-intercept of each line. 5. 3x + y = 2

6. 4x – 3y = 9

7. 2x + y = 4

18. An equation of a line is y = –x + b. Find the value of b if the line es through the point (1, –3). What is the value of m?

8. y = 5

Given the slope and y-intercept, write an equation of the line in the slope and y-intercept form. Then, write the equation in standard form. 1 10. m = –1; b = 2

9. m = 5; b = 2

2 11. m = ; b = –1 3

19. The equation 35n – t + 50 = 0 relates t, the total cost, in dollars, of boarding a dog in a kennel, with the number of nights, n, the dog is boarded, and the cost of medical insurance. a) Write the equation in the form y = mx + b. b) Graph the total cost versus the number of nights of boarding.

1 1 12. m = − ; b = − 4 3

300

200

13. Find the slope and y-intercept of the line through the points (–2, 11) and (10, –7).

100

0

2

4

6

8

Draw the graph of each line. 1 15. y = − x − 1 2

14. y = 3x + 1 y

c) What is the total cost to board a dog at the kennel for 7 nights?

y

4 –2

0

2

x

d) What is the slope of the line?

2

–2

0

2

–2

e) What does the slope represent?

–4

f) What is the cost of medical insurance?

x

–2



95

Name

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. Given the slopes of two lines, determine whether the lines are parallel, perpendicular, or neither. 1. m1 = 2, m2 = −

3 6

3. m1 = 5, m2 = –5

2 2.4 2. m1 = − , m2 = − 3 3.6 2 4. m1 = , m2 = –1.5 3

Identify whether each pair of lines is parallel, perpendicular, or neither. 1 13. 3x + y – 2 = 0 and y = − x + 2 3 14. 5x – y + 2 = 0 and 10x – 2y – 17 = 0 15. x + 2y – 2 = 0 and 2x – y + 3 = 0

Find the slope of a line perpendicular to a line with the given slope. 1 5. 2

6. –3

7. undefined

State the slope of a line a) parallel to each line b) perpendicular to each line

16. Determine an equation for the line ing through (–3, –2) and perpendicular to 2x – 3y = 3.

17. Determine an equation for the line parallel to 3x + y – 2 = 0 and having the same x-intercept as x – 2y + 5 = 0.

8. y = 3x – 1 Plot and the points in order. Classify each figure as a square, a rectangle, a parallelogram, or a trapezoid.

9. x + 4y = 5

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)

10. 2x – 6y – 3 = 0

y

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?

4 4 2 2

b) a parallelogram?

–4 –4

12. The slopes of two parallel lines are –3 and m . Find the value of m. 5 96

y

6


–2

0

2

–2

0

2 x

x

20. The equation of the path of a enger ship is given by 3x + y – 2 = 0. The equation of the path of a cargo ship is given by 6x + 2y + 10 = 0. If the ships remain on their given paths, are they likely to collide? Explain.


Name

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 scale or an equation, the equality is maintained. To solve an equation, isolate the variable on one side of the equation by adding or subtracting the same amount from both sides of the equation. To represent x as a variable, draw to represent 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 to represent each equation. 1. x – 2 = 6

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

=

2. x + 3 = 5

=

3. –4 = x – 3

=

25. y − 1 27. −

4. –5 = 4 + x

1 =7 2

1 +s=0 3

26.

11 2 =b+ 12 3

28. z −

7 1 =− 8 4

=

Solve and check. What number must be added to both sides of the equation to solve it? 5. b – 11 = 25

6. r – 3 = 8

7. –5 = –2 + k

8. x – 6 = –3

9. t −

1 =5 2

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

10. 1.3 = d – 1.4

What number must be subtracted from both sides of the 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


35. At one Winter Olympics, Canada won 2 gold medals. In the same year, Canada won 4 more gold medals in the Summer games than it did in the Winter games. Solve the equation x – 4 = 2 to determine the number of gold medals won in the Summer games.


97

Name

Solving Equations II (Using Division and Multiplication) To solve an equation, isolate the variable on one side of the equation by multiplying or dividing by the same amount on both sides of the equation. To represent x as a variable, draw to represent 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 to represent each equation. 1. 2x = 4

Solve. 14. 5t = –30

15. 3c = 24

16. –9r = 36

17. –56n = –8

=

18. −5 = 2. 3x = –6

3. −4 =

x 2

=

=

By what number must you divide both sides to solve each equation? 4. 6x = 60

5. 3y = –21

6. 15 = 5t

7. –36 = 9k

By what number must you multiply both sides to solve each equation? x k 8. 9. =5 = −2 5 7 y a 10. −8 = 11. 3 = 4 −8 Find the missing number. Then, check. m –y = 23 13. 12. − =8 4 –y × (–1) = 23 × (–1) m =8 y= −4 m (−4) × = −4 × 8 −4 m=

98


p −5

20. −10 =

k 15

19.

m = −3 3

21. 12 =

23. −

22. 8 = –y

s 7

x = 10 5

Solve and check. 24. 3x = 4.8

26.

p = 2.4 3

25.

s = −1.5 4

27. 2.5r = –7.5

Estimate. Then, use a calculator to solve. c 28. –291n = 20 661 29. − = 11 725

30. Circle the equation that represents the number of five-dollar bills in a pile that totals $60. x x a) 60 = b) = 5 c) 5x = 60 d) 60x = 5 5 60


Name

Solving Equations III (Multi-Step Equations) To solve equations involving more than one step, follow the process shown in the flow chart. Start Equation

Simplify both sides of the equation.

Add or Subtract the same value from both sides of the equation.

Draw a flow chart to show the solution steps for each equation. 1. 3x + 4 = 25

2. 4y – 3 = 21

Multiply or Divide both sides of the equation by the same value to isolate the variable.

Solve. x 16. =6+4 2

17.

Stop Solution

y = 11 − 3 4

18.

t −3=1 4

19. 5 +

20.

1 m −2 + = 5 5 5

21.

k = −2 3

n 3 −1 − = 2 4 4

Solve and check. 22. 3x + 2x + 3 = 13 3. 5 – x = 11

23. 6a – 3a + 5 = 14 24. 6 – 3c = 10 – 7

Solve. 4. 4t – 6 = –10

5. 5r = 9 + 6

25. 3b + 2b – b = 15 – 7 + 4 26. 3r + r + 2r – 6 = 11 + 2 – 7

6. 8m + 2m = 30

7. 4w + 3w = –28 Solve.

8. –3 = –2 – x

Solve and check. 10. 4b – 6b = 12

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

9. –5 – y = 6

11. 5 + 2n = –15

12. 4 + 8 = 3m

13. 3p – 7 = 14

14. –14 = 4y + 3y

15. 8a – 3a = 15


33. In 1992, the first year the Toronto Blue Jays won the World Series, the Jays played 162 regular-season games and 12 championship games. They won 34 more games than they lost. Solve the equation x + (x + 34) = 162 + 12 to find the number of games the Jays won and lost.


99

Name

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. Start Equation



Simplify both sides of the equation.

To represent x as a variable, draw to represent 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 to represent each equation. 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

Stop Solution

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 of energy. Doing yard work for 20 min uses approximately 2x kilojoules of energy. The number of kilojoules used doing yard work is equal to the number used washing a car plus 65. Solve the equation x + 65 = 2x to find the number of kilojoules of energy used to perform each 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.

Solve and check. 10. 3x + x = 5x – 6

b) Find the number.

11. 2t – 5t = t + 8 12. 3y + 2y = 3y + 6

22. Six more than 5 times a number is the same as 9 less than twice the number.

13. 4k – 6k = 6 – k

a) Write an equation to show this relationship.

14. 7m + 8m = –10 + 5m

b) Find the number.

15. 12 = 6b + 2b – 4 100



Name

Solving Equations V (With Brackets) To solve equations with brackets, follow the process shown in the flow chart.

Start Equation

Simplify both sides of the equation starting with the brackets.

Solve.



Stop Solution

Solve.

1. 3(x + 1) = 24

2. 2(x – 2) = 8

19. 4(x + 2) – 3(x + 1) = 2(x + 2)

3. –4(y + 3) = –16

4. 10 = 5(t + 2)

20. 3(n – 5) – (2n + 2) = 2(n – 1)

5. 2(z – 3) = –12

6. –15 = 3(k + 3)

21. 2(a – 4) – 3(a – 2) = 4(a + 1) + 4

Solve and check.

22. 5(c + 4) = 4(2c – 3) – 7

7. 4(2x – 2) = –16 8. 15 = 3(2y – 3) 9. 2(2k + 4) = 14 10. –12 = 3(2k + 2) Solve.

23. The largest flag in the world was presented to the city of Kao-hsiung, Taiwan, by the Republic of China. The diagram shows the dimensions of the flag. The equation 2(2x + 70) + 2(9x + 63) = 420 represents the perimeter of the flag, in centimetres. 9x + 63

11. 3(x + 2) = –9 – 2x P = 420 cm

2x + 70

12. 12 + 2(k + 3) = 3k – 6 13. 8 – 3y = 2(2y – 3) 14. 3(n – 2) – 19 = 5 + 2(n + 5)

a) Solve the equation.

Solve and check. 15. 3(x + 2) = 9 + 2(x + 4)

b) What are the dimensions of the flag?

16. 7 + 2(b – 3) = b + 4 17. 3(2 – 2z) = 1 – z 18. 3(4k – 1) + 2(5 – 3k) = 7k


24. The sides of a triangle, in centimetres, are given by the expressions 3(n – 2), 4(n + 3), and 2(n + 4). The perimeter of the triangle is 140 cm. Find the length of each side. Appendix A: Review of Prerequisite Skills

101

Name

Solving Equations VI (With Fractions and Decimals) To solve equations with fractions and decimals, follow the process shown in the flow chart.

Start Equation

Multiply each term of an equation with a) decimals by a power of 10. b) fractions by the lowest common denominator.

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

7. 0.5x = 1.5

9. 2 = –0.5d


10. 1.6y = –6.4

20.

2t t = −1 3 2

21.

3a 2a 5 − = +a 4 3 6

22.

(b + 1) = (b − 2)

23.

( 3 − k ) = ( k + 3)

24.

(x + 1) − x = 2

2

Solve and check. 26.

6

2

3

c+3 c+1 − 4 2

1− z z + 1 = −1 5 2

27. The mass of a snapping turtle is given by the expression (3x + 1.7) kg, while the mass of the case used to ship the turtle is given by the expression (x + 5.2) kg. The total mass of the turtle and the case is 39.9 kg.

12. –1.6 = 2 – 0.4y 13. 0.5a – 1 = 2 + 0.6a 14. 1.2 + 1.4y = 1.5y + 0.63 Solve. x 1 15. = 6 2

5

4

25. −4 = 11. 5x + 0.8 = 1.2

Stop Solution

Solve and check. x x 19. = +1 4 2

6. g – 3.4 = 1.65

8. 2.4s = –4.8


a) Find the mass of the turtle.

16.

y 1 =− 5 10

18.

−x 1 = 2 3

17.

−2 k = 5 2

102


b) Find the mass of the shipping case.


Name

Solving Proportions Ratios that make the same comparison are equivalent ratios or equal ratios. A statement that ratios are equal is a proportion.

For each ratio, write two equivalent ratios. 1. 1:4 3.

10 4

5. 1.5 to 4.5

2. 6 to 3 4. 3:7

6.

0.2 0.5

Write = or ≠ in each to make each statement true. 2 4 8. 3:5 9:15 7. – – 3 6 9. 5 to 2 10 to 5

Solve. y 1 11. = 16 4 13. 4:a = 2:5 15.

1.5 0.9 = d 1.8

10. 14:10 5:7

12.

3 75 = 4 x

20. The length of the shadow of a tree measures 6.2 m, and the shadow of a fence post measures 2.8 m. If the fence post is 1.4 m tall, how tall is the tree?

21. The ratio of uncooked rice to cooked rice, by volume, is 2 to 7. Complete the table. Uncooked Rice 1L 500 mL 250 mL

Cooked Rice

280 mL 1.4 L 2.45 L

14. 6:2 = c:6 16.

5 2 = 4.5 t

22. About 3 out of every 5 Canadians are far-sighted, while about 3 out of every 10 are near-sighted, and the rest have 20/20 vision.

17. To make concrete, 6 bags of cement are mixed with 4 bags of sand. How many bags of cement are needed to mix with 12 bags of sand?

a) Write the ratio of Canadians who have 20/20 vision to those who do not.

18. Scientists estimate that 8 out of every 9 people are right-handed. In a school of 360 students, how many students would you expect to be right-handed? left-handed?

b) How many more times likely is a Canadian to be far-sighted than near-sighted?

19. A basketball player makes 1 basket for every 2 shots missed. a) Write the ratio of baskets made to shots taken. b) How many baskets would you expect the player to make out of 138 shots?


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 using the following data: 8 out of 10 dentists recommend sugarless gum.


103

Name

Subtracting Polynomials To subtract polynomials, add the opposite of the polynomial that is being subtracted. Subtract expression B from expression A, represented by the algebra tiles. 1.

Subtract. 11. (b + 5) – (b + 2)

A

12. (2c – 3) – (c – 3)

B

13. (–2k – 4) – (–k – 2) 2.

A

14. (n2 + 3n + 2) – (n2 + 2n + 1)

B

15. (3w2 – w – 4) – (w2 – 3) Write the opposite of each polynomial. 3. 5ab + 6

16.

e2 + 4e + 6 e2 – 2e – 2

17.

–5f 2 – 2fg – g2 f 2 + fg + g2

18.

–3d2 + 4d – 2 –2d2 – d + 2

19.

2y2 – 5y – 4 –y2 – 3y – 2

4. z2 – z – 4

5. –2c + 3d + e

6. –4s2 – s + 5 20. a) Write an expression for the perimeter of this parking lot. 5m + 8

Model the expressions using algebra tiles or drawings on grid paper. Then, subtract. 7. (x + 4) – (x + 2)

3m – 1

3m + 7 7m – 1

8. (x2 + 3x + 2) – (x2 + x + 1)

9. (2x2 – 3x – 2) – (x2 – 2x – 1)

b) Find the difference in length for each pair of opposite sides.

10. (–2x2 – 2x – 4) – (–x2 – x – 3)

104



Name

Transformations I (Translations) A translation, or slide, is a motion that is described by length and direction.

y

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)

6

X′ 4

1. 3. 5. 7. 9.

[4, 1] 2. (x, y) → (x − 2, y + 1) 3 units right 4. [0, –2] 5 units left 6. (x, y) → (x + 1, y) [2, 2] 8. (x, y) → (x − 3, y − 1) 2 units right, 2 units down

Y

Z

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.

Draw an arrow on the grid to show each translation.

X

0

Y′ 2

4

Z′ 8 x

6

14. a) ____________________________________ b) _______________

c) ________________

15. a) ____________________________________ b) _______________

c) ________________

Complete the table. Original Point For questions 10–15, refer to the grid to a) describe each translation in words b) write the ordered pair that describes each translation c) write each translation as a mapping 11.

10. 14.

15.

13.

10. a) ____________________________________ b) _______________

c) ________________

11. a) ____________________________________ b) _______________

c) ________________

12. a) ____________________________________ b) _______________

(x, y) → (x, y – 2)

17. (1, 0)

4 units up

18. (–5, 0)

(x, y) → (x – 2, y + 3)

19. (–3, 4)

[3, 0]

Image Point

Complete the table. Image Point

20. (4, 2)

(–1, –3)

21. (5, –3)

(5, –5)

22. (–3, –2)

(–3, 1)

23. (0, 0)

(–1, 3)

Translation

24. ABC has vertices A(2, 2), B(4, 2), and C(5, 5). Draw ABC on the grid. Draw and label each translation image. a) [–4, –3] b) (x, y) → (x, y – 3) y

c) ________________

13. a) ____________________________________ b) _______________

16. (2, –3)

Original Point

12.

Translation

x

c) ________________



105

Name

Transformations II (Reflections) y

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.

6

4

X

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. Circle the pairs of figures that are reflections. Draw the reflection line for each pair. 2. F F

F 3. F

7. F F

FF

4. F

5.

9. F

10. FF

F

F

1. F F

l

X′

2

Z 0

Y

2

4

Y′

Z′

H 6

8

10

Draw and label the reflection images for each figure in the reflection lines l and m. 14. 15. Z y y m R

m

T

8.

F

6. F F

F

X l

0

D

B

x

l

y l

0

x

m

16. a) DEF has vertices D(–3, 1), E(0, 4), and F(–2, 3). Draw DEF on the grid.

n

A

S

Y

11. Draw and label the reflection image of figure ABCD in each reflection line. k

x

l

C m

12. a) Draw the reflection image of each line segment in the x-axis. b) Use a dotted line to draw the reflection image of each line segment in the y-axis. y C

2

D A B

–6

–4

–2

G

0

2

F 4

6

b)

(3, 1)

c)

(0, 3)

106

17. RST is reflected in the x-axis. The coordinates of the image R′S′T′ are R′(2, 2), S′(4, 3), and T′(3, 4). Write the coordinates of the original figure.

18. The word MOM has a vertical reflection line. The word BED has a horizontal reflection line.

13. Write the coordinates of the image of each point after a reflection in each axis. Reflection Line (–2, 3)

b) Reflect DEF in the line l.

x

H

a)

x

E –2

Point

0

x-axis

a) Write 3 other words that have a vertical reflection line.

y-axis


b) Write 3 other words that have a horizontal reflection line.


Name

Transformations III (Dilatations) y

A dilatation is a transformation that changes the size of an object. Dilatations are called enlargements or reductions, depending on the way 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. 1. a) Write the coordinates of each line segment and its image in the following diagram. D

y

E

A′ A

2

(2, 6)

6

m

4

(1, 3) (6, 2)

n

2

(3, 1) 0

2

4

6

8

x

10

3. A rectangle has vertices P(3, 2), Q(–1, 2), R(–1, –1), and S(3, –1). Write the coordinates of the vertices of the image of rectangle PQRS under the mapping (x, y) → (3x, 3y).

B′

E′ D′ –6

–4

–2

G

B G′

K 2

0

K′ 4

6

x

J –2

H′

4. Draw ABC with vertices A(1, 2), B(–1, –1), and C(2, –2). Draw the image A′B′C′ with vertices A′(2, 4), B′(–2, 2), and C′(4, –4). y

H

J′

AB

______________

A′B′ ___________

DE

______________

D′E′ ___________

GH ______________

G′H′

__________

JK _______________

J′K′

___________

x

0

a) What is the scale factor? b) What is the scale factor for each dilatation? AB _____

DE _____

GH _____

JK _____

2. Draw and label each dilatation image of XYZ.

b) How do the lengths of the sides of the original and the image compare? c) How do the measures of the angles of the original and the image compare?

a) with dilatation centre (0, 0) and scale factor 3 b) under the mapping ( x , y ) →  1 x , 1 y 2

5. Draw the image of each figure after a dilatation by the given scale factor.

2 

y

a) scale factor 2

b) scale factor 1 3

Y

X

x

0

Z



107

Answers

APPENDIX A: Review of Prerequisite Skills 6. Answers will vary. r = 105°, q = 105°, p = 105°

Adding Polynomials 3. –x2 – 2x – 2

7. s = 100°, r = 130°

8. x = 40°, y = 30°, z = 110°

1. 5x2 + 6x + 3

2. –2x2 – x + 5

4. 3x2 + 3x + 3

5. –3x2 + 3x – 2 6. x2 – 3x – 3

9. d = 65°, e = 65°, f = 50°

7. 3x2 – 2x – 2

8. –x2 – 4x + 1

10. a = 122°, b = 58°, c = 122°, d = 58°

9. 5y + 3z + 2

10. 5ab + 5bc

11. 2k2 – 4kj

11. v = 50°, w = 70°, x = 60°, y = 70°, z = 60°

12. s2 + 3s + 2t3 + t + 7

13. 6a + 3b – 3

14. 5m2 + 6mn + 3n2

15. 4r2 – 10r + 9

12. ∠RST = 120°, ∠SRU = 60°, ∠RUT = 120°, ∠STU = 60°, ∠RVW = ∠SVW = ∠UWV = ∠TWV = 90°

16. 4c2 + 2ac + 10 + a2

17. 2k2 + k – 2

18. 4x3 – 2y + 2

19. 4z3x – z + 4

20. a) 6x + 2

b) 50 cm

Common Factoring 1. 3

Polynomial

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°

4. 5.

GCF of Both

Other Factor

x

x+2

3y

2y – 3

xy

8 – 3xy

x2 + 2x 2

6y – 9y

6.

8xy – 3x y

7.

2m2n – 4mn2

8.

3. 2a2

2. 3b

2 2

2

2mn

2

6a + 9a – 12a

m – 2n 2

3a

2a + 3a – 4

6. d = 35°, e = 110°, f = 145° 7. h = 40°, j = 50°, k = 50°, m = 130°, n = 80°, p = 100°, q = 80°

9. 2(2y + 5)

10. 3m(2m + 3)

11. 2t2(2t – 3)

12. p2q2r(3pr – 4)

8. m = 72°, n = 116°, p = 64°

13. n is also a common factor. 4n(n – 3)

9. k = 128°

14. 4 is not the greatest common factor of the coefficients. 8y2z(2z + 3)

10. r = 90°, s = 72°, t = 121°, u = 77°, v = 103°

15. When multiplying powers, add the exponents. 5x2(5x4 – 1)

11. c = 60°, d = 125°, e = 50° 180o − ∠DBC 12. ∠BCD = ; 2

16. 5(a2 + 2ab – 3b2)

17. 3xy(3x2 + y + 5)

∠ACB = ∠ABC = 180° – ∠DBC;

18. 2st(s2t – 4st2 + 2)

19. 3y(2y – 3x + 4x2y)

∠DCA = ∠BCD + ∠ACB 360o − 90o − ∠S 13. ∠QRS = 2

20. 4c3(3cd + 2 – 4d) 21. a) x, 3 + 5y

b) x(3 + 5y)

22. a) l + l + w + w

b) 2l + 2w

Angle Properties II

23. a) 2x + 4y

1. ∠EDG = ∠FGD, ∠CDG = ∠HGD

24. a) 2πr(r + h)

2. ∠BGH = ∠GDE, ∠ADE = ∠DGH, ∠ADC = ∠DGF, ∠CDG = ∠FGB

Congruent Triangles

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°


c) 2(l + w)

b) 2(x + 2y) b) 471 cm2

c) πr(r + 2h)

1. 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


109

3. LM = PQ, LN = PR, MN = QR, ∠L = ∠P, ∠M = ∠Q, ∠N = ∠R

16. a) 14.5

b) 2.8

17. a) 9.9

b) 28.86

4. ST = XY, SV = XZ, TV = YZ, ∠S = ∠X, ∠T = ∠Y, ∠V = ∠Z

18. a) 1.5n

b) $12.00

5. 1 side

19. a) 6h, where h represents the height of a jump on Earth b) 16.8 m

6. 1 side or the angle between the two sides

20. a) 18 + 4g, where g represents the number of games rented b) $30.00

7. the side between the two angles

21. a) 1 l + 20, where l represents the length of the 2

8. BC = 5 m, EF = 4 m, FD = 3 m 9. ∠KLM = 95°, ∠JGH = 32°, ∠MKL = 53°

field in metres

b) width 75 m, perimeter 370 m

10. ∠TVS = 74°, ∠XYZ = 65°, ∠ZXY = 74°

Evaluating Expressions II

11. SAS, ∠X = ∠A, ∠Y = ∠B, ∠Z = ∠C, XY = AB, XZ = AC, YZ = BC

1. a) 9

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

c) –2

d) –12

e) –2

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

d) –16

e) 1

f) –6

4. a) 4

14. SAS, ∠QPR = ∠SPR, ∠Q = ∠S, ∠QRP = ∠SRP, PQ = PS, QR = SR, PR = PR

5.

m

3 + 2m

2

2

2

7

1

0

1

5

0

–2

0

3

–1

–4

–1

1

–2

–6

–2

1

s 2

s2 + 2s 8

4x + 5

1

2. x + 10

3. 9 × 12

4. 5y

6. m v

7. q + 8

8. t – 9

9. 2r

5. ab

7.

1 10. d or 0.5d 2

13.

x

8.

2

3

1

–2

0

0

–1

6

18

0

5

2

6

1

9

–1

–1

–1

2

–2

0

–2

7

3

3

17

4

12

10

45

9

27

1.5

11

0

0

6.2

29.8

10

30

8.3

38.2

30

90

t

3t – 4

3

r

y

2–r+y

5

3

4

3

2

2

0

1

3

5

11

2

5

5

2.3

2.9

1.1

3.2

4.1

4.1

8.3

0.8

0.9

2.1

7.8

19.4

8.3

20.9 b) 21

b) (4 – 2(–3))2 + 5(–3) = (4 + 6)2 – 15 = 102 – 15 = 100 – 15 = 85

9. a) 85

10. a) 422 m 14.

c) 8

f) –3

a2 – 2a – 1 –1

a

0

1

15. a) 3

110

12.

3s

6.

2x – 2

1. 12 + 7

s

b) –12 c) –6

x

Evaluating Expressions I

11.

b) 10

b) 142 m

11. a) 99 m; 75 m

c) 47 m

b) 0 m; rocket hits the ground

Evaluating Expressions III 1. a) 77.1 kg 2. a) 192

b) 102.9 kg

b) 8

c) Answers will vary.

3. a) 8.9 m/s

b) 1.4 m/s

4. 0.25q = 0.05n 5. a) $212.18 d) 15

e) 32


b) $689.79

c) $1276.28

f) 2 Copyright © 2001 McGraw-Hill Ryerson Limited

a

b) $12 400

b

c

s

A

5 cm

7 cm

8 cm

10 cm

17.3 cm2

11 cm

19 cm

20 cm

25 cm

102.5 cm2

5. a)

c) 48%

x

y 0

b)

100

8.4 m

3.6 m

6m

9m

9.4 m

0

1.5 m

1.5 m

2.1 m

2.55 m

1.12 m2

1

15.9

4

31.8

9

47.7

16

63.6

25

79.5

20

36

95.4

10

2

Evaluating Expressions IV 2

2

1. a) y = x – 2

2. a) y = x + 0.5

90 80

Speed (km/h)

6.

x

y

x

y

0

–2

0

0.5

1

–1

1

1.5

–1

–1

–1

1.5

2

2

2

4.5

Evaluating Radicals

–2

2

–2

4.5

1. 8, –8

3

7

3

9.5

–3

7

–3

9.5

b)

–2

50 40 30

0

3. 11, –11

6. 0.2, –0.2

9. 0.7

10. 0.3

13. 10.49

10 20 30 40 50

Skid Length (m)

2. 5, –5

5. 1.5, –1.5

y 6

60

c) about 70 km/h

y

b)

70

4. 0.9, –0.9

7. 4

8. –10

11. 7.87

12. –6.16

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

8

4

6

2

4

0

2

x

2

33. a) 20 cm –2 –2

0

2

b) 9.6 cm2 c) 3.6 cm

3. a) 20

x

b) 7 cm

34. a) 100 cm 35. a) 5 cm

c) 1.2 m

d) 1.5 m

b) 141.4 cm b) 7.14 m

c) 6 m

d) 13.0 cm

Area (cm2)

Expanding and Simplifying Expressions I 1. The length is (x + 3); the width is 2; and the area is 2(x + 3) or 2x + 6.

10

2. 2x + 4

0

2

4

3. 3x + 9

6. 4x + 8 7. 5x – 15 2 3x 10. x − 1 or −1 3 2 13. L 14. G

6

4. 4x + 2

5. 6x + 6

8. 0.3x + 1.5

9. 8x + 4

11. L

12. A

15. T

16. H

Diameter (cm)

17. R

Cost ($1000s)

4. a)

18. I

ALL RIGHT

18

20. 15 – 12n

21. 8h – 3

16 14

24. 8u + 2

25. 5x2 + 4x + 11

12 10 8

27. 3t2 – 10t – 5 30. a) 5x

6 4

19. 13x – 6

22. 12

23. 4 – 9k 26. –3y – 8

28. 4e + 20

29. x2 – 7x

b) –2x2 + 5x

2 0

10

20

30

40

50

60

70

80

Percent of Pollutants Removed



111

Expanding and Simplifying Expressions II 2

2

2

1. a + 3a

2. s – 5s

5. –6x + x2

6. –k2 + 3k

2

9. 6x – 2x

3. –y – 2y

4. 4b – b

7. 4r2 + 12r

8. 6m2 – 12m

10. –15y – 3y b) n – n + 1

12. 6n

Repeated Standard Multiplication Form (–3) × (–3) × (–3) × (–3) 81

8. 9.

(– 4)3

(–4) × (–4) × (– 4)

–64

3

(–5) × (–5) × (–5)

–125

10.

(– 4)3 ÷ (– 4)1

( − 4) × ( − 4) × ( − 4) ( − 4)

16

11.

(+5)4 ÷ (+5)2

( +5) × ( +5) × ( +5) × ( +5) ( +5) × ( +5)

25

12.

(+5)3 ÷ (+5)2

( +5) × ( +5) × ( +5) ( +5) × ( +5)

5

13.

(–3)5 ÷ (–3)2

( −3) × ( −3) × ( −3) × ( −3) × ( −3) ( −3) × ( −3)

14.

(–2)2 ÷ (–2)1

( −2) × ( −2) ( −2)

(–5)

14. n – 3n + 4 17. 4n2

16. –2n

18. 2n2 + 8n

2

A RHOMBUS

22. 3k2 – 5k

Exponential Form (–3)2 × (–3)2

2

13. n

15. 5n2 + n

2

7.

2

11. a) n(n + 1)

19. –n

2

2

20. 6a + 8a

23. 3d2 + 7d

21. r – 13r 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

–27 –2

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

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

Exponent Rules I

27. F

28. F

29. T

30. T

1. 25

2. 38

3. 46

4. 104

5. 97

6. 85

31. T

32. F

33. 8

34. 54

7. x7

8. y6

9. z5

10. 2

11. 3

12. 2

35. –23

36. –9

37. 18

38. –26

13. 1

14. 3

15. 4

16. 10 17. 1

18. 52

19. 43

20. 3

21. 93

22. 7

23. 22

24. m2

39. Radius (cm) 10 5

78.5

25. p2

26. a

27. 2

28. 2

29. 6

30. 4

2.5

19.6

31. 2

32. 7

33. 4

34. 8

35. 36

36. 28

1.3

5.3

37. 712 38. 68

39. 56

40. 415 41. x9

42. s4

6.2

120.8

43. r10

44. 3

45. 2

46. 4

48. 3

49. 3

50. 4

51. 3

52–63. PERFECT*WORK

47. 4

Exponent Rules II

1. 2. 3. 4. 5. 6.

112

Exponential Form (–2)3

Standard Form –8

Base –2

Exponent 3

31 5

3

1

3

5

1

5

(–3)3

–3

3

–27

(–2)5

–2

5

–32

–7

2

49

2

7


Area (cm2) 314.2

40. Negative bases that are multiplied an odd number 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 exponents of the variable, for example, (5n)(2n2) = 10n3. b) Multiply the coefficients and combine like variables 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


12. –6az3

13. –16r2s

13. 2a2b2

14. –20cde 15. 6x2y2

16. –6ab2m2

17. –5u2t2

17. −

18. –6a2b4c3d

19. –10r3s3t3

20. –60xyz

21. –18d2e

22. 8k3m2n5

23. 0.75s2

10. 8xyz

11. 12cde

b) 6 × 4b2 = 24b2

16. 2wx3

2 18. − c 4 d 5 3

21. 4x2y2

24. 5 d 3 f 3

25. x(x) + x(3x + 3x + x) = 8x2 26. a) (2b)(2b) = 4b2

3 2 3 f g h 2

20. 2a3b

24. 2x(2x) + x(4x) = 8x2

15. –2x3y2

14. 2q

25.

19. 2km2

3 22. − s 5 t 4 2

23.

8s 2t = 2s2t 4

26. 2x

27. 2r3s2

3 2 ef 2

28. 2a2b2

c) (2b)(2b)(2b) = 8b3

29. The length is 6xy and the width is 4xy.

27. a) 2n(n) = 2n2, 2n(4n) = 8n2, n(4n) = 4n2

30. front and back: 2x × 2y; two sides: 2x × 4xz2; top and bottom: 2y × 4xz2

b) 2(2n2) + 2(8n2) + 2(4n2) =28n2 28.

c) n(2n)(4n) = 8n3

29.

Graphing Equations I

4c

2a

1. y = –2x + 5

4c 4c

3a a

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, b) (–a)3(–a)3 = a6 a6. 2

4

4

y

x

y

–2

9

–2

4

–1

7

–1

2

0

5

0

0

1

3

1

–2

2

1

2

–4 4

8

6

3. x

4. s

5. c

6. –m15

7. f 18

8. s4t4

9. –x6y3

10. c2d4

11. y2z2

12. r3s3

13. –a6b6

y

y

6

2. y

2. 2x + y = 0

x

2

–2

0

2

x

2

x

4 6 4

14. f g ; FORTY WINKS 17. –8a6b3 5 3

21. y z

18. 9r6s2

15. 9t y

19. 25k6m4

3 3

5 5

22. –2a b

25. 18r4s4t3

2 2

23. –25s t 26. 32a5b6c6

3 3

16. –8x z

–2 2

20. –27q6r6 7

–4 9

24. 128k m

–2

0

2

x

27. –3m6n8p8

28. a) (xy2)2 = x2y4

b) (xy2)3 = x3y6

29. a) (2a2bc3)2 = 4a4b2c6

b) (2a2bc3)3 = 8a6b3c9

3. y = 2x + 1

30. The exponents are added, but should be multiplied. a6 31. The exponent of g is squared, but should be multiplied by 2. f 2g6 32. The square of a negative number is positive. 4x4y4

4. y = –x – 3

y

y

–2

0

2 –2 –2

0

2

x –4

–2

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



113

5. x + y = –1

6. 2x – y = 3

y

5.

6.

y

y

2

y

2

2 2

–2 –2

–2

0

0

–2

2

4

6

x

8

x + 4y = 8

–6

y + 1 = 2x

x

2

4 x

2

–4

x

2

0

–2 0 –2

–2 –2 –4

7. x-intercept –5; y-intercept 5; slope 1

y

–6

4

7. y = x + 4

y=x+5

8. y = 3x 1 9. y = x ; the domain is R 2 Depth of Heavy Wet Snow (cm)

10. a), b)

–4

8. x-intercept –4; y-intercept –10; 5 slope − 2

500 400 300

2

–2

x

0

y 2 –4 –2 0 –2

2

4

6

8 x

–4 –6

200

–8 100

5x + 2y = –20

–10

0

5

10

15

20

25

9. 5x + 4y – 20 = 0

Depth of Water (cm)

10. 4x – 7y + 28 = 0

The points can be ed because the measurements are continuous. c) 240 cm d) 18 cm

11. Answers may vary. For example, y = x – 1, y = x – 2, y = x – 3

Graphing Equations II

12. Answers may vary. For example, y = –x, y = –2x, 1 1 y = − – x, y = − – x 2 3

2.

y

2x + 3y = 6

2

y –2

0

4 x

2

14. slope undefined; y-intercept none; x-intercept 5

–2 –2

0

15. a)

4 x

2

–4

4x – y = 4

–2

3.

4

y

4.

y

3x + 5y – 15 = 0 –2

0

2

2

4

x

x+y+2=0 –4

114

2

x

60 50 40 30 20 10 0

–2 0

13. slope 0; y-intercept –3; x-intercept none

Number of Heartbeats

1.


10

20

30

Time (s)

b) slope 1.4; The slope represents the rate at which the heart beats. c) y = 1.4x d) 84 heartbeats per minute


Graphing Equations III 1. (1, 3)

(4, –1)

y 4

2. (5, –4)

3. (4, –1) 2

4.

y

 4 4 ,  3 3

2

0

2

x

4

–2

–2

0

x

2

–4

–2

10. If x = 0 for both planes as seen on the screen, then at x = 1, both planes will be at the point (1, 2) if the speeds allow. This assumes the plane y = 4 – 2x is going faster than the other. Diversionary tactics are immediately necessary to avoid a collision.

5. (2, 0) 6.

(3, 1)

y 2

–2

0

Greatest Common Factors

x

2

1. 2. 3. 4. 5. 6.

Number 20 30 18 54 150 252

7. 8. 9. 10. 11.

Expression 6x2 30st2 12a2bc3 10a2b3 24x2y2z

–2

7. a)

8.

x

y

x

y

0

5

b)

0

–8

5

10

5

2

13

18

13

18

x

y

x

y

0

1

0

3

1

3

–1

2

–1

–1

–2

1 (2, 5)

13. 3 × 17 × a × a

4

14. 2 × 2 × 19 × r × r × s

15. 4

16. 15

2

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

0

2

x

–2

9.

Prime Factors 2×3×x×x 2×3×5×s×t×t 2×2×3×a×a×b×c×c×c 2×5×a×a×b×b×b 2×2×2×3×x×x×y×y×z

12. 2 × 3 × m × m × n × n

y

–2

(13, 18)

Prime Factors 2×2×5 2×3×5 2×3×3 2×3×3×3 2×3×5×5 2×2×3×3×7

x

y

x

y

0

7

0

–5

1

5

5

0

2

3


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)


115

30. –1.3x5 + 0.2mx4 + 2.1x3 + 0.4m2 2 31. x 4 + b 3 x 2 + bx + 4b 32. x + a 3

b) ab 2b 4

2a

6

3a

33–35. Answers will vary.

2 2ab

36. a) 6x2 – 5

3ab 2a

2b

b) Yes; each has a degree of 2.

Slope I

3b

1. mAB =

2ab

a 4b

1 , mCD is undefined, mEF = 0, mGH = –1, 2

6b

5 2 mIJ = − , mKL = 3 3

b 4a

6a

2 3

Like 1. 4r, –r, 101r, r; and 5r2, –r2, (–r2); and r3, 2. a) 4

3

r , –2r3 2

b) coefficients –1, 5, –2; constant 3

3. 10t

4. –7b2

5. –13y

6. 21m3

7. 3p

8. 2c2

9. 4x

10. –12y

12. t2

11. d 15. 2b2

13. 15y – 2z

14. –8r + 3s

16. 7e3 – e2 17. 2s, 2

19. 5t – 3, –0.5 22. m + n + 3

23. 2 + 3z – 3w

5 4

3. –1

4. −

5.

3 4

6. 0

7. undefined

8. Answers may vary. For example, (0, –3), (2, –2), (6, 0), (8, 1) y

9.

10.

2

–4

–2

18. 4a2, 16

20. –6k + 2, 20

7 9

2.

21. c + 5d + 3

y 0

4

–2

2 x

0

2

–2

–4

–4

–6

24. 2 + 5x2 + y2

25. 3r + 2r + 2r, 7r, or 2 × 2r + 3r 11. Answers may vary. For example, (–1, 2) and (3, 5); (–2, 4) and (7, 11)

26. 1.5s + 0.5s + 1.5s + 0.5s, 4s 27–31. Answers will vary.

12. Answers may vary. For example, (–4, –3), (–7, –1), (2, –7)

Polynomials

13. Answers may vary. For example,

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

14. 2

15. 3

16. 4

19. 1

20. 2

21. 3

22. 3

23. 3n + 3, binomial, 1

17. 6

18. 4

1 bh, monomial, 2 24. 2

25. 2πr2 + 2πrh, binomial, 2 26. 2 + y + xy2 + y3 3 − y + y3 − y6 28. 4

116

b) (0, –8) and (8, 0); slope is 1 8 3

14. c = –6, d = 4

15.

16. non-collinear

17. non-collinear

12. 5

13–18. Find the greatest sum of the exponents in any one term. 13. 1

a) (0, –5) and (3, 1); slope is 2

18. collinear; m =

3 4

Slope II 1. m = 2, b = –5

1 2. m = − , b = 2 3

3. m = –1, b = 1

4. m =

1 5 , b=− 2 2

5. m = –3, b = 2

6. m =

4 , b = –3 3

27. x3 – 2x2y + 3xy3 29. x4 + 2x3 – 6x – 10



x

7. m = –2, b = 4

8. m = 0, b = 5

9. y = 5x + 2, 5x – y + 2 = 0 1 10. y = − x + , 2x + 2y – 1 = 0 2 2 11. y = x − 1, 2x – 3y – 3 = 0 3 1 1 12. y = − x − , 3x + 12y + 4 = 0 4 3

6.

7. 0

8. a) 3

b) −

9. a) −

1 4

b) 4

10. a)

1 3

b) –3

1 3

b) A, B, D, F or A, B, C, E

12. m = –15

13. neither

15. perpendicular

y

14.

5. –2

11. a) A, B, C, E

3 13. m = − ; b = 8 2

1 3

4. perpendicular

14. parallel

16. 3x + 2y + 13 = 0

17. 3x + y + 15 = 0

4

y = 3x + 1

18. trapezoid

y

2

E

6

D –2

0

4

4 x

2

C

–2

2

y

15.

F –2

0

–2

x

2

19. parallelogram

y

K

–2

1 y = – –x – 1 2

N M

3 1 3 1 ; b = ; y = x + ; 3x – 4y + 2 = 0 4 2 4 2

–4

1 1 17. m = − ; b = −2; y = − x − 2; x + 3y + 6 = 0 3 3 18. b = –2, m = –1

20. No. The paths are parallel.

19. a) t = 35n + 50

Solving Equations I

b)

1. G

Total Cost ($)

300

200

2. G

100

0

L

4

–4

16. m =

x

2

0

2

4

6

R R R R R R

=

R R R R R

=

G

=

R R R R

2 x

0

3.

8

Nights

c) $295

R R R

=

–2

2

d) 35

e) the cost per night to board the dog

f) $50

4.

G

Slope III 1. perpendicular

2. parallel

3. neither


1 2

5. 11

6. 3

7. 2

8. 6

9.

10. 1.4

11. 6

12. 3

13. 3

14. 2


117

15. 6

16. 0.4

17. 6

18. 15

19. 7

Solving Equations IV

20. –7

21. –7

22. –2

23. 0

24. 3

1.

1 25. 8 2

1 26. 4

1 27. 3

5 28. 8

29. 4.0

30. 11.8

31. 1.5

32. –1.4

33. 5.5

34. 1.8

G G G

=

G

R R R R

G G G

=

R R R R R R

G G G G

G G

=

G G G

2. R R

35. 6

Solving Equations II 1.

G G

2.

G G G

3.

=

3.

R R R R

=

=

G

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 kJ 21. a) 7x = 3x + 12 b) 3 22. a) 5x + 6 = 2x – 9 b) –5

4. 6

5. 3

6. 5

7. 9

8. 5

9. 7

10. 4

11. –8

12. –23

13. –32

Solving Equations V

18. 25

1. 7

2. 6

3. 1

4. 0

5. –3

14. –6

15. 8

16. –4

1 17. 7

19. –9

20. –150

21. 84

22. –8

23. –50

6. –8

7. –1

8. 4

9. 1.5

10. –3

24. 1.6

25. –6

26. 7.2

27. –3

28. –71

11. –3

12. 24

13. 2

14. 40

15. 11

16. 3

17. 1

18. 7

19. 1

20. –15

21. –2

22. 13

29. –7975

30. c) 5x = 60

Solving Equations III

23. a) x = 7

b) length = 126 cm, width = 84 cm

1.

START Divide by 3

3x + 4 x

Subtract 4 STOP

24. 36 cm, 68 cm, 36 cm

2.

START Divide by 4

4y – 3 y

Add 3 STOP

Solving Equations VI

3.

START Subtract 11

5–x x

Add x STOP,

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

1 2

17. −

2 3

19. –4

20. –6

10 11

22. –3

START 5–x Multiply or divide by –1

or

Subtract 5 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

118


16. − 21. − 26. 1

4 5

18. −

23. –1

27. a) 26.45 kg

24. –11 25. 17 b) 13.45 kg

Solving Proportions 1–6. Answers may vary. 2. 2 to 1, 4 to 2

1. 2:8, 3:12 20 30 6 30 , , 3. 4. 8 12 14 70


5. 1 to 3, 15 to 45 7. =

6. 9. ≠

8. =

c) (x, y) → (x + 2, y + 1)

2 20 , 5 50

13. a) 3 units up

10. ≠

11. y = 4

b) [0, 3]

c) (x, y) → (x, y + 3)

12. x = 100

13. a = 10

14. c = 18

14. a) 4 units right, 2 units down

15. d = 3.0

16. t = 1.8

17. 18

c) (x, y) → (x + 4, y – 2) 15. a) 2 units left, 3 units down

18. right-handed: 320, left-handed: 40 19. a) 1:3

b) 46

20. 3.1 m

21. Uncooked Rice 1L 500 mL 250 mL 80 mL 0.4 L 0.7 L

Cooked Rice 3.5 L 1.75 L 875 mL 280 mL 1.4 L 2.45 L

22. a) 1:10

b) twice

b) [4, –2]

b) [–2, –3]

c) (x, y) → (x – 2, y – 3) Original Point (2, –3)

Translation (x, y) → (x, y – 2)

Image Point (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)

20.

Original Point (4, 2)

Image Point (–1, –3)

Translation (x, y) → (x – 5, y – 5)

23. Answers will vary.

21.

(5, –3)

(5, –5)

(x, y) → (x, y – 2)

Subtracting Polynomials

22. 23.

(–3, –2)

(–3, 1)

(x, y) → (x, y + 3)

(0, 0)

(–1, 3)

(x, y) → (x – 1, y + 3)

16. 17. 18. 19.

c) Answers will vary.

1. 2x + 1

2. –x – 7

3. –5ab – 6

4. –z2 + z + 4

24.

y

C

6. 4s2 + s – 5

5. 2c – 3d – e

4

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

C′

2

–6

–4

C′′ A

0

–2

A′

–2

B

2

4

B′ A′′

6

x

B′′

2

19. 3y – 2y – 2 20. a) 18m + 13

Transformations II

b) For the long sides, the difference is 2m – 9; for the short sides, the difference is 8.

1.

Transformations I

11.

2.

1.

FF

3.

5.

6.

9. C′

7. 8.

10. a) 3 units left, 1 unit up

b) [–3, 1]

A′ A

D

FF

n

A′′ C

B′ B l

6.

C′′

A′′ k D′′

3.

FF

5.

B′′

D′ 4.

F F

D′′′ A′′′ C′′′

B′′′

m

IV

B

CIV IV

A

IV

D

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]



119

12.

D′′

G′ B′′ –6

A′′ C E′

F′′

–4

–2

G

Transformations III

D

y

H′

E C′

1. AB: (1, 2), (3, 1); A′B′: (2, 4), (6, 2); 2

A

2

C′′ F′ F B

0

2

4

6

B′ G′′

E′′

–2

DE: (–8, 4), (–4, 4); D′E′: (–2, 1), (–1, 1); x

GH: (–2, –2), (–2, –4); G′H′: (–1, –1), (–1, –2);

A′ H′′

H

a) b) c)

Point (–2, 3) (3, 1)

Reflection Line x-axis y-axis (–2, –3) (2, 3) (3, –1) (–3, 1)

(0, 3)

(0, –3)

2.

Y′ y

Y

(0, 3)

Y′′ X′

14.

Z

y

X X′′

x

Z′′

Z′′

m

1 2

JK: (0, –1), (1, 0); J′K′: (0, –4), (4, 0); 4

D′

13.

1 4

Z X

Y

X′′

Y′′

x

l

Z′ X′

Y′

3. P′(9, 6), Q′(–3, 6), R′(–3, –3), S′(9, –3) 4.

Z′

15.

A

y

R′ S′

A′ y

R

R′′ T T′ S′′

T′′

S

x

0

B x

B′

C C′

a) 2 m

l

16.

b) Images are twice as long as originals. c) same y

F

l

5. a)

E′

E

D

b)

F′

D′ x

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

120



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