MathPower 10 Ontario Edition Practice Masters

Transcription

™
MATHPOWER 10
ONTARIO EDITION
Practice Masters
Shirley Barrett
Richmond Hill, Ontario
Contributing Writer
Janice Nixon
Toronto, Ontario
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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
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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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C00-931907-7
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 Verifying 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
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 Terms
92
Polynomials
93
Slope I
(Using Points)
94
Slope II
(Linear Equations: Slope and y-Intercept
Form)
95
Slope III
(Parallel and Perpendicular Lines)
96
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
109
Exponent Rules II
To the Teacher
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
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.
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vi
To the Teacher
Name
Getting Started
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,
1
Name
Know Your Textbook and How to Use It
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 terms.
________________
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) verify 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
Name
Classroom Learning
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
Name
Homework
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
Name
Seeking Help
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
Name
Preparing for Tests and Exams
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.
6
Name
Writing Tests and Exams
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
Name
Problem Solving Skills
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 terms by dividing by 2 or multiplying by 1–.
2
How does this help?
6. What is the relationship?
7. Using the relationship, determine how many competitors play 21 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
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
Name
Goals
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 supported by actions.
I will always be prepared for class.
I will do my homework daily.
I will join a study group.
When I miss a class, I will catch up within one day.
2.
Write your actions.
A goal will have obstacles. Identify them up front.
I am on the soccer team. Two close friends are not. They will want me to do things with them, when 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
Name
1.1
Investigation: Ordered Pairs and Solutions
MATHPOWERTM 10, Ontario Edition, pp. 4–5
• To verify that an ordered pair satisfies an equation in two variables,
a) substitute the values for x and y in the original equation
b) evaluate the left side of the equation and the right side of the equation
c) check that L.S. = R.S.
• To verify that an ordered pair satisfies a system of equations,
a) substitute the values for x and y in each 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
Name
1.2
Solving Linear Systems Graphically
• 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
• 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
• 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
• To solve a linear system in two variables by elimination,
a) clear decimals and fractions, if necessary
b) rewrite the equations with like terms 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
• 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
• 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
pass 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
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.
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
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
• To find the length of a line segment joining (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
joining 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) Verify 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 joining (x1, y) and (x2, y), use the formula
 x1 + x 2

, y .
 2

• To find the midpoint, M, of a vertical line segment joining (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 joining 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 joining 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
• To find the midpoint, M, of a line segment joining (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
joining 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
Verifying Properties of Geometric Figures
• The following formulas can be used to determine characteristics of geometric figures and to verify
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). Verify 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) Verify 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.
Verify 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) Verify 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) Verify that ABCD is a parallelogram.
Name
2.5
Distance From a Point to a Line
• 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 passes 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 joining 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 Verifying 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).
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 joining 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
• To add polynomials, collect like terms.
• To subtract a polynomial, add its opposite.
• To multiply monomials, multiply the numerical coefficients. Then, multiply the variables
using the exponent rules for multiplication.
• To divide monomials, divide the numerical coefficients. Then, divide the variables using
the exponent rules for division.
• To multiply a polynomial by a monomial, use the distributive property to multiply each
term in the polynomial by the monomial.
1. Classify each polynomial by degree and by
number of terms.
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
terms 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
• 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 terms, Outside terms, Inside
terms, and Last terms
• Verify the product of two binomials by substituting a convenient value for the variable in the
original product and in the simplified expression.
1. Communication Explain how
the diagram models the product.
2x
x
+
y
+3
2x 2
3x
2xy
3y
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
• To square a binomial, use one of the following patterns.
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
• To find the product of the sum and difference of two terms, use the following pattern.
(a + b)(a – b) = a2 – b2
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
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)
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
• 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 terms 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
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
• 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 terms, which are two numbers whose sum is b and whose product is c
• When factoring a trinomial, first remove any common factors.
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
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
• To factor a trinomial in the form ax2 + bx + c, either use guess and check or break up the middle term.
• To factor by guess and check, list all the possible pairs of factors and expand to see which pair gives
the correct middle term of the trinomial.
• To factor by breaking up the middle term,
a) replace the middle term, bx, by two terms whose coefficients have a sum of b and a product of a × c
b) group pairs of terms and remove a common factor from each pair
c) remove the common binomial factor
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
• 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
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 terms
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
terms 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
• A function is a set of ordered pairs in which, for every x, there is only one y.
• If any vertical line passes through more than one point on the graph of a relation,
then the relation is not a function.
• The set of the first elements in a relation is called the domain. The set of the second
elements in a relation is called the range.
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 terms of the ordered pairs to get
(t, n). Is the new set of ordered pairs a function?
Explain.
10. Problem Solving The volume of a sphere is
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
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 passes through the
point (–1, –1). Find k.
1
b) y = –x2 and y = − x 2
2
6. The graph of y = ax2 + k passes 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 terms 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?
Name
4.5
Investigation: Sketching Parabolas in the Form y = ax(x – s) + t
• 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 terms:
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 passes 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 Verify 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
• 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. Verify 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
• 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
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?)
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
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)
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 terms 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 terms 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).
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
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
• 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
• 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
• 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
• 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
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
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)
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®
• 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 join 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
• join 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
• 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
• 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
• 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
• 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
• 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
• 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
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
• join 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
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
joining the rock and oak tree and the line joining
the rock and the birch tree is 25°. The angle
formed by the line joining the rock and the oak
tree and the line joining 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
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
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 terms 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 terms.
Then, divide all the terms by the GCF.
Determine the GCF of both terms.
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 Terms
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 terms.
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
Name
(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.
Name
(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 support 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.
77
Name
(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
account for a number of years. Compound
interest is paid on the amount in the account.
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)
Name
(Non-Linear Relations)
A non-linear relation has a graph that is not a straight line. When the equation includes
a power of x, such as y = x2 – 3, the points can be joined to form a smooth curve.
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
79
Name
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
(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
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.
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
____ ____ ____
____ ____ ____ ____ ____
81
Name
(Multiplying a Polynomial by a Monomial)
To multiply a polynomial by a monomial, use the distributive property.
When more than one expansion takes place, collect like terms and simplify.
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.
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
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
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
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
terms 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
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 Linear Equations)
To graph a linear equation, first make a table of values and substitute values of x to determine
the corresponding values of y. Then, plot the coordinates. Join the points if the domain is R.
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 join 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
(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
(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
To find the greatest common factor of two or more terms, first write all the factors of each term.
Then, determine which factors are common to all the terms, and
write the product of the common factors.
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 Terms
Like terms have the same variable raised to the same exponent. r, 4r, 101r
Unlike terms have different variables or the same variable, but different exponents. 7b, –3a, x2, x
1. List the like terms.
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 terms 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 terms, and a trinomial has three terms.
The degree of a polynomial in one variable is the highest power of the
variable. For example, 2x3 – 7x has degree 3.
The degree of a polynomial in two or more variables is the greatest sum of
the exponents in any one term. For example, 5m3n + m2n – mn2 has degree 4.
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 terms 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 terms 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 passing 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 passes
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 passes
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 passes 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 passing
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 join 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 passenger 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
(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
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
(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
(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
the same value to
Add or Subtract
the same value
from both sides
of the equation.
Simplify
both sides of
the equation.
to
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.
Add or Subtract
the same value
from both sides
of the equation.
the same value to
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.
101
Name
(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
Add or Subtract
the same value
from both sides
of the equation.
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
the same value to
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
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
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 Terms
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°
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
11.
b) 10
b) 142 m
11. a) 99 m; 75 m
c) 47 m
b) 0 m; rocket hits the ground
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
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
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)
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
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.
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 joined 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
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
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
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 Terms
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
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
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
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
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)
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

MathPower 10 Ontario Edition Practice Masters

Transcription

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