Mathematics: Journey from Basic Mathematics through Intermediate Algebra [1 ed.] 1285192362, 9781285192369

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Mathematics Journey from Basic Mathematics through Intermediate Algebra Annotated Instructor’s Edition

RICHARD N. AUFMANN PALOMAR COLLEGE

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Po Brief Contents AIM for Success: How to Succeed in This Course Whole Numbers Integers

Fractions Decimals and Percents Variable Expressions Introduction to Equations

General First-Degree Equations and Inequalities Linear Functions and Inequalities in Two Variables

Systems of Linear Equations in Two or Three Variables LD OONOOARWN—

10 11. 12 13 14 15 16 17 18 19 20 21

Polynomials Factoring Polynomials Rational Expressions Rational Exponents and Radicals Quadratic Equations More on Functions Exponential and Logarithmic Functions Conic Sections Sequences, Series, and the Binomial Theorem Measurement Geometry Statistics and Probability

TI-84 Plus Keystroke Guide for the TI-84 Plus Tables Index of Applications Index

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Contents

MODULE

A

AIM for Success: e e ¢ e e

How to Succeed in This Course

Get Ready 2 Motivate Yourself 2 Develop a “Can Do” Attitude Toward Math Strategies for Success 3 Time Management 4 Habits of Successful Students Use the Interactive Method 6

6

Use a Strategy to Solve Word Problems Ace the Test

8

9

Ready, Set, Succeed!

MODULE 1

3

9

Whole Numbers SECTION 1.1 Objective Objective Objective Objective

1.1A 1.1B 1.1C 1.1D

SECTION 1.2 Objective 1.2A Objective 1.2B Objective 1.2C

SECTION 1.3 Objective 1.3A Objective 1.3B Objective 1.3C

SECTION 1.4 Objective 1.4A Objective 1.4B

Introduction to Whole Numbers Identify the order relation between two numbers

2

Write whole numbers in words, in standard form, and in expanded form Round a whole number to a given place value 6 Solve application problems and use statistical graphs 7

Addition and Subtraction of Whole Numbers Add whole numbers

14

14

Subtract whole numbers

Solve application problems

17

20

Multiplication and Division of Whole Numbers

22

Multiply whole numbers 22 Divide whole numbers 25 Solve application problems 31

Exponential Notation and the Order of Operations Agreement 33 Simplify expressions that contain exponents 33 Use the Order of Operations Agreement to simplify expressions

Solutions to Check Your Understanding Solutions to Objective Practice Exercises

S-1 S-8

35

4

vi

Contents

MODULE

2

ntegers ‘7s

ECTION 2.1

Introduction to Integers

Objective 2.1A Objective 2.1B

Use inequality symbols with integers 2 Simplify expressions with absolute value

JECTION 2.2 Objective 2.2A Objective 2.2B Objective 2.2C

‘ECTION 2.3

‘n

Objective 2.3A Objective 2.3B Objective 2.3C

SECTION 2.4 Objective 2.4A Objective 2.4B

2 4

Addition and Subtraction of Integers Add integers 6 Subtract integers 8 Solve application problems

6

10

Multiplication and Division of Integers Multiply integers 11 Divide integers 13 Solve application problems

11

16

Exponents and the Order of Operations Agreement Simplify expressions containing exponents

17

17

Use the Order of Operations Agreement to simplify expressions

Solutions to Check Your Understanding

19

S-1

Solutions to Objective Practice Exercises

S-3

MODULE

Fractions SECTION 3.1 Objective 3.1A Objective 3.1B

Objective 3.1C

SECTION 3.2 Objective 3.2A Objective 3.2B

SECTION 3.3

The Least Common

Multiple and Greatest Common

Factor numbers and find the prime factorization of numbers Find the least common multiple (LCM) 4 Find the greatest common factor (GCF) 5

Introduction to Fractions

6

Write a fraction that represents part of

awhole

6

Write an improper fraction as a mixed number or a whole number, and a mixed number as an improper fraction 8

Writing Equivalent Fractions

9

Objective 3.3A Objective 3.3B

Write a fraction in simplest form 9 Find equivalent fractions by raising to higher terms

Objective 3.3C

Identify the order relation between two fractions

SECTION 3.4 Objective 3.4A Objective 3.4B Objective 3.4C

SECTION 3.5 Objective 3.5A Objective 3.5B Objective 3.5C

SECTION 3.6

Factor 2

11 12

Multiplication and Division of Fractions

13

Multiply fractions 13 Divide fractions 17 Solve application problems and use formulas

21

Addition and Subtraction of Fractions

22

Add fractions

22

Subtract fractions 26 Solve application problems

29

Operations on Positive and Negative Fractions

Objective 3.6A

Multiply and divide positive and negative fractions

Objective 3.6B

Add and subtract positive and negative fractions

31

34

31

2

Contents

SECTION 3.7 Objective 3.7A

Objective 3.7B

The Order of Operations Agreemerit and Complex Fractions 36 Use the Order of Operations Agreement to simplify expressions Simplify complex fractions 39

36

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-9

MODULE

4

Decimals and Percents SECTION 4.1 Objective 4.1A Objective 4.1B Objective 4.1C

SECTION 4.2

Introduction to Decimals

2

Write decimals in standard form and in

words

Round a decimal to a given place value Compare decimals 5

4

Adding and Subtracting Decimals

6

Objective 4.2A

Add and subtract decimals

Objective 4.2B

Solve application problems and use forjiulas

SECTION 4.3 Objective 4.3A Objective 4.3B Objective 4.3C

SECTION 4.4 Objective Objective Objective Objective

4.4A 4.4B 4.4C 4.4D

SECTION 4.5 Objective 4.5A Objective 4.5B

SECTION 4.6 Objective 4.6A Objective 4.6B Objective 4.6C

2

6

Multiplying and Dividing Decimal Multiply decimals 10 Divide decimals 13 Solve application problems and use for;

9

10

\ulas

17

Comparing and Converting Fract)}ons and Decimals Convert fractions to decimals 19 Convert decimals to fractions 21 Compare a fraction and adecimal Write ratios and rates

19

21

23

Introduction to Percents

26

Write a percent as a decimal or a fraction Write a decimal or a fraction as apercent

26 27

Radical Expressions and Real Numbers Find the square root of aperfect square 29 Approximate the square root of anatural number Solve application problems 33

Solutions to Check Your Understanding

Solutions to Objective Practice Exercises

29 31

S-1

S-7

MODULE

Variable Expressions SECTION 5.1 Objective 5.1A

SECTION 9.2 Objective 5.2A

Objective 5.2B Objective 5.2C Objective 5.2D

Evaluating Variable Expressions Evaluate a variable expression

2

2

Simplifying Variable Expressions

4

Simplify a variable expression using the Properties of Addition

4 Simplify a variable expression using the Properties of Multiplication Simplify a variable expression using the Distributive Property 9 Simplify general variable expressions 10

7

vii

viii

Contents

TION 5.3

n

dbjective 5.3A

Objective 5.3B

ipjective 5.3C

Translating Verbal Expressions into Variable Expressions Translate a verbal expression into a variable expression, given the variable 11 Translate a verbal expression into a variable expression and then simplify 13 Translate application problems

olutions to Check Your Understanding

S-1

»lutions to Objective Practice Exercises

o¢n tn

14

S-2

MODULE

ntroduction to Equations

—

i¢e)

=oTION 6.1 djective 6.1A Objective 6.1B o

jective 6.1C

bjective 6.1D

s ction 6.2 )bjective 6.2A Objective 6.2B

si cTion 6.3 Objective 6.3A Objective 6.3B

Objective 6.3C

SECTION 6.4

Introduction to Equations

Solve an equation of the formx+a=b 4

Solve basic uniform motion problems

6

Proportions

9

Solve proportions 9 Solve application problems using proportions

The Basic Percent Equation

Solve percent decrease problems

Objective 6.6A

16

Percent Increase and Percent Decrease

Objective 6.4B

SECTION 6.6

11

12

Solve the basic percent equation 12 Solve percent problems using proportions Solve application problems 18

Solve percent increase problems

Objective 6.5A Objective 6.5B

2

2

Solve an equation of the form ax=b

Objective 6.4A

SECTION 6.5

2

Determine whether a given number is a solution of an equation

Markup and Discount

22

24

Solve markup problems Solve discount problems

Simple Interest

21

21

24 26

28

Solve simple interest problems

28

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-7

MODULE

General First-Degree Equations and Inequalities SECTION 7.1 Objective Objective Objective Objective Objective

7.1A 7.1B 7.1C 7.1D 7.1E

SECTION 7.2 Objective 7.2A Objective 7.2B

General Equations Solve Solve Solve Solve Solve

2

an equation of the form ax +b=c 2 an equation of the form ax +b=cx+d an equation containing parentheses 6 a literal equation for one of the variables an absolute value equation 9

Translating Sentences into Equations Solve integer problems

4 7

11

11

Translate a sentence into an equation and solve

14

11

Contents

SECTION 7.3 Objective 7.3A Objective 7.3B Objective 7.3C

SECTION 7.4 Objective 7.4A Objective 7.4B Objective 7.4C

SECTION 7.5 Objective 7.5A Objective 7.5B

Mixture and Uniform Motion Problems Solve value mixture problems

15

15

Solve percent mixture problems Solve uniform motion problems

First-Degree Inequalities

17 19

22

Write sets of real numbers using set-builcer notation and interval notation 22 Solve an inequality in one variable 26 Solve application problems 29

Compound and Absolute Value Inequalities Solve a compound inequality 31 Solve an absolute value inequality

Solutions to Check Your Understanding

S-1

Solutions to Objective Practice Exercises

S-10

31

32

MODULE

8

Linear Functions and Inequalities SECTION 8.1 Objective 8.1A Objective 8.1B

SECTION 8.2 Objective 8.2A Objective 8.2B Objective 8.2C

SECTION 8.3 Objective Objective Objective Objective

8.3A 8.3B 8.3C 8.3D

SECTION 8.4 Objective 8.4A Objective 8.4B Objective 8.4C

SECTION 8.5 Objective 8.5A Objective 8.5B Objective 8.5C

SECTION 8.6 Objective 8.6A

SECTION 8.7 Objective 8.7A

in Two Variables

The Rectangular Coordinate Sys*2m Find the length and midpoint of a line

Graph an equation in two variables

Introduction to Functions Evaluate afunction 9 Graph a function 15 Apply the vertical line test

Linear Functions

Graph a linear function

2

segment

2

6

9

18

20

20

Graph an equation of the form Ax + By =C Find the x- and y-intercepts of a straight line Solve application problems 27

Slope of a Straight Line

22 25

28

Find the slope of a line given two points Find average rate of change 32 Graph a line given a point andthe slope

Finding Equations of Lines

28 34

36

Find the equation of a line given a point and the slope Find the equation of a line given two points 37 Solve application problems 39

Parallel and Perpendicular Lines Find parallel and perpendicular lines

Inequalities in Two Variables

40 40

43

Graph the solution set of an inequality in two variables

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-8

36

43

X

Contents

MODULE

9

42m =r

i?)

‘stems of Linear Equations Two or Three Variables TION 9.1

jective 9.1B

Solve a system of linear equations by graphing 2 Solve a system of linear equations by the substitution method

jective 9.1C

Solve investment problems

‘TION9.2

Solving Systems of Linear Equations by the Addition Method 10

jective 9.2A

Solve a system of two linear equations in two variables by the addition

bjective 9.1A

s

active 9.2B

si

Solving Systems of Linear Equations by Graphing and by the Substitution Method 2

method 10 Solve a system of three linear equations in three variables by the addition method 13

TION 9.3

Solving Systems of Equations by Using Determinants

jective 9.3A

Evaluate a determinant 18 Solve a system of equations by using Cramer’s Rule

jective 9.3B

Si cTION 9.4 »jective 9.4A jective 9.4B

s¢oTion 9.5 Objective 9.5A

Application Problems

21

24

Solve rate-of-wind or rate-of-current problems Solve application problems 26

24

Solving Systems of Linear Inequalities

29

Graph the solution set of a system of linear inequalities

Solutions to Check Your Understanding

S-1

Solutions to Objective Practice Exercises

S-8

MODULE

10

Polynomials SECTION 10.1

Addition and Subtraction of Polynomials

Objective 10.1A

Add polynomials

Objective 10.1B

Subtract polynomials

SECTION 10.2 Objective 10.2A

Objective 10.2B

SECTION 10.3 Objective 10.3A

Objective 10.3B Objective 10.3C Objective 10.3D

SECTION 10.4 Objective 10.4A Objective 10.4B

SECTION 10.5 Objective 10.5A Objective 10.5B

5

7

3

Multiplication of Monomials Multiply monomials 4 Simplify powers of monomials

4 6

Multiplication of Polynomials Multiply Multiply Multiply Multiply

2

2

7

a polynomial by amonomial 7 two polynomials 8 two binomials 9 binomials that have special products

10

Integer Exponents and Scientific Notation

12

Simplify expressions containing integer exponents Use scientific notation 16

12

Division of Polynomials

18

Divide a polynomial by amonomial Divide polynomials 19

Solutions to Check Your Understanding

Solutions to Objective Practice Exercises

S-1

S-4

18

29

18

Contents

MODULE

11

Factoring Polynomials SECTION 11.1 Objective 11.1A Objective 11.1B

SECTION 11.2

Common

Factors

2

Factor a monomial from a polynomial Factor by grouping 3

Factoring Polynomials of the Form x* + bx +c

Objective 11.2A

Factor trinomials of the form x* + bx

Objective 11.2B

Factor completely

SECTION 11.3

SECTION 11.4 Objective 11.4A

Objective 11.4B Objective 11.4C Objective 11.4D

SECTION 11.5 Objective 11.5A Objective 11.5B

5

5

8

Factoring Polynomials of the Form

ax?+bx+ec Objective 11.3A Objective 11.3B

+c

9

Factor trinomials of the form ax* + bx + c by using trial factors Factor trinomials of the form ax? + bx + c by grouping 13

Special Factoring

9

16

Factor the difference of two squares anc

perfect-square trinomials Factor the sum or difference of two periect cubes 20 Factor a trinomial that is quadratic in fo 21 Factor completely 23

Solving Equations

24

Solve equations by factoring 24 Solve application problems 26

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-6

MODULE

12

Rational Expressions SECTION 12.1 Objective 12.1A Objective 12.1B Objective 12.1C

SECTION 12.2 Objective 12.2A Objective 12.2B

SECTION 12.3 Objective 12.3A

SECTION 12.4 Objective 12.4A Objective 12.4B

SECTION 12.5 Objective 12.5A

SECTION 12.6

Multiplication and Division of Rational Expressions

2

Simplify rational expressions 2 Multiply rational expressions 4 Divide rational expressions 6

Addition and Subtraction of Rational Expressions Express two fractions in terms of acommon denominator Add and subtract rational expressions 10

Complex Fractions

13

Simplify complex fractions

13

Equations Containing Fractions

16

Solve equations containing fractions 16 Solve problems involving similar triangles

Variation

18

21

Solve direct and inverse variation problems

Application Problems

25

Objective 12.6A

Solve work problems

25

Objective 12.6B

Solve uniform motion problems

Solutions to Check Your Understanding Solutions to Objective Practice Exercises

S-1 S-7

27

21

8 8

16

Xi

xii

Contents

MODULE

13

ational Exponents and Radicals s

°TION 13.1

— Simplify numerical radical expressions

jective 13.1B

Simplify variable radical expressions

¢ ©TI0N 13.2 objective 13.2A

¢

©TION 13.3 ‘pjective 13.3A objective 13.3B

Ss:

s:

Introduction to Radical Expressions

pjective 13.1A

2

2 4

Addition and Subtraction of Radical Expressions Add and subtract radical expressions

6

6

— Multiplication and Division of Radical Expressions

8

Multiply radical expressions 8 Divide radical expressions 11

°oTION 13.4 — Solving Equations Containing Radical Expressions ‘objective 13.4A

Solve equations containing one or more radical expressions

bjective 13.4B

Solve application problems

16

TION 13.5 — Rational Exponents and Radical Expressions jective 13.54 ‘ojective 13.5B Ibjective 13.5C

13 13

19

Simplify expressions with rational exponents 19 Write exponential expressions as radical expressions and radical expressions as exponential expressions 21 Simplify radical expressions that are roots of perfect powers 22

Sciutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-5

MODULE

14

Quadratic Equations SECTION 14.1.

Solving Quadratic Equations by Factoring or by Taking Square Roots

Objective 14.1A Objective 14.1B

SECTION 14.2 Objective 14.2A

SECTION 14.3 Objective 14.3A

SECTION 14.4 Objective 14.4A

SECTION 14.5

2

Solve quadratic equations by factoring 2 Solve quadratic equations by taking square roots

— Solving Quadratic Equations by Completing the Square Solve quadratic equations by completing the square

Solve quadratic equations by using the quadratic formula

Applications of Quadratic Equations Solve application problems

Complex Numbers

9

11

11

16

Simplify complex numbers

Objective 14.5B

Add and subtract complex numbers

SECTION 14.6

6

6

— Solving Quadratic Equations by Using the Quadratic Formula

Objective 14.5A Objective 14.5C Objective 14.5D Objective 14.5E

3

16 18

= Multiply complex numbers 18 Divide complex numbers 20 Solve quadratic equations with complex number solutions

22

Equations That Are Reducible to Quadratic Equations

Objective 14.6A

Solve equations that are quadratic inform

Objective 14.6B

Solve radical equations

Objective 14.6C

Solve fractional equations

27

29

25

25

9

Contents

SECTION 14.7 Objective 14.7A

Nonlinear Inequalities

30

Solve nonlinear inequalities

Solutions to Check Your Understanding

30

S-1

Solutions to Objective Practice Exercises

S-9

MODULE

15

More on Functions SECTION 15.1 Objective 15.1A Objective 15.1B Objective 15.1C

SECTION 15.2 Objective 15.2A

Objective 15.2B

SECTION 15.3 Objective 15.3A Objective 15.3B

SECTION 15.4 Objective 15.4A Objective 15.4B

Properties of Quadratic Functions Graph a quadratic function

2

2

Find the x-intercepts of

aparabola 6 Find the zeros of a quadratic function

Applications of Quadratic Functions

10

Solve minimum and maximum problems 10 Solve applications of minimum and maximum

Algebra of Functions

13

Perform operations on functions 13 Find the composition of two functions

16

One-to-One and Inverse Functions

19

Determine whether a function is one-to-one Find the inverse of afunction

Solutions to Check Your Understanding

11

19

21

S-1

Solutions to Objective Practice Exercises

S-5

MODULE

16

Exponential and Logarithmic Functions SECTION 16.1 Objective 16.1A Objective 16.1B

SECTION 16.2 Objective 16.2A Objective 16.2B

SECTION 16.3 Objective 16.3A

SECTION 16.4 Objective 16.4A Objective 16.4B

SECTION 16.5 Objective 16.5A

Exponential Functions

2

Evaluate exponential functions 2 Graph exponential functions 4

Introduction to Logarithms

7

Write equivalent exponential and logarithmic equations Use the properties of logarithms

Graphs of Logarithmic Functions Graph logarithmic functions

17

17

Exponential and Logarithmic Equations Solve exponential equations Solve logarithmic equations

7

11

19

19 21

Applications of Exponential and Logarithmic Functions Solve application problems

22

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-7

22

Xili

Xiv_

Contents

MODULE

17

onic Sections

=~

s oTION17.1 bjective 17.1A n

cTlon17.2 \bjective 17.2A ojective 17.2B

s °T10N17.3 biective 17.3A \bjective 17.3B

s

TheParabola

2

Graphparabolas

2

TheCircle

5

—__Find the equation of a circle and then graph the circle 5 Write the equation of a circle in standard form and then graph the circle

The Ellipse and the Hyperbola

8

Graph an ellipse with center at the origin 8 | Graph a hyperbola with center at the origin 10

ction 17.4 — Solving Nonlinear Systems of Equations Dbjective 17.4A

— Solve nonlinear systems of equations

12

12

So'utions to Check Your Understanding S-1 Sc utions to Objective Practice Exercises S-3

MODULE

18

dp)

»quences, Series, and the Binomial Theorem

si TION 18.1

Introduction to Sequences and Series

Onjective 18.1A

Write the terms of

Objective 18.1B

Evaluate aseries

SECTION 18.2

asequence 4

Arithmetic Sequences and Series

6

Objective 18.2A

Find the nth term of an arithmetic sequence

Objective 18.2B

Evaluate an arithmetic series

Objective 18.2C

Solve application problems

SECTION 18.3 Objective Objective Objective Objective

18.3A 18.3B 18.3C 18.3D

SECTION 18.4

Objective 18.4A

2

2

6

8

10

Geometric Sequences and Series

10

— Find the nth term of a geometric sequence 10 Find the sum of a finite geometric series 13 — Find the sum of an infinite geometric series 14 Solve application problems 17

Binomial Expansions

17

Expand (a +b)" 17

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-6

MODULE 19

Measurement SECTION 19.1.

The U.S. Customary System

2

Objective 19.1A Objective 19.1B Objective 19.1C

Convert units of length in the U.S. Customary System 2 Convert units of weight in the U.S. Customary System 4 | Convert units of capacity in the U.S. Customary System 5

Objective 19.1D

| Convert units of time

6

7

Contents

SECTION 19.2 Objective 19.2A Objective 19.2B Objective 19.2C

SECTION 19.3 Objective 19.3A Objective 19.3B

The Metric System

8

Convert units of length in the metric system 8 Convert units of mass in the metric system 9 Convert units of capacity in the metric system 11

Conversion Between the U.S. Customary and the Metric Systems of Measurement 13 Convert U.S. Customary units to metric units Convert metric units to U.S. Customary units

13 14

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-4

MODULE

20

Geometry SECTION 20.1 Objective 20.1A Objective 20.1B Objective 20.1C

SECTION 20.2 Objective 20.2A Objective 20.2B Objective 20.2C

SECTION 20.3 Objective 20.3A Objective 20.3B Objective 20.3C

SECTION 20.4 Objective 20.4A

Angles, Lines, and Geometric Fic

ires

by intersecting lines

Perimeter of a Plane Geometric

“igure

11

Find the perimeter of a plane geometri:

‘igure

11

Find the perimeter of a composite geo! Solve application problems 19

‘etric figure

Area of a Plane Geometric Figure Find the area of a geometric figure

20

Find the area of a composite geometric figure

Solve application problems

Volume

17

2C 24

26

27

Find the volume of a geometric solid

27

Objective 20.4B

Find the volume of a composite geometric solid

Objective 20.4C

Solve application problems

SECTION 20.5

2

Define and describe lines and angles Define and describe geometric figures Solve problems involving angles forme

31

33

The Pythagorean Theorem

34

Objective 20.5A Objective 20.5B

Find the square root of anumber 34 Find the unknown side of a right triangle using the Pythagorean

Objective 20.5C

Theorem 35 Solve application problems

SECTION 20.6 Objective 20.6A

38

Similar and Congruent Triangles Solve similar and congruent triangles

Solutions to Check Your Understanding S-1 Solutions to Objective Practice Exercises S-7

39 39

9

XV

xvi

Contents

MODULE

21

Statistics and Probability s ction 21.1 Objective 21.1A Spjective 21.1B

SECTION 21.2 Objective 21.2A

n

Read apictograph

2

2

Read acircle graph

4

Bar Graphs and Broken-Line Graphs

)bjective 21.2B

Read abar graph 6 Read a broken-line graph

CTION 21.3

Organizing Data

10

Create frequency distributions

Ibjective 21.3C

Read frequency polygons

Read histograms

6

8

Ibjective 21.3A Objective 21.3B

SECTION 21.4

Ss

Pictographs and Circle Graphs

10

12

Statistical Measures

13 15

bjective 21.4A Objective 21.4B

Find the mean, median, and mode of a distribution

dbjective 21.4C

Find the standard deviation of a distribution

Draw a box-and-whiskers plot

19

oTion 21.5

Introduction to Probability

)bjective 21.5A

Calculate the probability of simple events Calculate the odds of an event 28

Dbjective 21.5B

Solutions to Check Your Understanding

Solutions to Objective Practice Exercises

TI-84 PLUS 12

INDEX OF APPLICATIONS INDEX

13

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KEYSTROKE GUIDE FOR THE TI-84 PLUS TABLES

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Letter to Instructor [,today’s multifaceted learning environment, math departments face the challenge of offering a curriculum that allows students easy access to quality courses that are affordable. These three elements represent what many call education’s “iron triangle.”! Iron Triangle

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When one element of the triangle is under stress, t!2 other elements are affected. For instance, as student access to courses increases, mat

departments experience increased

cost, which often puts a strain on quality as class size

increases.

To address these challenges, we offer Mathematics: Journey from Basic Mathematics through Intermediate Algebra. Our program is desig iced to assist mathematically underprepared students in making the transition to colleg»-level math courses. This comprehensive program is a single-source solution that (dresses the entire developmental mathematics curriculum. It provides access for all stu ents, and offers them a high quality learning environment at a modest cost. Mathematics: Journey from Basic Mathematics through Intermediate Algebra includes pre-built courses for basic mathematics, prealgebra, introductory algebra, and intermediate algebra. You can create a customized course to reflect both the needs of your department and your individual teaching style. The learning resources for students include a fully customizable eBook, multiple opportunities to practice skills and concepts, videos with practice questions, and a homework system—all designed to accommodate the student’s unique learning path. The student’s complete learning system is integrated into one online package contained within a single web environment. Through this new, innovative learning system, students can begin their journey with basic college mathematics or at any point in the curriculum that is appropriate to their skill level, and travel all the way through intermediate algebra. No matter where their journey begins, they will pay only one price for access to the entire digital program, a program designed by us, incorporating all of the proven pedagogical features that we use to promote student success.

Whether your course 1s lecture-based or lab-based—in fact, no matter how you model your course—Mathematics: Journey from Basic Mathematics through Intermediate Algebra is the perfectly flexible solution that provides mathematically diverse students access to a quality program at an affordable cost. It is a balanced approach to the “iron triangle.” Thank you for considering this innovative approach to the developmental mathematics curriculum. Richard N. Aufmann

'The National Center for Public Policy and Higher Education and Public Agenda

Joanne S. Lockwood

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Changing the

Equation...

Reimagining the Ok

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Imagine one, flexible solution suitab’ digital access. Mathematics: Journey Aufmann and Lockwood team, deliv: course curriculum covered in basic n streamlined learning path powered by eCourse was created with student suc

‘or the entire developmental mathematics sequence—all for one price for om Basic Mathematics through Intermediate Algebra, designed by the trusted s learning objectives organized by section into 21 modules that span the entire *‘hematics, prealgebra, introductory algebra, and intermediate algebra. With a “Unhanced WebAssign (EWA) and an integrated, fully customizable eBook, this ess and ease of use in mind.

Students can start with basic mathem «ics and travel through multiple courses to intermediate algebra, or they can begin somewhere in the middle. No matter :cre students begin their journey, they will not need to leave their lesson to find tools that will enhance their learning © perience: lecture videos with practice questions, eBook instruction, and various opportunities for practice. If you and / our students prefer a printed book, a custom-printed text including the Read It sections found in the integrated eBoo’ «an be arranged through your Learning Consultant. To further help you guide students along the learning path, we o. er a Guided Workbook based on the Aufmann Interactive Method (AIM), which encourages active participation along ic learning path. No matter your syllabus needs or cou) se environment—hybrid, lab-based, or online—you have everything you need to get started. From pre-built learning od homework assignments to module assessments and more, the Aufmann and Lockwood eCourse allows you to foci s on what matters most—helping your students succeed.

The Annotated Instructor’s E:\ition Is Your Teaching and Plannine Tool The Annotated Instructor’s Edition includes all 21 modules found in the eBook, so

AUEMANN & LOCKWOOD

that you (instructors and department curriculum developers) can examine the entire

MATHEMATICS

sequence of topics available. Choose the modules, sections, and learning objectives

THROUGH INTERMEDIATE ALGEBRA

§

that will enable you to build the course or courses that best fit your syllabi. Beyond some of the natural dependencies that are inherent in mathematics, these modules,

sections, and learning objectives are designed to be flexible, and can be organized by you to build the course sequence that fits your needs.

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approach to fit the rapidly changing needs of the developmental mathematics curriculum. This flexible and complete sequence of topics is designed to address any course type, from the lecturebased course to the flipped classroom. However you decide to

Whole Numbers

deliver your course, this carefully constructed set of learning

MODULE 1

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* Use the Interactive Method. 6 ssa oa

SECTION1.1 Introduction to Whole Numbers 2

objectives can be used to fit any model—traditionally paced or

Objective 1.1A

Identify the order relation between two numbers

Objective 1.1B Objective 1.1C Objective 1.1D

Write whole numbers in words, in standard form, and in expanded form Round a whole number to a given place value 6 Solve application problems and use statistical graphs 7

SECTION 1,2

Addition and Subtraction of Whole Numbers

Objective 1.2A Objective 1.2B

Add whole numbers 14 Subtract whole numbers

Objective 1.2C

Solve application problems

SECTION 1.3.

14

17

20

Multiplication and Division of Whole Numbers

Objective 1.3A

Multiply whole numbers

Objective 1.3B

Divide whole numbers

Objective 1.3C

Solve application problems

SECTION 1.4

2

22

22

25 31

Exponential Notation and the Order of Operations Agreement

33

accelerated,

lecture-based

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lab-based.

By Rethinking the Solution! A Clear Path to Success Mathematics: Journey from Basic Mathematics through Intermediate Algebra is an eCourse powered by Enhanced WebAssign. You have the option of creating your own course or choosing one of our pre-built Course Packs, each of which comes with a suggested Learning Path.

The Power of Enhanced WebAssign

Assighments

Available exclusively from Cengage Learning, Enhanced WebAssign combines exceptional mathematics content with the most powerful online homework solution, WebAssign. Enhanced WebAssign engages students with immediate feedback, rich tutorial content, and an interactive, fully customizable eBook, helping students to develop a deeper conceptual understanding of the subject matter. With Enhanced WebAssign, you can build a customized course that fits your specific needs. You can choose to show all 21 modules or to hide specific modules, and you can pick and choose from a variety of question types and assignments.

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The Students’ Learning Path Students will follow a learning path designed to guide them through assessment and learning activities that can be adjusted by you to fit the needs of your course. Each module begins with Learning Assignments correlated to each learning objective, through which students can review the lesson material, watch videos, and practice solving exercises in preparation for their homework or class. After completing each Learning Assignment for a section, students can begin their Homework. At the end of the module, students can take a Module Quiz to assess their readiness for a test.

My Assignments Current Assignments (16) Name

Due

LRN 1.1A — Identify the order relation be... LRN 1.1B —Write whole numbers in words,

Nov 8 2014 12:01 PM PST .

Nov 8 2014 12:02 PM PST

LRN 1.1C — Round a whole number to a give...

Nov 8 2014 12:03 PM PST

LRN 1.1D - Solve application problems and..

Nov 8 2014 12:04 PM PST

HIMK 1.1 - Introduction to Whole Numbers

Nov 8 2014 12:05 PM PST

LRN 1.2A - Add whole numbers LRN 1,26 - Subtract whole numbers

Nov 8 2014 12:06 PM PST Nov 8 2014 12:07 PM PST

LRN 1.2C - Solve application problems

Nov 8 2014 12:08 PM PST

HIMK 1.2 - Addition and Subtraction of Whole Num,

Nov 8 2014 12:09 PM PST

LRN 1.3A - Multiply whole numbers

LRN 1.3B - Divide whole numbers

Nov 8 2014 12:10 PM PST Nov 8 2014 12:11 PM PST

LRN 1.3C - Solve application problems

Nov 8 2014 12:12 PM PST

HMK 1,3 - Multiplication and Division of Whole ...

Noy 8 2014 12:13 PM PST

LRN 1.4A - Simplify expressions containing expo...

LRN 1.4B - Use the Order of Operations Agreemen...

Nov 8 2014 12:14 PM PST Nov 8 2014 12:15 PM PST

HMK 1.4 - Exponents and the Order of Operations...

Noy 8 2014 12:16 PM PST

CoursePack

XX

Tools for Learning

Tools for Learning Students will find many opportunities to practice with the support of videos and a fully customizable eBook integrated within the course.

“Read It” in the YouBook Each WebAssign question is accompanied by a “Read It” button that students can click on to access the Cengage YouBook (an interactive eBook). Here, they can access Aufmann and Lockwood’s written instruction, including examples for each learning objective. Each “Read It” button takes the student directly to the material that covers the learning objective he or she needs to learn.

What’s the YouBook?

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Once students have launched the YouBook via the “Read It” button, they are able to navigate the reading material for the entire modular course. In addition to the carefully crafted narrative, the YouBook contains the following features:

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Integrated Video Instruction Video instruction, by far the most popular way that students choose to learn and study, is integrated throughout the learning path in the form of “Watch It” videos and Video Example Questions. ‘“‘Watch It” videos accompany each problem to help students who may need additional instruction at that stage of the learning 5 Trelagite UieerBream | |< Ovemtion Dette Geach the cuceber en tha cumber bine 1. = Quenton Dette process. Video Example "Watch the video below then answer the ovastion (lek ond dragech pri a te arpraqvens peekicn Sorespotswiht be owned Questions ask students Rounding Whole Numbars totheNearest Ten to watch a video and then é answer a question about St the video. Videos are also A Wao integrated in a “just-iney Re time” manner into the YouBook “Read Its.”

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This flexible tool helps students identify which topics they have mastered and which require further study on their part. Comprised of quizzes that assess learning gaps, the PSP suggests learning tools that students can use to remediate the topics they have not yet mastered. Visual indicators enable students to easily view their performance in relation to a pre-determined mastery level.

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Tools for Teaching YouBook Features for Instructors The YouBook features a text-edit tool that allows you—the instructor—to modify the narrative as you see fit. With YouBook, you can quickly rearrange modules and sections or hide any content you don’t teach to create an eBook that perfectly matches your syllabus. You can further customize the eBook by adding links to your lecture notes, syllabi, video lectures, or files on a personal website. With this level of customization, the YouBook becomes a perfect class planning tool in addition to a learning tool.

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Resources for Teaching and Learning Within the Enhanced WebAssign course, the Resource Tab includes the Guided Workbook, lecture videos, a link to Paul Nolting’s Math Study Skills website, and additional instructor resources.

Cengage Learning Testing by Cognero Cengage Learning Testing, powered by Cognero, is a flexible online system that allows you to author, edit, and manage title-specific test bank content, create multiple test versions in an instant, and deliver tests from your LMS, your classroom, or wherever you want. This system is available online via the Instructor Companion site at login.cengage.com. 2% **

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Options to Suit Your Course Needs

Options to Suit Your Course Needs Along with your standalone, custom-built WebAssign course, you have the option of a print solution for you and your students.

Custom

Print Modules

All the material found in the YouBook (the interactive eBook) can be custom-printed. Contact your Learning Consultant to find out how you can choose the modules that fit your syllabus—the same ones you choose for your Enhanced WebAssign course—to create a custom-printed text.

Guided Workbook to Support Students Through the Learning Path The Guided Workbook is available automatically with digital access and can be found in the Resource Tab. Like the YouBook, this is also available for custom printing. Based on the Aufmann Interactive Method (AIM), this workbook encourages active participation as students work through the learning path. Working with your Learning Consultant, you can create a customized workbook that contains only the modules you need to accompany your Enhanced WebAssign course.

Aufmann Interactive Method (AIM) Available in the YouBook, Module A, titled “AIM for Success,” outlines the study skills used by students who have been successful in developmental mathematics courses. By making Module A the first module of the program, we have set the stage for a successful beginning to the student’s journey. A short quiz assessing comprehension of the module is offered in Enhanced WebAssign.

A

AIM for Success: How to Succeed in This Course

Get Ready

Motivate Yourself Develop a “Can Do” Attitude Toward Math

Strategies for Success Time Management

Paul Nolting’s Math Study Skills Workbook, 5e and Course Pack

The Nolting Math Study Skills, 5e, Course Pack is included within the Enhanced WebAssign course. The Nolting study skills strategy helps students identify their strengths, weaknesses, and personal learning styles. Then, it presents an easy-to-follow system to help them become more successful at learning mathematics. Sold separately, the Nolting Math Study Skills Workbook, 5e is key to getting the most out of the Course Pack. This best-selling math study skills workbook offers proven study tips, test-taking strategies, and recommendations for reducing math anxiety and improving grades.

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Acknowledgments

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Acknowledgments The authors thank the following mathematics educators for their insightful contributions to the development of Mathematics: Journey from Basic Mathematics through Intermediate Algebra. Wendy Ahrendsen, South Dakota State University Hamid Attarzadeh, Jefferson Community and Technical College Scott Barnett, Henry Ford Community College Michelle Beard, Ventura College Nicoleta Bila, Fayetteville State University Shane Brewer, Utah State University—College of Eastern Utah—Blanding Campus Michael Carr, Mott Community College Edie Carter, Amarillo College Linh Changaris, Jefferson Community and Technical College Fredrick Chapple, Baltimore City Community College Edward Ennels, Baltimore City Community College Ines Figueiras, Essex Community College Maggie Flint, Northeast State Community College Dot French, Community College of Philadelphia Asha Hill, Georgia Highlands College Steven Jackson, Jvy Tech Community College Kimberly Johnson, Mesa Community College Maryann Justinger, Erie Community College David Keller, Kirkwood Community College Alex Kolesnik, Ventura College Julianne Labbiento, Lehigh Carbon Community College Edith Lester, Volunteer State Community College Barbara Manley, Jackson State Community College Lola Marie McClendon, Pasadena City College Kim Minsu, Gainesville State College Carla Monticelli, Camden Community College Katie Motlow, Calhoun Community College Javad Moulai, Roxbury Community College Eli Nettles, Nashville State Community College Sandra Nieto, Santa Rosa Junior College Jon Odell, Richland Community College Priti Patel, Tarrant County College Betty Peterson, Mercer County Community College Edward Pierce, Ivy Tech Community College Matthew Pitassi, Rio Hondo College Michael Potter, University of Virginia’s College at Wise Harriette Roadman, New River Community College Elizabeth Rourke, College of Southern Maryland Christopher Sabino, Harold Washington College Daria Santerre, Norwalk Community College Barbara Savage, Roxbury Community College So Shing, University of Central Missouri Pam Stogsdill, Bossier Parish Community College M. Terry Simon, University of Toledo Chairsty Stewart, Montana State University—Billings Sean Stewart, Owens Community College Abolhassan Taghavy, Richard J. Daley College René Thompson, Vance-Granville Community College Mary Ann Teel, University of North Texas Rose Toering, Kilian Community College

Xxiv

Acknowledgments

Diane Veneziale, Burlington County College Lynette Wedig, New Mexico State University-Alamogordo Yoshi Yamoto, Pasadena City College Tzu-Yi Yang, Columbus State Community College Kim Yon, Passaic County Community College Jeffrey Zilahy, Montgomery County Community College Cathleen Zucco-Teveloff, Rider University

AIM for Success: How to Succeed in This Course Get Ready

Motivate Yourself Develop a “Can Do” Attitude Toward Math

Strategies for Success Time Management

Habits of Successful Students Use the Interactive Method Use a Strategy to Solve Word Problems

Ace the Test Ready, Set, Succeed!

2

Module A e AIM for Success:

How to Succeed in This Course

Get Ready We are committed to your success in learning mathematics and have developed many tools and resources to support you along the way.

DO YOU WANT TO EXCEL IN THIS COURSE? Read on to learn about the skills you’ll need and how best to use this text to get the results you want. We have written this text in an interactive style. More about this later but, in short, this means that you are supposed to interact with the text. Do not just read the text! Work along with it. Ready? Let’s begin!

WHY ARE YOU TAKING THIS COURSE? Did you interact with the text, or did you just read the last question? Get some paper and a pencil or pen and answer the question. Really—you will have more success in math and other courses you take if you actively participate. Now, interact. On your sheet of paper, write down one reason you are taking this course. Of course, we have no idea what you just wrote, but experience has shown us that many of you wrote something along the lines of “I have to take it to graduate” or “It is a prerequisite to another course I have to take” or “It is required for my major.” Those reasons are perfectly fine. Every teacher has had to take courses that were not directly related to his or her major.

WHY DO YOU WANT TO SUCCEED IN THIS COURSE? Think about why you want to succeed in this course. List the reasons (not in your head. . . on the sheet of paper’). One reason you may have listed is that math skills are important in order to be successful in your chosen career. That is certainly an important reason. Here are some other reasons.

e Math is a skill that applies across careers, which is certainly a benefit in our world of changing job requirements. A good foundation in math may enable you to more easily make a career change. ¢ Math can help you learn critical thinking skills, an attribute all employers want. e Math can help you see relationships between ideas and identify patterns. Take Note > Motivation alone won’t lead to success. For example, suppose a person who cannot swim is rowed out to the middle of a lake and thrown overboard. That person has a lot of motivation to swim, but most likely will drown without some help. You’ll need motivation and learning in order to succeed.

Motivate Yourself You'll find many real-life problems in this text, relating to sports, money, cars, music, and

more. We hope that these topics will help you understand how mathematics is used in eyeryday life. To learn all of the necessary skills and to understand how you can apply them to your life outside of this course, motivate yourself to learn. One of the reasons we asked you why you are taking this course was to provide motivation for you to succeed. When there is a reason to do something, that task is easier to accom-

plish. We understand that you may not want to be taking this course but, to achieve your career goal, this is a necessary step. Let your career goal be your motivation for success.

MAKE THE COMMITMENT TO SUCCEED! With practice, you will improve your math skills. Skeptical? Think about when you first learned to drive a car, ride a skateboard, dance, paint, surf, or any other talent that you now

have. You may have felt self-conscious or concerned that you might fail. But with time and practice, you learned the skill.

Write down a situation in which you accomplished your goal by spending time practicing and perfecting your skills (such as learning to play the piano or to play basketball).

Module A ¢ AIM for Success:

How to Succeed in This Course

3

You do not get “good” at something by doing it once a week. Practice is the backbone of any successful endeavor—including math!

Develop a “Can Do” Attitude Toward Math You can do math! When you first learned the skills you just listed, you may not have done them well. With practice, you got better. With practice, you will get better at math. Stay focused, motivated, and committed to success.

We cannot emphasize enough how important it is to overcome the “I Can’t Do Math” syndrome. If you listen to interviews of very successful athletes after a particularly bad performance, you will note that they focus on the positive aspects of what they did, not the negative. Sports psychologists encourage athletes always to be positive—to have a “can do” attitude. Develop this attitude toward math and you will succeed. Change your conversation about mathematics. Do not say “I can’t do math,” “I hate math,” or “Math is too hard.” These comments just give you an excuse to fail. You don’t want to fail, and we don’t want you to fail. Write it down now: | can do math!

Strategies for Success PREPARE TO SUCCEED There are a number of things that may be worrisome to you as you begin a new semester. List some of those things now. Here are some of the concerns that have been expressed by our students.

¢ Tuition Will I be able to afford school?

° Job I must work. Will my employer give me a schedule that will allow me to go to school? e Anxiety Will I succeed?

¢ Child care What will I do with my kids while I’m in class or when I need to study? ¢ Time Will I be able to find the time to attend class and study? Degree goals How long will it take me to finish school and earn my degree?

These are all important and valid concerns. Whatever your concerns, acknowledge them. Choose an education path that allows you to accommodate your concerns. Make sure they don’t prevent you from succeeding.

SELECT A COURSE Many schools offer math assessment tests. These tests evaluate your present math skills. They don’t evaluate how smart you are, so don’t worry about your score on the test. If you are unsure about where you should start in the math curriculum, these tests can show you where to begin. You are better off starting at a level that is appropriate for you than starting with a more advanced class and then dropping it because you can’t keep up. Dropping a class is a waste of time and money.

If you have difficulty with math, avoid short courses that compress the class into a few weeks. If you have struggled with math in the past, this environment does not give you the time to process math concepts. Similarly, avoid classes that meet once a week. The time delay between classes makes it difficult to make connections between concepts.

4

Module A « AIM for Success:

How to Succeed in This Course

Some career goals require a number of math courses. If that is true of your major, try to take a math course every semester until you complete the requirements. Think about it this way. If you take, say, French I, and then wait two semesters before taking French II, you may forget a lot of material. Math is much the same. You must keep the concepts fresh in your mind.

Time Management One of the most important requirements in completing any task is to acknowledge the amount of time it will take to finish the job successfully. Before a construction company starts to build a skyscraper, the company spends months looking at how much time each of the phases of construction will take. This is done so that resources can be allocated when appropriate. For instance, it would not make sense to schedule the electricians to run wiring until the walls are up.

MANAGE YOUR TIME! We know how busy you are outside of school. Do you have a full-time or part-time job? Do you have children? Do you visit your family often? Do you play school sports or participate in the school orchestra or theater company? It can be stressful to balance all of the important activities and responsibilities in your life. Creating a time management plan will help you schedule enough time to do everything you need to do. Let’s get started. First, you need a calendar. You can use a daily planner, a calendar for a smartphone, or an online calendar, such as the ones offered by Google, MSN, or Yahoo. It is best to have a calendar

on which you can fill in daily activities and be able to see a weekly or monthly view as well. Start filling in your calendar now, even if it means calendar. Some of the things you might include are: ¢ The hours each class meets

° Time to eat

e Time for driving to and from work or

e Your work schedule

school Take Note > Be realistic about how much time you have. One gauge is that working 10 hours per week is approximately equivalent to taking one three-unit course. If your college considers 15 units a full load and you are working 10 hours per week, you should consider taking 12 units. The more you work, the fewer units you should take.

stopping right here and finding a

e Leisure time, an important aspect of a

healthy lifestyle e Time for study. Plan at least one hour of study for each hour in class. oe Ee This is a minimum!

* Time for extracurricular activities such as sports, music lessons, or volunteer work

¢ Time for family and friends * Time for sleep ; es ¢ Time for exercise

We really hope you did this. If not, please reconsider. One of the best pathways to success is understanding how much time it takes to succeed. When you finish your calendar, if it does not allow you enough time to stay physically and emotionally healthy, rethink some of your school or work activities. We don’t want you to lose your job because you have to study math. On the other hand, we don’t want you to fail in math because of your job. If math is particularly difficult for you, consider taking fewer course units during the semesters you take math. This applies equally to any other subject that you may find difficult. There is no rule that you must finish college in four years. It is a myth—discard it now. Now extend your calendar for the entire semester. Many of the entries will repeat, such as the time a class meets. In your extended calendar, include significant events that may disrupt your normal routine. These might include holidays, family outings, birthdays, anniversaries, or special events such as a concert or a football game. In addition to these events, be sure to include the dates of tests, the date of the final exam, and dates that projects or

papers are due. These are all important semester events. Having them on your calendar will remind you that you need to make time for them.

Module A ¢ AIM for Success:

How to Succeed in This Course

5

CLASS TIME To be successful, attend class. You should consider your commitment to attend class as serious as your commitment to your job or to keeping an appointment with a close friend. It is difficult to overstate the importance of attending class. If you miss work, you don’t get paid. If you miss class, you are not getting the full benefit of your tuition dollar. You are losing money. If, by some unavoidable situation, you cannot attend class, find out as soon as possible what

was covered in class. You might: ° Ask a friend for notes and the assignment.

¢ Contact your instructor and get the assignment. Missing class is no excuse for not being prepared for the next class. ¢ Determine whether there are online resources that you can use to help you with the topics and concepts that were discussed in the class you missed.

Going to class is important. Once you are there, participate in class. Stay involved and active. When your instructor asks a question, try to at least mentally answer the question. If you have a question, ask. Your instructor expects questions and wants you to understand the concept being discussed.

HOMEWORK TIME In addition to attending class, you must do homework. Homework is the best way to reinforce the ideas presented in class. You should plan on at least one to two hours of homework and study for each hour you are in class. We’ve had many students tell us that one to two hours seems like a lot of time. That may be true, but if you want to attain your goals, you must be willing to devote the time to being successful in this math course.

You should schedule study time just as if it were class time. To do this, write down where and when you study best. For instance, do you study best at home, in the library, at the math center, under a tree, or somewhere else? Some psychologists who research successful study strategies suggest that just by varying where you study, you can increase the effectiveness of a study session. While you are considering where you prefer to study, also think about the time of day during which your study period will be most productive. Write down your thoughts. Look at what you have written, and be sure that you can consistently be in your favorite study environment at the time you have selected. Studying and homework are extremely important. Just as you should not miss class, do not miss study time. Before we leave this important topic, we have a few suggestions. If at all possible, create a study hour right after class. The material will be fresh in your mind, and the immediate review, along with your homework, will help reinforce the concepts you are learning. If you can’t study right after class, make sure that you set aside some time on the day of the class to review notes and begin the homework. The longer you wait, the more difficult it will be to recall some of the important points covered during class. Study math in small chunks—one hour a day (perhaps not enough for most of us), every day, is better than seven hours in one sit-

ting. If you are studying for an extended period of time, break up your study session by studying one subject for a while and then moving on to another subject. Try to alternate between similar or related courses. For instance, study math for a while, then science, and then back to

math. Or study history for a while, then political science, and then back to history.

Meet some of the people in your class and try to put together a study group. The group could meet two or three times a week. During those meetings, you could quiz each other, prepare for a test, try to explain a concept to someone else in the group, or get help on a topic that is difficult for you.

After reading these suggestions, you may want to rethink where and when you study best. If so, do that now. Remember, however, that it is your individual style that is important.

Choose what works for you, and stick to it.

6

Module A « AIM for Success:

How to Succeed in This Course

Habits of Successful Students There are a number of habits that successful students use. Think about what these might be, and write them down.

What you have written is very important. The habits you have listed are probably the things you know you must do to succeed. Here is a list of some responses from successful students we have known. ¢ Set priorities. You will encounter many distractions during the semester. Do not allow them to prevent you from reaching your goal. Take responsibility. Your instructor, this text, tutors, math centers, and other resources are there to help you succeed. Ultimately, however, you must choose to learn. You must choose success. Hang out with successful students. Success breeds success. When you work and study with successful students, you are in an environment that will help you succeed. Seek out people who are committed to their goals. e Study regularly. We have mentioned this before, but it is too important not to be repeated. Self test. Once every few days, select homework exercises from previous assignments and use them to test your understanding. Try to do these exercises without getting help from examples in the text. These self tests will help you gain confidence that you can do these types of problems on a test given in class. Try different strategies. If you read the text and are still having difficulty understanding a concept, consider going a step further. Contact the instructor or find a tutor. Many campuses have some free tutorial services. Go to the math or learning center. Consult another text. Be active and get the help you need. Make flash cards. This is one of the strategies that some math students do not think to try. Flash cards are a very important part of learning math. For instance, your instructor may use words or phrases such as linear, quadratic, exponent, base, rational, and many others. If you don’t know the meanings of these words, you will not know what is being discussed. In this text, you will see important concepts and rules set off in boxes. Here is one about multiplication. These rules are good candidates for flash cards.

Flash Card

Rule for Multiplying Exponential Expressions

RwWe for Multiplying Exponential E xpressiony If wand wave integers,

If m and n are integers, then x”"- x” = x m+n EXAMPLES

Examples:

In each example below, we are multiplying two exponential expressions with the same base. Simplify the expression by adding the exponents.

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2. y-y=y

then ur

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yee5s ba

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¢ Plod along. Your education is not a race. The primary goal is to finish. Taking too many classes and then dropping some does not get you to the end any faster. Take only as many classes as you can successfully manage.

Use the Interactive Method We want you to be actively involved in learning mathematics, and we have given you many suggestions for getting “hands-on” with this text.

Module A ¢ AIM for Success:

How to Succeed in This Course

7

Basic examples introduce a concept (in this case, solving a percent equation) and include a step-by-step solution of the type of exercise you will find in the homework problems. 60% of what number is 300? Percent

. base

—

amount

*

Use

0.60 -n

=

300

*

Percent

0.60n 0.60

=

300

——

the basic

percent

60%

equation

0.60,

base

* Solve for n by dividing each

= n, amount

300

side of the equation by 0.60

0.60 n = 500

60%

of 500 is 300.

Focus on/Check your understanding pairs help you get the practice you need to succeed.

_ Focus on solving the basic percent equation | a. Find 9.4% of 240. SOLUTION

| a. Use the basic percent equation.

Percent = 9.4% = 0.094, base = 240, amount = n Percent - base = amount

0.094 - 240 =n 22.56 =n 94%

of 240 is 22.56

| Check your understanding 1

a. Find 663% of 45. _ SOLUTION

See pages S-2-S-3.

You'll see that each Focus on is fully worked out. Study the Focus on example by carefully working through each step. Then, try to complete the Check your understanding. Use the solution to the Focus on as a model for solving the Check your understanding. If you get stuck, solutions to the Check your understandings are provided in an appendix. There is a page number directly following the Check your understanding exercises that shows you where you can find the completely-worked-out solution. Use the solution to get a hint for the step on which you are stuck. Then, try again! When you’ve arrived at your answer, check your work against the solution in the appendix. The solution to Check your understanding | above is given below.

Check your understanding 1 a. STRATEGY To find the amount, use the basic percent equation. Percent = 665% = 5 base = 45, amount = n SOLUTION

Percent : base = amount

=(45) =n

2 =(45)5—

30 =n

30 is 665% of 45.

8

Module A e AIM for Success:

How to Succeed in This Course

Use a Strategy to Solve Word Problems One of the reasons you are studying math is to learn to solve word problems. Solving word problems involves combining all of the critical thinking skills you have learned to solve practical problems. Try not to be intimidated by word problems. Basically, what you need is a strategy that will help you come up with the equation you will need to solve the problem. When you are looking at a word problem, try the following: Read the problem. This may seem pretty obvious, but we mean really read it. Don’t just scan it. Read the problem slowly and carefully. Write down what is known and what is unknown. Now that you have read the problem, go back and write down everything that is known. Next, write down what it is you are trying to find. Write this—don’t just think it! Be as specific as you can. For instance, if you are asked to find a distance, don’t just write “I need to find the distance.” Be specific and write “I need to find the distance between Earth and the moon.”

Think of a method to find the unknown. For instance, is there a formula that relates the known and unknown quantities? This is certainly the most difficult step. Eventually, you must write an equation to be solved. Solve the equation. Be careful as you solve the equation. There is no sense in getting to this point and then making a careless mistake. The unknown in most word problems will include a unit such as feet, dollars, or miles per hour. When you write your answer, include the unit. An answer such as 20 doesn’t mean much. Is it 20 feet, 20 dollars, 20 miles per hour, or something else?

Check your solution. Now that you have an answer, go back to the problem and ask yourself whether it makes sense. This is an important step. For instance, if, according to your answer, the cost of a car is $2.51, you know that something went wrong. In this text, the solution to every word problem is broken down into two steps, Strategy and Solution. The Strategy consists of the first three steps discussed above. The Solution is the last two steps.

| Focus on solving a percent concentration application To make a certain color of blue, 4 0z of cyan must be contained in | gal of paint. What is the percent concentration of cyan in the paint? STRATEGY

The amount of cyan is given in ounces and the amount of paint is given in gallons. We must convert ounces to gallons or gallons to ounces. For this problem, we will convert gallons to ounces: | gal = 128 oz. Solve Q = Ar for r, with Q = 4and A = 128. SOLUTION ‘@) =

Ar

4 = 128r

4

¢ Use the percent mixture equation

°-Q = 4,A

=

128

128r

128s iad28 0.03125 = 7

The percent concentration of cyan is 3.125%.

Check your understanding 5 The concentration of sugar in a certain breakfast cereal is 25%. If there are 2 oz of sugar contained in a bowl of cereal, how many ounces of cereal are in the bowl? | SOLUTION

See page S-4.

Module A ¢ AIM for Success:

How to Succeed in This Course

9

When you have finished studying a section, do the exercises your instructor has selected. Math is not a spectator sport. You must practice every day. Do the homework and do not get behind.

Ace the Test There are a number of features in this text that will help you prepare for a test. These features will help you even more if you do just one simple thing: When you are doing your homework, go back to each previous homework assignment for the current module and rework two exercises. That’s right—just two exercises. You will be surprised at how much better prepared you will be for a test by doing this.

Ready, Set, Succeed! It takes hard work and commitment to succeed, but we know you can do it! Doing well in mathematics is just one step you'll take on your path to success. Good luck. We wish you success.

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Whole Numbers

SECTION 1.1

Introduction to Whole Numbers

Objective 1.1A

Identify the order relation between two numbers

Objective 1.1B

Write whole numbers in words, in standard form, and in expanded form

Objective 1.1C

Round a whole number to a given place value

Objective 1.1D

Solve application problems and use statistical graphs

SECTION 1.2

Addition and Subtraction of Whole Numbers

Objective 1.2A

Add whole numbers

Objective 1.2B

Subtract whole numbers

Objective 1.2C

Solve application problems

SECTION

1.3

Objective 1.3A

Multiplication and Division of Whole Numbers Multiply whole numbers

Objective 1.3B

Divide whole numbers

Objective 1.3C

Solve application problems

SECTION 1.4

Exponential Notation and the Order of Operations Agreement

Objective 1.4A

Simplify expressions that contain exponents

Objective 1.4B

Use the Order of Operations Agreement to simplify expressions

2

Module 1 * Whole Numbers

SECTION

11

Introduction to Whole Numbers

Objective 1.1A

Identify the order relation between two numbers The natural numbers are |, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, The whole numbers are (), |. 2. 3. 4, 5, 6, 7. 8. 9, 10, 11, .... Note that the whole numbers include the natural numbers and zero.

Just as distances are associated with markings on the edge of a ruler, the whole numbers can be associated with points on a line. This line is called the number line and is shown below.

es 0

1

2

3

4

5

6

Wl

8

9

10

11

12

il}

14

The graph of a whole number is shown by placing a heavy dot on the number line directly above the number. Shown below is the graph of 6 on the number line.

| Focus on graphing a number on a number line _ Graph 4 on the number line.

| SOLUTION F-t+—tt-$ |

1

B45

ttt (5) 16. feos

te

OO

ois 12

|Check your understanding 1 | Graph 9 on the number line.

| SOLUTION

See page:S20)

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tester uae

race!

2

Ole

7

8.

SLO

het

On the number line, the numbers get larger as we move from left to right. The numbers get smaller as we move from right to left. Therefore, the number line can be used to visualize the order relation between two whole numbers.

_ Focus on finding a number on a number line |On the number line, what number is 3 units to the right of 4?

| SOLUTION

Lone: aia (OY til

st gh

dt

lal

5 6

oe

te

He EP AG) il

|7 is 3 units to the right of 4.

| Check your understanding 2 | On the number line, what number is 4 units to the left of 11? | SOLUTION ae

See page S-1.

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op

et2 2 4

®

7

i

Sa) Ril we

Section 1.1 © Introduction to Whole Numbers

3

A number that appears to the right of a given number is greater than the given number. The symbol for is greater than is >.

te

8 is to the right of 3. ; 8 is greater than 3. ip ae]

0

2

28

AS

Go

PS

10 iM 11s

44

A number that appears to the left of a given number is less than the given number. The symbol for is less than is

5 is to the left of 12.

An inequality symbol, < or >, points to the smaller number. The symbol opens toward the larger

Raetess than 19. Se

number.

An inequality expresses the relative order of two mathematical expressions. 8 > 3 and 5 < |2 are inequalities.

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| Focus on finding the order relation of two numbers | Place the correct symbol, < or >, between the two numbers.

Erm

Bho

es)

b.

0

54

\algesase 23

b.

O0
, between the two numbers.

a.

47

SOLUTION

19

bi

9265,

0

See page S-1.

a.

47

> 19

i

alo

(0)

tes

Focus on ordering a list of numbers Write the given numbers in order from smallest to largest.

16, 5, 47, 0, 83, 29 SOLUTION

055,16, 29,547,.83

Check your understanding 4 Write the given numbers in order from smallest to largest.

52, 17, 68, 0, 94, 3 SOLUTION

See page S-1.

0, 3, 17, 52, 68, 94

Objective 1.1A Practice 1. On a number line, what number is 3 units to the right of 2? 5 2. On a number line, what number is 7 units to the left of 7? 0 3. Place the correct symbol, < or >, between the two numbers:

Die

194278 = 94

4. Place the correct symbol, < or >, between the two numbers:

3827

6915

3827 < 6915

5. Arrange the numbers 440, 404, 400, 444, and 4000 in order from smallest to largest. 400, 404, 440, 444, 4000

Solutions on p. S-8.

4

Module 1 ¢ Whole Numbers

Objective

1.1B

Write whole numbers in words, in standard form, and in expanded form When a whole number is written using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, it is said to

be in standard form. The position of each digit in the number determines the digit’s place value. The diagram below shows a place-value chart naming the first twelve place values. The number 64,273 is in standard form and has been entered in the chart. In the number 64,273, the position of the digit 6 determines that its place value is ten-thousands.

When a number is written in standard form, each group of digits separated by a comma is called a period. The number 5.3!6,79.842 has four periods. The period names are shown in red in the place-value chart above. To write a number in words, start from the left. Name the number in each period. Then write the period name in place of the comma.

| Focus on writing a number in words

| Write 82,593,075 in words. | SOLUTION |

| |

s

~NY

ese

&

oas WO S

ar

| 82,593,075

*

| eighty-two million five hundred ninety-three

¢ Name the number in each period, replacing

thousand

|

There are three periods

each comma

seventy-five

by the period name

| Check your understanding 5

|Write 46,032,715 in words. | SOLUTION

See page S-1.

forty-six million thirty-two thousand seven hundred fifteen

To write a whole number in standard form, write the number named in each period, and

replace each period name with a comma.

|Focus on writing a number in standard form |

| Write six million fifty-one thousand eight hundred seventy-four in standard form.

|

| SOLUTION | six million fifty-one thousand eight hundred seventy-four

* Underline cach period name

| 6.051.874

e Write the number in each period, replacing each period name with a comma. Each period, except the first, must

contain three digits. Insert a zero as a placeholder for the hundred-thousands place

Section 1.1

¢ Introduction to Whole Numbers

Check your understanding 6 Write nine hundred twenty thousand eight in standard form.

| SOLUTION

See page S-1.

920,008

The whole number 37,286 can be written in expanded form as 30,000

+ 7000

+ 200 + 80

+ 6

The place-value chart can be used to find the expanded form of a number.

di

2

ae

Thousands

+

7000

a

8

Hundreds

a

6

Tens

8

Ones

thousands

30,000

200

80

Write the number 510,409 in expanded form.

5

0)

Hundred-

4 | Thousands

thousands

500,000

4

0

| 4 | Hundreds

| 1

9

Tens

33

Ones

0

}

9

thousands

+

10,000

+

0

+

= 500,000

+

10,000

400 + 400

+ 9

Focus on writing a number in expanded form Write 32,598 in expanded form. SOLUTION

30,000 += 2000) =- 500) == 90) +8

Check your understanding 7 Write 76,245 in expanded form.

SOLUTION

See page S-1.

70,000 + 6000 + 200 + 40 + 5

6

Module 1 ¢ Whole Numbers

Objective 1.1B Practice Write Write Write = 5BS Write

2801 in words. two thousand eight hundred one 356,943 in words. three hundred fifty-six thousand nine hundred forty-three three thousand four hundred fifty-six in standard form. 3456 seven million twenty-four thousand seven hundred nine in standard

form. 7,024,709 5. Write 403,705 in expanded form. 400,000 + 3000 + 700 + 5 Solutions on p. S-8.

Objective 1.1C

Round a whole number to a given place value When the distance to the sun is given as 93,000,000 mi, the number represents an approximation to the true distance. Giving an approximate value for an exact number is called rounding. A number is rounded to a given place value. 48 is closer to 50 than it is to 40. 48 rounded to the nearest ten is 50. 4872 rounded to the nearest ten is 5

4870.

4872 rounded to the nearest hundred is 4900.

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4880

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4900

We can round a number to a given place value without using the number line by looking at the first digit to the right of the given place value. If the digit to the right of the given place value is less than 5, replace that digit and all digits to the right of it by zeros.

Round 12,743 to the nearest hundred.

i

7

Given place value

12,743

4

8 12 1H: 2

SSUES

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Borrow | ten

Add the

subtract the

| Borrowing

from the tens

borrowed

1umbers in

_ is necessary.

column and

10 to 2

| 9tens =

write 10 in the

i 8 tens + | ten

ones column.

each column.

_ Focus on subtracting two numbers with borrowing

| Subtract 4392 — 678 and check. _SOLUTION

3

|

A

|| Check your understanding 4 | Subtract 3481 — 865 and check.

| SOLUTION

See page S-3.

2616

Subtraction may involve repeated borrowing.

| Subtract: 7325 — 4698 1

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j Borrow | ten _ (10 ones) from the - tens column and

Borrow | hundred (10 tens) from the hundreds column

Borrow 1 thousand (10 hundreds) from the thousands column

| add 10 to the 5 in

and add 10 to the |

and add 10 to the 2 in

_ the ones column. | Subtract 15 — 8.

in the tens column. Subtract 11 — 9.

the hundreds column. Subtract 12 — 6 and

j | The difference is 2627.

6 — 4.

_ Focus on subtracting two numbers with repeated borrowing | Find 23,954 less than 63,221 and check. |

SOLUTION

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2

ome reat

O

2B,

%

pe

2

3,9

5

&

BW Cle Det.

Check:

23,954 + 39,267

63,221

Section 1.2 ¢ Addition and Subtraction of Whole Numbers

Check your understanding 5 Find 54,562 decreased by 14,485 and check.

"SOLUTION

See page S-3.

40,077

When there is a zero in the minuend, subtraction involves repeated borrowing.

|

Subtract: 3904 — 1775

j

8

Wren cred!

10

8

Bas

fA

Shea!

Bye

5

Hi

14

BO WB fis

es OE es

i

Bp

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Borrow | ten from the tens column and add 10 to the

_ from the hundreds

4 in the ones

' column and write 10

column.

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Gi

14

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|There is a 0 in the i tens column. Borrow | 1 hundred (10 tens)

8

Ne

age

nC)

Subtract the numbers in each column.

| in the tens column.

Focus on subtracting two numbers with repeated borrowing Subtract 46,005 — 32,167 and check. SOLUTION 5 =

46 39

5

10 oF OS (ena

* There are two zeros in the minuend Borrow | thousand from the thousands column and write 10 in the hundreds column

2

9

5 4 6 7 Sey)

16 Or ie

10 30-5 You)

* Borrow | hundred from the hundreds column and write 10 in the tens column.

a

5 =

9

9

6

1

15

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5 9

Sh

ate!

4G fj 0 39) il

¢ Borrow

| ten from the tens column

and add 10 to the 5 in the ones column

y.

iE Si

stele

Gheck:=

3267,

+ 13,838 46,005

Check your understanding 6 Subtract 64,003 — 54,936 and check.

SOLUTION

See page S-3.

9067

19

20

Module 1 ¢ Whole Numbers

Take Note >

All of the phrases listed below indicate subtraction. An example is shown to the right of each phrase.

Note the order in which the numbers are subtracted when the phrase less than is used. Suppose that you

have $10 and I have $6 less than you do; then I have $6 less than

$10, or $10 — $6 = $4.

minus

10 minus 3

less

8 less 4

8 —4

2 less than9

Sh

the difference between

the difference between 6 and 1

6 — 1 zi

decreased by

7 decreased by 5

LSS

subtract... from

subtract 11 from 20

less than

:

i) = 3

PAD ce 1

| Focus on evaluating a variable expression | Evaluate c — d for c = 6183 and d = 2759. SOLUTION

C= Ul 6183 — 2759

* Replace c with 6183 and d with 2759

511) 7713)

6X83

* Arrange the numbers vertically and then subtract

| =2759

3424

| Check your understanding 7 | Evaluate 4007 — x when x = 389. | SOLUTION |

=

See page S-3.

3618

Objective 1.2B Practice 1. Subtract: 3526 — 387

3139

2. Subtract: 70,702 — 4239

66,463

3. Find the difference between 1003 and 447. 4. What is 29,797 less than 68,005? 38.208

5. Evaluate x —

556

y when x = 1605 and y = 839. 766

Solutions on pp. S-10-S-11.

Objective 1.2C

Solve application problems "iFocus on solving an application with a circle graph

|

The circle graph at the right shows the numbers of eggs produced in the United States in a recent year. The graph shows where the eggs that were produced went or how they were used. Use the graph to determine the sum of the number of cases of eggs sold by retail stores and the number of cases used for nonshell products.

Exported 2,000,000 Food Service Use 18,200,000

Non-shell Products 68,200,000 Retail Stores 125,500,000

Eggs Produced in the United States (in cases)

Section 1.2 ¢ Addition and Subtraction of Whole Numbers

21

STRATEGY

To find the sum of the number of cases of eggs sold by retail stores and the number of cases used for non-shell products, read the circle graph to find the number of each type. Then add the numbers. SOLUTION

125,500,000 cases of eggs were sold by retail stores. 68,200,000 cases of eggs were used for non-shell products.

+

125,500,000 68,200,000 193,700,000

193,700,000 cases of eggs were sold by retail stores or used for non-shell products.

Check your understanding 8 Use the circle graph given on the previous page to determine the total number of cases of eggs produced during the given year.

SOLUTION

See page S-3.

213,900,000 cases of eggs

Apply the Concept The graph in Figure 2 shows the actual or projected world energy consumption in quadrillion British

>

thermal units (Btu) for selected years. Find the dif-

ference between the projected world energy con-

3

sumption in 2020 and in 2030.

er

SOLUTION

~

%

6oy

§

&

2

.

To find the difference, subtract the projected world energy consumption in 2020 from the projected world energy consumption in 2030.

695 — 608

Thermal British quadrillions) (in Units

87

2005

The difference is 87 quadrillion Btu.

2010

2015

2020

2025

2030

Figure 2 World Energy Consumption (in quadrillion Btu) Source: http://ecopolitology.org

Focus on using a bar graph Use Figure 2 to find the difference between the world energy consumption in 2005 and that projected for 2020. STRATEGY

To find the difference in consumption between 2005 and 2020, read the graph to find the consumption in 2005 (462 quadrillion) and in 2020 (608 quadrillion). Then subtract the consumption in 2005 from the consumption in 2020. SOLUTION

608 — 462 146 The difference is 146 quadrillion Btu.

22

Module 1 ¢ Whole Numbers |

Check your understanding 9

| Use Figure 2 to find the difference between the world energy consumption projected for

|2015 and that projected for 2025. |

| SOLUTION

See page S-4.

89 quadrillion Btu

Objective 1.2C Practice

1. You eat an apple and one cup of cornflakes with one tablespoon of sugar and one cup of milk for breakfast. Find the total number of calories consumed if one apple contains 80 calories, one cup of cornflakes has 95 calories, one tablespoon of sugar has 45 calories, and one cup of milk has 150 calories. 370 calories 2. The odometer on your car read 58,376 at this time last year. It now reads 77,912. Estimate the number of miles your car has been driven during the past year. 20,000 mi 3. The graph at the right S : 500,000 Ss©

shows the projected :

:

Z S

sales of electric cars in

iS)

2015

a

; the United States from to 2020.

Between

5

2

C | z

200,000

increase the most?

3

100,000

Between 2016 and 2017

ice

~t

~

Ss

»

wy

;

: '

:

;

5

¥

i '

co)

|

/ 2015

OI

ee

oe S -===

oh

SE A eg

x

300,000

which two years shown A : is the number of electric cars sold projected to

SEER

s

9

400,000

2016

2017

a i

i

2018

2019

2020

Projected U.S. Electric Car Sales Source: An Analysis of Battery Electric Vehicle Production Projections by John Shamus Cunningham

SECTION

Multiplication and Division ORE inade lbs eta of Whole Numbers Objective 1.3A

Multiply whole numbers Multiplication is used to find the total number of objects in several groups when each group contains the same number of objects.

Sebastian purchased 8 six-packs of soda for a party. The total number of cans of soda he purchased can be found by adding 6 eight times. Sebastian purchased 48 cans of soda.

6

x

8

48

Multiplicand

x

Multiplier

Product

Section 1.3 © Multiplication and Division of Whole Numbers

23

The multiplicand is the number of objects in each group (6 cans in each six-pack); the multiplier is the number of groups (8 six-packs); the product is the total number of objects (48 cans). Frequently we will discuss the factors of a product. A factor is one of the numbers that are multiplied to obtain a product. 6 and 8 are factors of 48. The times sign “X” is only one symbol that is used to indicate multiplication. Each of the expressions that follow represents multiplication.

7X8

ea:

7(8)

(7)(8)

(7)8

The phrases below are used to indicate the operation of multiplication. An example is shown at the right of each phrase.

8 times 4

the product of

the product of 9 and 5

multiplied by

7 multiplied by 3 twice 6

Focus on finding the product of two numbers Find the product of 735 and 9. SOLUTION 34 735

°9x5=45 Write the 5 in the ones column.

re

Carry

the 4 to the tens column.

6615

9X

3=27,27+4

9X7=

63,63

+3

31 66

Check your understanding 1 Multiply: 648 < 7

SOLUTION

;i

i |i ia

See page S-4.

4536

Find the product of 47 and 23. Multiply by the ones digit.

Multiply by the tens digit.

47 nS

47 x23

141 (= 47 X 3)

EES ESP RR OEE

141

940 (= 47 X 20)

Add.

47 K 2B 141

940 1081

Writing the 0 keeps the columns aligned correctly.

The place-value chart on the right above illustrates the placement of the products.

24

Module 1 * Whole Numbers

| Focus on finding the product of larger numbers

Find 829 multiplied by 603. | SOLUTION

x

829

° 3 X 829 = 2487

603

* Write a zero in the tens

Pee

ee

column for 0 X 829

= 6X 829 = 4974

499,887

| Check your understanding 2

| Multiply: 756 x 305 | SOLUTION

See page S-4.

230,580

Apply the Concept Figure 1 shows the average weekly earnings of full-time workers in the United States. Using these figures, calculate the average earnings of a female full-time worker, age 27, for

working for 4 weeks. SOLUTION

asec stniaonend |7

Multiply the number of weeks (4) times the amount earned for one week ($649).

Weekly dollars) (in Earnings eeRce gnntemnnancaatebrannepte SCO

4(649) = 2596 The average earnings of a 27-yearold, female, full-time worker for working for 4 weeks are $2596.

[|

Source: Bureau of Labor Statistics

|SOLUTION

|510)

92° 7

x*y*zZ

* Replace each variable by its value.

|

=

100-7

¢ Multiply the first two numbers.

|

=

700

¢ Multiply the product by the next number.

| Check your understanding 3 Evaluate 5st for s =

SOLUTION

12 andt = 7.

See page S-4.

420

Years

Figure 1 Average Weekly Earnings of FullTime Workers

Evaluate xyz for x = 50, y = 2, and z = 7.

* xyz means

Tee.

Years

Years

Focus on evaluating a variable expression

|XYZ

Bae

Section 1.3 ¢ Multiplication and Division of Whole Numbers

Objective 1.3A Practice

1. 2. 3. 4. 5.

Multiply: 693 X 91 63,063 Multiply: 985-408 401,880 Find the product of 2, 19, and 34. 1292 What is 376 multiplied by 402? 151.152 Evaluate xyz when x = 5, y = 12, andz = 30.

1800

Solutions on pp. S-11-S-12.

Objective 1.3B

Divide whole numbers Division is used to separate objects into equal groups.

Apply the Concept Four friends want to share equally in the cost of a $24 birthday present for their friend Bianca. From the diagram below, each friend’s share of the cost is $6. Cost of the present

$24

Ne

ha

)

=}

4

se

lh

Gina’s share

Jason's share

Michelle’s share

Isaiah’s share

$6

$6

$6

$6

The solution of this division problem is written as follows: Each friend’s share 6

4)24 Ss

of the present Dividend

Note that the quotient multiplied by the divisor equals the dividend. 6

4)24

6

“

4

because | Quotient

=

Divisor

24

Dividend

6

9)54

because

6

x


(Number in each box)

3) 14—————__ Dividend al: (Total number of muffins)

(Number of boxes)

2 —_ | 12

||

-12

0

02

|

=O) ?

Check: |

* Think 4)2. Place 0 in the quotient.

* Multiply0 x 4. « Subtract:

2 —

0 =

2.

(630 X 4) +2 = 252 022

| Check your understanding 6 | Divide 6)5225 and check.

| SOLUTION

See page S-5.

870 15

When the divisor has more than one digit, estimate at each step by using the first digit of the divisor. If that product is too large, lower the guess by | and try again.

Section 1.3 ¢ Multiplication and Division of Whole Numbers

aa

Divide 34)1598 and check.

; 34) 1598

5 * Think 3)15

5

iy

i

4 34) 1598

— 136

* Multiply 5 x 34

| >) %£=170 fl f

* Multiply 4 x 34 ¢ Subtract 159 —

238

170 is too large. Lower the é guess by | and try again.

Bring down

136 = 23

the 8

|

47 y

29

Check:

34) 1598 136

a ence

* Think

238

¢ Multiply 7 x

—238

i

3)23

141

34

1598

¢ Subtract

0

The phrases below are used to indicate the operation of division. An example is shown at the right of each phrase.

the quotient of

the quotient of 9 and 3

divided by

6 divided by 2

Focus on dividing larger whole numbers Find 7077 divided by 34 and check. SOLUTION 208 r5

34) 7077 —68

ae27

_o

¢ Think 34)27. Place 0 in the quotient

etl) Dalal

¢ Multiply

0 X 34

¢ Subtract: 27 — 0 = 27. Bring down

Se

5 Check: (208 X 34) +5 = 7072) +5= 7077

Check your understanding 7 Divide 4578 + 42 and check. SOLUTION

See page S-5.

109

the 7

(cea

30

Module 1 ¢ Whole Numbers

_ Focus on dividing larger whole numbers Find the quotient of 21,312 and 56, and check. SOLUTION

380 r32 | 56) Des,

4 * Think

—168

5)21.4 X 56 is too large

Try 3

451 —448 32

aw 32 | Check: |

(380 X 56) + 32 = 2A 280)9 vate 32 21312.

|Check your understanding 8 |

| Divide 18,359 + 39 and check.

| SOLUTION

See page S-5.

470 129

|Focus on evaluating a variable expression |

| Evaluate

1

when x = 23.

| | SOLUTION

| 1081 |

x

1081 a

* Replace x by 23

23) 1081

+ Divide 1081 by 23

23

_ Check your understanding 9 | Evaluate w when x = 2173 and y = 53. Y

| SOLUTION

See page S-5.

4!

Objective 1.3B Practice

> Divides 2763)-3.9 307) . Divide: 7408 + 37 20018 . What is the quotient of 8172 and 9? 908 . Evaluate x + y when x = 39,200 and y = 4. x + y when x = Oandy = 23. 0 WN nh = . Evaluate Solutions on pp. S-12—S-13.

9800

Section 1.3 ¢ Multiplication and Division of Whole Numbers

Objective 1.3C

31

Solve application problems _ Focus on finding a monthly salary | A state park forester receives a salary of $1050 each week. How much does the forester earn in 4 weeks? | STRATEGY

| To find the forester’s earnings for 4 weeks, multiply the weekly salary (1050) by the | number of weeks (4).

| SOLUTION | xX

1050 4 4200

The forester earns $4200 in 4 weeks.

| Check your understanding 10 | An elephant will eat approximately 150 Ib of food each day. How many pounds of food | will an elephant eat in a 365-day year?

| SOLUTION

See pages S-5-S-6.

54,750 Ib

| Focus on finding a weekly salary with overtime | A pharmacist’s assistant earns $640 for working a 40-hour week. This week the assistant also worked 7 h of overtime at $26 an hour. Find the assistant’s total pay for the week. STRATEGY

To find the assistant’s total pay for the week: ¢ Find the overtime pay by multiplying the hours of overtime (7) by the overtime rate of

pay (26). ¢ Add the weekly salary (640) to the overtime pay.

| SOLUTION | X

26

640

i

ae ley

182

overtime pay

822

The assistant earned $822 this week.

Check your understanding 11 The buyer for Ross Department Store can buy 80 men’s suits for $7600. Each sports jacket will cost the store $62. The manager orders 80 men’s suits and 25 sports jackets. What is the total cost of the order? SOLUTION

See page S-6.

$9150

Focus on solving an application Ngan Hui, a freight supervisor, shipped 35,640 bushels of wheat in 9 railroad cars. Find the amount of wheat shipped in each car.

32

Module 1 * Whole Numbers STRATEGY To find the amount of wheat shipped in each car, divide the number of bushels (35,640) by the number of cars (9).

_ SOLUTION

3.960 9) 35,640

oer

86 —8] 54 mera. 0 Each car carried 3960 bushels of wheat.

Check your understanding 12 Suppose a Michelin retail outlet can store 270 tires on 15 shelves. How many tires can be stored on each shelf?

| SOLUTION

See page S-6.

18 tires

_ Focus on finding a monthly payment The used car you are buying costs $11,216. A down payment of $2000 is required. The remaining balance is paid in 48 equal monthly payments. What is the monthly payment?

| STRATEGY _ To find the monthly payment: | ¢ Find the remaining balance by subtracting the down payment (2000) from the total cost ) of the-car (11,216): ¢ Divide the remaining balance by the number of equal monthly payments (48).

- SOLUTION |

11,216

|

9,216

192

| — 2,000

48) 9216 * Remaining balance

macs

| |

44] — 432

|

96

=O

|

0 The monthly payment is $192.

Check your understanding 13 A soft-drink manufacturer produces 12,600 cans of soft drink each hour. Cans are packed 24 to a case. How many cases of soft drink are produced in 8 h? SOLUTION

See pages S-6-S-7.

4200 cases

Section 1.4 ¢ Exponential Notation and the Order of Operations Agreement

33

Objective 1.3C Practice

1. A computer analyst doing consulting work received $8064 for working 168 h ona project. Find the hourly rate the consultant charged. $48/h Use the table at the right for Exercises 2 and 3. 2. The owner of this company wants to provide the electrical installation for a new house. On the basis of the architectural plans, it is estimated

Electrician

$34

Plumber

$30

that the installation will require Clerk $16 three electricians, each working 50 h, to complete the job. What is the Bookkeeper $20 estimated cost for the electricians’ labor? $5100 Hourly Wages at a Small Construction 3. The owner of this company esti-mates Company that a kitchen remodel will require one electrician working 30 h and one plumber working 33 h. This project also requires 3 h of clerical work and 4 h of bookkeeping. What is the total cost for these four components of the remodel? $2138 Solutions on p. S-13.

SECTION

1.4 Objective 1.4A

Exponential Notation and the Order | of Operations Agreement Simplify expressions that contain exponents Repeated multiplication of the same factor can be written in two ways:

4:4-4-4-4

or 4°*——exponent at

base

The expression 4° is in exponential form. The exponent, 5, indicates how many times the base, 4, occurs as a factor in the multiplication.

It is important to be able to read numbers written in exponential form.

2=2!'

Read “two to the first power’ or just “two.” Usually the 1 is not written.

2°2=2?

2°2-2=23 Dei Dis

Da) 2

DODO wD) = 2°

Read “two squared” or “two to the second power.”

Read “two cubed” or “two to the third power.” Read “two to the fourth power.” Read “two to the fifth power.”

34

Module 1 * Whole Numbers

Focus on writing an expression in exponential form Write 7: 7+ 7-4-4 in exponential form.

| SOLUTION

LF.

7

4

= se

Check your understanding 1 | Write

2-2-2-3-3-3-3%in exponential form.

| SOLUTION

See page S-7.

PEEK

Variable expressions can contain exponents.

pS x

ats

x to the first power is usually written simply as x.

p05?

x° means x times x.

Phe JP OTROS

ieee

aie

Om

x° means x occurs as a factor 3 times.

nis

x? means x occurs as a factor 4 times.

Each place value in the place-value chart can be expressed as a power of 10. Ten =

10

=

10

= 10}

Hundred=

100

=

10-10

= 10°

Thousand =

1000

=

10-10-10

= 10°

Ten-thousand =

10,000

=

Hundred-thousand =

100,000

=

Million = 1,000,000

LOR OES OG

O

10-10-10-10-10

= 10-10-10: 10-10-10

= [0* =

10°

= 10°

Note that the exponent on 10 when the number is written in exponential form is the same as the number of zeros when the number is written in standard form. For example, 10° = 100,000; the exponent on 10 is 5, and the number 100,000 has 5 zeros.

To evaluate a numerical expression containing exponents, write each factor as many times as indicated by the exponent, and then multiply.

5 S55 25 5-5

22 6)

=e

(2 292) (62 6)

Se S612a8

| Focus on evaluating an exponential expression | Evaluate 8°. |

|

SOLUTION

8 =8-8-8

=64-8 = S12

Check your understanding 2 | |

Evaluate 6%.

| SOLUTION

See page S-7.

1296

Section 1.4 ¢ Exponential Notation and the Order of Operations Agreement

35

Focus on evaluating a power of 10 Evaluate 10’.

| SOLUTION 107 = 10,000,000 | (The exponent on 10 is 7. There are 7 zeros in 10,000,000.)

Check your understanding 3 Evaluate 10°. SOLUTION

See page S-7.

100,000,000

| Focus on evaluating a variable expression Evaluate x’y* for x = 4 and y = 2. SOLUTION

| x*y?

2.3 means x*eetimes y’.) 3 (xy?

Mele Aad) (0.2 62) = 16.8 = 198 | Check your understanding 4 Evaluate x*y? for x = 1 and y = 3.

SOLUTION

See page S-7.

9

Objective 1.4A Practice

Ve Wie 2

8 3. 355.5. 5 in exponential form:

2. Write a-a:b-b-b-b inexponential form. 3. Evaluate 2°. 64

4, Evaluate 24-37,

2° 2°25"

a»)

144

5. Evaluate x’y for x =

3andy = 4.

36

Solutions on p. S-14.

Objective 1.4B

Use the Order of Operations Agreement to simplify expressions More than one operation may occur in a numerical expression. For example, the expression

4 + 3(5) includes two arithmetic operations, addition and multiplication. The operations could be performed in different orders.

If we multiply first and then add, we have:

4 + 3(5) A 15 19

If we add first and then multiply, we have:

4 + 3(5) 7(5) 35

To prevent there being more than one answer to the same problem, an Order of Operations Agreement is followed. By this agreement, 19 is the only correct answer.

36

Module 1 ¢ Whole Numbers

The Order of Operations Agreement Step 1

Do all operations inside parentheses.

Step 2

Simplify any numerical expressions containing exponents.

Step 3

Do multiplication and division as they occur from left to right.

Step 4

Do addition and subtraction as they occur from left to right.

EXAMPLE Simplify: 5(6 + 4) — 23 5(6 + 4) — 23 = 5(10) — 2?

Perform operations inside parentheses.

5(10) — 8

Simplify expressions with exponents.

2) Us)

Do multiplication and division from left to right.

= 42

Do addition and subtraction from left to right.

One or more of the above steps may not be needed to simplify an expression. In that case, proceed to the next step in the Order of Operations Agreement.

|Focus on using the Order of Operations Agreement

| Simplify: 18 + (6 + 3)-9 — 4 | SOLUTION ae =

[eee

QoG)

=

|

|

18 +

(6 ate 3) “9 =

al

Se

ORO)

e—all(

4?

|

=2:-9—

|

= 18 — 16

|

=9) yt

16

* Perform operations inside parentheses. « Simplify expressions with exponents. ¢ Multiply or divide from left to right.

¢ Subtract.

|Check your understanding 5

| Simplify: 4-(8 — 3) +5 —2 | SOLUTION

See page S-7.

2

| Focus on using the Order of Operations Agreement

Simplify: 20 + 24(8 — 5) + 2? SOLUTION

| 20 + 24(8 F 5) +2?=20+

24(3) +2?

¢ Perform operations inside parentheses.

=

20 + 24(3) +4

¢ Simplify expressions with exponents.

=

20+

¢ Multiply or divide from left to right.

72-4

= 20 + 18

es Bias

* Add

Check your understanding 6

Simplify: 16 + 3(6 — 1)? + 5 SOLUTION

See page S-7.

3]

Section 1.4 * Exponential Notation and the Order of Operations Agreement

' Focus on evaluating a variable expression

Evaluate (a — b)? + 3c for a = 6, b = 4, andc = 1. SOLUTION (a ei b) + 3c =

(6 = 4)? IP 3(1)

=

3(1)

=4+

* Replace a with 6, b with 4, and c with 1. * Perform operations inside parentheses.

3(1)

* Simplify expressions with exponents.

=4+3

* Multiply or divide from left to right.

=7

* Add

Check your understanding 7

Evaluate (a — b)? + 5c fora = 7, b = 2, andc = 4. | SOLUTION

See page S-7.

45

Objective 1.4B Practice

. Simplify: 9 + (7+5)+6

. Simplify: 6(7) + 47-3? sounplifvel

Teoh

11

186 2)

4 S 14

. Evaluatex* + y + xforx = 2andy=8. wWn ah = . Evaluate°° +

Solutions on p. S-14.

8

4(¢ — y) + 2 forx = 8,y =6,andz=2.

66

37

ee

ayy A

Solutions to Module 1 S-1

SOLUTIONS TO MODULE

1

Solutions to Check Your Understanding Section 1.1

Check your understanding 1 he ORR

ZS eA

Ot

Or)

8

90a

1.12

Check your understanding 2 ey

ISS

Oppo

SSS

3 a.4 aed

aa

Sa

SOO

See

Ot

1D

7 is 4 units to the left of 11.

Check your understanding 3 a.

47 > 19

th,

Ay Se

Check your understanding 4 033, 17, 52,685.94

Check your understanding 5 forty-six million thirty-two thousand seven hundred fifteen

Check your understanding 6 920,008

Check your understanding 7 70,000 + 6000 + 200 + 40 + 5

Check your understanding 8 ae

Given place value

529,374

Poesy

529,374 rounded to the nearest ten-thousand is 530,000.

Check your understanding 9 mes

Given place value

7985 [Fein

Sie)

7985 rounded to the nearest hundred is 8000.

Check your understanding 10 STRATEGY

To find the sport named by the greatest number of people, find the largest number given in the circle graph. SOLUTION

The largest number given in the graph is 80. The sport named by the greatest number of people was football.

S-2

Solutions to Module 1

Check your understanding 11 STRATEGY

To find the shorter distance, compare the numbers 347 and 387. SOLUTION

347 < 387 The shorter distance is between Los Angeles and San Jose.

Check your understanding 12

STRATEGY To determine which state has fewer sanctioned league bowlers, compare the numbers 239,951 and 239,010. SOLUTION

239,010 < 239,951 Ohio has fewer sanctioned league bowlers.

Check your understanding 13

STRATEGY To find the land area to the nearest thousand square miles, round 3,851,809 to the nearest thousand. SOLUTION 3,851,809 rounded to the nearest thousand is 3,852,000. To the nearest thousand, the land area of Canada is 3,852,000 mi’.

Section 1.2

Check your understanding 1 11

347

2 743=40

+ 12,453

Write the 0 in the ones

Fp ae 12,800

column. Carry the | to the tens column.

1+4+5=10 Write the 0 in the tens

column.

Carry the | to the

hundreds column 14+3+4=8

347 increased by 12,453 is 12,800.

Check your understanding 2 Wl

PAR

392 4,079 89,035 + 4,992 98,498

Check your understanding 3 Neate AY, alae

1692 + 4783 + 5046 124

1692 4783 + 5046 ES 211

Solutions to Module 1 S-3

Check your understanding 4 2147

ZBABY = SOS

11

Check:

865 + 261 ale

481 =

P61 6

Check your understanding 5 15

Check:

5

14,485

+ 40,077

62 8 5

54,562

cle

Check your understanding 6 75

3)

10

o-@) 3

¢ There are two zeros in the minuend

maces to eg

Borrow

| thousand

from the thousands column and write 10 in the hundreds column 9

3

Ww

10

64,9

* Borrow

9 3

AOC

6

2

5

column and write

10 in the tens column

13

9

3

WW

9

13 6A, GO BB == Sy ah SY io) O20 67

Check:

| hundred

from the hundreds

© Borrow

| ten from the tens column and add 10 to the 3 in the ones column

54,936

+ 9,067 64,003

Check your understanding 7 4007 — x 4007 — 389

© Replace x by 389

Check your understanding 8 125,500,000 68,200,000 18,200,000 + 2,000,000 213,900,000

A total of 213,900,000 cases of eggs were produced during the year.

S-4

Solutions to Module 1

Check your understanding 9 2025: 652 quadrillion Btu 2015: 563 quadrillion Btu

652 =968 89 The difference is 89 quadrillion Btu.

Section 1.3

Check your understanding 1 Si

°7X8&=56

648 Tl

x

Write the 6 in the ones column Carry the 5 to the tens column.

1X4

4536

= PROBES —8

7X 6=42,42+3=45

Check your understanding 2 150! x

Fe 5

305

756:

3780

Write a zero in the tens column for 0 X 756

3780

3 X 756 = 2268

22680

230,580

Check your understanding 3 Sst So 227

¢ Replace s with 12 and f with 7.

=

60:7

= Multiply the first two numbers.

=

420

* Multiply the product by the next number

Check your understanding 4 453 9) 4077 =36

47 —45

7

=27 0

Check:

453 X 9 = 4077

Check your understanding 5 705

9)6345

eee)

04

—

0

* Think 9)4. Place 0 in the quotient.

0

¢ Multiply

a

¢ Bring down the 5.

—45

oO

Check: 705 X 9 = 6345

0 X 9. Subtract.

Solutions to Module 1

S-5

Check your understanding 6 870 r5 6) 5225 —48

29, =e)

0

05

¢ Think 6)5. Place 0 in the quotient

—()

* Multiply 0 x 6. Subtract

errs Check:

(870 X 6) + 5 = 5220 + 5 = 5225

Check your understanding 7 109 42) 4578

F

—42

* Think 42)37. Place

37

0 in the quotient

¢ Multiply 0 x 42. Subtract

378 —378 SPzU Check: (109 X 42) = 4578 Check your understanding 8 470 129

39) 18,359 1h DGS

Check:

-

* Think 3)18. 6 X 39 is too large. Try 5.5 large. Try 4

X 39 is too

(470 X 39) + 29 = 18,330 + 29 = 18,359

Check your understanding 9 xX

y PALI 53

Sa

* Replace x by 2173 and y by 53

41

53) 2173

* Divide 2173 by 53.

—212

PE =i)

0

When x = 2173 andy = 53,5 = 41. Check your understanding 10

STRATEGY To find the amount of food eaten in one year, multiply the amount eaten each day (150) by the number of days in one year (365).

S-6

Solutions to Module 1 SOLUTION

150 x 365 750 900 450

54,750 The elephant will eat 54,750 Ib of food in one year.

Check your understanding 11 STRATEGY

To find the total cost of the order: ¢ Find the cost of the sports jackets by multiplying the number of jackets (25) by the cost for each jacket (62). ¢ Add the product to the cost for the 80 suits (7600).

SOLUTION

62 eS

7600 ae lest)

310 124

9150

1550

© Cost for jackets

The total cost of the order is $9150.

Check your understanding 12 STRATEGY

To find the number of tires that can be stored on each shelf, divide the number of tires (270) by the number of shelves (15). SOLUTION

18

15) 270 = i115)

7120 = 120

eG) Each shelf can store 18 tires.

Check your understanding 13 STRATEGY

To find the number of cases produced in 8 h: ¢ Find the number of cases produced in 1 h by dividing the number of cans produced (12,600) by the number of cans to a case (24).

¢ Multiply the number of cases produced in | h by 8.

Solutions to Module 1

SOLUTION

525 24) 12,600 =[20 60 —48 THD

100 0 525 aan 4200 In 8 h, 4200 cases are produced.

Section 1.4

Check your understanding 1 2° BP

BiB oBiBesy2s Bt

Check your understanding 2 6' = 6°6°6°6 = 36:°6:'6 = 216-6 = 1296

Check your understanding 3 10° = 100,000,000

Check your understanding 4 x4

Sea (ie bet) 3.75) = 1-9 y)

Check your understanding 5

4-823) 45> 2 4egire5 = 2

¢ Perform operations inside parentheses.

= Aiea)

¢ Multiply or divide from left to right.

=2

¢ Subtract.

= 42

Check your understanding 6 16 + 3(6 — 1)? +5 = 16 + 3(5)? +5 = 16 + 3(25) +5

=16+75+5 16 + 15 rai

* Perform operations inside parentheses

¢ Simplify expressions with exponents. * Multiply or divide from left to right.

¢ Add.

Check your understanding 7 (a - b)? af Be

* Replace a with 7, b with 2, and c with 4.

(7 = De “bE 5(4) = 5? + 5(4) =

25+

5(4)

* Perform operations inside parentheses. ¢ Simplify expressions with exponents.

= 25 + 20

* Multiply or divide from left to right.

= 45

« Add.

S-7

S-8

Solutions to Module 1

Solutions to Objective Practice Exercises Objective 1.1A

1.

Pee ap (Qj WI Pe ey Ceo)

fet

2/33). OF I an

ay

5 is 3 units to the right of 2.

0 is 7 units to the left of 7.

3213194

4. 3827 < 6915 5. 400, 404, 440, 444, 4000

Objective 1.1B

1. Two thousand eight hundred one 2. Three hundred fifty-six thousand nine hundred forty-three

3. 3456 4. 7,024,709 5. 400,000 + 3000 + 700 + 5

Objective 1.1C

1.

ers Given place value 1638 Se 5 1638 rounded to the nearest hundred is 1600.

2.

ie Given place value 84,608 QS) 84,608 rounded to the nearest thousand is 85,000.

3:

i

Given place value

17,639 BIS5 17,639 rounded to the nearest hundred is 17,600.

4. aan

Given place value

5326 es

J=S

5326 rounded to the nearest thousand is 5000.

Sh

a

Given place value

746,898 QS 5 746,898 rounded to the nearest ten-thousand is 750,000.

Solutions to Module 1 S-9

Objective 1.1D

1. STRATEGY a. To find the annual per capita turkey consumption in the United States, use the pictograph to count each symbol to the right of “U.S.” as 2 Ib. b. To determine in which country the annual per capita turkey consumption is highest, use the pictograph to determine which country has the most symbols to its right. SOLUTION

a. The annual per capita turkey consumption in the United States is 18 Ib. b. The country with the highest annual per capita turkey consumption is Israel. STRATEGY To find the food that contains more calories, compare the numbers

190 and 114.

SOLUTION

190 > 114 Two tablespoons of peanut butter contain more calories. STRATEGY

a. To determine which was greater, the number of crashes in July or the number of crashes in October, use the Number of Crashes category of the bar graph to read which number above the corresponding bar is greater. b. To determine which was fewer, the number of vehicles involved in crashes in July or the number involved in December, use the Number of Vehicles category of the bar graph to read which number above the corresponding bar is smaller. SOLUTION

a. The number of crashes in July was 3459, and the number of crashes in October was 3344. The greater number of crashes was in July. b. The number of vehicles involved in crashes in July was 5210, and the number involved in December was 5242. There were fewer vehicles involved in crashes in July. STRATEGY

a. To determine during which school year enrollment was lowest, use the line graph to find the school year corresponding to the lowest point on the line.

b. To determine if enrollment increased or decreased between 1990 and 2000, read the graph at 1990 and at 2000. SOLUTION

a. Student enrollment was lowest during the 1980 school year. b. Enrollment was greater in 2000 than in 1990. Enrollment increased between 1990 and 2000. STRATEGY

a. To determine the most often mentioned complaint, find the complaint with the largest number of responses. b. To determine the least often mentioned complaint, find the complaint with the smallest number of responses. SOLUTION

a. The complaint with the largest number of responses is people talking. The most often mentioned complaint was people talking.

b. The complaint with the smallest number of responses is uncomfortable seats. The least often mentioned complaint was uncomfortable seats.

S-10

Solutions to Module 1 6. STRATEGY

To find the land area of Alaska to the nearest thousand square miles, round 570,833 to the nearest thousand. SOLUTION

570,833 rounded to the nearest thousand is 571,000. To the nearest thousand square miles, the land area of Alaska is 571,000 mi’. i ] Objective 1.2A

1.

Late 36.925

ODO 102,317

2.

3

111 20,958 3,218 +p 42 24,218

7988 +5678 13,666

‘“

111 45,872 + 7,894 53,766

Bh hear ay

38,229 + 51,671 11 38,229 +51,671 89,900 I i Objective 1.2B

1.

1

41 16 3576

= S3/ 3139 Pa

61076 yo12 7 O,FO2 — 4,239 66,463

3.

993 AA

9913

556

4, *

5179915 63,005 =D IST 38,208

Solutions to Module 1

S-11

5b, ge ys

1605 — 839 015915 1603S = O09 766

Objective 1.2C

1. STRATEGY To find the number of calories, add the number of calories in one apple (80), one cup of cornflakes (95), one tablespoon of sugar (45), and one cup of milk (150). SOLUTION

80 + 95 + 45 + 150 = 370 The breakfast contained 370 calories. STRATEGY

To estimate the number of miles driven: ¢ Round the two odometer readings. ¢ Subtract the rounded reading for the beginning of the year from the rounded reading at the

end of the year. SOLUTION

77,912 —

80,000

58,376 —

— 60,000 20,000

The car was driven approximately 20,000 mi during the year. .

STRATEGY

To determine between which two years electric car sales are projected to increase the most, find the difference between projected sales for each pair of consecutive years. SOLUTION

Between 2015 and 2016:

230,000 — 188,000 = 42,000 Between 2016 and 2017:

312,000 — 230,000 = 82,000 Between 2017 and 2018:

359,000 — 312,000 = 47,000 Between 2018 and 2019:

406,000 — 359,000 = 47,000 Between 2019 and 2020:

414,000 — 406,000 = 8000 Car sales are projected to increase the most between 2016 and 2017. The increase is projected to be 82,000 cars.

Objective 1.3A

1.

S-12

Solutions to Module 1

985 408

x

7 880 394 00 401,880 .

2X

19 = 38

38

x 34 152 4 1292

376 x 402 Ta2 1504

151,152 xyz

12-30

Objective 1.3B

it

2.

Solutions to Module 1

S-13

|e

39,200

4

9,800 4) 39,200 —36 32 oan 00 -0 00 -0

Objective 1.3C

1. STRATEGY To find the hourly rate, divide the total earnings (8064) by the number of hours worked (168). SOLUTION

168 The consultant received $48 per hour. 2.

STRATEGY

To find the estimated cost for the electricians’ labor, multiply the number of electricians (3), the number of hours each will work (50), and the wage per hour for each

($34). SOLUTION

3-50-34 = 5100 The estimated cost for the electricians’ labor is $5100. 3.

STRATEGY

To find the total cost, add the costs for the electrician, the plumber, clerical work, and

bookkeeping. To calculate the costs for each, multiply the number of hours for each type of work by the wage per hour. SOLUTION

Electrician: 30 - 34 = $1020 Plumber: 33 - 30 = $990

Clerical: 3 - 16 = $48 Bookkeeping: 4 - 20 = $80

The total cost is 1020 + 990 + 48 + 80 = $2138.

S-14

Solutions to Module 1

Objective 1.4A

1.

DMI

DORSEY Reon ae Oe ORY

2. a:a:b:b:b-b=a’b'

2 =2-2-2-2-2-2=64 a 25°37=

(2222-9) (3 23) = 169344

xy SecA == (Br3) a4 = 9-4 = 36 Objective 1.4B

9+(77+5)+6=9+12+6

= 9p) = {1 6(7) + 4°- 3° = 6(7) + 16-9

= 42+ 16-9 = 42 + 144 = 186 17+1-8:-2+4=17+1-16+4 =17+1-4 =18-4 =14 eae) oea

Pata ce

Aime cath i 9) =4+4

r+

4x -

ye 2

8? + 4(8 — 6) + 27 = 8° + 4-2+2? =64+4-2+4 =64+8+4 =644+2 = 66

MODULE

Integers SECTION 2.1

Introduction to Integers

Objective 2.1A

Use inequality symbols with integers

Objective 2.1B

Simplify expressions with absolute value

SECTION 2.2 Objective 2.2A

Addition and Subtraction of Integers Add integers

Objective 2.2B

Subtract integers

Objective 2.2C

Solve application problems

SECTION 2.3 Objective 2.3A

Multiplication and Division of Integers Multiply integers

Objective 2.3B

Divide integers

Objective 2.3C

Solve application problems

SECTION 2.4

Exponents and the Order of Operations Agreement

Objective 2.4A

Simplify expressions containing exponents

Objective 2.4B

Use the Order of Operations Agreement to simplify expressions

2

Module2 « Integers

SECTION

2.1 Objective 2.1A

Introduction to Integers Use inequality symbols with integers Mathematicians place objects with similar properties in groups called sets. A setis a collection of objects. The objects in a set are called the elements of the set.

The roster method of writing a set encloses a list of the elements in braces. Thus the set of sections within an orchestra is written {brass, percussion, string, woodwind}. When the elements of a set are listed, each element is listed only once. For instance, if the list of numbers 1, 2, 3, 2, 3 were placed in a set, the set would be {1, 2, 3}.

The symbol € means “is an element of.” 2 € B is read “2 is an element of set B.” Given C = {3, 5, 9}, then 3 € C,5 OnsenGe

€ C, and9 € C.7 € Cis read “7 is not an element

The numbers that we use to count objects, such as the students in a classroom or the horses on aranch, are the natural numbers.

Natural numbers = {1, 2. 3, 4, 5, 6, 7, 8, 9, 10, ...} The three dots mean that the list of natural numbers continues on and on, and that there is no largest natural number. The natural numbers alone do not provide all the numbers that are useful in applications. For instance, a meteorologist also needs the number zero and numbers below zero.

Integers = {.1.,

40,

>4).— 31-2

ee

eee

Each integer can be shown on a number line. The integers to the left of zero on the number line are called negative integers. The integers to the right of zero are called positive integers, or natural numbers. Zero is neither a positive nor a negative integer. Integers

EX SS

Sa we

5

4

3

ie

2

1

0

| \

Negative

integers

SSS 1

2

SS 3

G

4

5

Positive

Zero

integers

The graph of an integer is shown by placing a heavy dot on the number line directly above the number. The graphs of —3 and 4 are shown on the number line below.

Consider the following sentences.

The quarterback threw the football and the receiver caught it. A student purchased a computer and used if to write history papers.

Section 2.1 © Introduction to Integers

3

In the first sentence, it is used to mean the football; in the second sentence, it means the

computer. In language, the word it can stand for many different objects. Similarly, in mathematics, a letter of the alphabet can be used to stand for a number. Such a letter is called a variable. Variables are used in the following definition of inequality symbols.

Inequality Symbols If a and b are two numbers and a is to the left of b on the number line, then ais less than b. This is written a < b.

If a and b are two numbers and a is to the right of b on the number line, then

ais greater than b. This is written a > b. EXAMPLES 1.

4

—2 is to the left of 1 on the

, between the numbers below.

—14

16

14 < 16

2. Place the correct symbol, < or >, between the numbers below.

OLS 10 31 3. Use the roster method to write the set of positive integers less than 4. 4 = {1.2.3} 4. Given C = {—33, —24, —10, 0}, which elements of set C are less than —10?

— 33, —24

Solutions on pp. S-3—S-4.

Objective 2.1B

Simplify expressions with absolute value Two numbers that are the same distance from zero on the number line but are on opposite sides of zero are opposite numbers, or opposites. The opposite of a number is also called its additive inverse.

The opposite of 5 is —5.

The opposite of —5 is 5. The negative sign can be read “the opposite of.” =—(2) = 2

The opposite of 2 is —2.

=(—2)=2

The opposite of —2 is 2.

Focus on finding an additive inverse Given rss

A = {—12,0, 4}, find the additive inverse of each element of set A.

SOLUTION

—(—12) = 12 —0=0

* Zero is neither positive nor negative.

—(4) = -4 Check your understanding 3 Given B = {—11, 0, 8}, find the additive inverse of each element of set B. SOLUTION

See page S-1.

11,0,

—8

Section 2.1 © Introduction to Integers

5

The absolute value of a number is its distance from zero on the number line. Therefore, the absolute value of a number is a positive number or zero. The symbol for absolute value is two vertical bars, ||. The distance from 0 to 3 is 4. Therefore, the absolute value of 3 is 3.

j| Sat

itt

—5-4-3-2-1

|3| =3 The distance from 0 to —3 is 3. Therefore, the absolute value of —3 is 3.

a

ee

012345

|j}

|-3| =3

Absolute Value The absolute value of a positive number is the number itself. The absolute value of a negative number is the opposite of the number. The absolute value of zero is zero. EXAMPLES

1. |6| = 6

2 (sis

Using variables, the definition of absolute value is eee 0)

bel =

0,

x=0

=a)

Focus on evaluating an absolute value expression Evaluate |—4| and —|—10].

SOLUTION |—4| =4 =(—10) = —10 Check your understanding 4 Evaluate |—5| and —|—23]. SOLUTION

See page S-1.

55323

Objective 2.1B Practice . Find the additive inverse of 4. —4 . Find the additive inverse of —9. 9 . Find the additive inverse of —36. 36

. Evaluate —(—40). 40 . Evaluate |—74|. 74 —81 khwnds An = . Evaluate —|81|. Solutions on p. S-4.

6

Module2 @ Integers

SECTION

2.2 Objective 2.2A

J Addition and Subtraction of Integers Add integers A number can be represented anywhere along the number line by an arrow. A positive number is represented by an arrow pointing to the right, and a negative number is represented by an arrow pointing to the left. The size of the number is represented by the length of the arrow.

Gy

=)

8

i

AG

She teal

oie)

2

1

0

1

v

3°

4

5)

Oo

che

alto)

Addition is the process of finding the total of two numbers. The numbers being added are called addends. The total is called the sum. Addition of integers can be shown on the number line. To add integers, start at zero and draw, above the number line, an arrow representing the first number. At the tip of the first arrow, draw a second arrow representing the second number. The sum is below the tip of the second arrow.

44+2=6

4+ (2) = a

Sa ea

Syl

=4

Wf

4

+2

2

ee

2A fs A

5

—6

|

GOW

ie

afi

a

P=aja= p39) jl O it 2 g A

4+ (-2)=2 -4 |

+4 a

42

——s

SSS SS SS SS See 7-6-5 —-4-3-2-1 0 1 2 3 4

=2

we

SSS == A=f=2=l © 1 23 45 6 YF

The pattern for addition shown on the number lines above is summarized in the following rules for adding integers. Addition of Integers To add two numbers with the same sign, add the absolute values of the numbers. Then attach the sign of the addends. To add two numbers with different signs, find the absolute value of each number. Subtract the smaller of the absolute values from the larger. Then attach the sign of the number with the larger absolute value. EXAMPLES Adds

(—26)

VP se ( 26) =

21, INGE =D |ae 19| =

SP ee 19; |8| =8

19-8=11 —19

+38

—38

=

The signs are the same.

Add the absolute values of the numbers (12 + 26). Attach the sign of the addends.

The signs are different. Find the absolute value of each number.

Subtract the smaller absolute value from the larger. —{]

Attach the sign of the number with the larger absolute value.

Section 2.2 © Addition and Subtraction of Integers

Focus on adding integers

fPAdds 52 1 = 39) SOLUTION

—52 + (—39) =

—91

| Check your understanding 1 | Add:

100 + (—43)

| SOLUTION

Tips for Success >

See page S-1.

57

Focus on adding integers

One of the key instructional features

of this text is the Focus On/Check

Add: 37 + (—52) + (—14)

Your Understanding pairs. Each Focus On is completely worked.

SOLUTION

You are to solve the Check Your

37 + (—52) + (-14) = -15 + (-14)

Understanding problems. When you are ready, check your solution against the one given in the

Solutions section at the end of the dule.

pais

S65 |

4

| Check your understanding 2

Rad

s 1s 40.

_ SOLUTION

17 +1109)

See page S-1.

94

Apply the Concept Suppose you wake up in St. Paul, Minnesota, and the temperature is — 12°F. By noon, the temperature has risen 7°F. What is the temperature at noon? SOLUTION

To find the temperature at noon, add 7° to — 12°.

an ee

oe

The temperature at noon is —5°F.

( Find the sum of —23, 47, —18, and —10.

Recall that a sum is the answer to an addition problem. ; S73) sie A) Se (- 18) at (- 10) j = 94 (= 18) af (= 10)

* To add more than two numbers, add the first two numbers Then add the sum to the third number. Continue until all the numbers are added

= 6+ (-10) = —-4 All of the phrases below indicate addition.

added to

—6 added to 9

9+ (—6) = 3

more than

3 more than —8

Se}

the sum of

the sum of —2 and —8

increased by

—7 increased by 5

=i) st

the total of

the total of 4and —9

4+

plus

6 plus —10

6+ (—10) = —4

NO aera

2 + (-8) = k= 2

(—9) = —-5

-10

7

8

Module2

Integers

| Focus on translating an expression | Find 11 more than —23.

| SOLUTION = 23 oF

ie

| Check your understanding 3 |

| Find —8 increased by 7.

|SOLUTION

See page S-1.

—|

Objective 2.2A Practice

1. Add —6 + (—9).

15

Py POG! =i7/ ar WF, 0 30 Add —27 2 (42) (18)

87

4. Add —6 + (—8) + 14+ (—4). 5. What is 4 more than —8? 6. What is —8 added to —21?

~—4

4 ~—29

Solutions on p. S-4.

Objective 2.2B

Subtract integers Look at the expressions below. Note that each expression equals the same number.

8—3=5

8 minus 3is5.

8 + (—3) = 5

8plus the opposite of 3 is 5.

This example suggests the following.

Subtraction of Integers To subtract one number from another, add the opposite of the second number to the first number. EXAMPLES

1. Subtract: —21 — (—40) —21 — (—40) = —21 + 40 =

19

Rewrite the subtraction as addition of the opposite. Add.

5 mulomereie IS) = Sil 15 — 51 = 15 + (—5]) =

—36

Rewrite the subtraction as addition of the opposite. Add.

Section 2.2 ¢ Addition and Subtraction of Integers

Apply the Concept The table at the left shows the boiling point and the melting point, in degrees Celsius, of three chemical elements. Find the difference between the boiling point and the melting point of mercury. SOLUTION

To find the difference, subtract the melting point of mercury from the boiling point of mercury.

357 =| 39) = 307+ 39 = 396 The difference is 396°C.

Focus on subtracting integers Subtract:

—12 — (—21) — 15

SOLUTION =)

=

(—21) —15=-12+

21 + (-—15)

= 9+ (-—15) =

* Rewrite each subtraction as additionof the

-6

opposite. Then add

Check your understanding 4 Subtract:

—9 — (—12) - 17-4

SOLUTION

See page S-1.

—18

A difference is the answer to a subtraction problem. tFind the difference between —8 and 7. 8

l=

8+

( 7)

¢ Rewrite the subtraction as addition of the

Stee

opposite.

Then add

All of the following phrases indicate subtraction.

Take Note > Note the order in which numbers are subtracted when the phrase less than is used. If you have $10

—5 minus 11 a 3 less 5

and your friend has $6 less than you do, then your friend has $6

less than

—8 less than —2

ef

(3)

less than $10, or $10 — $6 = $4.

the difference between

the difference between —5 and 4

=

4 = —9

decreased by

—4 decreased by 9

subtract... from

subtract 8 from —3

Focus on translating an expression Find 9 less than —4. SOLUTION

-4-9=-4+ (-9) = -13

=6

9

10

Module2 « Integers |

Check your understanding 5

| Subtract |

| | Le

—12 from —11.

SOLUTION

See page S-1.

|

Objective 2.2B Practice

5 wllonere Ie = B12

. Subtract 6— (—12).

18

. Subtract -19 — (—19) — 18. —18 . What is 9 less than —12? —-21 . Find —21 decreased by 19. —40 WN ah Solutions on p. S-4.

Objective 2.2C

Solve application problems | Focus on solving an application The average temperature on Mercury’s sunlit side is 950°F. The average temperature | on Mercury’s dark side is —346°F. Find the difference between these two average

| temperatures. | STRATEGY

| To find the difference, subtract the average temperature on the dark side (—346) from the | average temperature on the sunlit side (950).

_ SOLUTION

950 — (—346)

950 + 346 = 1296

| The difference between the average temperatures is 1296°F.

| Check your understanding 6 _ The average daytime temperature on Mars is —17°F. The average nighttime temperature | on Mars is —130°F. Find the difference between these two average temperatures. | SOLUTION

See page S-2.

10358

Objective 2.2C Practice

1. The elevation, or height, of places on Earth is measured in relation to sea level, or the average level of the ocean’s surface. The table below shows height above sea level as a positive number and depth below sea level as a negative number. Use the table below to find the difference in elevation between Mt. Kilimanjaro and the Qattara Depression. 6028 m

Mt. Kilimanjaro

|=

Qattara Depression

Asia

Mt. Everest

Dead Sea

Europe

Mt. Elbrus

Caspian Sea

America

Mt. Aconcagua

Death Valley

Section 2.3 © Multiplication and Division of Integers

11

2. The elevation, or height, of places on Earth is measured in relation to sea level, or the average level of the ocean’s surface. The table below shows height above sea level as a positive number and depth below sea level as a negative number. Using the table below, for which continent shown is the difference between the highest and lowest elevations smallest? Purope

Africa ie Asia

|

Mt. Kilimanjaro

Qattara Depression

Mt. Everest

Dead Sea

Europe

Mt. Elbrus

5634

America

Mt. Aconcagua

—400

Caspian Sea Death Valley

—28 |

—86

i

3. The graph below shows the depths of Earth’s three deepest ocean trenches and the heights of its three tallest mountains. What is the difference between the depth of the Philippine Trench and the depth of the Mariana Trench? 980 m Qozir

9000

Kangchenjunga ban

8586

Mt. Everest

8611

Meters

Trench

Solutions on pp. S-4-S-S.

SECTION

2.3 Objective 2.3A

Multiplication and Division of Integers Multiply integers Several different symbols are used to indicate multiplication. The

3

numbers being multiplied are called factors; for instance, 3 and 2 are factors in each of the examples at the right. The result is called the product. Note that when parentheses are used and there is no operation symbol, the operation is multiplication.

Beli 6 6)(0) = 3)(2) = 6 3(2) = 6

Multiplication is repeated addition of the same number. The product 3 X 5 is shown on the number line below.

(3)2. = 6

5 5 eee SS

Gow

tt

shale Sei 7atg Voto

5

tia 1314 1s

D2 = 6

5 is added 3 times. ee a ahd aN

Si Gieke Osteo t=

-

15

12

Module2 « Integers

Now consider the product of a positive and a negative number. —5 is added 3 times.

3(— Sea

et eae)

=n

This example suggests that the product of a positive number and a negative number is negative. Here are a few more examples.

4(-7) = —28

—6:7 = —42

(—8)7 =

To find the product of two negative

These numbers

numbers, look at the pattern at the right. As —5 multiplies a sequence

LAE

These numbers

=

of decreasing integers, the products increase by 5.

—56

Y 4 oryee bye: 3 = 15

=) = 10) ACR

=
=1

The element 5 is greater than —1.

Check your understanding 3

Check your understanding 4

[=5} = 5 —|=23| = =23 Section 2.2

Check your understanding 1 100 + (—43) = 57

Check your understanding 2

—51 + 42 + 17 + (—102) = —9 + 17 + (—102) = 8 + (—102)

= —94

Check your understanding 3

-§+7=—-1

Check your understanding 4

9 — (-12) -17-4 0) Se Ei ete) eA) ll —14 + (—4)

er

= —18

Check your understanding 5

—11 — (-12)=-11+12=1

S-2

Solutions to Module 2

Check your understanding 6

STRATEGY To find the difference between the two average temperatures, subtract the smaller number

(—130) from the larger number (—17). SOLUTION

=),

(130)

ren

= ns

The difference is 113°F.

Section 2.3

Check your understanding 1 (=2)3(63)7 = 6(—8)9

= 48(7) = 336

Check your understanding 2 —9(34) = —306

Check your understanding 3

(—135) + (—9) = 15 Check your understanding 4 le)

4

= II

Check your understanding 5

Check your understanding 7 STRATEGY

To find the average daily low temperature: ¢ Add the seven temperature readings. ¢ Divide the sum by 7.

SOLUTION —6 + (~-7) + 0 + (—5) + (—8) + (-1) +

(-1) = -28

Iss = 7 = =A

The average daily low temperature was —4°C.

Check your understanding 8 STRATEGY

¢ Multiply each correct answer (23) by 5, multiply each incorrect answer (7) by —2, and multiply each question left blank (5) by 0. ¢ Add the results.

Solutions to Module 2 S-3 SOLUTION

There were 23 correct answers: 5(23) = 115 There were 7 incorrect answers: —2(7) = —14

There were 5 questions left blank: 0(5) = 0 115)+ (=14) +0=

101

The score for the student was 101.

Section 2.4

Check your understanding 1

a, 6 G2) = 3)G)G) -(=2)(—2)(—2) = 27(—8) = —216 b. The product of an odd number of negative factors is negative. Therefore,

(1)

=

ce. —2?-(—1)!- (-3)

=(2-2)- 1+ (=3) -(-3) — 2s | okey — eyes

Check your understanding 2

he 3

ne]

2 P

=7-

2[6 a

=f

2) 8 |:

=f

—

=

7 —

=

—12]1

14)

2[64] 128

¢ Perform operations inside grouping symbols

« Simplify

exponential expressions

¢ Multiply

or divide from left to right

* Subtract

Check your understanding 3

18 =5[8 — 2(2 —-5)] + 10 18 — 5[8 — 2(—3)] + 10

° Perform operations inside grouping symbols

= 18 — 5[8 + 6] + 10 = 18 — 5[14] + 10 =

18 —

70 +

10

* Multiply or divide from left to right

=18—7 —> a |

* Subtract

Check your understanding 4

36 + (8 — 5)? — (-3)*-2 =

36 +

(3)? =

=

36

~9-9-2

(—3)? -2

© Perform operations inside grouping symbols. * Multiply or divide from left to right

=4-9-2 =4-18 =-|4

* Subtract

Solutions to Objective Practice Exercises Objective 2.1A

1. —14
-31 ah 7 Rep

S-4

Solutions to Module 2

Oh

ops

=il0)

a. a NO} 0

0)

110)

10

The elements

—33 and —24 are less than —10.

Objective 2.1B

Objective 2.2A

6 + (-9) = =

-15

iat ean)

—27 + (—42) + (—18) = —69 + (—18) = —87 =6 (8) BYP

14

4)

id 4) 0 + (—4) = -4

5. -8+4=-4 6. —21 + (-8) = -29 Objective 2.2B

1. 16 — 8 = 16+

(-8) =8

2. 6— (-12) =6+ 12=18 32°19

(19)

1S

19019 18)

= 0 + (-18) = -18 4. =12-—9==12 + (9) = =21 5.

Objective 2.2C

=20

1921

E19) = —40

1. STRATEGY To find the difference, subtract the elevation of the Qattara Depression (— 133 m) from the elevation of Mt. Kilimanjaro (5895 m). SOLUTION

5895 — (—133) = 5895 + 133 = 6028 The difference in elevations is 6028 m. 2.

STRATEGY

To find which continent has the smallest difference in elevations, find the difference

for each continent and compare the differences.

Solutions to Module 2

S-5

SOLUTION

Africa:

5895 — (—133) = 5895 + 133 = 6028 m

Asia: 8850 — (—400) = 8850 + 400 = 9250 m Europe:

5634 — (—28) = 5634 + 28 = 5662 m

America:

6960 — (—86) = 6960 + 86 = 7046 m

The continent with the smallest difference between the highest and lowest elevation is Europe. STRATEGY To find the difference, subtract the depth of the Mariana Trench (—11,520 m) from the

depth of the Philippine Trench (— 10,540 m). SOLUTION

—10,540 — (—11,520) = —10,540 + 11,520 = 980 The difference is 980 m.

Objective 2.3A

1.

(—13)(—9) = 117

2.

17(—13) = —221

—12(—4)7(—2) = (48)7(—2) = 336(-2) = —672 ~2(—3)(—4)(—5) = 6(—4)(—5) = —24(—5) = 120 18 + (—3) = -6

Objective 2.3B

57 + (-3) = -19 =. =—17 .

0+ (-14) =0

ipsa al) eae Objective 2.3C

1. STRATEGY To find the average daily high temperature: ¢ Add the six temperature readings. ¢ Divide the sum by 6. SOLUTION

B29) (era 28) —156 + 6 = —26

(28)

The average daily high temperature was —26°F. STRATEGY

To find the average daily low temperature: ¢ Add the 10 temperature readings. * Divide the sum by 10.

(= 97)

156

S-6

Solutions to Module 2 SOLUTION

ee

Oe

5)

2)

(2) 1) 0-2)

= 30) = LOR 3 The average daily low temperature was —3° F. STRATEGY

To find the score: ¢ Multiply the number of correct answers by 7. ¢ Multiply the number of incorrect answers by —3.

¢ Multiply the number of blanks by —1. ¢ Add the results.

SOLUTION 17(7) = 119

8(—3) = —24

(1) = —2 119 + 24)

(2) = 955 2) = 93

The student’s score was 93.

Objective 2.4A

(—7) 47-3? = (-7)-4-4-3-3 = (-7)- 16-9 ll

—112:9 = —1008

(=2)(—3)( 1)

(2

2) 2) 5) (ea)

= —8(9)(—1) = —72(-1) = 72 Objective 2.4B

4—-8+2=4-4=0

24-183 +2= 24-642 S187

20

27 — 18

2 =

16+ 15

($37)

= 27

189)

= 2

+ (—-5) —-2 = 16 + (-3) -2= 13 -2=11

16-4 6+— De)

=)

12 = 4 + 12 =6+—-2 6 =6 +929 Se

=8-2

18 +2-4

—- (-3)=18+216-9 =9-16-9 =-7-9 = -16

=e

MODULE

Fractions

SECTION 3.1

The Least Common

Multiple and Greatest Common

Objective 3.1A

Factor numbers and find the prime factorization of numbers

Objective 3.1B

Find the least common

Objective 3.1C

Find the greatest common factor (GCF)

SECTION 3.2

Factor

multiple (LCM)

Introduction to Fractions

Objective 3.2A

Write a fraction that represents part of a whole

Objective 3.2B

Write an improper fraction as a mixed number or a whole number, and a mixed

number as an improper fraction

SECTION 3.3 Objective 3.3A

Writing Equivalent Fractions Write a fraction in simplest form

Objective 3.3B

Find equivalent fractions by raising to higher terms

Objective 3.3C

Identify the order relation between two fractions

SECTION 3.4

Multiplication and Division of Fractions

Objective 3.4A

Multiply fractions

Objective 3.4B

Divide fractions

Objective 3.4C

Solve application problems and use formulas

SECTION 3.5

Addition and Subtraction of Fractions

Objective 3.5A

Add fractions

Objective 3.5B

Subtract fractions

Objective 3.5C

Solve application problems

SECTION 3.6

Operations on Positive and Negative Fractions

Objective 3.6A

Multiply and divide positive and negative fractions

Objective 3.6B

Add and subtract positive and negative fractions

SECTION 3.7

The Order of Operations Agreement and Complex Fractions

Objective 3.7A

Use the Order of Operations Agreement to simplify expressions

Objective 3.7B

Simplify complex fractions

2

Module 3 e Fractions

SECTION

The Least Common Multiple and Greatest Common Objective 3.1A

Factor

| Factor numbers and find the prime factorization of numbers Natural number factors of a number divide that number evenly (there is no remainder).

|, 2, 4, and 6 are natural number factors of 6 because they divide 6 evenly. Note that both the divisor and the quotient are factors of the

dividend.

6

3

2

|

1)6 2)6 3)6 6)6

To find the factors of a number, try dividing the number by 1, 2, 3, 4,5, .... Those numbers that divide the number evenly are its factors. Continue this process until the factors start to repeat. The following rules are helpful in finding the factors of a number. 2 is a factor of a number

if the digit

in the ones place of the number 0.

2.4.6

31S a factor of sum

is

ors

a number

if the

of the digitsof the number

1s

divisibleby 3

436 ends in 6. Therefore, 2 is a factor of 436. (436 + 2 = 218)

The sum of the digits of 489 is 4 + 8 + 9 = 21. 21 is divisible by 3. Therefore, 3 is a factor of 489.

(489 + 3 = 163) 4 is

a factor

last two

of a

digits of

number

if the

the number

are

5 18 a factor

556 ends in 56.

56 is divisible by 4 (56 + 4 = 14). Therefore, 4 is a factor of 556. (556 + 4 = 139)

divisible by 4

of

a number

digit of the number is 0 0

if the ones r5

520 ends in 0. Therefore, 5 is a factor of 520.

(520 + 5 = 104) | Focus on finding the factors of a number

| Find all the factors of 42. |SOLUTION |Ae

{42

| and +2 are factors of 42.

| 42 +2 =21

2 and 2! are factors of42.

| 4D

+ and | are factors of 42.

3 —

4

|42 +4

4 will not divide 42 evenly.

| 42 + 5

5 will not divide 42 evenly.

| 4a

6 and / are factors of 42.

|42+7=6

/ and 6 are factors of 42.

| The factors are repeating. All the factors of 42 have been found.

| The factors of 42 are 1, 2, 3, 6, 7. 14, 21, and 42.

Section 3.1 © The Least Common

Multiple and Greatest Common

Factor

3

| Check your understanding 1 |Find all the factors of 30. | SOLUTION

See page S-1.

295. 2,0, 101. 30)

LE

A prime number is a natural number greater than | that has exactly two natural number factors, 1 and the number itself. 7 is prime because its only factors are | and 7. If a number is not prime, it is a composite number. Because 6 has factors of 2 and 3, 6 is a composite number. The prime numbers less than 50 are Deo

ett ls bin lO 2S, 2oc Sh, Slt), 43e47

The prime factorization of a number is the expression of the number as a product of its prime factors. To find the prime factors of 140, begin with the smallest prime number as a trial divisor and continue with prime numbers as trial divisors until the final quotient is prime.

_ Focus on finding the prime factorization of a number Find the prime factorization of 140.

| SOLUTION ~

|

pal

et76 2)140

35

5)35

2)70

2)70

2)140

| 140 is divisible by 2. Divide 140 by 2.

2)140

70 is divisible by 2. Divide 70 by 2.

| The prime factorizationof 140is

2:2-5-7

35 is not divisible by 2 or 3 but is divisible by 5. Divide 35 by 5. The quotient is 7, which is prime.

2

587,

Check your understanding 2 Find the prime factorization of 88. | SOLUTION

See page S-1.

Asay

Finding the prime factorization of larger numbers can be more difficult. Try each prime number as a trial divisor. Stop when the square of the trial divisor is greater than the number being factored.

Focus on finding the prime factorization of a number Find the prime factorization of 201. SOLUTION

201 is not divisible by 2 but is divisible by 3. Divide 201 by 3. 67

3)201

67 cannot be divided evenly by 2, 3, 5, 7, or 11. Prime

numbers greater than 11 need not be tried because 117 = 121 and 121 > 67.

The prime factorization of 201 is 3 - 67.

Check your understanding 3 Find the prime factorization of 295. SOLUTION

See page S-1.

5-59

4

Module3 « Fractions

Objective 3.1A Practice . Find all the factors of 12. . Find all the factors of 56. . Find all the factors of 48.

1, 2,3, 4,6, 12 1, 2, 4, 7, 8, 14, 28, 56 1, 2,3, 4, 6,8, 12, 16, 24, 48

. Find the prime factorization of 16. . Find the prime factorization of 40. Wn Auk = . Find the prime factorization of 37.

2 2-5 Prime

Solutions on pp. S-9-S-10.

Objective 3.1B

Find the least common

multiple (LCM)

The multiples of a number are the products of that number and the numbers

3X1=

3

32 =

10

3X8 3X 4= 12

1, 2, 3,

The multiples of 3 are 3, 6,9, 12, 15,....

3x5 = 15

A number that is a multiple of two or more numbers is a common multiple of those numbers. The multiples of4 are 4, 8, 12, 16, 20, 24, 28, 32, 36,....

The multiples of6 are 6, 12, 18, 24, 30, 36, 42,.... Some common multiples of 4 and 6 are |2, 24, and 30.

The least common more numbers.

multiple (LCM)

is the smallest common

multiple of two or

The least common multiple of 4 and 6 is | 2. Listing the multiples of each number is one way to find the LCM. Another way to find the LCM uses the prime factorization of each number. To find the LCM of 450 and 600, find the prime factorization of each number and write the factorization of each number in a table. Circle the greatest product in each column. The LCM is the product of the circled numbers.

2 2

450 =

soo= [2-2-2]

3 (5a)

3

5 (Ges)

equal. Circle just one product

The LCM is the product of the circled numbers. The

CM

2

283

13)

Focus on finding the LCM Find the LCM of 24, 36, and 50. SOLUTION

¢ In the column headed by 5, the products are

00!

Section 3.1 ¢ The Least Common

Multiple and Greatest Common

Factor

5

Check your understanding 4 Find the LCM of 12, 42, and 45. SOLUTION

See page S-1.

420

Objective 3.1B Practice . Find . Find . Find . Find . Find &WwYN Am = . Find

the the the the the the

LCM of 8 and 12.

24 56 9 and 36. 36 102 and 184. 9384 4, 8, and 12. 24 3,5, and 10. 30

LCM of 8 and 14.

LCM LCM LCM LCM

of of of of

Solutions on p. S-10.

Objective 3.1C

Find the greatest common factor (GCF) Recall that a number that divides another number evenly is a factor of that number. The number 64 can be evenly divided by 1, 2, 4, 8, 16, 32, and 64, so the numbers 1, 2, 4, 8, 16, 32, and 64 are factors of 64.

A number that is a factor of two or more numbers is a common factor of those numbers. The factors of 30 are |, 2, 3, 5, 6, 10, 15, and 30.

The factors of 105 are |, 3, 5, 7, 15, 21, 35, and 105. The common factors of 30 and 105 are |, 3, 5, and !5. The greatest common factor (GCF) is the largest common factor of two or more numbers.

The greatest common factor of 30 and 105 is |5. Listing the factors of each number is one way of finding the GCF. Another way to find the GCF is to use the prime factorization of each number. To find the GCF of 126 and 180, find the prime factorization of each number and write the factorization of each number in a table. Circle the least product in each column that does not have a blank. The GCF is the product of the circled numbers. * In the column headed by 3, the products are equal, Circle just one product. Columns 5 and 7 have a blank, so 5 and 7 are not common factors of 126 and 180. Do not circle any number in these columns,

The GCF is the product of the circled numbers. The GCF = 2-3-3 = 18.

|Focus on finding the GCF Find the GCF of 90, 168, and 420. SOLUTION

The GCF = 2 = 3'=6.

6

Module3 « Fractions

| Check your understanding 5 | Find the GCF of 36, 60, and 72. SOLUTION

See page S-1.

12

oe

Focus on finding the GCF Find the GCF of 7, 12, and 20.

SOLUTION

Because no numbers are circled, the GCF

=

1.

_ Check your understanding 6 | Find the GCF of 11, 24, and 30. |

| SOLUTION

See page S-1.

Objective 3.1C Practice . Find the GCF of 6and9. 3 . Find the GCF of 25 and 100. . Find the GCF of 32 and 51.

25 |!

. Find the GCF of 3,5, and 11. | . Find the GCF of 24, 40, and 72. & . Find the GCF DAnbhWYN

of 32,56, and 72.

&

Solutions on p. S-11.

Introduction to Fractions Objective 3.2A Take Note > The fraction bar separates the numerator from the denomina-

Write a fraction that represents part of a whole A fraction can represent the number of equal parts of a whole.

The shaded portion of the circle is represented by the 4

tor. The numerator is the part

fraction —. Four of the seven equal parts of the circle (that

of the fraction that appears above the fraction bar. The

is, four-sevenths of it) are shaded.

denominator is the part of the

Each part of a fraction has a name.

fraction that appears below the

Ractionibatt

4 < Numerator

UEDA Lee 7 < Denominator A proper fraction is a fraction less than 1. The numerator of a proper fraction is smaller than the denominator. The shaded portion of the circle can be represented by the proper fraction as

Section 3.2 ¢ Introduction to Fractions

7

A mixed number is a number greater than 1 with a whole-number part and a fractional part. The shaded portion of the circles can be represented by the mixed l number 2 |.

Focus on writing a mixed number Express the shaded portion of the circles as a mixed number.

OOO% SOLUTION »)

a

ole 5

Check your understanding 1

@000*

Express the shaded portion of the circles as a mixed number.

SOLUTION

See page S-2.

4

4

An improper fraction is a fraction greater than or equal to 1. The numerator of an improper fraction is greater than or equal to the denominator. The shaded portion of the circles can be represented by the improper

fraction re The shaded portion of the square can be rep-

:

4

resented by +.

Focus on writing an improper fraction Express the shaded portion of the circles as an improper fraction.

GSOOe SOLUTION

ly 5

Check your understanding 2

BREST

Express the shaded portion of the circles as an improper fraction.

SOLUTION

See page S-2.

3

4

4

8

Module 3 « Fractions

Objective 3.2A Practice

1. Express the shaded portion of the circle as a fraction.

2. Express the shaded portion of the circle as a fraction.

3. Express the shaded portion of the circles as a mixed number.

>:

,

b)

COOo@

=)

4. Express the shaded portion of the circles as an improper fraction.

9

Solutions on p. S-11.

Objective 3.2B

Write an improper fraction as a mixed number or a whole number, and a mixed number as an improper fraction a

Note from the diagram that the mixed number 23 and the improper fraction

- both represent the shaded

portion of the circles, so 2. —

=

|.

6

ee

66

An improper fraction can be written as a mixed number or a whole number.

ou

13

=

|Focus on writing an improper fraction as a mixed number Write a as a mixed number.

| SOLUTION |

We

;

:

:

_ Divide the numerator by the

To write the fractional part of

Write the

| denominator.

the mixed number, write the

answer.

remainder over the divisor.

|

|

2 =

Sais =I)

SRS 0

3

3

_ Check your understanding 3 | Write 2 as a mixed number.

SOLUTION

3 5k a5 z

See page S-2.

4

Mm

a)

13 as :

3 2.

Section 3.3 ¢ Writing Equivalent Fractions

9

Focus on writing an improper fraction as a whole number |

Write isas a whole number.

SOLUTION | 18 aera B er Oa D as) |Check your understanding 4 |

Write 2 as a whole number.

| SOLUTION

See page S-2.

4

To write a mixed number as an improper fraction, multiply the denominator of the fractional part by the whole-number part. The sum of this product and the numerator of the fractional part is the numerator of the improper fraction. The denominator remains the same.

| Focus on writing a mixed number as an improper fraction Pak ; | Write 73 as an improper fraction.

SOLUTION | 373). Se 8 | «8

3

SO ntioe i a22 8 8

eam 8

Be 8

| Check your understanding 5 | Write 143 as an improper fraction. _SOLUTION

i

See page S-2.

Q

Objective 3.2B Practice 1. Write raas a mixed or whole number.

2 ;

2. Write ieas a mixed or whole number.

°

Ju Write 65 as an improper fraction. : 4. Write 5% as an improper fraction.

58

Solutions on pp. S-11—S-12.

SECTION

3.3 Objective 3.3A

Writing Equivalent Fractions Write a fraction in simplest form Writing the simplest form of a fraction means writing it so that the numerator and denominator have no common factors other than 1. :

4

2

.

:

The fractions , and , are equivalent fractions.

Is ae Ce Cea 2 , 1s written in simplest form as ..

I+ eee :

;

=

2:

eae

7

i

:

|

10

Module3 ° Fractions

The Multiplication Property of One can be used to write fractions in simplest form. Write the numerator and denominator of the given fraction as a product of factors. Write factors common to both the numerator and denominator as an improper fraction equivalent to 1.

4 2252 }252|2 ey a

6 Feel

cos eae

The process of eliminating common factors is displayed with slashes through the common factors as

|

en ae (eae ae

shown at the right.

18

To write a fraction in simplest form, eliminate the

30

x. z- 3-3 3 “Gh Talo

common factors.

An improper fraction can be changed to a mixed

number. ' Focus on writing a fraction in simplest form a. Write R in simplest form.

_ b. Write & in simplest form.

| c. Write iin simplest form. | d. Write 4 as a fraction in simplest form.

| SOLUTION

15 40

|

Seng 2-2-2:

ed

Oia r yay ReySG 1

1

82 290s (Coes 0” 53-6 = e 8

io)S)

N= RH Nn

SS)

Nw

=N

—&

Al 2

|Check your understanding 1 | a. Write 6 in simplest form.

b. Write & in simplest form. c. Write B in simplest form.

_d. Write z as a fraction in simplest form. SOLUTION

=

6

See page S-2.

Objective 3.3A Practice

1. Write fain simplest form. 2. Write a in simplest form.

a.

35

1

m.

4. Write i in simplest form.

+

Solutions on p. S-12.

Objective 3.3B

_ Find equivalent fractions by raising to higher terms Equal fractions with different denominators are

called equivalent fractions.

ae

ce

4

_ 1s equivalent to |. Remember that the Multiplication Property of

One states that the product of a number and 1

p}

is the number. This is true for fractions as well as whole numbers. This property can be used to write equivalent fractions.

3 a

II

ll

x

wt

WIN

2 = XA

2 SSK

3

| hr to

2-4 =——=

3

2 3 Was

lI rye Wl

i

sip me

pac equivalent

to

2eC

>| &

|

8,

SPawny

12

; 2 scquvat to —

Se

3

Wee. ee fractions ¢ and 77.

+8

>= 8

=4

¢

5-4

20

ay

20

the larger

denominator

« Multiply the numerator

32

=

Divide

fraction by the

quotient

by

and denominator of the given (4)

5

_ Focus on writing equivalent fractions | Write ;as an equivalent fraction with a denominator of 42. SOLUTION

2

Sue

74

2-14 =

es

28 :

:

* Divide the larger denominator De

=

by the smaller

¢ Multiply the numerator and denominator of the given fraction by the quotient (14)

42 ~

2

42 1S equivalent to 3.

Check your understanding 2 | Write 2as an equivalent fraction with a denominator of 45. SOLUTION

See page S-2.

27 45

3

>|

Y

¢

ee ae EE

:

the smaller

39 1S equivalent to g.

4) = 3° =

+

a See

Write a fraction that is equivalent to 2and has a denominator of 32. 32

=

Ea — “ik

8

rewritten as the equivalent

3 x aay

||

=

12

Module3 « Fractions fi

Focus on writing equivalent fractions Write 4 as a fraction with a denominator of 12. | SOLUTION

| Write 4 as , ||

125

4

1

1e=s12

* Divide the larger denominator

AV? =

Pay,

48 =

by the smaller

¢ Multiply the numerator and denominator of the given fraction by the quotient (12)

1

48 | 4 1s equivalent to 75.

|Check your understanding 3 | Write 6 as a fraction with a denominator of 18. | SOLUTION

108

See page S-2.

18

Objective 3.3B Practice

1. Write ; as an equivalent fraction with a denominator of 16.

2. Write ;as an equivalent fraction with a denominator of 49. 3. Write 3as an equivalent fraction with a denominator of 18. 4. Write i as an equivalent fraction with a denominator of 64. Solutions on p. S-12.

Objective 3.3C

Identify the order relation between two fractions Recall that whole numbers can be graphed as points on the number line. Fractions can also be graphed as points on the number line. [pe The graph of 3 on the

é

0

1

4

mS

4

4

1

&

[Si

faa

©

“@

4

A

8

4

iO

4

4a mM

2B

4

1B

HM

RET

9

number line

The number line can be used to determine the order relation between two fractions. A fraction that appears to the left of a given fraction is less than the given fraction. A fraction that appears to the right of a given fraction is greater than the given fraction. 1 a>

3 8

6 g-

Bi 8

[+

++}

Ow

eB RO

A CeOm

++

& SG 7 Om MORES

+++ tt

2 MW Ss 3

ml )

+++ tm aS we fb Gh &

1g

To find the order relation between two fractions with the same denominator, compare the numerators. The fraction that has the smaller numerator is the smaller fraction. When the denominators are different, begin by writing equivalent fractions with a common denominator; then compare the numerators.

Focus on finding the order relation between two fractions _ Find the order relation between 7 and 2.

Section 3.4 © Multiplication and Division of Fractions SOLUTION

The LCD of the fractions is 72. 11 ies 44
, between the two numbers.

pee ek

Py

SOLUTION

See page S-2.

Objective 3.3C Practice

1. Place the correct symbol, < or >, between the two numbers. 1 ee)

40 40 2. Place the correct symbol, < or >, between the two numbers. 2S 5 tf 3. Place the correct symbol, < or >, between the two numbers. ey 7 Hey IL 4. Place the correct symbol, < or >, between the two numbers. 5 7 We

AIS

Solutions on p. S-12.

SECTION

3.4 Objective 3.4A

Multiplication and Division of Fractions Multiply fractions To multiply two fractions, multiply the numerators and multiply the denominators. Multiplication of Fractions

Take Note > Note that fractions do not need to have the same denominator in order to be

multiplied.

The product of two fractions is the product of the numerators over the product of the denominators. where

b#0O

and

d#0

13

14

Module3 e Fractions 2

i

:

5

The product 5 - 3 shown in example (1) on the previous page can be read 1»

‘

5

thu

gas

‘

9

CH

xe

”

"5 times i or

6

c

g

:

of 3. Reading the times sign as “of” is useful in diagramming the product of two fractions.

-of the bar at the right is shaded.

Po

We want to shade :of the i already shaded.

|

é of the bar is now shaded.

After multiplying two fractions, write the product in simplest form.

Multipl UCL DUS 3

eects ee

Simo

4 Ls B74

* Multiply the numerators

8 y 9 ae 8-9 = I

=

Multiply the denominators

Bo 2S 9)

Die PEI)

* Express the fraction in simplest form by first writing the prime factorization of each number

Sie}

|

« Divide by the common

6

factors and write the product in

simplest form

Sf WEG MAbeUiohys =| = }l|S— SN 3 SX

(2\(10 SNOT

=

3+2:10 hey Sl}

=

05

* Multiply the numerators. Multiply the denominators.

3°2°2'5

Bg

¢ Write the product in simplest form

cid 14

To multiply a whole number by a fraction or a mixed number, first write the whole number as a fraction with a denominator of 1.

5 _ Multiply: 3 3 Sa)

Sa

3

3-—-=—-—

a

* Write the whole number 3 as the fraction 7

oe)

¢ Multiply the fractions.

1:8

There are no common

15

=

=

W |—

factors in the numerator and denominator.

¢ The answer can be written as an improper fraction or a mixed number

Section 3.4 © Multiplication and Division of Fractions

Apply the Concept A seamstress is making 12 costumes for a dance recital. Each costume requires a yd

of fabric. How much fabric should the seamstress buy to make the 12 costumes? SOLUTION

To find the amount of fabric the seamstress should buy, multiply the amount of fabric needed for each costume (3)by the number of costumes (12).

B23 12-3 36 ei 4 4

2s

=9

The seamstress should buy 9 yd of fabric

Find the product of 4i and 2. ve. 2 Th i. i F =)

Gr

alOpns

* Write each mixed

number

as an improper

fraction

6. 10 Sy e707]

« Multiply

the fractions

6°10 BY COB

Syl)

a) =

| =>

the product in simplest form

5

45 =

* Write

«

1]

The answer can be written as an improper fraction or a mixed number

| Focus on multiplying fractions

| Multiply:

3/1\/8

—| — }{—

nie i(3)(5) SOLUTION H

sala) fh go (Pers

4\2/\9

=

* Multiply the numerators

49

Multiply the denominators

=

* Write the product in simplest form

Check your understanding 1

3(ra)las)

i | Multiply: — SOLUTION

See page S-3.

27

15

16

Module3 @ Fractions

Focus on multiplying a fraction and a whole number _ What is the product of é and 4? | SOLUTION

Ue

SS i

12

ges

A

SS eS

* Write 4 as >

aon ey

=

¢ Multiply

pee

is

UL?

=

PED 7

=

|

=)

“5

¢ Write the product in simplest form

8)? |

3

numerators and multiply denominators.

¢ The answer can be written as an improper fraction or a mixed number.

3

_ Check your understanding 2 | Find the product of 5and 6.

SOLUTION

See page S-3.

5

be

| Focus on multiplying mixed numbers

i

| Multiply: ee

5

| SOLUTION We

Gf

eee

=

iby

i

||

F

* Write the mixed numbers as improper fractions

3 =

15

+22

* Multiply numerators and multiply denominators

PRES |

= 375271

|

¢ Write the product in simplest form

2-5

=

|

be) = 33 1

| Check your understanding 3 Multipl Ultiplys ply 3== 7 A“9

| | SOLUTION

See page S-3.

3 oe

| Focus on evaluating a variable expression Evaluate the variable expression xy for x = 12 and y = 2. SOLUTION

xy

|Y le4S |=) \a= SoS Om 5

* Replace x by l= and y by 7. Write l= as an improper fraction

||

¢ Multiply numerators and multiply denominators.

| 8 ENO

|

»

;

©

9°5 a 5-6

=

Se eS SZ 3 |

= rycs

|||z

¢ Write the product in simplest form

F

Section 3.4 © Multiplication and Division of Fractions

17

Check your understanding 4 : 5 2 Evaluate the variable expression xy for x = 55 and y = 3. 5 SOLUTION See page S-3. 3 |> Objective 3.4A Practice

1. Multiply: 3-3 l

. Multiply: a ; 3 l

. Multiply: 3; ;ae

. Multiply: 35-55

19

WN & >

5. Find the product of is and =.

-

6. Evaluate xy for x = + and VS 6x.

~ or 3

Solutions on pp. S-12—S-13.

Objective 3.4B

Divide fractions The reciprocal of a fraction is that fraction with the numerator and denominator interchanged. The reciprocal of a number is also called the multiplicative inverse of the number. .

ire.

ae!

The reciprocal of | is ..

A

AY

Hoe

S)

The reciprocal of 5 is” 1

The product of a number and its reciprocal is 1.

The process of interchanging the numerator and denominator of a fraction is called inverting the fraction.

To find the reciprocal of a whole number, first rewrite the whole number as a fraction with a denominator of 1. Then invert the fraction.

6=

°

The reciprocal of 6 is i.

Reciprocals are used to rewrite division problems as related multiplication problems. Look at the following two problems: Oo

|

NM

\|

ios)

6 divided by 2 equals 3.

oO

II

1S)

6 times the reciprocal of 2 equals 3.

Division! isldefinted as/multiplication bythefeCiprocall Therefore, “divided by 2” s the same as

“times 5.” Fractions are divided by making this substitution.

18

Module3 e Fractions

Division of Fractions

To divide two fractions, multiply by the reciprocal of the divisor. where b # 0, c # 0, andd #0

EXAMPLE Cr

eS Divide: = + — oye Sie oyu. LL an pee S45 24 8 SS SS Dice

¢ Rewrite the division as multiplication by the reciprocal!

¢ Multiply the fractions

ie 7 14 IDinyitelos Se OES

ieee 100 15

= ue

=

=

=

* Rewrite the division as multiplication by the reciprocal

10 14 Pols

* Multiply the fractions

10-14 UI AS PRED) OI,©. Th

¢ Write the product in simplest form

7a Note in the next example that when we divide a fraction and a whole number, we first write the whole number as a fraction with a denominator of 1.

Take Note > Sie ie

6 =

#53

5 means that if

is divided into 6 equal parts, 1

If 6 people share 3 of a pizza, what fraction of the pizza does each person eat? SOLUTION

To find the fraction, find the quotient of ;and 6.

each equal part is 3

Therefore, if 6 people share 3 fiof a pizza, each person eats A of the pizza.

3 4 3 4 3 |

2

©

Write

6as 2

* Rewrite the division as multiplication by the reciprocal et ©

* Multiply the fractions

aS Nlwoa Orew

Each person eats ¢ of the pizza.

Section 3.4 ¢ Multiplication and Division of Fractions

19

When a number in a quotient is a mixed number, first write the mixed number as an im-

proper fraction. Then divide the fractions.

bay

ree!

l

PDivides= =, i 8) 4 2

:

i

Oy

=

2 aus 2)

2

4

=—-—-—

=

Se) 2:4

—_

8

=

SE

«

Write

the mixed

*

Rewrite

th

'

Multiply

number

vision

the

|; as

an improper

multiplication

by

fraction

the reciprocal

fractions

15

|Focus on dividing fractions Pog

A

8

(Divides | 5 15) |SOLUTION 4 ind 8

5

a

4 f 15

Thisaers!

AB

* Multiply

by the reciprocal

8

SySi)

_2-2-3°5 B) PLS POD) =

3 =

|

2 |

|

*

The

answer can be written

as an improper fractionor a mixed numbei

2

Check your understanding 5

5

| SOLUTION

10 See page S-3.

| Focus on dividing mixed numbers 4 | Divide: 3— + 2— 15

SOLUTION e Write the mixed numbers

as improper

¢ Write the product in simplest form

Check your understanding 6 3 1 Divide: ivide 4— 8 + 3— , SOLUTION

See page S-3.

|

|

fractions

20

Module3 ° Fractions

"Focus on dividing a whole number and a fraction | What is the quotient of 6 and 29 |

| SOLUTION +. OS

©

ae

3 eS

|

CED

| |

Sage1ae

|

wes

* Write 6 as 7

* Multiply by ultiply by the the reciprocal reciproca

IES) |

=

ae

¢ Write the product in simplest form

1-3

|

10

|

=—=

|

10

Check your understanding 7 _ Find the quotient of 4 and S. | SOLUTION

See page S-3.

4

_ Focus on evaluating a variable expression | Evaluate x + ytot x= 35 and y = 5. - SOLUTION lazy

| eee 5 ieee

a 25 at = RN PS Se 35°59 Be alih.

|Check your understanding 8 | Evaluate x + y for x = 2 and y = 9. |SOLUTION

|

See page S-4.

A

Objective 3.4B Practice

9 ‘e

1. Divide: ; = ;

2. Divide: 6+ 3 8 Sel woe. 3. Divide: 33 + 3

rele

We

3 e

WO 5

Tagen

4. Divide: 33 oo 26

yet

Se a

= or | e

5. Find the quotient of 25 and 2 Pye

2 or re 4

hl

5

6. Evaluate x + y when x = 43 and y = 7. Solutions on pp. S-13-S-14.

:

Section 3.4 ¢ Multiplication and Division of Fractions

Objective 3.4C

21

Solve application problems and use formulas Figure ABC is a triangle. AB is the base, b, of the triangle. The line segment from C that forms a right angle with the base is the height, /, of the triangle. The formula for the area of a triangle is given below.

Cc

Area of a Triangle The formula for the area of a triangle is A = tbh, where A is the area of the triangle, b is the base, and h is the height. EXAMPLE

eee 2 The area of the triangle at the right is 6 m

| Focus on finding the area of a triangle A riveter uses metal plates that are in the shape of a triangle and have a base of 12 cm and a height of 6 cm. Find the area of one metal plate. STRATEGY

To find the area, use the formula for the area of a triangle,

A = Shh. b=

12 andh=6

_ SOLUTION |

|

| A=-—bh | 2

(A=5 090 |

12cm

A = 36

| The area is 36 cm’.

Check your understanding 9 Find the amount of felt needed to make a banner that is in the shape of a triangle with a base of 18 in. and a height of 9 in.

ie

SOLUTION

See page S-4.

81 in?

| Focus on solving an application problem A 12-foot board is cut into pieces 25 ft long for use as bookshelves. What is the length of the remaining piece after as many shelves as possible are cut? | STRATEGY

_ To find the length of the remaining piece: | ¢ Divide the total length (12) by the length of each shelf (24).The quotient is the number |

of shelves cut, with a certain fraction of a shelf left over.

* Multiply the fraction left over by the length of a shelf.

22.

Module 3 © Fractions | SOLUTION

nasi g it, 52 2_2-2 aD. tO AS Modes

WA GS

4 5

_ 4 shelves, each 25 ft long, can be cut from the board. The piece remaining is Zof a 25-foot shelf.

4 t 6“3

45 _45 =2 6.9. 429

Phe length of the remaining piece is 2 ft.

_ Check your understanding 10 The Booster Club is making 22 sashes for the high school band members. Each sash requires | 2yd of materia] at a cost of $12 per yard. Find the total cost of the material.

SOLUTION

See page S-4.

$363

Objective 3.4C Practice

inthe NEWS!

>

Asteroid to Fly Within Orbit of

1. The Assyrian calendar was based on the phases of the moon. One lunation was 295 days long. There were 12 lunations in one year. Find the number of days in one year in the Assyrian calendar. 954 days 2. A car used 125 gal of gasoline on a 275-mile trip. How many miles can this car travel on J gal of gasoline? 22 m:

the Moon Asteroid GA®6 will zip past Earth on Thursday at 7:06 p.m, EDT. The asteroid, a space rock about 7\ ft wide, will fly within the orbit of the moon while it passes Parth,

3. Read the news clipping at the left. The distance between Earth and the moon is approximately 250,000 mi. At its closest point, asteroid GA6 was iiof that distance from Earth. Approximate the asteroid’s distance from Earth at its closest point. 5,000 mi 4. A vegetable garden is in the shape of a triangle with a base of 21 ft and a height of

Source:

Solutions on pp. S-14—S-15.,

news, yahoo.com

13 ft. Find the area of the vegetable garden.

4,

,,

SECTION

(35

Objective 3.5A

Addition and Subtraction of Fractions Add fractions Addition of Fractions with the Same Denominator

To add fractions with the same denominator, add the numerators and place the sum over the common denominator. EXAMPLES

litte Take Note > Ii Example 2 al the right, note that the answer is reduced to simplest form, AIWwiys

Hplest

Wile

fort

Your

atswer

in

=:

Section 3.5 ¢ Addition and Subtraction of Fractions

]

aL.

ay

18

=

18

18

|

¢

18

The denominators are the same. Add the numerators Place the sum over the common denominator

9) =

¢ Write the answer

in simplest form

To add fractions with different denominators, first rewrite the fractions as equivalent fractions with a common denominator. Then add the numerators and place the sum over the common denominator. The LCM of the denominators of the fractions is the least common denominator (LCD).

EXAMPLE

Adda =+ = 29.53 The LCM of the denominators 2 and 3 is ©. ¢ Write equivalent fractions with 6 as the denominator

«

Add the numerators

Focus on adding fractions INOW

oe # = se = eS)

SOLUTION 5 a y =

45

56

8

a)

WDD

9

¢ Write equivalent fractions using 72 (the denominators) as the common

LCM of the

denominator

AS + 56 =

72

101 =

=

92

¢«

Add the numerators

e

The answer

29 |

7p:

Check your understanding 1 INGO

TE eal Se Se a tee Ais)

SOLUTION

See page S-4.

23

can be written as a fraction or a mixed number

23

24

Module3 ° Fractions [|

Focus on adding fractions Find 5 more than 3.

| SOLUTION 3 a 8

7 = 12)

Z ak 14 24 24

* Write equivalent fractions using 24 (the LCM of the denominators) as the common denominator

9414

See page S-5.

ia

The sum of a whole number and a fraction is a mixed number.

The procedure below illus2 trates why 2 + — = De 3 3

Add:

2

6

2

D;

3

3

8

3

2 +

To add a whole number and a mixed number, write the fraction and then add the whole numbers.

2 Add: Te +4 2

Write the fraction.

UP

+4

Add the whole numbers.

VW

sets >

5

1] M|rN

Section 3.5 ¢ Addition and Subtraction of Fractions

25

To add two mixed numbers, add the fractional parts and then add the whole numbers. Remember to reduce the sum to simplest form.

Focus on adding mixed numbers

What is 673 added to 55? SOLUTION

The LCM of the denominators 9 and 15 is 45.

Add the fractional parts.

Add the whole numbers.

x4. 520 9 14

45 42

°15

O45

fa

20

9 14

45 42

ome

62 is

62 ig

2 eet

17 lg tga = 12

| Check your understanding 4

| Add: 7 ee + 13— eS 10 15 |

| SOLUTION

7

See page S-5.

2855

A pastry chef is making a blueberry cake that requires 13 cups of flour for the streusel topping and li cups of flour for the cake. To find the total amount of flour the chef needs, add 15 and |i SOLUTION

Objective 3.5A Practice

abageceet ; ra eae >)

todeee S35

2 ISG ESE ASE 16 165 48 SinAddasianeroleatt j

eee

aaa)

2

4. Find the sum of 3,2 and 3

5 5) 5. Add: 1 aF 35

i Le

Solutions on pp. S-15—S-16.

19

i

26

Module3 ¢ Fractions

Objective 3.5B

Subtract fractions

Subtraction of Fractions with the Same Denominator To subtract fractions with the same denominator, subtract the numerators and place the difference over the common denominator.

EXAMPLE =

2

11

I

PSUDthAC hei 18 1

—

18

7

=

= 18

Ws

18

*« The denominators are the same.

Subtract the numerators

18 =

4 ——

* Place the difference over the common

denominator

18 =

=

¢ Write

the answer

in simplest form

9

Subtraction of Fractions with Different Denominators

To subtract fractions with different denominators, first rewrite the fractions as equivalent fractions with a common denominator. Then subtract the numerators and place the difference over the common denominator. EXAMPLE

if

5

Subtract ubtrac 6

4

The LCD of the fractions is | 2.

10 6

4

3

* Rewrite each fraction as an equivalent fraction with 12 as

12

12

the

10-37 12

12

denominator

¢ Subtract the numerators and place the difference over the common

denominator

"Focus on subtracting fractions | Siloieeis

11

=

2

16

12

| SOLUTION 11

5

333)

AD)

16

12

48

48 48

=

—

* Write equivalent fractions using 48 (the LCM of the denominators) as the common denominator ¢ Subtract

the numerators

¢ Write the answer

in simplest form

Section 3.5 « Addition and Subtraction of Fractions

27

Check your understanding 5 im | Subtract Su 2 == ils

| SOLUTION

=g BA

See page S-5.

=

| Focus on subtracting fractions | Find the difference between iand =. |

SOLUTION | oT

5

eee

12

|

i

ZA a. 10

24

* Write

24

equivalent

lenominators)

fractions

as the

using

24 (the

LCM of the

common denominat

ap tegdal

24 |

=

|

Sree

1]

¢ Write

the

an

simple

24

| Check your understanding 6 |

Haters

5

8

| Find 6 less than 9:

| SOLUTION

See page S-5.

To subtract mixed numbers without borrowing, subtract the fractional parts and then subtract the whole numbers.

ubtrac SURE

5 3 B=6 — 2r

Subtract the fractional parts. 5

10

Ie eT

* The LCD of the

me

ee

ractions

fies Da 4

:

Subtract the whole numbers.

is

i

5

10

Fine tsOB

12

= =

=

ae

9

4

12

NW l

Pet

12

12

a

The difference is 375

Subtraction of mixed numbers sometimes involves borrowing.

Bese PeSUbiactay

P. 3

1 6 7=

Write equivalent fractions

Borrow | from 7. Add the 1 — Subtract the mixed

using the LCD, 24.

te 4. Write 1s ae 3

1 ==

4

memati = iS

ee eueeee)

15

| —2-=2— Pe C4

j

ae

28

Me

24

«24

5

15

15

Q— =)2— = 2 8 DA 04

:

13

| The difference is 457.

numbers.

i ok Go

Te

an

5

15

—2==2— Sua eod 13 4 24

28

Module 3 e Fractions

Apply the Concept

7

The inseam of a pant leg is 305 in. What is the length of the inseam after a tailor

cuts :in. from the pant leg? SOLUTION

To find the length of the inseam, subtract :from 305.

i)

3

ieee

3

7 :

3

Te eo ye

The length of the inseam is 293 in.

i)

SUbitactss))=- ar

_ Borrow | from 5.

Write | as a fraction so that the fractions have the

Subtract the mixed numbers.

same denominators. Nn

I] Na

ll aN

nn

nn

| i)

I] K

Il we)

cof

le)

|] 00 | (noo oo tw

The difference is 23.

_ Focus on subtracting mixed numbers |Find us decreased by 2.

| SOLUTION =

ie

(oO eeerR 11 33 Se

=

10

LCD = 48

48 33

ca

35 48

&-

|Check your understanding 7

| What is 215 minus 775? |

| SOLUTION

See page S-5.

E=3

ee 30

Objective 3.5B Practice

insane Oj

ubdtract:

0

Wl 2. Subtract: 9

ee 0

10

hui 6 ie

5 3. Subtract: gs dG

(Alyy

11

Section 3.5 e Addition and Subtraction of Fractions

28

4. Subtract:

ordbras

Oe =

85

75

Busine od6 e ubdtract: 9 6. Evaluate

29

as Z TS

x — y for x = 8 and y = 42. 3

Solutions on pp. S-16-S-17.

Objective 3.5C

Solve application problems -

| Focus on solving application problems involving fractions

| The

length of a regulation NCAA football must be no less than 10% in. and no more

"than ig in. What is the difference between the minimum and maximum lengths of an || NCAA regulation football? STRATEGY

| To find the difference, subtract the minimum length (103) from the maximum length

l 12) 16/° | SOLUTION Me

7

lO

16

poe;

7

lla

8

eye.

F

16

=

16

=

16

16

«16

Q

The difference is (6 in.

| Check your understanding 8 | The Heller Research Group conducted a survey to determine favorite doughnut flavors. :of the respondents named glazed doughnuts, 4 named filled doughnuts, and 3 named | frosted doughnuts. What fraction of the respondents did not name glazed, filled, or

| frosted as their favorite type of doughnut? 13 SOLUTION See page S-6. ‘0d

bes Focus on solving application problems involving fractions

A 23-inch piece is cut from a 62-inch board. How much of the board is left? | STRATEGY

To find the length remaining, subtract the length of the piece cut from the total length of the board. SOLUTION

5)

15

39

1

1

Cpe Ss

23

354 D3),

,

3s

354 in. of the board are left.

30

Module3 e Fractions

Check your understanding 9 A flight from New York to Los Angeles takes 5 5h. After the plane has been in the air for 23 h, how much flight time remains? 2

SOLUTION

See page S-6.

2h 4

Focus on solving application problems involving fractions Two painters are staining a house. In one day, one painter stains }of the house and the other stains tof the house. How much of the job remains to be done? STRATEGY

To find how much of the job remains: l

| ¢ Find the total amount of the house already stained (3ap 1). _ © Subtract the amount already stained from 1, which represents the complete job. SOLUTION 1

4

2

ora. i: 1 = 4

ae

3}

i a

7

1

12

iL?

5

5

12

12

> of the house remains to be stained

Check your understanding 10 A patient is put on a diet to lose 24 lb in 3 months. The patient loses 15 Ib the first month and 53 Ib the second month. How much weight must be lost during the third month to | achieve the goal?

| SOLUTION

See page S-6.

10; Ib

Objective 3.5C Practice 1. A roofer and an apprentice are roofing a newly constructed house. In one day, the roofer completes 3of the job and the apprentice completes f of the job. How much of the job remains to be done? Working at the same rate, can the roofer and the apprentice complete the job in one more day?

5

EYES

20

2. A student worked 43 h, 5 h, and 33 h this week at a part-time job. The student is paid

$9 an hour. How much did the student earn this week?

5117

3. A boxer is put on a diet to gain 15 lb in 4 weeks. The boxer gains 45 Ib the first week

and 33 Ib the second week. How much weight must the boxer gain during the third and fourth weeks in order to gain a total of 15 Ib? Solutions on pp. S-17-S-18.

6; Ib

Section 3.6 © Operations on Positive and Negative Fractions

31

SECTION

Operations on Positive and Negative Fractions Objective 3.6A

Multiply and divide positive and negative fractions The sign rules for multiplying positive and negative fractions are the same rules used to multiply integers.

The product of two numbers with the same sign is positive. The product of two numbers with different signs is negative.

3008 Multiply: Pere The signs are different. The product is negative.

3. 8 At 5 —

3-8 4: as PY cto ow) PP AD3C5

MIBLIEV® £0.00 rsiidin lite keno rebar bor *

Write

the

product

in

simy

lest

form

rs

The sign rules for dividing positive and negative fractions are the same rules used to divide integers.

The quotient of two numbers with the same sign is positive. The quotient of two numbers with different signs is negative.

The signs are the same. The quotient is positive.

ioe 108

LAN eee714 15 Hy BES) =

=

=

—-:—

10 14 els 10-14 Theis) Dies 2, 27

* Rewrite

¢

Multiply

the division as multiplication by the reciprocal

the fractions

* Write the product in simplest form

32

Module3 « Fractions

_ Focus on multiplying positive and negative fractions | Multiply: -2(5)(-5)

- SOLUTION 3/1

=

8

=

4\2

edhe

=

9

4-2-9 AONE So

* The product is positive Multiply the numerators Multiply the denominators

sD

2:2-2-3-3

one eed

¢ Write the

; product

in simplest form

| 3

_ Check your understanding 1

|

Weve

| SOLUTION

See page S-7.

| Multiply: aor Nees

|S

_ Focus on evaluating a variable expression :

;

4

5)

Evaluate the variable expression xy for x = 15 and y = —@. SOLUTION

(B22 eee pe if a

6

ee) :

as

4

« Replace x by 1= and y by

516 O55 = —— S26 Bee =-— JHZE8 3

=

=

—%.5 Write

154 as an improper fraction

* Multiply numerators and multiply denominators.

¢ Write the product in simplest form

|

|

*

The answer can be written as an improper fraction or a mixed number.

_ Check your understanding 2 | Evaluate the variable expression xy for x = —55 and y = 5.

| SOLUTION

See page S-7.

ae| iI)

|Focus on dividing positive and negative mixed numbers 1vide: soe 15 5 (-25] 10 Divide | |

| SOLUTION

sks

(24) =

VetSae

10

|

(2-2) 15

(49

a

10

* The signs are different. The quotient is negative. Write the mixed numbers as improper fractions.

10

¢ Write the product in simplest form

Section 3.6 * Operations on Positive and Negative Fractions

| Check your understanding 3

|

Dues IVIG

CS

| SOLUTION

a: nee

Ses

See page S-7.

15

| Focus on dividing positive and negative fractions | What is the quotient of 6 and —39

| SOLUTION Gah

3

;

5

Ss

6

j

3

it

¢ The signs are different. The quotient is negative

&

Write 6 as 7

HS

| |

Sa) ay

ey ee

¢ Multiply by the reciprocal

=—

* Write the product in simplest form

_| Check your understanding 4 |Find the quotient of 4 and 2

|

|| SOLUTION

See page S-7.

4

2)

Objective 3.6A Practice

1.. Multipl 26 ( | 1ply Bhkoa Verte: 5

3 ie)

2. Multiply:

:5

2 eo)

| ae

S) ull 3. Multiply: Pits

er

2 3 ultiply Si3 ( 24 INiuitiipohys ((SS4

ph

iy

55

ET

5. Evaluate xy for x = —jg andy =j75. 5

5)

a

- re

6

(O IDYaAivide: les ==~~ ae || ( :) 3 5

Ty JOXNRIGIER = 8

oh IDinasles ivide

(-5) 12 5 7 Sys [|4 5 (

¢

= 10

—8

9.% Find the quotient of 25 and 2 FBS

:

5

2 or te 4

|

ileal

10. Evaluate x + y for x= —g andy = —>5. Solutions on pp. S-18-S-19.

5

33

34

Module3 e Fractions

Objective 3.6B

Add and subtract positive and negative fractions To add a fraction with a negative sign, rewrite the fraction with the negative sign in the numerator. Then add the numerators and place the sum over the common denominator. i | |

Take Note > =|

1

12-12

3a

48S 4

Add: -> + = 6.

1

SoS

12

6

sls

OS a 6 + =lQ .

=

4

Although the sum could have

SSS 12

=i been left as Te all answers

Oa

in this text are written with the negative sign in front of the fraction.

SS

=

aa 12

* Rewrite each fraction as an equivalent fraction using the (12) of the fractions

Y

EE

12

Sa

* Rewrite the first fraction with the negative sign in the numerator

¢

|

ee

ee

12

Add the fractions

¢ Simplify the numerator and write the negative sign in front of the

12

2

4

3

5

LCD

fraction

Add: -= + (-= 2

+

{—

3

ENE = 5

en

+ 3

LO

ees2

es 15

— IS}

LOG

=

* Rewrite each negative fraction with the negative sign in the numerator

>

* Rewrite each fraction as an equivalent fraction using the LCD (15) of the fractions

Eat?

710 + (+12)

15

99 =

«

Add the fractions

7

=

=>

15

]

=

15

To subtract fractions with negative signs, first rewrite the fractions with the negative signs in the numerators.

By Simplify: Ee

2

See

9

12

5

=

* Rewrite the negative fraction with the negative sign in the

9 =

12

numerator

au — iS 36

=

8

15 =

36 23

=

¢ Write the fractions as equivalent fractions with a common

36

denominator

23 36

* Subtract the numerators and place the difference over the common denominator ¢ Write the negative sign in front of the fraction

36

:

2 P Subtracts— 3

4 (-2) 5

2 —— 13

2

a

3 NO, —

5

4 3)

{-—]=—+

=

=

* Rewrite subtraction as addition of the opposite

i

1S

— 15

IOs

12

———_

15

Te eh

¢ Write the fractions as equivalent fractions with a common denominator e Add the fractions

Section 3.6 ¢ Operations on Positive and Negative Fractions as

Focus on adding signed fractions Ad

|

ep Na) eee or

a

gene

3)

6

|

| SOLUTION le }

——+

|

8

||

gece)

5)

-

4

=

6

ee)

+

8

=

|

=

sao

ee

+

4

Omen

+

;

S90

24

24

24

=o

is

(20)

24

i |

* Rewrite each negative fraction with the negative

6

24

sign in the numerator

* Rewrite each fraction as an equivalent fraction using the LCD (24) of the fractions *

Add the fractions

1 24

| Check your understanding 5

:

leoANdGs

5. i2

ets “8

| ( ;)

+—+

| SOLUTION

|

See page S-8.

24

—

| Focus on subtracting signed fractions oe! | Subtract

SS2 6

|( SS4 8

| SOLUTION | |

lt

5

« 3 9 |-—_.—4 |) - 5 5.SSimplity.— Simplify 8 = (=

bei

Subriinys (7 (= = 5) =)) ===16. 6.» Simplify

ae

6A

Solutions on pp. S-21-S-22.

Objective 3.7B

Simplify complex fractions A complex fraction is a fraction whose numerator or denominator contains one or more fractions. Examples of complex fractions are shown below.

4

Main fraction bar ——~

10

coo|r|Ri[w

Look at the first example given above and recall that the fraction bar can be read “divided by.” =A )

Therefore, — can be read “* divided by 4

“ce 3

” and can be written | ~ ,. This is the division

8

of two fractions, which can be simplified by multiplying by the reciprocal, as shown below.

To simplify a complex fraction, first simplify the expression above the main fraction bar and the expression below the main fraction bar; the result is one number in the numerator and one number in the denominator. Then rewrite the complex fraction as a division problem by reading the main fraction bar as “divided by.”

| Focus on simplifying Simplify:

a complex fraction

i ———

2) SOLUTION a

pete1

= as 2)

2

¢ The numerator (4) is already simplified. Simplify the expression in the dene ( ) tenominator.

2

SS =

Note;

5 asD

Tes

o

ay)

a y; aa)

¢ Rewrite the complex fraction as division. I

a aoe >

¢ Divide.

12

sie

3 2}.

3 I 8

=—=

5

3

|—

5

;

¢ Write the answer in simplest form.

40

Module 3 « Fractions

Check your understanding 3 | Simplify: 7

i

|

Sel SOLUTION

See page S-9.

| Focus on simplifying

|

39

if

a complex fraction

nea

| 10:5 | Simplify: ——— i 4

SOLUTION =

Sie

SIs

10

ce:

3

5

10

¢ Simplify the expression in the numerator

5

lies

Note:

4 =

3

=|

5

==

= =

3

4

10

ll

anh y 3

4

Write the mixed number

as

in the denominator

* Rewrite the complex fraction as division. The quotient will be negative

¢ Divide by multiplying by the reciprocal

5 =

——

* Write the answer in simplest form

| Check your understanding 4

|

po se,

|

4

5

|

2 +

I

3).

&4

| Simplify:

| SOLUTION

See page S-9.

Objective 3.7B Practice

sack Fae LG 1. Simplify: ra 4 |

2. Simplify:

as

:

ile 3. Simplify:

ee

oes)

r

14° 7 43 4. Simplify: : 2 55 Sie SY

5. Evaluate = —

5 = : ad 4 Beye,

=o

eae lora 4, = 7,i= and @="95. 17

Solutions on pp. S-22—-S-23.

as an improper fraction,

Solutions to Module 3

SOLUTIONS TO MODULE

3

Solutions to Check Your Understanding Section 3.1

Check your understanding 1 S095 NO

0)

2 = Ip)

80% 3) — 10 30 + 4

* Does not divide evenly

30+5=6 30

a

6 =5

¢

The factors are repeating

The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30.

Check your understanding 2 11

2)22 2)44 2)88

88 =2:°2-2*11 =2?-11 Check your understanding 3 5g 5)295 DD ISy"5°Bh)

Check your understanding 4 7

The

LEM = 2 2+3-3-5-7 = 420.

Check your understanding 5

The GCF = 2° 2)°3 = 12,

Check your understanding 6 3

5

11

Because no numbers are circled, the GCF = 1.

S-1

S-2

Solutions to Module 3

Section 3.2

Check your understanding 1

“4

Check your understanding 2 17

oA Check your understanding 3

yo

4

ae

~20

2

Dy,

Check your understanding 4 4

pee

ee

93

0

Check your understanding 5

5 «88 Section 3.3

1124+ 5 8

8

Check your understanding 1 |

a

fOr 94 Tp ee 1

1 1

2-287

Rb

| Oe 1

“

1

1

8

ee eas il

Qnia

Oe 1

eae

1

ine 3-5 130.9 2-0 De 1

F AS

1

2-22

15 1

Die

Check your understanding 2 45+5=9 5 =

¢ Divide the larger denominator by the smaller

5-9 9 =

A5

¢ Multiply the numerator

and denominator of the given fraction by the quotient (9)

27;18 equivalent al 45 to 3=.

Check your understanding 3 Write 6 as $ 18+ C=

1=18

6-18

=

ire tN

¢ Divide the larger denominator

— 108 Se

by the smaller

* Multiply the numerator and denominator of the given fraction

18

by the quotient (18)

ra is equivalent to 6.

Check your understanding 4 9

27

1355

26

9 iS 13

14

42

7

ap)

14

21

Solutions to Module 3

Section 3.4

Check your understanding 1

(a \e) SII,

i538

=

15

Sel

* Multiply the numerators Multiply the denominators

als)

(S222 S}O DAS PROSSER)

=

* Write the product in simplest form

De pz|

Check your understanding 2 8 9

367 80 : = De le oe

-6

2s 2°2:2:2°-3

* Write 6 as ;.

== 16 =

3-3-1

5 I

3

¢

3

The

Then multiply

answer can be written as an improper

fraction or a mixed

number

Check your understanding 3

pl ab he. 3--Q- = S = YS

¢ Write the mixed numbers as improper

fractions.

qr ® es 3yo Bie Bye Pha il

+

(so)

Te ae ay

Then multiply

92

Pe)

Check your understanding 4 xy

Pale ae oe 3

eae

3

and y by

=. Then multiply

a [410-2 “88 abit Secv2 52-73

2 Se

Pigs

Bh 1)

Check your understanding 5

Oe ry 627 6 10 6-10 e UCESCaee

D305

* Multiply by the reciprocal

aa

4

A

Check your understanding 6

pate alee 33 ph 109522 3542 ReeeOee Sno Geko 837

See DDS PEDO GS

eh

4

Check your understanding 7

* Write the mixed numbers as improper fractions

Then multiply by the reciprocal

S-3

S-4

Solutions to Module 3

Check your understanding 8 IG => FY

,

1

5

eee eo ee 2 A 1 BAL ee _ 3235 eed 2725 3a8 2 a

4

Check your understanding 9 STRATEGY To find the amount of felt needed, use the formula for the area of a triangle, b = 18 andh = 9

1 A = i Dh.

SOLUTION

A=

Th 2

=ae5 18) (1809)(9 A=

8l

81 in’ of felt are needed.

Check your understanding 10 STRATEGY To find the total cost:

3

¢ Multiply the amount of material per sash (13)by the number of sashes (22) to find the total number of yards of material needed.

¢ Multiply the total number of yards of material needed by the cost per yard (12). SOLUTION

3 59 =lh 22s ee 8 a al 121 | = — = 30— 4 4 1 ON edible 30—-12= 4 40 4-1 11-11-2-2:3 = 363 Tl The total cost of the material is $363.

Section 3.5

Check your understanding 1 7 aie MI =

35 ae 33

9

45

15

45

_ 35 + 33 = =

45

68

eS

IG

¢ Write equivalent fractions using 45, the

LCM of the denominators

of the fractions

¢ Add the numerators

w3;

Ne

¢ Write the answer as a mixed number or an improper fraction.

Solutions to Module3

Check your understanding 2 5 i

9

12

16

20).

=

Pal

ae

48

¢ Write equivalent fractions using 48, the LCM of the denominators

48

of the fractions

_ 20.+.27 a

48

¢

Add the numerators

47 48

Check your understanding 3

2

OL ED

Sb

4

de Se

5

8

40

SUR —

ae

40

oa

Zo

40

87 40

ees

40

* Write equivalent fractions using 40, the

¢

Add the numerators

5 i

EN ae

aan

Spee

40

* Write the answer as a mixed number or an improper

Check your understanding 4 —74

a wee

=

5

°

30

7

is 3( 30 The e LCD is

PII

6—= 10

6—30

11 qr IBS] = 15

22 ISS 30 67 (j= SS Di 30 30

Check your understanding 5 13

V

eel

= 24 =

18 _

WD

TP

Dh mi

« Write equivalent fractions using 72, the denominatorsof the fractions

LCM of the

All

70

¢ Subtract the numerators

st G2,

Check your understanding 6 Sion asd ae Ol * Write equivalent 9

6

18

18

Dilcesis a

©

18

fractions using 18, the denominators of the fractions

¢ Subtract the numerators

a!

18

21]

=2)

LCM of the

denominators of the fractions

= 20

¢ The LCD is 36.

LCM of the

fraction

S-5

S-6

Solutions to Module 3

Check your understanding 8 STRATEGY

To find the fraction of the respondents who did not name glazed, filled, or frosted: ¢ Add the three fractions to find the fraction that named glazed, filled, or frosted.

¢ Subtract the fraction that named glazed, filled, or frosted from 1, the entire group surveyed.

SOLUTION yD, 8 3 +—+—= eee

20)

40

-

32

+

15

OO Rae OOOO

an

100

100

100

100 ~~100

ip of the respondents did not name glazed, filled, or frosted as their favorite type of doughnut.

Check your understanding 9 STRATEGY

To find the time remaining before the plane lands, subtract the number of hours already

in the air (23) from the total time of the trip (54). SOLUTION

eG 5— = 5= = 4— a 3 RG eee Aa 3 2h 4

The plane will land in 23 h. Check your understanding 10 STRATEGY

To find the amount of weight to be lost during the third month: ¢ Find the total weight loss during the first two months (73 + 53). e Subtract the total weight loss from the goal (24 Ib). SOLUTION

i gn 2 4

eee4

ee4 5 1 12> = 13—Ib lost 4 4 4

24 =23-4

Sig 4ae 4 4 10>4 Ib The patient must lose 103 Ib to achieve the goal.

Solutions to Module 3

Section 3.6

S-7

Check your understanding 1

(-3) S)-2.33 3

12

15

3.49

an

15

1€ product 18 positive.

eS See IP2F> ili 1hOS) OPES ByOO D2: NON ow) or.

775)

Check your understanding 2

xy 12

==

41

0 =

8

3

2

;

8

¢ The signs are different.

3

Improper

The answer ts negative.

|

Write 5g as an

traction

41-2 =

-(8-3

* Multiply the numerators.

Multiply the denominators

ey =

(5 ere

4i

5

_—

br

* Write the product in simplest form * Write the answer

ete)

as an improper

fraction or a mixed numbe:

Check your understanding 3 —q4—

3} 8

1

+ 3-—=

»

3)

—

8

:

* The signs are different.

»

oi 35 as improper

8 ; 5)

=

Soe

—|

——

2a

Be

=

a

Write 45 and

* Rewrite division as multiplication by the reciprocal

« Multiply the numerators.

SSS

5

The answer is negative.

fractions

Multiply the denominators

¢ Write the product in simplest form.

¢

Write the answer

as an improper

fraction or a mixed

number

4

Check your understanding 4 4

=

6 7

4

6

* The signs are different.

=

SS

arias

* Rewrite division as multiplication by the reciprocal.

ais

cael

¢ Multiply the numerators.

Lae) 4 7

easy

The answer is negative.

:

=

Write 4 =

4j

Multiply the denominators

LO) =

iM

2°2:7

* Write the product in simplest form.

re ANOS) 14

fe

¢ Write the answer as an improper fraction or a mixed number

S-8

Solutions to Module 3

Check your understanding 5 —

2

+ 2 +

d

I

&

6

=

2

ae 5 ae 71

(2 =

8

=10 24

* Rewrite each fraction with the negative sign in

6

the numerator.

ak 15 a. ea 24 24

* Write equivalent fractions using 24, the LCM of the denominators. Then add

_ 10 + 15 + (-4) 24

ane 24

SSS

SS

6

SS

9

9

Sj

SS Se 18 =

5

—;- with the negative sign in the

* Rewrite 6

numerator

aes

.

eee 18

slp

* Write equivalent fractions using 18, the LCM of the denominators.

14

¢ Subtract the fractions.

18 = 28 18

=

Section 3.7.

29

1

SSS SS 18

|

|= 18

¢ Write the difference as an improper fraction o1 a mixed number.

Check your understanding 1

C )Gra) 13

Bi( 2 4

13)

* Perform operations in parentheses

1 ¢ Simplify exponential expressions

:

* Multiply or divide from left to right

13 2 12

-13-= nee 13 1 O93 43.1925 a156 Check your understanding 2

( 7) =e 2 =

=

=

fa

LN

| ——]

D

wa

+

Ae

ad

-—

4+ —

=

¢ Simplify

5

als rae ae is

Bi) 5p 1 4 +

iO

S

—

kee 4

9

¢ Simplify (= 5

ae 1

8

5

IM

10

==

9 iI@

Solutions to Module 3

Check your understanding 3 2

2

Cees oe

a

sigy A

4

o +

l

3

84

2

2

—

SSS

=

cet

SoS

* Rewrite the complex fraction as division.

ee,

8?

323

9

15> 20°

12 3 520-20

SSS SS SS 8 4 &) |

2 =

* Simplify the expression in the denominator.

3

7

1]

=

12 3)

¢ Simplify the expression in the numerator and the expression in the denominator.

c

¢ Rewrite the complex fraction as division

Solutions to Objective Practice Exercises Objective 3.1A

1. 12S ID

112 2=6

12+3=4 12-4=3 Thefactors of 12 are 1, 2, 3, 4, 6, and 12.

I, Dose I= No) 56 52 = 28 56+4=14 56 = 1 =8 561-5 =F The factors of 56 are 1, 2, 4, 7, 8, 14, 28, and 56.

3. 48 = 1 = 48 48 +2 = 24 48 = 3'=16 48+4=12 48 -6=8 48 = 8=6 The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

S-9

S-10

Solutions to Module 3

2)16 16=2-2-2-2=2 5

ae

2)10 2)20 2)40 40 =2-2:2-5=23-5 6. 37 is a prime number.

Objective 3.1B

1.

2

3.

4.

5.

6

2

3 5=

10

KOM

3

3)

5

© 5

= Do Bos) = AO

Solutions to Module 3

Objective 3.1C

Objective 3.2A

Objective 3.2B

il.

S-11

S-12

Solutions to Module 3

33 6

2 {ee 11 Objective 3.3A

;

2

2 aes 11 11

a =

1.

=

10903

6

1

1

5

1

16° 2922-20

8

4

Dao ao eel

j 7

12a 22 15 5-7 Ce 36

1

2 35

1

1

“pba:

Objective 3.3B

1

1

1

1

Bi A 4-4

(0 27 See ji-7 5-2 139: =. : 3 2 15-4 4. 64 + 16 = 4; —— : 2 16:4 et:

Objective 3.4A

he 16

1. 16+ 4=4;——=— p

Objective 3.3C

1

PBI”. Sey 2-2-2-B

E49 10 IS =

60 — 64

ie ee 40 40 2 11635n Sees 2s =——-=—,—> Peeks bis ee 13-41-26) eae 8, = S; =—, = 18 36512 Bars ae 5 25 37. VOsae 7 4. =—, =—, < 12) 6015 60 noe 15 1. alae

1. >-1=-2.4

G2 Sa Se 322 VED 5

ey S

1504 " 16

10 loe4 16-9 375-22 DD Dates

3

Solutions to Module 3

“3 £4 8 9

2 3 PLE BOG SPL, OP GPLS Sie 3: a1

12

12

DADA e 27)

Objective 3.4B

1.

Di

ED

S-13

S-14

Solutions to Module 3

Objective 3.4C

1. STRATEGY

1 To find the number of days, multiply the number of days in one lunation (295) by 12. SOLUTION

Bele FO12 2D,

2)

=

539) o 112) Wo ||

= 354

There are 354 days in one year of the Assyrian calendar.

Solutions to Module3 -

S-15

STRATEGY

To find the number of miles traveled, divide the number of miles (275) by the

1 number of gallons used (125). SOLUTION

Il

21S

oD,

Zils).

Pes) + | 2 Pils) 72 se 25

The car can travel 22 mi on 1 gal of gas.

3. STRATEGY To approximate the asteroid’s distance from Earth at its closest point, multiply the distance between Earth and the moon in miles (250,000) by the fraction of that

9 distance that the asteroid was from Earth tl SOLUTION 2

!

.

eee

9 10

=

250,000

°

9

] 10 2,250,000 a 225,000 10

At its closest point, the asteroid was approximately 225,000 mi from Earth. 4.

STRATEGY

To find the area, use the formula below. Substitute 21 for b and 13 for h, and solve for A. SOLUTION

1

es

1

[ie D;ON 273 = S— = 2

es 3=

ik Bil

We

Dee.

|]

1 2

1

The area of the vegetable garden is 136, Ve

Objective 3.5A

1.

S-16

Solutions to Module 3

Bh

2_ 40 3 60 1 2 5 60 Feige2 12 60 Si 60 35459 aa 5 20 6 24

2 ee 60 20

plage gS iD 04 me Dy oh 5 en atS

(= f

ome 25

10,

Objective 3.5B

1

7

9 20 7 20 eal 20 +10 dele Om © als oes 6a ale 11 18 27 4 Sas To etl TO) beh Tee ee 5 45 45 4 20 20 ton aes 43

Solutions to Module 3

5.

40—

S-17

4 8 26 = 40— = 39 9 18 18 ]

Soap! Pye 6 18 18 see 18 (Ss Se

5h

5 6 §:=4-=7 6 6 ] = 3— 6

Objective 3.5C

4

5 6

1. STRATEGY To find the amount of roofing that remains: oa

al

¢ Add the amounts already done (Z3e i} ¢ Subtract the amount already done from the total job (1).

To determine if the roofers can finish in one more day, compare the amount already 1

done with —.

2

SOLUTION

2) i ay 54 9, 30 13.6201 2 13 20°20 20

eee 20 20

7 0 of the roofing job remains to be done. fl 1 Because 0 is less than 5° the roofing can be finished in one more day. 2.

STRATEGY

To find the amount earned: ¢ Add the hours worked (4;ar 8) ae 33). ¢ Multiply the hours worked by 9. SOLUTION

1 SE

z

aa

;

a

13

13-9 = 117

The student earned $117. 3.

STRATEGY

To find the amount of weight the boxer must gain: ¢ Add the amounts already gained (453F a) ¢ Subtract the amount gained from the goal amount (15).

S-18

Solutions to Module 3 SOLUTION

| Pa sceeeese3 4~+3—-=4-43 Pa ea ae 1S

Sa 4

4

Cee

8— =6 4 +

3 The boxer has or Ib left to gain.

Objective 3.6A

EQ

Solutions to Module 3 Nn

ON

nA rI|nr}n WN n|oa

NN

na ~

Aye We es 8 ( 4 ia G < a es

C4)

ey gee _ 340-233 7 3-220 Se’ y 16 neal 1) eo 5 10 ae 16 5> 7 SED DST eg Pe] 32-5 “ 5-7

=== -8 eee 5a 5) 5 12

3225

65"12 UE 5.55 WS 282.3 Sil

= 4 LOS

4

any,

Smeets.

\ae Oe.

-3+(-2)=3+4 ae

56

eer

815 3 5-2 7 239-9-3°5

S-19

S-20

Solutions to Module 3

Objective 3.6B

1.

sO dk 4

es Aes aes (res): ae _=9+88 =1 12 12 ik, + == + Pee orbs 8 =e 15 Ae, DA _-14+151 24 24 11 \ ° 2 ed + [=—+ 5 (-75) aS 2 Oppel! Ge erie =

—

1 12

See

6+(-11)

e + 8

—-5

15 15 1 hamec Case +—=—+ + 5) 41 Se Ope 12" 14 =—+ + 4. | 34) 9A O a(12) 14 4 ft 24

3G Se 3

a + 10

7 2

5 12

7 3. ay =—+ ieee iy aes O: =14 = ay eA _9+(-14) _ -5 30 30 _-1 3 ji 3 ee Se 8 8 _-4-3. -7 7 8 8 8 2\_-5 -2 3 oD aE hees: i sO

_ -5 = (-8) 12 See

en

i

Solutions to Module 3

-15+14

-1_

24

24

i571 36 Reis

Objective 3.7A

He

24

20 36 (—20)

36 156205 36

5 36

limeocd De 8 Ses 9 15

Se 45

45

ol

45

Suse Ono 4a —-=+-—+—-= EO

]

8 EE

5

21g Se

eo Sed 9 6 9 16m |— + —8 — — 1g 18 48

ee

=

ne,

Ee ie)

CTC

18 a

92 ee

Se, 1.9 38 Digue (Gtr == i 367) 36m

=

at

36.

eee 8

oil0r 35, Ty 55

72

aa

A

S-21

S-22

Solutions to Module 3

9 eis: 9 Be 10 \3 30510 wy 3 i i IG PS a3 24-26 10 ais 515 iris ~ 5

ae ecw Leb

a

to eye +3 BH19 iS Moa A394 = 1.0 re

Tie

G

aa

5) 16

pe

Ne

(; 2 Hi

glee p fond

one — 64—«W+6 ao 64. «64 é ~ 64 9

i 28

Objective 3.7B

3) Abn aa 4

Sx

eee-3 ~ 16 30nd me 4) ee ie AS). ik at a “86UF eee Ht 6-7 6 97 4 wy a 9° 1 Sir a 7 ei in eit 22 ad eg ie at

Solutions to Module 3

SI 4—-3=

8_ 8

Lots 8

7,

Ea

P4

pole Ca

Bo

Mes) 24 aay)

See

Bee

—-+—

ue 12

|7 a

nS 11)

Si 7

&ls 12

12

S-23

MODULE

Decimals and Percents

SECTION 4.1 Objective 4.1A

Introduction to Decimals Write decimals in standard form and in words

Objective 4.1B

Round a decimal to a given place value

Objective 4.1C

Compare decimals

SECTION 4.2

Adding and Subtracting Decimals

Objective 4.2A

Add and subtract decimals

Objective 4.2B

Solve application problems and use formulas

SECTION 4.3

Multiplying and Dividing Decimals

Objective 4.3A

Multiply decimals

Objective 4.3B

Divide decimals

Objective 4.3C

Solve application problems and use formulas

SECTION 4.4

Comparing and Converting Fractions and Decimals

Objective 4.4A

Convert fractions to decimals

Objective 4.4B

Convert decimals to fractions

Objective 4.4C

Compare a fraction and a decimal

Objective 4.4D

Write ratios and rates

SECTION 4.5

Introduction to Percents

Objective 4.5A

Write a percent as a decimal or a fraction

Objective 4.5B

Write a decimal or a fraction as a percent

SECTION 4.6

Radical Expressions and Real Numbers

Objective 4.6A

Find the square root of a perfect square

Objective 4.6B

Approximate the square root of a natural number

Objective 4.6C

Solve application problems

2

Module 4 e Decimals and Percents

SECTION

4.1 Objective 4.1A Take Note > In decimal notation, the part of the number that appears to

the left of the decimal point is the whole-number part. The part of the number that appears to the right of the decimal point is the decimal part. The decimal point separates the whole-number part from the decimal part.

| Introduction to Decimals Write decimals in standard form and in words The price tag on a sweater reads $61.88. The number 61.88 is in decimal notation. A number written in decimal notation is often called simply a decimal.

A number written in decimal notation has three parts.

61

:

Whole-number

Decimal

part

point

88 _—Decimal part

The decimal part of the number represents a number less than 1. For example, $.88 is less than $1. The decimal point (.) separates the whole-number part from the decimal part. The position of a digit in a decimal determines the digit’s place value. The place-value chart is extended to the right to show the place values of digits to the right of a decimal point.

In the decimal 458.302719, the position of the digit 7 determines that its place value is ten-thousandths.

To write a decimal in words, write the decimal part of the number as though it were a whole number, and then name the place value of the last digit.

0.9684

Nine thousand six hundred eighty-four ten-thousandths

The decimal point in a decimal is read as “and.”

372.516

Three hundred seventy-two and five hundred sixteen thousandths

To write a decimal in standard form when it is written in words, write the whole-number part, replace the word and with a decimal point, and write the decimal part so that the last digit is in the given place-value position. Four and twenty-three hundredths

3 is in the hundredths place.

4.23

Section 4.1

© Introduction to Decimals

3

Focus on naming a place value Name the place value of the digit 8 in the number 45.687. | SOLUTION

The digit 8 is in the hundredths place.

Check your understanding 1 | Name the place value of the digit 4 in the number 907.1342. SOLUTION

See page S-1.

Thousandths

Lo

When writing a decimal in standard form, you may need to insert zeros after the decimal point so that the last digit is in the given place-value position. Ninety-one and eight thousandths

8 is in the thousandths place.

91.008

Insert two zeros so that the 8 is in the thousandths place. Sixty-five ten-thousandths 5 is in the ten-thousandths place.

0.0065

Insert two zeros so that the 5 is in

the ten-thousandths place.

Focus on writing a number in words

| Write 293.50816 in words. SOLUTION | Two hundred ninety-three and fifty thousand eight hundred sixteen hundred-thousandths

Check your understanding 2 Write 55.6083 in words.

SOLUTION

See page S-1.

Fifty-five and six thousand eighty-three ten-thousandths

Focus on writing a number in standard form Write twenty-three and two hundred forty-seven millionths in standard form. SOLUTION 23.000247

¢ 7 is in the millionths place.

Check your understanding 3 Write eight hundred six and four hundred ninety-one hundred-thousandths in standard form.

SOLUTION

See page S-1.

806.00491

Objective 4.1A Practice

1. Name the place value of the digit 5 in the number 432.09157. 2. Write A asadecimal.

0.853

3. Write the number 25.6 in words.

Twenty-five and six tenths

—Ten-thousandths

4

Module 4 © Decimals and Percents 4. Write the number 1.004 in words. One and four thousandths 5. Write nine and four hundred seven ten-thousandths in standard form. 9.0407 6. Write six hundred twelve and seven hundred four thousandths in standard form.

612.704

Solutions on p. S-7.

Objective 4.1B Tips for Success > Have you considered joining a study group? Getting together regularly with other students in the class to go over material and quiz each other can be very beneficial. See AIM for Success.

Round a decimal to a given place value In general, rounding decimals is similar to rounding whole numbers except that the digits to the right of the given place value are dropped instead of being replaced by zeros. Round 6.9237 to the nearest hundredth. pany

jewncert place value (hundredths)

right are dropped.

6.9237 lid 3 5

Increase 3 by | and drop all digits to the right of 3.

12.385 rounded to the nearest tenth is 12.4.

_ Focus on rounding to a given place value | a. Round 0.9375 to the nearest thousandth. _ b. Round 2.5963 to the nearest hundredth. | ce. Round 72.416 to the nearest whole number. | SOLUTION

| a.

ee Given place value 0.9375 5=5 0.9375 rounded to the nearest thousandth is 0.938.

ca

ae

place value

2.5963 6>5 2.5963 rounded to the nearest hundredth is 2.60.

&

ee Given place value

72.416 4, between the numbers. 0.0135

0.02

| SOLUTION

0.02

0.0135

2

135

oe

100

10,000

200

135

10,000

10,000

200 10,000

O02

113)5%

=

10,000

==)

¢ Write the numbers as fractions

¢ Write the fractions with

a common

¢ Compare the fractions

O13

| Check your understanding 5 | Place the correct symbol, < or >, between the numbers.

0.15

SOLUTION

0.107

See page S-1.

0.15 > 0.107

Objective 4.1C Practice 1. Place the correct symbol, < or >, between the numbers.

0.278 > 0.203 2. Place the correct symbol, < or >, between the numbers.

0.045 > 0.038 3. Place the correct symbol, < or >, between the numbers.

0.037 < 0.13 4. Place the correct symbol, < or >, between the numbers. 0.031 > 0.00987 5. Place the correct symbol, < or >, between the numbers. 0.02883 < 0.0305 Solutions on p. S-8.

denominator.

27 ten-thousandths

.0027 eee!

=

PEA 10,000

6

Module 4 e Decimals and Percents

SECTION

4.2 Objective 4.2A

|

Adding and Subtracting Decimals Add and subtract decimals

| Add: 0.326 + 4.8 + 57.23

Note that placing the decimal points on a vertical line ensures that digits of the same place value

+

{5 ees

sis

are added.

| The sum is 62.356. _ Find the sum of 0.64, 8.731, 12, and 5.9.

Arrange the numbers vertically, placing the decimal points on a _ vertical line.

12

_ Add the numbers in each column.

0.64 Sa oll 12.

_ Write the decimal point in the sum directly below the decimal points in the addends.

te PITA

_ Subtract and check: 31.642 — 8.759

2 | 10

B\Y

|

By

Note that placing the decimal points on a vertical _ line ensures that digits of the same place value are subtracted. | The difference is 22.883.

Check:

Subtrahend + Difference = Minuend

8.759 + 22.883 31.642

y

212 elelsls:

Section 4.2 ¢ Adding and Subtracting Decimals

i Subtract and check: 5.4 —

1.6832

_ Insert zeros in the minuend so that it has the same

\

7

;

5.4000

number of decimal places as the subtrahend.

— 1.6832

iSubtract and then check. 4

_ Check:

139

910

1.6832

Z.AQOO

5.4000

3.7168

Apply the Concept Figure | shows the average price of a movie theater ticket in 1989, 1999, and 2009. Find the increase in price from 1989 to 2009.

= nn(=)

SOLUTION

To find the increase in price, subtract the price in 1989 ($3.99) from the price in 2009 ($7.50).

dollars) (in Price

7.50 SE) 3.51

1989

Figure 1

1999

2009

Average Price of a

From 1989 to 2009, the average price of a

Movie Theater Ticket

movie theater ticket increased by $3.5]

SOITESWWiePaLOORMAG CHE

The sign rules for adding and subtracting decimals are the same rules used to add and subtract integers.

Take Note >

| Simplify: —36.087 + 54.29

Recall that the absolute value

j

of a number is the distance

| The signs of the addends are different. Subtract the smaller absolute value from the larger

from zero to the number on the number line. The absolute

value of a number is a positive number or zero.

|54.29| = 54.29 |—36.087|

= 36.087

' absolute value.

{

| 54.29 — 36.087 = 18.203

' Attach the sign of the number with the larger absolute value.

| [54.29] > |-36.087| | The sum is positive.

—36.087 + 54.29 = 18.203

Recall that the opposite or additive inverse of n is —n, and the opposite of —n is n. To find the opposite of a number, change the sign of the number.

| Simplify: —2.86 — 10.3 iRewrite subtraction as addition of

Sesto =

NOS

|the opposite. The opposite of 10.3 _is = 03)

= —2.86 + (—10.3)

|The signs of the addends are the same.

_Add the absolute values of the numbers.

| Attach the sign of the addends.

1386

8

Module 4 e Decimals and Percents

Focus on adding and subtracting decimals a. Add: 35.8 + 182.406 + 71.0934 b. What is —251.49 more than —638.7? ce. Subtract and check: 73 — 8.16

| SOLUTION | a.

il, i

35.8 182.406

|

X 8 240

(32.41)(7.6). It is close to the

2 decimal places | decimal place

32.41

XO

S10)

240 1s an estimate of

11

3 decimal places

246.316

| Multiply: 0.061 (0.08)

actual product, 246.316. 0 i

a

0.061 x 0.08 0.00488

3 decimal places 2 decimal places 5 decimal places

* Insert two zeros between the 4 and the decimal point so that there are 5 decimal places in the product

To multiply a decimal by a power of 10 (10, 100, 1000, . . .), move the decimal point to the right the same number of places as there are zeros in the power of 10.

PrISIS 10

= 27.935

i

\A

| zero

| decimal place

PaJISP333) = 100

= 279.35

2 Zeros

ip Spy

2 decimal places

1000

= 2793.5

ae

Say

3 zeros

3 decimal places

DesIfSPBis\ 10,000

= 27,935.

a

ae}

4 zeros

4 decimal places

PIED © 100,000

—

ey

7 9350: SS

5 zeros

* A zero must be inserted before the decimal point.

5 decimal places

Note that if the power of 10 is written in exponential notation, the exponent indicates how many places to move the decimal point.

21935 -10' = 27.935 \A

| decimal place

2.195 ¢ 10? = 279.35 2 decimal places

251935." 10? = 2793.5 Se

3 decimal places

PAC Boe 10* = 27,935. al

4 decimal places

ZAD3S * 10° = 279,350. 7

5 decimal places

12

Module 4 « Decimals and Percents

| Find the product of 64.18 and 10°. | 64.18 - 10° =

64,180

¢ The exponent on 10 is 3. Move the decimal point in 64.18 three places to the right.

The sign rules for multiplying decimals are the same rules used to multiply integers. The product of two numbers with the same sign is positive. The product of two numbers with different signs is negative.

Multiply: (—3.2)(—0.008) (—3.2)(—0.008)

= 0.0256

* The signs are the same lhe product is positive. Multiply the absolute values of the numbers

|Focus on multiplying decimals

| Multiply: 0.00073(0.052) _ SOLUTION 0.00073

x

Recall that division is defined

as multiplication by the

To write a percent as a fraction, remove the percent sign and multiply by 799.

reciprocal. Therefore,

Sore

Pine

13%

multiplying by 799 is equiva-

1 ee ii) =100

13 X

=

100

lent to dividing by 100. ||

| Focus on writing a percent as a decimal and as a fraction | Write each percent as a decimal and as afraction.

a. 120%

| SOLUTION | 2s

|

IDG

=

120 « OO

120% = 120 x i bo 43%

=

11

hy AP Ok

00

100. 5

= 43" < 0:01=

010438

1 4.3% = 4.3 x —— 100 sepa 10. 100 =

43

10°.

x

1

Bene =

100

43

1000

10

* Multiply the fractions.

c. 0.45% = 0.45 X 0.01 = 0.0045 |

O45

i

7r—10'45

100

eee 20

9

2000

100

ase

100

20

¢ Multiply the fractions.

b. 4.3%

c. 0.45%

Section 4.5 ¢ Introduction to Percents

27

| Check your understanding 1 | Write each percent as a decimal and as afraction. | SOLUTION

See pages S-5-S-6.

pd

Re

eral

a. 125%

|

be 0035,

4

b. 8.5% 17

——

200

ec. 0.25%

2.00025:

l

——

400

-

| Focus on writing a percent as a fraction |

;

Q;

:

| Write 163% as a fraction.

| SOLUTION |

y)

D

iwh6=lo8—=

a

cie

x

sk #100

|

50

1

50

|

|Check your understanding 2 | Write 335% as a fraction. |

| SOLUTION

See page S-6.

L

]

Objective 4.5A Practice 1. Write 36% as a decimal and as a fraction. 2. Write 6.2% as a decimal and as a fraction.

0.36, = 0.062, ana

3. Write 0.25% as a decimal and as a fraction.

0.0025, Aa

4. Write 125% as a fraction. 2,

5. Write 4—% as a fraction. 7

70

Solutions on p. S-14.

Objective 4.5B

Write a decimal or a fraction as a percent A decimal or a fraction can be written as a percent by multiplying by 100%. | Write 0.37 as a percent.

037°

=

037 x 100%

=

37%

Move the decimal point two places to the right. Then write the percent sign.

When changing a fraction to a percent, if the fraction can be written as a terminating decimal, the percent is written in decimal form. If the decimal representation of the fraction is a repeating decimal, the answer is written with a fraction.

28

Module 4 ¢ Decimals and Percents

Take Note >

' Write gas a percent.

The decimal form of 4 terminates.

13 3 100% G22 vee 8.8 1 | 300% 8 = 37.5%

0.375 8) 3.000

=A

60

=56 40

3 . 3 =

(2.375 is a terminating decimal

¢ The answer is written in decimal form.

=40 0)

oy we ll

Write ¢ as a percent.

bd

Take Note >

0.166 6)1.000 6 40 =36 40 =36

4

2 100% xX ——

—=—

6

The decimal form of é repeats.

6 _ 100%

¢—

|

|

= 0.16 is a repeating decimal.

6

5 = 16-%

* The answer is written with a fraction

3

_ Focus on writing a decimal or a fraction as a percent | a. Write 0.015 and 2.3 as percents. | b. Write ioas a percent. ae) c. Write 3 as a percent.

| SOLUTION | a. 0.015 = 0.015 X 100%

|

= lat 2.3 = 2.3 X 100% == EY

19 . 100%

* 80

1900%

|

80 =

|

Des

ne?)

Claes

3

23.75%

* Write the answer in decimal form

100% a

3

_ 200%

1

3 =

2

007%

« Write the answer with a fraction

_ Check your understanding 3 |a. Write 0.048 and 3.6 as percents.

ess

’

| b. Write 7¢ as a percent. | c. Write 2 as a percent. |SOLUTION

See page S-6.

a. 4.8%, 360%

Objective 4.5B Practice

1. Write 0.73 as a percent. 73% 2. Write 1.012 as a percent. 101.2% 3. Write oe asapercent. 85%

b. 31.25%

]

Cumoomec

3

Section 4.6 ¢ Radical Expressions and Real Numbers 4. Write : asapercent.

29

225%

:

5. Write i as apercent.

|

58%

Solutions on p. S-14.

SECTION

Radical Expressions and Real Numbers Objective 4.6A

Find the square root of a perfect square Recall that the square of a number is equal to the number multiplied times itself.

3° =3:3=9 The square of an integer is called a perfect square.

9 is a perfect square because 9 is the square of 3: 37 = 9. The numbers

1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are perfect squares.

jee

a

3° = 9 A

1G

5? = 25 GAA A= AG

$= 64 Larger perfect squares can be found by squaring 11, squaring 12, squar-

ing 13, and so on.

O81

107 = 100

Note that squaring the negative integers results in the same

i=]

list of numbers.

(A2)= 4

(-3 =9 (—4)? = 16, and so on.

Perfect squares are used in simplifying square roots. The symbol for square root is a

Square Root A square root of a positive number x is a number whose square is x. If

ars then Vx =a.

EXAMPLE

The expression V/9, read “the square root of 9,” is equal to the number that, when squared, is equal to 9.

Since 3? = 9, V9 = 3.

30

Module 4 « Decimals and Percents

Every positive number has two square roots, one a positive number and one a negative

number. The symbol VV is used to indicate the positive square root of a number. When the negative square root of a number is to be found, a negative sign is placed in front of the square root symbol. For example,

/9=3

and

-V9=-3

The square root symbol, Wy is also called a radical. The number under the radical is

called the radicand. In the radical expression V9, 9 is the radicand.

| Simplify: V49 | V49 =7

Take Note >

¢ \/49 is equal to the number that, when squared, equals 49. 77 = 49.

| Simplify: 6 + 4/9

The radical is a grouping symbol. Therefore, when simplifying numerical expressions, simplify the radicand as part of Step | of the Order of Operations

LG

* Simplify V9

6c

AV 6

=6+

=

12

* Use the Order of Operations Agreement.

18

Agreement.

| Focus on simplifying perfect square radical expressions 4 |

Simplify: a. V121

bb. , ee

c. V36 — 9V/4

|

| SOLUTION

| a. Since 11? = 121,V121 = 11. Aran b. Since (2) OLE Wee Wo ei Gr |

Lc, \/46 ONeo) |

SoeGeilWe) |

= 6 + (-18)

|

—|2 a

Check your understanding 1

1 Simplify: a. —V/144—b. 4 |5 | | SOLUTION

See page S-6.

—1|2 aaa

c. 4V/16 — V9

_ Focus on evaluating an expression | Evaluate 6V/ab for a = 2 and b = 8. |

| | | | | |

SOLUTION 6V ab

6V 2-8 = 6V/16

= 6(4) = 04

iY

i =

10

ens

Section 4.6 ¢ Radical Expressions and Real Numbers

31

| Check your understanding 2 | Evaluate SV a + b for a = 17 and b = 19.

|

| SOLUTION

See page S-6.

30

Objective 4.6A Practice . Simplify: By

. Simplify: V81

|

9

. Simplify: aa

N/25, 7

. Simplify: V144 + 3V/9 21 . Evaluate 7V x + y for x = 34andy= 15. 49 by the square root of 16? 3

. What is 7 decreased = NY WwW & Nun Solutions on p. S-14.

Objective 4.6B

Approximate the square root of a natural number If the radicand is not a perfect square, the square root can only be approximated. For ex-

ample, the radicand in the radical expression V2 is 2, and 2 is not a perfect square. The square root of 2 can be approximated to any desired place value.

To the nearest tenth:

V2~

14

(1.4)? = 1.96

To the nearest hundredth:

V2 ~1A4l

(1.41)? = 1.9881

To the nearest thousandth:

V2 ~ 1.414

(1.414)? = 1.999396

To the nearest ten-thousandth:

V2 ~ 1.4142

(1.4142)? = 1.99996164

The square of each decimal approximation gets closer and closer to 2 as the number of place values in the approximation increases. But no matter how many place values are used to approximate Vo), the digits never terminate or repeat. In general, the square root of any number that is not a perfect square can only be approximated. Recall that a rational number has a decimal representation that terminates or repeats. A number such as V2 has a non-terminating, non-repeating decimal representation. Such a number is called an irrational number.

Irrational Numbers

An irrational number is a number whose decimal representation never terminates or repeats.

EXAMPLES OF IRRATIONAL NUMBERS (eee 3. 0.23239233323333...

The rational numbers and the irrational numbers taken together are called the real numbers.

32

Module 4 « Decimals and Percents

Real Numbers

The real numbers are all the rational numbers together with all the irrational numbers.

|

~ Whole Numbers Integers

—~—

'—

Zero

— ~~ Positive Integers (Natural Numbers)

Negative Integers

Rational Numbers

Terminating Decimals

Real Numbers ,

Irrational

Numbers

Fractions That Cannot

Be Reduced to Integers Repeating Decimals

| Focus on approximating a square root Approximate \/11 to the nearest ten-thousandth.

| SOLUTION 11 is not a perfect square. _ Use a calculator to approximate Witte

| VIL © 3.3166 | Check your understanding 3 | Approximate 3V/5 to the nearest ten-thousandth.

| SOLUTION

See page S-6.

6.7082

| Focus on finding a natural number inequality for a square root Between what two whole numbers is the value of V41?

| SOLUTION _ Since the number 41 is between the perfect squares 36 and 49, the value of V41 is | between 1/36 and V 49.

| Because \/36 = 6 and V49 = 7, the value of VAI is between the whole numbers 6 | and’7.

_ This can be written using inequality symbols as 6 < \/ 41 < 7, which is read “the square | root of 41 is greater than 6 and less than 7.” | Use a calculator to verify that Val ~ 6.4, which is between 6 and 7.

|Check your understanding 4 | Between what two whole numbers is the value of V572

| SOLUTION

See pageS-6.

7 < 57 0.203 2. 0.045 > 0.038

84 OWS < Ouls 4. 0.031 > 0.00987

5. 0.02883 < 0.0305

Objective 4.2A

1. —42.1 — 8.6 lI —42.1 + (—8.6) —50.7

2. —9.37 + 3.465 = —5.905 3. 382.9 + (—430.6) = —47.7 4.

—6.82 — 4.793 = —6.82 + (—4.793) AOI

Sy ab ar

—125.41 + 361.55 = 236.14 6. x =y¥

—3.69 — (—1.527) = =3.69 + 1.527 = —2.163

Objective 4.2B

1. sTRATEGY To find the new balance: e Add the amount of the deposit (189.53) to the old balance (347.80). ¢ Subtract the amount of the check (62.89) from the sum. SOLUTION

347.08 + 189.53 = 536.61

536.61 — 62.89 = 473.72 The new balance is $473.72. 2. STRATEGY

To find the markup, substitute 2231.81 for S and 1653.19 for C in the given formula and solve for M. SOLUTION

M=S—C M = 2231.81 — 1653.19 M = 578.62

The markup is $578.62.

Solutions to Module 4. S-9

oF

STRATEGY

To find the equity, replace V by 225,000 and L by 167,853.25 in the given formula and solve for E. SOLUTION

B=

VSL

E = 225,000 — 167,853.25 E = 57,146.75

The equity on the home is $57,146.75.

Objective 4.3A

1.

(269) (224) S15. 12

1.31(—0.006) = —0.00786 6.71210 =.67,100 (2.7) (— 16) (3.04)

i

(—43.2) (3.04)

= —131.328 ab

452(—0.86) = —388.72

Objective 4.3B

(—3.312) + (—0.8) = 4.14

84.66 + (—1.7) = —49.8 OSS 9.407 + 10° = 0.009407 so

eS y

26.22 ore =6.9

Objective 4.3C

20.22, 7

(0.9): = —3,8 ( )

if STRATEGY To find Ramon’s average yards per carry, divide the number of yards gained (162) by the number of carries (26). Round to the nearest hundredth. SOLUTION

162 + 26 ~ 6.23 Ramon averaged approximately 6.23 yards per carry.

S-10

Solutions to Module 4 2.

STRATEGY

To find the amount received for the 400 cans, multiply the weight of the cans (18.75) by the amount paid per pound (0.75). Round to the nearest cent. SOLUTION

18.75 - 0.75 ~ 14.06 The amount received for the 400 cans is $14.06. 3.

STRATEGY

To find the number of miles, divide the total distance traveled (295) by the number of

gallons (12.5). SOLUTION

2955

eet

28.6

You can travel 23.6 mi on one gallon of gas. 4.

STRATEGY

To find the area, substitute 7.8 for L and 4.6 for W in the formula below and solve for A. SOLUTION

A = LW

A = 7.8: 4.6 = 35.88 The area is 35.88 cm’.

Objective 4.4A

1.

D2,

0.55

O22 11)8.0000 = 30 22, 80 =i

30 =)

Solutions to Module

0.639639 111)71.000000 — 666 4 40 339 1 070 moo 710 — 666

440 = 333 1070 =e 71 71

—— = 0.639639... = 0.639 111

;

0.77272 22)17.00000 154 1 60 lot 60

—44 16 AS 0.77272... = 0.772 2p.

Ve

TE 40)3.000

=2 80 200 ~200 0 3 — = 0.075 40

9007.25 4)29.00

-28. 10 =8 20

4. S-11

S-12

Solutions to Module 4

ek

Objective 4.4B

A = %5) 7 S

1. 04=—=—

2. (ee '

100

eee 25

485 97 = —— = 1000 200 75 3 4,395.6 = : 100 an . 0485 psa

]

5. 0052=

1

5) 2-2°-13 13 = 2 1000 2:2-250 250 1

6. 0.00015=

1

15 100,000

gis

¥-20,000 1

Objective 4.4C

1.

0.7

iS 100 40 oy 8 200 26 —— —

40

:

3

20,000

Solutions to Module 4

12 =e 12) eo 55 100 240 © 242 1100 1100 240 242 7100 ~ 1100 ne < 0.22

55

055

> 9 Sh) ss 100 9 495 500 900 900 495 500 900~ 900 5

Oss < 5

wb)

=

3.14

22) 314 7 6.100 2200 2198 700 700. 2200_ 2198 700 ~ 700 224 7 Objective 4.4D

1.

Deine 28. 36in.36

9

28 in. :36 in. = 28:36 = 7:9 28 in. to 36 in. = 28to36 3202.

32

2

l6oz

16

1

= 7to9

320z:160z = 32:16 = 2:1 320z tol60z = 32 tol6 = 2 tol

10ft = = D Sill! 4s

$51,000 12 months $11.05 = 3.4 Ib

;

= $4250/month

$3.25/lb

639 mi = 42.6 mi/gal 15 gal

S-13

S-14

Solutions to Module 4

Objective 4.5A

1. 36% = 36 x 0.01 = 0.36

Aisi SOS

ee ee 100 100. 25 2. 6.2% = 6.2 X 0.01 = 0.062 in 6.2, ee Oe 6.2% = 6.2 X 100 100 1000 500 3. 0.25% = 0.25 X 0.01 = 0.0025

See a es 0 “100 100 10,000

‘pe oe 2 100 eon oes ~ 200° 8 5 pe 7 7. 100; 2 aes CO 200s Objective 4.5B

1. 0.73 = 0.73 X 100% = 73%

2. 1.012 = 1.012 x 100% = 101.2% 7 1700 3

. == 20 = —— 20 ws 100% 0 0 =

4. 2 = 2 x 100% = au 4 4 vf if 5 =pa

Objective 4.6A

100%

=

%‘O =

20

=

1. Since 1? = 1, - V1 = -1. 2. Since 9? = 81,V/81 = 9. 3. V144 — V25 = 12-5=7 4. V144 + 3V9 = 12 +3°3 =12+9=21

6. T= Vb= 7-423

Objective 4.6B

1. V/10 ~ 3.1623

2. 6V15 ~ 23.2379 3. 10/21 ~ 45.8258 4. —12\/53 ~ —87.3613

85%(4)

= 225%

700 ou

5. WVx+y 7V34 + 15 = 7V/49 =7-7=49

8

1 — 58%

400

Solutions to Module 4. S-15

5. 29 is between the perfect squares 25 and 36.

V/25 = 5 and V36 = 6 5 21/20 26 6. 130 is between the perfect squares 121 and 144.

V 121 = 11 and V 144 = 12 lil +n

|Check your understanding 2 | a. Translate “five times the difference between a number and sixty” into a variable expression. Then simplify. | b. Translate “negative four multiplied by the total of ten and the cube of a number” into a variable expression. Then simplify.

| SOLUTION

See page S-2.

a. 5(x —60) = 5x-—

300

ob, —4(10 + n3) = —40 — 4n3

Objective 5.3B Practice

1. Translate “twelve minus a number” into a variable expression. 12 — x 2. by 2. Translate “the quotient of twice a number and nine” into a variable expression. 9 3. Translate “the sum of five-eighths of a number and six” into a variable expression. —x + 6 4. Translate “a number added to the difference between twice the number and four” into a variable expression. Then simplify. (2x — 4) + x = 3x — 4 5. Translate “the sum of one-sixth of a number and four-ninths of the number” into a variable expression. Then simplify. by 4 f = . Solutions on p. S-4.

Objective 5.3C

6

9

18

Translate application problems Many applications in mathematics require that you identify the unknown quantity, assign a variable to that quantity, and then attempt to express other unknown quantities in terms of the variable. _ The height of a triangle is 10 ft longer than the base of the triangle. Express the height of _ the triangle in terms of the base of the triangle. the base of the triangle: b the height is 10 more than the base: ) +

¢ Assign a variable to the base of the triangle. 10

¢ Express the height of the triangle in terms of b.

Focus on translating an application problem a. The length of a swimming pool is 4 ft less than two times the width. Express the length of the pool in terms of the width. b. A banker divided $5000 between two accounts, one paying 10% annual interest and the second paying 8% annual interest. Express the amount invested in the 10% account in terms of the amount invested in the 8% account.

Section 5.3 © Translating Verbal Expressions into Variable Expressions

15

SOLUTION a.

the width of the pool: w the length is 4 ft less than two times the width: 2

— 4

. the amount invested at 8%: x the amount invested at 10%: 5000

—

x

Check your understanding 3 The speed of a new jet plane is twice the speed of an older model. Express the speed of the new model in terms of the speed of the older model. lb. A guitar string 6 ft long was cut into two pieces. Express the length of the shorter piece in terms of the length of the longer piece. a.

| SOLUTION

See page S-2.

a. Let s be the speed of the older model: 2s

b. Let y be the length of the longer piece: 6 — y

Objective 5.3C Practice In the News > U2 Concerts Top Annual Rankings in North America

The Irish rock band U2 performed the most popular concerts on the North American circuit this year. Bruce

Springsteen and the E Street Band came in second, with $28.5 million less in ticket sales. Source: new.music.yahoo.com

if See the news clipping at the left. Express Bruce Springsteen and the E Street Band’s ; ; F Bhieesrla eancals concert ticket sales in terms of U2’s concert ticket sales.'*' / be U's concert ucket sales: T — 28,500,000

. A recent survey conducted by Bankrate.com asked, “If you receive a tax refund, what will you do?” Thirty percent of respondents said they would pay down their debt. Express the number of people who would pay down their debt in terms of the number of people surveyed. Let N be the number of people surveyed: 0.30N . Ina triangle, the measure of the smallest angle is 10 degrees less than one-half the measure of the largest angle. Express the measure of the smallest angle in terms of the measure of the largest angle. Let L be the measure of the largest angle: = —7, — 19

Two cars are traveling in opposite directions at different rates. Two hours later, the cars are 200 mi apart. Express the distance traveled by the faster car in terms of the distance traveled by the slower car. Let x be the distance traveled by the slower car: 200 — x Solutions on p. S-5.

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Solutions to Module 5

SOLUTIONS

TO MODULE

5

Solutions to Check Your Understanding Section 5.1

Check your understanding 1 —4 is the constant term.

Check your understanding 2 a. 2xy + y’

2(—4)(2) + (2)?

era

4)

= 2(—4)(2) + 4 = (—8)(2) + 4 (16). 4 = —12 a+b Guat

Seb oer

5 +(-3)

25

9 C.F he — a el

5+ (-3) pe 2; =

17

ce xr —%xe+yt+2 (2)? — 2[2 + (—4)} + (-3)?

+ v= 2, y= -42=-3

Sg i=io(Sony =8§+4+9

=12+9 = 21

Section 5.2

Check your understanding 1 hy. Bi = 21D = Syke (Op

=

b. —3y? + 7+ 8y -— 14 =5y

a AD

-7

Check your understanding 2 a. —5(4y’) = —20y?

b. —7(—2a) = 14a =( fi ) 7 Ce

5)

9

Gl)

=

15

a

Check your understanding 3 a. —8(—2a + 7b) = 16a — 56b

b. 3(12x? — x + 8) = 36x” — 3x + 24 c. 3(-—a’ — 6a + 7) = —3a’* — 18a + 21

S-1

S-2

Solutions to Module 5

Check your understanding 4 Fae oii 2(y — 7x) = Sie Dy aitca lle =yt 14x

* Use the Distributive Property * Combine like terms.

b. —2(x — 2y) — (—x + 3y) = —2x + 4y + x - 3y =

S25 ar ay

c. 3y — 2[x — 4(2 — 3y)] = 3y — 2[x — 8 + 12y] = 3y — 2x + 16 — 24y ae = SA Yee Mey

Section 5.3

Check your understanding 1 a. the difference between twice n and the square of n 2n — n* b. the quotient of 7 less than b and 15

y= 1 115)

Check your understanding 2 a.

the unknown number: x the difference between the number and sixty: x — 60

5(x — 60) = 5x — 300 b. the unknown number: n the cube of the number: n° the total of ten and the cube of the number: 10 + n?

—4(10 + n’) = —40 — 4n3 Check your understanding 3 a. the the b. the the

speed speed length length

of the older model: 5 of the new jet plane is twice the speed of the older model: 2s of the longer piece: y of the shorter piece: 6 — y

Solutions to Objective Practice Exercises Objective 5.1A

1. 3a+ 2b

3(2) + 23) =6+6=

12

2. be + (2a)

3(-—4) + (2:2) = -12+4=-3 3.

(b — 2a)? + be

(3 — 2(2))? + 3(—4) = @ — 4)? + 3(—4) = (-1)? + 3(—4) = 1 + 3(-4)

=

(12)

11

4. (d- a) — 3c

Beg

ate Des 5

3h) (3128

Solutions to Module5 3)

Te

1

Ta Blac aie bd)

1 Sesl (= (9) +

3 ==(4))

Objective 5.2A

1 3 SE SSS +se ANB4@)] 9) = -F@ +52 + 12] 3 1 SS =(4) (74) + ae (14 5004)

2a)

6x + 8x = 14x

dg b= Bo —300

oe

=

-50D

Neo = 0

OS a Syn = Si7) Sp Sy) 9a 3 4

%

Sx

Objective 5.2B

1 3

%

7 8

o—

18 24

8

—%

24

x,

21 24

=

11 24

3

(—8y) — 10x + 4x = —3x — 8y

AV3x) = 12%

(4x)2 = 8x 1 =5(-2) = —0.2(10x) = —2x —0.5(—16y)

Objective 5.2C

= 8y

(5 — 3b)7 = 35 — 21b 3(5x° + 2x) = 15x? + 6x

(—3x — 6)5 = —15x — 30 40? — 3x + 5) = 4° — 12x + 20

3 3 9 =4x (20= Gy ey 6y + 8) 8) =k5? —(8b* — 6b + 9) = —8b? + 6b — 9

Objective 5.2D

4x — 2(3x + 8) = 4x — 6x — 16 = —2x —

16

5n — (7 — 2n) =5n —-7+2n=7n—-7

12(y — 2) + 37 = 3y) = 12y — 24 + 21 — 9y

=e 9/30 28 — 2x]

—2[3x + 2(4 — x)] ll

=—I|\x +8] = —2x — 16

S-3

S-4

Solutions to Module 6

§, —3[2x = & + 7)] = =3[2x =x = 7]

= =3[x = 7] = —3x + 21 6, 0,0Sy + 0,02(4 = x) = 0,08x + 0,08 = 0,02x = 0,03x + 0,08 Objective 5.3A

1, «less than 16

nm + (tQnm = 9) Objective 5.3B

1. the unkeowa pumber: x h2 = x

2. the unknown nepbder: ¢ twice the unkeownm number: Dy pe.

Q X

the unkeewn number: 4

=

ive-eighihs of the numer: 3 Ss

4

the unknewa number.» twice the number: 2s

the difference betweem (wae the mummies aint fine: Do — 4

Qe — 4)

+a =MB=— 44+ 5=B-4

SS the unknewr numder:o,

. : h One-kihOf The: nUMDeE a ; : : My four-ninthAs oF theemumberr: 5 I

+

—% 4 =

¢

&

x =

&

&

&

th =—%

Solutions to Module5

Objective 5.3C

1. U2’s concert ticket sales: T E Street Band’s concert ticket sales:

T — 28,500,000

number of people surveyed: NV number of people who would pay down their debt: 0.30N

measure of the largest angle: L

1 measure of the smallest angle: re = 0) distance traveled by the slower car: x

distance traveled by the faster car: 200 — x

S-5

rr -

a

7

>

=

che Viikneh UPer

A

|

7°

et

a) —_

-

Veoe) (aedaetabent 26

ais

2—_ , a

a

‘).

a

:

MODULE

Introduction to Equations SECTION 6.1

Introduction to Equations

Objective 6.1A

Determine whether a given number is a solution of an equation

Objective 6.1B

Solve an equation of the form x

Objective 6.1C

Solve an equation of the form ax = b

Objective 6.1D

Solve basic uniform motion problems

SECTION 6.2

+a=b

Proportions

Objective 6.2A

Solve proportions

Objective 6.2B

Solve application problems using proportions

SECTION 6.3

The Basic Percent Equation

Objective 6.3A

Solve the basic percent equation

Objective 6.3B

Solve percent problems using proportions

Objective 6.3C

Solve application problems

SECTION 6.4

Percent Increase and Percent Decrease

Objective 6.4A

Solve percent increase problems

Objective 6.4B

Solve percent decrease problems

SECTION 6.5

Markup and Discount

Objective 6.5A

Solve markup problems

Objective 6.5B

Solve discount problems

SECTION 6.6 Objective 6.6A

Simple Interest Solve simple interest problems

2

Module 6 e Introduction to Equations

SECTION

6.1 Objective 6.1A

Introduction to Equations Determine whether a given number is a solution of an equation An equation expresses the equality of two mathematical expressions. The expressions can be either numerical or variable expressions.

9 + 3 = 12 3x — 2 = 10 y+4=2y-1

lea

Lee The equation at the right is true if the variable is replaced by 5.

x + 8 = 13 5+ 8= 13

A true equation

The equation is false if the variable is replaced by 7.

7 + 8 = 13

A false equation

A solution of an equation is a number that, when substituted for the variable, results in a true equation. 5 is a solution of the equation x + 8 = 13. 7 is nota solution of the equation Xt Swiss

_ Focus on determining whether a number is a solution of an equation Is —2 a solution of 2x + 5 = x — 3? SOLUTION

Take Note > The Order of Operations Agreement applies when

evaluating 2(—2) + 5 and

x+5=x7r-3 | 2(=2) 5 5) (Gao Sil ae 5)

(2) a3.

(==

* Replace x by —2

AS

¢ Evaluate the numerical expressions

4]

¢ If the results are equal,

—2 is a solution of the

equation. If the results are not equal,

Yes, —2 1s a solution of

—2 is not a

solution of the equation

the equation.

Check your understanding 1 Is 5 a solution of 10x — x* = 3x — 10?

SOLUTION

See page S-1.

No

Objective 6.1A Practice 1. Is3 asolution of y+ 4=7?

Yes No 3. Is 4a solution of 3y — 4 = 2y? Yes 2. Is 2 asolution of 7 — 3n = 2?

4. Is 3 asolution of z7 + 1 = 4 + 3z? 5 lis 3 a solution of 8x — 1 =

12x + 3?

No No

Solutions on p. S-7.

Objective 6.1B

Solve an equation of the form x + a = b To solve an equation means to find a solution of the equation. The simplest equation to solve is an equation of the form variable = constant, because the constant is the solution. The solution of the equation x = 5 is 5 because 5 = 5 is a true equation.

Section 6.1 ¢ Introduction to Equations

Tips for Success > To learn mathematics, you must be an active participant. Listening and watching your professor do mathematics are not enough. Take notes in class, mentally think through every question your instructor asks, and try to answer

it even if you are not called on to do so. Ask questions when you have them. See AIM for Success.

The solution of the equation at the right is 7 because 7 + 2 = 9 isa true equation.

G29

Note that if 4 is added to each side of

3

i2 = 9

bao

theA ae equation x + 2 = 9,A the leat soluti is still7.

Galilee afae: x+6=13

If —5 is added to each side of the

Ma

equation x -+- 2 = 9, the solution is.

x -+ 2+

7+6=13

=,

(—5) = 9 + (—5)

Equations that have the same solution are called equivalent equations. The equations x+2=9,x

+ 6 =

13, and x — 3 = 4 are equivalent equations; each equation has 7 as

its solution. These examples suggest that adding the same number to each side of an equation produces an equivalent equation. This is called the Addition Property of Equations.

Addition Property of Equations The same number can be added to each side of an equation without changing its solution. In symbols, the equation a = b has the same solution as the equation at+c=bte. EXAMPLE

The equation x — 3 = 7 has the same solution as the equation x — 3 + 3 = 7 + 3.

In solving

an

equation,

the goal

is to rewrite

the

given

equation

in the

form

variable = constant. The Addition Property of Equations is used to Femovela erm from

one side of the equation bylldding theopposite OFthattermt6eachSide’ theequation. Focus on solving an equation Take Note > An equation has some properties that are similar to

Solve:

x — 4= 2

SOLUTION

those of a balance scale. For instance, if a balance scale is

x-4=2

in balance and equal weights

x—-—4+4=2+4

are added to each side of the scale, then the balance scale remains in balance. If an

x+0=6

equation is true, then adding

x=6

Check:

the same number to each

© Add4 to each side of the equation

¢ Simplify. © The equation is in the form variable = constant.

x —4=2 6-4

side of the equation produces another true equation.

« The goal is to rewrite the equation in the form variable = constant.

| 2 2=2

¢ A true equation

The solution is 6.

Check your understanding 2 Solve:

26 = y — 14

SOLUTION

See page S-1.

40

Because subtraction is defined in terms of addition, the Addition Property of Equations also makes it possible to subtract the same number from each side of an equation without changing the solution of the equation.

4

Module 6 @ Introduction to Equations

_ Focus on solving an equation with fractions

|| Solve:

SeeTEoeSs)

Ma

| SOLUTION |

See! Vie 4 =

y+

3)

7

2

3)

Ba

a =

y+O=

*

5

The goal is to rewrite the equation in the form variable = constant.

+ Subtract 73 from each side of the equation

4

wakes

cmt

a = A

* Simplify

| ses

A

¢ The equation is in the form variable

= constant

| The solution is - t You should check this solution.

_ Check your understanding 3 1 3 | Solve: (Cp =weiner + 4+ |

| SOLUTION

See page S-1.

=i

Objective 6.1B Practice

1 Solvecandicheck:yi-=o sn 2. Solve and check: -8 =n + 1 3. Solve and check: -9 =5 +x Qe~ 3

4s

Olverandichecke

ye 5

9 —14

=i

5 5. Solve and check: w + 2.932 = 4.801

1.869

Solutions on p. S-8.

Objective 6.1C

Solve an equation of the form ax = b The solution of the equation at the right is 3 because 2 - 3 = 6 is a true equation.

2x = 6 ona e

Note that if each side of 2x = 6 is multiplied by 5, the solution is still 3.

5(2x) = 5-6 10x = 30

: ‘ ee If each side of 2x = 6 is multiplied by ~—4, the solutionis still 3.

2x = 6 (Aion) = (eae eee oy

2°3=6

10-3 = 30 Mors Jetty

The equations 2x = 6, 10x = 30, and —8x = —24 are equivalent equations; each equation has 3 as its solution. These examples suggest that multiplying each side of an equation by the same nonzero number produces an equivalent equation.

Multiplication Property of Equations Each side of an equation can be multiplied by the same nonzero number without changing the solution of the equation. In symbols, if c # 0, then the equation a = b has the same solutions as the equation ac = be. EXAMPLE The equation 3x = 21 has the same solution as the equation 5“3x = i le

Section 6.1 ¢ Introduction to Equations

5

The Multiplication Property of Equations is used to coefficient remove a by multiplying

each side ofthe equation by the reciprocal ofthe coefficient.

t

3

j

ne =9

¢ The goal is to rewrite the equation in the form

Eee We 4 eas anaes

¢ Multiply each side of the equation by 7

i

1-z=12

* Simplity.

|3

z=

variable

= constant.

! :

=

12

* The equation is in the form variable

= constant

| The solution is 12. You should check this solution.

Because division is defined in terms of multiplication, each side of an equation can be divided by the same nonzero number without changing the solution of the equation. | Solve: Take Note > Remember to check the

solution. Check:

6x = 14

())

6

7

14

14 = 14

6x =

: 6x = i

14

Wise

meas

16x

|t

6

14 ¢ The goal is to rewrite the equation in the form variable = constant.

14

7.

* Divide each side of the equation by 6

6

H

7

hes

3

« Simplify.

The equation is in the form variable = constant

eae

} The solution is 3.

Focus on using the Multiplication Property of Equations to solve an equation a. Solve:

Bx —9 = ws

b. Solve:

5x — 9x = 12

SOLUTION

3x a.

—9=

5

a .

=(-9) = : or

4

eS =

ri

* Multiply each side by 5.

—| Pie? The solution is —12.

[ts ons — Whe —4x

—4x

ae

—4

117 =

a

12

12

ee

—4

6) The solution is —3.

* Combine like terms.

* Divide each side by —4.

6

Module6 « Introduction to Equations

Check your understanding 4 a. Solve:

2

ars =6

b. Solve: 4x — 8x = 16 SOLUTION

See page S-1.

ae

b.

—4

Objective 6.1C Practice 1. Solve and check: 2a =0

0

2. Solve and check: a

le)

Be

2

3. Solve and check: 5% =6 4. Solve and check: a

2.95

15 —7.88

—23.246

5. Solve and check: 7d — 4d =9

3

Solutions on pp. S-8—S-9.

Objective 6.1D Take Note > A car traveling in a circle at a constant speed of 45 mph is not in uniform motion because the direction of the car is always changing.

Solve basic uniform motion problems Any object that travels at a constant speed in a straight line is said to be in uniform motion. Uniform motion means that the speed and direction of an object do not change. For instance, a car traveling at a constant speed of 45 mph on a straight road is in uniform motion.

The solution of a uniform motion problem is based on the uniform motion equation d = rt, where d is the distance traveled, r is the rate of travel, and ¢ is the time

spent traveling. For instance, suppose a car travels at SO mph for 3 h. Because the rate (50 mph) and time (3 h) are known, we can find the distance traveled by solving the equation d = rt for d. d=

rt

d = 50(3)

or = 50,1

=3

d = 150 The car travels a distance of 150 mi.

A jogger runs 3 mi in 45 min. What is the rate of the jogger in miles per hour? To find the rate of the jogger, solve the equation d = rt for r. The answer must be in miles per hour and the time is given in minutes. Convert 45 min to hours: 45 min = 2 h= 3 h d= rt 3 =

(§)

4

ed

ai!

4

3 3= wi (=)3 =

t\3

( Fewer Students Take LSATs

This year, 137,444 people took the Law School Admission Test

(LSAT). Three years ago, the LSATs were administered to

2. Due to an increased number of service lines at a grocery store, the average amount of time a customer waits in line has decreased from 3.8 min to 2.5 min. Find the percent decrease. 34.2% 3. Use the news clipping at the left to find the percent decrease in the number of people who took the LSATs in the last three years. 7.1% 4. It is estimated that the value of a new car is reduced by 30% after one year of ownership. Find the value after one year of a car that cost $21,900 new. $15,330 Solutions on pp. S-16—S-17.

148,014 people. Source: Law School Admission Council

Markup and Discount Objective 6.5A

Solve markup problems Cost is the amount a merchandising business or retailer pays for a product. Selling price, or retail price, is the price for which a merchandising business or retailer sells a product to a customer. The difference between selling price and cost is called markup. Markup is added to cost to cover the expenses of operating a business and to provide a profit to the owners.

Markup can be expressed as a percent of the cost, or it can be expressed as a percent of the selling price. Here we present markup as a percent of the cost, which is the most common practice. wae]

| | Selling Price

Cost

The percent markup is called the markup rate, and it is expressed as the markup based on the cost. A diagram is useful for expressing the markup equations. In the diagram at the left, the total length is the selling price. One part of the diagram is the cost, and the other part is the markup.

The Markup Equations Take Note > If C is added to both

sides of the first equation, M = P — C the result is the second equation,

Mi=