1,001 Pre-Calculus

Practice Problems

by Mary Jane Sterling

1,001 Pre-Calculus Practice Problems For Dummies®

Published by: John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, www.wiley.com

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10 9 8 7 6 5 4 3 2 1

Contents at a Glance

Introduction ............................................................................ 1

Part I: The Questions ................................................................ 7

Chapter 1: Getting Started with Algebra Basics ................................................................................ 9

Chapter 2: Solving Some Equations and Inequalities ..................................................................... 15

Chapter 3: Function Basics ................................................................................................................ 21

Chapter 4: Graphing and Transforming Functions ......................................................................... 29

Chapter 5: Polynomials ...................................................................................................................... 37

Chapter 6: Exponential and Logarithmic Functions ....................................................................... 45

Chapter 7: Trigonometry Basics ....................................................................................................... 53

Chapter 8: Graphing Trig Functions ................................................................................................. 61

Chapter 9: Get ting Star ted with Trig Identities .............................................................................. 67

Chapter 10: Continuing with Trig Identities .................................................................................... 73

Chapter 11: Working with Triangles and Trigonometry ................................................................ 79

Chapter 12: Complex Numbers and Polar Coordinates ................................................................. 89

Chapter 13: Conic Sections ................................................................................................................ 97

Chapter 14: Systems of Equations and Inequalities ..................................................................... 103

Chapter 15: Sequences and Series .................................................................................................. 111

Chapter 16: Introducing Limits and Continuity ............................................................................ 117

Part II: The Answers ............................................................ 125

Chapter 17: Answers......................................................................................................................... 127

Index .................................................................................. 527

Table of Contents

Introduction ................................................................. 1

What You’ll Find .............................................................................................. 1

How This Workbook Is Organized ................................................................. 2

Part I: The Questions ............................................................................. 2

Part II: The Answers .............................................................................. 3

Beyond the Book ............................................................................................. 3

What you’ll find online .......................................................................... 3

How to register ....................................................................................... 4

Where to Go for Additional Help ................................................................... 4

Part I: The Questions .................................................... 7

Chapter 1: Getting Started with Algebra Basics . . . . . . . . . . . . . . . . . . .9

The Problems You’ll Work On ....................................................................... 9

What to Watch Out For ................................................................................... 9

Identifying Which System or Systems a Number Belongs To .................. 10

Recognizing Properties of Number Systems .............................................. 10

Simplifying Expressions with the Order of Operations ............................ 11

Graphing Inequalities ....................................................................................12

Using Graphing Formulas .............................................................................13

Applying Graphing Formulas ....................................................................... 13

Chapter 2: Solving Some Equations and Inequalities . . . . . . . . . . . . . .15

The Problems You’ll Work On ..................................................................... 15

What to Watch Out For ................................................................................. 15

Using Interval and Inequality Notation ....................................................... 16

Solving Linear Inequalities ........................................................................... 17

Solving Quadratic Inequalities ..................................................................... 17

Solving Absolute Value Inequalities ............................................................17

Working with Radicals and Fractional Notation ........................................18

Performing Operations Using Fractional Exponents ................................ 18

Factoring Using Fractional Notation ........................................................... 19

Solving Radical Equations ............................................................................ 19

Rationalizing Denominators .........................................................................20

Chapter 3: Function Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21

The Problems You’ll Work On ..................................................................... 21

What to Watch Out For ................................................................................. 21

Using Function Notation to Evaluate Function Values ............................. 22

Determining the Domain and Range of a Function .................................... 22

Recognizing Even Functions ........................................................................ 23

Identifying Odd Functions ............................................................................ 23

1,001 Pre-Calculus Practice Problems For Dummies

vi

Ruling Out Even and Odd Functions ........................................................... 23

Recognizing One-to-One Functions from Given Relations ........................23

Identifying One-to-One Functions from Equations .................................... 25

Recognizing a Function’s Inverse ................................................................25

Determining a Function’s Inverse ................................................................26

Executing Operations on Functions ............................................................26

Performing Function Composition .............................................................. 27

Doing More Function Composition ............................................................. 27

Using the Difference Quotient ......................................................................28

Chapter 4: Graphing and Transforming Functions . . . . . . . . . . . . . . . . .29

The Problems You’ll Work On ..................................................................... 29

What to Watch Out For ................................................................................. 29

Functions and Their Inverses ......................................................................30

Sketching Quadratic Functions from Their Equations ............................. 30

Writing Equations from Graphs of Parabolas ............................................31

Investigating and Graphing Radical Functions .......................................... 32

Investigating Absolute Value Functions .....................................................33

Investigating the Graphs of Polynomial Functions ................................... 33

Investigating Rational Functions ................................................................. 34

Transformation of Functions ....................................................................... 34

Transforming Selected Points Using Functions ......................................... 34

Sketching Graphs Using Basic Functions and Transformations ..............35

Sketching More Graphs Using Basic Functions and Transformations ...... 35

Chapter 5: Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37

The Problems You’ll Work On ..................................................................... 37

What to Watch Out For ................................................................................. 37

Using Factoring to Solve Quadratic Equations .......................................... 38

Solving Quadratic Equations by Using the Quadratic Formula ............... 38

Using Completing the Square to Solve Quadratic Equations ................... 39

Solving Polynomial Equations for Intercepts ............................................. 39

Using Factoring by Grouping to Solve Polynomial Equations .................40

Applying Descartes’s Rule of Signs ............................................................. 40

Listing Possible Roots of a Polynomial Equation ...................................... 40

Dividing Polynomials ....................................................................................41

Using Synthetic Division to Divide Polynomials ........................................41

Checking for Roots of a Polynomial by Using Synthetic Division .............41

Writing Polynomial Expressions from Given Roots ..................................42

Writing Polynomial Expressions When Given Roots and a Point ............ 42

Graphing Polynomials ...................................................................................43

Writing Equations from Graphs of Polynomials ........................................43

Chapter 6: Exponential and Logarithmic Functions . . . . . . . . . . . . . . . .45

The Problems You’ll Work On ..................................................................... 45

What to Watch Out For ................................................................................. 45

Understanding Function Notation ...............................................................46

Graphing Exponential Functions ................................................................. 46

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vii

Solving Exponential Equations .................................................................... 47

Using the Equivalence bx = y ⇔ log y = x to Rewrite Expressions .........48

b

Using the Equivalence log y = x ⇔ bx = y to Rewrite Expressions .........48

b

Rewriting Logarithmic Expressions ............................................................48

Rewriting Logs of Products and Quotients as Sums and Differences ..... 49

Solving Logarithmic Equations .................................................................... 49

Applying Function Transformations to Log Functions ............................. 50

Applying Logarithms to Everyday Life ....................................................... 51

Chapter 7: Trigonometry Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53

The Problems You’ll Work On ..................................................................... 53

What to Watch Out For ................................................................................. 53

Using Right Triangles to Determine Trig Functions ..................................54

Solving Problems by Using Right Triangles and Their Functions ........... 55

Working with Special Right Triangles .........................................................56

Changing Radians to Degrees ......................................................................57

Changing Degrees to Radians ......................................................................57

Finding Angle Measures (in Degrees) in Standard Position ..................... 57

Determining Angle Measures (in Radians) in Standard Position ............ 58

Identifying Reference Angles ........................................................................ 58

Determining Trig Functions by Using the Unit Circle ............................... 58

Calculating Trig Functions by Using Other Functions

and Terminal Side Positions ..................................................................... 59

Using the Arc Length Formula ..................................................................... 59

Evaluating Inverse Functions ....................................................................... 60

Solving Trig Equations for x in Degrees .....................................................60

Calculating Trig Equations for x in Radians ...............................................60

Chapter 8: Graphing Trig Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61

The Problems You’ll Work On ..................................................................... 61

What to Watch Out For ................................................................................. 61

Recognizing Basic Trig Graphs .................................................................... 62

Graphing Sine and Cosine .............................................................................64

Applying Function Transformations to Graphs of Trig Functions ..........64

Writing New Trig Functions Using Transformations ................................ 64

Graphing Tangent and Cotangent ...............................................................65

Interpreting Transformations of Trig Functions ....................................... 65

Graphing Secant and Cosecant .................................................................... 66

Interpreting Transformations from Function Rules ..................................66

Chapter 9: Get ting Star ted with Trig Identities . . . . . . . . . . . . . . . . . . .67

The Problems You’ll Work On ..................................................................... 67

What to Watch Out For ................................................................................. 67

Proving Basic Trig Identities ........................................................................ 68

Returning to Basic Sine and Cosine to Solve Identities ............................ 69

Using Multiplication by a Conjugate to Solve Identities ...........................70

Solving Identities After Raising a Binomial to a Power ............................. 70

Solving Identities After Factoring out a Common Function ..................... 70

1,001 Pre-Calculus Practice Problems For Dummies

viii

Solving Identities After Combining Fractions ............................................71

Performing Algebraic Processes to Make Identities More Solvable .......71

Chapter 10: Continuing with Trig Identities . . . . . . . . . . . . . . . . . . . . . . .73

The Problems You’ll Work On ..................................................................... 73

What to Watch Out For ................................................................................. 73

Using Identities That Add or Subtract Angle Measures ...........................74

Confirming Double-Angle Identities ............................................................74

Using Identities That Double the Size of the Angle ................................... 74

Confirming the Statements of Multiple-Angle Identities ...........................74

Creating Half-Angle Identities from Double-Angle Identities ..................... 75

Creating a Half-Angle Identity for Tangent ................................................. 75

Using Half-Angle Identities to Simplify Expressions ..................................75

Creating Products of Trig Functions from Sums and Differences ........... 75

Using Product-to-Sum Identities to Evaluate Expressions .......................75

Using Sum-to-Product Identities to Evaluate Expressions .......................76

Applying Power-Reducing Identities ........................................................... 76

Using Identities to Determine Values of Functions at Various Angles ...... 76

Working through Identities Using Multiple Methods ................................ 77

Chapter 11: Working with Triangles and Trigonometry . . . . . . . . . . . .79

The Problems You’ll Work On ..................................................................... 79

What to Watch Out For ................................................................................. 79

Applying the Law of Sines to Find Sides .....................................................80

Utilizing the Law of Sines to Find Angles .................................................... 80

Using the Law of Sines for Practical Applications .....................................81

Investigating the Ambiguous Case of the Law of Sines ............................. 81

Determining All Angles and Sides of a Triangle ......................................... 82

Finding Side Measures by Using the Law of Cosines ................................ 82

Using the Law of Cosines to Determine an Angle ...................................... 82

Applying the Law of Cosines to Real-World Situations ............................. 83

Finding Areas of Triangles by Using the Sine ............................................. 83

Applying the Trig Formula for Area of a Triangle .....................................84

Using the Trig Formula for Area in Various Situations ............................. 84

Solving Area Problems Needing Additional Computations ...................... 85

Finding Areas of Triangles by Using Heron’s Formula ..............................86

Applying Heron’s Formula ............................................................................ 86

Practical Applications Using Heron’s Formula .......................................... 87

Tackling Practical Applications by Using Triangular Formulas ..............87

Chapter 12: Complex Numbers and Polar Coordinates . . . . . . . . . . . . .89

The Problems You’ll Work On ..................................................................... 89

What to Watch Out For ................................................................................. 89

Writing Powers of i in Their Simplest Form ...............................................90

Adding and Subtracting Complex Numbers ...............................................90

Multiplying Complex Numbers .................................................................... 91

Using Multiplication to Divide Complex Numbers .................................... 91

Solving Quadratic Equations with Complex Solutions ............................. 92

Graphing Complex Numbers ........................................................................ 92

Identifying Points with Polar Coordinates ................................................. 94

Table of Contents

ix

Identifying Points Whose Angles Have Negative Measures ..................... 94

Converting Polar to Rectangular Coordinates ...........................................95

Converting Rectangular to Polar Coordinates ...........................................95

Recognizing Polar Curves .............................................................................96

Chapter 13: Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .97

The Problems You’ll Work On ..................................................................... 97

What to Watch Out For ................................................................................. 97

Identifying Conics from Their Equations .................................................... 98

Rewriting Conic Equations in Standard Form ............................................98

Writing Equations for Circles ....................................................................... 98

Determining Foci and Axes of Symmetry of Parabolas ............................. 99

Finding the Vertices and Directrixes of Parabolas .................................... 99

Writing Equations of Parabolas ................................................................. 100

Determining Centers and Foci of Ellipses ................................................. 100

Writing Equations of Ellipses ..................................................................... 100

Determining Asymptotes of Hyperbolas .................................................. 101

Writing Equations of Hyperbolas .............................................................. 101

Changing Equation Format from Trig Functions to Algebraic ............... 101

Changing Equation Format from Algebraic to Trig ................................. 102

Chapter 14: Systems of Equations and Inequalities . . . . . . . . . . . . . . .103

The Problems You’ll Work On ................................................................... 103

What to Watch Out For ............................................................................... 104

Using Substitution to Solve Systems of Linear Equations

with Two Variables .................................................................................. 104

Using Elimination to Solve Systems of Linear Equations

with Two Variables .................................................................................. 104

Solving Systems of Equations Involving Nonlinear Functions ............... 105

Solving Systems of Linear Equations ........................................................ 105

Solving Systems of Linear Equations with Four Variables ..................... 106

Graphing Systems of Inequalities .............................................................. 106

Decomposition of Fractions ....................................................................... 107

Operating on Matrices ................................................................................ 107

Changing Matrices to the Echelon Form .................................................. 108

Solving Systems of Equations Using Augmented Matrices .................... 108

Solving Systems of Equations Using the Inverse of the

Coefficient Matrix ..................................................................................... 109

Applying Cramer’s Rule to Solve Systems of Equations ......................... 110

Chapter 15: Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . .111

The Problems You’ll Work On ................................................................... 111

What to Watch Out For ............................................................................... 111

Finding Terms of Sequences ...................................................................... 112

Determining Rules for Sequences .............................................................. 112

Working with Recursively Defined Sequences ......................................... 112

Adding Terms in an Arithmetic Series ...................................................... 113

Summing Terms of a Series ........................................................................ 113

Finding Rules and Summing Terms of a Series ........................................ 113

Calculating the Sum of a Geometric Series .............................................. 114

1,001 Pre-Calculus Practice Problems For Dummies

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Determining Formulas and Finding Sums ................................................. 114

Counting Items by Using Combinations ................................................... 114

Constructing Pascal’s Triangle .................................................................. 115

Applying Pascal’s Triangle ......................................................................... 115

Utilizing the Binomial Theorem ................................................................. 115

Chapter 16: Introducing Limits and Continuity . . . . . . . . . . . . . . . . . . .117

The Problems You’ll Work On ................................................................... 117

What to Watch Out For ............................................................................... 117

Determining Limits from Graphs ............................................................... 118

Determining One-Sided Limits ................................................................... 119

Determining Limits from Function Values ................................................ 120

Determining Limits from Function Rules .................................................. 121

Applying Laws of Limits .............................................................................. 122

Investigating Continuity .............................................................................. 123

Part II: The Answers ................................................. 125

Chapter 17: Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .127

Index ....................................................................... 527

Introduction

P re-calculus is a rather difficult topic to define or describe. There’s a little bit of this, a lot of that, and a smattering of something else. But you need the mathematics considered to be pre-calculus to proceed to what changed me into a math major: calculus! Yes, believe it or not, I started out as a biology major — inspired by my high school biology teacher. Then I got to the semester where I was taking invertebrate zoology, chemistry, and calculus (yes, all at the same time). All of a sudden, there was a bright light! An awakening! “So this is what mathematics can be!” Haven’t turned back since. Calculus did it for me, and my great preparation for calculus made the adventure wonderful.

Pre-calculus contains a lot of algebra, some trigonometry, some geometry, and some analytic geometry. These topics all get tied together, mixed up, and realigned until out pops the mathematics you’ll use when working with calculus. I keep telling my calculus students that

“calculus is 60 percent algebra.” Maybe my figures are off a bit, but believe me, you can’t succeed in calculus without a good background in algebra (and trigonometry). The geometry is very helpful, too.

Why would you do 1,001 pre-calculus problems? Because practice makes perfect. Unlike other subjects where you can just read or listen and absorb the information sufficiently, mathematics takes practice. The only way to figure out how the different algebraic and trigonometric rules work and interact with one another, or how measurements in degrees and radians fit into the big picture, is to get into the problems — get your hands dirty, so to speak. Many problems given here may appear to be the same on the surface, but different aspects and challenges have been inserted to make them unique. The concepts become more set in your mind when you work with the problems and have your solutions confirm the properties.

What You’ll Find

This book contains 1,001 pre-calculus problems, their answers, and complete solutions to each. There are 16 problem chapters, and each chapter has many different sets of questions. The sets of questions are sometimes in a logical, sequential order, going from one part of a topic to the next and then to the next. Or sometimes the sets of questions represent the different ways a topic can be presented. In any case, you’ll get instructions on doing the problems. And sometimes you’ll get a particular formula or format to use. Feel free to refer to other mathematics books, such as Yang Kuang and Elleyne Kase’s Pre-Calculus For Dummies, my Algebra II For Dummies, or my Trigonometry For Dummies (all published by Wiley) for even more ideas on how to solve some of the problems.

Instead of just having answers to the problems, you’ll find a worked-out solution for each and every one. Flip to the last chapter of this book for the step-by-step processes needed to solve the problems. The solutions include verbal explanations inserted in the work where necessary. Sometimes, the explanation may offer an alternate procedure. Not everyone does algebra and trigonometry problems exactly the same way, but this book tries to provide the most understandable and success-promoting process to use when solving the problems presented.

2 1,001 Pre-Calculus Practice Problems For Dummies

How This Workbook Is Organized

This workbook is divided into two main parts: questions and answers. But you probably figured that out already.

Part I: The Questions

The chapters containing the questions cover many different topics:

✓ Review of basic algebraic processes: Chapters 1 and 2 contain problems on basic algebraic rules and formulas, solving many types of equations and inequalities, and interpreting and using very specific mathematical notation correctly. They thoroughly cover functions and function properties, with a segue into trigonometric functions.

✓ Graphing functions and transformations of functions: Functions and properties of functions are a big part of pre-calculus and calculus. You work with operations on functions, including compositions. These operations translate into transformations. And all this comes together when you look at the graphs of the functions. Transformations of functions help you see the similarities and differences in basic mathematical models — and the practice problems help you see how all this can save you a lot of time in the end.

✓ Polynomial functions: Some of the more familiar algebraic functions are the polynomials. The graphs of polynomials are smooth, rolling curves. Their characteristics include where they cross the axes and where they make their turns from moving upward to moving downward or vice versa. You get to practice your equation-solving techniques when determining the x-intercepts and y-intercept of polynomial functions.

✓ Exponential and logarithmic functions: You’re not in Kansas anymore, so it’s time to leave the world of algebraic functions and open your eyes to other types: exponential and logarithmic, to name two. You practice the operations specific to these types of functions and see how one is the inverse of the other. The applications of these functions are closer to real-world than most others in earlier chapters.

✓ Trigonometric functions: Trigonometric functions take being different one step further. You’ll see how the input values for these functions have to be angle measures, not just any old numbers. The trig functions have their own rules, too, and lots of ways to interact, called identities. Solving trig identities helps you prepare for that most exciting process in calculus, where you get to find the area under a trigonometric curve. So keep your eye on that goal! And the trig applications are some of my favorite — all so easy to picture (and draw) and to solve.

✓ Complex numbers and polar coordinates: Complex numbers were created; no, they aren’t real or natural. Mathematicians needed to solve problems whose solutions were the square roots of negative numbers, so they adopted the imaginary number i to accomplish this task. Performing operations on complex numbers and finding complex solutions are a part of this general arena. Polar coordinates are a way of graphing curves by using angle measures and radii. You open up a whole new world of curves when you practice with these problems dealing with polar graphs.

✓ Conic sections: A big family of curves belongs in the classification of conics. You find the similarities and differences between circles, ellipses, hyperbolas, and parabolas.

Exercises have you write the standard forms of the equations so you can better determine individual characteristics and create reasonable sketches of the graphs of the curves.

Introduction

3

✓ Systems of equations and inequalities: When you have two or more statements or equations and want to know whether any solutions are common to both or all of them at the same time, you’re talking about solving systems. The equations can be linear, quadratic, exponential, and so on. You’ll use algebraic techniques and also use matrices to solve some of the linear systems.

✓ Sequences and series: Some problems cover the basic arithmetic and geometric series. And, as a huge bonus, you’ll use the binomial theorem and Pascal’s triangle to expand binomials to fairly high powers.

✓ Limits and continuity: The basics of limits and continuity are covered — analytically and graphically. This point is actually the launching spot for calculus — where precalculus finishes, calculus begins.

Part II: The Answers

This part provides not only the answers to all the questions but also explanations of the answers. So you get the solution, and you see how to arrive at that solution.

Beyond the Book

This book is chock-full of pre-calculus goodness, but maybe you want to track your progress as you tackle the problems. Or maybe you’re stuck on a few particularly challenging types of pre-calculus problems and wish they were all presented in one place where you could methodically make your way through them. No problem! Your book purchase comes with a free one-year subscription to all 1,001 practice problems online. Track your progress and view personalized reports that show where you need to study the most. And then do it.

Study what, where, when, and how you want.

What you’ll find online

The online practice that comes free with this book offers you the same 1,001 questions and answers that are available here, presented in a multiple-choice format. The beauty of the online problems is that you can customize your online practice to focus on the topic areas that give you the most trouble. So if you aren’t yet a whiz at exponential and logarithmic functions, you can select these problem types and BAM! — just those types of problems appear for your solving pleasure. Or, if you’re short on time but want to get a mixed bag of a limited number of problems, you can plug in the quantity of problems you want to practice and that many — or few — of a variety of pre-calculus problems appears. Whether you practice a couple hundred problems in one sitting or a couple dozen, or whether you focus on a few types of problems or practice every type, the online program keeps track of the questions you get right and wrong so that you can monitor your progress and spend time studying exactly what you need.

You can access this online tool by using a PIN code, as described in the next section. Keep in mind that you can create only one login with your PIN. After the PIN is used, it’s no longer valid and is nontransferable. So you can’t share your PIN with other users after you’ve established your login credentials.

4 1,001 Pre-Calculus Practice Problems For Dummies

This book also comes with an online Cheat Sheet full of frequently used formulas and more goodies. Check it out for free at www.dummies.com/cheatsheet/1001precalculus. (No PIN is required. You can access this info before you even register.) How to register

Purchasing this book entitles you to one year of free access to the online, multiple-choice version of all 1,001 of this book’s practice problems. All you have to do is register. Just follow these simple steps:

1. Find your PIN code.

• Print book users: If you purchased a hard copy of this book, turn to the front of the book to find your PIN.

• E-book users: If you purchased this book as an e-book, you can get your PIN by registering your e-book at www.dummies.com/go/getaccess. Go to this website, find your book and click it, and then answer the security question to verify your purchase. Then you’ll receive an e-mail with your PIN.

2. Go to http://learn.dummies.com.

3. Enter your PIN.

4. Follow the instructions to create an account and establish your own login information.

That’s all there is to it! You can come back to the online program again and again — simply log in with the username and password you choose during your initial login. No need to use the PIN a second time.

If you have trouble with the PIN or can’t find it, please contact Wiley Product Technical Support at 877-762-2974 or http://support.wiley.com.

Your registration is good for one year from the day you activate your PIN. After that time frame has passed, you can renew your registration for a fee. The website gives you all the important details about how to do so.

Where to Go for Additional Help

The written directions given with the individual problems are designed to tell you what you need to do to get the correct answer. Sometimes the directions may seem vague if you aren’t familiar with the words or the context of the words. Go ahead and look at the solution to see whether it helps you with the meaning. But if the vocabulary is still unrecognizable, you may want to refer to Pre-Calculus For Dummies, Algebra II For Dummies, or Trigonometry For Dummies, all published by the fine folks at Wiley.

You may not be able to follow a particular solution from one step to the next. Is something missing? This book is designed to provide you with enough practice to become very efficient in pre-calculus topics, but it isn’t intended to give the step-by-step explanation of how and why each step is necessary. You may need to refer to the books listed in the preceding Introduction

5

paragraph or their corresponding workbooks to get more background on a problem or to understand why a particular step is taken in the solution of the problem.

Some pre-calculus topics are sometimes seen as being a bunch of rules without a particular purpose. Why do you have to solve for the exponent of that equation? Where will you use the fact that tan2 x + 1 = sec2 x? All these questions are more apparent when you see them tied together and when more background information is available. Don’t be shy about seek-ing out that kind of information. And all this practice will pay off when you begin your first calculus experience. It may even be with Mark Ryan’s Calculus For Dummies!

6 1,001 Pre-Calculus Practice Problems For Dummies

Part I

The Questions

Visit www.dummies.com for great (and free!) Dummies content online.

In this part . . .

Y ou find 1,001 pre-calculus problems — many different types

in three different difficulty levels. The types of problems

you’ll find are

✓ Basic algebraic rules and graphs as well as solving

algebraic equations and inequalities (Chapters 1 through 5)

✓ Properties of exponential and logarithmic functions and their

equations (Chapter 6)

✓ Trigonometry basics and solving trig identities (Chapters 7

through 11)

✓ Complex numbers, polar coordinates, and conic sections

(Chapters 12 through 13)

✓ Systems of equations, sequences, and series (Chapters 14

and 15)

✓ Limits and continuity (Chapter 16)

Chapter 1

Getting Started with Algebra Basics

T he basics of pre-calculus consist of reviewing number systems, properties of the number systems, order of operations, notation, and some essential formulas used in coordinate graphs. Vocabulary is important in mathematics because you have to relate a number or process to its exact description. The problems in this chapter reacquaint you with many old friends from previous mathematics courses.

The Problems You’ll Work On

In this chapter, you’ll work with simplifying expressions and writing answers in the following ways:

✓ Identifying which are whole numbers, integers, and rational and irrational numbers

✓ Applying the commutative, associative, distributive, inverse, and identity properties

✓ Computing correctly using the order of operations (parentheses, exponents/powers and roots, multiplication and division, and then addition and subtraction)

✓ Graphing inequalities for the full solution

✓ Using formulas for slope, distance, and midpoint

✓ Applying coordinate system formulas to characterize geometric figures What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working with simplifying expressions and communicating answers, your challenges will be

✓ Distributing the factor over every term in the parentheses

✓ Changing the signs of all the terms when distributing a negative factor

✓ Working from left to right when applying operations at the same level

✓ Assigning points to the number line in the correct order

✓ Placing the change in y over the change in x when using the slope formula

✓ Satisfying the correct geometric properties when characterizing figures 10 Part I: The Questions

7. Which is not an irrational number?

Identifying Which System or

Systems a Number Belongs To

1–10 Identify which number doesn’t belong to the

number system.

8. Which is not an irrational number?

1. Which is not a rational number?

9. Which is not an imaginary number?

2. Which is not a rational number?

10. Which is not an imaginary number?

3. Which is not a natural number?

Recognizing Properties of

4. Which is not a natural number?

Number Systems

11–20 Identify which property of numbers the

equation illustrates.

11.

5. Which is not an integer?

12.

6. Which is not an integer?

Chapter 1: Getting Started with Algebra Basics

11

13.

Simplifying Expressions with

the Order of Operations

21–30 Simplify the expression by using the order of

operations.

14.

21.

15.

22.

16.

23.

17.

24.

18.

19. If

x = 3 and y = x, then y = 3.

25.

20.

26.

12 Part I: The Questions

27.

33.

34.

28.

35.

29.

36.

30.

37.

38.

Graphing Inequalities

31–40 Graph the inequality.

31.

39.

32.

40.

Chapter 1: Getting Started with Algebra Basics

13

48. Find the midpoint of the segment between

Using Graphing Formulas

the points (−5, 2) and (7, −8).

41–50 Solve by using the necessary formula.

41. Find the slope of the line through the

points (−2, 3) and (4, 9).

49. Find the midpoint of the segment between

the points (6, 3) and (−4, −4).

42. Find the slope of the line through the

points (−4, −3) and (−6, 2).

50. Find the midpoint of the segment between

the points

and

.

43. Find the slope of the line through the

points (4, −3) and (4, −7).

Applying Graphing Formulas

51–60 Use an appropriate formula to compute the

indicated value.

44. Find the slope of the line through the

points (−2, −9) and (2, −9).

51. Find the perimeter of triangle ABC, whose

vertices are A (1, 1), B (1, 4), and C (5, 1).

45. Find the distance between the points

(−8, −1) and (−2, 7).

52. Find the perimeter of the parallelogram

DEFG, whose vertices are D (0, 10),

E (9, 13), F (11, 7), and G (2, 4).

46. Find the distance between the points

(0, 16) and (7, −8).

53. Find the center of the rhombus HJKL,

whose vertices are H (0, 3), J (4, 6), K (8, 3),

and L (4, 0).

47. Find the distance between the points

(6, −5) and (−4, 3).

14 Part I: The Questions

54. Determine which type(s) of triangle ABC is 58. Find the area of the parallelogram PQRS

if the vertices are A (1, 1), B (4, 5), and

with vertices P (4, 7), Q (7, 12), R (15, 12),

C (9, −5).

and S (12, 7).

55. Determine which type of triangle ABC is if

59. Find the area of circle A if the endpoints of

the vertices are A (0, 0), B (0, 12), and

its diameter are at (6, 13) and (−8, 21).

C

.

60. Find the area of triangle ABC with vertices

A (0, 0), B (5, 12), and C (14, 0).

56. Find the length of the altitude of triangle

ABC, drawn to side AC, with vertices

A (0, 0), B (5, 12), and C (21, 0).

57. Find the length of the altitude of triangle

DEF, drawn to side DF, with vertices

D (2, 3), E (2, 12), and F (42, 3).

Chapter 2

Solving Some Equations

and Inequalities

T he object of solving equations and inequalities is to discover which number or numbers will create a true statement in the given expression. The main techniques you use to find such solutions include factoring, applying the multiplication property of zero, creating sign lines, finding common denominators, and squaring both sides of an equation. Your challenge is to determine which techniques work in a particular problem and whether you have a correct solution after applying those techniques.

The Problems You’ll Work On

In this chapter, you’ll work with equations and inequalities in the following ways:

✓ Writing inequality solutions using both inequality notation and interval notation

✓ Solving linear and quadratic inequalities using a sign line

✓ Determining the solutions of absolute value equations and inequalities

✓ Taking on radical equations and checking for extraneous roots

✓ Rationalizing denominators as a method for finding solutions

What to Watch Out For

Keep in mind that when solving equations and inequalities, your challenges will include

✓ Factoring trinomials correctly when solving quadratic equations and inequalities

✓ Choosing the smallest exponent when the choices include fractions and negative numbers

✓ Assigning the correct signs in intervals on the sign line

✓ Recognizing the solution x = 0 when a factor of the expression is x

✓ Determining which solutions are viable and which are extraneous

✓ Squaring binomials correctly when working with radical equations

✓ Finding the correct format for a binomial’s conjugate

16 Part I: The Questions

Using Interval and Inequality 66. Describe the graph using inequality notation.

Notation

61–70 Write the expression using the indicated

Illustration by Thomson Digital

notation.

61. Write the expression

in interval

notation.

67. Describe the graph using inequality

notation.

62. Write the expression

in interval

Illustration by Thomson Digital

notation.

68. Describe the graph using interval notation.

63. Write the interval notation

as an

inequality expression.

Illustration by Thomson Digital

64. Write the interval notation

as an

inequality expression.

69. Describe the graph using interval notation.

Illustration by Thomson Digital

65. Describe the graph using inequality

notation.

70. Describe the graph using interval notation.

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 2: Solving Some Equations and Inequalities

17

77.

Solving Linear Inequalities

71–74 Solve the inequality for x.

71.

78.

72.

79.

73.

80.

74.

Solving Absolute Value

Inequalities

81–85 Solve the absolute value inequality for x.

Solving Quadratic Inequalities

81.

75–80 Solve the nonlinear inequality for x.

75.

82.

76.

83.

18 Part I: The Questions

84.

Performing Operations Using

Fractional Exponents

90–94 Simplify the expression.

85.

90.

Working with Radicals and

91.

Fractional Notation

86–89 Change the expression using rules for

fractional exponents.

86. Rewrite the radical expression

with a

92.

fractional exponent.

87. Rewrite the radical expression

with a

93.

fractional exponent.

88. Rewrite the fractional exponent

as a

94.

radical expression.

89. Rewrite the fractional exponent

as a

radical expression.

Chapter 2: Solving Some Equations and Inequalities

19

Factoring Using Fractional

Solving Radical Equations

Notation

101–110 Solve for x. Check for any extraneous

solutions.

95–100 Factor the greatest common factor (GCF)

from each term and write the factored form.

101.

95.

102.

96.

103.

97.

104.

98.

105.

99.

106.

100.

20 Part I: The Questions

107.

114.

108.

115.

109.

116.

110.

117.

Rationalizing Denominators

111–120 Simplify by rationalizing the denominator.

111.

118.

112.

119.

113.

120.

Chapter 3

Function Basics

A function is a special type of rule or relationship. The difference between a function and a relation is that a function has exactly one output value (from the range) for every input value (from the domain). Functions are very useful when you’re describing trends in business, heights of objects shot from a cannon, times required to complete a task, and so on. Functions have some special properties and operations that allow for investigation into what happens when you change the rule.

The Problems You’ll Work On

In this chapter, you’ll work with functions and function operations in the following ways:

✓ Writing and using function notation

✓ Determining the domain and range of different types of functions

✓ Recognizing even and odd functions

✓ Checking on whether a function is one-to-one

✓ Finding inverses of one-to-one functions

✓ Performing the basic operations on functions and function rules

✓ Working with the composition of functions and the difference quotient What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working with functions, your challenges will include

✓ Following the order of operations when evaluating functions

✓ Determining which values need to be excluded from a function’s domain

✓ Working with negative signs correctly when checking for even and odd functions

✓ Being sure a function is one-to-one before trying to determine an inverse

✓ Correctly applying function rules when performing function composition

✓ Raising binomials to higher powers and including all the terms

22 Part I: The Questions

Using Function Notation to

Determining the Domain and

Evaluate Function Values

Range of a Function

121−125 Evaluate the function for the given value.

126−135 Find the domain and range for the

function.

121. Given

, find

.

126.

122. Given

, find

.

127.

123. Given

, find

.

124. Given

,

Illustration by Thomson Digital

find

.

128.

125. Given

, find

.

129.

130.

Chapter 3: Function Basics

23

131.

Identifying Odd Functions

138−139 Determine which function is odd.

138.

,

,

132.

,

,

133.

139.

,

,

,

,

134.

Ruling Out Even and

Odd Functions

135.

140 Determine which function is neither even

nor odd.

140.

,

,

,

,

Recognizing Even Functions

136−137 Determine which function is even.

136.

,

,

,

,

Recognizing One-to-One

Functions from Given

Relations

141−143 Determine which of the given relations is a

137.

,

one-to-one function over its domain.

,

,

141.

,

,

,

,

,

24 Part I: The Questions

142.

,

,

,

,

143.

Illustration by Thomson Digital

Illustration by Thomson Digital

Illustration by Thomson Digital

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 3: Function Basics

25

147.

,

Identifying One-to-One

,

Functions from Equations

,

144−145 Determine which of the functions is one-to-

one over its domain.

,

144.

,

,

,

,

148.

,

145.

,

,

,

,

,

,

,

Recognizing a Function’s

Inverse

146−150 Determine which pair of functions are

149.

,

inverses of each other.

,

146.

,

,

,

,

,

,

150.

,

,

,

,

26 Part I: The Questions

158.

Determining a Function’s

Inverse

151−160 Find the inverse of the function.

159.

151.

160.

152.

153.

Executing Operations on

Functions

161−165 Perform the indicated operation using the

given functions.

154.

161.

Find

.

155.

162.

Find

.

156.

163.

157.

Find

.

Chapter 3: Function Basics

27

164.

169.

Find

.

Doing More Function

Composition

165.

170−176 Perform the indicated operation.

170.

Find

.

Find .

Performing Function

Composition

171.

166−169 Find

.

Find .

166.

172.

167.

Find .

173.

168.

Find

.

28 Part I: The Questions

174.

Using the Difference Quotient

177−180 Evaluate the difference quotient,

Find

.

, for the given function

Assume

.

177.

175.

Find

.

178.

176.

179.

Find

.

180.

Chapter 4

Graphing and Transforming Functions

Y ou can graph functions fairly handily using a graphing calculator, but you’ll be frustrated using this technology if you don’t have a good idea of what you’ll find and where you’ll find it. You need to have a fairly good idea of how high or how low and how far left and right the graph extends. You get information on these aspects of a graph from the intercepts (where the curve crosses the axes), from any asymptotes (in rational functions), and, of course, from the domain and range of the function. A good working knowledge of the characteristics of different types of functions goes a long way toward making your graphing experience a success.

Another way of graphing functions is to recognize any transformations performed on basic function definitions. Just sliding a graph to the left or right or flipping the graph over a line is a lot easier than starting from scratch.

The Problems You’ll Work On

In this chapter, you’ll work with function graphs in the following ways:

✓ Graphing both a function and its inverse

✓ Determining the vertices of quadratic functions (parabolas)

✓ Recognizing the limits of some radical functions when graphing

✓ Pointing out the top or bottom point of an absolute-value function graph to establish its range

✓ Solving polynomial equations for intercepts

✓ Writing equations of the asymptotes of rational functions

✓ Using function transformations to quickly graph variations on functions What to Watch Out For

When graphing functions, your challenges include the following:

✓ Taking advantage of alternate formats of function equations (slope-intercept form, factored polynomial or rational functions, and so on)

✓ Determining whether a parabola opens upward or downward and how steeply 30 Part I: The Questions

✓ Graphing radical functions with odd-numbered roots and recognizing the unlimited domain

✓ Recognizing when polynomial functions don’t cross the x-axis at an intercept

✓ Using asymptotes correctly as a guide in graphing

✓ Reflecting functions vertically or horizontally, depending on the function transformation

Functions and Their Inverses 187.

181–190 Find the inverse of the function. Then

graph both the function and its inverse.

181.

188.

182.

189.

183.

190.

184.

Sketching Quadratic Functions

from Their Equations

191–195 Sketch the graph of the quadratic function.

185.

191.

186.

192.

Chapter 4: Graphing and Transforming Functions

31

193.

197.

194.

195.

Illustration by Thomson Digital

Writing Equations from

198.

Graphs of Parabolas

196–200 Given the graph of a quadratic function,

write its function equation in vertex form,

.

196.

Illustration by Thomson Digital

Illustration by Thomson Digital

32 Part I: The Questions

199.

Investigating and Graphing

Radical Functions

201–210 Identify the domain of the radical function

and any intercepts. Then sketch a graph of the

function.

201.

202.

Illustration by Thomson Digital

203.

200.

204.

205.

206.

Illustration by Thomson Digital

207.

Chapter 4: Graphing and Transforming Functions

33

208.

Investigating the Graphs of

Polynomial Functions

216–223 Determine the intercepts of the graph of the

polynomial. Then sketch the graph.

209.

216.

210.

217.

Investigating Absolute Value

218.

Functions

211–215 Find the range of the absolute value

function. Then sketch the function’s graph.

211.

219.

212.

220.

213.

221.

214.

222.

215.

223.

34 Part I: The Questions

Investigating Rational

Transformation of Functions

Functions

231–235 Describe the transformations from the

f function to the g function.

224–230 Determine any asymptotes of the rational

function. Then graph the function.

231.

,

224.

232.

,

225.

233.

,

226.

234.

,

227.

235.

,

228.

Transforming Selected Points

229.

Using Functions

236–240 Transform the points (3, 4), (‒ 2, 6),

and (5, ‒ 1) using the function description of the

transformations.

230.

236.

Chapter 4: Graphing and Transforming Functions

35

237.

244.

238.

245.

239.

Sketching More Graphs

Using Basic Functions

and Transformations

246–250 Find the vertical asymptote of the given

240.

function, which is a transformation of

. Then

sketch the graph of the new function.

246.

Sketching Graphs Using Basic

Functions and Transformations

241–245 Find the vertex of the given function,

which is a transformation of

. Then sketch

247.

the graph of the new function.

241.

248.

242.

249.

243.

250.

36 Part I: The Questions

Chapter 5

Polynomials

P olynomial functions have graphs that are smooth curves. They go from negative infinity to positive infinity in a nice, flowing fashion with no abrupt changes of direction. Pieces of polynomial functions are helpful when modeling physical situations, such as the height of a rocket shot in the air or the time a person takes to swim a lap depending on his or her age.

Most of the focus on polynomial functions is in determining when the function changes from negative values to positive values or vice versa. Also of interest is when the curve hits a relatively high point or relatively low point. Some good algebra techniques go a long way toward studying these characteristics of polynomial functions.

The Problems You’ll Work On

In this chapter, you’ll work with polynomial functions in the following ways:

✓ Solving quadratic equations by factoring or using the quadratic formula

✓ Rewriting quadratic equations by completing the square

✓ Factoring polynomials by using grouping

✓ Looking for rational roots of polynomials by using the rational root theorem

✓ Counting real roots with Descartes’s rule of signs

✓ Using synthetic division to quickly compute factors

✓ Writing equations of polynomials given roots and other information

✓ Graphing polynomials by using end-behavior and the factored form

What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working with polynomial functions, your challenges will include

✓ Watching the order of operations when using the quadratic formula

✓ Adding to both sides when completing the square

✓ Remembering to insert zeros for missing terms when using synthetic division

✓ Recognizing the effect of imaginary roots on the graph of a polynomial 38 Part I: The Questions

258.

Using Factoring to Solve

Quadratic Equations

251–260 Find the solution set for the equation by

factoring.

259.

251.

260.

252.

Solving Quadratic Equations

253.

by Using the Quadratic

Formula

261–265 Use the quadratic formula to solve the

equation.

254.

261.

255.

262.

256.

263.

257.

Chapter 5: Polynomials

39

264.

Solving Polynomial Equations

for Intercepts

271–276 Find the real roots (x-intercepts) of the

polynomial.

265.

271.

Using Completing the Square

to Solve Quadratic Equations

272.

266–270 Solve the equation by completing the

square.

266.

273.

267.

274.

268.

275.

269.

276.

270.

40 Part I: The Questions

283.

Using Factoring by Grouping

to Solve Polynomial Equations

277–280 Find the real roots (x-intercepts) of the

polynomial by using factoring by grouping.

284.

277.

285.

278.

Listing Possible Roots of a

279.

Polynomial Equation

286–290 Use the rational root theorem to list the

possible rational roots of the function.

286.

280.

287.

Applying Descartes’s Rule

of Signs

281–285 Use Descartes’s rule of signs to determine

the possible number of positive and negative real

288.

roots of the function.

281.

289.

282.

290.

Chapter 5: Polynomials

41

Dividing Polynomials

Using Synthetic Division to

291–295 Perform the division by using long division. Divide Polynomials Write the remainder as a fraction.

296–298 Use synthetic division to find the quotient

and remainder.

291.

296.

292.

297.

293.

298.

294.

Checking for Roots of a

Polynomial by Using

295.

Synthetic Division

299–300 Use synthetic division to determine which

of the given values is a zero of the function.

299.

Possible zeros: –3, 3, –6, 6, 1

42 Part I: The Questions

300.

Writing Polynomial

Possible zeros: –8, –2, –1, 4, 8

Expressions When Given

Roots and a Point

306–310 Find the lowest-order polynomial that

has the listed values as its zeros and whose graph

passes through the given point.

Writing Polynomial

Expressions from

306. Zeros:

–3, 1, 2

Given Roots

Point: (–1, 24)

301–305 Find the lowest order polynomial with 1 as

the leading coefficient with the listed values as its

zeros.

301. –3, 2, 3

307. Zeros:

–4, –2, 3

Point: (2, 48)

302. –4, –3, 0, 1, 2

308. Zeros:

–3, –2, 1, 2

Point: (0, 36)

303. 1

+ i,1 – i, –1, 2

309. Zeros:

,

, –2, 1

304.

,

, –1, and –4

Point: (2, 16)

305.

,

, 1, 2

310. Zeros: 2 + i, 2 – i, –3, 2

Point: (–1, 120)

Chapter 5: Polynomials

43

Graphing Polynomials

Writing Equations from

311–316 Graph the polynomial by determining the

Graphs of Polynomials

end behavior, finding the x- and y-intercepts, and

using test points between the x-intercepts.

317−320 Write an equation for the given

polynomial graph.

311.

317.

312.

313.

Illustration by Thomson Digital

314.

318.

315.

316.

Illustration by Thomson Digital

44 Part I: The Questions

319.

320.

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 6

Exponential and Logarithmic Functions

E xponential and logarithmic functions go together. You wouldn’t think so at first glance, because exponential functions can look like

, and logarithmic (log) functions

can look like

. What joins them together is that exponential functions and log

functions are inverses of each other.

Exponential and logarithmic functions can have bases that are any positive number except the number 1. The special cases are those with base 10 (common logarithms) and base e (natural logarithms), which go along with their exponential counterparts.

The whole point of these functions is to tell you how large something is when you use a particular exponent or how big of an exponent you need in order to create a particular number. These functions are heavily used in the sciences and finance, so studying them here can pay off big time in later studies.

The Problems You’ll Work On

In this chapter, you’ll work with exponential and logarithmic functions in the following ways:

✓ Evaluating exponential and log functions using the function rule

✓ Simplifying expressions involving exponential and log functions

✓ Solving exponential equations using rules involving exponents

✓ Solving logarithmic equations using laws of logarithms

✓ Graphing exponential and logarithmic functions for a better view of their powers

✓ Applying exponential and logarithmic functions to real-life situations What to Watch Out For

Don’t let common mistakes trip you up. Here are some of the challenges you’ll face when working with exponential and logarithmic functions:

✓ Using the rules for exponents in various operations correctly

✓ Applying the laws of logarithms to denominators of fractions

46 Part I: The Questions

✓ Remembering the order of operations when simplifying exponential and log expressions

✓ Checking for extraneous roots when solving logarithmic equations

Understanding Function

Graphing Exponential

Notation

Functions

321–325 Evaluate the function at the indicated

326–330 Graph the exponential function.

points.

326.

321. Evaluate the function

at

and

.

327.

322. Evaluate the function

at

and

.

328.

323. Evaluate the function

at

and

.

329.

324. Evaluate the function

at

and

.

330.

325. Evaluate the function

at

and

.

Chapter 6: Exponential and Logarithmic Functions

47

339.

Solving Exponential Equations

331–345 Solve the exponential equation for x.

331.

340.

332.

341.

333.

342.

334.

343.

335.

344.

336.

345.

337.

338.

48 Part I: The Questions

Using the Equivalence

Rewriting Logarithmic

to Rewrite

Expressions

Expressions

351–358 Write the expression in a new form using

the laws of logarithms.

346–348 Rewrite the exponential expression

as a logarithmic expression.

351.

346.

352.

347.

353.

348.

354.

Using the Equivalence

to Rewrite

Expressions

355.

349–350 Rewrite the logarithmic expression as an

exponential expression.

349.

356.

350.

357.

358.

Chapter 6: Exponential and Logarithmic Functions

49

Rewriting Logs of Products

Solving Logarithmic Equations

and Quotients as Sums and

366–375 Solve the logarithmic equation for x.

Differences

366.

359–365 Write the expression as an expanded

logarithm.

359.

367.

360.

368.

361.

369.

362.

370.

363.

371.

364.

372.

365.

373.

50 Part I: The Questions

374.

377.

375.

Applying Function

Transformations to Log

Illustration by Thomson Digital

Functions

(A)

376–380 Find the best choice for a function rule

of the transformation of

shown in

(B)

the graph.

(C)

(D)

376.

(E)

378.

Illustration by Thomson Digital

(A)

Illustration by Thomson Digital

(B)

(C)

(A)

(D)

(B)

(E)

(C)

(D)

(E)

Chapter 6: Exponential and Logarithmic Functions

51

379.

Applying Logarithms

to Everyday Life

381‒ 390 Use the given formula and properties of logarithms to answer the question.

381. Orange juice has a hydrogen ion concentra-

tion of 0.001 moles/liter. What is the pH

of orange juice? Use the pH formula

, where H + is the hydrogen

ion concentration in moles/liter.

Illustration by Thomson Digital

(A)

382. Ammonia water has a pH of 11.6. What is

(B)

the hydrogen ion concentration (in moles/

liter)? Use the pH formula

,

(C)

where H + is the hydrogen ion concentra-

(D)

tion in moles/liter.

(E)

380.

383. You invest $5,000 in an account at an annual

rate of 1.475% with interest compounding

quarterly. Find out how much is in the

account (to the nearest cent) after 3 years

by using the compound interest formula

, where A is the total amount

in the account in dollars, P is the principal

(amount invested), r is the annual interest

rate , n is the number of times compounded

per year, and t is the time in years .

Illustration by Thomson Digital

(A)

(B)

(C)

(D)

(E)

52 Part I: The Questions

384. If you invest $1,000 at a rate of 3.25%, com- 388. A town had a population of 6,250 in 1975 and pounded continuously, how many years

a population of 8,125 in 2010. If the rate of

would it take for your money to double?

growth is exponential, what will the popula-

Round to the nearest hundredth of a year.

tion be in 2040? Use the exponential model

Use the continuous compound interest

, where y is the resulting population,

formula

, where A is the amount in

C is the initial amount, k is the growth or

the account, P is the principal, r is the inter-

decay rate, and t is the time in years.

est rate, and t is the time in years .

389. In 2000, a house was appraised at $179,900.

385. You buy a car that costs $23,495, and it loses

In 2013, it was reappraised at $138,000. If the

$4,500 in value when you drive it off the car

decline rate is exponential, what will the

lot. The car then loses 10.785% in value each

value of the house be in 2020? Round the

year (exponential depreciation). Rounded to

value to the nearest dollar. Use the exponen-

the nearest cent, what is the value of the car

tial model

, where y is the resulting

after 5 years? Use the exponential equation

value, C is the initial amount, k is the growth

, where y is the resulting value, C is

or decay rate, and t is the time in years.

the initial amount, k is the growth or decay

rate, and t is the time in years.

390. One student with a contagious flu virus goes

to school when he’s sick. The school district

386. A normal conversation is 60 decibels (dB).

includes 7,500 students and 1,000 staff

What is the intensity of the sound of the

members. The growth of the virus is mod-

conversation (in watts per square meter)?

eled by the following equation, where y is

Use the following formula, where L is deci-

the number infected and t is time in days:

bels and I is the intensity of the sound:

. When 45% of the students

.

and staff members fall ill, the schools will

close. How many days before the schools

are closed? Round to the nearest whole day.

387. Two earthquakes occurred in different

parts of the country. The first earthquake

had a magnitude of 8.0, and the second had

a magnitude of 4.0. Compare the intensity

of the earthquakes using the equation

, where M is the magnitude, I is

the intensity measured from the epicenter,

and I is a constant.

0

Chapter 7

Trigonometry Basics

T rigonometric functions are special in several ways. The first characteristic that separates them from all the other types of functions is that input values are always angle measures.

You input an angle measure, and the output is some real number. The angle measures can be in degrees or radians — a degree being one-360th of a slice of a circle, and a radian being about one-sixth of a circle. Each type has its place and use in the study of trigonometry.

Another special feature of trig functions is their periodicity; the function values repeat over and over and over, infinitely. This predictability works well with many types of physical phenomena, so trigonometric functions serve as models for many naturally occurring observations.

The Problems You’ll Work On

In this chapter, you’ll work with trigonometric functions and their properties in the following ways:

✓ Defining the basic trig functions using the sides of a right triangle

✓ Expanding the input values of trig functions by using the unit circle

✓ Exploring the right triangle and trig functions to solve practical problems

✓ Working with special right triangles and their unique ratios

✓ Changing angle measures from degrees to radians and vice versa

✓ Determining arc length of pieces of circles

✓ Using inverse trig functions to solve for angle measures

✓ Solving equations involving trig functions

What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working with trigonometric functions, some challenges will include the following:

✓ Setting up ratios for the basic trig functions correctly

✓ Recognizing the corresponding sides of right triangles when doing applications 54 Part I: The Questions

✓ Remembering the counterclockwise rotation in the standard position of angles

✓ Measuring from the terminal side to the x-axis when determining reference angles

✓ Keeping the trig functions and their inverses straight from the functions and their reciprocals

Using Right Triangles to

393.

Determine Trig Functions

391−396 Use the triangle to find the trig ratio.

391.

Illustration by Thomson Digital

Illustration by Thomson Digital

394.

392.

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 7: Trigonometry Basics

55

395.

398. A boy flying a kite lets out 300 feet of string,

which makes an angle of 35° with the

ground. Assuming that the string is straight,

how high above the ground is the kite?

Illustration by Thomson Digital

399. An airplane climbs at an angle of 9° with

the ground. Find the ground distance it has

traveled when it reaches an altitude of

400 feet.

396.

400. A car is traveling up a grade with an angle

of elevation of 4°. After traveling 1 mile,

what is the vertical change in feet?

(

)

401. Jase is in a hot air balloon that is 600 feet

Illustration by Thomson Digital

above the ground, where he can see his

brother Willie. The angle of depression

from Jase’s line of sight to Willie is 25°. How

far is Willie from the point on the ground

directly below the hot air balloon?

Solving Problems by Using

Right Triangles and Their

Functions

402. A bird is flying at a height of 36 feet and

spots a windowsill 8 feet off the ground on

397−406 Solve the problem. Round your answer to

which to perch. If the windowsill is at a 28°

the nearest tenth unless otherwise indicated.

angle of depression from the bird, how far

must the bird fly before it can land?

397. A ladder leaning against a house makes an

angle of 30° with the ground. The foot of

the ladder is 7 feet from the foot of the

house. How long is the ladder?

56 Part I: The Questions

403. A submarine traveling 8 mph is descending

at an angle of depression of 5°. How many

Working with Special Right

minutes does the submarine take to reach

a depth of 90 feet?

Triangles

407−410 Find the length of the side indicated by x.

407.

404. You’re a block away from a building that is

820 feet tall. Your friend is between you

and the building. The angle of elevation

from your position to the top of the build-

Illustration by Thomson Digital

ing is 42°. The angle of elevation from your

friend’s position to the top of the building

is 71°. To the nearest foot, how far are you

from your friend?

408.

405. A lighthouse is located at the top of a hill

100 feet tall. From a point P on the ground,

the angle of elevation to the top of the hill

Illustration by Thomson Digital

is 25°. From the same point P, the angle of

elevation to the top of the lighthouse is 50°.

How tall is the lighthouse?

409.

406. A surveyor needs to find the distance BC

across a lake as part of a project to build a

bridge. The distance from point A to point

Illustration by Thomson Digital

B is 250 feet. The measurement of angle A

is 40°, and the measurement of angle B is

112°. What is the distance BC across the

lake to the nearest foot?

410.

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 7: Trigonometry Basics

57

417. 60°

Changing Radians to Degrees

411−415 Convert the radian measure to degrees.

Round any decimals to two places.

411.

418. −405°

412.

419. 36°

413.

420. 167°

414. 3

Finding Angle Measures (in

Degrees) in Standard Position

421−422 The terminal side of an angle in standard

position on the unit circle contains the given point.

Give the degree measure of .

415. 0.93

421.

Changing Degrees to Radians

416−420 Convert the degree measure to radians.

422.

Give the exact measure in terms of .

416. 225°

58 Part I: The Questions

429. 792°

Determining Angle Measures

(in Radians) in Standard

Position

423

430.

−425 The terminal side of an angle in standard

−748°

position on the unit circle contains the given point.

Give the radian measure of .

423.

Determining Trig Functions

by Using the Unit Circle

431−434 Find the exact value of the trigonometric

424.

function. If any are not defined, write undefined .

431. sin 225°

425.

432. tan 330°

Identifying Reference Angles

426−430 Find the reference angle for the angle

433. cos 405°

measure. (Recall that the quadrants in standard

position are numbered counterclockwise, starting in

the upper right-hand corner.)

426. 167°

434.

427. 342°

428. 265°

Chapter 7: Trigonometry Basics

59

Calculating Trig Functions by Using the Arc Length Formula

Using Other Functions and

441−445 Solve the problem using the arc length

formula

, where r is the radius of the circle

Terminal Side Positions

and is in radians.

435−440 Use the given information to find the

exact value of the trigonometric function.

441. The central angle in a circle with a radius

of 20 cm is

Find the exact length of the

435. Given:

; is in quadrant I

intercepted arc.

Find:

442. The central angle in a circle of radius 6 cm

is 85°. Find the exact length of the inter-

436. Given:

; is in quadrant II

cepted arc.

Find:

443. Find the radian measure of the central angle

that intercepts an arc with a length of

437. Given:

; is in quadrant IV

inches in a circle with a radius of 13 inches.

Find:

444. The second hand of a clock is 18 inches

long. In 25 seconds, it sweeps through an

438. Given:

; is in quadrant III.

angle of 150°. How far does the tip of the

second hand travel in 25 seconds?

Find:

445. How far does the tip of a 15 cm long minute

439. Given:

;

hand on a clock move in 10 minutes?

Find:

440. Given:

,

Find:

60 Part I: The Questions

453.

Evaluating Inverse Functions

446−450 Find an exact value of y.

446.

; give y in radians.

454.

447.

; give y in radians.

455.

448.

; give y in radians.

Calculating Trig Equations for

x in Radians

456−460 Find all solutions of the equation in the

interval

. (Recall that the quadrants in standard

449.

; give y in radians.

position are numbered counterclockwise, starting in

the upper right-hand corner.)

456.

450.

457.

Solving Trig Equations for x

in Degrees

458.

451−455 Find all solutions of the equation in the

interval

. (Recall that the quadrants in

standard position are numbered counterclockwise,

starting in the upper right-hand corner.)

451.

459.

452.

460.

Chapter 8

Graphing Trig Functions

T he graphs of trigonometric functions are usually easily recognizable — after you become familiar with the basic graph for each function and the possibilities for transformations of the basic graphs.

Trig functions are periodic. That is, they repeat the same function values over and over, so their graphs repeat the same curve over and over. The trick is to recognize how often this curve repeats and where one of the basic graphs starts for a particular function.

An interesting feature of four of the trig functions is that they have asymptotes — those not-really-there lines used as guides to the shape of a curve. The sine and cosine functions don’t have asymptotes, because their domains are all real numbers. The other four functions have vertical asymptotes to mark where their domains have gaps.

The Problems You’ll Work On

In this chapter, you’ll work with the graphs of trigonometric functions in the following ways:

✓ Marking any intercepts on the x-axis to help graph the curves

✓ Locating and drawing in vertical asymptotes for the tangent, cotangent, secant, and cosecant functions

✓ Computing the change in the period of a function based on some transformation

✓ Adjusting the amplitude of the sine or cosine when the basic curve has a multiplier

✓ Making sideways moves when transformations involve horizontal translations

✓ Moving trig functions upward or downward with vertical translations What to Watch Out For

When graphing trigonometric functions, some challenges will include

✓ Not misreading the period of the trig function when a transformation involves a fraction

✓ Drawing enough full cycles of the curve to show its characteristics properly 62 Part I: The Questions

✓ Marking the axes appropriately for the situation

✓ Making use of intercepts when they’re helpful in a graph

Recognizing Basic Trig Graphs

461– 465 Determine which trig function equation matches the graph.

461.

Illustration by Thomson Digital

(A)

(B)

(C)

(D)

(E)

Chapter 8: Graphing Trig Functions

63

462.

464.

Illustration by Thomson Digital

Illustration by Thomson Digital

(A)

(B)

(A)

(C)

(B)

(D)

(C)

(E)

(D)

463.

(E)

465.

Illustration by Thomson Digital

Illustration by Thomson Digital

(A)

(A)

(B)

(B)

(C)

(C)

(D)

(D)

(E)

(E)

64 Part I: The Questions

Graphing Sine and Cosine

473.

to

466– 470 Sketch the graph of the function.

466.

474.

to

467.

475.

to

468.

Writing New Trig Functions

Using Transformations

476– 480 Given the description of the transformation

from function f to function g, write the function

469.

equation for g.

476. Shift

left units and up 5 units.

470.

477. Shift

right units and down units.

Applying Function

Transformations to Graphs

478. Shift

right units and down

of Trig Functions

units.

471– 475 Describe the transformation from function

f to function g.

471.

to

479. Shift

left units and down 3 units.

480. Shift

right units and up units.

472.

to

Chapter 8: Graphing Trig Functions

65

487.

to

Graphing Tangent and

Cotangent

481– 484 Sketch the graph of the function.

488.

to

481.

489.

to

482.

490.

to

483.

491.

to

484.

492.

to

Interpreting Transformations

of Trig Functions

485– 495 Describe the transformations from function

f to function g.

493.

to

485.

to

494.

to

486.

to

495.

to

66 Part I: The Questions

Graphing Secant and

Interpreting Transformations

Cosecant

from Function Rules

496–503 Give a rule for the equations of the

504–510 Describe the transformation from function

asymptotes. Then sketch the graph of the function.

f to function g.

496.

504.

to

497.

505.

to

498.

506.

to

499.

507.

to

500.

508.

to

501.

509.

to

502.

510.

to

503.

Chapter 9

Get ting Star ted with Trig Identities

D on’t have an identity crisis! In this chapter, you become more familiar with the possibilities for rewriting trigonometric expressions. A trig identity is really an equivalent expression or form of a function that you can use in place of the original. The equivalent format may make factoring easier, solving an application possible, and (later) performing an operation in calculus more manageable.

The trigonometric identities are divided into many different classifications. These groupings help you remember the identities and make determining which identity to use in a particular substitution easier. In a classic trig identity problem, you try to make one side of the equation match the other side. The best way to do so is to work on just one side — the left or the right — but sometimes you need to work on both sides to see just how to work the problem to the end.

The Problems You’ll Work On

In this chapter, you’ll work with the basic trigonometric identities in the following ways:

✓ Determining which trig functions are reciprocals of one another

✓ Creating Pythagorean identities from a right triangle whose hypotenuse measures 1 unit

✓ Determining the sign of identities whose angle measure is negated

✓ Matching up trig functions and their co-functions

✓ Using the periods of functions in identities

✓ Making the most of selected substitutions into identities

✓ Working on only one side of the identity

✓ Figuring out where to go with an identity by working both sides at once What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working on trigonometric identities, some challenges will include the following:

✓ Keeping track of where the 1 goes in the Pythagorean identities

✓ Remembering the middle term when squaring binomials involving trig functions 68 Part I: The Questions

✓ Correctly rewriting Pythagorean identities when solving for a squared term

✓ Recognizing the exponent notation

518.

Proving Basic Trig Identities

511–535 Prove the trig identity. Indicate your first

identity substitution.

511.

519.

512.

520.

513.

521.

514.

522.

515.

523.

516.

524.

517.

525.

Chapter 9: Getting Started with Trig Identities

69

526.

534.

527.

535.

528.

Returning to Basic Sine and

Cosine to Solve Identities

536–540 Determine the missing term or factor in

the identity by changing all functions to those using

529.

sine or cosine.

536.

530.

537.

531.

538.

532.

539.

533.

540.

70 Part I: The Questions

547.

Using Multiplication by a

Conjugate to Solve Identities

541–545 Determine the missing term or factor in

the identity by multiplying by a conjugate.

548.

541.

549.

542.

550.

543.

Solving Identities After

544.

Factoring out a Common

Function

551–555 Determine the missing term or factor in the

identity after factoring.

545.

551.

Solving Identities After

Raising a Binomial to

552.

a Power

546–550 Determine the missing term or terms in the

identity after computing the power of the binomial.

553.

546.

Chapter 9: Getting Started with Trig Identities

71

554.

Performing Algebraic

Processes to Make Identities

More Solvable

555.

561–570 Give the rewritten identity after performing

the action. Then complete the solution of the identity.

561. Distribute on the right and move all terms

to the left.

Solving Identities After

Combining Fractions

556–560 Determine the missing term or factor in

the identity after adding the fractions.

562. Split up the first fraction on the left and

distribute on the right.

556.

557.

563. Cross-multiply.

558.

564. Cross-multiply.

559.

565. Factor the numerator on the left and

reduce the fraction.

560.

72 Part I: The Questions

566. Cross-multiply.

569. Subtract the two terms on the left.

567. Add the middle fraction to both sides, and 570. Square both sides of the equation.

then add the two fractions on the right.

Simplify by using reciprocal identities.

568. Combine the two terms on the right.

Chapter 10

Continuing with Trig Identities

T he basic trig identities (see Chapter 9) will get you through most problems and applications involving trigonometry. But if you’re going to broaden your horizons and study more and more mathematics, you’ll find the additional identities in this chapter crucial to your success. Also, in some of the sciences, especially physics, these specialized identities come up in the most unlikely (and likely) places.

These trigonometric identities are broken into groups, depending on whether you’re trying to combine angles or split them apart, increase exponents or reduce them, and so on.

The groupings can help you to decide which identity to use in which situation. Keep a list of these identities handy, because you’ll want to refer to them as you work through the problems.

The Problems You’ll Work On

You’ll work with the more-advanced trig identities in the following ways:

✓ Using the function values of two angles to determine the function value of the sum of the angles

✓ Applying the identities for the difference between two angles

✓ Making use of the half-angle identities

✓ Working from product-to-sum and sum-to-product identities

✓ Using the periods of functions in identities

✓ Applying power-reducing identities

✓ Deciding which identity to use first

What to Watch Out For

When you’re working on these particular trig identities, some challenges will include the following:

✓ Applying the identities using the correct order of operations

✓ Simplifying the radicals correctly in half-angle identities

✓ Making the correct choices between positive and negative identities 74 Part I: The Questions

Using Identities That Add or

Using Identities That Double

Subtract Angle Measures

the Size of the Angle

571–573 Use a sum or difference identity to

576–578 Use a double-angle identity to determine

determine the missing term in the identity.

the missing term in the identity.

576.

571.

577.

572.

578.

573.

Confirming the Statements of

Confirming Double-Angle

Multiple-Angle Identities

Identities

579–580 Use sum and double-angle identities to

574–575 Use a sum identity to determine the

determine the missing term or factor in the multiple-

missing factor or term in the double-angle identity.

angle identity.

574.

579.

575.

580.

Chapter 10: Continuing with Trig Identities

75

Creating Half-Angle Identities 585.

from Double-Angle Identities

581–582 Determine the missing term in the half-

angle identity by starting with an identity for

.

581.

Creating Products of Trig

Functions from Sums and

Differences

586–588 Determine the missing factor in the

product-to-sum identity.

582.

586.

Creating a Half-Angle

587.

Identity for Tangent

583 Determine the missing factor in the half-angle

identity by using the ratio identity for tangent and

the half-angle identities for sine and cosine.

588.

583.

Using Product-to-Sum

Identities to Evaluate

Using Half-Angle Identities

Expressions

to Simplify Expressions

589–590 Write the product as a sum using a

product-to-sum identity. Then evaluate.

584–585 Determine the missing term in the identity.

589.

584.

76 Part I: The Questions

590.

Applying Power-Reducing

Identities

596–600 Use a power-reducing identity to determine

the missing term in the identity.

Using Sum-to-Product

Identities to Evaluate

596.

Expressions

591–595 Write the sum as a product using a

sum-to-product identity. Then evaluate.

597.

591.

598.

592.

599.

593.

600.

594.

Using Identities to Determine

595.

Values of Functions at

Various Angles

601–610 Evaluate the angle’s function using an

appropriate identity.

601.

Chapter 10: Continuing with Trig Identities

77

602.

Working through Identities

Using Multiple Methods

611–630 Determine the missing term or factor

in the identity.

603.

611.

604.

612.

605.

613.

606.

614.

607.

615.

608.

616.

609.

617.

610.

78 Part I: The Questions

618.

625.

619.

626.

620.

627.

621.

628.

622.

629.

623.

630.

624.

Chapter 11

Working with Triangles

and Trigonometry

Y ou may think that you’re “not in Kansas anymore” when you leave the familiar world of right triangles and Mr. Pythagoras to enter this new world of oblique triangles.

Trigonometry allows for some calculations that aren’t possible with the geometric formulas and other types of measurement. The Law of Sines and Law of Cosines are relationships between the sides and angles of triangles that aren’t right triangles.

The applications you can solve using these new laws are many and varied. Whole new worlds are opened to you now that you don’t have triangle-type restrictions. The biggest challenge is in deciding which law to use, but even that is pretty straightforward.

The Problems You’ll Work On

In this chapter, you’ll work on solving for parts of triangles in the following ways:

✓ Using the Law of Sines and choosing the pair of ratios

✓ Applying the Law of Cosines and determining which version works

✓ Working with the ambiguous case and deciding which angle applies

✓ Finding the missing values of sides and angles of a triangle

✓ Computing areas of triangles by using a formula involving a trig function

✓ Using Heron’s formula for the area of a triangle

✓ Practicing with practical applications

What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working with these formulas for triangles, some challenges will include

✓ Choosing the correct pair of ratios when using the Law of Sines

✓ Performing the order of operations correctly when applying the Law of Cosines 80 Part I: The Questions

✓ Identifying the correct parts of the triangle when finding the area

✓ Writing the correct relationships between triangle parts when working on applications Applying the Law of Sines

Utilizing the Law of Sines

to Find Sides

to Find Angles

631–633 Use the Law of Sines to find the indicated

634–636 Use the Law of Sines to find the indicated

side. Round your answer to the nearest tenth.

angle. Round your answer to the nearest tenth of a

degree.

631.

634.

Illustration by Thomson Digital

Find

c.

Illustration by Thomson Digital

Find .

632.

635.

Illustration by Thomson Digital

Find

b.

Illustration by Thomson Digital

633.

Find .

Illustration by Thomson Digital

Find

c.

Chapter 11: Working with Triangles and Trigonometry

81

636.

639. Jake is taking a walk along a straight road.

He decides to take a detour through some

woods, so he walks on a path that makes

an angle of 50° with the road. After walking

for 1,450 feet, he turns 105° and heads

back to the road. How far does Jake have

to walk to get back to the road? Round

your answer to the nearest foot.

Illustration by Thomson Digital

Find .

640. Carlos measures the angle of elevation of

the top of a tree as 75°. Elena (who is

20 feet farther from the tree on a straight,

level path) measures the angle of elevation

Using the Law of Sines for

as 45°. How tall is the tree to the nearest

foot?

Practical Applications

637–640 Use the Law of Sines to solve the problem.

637. Points

A and B are on one side of a river,

100 feet apart, with C on the opposite side.

Investigating the Ambiguous

The angles A and B measure 70° and 60°,

respectively. What is the distance from

Case of the Law of Sines

point A to point C, to the nearest foot?

641–643 Determine the number of triangles that

can be formed using the given measurements. If the

answer is unknown, write, “Cannot be determined.”

641.

,

,

638. A 14-foot pole is leaning. A wire attached

to the top of the pole is anchored in the

ground. The wire is 14.5 feet long and

makes a 74.2° angle with the ground. What

angle does the pole make with the ground?

Round your answer to the nearest tenth of

642.

,

,

a degree.

643.

,

,

82 Part I: The Questions

648.

Determining All Angles and

Sides of a Triangle

644–645 Use the Law of Sines to solve the triangle.

Round your answers to the nearest tenth. If no

triangle exists, write, “No solution.”

644.

,

,

Illustration by Thomson Digital

645.

,

,

Using the Law of Cosines to

Determine an Angle

649–651 Use the Law of Cosines to find the indicated

Finding Side Measures by

angle. Round your answer to the nearest tenth of a

degree.

Using the Law of Cosines

646–648 Use the Law of Cosines to find the length

649.

of the missing side. Round your answer to the

nearest tenth.

646.

Illustration by Thomson Digital

Illustration by Thomson Digital

Find .

647.

Illustration by Thomson Digital

Chapter 11: Working with Triangles and Trigonometry

83

650.

653. In a rhombus with a side length of 24, the

longer diagonal has a length of 35. Find the

larger angle of the rhombus. Round your

answer to the nearest degree.

Illustration by Thomson Digital

Find .

654. A baseball player in center field is playing

350 feet from a television camera that’s

behind home plate. The batter hits a fly ball

that travels a horizontal distance of 420 feet

from the camera. About how many feet does

the center fielder run to catch the ball if the

camera turns 8° while following the play?

651.

655. A bicycle race follows a triangular course.

Illustration by Thomson Digital

The three legs of the race, in order, are

10.4 kilometers, 6.3 kilometers, and

Find .

8.3 kilometers. Find the angle between

the first two legs of the race to the nearest

degree.

Applying the Law of Cosines

to Real-World Situations

Finding Areas of Triangles

652–655 Use the Law of Cosines to solve the

by Using the Sine

problem.

656–658 Find the area of the triangle. Round your

answer to the nearest tenth.

652. To approximate the length of a lake, a

surveyor starts at one end of the lake

and walks northeast 145 yards. He then

656.

turns 110° to the right and walks 270 yards

until he arrives at the other end of the

lake. What is the length of the lake to the

nearest yard?

Illustration by Thomson Digital

84 Part I: The Questions

657.

660.

,

, and

Illustration by Thomson Digital

Illustration by Thomson Digital

658.

661.

,

, and

Illustration by Thomson Digital

Illustration by Thomson Digital

Applying the Trig Formula

for Area of a Triangle

Using the Trig Formula for

659–661 Use the figure and the given information

to find the area of the triangle. (The triangle isn’t

Area in Various Situations

necessarily drawn to scale.) Round your answer to

the nearest tenth.

662–663 Solve the problem.

659.

,

, and

662. In an isosceles triangle, the two equal sides

each measure 20 meters; they include an

angle of 30°. Find the exact area of the

isosceles triangle.

Illustration by Thomson Digital

663. In a rhombus, each side is 15 inches, and

one angle is 96°. Find the area of the rhom-

bus to the nearest square inch.

Chapter 11: Working with Triangles and Trigonometry

85

667.

Solving Area Problems

Needing Additional

Computations

664–669 Find the area of the triangle. Round your

answer to the nearest tenth.

Illustration by Thomson Digital

664.

668.

Illustration by Thomson Digital

Illustration by Thomson Digital

665.

669.

Illustration by Thomson Digital

Illustration by Thomson Digital

666.

670. Find the area of triangle BCD.

Illustration by Thomson Digital

Illustration by Thomson Digital

86 Part I: The Questions

Finding Areas of Triangles by Applying Heron’s Formula

Using Heron’s Formula

674–676 Use the figure and the given information

to find the area of the triangle with Heron’s formula.

671–673 Use Heron’s formula to find the exact area

(The triangle isn’t necessarily drawn to scale.)

of the triangle. Simplify all radicals.

Round your answer to the nearest tenth.

671.

674.

,

,

Illustration by Thomson Digital

Illustration by Thomson Digital

672.

675.

,

,

Illustration by Thomson Digital

Illustration by Thomson Digital

673.

676.

,

,

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 11: Working with Triangles and Trigonometry

87

680. Find the area of the triangle on the coordi-

Practical Applications Using

nate plane with vertices at A (–3, 1), B (1, 3),

and C (3, –2).

Heron’s Formula

677–680 Use Heron’s formula to solve the problem.

677. If a triangular plot of land has sides that

measure 375 feet, 250 feet, and 300 feet,

Tackling Practical

what is the area of the plot? Round your

answer to the nearest whole number.

Applications by Using

Triangular Formulas

681–690 Solve the problem using the Law of Sines,

the Law of Cosines, or an area formula.

678. The Bermuda Triangle is an imaginary area

located off the southeastern Atlantic coast 681. A man and a woman are walking toward of the United States of America; it’s noted

each other on an east-west path through

for a supposedly high incidence of unex-

the park. A hot air balloon is directly

plained disappearances of ships and

above the path between them. The woman

aircraft. The vertices of the triangle are

sees the balloon while looking east at an

generally believed to be Bermuda; Miami,

angle of elevation of 46°. The man sees the

Florida; and San Juan, Puerto Rico. The

balloon while looking west at an angle of

distance from Bermuda to Miami is about

elevation of 58°. If the walkers are 75 yards

1,040 miles. The distance from Miami to

apart, how far is the balloon from the man

San Juan is about 1,000 miles, and the

to the nearest tenth of a yard?

distance from San Juan to Bermuda is

about 960 miles. Find the area of the

Bermuda Triangle to the nearest one hun-

dred square miles.

682. A surveyor has to stake the lot markers for

a new public park. The diagram shows

some of the dimensions of the park. How

much fencing is necessary to enclose the

679. Find the area of triangle BCD to the nearest

entire park? Round your answer to the

whole number.

nearest meter.

Illustration by Thomson Digital

Illustration by Thomson Digital

88 Part I: The Questions

683. Two tracking stations, A and B, are 20 miles 687. A perfect triangle is a triangle whose side

apart. At noon, each station measures the

lengths are integers and whose area is

angle of elevation of a rocket. The rocket

numerically equal to its perimeter. If a par-

isn’t between the two stations. If the angle

ticular perfect triangle has two sides mea-

of elevation from station A is 45° and the

suring 9 and 17, what is the measure of the

angle of elevation from station B is 72°, find

third side?

the altitude of the rocket to the nearest

tenth of a mile. Assume that D, the point

directly beneath the rocket, is on the same

line as A and B.

688. The diagram shows a field. Find the area of

the field to the nearest square meter.

684. A boat leaves port and heads due east for

25 miles. It then makes a turn 35° southeast

and travels 15 miles. How far is the boat

from port? Round your answer to the near-

est tenth.

685. The lengths of the diagonals of a parallelo-

Illustration by Thomson Digital

gram are 8 centimeters and 12 centimeters.

Find the lengths of the short sides of the

parallelogram if the diagonals intersect at

an angle of 36°. Round your answer to the

nearest tenth.

689. Two hikers, Don and Marie, leave the same

point at the same time. Marie walks due

north at a rate of 3 miles per hour, and Don

walks 62° northeast at a rate of 4 miles per

hour. How far apart are they after three

686. The legs of a stepladder usually have dif-

hours? Round your answer to the nearest

ferent lengths. One stepladder has a front

tenth.

leg (the leg with the steps) 60 inches long

and a back leg of 57 inches. When the

ladder is open, the distance between the

legs is 40 inches. To the nearest degree,

what is the angle between the legs when

the ladder is open?

690. You’re constructing an octagonal play-

ground with sides that measure 8 yards.

This shape is a regular octagon. What is the

total area of the playground, rounded to the

nearest tenth?

Chapter 12

Complex Numbers and

Polar Coordinates

C omplex numbers are unreal. Yes, that’s the truth. A complex number has a term with a multiple of i, and i is the imaginary number equal to

. Many of the algebraic rules

that apply to real numbers also apply to complex numbers, but you have to be careful because many rules are different for these numbers. The problems in this chapter will help you understand their particular properties.

You’ll also work on graphing complex numbers. Polar coordinates are quite different from the usual ( x, y) points on the Cartesian coordinate system. Polar coordinates bring together both angle measures and distances, all in one neat package. With the polar coordinate system, you can graph curves that resemble flowers and hearts and other elegant shapes.

The Problems You’ll Work On

In this chapter, you’ll work on complex numbers and polar coordinates in the following ways:

✓ Simplifying powers of i into one of four values

✓ Adding and subtracting complex numbers by combining like parts

✓ Multiplying complex numbers and simplifying resulting powers of i

✓ Dividing complex numbers by multiplying by a conjugate

✓ Interpreting graphs of basic polar coordinates

✓ Graphing polar equations such as cardioids and lemniscates

What to Watch Out For

When working with complex numbers and polar coordinates, some challenges will include the following:

✓ Multiplying imaginary numbers correctly

✓ Choosing the correct conjugate and simplifying the difference of squares correctly when dividing complex numbers

90 Part I: The Questions

✓ Moving in a counterclockwise direction when graphing polar coordinates

✓ Recognizing which ray to use when graphing negative and multiple angle measures

Writing Powers of i in Their

Adding and Subtracting

Simplest Form

Complex Numbers

691–695 Write the power of i in its simplest form.

696–705 Perform the operations. Write your answer

in the form

.

691.

696.

692.

697.

693.

698.

694.

699.

695.

700.

701.

Chapter 12: Complex Numbers and Polar Coordinates

91

702.

710.

703.

711.

704.

712.

705.

713.

714.

Multiplying Complex Numbers

706–715 Multiply. Write your answer in

form.

706.

715.

707.

Using Multiplication to Divide

Complex Numbers

716–720 Perform the division by multiplying by a

708.

conjugate. Write your answer in the form

.

716.

709.

92 Part I: The Questions

717.

723.

718.

724.

719.

725.

720.

Graphing Complex Numbers

726–730 Identify the point on the complex

coordinate plane. Give your answer in the

form

.

Solving Quadratic Equations 726.

with Complex Solutions

721–725 Solve the equation. Write your answer in

the form

.

721.

722.

Illustration by Thomson Digital

Chapter 12: Complex Numbers and Polar Coordinates

93

727.

729.

Illustration by Thomson Digital

Illustration by Thomson Digital

728.

730.

Illustration by Thomson Digital

Illustration by Thomson Digital

94 Part I: The Questions

733.

Identifying Points with

Polar Coordinates

731–733 Identify the point on the polar coordinate

plane. Give your answer in

form, where is in

radians.

731.

Illustration by Thomson Digital

Identifying Points Whose

Angles Have Negative

Illustration by Thomson Digital

Measures

734–735 Identify the point on the polar coordinate

plane. Give your answer in

form, where is a

negative angle in radians.

732.

734.

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 12: Complex Numbers and Polar Coordinates

95

735.

740.

Converting Rectangular to

Polar Coordinates

741–745 Change the rectangular coordinates

to polar coordinates.

741.

Illustration by Thomson Digital

742. (–1, 1)

Converting Polar to

Rectangular Coordinates

736–740 Change the polar coordinates to

rectangular coordinates.

743.

736.

744. (3, 0)

737.

745.

738.

Note: Round your answer to three decimal

places.

739.

96 Part I: The Questions

748.

Recognizing Polar Curves

746–750 Identify the polar curve and sketch

its graph.

746.

for

749.

747.

750.

Chapter 13

Conic Sections

C onic sections can be described or illustrated with exactly what their name suggests: cones. Imagine an orange cone in the street, steering you in the right direction. Then picture some clever highway engineer placing one cone on top of the other, tip to tip. That engineer is trying to demonstrate how you can create conic sections (well, perhaps I stretch this a bit).

If you come along and slice one of those cones parallel to the ground, the cut edges form a circle. Slice the cone on an angle, and you have an ellipse. Slice the cone parallel to one of the sides, and you have a parabola. And, finally, slice through both cones together, perpendicular to the ground, and you have a hyperbola. If these descriptions just don’t work for you, the problems in this chapter will certainly do the trick.

The Problems You’ll Work On

In this chapter, you’ll work on conic sections in the following ways:

✓ Recognizing which conic you have from the general equation

✓ Finding the centers of circles and ellipses

✓ Determining the foci of circles, ellipses, and parabolas

✓ Using the directrix of a parabola to complete the sketch

✓ Writing the equations of a hyperbola’s asymptotes

✓ Changing basic conic section equations from parametric to rectangular What to Watch Out For

When working with conic sections, some challenges will include the following:

✓ Determining the major axis of an ellipse

✓ Sketching the graph of a parabola in the correct direction

✓ Using the asymptotes of a hyperbola correctly in a graph

✓ Finding the square root in the equation of a circle when finding the radius 98 Part I: The Questions

757.

Identifying Conics from

Their Equations

751–755 Name the conic and its center.

758.

751.

759.

752.

760.

753.

Writing Equations for Circles

754.

761–770 Write the equation of the circle described.

Then graph the circle.

761. center: (0, 0); radius: 4

755.

762. center: (4, 3); radius: 5

Rewriting Conic Equations

in Standard Form

756–760 Write the equation of the conic in standard

form.

763. center: (–2, 1); radius: 1

756.

764. center: (0, –2); radius:

Chapter 13: Conic Sections

99

765. center: (–5, 0); radius:

772.

766. endpoints of diameter: (4, 7) and (–2, –1)

773.

767. endpoints of diameter: (–3, –4) and (–1, 0)

774.

768. tangent to both axes in third quadrant;

Finding the Vertices and

radius: 3

Directrixes of Parabolas

775–777 Identify the vertex and directrix of the

parabola.

769. tangent to

and

with center in

775.

third quadrant; radius: 7

776.

770. tangent to both axes with center on the

line

; radius: 4

777.

Determining Foci and Axes

of Symmetry of Parabolas

771–774 Identify the focus and axis of symmetry of

the parabola.

771.

100 Part I: The Questions

784.

Writing Equations

of Parabolas

778–780 Write the equation of the parabola

described.

785.

778. vertex: (–3, 2); parabola goes through

(0, 20); axis of symmetry:

786.

779. vertex: (4, 1); parabola goes through (6, 3);

axis of symmetry:

787.

780. vertex: (3, 0); axis of symmetry:

;

directrix:

Writing Equations of Ellipses

788–790 Use the given information to write the

equation of the ellipse. Then sketch its graph.

Determining Centers and Foci 788. center: (–3, 2); vertex: (2, 2); of Ellipses

co-vertex: (–3, 3)

781–787 Identify the center and foci of the ellipse.

781.

789. vertices: (1, 1), (1, –9); co-vertices:

(–3, –4), (5, –4)

782.

790. vertices: (–6, 5), (0, 5); foci: (–5, 5), (–1, 5)

783.

Chapter 13: Conic Sections

101

Determining Asymptotes of

Writing Equations of

Hyperbolas

Hyperbolas

791–797 Identify the asymptotes of the hyperbola.

798–800 Write the standard equation of the hyperbola

given the description. Then sketch its graph.

791.

798. center: (3, –2); foci: (–1, –2), (7, –2);

792.

799. center: (0, 7); foci: (0, 1), (0, 13);

793.

800. foci:

,

; asymptotes:

,

794.

Changing Equation Format

from Trig Functions to

795.

Algebraic

801–805 Write the equation of the curve in

rectangular form (algebraic terms) by eliminating

the parameter .

796.

801.

,

797.

802.

,

102 Part I: The Questions

803.

,

807.

804.

,

808.

805.

,

809.

810.

Changing Equation Format

from Algebraic to Trig

806–810 Change the equation of the conic from

rectangular form (algebraic terms) to parametric

form (using trig functions).

806.

Chapter 14

Systems of Equations

and Inequalities

A system of equations is a collection of two or more equations involving two or more variables. If the number of equations is equal to the number of different variables, then you may be able to find a unique solution that’s common to all the equations. Having the correct number of variables isn’t a guarantee that you’ll have that solution, and it’s not terrible if a unique solution doesn’t exist; sometimes you just write a rule to represent the many solutions shared by the equations in the collection.

A system of inequalities has an infinite number of solutions (unless it has none at all). You solve these systems using graphs of the separate statements.

The techniques for solving these systems involve algebraic manipulation and/or matrices and matrix mathematics. Lots of practice using these techniques is available here. When you’re solving your own systems, the approach you use is up to you.

The Problems You’ll Work On

In this chapter, you’ll work on solving systems of equations and inequalities in the following ways:

✓ Using substitution to solve linear and nonlinear systems of equations

✓ Applying the elimination method when solving systems of linear equations

✓ Writing a rule for multiple solutions of systems of equations

✓ Graphing systems of inequalities

✓ Creating partial fractions using fraction decomposition

✓ Writing coefficient matrices and constant matrices to use in matrix solutions of systems

✓ Determining matrix inverses to use in solving systems of linear equations 104 Part I: The Questions

What to Watch Out For

When you’re working with systems of equations and inequalities, some challenges will include

✓ Recognizing that the answer may be no solution

✓ Distributing correctly when using substitution for solving systems

✓ Performing matrix operations correctly when doing row reductions and eliminating terms

✓ Writing solutions from resulting variable matrices

Using Substitution to Solve

Using Elimination to Solve

Systems of Linear Equations

Systems of Linear Equations

with Two Variables

with Two Variables

811–815 Solve the system by using substitution.

816–820 Solve the system using elimination. Write

Write your answer as an ordered pair,

.

your answer as an ordered pair,

.

811.

,

816.

,

812.

,

817.

,

813.

,

818.

,

814.

,

819.

,

815.

,

820.

,

Chapter 14: Systems of Equations and Inequalities

105

829.

Solving Systems of Equations

Involving Nonlinear Functions

821–830 Solve the system of equations. Write your

answers as order pairs,

.

830.

821.

,

Solving Systems of Linear

822.

,

Equations

831–838 Solve each system of equations. Write the

solution as an ordered triple,

.

823.

,

831.

824.

,

832.

825.

,

826.

,

833.

827.

,

834.

828.

,

106 Part I: The Questions

835.

840.

836.

Graphing Systems

of Inequalities

841–845 Graph the system of inequalities.

841.

837.

842.

838.

843.

Solving Systems of Linear

Equations with Four Variables

839–840 Solve the system of equations. Write the

solution as

.

844.

839.

Chapter 14: Systems of Equations and Inequalities

107

Operating on Matrices

845.

851–855 Perform the matrix operation.

851. Given

and

,

find

.

Decomposition of Fractions

852. Given

and

,

846–850 Write the fraction as the sum or difference

of fractions using decomposition.

find

.

846.

853. Given

and

,

find

.

847.

854. Given

and

,

848.

find

.

855. Given

and

849.

, find

.

850.

108 Part I: The Questions

Changing Matrices to the

Solving Systems of Equations

Echelon Form

Using Augmented Matrices

856–860 Use matrix operations to rewrite the matrix

861–870 Solve the system of equations using an

in echelon form.

augmented matrix.

861.

856.

862.

857.

863.

858.

864.

859.

865.

860.

Chapter 14: Systems of Equations and Inequalities

109

Solving Systems of Equations

866.

Using the Inverse of the

Coefficient Matrix

871–875 Solve the system of equations using the

inverse of the coefficient matrix.

871.

867.

872.

868.

873.

869.

874.

870.

875.

110 Part I: The Questions

Applying Cramer’s Rule to

879.

Solve Systems of Equations

876–880 Use Cramer’s Rule to solve the system of

equations.

876.

880.

877.

878.

Chapter 15

Sequences and Series

A sequence is a list of items; in mathematics, a sequence usually consists of numbers such as 1, 2, 4, 8, . . . or 1, 1, 2, 3, 5, 8, 13, . . . See the patterns in these two sequences?

That’s part of what this chapter is about: finding patterns in lists of numbers so you can write the rest of the numbers in the list.

A series is the sum of a list of numbers, such as 1 + 2 + 4 + 8. Many times, you can find a formula to help you add up the numbers in a series. Formulas are especially helpful when you have a lot of numbers to add or if they’re fractions or alternating negative and positive.

You can classify many sequences and series by type, which helps you determine particular terms or sums. Recognize the type, and you’re pretty much home free when doing your calculations.

The Problems You’ll Work On

In this chapter, you’ll work on sequences and series in the following ways:

✓ Finding a particular term in a sequence when given the general formula

✓ Determining a formula or standard expression for all the terms in a sequence

✓ Writing terms in a recursively formed sequence

✓ Becoming familiar with summation notation

✓ Computing the sum of terms in arithmetic and geometric series

✓ Generating terms by using the binomial theorem

What to Watch Out For

Don’t let common mistakes trip you up; keep in mind that when working with sequences and series, some challenges will include

✓ Correctly formulating terms in a sequence by using the order of operations

✓ Recognizing when a sequence or series is arithmetic or geometric

✓ Using the infinite series sum only when the ratio of consecutive terms has an absolute value less than 1

✓ Assigning the correct signs in the terms of an alternating sequence 112 Part I: The Questions

Finding Terms of Sequences

888.

881–885 Given the general term of the sequence,

find the fifth term.

881.

889. 2, 5, 10, 17, 26, ...

882.

890.

883.

Working with Recursively

Defined Sequences

891–895 Find the third and fourth terms of the

sequence.

884.

891.

,

,

885.

892.

,

,

Determining Rules for

Sequences

893.

,

,

886–890 Write an expression for the general term

of the sequence.

886. 1, 3, 9, 27, 81, ...

894.

,

,

887. 1, 2, 6, 24, 120, ...

895.

,

,

Chapter 15: Sequences and Series

113

903.

Adding Terms in an

Arithmetic Series

896–900 Find the sum of the arithmetic series.

904.

896.

905.

897.

898.

Finding Rules and Summing

Terms of a Series

906–910 Find the sum of the first ten terms

of the series.

899.

906. 1, 3, 5, 7, ...

900.

907. –8, –3, 2, 7, ...

Summing Terms of a Series

901–905 Find the sum of the series.

908.

901.

909. 12, 9, 6, 3, ...

902.

910. ‒2, ‒10, ‒18, ‒26, ...

114 Part I: The Questions

Calculating the Sum of a

918.

Geometric Series

911–915 Find the sum of the geometric series.

919. 64

‒ 16 + 4 ‒ 1 + ...

911.

920.

912.

Counting Items by Using

913.

Combinations

921–925 Compute the value of the combination.

914.

921.

915.

922.

Determining Formulas and

923.

Finding Sums

916–920 Find the sum of the infinite series.

924.

916.

925.

917. 27

+ 18 + 12 + 8 + ...

Chapter 15: Sequences and Series

115

933.

Constructing Pascal’s Triangle

926–930 Construct Pascal’s triangle to include the

coefficients of the expansion of the binomial.

926.

934.

927.

935.

928.

Utilizing the Binomial

Theorem

936–940 Use the binomial theorem to find the

coefficient of the term in the expansion of the

binomial.

929.

936.

term in

930.

937.

term in

Applying Pascal’s Triangle

931–935 Expand the binomial by using Pascal’s

938.

term in

triangle.

931.

939.

term in

932.

940.

term in

116 Part I: The Questions

Chapter 16

Introducing Limits and Continuity

“W hat’s the limit on this road?” “The sky’s the limit.” “We need to introduce term limits.” “Have you reached your limit?” People talk about limits in many ways.

In mathematics, a limit suggests that you’re approaching some value. Some functions, such as a rational function with a horizontal asymptote, have a limit as the x values move toward positive or negative infinity — that is, as the value of x gets very small or very large. Limits are another way of describing the characteristics of particular functions.

Although limits are often demonstrated graphically (a picture is worth a thousand words?), you can describe limits more precisely using algebra. In this chapter, you find examples of both techniques.

Coupled with limits is the concept of continuity — whether a function is defined for all real numbers or not. Investigating a function’s continuity (or lack thereof) is part of this chapter.

The Problems You’ll Work On

In this chapter, you’ll work on limits and continuity in the following ways:

✓ Looking at graphs for information on a function’s limits

✓ Using analytic techniques to investigate limits

✓ Performing algebraic operations to solve for a function’s limits

✓ Determining where a function is continuous

✓ Searching for any removable discontinuities

What to Watch Out For

When you’re working with limits and continuity, some challenges include the following:

✓ Recognizing a function’s behavior at negative infinity or positive infinity

✓ Using the correct technique for an analytic look at limits

✓ Factoring correctly when investigating limits algebraically

✓ Using the correct conjugates in algebraic procedures

✓ Forgetting that the “removable” part of a removable discontinuity doesn’t really change a function’s continuity; a function with a removable discontinuity is not continuous 118 Part I: The Questions

Determining Limits from

943.

Graphs

941–945 Given the graph of

, find

.

941.

Illustration by Thomson Digital

944.

Illustration by Thomson Digital

942.

Illustration by Thomson Digital

Illustration by Thomson Digital

Chapter 16: Introducing Limits and Continuity

119

945.

947.

Illustration by Thomson Digital

Illustration by Thomson Digital

948.

Determining One-Sided Limits

946–950 Given the graph of

, find the one-sided

limit.

946.

Illustration by Thomson Digital

Illustration by Thomson Digital

120 Part I: The Questions

949.

950.

Illustration by Thomson Digital

Illustration by Thomson Digital

Determining Limits from Function Values

951–955 Determine the limit using the values given in the chart.

951.

Illustration by Thomson Digital

952.

Illustration by Thomson Digital

953.

Illustration by Thomson Digital

Chapter 16: Introducing Limits and Continuity

121

954.

Illustration by Thomson Digital

955.

Illustration by Thomson Digital

Determining Limits from

961.

Function Rules

956–980 Find the limit.

962.

956.

963.

957.

964.

958.

965.

959.

966.

960.

122 Part I: The Questions

967.

976.

where

968.

977.

where

969.

978.

where

970.

979.

where

971.

972.

980.

where

973.

Applying Laws of Limits

974.

981–990 Use the limit laws to evaluate the

expression.

981. If

and

, then

975.

Chapter 16: Introducing Limits and Continuity

123

982. If

and

, then

989. If

and

, then

983. If

and

, then

990. If

and

, then

984. If

, then

Investigating Continuity

991–1,001 Determine for which values of x the

function is not continuous.

985. If

, then

991.

986. If

and

, then

992.

987. If

and

, then

993.

994.

988. If

and

, then

124 Part I: The Questions

995.

on the interval

999.

996.

on the interval

1,000.

997.

on the interval

1,001.

998.

Part II

The Answers

To access the free Cheat Sheet created specifically for this

book, go to www.dummies.com/cheatsheet/

1001precalculus.

In this part . . .

Y ou find answers and explanations for all 1,001 problems. You

already know that you can often solve a particular math-

ematics problem in more than one way, and you have several

more resources that you can refer to for those options (all pub-

lished by Wiley):

✓ Algebra I For Dummies (Mary Jane Sterling)

✓ Algebra II For Dummies (Mary Jane Sterling)

✓ Trigonometry For Dummies (Mary Jane Sterling)

✓ Pre-Calculus For Dummies (Yang Kuang, PhD, and

Elleyne Kase)

And after you’ve mastered pre-calculus and are ready to move on,

you find

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Chapter 17

Answers 1–100

Answers

1.

You can write a rational number as a fraction with an integer in both the numerator and denominator. You can’t write the square root of 3 as an integer.

2.

You can write a rational number as a fraction with an integer in both the numerator and denominator. The number

is an imaginary number with

, which you

can’t write as an integer.

3. 0

A natural number (also referred to as a counting number) is a positive integer. The number 0 isn’t positive.

4. 1.9

A natural number (also referred to as a counting number) is a positive integer. The number 1.9 is actually a fraction (rational number) and isn’t a natural number.

5.

An integer is a positive or negative counting number or 0. The number is

irrational.

6.

An integer is a positive or negative counting number or 0. You can reduce the number to

, but that’s still a rational number, which you can’t write as an integer.

7.

You can write the square root of 25 as 5. It is rational and not irrational.

128 Part II: The Answers

8.

Answers 1–100

The square root of 0 is 0. It isn’t irrational.

9.

Because

, you can write it as an integer.

10. 2 + i 4

Because

, you can write the expression as

.

11. commutative

The commutative property of addition says that changing the order of the terms in addition doesn’t change the result.

12. associative

The associative property of addition says that changing the grouping of the terms in addition doesn’t change the result.

13. associative

The associative property of multiplication says that changing the grouping of the factors in multiplication doesn’t change the result.

14. identity

When you add the additive identity, 0, to a term, it doesn’t change that term.

15. distributive

The distributive property of multiplication over addition says that multiplying each term in the parentheses gives the same result as the factored form.

16. inverse

Multiplying a number by its multiplicative inverse results in the multiplicative identity, 1.

17. distributive

The distributive property of multiplication over addition says that multiplying each term in the parentheses gives the same result as the factored form.

Chapter 17: Answers 129

18. commutative

Answers 1–100

The commutative property of multiplication says that changing the order of the factors in a multiplication problem doesn’t change the result.

19. transitive

The transitive property says that if one value is equal to both a second value and a third, then the second and third values are equal to one another.

20. inverse

Adding a number to its additive inverse results in 0.

21. 36

The order of operations says to perform powers (exponents) before multiplication: 22. 4

The order of operations says to first perform the addition in the numerator of the fraction because the fraction bar acts like a grouping symbol:

An alternate way to simplify is to factor the terms in the numerator and then divide: 23.

First, find common denominators for the fractions in the numerator and denominator; then add the fractions and reduce the results:

Multiply the numerator by the reciprocal of the denominator:

130 Part II: The Answers

24.

Answers 1–100

First, find common denominators for the fractions in the numerator and denominator; then subtract and add the fractions and reduce the results:

Multiply the numerator by the reciprocal of the denominator:

25.

Using the order of operations, you first simplify what’s under the radical. Applying the order of operations there, first multiply the three numbers together and then subtract the product from the 4:

Find the square root of 100 and then add the two numbers in the numerator together before reducing the fraction:

26. 3

Using the order of operations, you first simplify what’s under the radical. Applying the order of operations there, first multiply the three numbers together and then subtract the product from the 400:

Find the square root of 100 and then subtract the two numbers in the numerator before reducing the fraction:

27. 5

Using the order of operations, first square the 18 and then multiply the last two numbers under the radical. Add their results. In the denominator, add the two numbers in the absolute value.

Chapter 17: Answers 131

Add the numbers under the radical, find the square root, and then apply the absolute value before reducing the fraction:

Answers 1–100

28.

Perform the operations under each of the radicals, using the order of operations: Find the roots and then add the terms in the numerator:

29.

Complete the computations within the absolute value symbols before applying that operation. In the denominator, square the 3 and subtract.

Find the absolute value in the numerator and multiply in the denominator; then reduce the fraction:

30.

Square each of the terms in the absolute value and parentheses. Then find their differences.

Find the absolute values in the numerator and the square in the denominator; then find the difference in the numerator:

132 Part II: The Answers

31.

Answers 1–100

Illustration by Thomson Digital

First, find the intercepts of the line

and draw the line through those two inter-

cepts. You find the x-intercept by letting y = 0 in the equation

, giving you (7, 0).

You find the y-intercept by letting x = 0 in the same equation, giving you (0, 7).

Next, use the test point (0, 0) to see whether it’s part of the solution. Because 0 + 0 isn’t greater than 7, the origin isn’t in the solution, so shade on the other side of the line.

32.

Illustration by Thomson Digital

First, find the intercepts of the line

and draw the line through those two inter-

cepts. You find the x-intercept by letting y = 0 in the equation

, giving you

(2, 0). You find the y-intercept by letting x = 0 in the same equation, giving you (0, ‒4).

Chapter 17: Answers 133

Next, use the test point (0, 0) to see whether it’s part of the solution. Because 0 ‒ 0 is less than 4, the origin is part of the solution, so shade on that side of the line.

Answers 1–100

33.

Illustration by Thomson Digital

First, find the intercepts of the line

and draw the line through those two

intercepts; use a dashed line because the points on the line aren’t included in the solution. You find the x-intercept by letting y = 0 in the equation

, giving you

(5, 0). You find the y-intercept by letting x = 0 in the same equation, giving you (0, 4).

Next, use the test point (0, 0) to see whether it’s part of the solution. Because 0 + 0 isn’t greater than 20, the origin isn’t part of the solution, so shade on the other side of the line.

34.

Illustration by Thomson Digital

134 Part II: The Answers

First, find the intercepts of the line

and draw the line through those two

intercepts; use a dashed line because the points on the line aren’t included in the solution. You find the x-intercept by letting y = 0 in the equation

, giving you

Answers 1–100

(4, 0). You find the y-intercept by letting x = 0 in the same equation, giving you (0, ‒6).

Next, use the test point (0, 0) to see whether it’s part of the solution. Because 0 ‒ 0 is less than 12, the origin is part of the solution, so shade on that side of the line.

35.

Illustration by Thomson Digital

First, draw the vertical line through the intercept on the x-axis, (6, 0); make the line dashed because the points on the line aren’t part of the solution. Because all the x values must be greater than 6, shade to the right of the vertical line.

36.

Illustration by Thomson Digital

Chapter 17: Answers 135

First, draw the vertical line through the intercept on the x-axis, (‒2, 0). Because all the x values must be less than ‒2, shade to the left of the vertical line.

Answers 1–100

37.

Illustration by Thomson Digital

First, draw the horizontal line through the intercept on the y-axis, (0, ‒4). Because all the y values must be greater than ‒4, shade above the horizontal line.

38.

Illustration by Thomson Digital

First, draw the horizontal line through the origin, along the x-axis. Because all the y values must be less than 0, shade below the axis.

136 Part II: The Answers

39.

Answers 1–100

Illustration by Thomson Digital

First, find the intercepts of the line

. Letting y = 0, you get the x-intercept

.

Then, letting x = 0, you get the y-intercept (0, ‒7). Draw a line through those two points.

Use the test point (0, 0) to see whether it’s a solution. Because 0 isn’t less than 0 ‒ 7, the origin isn’t part of the solution, so you shade on the other side of the line.

40.

Illustration by Thomson Digital

First, find the intercepts of the line

. Letting y = 0, you get the x-intercept (1, 0);

and letting x = 0, you get the y-intercept

. Draw a line through those two points.

Use the test point (0, 0) to see whether it’s a solution. Because 0 isn’t greater than 0 + 1, the origin isn’t part of the solution, so you shade on the other side of the line.

Chapter 17: Answers 137

41. 1

Answers 1–100

Use the slope formula,

:

42.

Use the slope formula,

:

43. No slope

Use the slope formula,

:

You can’t divide by 0, so the line has no slope. The line is vertical.

44. 0

Use the slope formula,

:

The line is horizontal.

45. 10 units

Use the distance formula,

:

46. 25 units

Use the distance formula,

:

138 Part II: The Answers

47.

units

Answers 1–100

Use the distance formula,

:

You can simplify the radical:

48. (1, ‒3)

Use the midpoint formula,

:

49.

Use the midpoint formula,

:

50.

Use the midpoint formula,

:

51. 12 units

Find the lengths of the sides of the triangle by using the distance formula,

, for each side:

The perimeter of a triangle is the sum of the lengths of the sides, so

.

Chapter 17: Answers 139

52.

units

Answers 1–100

The opposite sides of a parallelogram are congruent, and you find the perimeter of a parallelogram with

, so you need to find the lengths of two adjacent sides

by using the distance formula,

, and put those lengths in the

formula:

So

.

53. (4, 3)

The center of a rhombus is at the intersection of the diagonals, which is also the midpoint of each diagonal. Find the midpoint of either diagonal by using

to

find the midpoint of the rhombus. Using the diagonal HK, the midpoint is 54. right and scalene

Find the lengths of the three sides by using the distance formula,

:

The triangle isn’t isosceles or equilateral. Use the Pythagorean theorem to

see whether it’s a right triangle:

This triangle is a right triangle; because the measures of the sides are all different, it’s also scalene.

55. equilateral

Find the lengths of the three sides by using the distance formula,

:

140 Part II: The Answers

Answers 1–100

The three lengths are the same, so the triangle is equilateral.

56. 12 units

The altitude to side AC is drawn from vertex B perpendicular to AC. Because AC lies along the x-axis, the altitude is a vertical line through point B and intersects the x-axis at the point (5, 0).

Illustration by Thomson Digital

The distance from (5, 0) to (5, 12) is 12 units.

57. 9 units

The triangle is a right triangle because sides DE and DF are parallel to the y-axis and x-axis, respectively, making DE perpendicular to DF.

Illustration by Thomson Digital

So the altitude to DF is side DE, which has a length of

units.

58. 40 square units

You find the area of a parallelogram by multiplying the length of one side by the distance between that side and the other side parallel to it; another form for area is

.

Chapter 17: Answers 141

The sides QR and PS are both parallel to the x-axis, and the distance between them is

.

Answers 1–100

Illustration by Thomson Digital

The lengths of QR and PS are 8 units. So the area of the parallelogram is square units.

59. 65π square units

The radius of a circle is half its diameter, so find the length of the diameter by using and halve that amount.

Half of

is

, which is the length of the radius.

You find the area of a circle with the formula

, so this circle has area

square units.

60. 84 square units

The triangle has its base along the x-axis, so the altitude drawn to the base from point B

has a length of 12 units.

Illustration by Thomson Digital

You find the area of a triangle with the formula

. The base of the triangle, AC, has

a length of

units, so the area of the triangle is

square units.

142 Part II: The Answers

61.

Answers 1–100

Use a parenthesis next to the 6 to show that it isn’t included in the solution.

62.

Write the numbers in order, from left to right, with the smaller number to the left. Use a bracket next to the ‒5 to show that it’s included in the solution. Use a parenthesis next to the 8 to show that it isn’t included in the solution.

63.

The parenthesis indicates that the 4 isn’t included in the solution; the bracket indicates that the 16 is included in the solution.

64.

The x indicates all numbers starting with ‒8 and larger. You use the greater-than-or-equal-to symbol to include the ‒8.

65.

The 3 is included in the solution, but the 7 is not.

66.

Neither the 0 nor the 11 is a part of the solution.

67.

or

The numbers between ‒5 and 7, including the 7, are in the solution. Also, the numbers greater than 11 are in the solution.

68.

Use brackets to show that the ‒6 and the –2 are included in the solution.

69.

The points between ‒10 and 0 are in the solution, but neither the ‒10 nor the 0 is included. Many people mistake this notation for the coordinates of a point, so make your intention clear when using it.

70.

All the numbers smaller than 3 are in the solution, but the number 3 isn’t. Also, the number 5 and all the numbers bigger than 5 are in the solution.

Chapter 17: Answers 143

71.

Answers 1–100

Add 3 to each side of the inequality:

Divide each side by 4:

In interval notation, you write the solution as

.

72.

Subtract 7 x from each side of the inequality:

Add 6 to each side:

Divide each side by ‒2, reversing the direction of the inequality:

In interval notation, you write the solution as

.

73.

Add ‒4 x to each side of the inequality:

Subtract 16 from each side:

Divide each side by ‒7, reversing the direction of the inequality.

In interval notation, you write the solution as

.

74.

Multiply each side of the inequality by ‒3, reversing the direction of the inequality: Subtract x from each side:

144 Part II: The Answers

Subtract 9 from each side:

Answers 1–100

Divide each side by 4:

In interval notation, you write the solution as

.

75.

or

First factor the quadratic and set it equal to 0 to find the critical numbers: Use a sign line to find the sign of each factor in the intervals determined by the critical numbers:

You want the product to be positive or zero, so all x less than ‒1 or greater than 8 will give you that product, as will ‒1 and 8:

In interval notation, you write the solution as

.

76.

or

First factor the quadratic and set it equal to 0 to find the critical numbers: Use a sign line to find the sign of each factor in the intervals determined by the critical numbers:

You want the product to be positive, so all x less than 4 or greater than 6 will give you that product:

In interval notation, you write the solution as

.

Chapter 17: Answers 145

77.

or

Answers 1–100

First factor the polynomial and set it equal to 0 to find the critical numbers: Use a sign line to find the sign of each factor in the intervals determined by the critical numbers:

You want the product to be negative or zero, so all x less than ‒3 or between 0 and 3

will give you that product, as will ‒3, 0, and 3:

In interval notation, you write the solution as

.

78.

or

First use grouping to factor the polynomial. Set the function equal to 0 to find the critical numbers:

Use a sign line to find the sign of each factor in the intervals determined by the critical numbers:

You want the product to be negative, so all x less than ‒1 or between 1 and 6 will give you that product:

In interval notation, you write the solution as

.

79.

or

First determine the critical numbers by setting the numerator and denominator equal to 0: 146 Part II: The Answers

Use a sign line to find the sign of each factor in the intervals determined by the critical numbers:

Answers 1–100

You want the quotient to be positive or zero, so all x less than ‒4 or greater than 3 will give you that quotient; also, 3 will give you a 0.

In interval notation, you write the solution as

.

80.

or

First determine the critical numbers by setting the numerator and denominator equal to 0:

Use a sign line to find the sign of each factor in the different intervals determined by the critical numbers:

You want the quotient to be negative, so all x less than ‒7 or between ‒5 and 5 will give you that quotient:

In interval notation, you write the solution as

.

81.

Rewrite the absolute value inequality as

Then add 4 to each expression:

In interval notation, you write the solution as

.

Chapter 17: Answers 147

82.

or

Answers 1–100

Rewrite the absolute value inequality as

Then add 3 to each side of each expression:

Divide each side by 2:

In interval notation, you write the solution as

.

83.

First divide each side of the inequality by 4:

Now rewrite the absolute value inequality as

Subtract 1 from each expression:

Divide each expression by 2:

In interval notation, you write the solution as

.

84.

First subtract 3 from each side of the inequality:

Next, multiply each side by ‒1, reversing the direction of the inequality sign: Now rewrite the absolute value inequality as

Subtract 5 from each expression:

In interval notation, you write the solution as

.

85. no solution

First subtract 5 from each side of the inequality:

You can stop right there. The absolute value expression is either a positive number or 0.

It can’t be negative, so there’s no solution.

148 Part II: The Answers

86.

Answers 1–100

The root outside the radical, 3, goes in the denominator of the fraction. The power, 4, goes in the numerator.

87.

Remember that when you don’t see a root indicated outside the radical, you assume that it’s the square root. That means you can then write the fraction as

. When you

bring the denominator up to the numerator, the sign of the exponent becomes negative:

.

88.

The denominator of the fraction, 3, becomes the index of the radical. The numerator of the fraction, 5, is the power of x under the radical.

89.

First rewrite the expression as a fraction with the power of x in the denominator:

.

The 4 in the denominator of the exponent means that the root is 4; the 3 in the numerator of the exponent means that the x is raised to the third power.

90.

When multiplying expressions with the same base, you add the exponents: 91.

When multiplying expressions with the same base, you add the exponents: 92.

When dividing expressions with the same base, you subtract the exponents: 93. x

When dividing expressions with the same base, you subtract the exponents.

Chapter 17: Answers 149

94.

Answers 1–100

When dividing expressions with the same base, you subtract the exponents: Find a common denominator before combining the fractions:

95.

The GCF is

. Factoring gives you

96.

The GCF is

. Factoring gives you

97.

The GCF is

. Factoring gives you

Remember that when dividing, you subtract exponents:

98.

The GCF is

. Remember, the factor with the lowest power of the variable is the

GCF. Factoring gives you

99.

The GCF is

. Dividing each term by that factor gives you

Therefore,

.

150 Part II: The Answers

100.

The GCF is

. The GCF could also be

if you prefer a positive factor in

front. The exponent

is the smallest of the three exponents on y.

Factoring gives you

Answers 101–200

101. x = 5

Square both sides of the equation:

Now solve for x:

Check the solution in the original equation:

It checks.

102. x = ‒3

Square both sides of the equation:

Now solve for x:

Check the solution in the original equation:

It checks.

Chapter 17: Answers 151

103. x = 9

First subtract 4 from both sides of the equation:

Next, square both sides of the equation:

Answers 101–200

Solve for x:

Check the solution in the original equation:

It checks.

104. x = ‒3

First add 3 to both sides of the equation:

Next, square both sides of the equation:

Solve for x:

Check the solution in the original equation:

It checks.

152 Part II: The Answers

105.

x = 5

Square both sides of the equation. Be careful to square the binomial correctly.

Answers 101–200

Set the quadratic equation equal to 0 and solve for x:

The solutions of the quadratic equation are

and x = 5.

Check

in the original equation:

This solution is extraneous.

Now check x = 5 in the original equation:

This solution checks.

106. x = ‒1

Square both sides of the equation. Be careful to square the binomial correctly.

Set the quadratic equation equal to 0 and solve for x:

The solutions of the quadratic equation are x = ‒1 and x = ‒9.

Check x = ‒1 in the original equation:

This solution works.

Chapter 17: Answers 153

Now check x = ‒9 in the original equation:

This solution is extraneous.

Answers 101–200

107.

,

First add 6 to each side of the equation:

Now square both sides of the equation. Be careful to square the binomial correctly.

Set the quadratic equation equal to 0 and solve for x:

The solutions of the quadratic equation are

and x = ‒2.

Check

in the original equation:

This solution works.

Now check x = ‒2 in the original equation:

This solution also works.

108. x = ‒2, x = ‒3

First add x to each side of the equation:

154 Part II: The Answers

Now square both sides of the equation. Be careful to square the binomial correctly.

Set the quadratic equation equal to 0 and solve for x:

Answers 101–200

The solutions of the quadratic equation are x = ‒3 and x = ‒2.

Check x = ‒3 in the original equation:

This solution works.

Now check x = ‒2 in the original equation:

This solution also works.

109. x = 3, x = ‒1

First add

to each side of the equation:

Now square both sides of the equation. Be careful to square the binomial with the radical correctly.

Now subtract x and 2 from each side to isolate the radical on the right: Square both sides of the equation. Then set the quadratic equation equal to 0 and solve for x:

Chapter 17: Answers 155

The two solutions of the quadratic are x = 3 and x = ‒1.

Check x = 3 in the original equation:

Answers 101–200

This solution works.

Now check x = ‒1 in the original equation:

This solution works.

110.

,

First subtract

from each side of the equation:

Now square both sides of the equation. Be careful to square the binomial with the radical correctly.

Now subtract x and 55 (that is, 49 + 6) from each side to isolate the radical on the right: Before squaring both sides of the equation, divide each term by 2. Then set the quadratic equation equal to 0 and solve for x:

The two solutions of the quadratic are x = ‒5 and

.

156 Part II: The Answers

Check x = ‒5 in the original equation:

Answers 101–200

This solution works.

Now check

in the original equation:

This solution also works.

111.

Multiply both the numerator and the denominator by

:

Reduce the fraction:

112.

Multiply both the numerator and the denominator by

:

Reduce the fraction:

Chapter 17: Answers 157

113.

Multiply both the numerator and the denominator by

:

Answers 101–200

114.

Multiply both the numerator and the denominator by

:

In the numerator, simplify the radical on the right:

Reduce the fraction:

115.

Multiply both the numerator and the denominator by the conjugate of the denominator: Simplify the fraction:

116.

Multiply both the numerator and the denominator by the conjugate of the denominator: Simplify the fraction:

158 Part II: The Answers

117.

Multiply both the numerator and the denominator by the conjugate of the denominator: Answers 101–200

Simplify the fraction:

118.

Multiply both the numerator and the denominator by the conjugate of the denominator: In the numerator, simplify the radical on the right. Then simplify and reduce the fraction: 119.

Multiply both the numerator and the denominator by the conjugate of the denominator: Simplify the radicals in the numerator; then subtract in the denominator: None of the radicals combine.

120.

Multiply both numerator and denominator by the conjugate of the denominator: Simplify the radical in the numerator; then subtract in the denominator: Chapter 17: Answers 159

121. −79

Substitute –3 for n in the expression for

:

Answers 101–200

122. 11

Substitute –6 for x in the expression for

:

123. −256

Substitute 3 for y in the expression for

:

Next, find the 8th power of 2. Using the order of operations, you perform the power before performing multiplication or division. The negative sign in front of the 2 acts as a ‒1 multiplier, so you perform it after the power.

124. 27

This function is a piecewise function, so evaluate it by using the portion of the function defined for

. In this case, because

, use

125.

Replace x with

and then simplify the expression.

126. domain: {−2, 0, 1, 2, 5}; range: {2, 3, 4, 6}

The domain is the set of x values, and the range is the set of y values. f is a function because each x value maps to only one y value.

160 Part II: The Answers

127. domain:

; range:

or

The domain is the set of x values, and the range is the set of y values for which the function is defined. In this case, x is defined for all real numbers greater than −5. You don’t include −5 because there’s an open dot on −5 in the graph. If

, then the

range is

. If

, then the range is all real numbers less than −1. So the range

is

or

.

Answers 101–200 128. domain:

; range:

The domain and range of a linear function are all real numbers, written

. The

graph of the function makes this fact clear.

Illustration by Thomson Digital

129. domain:

; range:

The domain of any polynomial function is all real numbers. However, the value of the function can only be a positive real number because squaring any number gives a positive result. Letting x = 0 gives

which is the minimum value of the function, so

the range is the set of all real numbers greater than or equal to 1. The graph of the function makes this fact clear.

Illustration by Thomson Digital

130. domain:

; range:

The definition of the function restricts the values of t to the real numbers between 0

and 8, inclusive. Look at a factored version of the function equation:

.

Chapter 17: Answers 161

Use the TABLE function of a graphing calculator to find some values of the function.

Illustration by Thomson Digital

Answers 101–200

The minimum value of the function is 0 and the maximum value of the function is 256, so the range is [0, 256]. The graph of the function makes this fact clear.

Illustration by Thomson Digital

131. domain:

; range:

The domain of any polynomial function is all real numbers. The range of an odd degree polynomial is always all real numbers. The graph of the function makes this fact clear.

Illustration by Thomson Digital

132. domain:

; range:

The argument of the square root must be greater than or equal to zero because the square root of a negative number is not defined. So solve

162 Part II: The Answers

Therefore, the domain is all real numbers less than or equal to 2.5.

Square roots are always positive or zero. Therefore, the range is all real numbers greater than or equal to zero. The graph of the function makes this fact clear.

Answers 101–200

Illustration by Thomson Digital

133. domain:

; range:

The domain and range of the cube root function (and any odd-numbered integer root) is all real numbers. The graph of the function makes this fact clear.

Illustration by Thomson Digital

134. domain:

; range:

The domain of the function is restricted to those values that don’t make the denominator equal zero.

Thus, the domain is

.

Use the TABLE function of a graphing calculator to find some values of the function.

You see that the values of Y approach 0.5 as x becomes very small or very large.

1

Chapter 17: Answers 163

However, the value of the function will never equal 0.5. (Try to solve You

end up with

, which isn’t true!) Therefore, 0.5 is not in the range of

Answers 101–200

Illustration by Thomson Digital

The graph of the function makes this fact clear.

Illustration by Thomson Digital

135. domain:

; range:

The domain of an absolute value function is all real numbers. The absolute value part of the function will always be greater than or equal to zero. (In this case, you know the absolute value will equal zero if x = 4.) Because 3 is being added to the absolute value, the range is

The graph of the function makes this fact clear.

Illustration by Thomson Digital

164 Part II: The Answers

136.

A function is even if

for any value of x. In the case of a polynomial function,

it’s even if the exponent of every variable is an even integer.

Therefore,

is an even function.

Answers 101–200 137.

A function is even if

for any value of x. In the case of a polynomial function,

it’s even if the exponent of every variable is an even integer.

Therefore,

is an even function.

138.

A function is odd if

for any value of x. In the case of a polynomial function,

it’s odd if the exponent of every variable is an odd integer and there are no constant terms.

Therefore,

is an odd function.

139.

A function is odd if

for any value of x. In the case of a polynomial function,

it’s odd if the exponent of every variable is an odd integer and there are no constant terms.

Therefore,

is an odd function.

140.

Test

to see whether it’s an even function or an odd function:

Chapter 17: Answers 165

Therefore,

is neither even nor odd. Note that the exponents of the vari-

ables are a mixture of even and odd integers.

141.

Answers 101–200

This relation is a one-to-one function because each x value maps to just one y value and each y value maps to just one x value.

142.

This relation is a one-to-one function because each x value maps to just one y value and each y value maps to just one x value.

143.

Illustration by Thomson Digital

You can use two tests to determine whether a graph is a one-to-one function. The horizontal line test states that if a horizontal line intersects a function’s graph more than once, then the function is not one-to-one. The vertical line test states that a relation is a function if there are no vertical lines that intersect the graph at more than one point.

This graph passes both the horizontal line test and the vertical line test, so it’s a one-to-one function.

144.

Suppose that a and b are real numbers and that

To prove that

is one-to-

one, show that a = b:

Let

Then, substituting, you get

and

.

Taking the cube root of each side, a = b, so the function is one-to-one over its domain.

166 Part II: The Answers

145.

Suppose that a and b are real numbers and that

To prove that

is one-to-

one, show that a = b:

Answers 101–200

Let

Then, substituting, you get

Cross-multiply and distribute:

Subtract 2 ab from each side to get b = a. Therefore, is one-to-one over its domain.

146.

Show that f and g are inverses of each other by verifying that and

:

147.

Show that f and g are inverses of each other by verifying that and

:

148.

Show that f and g are inverses of each other by verifying that and

:

Chapter 17: Answers 167

149.

Show that f and g are inverses of each other by verifying that and

:

Answers 101–200

150.

Show that f and g are inverses of each other by verifying that and

:

151.

Find the inverse by interchanging the x and y values: 152.

Change

to y:

Interchange x and y:

168 Part II: The Answers

Now solve for y:

Rename y as

:

Answers 101–200

153.

Change

to y:

Interchange x and y:

Now solve for y:

Rename y as

:

154.

Change

to y:

Interchange x and y:

Now solve for y by first multiplying each term in the equation by y: Chapter 17: Answers 169

Rename y as

:

155.

Change

to y:

Answers 101–200

Interchange x and y:

Now solve for y:

Rename y as

:

156.

Change

to y:

Interchange x and y:

Now solve for y:

Rename y as

:

157.

Change

to y:

170 Part II: The Answers

Interchange x and y:

Now solve for y:

Answers 101–200

Rename y as

:

158.

Change

to y:

Interchange x and y:

Now solve for y:

Rename y as

:

159.

Change

to y:

Chapter 17: Answers 171

Interchange x and y:

Now solve for y:

Answers 101–200

Rename y as

:

160.

Note that the domain of f is restricted to those values of x between 0 and 1, inclusive.

The domain is restricted so that f is one-to-one and an inverse exists. Also note that the range of f is restricted to those values of y between 0 and 1, inclusive. The range of f becomes the domain of .

Change

to y:

Interchange x and y:

Now solve for y:

Rename y as

:

The domain of the same as the range of f, so it’s restricted to those values of y between 0 and 1, inclusive. This function, on its restricted domain, is its own inverse.

172 Part II: The Answers

161.

Find

by adding the expressions for

and

and simplifying:

Answers 101–200 162.

Find

by subtracting the expression for

from the expression for

and

simplifying:

163.

Find

by multiplying the expressions for

and

. Use the FOIL method to

multiply the binomials and then simplify:

164.

Find

by dividing the expression for

by the expression for

. Simplify the

quotient if possible.

The value of x cannot be –1.

Chapter 17: Answers 173

165. 112

First find

and

by substituting –2 for x in the expressions for

and

.

Then multiply the results.

Answers 101–200

Alternatively, find

by multiplying the expressions for

and

. Then substi-

tute –2 for x in the resulting expression to find

.

166.

The notation

means the same thing as

. So substitute for x in

:

167.

The notation

means the same thing as

. So substitute

for x in

.

Then simplify the resulting expression.

168.

The notation

means the same thing as

. So substitute

for each x in

. Then simplify the resulting expression:

174 Part II: The Answers

169. x

The notation

means the same thing as

. So substitute

for x in

.

Then simplify the resulting expression:

Answers 101–200

170.

The notation

means the same thing as

. So substitute 3 x for x in

:

171.

The notation

means the same thing as

. So substitute

for x in

. Then simplify the resulting expression.

172.

The notation

means the same thing as

. So substitute

for each x in

. Then simplify the resulting expression.

173.

The notation

means the same thing as

. So substitute

for each x

in

. Then simplify the resulting expression.

Chapter 17: Answers 175

174. −159

The notation

means the same thing as

. First, find

by substitut-

ing 12 for x in the expression for

. Next, find

by substituting the value of

for x in the expression for

. Finally, simplify the resulting expression.

Answers 101–200

175. 4

The notation

means the same thing as

. First, find

by substitut-

ing –3 for x in the expression for

. Next, find

by substituting the value of

for x in the expression for

. Finally, simplify the resulting expression.

176.

The notation

means the same thing as

. Work from the inside out

to evaluate the function. First, substitute

, the expression for

, for x in the

expression for

:

Next, evaluate

:

Now evaluate

by substituting

for x in the expression for

:

177. 4

To find

, substitute

for x in the expression for

Then simplify the result-

ing expression.

Now plug this result and the expression for

into the difference quotient and

simplify:

176 Part II: The Answers

178.

To find

, substitute

for x in the expression for

Then simplify the result-

ing expression.

Answers 101–200

Now plug this result and the expression for

into the difference quotient and

simplify:

179.

To find

, substitute

for x in the expression for

Then simplify the result-

ing expression.

To make the computation a little easier, work on the numerator of the difference quotient next. Find

, the numerator of the difference quotient, by using the

result you found for

and subtracting the expression for

from it. Be sure to

simplify the result.

Finally, plug this expression into the numerator of the difference quotient and simplify: 180.

To find

, substitute

for x in the expression for

To make the computation a little easier, work on the numerator of the difference quotient next. Find

, the numerator of the difference quotient, by using the

result you found for

and subtracting the expression for

from it. Be sure to

simplify the result; the lowest common denominator for the two fractions in the second step is

.

Chapter 17: Answers 177

Answers 101–200

Finally, plug this expression into the numerator of the difference quotient and simplify: 181.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y:

Finally, rewrite the function, replacing the y with

:

Here are the graphs of

and

:

Illustration by Thomson Digital

178 Part II: The Answers

Note that the graph of

is the reflection of

over the line

. After using the

slopes and y-intercepts to help you sketch both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

182.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Answers 101–200

Solve for y:

Finally, rewrite the function, replacing the y with

:

In slope-intercept form, that’s

.

Here are the graphs of

and

:

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After using the

slopes and y-intercepts to help you sketch both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

Chapter 17: Answers 179

183.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y:

Answers 101–200

Finally, rewrite the function, replacing the y with

:

Here are the graphs of

and

, which have a common intercept at (0, 0) and

shared points at (1, 1) and (‒1, ‒1):

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After sketching

both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

184.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y:

Finally, rewrite the function, replacing the y with

:

180 Part II: The Answers

Here are the graphs of

and

, which have a common intercept at (0, 0) and

shared points at (1, 1) and (‒1, ‒1):

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After sketching

both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

185.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y:

Finally, rewrite the function, replacing the y with

:

Chapter 17: Answers 181

Here are the graphs of

and

:

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After sketching

both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

Also note that the functions give you the “opposite operations in the opposite order”: In

, you cube x and then add 3; in

, you subtract 3 and then take the cube root

of the result.

186.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y:

Finally, rewrite the function, replacing the y with

:

182 Part II: The Answers

Here are the graphs of

and

:

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After sketching

both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

Also note that the functions give you the “opposite operations in the opposite order”: In

, you subtract 4, then take a root, and then add 7; in

, you subtract 7, and

then cube the result, and then add 4.

187.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y:

Finally, rewrite the function, replacing the y with

:

Chapter 17: Answers 183

Here are the graphs of

and

:

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After using the

intercepts and asymptotes to help you sketch both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

When the function and its inverse intersect, they do so at a point on the line

.

188.

To find the inverse, first rewrite the function as

. Then exchange the y and x’s:

Solve for y:

Finally, rewrite the function, replacing the y with

:

184 Part II: The Answers

Here are the graphs of

and

:

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After using the

intercepts and asymptotes to help you sketch both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

.

When the function and its inverse intersect, they do so at a point on the line

.

189.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y, starting by multiplying each side by the denominator on the right: Finally, rewrite the function, replacing the y with

:

Chapter 17: Answers 185

Here are the graphs of

and

:

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After sketching

both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

. When the function and its inverse intersect,

they do so at a point on the line

.

190.

To find the inverse, first rewrite the function as

. Then exchange the y and x:

Solve for y, starting by multiplying each side by the denominator on the right: 186 Part II: The Answers

Finally, rewrite the function, replacing the y with

:

Here are the graphs of

and

:

Answers 101–200

Illustration by Thomson Digital

Note that the graph of

is the reflection of

over the line

. After sketching

both the function and its inverse on the same graph, you should see the symmetry (mirror image) over the diagonal

. When the function and its inverse intersect,

they do so at a point on the line

.

191.

Illustration by Thomson Digital

Rewrite the equation in vertex form,

:

Chapter 17: Answers 187

The vertex is ( h, k), so in this parabola, you see the vertex at (0, ‒1).

The parabola opens upward if a is positive and downward if a is negative, and if

,

then the curve is steep. Therefore, this parabola opens upward and is flattened rather than steep.

The y-intercept is (0, ‒1), the vertex. Find the x-intercepts by setting y = 0 and solving for x:

Answers 101–200

So the x- intercepts are

and

.

192.

Illustration by Thomson Digital

Rewrite the equation in vertex form,

:

The vertex is ( h, k), so in this parabola, you see the vertex at (‒3, ‒1).

The parabola opens upward if a is positive and downward if a is negative, and if

,

then the curve is steep. Because a = 1, this parabola opens upward.

To find the y-intercept, let x = 0 and solve for y; the y-intercept is (0, 8). Find the x-intercepts by setting y = 0 and solving for x: Therefore, the x- intercepts are (‒2, 0) and (‒4, 0).

188 Part II: The Answers

193.

Answers 101–200

Illustration by Thomson Digital

Start with the equation in vertex form,

:

The vertex is ( h, k), so in this parabola, you see the vertex at (‒2, 9).

The parabola opens upward if a is positive and downward if a is negative, and if

,

then the curve is steep. Because a = –1, this parabola opens downward.

To find the y-intercept, let x = 0 and solve for y: So the y-intercept is (0, 5). Find the x-intercepts by setting y = 0 and solving for x: Therefore, the x-intercepts are (‒5, 0) and (1, 0).

Chapter 17: Answers 189

194.

Answers 101–200

Illustration by Thomson Digital

Start with the equation in vertex form,

:

The vertex is ( h, k), so in this parabola, you see the vertex at

.

The parabola opens upward if a is positive and downward if a is negative, and if

,

then the curve is steep. Because a = 4, this parabola is steep and opens upward.

To find the y-intercept, let x = 0 and solve for y: So the y-intercept is (0, 2). Find the x-intercepts by setting y = 0 and solving for x: Use the quadratic formula to solve for x:

There are no real roots, so there are no x-intercepts.

190 Part II: The Answers

195.

Answers 101–200

Illustration by Thomson Digital

Rewrite the equation in vertex form,

:

The vertex is ( h, k), so in this parabola, you see the vertex at (1, 7).

The parabola opens upward if a is positive and downward if a is negative, and if

,

then the curve is steep. Because a = –7, this parabola is steep and opens downward.

To find the y-intercept, let x = 0 and solve for y: So the y-intercept is (0, 0). Find the x-intercepts by setting y = 0 and solving for x: Therefore, the x-intercepts are (0, 0) and (2, 0).

196.

The vertex is (0, 9), and the graph opens downward. Using the vertex form of a quadratic equation,

, these characteristics are represented by

, where

a is a negative number.

The x-intercepts are (‒3, 0) and (3, 0). Substitute (3, 0) into the equation and solve for a: So the equation of the parabola is

.

197.

The vertex is (‒1, ‒4), and the graph opens upward. Using the vertex form of a quadratic equation,

, these characteristics are represented by

,

where a is a positive number.

Chapter 17: Answers 191

The x-intercepts are (‒3, 0) and (1, 0). Choosing one of the intercepts, (1, 0), substitute the values into the equation and solve for a:

So the equation of the parabola is

.

Answers 101–200

198.

The vertex is (‒3, 8), and the graph opens downward. Using the vertex form of a quadratic equation,

, these characteristics are represented by

, where

a is a negative number.

The x-intercepts are (‒5, 0) and (‒1, 0). Substitute (‒1, 0) into the equation and solve for a: So the equation of the parabola is

.

199.

It’s hard to tell exactly where the vertex is, because it isn’t at an intersection on the grid. But the x-intercepts are easy to determine; they’re at (1, 0) and (7, 0). Also, the curve goes through the point (3, 12).

An equation representing the intercepts is

, where a is a negative

number. Substituting 3 for x and 12 for y from the point on the curve, you can solve for a: Now replace the a in the equation that uses the intercepts:

To put this equation in vertex form, first multiply the 7 in the parentheses by and

move the product outside the parentheses. Then complete the square using the two remaining terms in the parentheses.

Recall that to complete the square of a quadratic ax 2 + bx, add and subtract the square of half b to the variable terms

. The first three terms should be a

perfect square trinomial. Multiply the last term by the

and move it outside the

parentheses.

192 Part II: The Answers

Here are the calculations:

Answers 101–200

This equation, which is in vertex form, tells you that the vertex is

.

200.

It’s hard to tell exactly where the vertex is, because it isn’t at an intersection on the grid. But the x-intercepts are easy to determine; they’re at (‒6, 0) and (8, 0). Also, the curve has a y-intercept of (0, ‒6).

An equation representing the x-intercepts is

, where a is a positive

number. Substituting 0 for x and ‒6 for y from the point on the curve, you can solve for a: Now replace the a in the equation using the intercepts:

To put this equation in the standard form, multiply the binomials, and complete the square.

Recall that to complete the square of a quadratic ax 2 + bx, add and subtract the square of half b to the variable terms

. The first three terms should be a perfect

square trinomial. Multiply the last term by and move it outside the parentheses.

Here are the calculations:

This equation, which is in vertex form, tells you that the vertex is

.

Chapter 17: Answers 193

201. domain:

; intercepts: (‒4, 0), (0, 2)

The domain begins at x = ‒4 with the x-intercept (‒4, 0) and the curve rises as x goes to positive infinity. The y-intercept is (0, 2).

The graph of

looks like half of a parabola:

Answers 201–300

Illustration by Thomson Digital

202. domain:

; intercept: (2, 0)

The domain begins at with the point

and the curve falls as x goes toward posi-

tive infinity. The function has an x-intercept at (2, 0) and no y-intercept.

The graph of

looks like half of a parabola:

Illustration by Thomson Digital

194 Part II: The Answers

203. domain:

; intercepts: (0, ‒6), (4, 0)

The domain begins at 0 with (0, ‒6), which is the y-intercept, and the curve rises as x goes to positive infinity. The x-intercept is (4, 0).

The graph of

looks like half of a parabola:

Answers 201–300

Illustration by Thomson Digital

204. domain:

; intercepts:

, (0, ‒8)

The domain begins at 0 with (0, ‒8), which is the y-intercept, and the curve rises as x goes to positive infinity. The x-intercept is

.

The graph of

looks like half of a parabola:

Illustration by Thomson Digital

Chapter 17: Answers 195

205. domain:

; intercept:

The domain begins at with

and the curve falls as x goes to positive infinity.

There is no y-intercept. The x-intercept is

. The graph looks like half a parabola.

The graph of

looks like half of a parabola:

Answers 201–300

Illustration by Thomson Digital

206. domain:

; intercepts: (‒3, 0),

The domain is all real numbers. The curve falls moving from left to right. The x-intercept is (‒3, 0), and the y-intercept is

.

The graph of

looks like two halves of a parabola, with one half rotated

around 180 degrees from the other:

Illustration by Thomson Digital

196 Part II: The Answers

207. domain:

; intercepts:

,

The domain is all real numbers. The curve rises moving from left to right. The x-intercept is

, and the y-intercept is

which is about (0, ‒4.6).

The graph of

looks like two halves of a parabola, with one half rotated

around 180 degrees from the other:

Answers 201–300

Illustration by Thomson Digital

208. domain:

; intercepts:

, (‒167, 0)

The domain is all real numbers. The curve rises moving from left to right. The y-intercept is

, and the x-intercept is (‒167, 0).

The graph of

looks like two halves of a parabola, with one half

rotated around 180 degrees from the other:

Illustration by Thomson Digital

Chapter 17: Answers 197

209. domain:

; intercepts: (0, ‒7),

The domain is all real numbers. The curve rises moving from left to right. The y-intercept is (0, ‒7), and the x-intercept is

.

The graph of

looks like two halves of a parabola, with one half rotated

around 180 degrees from the other:

Answers 201–300

Illustration by Thomson Digital

210. domain:

; intercepts:

,

The domain is all real numbers. The curve falls moving from left to right. The y-intercept is

, and the x-intercept is

.

The graph of

looks like two halves of a parabola, with one half

rotated around 180 degrees from the other:

Illustration by Thomson Digital

198 Part II: The Answers

211. range:

The domain of the function is all real numbers, and the range is from 3 to positive infinity (including 3). The graph has the characteristic V shape with the lowest point at (3, 3).

There is no x-intercept; the y-intercept is (0, 9).

Here’s the graph of

:

Answers 201–300

Illustration by Thomson Digital

212. range:

The domain of the function is all real numbers, and the range is from ‒7 to positive infinity (including ‒7). The graph has the characteristic V shape, made more steep with the multiplier of 3; the lowest point is (0, ‒7), the y-intercept. The x-intercepts are and

.

Here’s the graph of

:

Illustration by Thomson Digital

Chapter 17: Answers 199

213. range:

The domain is all real numbers, and the range goes from 0 to positive infinity. You can draw the curve starting with the graph of the parabola

with the portion

between x = ‒2 and x = 2 reflected over the x-axis. The y-intercept is (0, 4), and the x-intercepts are (‒2, 0) and (2, 0).

Here’s the graph of

:

Answers 201–300

Illustration by Thomson Digital

214. range:

The domain is all real numbers, and the range goes from 0 to positive infinity. You can draw the curve starting with the graph of the cubic

with the portion from

to ‒1 reflected over the x-axis. The y-intercept is (0, 1), and the x-intercept is (‒1, 0).

Here’s the graph of

:

Illustration by Thomson Digital

200 Part II: The Answers

215. range:

The domain is all real numbers, and the range goes from 0 to positive infinity. You can draw the curve starting with the graph of the cube root

with the portion

from

to 3 reflected over the x-axis. The curve comes down to the point (3, 0) with a vertical tangent. The y-intercept is

, and the x-intercept is (3, 0).

Here’s the graph of

:

Answers 201–300

Illustration by Thomson Digital

216. intercepts: (0, 0), (4, 0), (‒5, 0)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (0, 0), (4, 0), and (‒5, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 0).

To sketch the graph, note that the exponents on the factors are odd numbers, so the curve crosses the x-axis at each x-intercept. The curve rises to the right as x approaches positive infinity, as determined when you test an x value greater than the right-most intercept. Here’s the graph:

Illustration by Thomson Digital

Chapter 17: Answers 201

217. intercepts: (3,0), (‒2,0), (0, 36)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (3, 0) and (‒2, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 36).

To sketch the graph, note that the exponents on the factors are even numbers, so the curve just touches the x-axis at each x-intercept. The curve rises to the right as x approaches positive infinity, as determined when you test an x value greater than the right-most intercept. Here’s the graph:

Answers 201–300

Illustration by Thomson Digital

218. intercepts: (‒4, 0), (3, 0), (0, ‒48)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (‒4, 0) and (3, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, ‒48).

To sketch the graph, note that the curve crosses the x-axis at the x-intercept (3, 0) because the exponent on

is odd, and the curve just touches the x-axis at (‒4, 0)

because the exponent on

is even. The curve rises to the right as x approaches

positive infinity, as determined when you test an x value greater than the right-most intercept. Here’s the graph:

202 Part II: The Answers

Answers 201–300

Illustration by Thomson Digital

219. intercepts: (0, 0), (2, 0)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (0, 0) and (2, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 0).

To sketch the graph, note that the exponents on the factors are odd numbers, so the curve crosses the x-axis at both x-intercepts. The curve also flattens out at each x-intercept, with a low point between the intercepts at x = 1. The curve rises to the right as x approaches positive infinity, as determined when you test an x value greater than the right-most intercept. Here’s the graph:

Illustration by Thomson Digital

Chapter 17: Answers 203

220. intercepts: (‒5, 0), (‒2, 0), (2, 0), (5, 0), (0, 100)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (‒2, 0), (2, 0), (‒5, 0), and (5, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 100).

To sketch the graph, note that the exponent on each factor is odd, so the curve crosses the x-axis at each intercept. The function is symmetric about the y-axis, because replacing x with ‒ x doesn’t change the equation. The curve rises to the right as x approaches positive infinity, as determined when you test an x value greater than the right-most intercept. Here’s the graph:

Answers 201–300

Illustration by Thomson Digital

221. intercepts: (0, 0), (‒3, 0), (3, 0), (‒6, 0), (6, 0)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (0, 0), (‒3, 0), (3, 0), (‒6, 0), and (6, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 0).

To sketch the graph, note that the exponents on the factors are all odd, so the curve crosses the x-axis at each intercept. The function is symmetric about the origin, because replacing each x with ‒ x changes

to

. The curve rises to the right as x

approaches positive infinity, as determined when you test an x value greater than the right-most intercept. Here’s the graph:

204 Part II: The Answers

Answers 201–300

Illustration by Thomson Digital

222. intercepts: (0, 0), (‒4, 0), (4, 0)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (0, 0), (‒4, 0), (4, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 0).

To sketch the graph, note that the curve crosses the x-axis at the intercepts (‒4, 0) and (4, 0), because the exponents on those factors are odd, and the curve touches the axis at (0, 0), because the exponent on the factor x is even. The function is symmetric about the y-axis, because replacing each x with ‒ x doesn’t change the function equation. The curve rises to the right as x approaches positive infinity, as determined when you test an x-value greater than the right-most intercept. Here’s the graph: Illustration by Thomson Digital

Chapter 17: Answers 205

223. intercepts: (‒5, 0), (5, 0), (‒1, 0), (0, ‒25)

Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (‒5, 0), (5, 0), and (‒1, 0).

Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, ‒25).

To sketch the graph, note that the curve crosses the x-axis at the intercepts (‒5, 0) and (5, 0), because the exponents on those factors are odd, and the curve touches the axis at (‒1, 0), because the exponent on that factor is even. The curve rises to the right as x approaches positive infinity. Here’s the graph:

Answers 201–300

Illustration by Thomson Digital

224. asymptotes: x = ‒3, y = 1

Find the vertical asymptote of

by setting the denominator equal to 0 and solv-

ing for x. The vertical asymptote is x = ‒3.

The horizontal asymptote is y = 1, because the highest power in both the numerator and denominator is 1, and the ratio formed by the two variables raised to the first power is

.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercept of

is (2, 0). Find the y-intercept by let-

ting x = 0 and solving for y. The y-intercept is

. Here’s the graph:

206 Part II: The Answers

Answers 201–300

Illustration by Thomson Digital

225. asymptotes: x = 1, x = ‒4, y = 1

Factor the numerator and denominator to determine whether there are any removable discontinuities and to aid in finding intercepts and asymptotes:

No common factors appear in the numerator and denominator, so there’s no removable discontinuity.

Find the vertical asymptotes by setting the denominator equal to 0 and solving for x.

The vertical asymptotes are x = 1 and x = ‒4.

The horizontal asymptote is y = 1, because the highest power in both the numerator and denominator is 2, and the ratio formed by the two variables raised to the second power is

.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts are (‒3, 0) and (3, 0). Find the y-intercept by letting x = 0 and solving for y. The y-intercept is

. Here’s the graph:

Chapter 17: Answers 207

Answers 201–300

Illustration by Thomson Digital

226. asymptotes: x = 5, x = ‒5, y = 0

Factor the denominator to determine whether there are any removable discontinuities and to aid in finding intercepts and asymptotes:

No common factors appear in the numerator and denominator, so there’s no removable discontinuity.

Find the vertical asymptotes by setting the denominator equal to 0 and solving for x.

The vertical asymptotes are x = 5 and x = ‒5.

The horizontal asymptote is y = 0, because the highest power in the numerator is smaller than the highest power in the denominator.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercept is (‒3, 0). Find the y-intercept by letting x = 0

and solving for y. The y-intercept is

. Here’s the graph:

208 Part II: The Answers

Answers 201–300

Illustration by Thomson Digital

227. asymptotes: x = ‒5, y = 1

Factor the numerator and the denominator to determine whether there are any removable discontinuities and to aid in finding intercepts and asymptotes: The common factor of ( x – 5) in the numerator and denominator produces a removable discontinuity at x = 5.

After factoring out the common factor, find the asymptotes. The vertical asymptote is x = ‒5; you find that asymptote by setting the denominator equal to 0 and solving for x.

The horizontal asymptote is y = 1, because the highest power in both the numerator and denominator of the reduced fraction is 1, and the ratio formed by the two variables raised to the first power is

.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercept is (0, 0). Find the y-intercept by letting x = 0 in the factored version and solving for y. The y-intercept is also (0, 0). Here’s the graph: Chapter 17: Answers 209

Answers 201–300

Illustration by Thomson Digital

228. asymptotes: x = 2, y = 1

Factor the numerator and the denominator to determine whether there are any removable discontinuities and to aid in finding intercepts and asymptotes: The common factor of ( x – 5) in the numerator and denominator produces a removable discontinuity at x = 5.

After factoring out the common factor, find the asymptotes. The vertical asymptote is x = 2; you find that asymptote by setting the denominator equal to 0 and solving for x.

The horizontal asymptote is y = 1, because the highest power in both the numerator and denominator of the reduced fraction is 1, and the ratio formed by the two variables raised to the first power is

.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercept is (‒3, 0). Find the y-intercept by letting x = 0 in the factored version and solving for y. The y-intercept is

. Here’s the graph:

210 Part II: The Answers

Answers 201–300

Illustration by Thomson Digital

229. asymptotes: x = 1, y = x + 1

The vertical asymptote of

is x = 1; you find that asymptote by setting the

denominator equal to 0 and solving for x.

Find the slant/oblique asymptote by dividing the numerator by the denominator and writing an equation using the quotient without the remainder. The slant/oblique asymptote is

.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercepts of are (‒2, 0) and (2, 0). Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 4). Here’s the graph: Illustration by Thomson Digital

Chapter 17: Answers 211

230. asymptotes: ,

,

Find the vertical asymptotes of

by setting the denominator equal to 0

and solving for x. The vertical asymptotes are x = 1 and x = ‒1.

Find the slant/oblique asymptote by dividing the numerator by the denominator and writing an equation using the quotient without the remainder. The slant/oblique asymptote is

.

For the graph, you also need to identify the intercepts. Find the x-intercepts by letting y = 0 and solving for x. The x-intercept is (2, 0). Find the y-intercept by letting x = 0 and solving for y. The y-intercept is (0, 2). Here’s the graph: Answers 201–300

Illustration by Thomson Digital

231. reflection over the x-axis, right 5 units, down 7 units The negative sign in front of the parentheses in the g function represents a reflection over the x-axis. The –5 inside the parentheses represents a shift right 5 units, and the –7 on the outside represents a shift down 7 units.

232. shrink of , left 4 units

The in front of the parentheses in the g function represents a shrink of (the graph gets flatter). The +4 inside the parentheses represents a shift left 4 units.

233. reflection over the x-axis, left 3 units, up 2 units

The negative sign in front of the absolute value in the g function represents a reflection over the x-axis. The +3 inside the absolute value represents a shift left 3 units, and the +2 on the outside represents a shift up 2 units.

212 Part II: The Answers

234. stretch of 3, down 8 units

The 3 in front of the x term in the g function represents a stretch of 3 (the graph gets steeper). The –8 after the x term represents a shift down 8 units.

235. stretch of 4, reflection over the y-axis, up 3 units

The 4 in front of the square root sign in the g function represents a stretch of 4; the graph gets steeper. The negative sign under the square root sign represents a reflection over the y-axis. The +3 after the square root sign represents a shift up 3 units.

236. (1, 0), (–4, 2), (3, –5)

Answers 201–300

The transformation for the function is left 2 units (subtract 2 from the x coordinates) and down 4 units (subtract 4 from the y coordinates).

237. (3, 26), (–2, 36), (5, 1)

The transformations for the function are a vertical stretch of 5 (5 times the y coordinates) and up 6 units (add 6 to the y coordinates).

238. (6, 1), (1, –1), (8, 6)

The transformations for the function are a reflection over the x-axis (the signs of the y coordinates change), up 5 units (add 5 to the y coordinates after the sign change), and right 3 units (add 3 to the x coordinates).

239. (–2, –6), (–7, –5), (0, –8.5)

The transformations for the function are a shrink of ( times the y coordinates), left 5 units (subtract 5 from the x coordinates), and down 8 units (subtract 8 from the y coordinates after finding half).

240. (–3, 1), (2, 3), (–5, –4)

The transformations for the function are a reflection over the y-axis (the signs of the x coordinates change) and down 3 units (subtract 3 from the y coordinates).

Chapter 17: Answers 213

241. vertex: (0, ‒3)

The parabola is shifted 3 units down from the basic graph’s vertex of (0, 0). It’s steeper because of the 2 multiplier. Here’s the graph:

Answers 201–300

Illustration by Thomson Digital

242. vertex: (2, 0)

The parabola is shifted 2 units right from the basic graph’s vertex of (0, 0). It’s flatter because of the multiplier. Here’s the graph:

Illustration by Thomson Digital

214 Part II: The Answers

243. vertex: (3, 4)

The parabola is shifted 3 units right and 4 units up from the basic graph’s vertex of (0, 0).

Here’s the graph:

Answers 201–300

Illustration by Thomson Digital

244. vertex: (‒1, 6)

The parabola is shifted 1 unit left and 6 units up from the basic graph’s vertex of (0, 0).

It’s steeper and opens downward because of the ‒3 multiplier. Here’s the graph: Illustration by Thomson Digital

Chapter 17: Answers 215

245. vertex: (4, 2)

The parabola is shifted 4 units right and 2 units up from the basic graph’s vertex of (0, 0). It’s flatter and opens downward because of the

multiplier.

Answers 201–300

Illustration by Thomson Digital

246. vertical asymptote: x = 3

From the basic graph’s intersection of vertical and horizontal asymptotes at (0, 0), the curve is shifted to 3 units the right. The vertical asymptote, which you find by setting the denominator equal to 0 and solving for x, is x = 3.

The horizontal asymptote is y = 0, because the highest power of the variable in the numerator is smaller than that in the denominator. Here’s the graph of

:

Illustration by Thomson Digital

216 Part II: The Answers

247. vertical asymptote: x = ‒4

From the basic graph’s intersection of vertical and horizontal asymptotes at (0, 0), the curve is shifted 4 units to the left. The vertical asymptote, which you find by setting the denominator equal to 0 and solving for x, is x = ‒4.

The horizontal asymptote is y = 0, because the highest power of the variable in the numerator is smaller than that in the denominator. The numerator of 2 stretches the graph vertically. Here’s the graph of

:

Answers 201–300

Illustration by Thomson Digital

248. vertical asymptote: x = ‒1

From the basic graph’s intersection of vertical and horizontal asymptotes at (0, 0), the curve is shifted 1 unit to the left. The vertical asymptote, which you find by setting the denominator equal to 0 and solving for x, is x = ‒1.

The horizontal asymptote is y = 0, because the highest power of the variable in the numerator is smaller than that in the denominator. The numerator of 3 stretches the graph vertically, and the negative sign flips the graph over the x-axis. Here’s the graph of

:

Chapter 17: Answers 217

Answers 201–300

Illustration by Thomson Digital

249. vertical asymptote: x = 5

Rewrite the function equation by factoring ‒1 from the terms in the denominator: This puts the function in the general format of the basic graph,

.

From the basic graph’s intersection of vertical and horizontal asymptotes at (0, 0), the curve is shifted 5 units to the right and 1 unit up. The vertical asymptote, which you find by setting the denominator equal to 0 and solving for x, is x = 5.

The horizontal asymptote is y = 1, because the +1 raises the basic graph 1 unit up off the x-axis. The negative sign flips the graph over the horizontal asymptote. Here’s the graph: Illustration by Thomson Digital

218 Part II: The Answers

250. vertical asymptote:

Rewrite the function equation by factoring ‒3 from the terms in the denominator: This puts the function in the general format of the basic graph,

.

From the basic graph’s intersection of vertical and horizontal asymptotes at (0, 0), the curve is shifted unit to the right and 2 units down. The vertical asymptote, which you find by setting the denominator equal to 0 and solving for x, is

.

Answers 201–300

The horizontal asymptote is y = ‒2, because the ‒2 lowers the basic graph 2 units below the x-axis. The negative sign flips the graph over the horizontal asymptote.

Here’s the graph:

Illustration by Thomson Digital

251. {–2, 0}

Each term on the left side of the equation contains the common factor x, so factor it out:

, so you can rewrite the given equation as

.

Now use the zero product property to solve the equation:

The solution set is {–2, 0}.

Chapter 17: Answers 219

252. {–3, 4}

Subtract 9 from each side to set the equation equal to 0:

Now factor the left side:

Then use the zero product property to solve the equation:

Answers 201–300

The solution set is {–3, 4}.

253.

The expression

is a difference of squares, so you can factor it as

. Rewrite the given equation as

and use the zero product property to solve for x:

The solution set is

254.

The expression

is a perfect square trinomial, so there’s just one bino-

mial factor and one solution:

The solution set is

220 Part II: The Answers

255. {–4, 7}

First, subtract 24 from each side to set the equation equal to zero: All three terms on the left side of the equation have a common factor of 3, so factor it out: Now divide both sides by 3:

Answers 201–300

Factor the left side of the equation to rewrite it as

and then use the zero

product property to solve for x:

The solution set is {–4, 7}.

256.

First, factor the left side of the equation:

Now use the zero product property to solve for x:

The solution set is

257.

First, factor the left side of the equation:

Chapter 17: Answers 221

Now use the zero product property to solve for x:

The solution set is

258.

First, factor the left side of the equation:

Answers 201–300

Now use the zero product property to solve for x:

The solution set is

259.

You can write

as a quadratic equation by letting

(because x 3 is

a factor of both x 6 and x 3). Then

. When you substitute u for x 3, you get

. Factor the left side of the equation:

Use the zero product property to solve for u:

You’re almost there, but you aren’t looking for the value of u; what you need is the value of x. Substitute the u values into the equation to solve for x:

The solution set is

222 Part II: The Answers

260.

You can write

as a quadratic equation by letting

(because x 2 is a

factor of both x 4 and x 2). Then

When you substitute u for x 2, you get

. Factor the left side of the equation:

Use the zero product property to solve for u:

Answers 201–300

You’re almost there, but you aren’t looking for the value of u; what you need is the value of x. Substitute the u values into the equation to solve for x:

The solution set is

261.

The given equation is a quadratic equation already written in standard form

, so you know that a = 4, b = –7, and c = –15. Plug the values of a, b, and c into the quadratic formula,

, and solve:

Find both x values:

The solution set is

Chapter 17: Answers 223

262.

First write the equation in standard form

in order to determine a, b,

and c:

So a = 1, b = –1, and c = –4.

Then plug the values of a, b, and c into the quadratic formula,

, and

solve:

Answers 201–300

The solution set is

263.

First write the equation in standard form

in order to determine a, b,

and c:

So a = 4, b = 8, and c = –1. Plug the values of a, b, and c into the quadratic formula,

, and solve:

The solution set is

224 Part II: The Answers

264.

The equation is already written in standard form

, so you know that

a = 2, b = –5, and c = –5. Plug the values of a, b, and c into the quadratic formula,

, and solve:

Answers 201–300

The solution set is

265.

First, expand the left side of the equation by multiplying

, and then write

the equation in standard form

in order to determine a, b, and c:

So a = 1, b = –8, and c = 5. Plug the values of a, b, and c into the quadratic formula,

, and solve:

The solution set is

Chapter 17: Answers 225

266. {‒4, 2}

First, add the constant term, 8, to both sides of the equation so that you have only terms with the variable on the left side of the equation:

Next, complete the square by taking half of the linear coefficient (the b value from the quadratic equation — in this case, 2) and squaring the result:

. Add that value

to both sides of the equation:

Answers 201–300

Now simplify the equation, factor the left side, and take the square root of both sides: Isolate the variable to solve the equation:

The expression

equals –4 or 2, so the solution set is {–4, 2}.

267. {‒9, ‒2}

In this equation, terms containing the variable are only on the left side. So complete the square by taking half of the linear coefficient (what would be the b value if you made this a quadratic equation — in this case, 11) and squaring the result. Add that value,

, to both sides of the equation:

Next, simplify the equation, factor the left side, and take the square root of both sides: Isolate the variable to solve the equation:

The expression

equals

or

, so the solution set is {‒9, ‒2}.

226 Part II: The Answers

268.

In this equation, terms containing the variable are only on the left side. So complete the square by taking half of the linear coefficient (what would be the b value if you made this a quadratic equation — in this case, 4) and squaring the result:

.

Add that value to both sides of the equation:

Next, simplify the equation, factor the left side, and take the square root of both sides: Answers 201–300

Isolate the variable to solve the equation:

The solution set is

269.

First, divide every term by the leading coefficient, 5, so that a = 1: Now complete the square by taking half of the linear coefficient (what would be the b value if you made this a quadratic equation — in this case, ) and squaring the result.

Add that value to both sides of the equation:

Simplify the equation, factor the left side, and then take the square root of both sides: Chapter 17: Answers 227

Isolate the variable to solve the equation:

The expression

equals –1 or , so the solution set is

270.

First you have to get only terms with the variable on the left side of the equation, so add the constant term, 2, to both sides of the equation:

Answers 201–300

Next, divide every term by the leading coefficient, 3, so that a = 1: Now complete the square by taking half of the linear coefficient (the b value from the quadratic equation — in this case,

) and squaring the result. Add that value to

both sides of the equation:

Simplify the equation, factor the left side, and then take the square root of both sides: Isolate the variable to solve the equation:

The expression

equals or 5, so the solution set is

228 Part II: The Answers

271.

Both left-side terms have a common factor of ‒3 x 4, so factor it out of each term: Set each factor equal to zero and solve for x to find the x-intercepts: Answers 201–300 272.

All left-side terms have a common factor of 2 x, so factor it out of each term: Set each factor equal to zero and solve for x to find the x-intercepts: 273.

All left-side terms have a common factor of x 2, so factor it out of each term: Then factor the expression inside the parentheses to get

Set each factor equal to zero and solve for x to find the x-intercepts: Chapter 17: Answers 229

274. –2, 2

This equation is in the form of the difference of squares. The formula for factoring the difference of squares is

, so you can rewrite the given equation as follows:

Set each factor equal to zero and solve for x to find the x-intercepts. Start with the first factor,

:

Answers 201–300

Remember that

, so

. These aren’t real roots or intercepts,

so you disregard this solution when choosing your answer.

On to the second factor. Notice that

is also the difference of squares. Factor it

further to get

Then set those factors equal to zero and solve for x:

These are real roots, so the x-intercepts are –2 and 2.

275.

All the left-side terms have a common factor of 2 x 2, so factor it out of each term: Now set each factor equal to zero and solve for x to find the x-intercepts. The first factor is pretty straightforward:

You need to use the quadratic formula,

, to solve

because it isn’t factorable. It is in standard form

, though, so you can easily

determine that a = 1, b = 10, and c = 3. Plug those values into the quadratic formula: 230 Part II: The Answers

Answers 201–300 276.

You can use the sum of cubes to solve this equation. The formula for factoring the sum of cubes is

. Plug in the values from the given equation,

noting that both 27 and 8 are perfect cubes:

Now set each factor equal to zero and solve for x to find the x-intercepts. The first factor is pretty straightforward:

So

is an x-intercept.

You need to use the quadratic formula,

, to solve

because it isn’t factorable. It is in standard form

, though, so you can

easily determine that a = 9, b = –6, and c = 4. Plug those values into the quadratic formula: Chapter 17: Answers 231

(Remember that

, so

.) These roots aren’t real, so

they aren’t intercepts.

277.

First, factor by grouping. Break up the polynomial into sets of two and then find the greatest common factor of each set and factor it out. Finally, factor again.

Answers 201–300

Next, set each factor equal to zero and solve for x to find the x-intercepts: 278.

First, factor by grouping. Break up the polynomial into sets of two and then find the greatest common factor of each set and factor it out. Finally, factor again.

Next, set each factor equal to zero and solve for x to find the x-intercepts: 279.

First, factor by grouping. Break up the polynomial into sets of two and then find the greatest common factor of each set and factor it out. Finally, factor again.

232 Part II: The Answers

Next, set each factor equal to zero and solve for x to find the x-intercepts: 280.

First, factor by grouping. Break up the polynomial into sets of two and then find the greatest common factor of each set and factor it out. Finally, factor again.

Answers 201–300

Next, set each factor equal to zero and solve for x to find the x-intercepts: 281. possible number of positive real roots: 2 or 0; possible number of negative real roots: 1

Descartes’s rule of signs says that the number of positive roots is equal to the number of changes in sign in

or is less than that number by an even number (so you keep

subtracting 2 until you get either 1 or 0).

In

, there’s a change in sign between the first and second terms

and between the third and fourth terms, so the possible number of positive real roots is 2 or 0. (Remember, 6 x 3 has a positive sign even though the plus sign isn’t written.) To find the possible number of negative real roots, find

and count the number of

sign changes from term to term. The number of negative roots is equal to the changes in sign for

or must be less than that number by an even number.

There’s one sign change between the second and third terms, so the possible number of negative real roots is 1.

282. possible number of positive real roots: 2 or 0; possible number of negative real roots: 2 or 0

Descartes’s rule of signs says that the number of positive roots is equal to the number of changes in sign in

or is less than that number by an even number (so you keep

subtracting 2 until you get either 1 or 0).

Chapter 17: Answers 233

In

, there’s a change in sign between the first and second

terms and between the third and fourth terms, so the possible number of positive real roots is 2 or 0. (Remember, 18 x 6 has a positive sign even though the plus sign isn’t written.)

To find the possible number of negative real roots, find

and count the number of

sign changes from term to term. The number of negative roots is equal to the changes in sign for

or must be less than that number by an even number.

There’s a sign change between the first and second terms and between the third and fourth terms, so the possible number of negative real roots is 2 or 0.

Answers 201–300

283. possible number of positive real roots: 3 or 1; possible number of negative real roots: 1

Descartes’s rule of signs says that the number of positive roots is equal to the number of changes in sign in

or is less than that number by an even number (so you keep

subtracting 2 until you get either 1 or 0).

In

, there’s a change in sign between the first and second terms,

between the second and third terms, and between the third and fourth terms, so the possible number of positive real roots is 3 or 1. (Remember, 7 x 6 has a positive sign even though the plus sign isn’t written.)

To find the possible number of negative real roots, find

and count the number of

sign changes from term to term. The number of negative roots is equal to the changes in sign for

or must be less than that number by an even number.

There’s a sign change between the first and second terms, so the possible number of negative real roots is 1.

284. possible number of positive real roots: 4 or 2 or 0; possible number of negative real roots: 0

Descartes’s rule of signs says that the number of positive roots is equal to the number of changes in sign in

or is less than that number by an even number (so you keep

subtracting 2 until you get either 1 or 0).

In

, there’s a change in sign between the first and second

terms, between the second and third terms, between the third and fourth terms, and between the fourth and fifth terms, so the possible number of positive real roots is 4 or 2 or 0. (Remember, 2 x 4 has a positive sign even though the plus sign isn’t written.) To find the possible number of negative real roots, find

and count the number of

sign changes from term to term. The number of negative roots is equal to the changes in sign for

or must be less than that number by an even number.

There are no sign changes, so the possible number of negative real roots is 0.

234 Part II: The Answers

285. possible number of positive real roots: 0; possible number of negative real roots: 5 or 3 or 1

Descartes’s rule of signs says that the number of positive roots is equal to the number of changes in sign in

or is less than that number by an even number (so you keep

subtracting 2 until you get either 1 or 0).

In

, there are no changes in sign, so the possible number of

positive real roots is 0. (Remember, x 5 has a positive sign even though the plus sign isn’t written.)

To find the possible number of negative real roots, find

and count the number of

sign changes from term to term. The number of negative roots is equal to the changes in sign for

or must be less than that number by an even number.

Answers 201–300

There’s a sign change between each pair of terms, so there are five changes in sign.

The possible number of negative real roots is 5 or 3 or 1.

286.

The rational root theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial.

In

, the constant term is −7, and its factors are and

The

leading coefficient is 7, and its factors are and

Now you can divide. The possible

rational roots are

,

, and

.

287.

The rational root theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial.

In

, the constant term is −5, and its factors are and

The leading

coefficient is 3, and its factors are and

Now you can divide. The possible rational

roots are

,

,

, and

.

288.

The rational root theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial.

In

, the constant term is −1, and its factors are

The leading

coefficient is 4, and its factors are , , and

Now you can divide. The possible

rational roots are

,

, and

.

Chapter 17: Answers 235

289.

The rational root theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial.

In

, the constant term is 8, and its factors are , , , and

The leading coefficient is 4, and its factors are , , and

Now you can divide:

Eliminating the duplicates, you have , , , ,

, and

.

Answers 201–300

290.

The rational root theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial.

In

, the constant term is −8, and its factors are , , , and

The leading coefficient is 5, and its factors are and

Now you can divide.

The possible rational roots are

,

,

,

,

,

,

, and

.

291.

When dividing one polynomial by another, write the divisor outside the division symbol and the dividend inside. The terms should be in descending order of the exponents. If there are any missing terms in the dividend, put zeros in as placeholders. You find the terms in the quotient one at a time by multiplying the divisor by an appropriate term.

To write the remainder as a fraction, place it over the divisor. Written this way, the quotient is

236 Part II: The Answers

292.

When dividing one polynomial by another, write the divisor outside the division symbol and the dividend inside. The terms should be in descending order of the exponents. If there are any missing terms in the dividend, put zeros in as placeholders. You find the terms in the quotient one at a time by multiplying the divisor by an appropriate term.

Answers 201–300

To write the remainder as a fraction, place it over the divisor. Written this way, the quotient is

293.

When dividing one polynomial by another, write the divisor outside the division symbol and the dividend inside. The terms should be in descending order of the exponents. The linear term, x, is missing from the dividend, so insert 0 x in order to leave space for an x term column. You find the terms in the quotient one at a time by multiplying the divisor by an appropriate term.

To write the remainder as a fraction, place it over the divisor. Written this way, the quotient is

Chapter 17: Answers 237

294.

When dividing one polynomial by another, write the divisor outside the division symbol and the dividend inside. The terms should be in descending order of the exponents. The square term, x 2, is missing from the dividend, so insert 0 x 2 in order to leave space for an x 2 term column. You find the terms in the quotient one at a time by multiplying the divisor by an appropriate term.

Answers 201–300

To write the remainder as a fraction, place it over the divisor. Written this way, the quotient is

295.

When dividing one polynomial by another, write the divisor outside the division symbol and the dividend inside. The terms should be in descending order of the exponents. If there are any missing terms in the dividend, put zeros in as placeholders. You find the terms in the quotient one at a time by multiplying the divisor by an appropriate term.

To write the remainder as a fraction, place it over the divisor. Written this way, the quotient is

238 Part II: The Answers

296. quotient: ; remainder: −20

Synthetic division works only when a polynomial is divided by a linear factor and when the coefficient of the linear factor is 1. If

is a factor of the polynomial, then x = n is

a zero of the polynomial, and if x = n is a zero, then

is a factor.

In this case,

may be a factor of

, so x = –5 may be a zero. Therefore,

write –5 on the outside of the synthetic division sign. Write the coefficients of the polynomial on the inside. Here’s the process, step by step:

1. Drop the 1 from the coefficient line to below the division line.

2. Multiply 1 by –5 to get –5 and write that under the 6.

3. Add

to get 1. Write that under the line in the same column.

Answers 201–300

4. Multiply 1 by –5 to get –5 and write that under the –15.

5. Add

to get –20. Write that under the line in the same column. This

amount is the remainder.

The first two numbers under the line are the coefficients of the quotient. The answer is always one degree lower than the original polynomial. In this case, the quotient is

,

and the remainder is –20.

297. quotient:

; remainder: 48

Synthetic division works only when a polynomial is divided by a linear factor and when the coefficient of the linear factor is 1. If

is a factor of the polynomial, then x = n is

a zero of the polynomial, and if x = n is a zero, then

is a factor.

In this case,

may be a factor of

, so x = 4 may be a zero.

Therefore, write 4 on the outside of the synthetic division sign. Write the coefficients of the polynomial on the inside. Here’s the process, step by step:

1. Drop the 2 from the coefficient line to below the division line.

2. Multiply 4 by 2 to get 8 and write that under the –13.

3. Add

to get –5. Write that under the line in the same column.

4. Multiply 4 by –5 to get –20 and write that under the 27.

5. Add

to get 7. Write that under the line in the same column.

6. Multiply 4 by 7 to get 28 and write that under the 20.

7. Add

to get 48. This amount is the remainder.

The first three numbers under the line are the coefficients of the quotient. The answer is always one degree lower than the original polynomial. In this case, the quotient is

, and the remainder is 48.

Chapter 17: Answers 239

298. quotient:

; remainder: –47

Synthetic division works only when a polynomial is divided by a linear factor and when the coefficient of the linear factor is 1. If

is a factor of the polynomial, then x = n is

a zero of the polynomial, and if x = n is a zero, then

is a factor.

In this case,

may be a factor of

, so x = –3 may be a zero. Therefore, write –3

on the outside of the synthetic division sign. Write the coefficients of the polynomial on the inside, using 0 as the coefficients of the missing and x terms. Here’s the process, step by step:

1. Drop the 1 from the coefficient line to below the division line.

2. Multiply −3 by 1 to get –3 and write that under the first 0.

3. Add

to get –3. Write that under the line in the same column.

Answers 201–300

4. Multiply –3 by –3 to get 9 and write that under the next 0.

5. Add

to get 9. Write that under the line in the same column.

6. Multiply –3 by 9 to get –27 and write that under the –20.

7. Add

to get –47. This amount is the remainder.

The first three numbers under the line are the coefficients of the quotient. The answer is always one degree lower than the original polynomial. In this case, the quotient is

, and the remainder is –47.

299. 3

The remainder is 0 when

is divided by

, where a is a zero of the function. So

use synthetic division to test each value.

The remainder equals 66, so x = –3 isn’t a zero.

The remainder equals 0, so x = 3 is a zero. From this point, you can use a “reduced polynomial” and not the original. The following steps show you the work with the original coefficients.

240 Part II: The Answers

The remainder equals 168, so x = –6 isn’t a zero.

The remainder equals 48, so x = 6 isn’t a zero.

The remainder equals –2, so x = 1 isn’t a zero.

300. –1

Answers 201–300

The remainder is 0 when

is divided by

, where a is a zero of the function. So

use synthetic division to test each value.

The remainder equals 2,520, so x = –8 isn’t a zero.

The remainder equals −24, so x = –8 isn’t a zero.

The remainder equals 0, so x = –1 is a zero. From this point, you can use a “reduced polynomial” and not the original. The following steps show you the work with the original coefficients.

The remainder equals 240, so x = 4 isn’t a zero.

The remainder equals 4,536, so x = 8 isn’t a zero.

Chapter 17: Answers 241

301.

The zeros of the polynomial are –3, 2, and 3, so the factors of the polynomial are

,

, and

Multiply the factors to find the polynomial:

302.

The zeros of the polynomial are –4, –3, 0, 1, and 2, so the factors of the polynomial are

,

,

,

, and

. Multiply the factors to find the polynomial:

303.

The zeros of the polynomial are

,

, –1, and 2, so the factors of the polynomial are

,

,

, and

.

Answers 301–400

Remember that

and that imaginary roots occur when a polynomial has a factor

of the form

. When inserting those imaginary roots back into the polynomial form, you’ll end up with i 2 terms that you can replace with ‒1.

Multiply the factors to find the polynomial:

304.

The zeros of the polynomial are

,

, –1, and –4, so the factors of the polynomial

are

,

,

, and

. Multiply the factors to find the polynomial:

242 Part II: The Answers

305.

The zeros of the polynomial are

,

, 1, and 2, so the factors of the polyno-

mial are

,

,

, and

. Multiply the factors to find the

polynomial:

306.

The zeros of the polynomial are

Answers 301–400

−3, 1, and 2, so the factors are

,

, and

.

The polynomial also has a constant factor, a. Therefore, the polynomial is given by The value of the polynomial at x = −1 is 24, so plug that in for x and solve for a: Therefore, the polynomial is

307.

The zeros of the polynomial are −4, −2, and 3, so the factors are

,

, and

. The polynomial also has a constant factor, a. Therefore, the polynomial is given by Chapter 17: Answers 243

The value of the polynomial at x = 2 is 48, so plug that in for x and solve for a: Therefore, the polynomial is

308.

The zeros of the polynomial are −3, −2, 1, and 2, so the factors are

,

,

, and

. The polynomial also has a constant factor, a. Therefore, the polynomial is given by

Answers 301–400

The value of the polynomial at x = 0 is 36, so plug that in for x and solve for a: Therefore, the polynomial is

309.

The zeros of the polynomial are

,

, –2, and 1, so the factors are

,

,

, and

. The polynomial also has a constant factor, a. Therefore, the polynomial is given by

244 Part II: The Answers

The value of the polynomial at x = 2 is 16, so plug that in for x and solve for a: Therefore, the polynomial is

310.

The zeros of the polynomial are

,

, –3, and 2, so the factors are

,

,

, and

.

Remember that

and that imaginary roots occur when a polynomial has a factor

of the form

. When inserting those imaginary roots back into the polynomial form, Answers 301–400

you’ll end up with i 2 terms that you can replace with ‒1.

The polynomial also has a constant factor, a. Therefore, the polynomial is given by The value of the polynomial at x = –1 is 120, so plug that in for x and solve for a: Therefore, the polynomial is

Chapter 17: Answers 245

311.

Illustration by Thomson Digital

End behavior: The leading coefficient test says that if the degree of the polynomial is odd and the leading coefficient is negative, then the left side of the graph points up and Answers 301–400

the right side points down.

Intercepts: Solve

to find the x-intercepts:

Each zero has multiplicity 1 (meaning the exponent on each factor is 1, which is an odd number), so the graph crosses the x-axis at each zero.

Find the y-intercept by finding

:

Test points: Test points in the intervals

,

,

, and

.

Illustration by Thomson Digital

246 Part II: The Answers

312.

Illustration by Thomson Digital

End behavior: The leading coefficient test says that if the degree of the polynomial is odd and the leading coefficient is positive, then the left side of the graph points down and the right side points up.

Answers 301–400

Intercepts: Solve

to find the x-intercepts. The rational root theo-

rem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial. Therefore, the possible rational zeros are Use synthetic division to find that x = 2 is one of the zeros: Now solve the resulting equation,

, to find the remaining zeros:

Each zero has multiplicity 1 (the exponent on each factor is 1, which is an odd number), so the graph crosses the x-axis at each zero.

Find the y-intercept by finding

:

Chapter 17: Answers 247

Test points: Test points in the intervals

and

.

Illustration by Thomson Digital

313.

Answers 301–400

Illustration by Thomson Digital

End behavior: The leading coefficient test says that if the degree of the polynomial is even and the leading coefficient is positive, then both sides of the graph point up.

Intercepts: Solve

to find the x-intercepts:

The zeros at x = –2 and x = 3 have multiplicity 1 (the exponent on each factor is 1, which is an odd number), so the graph crosses the x-axis at –2 and 3. The zero at x = 0

has multiplicity 2 (the exponent of x is an even number, meaning there’s no change of sign on either side), so the graph touches the x-axis at x = 0 but doesn’t cross it.

Find the y-intercept by finding

:

248 Part II: The Answers

Test points: Test points in the intervals

,

,

, and

.

Illustration by Thomson Digital

314.

Answers 301–400

Illustration by Thomson Digital

End behavior: The leading coefficient test says that if the degree of the polynomial is even and the leading coefficient is positive, then both sides of the graph point up.

Intercepts: Solve

to find the x-intercepts. The rational root

theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial. Therefore, the possible rational zeros are Use synthetic division to find that x = 1 is one of the zeros: Now solve the resulting equation,

, to find another zero. The possible

rational zeros are

Chapter 17: Answers 249

Use synthetic division to find that x = 1 is one of the zeros: Now solve this resulting equation,

, to find the remaining zeros:

The zero at x = –4 has multiplicity 1 (the exponent on the factor is 1, which is an odd number), so the graph crosses the x-axis at –4. The zero at x = 1 has multiplicity 3 (the exponent is 3, which is an odd number), so the graph crosses the x-axis at x = 1.

Find the y-intercept by finding

:

Test points: Test points in the intervals

,

, and

.

Answers 301–400

Illustration by Thomson Digital

315.

Illustration by Thomson Digital

End behavior: The leading coefficient test says that if the degree of the polynomial is even and the leading coefficient is negative, then both sides of the graph point down.

Intercepts: Solve

to find the x-intercepts. The rational root

theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial. Therefore, the possible rational zeros are 250 Part II: The Answers

Use synthetic division to find that x = 2 is one of the zeros: Solve the resulting equation,

, to find another zero. The possible

rational zeros are

Use synthetic division to find that x = 3 is one of the zeros: Answers 301–400

Now solve the resulting equation,

, to find the remaining zeros:

Therefore, two zeros are imaginary and can’t be graphed. The zeros at x = 2 and x = 3

have multiplicity 1 (the exponent on each factor is 1, an odd number), so the graph crosses the x-axis at 2 and 3.

Find the y-intercept by finding

:

Test points: Test points in the intervals

,

, and

.

Illustration by Thomson Digital

Chapter 17: Answers 251

316.

Illustration by Thomson Digital

End behavior: The leading coefficient test says that if the degree of the polynomial is odd and the leading coefficient is negative, then the left side of the graph points up and Answers 301–400

the right side points down.

Intercepts: Solve

to find the x-intercepts. The rational

root theorem says that if you take all the factors of the constant term in a polynomial and divide them by all the factors of the leading coefficient, you produce a list of all the possible rational roots of the polynomial. Therefore, the possible rational zeros are Use synthetic division to find that x = 2 is one of the zeros: Solve the resulting equation,

, to find another zero. The pos-

sible rational zeros are

Use synthetic division to find that x = 1 is one of the zeros: Solve the resulting equation,

, to find another zero. The possible

rational zeros are

252 Part II: The Answers

Use synthetic division to find that x = −1 is one of the zeros: Now solve the resulting equation,

, to find the remaining zeros:

Thus, the zeros are −2, −1, 1, and 2. The zeros at x = −2, x = 1, and x = 2 have multiplicity 1 (the exponent on each factor is 1, an odd number), so the graph crosses the x-axis at these points. The zero at x = −1 has multiplicity 2 (the exponent on the factor is 2, an even number, meaning there’s no sign change at the root), so the graph touches but doesn’t cross the x-axis at this point.

Find the y-intercept by finding

:

Answers 301–400

Test points: Test points in the intervals

,

,

,

, and

.

Illustration by Thomson Digital

317.

The graph crosses the x-axis at x = –1, x = 2, and x = 4, so the function is given by where a is a constant. The y-intercept is (0, 8), so find a by plugging in these values and solving:

Therefore, the function is

Chapter 17: Answers 253

318.

The graph crosses the x-axis at x = –3, x = –2, x = 2, and x = 3, so the function is given by where a is a constant. The y-intercept is (0, –72), so find a by plugging in these values and solving:

Therefore, the function is

319.

The graph crosses the x-axis at x = 0 and x = –3. This means that two factors of are

Answers 301–400

x and

. The graph touches the x-axis but doesn’t cross it at x = –5, so has an

even number of factors of

. Therefore, try the following as an equation of the

graph:

To find the constant a, substitute the coordinates of the given point, (–1, 32), for x and y and solve:

Therefore, the function is

320.

The graph crosses the x-axis at x = –3 and x = –2, so two factors of are

and

. The graph also crosses the x-axis at x = 1, so

is also a factor. If

, then

is a cubic polynomial. But

must be a factor of

an odd number of times, so try the following as an equation of the graph: The y-intercept is (0, 6), so find the constant a by plugging in these values and solving: Therefore, the function is

254 Part II: The Answers

321.

,

To evaluate the function, plug in the given values for x and solve. At

, the function

equals

And at

, the function equals

322.

,

To evaluate the function, plug in the given values for x and solve. At

, the function

equals

Answers 301–400

And at

, the function equals

323.

,

To evaluate the function, plug in the given values for x and solve. At

, the function

equals

And at

, the function equals

324.

,

To evaluate the function, plug in the given values for x and solve. At

, the function

equals

Chapter 17: Answers 255

And at

, the function equals

325.

,

To evaluate the function, plug in the given values for x and solve. At

, the function

equals

Answers 301–400

And at

, the function equals

326.

Illustration by Thomson Digital

You find the x-intercepts by solving for

. No values of x make the equation true,

so there are no x-intercepts.

256 Part II: The Answers

You find the y-intercept by substituting 0 for x:

So the y-intercept is (0, –1).

There’s a horizontal asymptote at y = 0 because the limit as x approaches is 0. The

function is decreasing as x approaches

because the values of the function are get-

ting smaller and smaller, and the function approaches 0 as x approaches because

of the horizontal asymptote.

327.

Answers 301–400

Illustration by Thomson Digital

You find the x-intercepts by solving for

:

Taking the log of each side, you get

. So when x is about 1.4, you have an

x-intercept.

You find the y-intercept by substituting 0 for x:

So the y-intercept is (0, 9).

There’s a horizontal asymptote at y = 10 because the limit as x approaches is 10.

The function is decreasing as x approaches

because the function values are getting

smaller and smaller, and the function approaches 10 as x approaches because of

the horizontal asymptote.

Chapter 17: Answers 257

328.

Illustration by Thomson Digital

You find the x-intercepts by solving for

. No values of x make the equation true,

Answers 301–400

so there are no x-intercepts.

You find the y-intercept by substituting 0 in for x:

So the y-intercept is (0, 6).

There’s a horizontal asymptote at y = 0 because the limit as x approaches is 0.

The function is increasing as x approaches

because the values of the function keep

getting larger and larger, and the function approaches 0 as x approaches because

of the horizontal asymptote.

329.

Illustration by Thomson Digital

258 Part II: The Answers

You find the x-intercepts by solving for

. No values of x make the equation true,

so there are no x-intercepts.

You find the y-intercept by substituting 0 in for x:

So the y-intercept is (0, –7).

There’s a horizontal asymptote at y = –3 because the limit as x approaches is ‒3.

The function is decreasing as x approaches

, as you see by finding values of the

function as x gets smaller. The function approaches –3 as x approaches because of

the horizontal asymptote.

330.

Answers 301–400

Illustration by Thomson Digital

You find the x-intercepts by solving for

. No values of x make the equation true,

so there are no x-intercepts.

You can find the y-intercept by substituting 0 in for x: So the y-intercept is (0, 5.75).

There’s a horizontal asymptote at y = 5 because the limit as x approaches is 5.

The function is increasing as x approaches

because of the increasing values of the

function, and the function approaches 5 as x approaches

because of the horizontal

asymptote.

331. x = 6

Because the bases are already the same, you can set the exponents on each side equal to one another and solve:

Chapter 17: Answers 259

332. x = ‒5

First, rewrite the right side of the equation so that it has the same base as the left: The bases are now the same, so set the exponents on each side equal to each other: 333.

First, rewrite the left side to have the same base as the right side: Answers 301–400

When the bases are the same, you can set the exponents on each side equal to each other and solve for x:

334.

First, rewrite the right side of the equation so that it has the same base as the left side: When the bases are the same, set the exponents on each side equal to each other and solve for x:

260 Part II: The Answers

335.

First, rewrite the left and right sides of the equation so they have the same base: When the bases are the same, set the exponents on each side equal to each other and solve for x:

336.

Answers 301–400

The bases are the same, and so are the coefficients:

Set the exponents on each side of the equation equal to each other and solve for x: 337. x = 0

First, add 44 to each side and rewrite the left and right sides of the equation so that they have the same base:

When the bases are the same, set the exponents on each side equal to each other and solve for x:

Chapter 17: Answers 261

338. x = ‒12

First, subtract 80 from each side and rewrite the left and right sides of the equation so that they have the same base:

When the bases are the same, set the exponents on each side equal to each other and solve for x:

339.

First, add to each side and rewrite the left and right sides of the equation so that Answers 301–400

they have the same base:

When the bases are the same, set the exponents on each side equal to each other and solve for x:

340. x = ‒3

First, divide each side by 6 and rewrite the left and right sides of the equation so that they have the same base:

When the bases are the same, set the exponents on each side equal to each other and solve for x:

262 Part II: The Answers

341. x = 1

First, subtract 24 from each side and divide each side of the resulting equation by 2.

Then rewrite the right and left sides so that the bases are the same: Next, set the exponents on each side equal to each other and solve for x: 342.

Answers 301–400

First, subtract 25 from each side of the equation and rewrite the right and left sides so that the bases are the same:

Then set the exponents on each side equal to each other and solve for x: 343. x = –2

First, subtract from each side and rewrite the right and left sides so that the bases are the same:

Then set the exponents on each side equal to each other and solve for x: Chapter 17: Answers 263

344. x = –8, 5

First, rewrite the right side so that the bases are the same:

Then you can set the exponents on each side equal to each other and solve for x: 345. x = –5, 5

Answers 301–400

First, rewrite the right and left sides of the equation so that the bases are the same: Then set the exponents on each side equal to each other and solve for x: 346.

To convert an exponential expression to a logarithm, use

Therefore,

becomes

.

347.

To convert an exponential expression to a logarithm, use

Therefore,

becomes

.

264 Part II: The Answers

348.

To convert an exponential expression to a logarithm, use

Therefore,

becomes

.

349.

To convert a logarithmic expression to an exponential expression, use Therefore,

becomes

.

350.

To convert a logarithmic expression to an exponential expression, use Answers 301–400

Therefore,

becomes

.

351.

Using the logarithm product rule,

rewrite the logarithm as a sum:

352.

Using the logarithm quotient rule,

rewrite the logarithm as a difference:

353.

Using the logarithm power rule and the logarithm of a base rule,

rewrite each term in the sum:

Chapter 17: Answers 265

354. 2

Using the logarithm of a base rule and the logarithm of 1 rule,

rewrite each term in the sum:

355.

Using the power and product rules of logarithms,

condense the expression:

Answers 301–400

356.

Using the power and product rules of logarithms,

condense the expression:

357. –1

Using the power rule and logarithm of a base rule,

condense the expression:

266 Part II: The Answers

358.

Using the power and quotient rules of logarithms,

condense the expression:

359.

Using the product, quotient, and power rules of logarithms, as well as the logarithm of a base rule,

Answers 301–400

expand the expression:

360.

Using the product, quotient, and power rules of logarithms, as well as the logarithm of a base rule,

Chapter 17: Answers 267

expand the expression:

361.

Using the product and power rules of logarithms, as well as the logarithm of a base rule,

expand the expression:

Answers 301–400

362.

Using the quotient and power rules of logarithms, as well as the logarithm of a base rule and the logarithm of 1 rule,

expand the expression:

268 Part II: The Answers

363.

Using the product, quotient, and power rules of logarithms,

expand the expression:

Answers 301–400 364.

Using the product, quotient, and power rules of logarithms, as well as the logarithm of a base rule,

expand the expression:

365.

Using the product, quotient, and power rules of logarithms, as well as the logarithm of a base rule,

Chapter 17: Answers 269

expand the expression:

366. 11

First, convert the equation to an exponential format. Then solve for x: Answers 301–400

367.

First, convert the equation to an exponential format (recall that e is the base of the natural log). Then solve for x:

368.

First, convert the equation to an exponential format (remember that if the base is unlisted, it’s equal to 10). Then solve for x:

270 Part II: The Answers

369. 5

First, rewrite the equation as one logarithm:

Next, change the logarithmic equation to an exponential format and solve for x: The solution

is extraneous for the equation because it creates a negative argument Answers 301–400

for the second log term.

370.

First, rewrite the equation as one logarithm:

Then convert the equation to an exponential format and solve for x: 371.

First, rewrite the equation as one logarithm.

Chapter 17: Answers 271

Then convert the equation to an exponential format and solve for x (recall that the base of the natural log is e):

372. 3

First, rewrite the equation as one logarithm:

Then convert the equation to an exponential format and solve for x: Answers 301–400

The solution –7 is extraneous for the equation because it makes both of the arguments negative in the original log terms.

373. –3, 15

Because the two logarithms have the same base (base 10), their arguments are equal.

Set those two arguments equal to one another. Then solve the equation by using inverse operations.

272 Part II: The Answers

374. no solution

First, write each side of the equation as a power of e to eliminate the natural log. Then solve the equation by using inverse operations.

The solutions –4 and –8 are extraneous for the equation because the arguments of logs have to be positive. These numbers create negative and zero arguments, so this equation has no solution.

375. 10

Answers 301–400

First, rewrite the left side of the equation as one logarithm:

Next, write each side of the equation as a power of 3 to eliminate the log. Then use inverse operations to solve the equation.

376.

The graph appears to be raised 4 units above where the graph of the original function would be. The function transformation

raises the original graph by k units. So

in this case, using

, the point (1, 0) in the original graph moves to (1, 4).

The x-intercept of

is found by setting the function equal to 0 and solving

for x:

Then change the log expression to its equivalent exponential expression: So the x-intercept is (0.0001, 0).

The vertical asymptote is still x = 0, because this is a vertical slide.

Chapter 17: Answers 273

377.

The graph appears to have slid 1 unit to the left of where the graph of the original function would be.

The function transformation

slides the original graph k units to the left. Using

, the point (1, 0) on the original graph moves to (0, 0). The vertical asymptote also slides to the left and is now x = ‒1.

378.

The graph appears to be a reflection of the original graph over a vertical line, and it seems slightly steeper.

The function transformation

reflects the original graph over a vertical line. Using

, the point (1, 0) on the original graph flips over the vertical axis to (‒1, 0), and the vertical asymptote stays the same because it’s the line of reflection.

The multiplier of 3 makes the graph slightly steeper than the original. For example, when x = −5 on the graph of

, then

. Multiplying that

Answers 301–400

by 3, the point on the new graph becomes about (‒5, 2.09691).

379.

The graph appears to be a reflection of the original graph over a horizontal line.

The function transformation

reflects the original graph over a horizontal line, the

x-axis. Using

, the point (1, 0) on the original graph stays in place because

it’s on the line of reflection; also, the vertical asymptote stays the same. The signs of the function values are all reversed — positive to negative and vice versa. However, this function doesn’t appear to be one of the choices.

Investigating the choice

, first apply the law of logarithms involving a

quotient and rewrite the function rule as

, which simplifies to

. This is the desired function.

380.

The graph appears to have slid 1 unit down from where the graph of the original function would be.

The function transformation

slides the original graph k units down. Using

, the point (1, 0) on the original graph moves down to (1, ‒1). The vertical asymptote stays the same. However, this function doesn’t appear to be one of the choices.

Investigating the choice

, first apply the law of logarithms involving a

quotient and rewrite the function rule as

, which simplifies to

. This is the desired function.

274 Part II: The Answers

381. 3

To find the pH, substitute 0.001 into the formula and simplify:

382. 10‒11.6

To find the H +, plug 11.6 into the formula for pH:

Using the log equivalence,

becomes

, so you have

Answers 301–400 383. $5,225.79

Substitute the following values into the equation: P = $5,000, r = 0.01475, n = 4, and t = 3. Then simplify the equation to find the amount in the account: 384. 21.33 years

Substitute the following values into the equation: P = $1000, r = 0.0325, and A = $2,000.

Then simplify the equation:

To eliminate the e, take the natural log of both sides of the equation; then solve for t: 385. $11,077.61

Use the exponential model,

. Substitute the following values into the equation:

C = $18,995 (the value of the car when you drive off the lot), k = ‒0.10785, and t = 5.

Then simplify the equation to find the value of the car:

Chapter 17: Answers 275

386. 10‒6

Substitute L = 60 into the decibel equation. Then simplify the equation and expand the right side into two logs.

Answers 301–400

Using the log equivalence,

becomes

, so you have

387.

You have two equations to use for the two earthquakes. Earthquake 1 has M = 8.0, and Earthquake 2 has M = 4.0. Substitute these values into each equation and then expand the log:

Solve the resulting system of equations to eliminate the log I : 0

Using the log equivalence,

becomes

, so you have

The intensity of the first earthquake is 10,000 times that of the second earthquake.

276 Part II: The Answers

388. 10,174

Substitute the following values into the equation: C = 6,250, y = 8,125, and t = 2010 –

1975 = 35. Then simplify the equation to find the growth rate:

To eliminate the e, take the natural log on both sides of the equation. Then solve the problem for k:

Now, to find the population in the year 2040, substitute the values C = 6,250, k = 0.007496, and t = 2040 – 1975 = 65 into the equation: Answers 301–400

389. $119,639

Substitute the following values into the equation: C = 179,900, y = 138,800, and t = 2013 –

2000 = 13. Then solve for k to find the decay rate:

To eliminate the e, take the natural log on both sides of the equation. Then solve the problem for k:

Chapter 17: Answers 277

Now, to find the value of the house in the year 2020, plug the values C = 179,900, k = ‒0.020396, and t = 2020 – 2000 = 20 into the exponential growth equation: 390. 11

First find y, the required number of sick students and employees, by taking 45%

of 8,500:

Plug 8,500 into the equation for y. Then solve for t:

Answers 301–400

To eliminate the e, take the natural log on both sides of the equation. Solve for t, the time in days:

391.

Using the right-triangle definition for the sine of an angle, determine the measure of the side opposite the angle in question and divide that measure by the length of the hypotenuse of the triangle:

278 Part II: The Answers

392.

Using the right-triangle definition for the cosine of an angle, determine the measure of the side adjacent to the angle in question (not the hypotenuse) and divide that measure by the length of the hypotenuse of the triangle:

393. 1

Using the right-triangle definition for the tangent of an angle, determine the measure of the side opposite the angle in question and divide that measure by the length of the side adjacent to the angle (not the hypotenuse):

Answers 301–400 394.

Using the right-triangle definition for the cotangent of an angle, determine the measure of the side adjacent to the angle in question and divide that measure by the length of the side opposite the angle:

395.

Using the right-triangle definition for the secant of an angle, determine the measure of the hypotenuse and divide that measure by the length of the side adjacent to the angle: 396.

Using the right-triangle definition for the cosecant of an angle, determine the measure of the hypotenuse and divide that measure by the length of the side opposite the angle.

Chapter 17: Answers 279

397. 8.1 ft

The diagram represents the situation. You’re looking for the length of the ladder, which is the hypotenuse of the right triangle. You’re given an angle and the length of the adjacent side, so use

:

Illustration by Thomson Digital

398. 172.1 ft

Answers 301–400

The diagram shows the situation. You’re given an angle and the length of the hypotenuse of the right triangle. You’re looking for the length of the opposite side, so use

:

Illustration by Thomson Digital

399. 2,525.5 ft

The diagram shows the situation. You’re given an angle and the height of the right triangle. You’re looking for the length of the adjacent side, so use

:

Illustration by Thomson Digital

280 Part II: The Answers

400. 368.3 ft

The diagram shows the situation. You’re given an angle and the length of the hypotenuse.

You’re looking for the length of the opposite side, so use

:

Illustration by Thomson Digital

The vertical change in feet is about 368.3 feet.

401. 1,286.7 ft

The diagram shows the situation. The angle of depression is outside the triangle.

However, the angle of depression is defined as the angle between a horizontal line and the observer’s line of sight. The horizontal line is parallel to the ground, so the angle of Answers 401–500

depression is equal to the angle formed by the line of sight and the ground.

Illustration by Thomson Digital

The side you’re looking for is adjacent to the given angle and the given side is opposite the angle, so use

:

Willie is about 1,286.7 feet from the point on the ground directly below the hot air balloon.

Chapter 17: Answers 281

402. 59.6 ft

The diagram shows the situation.

Illustration by Thomson Digital

Because the bird is landing 8 feet above the ground, its height is 28 feet above the windowsill.

Illustration by Thomson Digital

Answers 401–500

The angle of depression is outside the triangle. However, the angle of depression is defined as the angle between a horizontal line and the observer’s line of sight. The horizontal line is parallel to the ground, so the angle of depression is equal to the angle formed by the line of sight and the ground.

To find how far the bird must fly, use

:

The bird must fly about 59.6 feet before it can land on the windowsill.

403. 1.5 min

The diagram shows the situation. You’re given an angle and the height of the right triangle. The height represents the depth of the submarine, and the hypotenuse of the triangle is the distance the sub travels to reach that depth.

Illustration by Thomson Digital

282 Part II: The Answers

You’re looking for the length of the hypotenuse, so use

:

The submarine is traveling at 8 miles per hour, which is

At that rate, to find out how long it takes to travel 1,032.6 feet, divide: The submarine will take about 1.5 minutes to reach a depth of 90 feet.

404. 628 ft

The diagram shows the situation. Let x represent the distance between you and your friend and let y represent the distance between your friend and the building. Then the distance between you and the building is

.

Answers 401–500

Illustration by Thomson Digital

The side represented by y is adjacent to the 71° angle. You know the length of the side opposite the 71° angle, so use

:

The side represented by

is adjacent to the 42° angle, so use

to find

:

Chapter 17: Answers 283

Now substitute

into the expression for

and solve for x:

You’re about 628 feet from your friend.

405. 155.6 ft

The diagram shows the situation.

Answers 401–500

Illustration by Thomson Digital

First, find the length of PC in triangle APC by using

:

Next, find AB by finding the length of CB in triangle PCB. Again, use

:

Now, subtract 100 from the length of CB to find the height of the lighthouse, AB: The lighthouse is about 155.6 feet tall.

284 Part II: The Answers

406. 342 ft

Draw BD perpendicular to AC. In triangle ABC, Illustration by Thomson Digital

In triangle ABD, find BD by using

:

Answers 401–500

Now use BD and

to find BC:

The distance across the lake is about 342 feet.

407.

In a 45°-45°-90° triangle, the lengths of the two legs are equal, and the length of the hypotenuse equals

times the length of a leg.

408.

In a 45°-45°-90° triangle, the lengths of the two legs are equal, and the length of the hypotenuse equals

times the length of a leg. The length of the hypotenuse is given,

so you find the length of a leg as follows:

Chapter 17: Answers 285

409.

In a 30°-60°-90° triangle where the length of the shorter leg is a, the length of the longer leg is

, and the length of the hypotenuse is 2 a. In this case, you’re given the length of the shorter leg and are looking for the length of the longer leg. Substitute 15 for a in to get

.

410.

In a 30°-60°-90° triangle where the length of the shorter leg is a, the length of the longer leg is

, and the length of the hypotenuse is 2 a. In this case, you’re given the length of the hypotenuse and are looking for the length of the longer leg. First, find the length of the shorter leg, a:

Now find the length of the longer leg by substituting 4 for a in to get

.

411. 120°

To convert radians to degrees, multiply the radian measure by

:

Answers 401–500

412. 2,160°

To convert radians to degrees, multiply the radian measure by

:

413. −330°

To convert radians to degrees, multiply the radian measure by

:

414. 171.89°

To convert radians to degrees, multiply the radian measure by

:

415. 53.29°

To convert radians to degrees, multiply the radian measure by

286 Part II: The Answers

416.

To convert degrees to radians, multiply the degree measure by

:

417.

To convert degrees to radians, multiply the degree measure by

:

418.

To convert degrees to radians, multiply the degree measure by

:

419. 0.63

To convert degrees to radians, multiply the degree measure by

:

Answers 401–500

420. 2.91

To convert degrees to radians, multiply the degree measure by

:

421. 60°

This is a special value that you should memorize. Refer to the unit circle.

Illustration by Thomson Digital

Chapter 17: Answers 287

422. 210°

This is a special value that you should memorize. Refer to the unit circle.

Illustration by Thomson Digital

423.

This is a special value that you should memorize. Refer to the unit circle.

Answers 401–500

Illustration by Thomson Digital

288 Part II: The Answers

424.

This is a special value that you should memorize. Refer to the unit circle.

Illustration by Thomson Digital

425.

Answers 401–500

This is a special value that you should memorize. Refer to the unit circle.

Illustration by Thomson Digital

426. 13°

Because 167° lies in quadrant II, its reference angle is

.

Chapter 17: Answers 289

427. 18°

Because 342° lies in quadrant IV, its reference angle is

.

428. 85°

Because 265° lies in quadrant III, its reference angle is

.

429. 72°

Because

, subtract 360° from 792° until the result is less than 360°:

An angle with a measure of 72° lies in quadrant I, so the reference angle is 72°.

430. 28°

Because −748° is less than 360°, add 360° to –748° until the result is greater than 0°.

Answers 401–500

An angle with a measure of 332° lies in quadrant IV, so the reference angle is 360° – 332° = 28°.

431.

The terminal side of a 225° angle in standard position on the unit circle contains the point

:

290 Part II: The Answers

432.

The terminal side of a 330° angle in standard position on the unit circle contains the point

:

433.

First, find the reference angle. Because 405° > 360°, subtract 360° from 405° until the result is less than 360°.

Answers 401–500

The reference angle is 45°.The terminal side of a 45° angle in standard position on the unit circle contains the point

:

434. undefined

The terminal side of an angle with a measure of in standard position on the unit circle contains the point (0, −1)

which is undefined.

435.

Recall that

. Use the Pythagorean theorem to find the length of the missing

side, y:

Chapter 17: Answers 291

Because is in quadrant I,

. Draw a right triangle on the coordinate plane with

the vertex of located at the origin.

Illustration by Thomson Digital

436.

Answers 401–500

Recall that

. Use the Pythagorean theorem to find the length of the missing

side, x:

Because is in quadrant II,

Draw a right triangle on the coordinate plane

with the vertex of located at the origin:

Illustration by Thomson Digital

292 Part II: The Answers

437.

Recall that

. Use the Pythagorean theorem to find the length of the missing

side, r. Angle is in quadrant IV, so y is negative.

By definition, r is always positive.

Draw a right triangle on the coordinate plane with the vertex of located at the origin: Answers 401–500

Illustration by Thomson Digital

438.

Recall that

Use the Pythagorean theorem to find the length of the missing

side, y. Angle is in quadrant III, so x and y are both negative.

Draw a right triangle on the coordinate plane with the vertex of located at the origin: Chapter 17: Answers 293

Illustration by Thomson Digital

439.

Recall that

. Use the Pythagorean theorem to find the length of the missing

side, x. If

is positive and

is negative, then lies in quadrant II, and x is nega-

tive while y is positive.

Answers 401–500

Draw a right triangle on the coordinate plane with the vertex of located at the origin.

Illustration by Thomson Digital

294 Part II: The Answers

440.

Recall that

. Use the Pythagorean theorem to find the length of the missing

side, x. If

is negative and

is negative, then lies in quadrant IV, and x is posi-

tive while y is negative.

By definition, r is always positive.

Draw a right triangle on the coordinate plane with the vertex of located at the origin.

Answers 401–500

Illustration by Thomson Digital

441.

Plug the given values into the arc length formula. Replace r with 20 and with Simplify and express the answer as a multiple of .

442.

First change 85° to radians:

Plug the given values into the arc length formula. Replace r with 6 and with Simplify and express the answer as a multiple of π.

Chapter 17: Answers 295

443.

Plug the given values into the arc length formula. Replace s with and r with 13.

Then solve for .

444.

inches

Plug the given values into the arc length formula:

445. cm

Answers 401–500

Ten minutes is of an hour, so the minute hand moves through an angle with radian measure

. Plug this value and the given length into the arc length formula:

446.

You solve the equation

by writing the corresponding equation

and evaluating by using the domain of the inverse function:

The domain of the inverse sine function is

, and the range is

. Therefore,

So

.

296 Part II: The Answers

447.

You solve the equation

by writing the corresponding equation

and evaluating by using the domain of the inverse function:

The domain of the inverse cosine function is

, and the range is

. Therefore,

. So

448.

You solve the equation

by writing the corresponding equation

and evaluating by using the domain of the inverse function:

Answers 401–500

The domain of the inverse tangent function is

, and the range is

Therefore,

So

, or

.

449. 0

You solve the equation

by writing the corresponding equation

and

evaluating by using the domain of the inverse function. Because

, you know that

The domain of the inverse sine function is

, and the range is

. Therefore,

So

.

450.

You solve the equation y = tan–1( x) by writing the corresponding equation tan( y) = x and evaluating by using the domain of the inverse function:

Chapter 17: Answers 297

The domain of the inverse tangent function is

and the range is

Therefore,

So

, and you can plug that into the given equation:

451. {120°, 240°}

Solve the equation for cos x by isolating that term on the left: The two angles for which the cosine is

on the interval are in the second and third

quadrants: 120° and 240°.

Answers 401–500

452. {30°, 210°}

Solve the equation for tan x by isolating that term on the left and then simplifying on the right:

The two angles for which the tangent is

on the interval are in the first and third

quadrants: 30° and 210°.

298 Part II: The Answers

453. {30°, 150°, 210°, 330°}

Solve the equation for sin x by isolating the trig term on the left and then using the square-root rule for quadratic functions:

The four angles for which the sine is

on the interval are 30°, 150°, 210°, and 330°.

454. {90°, 210°, 330°}

Solve the equation for sin x. The first thing you have to do is factor the given equation: Then set each factor equal to zero and solve. First factor:

Answers 401–500

Second factor:

So the complete set of solutions includes 90°, 210°, and 330°.

455. {120°, 240°}

Solve the equation for cos x. First, rearrange the given equation as a quadratic equation and then factor it:

Then set each factor equal to zero and solve. First factor:

Chapter 17: Answers 299

Second factor:

There are no solutions for the second factor because

. So the complete set

of solutions includes just 120° and 240°.

456.

First, solve the equation for cos x:

Because cos x is positive in quadrants I and IV, there are two solutions in the interval

:

457.

First, solve the equation for tan x:

Answers 401–500

Because the period of

is and the instructions ask for solutions in the interval

, there are four solutions. If

, then

and

. If

, then

and

.

458.

First, solve the equation for cos x by rearranging the given equation as a quadratic equation and factoring it:

300 Part II: The Answers

Then set each factor equal to zero and solve. First factor:

Second factor:

If

, then there are two solutions in the interval

,

and

.

If cos x = 1, then x = 0.

459.

First solve the equation for sin x by factoring (factor by grouping): Answers 401–500

Then set each factor equal to zero and solve. First factor:

Second factor:

If

, then there are two solutions in the interval

,

and

. If

, then there are two solutions in the interval

,

and

. If

, then there are two solutions in the interval

,

and

.

Chapter 17: Answers 301

460.

Factor the left side of the equation:

Then use the zero product property to solve for

and

. (The zero product

property says that if

, then either a or b is equal to 0.)

If

, then

and

. If

, then there are two solutions in the

interval

,

and

.

461.

Use the standard equation

, where A is the amplitude and

is the

period. The trigonometric function has a maximum value of 2 and a minimum value of

–2, for a difference of

. This means that the amplitude of the function is half

Answers 401–500

that difference, or 2, so A = 2. The period of the function is about 1.57 units, or , so B = 4.

462.

Use the standard equation

, where A is the amplitude and

is the

period. The trigonometric function has a maximum value of 3 and a minimum value of

−3, for a difference of

. This means that the amplitude of the function is half

that difference, or 3, so A = 3. The period of the function is about 3.14 units, or exactly , so B = 2.

463.

Use the standard equation

, where A is the amplitude and

is the period.

The trigonometric function has a maximum value of and a minimum value of

, for

a difference of

. This means that the amplitude of the function is half that

difference, or , so

. Because the standard sine graph is flipped over the x-axis, the function is multiplied by ‒1. The period of the function is exactly 2 units, so

.

302 Part II: The Answers

464.

Use the standard equation

, where A is the amplitude and

is the

period. The trigonometric function has a maximum value of and a minimum value of

, for a difference of

. This means that the amplitude of the function is

half the difference, or , so

. The period of the function is 4 units, so

.

465.

Use the standard equation

, where A is the amplitude and

is the

period. The trigonometric function has a maximum value of 8 and a minimum value of

−8, for a difference of

. This means that the amplitude of the function is half

that difference, or 8, so A = 8. Because the standard sine graph is flipped over the x-axis, the function is multiplied by ‒1. The period of the function is exactly 8 units, so

.

466.

Answers 401–500

Illustration by Thomson Digital

The given function is

Using

, the amplitude of the function is A. Therefore, the amplitude of the graph is . The period of the function is determined by

;

, so the period of the

function is

.

Chapter 17: Answers 303

467.

Illustration by Thomson Digital

The given function is

Using

, the amplitude of the function is A. Therefore, the amplitude of the graph is 5. The period of the function is determined by

;

, so the period of the

Answers 401–500

function is 1. Because the function is multiplied by ‒1, the standard sine graph is reflected over the x-axis.

468.

Illustration by Thomson Digital

The given function is

304 Part II: The Answers

Using

, the amplitude of the function is A. Therefore, the amplitude of the graph is . Because the function is multiplied by ‒1, the standard cosine graph is reflected over the x-axis. The period of the function is determined by

; B = 6, so the

period of the function is

.

469.

Illustration by Thomson Digital

The given function is

Using

, the amplitude of the function is A. Therefore, the amplitude of the Answers 401–500

graph is . The period of the function is determined by

;

, so the period of

the function is

. The negative multiplier of the angle reflects the

graph over the y-axis.

470.

Illustration by Thomson Digital

The given function is

Chapter 17: Answers 305

Using

, the amplitude of the function is A. Therefore, the amplitude of the graph is

. Because the function is multiplied by ‒1, the standard cosine graph is reflected over the x-axis. The period of the function is determined by

;

, so the

period of the function is

.

471. right , up 3

The transformed function is

Using

, C represents a horizontal shift and D represents a vertical shift. In this function, C is

, which is a shift to the right, and D is 3, which is a

shift upward.

472. left , down 2

The transformed function is

Using

, C represents a horizontal shift and D represents a verti-Answers 401–500

cal shift. In this function, C is

, which is a shift to the left, and D is ‒2, which is a shift downward.

473. right , down ; period:

Using

, C represents a horizontal shift and D represents a vertical shift. Rewrite

to match the standard form:

In this function, C is

, which is a shift to the right, and D is

, which is a shift down-

ward. In the general form

, the period of sine is

; therefore,

the 4 multiplier changes the period to because

.

474. right , up ; reflected over y-axis; period:

The transformed function is

Using

, C represents a horizontal shift and D represents a vertical shift. Rewrite

to match the standard form:

306 Part II: The Answers

In this function, C is

, which is a shift to the right, and D is , which is a shift upward. The negative multiplier on the angle reflects the graph over the y-axis. In the general form

, the period of cosine is

; therefore, the 2 multi-

plier changes the period to because

.

475. left , up 8; period:

The transformed function is

Using

, C represents a horizontal shift and D represents a vertical shift. Rewrite

to match the standard form:

In this function, C is

, which is a shift to the left, and D is 8, which is a shift upward.

In the general form

, the period of sine is

; therefore, the

4 multiplier changes the period to

.

476.

Use

, where C represents a horizontal shift and D represents a Answers 401–500

vertical shift. In this case,

and

. Inserting the values, you have

477.

Use

, where C represents a horizontal shift and D represents a vertical shift. In this case,

and

. Inserting the values, you have

478.

Use

, where C represents a horizontal shift and D represents a vertical shift. In this case,

and

. Inserting the values, you have

479.

Use

, where C represents a horizontal shift and D represents a vertical shift. In this case,

and

. Inserting the values, you have

Chapter 17: Answers 307

480.

Use

, where C represents a horizontal shift and D represents a vertical shift. In this case,

and

. Inserting the values, you have

481.

Answers 401–500

Illustration by Thomson Digital

The given function is

Using

, the period of the function is determined by

; therefore, the

graph is the standard tangent function, except with a period of .

482.

Illustration by Thomson Digital

308 Part II: The Answers

The given function is

Using

, the period of the function is determined by

; therefore, the

graph is the standard cotangent function, except with a period of .

483.

Illustration by Thomson Digital

Answers 401–500

The given function is

Using

, the multiplier A = 2 makes the graph steeper, and the negative sign reflects the graph over the x-axis. The period of the function is determined by

;

because B = π, the graph has a period of 1.

484.

Illustration by Thomson Digital

The given function is

Chapter 17: Answers 309

Using

, the multiplier A = 3 makes the graph steeper. The period of the function is determined by

, so the graph has a period of 2. The negative multiplier

on the angle reflects the graph over the y-axis.

485. period: ; amplitude: ; reflected over the x-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

With

, the curve flattens, because is smaller than 1. The negative multiplier creates a reflection over the x-axis, and B = 6 results in a period of . The period of tangent or cotangent is found by dividing π by the absolute value of B.

486. period: ; amplitude: 2

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical Answers 401–500

shift.

With A = 2, the curve steepens. You find the period of tangent or cotangent by dividing π by the absolute value of B, so

results in a period of

.

487. period: 3; amplitude:

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

With

, the curve steepens, because

is greater than 1.

results in a period

of

.

488. period: 6; amplitude: ; reflected over the y-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

310 Part II: The Answers

With

, the curve flattens, because is smaller than 1. You find the period of tangent or cotangent by dividing π by the absolute value of B, so results in a period

of

. The negative sign on B results in a reflection over the y-axis.

489. right , up 3; period:

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

You find the period of tangent or cotangent by dividing by the absolute value of B, so with B = 4, the period is . The value

means a shift to the right, and with D = 3,

the curve moves up 3 units.

490. left , down 1; period:

Answers 401–500

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

You find the period of tangent or cotangent by dividing by the absolute value of B, so with B = 3, the period is . The value

means a shift to the left, and with D = ‒1,

the curve moves down 1 unit.

491. right 1, up ; period: ; reflected over the y-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

Chapter 17: Answers 311

You find the period of tangent or cotangent by dividing π by the absolute value of B, so with B = ‒2, the period is . The value

means a shift to the right, and with

, the

curve moves up units. The negative sign on B = ‒2 indicates a reflection over the y-axis.

492. left , up ; period: ; reflected over the y-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

You find the period of tangent or cotangent by dividing by the absolute value of B, so with B = ‒3, the period is . Because B is negative, the function is reflected over the y-axis. The value

means a shift to the left, and with

, the curve moves up

units.

493. left , down ; period: ; reflected over the x-axis

Answers 401–500

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

With A = ‒3, the curve is steeper, because 3 is greater than 1. There’s also a reflection over a horizontal axis — in this case,

— because of the negative sign.

You find the period of tangent or cotangent by dividing by the absolute value of B, so with B = 2, the period is . The value

means a shift to the left, and with

,

the curve moves down units.

494. right , up 4; period: ; reflected over the x-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

312 Part II: The Answers

First, rewrite the function equation as

With

, the curve is flatter, because is less than 1. There’s also a reflection over a horizontal axis — in this case, y = 4 — because of the negative sign.

You find the period of tangent or cotangent by dividing by the absolute value of B, so with B = 5, the period is . The value

means a shift to the right, and with

D = 4, the curve moves up 4 units.

495. right , down ; period:

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

Answers 401–500

With

, the curve is flatter, because is less than 1.

You find the period of tangent or cotangent by dividing by the absolute value of B, so with

, the period is . The value

means a shift to the right, and with

, the curve moves down units.

496.

Illustration by Thomson Digital

Chapter 17: Answers 313

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

.

The asymptotes are found where the reciprocal of the secant,

, is equal

to 0:

when

Solving for x, you divide each term by to get

Letting k be an integer, the general rule for the equations of the asymptotes is 497.

Answers 401–500

Illustration by Thomson Digital

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

.

The asymptotes are found where the reciprocal of the cosecant,

, is equal

to 0:

when

314 Part II: The Answers

Solving for x, you divide each term by to get

Letting k be an integer, the general rule for the equations of the asymptotes is 498.

Answers 401–500

Illustration by Thomson Digital

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

.

The asymptotes are found where the reciprocal of the secant,

, is equal

to 0:

when

Solving for x, you divide each term by ‒3 to get

Letting k be an integer, the general rule for the equations of the asymptotes is Chapter 17: Answers 315

499.

Answers 401–500

Illustration by Thomson Digital

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

. The multiplier 4 brings the upper curves down to 4 and the lower curves up to ‒4.

The asymptotes are found where the reciprocal of the secant,

, is equal

to 0:

when

Solving for x, you divide each term by 5 to get

Letting k be an integer, the general rule for the equations of the asymptotes is 316 Part II: The Answers

500.

Illustration by Thomson Digital

Answers 401–500

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

. The multiplier brings the upper curves down to and the lower curves up to

.

The asymptotes are found where the reciprocal of the secant,

, is equal

to 0:

when

Solving for x, you divide each term by to get

Letting k be an integer, the general rule for the equations of the asymptotes is Chapter 17: Answers 317

501.

Illustration by Thomson Digital

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

. The multiplier ‒2 brings the upper curves down to 2 and the lower

Answers 501–600

curves up to ‒2. The negative multiplier of the angle reflects the standard curve over the y-axis.

The asymptotes are found where the reciprocal of the cosecant,

, is

equal to 0:

when

Solving for x, you divide each term by ‒8 to get

Letting k be an integer, the general rule for the equations of the asymptotes is 318 Part II: The Answers

502.

Illustration by Thomson Digital

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period

Answers 501–600

is

. The multiplier

brings the upper curves down to

and the lower

curves up to

. The negative multiplier reflects the standard curve over the x-axis.

The asymptotes are found where the reciprocal of the secant,

, is equal

to 0:

when

Solving for x, you divide each term by to get

Letting k be an integer, the general rule for the equations of the asymptotes is Chapter 17: Answers 319

503.

Illustration by Thomson Digital

Use

, where A is the amplitude,

is the period, C represents a

horizontal shift, and D represents a vertical shift. For

, the period is

. The multiplier brings the upper curves down to and the lower

Answers 501–600

curves up to

. The negative multiplier of the angle reflects the standard curve over the y-axis.

The asymptotes are found where the reciprocal of the cosecant,

, is

equal to 0:

when

Solving for x, you divide each term by

to get

Letting k be an integer, the general rule for the equations of the asymptotes is 320 Part II: The Answers

504. right , up 5; period:

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

With B = 3, you have a period of , because the period is found with

.

, so

the curve shifts to the right. D = 5 moves the curve up 5 units.

505. left , up 3; period

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

Answers 501–600

With B = 2, you have a period of

, because the period is found with

.

,

so the curve shifts to the left. D = 3 moves the curve up 3 units.

506. right 8, down

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

C = ‒8, so the curve shifts to the right.

moves the curve down units.

Chapter 17: Answers 321

507. right , up ; period: ; reflected over the y-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

With B = ‒1, you have a period of , because the period is found with

. The nega-

tive sign reflects the curve over the y-axis.

, so the curve shifts to the right.

moves the curve up units.

508. right , down ; period: ; reflected over the x-axis

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

Answers 501–600

With A = ‒2, the curve is steeper, and the distance between the upper and lower curves is greater because of the 2 multiplier. The negative sign reflects the curve over the x-axis. With B = ‒2, you have a period of

, because the period is found with

. The negative sign on B also reflects the graph over the y-axis.

, so the curve

shifts to the right.

moves the curve down units.

509. left , down 5; period: 2

The transformed function is

322 Part II: The Answers

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

With

, the curve is flatter, and the distance between the upper and lower curves is smaller because of the multiplier. With

, you have a period of 2, because the

period is found with

.

, so the curve shifts to the left. D = ‒5 moves the curve

down 5 units.

510. left 3, down ; period:

The transformed function is

Using

, A represents a multiplier that steepens or flattens the

curve and changes its low points and high points, B affects the period, C represents a horizontal shift, and D represents a vertical shift.

First, rewrite the function equation as

With

, the curve is slightly flatter, and the distance between the upper and lower Answers 501–600

curves is smaller because of the multiplier. With

, you have a period of

. C = +3, so the curve shifts to the left.

moves the curve down units.

511. use the reciprocal identity

Because each term contains a function and its reciprocal, using reciprocal identities will simplify the terms quickly.

Replace

with its reciprocal identity,

, and

with its reciprocal identity,

:

Multiply and simplify:

Chapter 17: Answers 323

512. use the reciprocal identity

Because each term contains a function and its reciprocal, using reciprocal identities will simplify the terms quickly.

First, replace

with its reciprocal identity,

, and

with its reciprocal

identity,

. Then simplify the complex fractions.

Finally, replace

with 1, using the Pythagorean identity:

513. use the ratio identity

The term on the left can actually be rewritten using the ratio identity directly, but the following shows how to change to positive angles first.

Replace

with

and

with

using the two even-odd identities:

Answers 501–600

Use the ratio identity to replace

with

:

Replace

with

using the even-odd identity:

514. use the co-function identity

You want both trig terms to use the same angle measure, so first apply the co-function identity.

Rewrite the first term on the left as the power of the sine:

Part II: The Answers

324

Replace

with

using the co-function identity:

Rewrite the first term on the left:

Replace

with 1, using the Pythagorean identity:

515. use the periodicity identity

You want the term on the left to have the same angle as that on the right (simplifying the more complex expression), so apply the periodicity identity.

Rewrite

as the sine of a negative angle:

Use the even-odd identity to replace

with

:

Replace

with

using the periodicity identity:

Another method would be to use the difference formula, although it might not be as intuitive that it will work with two different angle measures.

Answers 501–600

Replace the left side using the difference formula and substitute in the function values.

516. use the reciprocal identity

The more complex term in this identity is the fraction on the left. Because it contains a function and its reciprocal, apply a reciprocal identity.

Replace

with its reciprocal identity,

. Then simplify the complex fraction.

Chapter 17: Answers 325

Use the Pythagorean identity to replace

with

:

517. use the Pythagorean identity

Even though you could split the term on the left into two fractions and simplify, a quicker move is to apply the Pythagorean identity to the numerator to create one term.

Replace

with 1 using the Pythagorean identity:

Replace

with

using the reciprocal identity:

518. use the even-odd identity

You want the angle measures in the trig terms to be the same, so apply the even-odd identity to the first term.

Rewrite the first term on the left as the power of the sine:

Use the even-odd identity to replace

with

:

Answers 501–600

Square the first term:

Replace

with 1 using the Pythagorean identity:

519. use the co-function identity

You want the angle measures in the trig terms to be the same, so apply the co-function identity to the first term on the left.

Rewrite the first term on the left as the power of the cosine:

Part II: The Answers

326

Replace

with

using the co-function identity:

Factor the terms on the left:

520. use the ratio identity

Before distributing through the terms on the left, rewrite the factor outside the parentheses by using the ratio identity.

Use the ratio identity to replace

with

:

Replace

with

using the reciprocal identity. Then distribute

.

Replace

with

using the Pythagorean identity:

521. use the ratio identity

Answers 501–600

This equation contains three different trig functions, so you want to rewrite it with trig functions that are more closely related. Taking advantage of the fact that cotangent is cosine over sine — and planning on using that ratio in a future multiplication — use the ratio identity first.

Replace

with

using the ratio identity, and replace

with

using the

reciprocal identity. Multiply on the left.

Chapter 17: Answers 327

522. use the Pythagorean identity

Looking ahead to the product you’ll get on the right after multiplying the binomials, use the Pythagorean identity to rewrite the single term on the left.

Take the Pythagorean identity,

, and subtract

from each side:

Replace

with

in the identity you’re proving:

Factor the binomial as the difference of two squares:

Another option is to first multiply the two terms on the right, add

to each side,

and then apply the Pythagorean identity, as follows. It all depends on how you see the problem.

523. use the reciprocal identity

To change the ratio on the left to a single term, first use the reciprocals of each factor and simplify.

Answers 501–600

Replace

and

using reciprocal identities:

Simplify the complex fraction:

Part II: The Answers

328

Use the even-odd identities to replace

with

and

with

:

Now use the ratio identity to replace

with

:

524. use the even-odd identity

You want the angles of the two terms to be the same so you can apply the even-odd identity.

Factor –1 from the two terms of the tangent’s angle:

Use the even-odd identity to replace the expression on the left:

525. use the reciprocal identity

The two terms in the parentheses on the left are reciprocals of the multiplier outside Answers 501–600

the parentheses, so first rewrite the tangent by using the reciprocal identity.

Replace

on the left with

using the reciprocal identity:

Distribute

through the parentheses. Then multiply and simplify terms.

Use the Pythagorean identity to replace

with

:

Chapter 17: Answers 329

526. use the ratio identity

The equation has four different trig identities with no apparent quick connections, so first distribute the tangent over the two terms and then use the ratio identity on the tangent to simplify the terms.

Distribute

:

Use the ratio identity to replace

with

; then use the reciprocal identity to

replace

with

. Multiply and simplify.

527. use the Pythagorean identity

When one side of an identity is more complex because of products or quotients, it’s a good idea to multiply or divide to see the results. In this case, the product produces terms that belong in a Pythagorean identity.

Multiply the two binomials:

Use the Pythagorean identity to replace

with

:

Answers 501–600

528. use the even-odd identity

You want the angles in the four trig functions to be the same, so first deal with the by using an even-odd identity.

Replace

with

using the even-odd identity:

Part II: The Answers

330

Use the ratio identity to replace

with

:

Multiply and simplify:

529. use the co-function identity

You want the angles in the three trig functions to be the same, so first deal with the tangent by using a co-function identity.

Rewrite the second term on the left as a power of the tangent:

Use the co-function identity to replace

with

:

Rewrite the second term and combine the terms on the left:

Rewrite the Pythagorean identity,

, by subtracting 1 from each side to

get

. Replace

in the identity being proven with

.

Answers 501–600

530. use the ratio and reciprocal identities

You have four different trig functions and need to distribute over the two terms on the left. Before doing that, change the two terms in the parentheses to functions involving just sine and cosine.

Replace

with

using the ratio identity, and replace

with

using the

reciprocal identity:

Distribute

. Then multiply and simplify.

Chapter 17: Answers 331

531. use the reciprocal identity

You have four different trig functions in the identity, two of which are reciprocals of sine and cosine.

Use the reciprocal identities to replace

with

and

with

.

Then simplify the complex fractions and multiply.

Replace

with

using the ratio identity and then simplify:

532. use the ratio identity

The left side of the identity is more complicated than the right. To simplify on the left, distribute through the two terms and then apply a ratio identity to rewrite the first term in the result.

Distribute

. Then multiply and simplify.

Answers 501–600

Write the first term as the power of a fraction. Then use the ratio identity to replace the fraction with tan x.

Finally, use the Pythagorean identity to replace

with

:

Part II: The Answers

332

533. use the co-function identity

You want the angles of the trig functions to be the same, so use a co-function identity to rewrite the angle of the sine function on the left.

Use the co-function identity to replace the sine of the difference:

Now use the ratio identity to replace

with

. Multiply and simplify.

534. use the even-odd identity

You want the angles of the four trig functions to be the same, so rewrite the functions with negative angles by using even-odd identities.

Rewrite each function with a negative angle:

Use the reciprocal identities to replace

with

and

with

. Multiply

and simplify.

Answers 501–600

535. use the Pythagorean identity

Usually, you want to change two terms to one term. But in this case, because of the shared functions on both sides of the equation, a good move is to rewrite the cosecant term on the left by using a Pythagorean identity and then to distribute to create two terms.

Use the Pythagorean identity to replace

with

. Then distribute

.

Replace the

in the first term using the reciprocal identity, and replace the

using the ratio identity. Multiply and simplify.

Chapter 17: Answers 333

Finally, use the reciprocal identity to replace

with

:

536.

Use the reciprocal identities to replace

and

. Use the ratio identities to

replace

and

.

Reduce the fractions and simplify:

The missing term is

.

537.

Use the reciprocal identities to replace

and

. Use the ratio identities to

replace

and

.

Answers 501–600

Simplify the complex fractions by multiplying the numerator by the reciprocal of the denominator:

The missing term is

.

538. 1

Use the reciprocal identity to replace

, and use the ratio identity to

replace

:

Part II: The Answers

334

Distribute

, simplify, and then combine the two terms:

Rewrite the Pythagorean identity,

, by subtracting

from each

side to get

. Replace the numerator of the fraction in the identity

with

:

The missing term is 1.

539.

Use the reciprocal identity to replace sec x. Use the ratio identities to replace and

.

Combine the two fractions in the parentheses; multiply by the first factor: Answers 501–600

The missing factor is

.

540.

Use the reciprocal identities to replace

and

. Use the ratio identity to

replace

.

Add the two fractions in the parentheses after finding a common denominator: Chapter 17: Answers 335

Divide out the

factor; then multiply:

Write the fraction as two separate fractions. Reduce the fraction on the left, and write the fraction on the right as a product.

Use the reciprocal identity to replace

. Use the ratio identity to replace

.

Finally, factor out

.

The missing factor is

.

541.

Multiply both the numerator and the denominator by

:

Answers 501–600

Distribute

in the numerator and multiply in the denominator:

Replace

in the numerator by using the reciprocal identity. Replace

in

the denominator by using the Pythagorean identity. Simplify.

Multiply numerator and denominator by –1:

The missing factor is

.

Part II: The Answers

336

542.

Multiply both the numerator and the denominator by

:

Distribute

in the numerator and multiply in the denominator:

Replace

in the numerator using the reciprocal identity. Replace

in the

denominator with

after subtracting 1 from each side of the Pythagorean

identity,

. Simplify.

The missing factor is

.

543.

Multiply both the numerator and the denominator by

:

Answers 501–600

Multiply in the denominator:

Rewrite the Pythagorean identity,

, by subtracting

from each

side. Then substitute

for

in the denominator. Simplify.

The missing factor is

.

Chapter 17: Answers 337

544.

Multiply both the numerator and the denominator of the first fraction by

, and

multiply the numerator and denominator of the second fraction by

:

Add the two fractions together. Then replace the 1 in the denominator with from the Pythagorean identity.

Replace

with its reciprocal and then simplify:

The missing factor is

.

Answers 501–600

545.

Multiply both the numerator and the denominator of the first fraction by

,

and multiply the numerator and denominator of the second fraction by

:

Leave the numerators factored. Multiply in the denominators.

Part II: The Answers

338

Use the Pythagorean identity to replace each 1 in the denominators with

.

Simplify.

Reduce each term:

The missing term is

.

546.

Square the binomial:

Take the Pythagorean identity,

, and subtract

from each side.

Then replace

in the problem with

. Simplify.

The missing term is

.

547.

Answers 501–600

Square the binomial:

Use the reciprocal identity to replace

with

. Simplify.

Take the Pythagorean identity,

, and subtract 1 from each side of the

equation. Then replace

in the problem with

. Simplify.

Using the Pythagorean identity, replace

with

:

The missing term is

.

Chapter 17: Answers 339

548. 1

Square the binomial:

Using the Pythagorean identity, replace

with 1:

The missing term is 1.

549.

Cube the binomial:

Replace

with its reciprocal,

, and

with its reciprocal,

, in the

second and third terms. Multiply and simplify.

The missing term is

.

550.

Raise the binomial to the fourth power. Using Pascal’s triangle for the coefficients of Answers 501–600

the terms, you’ll have 1, 4, 6, 4, and 1 for the coefficients, decreasing powers of sine, and increasing powers of cosine:

The problem becomes

Replace

factor with

, obtained from subtracting

from each side of

the Pythagorean identity:

Multiply and simplify:

Part II: The Answers

340

Now combine like terms:

The missing term is

.

551.

Factor

from the two terms:

Replace

with

using the Pythagorean identity:

Replace

with its reciprocal,

; then multiply and rewrite:

Using the ratio identity, replace

with

, and change

back to its

reciprocal:

The missing factor is

.

552.

Answers 501–600

Factor

from each term:

Use the Pythagorean identity to replace

with 1:

The missing term is

.

553.

Factor the numerator as the square of a binomial:

Chapter 17: Answers 341

Reduce the fraction:

The missing term is

.

554.

Factor the numerator of the fraction as the difference of squares. Then reduce the fraction.

The missing term is

.

555. 1

Factor the terms by grouping:

Factor

from each term:

Answers 501–600

Using the Pythagorean identity, replace

with 1:

The missing term is 1.

556.

The common denominator is

. Write both fractions with that denominator:

Part II: The Answers

342

Rewrite

and

using their ratio identities. Multiply and simplify.

Reduce each fraction. Then replace

and

with their respective

reciprocal identities.

The missing term is

. As it turns out, this problem would be much shorter if you just worked on the two fractions separately and didn’t add them together. There are many ways to solve most identities.

557.

The common denominator is

. Write both fractions with that denominator:

Add the fractions. Then replace

with 1 from the Pythagorean identity.

Answers 501–600

Write the fraction as the product of two fractions. Then replace

with its recipro-

cal identity and

with its reciprocal identity.

The missing factor is

.

558.

The common denominator is

. Write both fractions with that denominator:

Chapter 17: Answers 343

Write the fraction as the product of two fractions. Then replace

with its recipro-

cal identity and

with its reciprocal identity. Simplify the numerators.

Add the two fractions. Then rewrite the fraction as a product.

Replace

with its reciprocal identity and

with its reciprocal identity:

The missing factor is

.

559. 3

The common denominator is

. Write both fractions with that denominator:

Answers 501–600

Replace

with its reciprocal and

with its reciprocal. Multiply. Then add

the fractions together.

Replace

with 1 from the Pythagorean identity:

The missing factor is 3.

Part II: The Answers

344

560.

The common denominator is

. Write both fractions with that

denominator:

Because

and

are reciprocal functions, their product is 1. Both denominators

have products of 1:

Rewrite all the functions in terms of

and

:

Now use the ratio identities to rewrite

and

:

Answers 501–600

The missing term is

.

Another way to solve the problem is to apply reciprocal identities to and

in

the original fractions:

Then use the reciprocal identities on the two terms on the left to get

.

Chapter 17: Answers 345

561.

The directions say to distribute on the right and move all terms to the left. On the right side, distribute the

:

Subtract

from both sides and add

to both sides:

To finish solving the identity, factor

from each term:

Replace

with

using the Pythagorean identity:

562.

Split up the first fraction on the left and distribute on the right: To finish solving the identity, replace

with its reciprocal identity,

.

Then subtract

from both sides of the equation:

Answers 501–600

Find a common denominator on the left and subtract; replace

by using the

reciprocal identity, replace

by using the reciprocal identity, and then multiply:

Replace

with

using the Pythagorean identity. Simplify the numerator.

Part II: The Answers

346

563.

Cross-multiply:

To finish solving the identity, move all the terms to the right by subtracting

, adding

, and subtracting

:

Replace

with 1 using the Pythagorean identity. Simplify.

564.

Cross-multiply:

To finish solving the identity, add

to each side. Then replace

with the

ratio identity.

Answers 501–600

565.

Factor the numerator on the left and reduce the fraction:

To finish solving the identity, cross-multiply:

Chapter 17: Answers 347

Replace the 1 with

from the Pythagorean identity:

566.

Cross-multiply:

To finish solving the identity, note that the product of

is equal to 1 because

is the reciprocal function of

. Replace

with 1 and then simplify:

567.

Add the middle fraction to both sides, and then add the two fractions on the right: Answers 501–600

Replace the two

terms in the numerator with their reciprocal, multiply, and

reduce the fractions:

Part II: The Answers

348

To finish solving the identity, cross-multiply:

Replace

with its ratio identity and

with its reciprocal identity. Multiply

and simplify.

568.

The directions say to combine the two terms on the right. First, replace with

its reciprocal identity and

with its ratio identity; then subtract:

To finish solving the identity, cross-multiply:

Answers 501–600

Replace the 1 with

from the Pythagorean identity:

569.

The directions say to subtract the two terms on the left. Find a common denominator and then subtract:

To finish solving the identity, replace

with

from the Pythagorean iden-

tity. On the right, replace

with its reciprocal identity.

Chapter 17: Answers 349

Reduce the fraction on the left:

570.

Square both sides of the equation:

To finish solving the identity, replace

with its reciprocal identity. Then simplify

the complex fraction.

Answers 501–600

Replace

with its ratio identity:

Replace

with 1 from the Pythagorean identity:

571.

Use the sine-of-a-sum identity:

Part II: The Answers

350

572.

Use the tangent-of-a-sum identity:

573.

Use the cosine-of-a-difference identity:

574.

Change the double angle to the sum of two angles:

Answers 501–600

Now apply the sine-of-a-sum identity:

575.

Change the double angle to the sum of two angles:

Now apply the tangent-of-a-sum identity:

Chapter 17: Answers 351

576.

Replace

with the double-angle identity:

Factor:

577. 0

Replace

with the double-angle identity involving the cosine:

Replace the 1 with

from the Pythagorean identity:

578. 1

Answers 501–600

Rewrite the Pythagorean identity

as

by

subtracting

from each side of the equation. Then replace

with

:

Rewrite

as the square of

. Then replace

with its double-angle

identity. Square and simplify:

Part II: The Answers

352

579. 1

Replace

with the cosine of a sum:

Replace the cosine of the sum with its identity:

Now replace the cosine double angles with the identity involving cosine, and replace the sine double angles with their identity:

Multiply and simplify:

Replace

with

using the Pythagorean identity:

580.

Replace

with the sine of a sum:

Answers 501–600

Replace the sine of the sum with its identity:

Now replace

and

with their respective double-angle identities; for the

cosine, use the identity involving sine:

581.

Begin with the identity

. Replace the 2 x with and replace the x

with :

Chapter 17: Answers 353

Add

to each side and subtract

from each side of the equation:

Divide each side by 2:

Take the square root of each side:

582.

Begin with the identity

. Replace the 2 x with and replace the x

with :

Add 1 to each side of the equation:

Divide each side by 2:

Answers 501–600

Take the square root of each side:

583.

Use the ratio identity to replace

with identities in sine and cosine:

Part II: The Answers

354

Replace

and

with their respective identities; then simplify:

Inside the radical, multiply the numerator and denominator by

:

Replace the 1 in the denominator with

from the Pythagorean identity;

then simplify:

You can drop the sign because the numerator will always be a positive number or 0

and the denominator will be the sign of

. Remember: When is an angle between 0°

and 180°, the sine is positive — corresponding to the tangent positive between 0° and Answers 501–600

90° (half 0° to 180°). When is an angle between 180° and 360°, the sine is negative —

corresponding to the tangent negative between 90° and 180° (half 180° to 360°). The sign will be determined by the angle being evaluated.

For the other form of the identity, begin with

and, inside the radical,

multiply the numerator and the denominator by

:

Replace the 1 in the numerator with

from the Pythagorean identity; then

simplify:

Chapter 17: Answers 355

As before, you can drop the sign:

584.

Rewrite the denominator as

using the Pythagorean identity. Then replace

with its reciprocal identity:

Multiply the numerator by the reciprocal of the denominator:

Replace the tangent term using its ratio identity; then multiply:

Answers 501–600

Use the Pythagorean identity to replace

with

:

The expression on the right is one of the double-angle forms for

. So if

, then twice is equal to and the left side of the identity equals

:

Part II: The Answers

356

585.

Replace

and

with their respective half-angle identities, using the

positive version of each identity:

Multiply numerator and denominator by

:

Multiply the terms under the radical. Replace the result in the numerator with using the Pythagorean identity. Then simplify:

Answers 501–600

Chapter 17: Answers 357

Multiply the numerator and denominator by 2 and simplify:

Rewrite the fraction as the sum of two fractions. Then replace the first term with its reciprocal identity and the second term with its ratio identity:

586. cos b

Working on the right side of the identity, replace the sine of the sum and sine of the difference with their respective identities. Then simplify:

587. cos b

Answers 501–600

Working on the right side of the identity, replace the cosine of the sum and cosine of the difference with their respective identities. Then simplify:

588. sin b

Working on the right side of the identity, replace the cosine of the difference and cosine of the sum with their respective identities. Then simplify:

Part II: The Answers

358

589.

Use the identity

, letting a = 75 and b = 15:

590.

Use the identity

, letting a = 120 and b = 60:

591.

Use the sum-to-product identity

:

Answers 501–600

592.

Use the sum-to-product identity

:

Chapter 17: Answers 359

593.

Use the sum-to-product identity

:

594.

Use the sum-to-product identity

:

595.

Answers 501–600

Use the sum-to-product identity

:

Now use the cosine-of-a-sum identity to find

:

And use the cosine-of-a-difference identity to find

:

Part II: The Answers

360

Now use these values to finish the problem:

596.

Use the power-reducing identity

:

597.

Use the power-reducing identities

and

, replacing x

with and 2 x with . Then simplify:

Answers 501–600

598. 1

Use the power-reducing identity

:

Now use the double-angle identity involving the cosine to replace

:

Chapter 17: Answers 361

Rewrite the fraction as the difference of two fractions. Then use the reciprocal identity for

:

599. 1

Use the reciprocal of the power-reducing identity

and also use

:

As an alternative, if you prefer not to use power-reducing identities, you can use the Pythagorean identity

in the original statement and get to the result

even more quickly.

600.

Use the reciprocal of the power-reducing identity

and also use

:

Answers 501–600

Find a common denominator, change the fractions, and subtract:

Part II: The Answers

362

601.

Use the half-angle identity:

Because an angle of 75° is in the first quadrant, the sine is positive.

You can also solve this problem using the angle-sum identity with angles of 45° and 30°.

The two answers are equivalent, even though they don’t look the same. Here’s the math:

Answers 601–700

Chapter 17: Answers 363

602.

Use the half-angle identity:

Because an angle of 15° is in the first quadrant, the cosine is positive.

You can also solve this problem using the angle-difference identity with angles of 45°

and 30°.

603.

Use the angle-sum identity with angles of 120° and 45°:

Then rationalize the denominator:

You can also solve this problem using the half-angle identity with an angle of 330°.

Answers 601–700

604.

Use the half-angle identity:

Part II: The Answers

364

Because an angle of radians is in the first quadrant, the sine is positive.

You can also solve this problem using the angle-difference identity with angles of radians and radians:

Showing that the two answers are equivalent:

605.

Use the angle-sum identity:

606.

Answers 601–700

Use the angle-difference identity to find

:

Then rationalize the denominator:

Chapter 17: Answers 365

607.

First find

using the angle-sum identity:

Before rationalizing the denominator, use the reciprocal identity to determine

:

Now rationalize the denominator:

608.

First find

using the angle-sum identity with angles of 120° and 45°:

Now use the reciprocal identity with the value you just determined:

Rationalize the denominator:

Answers 601–700

Part II: The Answers

366

609.

First, find

using the angle-sum identity and angles and :

Now use the half-angle identity to find

:

Next, use the value you just determined and the positive radical (because the angle is in the first quadrant):

Finally, use the reciprocal identity:

Rationalizing this answer requires multiplying by three different conjugates — and it Answers 601–700

gets pretty messy.

610.

First, find

using the half-angle identity:

Chapter 17: Answers 367

Because is in the first quadrant, the sine is positive.

Now solve for

using

and the Pythagorean identity:

Use the positive radical because the angle is in the first quadrant.

Now, armed with the sine and cosine of , use the half-angle identity for the tangent: Yes, these radicals within radicals are messy. You can get a simpler answer using angle sum formulas — after using half-angles on sine and cosine. It’s your choice.

611.

Answers 601–700

Replace

with

using the Pythagorean identity. Then break the fraction

into two terms:

Part II: The Answers

368

Replace

using the reciprocal identity. Then simplify each complex fraction:

Replace

using the ratio identity, multiply, and simplify:

Replace

with

using the Pythagorean identity:

The resulting expression is a version of the double-angle identity for cosine.

612.

Rewrite the fractions with a common denominator, add, and simplify:

Replace

with 1 from the Pythagorean identity. Then simplify:

Answers 601–700

Replace

with the reciprocal identity:

Chapter 17: Answers 369

613.

Multiply the two binomials together:

Replace

with its reciprocal identity and

with its ratio identity. Multiply, com-

bine terms, and simplify:

Replace the 1 with

from the Pythagorean identity:

614.

Factor the trinomial as the square of a binomial:

Replace

with

from the Pythagorean identity:

Answers 601–700

615. 0

Use the double-angle identity,

, letting

:

Part II: The Answers

370

616.

Replace

with its reciprocal identity. Combine the terms in the denominator; then simplify the complex fraction:

617.

Factor the terms in the denominator on the left:

Replace the first

in the denominator with its reciprocal identity:

Simplify the complex fraction:

618. 1

Factor the terms by grouping:

Answers 601–700

Use the Pythagorean identity to replace

with

. Also, replace

with

, using its Pythagorean identity:

Replace

with its reciprocal identity:

619.

Replace

in the denominator with its ratio identity and

in the denominator

with its reciprocal identity. Multiply and simplify:

Chapter 17: Answers 371

Split the term on the left into two fractions:

Replace

with its ratio identity and

with its reciprocal identity. Simplify both

complex fractions:

Replace

with its reciprocal identity:

620.

Replace

with its reciprocal identity. Then find a common denominator and

combine the terms in the denominator:

Replace

with its ratio identity. Also replace

with

using the

Pythagorean identity:

Answers 601–700

Simplify the complex fraction:

621.

Use the cosine angle-sum identity with angles of and :

Part II: The Answers

372

Now use the cosine double-angle identity (cosine version) and the sine double-angle identity to replace

and

:

Replace

with

using the Pythagorean identity:

622. 1

Use the cosine’s double-angle identity involving sine to replace

:

Now replace

using the double-angle identity for sine after rewriting the factor

as

. Then simplify:

Use the Pythagorean identity to replace

with

:

623. 1

Answers 601–700

Factor the numerator. Then reduce the fraction:

Replace

with 1:

Replace

using the double-angle identity for sine. Recall that

.

Dividing both sides of the equation by 2 gives you

, so

Chapter 17: Answers 373

624. 0

Subtract the two fractions after finding a common denominator. Then simplify: Multiply the binomials in the denominator:

Use the Pythagorean identity to replace

with

:

The denominator is one of the double-angle identities for cosine, so replace the denominator with

:

Finally, use the reciprocal identity to replace

:

If you recognized that

is the double-angle identity for cosine, you could’ve

Answers 601–700

used that directly to replace the denominator instead of going through the Pythagorean identity.

625. 8

Square the binomials and combine like terms:

Part II: The Answers

374

Replace

with its reciprocal identity:

626. 2

Replace

and

with their double-angle identities. It doesn’t really matter

which double-angle identity you use for cosine.

Replace the 1 with

using the Pythagorean identity:

Break up the left side into three fractions. Then reduce:

Use ratio identities to replace the two fractions:

627. 1

Replace

and

with their double-angle identities. Then simplify:

Answers 601–700

Replace

with

using the Pythagorean identity. Then simplify:

Chapter 17: Answers 375

628. 0

Replace

and

with their double-angle identities. Then simplify:

Rewrite the fraction as the product of two fractions and then reduce: Replace

with its reciprocal identity and

with its ratio identity:

629.

Multiply both numerator and denominator by

and then simplify:

Answers 601–700

Replace

with

using the Pythagorean identity. Then simplify:

Part II: The Answers

376

Write the fraction as the sum of two fractions:

Replace

with its reciprocal identity and

with its ratio identity:

630. 1

Square the binomial and simplify:

In the first term, replace

with its reciprocal identity and then multiply. In the

second term, replace

with

using the Pythagorean identity:

Replace

with its ratio identity:

Factor the binomial:

Answers 601–700

631. 16.4

You know two angles and a consecutive side, so this is an angle-angle-side (AAS) problem. Find the missing side by using the Law of Sines,

:

Chapter 17: Answers 377

632. 14.0

You know two angles and a consecutive side, so this is an angle-angle-side (AAS) problem. Find the missing side by using the Law of Sines,

.

You’re looking for side b, so you need to know the measure of angle B. You know the measures of angles A and C, so find the measure of angle B by subtracting the sum of A and C from 180°:

Now use the Law of Sines to solve for b:

633. 24.1

You know two angles and the side between them, so this is an angle-side-angle (ASA) problem. Find the missing side by using the Law of Sines,

.

You know the length of side a but not the measure of angle A. You know the measures of angles B and C, so find the measure of angle A by subtracting the sum of B and C

from 180°:

Now use the Law of Sines to solve for c:

Answers 601–700

634. 43.1°

You know two sides and a consecutive angle, so this is a side-side-angle (SSA) problem.

Find the measure of angle A by using the Law of Sines,

:

Part II: The Answers

378

635. 52.4°

You know two sides and a consecutive angle, so this is a side-side-angle (SSA) problem.

Find the measure of angle C by using the Law of Sines,

:

636. 15.0°

You know two sides and a consecutive angle, so this is a side-side-angle (SSA) problem.

Find the measure of angle C by using the Law of Sines,

:

637. 113 feet

Always start by drawing and labeling a diagram to represent the problem: Answers 601–700

Illustration by Thomson Digital

You can see from the diagram that you know two angles and the side between them, so this is an angle-side-angle (ASA) problem.

You know the length of side c, so you need to find the measure of angle C. You know the measures of angles A and B, so find the measure of angle C by subtracting the sum of A and B from 180°:

Chapter 17: Answers 379

The side from point A to point C is b, so use the Law of Sines to solve for b: The distance from point A to point C is approximately 113 feet.

638. 85.3°

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

You can see from the diagram that you know two sides and a consecutive angle, so this is a side-side-angle (SSA) problem. You know the lengths of sides a and b, and you need to find the measure of angle A. Use the Law of Sines:

Answers 601–700

The pole makes an angle of about 85.3° with the ground.

Part II: The Answers

380

639. 1,356 feet

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

The arrows show the direction that Jake is walking. At A, Jake turns and walks along a path that makes an angle of 50° with the road. At B, he turns 105° to walk back to the road. This redirection makes an angle of

with the path, so the measure

of angle

. You want to find the length of the path from B to C, which is a.

You can see from the diagram that you know two angles and a side between them, so this is an angle-side-angle (ASA) problem. You know the length of side c, so you need to find the measure of angle C. You know the measures of angles A and B, so find the measure of angle C by subtracting the sum of A and B from 180°: Now use the Law of Sines to solve for a:

Answers 601–700

Jake has to walk about 1,356 feet to get back to the road. He takes more than a half-mile detour!

Chapter 17: Answers 381

640. 27 feet

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

The tree is labeled BT, Elena is standing at E, and Carlos is standing at C. There are three triangles in the diagram: triangle BET, triangle CET, and triangle BCT. You need to find length BT, but neither triangle containing this side has enough information .

Triangles BCT and BET are both right triangles with one acute angle labeled. If you can find the length of the hypotenuse in either triangle, you’ll have enough information to find BT.

Use triangle CET. In this triangle, the measure of angle ECT is

. Then the

measure of angle ETC is

. Use the Law of Sines to find the length

of

, which is the hypotenuse of triangle BET:

Because you need to use this result as-is in later calculations, don’t approximate it yet.

Now look at triangle BET:

Answers 601–700

Illustration by Thomson Digital

Part II: The Answers

382

Because you now have the length of the hypotenuse in triangle BET, you have enough information to solve for the length BT by using the ratio definition of sine: Now, replace

with

:

The tree is about 27 feet tall.

641. 1

The following chart summarizes the ambiguous case of the Law of Sines. When you’re given the lengths of two sides of a triangle and the angle opposite one of them — SSA — various possibilities exist: Two triangles may be possible, one triangle may be possible, or no triangles at all may be possible.

Illustration by Thomson Digital

Answers 601–700

This figure shows the situation:

Illustration by Thomson Digital

Chapter 17: Answers 383

Drawing an altitude from vertex C, you form a right triangle with an acute angle of 85°.

You get the 85° by subtracting

because the two angles along the base are

supplementary. Then, in that small right triangle formed to the left of ABC, Because A is obtuse and the length of the altitude is less than that of the opposite side, only one triangle is possible.

642. 0

The following chart summarizes the ambiguous case of the Law of Sines. When you’re given the lengths of two sides of a triangle and the angle opposite one of them — SSA — various possibilities exist: Two triangles may be possible, one triangle may be possible, or no triangles at all may be possible.

Illustration by Thomson Digital

This figure shows the situation. Drawing an altitude from angle B, you’re not sure whether angle A is obtuse (making side c fold left toward C) or acute (making side c fold right). Or maybe neither works.

Answers 601–700

Illustration by Thomson Digital

To compute the altitude, use the following formula.

This situation is impossible because the opposite side can’t be shorter than the altitude. Therefore, no triangles are possible.

Part II: The Answers

384

643. 2

The following chart summarizes the ambiguous case of the Law of Sines. When you’re given the lengths of two sides of a triangle and the angle opposite one of them — SSA — various possibilities exist: Two triangles may be possible, one triangle may be possible, or no triangles at all may be possible.

Illustration by Thomson Digital

The figure shows the situation:

Illustration by Thomson Digital

Drawing an altitude from angle C, you’re not sure whether angle B is obtuse (making side a fold left toward A) or if angle B is acute (making side a fold right). Or maybe neither works. To compute the altitude, use the following formula:

Answers 601–700

Because the altitude is less than the opposite side, you have the possibility of two different triangles.

644. Triangle 1:

,

,

; Triangle 2:

,

,

First find the measure of angle C by using the Law of Sines: Chapter 17: Answers 385

Now find the measure of angle B:

Finally, find b using the Law of Sines. Always use as much of the given information as possible just in case you’ve made an error elsewhere.

Because

, there’s another possible angle, C ′ :

This solution creates a second triangle, so find

:

Now find b′ by using the Law of Sines:

Here are the two solutions, shown in the following diagram:

Answers 601–700

Illustration by Thomson Digital

Part II: The Answers

386

645. no solution

First, find the measure of angle B by using the Law of Sines: If you plug

into your calculator, you get an error message because

, and sin B must be between −1 and 1. Therefore, no triangle exists, and no solution is possible.

646. 17.4

This is an SAS problem (you’re given two sides and the included angle), so find the missing side, a, by using the Law of Cosines:

647. 38.1

This is an SAS problem (you’re given two sides and the included angle), so find the missing side, c, by using the Law of Cosines:

Answers 601–700

648. 19.0

This is an SAS problem (you’re given two sides and the included angle), so find the missing side, b, by using the Law of Cosines:

Chapter 17: Answers 387

649. 41.7°

This is an SSS problem (you’re given all three sides), so find the measure of angle C by using the Law of Cosines:

650. 137.0°

This is an SSS problem (you’re given all three sides), so find the measure of angle A by using the Law of Cosines:

651. 72.1°

This is an SSS problem (you’re given all three sides), so find the measure of angle B by using the Law of Cosines:

Answers 601–700

652. 347 yards

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

Part II: The Answers

388

The arrows show the direction that the surveyor is walking. He starts at A, walks 145 yards to B, turns right 110°, and then walks 270 yards to C.

The diagram shows that this is an SAS problem (you’re given two sides and the included angle), so use the Law of Cosines to find AC, the length of the lake. Note that AC is opposite angle B, so it’s b in the formula for the Law of Cosines.

653. 94°

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

The diagram shows that this is an SSS problem (you’re given all three sides), so use the Law of Cosines to find the measure of angle B, which is the angle opposite the diagonal AC: Answers 601–700

Because 94° is an obtuse angle, it’s the larger angle of the rhombus.

654. 88 feet

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

Chapter 17: Answers 389

The television camera is located at T, the player is originally located at P, and he catches the ball at C.

The diagram shows that this is an SAS problem (you’re given two sides and the included angle), so use the Law of Cosines to find PC, the distance the center fielder has to run to catch the ball:

The center fielder runs about 88 feet in order to catch the ball.

655. 53°

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

The race starts at A, the next leg starts at B, and the final leg starts at C.

The diagram shows that this is an SSS problem (you’re given all three sides), so use the Law of Cosines to find the measure of angle B, which is the angle between the first two legs of the race:

Answers 601–700

The angle between the first two legs of the race is about 53°.

656. 106.9

You know the measures of two sides and the included angle, so use the formula to find the area:

Part II: The Answers

390

657. 120.6

You know the measures of two sides and the included angle, so use the formula to find the area:

658. 102.1

You know the measures of two sides and the included angle, so use the formula to find the area:

659. 24.0

The problem tells you the measures of c and a, which it calls AB and BC, respectively.

So you know that

and

.

Because you know the measures of two sides and the included angle, use the formula to find the area:

660. 13.7

The problem tells you the measures of b and c, which it calls AC and AB, respectively.

Answers 601–700

So you know that

and

.

Because you know the measures of two sides and the included angle, use the formula to find the area:

661. 173.9

The problem tells you the measures of b and a; it just calls them AC and BC, respectively. So you know that

and

.

Chapter 17: Answers 391

Because you know the measures of two sides and the included angle, use the formula to find the area:

662. 100 square meters

You know the measures of two sides and the included angle, so use the formula to find the area:

The area of the triangle is 100 square meters.

663. 224 square inches

Start by drawing and labeling a diagram to represent the problem:

Answers 601–700

Illustration by Thomson Digital

When you draw the diagonal AC, you get two congruent triangles. To find the area of the rhombus, find the area of one of the triangles and then double it.

In triangle ABC, you know the measures of two sides and the included angle, so use the formula

to find the triangle’s area:

The area of the rhombus is about

square inches.

Part II: The Answers

392

664. 381.4

To use the area formula involving an angle and the two sides forming that angle, you need the measure of one more side. You have angle A, so you need the measures of sides

and

. You’re missing AB, but you can determine this measure by using the Law of Sines.

First, find

, the only angle you don’t have yet:

Now use the Law of Sines to find AB:

Don’t bother computing c. Just substitute

for c in the area formula. That way,

you carry all the decimal places through the calculation so that the answer is as accurate as possible.

Now find the area of the triangle by using

:

665. 72.4

To use the area formula involving an angle and the two sides forming that angle, you need the measure of one more side. You can use either angle B or angle C — your choice.

To use angle B in the area formula, you need the measures of sides and

.

Answers 601–700

You’re missing AB, but you can determine this measure by using the Law of Sines.

First, find

:

Now use the Law of Sines to find AB:

Don’t bother computing c. Just substitute

for c in the area formula. That way,

you carry all the decimal places through the calculation so that the answer is as accurate as possible.

Chapter 17: Answers 393

Now find the area of the triangle by using

:

666. 194.4

To use the area formula involving an angle and the two sides forming that angle, you need the measure of one more side. You can use either angle B or angle C — your choice.

To use angle C in the area formula, you need the measures of sides and

.

You’re missing AC, but you can determine this measure by using the Law of Sines.

First, find

:

Now use the Law of Sines to find AC:

Don’t bother computing c. Just substitute

for b in the area formula. That way,

you carry all the decimal places through the calculation so that the answer is as accurate as possible.

Now find the area of the triangle by using

:

667. 50.7

Answers 601–700

To use the area formula involving an angle and the two sides forming that angle, you need the measure of one more side. You can use either angle A or angle C — your choice. Here I use angle C.

To use angle C in the area formula, you need the measures of sides and

.

You’re missing AC, but you can determine this measure by using the Law of Sines.

First, find

:

Now use the Law of Sines to find AC:

Part II: The Answers

394

Don’t bother computing b. Just substitute

for b in the area formula. That way,

you carry all the decimal places through the calculation so that the answer is as accurate as possible.

Now find the area of the triangle by using

:

668. 219.2

To use the area formula involving an angle and the two sides forming that angle, you need the measure of angle B, the angle between the two given sides. You can do so by first solving for the measure of angle C — using the Law of Sines — and then subtracting the sum of the measures of angles A and C from 180.

Solve for the measure of angle C:

Don’t bother computing C. Just plug

into the formula to find the mea-

sure of angle B:

Now find the area of the triangle by using

:

Answers 601–700

669. 279.0

To use the area formula involving an angle and the two sides forming that angle, you need the measure of angle C, the angle between the two given sides. You can do so by first solving for the measure of angle A — using the Law of Sines — and then subtracting the sum of the measures of angles A and B from 180.

Chapter 17: Answers 395

Find the measure of angle A:

Don’t bother computing A. Just plug

into the formula to find the

measure of angle C:

Now find the area of the triangle by using

:

670. 31.3 square inches

In triangle BCD, you’re given an angle and an adjacent side. You need to find the length of the other adjacent side, BD, to find the area of the triangle.

Look at triangle ABD. It’s a 30°-60°-90° triangle, and you know the length of the hypotenuse. BD is the shorter leg, so

Now find the area of the triangle by using

:

Answers 601–700

The area of triangle BCD is about 31.3 square inches

671.

This is an SSS problem (you’re given all three sides of the triangle), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of triangle PQR is

Now substitute the value for s and the lengths of the sides into the area formula: Part II: The Answers

396

672.

This is an SSS problem (you’re given all three sides of the triangle), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of triangle ABC is

Now substitute the value for s and the lengths of the sides into the formula: 673.

This is an SSS problem (you’re given all three sides of the triangle), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of triangle XYZ is

Now substitute the value for s and the lengths of the sides into the formula: 674. 21.3

This is an SSS problem (you’re given all three sides of the triangle), so use Heron’s Answers 601–700

formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of triangle ABC is

Now substitute the value for s and the lengths of the sides into the formula: 675. 25.7

This is an SSS problem (you’re given all three sides of the triangle), so use Heron’s formula,

, to find the area.

Chapter 17: Answers 397

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of triangle ABC is

Now substitute the value for s and the lengths of the sides into the formula: 676. 26.5

This is an SSS problem (you’re given all three sides of the triangle), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of triangle ABC is

Now substitute the value for s and the lengths of the sides into the formula: 677. 37,382 square feet

This triangle is an SSS problem (you’re given all three sides), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of the given triangle is

Answers 601–700

Now substitute the value for s and the lengths of the sides into the formula: The area of the plot is about 37,382 square feet.

678. 431,600 square miles

This triangle is an SSS problem (you’re given all three sides), so use Heron’s formula,

, to find the area.

Part II: The Answers

398

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of the given triangle is

Now substitute the value for s and the lengths of the sides into the formula: The area of the Bermuda Triangle is about 431,600 square miles.

Note: When you multiply

on a calculator, you may get an answer

like 1.863E11. The product just has more digits than the calculator can handle. One way to deal with this issue is to simplify the radical first:

679. 32 square inches

For triangle BCD, the diagram shows that

and

inches. Triangle ABD is a

30°-60°-90° triangle with hypotenuse

in. In a 30°-60°-90° triangle, the length of

the side opposite the 30° angle is half the length of the hypotenuse, so inches.

That means that

inches as well.

Triangle BCD is an SSS problem (you’re given all three sides), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of the given triangle is

Now substitute the value for s and plug the lengths of the sides into the formula: Answers 601–700

The area of triangle BCD is about 32 square inches.

680. 12 square units

First, find the length of each side of the triangle using the distance formula

:

Chapter 17: Answers 399

This triangle is an SSS problem (you’re given all three sides), so use Heron’s formula,

, to find the area.

In Heron’s formula, s is the semiperimeter (half the perimeter), represented by

. The semiperimeter of the given triangle is

Now substitute the value for s and the lengths of the sides into the formula: Answers 601–700

This result looks really complicated, but it simplifies as follows:

Part II: The Answers

400

The numerators have the form

. This expression is the special product

, so you can rewrite the equation as follows:

The area of the triangle is 12 square units. (Of course, you can just use a calculator!) 681. 55.6 yards

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

The woman is located at F, the man is located at M, and the balloon is located at B.

You can see from the diagram that you know two angles and the side between them, so this is an ASA problem. You know the length of side b, so you need to find the measure of angle B:

Answers 601–700

The side from point M to point B is f, so use the Law of Sines and solve for f: The distance from the man to the balloon is approximately 55.6 yards.

Chapter 17: Answers 401

682. 191 meters

The diagram shows that you know two sides and a consecutive angle, so this is an SSA problem. First, find the measure of angle A:

Next, find the measure of angle B:

Now use the Law of Sines to find the third side, AC:

The question asks for the amount of fencing needed to enclose the entire park, so find the perimeter of the park:

The park needs 191 meters of fencing.

683. 29.6 miles

Always start by drawing and labeling a diagram to represent the problem: Answers 601–700

Illustration by Thomson Digital

The tracking stations are located at A and B, and the rocket is located at C. C is located where shown in the diagram because the problem specifically says that the rocket isn’t between the two stations. The altitude of the rocket is CD. Triangle ADC is a 45°-45°-90°

triangle, so when you know the length of the hypotenuse, AC, you can easily find CD.

Part II: The Answers

402

Angle ABC is supplementary to the 72° angle in triangle BDC, so its measure is Now you know two of the three angles in triangle ABC, so you can subtract them from 180° to find the remaining angle:

With this information, you can use the Law of Sines to find AC in triangle ABC: Because you’ll need to use this result to find the hypotenuse of triangle ADC, don’t compute the result yet. (Not simplifying yet ensures a more accurate answer later because you don’t lose any decimal places.)

Remember that in a 45°-45°-90° triangle, the length of a leg is equal to the length of the hypotenuse divided by

:

The rocket is about 29.6 miles above the earth.

684. 38.3 miles

Always start by drawing and labeling a diagram to represent the problem: Answers 601–700

Illustration by Thomson Digital

First, find

. It’s the supplement of the 35° angle, so

.

The diagram shows that this is an SAS problem (you’re given two sides and the included angle).

You could use the Law of Sines, but using the Law of Cosines is less work: The boat is about 38.3 miles from port.

Chapter 17: Answers 403

685. 3.6 centimeters

Always start by drawing and labeling a diagram to represent the problem: Illustration by Thomson Digital

Remember that the opposite sides of a parallelogram are equal and that the diagonals of a parallelogram bisect each other. So

centimeters and

centimeters.

You’re looking for BC.

The diagram shows that this is an SAS problem (you’re given two sides and the included angle). You could use the Law of Sines, but using the Law of Cosines is less work:

The short sides are about 3.6 centimeters.

686. 40°

Always start by drawing and labeling a diagram to represent the problem: Answers 601–700

Illustration by Thomson Digital

Part II: The Answers

404

The diagram shows that this is an SSS problem (you’re given all three sides). You’re looking for the measure of angle C, so use the Law of Cosines: The angle between the legs when the ladder is open measures about 40°.

687. 10

To solve this problem, you set the triangle’s perimeter equal to its area, using Heron’s formula,

, to find the area. In Heron’s formula, s is the semi-

perimeter (half the perimeter). The perimeter of the triangle is

, and half of

that is

.

Here’s the equation to solve:

Notice the nice pairs of binomials under the radical. Multiply them together and then square both sides of the equation:

Answers 601–700

Now multiply each term by 16 to eliminate the fractions and then simplify: Yes, this is a fourth-degree polynomial that you can solve using the rational root theorem, but you’re interested only in finding a solution that could be the side of a triangle.

The triangle inequality theorem says that any side of a triangle must be shorter than the other two sides added together. So if two of the sides of the triangle are 9 and 17, the third side can be any length between and including 9 and 25. This is assuming, of course, that the measure is an integer.

Chapter 17: Answers 405

Looking at the polynomial equation, you see from the coefficient of 1 on the term that any rational roots will be integers. Now look at the constant term. The factors of 54,080

include 1, 2, 4, 5, 8, 10, 13, 16, 20, and so on. These are some of the possible roots/

solutions of the equation. The four numbers between 9 and 25 are 10, 13, 16, and 20, so check those four factors. Replacing the x’s with 10 in the equation, you have a solution!

Neither 13, 16, nor 20 works. So the third side of the triangle measures 10 units.

688. 305,782 square meters

The field consists of two triangles, so find the area of each and add them together to find the area of the field.

In triangle PQR, you know the length of one side and the measures of two angles, but the given side isn’t between the two angles, and that’s the side you need to know to use the area formula. Because you know PQ and the measures of and

, you

need to find PR to find the area of the triangle. First, find

:

Now use the Law of Sines to find PR:

Don’t bother computing PR. Just substitute

for PR in the area formula. That

way, you carry all the decimal places through the calculation so that the answer is as accurate as possible.

Now find the area of the triangle:

Answers 601–700

Now you have to find the area of triangle PSR. You know the measures of two sides, meters and

, as well as the included angle

:

The total area of the field is

square meters.

689. 5.9 miles

Remember that

. So after three hours, Marie has walked

miles

and Don has walked

miles. Now draw a diagram to represent the problem. S rep-

resents the starting point of both Don and Marie, M is Marie’s destination, and D is Don’s stopping point.

Part II: The Answers

406

Illustration by Thomson Digital

You need to find length MD. In order to do that, first find

. Marie is walking due

north, which forms a 90° angle with the baseline. Don is walking 62° northeast, so

. Now you have an SAS problem (you know two sides and the

included angle), so use the Law of Cosines to find MD:

They’re about 5.9 miles apart after three hours.

690. 309.0 square yards

You can divide the regular octagon into eight congruent triangles. Find the area of one of the triangles and multiply by 8.

Each triangle in the octagon has an angle at the center measuring 45°. Dividing a triangle in half at that center, you have a right triangle with an angle of 22.5° and opposite side measuring 4 yards. Use the tangent of 22.5° to solve for the adjacent side of the triangle, h:

Answers 601–700

Now find the area of one of the eight triangles, which has a base of 8 yards and a height of 9.6569 yards:

So the area of the entire octagon is 8 times that area:

You could do this problem using Heron’s formula, which involves finding the length of one of the segments drawn from the center to a vertex of the octagon. That process would involve using the square root. You could also use the trig formula involving the sine of the angle and measures of two sides — the height and the segment to the vertex.

That one is too much work. The method that uses right triangles seems to be the easi-est. And, finally, if you have the formula in your back pocket, you can find the area of any regular polygon using

, where the apothem is the

height of the triangle. So many choices!

Chapter 17: Answers 407

691.

i

Rewrite the exponent as the sum of a multiple of 4 and a number between 0 and 3: Now write the power of i as the product of two powers:

The value of is 1, so is

.

692. –1

Rewrite the exponent as the sum of a multiple of 4 and a number between 0 and 3: Now write the power of i as the product of two powers:

The value of is 1, and

, so is

.

693. 1

Rewrite the exponent as the sum of a multiple of 4 and a number between 0 and 3: Now write the power of i as the product of two powers:

The value of is 1, and

, so

is

.

694.

i

Rewrite the exponent as the sum of a multiple of 4 and a number between 0 and 3: Answers 601–700

Now write the power of i as the product of two powers:

The value of is 1, so

.

695. – i

Rewrite the exponent as the sum of a multiple of 4 and a number between 0 and 3: Now write the power of i as the product of two powers:

The value of is 1, and

, so

is

Part II: The Answers

408

696.

Add the two real numbers together, and add the two imaginary numbers together: The answer is

.

697.

Subtract the two real numbers, and subtract the two imaginary numbers: The answer is

.

698. 3

Add the first two real numbers and subtract the third from the sum:

Also add the first two imaginary numbers together and subtract the third from the sum:

The answer is

.

699.

Add the first two real numbers and add the third to the sum:

Answers 601–700

Also add the first two imaginary numbers together:

The answer is

.

700.

Subtract the second real number from the first:

Subtract the second imaginary from the first and then subtract the third imaginary number from the result:

The answer is

.

Chapter 17: Answers 409

701.

Add the two real numbers together, and add the two imaginary numbers together: The answer is

.

702.

You can’t combine the i terms unless the radicals match, so first simplify

:

The problem now reads

Subtract the two real numbers, and subtract the two imaginary numbers: The answer is

.

703.

First simplify

and

so you can combine the i terms:

The problem now reads

Add the two real numbers together, and add the two imaginary numbers together: Answers 701–800

The answer is

.

410 Part II: The Answers

704.

Subtract the two real numbers, and subtract the two imaginary numbers: The answer is

.

705.

Add the two real numbers together, and add the two imaginary numbers together: The answer is

.

706.

Multiply each term in the parentheses by 6:

The answer is

.

707.

Multiply each term in the parentheses by –5:

The answer is

.

708.

Multiply each term in the parentheses by 2 i:

Answers 701–800

The answer is

.

Chapter 17: Answers 411

709.

Multiply each term in the parentheses by –3 i:

The answer is

.

710. 13

Use FOIL to multiply the binomials:

711. 68

Use FOIL to multiply the binomials:

712.

Use FOIL to multiply the binomials:

Answers 701–800

412 Part II: The Answers

713.

Rewrite the square of the binomial as the product of two binomials; then multiply: 714.

To find the cube of the binomial, use Pascal’s triangle:

Recall that when using the triangle to determine the coefficients of the power of a binomial, you use the row whose second number is the same as the power.

First write the coefficients from the row whose second number is 3:

The first term in

is 2, so write in decreasing powers of 2:

Now write in increasing powers of the second term, 3 i:

Simplify and write the sum of the terms:

Answers 701–800

Chapter 17: Answers 413

715.

To find the fourth power of the binomial, use Pascal’s triangle:

Recall that when using the triangle to determine the coefficients of the power of a binomial, you use the row whose second number is the same as the power.

First write the coefficients from the row whose second term is 4:

The first term in

is –1, so write in decreasing powers of –1:

Now write in increasing powers of the second term, –4 i:

Simplify and write the sum of the terms:

716.

Multiply both the numerator and the denominator by the conjugate of the denominator (the conjugate of a binomial is the same two terms with the opposite sign between them): Answers 701–800

717.

Multiply both the numerator and the denominator by the conjugate of the denominator (the conjugate of a binomial is the same two terms with the opposite sign between them): 414 Part II: The Answers

718.

Multiply both the numerator and the denominator by the conjugate of the denominator (the conjugate of a binomial is the same two terms with the opposite sign between them): 719. –3 – 3 i

Multiply both the numerator and the denominator by the conjugate of the denominator (the conjugate of a binomial is the same two terms with the opposite sign between them): 720. 7 – 4 i

Multiply both the numerator and the denominator by the conjugate of the denominator (the conjugate of a binomial is the same two terms with the opposite sign between them): 721.

The quadratic doesn’t factor, so use the quadratic formula,

. In this

problem, a = 1, b = –3, and c = 4:

722.

Answers 701–800

The quadratic doesn’t factor, so use the quadratic formula,

. In this

problem, a = 1, b = 4, and c = 8:

Chapter 17: Answers 415

723.

The quadratic doesn’t factor, so use the quadratic formula,

. In this

problem, a = 2, b = –5, and c = 25:

724.

The quadratic doesn’t factor, so use the quadratic formula,

. In this

problem, a = 4, b = 1, and c = 4:

725.

The quadratic doesn’t factor, so use the quadratic formula,

. In this

problem, a = 5, b = 2, and c = 3:

726.

The point is 3 units to the right on the real axis and 2 units up on the imaginary axis.

727.

Answers 701–800

The point is 5 units to the left on the real axis and 1 unit up on the imaginary axis.

416 Part II: The Answers

728.

The point is 4 units to the left on the real axis and 4 units down on the imaginary axis.

729. –3

The point is 3 units to the left on the real axis, and it sits on that axis with an imaginary coordinate of 0.

730. 3 i

The point is 3 units up on the imaginary axis, and it sits on that axis with a real coordinate of 0.

731.

The point is on the 2-unit ring, so the radius is 2. You measure the angle from the positive x-axis in a counterclockwise direction. Each ray represents 15° (or ), so this point is on the ray representing . The radius is 2, so the point is 2 units out from the origin.

732.

The point is on the 3-unit ring, so the radius is 3. You measure the angle from the positive x-axis in a counterclockwise direction. Each ray represents 15° (or ), so this point is on the ray representing .

733.

The point is on the 2-unit ring, so the radius is 2. You measure the angle from the positive x-axis in a counterclockwise direction. Each ray represents 15° (or ), so this point is on the ray representing .

734.

The point is on the 1-unit ring, so the radius is 1. You’re measuring the angle from the Answers 701–800

positive x-axis in a clockwise direction. Each ray represents 15° (or ), so this point is on the ray representing

. Therefore, the answer is

. If you were to use a posi-

tive angle instead, this point would be

.

Chapter 17: Answers 417

735.

The point is on the 3-unit ring, so the radius is 3. You’re measuring the angle from the positive x-axis in a clockwise direction. Each ray represents 15° (or ), so this point is on the ray representing

. Therefore, the answer is

. If you were to use a

positive angle instead, this point would be

.

736. (0, –2)

You find the x coordinate with

, so

You find the y coordinate with

, so

737.

You find the x coordinate with

, so

You find the y coordinate with

, so

738.

You find the x coordinate with

, so

You find the y coordinate with

, so

Answers 701–800

418 Part II: The Answers

739.

You find the x coordinate with

, so

An angle measuring

is coterminal with the angle (they have the same terminal

ray when graphed in standard position), so

You find the y coordinate with

, so

740. (–2, 0)

You find the x coordinate with

, so

You find the y coordinate with

, so

741.

You find the radius with

, so

You find the angle with

, so

The tangent of an angle is negative in the second and fourth quadrants. The rectangular coordinates are those of a point in the fourth quadrant, so you want an angle between Answers 701–800

and . Add to

to get

Chapter 17: Answers 419

742.

You find the radius with

, so

You find the angle with

, so

The tangent of an angle is negative in the second and fourth quadrants. The rectangular coordinates are those of a point in the second quadrant, so you want an angle between and . Add to

to get

743.

You find the radius with

, so

You find the angle with

, so

The tangent of an angle is positive in the first and third quadrants. The rectangular coordinates are those of a point in the first quadrant, so

744. (3, 0)

You find the radius with

, so

Answers 701–800

You find with

, so

420 Part II: The Answers

The tangent of an angle is 0 on the x-axis. The rectangular coordinates are those of a point on the positive x-axis, so

745. (1, 4.429)

You find the radius with

, so

You find the angle with

, so

The tangent of an angle is positive in the first and third quadrants. The rectangular coordinates are those of a point in the third quadrant, so

746. Archimedean spiral

Illustration by Thomson Digital

Answers 701–800

The equation has the general form for an Archimedean spiral,

. In this case,

a = 0 and b = 2.

Chapter 17: Answers 421

747. cardioid

Illustration by Thomson Digital

The equation has the general form for a cardioid,

, where

. In this

case, a = 2 and b = 2.

748. lemniscate

Illustration by Thomson Digital

The equation has the general form for a lemniscate,

. In this case, a = 2.

Answers 701–800

422 Part II: The Answers

749. rose

Illustration by Thomson Digital

The equation has the general form for a rose,

, In this case, a = 3 and n = 5.

750. limaçon

Illustration by Thomson Digital

The equation has the general form for a limaçon,

. In this case, a = 2

Answers 701–800

and b = 3.

Chapter 17: Answers 423

751. hyperbola; center: (3, –2)

The standard form for the equation of a hyperbola with center ( h, k) is The given equation is already in this form, so you can identify the center coordinates by looking at the values substituted for h and k.

752. ellipse; center: (–4, 1)

The standard form for the equation of an ellipse with center ( h, k) is The given equation is already in this form, so you can identify the center coordinates by looking at the values substituted for h and k.

753. circle; center: (5, 0)

The standard form for the equation of a circle with center ( h, k) is The given equation is already in this form, with a bit of a twist: No k value is given, so you have to recognize that k is 0 (because

). You can identify the x coordi-

nate more easily by looking at the values substituted for h.

754. ellipse; center: (0, –3)

The standard form for the equation of an ellipse with center ( h, k) is To put the given equation in the standard form, divide each term in the equation by 36: Answers 701–800

Now you can identify the center coordinates by looking at the values substituted for h and k. In this case, no h value is given, so you have to recognize that h is 0 (because

).

424 Part II: The Answers

755. hyperbola; center: (–3, 8)

The standard form for the equation of a hyperbola with center ( h, k) is To put the equation in the standard form, divide each term in the equation by 54: This last form of the equation is an alternate that lets you see, from the –1 on the right, that the hyperbola will open upward and downward.

Now you can identify the center coordinates by looking at the values substituted for h and k.

756.

Rewrite the equation by grouping the x terms and the y term and moving the constant to the right side:

Complete the square with the x terms, adding 9 to both sides: Factor the perfect square trinomial and simplify the equation:

This is the equation of a circle with center (3, 0) and a radius of 4.

757.

Rewrite the equation by grouping the x terms and the y terms and moving the constant to the right side:

Factor 25 from each of the x terms and 9 from each of the y terms: Answers 701–800

Complete the square with the x terms by adding 4 in the parentheses; add 100 to the right side (because

):

Complete the square with the y terms by adding 9 in the parentheses; add 81 to the right side (because

).

Chapter 17: Answers 425

Factor the perfect square trinomials and simplify the equation:

Divide each term by 225:

This is the equation of an ellipse with its center at (–2, 3).

758.

Rewrite the equation by grouping the y terms and moving the constant to the right side:

Factor 25 from each of the y terms:

Complete the square with the y terms by adding 9 in the parentheses; add 225 to the right side (because

):

Factor the perfect square trinomial and simplify the equation:

Divide each term by 25:

This last form of the equation is an alternate that lets you see, from the –1 on the right, that the hyperbola will open upward and downward.

This is the equation of a hyperbola with its center at (0, 3).

759.

Answers 701–800

Rewrite the equation by grouping the x terms and the y terms and moving the constant to the right side:

Factor 4 from each of the x terms and 4 from each of the y terms: Complete the square with the x terms by adding 16 in the parentheses; add 64 to the right side (because

):

426 Part II: The Answers

Complete the square with the y terms by adding 16 in the parentheses; add another 64

to the right side (because

):

Factor the perfect square trinomials and simplify the equation:

Divide each term by 4:

This is the equation of a circle with its center at (4, –4).

760.

Rewrite the equation by grouping the x terms and the y terms and moving the constant to the right side:

Factor 4 from each of the x terms and 3 from each of the y terms: Complete the square with the x terms by adding 1 in the parentheses; add 4 to the right side (because

):

Complete the square with the y terms by adding 4 in the parentheses; add 12 to the right side (because

):

Factor the perfect square trinomials and simplify the equation:

Divide each term by 3. Then multiply the first term by divided by to create a coefficient of 1 in the numerator:

Answers 701–800

This is the equation of an ellipse with its center at (1, 2).

Chapter 17: Answers 427

761.

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the given point (0, 0) for the ( h, k) and square the 4: Illustration by Thomson Digital

762.

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the given point (4, 3) for the ( h, k) and square the 5: Answers 701–800

Illustration by Thomson Digital

428 Part II: The Answers

763.

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the given point (–2, 1) for the ( h, k) and square the 1: Illustration by Thomson Digital

764.

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the given point (0, –2) for the ( h, k) and square the .

Illustration by Thomson Digital

765.

Answers 701–800

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the given point (–5, 0) for the ( h, k) and square the

:

Chapter 17: Answers 429

Illustration by Thomson Digital

766.

Find the center of the diameter (and therefore the center of the circle) by using the midpoint formula:

To find the radius, find the distance from the center to one of the endpoints of the diameter by using the distance formula:

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the center point (1, 3) for the ( h, k) and square the 5: Answers 701–800

Illustration by Thomson Digital

430 Part II: The Answers

767.

Find the center of the diameter (and therefore the center of the circle) by using the midpoint formula:

To find the radius, find the distance from the center to one of the endpoints of the diameter by using the distance formula:

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the center point (–2, –2) for the ( h, k) and square the

:

Illustration by Thomson Digital

768.

Look at a sketch of possible circles first. Each circle shown is tangent to both axes and lies in the third quadrant.

Answers 701–800

Illustration by Thomson Digital

Chapter 17: Answers 431

For the circle you want, the points of tangency of the circle and axes are at (–3, 0) and (0, –3). The center of the circle is at (–3, –3).

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the center point (–3, –3) for the ( h, k) and square the 3: Now you can sketch the exact graph:

Illustration by Thomson Digital

769.

Look at a sketch of some possible circles first. Each circle is tangent to the lines and

and lies in the third quadrant.

Answers 701–800

Illustration by Thomson Digital

The center of the circle is 7 units to the left of the vertical line

, which means it’s

at

. The center is also 7 units below the horizontal line

, putting it at

.

So the center has coordinates (–8, –9).

432 Part II: The Answers

The standard equation of a circle with radius ( h, k) and radius r is

.

Substitute the center point (–8, –9) for the ( h, k) and square the 7: Now you can sketch the exact graph:

Illustration by Thomson Digital

770.

and

Look at a sketch of some possible circles first. Each is tangent to both axes and has its center on the line

.

Answers 701–800

Illustration by Thomson Digital

Chapter 17: Answers 433

Two circles fit this description: one in the second quadrant and the other in the fourth quadrant. With the centers on the line

, the coordinates of the centers are oppo-

sites: (–4, 4) and (4, –4).

The standard equation of a circle with radius ( h, k) and radius r is

.

First, substitute one of the center points, (–4, 4), for the ( h, k) and square the 4: Now do the same for the second center point, (4, –4):

Now you can sketch the exact figures:

Illustration by Thomson Digital

771. focus:

; axis of symmetry:

The standard equation of a parabola that opens upward or downward is

,

Answers 701–800

where ( h, k) are the coordinates of the vertex and

. The value of p gives you the

distance from the vertex to the focus and directrix.

In this parabola,

, so

. The focus is unit above the vertex at

, and the

directrix is unit below the vertex at

. The axis of symmetry goes through the

vertex and is perpendicular to the directrix; its equation is

.

434 Part II: The Answers

772. focus: (–1, 1); axis of symmetry:

The standard equation of a parabola that opens upward or downward is

,

where ( h, k) are the coordinates of the vertex and

. The value of p gives you the

distance from the vertex to the focus and directrix.

Pull the values for h and k from the given equation. The vertex is (–1, 2).

In this parabola,

, so p = –1. The focus is 1 unit below the vertex at (–1, 1), and the directrix is 1 unit above the vertex at

. The axis of symmetry goes through the

vertex and is perpendicular to the directrix; its equation is

.

773. focus:

; axis of symmetry:

The standard equation of a parabola that opens to the left or right is

,

where ( h, k) are the coordinates of the vertex and

. The value of p gives you the

distance from the vertex to the focus and directrix.

Pull the values for h and k from the given equation. The vertex is (–1, 4).

In this parabola,

, so

. The focus is unit to the right of the vertex at

,

and the directrix is unit to the left the vertex at

. The axis of symmetry goes

through the vertex and is perpendicular to the directrix; its equation is

.

774. focus:

; axis of symmetry:

The standard equation of a parabola that opens to the left or right is

,

where ( h, k) are the coordinates of the vertex and

. The value of p gives you the

distance from the vertex to the focus and directrix.

Pull the values for h and k from the given equation. The vertex is (6, –1).

In this parabola,

, so

. The focus is unit to the left of the vertex at

, and the directrix is unit to the right the vertex at

. The axis of symme-

try goes through the vertex and is perpendicular to the directrix; its equation is

.

775. vertex: (0, –2); directrix:

Answers 701–800

The standard equation of a parabola that opens upward or downward is

,

where ( h, k) are the coordinates of the vertex and

. The value of p gives you the

distance from the vertex to the focus and directrix.

Pull the values for h and k from the given equation. No h term is given, so you know the x coordinate is 0. The vertex is (0, –2).

In this parabola,

, so

. The focus is 2 units above the vertex at (0, 0), and the

directrix is 2 units below the vertex at

. The axis of symmetry goes through the

vertex and is perpendicular to the directrix; its equation is

.

Chapter 17: Answers 435

776. vertex:

; directrix:

The standard equation of a parabola that opens to the left or right is

,

where ( h, k) are the coordinates of the vertex and

. The value of p gives you the

distance from the vertex to the focus and directrix.

Pull the values for h and k from the given equation. No h term is given, so you know the x coordinate is 0. The vertex is (0, 3).

In this parabola,

, so

. The focus is unit to the left of the vertex at

, and the directrix is unit to the right the vertex at

. The axis of symme-

try goes through the vertex and is perpendicular to the directrix; its equation is

.

777. vertex: (2, 2); directrix:

The standard equation of a parabola that opens to the left or right is

,

where ( h, k) are the coordinates of the vertex and

.

To put the equation in the standard form, multiply each term by 2:

The value of p gives you the distance from the vertex to the focus and directrix.

Pull the values for h and k from the rewritten equation. The vertex is (2, 2).

In this parabola,

, so

. The focus is 1 unit to the right of the vertex at (3, 2),

and the directrix is 1 unit to the left the vertex at

. The axis of symmetry goes

through the vertex and is perpendicular to the directrix; its equation is

.

778.

The axis of symmetry is a vertical line, so the standard form of the parabola is

. Replace the h and k in the equation with the coordinates of the vertex: Solve for a by replacing the x and y in the equation with the coordinates of the given Answers 701–800

point, (0, 20):

436 Part II: The Answers

The equation of the parabola is

.

Illustration by Thomson Digital

779.

The axis of symmetry is a horizontal line, so the standard form of the parabola is

. Replace the h and k in the equation with the coordinates of the vertex: Solve for a by replacing the x and y in the equation with the coordinates of the given point, (6, 3):

Answers 701–800

The equation of the parabola is

.

Chapter 17: Answers 437

Illustration by Thomson Digital

780.

The axis of symmetry is a vertical line, so the standard form of the parabola is

. Replace the h and k in the equation with the coordinates of the vertex: The directrix,

, is 1 unit below the vertex. Also,

, where the value of p gives

you the distance from the vertex to the focus and directrix. The value of p here is 1, so

.

The equation of the parabola is

.

Answers 701–800

Illustration by Thomson Digital

438 Part II: The Answers

781. center: (0, 0); foci: (–1, 0), (1, 0)

The standard form for the equation of an ellipse with center at ( h, k) is if the ellipse is wider horizontally and

if the ellipse is taller vertically.

This ellipse uses the first version because the larger square is under the x term.

The given equation has no h or k terms, so in this ellipse, the center is (0, 0). The value of a and b are

and

.

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

The coordinates of the foci are 1 unit to the left and the right of the center, at (–1, 0) and (1, 0).

782. center: (0, 0); foci: (0, –8), (0, 8).

The standard form for the equation of an ellipse with center at ( h, k) is if the ellipse is wider horizontally and

if the

ellipse is taller vertically. This ellipse uses the second version because the larger square is under the y term.

The given equation has no h or k terms, so in this ellipse, the center is (0, 0). The value of a and b are a = 6 and b = 10.

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

Answers 701–800

The coordinates of the foci are 8 units below and above the center, at (0, –8) and (0, 8).

783. center: (1, –2); foci: (1, –6), (1, 2).

The standard form for the equation of an ellipse with center at ( h, k) is if the ellipse is wider horizontally and

if the

Chapter 17: Answers 439

ellipse is taller vertically. This ellipse uses the second version because the larger square is under the y term.

In this ellipse, the center is (1, –2). Also, a = 3 and b = 5.

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

The coordinates of the foci are 4 units below and above the center, at (1, –6) and (1, 2).

784. center: (–3, 0); foci: (–3, –12), (–3, 12).

The standard form for the equation of an ellipse with center at ( h, k) is if the ellipse is wider horizontally and

if the

ellipse is taller vertically. This ellipse uses the second version because the larger square is under the y term.

In this ellipse, the center is (–3, 0). The values of a and b are a = 5 and b = 13.

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

The coordinates of the foci are 12 units below and above the center, at (–3, –12) and (–3, 12).

785. center: (0, 1); foci: (–6, 1), (6, 1)

The standard form for the equation of an ellipse with center at ( h, k) is Answers 701–800

if the ellipse is wider horizontally and

if the

ellipse is taller vertically. This ellipse uses the first version because the larger square is under the x term.

In this ellipse, the center is (0, 1). Also, a = 10 and b = 8.

440 Part II: The Answers

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

The coordinates of the foci are 6 units to the left and the right of the center, at (–6, 1) and (6, 1).

786. center: (7, –7); foci: (6, –7), (8, –7)

The standard form for the equation of an ellipse with center at ( h, k) is if the ellipse is wider horizontally and

if the

ellipse is taller vertically. This ellipse uses the first version because the larger square is under the x term.

In this ellipse, the center is (7, –7). The values of a and b are and

.

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

The coordinates of the foci are 1 unit to the left and the right of the center, at (6, –7) and (8, –7).

787. center: (2, 2); foci: (2, 0), (2, 4).

The standard form for the equation of an ellipse with center at ( h, k) is if the ellipse is wider horizontally and

if the

Answers 701–800

ellipse is taller vertically. This ellipse uses the second version because the larger square is under the y term.

In this ellipse, the center is (2, 2). The values of a and b are a = 4 and

.

Chapter 17: Answers 441

To find the foci, you use the equation

, where c is the distance from the

center to a focus:

The coordinates of the foci are 2 units below and above the center, at (2, 0) and (2, 4).

788.

The distance between the center and the vertex is

units. This is a horizontal

distance, so

. The distance between the center and the co-vertex is

; this is

a vertical distance, so

.

The standard form for the equation of an ellipse with its vertices along a horizontal line and with center at ( h, k) is

. You know that this is a wide (rather than

tall) ellipse because the given vertex and center both have y coordinates of 2. Replacing h with –3, k with 2, a with 5, and b with 1, you get the equation

.

Illustration by Thomson Digital

789.

The center of the ellipse is at the midpoint, M, of the major axis between the vertices: Answers 701–800

The distance from the center to the vertex (1, 1) is

. This is a vertical

distance, so

. The distance between the center and the co-vertex (–3, –4) is

; this is a horizontal distance, so

.

The standard form for the equation of an ellipse with its vertices along a vertical line and center at ( h, k) is

. You know that this is a tall (rather than

wide) ellipse because the given vertices both have x coordinates of 1. Replacing h with 1, k with –4, a with 4, and b with 5, you get the equation

.

442 Part II: The Answers

Illustration by Thomson Digital

790.

The center of the ellipse is at the midpoint, M, of the major axis between the vertices: The relationship between the foci and lengths of the axes is

because the

major axis is horizontal. Find c by finding the distance between the center and the right focus:

. Then find a by finding the distance between the right

vertex and the center:

. Now you can find b using

:

The standard form for the equation of an ellipse with its vertices along a horizontal line and center at ( h, k) is

. You know that this is a wide (rather than

tall) ellipse because the given vertices both have y coordinates of 5. Replacing h with Answers 701–800

–3, k with 5, a with 3 and b 2 with 5, you get the equation

.

Chapter 17: Answers 443

Illustration by Thomson Digital

791. asymptotes:

,

The standard equation for this hyperbola is

where ( h, k) are the

coordinates of the center. Because the given equation has no h or k values, the center of this hyperbola is (0, 0).

To find the foci, use

, where c is the distance from the center to a focus. Both

a and b are 1, so

The foci are

units to the left and the right of the center, at

and

.

Solve for the asymptotes using the equation

:

792. asymptotes:

,

Answers 701–800

The standard equation for this hyperbola is

, where ( h, k) are the

coordinates of the center. Because the given equation has no h or k values, the center of this hyperbola is (0, 0).

444 Part II: The Answers

To find the foci, use

, where c is the distance from the center to a focus:

The foci are 5 units to the left and the right of the center, at (–5, 0) and (5, 0).

Solve for the asymptotes using the equation

:

793. asymptotes:

,

The standard equation for this hyperbola is

, where ( h, k) are the

coordinates of the center. Because the given equation has no h or k values, the center of this hyperbola is (0, 0).

To find the foci, use

, where c is the distance from the center to a focus:

The foci are 25 units above and below the center, at (0, –25) and (0, 25).

Solve for the asymptotes using the equation

:

Answers 701–800

794. asymptotes:

,

The standard equation for this hyperbola is

, where ( h, k) are the

coordinates of the center. The center of this hyperbola is (1, 4).

Chapter 17: Answers 445

To find the foci, use

, where c is the distance from the center to a focus:

The foci are 4 units above and below the center, at (1, 0) and (1, 8).

Solve for the asymptotes using the equation

:

The asymptotes are

and

.

795. asymptotes:

,

The standard equation for this hyperbola is

, where ( h, k) are the

coordinates of the center. The center of this hyperbola is (–5, 0).

To find the foci, use

, where c is the distance from the center to a focus:

The foci are 5 units above and below the center, at (–5, –5) and (–5, 5).

Solve for the asymptotes using the equation

:

Answers 701–800

The asymptotes are

and

.

796. asymptotes:

,

The standard equation for this hyperbola is

, where ( h, k) are the

coordinates of the center. The center of this hyperbola is (–3, 3).

446 Part II: The Answers

To find the foci, use

, where c is the distance from the center to a focus:

The foci are 5 units to the left and right of the center, at

and

.

Solve for the asymptotes using the equation

:

Distribute the terms on the right side and then add 3 to both sides to isolate y. The equations of the asymptotes are

and

.

797. asymptotes:

,

The standard equation for this hyperbola is

, where ( h, k) are the

coordinates of the center. The center of this hyperbola is (4, 4).

To find the foci, use

, where c is the distance from the center to a focus:

Answers 701–800

The foci are 5 units above and below the center, at

and

.

Solve for the asymptotes using the equation

:

The asymptotes are

and

.

Chapter 17: Answers 447

798.

The two foci are to the left and right of the center, so the hyperbola has this standard form:

The foci are 4 units on either side of the center. You determine this by finding the difference between the x coordinate of the center and the x coordinate of a focus:

.

The foci have the relation

, where c is the distance from the center to a focus.

Substitute 4 for c and 3 for b:

Plug these values into the standard form equation, replacing h with 3, k with –2, a 2

with 7, and b 2 with 9:

Illustration by Thomson Digital

Answers 701–800

799.

The two foci are above and below the center, so the hyperbola has this standard form: The foci are 6 units above and below the center. You determine this by finding the difference between the y coordinate of a focus and the y coordinate of the center:

.

448 Part II: The Answers

The foci have the relation

, where c is the distance from the center to a focus.

Substitute 6 for c and 5 for b:

Plug these values into the standard form equation, replacing h with 0, k with 7, a 2

with 11, and b 2 with 25:

Illustration by Thomson Digital

800.

The relationship between the foci and the axes is

, where c is the distance from

Answers 701–800

the hyperbola’s center to a focus. The foci are on either side of the center, which lies on the x-axis (because the y coordinates are 0). Halfway between the foci is the origin, so the center of the hyperbola is (0, 0). Because the foci are

and

, you

know that

and that

, giving you

.

Chapter 17: Answers 449

The asymptotes are

and

. Work backward to start putting these equations

into standard form:

Because the foci lie along a horizontal line, the square under the x term is larger than that under the y term. Using

and

, those values satisfy the equation

.

The equation of the hyperbola has the form

, because the foci are

along a horizontal axis, so

becomes

.

Illustration by Thomson Digital

801.

Solve each equation for a function of :

Square both sides of each equation:

Substitute the squares of the trig expressions into

:

Answers 801–900

450 Part II: The Answers

Multiply each side of the equation by 16 and simplify:

This curve is a circle with its center at the origin and a radius of 4.

802.

Solve each equation for a function of :

Square both sides of each equation:

Substitute the squares of the trig expressions into

.

This curve is an ellipse with its center at (3, 2).

803.

Solve each equation for a function of :

Square both sides of each equation:

Substitute the squares of the trig expressions into

:

This curve is a hyperbola with its center at (–4, 0).

804.

Solve each equation for a function of :

Answers 801–900

Square both sides of each equation.

Chapter 17: Answers 451

Substitute the squares of the trig expressions into

.

Multiply both sides of the equation by 9 and simplify:

This curve is a circle with its center at (5, –2) and radius 3.

805.

Solve each equation for a function of :

Square both sides of each equation:

Substitute the squares of the trig expressions into

.

This curve is a hyperbola with its center at (3, 6).

806.

,

Your goal is to substitute terms into a version of the Pythagorean theorem,

, so take advantage of the fact that each term is a perfect square.

Divide each term by 100 to create a 1 on the right:

This equation looks similar to the identity. Let

and

, corresponding

to

. Now determine what x and y can be in this new format — how they work with the trig functions. Solving for x and y,

Answers 801–900

If you pick an angle and substitute it into the two equations for x and y, you have a point on the curve, now written in parametric form.

452 Part II: The Answers

807.

,

First put the conic in standard form by completing the square:

Your goal is to substitute terms into a version of the Pythagorean theorem,

, so take advantage of the fact that each term is a perfect square.

Divide each term by 9 to create a 1 on the right:

This equation looks similar to the identity. Let

and

,

corresponding to

. Now determine what x and y can be in this new

format — how they work with the trig functions. Solving for x and y, If you pick an angle and substitute it into the two equations for x and y, you have a point on the curve, now written in parametric form.

808.

,

First put the conic in standard form by completing the square:

Answers 801–900

Your goal is to substitute terms into a version of the Pythagorean theorem,

. Don’t let the fact that the denominators aren’t perfect squares

throw you; radicals work just fine. Let

and

,

Chapter 17: Answers 453

corresponding to

. Now determine what x and y can be in this new

format — how they work with the trig functions. Solving for x and y, If you pick an angle and substitute it into the two equations for x and y, you have a point on the curve, now written in parametric form.

809.

,

First put the conic in standard form by using completing the square: Your goal is to substitute terms into a version of the Pythagorean theorem,

, so take advantage of the fact that each term is a perfect square. Let and

, corresponding to

. Now determine

what x and y can be in this new format — how they work with the trig functions.

Solving for x and y,

Answers 801–900

If you pick an angle and substitute it into the two equations for x and y, you have a point on the curve, now written in parametric form.

454 Part II: The Answers

810.

,

First put the conic in standard form by using completing the square: Your goal is to substitute terms into a version of the Pythagorean theorem,

. Don’t let the fact that the denominators aren’t perfect squares throw you; radicals work just fine. Let

and

, corresponding to

. Now determine what x and y can be in this new format — how they work with the trig functions. Solving for x and y,

If you pick an angle and substitute it into the two equations for x and y, you have a point on the curve, now written in parametric form.

811. (–1, 10)

Solve for y in the first equation,

:

Substitute the expression equal to y into the second equation; then solve for x: Answers 801–900

Solve for y:

The answer is (–1, 10).

Chapter 17: Answers 455

812. (4, 2)

Solve for x in the second equation,

:

Substitute the expression equal to x into the first equation; then solve for y: Solve for x:

The answer is (4, 2).

813. (–7, 1)

Solve for x in the first equation,

:

Substitute the expression equal to x into the second equation; then solve for y: Solve for x:

The answer is (–7, 1).

814. no solution

Solve for y in the first equation,

:

Substitute the expression equal to y into the second equation; then solve for x: Answers 801–900

The statement

is never true, so the system has no solution.

456 Part II: The Answers

815.

Solve for y in the first equation,

:

Substitute the expression equal to y into the second equation; then solve for x: This statement is always true, so the system has an infinite number of solutions.

Choose the parameter k for x. You can write the solutions as the ordered pair

.

816. (2, –3)

Multiply the terms in the second equation

by 3 to eliminate the y term. Then

add the two equations together:

Divide by 10 to get x = 2.

To find y, substitute 2 for x in the second equation: The answer is (2, –3).

817. (0, 4)

Multiply the terms in the second equation

by –1 to eliminate the x term.

Then add the two equations together:

To find x, substitute 4 for y in the first equation: Answers 801–900

The answer is (0, 4).

Chapter 17: Answers 457

818. (–2, 3)

Multiply the terms in the second equation

by –2 to eliminate the x term.

Then add the two equations together:

Divide both side of the equation by –9 to get y = 3.

To find x, substitute 3 for y in the first equation: The answer is (–2, 3).

819. no solution

Rearrange the terms in the second equation

so they align with those in

the first equation:

Multiply the terms in the first equation

by 2. Then add the two equations

together:

This statement is false, so there’s no solution.

820.

Rearrange the terms in the second equation

so they align with those in

the first equation:

Multiply the terms in the first equation

by –2. Then add the two equations

together:

Answers 801–900

458 Part II: The Answers

This statement is always true, so the system has an infinite number of solutions.

Choose the parameter k for y. Then solve for x in the first equation: The answer is

.

821. (0, –2), (4, 50)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Set this equation equal to 0 by adding –13 x and 2 to each side of the equation. Then factor and solve for x:

Use the system’s second equation,

, to find the corresponding y value for

each x:

822. (–3, –33), (1, –1)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Set this equation equal to 0 by adding –8 x and 9 to each side of the equation. Then factor and solve for x:

Use the system’s second equation,

, to find the corresponding y value for each x:

Answers 801–900

Chapter 17: Answers 459

823. (0, 0), (1, 1)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Square both sides of the equation:

Set this equation equal to 0 by adding – x to each side of the equation. Then factor and solve for x. ( Warning: Don’t divide both sides of the equation by x; you lose a solution if you do that.)

Use the system’s second equation,

, to find the corresponding y value for each x: 824. (0, –4), (3, 2)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Set this equation equal to 0 by adding 4 and –2 x to each side of the equation. Then factor and solve for x:

Use the system’s second equation,

, to find the corresponding y value for each x:

825. (1, 0), (–2, –12), (3, 8)

Using substitution, replace the y in the first equation with its equivalent in the second Answers 801–900

equation:

460 Part II: The Answers

Set this equation equal to 0 by adding 4 and –4 x to each side of the equation. Then factor.

You can find the first two factorizations using synthetic division:

Then factor the quadratic and solve for x:

Use the system’s second equation,

, to find the corresponding y value for each x:

826. (1, –3), (–2, –6), (3, –11)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Set this equation equal to 0 by adding and 2 to each side of the equation. Then factor. You can do the first factorization using synthetic division: Then factor the quadratic and solve for x:

Answers 801–900

Use the system’s second equation,

, to find the corresponding y value for

each x:

Chapter 17: Answers 461

827. (–3, 61), (5, –83), (1, 5)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Set this equation equal to 0 by adding , 16 x, and –22 to each side of the equation.

Then factor. You can do the first factorization using synthetic division: Then factor the quadratic and solve for x:

Use the system’s second equation to find the corresponding y value for each x: 828. (8, 843), (–4, –237), (–1, –21)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Set this equation equal to 0 by adding

, –82 x, and –59 to each side of the equation;

then factor. First take out the common factor of 2. You can then use synthetic division for the next factorization:

Answers 801–900

Then factor the quadratic and solve for x:

462 Part II: The Answers

Use the system’s second equation,

, to find the corresponding y value

for each x:

829. (–3, –6), (–5, 8)

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Multiply each side of the equation by

, the denominator of the fraction on the

right:

Set this equation equal to 0 by adding – x and 3 to each side of the equation. Then factor and solve for x:

Use the system’s first equation,

, to find the corresponding y value for each x:

830. (2, –2),

, (–1, 7)

Answers 801–900

Using substitution, replace the y in the first equation with its equivalent in the second equation:

Chapter 17: Answers 463

Multiply each side of the equation by

, the denominator of the fraction on the right:

Set this equation equal to 0 by adding – x and 6 to each side of the equation. Then factor. First take out the common factor of –1. You can then use synthetic division for the next factorization:

Then factor the quadratic and solve for x:

Use the system’s second equation,

, to find the corresponding y value for

each x:

831. (2, –1, 3)

Eliminate z in the third equation. To do so, multiply the second equation by 4 and add it to the third equation.

Divide each side of the equation by 9 to get

.

Substitute –1 for y in the original second equation to solve for z: Now substitute –1 for y and 3 for z in the original first equation to solve for x: Answers 801–900

In

form, the answer is

.

464 Part II: The Answers

832. (0, 4, 2)

Eliminate x in the first equation. To do so, multiply the second equation by –4 and add it to the first equation:

Now use this new equation and the original third equation to eliminate y. Multiply the third equation

by –4 and add it to the new equation:

Multiply each side of the equation by –1 to get z = 2.

Substitute 2 for z in the original third equation to solve for y: To solve for x, substitute 2 for z in the original first equation: In

form, the answer is

.

833. (4, –1, –2)

Eliminate the x term in the second and third equations. To do so, multiply the first equation

by –3 and add it to the second equation:

Then multiply the original first equation by –2 and add it to the original third equation: Now eliminate the y term in the first new equation

by adding it to twice

Answers 801–900

the second new equation

:

Divide each side of the equation by –9 to get

.

Chapter 17: Answers 465

Now substitute –2 for z in the second new equation to solve for y: Now substitute –1 for y and –2 for z in the original first equation to solve for x: In

form, the answer is

.

834. (–3, 0, 2)

Eliminate the z term in the second and third equations. To do so, add the second equation to the first equation:

Then multiply the original first equation

by –2 and add it to the third

equation:

Now eliminate the y in the new equations. To do this, multiply the first new equation by –5 and add it to the second new equation:

Divide each side of the equation by –48, and you have

.

Substitute –3 for x in the first new equation to solve for y: Now substitute 0 for y and –3 for x in the original first equation to solve for z: Answers 801–900

5

In

form, the answer is

.

466 Part II: The Answers

835.

Multiply the first equation

by –2 and add it to the second equation to elimi-

nate the x term:

Eliminate the z term in this new equation and the original third equation. To do so, multiply the terms in the new equation

by 3 and the terms in the original

third equation

by 5 before adding the equations:

Divide each side of the equation by 43 to get

.

Substitute 4 for y in the original first equation to solve for x: Now substitute 4 for y in the original third equation to solve for z: In

form, the answer is

.

836.

Add the first and second equations to eliminate the x and z terms: Substitute

for y in the original third equation to solve for z:

Answers 801–900

Chapter 17: Answers 467

Substitute

for both y and z in the original first equation to solve for x: In

form, the answer is

.

837.

Eliminate the x term. To do so, first multiply the terms in the first equation by –4 and add them to the second equation:

Now multiply the terms in the original first equation

by –6 and add

them to the third equation:

Next, eliminate the y term by multiplying the first new equation by –1 and adding it to the second new equation:

This statement is always true, so there are an infinite number of solutions for this system.

Let z be represented by the parameter k. Use the second new equation to solve for y in terms of k:

Now solve for x in terms of k using the original first equation, the expression for y in terms of k, and z = k:

Answers 801–900

In

form, the answer is

.

468 Part II: The Answers

838.

Eliminate the z term in the second and third equations. To do so, first multiply the terms in the second equation

by –1 and add them to the third equation:

Divide each term in this new equation by 2 and add it to the original first equation: This statement is always true, so there are an infinite number of solutions for this system.

Let x be represented by the parameter k. Use the original first equation to solve for y in terms of k:

Now substitute k for x and

for y in the original second equation to solve for z

in terms of k:

In

form, the answer is

.

839.

Even though the coefficients on the three y terms offer a challenge, eliminating the y terms requires only two processes, whereas eliminating one of the other terms takes three. First, multiply the first equation

by 3 and the second equa-

tion

by 2 and add them together:

Now multiply the original first equation

by 2 and add it to the

third equation:

Answers 801–900

Chapter 17: Answers 469

The new system equations, without the y term, consist of these two new equations and the original fourth equation:

The next steps involve eliminating the w term. First, multiply the top equation by 5 and add it to the middle equation:

Now multiply the bottom equation from that new system

by 5 and add

it to the middle equation:

Each of these terms is divisible by 2:

You now have a new system of two equations:

The numbers still aren’t very pretty, but you can solve this newest system using elimination one more time. Multiply the terms in the second equation by 43 and add it to the first equation.

Divide each side by 425, and you have

.

Now back-substitute to find the values of the other variables:

Answers 801–900

In

form, the answer is

.

470 Part II: The Answers

840.

Start by eliminating the w term. Multiply the second equation by 2

and add it to the third equation:

Next, multiply the fourth equation

by 2 and add it to the third equation:

The new system of equations, without the y term, consists of these two new equations and the original first equation:

The next step involves eliminating the y term. Add the first two equations of the new system together:

Each term in the new equation is divisible by 2, giving you

. Multiply the terms

in this equation by –3 and add it to the last equation in the new system: Dividing by –4, you have x = 1. Now back-solve to find the values of the rest of the variables:

Answers 801–900

In

form, the answer is

.

Chapter 17: Answers 471

841.

Illustration by Thomson Digital

To graph

, first graph the line

(or

in slope-intercept form).

Choose a test point to determine which side of the line to shade. Using (0, 0) as a test point,

, so shade the area below and to the left of the line.

Then graph the line

(or

). Use a dashed line to show that the points

on the line are not in the solution. Choose a test point. Using (0, 0) as a test point,

. The point is not in the solution, so shade below and to the right of the line.

The solution is the intersection of the two shadings.

842.

Illustration by Thomson Digital

Answers 801–900

To graph

, first graph the line

(or

in slope-intercept form).

Choose a test point to determine which side of the line to shade. Using (0, 0) as a test point,

, so shade the area above and to the right of the line.

472 Part II: The Answers

Then graph the line

. Shade to the left of the line. Finally, graph the line

and

shade below that line.

The solution is the intersection of the three shadings.

843.

Illustration by Thomson Digital

To graph

, first graph the line

(or

in slope-intercept

form). Choose a test point to determine which side of the line to shade. Using (0, 0) as a test point,

, so shade the area below and to left of the line.

Next, graph the line

(or

). Choose a test point to determine which

side of the line to shade. Using (0, 0) as a test point,

, so shade the area

below and to left of the line.

Then graph the line

(the y-axis) and the line

(the x-axis). The solution will

contain only points in the first quadrant and on those axes adjacent to the first quadrant.

The solution is the intersection of the shadings.

844.

Answers 801–900

Illustration by Thomson Digital

Chapter 17: Answers 473

First graph the line

(or

in slope-intercept form). Choose a test

point to determine which side of the line to shade. Using (0, 0) as a test point,

, so shade the area above and to right of the line.

Next, graph the line

(or

). Choose a test point to determine

which side of the line to shade. Using (0, 0) as a test point,

, so shade

the area above and to right of the line.

Then graph the line

(the y-axis) and the line

(the x-axis). The solution will

contain only points in the first quadrant and on those axes adjacent to the first quadrant.

The solution is the intersection of the shadings.

845.

Illustration by Thomson Digital

First graph the line

(or

in slope-intercept form). Choose a test point

to determine which side of the line to shade. Using (0, 0) as a test point,

, so

shade the area above and to right of the line.

Next, graph the line

(or

). Choose a test point to determine which

side of the line to shade. Using (0, 0) as a test point,

, so shade the area

below and to the left of the line.

Then graph the line

(or

). Choose a test point to determine which

side of the line to shade. Using (0, 0) as a test point,

, so shade the area

below and to the left of the line.

Then graph the line

(the y-axis) and the line

(the x-axis). The solution will

contain only points in the first quadrant and on those axes adjacent to the first quadrant.

The solution is the intersection of the shadings.

846.

Answers 801–900

You want to decompose

. Factor the denominator of the fraction:

Next, set the original fraction equal to the sum of two fractions whose denominators are those factors you found. Let A and B represent the numerators of those fractions.

474 Part II: The Answers

The common denominator of the fractions on the right is, of course, the fraction you started with. Rewrite the fractions on the right so they all have that common denominator. You’ll be multiplying the numerator and denominator by a missing factor, which is the denominator of the other fraction.

Write a new equation, setting the numerator on the left equal to the numerator on the right:

Distribute over the binomials on the right; then regroup the terms into like terms.

Factor the x from the terms containing the variable:

The coefficient of x on the left is equal to the coefficient of x on the right: The constant term on the left is equal to the constant term on the right: The two equations in A and B form a system of equations whose solution gives you the numerators of the fractions.

Adding the two equations together, you get

, or

. Substituting 5 for A in the

first equation, you get

, or

. Therefore, the decomposed fraction is

847.

Factor the denominator of the fraction:

Next, set the original fraction equal to the sum of two fractions whose denominators are those factors you found. Let A and B represent the numerators of those fractions.

Answers 801–900

The common denominator of the fractions on the right is, of course, the fraction you started with. Rewrite the fractions on the right so they all have that common denominator. You’ll be multiplying the numerator and denominator by a missing factor, which is the denominator of the other fraction.

Chapter 17: Answers 475

Write a new equation, setting the numerator on the left equal to the numerator on the right:

Distribute over the binomials on the right; then regroup the terms into like terms.

Factor the x from the terms containing the variable:

The coefficient of x on the left is equal to the coefficient of x on the right: The constant term on the left is equal to the constant term on the right: The two equations in A and B form a system of equations whose solution gives you the numerators of the fractions.

Multiply the terms in the first equation

by 2 and add them to the second

equation:

Divide each side by 5, and you have

. Replace A with 6 in the original first equa-

tion, and you have

, so

. Therefore, the decomposed fraction is

848.

Factor the denominator of the fraction:

Next, set the original fraction equal to the sum of two fractions whose denominators are those factors you found. Let A and B represent the numerators of those fractions.

Answers 801–900

The common denominator of the fractions on the right is, of course, the fraction you started with. Rewrite the fractions on the right so they all have that common denominator. You’ll be multiplying the numerator and denominator by a missing factor, which is the denominator of the other fraction.

476 Part II: The Answers

Write a new equation, setting the numerator on the left equal to the numerator on the right:

Distribute over the binomial on the right; then regroup the terms into like terms.

Factor the x from the terms containing the variable:

The coefficient of x on the left is equal to the coefficient of x on the right: The constant term on the left is equal to the constant term on the right: Divide each side by 7, and you have

. Replace A with 4 in the original first equa-

tion, and you have

, so

. Therefore, the decomposed fraction is

849.

Factor the denominator of the fraction:

Next, set the original fraction equal to the sum of three fractions whose denominators are those factors you found. Let A, B, and C represent the numerators of the fractions.

The common denominator of the fractions on the right is, of course, the fraction you started with. Rewrite the fractions on the right so they all have that common denominator. You’ll be multiplying the numerator and the denominator by missing factors, which are the denominators of the other fractions.

Answers 801–900

Chapter 17: Answers 477

Write a new equation, setting the numerator on the left equal to the numerator on the right:

Distribute over the trinomial and binomials on the right; then regroup the terms into like terms. Factor the x from the terms containing the variable: The coefficient of on the left is equal to the coefficient of on the right: The coefficient of x on the left is equal to the coefficient of x on the right: The constant term on the left is equal to the constant term on the right: Divide each side by –10, and you have A = 3. Replace A with 3 in the equations for and x:

You now have the following system of equations:

Solve the system by multiplying each term in the first equation by 5 and adding the two equations together:

Divide each side of the equation by 7 to get

. Then solve for B using the first equa-

tion:

, so

. Therefore, the decomposed fraction is

850.

Answers 801–900

Factor the denominator of the fraction:

478 Part II: The Answers

Next, set the original fraction equal to the sum of three fractions whose denominators are those factors you found. Let A, B, and C represent the numerators of those fractions.

The common denominator of the fractions on the right is, of course, the fraction you started with. Rewrite the fractions on the right so they all have that common denominator. You’ll be multiplying the numerator and the denominator by missing factors, which are the denominators of the other fractions.

Write a new equation, setting the numerator on the left equal to the numerator on the right:

Distribute over the trinomial and binomials on the right; then regroup the terms into like terms. Factor the x from the terms containing the variable: The coefficient of on the left is equal to the coefficient of on the right: The coefficient of x on the left is equal to the coefficient of x on the right: The constant term on the left is equal to the constant term on the right: Divide each side by –16, and you have

. Replace A with –4 in the equation for :

You now have the following system of equations:

Solve the system by multiplying each term in the first equation by 4 and adding the two equations together:

Answers 801–900

Divide each side of the equation by 8 to get

. Then solve for C using the first equa-

tion:

, so

. Therefore, the decomposed fraction is

Chapter 17: Answers 479

851.

Find the sum of each pair of corresponding elements:

852.

Find the difference between each pair of corresponding elements:

853.

You perform scalar multiplication by multiplying each element in the matrix by the multiplier outside the matrix. So first perform the scalar multiplication on A and B: Subtract the results by subtracting the corresponding elements:

854.

Multiply the elements in each row of matrix A by the elements in the columns of matrix B; then add the products:

Answers 801–900

480 Part II: The Answers

855.

Multiply the elements in each row of matrix B by the elements in the columns of matrix C; then add the products:

856.

The echelon form of a matrix has only zeros as elements below the main diagonal and only ones as elements along that diagonal. You change a matrix to echelon form by performing various row operations: exchanging rows, multiplying a row by a constant, and adding a multiple of one row to another. You usually start with the first row, perform operations to change it to what you want, and then work your way down.

Answers 801–900

Chapter 17: Answers 481

857.

858.

Answers 801–900

482 Part II: The Answers

859.

Answers 801–900

Chapter 17: Answers 483

860.

Answers 801–900

484 Part II: The Answers

861. (2, –3)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

From the bottom row, you have

. Substitute –3 for y in the equation represented

by the new first row to solve for x:

862. (–1, –4)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

Answers 801–900

From the bottom row, you have

. Substitute –4 for y in the equation represented

by the new first row to solve for x:

Chapter 17: Answers 485

863. (2, 3)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

From the bottom row, you have

. Substitute 3 for y in the equation represented by

the new first row to solve for x:

864.

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

Answers 801–900

486 Part II: The Answers

From the bottom row, you have

. Substitute –3 for y in the equation represented

by the new first row to solve for x:

865. (–1, –2, 3)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

From the bottom row, you have

. Substitute 3 for z in the equation represented by

the new second row to solve for y:

Now substitute the values for y and z into the new first row to solve for x: Answers 801–900

Chapter 17: Answers 487

866. (2, –1, 0)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

From the bottom row, you have

. Substitute 0 for z in the equation represented by

the new second row to solve for y:

Now substitute the values for y and z into the new first row to solve for x: Answers 801–900

488 Part II: The Answers

867. (4, –3, 5)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

From the bottom row, you have

. Substitute 5 for z in the equation represented by

the new second row to solve for y:

Now substitute the values for y and z into the new first row to solve for x: 868. (7, –3, 2)

First write the augmented matrix for the system. The augmented matrix consists of the Answers 801–900

coefficient matrix separated from a column matrix of the constants by a bar: Chapter 17: Answers 489

Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

From the bottom row, you have

. Substitute 2 for z in the equation represented by

the new second row to solve for y:

Now substitute the value for z into the first row to solve for x: 869. (1, –1, 2, –1)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

Answers 801–900

490 Part II: The Answers

From the bottom row, you have

. Substitute –1 for w in the equation represented

by the new third row to solve for z:

Now substitute the values for z and w into the new second row to solve for y: Answers 801–900

Chapter 17: Answers 491

Finally, solve for x by substituting the known values into the new first row: 870. (0, 3, –3, 2)

First write the augmented matrix for the system. The augmented matrix consists of the coefficient matrix separated from a column matrix of the constants by a bar: Perform row operations to put the matrix in echelon form. You perform row operations on the entire row — coefficients and constants. The ones along the main diagonal in the resulting echelon form correspond to the variables in the equation.

Answers 801–900

492 Part II: The Answers

From the bottom row, you have

. Substitute 2 for w in the equation represented by

the new third row to solve for z:

Now substitute the values for z and w into the second new row to solve for y: Finally, solve for x by substituting the known values into the first new row: 871. (–4, 5)

To determine the inverse of the coefficient matrix corresponding to the system of equations, use the formula for the inverse of a

matrix. (Recall that the determinant

of a

matrix is

.)

The coefficient matrix for the system

is

, so the inverse is

Multiply the inverse of the coefficient matrix by the constant matrix: Answers 801–900

The solution is x = –4, y = 5.

Chapter 17: Answers 493

872. (2, –3)

To determine the inverse of the coefficient matrix corresponding to the system of equations, use the formula for the inverse of a

matrix. (Recall that the determinant

of a

matrix is

.)

The coefficient matrix for the system

is

, so the inverse is

This matrix is its own inverse! Multiply the inverse of the coefficient matrix by the constant matrix:

The solution is

,

.

873. (–1, 3)

To determine the inverse of the coefficient matrix corresponding to the system of equations, use the formula for the inverse of a

matrix. (Recall that the determinant

of a

matrix is

.)

The coefficient matrix for the system

is

, so the inverse is

Answers 801–900

Multiply the inverse of the coefficient matrix by the constant matrix: The solution is

,

.

494 Part II: The Answers

874. (1, 0, –1)

The system is

Find the inverse of the coefficient matrix. To do so, perform row operations that change the matrix on the left of the bar to an identity matrix like the one on the right.

Multiply the inverse matrix for the coefficients by the constant matrix: 875. (–1, 2, –4)

The system is

Find the inverse of the coefficient matrix: To do so, perform row operations that change the matrix on the left of the bar to an identity matrix like the one on the right.

Answers 801–900

Multiply the inverse matrix for the coefficients by the constant matrix: Chapter 17: Answers 495

876. (–2, 4)

Cramer’s Rule uses determinants to find x and y:

Recall that the determinant of a

matrix is

With the system of

and

, you have the following:

877. (5, 6)

Cramer’s Rule uses determinants to find x and y:

Recall that the determinant of a

matrix is

With the system of

and

, you have the following:

Answers 801–900

496 Part II: The Answers

878.

Cramer’s Rule uses determinants to find x and y:

Recall that the determinant of a

matrix is

With the system of

and

, you have the following:

879. (7, –3)

Cramer’s Rule uses determinants to find x and y:

Recall that the determinant of a

matrix is

With the system of

and

, you have the following:

Answers 801–900

Chapter 17: Answers 497

880. (8, 2, 3)

Cramer’s Rule uses determinants to find x, y, and z: Recall that you find the determinant of a

matrix by multiplying along the diagonals

and finding a difference:

With the system of

,

, and

, you have the

following:

881. 31

You find the fifth term, , by replacing the n with 5 in the general term and simplifying: Answers 801–900

498 Part II: The Answers

882.

You find the fifth term, , by replacing each n with 5 in the general term and simplifying:

883. 10

You find the fifth term, , by replacing each n with 5 in the general term and simplifying:

884.

You find the fifth term, , by replacing each n with 5 in the general term and simplifying: 885. 3

You find the fifth term, , by replacing the n with 5 in the general term and simplifying: 886.

Each term is a power of 3. The number 1 is 30, so you subtract 1 from n to create a 0

exponent for the first term.

887.

Each term is the result of performing the factorial operation on the number of the term.

888.

The coefficient of each term is twice the number of the term, and the power of the term is the number of the term.

Answers 801–900 889.

Each term is one more than a perfect square. The number of the term is squared.

Chapter 17: Answers 499

890.

The pattern is easier to see if you write the first term as a fraction and the third term with a denominator of 9:

The numerators are all 3 more than the number of the term, and the denominators are the number of the term squared.

891. ‒1, 5

A sequence has a recursively defined general term if the rule involves one or more previous terms in the computation of a given term. In this recursively defined sequence, the next two terms are as follows:

892.

A sequence has a recursively defined general term if the rule involves one or more previous terms in the computation of a given term. In this recursively defined sequence, the next two terms are as follows:

893. 3, 10

A sequence has a recursively defined general term if the rule involves one or more previous terms in the computation of a given term. In this recursively defined sequence the next two terms are as follows:

894.

A sequence has a recursively defined general term if the rule involves one or more previous terms in the computation of a given term. In this recursively defined sequence, the next two terms are as follows:

Answers 801–900

500 Part II: The Answers

895. 2, 3

A sequence has a recursively defined general term if the rule involves one or more previous terms in the computation of a given term. In this recursively defined sequence, the next two terms are as follows:

This sequence is, of course, the famous Fibonacci sequence.

896. 205

You find the sum of an arithmetic series with the formula

, where

you’re averaging the first and last terms and multiplying by n, the number of terms.

In this series,

, the first term is

, and the last term is

:

897. ‒24

You find the sum of an arithmetic series with the formula

, where

you’re averaging the first and last terms and multiplying by n, the number of terms.

In this series,

, the first term is

, and the last term is

:

898. 190

You find the sum of an arithmetic series with the formula

, where

you’re averaging the first and last terms and multiplying by n, the number of terms.

In this series, you’re actually going from the eighth to the eleventh term, which means you’re adding four terms. Using the formula,

, the eighth term is 40, and the elev-

enth term is 55:

899.

Answers 801–900

28

You find the sum of an arithmetic series with the formula

, where

you’re averaging the first and last terms and multiplying by n, the number of terms.

Chapter 17: Answers 501

In this series, you’re actually going from the third to the ninth term, which means you’re adding seven terms. Using the formula,

, the third term is , and the ninth term is :

900.

You find the sum of an arithmetic series with the formula

, where

you’re averaging the first and last terms and multiplying by n, the number of terms.

In this series, you’re actually going from the zeroth to the sixth term, which means you’re adding seven terms. (It’s often more convenient to number the terms starting with 0

rather than 1; you can write the formula more simply or combine terms of different series more easily.) Using the formula,

, the zeroth term is , and the sixth term is :

901. 37

Write out the four terms and add them:

902. 11

Write out the five terms, simplify, and add them. The signs will alternate with the powers of ‒1.

903. 0

Write out the six terms, compute the value of the trig function, and then add the terms: Answers 901–1001

502 Part II: The Answers

904.

Write out the ten terms and add them together:

905.

Write out the six terms and add them:

906. 100

This series is an arithmetic series with

and a difference of 2. First, find the tenth

term by using the formula

:

Now find the sum of the first ten terms by using the following formula: 907. 145

This series is an arithmetic series with

and a difference of 5. First, find the tenth

term by using the formula

:

Answers 901–1001

Chapter 17: Answers 503

Now find the sum of the first ten terms by using the following formula: 908. 25

This series is an arithmetic series with

and a difference of . First, find the tenth

term by using the formula

:

Now find the sum of the first ten terms by using the following formula: 909. ‒15

This series is an arithmetic series with

and a difference of ‒3. First, find the

tenth term by using the formula

:

Now find the sum of the first ten terms by using the following formula: 910. ‒380

This series is an arithmetic series with

and a difference of ‒8. Find the tenth

term by using the formula

:

Now find the sum of the first ten terms by using the following formula: Answers 901–1001

504 Part II: The Answers

911. 63

You find the sum of the first n terms of a geometric series by using the following formula:

The sum of the terms in this series with

is

912. 2,188

You find the sum of the first n terms of a geometric series by using the following formula:

The sum of the terms in this series with

and

is

913.

You find the sum of the first n terms of a geometric series by using the following formula:

The sum of the terms in this series with

and

is

914.

You find the sum of the first n terms of a geometric series by using the following formula:

Answers 901–1001

Chapter 17: Answers 505

To use the formula for this series, you need to rewrite the general term with an exponent of i:

You can now rewrite the problem as

This version has the correct format for the formula, letting

and

:

915.

You find the sum of the first n terms of a geometric series by using the following formula:

To use the formula for this series, you need to rewrite the problem from 0 to n. The first term in this series is , so to begin the summation with

, write the general

term as

and change n from 6 to 5. You can now rewrite the problem as This version has the correct format for the formula, letting

and

:

916. 32

You find the sum of the terms of an infinite geometric series with

by using the

formula

. In this series, the first term, a, is 16, and the ratio, r, is . You find the ratio by dividing any term by the term immediately preceding it.

Answers 901–1001

506 Part II: The Answers

917. 81

You find the sum of the terms of an infinite geometric series with

by using the

formula

. In this series, the first term, a, is 27, and the ratio, r, is . You find the ratio by dividing any term by the term immediately preceding it.

918.

You find the sum of the terms of an infinite geometric series with

by using the for-

mula

. In this series, the first term, a, is , and the ratio, r, is . You find the ratio by dividing any term by the term immediately preceding it.

919.

You find the sum of the terms of an infinite geometric series with

by using the

formula

. In this series, the first term, a, is 64, and the ratio, r, is

. You

find the ratio by dividing any term by the term immediately preceding it.

920.

You find the sum of the terms of an infinite geometric series with

by using the

following formula:

.

In this series, the first term, a, is and the ratio, r, is . You find the ratio by dividing any term by the term immediately preceding it.

Answers 901–1001

Chapter 17: Answers 507

921. 15

You find the combination of n things taken r at a time by using the formula

, so

Instead of writing 6! as

, you can use 4! for the last four factors to simplify

the fraction reduction:

922. 9

You find the combination of n things taken r at a time by using the formula

, so

Instead of writing 9! as

, you can use 8! for the last eight factors to

simplify the fraction reduction:

923. 1

You find the combination of n things taken r at a time by using the formula

, so

By special definition, 0! = 1:

924. 1

You find the combination of n things taken r at a time by using the formula

, so

Answers 901–1001

508 Part II: The Answers

925. 25,827,165

You find the combination of n things taken r at a time by using the formula

, so

Instead of writing 54! as

and so on, you can use 48! for the last 48 factors to

simplify the fraction reduction:

926. bottom row is

You form each new row by placing the sums of the two adjacent numbers from the row above diagonally below. The beginning and ending numbers of each row are 1.

927. bottom row is

You form each new row by placing the sums of the two adjacent numbers from the row above diagonally below. The beginning and ending numbers of each row are 1.

928. bottom row is

You form each new row by placing the sums of the two adjacent numbers from the row above diagonally below. The beginning and ending numbers of each row are 1.

Answers 901–1001

Chapter 17: Answers 509

929. bottom row is

You form each new row by placing the sums of the two adjacent numbers from the row above diagonally below. The beginning and ending numbers of each row are 1.

930. bottom row is

You form each new row by placing the sums of the two adjacent numbers from the row above diagonally below. The beginning and ending numbers of each row are 1.

931.

First, write the coefficients from the fourth row of the triangle:

Next, write in decreasing powers of x:

Now write in increasing powers of y:

Simplify and write the sum of the terms:

932.

First, write the coefficients from the fifth row of the triangle:

Next, write in decreasing powers of x:

Now write in increasing powers of 2:

Simplify and write the sum of the terms:

Answers 901–1001

510 Part II: The Answers

933.

First, write the coefficients from the sixth row of the triangle:

Next, write in decreasing powers of y:

Now write in increasing powers of ‒3:

Simplify and write the sum of the terms:

934.

First, write the coefficients from the fourth row of the triangle:

Next, write in decreasing powers of 2 x:

Now write in increasing powers of ‒5:

Simplify and write the sum of the terms:

935.

First, write the coefficients from the fifth row of the triangle:

Next, write in decreasing powers of 4 x:

Now write in increasing powers of ‒3 y:

Simplify and write the sum of the terms:

936.

The binomial theorem produces the expansion of the binomial

by using the

following formula:

Answers 901–1001

Chapter 17: Answers 511

You can find the

term in

by plugging in the given information:

Instead of writing 9! as

, you can use 7! for the last seven factors to

simplify the fraction reduction:

937.

The binomial theorem produces the expansion of the binomial

by using the

following formula:

You can find the

term in

by plugging in the given information:

Instead of writing 8! as

, you can use 5! for the last five factors to sim-

plify the fraction reduction:

938.

The binomial theorem produces the expansion of the binomial

by using the

following formula:

You can find the term in

by plugging in the given information:

Instead of writing 6! as

, you can use 3! for the last three factors to simplify

the fraction reduction:

939.

The binomial theorem produces the expansion of the binomial

by using the

following formula:

Answers 901–1001

512 Part II: The Answers

You can find the

term in

by plugging in the given information:

Instead of writing 7! as

, you can use 4! for the last four factors to sim-

plify the fraction reduction:

940.

The binomial theorem produces the expansion of the binomial

by using the

following formula:

You can find the

term in

by plugging in the given information:

Instead of writing 5! as

, you can use 3! for the last three factors to simplify

the fraction reduction:

941. 3

The function is continuous, and it has y values that approach 3 when x is approaching 2.

942. 3

The function has a hole at (2, 3). The limit as x approaches 2 from the left is 3, and the limit as x approaches 2 from the right is 3.

943. does not exist

The limit as x approaches 2 from the left is 3, and the limit as x approaches 2 from the right is ‒2. They aren’t the same, so the limit doesn’t exist.

944. 3

The function has a hole at (2, 3). The limit as x approaches 2 from the left is 3, and the limit as x approaches 2 from the right is 3. The asymptotes aren’t involved in this limit.

Answers 901–1001

Chapter 17: Answers 513

945. does not exist

The function has a vertical asymptote when

. The function approaches

when x

approaches 2 from the left and

when x approaches 2 from the right. The left- and

right-hand limits don’t match, so the limit doesn’t exist.

946. ‒3

The function is approaching a y value of ‒3 as x moves toward ‒1 from the right.

947. 0

The function is approaching a y value of 0 as x moves toward ‒3 from the right.

948.

The function is rising and approaching

as x gets closer and closer to ‒3 from the

left.

949. ‒2

The function is approaching a y value of ‒2 as x moves toward 2 from the right.

950.

The function is dropping and approaching

as x gets closer and closer to 3 from the

left.

951. 8

The y values are getting closer and closer to 8 as x approaches 3 from the left and from the right.

952. 728

The y values are getting closer and closer to 728 as x approaches 9 from the left and from the right.

953. ‒9

The y values are getting closer and closer to ‒9 as x approaches ‒2 from the left and from the right.

954. does not exist

The y values are getting closer and closer to

as x approaches 3 from the left and

to

as x approaches 3 from the right.

Answers 901–1001

514 Part II: The Answers

955.

The y values are getting closer and closer to

as x approaches 5 from the left and

from the right.

956. 1

Replace each x in the function rule with 3; then simplify:

957. 0

Replace each x in the function rule with ‒2; then simplify:

958.

Replace each x in the function rule with 1; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor and reduce the rational expression before replacing each x with 1:

The function has a removable discontinuity when

. A removable discontinuity

occurs when you can rewrite a function rule by factoring out a common factor. There’s still a discontinuity at the point identified by the removed factor, but this discontinuity behaves differently from other discontinuities.

959.

Replace each x in the function rule with ‒3; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor and reduce the rational expression before replacing each x with –3:

The function has a removable discontinuity when

.

Answers 901–1001

Chapter 17: Answers 515

960. does not exist

Replace each x in the function rule with 5; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor and reduce the rational expression before replacing each x with 5:

The resulting expression still has no limit.

961. does not exist

Replace each x in the function rule with 5; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor the rational expression before replacing each x with 5:

The resulting expression still has no limit.

962.

Replace each x in the function rule with 3; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor and reduce the rational expression before replacing each x with 3:

The function has a removable discontinuity when

.

Answers 901–1001

516 Part II: The Answers

963. ‒3

Replace each x in the function rule with ‒2; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor and reduce the rational expression before replacing each x with –2:

The function has a removable discontinuity when

.

964.

Replace each x in the function rule with ‒3; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor and reduce the rational expression before replacing each x with –3:

The function has a removable discontinuity when

.

965. does not exist

Replace each x in the function rule with ‒2; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor the rational expression before replacing each x with –2:

Answers 901–1001

Chapter 17: Answers 517

The resulting expression still has no limit.

966. 1

Replace each x in the function rule with ; then simplify:

967. does not exist

Replace each x in the function rule with ; then simplify:

Division by 0 is undefined, so the limit doesn’t exist.

968.

Replace each x in the function rule with ; then simplify:

969. ‒2

Replace each x in the function rule with 0; then simplify:

Division by 0 is undefined, so go back to the original problem. Factor the numerator and reduce the rational expression before replacing each x with 0: The function has a removable discontinuity when

.

Answers 901–1001

518 Part II: The Answers

970. does not exist

Replace each x in the function rule with

; then simplify:

Division by 0 is undefined, so the limit doesn’t exist.

971.

Replace each x in the function rule with 9; then simplify:

Division by 0 is undefined, so go back to the original problem. Multiply both the numerator and the denominator of the rational expression by the conjugate of the numerator before replacing each x with 9:

The function has a removable discontinuity when

.

972. does not exist

Replace each x in the function rule with 4; then simplify:

The result is not the fraction , so rationalizing won’t change the value. The limit doesn’t exist.

973.

Replace each x in the function rule with 3; then simplify:

Answers 901–1001

Chapter 17: Answers 519

The fraction is undefined, so go back to the original problem. Multiply both the numerator and the denominator of the rational expression by the conjugate of the numerator before replacing each x with 3:

The function has a removable discontinuity when

.

974.

Replace each x in the function rule with 0; then simplify:

The fraction is undefined, so go back to the original problem. Add the two fractions in the numerator:

Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator; then replace each x with 0:

The function has a removable discontinuity when

.

Answers 901–1001

520 Part II: The Answers

975.

Replace each x in the function rule with ; then simplify:

The fraction is undefined, so go back to the original problem. Rewrite the expression using the ratio identity for the tangent. Combine the fractions in the numerator and denominator.

Replace the 1 in the expression

with

from the Pythagorean

identity. Simplify. Then multiply the numerator of the rational expression by the reciprocal of the denominator to simplify the complex fraction.

Factor the numerator of the first fraction and reduce before replacing each x with : The function has a removable discontinuity when

.

Answers 901–1001

Chapter 17: Answers 521

976. 1

, and as x approaches 0 from the left,

.

Remember: When working with piece-wise functions, you determine the function value for a given x by using the rule corresponding to that x value. When evaluating a limit at an x value that occurs where the rule changes, you find the function values for both rules — coming from both the left and the right — to determine whether they give you the same y value.

977. ‒4

, and as x approaches ‒2 from the left,

.

978. ‒3

, and as x approaches 0 from the left,

.

979. does not exist

, and as x approaches ‒1 from the left,

.

980. 2

, and as x approaches 3 from the left,

.

981. 13

The addition law states that if

and

, then

.

Using the law,

982. ‒144

The multiplication law states that if

and

, then

.

Using the law,

983. does not exist

The division law states that if

and

, then

as long

as

. In

, the limit doesn’t exist because

.

Answers 901–1001

522 Part II: The Answers

984. ‒27

The power law states that if

, then

. Using the law,

985. ‒5

The constant factor law states that if

, then

. Using the law,

986. ‒21

The constant factor law states that if

, then

. The power law

states that if

, then

. The subtraction law states that if

and

, then

. Using these laws,

987.

The constant factor law states that if

, then

. The subtrac-

tion law states that if

and

, then

. The

power law states that if

, then

. The division law states that if

and

, then

as long as

. Using these laws,

988.

The power law states that if

, then

. The constant factor law

states that if

, then

. The addition law states that if

and

, then

. The division law states that if

and

, then

as long as

. Using these laws,

Answers 901–1001

Chapter 17: Answers 523

989. ‒3

The power law states that if

, then

. The subtraction law states

that if

and

, then

. The division law states

that if

and

, then

as long as

. Using these laws,

990. ‒5

The constant factor law states that if

, then

. The power law

states that if

, then

. The addition law states that if

and

, then

. The subtraction law states that if

and

, then

. The division law states that if

and

, then

as long as

. Using these laws,

991. 3

The function

is not continuous when

, because

. When

working with rational functions, you determine discontinuities by setting the denominator equal to 0 and solving for x. The domain consists of all real numbers except 3.

992. 4

The function

is not continuous when

, because

.

When working with rational functions, you determine discontinuities by setting the denominator equal to 0 and solving for x. The domain consists of all real numbers except 4.

993.

or

4

The function

is not continuous when x is greater than 4 or x is less than ‒4. When the root is an even number, you can’t have a negative value under the radical. Solving the inequality

, you have

This is true when

, so the domain consists of all x such that

.

Therefore, x values smaller than ‒4 or larger than 4 are not in the domain.

Answers 901–1001

524 Part II: The Answers

994. 3, ‒3

The function

is not continuous when

, because

. Also,

, which is undefined. When working with

rational functions, you determine discontinuities by setting the denominator equal to 0

and solving for x. The domain consists of all real numbers except 3 and ‒3.

995.

The function

is not continuous when

, because

. The domain consists of all real numbers except those

that create a 0 in the denominator, so set the denominator equal to 0 and solve for x: This equation is true for all x such that

— all odd multiples of .

996.

The function

is not continuous when

, because

,

and

is not in the domain of the tangent. The domain consists of all real numbers except those that are odd multiples of .

997.

The function

is not continuous when

, because

.

The domain consists of all real numbers except those that create a 0 in the denominator, so set the denominator equal to 0 and solve for x:

This is true for all x such that

, or

.

998. 0

The function

is not continuous when

, because

.

When working with rational functions, you determine discontinuities by setting the denominator equal to 0 and solving for x. In this case,

when

, which is

when

. The domain consists of all real numbers except 0.

Answers 901–1001

Chapter 17: Answers 525

999. 1

The function

is not continuous when

, because

. When working with rational functions, you determine discontinuities by setting the denominator equal to 0 and solving for x. In this case, when

, which is when

. The domain consists of all real numbers except 1.

1,000. 0

, and as x approaches 0 from the left,

. The sine and cosine

functions are continuous for all x, so the function is continuous everywhere except at 0.

1,001. 1

, and as x approaches 1 from the right,

. All the func-

tion rules used in the piecewise function are continuous for all real numbers, but the values when

aren’t the same.

Answers 901–1001

526 Part II: The Answers

Index

• A •

• B •

absolute value functions, 33, 109–200

Binomial theorem, 115

absolute value inequalities, 17–18, 146–147

binomials, 70, 338–340

adding

angle measures, 74, 349–350

• C •

complex numbers, 90–91, 408–410

terms in arithmetic series, 113, 500–501

calculating. See also finding; solving

algebra

sum of geometric series, 114, 504–505

about, 9

trig equations, 60, 299–301

graphing formulas, 13–14, 137–141

trig functions, 59, 290–294

graphing inequalities, 12, 132–136

Calculus For Dummies (Ryan), 5

identifying systems, 10, 127–128

centers of ellipses, finding, 100, 438–441

recognizing properties of number systems,

Cheat Sheet (website), 4

10–11, 128–129

circles, writing equations for, 98–99, 427–433

simplifying expressions, 11–12, 75, 129–131,

coefficient matrix, 109, 492–494

355–357

completing the square, 39, 225–227

Algebra II For Dummies (Sterling), 1, 4

complex numbers

algebraic processes, performing, 71–72,

about, 89–90

345–349

adding, 90–91, 408–410

angles

dividing using multiplication, 91–92, 413–414

adding measures, 74, 349–350

graphing, 92–93, 415–416

doubling size, 74, 350–351

multiplying, 91, 410–413

finding, 57–58, 80–83, 286–288, 377–378,

solving quadratic equations with complex

384–387

solutions, 92, 414–415

subtracting measures, 74, 349–350

subtracting, 90–91, 408–410

answers to practice problems. See specific topics

composition of functions, 27–28, 173–175

applying

confirming statements of multiple-angle

Cramer’s rule, 110, 495–497

identities, 74, 352

Descartes’s rule, 40, 232–234

conic sections

function transformations, 50–51, 64, 272–273,

about, 97

305–306

changing equation format, 101–102, 449–454

graphing formulas, 13–14, 137–141

finding asymptotes of hyperbolas, 101, 443–446

Heron’s formula, 86, 396–397

finding centers and foci of ellipses, 100,

laws of limits, 122–123, 521–523

438–441

logarithms to everyday life, 51–52, 274–277

finding foci and axes of symmetry of parabolas,

Pascal’s triangle, 115, 509–510

99, 433–434

power-reducing identities, 76, 360–361

finding vertices and directrixes of parabolas,

trig formula for area of triangles, 84, 390–391

99, 434–435

arc length formula, 59, 294–295

identifying from their equations, 98, 423–424

area, finding for triangles, 83–84, 86, 389–391,

rewriting conic equations in standard form, 98,

395–396

424–426

asymptotes, finding of hyperbolas, 101,

writing equations for circles, 98–99, 427–433

443–446

writing equations for ellipses, 100, 441–443

augmented matrices, 108–109, 484–492

writing equations for hyperbolas, 101, 447–449

axes, finding, 99, 433–434

writing equations for parabolas, 100, 435–437

528 1,001 Pre-Calculus Practice Problems For Dummies

conjugate, solving trig identities with

writing from graphs, 31–32, 43–44, 190–192,

multiplication by a, 70, 335–338

252–253

continuity, 117, 123–124, 523–525

writing for ellipses, 100, 441–443

converting coordinates, 20, 417–418

writing for hyperbolas, 101, 447–449

cosecant, graphing, 66, 312–319

writing for parabolas, 100, 435–437

cosine

equivalence log, 48, 263–264

graphing, 64, 302–305

evaluating

solving trig identities with, 69, 333–335

expressions, 75–76, 358–360

cotangent, graphing, 65, 307–309

function values with function notation,

counting items using combinations, 114,

22, 159

507–508

inverse functions, 60, 295–297

Cramer’s rule, 110, 495–497

even functions, 23, 164

creating

exponential and logarithmic functions

half-angle identities, 75, 352–353, 353–355

about, 45–46

products of trig functions from sums and

applying function transformations to log

differences, 75, 357

functions, 50–51, 272–273

applying logarithms to everyday life, 50–52,

• D •

272–277

function notation, 46, 254–255

decomposition of fractions, 107, 473–478

graphing exponential functions, 46, 255–258

degrees

rewriting expressions, 48, 263–266

changing radians to, 57, 285

rewriting logs of products and quotients as

changing to radians, 57, 286

sums and differences, 49, 266–269

solving trig equations, 60, 297–299

solving exponential equations, 47, 258–263

denominators, rationalizing, 20, 156–158

solving logarithmic equations, 49–50,

Descartes’s rule, 40, 232–234

269–272

difference quotient, 28, 175–177

exponential equations, solving, 47, 258–263

differences, 49, 75, 266–269, 357

expressions

directrixes, 99, 434–435

evaluating, 75–76, 358–360

dividing

rewriting, 48, 263–266

complex numbers using multiplication, 91–92,

simplifying, 11–12, 75, 129–131, 355–357

413–414

polynomials, 41, 235–239

• F •

domain, finding, 22–23, 159–163

double-angle identities, 75, 352–353

factoring, 19, 38, 149–150, 218–222

doubling angle size, 74, 350–351

factoring by grouping, solving polynomial

equations using, 40, 231–232

• E •

finding. See also calculating; solving

angle measures, 57–58, 286–288

Echelon form, changing matrices to, 108,

area of triangles, 83–84, 86, 389–390,

480–483

395–396

elimination, solving systems of linear equations

asymptotes of hyperbolas, 101, 443–446

using, 104, 456–458

axes of symmetry of parabolas, 99,

ellipses, 100, 438–441

433–434

equations

centers of ellipses, 100, 438–441

about, 15

directrixes of parabolas, 99, 434–435

changing format of, 101–102, 449–454

domain of functions, 22–23, 159–163

identifying conics from their, 98, 423–424

foci of ellipses, 100, 438–441

sketching quadratic functions from their,

foci of symmetry of parabolas, 99, 433–434

30–31, 186–190

formulas, 114, 505–507

writing for circles, 98–99, 427–433

inverse of functions, 26, 167–171

Index 529

limits from function rules, 121–122,

recognizing, 23–25, 164–167

514–521

ruling out even and odd, 23, 164

limits from function values, 120–121,

solving trig identities, 70–71, 340–341

513–514

transforming selected points using, 34–35,

limits from graphs, 118–119, 512–513

213–215

one-sided limits, 119–120, 513

range of functions, 22–23, 159–163

• G •

rules for sequences, 112, 498–499

rules in series, 113, 502–503

graphing trig functions

sides of triangles, 80, 82, 376–377, 384–386

about, 61–62

sums, 114, 505–507

applying function transformations, 64,

terms of sequences, 112, 497–498

305–306

trig functions, 54–55, 58, 277–278, 289–290

basic graphs, 62–63, 301–302

values of functions using identities, 76–77,

cosecant, 66, 312–319

362–367

cosine, 64, 302–305

vertices of parabolas, 99, 434–435

cotangent, 65, 307–309

foci, 99, 100, 433–434, 438–411

interpreting transformations, 65, 66, 309–312,

formulas

320–322

finding, 114, 505–507

secant, 66, 312–319

graphing, 13–14, 137–141

sine, 64, 302–305

fractional exponents, 18, 148–149

tangent, 65, 307–309

fractional notation, 18, 19, 148–150

transformations, 64, 306–307

fractions

graphs and graphing

decomposition of, 107, 473–478

complex numbers, 92–93, 415–416

solving trig identities after combining, 71,

exponential functions, 46, 255–258

341–344

finding limits from, 118–119, 512–513

function notation, 46, 254–255

formulas, 13–14, 137–141

function rules, 66, 121–122, 320–322, 514–521

functions, 29–30

function transformations, applying, 50–51, 64,

inequalities, 12, 132–136

272–273, 305–306

of polynomial functions, 33, 200–205

function values, finding limits from, 120–121,

polynomials, 43, 245–252

513–514

radical functions, 32–33, 193–197

functions

sketching, 35, 215–218

about, 21

systems of inequalities, 106–107, 471–473

absolute value, 33, 198–200

writing equations from graphs of parabolas,

calculating trig functions, 59, 290–294

31–32, 190–192

composition of, 27–28, 173–175

writing equations from graphs of polynomials,

difference quotient, 28, 175–177

43–44, 252–253

evaluating function values, 22, 159

finding domain and range of functions, 22–23,

• H •

159–163

finding inverse of, 26, 167–171

half-angle identities, 75, 352–357

graphing, 29–30. See also graphing trig

Heron’s formula, 86, 87, 395–400

functions

hyperbolas, 101, 443–449

identifying, 23, 25, 164, 165–166

inverses of, 30, 177–186

performing operations on, 26–27, 172–173

• I •

polynomial, 33, 200–205

radical, 32–33, 193–197

i, powers of, 90, 407

rational, 34, 205–211

icons, explained, 4

530 1,001 Pre-Calculus Practice Problems For Dummies

identifying

• N •

conics from their equations, 98, 423–424

number systems, 10, 127–128

nonlinear functions, solving systems of

odd functions, 23, 164

equations involving, 105, 458–463

one-to-one functions, 25, 166–167

number systems, 10–11, 127–129

points, 94–95, 416–417

reference angles, 58, 288–289

inequalities

• O •

about, 15

absolute value, 17–18, 146–147

odd functions, 23, 164

graphing, 12, 132–136

one-sided limits, 119–120, 513

linear, 17, 143

one-to-one functions, 23–24, 25, 165–167

quadratic, 17, 144–146

operating on matrices, 107, 479–480

using notation, 16, 142

operations, performing using fractional

inequality notation, 16, 142

exponents, 18, 148–149

intercepts, solving polynomial equations for,

order of operations, 11–12, 129–131

39, 228–231

interpreting transformations, 65, 66, 309–312,

• P •

320–322

interval notation, 16, 142

parabolas

inverse of functions, 25, 26, 30, 60, 166–171,

finding foci and axes of symmetry of, 99,

177–186, 295–297

433–434

finding vertices and directrixes of, 99, 434–435

• K•

writing equations from graphs of, 31–32,

190–192

Kase, Elleyne (author)

writing equations of, 100, 435–437

Pre-Calculus For Dummies, 1, 4

Pascal’s triangle, 115, 508–510

Kuang, Yang (author)

PIN code, 3–4

Pre-Calculus For Dummies, 1, 4

points, identifying, 94–95, 416–417

polar coordinates

• L •

about, 89–90

converting rectangular coordinates to, 95,

418–420

law of cosines, 82–83, 386, 387–389

converting to rectangular coordinates, 95,

law of sines, 80–81, 376–384

417–418

laws of limits, applying, 122–123, 521–523

identifying points with, 94, 416

limits

recognizing polar curves, 96, 420–422

about, 117

polar curves, recognizing, 96, 420–422

applying laws of, 122–123, 521–523

polynomial equations, 39, 40, 228–232, 234–235

finding, 118–122, 512–521

polynomial expressions, 42, 241–244

linear inequalities, solving, 17, 143

polynomial functions, 33, 200–205

listing possible roots of polynomial equations,

polynomials

40, 234–235

about, 37

logarithmic functions. See exponential and

checking for roots of using synthetic division,

logarithmic functions

41–42, 239–240

logs of products/quotients, 49, 266–269

Descartes’s rule, 40, 232–234

• M •

dividing, 41, 235–239

graphing, 43, 245–252

listing possible roots of polynomial equations,

matrices, 107, 108, 479–483

40, 234–235

multiplication, 91–92, 410–414

solving polynomial equations, 39, 40, 228–232

Index 531

solving quadratic equations, 38–39, 218–227

sine

writing equations from graphs of, 43–44,

finding area of triangles using, 83–84,

252–253

389–390

writing polynomial expressions, 42, 241–244

graphing, 64, 302–305

power-reducing identities, 76, 360–361

solving trig identities with, 69,

powers of i, writing in simplest form, 90, 407

333–335

Pre-Calculus For Dummies (Kuang and Kase),

sketching

1, 4

graphs using basic functions and

products of trig functions, 75, 357

transformations, 35, 215–218

product-to-sum identities, 75–76, 358

quadratic functions, 30–31, 186–190

properties, recognizing of number systems,

solving. See also calculating; finding

10–11, 128–129

absolute value inequalities, 17–18, 146–147

proving basic trig identities, 68–69, 322–333

exponential equations, 47, 258–263

linear inequalities, 17, 143

• Q •

logarithmic equations, 49–50, 269–272

polynomial equations, 39, 40, 228–232

quadratic equations, 38–39, 92, 218–227,

quadratic equations, 38–39, 92, 218–227,

414–415

414–415

quadratic formula, 38–39, 222–224

quadratic inequalities, 17, 144–146

quadratic functions, 30–31, 186–190

radical equations, 19–20, 150–156

quadratic inequalities, 17, 144–146

systems of equations, 105, 108–110, 458–463,

484–497

• R •

systems of linear equations, 104–106, 454–458,

463–470

radians

trig equations, 60, 297–299

calculating trig equations, 60, 299–301

trig identities, 69, 333–344

changing degrees to, 57, 286

by using right triangles and their functions,

changing to degrees, 57, 285

55–56, 279–284

radical equations, 19–20, 150–156

special right triangles, 56, 284–285

radical functions, 32–33, 193–197

standard form, rewriting conic equations in,

radicals, working with, 18, 148

98, 424–426

range, finding, 22–23, 159–163

standard position, finding angle measures in,

rational functions, 34, 205–211

57–58, 286–288

rationalizing denominators, 20, 156–158

statements of multiple-angle identities,

rectangular coordinates, 95, 418–420

confirming, 74, 352

recursively defined sequences, 112, 499–500

Sterling, Mary Jane (author)

reference angles, identifying, 58, 288–289

Algebra II For Dummies, 1, 4

registering online, 4

Trigonometry For Dummies, 1, 4

Remember icon, 4

substitution, solving systems of linear

right triangles, 54–56, 277–284

equations with two variables using,

rules, 112, 113, 498–499, 502–503

104, 454–456

ruling out even/odd functions, 23, 164

subtracting

Ryan, Mark (author)

angle measures, 74, 349–350

Calculus For Dummies, 5

complex numbers, 90–91, 408–410

sum of geometric series, 114, 504–505

• S •

summing terms in series, 113, 501–502

sums

creating products of trig functions from,

secant, graphing, 66, 312–319

75, 357

sequences, 111, 112, 497–500

finding, 114, 505–507

series, 111, 113, 114, 500–505

rewriting logs of products/quotients as,

sides of triangles, finding, 82, 384–386

49, 266–269

simplifying expressions, 11–12, 75, 129–131,

sum-to-product identities, 76, 358–360

355–357

532 1,001 Pre-Calculus Practice Problems For Dummies

symmetry of parabolas, 99, 433–434

triangular formulas, practical applications of,

synthetic division, 41–42, 238–240

87–88, 400–406

system of equations

trig functions

about, 103–104

calculating using functions and terminal side

solving linear equations, 104–106, 454–458,

positions, 59, 290–294

463–470

finding using right triangles, 54–55,

solving using augmented matrices, 108–109,

277–278

484–492

finding using unit circles, 58, 289–290

solving using Cramer’s rule, 110, 495–497

trig identities

solving using inverse of coefficient matrix, 109,

about, 67–68, 73

492–494

adding angle measures, 74, 349–350

solving using nonlinear functions, 105, 458–463

applying power-reducing identities, 76,

system of inequalities, 103–104, 106–107,

360–361

471–473

confirming statements of multiple-angle

identities, 74, 352

• T •

creating half-angle, 75, 352–355

creating products of trig functions from sums

tangent, 65, 75, 307–309, 353–355

and differences, 75, 357

terminal side positions, calculating trig

doubling angle size, 74, 350–351

functions using functions and, 59, 290–294

evaluating expressions, 75–76, 358–360

terms

finding values of functions using, 76–77,

adding in arithmetic series, 113, 500–501

362–367

finding of sequences, 112, 497–498

multiple methods for working through, 77–78,

summing in series, 113, 501–502

367–376

Tip icon, 4

performing algebraic processes for, 71–72,

transformations

345–349

interpreting, 65, 66, 309–312, 320–322

proving basic, 68–69, 322–333

selected points using functions, 34–35, 213–215

simplifying expressions with half-angle, 75,

sketching graphs using basic, 35, 215–218

355–357

writing new trig functions using, 64, 306–307

solving, 69–71, 333–344

transforming functions, 29–30, 34, 211–212

subtracting angle measures, 74, 349–350

triangles

trigonometry. See also triangles

about, 79–80

about, 53–54

ambiguous case of law of sines, 81, 382–384

arc length formula, 59, 294–295

applying Heron’s formula, 86, 396–397

calculating trig functions using functions

applying trig formula for area of triangles, 84,

and terminal side positions, 59,

390–391

290–294

finding angles of, 80–83, 377–378, 384–387

changing degrees to radians, 57, 286

finding area, 83–84, 86, 389–390, 395–396

changing radians to degrees, 57, 285

finding sides, 80, 82, 376–377, 384–386

evaluating inverse functions, 60

practical applications of Heron’s formula, 87,

finding angle measures in standard position,

397–400

57–58, 286–288

practical applications of law of cosines, 83,

finding trig functions, 54–55, 58, 277–278,

387–389

289–290

practical applications of law of sines, 81,

identifying reference angles, 58, 288–289

378–382

solving, 55–56, 279–284

practical applications of triangular formulas,

solving for x in degrees, 60, 297–299

87–88, 400–406

solving for x in radians, 60, 299–301

solving area with additional computations, 85,

special right triangles, 56, 284–285

392–395

Trigonometry For Dummies (Sterling), 1, 4

Index 533

• U •

• W •

unit circles, finding trig functions using, 58,

websites. See specific websites

289–290

Wiley Product Technical Support (website), 4

writing

• V •

equations for circles, 98–99, 427–433

equations from graphs, 31–32, 43–44, 190–192,

values of functions, finding using identities,

252–253

76–77, 362–367

equations of ellipses, 100, 441–443

variables, solving systems of linear equations

equations of hyperbolas, 101, 447–449

with multiple, 104, 106, 454–458,

equations of parabolas, 100, 435–437

468–470

new trig functions using transformations, 64,

vertices, finding of parabolas, 99,

306–307

434–435

powers of i in simplest form, 90, 407

Workspace

Workspace

Workspace

About the Author

Mary Jane Sterling is the author of six For Dummies titles: Algebra I For Dummies, Algebra II For Dummies, Trigonometry For Dummies, Math Word Problems For Dummies, Business Math For Dummies, and Linear Algebra For Dummies (all published by Wiley) .

She has also written many supplements — workbooks and study aids.

Mary Jane has been teaching at Bradley University in Peoria, Illinois, for 35 years and loves hearing from former students. She still remembers her favorite student evaluation remark: “Mrs. Sterling is way too excited about mathematics.” Yes!

Dedication

I dedicate this book to my friends and colleagues — current and former — at Bradley University. The feeling of community at this institution is unique and has made my tenure there a delight.

Author’s Acknowledgments

I issue a big thank you to project editor Georgette Beatty, who has taken on the huge challenge of pulling together this project. She has been a delight to work with — always upbeat and helpful. Thank you so much for your hard work and patience. Also, another big salute with sincere appreciation goes to the copy editors, Megan Knoll and Danielle Voirol. Their thoroughness and attention to detail help make for a polished product.

And, of course, a heartfelt thank you to the technical editors, Mark Kannowski and Becky Moening. As much as I try to check the problems carefully, there is always that chance of a silly error. The editors keep me honest!

As always, a grateful thank you to acquisitions editor Lindsay Lefevere, who again found me another interesting project.

Publisher’s Acknowledgments

Executive Editor: Lindsay Sandman Lefevere

Project Coordinator: Lauren Buroker

Senior Project Editor: Georgette Beatty

Cover Image: ©iStock.com/ngkaki

Senior Copy Editor: Danielle Voirol

Copy Editor: Megan Knoll

Technical Editors: Mark Kannowski, Becky Moening

Art Coordinator: Alicia B. South

WILEY END USER LICENSE AGREEMENT

Go to www.wiley.com/go/eula to access Wiley’s ebook EULA.

Document Outline

- Title Page

- Copyright Page

- Contents at a Glance

- Table of Contents

- Introduction

- What You�ll Find

- How This Workbook Is Organized

- Part I: The Questions

- Part II: The Answers

- Beyond the Book

- What you�ll find online

- How to register

- Where to Go for Additional Help

- Part I: The Questions

- Chapter 1: Getting Started with Algebra Basics

- The Problems You�ll Work On

- What to Watch Out For

- Identifying Which System or Systems a Number Belongs To

- Recognizing Properties of Number Systems

- Simplifying Expressions with the Order of Operations

- Graphing Inequalities

- Using Graphing Formulas

- Applying Graphing Formulas

- Chapter 2: Solving Some Equations and Inequalities

- The Problems You�ll Work On

- What to Watch Out For

- Using Interval and Inequality Notation

- Solving Linear Inequalities

- Solving Quadratic Inequalities

- Solving Absolute Value Inequalities

- Working with Radicals and Fractional Notation

- Performing Operations Using Fractional Exponents

- Factoring Using Fractional Notation

- Solving Radical Equations

- Rationalizing Denominators

- Chapter 3: Function Basics

- The Problems You�ll Work On

- What to Watch Out For

- Using Function Notation to Evaluate Function Values

- Determining the Domain and Range of a Function

- Recognizing Even Functions

- Identifying Odd Functions

- Ruling Out Even and Odd Functions

- Recognizing One-to-One Functions from Given Relations

- Identifying One-to-One Functions from Equations

- Recognizing a Function�s Inverse

- Determining a Function�s Inverse

- Executing Operations on Functions

- Performing Function Composition

- Doing More Function Composition

- Using the Difference Quotient

- Chapter 4: Graphing and Transforming Functions

- The Problems You�ll Work On

- What to Watch Out For

- Functions and Their Inverses

- Sketching Quadratic Functions from Their Equations

- Writing Equations from Graphs of Parabolas

- Investigating and Graphing Radical Functions

- Investigating Absolute Value Functions

- Investigating the Graphs of Polynomial Functions

- Investigating Rational Functions

- Transformation of Functions

- Transforming Selected Points Using Functions

- Sketching Graphs Using Basic Functions and Transformations

- Sketching More Graphs Using Basic Functions and Transformations

- Chapter 5: Polynomials

- The Problems You�ll Work On

- What to Watch Out For

- Using Factoring to Solve Quadratic Equations

- Solving Quadratic Equations by Using the Quadratic Formula

- Using Completing the Square to Solve Quadratic Equations

- Solving Polynomial Equations for Intercepts

- Using Factoring by Grouping to Solve Polynomial Equations

- Applying Descartes�s Rule of Signs

- Listing Possible Roots of a Polynomial Equation

- Dividing Polynomials

- Using Synthetic Division to Divide Polynomials

- Checking for Roots of a Polynomial by Using Synthetic Division

- Writing Polynomial Expressions from Given Roots

- Writing Polynomial Expressions When Given Roots and a Point

- Graphing Polynomials

- Writing Equations from Graphs of Polynomials

- Chapter 6: Exponential and Logarithmic Functions

- The Problems You�ll Work On

- What to Watch Out For

- Understanding Function Notation

- Graphing Exponential Functions

- Solving Exponential Equations

- ��

- ��

- Rewriting Logarithmic Expressions

- Rewriting Logs of Products and Quotients as Sums and Differences

- Solving Logarithmic Equations

- Applying Function Transformations to Log Functions

- Applying Logarithms to Everyday Life

- Chapter 7: Trigonometry Basics

- The Problems You�ll Work On

- What to Watch Out For

- Using Right Triangles to Determine Trig Functions

- Solving Problems by Using Right Triangles and Their Functions

- Working with Special Right Triangles

- Changing Radians to Degrees

- Changing Degrees to Radians

- Finding Angle Measures (in Degrees) in Standard Position

- Determining Angle Measures (in Radians) in Standard Position

- Identifying Reference Angles

- Determining Trig Functions by Using the Unit Circle

- Calculating Trig Functions by Using Other Functions and Terminal Side Positions

- Using the Arc Length Formula

- Evaluating Inverse Functions

- Solving Trig Equations for x in Degrees

- Calculating Trig Equations for x in Radians

- Chapter 8: Graphing Trig Functions

- The Problems You�ll Work On

- What to Watch Out For

- Recognizing Basic Trig Graphs

- Graphing Sine and Cosine

- Applying Function Transformations to Graphs of Trig Functions

- Writing New Trig Functions Using Transformations

- Graphing Tangent and Cotangent

- Interpreting Transformations of Trig Functions

- Graphing Secant and Cosecant

- Interpreting Transformations from Function Rules

- Chapter 9: Getting Started with Trig Identities

- The Problems You�ll Work On

- What to Watch Out For

- Proving Basic Trig Identities

- Returning to Basic Sine and Cosine to Solve Identities

- Using Multiplication by a Conjugate to Solve Identities

- Solving Identities After Raising a Binomial to a Power

- Solving Identities After Factoring out a Common Function

- Solving Identities After Combining Fractions

- Performing Algebraic Processes to Make Identities More Solvable

- Chapter 10: Continuing with Trig Identities

- The Problems You�ll Work On

- What to Watch Out For

- Using Identities That Add or Subtract Angle Measures

- Confirming Double-Angle Identities

- Using Identities That Double the Size of the Angle

- Confirming the Statements of Multiple-Angle Identities

- Creating Half-Angle Identities from Double-Angle Identities

- Creating a Half-Angle Identity for Tangent

- Using Half-Angle Identities to Simplify Expressions

- Creating Products of Trig Functions from Sums and Differences

- Using Product-to-Sum Identities to Evaluate Expressions

- Using Sum-to-Product Identities to Evaluate Expressions

- Applying Power-Reducing Identities

- Using Identities to Determine Values of Functions at Various Angles

- Working through Identities Using Multiple Methods

- Chapter 11: Working with Triangles and Trigonometry

- The Problems You�ll Work On

- What to Watch Out For

- Applying the Law of Sines to Find Sides

- Utilizing the Law of Sines to Find Angles

- Using the Law of Sines for Practical Applications

- Investigating the Ambiguous Case of the Law of Sines

- Determining All Angles and Sides of a Triangle

- Finding Side Measures by Using the Law of Cosines

- Using the Law of Cosines to Determine an Angle

- Applying the Law of Cosines to Real-World Situations

- Finding Areas of Triangles by Using the Sine

- Applying the Trig Formula for Area of a Triangle

- Using the Trig Formula for Area in Various Situations

- Solving Area Problems Needing Additional Computations

- Finding Areas of Triangles by Using Heron�s Formula

- Applying Heron�s Formula

- Practical Applications Using Heron�s Formula

- Tackling Practical Applications by Using Triangular Formulas

- Chapter 12: Complex Numbers and Polar Coordinates

- The Problems You�ll Work On

- What to Watch Out For

- Writing Powers of i in Their Simplest Form

- Adding and Subtracting Complex Numbers

- Multiplying Complex Numbers

- Using Multiplication to Divide Complex Numbers

- Solving Quadratic Equations with Complex Solutions

- Graphing Complex Numbers

- Identifying Points with Polar Coordinates

- Identifying Points Whose Angles Have Negative Measures

- Converting Polar to Rectangular Coordinates

- Converting Rectangular to Polar Coordinates

- Recognizing Polar Curves

- Chapter 13: Conic Sections

- The Problems You�ll Work On

- What to Watch Out For

- Identifying Conics from Their Equations

- Rewriting Conic Equations in Standard Form

- Writing Equations for Circles

- Determining Foci and Axes of Symmetry of Parabolas

- Finding the Vertices and Directrixes of Parabolas

- Writing Equations of Parabolas

- Determining Centers and Foci of Ellipses

- Writing Equations of Ellipses

- Determining Asymptotes of Hyperbolas

- Writing Equations of Hyperbolas

- Changing Equation Format from Trig Functions to Algebraic

- Changing Equation Format from Algebraic to Trig

- Chapter 14: Systems of Equations and Inequalities

- The Problems You�ll Work On

- What to Watch Out For

- Using Substitution to Solve Systems of Linear Equations with Two Variables

- Using Elimination to Solve Systems of Linear Equations with Two Variables

- Solving Systems of Equations Involving Nonlinear Functions

- Solving Systems of Linear Equations

- Solving Systems of Linear Equations with Four Variables

- Graphing Systems of Inequalities

- Decomposition of Fractions

- Operating on Matrices

- Changing Matrices to the Echelon Form

- Solving Systems of Equations Using Augmented Matrices

- Solving Systems of Equations Using the Inverse of the Coefficient Matrix

- Applying Cramer�s Rule to Solve Systems of Equations

- Chapter 15: Sequences and Series

- The Problems You�ll Work On

- What to Watch Out For

- Finding Terms of Sequences

- Determining Rules for Sequences

- Working with Recursively Defined Sequences

- Adding Terms in an Arithmetic Series

- Summing Terms of a Series

- Finding Rules and Summing Terms of a Series

- Calculating the Sum of a Geometric Series

- Determining Formulas and Finding Sums

- Counting Items by Using Combinations

- Constructing Pascal�s Triangle

- Applying Pascal�s Triangle

- Utilizing the Binomial Theorem

- Chapter 16: Introducing Limits and Continuity

- The Problems You�ll Work On

- What to Watch Out For

- Determining Limits from Graphs

- Determining One-Sided Limits

- Determining Limits from Function Values

- Determining Limits from Function Rules

- Applying Laws of Limits

- Investigating Continuity

- Part II: The Answers

- Chapter 17: Answers

- Index

- About the Author

- Wiley End User License Agreement

Table of Contents

Introduction ............................................................................ 1

Part I: The Questions ................................................................ 7

Chapter 1: Getting Started with Algebra Basics .....................................................

Chapter 2: Solving Some Equations and Inequalities .................................................

Chapter 3: Function Basics .........................................................................

Chapter 4: Graphing and Transforming Functions .....................................................

Chapter 5: Polynomials .............................................................................

Chapter 6: Exponential and Logarithmic Functions ...................................................

Chapter 7: Trigonometry Basics .....................................................................

Chapter 8: Graphing Trig Functions .................................................................

Chapter 9: Get ting Star ted with Trig Identities ............................................

Chapter 10: Continuing with Trig Identities ........................................................

Chapter 11: Working with Triangles and Trigonometry ................................................

Chapter 12: Complex Numbers and Polar Coordinates ..................................................

Chapter 13: Conic Sections .........................................................................

Chapter 14: Systems of Equations and Inequalities ..................................................

Chapter 15: Sequences and Series ...................................................................

Chapter 16: Introducing Limits and Continuity ......................................................

Part II: The Answers ............................................................ 125

Chapter 17: Answers.................................................................................

Index .................................................................................. 527

How This Workbook Is Organized ................................................................. 2

Part II: The Answers ..............................................................................

How to register ....................................................................................

Identifying Which System or Systems a Number Belongs To .................. 10

Simplifying Expressions with the Order of Operations ............................ 11

Graphing Inequalities ..............................................................................

Using Graphing Formulas ............................................................................

Using Interval and Inequality Notation ....................................................... 16

Solving Linear Inequalities ........................................................................

Working with Radicals and Fractional Notation ........................................18

Factoring Using Fractional Notation ........................................................... 19

Rationalizing Denominators .........................................................................

Using Function Notation to Evaluate Function Values ............................. 22

Recognizing Even Functions ........................................................................

Identifying One-to-One Functions from Equations .................................... 25

Determining a Function’s Inverse ................................................................26

Performing Function Composition .............................................................. 27

Using the Difference Quotient ......................................................................

Functions and Their Inverses ......................................................................3

Writing Equations from Graphs of Parabolas ............................................31

Investigating and Graphing Radical Functions .......................................... 32

Investigating Absolute Value Functions .....................................................33

Investigating Rational Functions ................................................................. 3

Sketching Graphs Using Basic Functions and Transformations ..............35

Using Factoring to Solve Quadratic Equations .......................................... 38

Using Completing the Square to Solve Quadratic Equations ................... 39

Using Factoring by Grouping to Solve Polynomial Equations .................40

Dividing Polynomials ...............................................................................

Writing Polynomial Expressions from Given Roots ..................................42