chapter 11 resource masters - commack school district 11 worksheets.pdf · chapter 11 resource...

Chapter 11 Resource Masters

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All rights reserved. The contents, or parts thereof, may be reproduced in print form for non-profit educational use with Glencoe Algebra 1, provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, network storage or transmission, or broadcast for distance learning.

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ISBN: 978-0-07-660285-8MHID: 0-07-660285-0

Printed in the United States of America.

1 2 3 4 5 6 7 8 9 10 DOH 16 15 14 13 12 11

CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters booklets are available as consumable workbooks in both English and Spanish.

MHID ISBNStudy Guide and Intervention Workbook 0-07-660292-3 978-0-07-660292-6Homework Practice Workbook 0-07-660291-5 978-0-07-660291-9

Spanish VersionHomework Practice Workbook 0-07-660294-X 978-0-07-660294-0

Answers For Workbooks The answers for Chapter 11 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

ConnectED All of the materials found in this booklet are included for viewing, printing, and editing at connected.mcgraw-hill.com.

Spanish Assessment Masters (MHID: 0-07-660289-3, ISBN: 978-0-07-660289-6) These masters contain a Spanish version of Chapter 11 Test Form 2A and Form 2C.

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Teacher’s Guide to Using the Chapter 11 Resource Masters ..........................................iv

Chapter Resources Chapter 11 Student-Built Glossary .................... 1Chapter 11 Anticipation Guide (English) ............ 3Chapter 11 Anticipation Guide (Spanish) .......... 4

Lesson 11-1Inverse Variation Study Guide and Intervention ............................ 5Skills Practice .................................................... 7Practice .............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10

Lesson 11-2Rational Functions Study Guide and Intervention ...........................11Skills Practice .................................................. 13Practice ............................................................ 14Word Problem Practice .................................... 15Enrichment ...................................................... 16

Lesson 11-3Simplifying Rational Expressions Study Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice ............................................................ 20Word Problem Practice .................................... 21Enrichment ...................................................... 22

Lesson 11-4Multiplying and Dividing Rational Expressions Study Guide and Intervention .......................... 23Skills Practice .................................................. 25Practice ............................................................ 26Word Problem Practice .................................... 27Enrichment ...................................................... 28Spreadsheet Activity ........................................ 29

Lesson 11-5Dividing PolynomialsStudy Guide and Intervention .......................... 30Skills Practice .................................................. 32Practice ............................................................ 33Word Problem Practice .................................... 34Enrichment ...................................................... 35TI-Nspire® Activity ........................................... 36

Lesson 11-6Adding and Subtracting Rational ExpressionsStudy Guide and Intervention .......................... 37Skills Practice .................................................. 39Practice ............................................................ 40Word Problem Practice .................................... 41Enrichment ...................................................... 42

Lesson 11-7Mixed Expressions and Complex FractionsStudy Guide and Intervention .......................... 43Skills Practice .................................................. 45Practice ............................................................ 46Word Problem Practice .................................... 47Enrichment ...................................................... 48

Lesson 11-8Rational Equations

Study Guide and Intervention .......................... 49Skills Practice .................................................. 51Practice ............................................................ 52Word Problem Practice .................................... 53Enrichment ...................................................... 54

AssessmentStudent Recording Sheet ............................... 55Rubric for Scoring Extended Response .......... 56Chapter 11 Quizzes 1 and 2 ............................ 57Chapter 11 Quizzes 3 and 4 ............................ 58Chapter 11 Mid-Chapter Test ........................... 59Chapter 11 Vocabulary Test ............................. 60Chapter 11 Test, Form 1 .................................. 61Chapter 11 Test, Form 2A ................................ 63Chapter 11 Test, Form 2B ................................ 65Chapter 11 Test, Form 2C ............................... 67

Chapter 11 Test, Form 2D ............................... 69Chapter 11 Test, Form 3 .................................. 71Chapter 11 Extended-Response Test .............. 73Standardized Test Practice ...............................74

Answers ........................................... A1–A37

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iv

Teacher’s Guide to Using the Chapter 11 Resource Masters

The Chapter 11 Resource Masters includes the core materials needed for Chapter 11. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing, printing, and editing at connectED.mcgraw-hill.com.

Chapter Resources

Student-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 11-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson Resources

Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI-Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

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Assessment Options

The assessment masters in the Chapter 11 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master cor-responds with the standardized test practice at the end of the chapter.

Extended Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 12 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on-grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers

• The answers for the Anticipation Guide and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

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Chapter 11 1 Glencoe Algebra 1

Student-Built Glossary

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 11. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.

11

Vocabulary TermFound

on PageDefi nition/Description/Example

asymptote

complex fraction

excluded values

extraneous solutionsehk·STRAY·nee·uhs

inverse variationihn·VUHRS

least common denominator

least common multiple

mixed expression

(continued on the next page)

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Student-Built Glossary (continued)11

Vocabulary TermFound

on PageDefi nition/Description/Example

product rule

rate problems

rational equations

rational expression

rational function

work problems


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Anticipation GuideRational Expressions and Equations

Before you begin Chapter 11

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

After you complete Chapter 11

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

11

Step 1

STEP 1A, D, or NS

StatementSTEP 2A or D

1. Since a direct variation can be written as y = kx, an inversevariation can be written as y = x −

k .

2. A rational expression is an algebraic fraction that containsa radical.

3. To multiply two rational expressions, such as 2xy2

−

3c and 3 c 2 −

5y ,

multiply the numerators and the denominators.

4. When solving problems involving units of measure, dimensional analysis is the process of determining the units of the final answer so that the units can be ignored while performing calculations.

5. To divide (4x2 + 12x) by 2x, divide 4x2 by 2x and 12x by 2x.

6. To find the sum of 2a −

(3a - 4) and 5 −

(3a - 4) , first add the

numerators and then the denominators.

7. The least common denominator of two rational expressions will be the least common multiple of the denominators.

8. A complex fraction contains a fraction in its numerator or denominator.

9. The fraction ( a −

b ) −

( c −

d ) can be rewritten as ac −

bd .

10. Extraneous solutions are solutions that can be eliminated because they are extremely high or low.

Step 2


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Ejercicios preparatoriosExpresiones y ecuaciones racionales

Antes de comenzar el Capítulo 11

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a)).

Después de completar el Capítulo 11

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

11

Paso 1

PASO 1A, D, o NS

EnunciadoPASO 2A o D

1. Dado que una variación directa se puede escribir como y = kx, una variación inversa se puede escribir como y = x −

k .

2. Una expresión racional es una fracción algebraica que contiene un radical.

3. Para multiplicar dos expresiones racionales, como 2xy2

−

3c y 3 c 2 −

5y ,

multiplica los numeradores y los denominadores.

4. Al resolver problemas con unidades de medida, el análisis dimensional es el proceso de determinar las unidades de la respuesta final, de manera que las unidades pueden ignorarse mientras se desarrollan los cálculos.

5. Para dividir (4x2 + 12x) entre 2x, divide 4x2 entre 2x y 12x entre 2x.

6. Para sumar 2a −

(3a - 4) y 5 −

(3a - 4) , primero suma los

numeradores y luego los denominadores.

7. El mínimo común denominador de dos expresiones racionales será el mínimo común múltiplo de los denominadores.

8. Una fracción compleja contiene una fracción en su numerador o denominador.

9. La fracción ( a −

b ) −

( c −

d ) se puede volver a plantear como ac −

bd .

10. Las soluciones extrínsecas son soluciones que se pueden eliminar porque son extremadamente altas o bajas.

Paso 2

Capítulo 11 4 Álgebra 1 de Glencoe


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Both methods show that x2 = 8 −

3 when y = 18.

ExercisesDetermine whether each table or equation represents an inverse or a direct variation. Explain.

1. x y

3 6

5 10

8 16

12 24

2. y = 6x 3. xy = 15

Assume that y varies inversely as x. Write an inverse variation equation that relates x and y. Then solve.

4. If y = 10 when x = 5, 5. If y = 8 when x = -2, find y when x = 2. find y when x = 4.

6. If y = 100 when x = 120, 7. If y = -16 when x = 4, find x when y = 20. find x when y = 32.

8. If y = -7.5 when x = 25, find y when x = 5.

9. DRIVING The Gerardi family can travel to Oshkosh, Wisconsin, from Chicago, Illinois, in 4 hours if they drive an average of 45 miles per hour. How long would it take them if they increased their average speed to 50 miles per hour?

10. GEOMETRY For a rectangle with given area, the width of the rectangle varies inversely as the length. If the width of the rectangle is 40 meters when the length is 5 meters, find the width of the rectangle when the length is 20 meters.

Study Guide and InterventionInverse Variation

Identify and Use Inverse Variations An inverse variation is an equation in the form of y = k − x or xy = k. If two points (x1, y1) and (x2, y2) are solutions of an inverse variation, then x1 ∙ y1 = k and x2 ∙ y2 = k.

Product Rule for Inverse Variation x1 ∙ y1

= x2 ∙ y2

From the product rule, you can form the proportion x1 − x2

= y2 − y1

.

If y varies inversely as x and y = 12 when x = 4, find x when y = 18.

Method 1 Use the product rule. x1 ∙ y1 = x2 ∙ y2 Product rule for inverse variation

4 ∙ 12 = x2 ∙ 18 x1 = 4, y

1 = 12, y

2 = 18

48 −

18 = x 2 Divide each side by 18.

8 −

3 = x 2 Simplify.

Method 2 Use a proportion.

x 1 − x 2

= y 2 − y 1

Proportion for inverse variation

4 − x 2 = 18 −

12 x

1 = 4, y

1 = 12, y

2 = 18

48 = 18x2 Cross multiply.

8 −

3 = x2 Simplify.

11-1

Example


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Study Guide and Intervention (continued)

Inverse Variation

Graph Inverse Variations Situations in which the values of y decrease as the values of x increase are examples of inverse variation. We say that y varies inversely as x, or y is inversely proportional to x.

Suppose you drive 200 miles without stopping. The time it takes to travel a distance varies inversely as the rate at which you travel. Let x = speed in miles per hour and y = time in hours. Graph the variation.

The equation xy = 200 can be used to represent the situation. Use various speeds to make a table.

x

y

O 20 40 60

30

20

10

Graph an inverse variation in which y varies inversely as x and y = 3 when x = 12.

Solve for k. xy = k Inverse variation equation

12(3) = k x = 12 and y = 3

36 = k Simplify.

Choose values for x and y, which have a product of 36.

x

y

O 12 24

24

12

ExercisesGraph each variation if y varies inversely as x.

1. y = 9 when x = -3 2. y = 12 when x = 4 3. y = -25 when x = 5

x

y

O

24

12

-12

-24

-12-24 12 24

x

y

O

32

16

-16

-32

-16-32 16 32

x

y

O 50-50-100 100

100

50

-50

-100

4. y = 4 when x = 5 5. y = -18 when x = -9 6. y = 4.8 when x = 5.4

x

y

O

20

10

-10

-20

-10-20 10 20

x

y

O

36

18

-18

-36

-18-36 18 36

x

y

O

7.2

3.6

-3.6

-7.2

-3.6-7.2 3.6 7.2

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Example 1 Example 2

x y

10 20

20 10

30 6.7

40 5

50 4

60 3.3

x y

−6 −6

−3 −12

−2 −18

2 18

3 12

6 6

Inverse Variation Equation an equation of the form xy = k, where k ≠ 0


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Skills PracticeInverse Variation

Determine whether each table or equation represents an inverse or a direct variation. Explain.

1. x y

0.5 8

1 4

2 2

4 1

2. xy = 2 −

3 3. -2x + y = 0

Assume that y varies inversely as x. Write an inverse variation equation that relates x and y. Then graph the equation.

4. y = 2 when x = 5 5. y = -6 when x = -6

6. y = -4 when x = -12 7. y = 15 when x = 3

Solve. Assume that y varies inversely as x.

8. If y = 4 when x = 8, 9. If y = -7 when x = 3, find y when x = 2. find y when x = -3.

10. If y = -6 when x = -2, 11. If y = -24 when x = -3, find y when x = 4. find x when y = -6.

12. If y = 15 when x = 1, 13. If y = 48 when x = -4, find x when y = -3. find y when x = 6.

14. If y = -4 when x = 1 −

2 , find x when y = 2.

11-1

x

y

O 8-8-16 16

16

8

-8

-16

x

y

O 8-8-16

16

16

8

-8

-16

x

y

O 10-10-20

20

20

10

-10

-20

x

y

O 4-4-8

8

8

4

-4

-8


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Practice Inverse Variation

Determine whether each table or equation represents an inverse or a direct variation. Explain.

1. x y

0.25 40

0.5 20

2 5

8 1.25

2. x y

-2 8

0 0

2 -8

4 -16

3. y − x = -3 4. y = 7 − x

Asssume that y varies inversely as x. Write an inverse variation equation that relates x and y. Then graph the equation.

5. y = -2 when x = -12 6. y = -6 when x = -5 7. y = 2.5 when x = 2

Write an inverse variation equation that relates x and y. Assume that y varies inversely as x. Then solve.

8. If y = 124 when x = 12, find y when x = -24.

9. If y = -8.5 when x = 6, find y when x = -2.5.

10. If y = 3.2 when x = -5.5, find y when x = 6.4.

11. If y = 0.6 when x = 7.5, find y when x = -1.25.

12. EMPLOYMENT The manager of a lumber store schedules 6 employees to take inventory in an 8-hour work period. The manager assumes all employees work at the same rate.a. Suppose 2 employees call in sick. How many hours will 4 employees need to take

inventory? b. If the district supervisor calls in and says she needs the inventory finished in 6 hours,

how many employees should the manager assign to take inventory?

13. TRAVEL Jesse and Joaquin can drive to their grandparents’ home in 3 hours if they average 50 miles per hour. Since the road between the homes is winding and mountainous, their parents prefer they average between 40 and 45 miles per hour. How long will it take to drive to the grandparents’ home at the reduced speed?

11-1

x

y

O

24

12

-12

-24

-24 -12 12 24 x

y

Ox

y

O

16

8

-8

-16

-8-16 8 16


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1. PHYSICAL SCIENCE The illumination Iproduced by a light source varies inversely as the square of the distance dfrom the source. The illumination produced 5 feet from the light source is 80 foot-candles.

Id2= k

80(5)2= k

2000 = k

Find the illumination produced 8 feet from the same source.

2. MONEY A formula called the Rule of 72 approximates how fast money will double in a savings account. It is based on the relation that the number of years it takes for money to double varies inversely as the annual interest rate. Use the information in the table to write the Rule of 72 formula.

3. ELECTRICITY The resistance, in ohms, of a certain length of electric wire varies inversely as the square of the diameter of the wire. If a wire 0.04 centimeter in diameter has a resistance of 0.60 ohm, what is the resistance of a wire of the same length and material that is 0.08 centimeter in diameter?

4. BUSINESS In the manufacturing of a certain digital camera, the cost of producing the camera varies inversely as the number produced. If 15,000 cameras are produced, the cost is $80 per unit. Graph the relationship and label the point that represents the cost per unit to produce 25,000 cameras.

5. SOUND The sound produced by a string inside a piano depends on its length. The frequency of a vibrating string varies inversely as its length.

a. Write an equation that represents the relationship between frequency f and length �. Use k for the constant of variation.

b. If you have two different length strings, which one vibrates more quickly (that is, which string has a greater frequency)?

c. Suppose a piano string 2 feet long vibrates 300 cycles per second. What would be the frequency of a string 4 feet long?

Word Problem Practice Inverse Variation

11-1

Years

to Double

Money

Annual Interest

Rate

(percent)

18 4

14.4 5

12 6

10.29 7

Pric

e p

er U

nit

($)

200

300

100

0Units Produced

(thousands)

10 20 30

y

x


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Enrichment

Direct or Indirect Variation

Fill in each table below. Then write inversely, or directly to complete each conclusion.

1. � 2 4 8 16 32

W 4 4 4 4 4

A

2. Hours 2 4 5 6

Speed 55 55 55 55

Distance

For a set of rectangles with a width For a car traveling at 55 mi/h, the of 4, the area varies distance covered varies

as the length. as the hours driven.

3. Oat Bran 1 −

3 cup 2 −

3 cup 1 cup

Water 1 cup 2 cup 3 cup

Servings 1 2

4. Hours of Work 128 128 128

People Working 2 4 8

Hours per Person

The number of servings of oat bran A job requires 128 hours of work. The varies as the number number of hours each person works of cups of oat bran. varies as the number

of people working.

5. Miles 100 100 100 100

Rate 20 25 50 100

Hours 5

6. b 3 4 5 6

h 10 10 10 10

A 15

For a 100-mile car trip, the time the For a set of right triangles with a height trip takes varies as the of 10, the area varies

average rate of speed the car travels. as the base.

Use the table at the right.

7. x varies as y.

8. z varies as y.

9. x varies as z.

11-1

x 1 1.5 2 2.5 3

y 2 3 4 5 6

z 60 40 30 24 20


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Identify Excluded Values The function y = 10 − x is an example of a rational function. Because division by zero is undefined, any value of a variable that results in a denominator of zero must be excluded from the domain of that variable. These are called excluded values of the rational function.

State the excluded value for each function.

Exercises


1. y = 2 −

x 2. y = 1 −

x - 4 3. y = x - 3 −

x + 1

4. y = 4 −

x - 2 5. y = x −

2x - 4 6. y = - 5 −

3x

7. y = 3x - 2 −

x + 3 8. y = x - 1 −

5x + 10 9. y = x + 1 − x

10. y = x - 7 −

2x + 8 11. y = x - 5 −

6x 12. y = x - 2 −

x + 11

13. y = 7 −

3x + 21 14. y = 3x - 4 −

x + 4 15. y = x −

7x - 35

16. DINING Mya and her friends are eating at a restaurant. The total bill of $36 is split among x friends. The amount each person pays y is given by y = 36 −

x , where x is the

number of people. Graph the function.

11-2 Study Guide and InterventionRational Functions

Example

a. y = 3 − x The denominator cannot equal zero. The excluded value is x = 0.

b. y = 4 −

x - 5

x - 5 = 0 Set the denominator equal to 0.

x = 5 Add 5 to each side.

The excluded value is x = 5.

Bill

per P

erso

n ($

)

4

0

8

12

16

20

24

28

32

36

Number of People1 2 3 4 5 6 7 8


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Identify and Use Asymptotes Because excluded vales are undefined, they affect the graph of the function. An asymptote is a line that the graph of a function approaches.

A rational function in the form y = a −

x - b + c has a vertical asymptote at the x-value that

makes the denominator equal zero, x = b. It has a horizontal asymptote at y = c.

Identify the asymptotes of y = 1 −

x - 1 + 2 . Then graph the function.

Step 1 Identify and graph the asymptotes using dashed lines. vertical asymptote: x = 1 horizontal asymptote: y = 2

Step 2 Make a table of values and plot the points. Then connect them.

x –1 0 2 3

y 1.5 1 3 2.5

ExercisesIdentify the asymptotes of each function. Then graph the function.

1. y = 3 − x 2. y = -2 −

x 3. y = 4 − x + 1

4. y = 2 −

x - 3 5. y = 2 −

x + 1 6. y = -2 −

x - 3

11-2 Study Guide and Intervention (continued)

Rational Functions

Example

y

x

y = + 2

y = 2

x = 1

1x-1

y

x

y

x

y

x

y

x

y

x

y

x


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1. y = 6 − x 2. y = 2 −

x - 2 3. y = x −

x + 6

4. y = x - 3 −

x + 4 5. y = 3x - 5 −

x + 8 6. y = -5 −

2x - 14

7. y = x −

3x + 21 8. y = x - 1 −

9x - 36 9. y = 9 −

5x + 40

Identify the asymptotes of each function. Then graph the function.

10. y = 1 − x 11. y = 3 − x 12. y = 2 −

x + 1

13. y = 3 −

x - 2 14. y = 2 −

x + 1 - 1 15. y = 1 −

x - 2 + 3

11-2 Skills PracticeRational Functions

y

x

y

x

y

x

y

x

y

x

y

x


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1. y = -1

−

x 2. y = 3 −

x + 5 3. y = 2x −

x - 5

4. y = x - 1 −

12x + 36 5. y = x + 1 −

2x + 3 6. y = 1 −

5x - 2

Identify the asymptotes of each function. Then graph the function.

7. y = 1 − x 8. y = 3 − x 9. y = 2 −

x - 1

10. y = 2 −

x + 2 11. y = 1 −

x - 3 + 2 12. y = 2 −

x + 1 -1

13. AIR TRAVEL Denver, Colorado, is located approximately

1000 miles from Indianapolis, Indiana. The average speed of a

plane traveling between the two cities is given by y = 1000 −

x ,

where x is the total flight time. Graph the function.

11-2 PracticeRational Functions

Aver

age

Spee

d (m

ph)

400

600

200

0 1

800

1000

Total Flight Time

2 3 4 5

y

x

y

x

y

x

y

x

y

x

y

x


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1. BULLET TRAINS The Shinkansen, or Japanese bullet train network, provides high-speed transportation throughout Japan. Trains regularly operate at speeds in excess of 200 kilometers per hour. The average speed of a bullet train traveling between Tokyo and Kyoto is given by y =

515−x , where x is the total travel time

in hours. Graph the function.

2. DRIVING Peter is driving to his

grandparents’ house 40 miles away.

During the trip, Peter makes a 30-minute

stop for lunch. The average speed of

Peter’s trip is given by y =

40−

x + 0.5, where

x is the total time spent in the car. What

are the asymptotes of the function?

3. ERROR ANALYSIS Nicolas is graphing

the equation y =

20−

x + 3 - 6 and draws a

graph with asymptotes at y = 3 and

x = – 6. Explain the error that Nicolas

made in his graph.

11-2 Word Problem Practice Rational Functions

Aver

age

Spee

d (k

m/h

)

100

150

50

0 1

200

300250

Time (hours)2 3 4

4. USED CARS While researching cars to purchase online, Ms. Jacobs found that the value of a used car is inversely proportional to the age of the car. The average price of a used car is given by

y =

17,900−

x + 1.2+ 100, where x is the age of

the car. What are the asymptotes of the function? Explain why x = 0 cannot be an asymptote.

5. FAMILY REUNION The Gaudet family is holding their annual reunion at Watkins Park. It costs $50 to get a permit to hold the reunion at the park, and the family is spending $8 per person on food. The Gaudets have agreed to split the cost of the event evenly among all those attending.

a. Write an equation showing the cost per person y if x people attend the reunion.

b. What are the asymptotes of the equation?

c. Now assume that the family wants to let a long-lost cousin attend for free. Rewrite the equation to find the new cost per paying person y.

d. What are the asymptotes for the new equation?


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Inequalities involving rational functions can be graphed much like those involving linear functions.

Graph y ≥ 1 − x .

Step 1 Plot points and draw a smooth solid curve. Because the inequality involves a greater than or equal to sign, solutions that satisfy y = 1 − x will be a part of the graph.

Step 2 Plot the asymptotes, x = 0 and y = 0, as dashed lines.

Step 3 Begin testing values. A value must be tested between each set of lines, including asymptotes.Region 1 Test (–1, 1). This returns a true value for

the inequality.Region 2 Test (–1, –0.5). This returns a true value

for the inequality.Region 3 Test (–1, –2). This returns a false value for

the inequality.Region 4 Test (1, 2). This returns a true value for

the inequality.Region 5 Test (1, 0.5). This returns a false value for

the inequality.Region 6 Test (1, –1). This returns a false value for

the inequality.

Step 4 Shade the regions where the inequality is true.

ExercisesGraph each inequality.

1. y ≤ 1 −

x + 1 2. y > 2 − x 3. y ≤ 1 −

x + 1 -1

11-2 EnrichmentInequalities involving Rational Functions

Exampley

x

y

x

y

x

y

x

y

x


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Study Guide and InterventionSimplifying Rational Expressions

Identify Excluded Values

Because a rational expression involves division, the denominator cannot equal zero. Any value of the denominator that results in division by zero is called an excluded value of the denominator.

State the excluded

value of 4m - 8 −

m + 2 .

Exclude the values for which m + 2 = 0. m + 2 = 0 The denominator cannot equal 0.

m + 2 - 2 = 0 - 2 Subtract 2 from each side.

m = -2 Simplify.

Therefore, m cannot equal -2.

State the excluded

values of x 2 + 1 −

x 2 - 9 .

Exclude the values for which x2 - 9 = 0. x2 - 9 = 0 The denominator cannot equal 0.

(x + 3)(x - 3) = 0 Factor.

x + 3 = 0 or x - 3 = 0 Zero Product Property

= -3 = 3Therefore, x cannot equal -3 or 3.

ExercisesState the excluded values for each rational expression.

1. 2b −

b2 - 8 2. 12 - a −

32 + a

3. x 2 - 2 −

x2 + 4 4. m 2 - 4 −

2m2 - 8

5. 2n - 12 −

n 2 - 4 6. 2x + 18 −

x 2 - 16

7. x 2 - 4 −

x 2 + 4x + 4 8. a - 1 −

a 2 + 5a + 6

9. k 2 - 2k + 1 −

k 2 + 4k + 3 10. m 2 - 1 −

2 m 2 - m - 1

11. 25 - n 2 −

n 2 - 4n - 5 12. 2 x 2 + 5x + 1 −

x 2 - 10x + 16

13. n 2 - 2n - 3 −

n 2 + 4n - 5 14.

y 2 - y - 2 −

3 y 2 - 12

15. k 2 + 2k - 3 −

k 2 - 20k + 64 16. x 2 + 4x + 4 −

4 x 2 + 11x - 3

11-3

Rational

Expression

an algebraic fraction with numerator and

denominator that are polynomialsExample:

x 2 + 1 −

y 2

Example 2Example 1

017_035_ALG1_A_CRM_C11_CR_660285.indd 17017_035_ALG1_A_CRM_C11_CR_660285.indd 17 12/20/10 7:40 PM12/20/10 7:40 PM

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Simplifying Rational Expressions

Simplify Expressions Factoring polynomials is a useful tool for simplifying rational expressions. To simplify a rational expression, first factor the numerator and denominator. Then divide each by the greatest common factor.

Simplify 54 z 3 −

24yz .

54 z 3 −

24yz = (6z)(9 z 2 )

−

(6z)(4y) The GCF of the numerator and the denominator is 6z.

= 1 (6z)(9 z 2 )

−

1 (6z)(4y) Divide the numerator and denominator by 6z.

= 9 z 2 −

4y Simplify.

Simplify 3x - 9 −

x 2 - 5x + 6 . State the excluded values of x.

3x - 9 −

x 2 - 5x + 6 = 3(x - 3)

−

(x - 2)(x - 3) Factor.

= 3 (x - 3) 1 −

(x - 2) (x - 3) 1 Divide by the GCF, x – 3.

= 3 −

x - 2 Simplify.

Exclude the values for which x2 - 5x + 6 = 0. x2 - 5x + 6 = 0 (x - 2)(x - 3) = 0 x = 2 or x = 3Therefore, x ≠ 2 and x ≠ 3.

ExercisesSimplify each expression. State the excluded values of the variables.

1. 12ab −

a 2 b 2 2. 7 n 3 −

21 n 8

3. x + 2 −

x 2 - 4 4. m 2 - 4 −

m 2 + 6m + 8

5. 2n - 8 −

n 2 - 16 6. x 2 + 2x + 1 −

x 2 - 1

7. x 2 - 4 −

x 2 + 4x + 4 8. a 2 + 3a + 2 −

a 2 + 5a + 6

9. k 2 - 1 −

k 2 + 4k + 3 10. m 2 - 2m + 1 −

2 m 2 - m - 1

11. n 2 - 25 −

n 2 - 4n - 5 12. x 2 + x - 6 −

2 x 2 - 8

13. n 2 + 7n + 12 −

n 2 + 2n - 8 14.

y 2 - y - 2 −

y 2 - 10y + 16

11-3

Example 1

Example 2


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Skills PracticeSimplifying Rational Expressions

State the excluded values for each rational expression.

1. 2p −

p - 7 2. 4n + 1 −

n + 4

3. k + 2 −

k 2 - 4 4. 3x + 15 −

x 2 - 25

5. y 2 - 9 −

y 2 + 3y - 18 6. b 2 - 2b - 8 −

b 2 + 7b + 10

Simplify each expression. State the excluded values of the variables.

7. 21bc −

28 bc 2 8. 12 m 2 r −

24 mr 3

9. 16 x 3 y 2

−

36 x 5 y 3 10. 8 a 2 b 3 −

40 a 3 b

11. n + 6 −

3n + 18 12. 4x - 4 −

4x + 4

13. y 2 - 64

−

y + 8 14.

y 2 - 7y - 18 −

y - 9

15. z + 1 −

z 2 - 1 16. x + 6 −

x 2 + 2x - 24

17. 2d + 10 −

d 2 - 2d - 35 18. 3h - 9 −

h 2 - 7h + 12

19. t 2 + 5t + 6 −

t 2 + 6t + 8 20. a 2 + 3a - 4 −

a 2 + 2a - 8

21. x 2 + 10x + 24 −

x 2 - 2x - 24 22. b 2 - 6b + 9 −

b 2 - 9b + 18

11-3


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Practice Simplifying Rational Expressions

State the excluded values for each rational expression.

1. 4n - 28 −

n 2 - 49 2.

p 2 - 16 −

p 2 - 13p + 36 3. a 2 - 2a - 15 −

a 2 + 8a + 15

Simplify each expression. State the excluded values of the variables.

4. 12a −

48 a 3 5.

6xy z 3 −

3 x 2 y 2 z 6.

36 k 3 n p 2 −

20 k 2 n p 5

7. 5 c 3 d 4 −

40 cd 2 + 5 c 4 d 2 8.

p 2 - 8p + 12 −

p - 2

9. m 2 - 4m - 12 −

m - 6 10. m + 3 −

m 2 - 9

11. 2b - 14 −

b 2 - 9b + 14 12. x 2 - 7x + 10 −

x 2 - 2x - 15

13. y 2 + 6y - 16

−

y 2 - 4y + 4 14. r 2 - 7r + 6 −

r 2 + 6r - 7

15. t 2 - 81 −

t 2 - 12t + 27 16. r 2 + r - 6 −

r 2 + 4r - 12

17. 2 x 2 + 18x + 36 −

3 x 2 - 3x - 36 18.

2 y 2 + 9y + 4 −

4 y 2 - 4y - 3

19. ENTERTAINMENT Fairfield High spent d dollars for refreshments, decorations, and advertising for a dance. In addition, they hired a band for $550.

a. Write an expression that represents the cost of the band as a fraction of the total amount spent for the school dance.

b. If d is $1650, what percent of the budget did the band account for?

20. PHYSICAL SCIENCE Mr. Kaminksi plans to dislodge a tree stump in his yard by using a 6-foot bar as a lever. He places the bar so that 0.5 foot extends from the fulcrum to the end of the bar under the tree stump. In the diagram, b represents the total length of the bar and t represents the portion of the bar beyond the fulcrum.

a. Write an equation that can be used to calculate the mechanical advantage.

b. What is the mechanical advantage?

c. If a force of 200 pounds is applied to the end of the lever, what is the force placed on the tree stump?

b

fulcrum

tree stumpt

11-3


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Word Problem Practice Simplifying Rational Expressions

1. PHYSICAL SCIENCE Pressure is equal to the magnitude of a force divided by the area over which the force acts.

P =

F−

A

Gabe and Shelby each push open a door with one hand. In order to open, the door requires 20 pounds of force. The surface area of Gabe’s hand is 10 square inches, and the surface area of Shelby’s hand is 8 square inches. Whose hand feels the greater pressure?

2. GRAPHING Recall that the slope of a line is a ratio of the vertical change to the horizontal change in coordinates for two given points. Write a rational expression that represents the slope of the line containing the points at (p, r) and (7, -3).

3. AUTOMOBILES The force needed to keep a car from skidding out of a turn on a particular road is given by the formula below. What force is required to keep a 2000-pound car traveling at 50 miles per hour on a curve with radius of 750 feet on the road? What value of r is excluded?

f = 0.0672w s 2 − r

f = force in pounds

w = weight in pounds

s = speed in mph

r = radius in feet

4. PACKAGING In order to safely ship a new electronic device, the distribution manager at Data Products Company determines that the package must contain a certain amount of cushioning on each side of the device. The device is shaped like a cube with side length x, and some sides need more cushioning than others because of the device’s design. The volume of a shipping container is represented by the expression (x2 + 6x + 8)(x + 6). Find the polynomial that represents the area of the top of the box if the height of the box is x + 2.

5. SCHOOL CHOICE During a recent school year, the ratio of public school students to private school students in the United States was approximately 7.6 to 1. The students attending public school outnumbered those attending private schools by 42,240,000.

a. Write a rational expression to express the ratio of public school students to x private school students.

b. How many students attended private school?

11-3

x + 2


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Shannon’s Juggling TheoremMathematicians look at various mathematical ways to represent juggling. One way they have found to represent juggling is Shannon’s Juggling Theorem. Shannon’s Juggling Theorem uses the rational equation

f + d

−

v + d = b −

h

where f is the flight time, or how long a ball is in the air, d is the dwell time, or how long a ball is in a hand, v is the vacant time, or how long a hand is empty, b is the number of balls, and h is the number of hands (either 1 or 2 for a real-life situation, possibly more for a computer simulation).So, given the values for f, d, v, and h, it is possible to determine the number of balls being juggled. If the flight time is 9 seconds, the dwell time is 3 seconds, the vacant time is 1 second, and the number of hands is 2, how many balls are being juggled?

f + d

−

v + d = b −

h Original equation

9 + 3 −

1 + 3 = b −

2 f = 9, d = 3, v = 1, and h = 2

12 −

4 = b −

2 Simplify.

24 = 4b Cross multiply.

6 = b Divide.

So, the number of balls being juggled is 6.

Given the following information, determine the number of balls being juggled.

1. flight time = 6 seconds, vacant time = 1 second, dwell time = 1 second, number of hands = 2

2. flight time = 13 seconds, vacant time = 1 second, dwell time = 5 seconds, number of hands = 1

3. flight time = 4 seconds, vacant time = 1 second, dwell time = 1 second, number of hands = 2

4. flight time = 16 seconds, vacant time = 1 second, dwell time = 2 seconds, number of hands = 2

5. flight time = 18 seconds, vacant time = 3 seconds, dwell time = 2 seconds, number of hands = 1

Enrichment 11-3


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Study Guide and Intervention Multiplying and Dividing Rational Expressions

Multiply Rational Expressions To multiply rational expressions, you multiply the numerators and multiply the denominators. Then simplify.

Find 2 c 2 f

−

5a b 2 · a 2 b −

3cf .

2 c 2 f

−

5a b 2 · a 2 b −

3cf =

2 a 2 b c 2 f −

15a b 2 cf Multiply.

= 1(abcf)(2ac)

− 1(abcf)(15b)

Simplify.

= 2ac −

15b Simplify.

Find x 2 - 16 −

2x + 8 · x + 4

−

x 2 + 8x + 16 .

x 2 - 16 −

2x + 8 · x + 4 −

x 2 + 8x + 16 = (x - 4)(x + 4)

−

2(x + 4) · x + 4 −

(x + 4)(x + 4) Factor.

= (x - 4)(x + 4)1

−

2(x + 4)1

· x + 41

−

(x + 4)(x + 4)1

Simplify.

= x - 4 −

2x + 8 Multiply.

ExercisesFind each product.

1. 6ab −

a 2 b 2 · a 2 −

b 2 2.

m p 2 −

3 · 4 − mp

3. x + 2 −

x - 4 · x - 4 −

x - 1 4. m - 5 −

8 · 16 −

m - 5

5. 2n - 8 −

n + 2 · 2n + 4 −

n - 4 6. x 2 - 64 −

2x + 16 · x + 8 −

x 2 + 16x + 64

7. 8x + 8 −

x 2 - 2x + 1 · x - 1 −

2x + 2 8. a 2 - 25 −

a + 2 · a 2 - 4 −

a - 5

9. x 2 + 6x + 8 −

2 x 2 + 9x + 4 · 2 x 2 - x - 1 −

x 2 - 3x + 2 10. m 2 - 1 −

2 m 2 - m - 1 · 2m + 1 −

m 2 - 2m + 1

11. n 2 - 1 −

n 2 - 7n + 10 · n2 - 25 −

n2 + 6n + 5 12.

3p - 3r −

10pr ·

20 p 2 r 2 −

p 2 - r 2

13. a 2 + 7a + 12 −

a 2 + 2a - 8 · a 2 + 3a - 10 −

a 2 + 2a - 8 14. v 2 - 4v - 21 −

3 v 2 + 6v · v 2 + 8v −

v 2 + 11v + 24

11-4

Example 1

Example 2


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Multiplying and Dividing Rational Expressions

Divide Rational Expressions To divide rational expressions, multiply by the reciprocal of the divisor. Then simplify.

Find 12 c 2 f

−

5 a 2 b 2 ÷

c 2 f 2 −

10ab .

12 c 2 f

−

5 a 2 b 2 ÷

c 2 f 2 −

10ab =

12 c 2 f −

5 a 2 b 2 × 10ab −

c 2 f 2

= 121c2f 1

−

15aa2b2

b

× 2 10 1 ab 1 −

1 c 2 f 2 f

= 24 −

abf

Find x 2 + 6x - 27 −

x 2 + 11x + 18 ÷ x - 3 −

x 2 + x - 2 .

x 2 + 6x - 27 −

x 2 + 11x + 18 ÷ x - 3 −

x 2 + x - 2 = x 2 + 6x - 27 −

x 2 + 11x + 18 × x 2 + x - 2 −

x - 3

= (x + 9)(x - 3) −

(x + 9)(x + 2) × (x + 2)(x - 1)

−

x - 3

= 1(x + 9)(x - 3)1

−

1(x + 9)(x + 2)1

× 1(x + 2)(x - 1)

−

x - 31

= x - 1

ExercisesFind each quotient.

1. 12ab −

a 2 b 2 ÷ b − a 2. n −

4 ÷ n − p

3. 3 xy 2

−

8 ÷ 6xy 4. m - 5 −

8 ÷ m - 5 −

16

5. 2n - 4 −

2n ÷ n 2 - 4 − n 6.

y 2 - 36 −

y 2 - 49 ÷

y + 6 −

y + 7

7. x 2 - 5x + 6 −

5 ÷ x - 3 −

15 8. a 2 b 3 c −

3 r 2 t ÷ 6 a 2 bc −

8r t 2 u

9. x 2 + 6x + 8 −

x 2 + 4x + 4 ÷ x + 4 −

x + 2 10. m 2 - 49 − m ÷ m 2 - 13m + 42 −

3 m 2

11. n 2 - 5n + 6 −

n 2 + 3n ÷ 3 - n −

4n + 12 12.

p 2 - 2pr + r 2 − p + r ÷

p 2 - r 2 − p + r

13. a 2 + 7a + 12 −

a 2 + 3a - 10 ÷ a 2 - 9 −

a 2 - 25 14. a 2 - 9 −

2 a 2 + 13a - 7 ÷ a + 3 −

4 a 2 - 1

11-4

Example 1

Example 2


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Skills PracticeMultiplying and Dividing Rational Expressions

Find each product.

1. 14 −

c 2 · c 5 −

2c 2. 3 m 2 −

2t · t

2 −

12

3. 2 a 2 b −

b 2 c · b − a 4.

2 x 2 y −

3 x 2 y ·

3xy −

4y

5. 3(4m - 6) −

18r · 9 r 2 −

2(4m - 6) 6. 4(n + 2)

−

n(n - 2) · n - 2 −

n + 2

7. (y - 3)(y + 3)

−

4 · 8 −

y + 3 8. (x - 2)(x + 2)

−

x(8x + 3) · 2(8x + 3)

−

x - 2

9. (a - 7)(a + 7) −

a(a + 5) · a + 5 −

a + 7 10. 4(b + 4)

−

(b - 4)(b - 3) · b - 3 −

b + 4

Find each quotient.

11. c3 −

d3 ÷ d

3 −

c3 12. x

3 −

y2 ÷ x

3 − y

13. 6a3 −

4f 2 ÷ 2a2

−

12f 2 14. 4m3

−

rp2 ÷ 2m − rp

15. 3b + 3 −

b + 2 ÷ (b + 1) 16. x - 5 −

x + 3 ÷ (x - 5)

17. x2 - x - 12 −

6 ÷ x + 3 −

x - 4 18. a

2 - 5a - 6 −

3 ÷ a - 6 −

a + 1

19. m2 + 2m + 1 −

10m - 10 ÷ m + 1 −

20 20.

y2 + 10y + 25 −

3y - 9 ÷

y + 5 −

y - 3

21. b + 4 −

b2 - 8b + 16 ÷ 2b + 8 −

b - 8 22. 6x + 6 −

x - 1 ÷ x

2 + 3x + 2 −

2x - 2

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Practice Multiplying and Dividing Rational Expressions

Find each product or quotient.

1. 18 x 2 −

10 y 2 ·

15 y 3 −

24x 2. 24 rt 2 −

8 r 4 t 3 · 12 r 3 t 2 −

36 r 2 t

3. (x + 2)(x + 2) −

8 · 72 −

(x + 2)(x - 2) 4. m + 7 −

(m - 6)(m + 2) · (m - 6)(m + 4)

−

(m + 7)

5. a - 4 −

a 2 - a - 12 · a + 3 −

a - 6 6. 4x + 8 −

x 2 · x −

x 2 - 5x - 14

7. n 2 + 10n + 16 −

5n - 10 · n - 2 −

n 2 + 9n + 8 8.

3y - 9 −

y 2 - 9y + 20 ·

y 2 - 8y + 16 −

y - 3

9. b 2 + 5b + 4 −

b 2 - 36 · b 2 + 5b - 6 −

b 2 + 2b - 8 10. t 2 + 6t + 9 −

t 2 - 10t + 25 · t

2 - t - 20 −

t 2 + 7t + 12

11. 28a2 −

7b2 ÷ 21a3

−

35b 12.

mn2p3

−

x4y2 ÷

mnp2

−

x3y

13. 2a −

a - 1 ÷ (a + 1) 14. z

2 - 16 −

3z ÷ (z - 4)

15. 4y + 20

−

y - 3 ÷

y + 5 −

2y - 6 16. 4x + 12 −

6x - 24 ÷ 2x + 6 −

x + 3

17. b2 + 2b - 8 −

b2 - 11b + 18 ÷ 2b - 8 −

2b - 18 18. 3x - 3 −

x2 - 6x + 9 ÷ 6x - 6 −

x2 - 5x + 6

19. a2 + 8a + 12 −

a2 - 7a + 10 ÷ a

2 - 4a - 12 −

a2 + 3a - 10 20.

y2 + 6y - 7 −

y2 + 8y - 9 ÷

y2 + 9y + 14 −

y2 + 7y - 18

21. BIOLOGY The heart of an average person pumps about 9000 liters of blood per day. How many quarts of blood does the heart pump per hour? (Hint: One quart is equal to 0.946 liter.) Round to the nearest whole number.

22. TRAFFIC On Saturday, it took Ms. Torres 24 minutes to drive 20 miles from her home to her office. During Friday’s rush hour, it took 75 minutes to drive the same distance.

a. What was Ms. Torres’s average speed in miles per hour on Saturday?

b. What was her average speed in miles per hour on Friday?

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Word Problem Practice Multiplying and Dividing Rational Expressions

1. JOBS Rosa earned $26.25 for babysitting for 3 1−

2 hours. At this rate, how much will

she earn babysitting for 5 hours?

2. HOMEWORK Alejandro and Ander were working on the following homework problem.

Find n - 10 −

n + 3 · 2n + 6 −

n + 3 .

Alejandro’s Solution

n - 10 −

n + 3 · 2n + 6 −

n + 3

= 2(n - 10)(n + 3 ) 1 −−

(n + 3)(n + 3 ) 1

= 2n - 20 −

n + 3

Ander’s Solution

n - 10 −

n + 3 · 2n + 6 −

n + 3

= 2(n - 10)(n + 3 ) 1 −−

n + 3 1

= 2n - 20

Is either of them correct? Explain.

3. GEOMETRY Suppose the rational

expression 5k2m3 −

2ab represents the area

of a section in a tiled floor and 2km − a

represents the section’s length. Write a rational expression to represent the section’s width.

4. TRAVEL Helene travels 800 miles from Amarillo to Brownsville at an average speed of 40 miles per hour. She makes the return trip driving an average of 60 miles per hour. What is the average rate for the entire trip? (Hint: Recall that t = d ÷ r.)

5. MANUFACTURING India works in a metal shop and needs to drill equally spaced holes along a strip of metal. The centers of the holes on the ends of the strip must be exactly 1 inch from each end. The remaining holes will be equally spaced.

a. If there are x equally spaced holes, write an expression for the number of equal spaces there are between holes.

b. Write an expression for the distance between the end screws if the length is ℓ.

c. Write a rational equation that represents the distance between the holes on a piece of metal that is� inches long and must have x equally spaced holes.

d. How many holes will be drilled in a metal strip that is 6 feet long with a distance of 7 inches between the centers of each screw?

ddddd

1inch

1inch

�

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Geometric SeriesA geometric series is a sum of the terms in a geometric sequence. Each term of a geometric sequence is formed by multiplying the previous term by a constant term called the common ratio.

1 + 1 −

2 + 1 −

4 +

1 −

8 ← geometric sequence where the common ratio is 1 −

2

The sum of a geometric series can be represented by the rational expressionx0

(r)n - 1 −

r - 1 , where x0 is the first term of the series, r is the common ratio, and n is the

number of terms.

In the example above, 1 + 1 −

2 + 1 −

4 + 1 −

8 = 1 ․

( 1 −

2 )

4 - 1 −

1 −

2 - 1

or 15 −

8 .

You can check this by entering 1 + 1 −

2 + 1 −

4 + 1 −

8 a calculator. The result is the same.

Rewrite each sum as a rational expression and simplify.

1. 9 + 3 + 1 + 1 −

3 + 1 −

9

2. 500 + 250 + 125 + 62 1 −

2

3. 6 + 1 + 1 −

6 + 1 −

36

4. 100 + 20 + 4 + 4 −

5

5. 1000 + 100 + 10 + 1 + 1 −

10 + 1 −

100 + 1 −

1000

6. 55 + 5 + 5 −

11 + 5 −

121

Enrichment 11-4


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Spreadsheet ActivityRevolutions per Minute

One of the characteristics that makes a spreadsheet powerful is the ability to recalculate values in formulas automatically. You can use this ability to investigate real-world situations.

Exercises

Use the spreadsheet of revolutions per minute.

1. How is the number of revolutions affected if the speed of a wheel of a given diameter is doubled?

2. Name two ways that you can double the RPM of a bicycle wheel.

Use a spreadsheet to investigate the effect of doubling the diameter of a tire on the number of revolutions the tire makes at a given speed.

Use dimensional analysis to find the formula for the revolutions per minute of a tire with diameter of x inches traveling at y miles per hour.

1 revolution −

π · x × y

−

1 × 1 −

60 minutes ×

63,360 −

1 =

1056y revolutions −−

π · x minutes

Step 1 Use Column A of the spreadsheet for diameter of the tire in inches. Use Column B for the speed in miles per hour.

Step 2 Column C contains the formula for the number of rotations per minute. Notice that in Excel, π is entered as PI( ).

Step 3 Choose values for the diameter and speed and study the results shown in the spreadsheet. To compare the revolutions per minute for doubled diameters, keep the speed constant and change the diameters. It appears that when the diameter is doubled, the number of revolutions per minute is halved.

A1

3456

2

B CDiameter (in.) Speed (mph) RPM

=(1056*B2)/(A2*PI())=(1056*B3)/(A3*PI())=(1056*B4)/(A4*PI())=(1056*B5)/(A5*PI())

Sheet 1 Sheet 2 Sheet 3

A1

3456

2

B CDiameter (in.) Speed (mph) RPM

18487.449243.724621.862310.93

1248

55555555

Sheet 1 Sheet 2 Sheet 3

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Example


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Study Guide and Intervention Dividing Polynomials

Divide Polynomials by Monomials To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

Find (4r2 - 12r) ÷ (2r).

(4r2 - 12r) ÷ 2r = 4r2 - 12r −

2r

= 4r2 −

2r - 12r −

2r Divide each term.

=

2r4r2

−

1

2r -

12r 6

−

2r 1

Simplify.

= 2r - 6 Simplify.

Find (3x2 - 8x + 4) ÷ (4x).

(3x2 - 8x + 4) ÷ 4x = 3x2 - 8x + 4 −

4x

= 3x2 −

4x - 8x −

4x + 4 −

4x

= 3x4 3x2

−

4x - 8x −

4x + 4 −

4x

= 3x −

4 - 2 + 1 − x


1. (x3 + 2x2 - x) ÷ x 2. (2x3 + 12x2 - 8x) ÷ (2x)

3. (x2 + 3x - 4) ÷ x 4. (4m2 + 6m - 8) ÷ (2m2)

5. (3x3 + 15x2 - 21x) ÷ (3x) 6. (8m2p2 + 4mp - 8p) ÷ p

7. (8y4 + 16y2 - 4) ÷ (4y2) 8. (16x4y2

+ 24xy + 5) ÷ (xy)

9. 15x2 - 25x + 30 −

5 10. 10a2b + 12ab - 8b −−

2a

11. 6x3 + 9x2 + 9 −

3x 12. m

2 - 12m + 42 −

3m2

13. m2p2 - 5mp + 6

−

m2p2 14.

p2 - 4pr + 6r2

− pr

15. 6a2b2 - 8ab + 12 −−

2a2 16.

2x2y3 - 4x2y2 - 8xy −−

2xy

17. 9x2y2z - 2xyz + 12x

−− xy 18. 2a3b3 + 8a2b2 - 10ab + 12 −−

2a2b2

11-5

Example 1 Example 2

2

1


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Dividing Polynomials

Divide Polynomials by Binomials To divide a polynomial by a binomial, factor the dividend if possible and divide both dividend and divisor by the GCF. If the polynomial cannot be factored, use long division.

Find (x2 + 7x + 10) ÷ (x + 3).

Step 1 Divide the first term of the dividend, x2 by the first term of the divisor, x. x

x + 3 � �� x2 + 7x + 10 (-) x2 + 3x Multiply x and x + 3.

4x Subtract.

Step 2 Bring down the next term, 10. Divide the first term of 4x + 10 by x. x + 4

x + 3 � �� x2 + 7x + 10 x2 + 3x 4x + 10 (-) 4x + 12 Multiply 4 and x + 3.

-2 Subtract.

The quotient is x + 4 with remainder -2. The quotient can be written as x + 4 + -2 −

x + 3 .


1. (b2 - 5b + 6) ÷ (b - 2) 2. (x2 - x - 6) ÷ (x - 3)

3. (x2 + 3x - 4) ÷ (x - 1) 4. (m2 + 2m - 8) ÷ (m + 4)

5. (x2 + 5x + 6) ÷ (x + 2) 6. (m2 + 4m + 4) ÷ (m + 2)

7. (2y2 + 5y + 2) ÷ ( y + 2) 8. (8y2 - 15y - 2) ÷ ( y - 2)

9. 8x2 - 6x - 9 −

4x + 3 10. m

2 - 5m - 6 −

m - 6

11. x3 + 1 −

x - 2 12. 6m3 + 11m2 + 4m + 35 −−

2m + 5

13. 6a2 + 7a + 5 −

2a + 5 14.

8p3 + 27 −

2p + 3

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Example


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Skills PracticeDividing Polynomials

Find each quotient.

1. (20x2 + 12x) ÷ 4x 2. (18n2 + 6n) ÷ 3n

3. (b2 - 12b + 5) ÷ 2b 4. (8r2 + 5r - 20) ÷ 4r

5. 12p3r2 + 18p2r - 6pr

−−

6p2r 6. 15k2u - 10ku + 25u2

−−

5ku

7. (x2 - 5x - 6) ÷ (x - 6) 8. (a2 - 10a + 16) ÷ (a - 2)

9. (n2 - n - 20) ÷ (n + 4) 10. ( y2 + 4y - 21) ÷ ( y - 3)

11. (h2 - 6h + 9) ÷ (h - 2) 12. (b2 + 5b - 2) ÷ (b + 6)

13. ( y2 + 6y + 1) ÷ ( y + 2) 14. (m2 - 2m - 5) ÷ (m - 3)

15. 2c2 - 5c - 3 −

2c + 1 16. 2r2 + 6r - 20 −

2r - 4

17. x3 - 3x2 - 6x - 20 −−

x - 5 18.

p3 - 4p2 + p + 6 −

p - 2

19. n3 - 6n - 2 −

n + 1 20.

y3 - y2 - 40 −

y - 4

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Practice Dividing Polynomials

Find each quotient.

1. (6q2 - 18q - 9) ÷ 9q 2. (y2 + 6y + 2) ÷ 3y 3. 12a2b - 3ab2 + 42ab −−

6a2b

4. 2m3p2 + 56mp - 4m2p3

−−

8m3p 5. (x2 - 3x - 40) ÷ (x + 5) 6. (3m2 - 20m + 12) ÷ (m - 6)

7. (a2 + 5a + 20) ÷ (a - 3) 8. (x2 - 3x - 2) ÷ (x + 7) 9. (t2 + 9t + 28) ÷ (t + 3)

10. (n2 - 9n + 25) ÷ (n - 4) 11. 6r2 - 5r - 56 −

3r + 8 12. 20w2 + 39w + 18 −−

5w + 6

13. (x3 + 2x2 - 16) ÷ (x - 2) 14. (t3 - 11t - 6) ÷ (t + 3)

15. x3 + 6x2 + 3x + 1 −−

x - 2 16. 6d3 + d2 - 2d + 17 −−

2d + 3

17. 2k3 + k2 - 12k + 11 −−

2k - 3 18.

9y3 - y - 1 −

3y + 2

19. LANDSCAPING Jocelyn is designing a bed for cactus specimens at a botanical garden. The total area can be modeled by the expression 2x2 + 7x + 3, where x is in feet.

a. Suppose in one design the length of the cactus bed is 4x, and in another, the length is 2x + 1. What are the widths of the two designs?

b. If x = 3 feet, what will be the dimensions of the cactus bed in each of the designs?

20. FURNITURE Teri is upholstering the seats of four chairs and a bench. She needs 1 −

4 square yard of fabric for each chair, and 1 −

2 square yard for the bench. If the fabric at

the store is 45 inches wide, how many yards of fabric will Teri need to cover the chairs and the bench if there is no waste?

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Word Problem Practice Dividing Polynomials

1. TECHNOLOGY The surface area in square millimeters of a rectangular computer microchip is represented by the expression x2 - 12x + 35, where x is the number of circuits. If the width of the chip is x - 5 millimeters, write a polynomial that represents the length.

2. HOMEWORK Your classmate Ava writes her answer to a homework problem on the chalkboard. She has

simplified 6x2 - 12x −

6 as x2 - 12x. Is this

correct? If not, what is the correct simplification?

3. CIVIL ENGINEERING Suppose 5400 tons of concrete costs (500 + d) dollars. Write a formula that gives the cost C of t tons

of concrete.

4. SHIPPING The Overseas Shipping Company loads cargo into a container to be shipped around the world. The volume of their shipping containers is determined by the following equation.

x3+ 21x2

+ 99x + 135

The container’s height is x + 3. Write an expression that represents the area of the base of the shipping container.

5. CIVIL ENGINEERING Greenshield’s Formula can be used to determine the amount of time a traffic light at an intersection should remain green.

G = 2.1n + 3.7

G = green time in seconds n = average number of vehicles traveling

in each lane per light cycle Write a simplified expression to

represent the average green light time per vehicle.

6. SOLID GEOMETRY The surface area of a right cylinder is given by the formula S = 2πr2 + 2πrh.

a. Write a simplified rational expression that represents the ratio of the surface area to the circumference of the cylinder.

b. Write a simplified rational expression that represents the ratio of the surface area to the area of the base.

11-5


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Enrichment

Synthetic Division

You can divide a polynomial such as 3x3 - 4x2 - 3x - 2 by a binomial such as x - 3 by a

process called synthetic division. Compare the process with long division in the following explanation.

Divide (3x 3 - 4x 2 - 3x - 2) by (x - 3) using synthetic division.

1. Show the coefficients of the terms in descending order.

2. The divisor is x - 3. Since 3 is to besubtracted, write 3 in the corner .

3. Bring down the first coefficient, 3.

4. Multiply. 3 . 3 = 9

5. Add. -4 + 9 = 5

6. Multiply. 3 . 5 = 15

7. Add. -3 + 15 = 12

8. Multiply. 3 . 12 = 36

9. Add. -2 + 36 = 34

Check Use long division.3x2 + 5x + 12

x - 3 � �� 3x3 - 4x2 - 3x - 2 3x3 - 9x2

5x2 - 3x 5x2 - 15x

12x - 212x - 36

34 The result is 3x2 + 5x + 12 + 34 −

x - 3 .

Divide by using synthetic division. Check your result using long division.

1. (x3 + 6x2 + 3x + 1) ÷ (x - 2) 2. (x3 - 3x2 - 6x - 20) ÷ (x - 5)

3. (2x3 - 5x + 1) ÷ (x + 1) 4. (3x3 - 7x2 + 4) ÷ (x - 2)

5. (x3 + 2x2 - x + 4) ÷ (x + 3) 6. (x3 + 4x2 - 3x - 11) ÷ (x - 4)

3 3 -4 -3 -2 9 15 36 3 5 12 34

3x2 + 5x + 12, remainder 34

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Example

017_035_ALG1_A_CRM_C11_CR_660285.indd 35017_035_ALG1_A_CRM_C11_CR_660285.indd 35 12/23/10 11:30 AM12/23/10 11:30 AM

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11-5 TI-Nspire® ActivityDividing Polynomials

You can use a graphing calculator with a computer algebra system (CAS) to divide polynomials with any divisor. One function is used to find the quotient, while another function is used to find the remainder.

Use CAS to find ( x 4 + x 3 – 3 x 2 + x) ÷ ( x 2 + 4x + 5).

Step 1 Add a new Calculator page on the TI-Nspire.

Step 2 From the menu, select Algebra, Polynomial Tools and Quotient of Polynomial.

Step 3 Type the dividend, a comma, and the divisor.

The CAS indicates that ( x 4 + x 3 – 3 x 2 + x) ÷ ( x 2 + 4x + 5) is x 2 − 3x + 4. We need to determine whether there is a remainder.

Step 4 Use the Remainder of a Polynomial option from the Algebra, Polynomial Tools menu to determine the remainder. Then type the dividend, a comma, and the divisor.

The remainder is −20. Therefore, ( x 4 + x 3 − 3 x 2 + x) ÷ ( x 2 + 4x + 5) is x 2 − 3x + 4 − 20 −

x 2 + 4x + 5.

Check Use the Expand option from the Algebra menu to confirm your answer. Type the quotient multiplied by the sum of the divisor and the remainder over the divisor. The CAS confirms that the division was correct.

Exercises

Find each quotient.

1. ( x 4 + 2 x 3 – 19 x 2 + 28x – 12) ÷ ( x 2 + 5x – 6) 2. (2 x 4 + 3 x 2 + x + 9) ÷ ( x 2 − 2x + 3)

3. (3 x 4 − 20 x 3 + 58 x 2 − 30x + 8) ÷ (3 x 2 − 2x + 1) 4. (4 x 4 − 7 x 3 + 29 x 2 − 3x − 40) ÷ (4 x 2 + x + 5)

5. (2 x 5 + 15 x 4 + 10 x 3 + 4 x 2 + 15x + 16) ÷ ( x 2 + 3x + 5)

6. ( x 5 + 4 x 4 − 5 x 3 + 12 x 2 + 40x + 60) ÷ ( x 3 + x + 8)

7. Think About It When a polynomial is divided by ( x 2 + 3x + 5), the quotient is ( x 2 + 6x − 8) with a remainder of 4x + 10. What is the polynomial?

Example


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11-6 Study Guide and InterventionAdding and Subtracting Rational Expressions

Add and Subtract Rational Expressions with Like Denominators To add rational expressions with like denominators, add the numerators and then write the sum over the common denominator. To subtract fractions with like denominators, subtract the numerators. If possible, simplify the resulting rational expression.

Find 5n −

15 + 7n −

15 .

5n −

15 + 7n −

15 = 5n + 7n −

15 Add the numerators.

= 12n −

15 Simplify.

= 12n −

15

5 4n

Divide by 3.

= 4n −

5 Simplify.

Find 3x + 2 −

x - 2 - 4x −

x - 2 .

3x + 2 −

x - 2 - 4x −

x - 2 = 3x + 2 - 4x −

x - 2

The common

denominator is x - 2.

= 2 - x −

x - 2 Subtract.

= -1(x - 2) −

x - 2 2 - x = -1(x - 2)

= -1(x - 2) −

x - 2

=

-1 −

1 Simplify.

= -1

ExercisesFind each sum or difference.

1. 3 − a + 4 − a 2. x2 −

8 + x −

8

3. 5x −

9 - x −

9 4. 11x −

15y - x −

15y

5. 2a - 4 −

a - 4 + -a −

a - 4 6. m + 1 −

2m - 1 + 3m - 3 −

2m - 1

7. y + 7

−

y + 6 - 1 −

y + 6 8.

3y + 5 −

5 -

2y −

5

9. x + 1 −

x - 2 + x - 5 −

x - 2 10. 5a −

3b2 + 10a −

3b2

11. x2 + x − x - x

2 + 5x − x 12. 5a + 2 −

a2 - 4a + 2 −

a2

13. 3x + 2 −

x + 2 + x + 6 −

x + 2 14. a - 4 −

a + 1 + a + 6 −

a + 1

Example 1 Example 2

1

1


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Adding and Subtracting Rational Expressions

Add and Subtract Rational Expressions with Unlike Denominators Adding or subtracting rational expressions with unlike denominators is similar to adding and subtracting fractions with unlike denominators.

Adding and Subtracting

Rational Expressions

Step 1 Find the LCD of the expressions.

Step 2 Change each expression into an equivalent expression with the LCD as

the denominator.

Step 3 Add or subtract just as with expressions with like denominators.

Step 4 Simplify if necessary.

Find n + 3 − n + 8n - 4 −

4n .

Factor each denominator. n = n 4n = 4 . n LCD = 4nSince the denominator of 8n - 4 −

4n is already

4n, only n + 3 − n needs to be renamed.

n + 3 − n + 8n - 4 −

4n = 4(n + 3)

−

4n + 8n - 4 −

4n

= 4n + 12 −

4n + 8n - 4 −

4n

= 12n + 8 −

4n

= 3n + 2 − n

Find 3x −

x2 - 4x - 1 −

x - 4 .

3x −

x2 - 4x - 1 −

x - 4 = 3x −

x(x - 4) - 1 −

x - 4

= 3x −

x(x - 4) - 1 −

x - 4 · x − x

= 3x −

x(x - 4) - x −

x(x - 4)

= 2x −

x(x - 4)

= 2 −

x - 4

ExercisesFind each sum or difference.

1. 1 − a + 7 −

3a 2. 1 −

6x + 3 −

8

3. 5 −

9x - 1 −

x2 4. 6 −

x2 - 3 −

x3

5. 8 −

4a2 + 6 −

3a 6. 4 −

h + 1 + 2 −

h + 2

7. y −

y - 3 - 3 −

y + 3 8.

y −

y - 7 -

y + 3 −

y2 - 4y - 21

9. a −

a + 4 + 4 −

a - 4 10. 6 −

3(m + 1) + 2 −

3(m - 1)

11. 4 −

x - 2y - 2 −

x + 2y 12. a - 6b

−

2a2 - 5ab + 2b2 - 7 −

a - 2b

13. y + 2 −

y2 + 5y + 6 +

2 - y −

y2 + y - 6 14.

q −

q2 - 16 +

q + 1 −

q2 + 5q + 4

Example 1 Example 2

Factor the

denominator.

The LCD is

x(x - 4).

1 · x = x

Subtract

numerators.

Simplify.


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11-6 Skills PracticeAdding and Subtracting Rational Expressions

Find each sum or difference.

1. 2y

−

5 +

y −

5 2. 4r −

9 + 5r −

9

3. t + 3 −

7 - t −

7 4. c + 8 −

4 - c + 6 −

4

5. x + 2 −

3 + x + 5 −

3 6.

g + 2 −

4 +

g - 8 −

4

7. x −

x - 1 - 1 −

x - 1 8. 3r −

r + 3 - r −

r + 3

Find the LCM of each pair of polynomials.

9. 4x2y, 12xy2 10. n + 2, n - 3

11. 2r - 1, r + 4 12. t + 4, 4t + 16


13. 5 −

4r - 2 −

r2 14. 5x −

3y2 - 2x −

9y

15. x −

x + 2 - 4 −

x - 1 16. d - 1 −

d - 2 - 3 −

d + 5

17. b −

b - 1 + 2 −

b - 4 18. k −

k - 5 + k - 1 −

k + 5

19. 3x + 15 −

x2 - 25 + x −

x + 5 20. x - 3 −

x2 - 4x + 4 + x + 2 −

x - 2


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11-6 Practice Adding and Subtracting Rational Expressions


1. n −

8 + 3n −

8 2. 7u −

16 + 5u −

16 3. w + 9 −

9 + w + 4 −

9

4. x - 6 −

2 - x - 7 −

2 5. n + 14 −

5 - n - 14 −

5 6. 6 −

c - 1 - -2 −

c - 1

7. x - 5 −

x + 2 + -2 −

x + 2 8. r + 5 −

r - 5 + 2r - 1 −

r - 5 9.

4p + 14 −

p + 4 +

2p + 10 −

p + 4

Find the LCM of each pair of polynomials.

10. 3a3b2, 18ab3 11. w - 4, w + 2 12. 5d - 20, d - 4

13. 6p + 1, p - 1 14. x2 + 5x + 4, (x + 1)2 15. m2 + 3m - 10, m2 - 4


16. 6p

−

5x2 -

2p −

3x 17. m + 4 −

m - 3 - 2 −

m - 6

18. y + 3

−

y2 - 16 +

3y - 2 −

y2 + 8y + 16 19.

p + 1 −

p2 + 3p - 4 +

p −

p + 4

20. t + 3 −

t2 - 3t - 10 - 4t - 8 −

t2 - 10t + 25 21.

4y −

y2 - y - 6 -

3y + 3 −

y2 - 4

22. SERVICE Members of the ninth grade class at Pine Ridge High School are organizing into service groups. What is the minimum number of students who must participate for all students to be divided into groups of 4, 6, or 9 students with no one left out?

23. GEOMETRY Find an expression for the perimeter of rectangle 5a + 4b2a + b

BA

CD

3a + 2b2a + b

ABCD. Use the formula P = 2� + 2w.


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11-6 Word Problem Practice Rational Expressions with Unlike Denominators

1. TEXAS Of the 254 counties in Texas, 4 are larger than 6000 square miles. Another 21 counties are smaller than 300 square miles. What fraction of the counties are 300 to 6000 square miles in size?

2. SWIMMING Power Pools installs swimming pools. To determine the appropriate size of pool for a yard, they measure the length of the yard in meters and call that value x. The length and width of the pool are calculated with the diagram below. Write an expression in simplest form for the perimeter of a rectangular pool for the given variable dimensions.

m x 4

m 2x 5

3. EGYPTIAN FRACTIONS Ancient Egyptians used only unit fractions, which are fractions in the form 1 − n . Their mathematical notation only allowed for a numerator of 1. When they needed to express a fraction with a numerator other than 1, they wrote it as a sum of unit fractions. An example is shown below. 5 −

6 = 1 −

3 + 1 −

2

Simplify the following expression so it is a sum of unit fractions.

5x + 6 −

10x2 + 12x + 2x −

8x2

4. INSURANCE For a hospital stay, Paul’s health insurance plan requires him to pay 2−

5 the cost of the first day in the

hospital and 1−5

the cost of the second and

third days. If Paul’s hospital stay is 3 days and cost him $420, what was the full daily cost?

5. PACKAGE DELIVERY Fredricksburg Parcel Express delivered a total of 498 packages on Monday, Tuesday, and Wednesday. On Tuesday, they delivered 7 less than 2 times the number of packages delivered on Monday. On Wednesday, they delivered the average number delivered on Monday and Tuesday.

a. Write a rational equation that represents the sum of the numbers of packages delivered on Monday, Tuesday, and Wednesday.

b. How many packages were delivered on Monday?


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11-6 Enrichment

Sum and Difference of Any Two Like PowersThe sum of any two like powers can be written an + bn, where n is a positive integer. The difference of like powers is an - bn. Under what conditions are these expressions exactly divisible by (a + b) or (a - b)? The answer depends on whether n is an odd or even number.

Use long division to find the following quotients. (Hint: Write a3 + b3 as a3 + 0a2 + 0a + b3.) Is the numerator exactly divisible by the denominator? Write yes or no.

1. a3 + b3

−

a + b 2. a

3 + b3

−

a - b 3. a

3 - b3 −

a + b 4. a

3 - b3 −

a - b

5. a4 + b4

−

a + b 6. a

4 + b4

−

a - b 7. a

4 - b4 −

a + b 8. a

4 - b4 −

a - b

9. a5 + b5

−

a + b 10. a

5 + b5

−

a - b 11. a

5 - b5 −

a + b 12. a

5 - b5 −

a - b

13. Use the words odd and even to complete these two statements.

a. an + bn is divisible by a + b if n is , and by neither a + b nor a - b if n is .

b. an - bn is divisible by a - b if n is , and by both a + b and a - b if n is .

14. Describe the signs of the terms of the quotients when the divisor is a - b.

15. Describe the signs of the terms of the quotient when the divisor is a + b.


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11-7 Study Guide and InterventionMixed Expressions and Complex Fractions

Simplify Mixed Expressions Algebraic expressions such as a + b − c and 5 + x + y

−

x + 3 are

called mixed expressions. Changing mixed expressions to rational expressions is similar to changing mixed numbers to improper fractions.

Simplify 5 + 2 − n .

5 + 2 − n = 5 · n − n + 2 − n LCD is n.

= 5n + 2 − n Add the numerators.

Therefore, 5 + 2 − n = 5n + 2 − n .

Simplify 2 + 3 −

n + 3 .

2 + 3 −

n + 3 = 2(n + 3)

−

n + 3 + 3 −

n + 3

= 2n + 6 −

n + 3 + 3 −

n + 3

= 2n + 6 + 3 −

n + 3

= 2n + 9 −

n + 3

Therefore, 2 + 3 −

n + 3 = 2n + 9 −

n + 3 .

ExercisesWrite each mixed expression as a rational expression.

1. 4 + 6 − a 2. 1 −

9x - 3

3. 3x - 1 −

x2 4. 4 −

x2 - 2

5. 10 + 60 −

x + 5 6. h −

h + 4 + 2

7. y −

y - 2 + y2 8. 4 - 4 −

2x + 1

9. 1 + 1 − x 10. 4 −

m - 2 - 2m

11. x2 + x + 2 −

x - 3 12. a - 3 + a - 2 −

a + 3

13. 4m + 3p

−

2t 14. 2q2 +

q − p + q

15. 2 −

y2 - 1 - 4y2 16. t2 +

p + t − p - t

Example 1 Example 2


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Mixed Expressions and Complex Fractions

Simplify Complex Fractions If a fraction has one or more fractions in the numerator or denominator, it is called a complex fraction.

Simplifying a Complex Fraction

Any complex fraction a −

b −

c −

d where b ≠ 0, c ≠ 0, and d ≠ 0, can be expressed as ad −

bc .

Simplify 2 + 4 − a

−

a + 2 −

3

.

2 + 4 − a

−

a + 2 −

3

=

2a − a + 4 − a −

a + 2 −

3

Find the LCD for the numerator and rewrite as like fractions.

=

2a + 4 − a −

a + 2 −

3

Simplify the numerator.

= 2a + 4 − a · 3 −

a + 2 Rewrite as the product of the numerator and the reciprocal of the denominator.

= 2(a + 2) − a · 3 −

a + 2 Factor.

= 6 − a Divide and simplify.

Exercises

Simplify each expression.

1. 2 2 −

5 −

3 3 −

4 2.

3 − x −

4 − y 3.

x −

y3 −

x3 −

y2

4. 1 - 1 − x

−

1 + 1 − x 5.

1 - 1 − x −

1 - 1 −

x2 6.

1 −

x - 3 −

2 −

x2 - 9

7. x

2 - 25 − y −

x3 - 5x2 8.

x - 12 −

x - 1 −

x - 8 −

x - 2 9.

3 −

y + 2 - 2 −

y - 2 −

1 −

y + 2 - 2 −

y - 2

Example


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11-7 Skills PracticeMixed Expressions and Complex Fractions

Write each mixed expression as a rational expression.

1. 6 + 4 −

h 2. 7 + 6 − p

3. 4b + b − c 4. 8q - 2q

− r

5. 2 + 4 −

d - 5 6. 5 - 6 −

f + 2

7. b2 + 12 −

b + 3 8. m - 6 −

m - 7

9. 2a + a - 2 − a 10. 4r - r + 9 −

2r


11. 2 1 −

2 −

4 3 −

4 12.

3 2 −

3 −

5 2 −

5 13.

r −

n2 −

r2 − n

14. a

2 −

b3 −

a −

b 15.

x2y

− c −

xy3

−

c2 16.

r - 2 −

r + 3 −

r - 2 −

3

17. w + 4 − w

−

w2 - 16 − w

18. x

2 - 1 − x −

x - 1 −

x2

19. b2 - 4 −

b2 + 7b + 10 −

b - 2

20. k

2 + 5k + 6 −

k2 - 9 −

k + 2 21.

g + 12 −

g + 8 −

g + 6 22.

p + 9 −

p - 6 −

p - 3


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11-7 PracticeMixed Expressions and Complex Fractions

Write each mixed expression as a rational expression.

1. 14 - 9 − u 2. 7d + 4d − c 3. 3n + 6 - n − n

4. 5b - b + 3 −

2b 5. 3 + t + 5 −

t2 - 1 6. 2a + a - 1 −

a + 1

7. 2p + p + 1

−

p - 3 8. 4n2 + n - 1 −

n2 - 1 9. (t + 1) + 4 −

t + 5


10. 3 2 −

5 −

2 5 −

6 11.

m2 −

6p −

3m −

p2 12.

x2 - y2

−

x2 −

x + y

−

3x

13. a - 4 −

a2 −

a2 - 16 − a

14. q2 - 7q + 12

−

q2 - 16 −

q - 3 15.

k2 + 6k −

k2 + 4k - 5 −

k - 8 −

k2 - 9k + 8

16. b

2 + b - 12 −

b2 + 3b - 4 −

b - 3 −

b2 - b 17.

g - 10 −

g + 9 −

g - 5 −

g + 4 18.

y + 6 −

y - 7 −

y - 7 −

y + 6

19. TRAVEL Ray and Jan are on a 12 1 −

2 -hour drive from Springfield, Missouri, to Chicago,

Illinois. They stop for a break every 3 1 −

4 hours.

a. Write an expression to model this situation.

b. How many stops will Ray and Jan make before arriving in Chicago?

20. CARPENTRY Tai needs several 2 1 −

4 -inch wooden rods to reinforce the frame on a futon.

She can cut the rods from a 24 1 −

2 -inch dowel purchased from a hardware store. How many

wooden rods can she cut from the dowel?


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11-7

1. CYCLING Natalie rode in a bicycle event

for charity on Saturday. It took her 2−3

of an hour to complete the 18-mile race.What was her average speed in miles per hour?

2. QUILTING Mrs. Tantora sews and sells

Amish baby quilts. She bought 42 3 −

4 yards

of backing fabric, and 2 1 −

4 yards are

needed for each quilt she sews. How many quilts can she make with the backing fabric she bought?

3. TRAVEL The Franz family traveled from Galveston to Waco for a family reunion. Driving their minivan, they averaged 30 miles per hour on the way to Waco and 45 miles per hour on the return trip home to Galveston. What is their average rate for the entire trip? (Hint: Remember that average rate equals total distance divided by total time and that time can be represented as a ratio of distance x to rate.)

4. PHYSICAL SCIENCE The volume of a gas varies directly as the Kelvin temperature T and inversely as the pressure P, where k is the constant of variation.

V = k ( T −

P )

If k = 13 −

157 , find the volume in liters of

helium gas at 273 degrees Kelvin and13 −

3 atmospheres of pressure. Round your

answer to the nearest hundredth.

5. SAFETY The Occupational Safety and Health Administration provides safety standards in the workplace to keep workers free from dangerous working conditions. OSHA recommends that for general construction there be 5 foot-candles of illumination in which to work. A foreman using a light meter finds that the illumination of a construction light on a surface 8 feet from the source is 11 foot-candles. The illumination produced by a light source varies inversely as the square of the distance from the source.

I is illumination (in foot-candles). I = k −

d2 d is the distance from the source

(in feet).k is a constant.

a. Find the illumination of the same

light at a distance of 15 3 −

4 feet. Round

your answer to the nearest hundredth.

b. Is there enough illumination at this distance to meet OSHA requirements for lighting?

c. In order to comply with OSHA, what is the maximum allowable working distance from this light source? Round your decimal answer to nearest tenth.

Word Problem Practice Mixed Expressions and Complex Fractions


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11-7 Enrichment

Continued Fractions

Continued fractions are a special type of complex fraction. Each fraction in a continued fraction has a numerator of 1.

1 + 1 −

2 + 1 −

3 + 1 −

4

Evaluate the continued fraction above. Start at the bottom and work your way up.

Step 1 3 + 1 −

4 =

12 −

4 + 1 −

4 = 13 −

4

Step 2 1 −

3 + 1 −

4 = 4 −

13

Step 3 2 + 4 −

13 = 26 −

13 + 4 −

13 = 30 −

13

Step 4 1 −

30 −

13 = 13 −

30

Step 5 1 + 13 −

30 = 1 13 −

30

Change 13 −

7 into a

continued fraction.

Step 1 13 −

7 = 7 −

7 + 6 −

7 = 1 + 6 −

7

Step 2 6 −

7 = 1 −

7 −

6

Step 3 7 −

6 = 1 + 1 −

6

Stop, because the numerator is 1.

Thus, 13 −

7 can be written as 1 + 1 −

1 + 1 −

6 .

Example 1 Example 2

Evaluate each continued fraction.

1. 0 + 1 −

2 + 1 −

4 + 1 −

6 + 1 −

8 2. 0 + 1 −

1 + 1 −

2 + 1 −

6

3. 3 + 1 −

1 + 1 −

1 + 1 −

1 + 1 −

2 4. 1 +

1 −

3 + 1 −

5 + 1 −

7

Change each fraction into a continued fraction.

5. 71 −

26

6. 9 −

56 7. 4626 −

2065


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11-8 Study Guide and InterventionRational Equations

Solve Rational Equations Rational equations are equations that contain rational expressions. To solve equations containing rational expressions, multiply each side of the equation by the least common denominator.Rational equations can be used to solve work problems and rate problems.

ExercisesSolve each equation. State any extraneous solutions.

1. x - 5 −

5 + x −

4 = 8 2. 3 − x = 6 −

x + 1 3. x - 1 −

5 = 2x - 2 −

15

4. 8 −

n - 1 = 10 −

n + 1 5. t - 4 −

t + 3 = t + 3 6. m + 4 − m + m −

3 = m −

3

7. q + 4

−

q - 1 +

q −

q + 1 = 2 8. 5 - 2x −

2 - 4x + 3 −

6 = 7x + 2 −

6

9. m + 1 −

m - 1 - m −

1 - m = 1 10. x 2 - 9 −

x - 3 + x2 = 9

11. 2 −

x 2 - 36 - 1 −

x - 6 = 0 12. 4z −

z 2 + 4z + 3 = 6 −

z + 3 + 4 −

z + 1

13. 4 −

4 - p -

p2

−

p - 4 = 4 14. x

2 - 16 −

x - 4 + x 2 = 16

Solve x - 3 −

3 + x −

2 = 4.

x - 3 −

3 + x −

2 = 4

6 ( x - 3 −

3 + x −

2 ) = 6(4) The LCD is 6.

2(x - 3) + 3x = 24 Distributive Property

2x - 6 + 3x = 24 Distributive Property

5x = 30 Simplify.

x = 6 Divide each side by 5.

The solution is 6.

Solve 15 −

x 2 - 1 = 5 −

2(x - 1) . State

any extraneous solutions.

15 −

x 2 - 1 = 5 −

2(x - 1) Original equation

30(x - 1) = 5(x2 - 1) Cross multiply.

30x - 30 = 5x2 - 5 Distributive Property

0 = 5x2 - 30x + 30 - 5 Add -30x + 30 to

each side.

0 = 5x2 - 30x + 25 Simplify.

0 = 5(x2 - 6x + 5) Factor.

0 = 5(x - 1)(x - 5) Factor.

x = 1 or x = 5 Zero Product Property

The number 1 is an extraneous solution, since 1 is an excluded value for x. So, 5 is the solution of the equation.

Example 1 Example 2


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Rational Equations

Use Rational Equations to Solve Problems Rational equation can be used to solve work problems and rate problems.

WORK PROBLEM Marla can paint Percy’s kitchen in 3 hours. Percy can paint it in 2 hours. Working together, how long will it take Marla and Percy to paint the kitchen?

In t hours, Marla completes t · 1 −

3 of the job and Percy completes t · 1 −

2 of the job. So an

equation for completing the whole job is t −

3 + t −

2 = 1.

t −

3 + t −

2 = 1

2t + 3t = 6 Multiply each term by 6.

5t = 6 Add like terms.

t = 6 −

5 Solve.

So it will take Marla and Percy 1 1 −

5 hours to paint the room if they work together.

Example

Exercises1. GREETING CARDS It takes Kenesha 45 minutes to prepare 20 greeting cards. It takes

Paula 30 minutes to prepare the same number of cards. Working together at this rate, how long will it take them to prepare the cards?

2. BOATING A motorboat went upstream at 15 miles per hour and returned downstream at 20 miles per hour. How far did the boat travel one way if the round trip took 3.5 hours?

3. FLOORING Maya and Reginald are installing hardwood flooring. Maya can install flooring in a room in 4 hours. Reginald can install flooring in a room in 3 hours. How long would it take them if they worked together?

4. BICYCLING Stefan is bicyling on a bike trail at an average of 10 miles per hour. Erik starts bicycling on the same trail 30 minutes later. If Erik averages 16 miles per hour, how long will it take him to pass Stefan?


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11-8 Skills PracticeRational Equations

Solve each equation. State any extraneous solutions.

1. 5

−

c = 2 −

c + 3 2. 3 − q =

5 −

q + 4

3. 7 −

m + 1 = 12 −

m + 2 4. 3 −

x + 2 = 5 −

x + 8

5. y −

y - 2 =

y + 1 −

y - 5 6. b - 2 −

b = b + 4 −

b + 2

7. 3m −

2 - 1 −

4 = 10m −

8 8.

7g −

9 + 1 −

3 =

5g −

6

9. 2a + 5 −

6 - 2a −

3 = -

1 −

2 10. n - 3 −

10 + n - 5 −

5 = 1 −

2

11. c + 2 − c + c + 3 − c = 7 12. 3b - 4 −

b - b - 7 −

b = 1

13. m - 4 − m -

m – 11 −

m + 4 = 1 − m 14.

f + 2 −

f -

f + 1 −

f + 5 = 1 −

f

15. r + 3 −

r - 1 - r −

r - 3 = 0 16. u + 1 −

u - 2 - u −

u + 1 = 0

17. - 2 −

x + 1 + 2 − x = 1 18. 5 −

m – 4 - m −

2m – 8 = 1

19. ACTIVISM Maury and Tyra are making phone calls to state representatives’ offices to lobby for an issue. Maury can call all 120 state representatives in 10 hours. Tyra can call all 120 state representatives in 8 hours. How long would it take them to call all 120 state representatives together?


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11-8 Practice Rational Equations


1. 5 −

n + 2 = 7 −

n + 6 2. x −

x - 5 = x + 4 −

x - 6 3. k + 5 −

k = k - 1 −

k + 9

4. 2h −

h - 1 = 2h + 1 −

h + 2 5.

4y −

3 + 1 −

2 =

5y −

6 6.

y - 2 −

4 -

y + 2 −

5 = -1

7. 2q - 1

−

6 -

q −

3 =

q + 4 −

18 8. 5 −

p - 1 - 3 −

p + 2 = 0 9. 3t −

3t - 3 - 1 −

9t + 3 = 1

10. 4x −

2x + 1 - 2x −

2x + 3 = 1 11. d - 3 −

d - d - 4 −

d - 2 = 1 −

d 12.

3y - 2 −

y - 2 +

y 2 −

2 - y = -3

13. 2 −

m + 2 - m + 2 −

m - 2 = 7 −

3 14. n + 2 − n + n + 5 −

n + 3 = -

1 − n 15. 1 −

z + 1 - 6 - z −

6z = 0

16. 2p −

p - 2 +

p + 2 −

p 2 - 4 = 1 17. x + 7 −

x 2 - 9 - x −

x + 3 = 1 18. 2n −

n - 4 - n + 6 −

n 2 - 16 = 1

19. PUBLISHING Tracey and Alan publish a 10-page independent newspaper once a month. At production, Alan usually spends 6 hours on the layout of the paper. When Tracey helps, layout takes 3 hours and 20 minutes.

a. Write an equation that could be used to determine how long it would take Tracey to do the layout by herself.

b. How long would it take Tracey to do the job alone?

20. TRAVEL Emilio made arrangements to have Lynda pick him up from an auto repair shop after he dropped his car off. He called Lynda to tell her he would start walking and to look for him on the way. Emilio and Lynda live 10 miles from the auto shop. It takes Emilio 2 1 −

4 hours to walk the distance and Lynda 15 minutes to drive the distance.

a. If Emilio and Lynda leave at the same time, when should Lynda expect to spot Emilio on the road?

b. How far will Emilio have walked when Lynda picks him up?


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11-8 Word Problem PracticeRational Equations

1. ELECTRICITY The current in a simple electric circuit varies inversely as the resistance. If the current is 20 amps when the resistance is 5 ohms, find the current when the resistance is 8 ohms.

2. MASONRY Sam and Belai are masons who are working to build a stone wall that will be 120 feet long. Sam works from one end and is able to build one ten-foot section in 5 hours. Belai works from the other end and is able to finish a ten-foot section in 4 hours. How long will it take Sam and Belai to finish building the wall?

3. NUMBERS The formula to find the sum of the first n whole numbers is

sum = n2 + n −

2 . In order to encourage

students to show up early to a school dance, the dance committee decides to charge less for those who come to the dance early. Their plan is to charge the first student to arrive 1 penny. The second student through the door is charged 2 pennies; the third student through the door is charged 3 pennies, and so on. How much money, in total, would paid by the first 150 students?

4. NAUTICAL A ferry captain keeps track of the progress of his ship in the ship’s log. One day, he records the following entry.

With the recent spring snow melt, the current is running strong today. The six-mile trip downstream to Whyte’s landing was very quick. However, we only covered two miles in the same amount of time when we headed back upstream.

Write a rational equation using b for the speed of the boat and c for the speed of the stream and solve for b in terms of c.

5. HEALTH CARE The total number of Americans waiting for kidney and heart transplants is approximately 66,500. The ratio of those awaiting a kidney transplant to those awaiting a heart transplant is about 20 to 1.

a. How many people are on each of the waiting lists? Round your answers to the nearest hundred.

b. These two groups make up about 3 −

4 of

the transplant candidates for all organs. About how many organ transplant candidates are there altogether? Round your answer to the nearest thousand.


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11-8 Enrichment

Winning DistancesIn 1999, Hicham El Guerrouj set a world record for the mile run with a time of 3:43.13 (3 min 43.13 s). In 1954, Roger Bannister ran the first mile under 4 minutes at 3:59.4. Had they run those times in the same race, how far in front of Bannister would El Guerrouj have been at the finish?Use d − t = r. Since 3 min 43.13 s = 223.13 s, and 3 min 59.4 s = 239.4 s,

El Guerrouj’s rate was 5280 ft −

223.13 s and Bannister’s rate was 5280 ft −

239.4 s .

r t d

El Guerrouj 5280 −

223.13 223.13 5280 feet

Bannister 5280 −

239.4 223.13 5280

−

239.4 . 223.13 or 4921.2 feet

Therefore, when El Guerrouj hit the tape, he would be 5280 - 4921.2, or 358.8 feet, ahead of Bannister. Let’s see whether we can develop a formula for this type of problem. Let D = the distance raced, W = the winner’s time, and L = the loser’s time.Following the same pattern, you obtain the results shown in the table at the right.The winning distance will be D - DW −

L .

1. Show that the expression for the winning distance is equivalent to D(L - W) −

L .

Use the formula winning distance = D(L - W) −

L to find the winning

distance to the nearest tenth for each of the following Olympic races.

2. women’s 400 meter relay: East Germany 41.6 s (1980);Canada 48.4 s (1928)

3. men’s 200 meter freestyle swimming: Michael Phelps 1 min 42.96 s (2008); Yevgeny Sadovyi 1 min 46.70 s (1992)

4. men’s 50,000 meter walk: Alex Schwazer 3 h 37 min 9 s (2008); Hartwig Gauter 3 h 49 min 24 s (1980)

5. women’s 400 meter freestyle swimming relay: Netherlands 3 min 33.76 s (2008); United States 3 min 39.29 s (1996)

r t d

Winner D −

W W D −

W · W = D

Loser D −

L W D −

L · W = DW −

L


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11 Student Recording SheetUse this recording sheet with pages 742–743 of the Student Edition.

7a. ————————

7b. ————————

7c. ————————

8. ———————— (grid in)

9a. 9b.

10. ————————

11. ————————

8.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

Record your answers for Question 12 on the back of this paper.

Extended Response

y

x

————————

Read each question. Then fill in the correct answer.

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

Multiple Choice

Short Response/Gridded Response

Record your answer in the blank.

For gridded response questions, also enter your answer in the grid by writing each number or symbol in a box. Then fill in the corresponding circle for that number or symbol.

x y123456

055-065_ALG1_A_CRM_C11_AS_660285.indd 55055-065_ALG1_A_CRM_C11_AS_660285.indd 55 12/20/10 10:25 PM12/20/10 10:25 PM

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11

General Scoring Guidelines

• If a student gives only a correct numerical answer to a problem but does not show how he or she arrived at the answer, the student will be awarded only 1 credit. All extended response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for all parts of the question. For example, if a question has three parts, the correct response to one or two parts of the question that required work to be shown is not considered a fully correct response.

• Students who use trial and error to solve a problem must show their method. Merely showing that the answer checks or is correct is not considered a complete response for full credit.

Exercise 12 Rubric

Score Specific Criteria

4 A correct solution that is supported by well-developed, accurate explanations. The sketch in part a should be clearly labeled with width w, height w - 1, and length w + 2. In part b, students should simplify the expression w(w - 1)(w + 2) as w3

+ w2 - 2w. In part c,

the student will set the expression obtained in part b equal to 30 and solve for w to get w = 3 ft. This value is then used to find the length (5 ft) and the height (2 ft).

3 A generally correct solution, but may contain minor flaws in reasoning or computation.

2 A partially correct interpretation and/or solution to the problem.1 A correct solution with no evidence or explanation.0 An incorrect solution indicating no mathematical understanding of

the concept or task, or no solution is given.

Rubric for Scoring Extended Response


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Chapter 11 Quiz 2 (Lessons 11-3 and 11-4)

11

PDF Pass


11 SCORE

1. MULTIPLE CHOICE Which equation does not represent an inverse variation?

A y = 3 − x B 4xy = 12 C x = 5 − y D x = 2y

2. Graph an inverse variation in which y varies inversely as x and y = 2 when x = 7.

3. Write an inverse variation equation that relates x and y if y = 6 when x = -3. Then find y when x = -9.

4. State the excluded value for the function y =

5 −

x + 2 .

5. Identify the asymptotes of y = 2 −

x + 1 + 2 .


1. MULTIPLE CHOICE Simplify 5x - 10−

x 2 - 4x + 4 .

State the excluded value of x.

A 5 −

x - 2 ; 2 B 5 − x - 2; -2 C 5 −

x - 4 ; 4 D 5 −

x + 4 ; -4

Find each product or quotient.

2. 10 b 2 −

4 c 5 · 32 c 4 −

25 b 2 3. 4r2

− q · 3q2

−

8r4

4. x 2 - x - 12 −

x 2 - 3x - 4 · x 2 - 1 −

x 2 + 2x - 3

5. xy 2

−

wz 4 ÷

x 4 y −

w 3 z 6. 8k4

−

j2 ÷ 2k4

−

5j3

7. m 2 + 3m + 2 −

m 3 ÷ m 2 + m - 2 −

m 2

8. b 2 + 7b + 6 −

b 2 + b - 2 ÷ b 2 - 36 −

b 2 - 7b + 6

Find the roots of each function.

9. f(x) =

x2 + 2x -8 −

x - 2 10. f(x) =

x2 - 8x + 7 −

x2 + x - 2

11

1.

2. y

x

O

10

10 20-10-20

20

-10

-20

3.

4.

5.

1.

2.

3.

4.

5.

6.

7.

8.

9. 10.

SCORE


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PDF 2nd


11 SCORE Chapter 11 Quiz 3(Lessons 11-5 and 11-6)

Find each quotient.

1. (x2 + 6x - 5) ÷ (2x) 2. (5x2 + 33x - 14) ÷ (x + 7)

3. (6x2 - x - 12) ÷ (2x - 3)

Find the LCM for each pair of polynomials.

4. 8rt, 6r2t 5. x2 + 10x + 21, (x + 3)2


6. 12 −

n 2 - 16 + 3 −

4 - n 7. x + 16 −

x - 5 + 3x - 2 −

x - 5

8. 11a −

6 - 7a −

6 9. 4x −

x - 3 - 7 −

3 - x

10. MULTIPLE CHOICE Find m - 3 −

m 2 + 6m + 9 - m - 2 −

m 2 - 9 .

A -5m + 10 −

(m + 3)(m - 3) C 5m + 10 −

(m + 3 ) 2 (m - 3)

B -7m + 15 −

(m + 3)(m - 3) D -7m + 15 −

(m + 3 ) 2 (m - 3)

1. Write 3m + 1 + m − m as a rational expression.


2. x 2 y

−

5 n 3 −

x 3 y 5

−

15n 3.

x - 1 −

x + 3 −

x + 2 −

x + 3

4. Solve 5 −

3 + 3 −

2x = 1 −

6 . State any extraneous solutions.

5. MULTIPLE CHOICE Byron can paint a room in 3 hours. Latoya can paint the same room in 2 hours 15 minutes. How long would it take them if they worked together?

A 4 −

21 hour B 1 2 −

7 hours C 2 5 −

8 hours D 5 1 −

4 hours


11

11

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1.

2.

3.

4.

5.

SCORE

055-065_ALG1_A_CRM_C11_AS_660285.indd 58055-065_ALG1_A_CRM_C11_AS_660285.indd 58 12/23/10 11:50 AM12/23/10 11:50 AM

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11 SCORE

Part I Write the letter for the correct answer in the blank at the right of each question.

1. If y varies inversely as x and y = -9 when x = -11, find y when x = 66.

A 3 B 6 C 11 D 3 −

2

2. State the excluded values of a 2 - 3a - 28 −

a 2 + 3a - 4 .

F -4, 7 G -4, 1 H -4, -1 J -4, 1, 7

3. Simplify 3 x 2 - 5x + 2 −

x 2 - 3x + 2 .

A 3x - 2 −

x - 2 B 3x + 1 −

x - 1 C 3x - 5 −

x - 3 D 2 −

3

4. Find m 2 −

8 m 2 - 8 p 2 ·

m + p −

m 3 + m2 .

F 0 G 1 −

8(m - p)(m + 1) H 1 −

8m(m - p) J 1 −

(m - n)(m + 1)

5. Find 4 m 2 c −

p 3 ÷ mc −

6 p 4 .

A 24mp B 1 −

24mp C 2 m 3 c 2 −

3 p 7 D 1 −

10mp

6. Find 6 x 2 + 5x - 4 −

2x - 1 .

F 6x2 + 3x G 3x + 5 −

2 H 3x + 4 J 4x + 4

Part II

7. Assume that y varies inversely as x. If y = 7 when x = 3, write an inverse variation equation that relates x and y.

8. State the excluded value of the function y = 9 −

x - 9 .

9. Identify the asymptotes of y = 3 − x + 2.

10. Find the roots of f(x) = x2

- x - 12 −

x - 4 .

11. Find x 2 - 6x - 7 −

x 2 - 3x - 10 · x + 2 −

x - 7 .

12. Find 3 x 2 - 12x - 36 −

6 x 2 + 12x - 18 ÷ x 2 - 2x - 24 −

x 2 + 3x - 4 .

Chapter 11 Mid-Chapter Test (Lessons 11-1 through 11-4 )

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.


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11 SCORE Chapter 11 Vocabulary Test

asymptote

complex fraction

excluded values

extraneous

solutions

inverse variation

least common

denominator

least common

multiple

mixed expression

product rule

rate problem

rational equation

rational expression

work problem

Underline or circle the correct term or phrase to complete each sentence. 1. To add rational expressions with unlike denominators,

rename them using their (excluded values, least common denominator, least common multiple).

2. 7x + 6 −

x - 3 is an example of a(n) (complex fraction, mixed

expression, rational expression).

3. Suppose it takes you 3 hours to stock grocery shelves and you and your friend 1 1 −

4 hours. To find how long it would .solve a(n)

(inverse variation, complex fraction, work problem).

4. When you solve an equation and some of the numbers you get are not solutions of the original equation, these numbers are called (excluded values, extraneous solutions, work problems).

5. The equation y = 10 − x is an example of a(n) (inverse variation, rate problem, rational expression).

6. A fraction that contains one or more fractions in the numerator or the denominator is a (complex fraction, mixed expression, rational expression).

7. Values of the variable that make the denominator of a rational expression 0 are called (excluded values, extraneous solutions, least common denominator).

8. An expression that is the sum of a monomial and a rational expression is a (complex fraction, mixed expression, rational expression).

9. If you write a product using each prime factor of two polynomials the greatest number of times it occurs in the polynomial, you get the (least common denominator, least common multiple, mixed expression) for the polynomials.

10. A problem that involves uniform motion is a(n) (inverse variation, rate problem, work problem).

Define each term in your own words.

11. product rule

12. rational equation

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.


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11 SCORE Chapter 11 Test, Form 1

Write the letter for the correct answer in the blank at the right of each question.

1. Identify the inverse variation that is graphed.

A y = 15x C xy = -15

B xy = 15 D y = -15x

2. If y varies inversely as x and y = 18 when x = 2, find y when x = 4.

F 144 G 36 H 9 J 9 −

4

3. State the excluded value(s) of x - 2 −

x - 4 .

A 2 B -2, 2 C 2, 4 D 4

4. Simplify 18 r 2 n −

30 rn 2 .

F 90r3n3 G 3r −

5n H 6rn J 9rn −

15rn

5. Find 3 r 2 −

2 p 3 ·

4 p 4 −

9r .

A 2rp

−

3 B 2 r 2 −

3p C 27 r 3 −

8 p 7 D

4p −

6r

6. Find (t + 2)(t + 3) −

(t - 1)(t + 2) · (t - 1)(t + 7)

−

(t + 7)(t - 6) .

F t + 3 −

t - 6 G t + 2 −

t + 7 H (t + 3)(t - 1)

−

(t - 1)(t - 6) J -

1 −

2

7. Find 4r −

r + 1 ÷ 3 −

r + 1 .

A 12r −

(r + 1 ) 2 B 3 −

4r C 4r −

3 D 4r −

3(r + 1)

8. Find (a2 + 4a - 5) ÷ (5a).

F a 2 + 4a - 1 − a G a2 + 3a H 5a + 20 - 1 − a J a −

5 + 4 −

5 - 1 − a

9. Find (x2 - 5x - 14) ÷ (x + 2).

A x - 7 B -5x - 7 C x + 7 D x2 - 6x - 12

10. Identify the asymptotes of y = 3 − x + 2.

F x = 0, y = 2 G x = 0, y = -2 H x = 2, y = 0 J x = -2, y = 0

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

y

xO

5

5 10 15

1015

-5

-10

-10

-15

-15


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11 Chapter 11 Test, Form 1 (continued)

11. Find the roots of f(x) = x2

+ x - 6 −

x - 2 .

A -3 B 2 C 2 and -3 D 3

12. Find 4 −

x + 1 + 3x −

x + 1 .

F 7x −

x + 1 G 3x + 4 −

2x + 2 H 3x + 4 −

x + 1

13. Find x −

x + 1 + 2 −

x + 3 .

A 3x + 4 −

(x + 1)(x + 3) B x + 2 −

2x + 4 C x 2 + 2x + 4 −

(x + 1)(x + 3) D x 2 + 5x + 2 −

(x + 1)(x + 3)

14. Find x - 4 −

8 - x - 7 −

8 .

F -

11 −

8 G 3 −

8 H x - 11 −

8 J x + 3 −

8

15. Find x - 1 −

x + 2 - 3 −

x + 1 .

A x 2 - 3x - 7 −

(x + 2)(x + 1) B x - 4 −

(x + 1)(x + 2) C -3 −

x + 2 D x 2 - 3x + 5 −

(x + 2)(x + 1)

16. Write 9 + 6 − a as a rational expression.

F 9a + 6 − a G 15 − a H 9a + 6a − a J 15a

17. Simplify u 4 −

x 3 −

u −

x 2 .

A u 5 −

x 5 B u − x C u 3 − x D u 2 −

x 2

18. Solve 3 −

x - 2 = 6 −

x + 1 .

F 1 G 5 −

3 H 5 J -

15 −

2

19. Which value is an extraneous solution of x −

x - 2 + x - 4 −

x - 2 = 5?

A 0 B 4 C 1 D 2

20. Find 200 m −

1 min · 60 min −

1 hr · 1 km −

1000 m .

F 300 km/hr G 12 km/hr H 300 m/hr J 12 m/hr

Bonus Simplify 1 −

1 + 3 −

4 .

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

J 7x −

2x + 2


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11 SCORE


1. Identify the inverse variation that is graphed.

A xy = 30 C y = -30x B xy = -30 D y = 30x

2. If y varies inversely as x and y = 0.8 when x = 1.8, find y when x = 4.8.

F 0.1 G 0.3 H 2.13 J 10.8

3. State the excluded value(s) of n 2 - 3n - 28 −

n 2 + 3n - 4 .

A -4, 7 B -4, 1, 7 C 1 D -4, 1

4. Simplify k2 - 8k + 16 −

k2 - 16 .

F k - 4 G -1 H k - 4 −

k + 4 J k + 4 −

k - 4

5. Find 3 a 2 b −

6 b 2 c 3 · 12 bc 2 −

15a .

A ab −

3c B 2 a 2 b −

5c C 2a −

5c D a 2 −

3 c 2

6. Find y 2 + 4y + 4

− y · 9y −

y 2 - 4 .

F 9(y + 2)

−

y - 2 G -9 −

y + 2 H 9 −

y - 2 J -9 −

8(y - 2)

7. Find n 2 + 3n - 10 −

n 2 + 6n + 8 ÷ n - 2 −

n 2 + 2n .

A n(n + 5) −

n + 4 B (n + 5)(n - 2 ) 2

−

(n + 3 ) 3 C n + 5 −

n + 2 D n(n - 5)

−

(n - 2 ) 2

8. Identify the asymptotes of y = 4 − x - 1.

F x = 4, y = -1 G x = 0, y = -1 H x = 0, y = 1 J x = -1, y = 0

9. Find 10 x 3 + 15 x 2 - 8x - 12 −−

2x + 3 .

A 5 x 2 + 15x + 37 −

2 C 5 x 2 + 5x - 13 −

2

B 5x2 - 4 D 5 x 2 - 4 - 15 −

2x + 3

10. Find 3t - 2 −

3t - 1 + 5t + 4 −

3t - 1 .

F 8t + 2 −

6t - 2 G 8t + 2 −

3t - 1 H 8t - 8 −

3t - 1 J 10t −

4t

Chapter 11 Test, Form 2A

y

x5

-5-10-15-5

-10

10

10 15

-15

O

11

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


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11

11. Find 2x + 3 −

(x - 4 ) 2 + x + 4 −

4 - x .

A x 2 + 2x - 13 −

(x - 4 ) 2 B x - 1 −

x - 4 C x - 1 −

(x - 4 ) 2 D - x 2 + 2x + 19 −

(x - 4 ) 2

12. Find 6x −

x - 5 - 4 −

x - 5 .

F 2x −

x - 5 G 6x - 4 −

x - 5 H 6x - 4 −

5 J 6x - 4 −

2x - 10

13. Find 10 −

a - b - 6b −

a 2 - b 2 .

A 4 −

a - b B 10a - 6b −

a 2 - b 2 C 4b −

a - b D 10a + 4b −

a 2 - b 2

14. Write x + x + 1 −

x + 8 as a rational expression.

F x 2 + 9x + 1 −

x + 8 G 2x + 1 H 2x + 1 −

x + 8 J x 2 + x + 1 −

x + 8

15. Simplify r 2 + 2r - 3 − r

−

r 2 + 3r −

r 2 - r .

A -1(r - 1) B (r - 1 ) 2 − r C (r + 3 ) 2

− r D 3(r + 3)

16. Solve 5x −

3x + 1 - 10 −

9x + 3 = 7 −

6 .

F 3 G -

27 −

11 H 67 −

9 J 7 −

3


x + 1 - 6 −

x 2 - 4x - 5 = 1?

A 5 B 0 C -1 D 6

18. Find 60 miles −

1 hour · 5280 feet −

1 mile · 1 hour −

60 minutes · 1 minute −

60 seconds .

F 22 ft −

15 s G 68 ft/s H 5280 mi/min J 88 ft/s

19. ART Vanesa is sculpting statues which require about 2 −

9 cubic foot of clay.

If she has 2 cubic yards of clay, how many statues can Vanesa make?

A 9 B 243 C 27 D 729

20. PHYSICS The formula 1 −

R T = 1 −

R 1 + 1 −

R 2 + 1 −

R 3 represents the total resistance

of a circuit. What is R1 if RT = 2.5 ohms, R2 = 10 ohms, and R3 = 5 ohms?

F 17.5 ohms G 2.5 ohms H 10 ohms J 7 ohms

Bonus Find the value of k so that x + 1 is a factor of 3x3 - 2x2 + x + k.

Chapter 11 Test, Form 2A (continued)

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:


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11 SCORE


1. Identify the inverse variation that is graphed. A xy = -42 C y = -42x B xy = 42 D y = 42x

2. If y varies inversely as x and y = 0.4 when x = 2.7, find y when x = 1.2.

F 8.1 G 0.9 H 0.1 J 0.18

3. State the excluded values of n 2 - 2n - 24 −

n 2 + 2n - 8 .

A -4, 6 B -4, 2, 6 C 2 D -4, 2

4. Simplify y 2 + 2y - 24

−

y 2 - 16 .

F 3 −

2 G

y - 3 −

-1 H

y - 6 −

y - 4 J

y + 6 −

y + 4

5. Find 2 n 2 y 3

−

6 c 4 · 30nc −

14 y 5 .

A 5 n 3 −

7 y 2 c 3 B

5 n 3 c 3 y 2 −

7 C

n 3 c 3 y 2 −

2 D n 2 −

2 y 2 c 4

6. Find y 2 + 8y + 16

− y · 4y −

y 2 - 16 .

F -4 −

y - 4 G 4 −

y + 4 H

4(y + 4) −

y - 4 J

64(y + 4 ) 2 −

y 2

7. Find m 2 - 4m - 12 −

m 2 + 6m + 8 ÷ m - 8 −

m 2 + 4m .

A m(m - 6) −

m - 8 B (m - 6)(m - 8)

−

m(m + 4) C m(m + 6)

−

m - 8 D m - 6 −

8

8. Find 16 u 2 x 2 - 12 u 3 x - 20 x 3 −−

4 u 2 x 2 .

F 16 - 12u − x - 20x −

u 2 H 4 - 3u − x - 5x −

u2

G 4 - 3ux - 5u2x J 16u4x4 - 12u5x3 - 20u2x5

9. Find 20 x 3 - 8 x 2 + 15x - 6 −−

5x - 2 .

A 4 x 2 - 16 −

5 x + 43 −

25 C 4 x 2 - 12 −

5 x + 87 −

25

B 4 x 2 + 3 - 4 −

5x - 2 D 4x2 + 3

10. Find 2k + 5 −

5k + 2 + 3k - 7 −

5k + 2 .

F 5k - 2 −

10k + 4 G 5k - 35 −

5k + 2 H -28k −

49k J 5k - 2 −

5k + 2

Chapter 11 Test, Form 2B

y

x

O-10

10

10

-10

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


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11

11. Find 3m + 2 −

(m - 3 ) 2 + m + 3 −

3 - m .

A m 2 + 3m - 6 −

(m - 3 ) 2 B - m 2 + 3m + 11 −

(m - 3 ) 2 C -3 m 2 + 7m + 6 −

(m - 3 ) 2 D 2m - 1 −

(m - 3 ) 2

12. Find 5 −

x + 9 - 4x −

x + 9 .

F 5 - 4x −

x + 9 G 5 - 4x −

2x + 18 H x −

x + 9 J 5 −

9 - 4x

13. Find x −

x - 3 - 4 −

x 2 - 9 .

A x - 4 −

x - 3 B x - 4 −

x 2 - 9 C (x + 4)(x - 1)

−

(x + 3)(x - 3) D (x + 1)(x - 1)

−

(x + 3)(x - 3)

14. Write x + x + 2 −

x + 5 as a rational expression.

F x 2 + x + 2 −

x + 5 G 2x + 2 H x 2 + 6x + 2 −

x + 5 J 2x + 2 −

x + 5

15. Simplify x 2 - 4x - 5 − x

−

x 2 - 5x −

x 2 + x .

A 1 − x B 5(x - 5) C (x + 1 ) 2 − x D (x - 5 ) 2

− x

16. Solve 5x −

2x + 1 + 8 −

4x + 2 = 8 −

3 .

F 4 G 8 −

7 H 8 −

17 J 7 −

3


x + 2 - 10 −

x 2 - x - 6 = 1?

A 3 B -2 C 0 D 10

18. Find 126 kilometers −

1 hour · 1000 meters −

1 kilometer · 1 hour −

60 minutes ·

1 minute −

60 seconds .

F 2100 m/s G 5 m −

18 s H 35 m/s J 28 m/s

19. GARDENING Yong is filling pots with soil. Each pot has a volume of 3 −

8 cubic foot. If he has 3 cubic yards of soil, how many pots can Yong fill?

A 8 B 24 C 648 D 216

20. PHYSICS The formula 1 −

R T = 1 −

R 1 + 1 −

R 2 + 1 −

R 3 represents the total resistance

of a circuit. What is R2 if RT = 1.5 ohms, R1 = 6 ohms, and R3 = 10 ohms?

F 27 ohms G 2.5 ohms H 3 ohms J 6 ohms

Bonus One factor of x3 + x2

- 14x - 24 is x - 4. Find the other two factors.

Chapter 11 Test, Form 2B (continued)

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

066_076_ALG1_A_CRM_C11_AS_660285.indd 66066_076_ALG1_A_CRM_C11_AS_660285.indd 66 12/20/10 10:25 PM12/20/10 10:25 PM

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11 SCORE

1. Graph an inverse variation in which y varies inversely as x and y = -16 when x = -2.

2. Write an inverse variation equation that relates x and y if y varies inversely as x and y = 3.2 when x = 39.5. Find y when x = 31.6.

3. State the excluded value(s) of y 2 + 4y + 3

−

y 2 - 2y - 15 .

4. Simplify a 2 - 3a - 28 −

a 2 + 3a - 4 . State the excluded value(s) of a.

Find each product.

5. y 2 - 9

−

4 · 8 −

y + 3

6. r 2 + 2r - 8 −

r 2 + 3r - 10 · r + 5 −

r 2 - 16

Find each quotient.

7. 5m −

m + 1 ÷ 25 m 2 −

m 2 + 6m + 5

8. x 2 + 5x - 14 −

x 2 - 2x - 15 ÷ 7x - 14 −

6x + 18

9. (6rt2 + 4r2t2 - 9r2t) ÷ (3rt)

10. (3b2 - 17b + 22) ÷ (b - 4)

Find each sum.

11. 2r - 3 −

r - 5 + 6r + 7 −

r - 5

12. 2a + 6 −

a 2 + 6a + 9 + 1 −

a + 3

13. -2 −

6 - n + 13 −

n 2 - 36

Chapter 11 Test, Form 2C

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

y

xO

10

-10

-10

-20

-2010 20

20


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11

Find each difference.

14. 15 −

2x - 5 - 6x −

2x - 5

15. 4 n 2 −

2n - 3 - 9 −

3 - 2n

16. -9 −

5y - 35 - -4 −

y 2 - 7y

17. Write a + a + 7 −

a + 2 as a rational expression.


18. r 2 + r - 6 − r

−

3r + 9 −

r + 2

19. x - 15 −

x + 2 −

x + 5 −

x + 6


20. 2 −

x + 2 = 9 −

8 - 5x −

4x + 8

21. x 2 −

x - 4 + 3 = 16 −

x - 4

22. Find 0.4 inch −

1 day · 1 foot −

12 inches ·

1 yard −

3 feet ·

365 days −

1 year .

23. FITNESS If Nate maintains a pace of 2 −

13 mile per minute,

how long will it take for him to run 8 miles?

24. CONSTRUCTION Seth can build a doghouse in 6 hours. Violet can build a doghouse in 5 hours. How long will it take them to build a doghouse if they work together?

25. PHYSICS Given the formula 1 −

R T = 1 −

R 1 + 1 −

R 2 + 1 −

R 3 for the

resistance of a circuit, determine the resistance R2 if RT = 2 ohms, R1 = 4 ohms, and R3 = 3 ohms.

Bonus Simplify [ ( 45 miles −


1 mile ) ÷ 60 minutes −

1 hour

] ÷ 60 seconds −

1 minute .

Chapter 11 Test, Form 2C (continued)

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B:


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11 SCORE

1. Graph an inverse variation in which y varies inversely as x and y = -15 when x = -6.

2. Write an inverse variation equation that relates x and y if y varies inversely as x and y = 34.5 when x = 3.2. Find y when x = 13.8.

3. State the excluded value(s) of x 2 + x - 12 −

x 2 - 7x + 12 .

4. Simplify k 2 + 2k - 15 −

k 2 - 4k + 3 . State the excluded value(s) of k.

Find each product.

5. 5 −

m 2 - 4 · m + 2 −

10

6. r 2 + 2r - 3 −

r 2 + 5r + 6 · r + 2 −

r 2 - 1

Find each quotient.

7. 2x −

x + 7 ÷ 8 x 2 −

x 2 + 8x + 7

8. x 2 - 4x - 21 −

x 2 - 7x - 18 ÷ 2x - 14 −

5x + 10

9. (12a4b2 - 5a2b3 - 15a3b2) ÷ (3a2b)

10. (5r2 - 13r + 2) ÷ (r - 3)

Find each sum.

11. 4t - 5 −

t + 6 + 5t + 3 −

t + 6

12. 3v - 6 −

v 2 - 4v + 4 + 1 −

v - 2

13. -3 −

7 - p + 11 −

p 2 - 49

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

y

xO

10

10 20

20

-10-20

-10

-20

Chapter 11 Test, Form 2D11


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11


14. 14 −

4x - 7 - 8x −

4x - 7

15. 25 n 2 −

5n - 4 - 16 −

4 - 5n

16. -16 −

6x - 30 - -1 −

x 2 - 5x

17. Write n + n + 7 −

n + 3 as a rational expression.


18. 3x + 3 −

5 x 2 - 5x - 10 −

x - 2 −

15

19. x - 12 −

x + 1 −

x - 8 −

x + 2


20. 2 −

m + 1 - 1 −

3m + 3 = -

5 −

9

21. x 2 −

x - 3 - 9 −

x - 3 = -2

22. Find 1.2 ounces −

1 minute ·

1 quart −

16 ounces · 60 minutes −

1 hour · 24 hours −

1 day .

23. FITNESS If Rozene maintains a pace of 2 −

45 mile per minute,

how long will it take for her to swim 3 miles?

24. CONSTRUCTION Raquel can build a chair in 3 hours. Dekota can build a chair in 2 hours. How long will it take them to build a chair if they work together?

25. PHYSICS Given the formula 1 −

R T = 1 −

R 1 + 1 −

R 2 + 1 −

R 3 for the

resistance of a circuit, determine the resistance R3 if RT = 4 ohms, R1 = 6 ohms, and R2 = 3 ohms.

Bonus Simplify [ ( 30 miles −



1 hour

] ÷ 60 seconds −

1 minute .

Chapter 11 Test, Form 2D (continued)

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

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11 SCORE

1. Graph an inverse variation in which y varies inversely as x and y = 3.6 when x = -1.2.

2. Write an inverse variation equation that relates x and y

if y varies inversely as x and y = 6 when x = 2 −

5 . Find y

when x = 4.

3. Identify the asymptotes of y = 5 −

2x - 1 .

4. State the excluded values of x 4 - 10 x 2 + 9 −

x 3 + 4 x 2 + 3x .

5. Simplify 6 x 2 - 7x - 3 −

6 x 2 - x - 1 . State the excluded values of x.

Find each product.

6. a 2 - 5a - 14 −

48 a 2 · 60 a 3 −

a 2 - 4a - 21

7. 2 b 2 - b - 15 −

6 b 2 + 7b - 3 · 3 b 2 + 5b - 2 −

b 2 - b - 6

Find each quotient.

8. n 2 - 5n + 6 −

w 2 + 9w + 14 ÷ (2 - n)(3 - n)

−

w + 7

9. 3 x 5 y 3

−

x 2 - 9 ÷

12 x 2 y 5 −

x + 3

10. (3x3 - 2x2 - 48x + 32) ÷ (3x - 2)

Find each sum.

11. 3 - 2w −

5 w 2 - 1 + 4w + 1 −

5 w 2 - 1

12. 2 −

(2x + 7 ) 2 + 1 −

4 x 2 - 49

Chapter 11 Test, Form 3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

y

xO

2

2 4

4

-2-4

-2

-4


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11


13. 12x + 1 −

4x - 1 - 7x + 3 −

4x - 1

14. 5y −

y 2 - 36 -

2y + 5 −

y + 6

15. Find 2w + 7 −

w + 5 - 3w −

5 + w + 8w + 5 −

w + 5 .

16. Find 2 −

y - 7 -

3y −

y - 7 +

-5y −

7 - y .

17. Write (p + 2) + p + 1

−

p - 4 as a rational expression.

18. Simplify x + 5 −

x 2 + 2x - 8 −

x 2 + 2x - 15 −

x 2 - 5x + 6 .

19. What is the quotient of 4 u 2 −

9 r 3 and 6 u 4 −

5r ?


20. p 2 −

p - 3 + 5 −

3 - p = 2

21. x - 1 −

x - 2 - 7 −

x + 3 = 5 −

x 2 + x - 6

22. CLEANING James can clean a car interior in 20 minutes, and Sally can clean the same car in 15 minutes. If they work together, how long will it take to clean 15 cars?

23. POOL Bill’s pool has a hole and is losing water at a rate of

4 −

9 gallon per minute. How many hours will it take for Bill

to lose 100 gallons?

24. Simplify [ ( 20 miles −



1 hour

] ÷ 60 seconds −

1 minute .

25. GEOMETRY The volume of a box is 2x3 + 11x2 + 18x + 9. One dimension of the box is x + 1. What are the other two dimensions if they are polynomials in x with integer coefficients?

Bonus Find the constants k and c so that

7x + 17 −

x 2 + 4x + 3 = k −

x + 3 + c −

x + 1 .

Chapter 11 Test, Form 3 (continued)

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

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11 SCORE Chapter 11 Extended-Response Test

Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.

1. The time it takes to travel a certain distance varies inversely as the rate at which you travel. The equation rt = D can be used to represent a person traveling D miles. Choose a distance you might travel to go on vacation, and substitute this value in the formula for D. Choose 5 possible values for r and find the corresponding values for t. Draw a graph of this relation.

2. a. Write a rational expression that has a common factor of the form x + n in the numerator and denominator. Assume that n is an integer and n ≠ 0. Simplify the expression. State the excluded values of the variable.

b. Write two rational expressions whose product is x - 2 −

x + 3 . Find the

quotient of the two expressions.

c. Write two rational expressions whose difference is 3x + 1 −

4x - 7 . Find the

sum of the two expressions.

3. a. Write a rational expression in the form (x + m)(ax + n) − x + m where

a, m, and n are integers and a ≠ 0. Write the expression in the form

ax 2 + bx + c − x + m by multiplying the two factors in the numerator. Use

long division to simplify the expression.

b. Compare and contrast long division of a polynomial by a binomial with simplifying a rational expression whose denominator is a binomial.

4. FISHING Roshanda uses an average of 14 1 −

5 inches of thread to tie a

fishing fly. She has 18 1 −

2 yards of thread to use for her flies. Describe

how division can be used to find out how many flies Roshanda can make. How many flies can she make?

5. FARMING Pat and Harold own a farm. Working alone, Pat can irrigate their fields in x hours. Harold can irrigate their fields in two hours less time than Pat. When working together they can irrigate their fields in y hours.

The equation 1 − x + 1 −

x - 2 = 1 − y represents this situation.

a. Determine if x and y can be equal. b. Choose a value for x, and solve the equation for y. For this value of x,

how long does it take for Pat and Harold to irrigate the fields together? c. Choose a value for y, and solve the equation for x. For this value of y,

how long does it take for Pat to irrigate the fields when working alone?


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11 SCORE Standardized Test Practice

1. Which point lies on the line given by y = 2x - 3? (Lesson 3-1)

A (2, 7) B (-1, -5) C (0, -1) D (-4, -9)

2. Solve x3 - 4x = 0. (Lesson 8-6)

F {-4, 0, 4} G {-2, 0, 2} H {-4, 4} J {-2, 2}

3. Solve x2 - 8x + 16 = 25 by taking the square root of each side. (Lesson 9-4)

A {-9, 1} B {-1, 1} C {-9, 9} D {-1, 9}

4. The District of Columbia has been experiencing a 1.6% annual decrease in population. In 1999, its population was 519,000. If the trend continues, predict the District of Columbia’s population in 2015. (Lesson 7-7)

F 441,691 G 1883 H 427,656 J 400,949

5. Find (5 - √ �� 14 ) 2 . (Lesson 10-3)

A 11 B 29 - 29 √ � 7 C 39 - 10 √ �� 14 D 29 √ �� 14

6. Solve √ �� x + 3 = x - 3. (Lesson 10-4)

F 6 G 1, 6 H 1, 3, 6 J 1

7. A water ski ramp is 6 feet tall and 9 feet long. Find the measure of the angle the ramp makes with the water. (Lesson 10-6)

A about 33.7°

B about 41.8°

C about 48.2°

D about 56.3°

8. Find (12x2 - 16x) ÷ (4x). (Lesson 11-5)

F 3x - 4 G -1 H 3x - 4 −

4 J 3x - 1

9. The dimensions of a rectangle are 5 −

3 - y and 7 −

y 2 - 9 . Find the

perimeter of the rectangle. (Lesson 11-6)

A 10y + 44

−

y 2 - 9 B

-10y - 16 −

y 2 - 9 C

10y + 16 −

2 y 2 - 18 D

-5y - 16 −

2 y 2 - 18

10. Simplify x 2 - 9 −

x 2 + 5x + 6 −

x - 3 . (Lesson 11-7)

F -3 −

(5x + 2)(-3) G 1 −

x + 2 H (x - 3 ) 2

−

x + 2 J x + 3 −

x 2 + 5x + 6

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. F G H J

8. F G H J

9. A B C D

10. F G H J


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NAME DATE PERIOD

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11 Standardized Test Practice (continued)

11. Evaluate a + bc if a = 5, b = 2.4, and c = 1.2. (Lesson 1-3)

A 2.88 B 14.4 C 7.88 D 8.6

12. The formula A = P(1 + r) gives the amount A in an account with simple interest rate r and principal P. What is the interest rate as a decimal if the amount in the account is $1080 and the principal was $900? (Lesson 2-8)

F 0.1 G 0.2 H 0.3 J 0.4

13. Simplify √ ��

12 x 2 y . (Lesson 10-2)

A 2x √ � 3y B 6x √ �

y C 3y √ � 2x D y √ � 6x

14. Simplify 3 x 2 - 6x −

x 2 - 4 . State the excluded values of the variables. (Lesson 11-3)

F -1, 1 G -2, 2 H -3, 3 J -4, 4

15. Find (x3 + 7x + 22) ÷ (x + 2). (Lesson 11-5)

A x2 - 2x + 11 B x2 + 2x - 11 C x2 + 11 D x2 - 11

16. Solve 2 −

x + 2 + 5 −

3x = 4 − x . (Lesson 11-8)

F 12 G -12 H 14 J -14

17. The voltage V required for a circuit is given by V = √ �� PR , where P is the power in watts and R is the resistance in ohms. What is the voltage of a 65-watt bulb if the resistance is 110 ohms? Round your answer to the nearest tenth. (Lesson 10-2)

18. The Belmont Stakes is a 1.5-mile race for horses. The winning time in 2000 was about 2 minutes, 30 seconds. What was the average speed for the winning horse in miles per hour? (Lesson 11-4)

Part 2: Gridded Response

Instructions: Enter your answer by writing each digit of the answer in a column box

and then shading in the appropriate circle that corresponds to that entry.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

NAME DATE PERIOD

PDF Pass


11 Standardized Test Practice (continued)

19. Determine whether x −

3 = 5 +

3y −

4 is a linear equation. If so,

write the equation in standard form. (Lesson 3-1)

20. A lighthouse in North Carolina was 1500 feet from the shoreline in 1870. About 1988, the lighthouse was 160 feet from the shoreline and had to be moved. Write the point-slope form of an equation to find the approximate distance from the shoreline for any given year x between 1870 and 1988. Use x = 0 to correspond with the year 1870. (Lesson 4-3)

21. Solve x - 10 > -2 and 2x - 5 ≤ 10. Then graph the solution set. (Lesson 5-4)

22. Solve x2 + 4x + 3 - x(x - 5) = 3(2x - 1). (Lesson 8-2)

23. Six times a number subtracted from the number squared is 0. Find the number. (Lesson 8-5)

24. Factor x2 + 5x - 24. (Lesson 8-6)

25. Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of the equation y = -2x2 + 4x - 5. (Lesson 9-1)

26. Graph y = 4(2x). State the y-intercept. (Lesson 7-5)

27. Determine whether the side measures 5, 17, and 19 form a right triangle. Justify your answer. (Lesson 10-5)

28. Find AB. (Lesson 10-5)

29. Graph an inverse variation in which y varies inversely as x and y = -12 when x = 4. (Lesson 11-1)

30. A square has an area of 121 square inches. The formula for the area A of a square with side length s is A = s2. (Lesson 10-5)

a. Find the length of one side of the square.

b. Find the length of the diagonal.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30a.

30b.

y

xO 8 16-8-16

-8

-16

8

16

y

xO 2-2

4

8

12

16

-1-2-3-5-4 0 1 2 63 4 5

A

C B

9 cm

21 cm

Part 3: Short Response

Instructions: Write your anwers in the space provided.


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass

Chapter 11 A1 Glencoe Algebra 1Chapter 11 A1 Glencoe Algebra 1

Answers (Anticipation Guide and Lesson 11-1)

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

Chapter Resources

NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

3

Gle

ncoe

Alg

ebra

1

Ant

icip

atio

n G

uide

Rati

on

al

Exp

ressio

ns a

nd

Eq

uati

on

s

B

efor

e yo

u b

egin

Ch

ap

ter

11

•

Rea

d ea

ch s

tate

men

t.

•

Dec

ide

wh

eth

er y

ou A

gree

(A

) or

Dis

agre

e (D

) w

ith

th

e st

atem

ent.

•

Wri

te A

or

D i

n t

he

firs

t co

lum

n O

R i

f yo

u a

re n

ot s

ure

wh

eth

er y

ou a

gree

or

disa

gree

, wri

te N

S (

Not

Su

re).

Aft

er y

ou c

omp

lete

Ch

ap

ter

11

•

Rer

ead

each

sta

tem

ent

and

com

plet

e th

e la

st c

olu

mn

by

ente

rin

g an

A o

r a

D.

•

Did

an

y of

you

r op

inio

ns

abou

t th

e st

atem

ents

ch

ange

fro

m t

he

firs

t co

lum

n?

•

For

th

ose

stat

emen

ts t

hat

you

mar

k w

ith

a D

, use

a p

iece

of

pape

r to

wri

te a

n

exam

ple

of w

hy

you

dis

agre

e.

11 Step

1

ST

EP

1A

, D, o

r N

SS

tate

men

tS

TE

P 2

A o

r D

1.

Sin

ce a

dir

ect

vari

atio

n c

an b

e w

ritt

en a

s y

= k

x, a

n i

nve

rse

vari

atio

n c

an b

e w

ritt

en a

s y

= x −

k .

2.

A r

atio

nal

exp

ress

ion

is

an a

lgeb

raic

fra

ctio

n t

hat

con

tain

sa

radi

cal.

3.

To

mu

ltip

ly t

wo

rati

onal

exp

ress

ion

s, s

uch

as

2xy2

−

3c

and

3 c 2

−

5y ,

mu

ltip

ly t

he

nu

mer

ator

s an

d th

e de

nom

inat

ors.

4.

Wh

en s

olvi

ng

prob

lem

s in

volv

ing

un

its

of m

easu

re,

dim

ensi

onal

an

alys

is i

s th

e pr

oces

s of

det

erm

inin

g th

e u

nit

s

of t

he

fin

al a

nsw

er s

o th

at t

he

un

its

can

be

ign

ored

wh

ile

perf

orm

ing

calc

ula

tion

s.

5.

To

divi

de (

4x2

+ 12

x) b

y 2x

, div

ide

4x2

by 2

x an

d 12

x by

2x.

6.

To

fin

d th

e su

m o

f

2a

−

(3a

- 4

) an

d

5 −

(3a

- 4

) , fi

rst

add

the

nu

mer

ator

s an

d th

en t

he

den

omin

ator

s.

7.

Th

e le

ast

com

mon

den

omin

ator

of

two

rati

onal

exp

ress

ion

s

wil

l be

th

e le

ast

com

mon

mu

ltip

le o

f th

e de

nom

inat

ors.

8.

A c

ompl

ex f

ract

ion

con

tain

s a

frac

tion

in

its

nu

mer

ator

or

den

omin

ator

.

9.

Th

e fr

acti

on ( a

−

b ) −

( c −

d ) c

an b

e re

wri

tten

as

ac

−

bd .

10.

Ext

ran

eou

s so

luti

ons

are

solu

tion

s th

at c

an b

e el

imin

ated

be

cau

se t

hey

are

ext

rem

ely

hig

h o

r lo

w.

Step

2

D D A D A D A A D D

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

312

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-1

Cha

pte

r 11

5

Gle

ncoe

Alg

ebra

1

Bot

h m

eth

ods

show

th

at x

2 =

8 −

3 wh

en y

= 1

8.

Exer

cise

sD

eter

min

e w

het

her

eac

h t

able

or

equ

atio

n r

epre

sen

ts a

n i

nve

rse

or a

dir

ect

vari

atio

n. E

xpla

in.

1.

xy

36

510

816

12

24

2.

y =

6x

3. x

y =

1

5

Ass

um

e th

at y

var

ies

inve

rsel

y as

x. W

rite

an

in

vers

e va

riat

ion

eq

uat

ion

th

at

rela

tes

x an

d y

. Th

en s

olve

.

4. I

f y

= 1

0 w

hen

x =

5,

5. I

f y

= 8

wh

en x

= -

2,

fin

d y

wh

en x

= 2

. xy =

50;

25

fi

nd

y w

hen

x =

4.

xy =

-16

; -

4

6. I

f y

= 1

00 w

hen

x =

120

, 7.

If

y =

-16

wh

en x

= 4

, fi

nd

x w

hen

y =

20.

xy =

12,

00

0; 6

00

fi

nd

x w

hen

y =

32.

xy =

-64

;-2

8. I

f y

= -

7.5

wh

en x

= 2

5, f

ind

y w

hen

x =

5.

xy =

-18

7.5;

-37

.5

9. D

RIV

ING

Th

e G

erar

di f

amil

y ca

n t

rave

l to

Osh

kosh

, Wis

con

sin

, fro

m C

hic

ago,

Ill

inoi

s,

in 4

hou

rs i

f th

ey d

rive

an

ave

rage

of

45 m

iles

per

hou

r. H

ow l

ong

wou

ld i

t ta

ke t

hem

if

they

in

crea

sed

thei

r av

erag

e sp

eed

to 5

0 m

iles

per

hou

r? 3

.6 h

10. G

EOM

ETRY

For

a r

ecta

ngl

e w

ith

giv

en a

rea,

th

e w

idth

of

the

rect

angl

e va

ries

in

vers

ely

as t

he

len

gth

. If

the

wid

th o

f th

e re

ctan

gle

is 4

0 m

eter

s w

hen

th

e le

ngt

h i

s 5

met

ers,

fin

d th

e w

idth

of

the

rect

angl

e w

hen

th

e le

ngt

h i

s 20

met

ers.

10

m

Stud

y G

uide

and

Inte

rven

tion

Invers

e V

ari

ati

on

Iden

tify

an

d U

se In

vers

e V

aria

tio

ns

An

in

vers

e va

riat

ion

is

an e

quat

ion

in

th

e fo

rm o

f y

= k −

x o

r xy

= k

. If

two

poin

ts (

x 1, y 1)

an

d (x

2, y 2)

are

sol

uti

ons

of a

n i

nve

rse

vari

atio

n,

then

x1

∙ y1

= k

an

d x 2

∙ y2

= k

.

Pro

du

ct R

ule

fo

r In

vers

e V

aria

tio

nx 1

∙ y1

= x

2 ∙ y

2

Fro

m t

he

prod

uct

ru

le, y

ou c

an f

orm

th

e pr

opor

tion

x 1 −

x 2 = y 2 −

y 1 .

If

y v

arie

s in

vers

ely

as x

an

d y

= 1

2 w

hen

x =

4, f

ind

x w

hen

y =

18.

Met

hod

1 U

se t

he

prod

uct

ru

le.

x1

∙ y1

= x

2 ∙ y

2 P

roduct

rule

for

inve

rse v

ari

ation

4 ∙

12 =

x2

∙ 18

x 1 =

4,

y 1 =

12,

y 2 =

18

48

−

18 =

x 2

Div

ide e

ach s

ide b

y 1

8.

8 −

3 = x

2 S

implif

y.

Met

hod

2 U

se a

pro

port

ion

.

x 1 −

x 2 =

y 2 −

y 1

Pro

port

ion for

inve

rse v

ari

ation

4 −

x 2 =

18

−

12

x 1 =

4,

y 1 =

12,

y 2 =

18

48

= 1

8x2

Cro

ss m

ultip

ly.

8 −

3 = x

2 S

implif

y.

11-1

Exam

ple

dir

ect

vari

atio

n;

of

the

form

y =

kx

dir

ect

vari

atio

n;

of

the

form

y =

kx

inve

rse

vari

atio

n;

of

the

form

xy =

k

001-

016_

ALG

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M_C

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5.in

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10

7:40

PM

A01_A14_ALG1_A_CRM_C11_AN_660285.indd A1A01_A14_ALG1_A_CRM_C11_AN_660285.indd A1 12/20/10 10:19 PM12/20/10 10:19 PM

Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass


An swers (Lesson 11-1)


NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

6

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Invers

e V

ari

ati

on

Gra

ph

Inve

rse

Var

iati

on

s S

itu

atio

ns

in w

hic

h t

he

valu

es o

f y

decr

ease

as

the

valu

es

of x

in

crea

se a

re e

xam

ples

of

inve

rse

vari

atio

n. W

e sa

y th

at y

var

ies

inve

rsel

y as

x, o

r y

is

inve

rsel

y pr

opor

tion

al t

o x.

S

up

pos

e yo

u d

rive

20

0 m

iles

wit

hou

t st

opp

ing.

Th

e ti

me

it t

akes

to

trav

el a

dis

tan

ce v

arie

s in

vers

ely

as t

he

rate

at

wh

ich

you

tr

avel

. Let

x =

sp

eed

in

mil

es p

er h

our

and

y =

tim

e in

hou

rs. G

rap

h t

he

vari

atio

n.

Th

e eq

uat

ion

xy

= 2

00 c

an b

e u

sed

to

repr

esen

t th

e si

tuat

ion

. Use

var

iou

s sp

eeds

to

mak

e a

tabl

e.

x

y O20

4060

30 20 10

G

rap

h a

n i

nve

rse

vari

atio

n i

n w

hic

h y

var

ies

inve

rsel

y as

x

and

y =

3 w

hen

x =

12.

Sol

ve f

or k

.

xy =

k

Inve

rse v

ari

ation e

quation

12(

3) =

k

x =

12 a

nd y

= 3

36

= k

S

implif

y.

Ch

oose

val

ues

for

x a

nd

y, w

hic

h h

ave

a pr

odu

ct o

f 36

.

x

y O12

24

24 12

Exer

cise

sG

rap

h e

ach

var

iati

on i

f y

vari

es i

nve

rsel

y as

x.

1. y

= 9

wh

en x

= -

3 2.

y =

12

wh

en x

= 4

3.

y =

-25

wh

en x

= 5

x

y

O24 12

-12

-24

-12

-24

1224

x

y

O32 16

-16

-32

-16

-32

1632

x

y

O50

-50

-10

010

0

100 50

-50

-10

0

4. y

= 4

wh

en x

= 5

5.

y =

-18

wh

en x

= -

9 6.

y =

4.8

wh

en x

= 5

.4

x

y

O20 10

-10

-20

-10

-20

1020

x

y

O36 18

-18

-36

-18

-36

1836

x

y

O7.2

3.6

-3.

6

-7.

2

-3.

6-

7.2

3.6

7.2

11-1

Exam

ple

1Ex

amp

le 2

xy

10

20

20

10

30

6.7

40

5

50

4

60

3.3

xy

−

6−

6

−

3−

12

−

2−

18

218

312

66

Inve

rse

Var

iati

on

Eq

uat

ion

an

eq

ua

tio

n o

f th

e fo

rm x

y =

k,

wh

ere

k ≠

0

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

612

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-1

Cha

pte

r 11

7

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceIn

vers

e V

ari

ati

on

Det

erm

ine

wh

eth

er e

ach

tab

le o

r eq

uat

ion

rep

rese

nts

an

in

vers

e or

a d

irec

t va

riat

ion

. Exp

lain

.

1.

xy

0.5

8

14

22

41

2.

xy

= 2 −

3 3.

-2x

+ y

= 0

Ass

um

e th

at y

var

ies

inve

rsel

y as

x. W

rite

an

in

vers

e va

riat

ion

eq

uat

ion

th

at

rela

tes

x an

d y

. Th

en g

rap

h t

he

equ

atio

n.

4.

y =

2 w

hen

x =

5

5. y

= -

6 w

hen

x =

-6

6.

y =

-4

wh

en x

= -

12

7. y

= 1

5 w

hen

x =

3

Sol

ve. A

ssu

me

that

y v

arie

s in

vers

ely

as x

.

8. I

f y

= 4

wh

en x

= 8

, 9.

If

y =

-7

wh

en x

= 3

, fi

nd

y w

hen

x =

2.

xy =

32;

16

fin

d y

wh

en x

= -

3. x

y =

-21

; 7

10. I

f y

= -

6 w

hen

x =

-2,

11

. If

y =

-24

wh

en x

= -

3,

fin

d y

wh

en x

= 4

. xy =

12;

3

fin

d x

wh

en y

= -

6. x

y =

72;

-12

12. I

f y

= 1

5 w

hen

x =

1,

13. I

f y

= 4

8 w

hen

x =

-4,

fi

nd

x w

hen

y =

-3.

xy =

15;

-5

fin

d y

wh

en x

= 6

. xy =

-19

2; -

32

14. I

f y

= -

4 w

hen

x =

1 −

2 , fi

nd

x w

hen

y =

2.

xy =

-2;

-1

11-1

x

y

O8

-8

-16

16

16 8

-8

-16

x

y

O8

-8

-16

16

16 8

-8

-16

x

y

O10

-10

-20

20

20 10

-10

-20

x

y

O4

-4

-8

8

8 4

-4

-8

inve

rse,

xy =

4

inve

rse,

xy =

2 −

3 d

irec

t, y =

2x

xy =

10

xy =

36

xy =

48

xy =

45

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

712

/20/

10

7:40

PM


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF 2nd


Answers (Lesson 11-1)


NA

ME

DAT

E

P

ER

IOD

Lesson 11-1

Cha

pte

r 11

9

Gle

ncoe

Alg

ebra

1

1.PH

YSI

CA

L SC

IEN

CE

Th

e il

lum

inat

ion

Ipr

odu

ced

by a

lig

ht

sou

rce

vari

es

inve

rsel

y as

th

e sq

uar

e of

th

e di

stan

ce d

from

th

e so

urc

e. T

he

illu

min

atio

n

prod

uce

d 5

feet

fro

m t

he

ligh

t so

urc

e is

80

foo

t-ca

ndl

es.

Id2

=k

80

(5)2

=k

20

00=

k

Fin

d th

e il

lum

inat

ion

pro

duce

d 8

feet

fr

om t

he

sam

e so

urc

e.

31.2

5 fo

ot-

can

dle

s

2. M

ON

EY A

for

mu

la c

alle

d th

e R

ule

of

72 a

ppro

xim

ates

how

fas

t m

oney

wil

l do

ubl

e in

a s

avin

gs a

ccou

nt.

It

is b

ased

on

th

e re

lati

on t

hat

th

e n

um

ber

of y

ears

it

tak

es f

or m

oney

to

dou

ble

vari

es

inve

rsel

y as

th

e an

nu

al i

nte

rest

rat

e.

Use

th

e in

form

atio

n i

n t

he

tabl

e to

wri

te

the

Ru

le o

f 72

for

mu

la.

yr

= 7

2

3. E

LEC

TRIC

ITY

Th

e re

sist

ance

, in

oh

ms,

of

a c

erta

in l

engt

h o

f el

ectr

ic w

ire

vari

es

inve

rsel

y as

th

e sq

uar

e of

th

e di

amet

er

of t

he

wir

e. I

f a

wir

e 0.

04 c

enti

met

er i

n

diam

eter

has

a r

esis

tan

ce o

f 0.

60 o

hm

, w

hat

is

the

resi

stan

ce o

f a

wir

e of

th

e sa

me

len

gth

an

d m

ater

ial

that

is

0.08

cen

tim

eter

in

dia

met

er?

0.

15 o

hm

4.B

USI

NES

S In

th

e m

anu

fact

uri

ng

of a

ce

rtai

n d

igit

al c

amer

a, t

he

cost

of

prod

uci

ng

the

cam

era

vari

es i

nve

rsel

y as

th

e n

um

ber

prod

uce

d. I

f 15

,000

cam

eras

ar

e pr

odu

ced,

th

e co

st i

s $8

0 pe

r u

nit

. G

raph

th

e re

lati

onsh

ip a

nd

labe

l th

e po

int

that

rep

rese

nts

th

e co

st p

er u

nit

to

prod

uce

25,

000

cam

eras

. $4

8

5. S

OU

ND

Th

e so

un

d pr

odu

ced

by a

str

ing

insi

de a

pia

no

depe

nds

on

its

len

gth

. Th

e fr

equ

ency

of

a vi

brat

ing

stri

ng

vari

es

inve

rsel

y as

its

len

gth

.

a. W

rite

an

equ

atio

n t

hat

rep

rese

nts

th

e re

lati

onsh

ip b

etw

een

fre

quen

cy f

an

d le

ngt

h �

. Use

k f

or t

he

con

stan

t of

var

iati

on.

b.

If y

ou h

ave

two

diff

eren

t le

ngt

h

stri

ngs

, wh

ich

on

e vi

brat

es m

ore

quic

kly

(th

at i

s, w

hic

h s

trin

g h

as a

gr

eate

r fr

equ

ency

)?

T

he

sho

rter

str

ing

vib

rate

s m

ore

qu

ickl

y th

an t

he

lon

ger

st

rin

g.

c. S

upp

ose

a pi

ano

stri

ng

2 fe

et l

ong

vibr

ates

300

cyc

les

per

seco

nd.

Wh

at

wou

ld b

e th

e fr

equ

ency

of

a st

rin

g 4

feet

lon

g?

15

0 cy

cles

per

sec

on

d

Wor

d Pr

oble

m P

ract

ice

Invers

e V

ari

ati

on

11-1

Year

s to

Do

ub

le

Mo

ney

An

nu

al In

tere

st

Rat

e (p

erce

nt)

18

4

14

.45

12

6

10

.29

7

Price per Unit ($)

200

300

100 0

Un

its

Pro

du

ced

(th

ou

san

ds)

1020

30

y

x

25,0

00,4

8

f ×

ℓ =

k o

r f

= k

−

�

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

912

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

8

Gle

ncoe

Alg

ebra

1

Prac

tice

In

vers

e V

ari

ati

on

Det

erm

ine

wh

eth

er e

ach

tab

le o

r eq

uat

ion

rep

rese

nts

an

in

vers

e or

a d

irec

t va

riat

ion

. Exp

lain

.

1.

xy

0.2

54

0

0.5

20

25

81.

25

2.

x

y

-2

8

00

2-

8

4-

16

3.

y −

x =

-3

4. y

= 7 −

x

Ass

sum

e th

at y

var

ies

inve

rsel

y as

x. W

rite

an

in

vers

e va

riat

ion

eq

uat

ion

th

at

rela

tes

x an

d y

. Th

en g

rap

h t

he

equ

atio

n.

5. y

= -

2 w

hen

x =

-12

6.

y =

-6

wh

en x

= -

5 7.

y =

2.5

wh

en x

= 2

Wri

te a

n i

nve

rse

vari

atio

n e

qu

atio

n t

hat

rel

ates

x a

nd

y. A

ssu

me

that

y v

arie

s in

vers

ely

as x

. Th

en s

olve

.

8. I

f y

= 1

24 w

hen

x =

12,

fin

d y

wh

en x

= -

24.

xy =

148

8; -

62

9. I

f y

= -

8.5

wh

en x

= 6

, fin

d y

wh

en x

= -

2.5.

xy =

-51

; 20

.4

10. I

f y

= 3

.2 w

hen

x =

-5.

5, f

ind

y w

hen

x =

6.4

. xy =

-17

.6;

-2.

75

11. I

f y

= 0

.6 w

hen

x =

7.5

, fin

d y

wh

en x

= -

1.25

. xy =

4.5

; -

3.6

12. E

MPL

OY

MEN

T T

he

man

ager

of

a lu

mbe

r st

ore

sch

edu

les

6 em

ploy

ees

to t

ake

inve

nto

ry

in a

n 8

-hou

r w

ork

peri

od. T

he

man

ager

ass

um

es a

ll e

mpl

oyee

s w

ork

at t

he

sam

e ra

te.

a. S

upp

ose

2 em

ploy

ees

call

in

sic

k. H

ow m

any

hou

rs w

ill

4 em

ploy

ees

nee

d to

tak

e in

ven

tory

? 12

hb

. If

th

e di

stri

ct s

upe

rvis

or c

alls

in

an

d sa

ys s

he

nee

ds t

he

inve

nto

ry f

inis

hed

in

6 h

ours

, h

ow m

any

empl

oyee

s sh

ould

th

e m

anag

er a

ssig

n t

o ta

ke i

nve

nto

ry?

8

13. T

RA

VEL

Jes

se a

nd

Joaq

uin

can

dri

ve t

o th

eir

gran

dpar

ents

’ hom

e in

3 h

ours

if

they

av

erag

e 50

mil

es p

er h

our.

Sin

ce t

he

road

bet

wee

n t

he

hom

es i

s w

indi

ng

and

mou

nta

inou

s, t

hei

r pa

ren

ts p

refe

r th

ey a

vera

ge b

etw

een

40

and

45 m

iles

per

hou

r. H

ow l

ong

wil

l it

tak

e to

dri

ve t

o th

e gr

andp

aren

ts’ h

ome

at t

he

redu

ced

spee

d?

bet

wee

n 3

h 2

0 m

in a

nd

3 h

45

min

11-1

x

y

O24 12

-12

-24

-24

-12

1224

x

y

Ox

y

O16 8

-8

-16

-8

-16

816

inve

rse;

xy =

kd

irec

t; y

= k

x

dir

ect;

y =

kx

inve

rse;

xy =

k

xy =

24

xy =

30

xy =

5

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

812

/23/

10

2:01

PM


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass


Answers (Lesson 11-1 and Lesson 11-2)


NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

10

G

lenc

oe A

lgeb

ra 1

Enri

chm

ent

Dir

ect

or

Ind

irect

Vari

ati

on

Fil

l in

eac

h t

able

bel

ow. T

hen

wri

te i

nve

rsel

y, o

r d

irec

tly

to c

omp

lete

eac

h

con

clu

sion

.

1.

�2

48

16

32

W4

44

4

4

A8

1632

6412

8

2.

H

ou

rs2

45

6

Sp

eed

55

55

55

55

Dis

tan

ce11

022

027

533

0

F

or a

set

of

rect

angl

es w

ith

a w

idth

F

or a

car

tra

veli

ng

at 5

5 m

i/h, t

he

of

4, t

he

area

var

ies

d

ista

nce

cov

ered

var

ies

as t

he

len

gth

. a

s th

e h

ours

dri

ven

.

3.

Oat

Bra

n 1

−

3 c

up

2

−

3 c

up

1 c

up

Wat

er1

cu

p2

cu

p3

cu

p

Ser

vin

gs

12

3

4.

H

ou

rs o

f Wo

rk12

812

812

8

Peo

ple

Wo

rkin

g

24

8

Ho

urs

per

Per

son

6432

16

T

he

nu

mbe

r of

ser

vin

gs o

f oa

t br

an

A jo

b re

quir

es 1

28 h

ours

of

wor

k. T

he

v

arie

s a

s th

e n

um

ber

nu

mbe

r of

hou

rs e

ach

per

son

wor

ks

o

f cu

ps o

f oa

t br

an.

var

ies

as

the

nu

mbe

r

o

f pe

ople

wor

kin

g.

5.

Mile

s10

010

010

010

0

Rat

e

20

25

50

10

0

Ho

urs

5

42

1

6.

b

34

56

h10

10

10

10

A15

2025

30

F

or a

100

-mil

e ca

r tr

ip, t

he

tim

e th

e F

or a

set

of

righ

t tr

ian

gles

wit

h a

hei

ght

tri

p ta

kes

vari

es

as

the

of

10, t

he

area

var

ies

av

erag

e ra

te o

f sp

eed

the

car

trav

els.

a

s th

e ba

se.

Use

th

e ta

ble

at

the

righ

t.

7. x

var

ies

as

y.

8. z

var

ies

as

y.

9. x

var

ies

as

z.

inve

rsel

y

11-1

dir

ectl

yd

irec

tly

dir

ectl

y

dir

ectl

yinve

rsel

y

x1

1.5

22

.53

y2

34

56

z6

04

03

02

42

0

inve

rsel

y

inve

rsel

y

dir

ectl

y

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1012

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-2

Cha

pte

r 11

11

G

lenc

oe A

lgeb

ra 1

Iden

tify

Exc

lud

ed V

alu

es T

he

fun

ctio

n y

= 10

−

x i

s an

exa

mpl

e of

a r

atio

nal

fu

nct

ion

. B

ecau

se d

ivis

ion

by

zero

is

un

defi

ned

, an

y va

lue

of a

var

iabl

e th

at r

esu

lts

in a

den

omin

ator

of

zer

o m

ust

be

excl

ude

d fr

om t

he

dom

ain

of

that

var

iabl

e. T

hes

e ar

e ca

lled

exc

lud

ed

valu

es o

f th

e ra

tion

al f

un

ctio

n.

S

tate

th

e ex

clu

ded

val

ue

for

each

fu

nct

ion

.

Exer

cise

s

Sta

te t

he

excl

ud

ed v

alu

e fo

r ea

ch f

un

ctio

n.

1. y

= 2 −

x

x =

0

2. y

=

1 −

x -

4 x

= 4

3.

y =

x -

3

−

x +

1

x =

-1

4. y

=

4 −

x -

2 x

= 2

5.

y =

x −

2x -

4 x

= 2

6.

y =

-

5 −

3x x

= 0

7. y

= 3x

- 2

−

x +

3

x =

-3

8. y

= x

- 1

−

5x +

10

x =

-2

9. y

= x

+ 1

−

x x

= 0

10. y

= x

- 7

−

2x +

8 x

= -

4 11

. y =

x

- 5

−

6x

x =

0

12. y

= x

- 2

−

x +

11

x =

-11

13. y

=

7 −

3x +

21

x =

-7

14. y

= 3x

- 4

−

x +

4

x =

-4

15

. y =

x −

7x -

35

x =

5

16. D

ININ

G M

ya a

nd

her

fri

ends

are

eat

ing

at a

res

tau

ran

t. T

he

tota

l bi

ll o

f $3

6 is

spl

it

amon

g x

frie

nds

. Th

e am

oun

t ea

ch p

erso

n p

ays

y is

giv

en b

y y

= 36

−

x

, w

her

e x

is t

he

nu

mbe

r of

peo

ple.

Gra

ph t

he

fun

ctio

n.

11-2

Stud

y G

uide

and

Inte

rven

tion

Rati

on

al

Fu

ncti

on

s

Exam

ple

a. y

= 3 −

x

T

he

den

omin

ator

can

not

equ

al z

ero.

Th

e ex

clu

ded

valu

e is

x =

0.

b.

y =

4

−

x -

5

x

- 5

= 0

S

et

the d

enom

inato

r equal to

0.

x =

5

Add 5

to e

ach s

ide.

T

he

excl

ude

d va

lue

is x

= 5

.

Bill per Person ($)

4 0812162024283236

Num

bero

fPeo

ple

12

34

56

78

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1112

/20/

10

7:40

PM


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

12

G

lenc

oe A

lgeb

ra 1

Iden

tify

an

d U

se A

sym

pto

tes

Bec

ause

exc

lude

d va

les

are

un

defi

ned

, th

ey a

ffec

t th

e gr

aph

of

the

fun

ctio

n. A

n a

sym

pto

te i

s a

lin

e th

at t

he

grap

h o

f a

fun

ctio

n a

ppro

ach

es.

A r

atio

nal

fu

nct

ion

in

th

e fo

rm y

=

a −

x -

b

+ c

has

a v

erti

cal

asym

ptot

e at

th

e x-

valu

e th

at

mak

es t

he

den

omin

ator

equ

al z

ero,

x =

b. I

t h

as a

hor

izon

tal

asym

ptot

e at

y =

c.

Id

enti

fy t

he

asym

pto

tes

of y

=

1 −

x -

1 +

2 .

Th

en g

rap

h t

he

fun

ctio

n.

Ste

p 1

Id

enti

fy a

nd

grap

h t

he

asym

ptot

es u

sin

g da

shed

lin

es.

ve

rtic

al a

sym

ptot

e: x

= 1

h

oriz

onta

l as

ympt

ote:

y =

2

Ste

p 2

M

ake

a ta

ble

of v

alu

es a

nd

plot

th

e po

ints

. T

hen

con

nec

t th

em.

x–1

02

3

y1.

51

32

.5

Exer

cise

sId

enti

fy t

he

asym

pto

tes

of e

ach

fu

nct

ion

. Th

en g

rap

h t

he

fun

ctio

n.

1. y

= 3 −

x x

= 0

; y =

0

2. y

= -

2 −

x

x

= 0

; y =

0

3. y

= 4 −

x +

1 x

= 0

; y =

1

4. y

= 2 −

x

- 3

x =

0;

y =

3

5. y

=

2 −

x +

1 x

= 1

; y =

0

6. y

= -

2 −

x -

3 x

= 3

; y =

0

11-2

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Rati

on

al

Fu

ncti

on

s

Exam

ple

y

x

y =

+

2

y =

2

x =

1

1 x-1

y

x

y

x

y

x

y

x

y

x

y

x

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1212

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-2

Cha

pte

r 11

13

G

lenc

oe A

lgeb

ra 1

Sta

te t

he

excl

ud

ed v

alu

e fo

r ea

ch f

un

ctio

n.

1. y

= 6 −

x x

= 0

2.

y =

2

−

x -

2 x

= 2

3.

y =

x

−

x +

6 x

= -

6

4. y

= x

- 3

−

x +

4

x =

-4

5. y

= 3x

- 5

−

x +

8

x =

-8

6. y

=

-5

−

2x -

14

x =

7

7. y

=

x −

3x +

21

x =

-7

8. y

= x

- 1

−

9x -

36

x =

4

9. y

=

9 −

5x +

4

0

x =

-8

Iden

tify

th

e as

ymp

tote

s of

eac

h f

un

ctio

n. T

hen

gra

ph

th

e fu

nct

ion

.

10. y

= 1 −

x

11. y

= 3 −

x

12. y

=

2 −

x +

1

x

= 0

, y =

0

x =

0, y

= 0

x

= -

1, y

= 0

13. y

=

3 −

x -

2

14. y

=

2 −

x +

1 -

1

15. y

=

1 −

x -

2 +

3

x

= 2

, y =

0

x =

-1,

y =

-1

x =

2, y

= 3

11-2

Skill

s Pr

acti

ceR

ati

on

al

Fu

ncti

on

s

y

x

y

x

y

x

y

x

y

x

y

x

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1312

/20/

10

7:40

PM


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

14

G

lenc

oe A

lgeb

ra 1

Sta

te t

he

excl

ud

ed v

alu

e fo

r ea

ch f

un

ctio

n.

1. y

= -

1

−

x

x

= 0

2.

y =

3 −

x +

5 x

= -

5 3.

y =

2x

−

x -

5 x

= 5

4. y

=

x -

1

−

12x

+ 36

x =

-3

5. y

= x

+ 1

−

2x +

3 x

= -

3 −

2 6.

y =

1 −

5x -

2

x =

2

−

5

Iden

tify

th

e as

ymp

tote

s of

eac

h f

un

ctio

n. T

hen

gra

ph

th

e fu

nct

ion

.

7. y

= 1 −

x

8. y

= 3 −

x

9. y

=

2 −

x -

1

x

= 0

, y =

0

x

= 0

, y =

0

x =

1, y

= 0

10. y

=

2 −

x +

2

11. y

=

1 −

x -

3 +

2

12. y

=

2 −

x +

1 -

1

x

= -

2, y

= 0

x

= 3

, y =

2

x =

-1,

y =

-1

13. A

IR T

RA

VEL

Den

ver,

Col

orad

o, i

s lo

cate

d ap

prox

imat

ely

1000

mil

es f

rom

In

dian

apol

is, I

ndi

ana.

Th

e av

erag

e sp

eed

of a

plan

e tr

avel

ing

betw

een

th

e tw

o ci

ties

is

give

n b

y y

= 10

00

−

x

,

wh

ere

x is

th

e to

tal

flig

ht

tim

e. G

raph

th

e fu

nct

ion

.

11-2

Prac

tice

Rati

on

al

Fu

ncti

on

s

Average Speed (mph)

400

600

200 0

1

800

1000

Tota

l Flig

ht T

ime

23

45

y

x

y

x

y

x

y

x

y

x

y

x

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1412

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-2

Cha

pte

r 11

15

G

lenc

oe A

lgeb

ra 1

1. B

ULL

ET T

RA

INS

Th

e S

hin

kan

sen

, or

Japa

nes

e bu

llet

tra

in n

etw

ork,

pro

vide

s h

igh

-spe

ed t

ran

spor

tati

on t

hro

ugh

out

Japa

n. T

rain

s re

gula

rly

oper

ate

at s

peed

s in

exc

ess

of 2

00 k

ilom

eter

s pe

r h

our.

Th

e av

erag

e sp

eed

of a

bu

llet

tra

in t

rave

lin

g be

twee

n T

okyo

an

d K

yoto

is

give

n b

y y

=

515

−

x, w

her

e x

is t

he

tota

l tr

avel

tim

e in

hou

rs. G

raph

th

e fu

nct

ion

.

2. D

RIV

ING

Pet

er i

s dr

ivin

g to

his

gran

dpar

ents

’ hou

se 4

0 m

iles

aw

ay.

Du

rin

g th

e tr

ip, P

eter

mak

es a

30-

min

ute

stop

for

lu

nch

. Th

e av

erag

e sp

eed

of

Pet

er’s

tri

p is

giv

en b

y y

=

40−

x +

0.

5, w

her

e

x is

th

e to

tal

tim

e sp

ent

in t

he

car.

Wh

at

are

the

asym

ptot

es o

f th

e fu

nct

ion

?x

=-

0.5,

y=

0

3. E

RR

OR

AN

ALY

SIS

Nic

olas

is

grap

hin

g

the

equ

atio

n y

=

20−

x +

3

-

6

and

draw

s a

grap

h w

ith

asym

ptot

es a

t y

= 3

and

x=

– 6.

Exp

lain

the

err

or t

hat

Nic

olas

mad

e in

his

gra

ph.

Th

e as

ymp

tote

s sh

ou

ld b

e x

=-

3, y

=–6.

11-2

Wor

d Pr

oble

m P

ract

ice

Rati

on

al

Fu

ncti

on

s

Average Speed (km/h)

100

150 50 0

1

200

300

250

Tim

e (h

ours

)2

34

4. U

SED

CA

RS

Wh

ile

rese

arch

ing

cars

to

purc

has

e on

lin

e, M

s. J

acob

s fo

un

d th

at

the

valu

e of

a u

sed

car

is i

nve

rsel

y pr

opor

tion

al t

o th

e ag

e of

th

e ca

r. T

he

aver

age

pric

e of

a u

sed

car

is g

iven

by

y =

17,9

00−

x +

1.

2+

10

0, w

her

e x

is t

he

age

of

the

car.

Wh

at a

re t

he

asym

ptot

es o

f th

e fu

nct

ion

? E

xpla

in w

hy

x=

0 c

ann

ot b

e an

asy

mpt

ote.

x=

-1.

2, y

= 1

00;

An

asy

mp

tote

at

x=

0 w

ou

ld g

ive

y, t

he

cost

of

the

car,

an in

fi n

ite

valu

e w

hen

th

e ca

r is

bra

nd

new

. Th

e as

ymp

tote

n

eed

s to

be

loca

ted

at

x=

a,

wh

ere

a<

0.

5. F

AM

ILY

REU

NIO

N T

he

Gau

det

fam

ily

is h

oldi

ng

thei

r an

nu

al r

eun

ion

at

Wat

kin

s P

ark.

It

cost

s $5

0 to

get

a

perm

it t

o h

old

the

reu

nio

n a

t th

e pa

rk,

and

the

fam

ily

is s

pen

din

g $8

per

per

son

on

foo

d. T

he

Gau

dets

hav

e ag

reed

to

spli

t th

e co

st o

f th

e ev

ent

even

ly a

mon

g al

l th

ose

atte

ndi

ng.

a. W

rite

an

equ

atio

n s

how

ing

the

cost

pe

r pe

rson

y i

f x

peop

le a

tten

d th

e re

un

ion

.

b.

Wh

at a

re t

he

asym

ptot

es o

f th

e eq

uat

ion

?

c. N

ow a

ssu

me

that

th

e fa

mil

y w

ants

to

let

a lo

ng-

lost

cou

sin

att

end

for

free

. R

ewri

te t

he

equ

atio

n t

o fi

nd

the

new

co

st p

er p

ayin

g pe

rson

y.

y =

8 +

58

−

x -

1

d.

Wh

at a

re t

he

asym

ptot

es f

or t

he

new

eq

uat

ion

?

y =

8 +

50

−

x

x =

0, y

= 8

x =

1, y

= 8

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1512

/20/

10

7:40

PM


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

16

G

lenc

oe A

lgeb

ra 1

Ineq

ual

itie

s in

volv

ing

rati

onal

fu

nct

ion

s ca

n b

e gr

aph

ed m

uch

lik

e th

ose

invo

lvin

g li

nea

r fu

nct

ion

s.

Gra

ph

y ≥

1 −

x .

Ste

p 1

P

lot

poin

ts a

nd

draw

a s

moo

th s

olid

cu

rve.

Bec

ause

th

e in

equ

alit

y in

volv

es a

gre

ater

th

an o

r eq

ual

to

sign

, so

luti

ons

that

sat

isfy

y =

1 −

x w

ill

be a

par

t of

th

e gr

aph

.

Ste

p 2

P

lot

the

asym

ptot

es, x

= 0

an

d y

= 0

, as

dash

ed l

ines

.

Ste

p 3

B

egin

tes

tin

g va

lues

. A v

alu

e m

ust

be

test

ed b

etw

een

ea

ch s

et o

f li

nes

, in

clu

din

g as

ympt

otes

.R

egio

n 1

T

est

(–1,

1).

Th

is r

etu

rns

a tr

ue

valu

e fo

r

th

e in

equ

alit

y.R

egio

n 2

T

est

(–1,

–0.

5). T

his

ret

urn

s a

tru

e va

lue

for

the

ineq

ual

ity.

Reg

ion

3

Tes

t (–

1, –

2). T

his

ret

urn

s a

fals

e va

lue

for

the

ineq

ual

ity.

Reg

ion

4

Tes

t (1

, 2).

Th

is r

etu

rns

a tr

ue

valu

e fo

r

th

e in

equ

alit

y.R

egio

n 5

T

est

(1, 0

.5).

Th

is r

etu

rns

a fa

lse

valu

e fo

r

th

e in

equ

alit

y.R

egio

n 6

T

est

(1, –

1). T

his

ret

urn

s a

fals

e va

lue

for

the

ineq

ual

ity.

Ste

p 4

S

had

e th

e re

gion

s w

her

e th

e in

equ

alit

y is

tru

e.

Exer

cise

sG

rap

h e

ach

in

equ

alit

y.

1. y

≤

1 −

x +

1

2. y

> 2 −

x

3. y

≤

1 −

x +

1 -1

11-2

Enri

chm

ent

Ineq

ualiti

es i

nvo

lvin

g R

ati

on

al

Fu

ncti

on

s

Exam

ple

y

x

y

x

y

x

y

x

y

x

001-

016_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1612

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-3

Cha

pte

r 11

17

G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

Sim

plify

ing

Rati

on

al

Exp

ressio

ns

Iden

tify

Exc

lud

ed V

alu

es

Bec

ause

a r

atio

nal

exp

ress

ion

in

volv

es d

ivis

ion

, th

e de

nom

inat

or c

ann

ot e

qual

zer

o. A

ny

valu

e of

th

e de

nom

inat

or t

hat

res

ult

s in

div

isio

n b

y ze

ro i

s ca

lled

an

exc

lud

ed v

alu

e of

th

e de

nom

inat

or.

S

tate

th

e ex

clu

ded

valu

e of

4m

- 8

−

m +

2 .

Exc

lude

th

e va

lues

for

wh

ich

m +

2 =

0.

m

+ 2

= 0

T

he d

enom

inato

r cannot

equal 0.

m +

2 -

2 =

0 -

2 S

ubtr

act

2 f

rom

each s

ide.

m

= -

2 S

implif

y.

Th

eref

ore,

m c

ann

ot e

qual

-2.

S

tate

th

e ex

clu

ded

valu

es o

f x

2 +

1

−

x 2

- 9

.

Exc

lude

th

e va

lues

for

wh

ich

x2

- 9

= 0

.

x2 -

9 =

0

The d

enom

inato

r cannot

equal 0.

(x

+ 3

)(x

- 3

) =

0

Facto

r.

x +

3 =

0

or

x -

3 =

0

Zero

Pro

duct

Pro

pert

y

= -

3 =

3T

her

efor

e,

x ca

nn

ot e

qual

-3

or 3

.

Exer

cise

sS

tate

th

e ex

clu

ded

val

ues

for

eac

h r

atio

nal

exp

ress

ion

.

1.

2b

−

b2 -

8

√ �

8

, - √

�

8

2.

12 -

a

−

32 +

a

-32

3.

x 2 -

2

−

x2 +

4

2, -

2 4.

m

2 -

4

−

2m2

- 8

2,

-2

5.

2n -

12

−

n 2

- 4

-

2, 2

6.

2x

+ 1

8 −

x 2 -

16

-4,

4

7.

x 2

- 4

−

x 2 +

4x

+ 4

-

2 8.

a

- 1

−

a 2

+ 5

a +

6 -

3, -

2

9.

k 2 -

2k

+ 1

−

k 2 +

4k

+ 3

-

3, -

1 10

.

m 2

- 1

−

2 m 2

- m

- 1

-

1 −

2 , 1

11.

25

- n

2 −

n 2

- 4

n -

5

-1,

5

12.

2 x 2

+ 5

x +

1

−

x 2 -

10x

+ 1

6 2,

8

13.

n 2

- 2

n -

3

−

n 2

+ 4

n -

5

-5,

1

14.

y 2 -

y -

2

−

3 y 2

- 1

2 -

2, 2

15.

k 2 +

2k

- 3

−

k 2 -

20k

+ 6

4 4,

16

16.

x 2 +

4x

+ 4

−

4 x 2

+ 1

1x -

3

-3,

1 −

4

11-3

Rat

ion

alE

xpre

ssio

na

n a

lge

bra

ic f

ractio

n w

ith

nu

me

rato

r a

nd

de

no

min

ato

r th

at

are

po

lyn

om

ials

Exa

mp

le:

x 2 +

1

−

y 2

Exam

ple

2Ex

amp

le 1

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1712

/20/

10

7:40

PM


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

18

G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Sim

plify

ing

Rati

on

al

Exp

ressio

ns

Sim

plif

y Ex

pre

ssio

ns

Fact

orin

g po

lyn

omia

ls i

s a

use

ful

tool

for

sim

plif

yin

g ra

tion

al

expr

essi

ons.

To

sim

plif

y a

rati

onal

exp

ress

ion

, fir

st f

acto

r th

e n

um

erat

or a

nd

den

omin

ator

. T

hen

div

ide

each

by

the

grea

test

com

mon

fac

tor.

S

imp

lify

54 z 3

−

24yz

.

54 z 3

−

24yz

= (6

z)(9

z 2 )

−

(6z)

(4y)

T

he G

CF

of

the n

um

era

tor

and t

he d

enom

inato

r is

6z.

= 1 (

6z)(

9 z 2 )

−

1 (6z

)(4y

)

Div

ide t

he n

um

era

tor

and d

enom

inato

r by 6

z.

=

9 z 2

−

4y

S

implif

y.

S

imp

lify

3x

- 9

−

x 2

- 5

x +

6 .

Sta

te t

he

excl

ud

ed v

alu

es o

f x.

3x -

9

−

x 2 - 5

x +

6

=

3(x

- 3

) −

(x -

2)(

x -

3)

Facto

r.

=

3 (x

- 3

) 1 −

(x -

2) (

x -

3) 1

D

ivid

e b

y t

he G

CF,

x –

3.

=

3 −

x -

2

Sim

plif

y.

Exc

lude

th

e va

lues

for

wh

ich

x2

- 5

x +

6 =

0.

x2

- 5

x +

6 =

0 (

x -

2)(

x -

3)

= 0

x =

2

or

x =

3T

her

efor

e, x

≠ 2

an

d x

≠ 3

.

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n. S

tate

th

e ex

clu

ded

val

ues

of

the

vari

able

s.

1.

12ab

−

a 2 b

2

12

−

ab ;

a ≠

0;

b ≠

0

2.

7 n 3

−

21 n

8

1 −

3 n 5 ;

n ≠

0

3.

x +

2

−

x 2 -

4

1

−

x -

2 ;

x ≠

-2

or

2 4.

m

2 -

4

−

m 2

+ 6

m +

8

m -

2

−

m +

4 ;

m ≠

-4

or

-2

5.

2n -

8

−

n 2

- 1

6

2 −

n +

4 ;

n ≠

-4

or

4 6.

x 2

+ 2

x +

1

−

x 2 -

1

x

+ 1

−

x -

1 ;

x ≠

-1

or

1

7.

x 2

- 4

−

x 2 +

4x

+ 4

x

- 2

−

x +

2 ;

x ≠

-2

8.

a 2

+ 3

a +

2

−

a 2

+ 5

a +

6

a +

1

−

a +

3 ;

a ≠

-3

or

-2

9.

k 2

- 1

−

k 2 +

4k

+ 3

k

- 1

−

k +

3 ;

k ≠

-3

or

-1

10.

m 2

- 2

m +

1

−

2 m 2

- m

- 1

m

- 1

−

2m +

1 ;

m ≠

-

1 −

2 or

1

11.

n

2 -

25

−

n 2

- 4

n -

5

n +

5

−

n +

1 ;

n ≠

-1

or

5 12

. x 2

+ x

- 6

−

2 x 2

- 8

x +

3

−

2x +

4 ;

x ≠

-2

or

2

13.

n 2

+ 7

n +

12

−

n 2

+ 2

n -

8

n +

3

−

n -

2 ;

n ≠

-4

or

2 14

.

y 2 -

y -

2

−

y 2 -

10y

+ 1

6 y

+ 1

−

y -

8 ;

y ≠

2 o

r 8

11-3

Exam

ple

1

Exam

ple

2

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1812

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-3

Cha

pte

r 11

19

G

lenc

oe A

lgeb

ra 1

Skill

s Pr

acti

ceS

imp

lify

ing

Rati

on

al

Exp

ressio

ns

Sta

te t

he

excl

ud

ed v

alu

es f

or e

ach

rat

ion

al e

xpre

ssio

n.

1.

2p

−

p -

7 7

2.

4n +

1

−

n +

4

-4

3.

k +

2

−

k 2 -

4 -

2, 2

4.

3x +

15

−

x 2 -

25

-5,

5

5.

y 2

- 9

−

y 2 +

3y

- 1

8 -6,

3

6.

b 2 -

2b

- 8

−

b 2 +

7b

+ 1

0 -5,

-2

Sim

pli

fy e

ach

exp

ress

ion

. Sta

te t

he

excl

ud

ed v

alu

es o

f th

e va

riab

les.

7.

21bc

−

28 bc

2 3 −

4c ;

0, 0

8.

12 m

2 r

−

24 m

r 3 m

−

2 r 2 ;

0, 0

9.

16 x 3 y

2 −

36 x 5 y

3

4 −

9 x 2 y ;

0, 0

10

. 8 a

2 b 3

−

40 a

3 b

b 2

−

5a ;

0, 0

11.

n +

6

−

3n +

18

1 −

3 ; -

6 12

. 4x

- 4

−

4x +

4

x –

1

−

x +

1 ;

-1

13.

y 2 -

64

−

y +

8

y -

8;

-8

14.

y 2 -

7y

- 1

8

−

y -

9

y +

2;

9

15.

z +

1

−

z 2 -

1

1

−

z -

1 ;

-1,

1

16.

x

+ 6

−

x 2 +

2x

- 2

4

1 −

x –

4 ;

-6,

4

17.

2d

+ 1

0 −

d 2

- 2

d -

35

2

−

d -

7 ;

-5,

7

18.

3h

- 9

−

h 2

- 7

h +

12

3

−

h -

4 ;

3, 4

19.

t 2 +

5t

+ 6

−

t 2 +

6t

+ 8

t

+ 3

−

t +

4 ;

-4,

-2

20.

a 2

+ 3

a -

4

−

a 2

+ 2

a -

8

a -

1

−

a -

2 ;

-4,

2

21.

x 2 +

10x

+ 2

4

−

x 2 -

2x

- 2

4 x

+ 6

−

x -

6 ;

-4,

6

22.

b 2 -

6b

+ 9

−

b 2 -

9b

+ 1

8 b

- 3

−

b -

6 ;

3, 6

11-3

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

1912

/20/

10

7:40

PM


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

20

G

lenc

oe A

lgeb

ra 1

Prac

tice

S

imp

lify

ing

Rati

on

al

Exp

ressio

ns

Sta

te t

he

excl

ud

ed v

alu

es f

or e

ach

rat

ion

al e

xpre

ssio

n.

1.

4n -

28

−

n 2

- 4

9 -

7, 7

2.

p

2 -

16

−

p 2

- 1

3p +

36

4, 9

3.

a

2 -

2a

- 1

5

−

a 2

+ 8

a +

15

-5,

-3

Sim

pli

fy e

ach

exp

ress

ion

. Sta

te t

he

excl

ud

ed v

alu

es o

f th

e va

riab

les.

4.

12a

−

48 a

3 1 −

4 a 2 ;

0 5.

6x

y z 3

−

3 x 2 y

2 z

2 z 2

−

xy ;

0, 0

, 0

6.

36 k 3 n

p 2

−

20 k 2 n

p 5

9k

−

5 p 3 ;

0, 0

, 0

7.

5 c

3 d 4

−

40 cd

2 +

5 c 4 d

2 c

2 d 2

−

8 +

c 3 ;

c:

0, -

2, d

: 0

8.

p 2

- 8

p +

12

−

p -

2

p -

6;

2

9.

m 2

- 4

m -

12

−

m -

6

m

+ 2

; 6

10.

m +

3

−

m 2

- 9

1 −

m -

3 ;

-3,

3

11.

2b

- 1

4 −

b 2 -

9b

+ 1

4

2 −

b -

2 ;

2, 7

12

. x 2

- 7

x +

10

−

x 2 -

2x

- 1

5 x

- 2

−

x +

3 ;

-3,

5

13.

y 2 +

6y

- 1

6

−

y 2 -

4y

+ 4

y

+ 8

−

y -

2 ;

2 14

. r 2

- 7

r +

6

−

r 2 +

6r

- 7

r

– 6

−

r +

7 ;

-7,

1

15.

t 2

- 8

1 −

t 2 -

12t

+ 2

7 t

+ 9

−

t -

3 ;

3, 9

16

. r 2

+ r

- 6

−

r 2 +

4r

- 1

2 r

+ 3

−

r +

6 ;

-6,

2

17.

2 x 2

+ 1

8x +

36

−

3 x 2

- 3

x -

36

2(x +

6)

−

3(x -

4) ;

-3,

4

18.

2 y 2

+ 9

y +

4

−

4 y 2

- 4

y -

3

y +

4

−

2y -

3 ;

-

1 −

2 , 3 −

2

19. E

NTE

RTA

INM

ENT

Fair

fiel

d H

igh

spe

nt

d d

olla

rs f

or r

efre

shm

ents

, dec

orat

ion

s, a

nd

adve

rtis

ing

for

a da

nce

. In

add

itio

n, t

hey

hir

ed a

ban

d fo

r $5

50.

a. W

rite

an

exp

ress

ion

th

at r

epre

sen

ts t

he

cost

of

the

ban

d as

a

55

0 −

d +

550

fr

acti

on o

f th

e to

tal

amou

nt

spen

t fo

r th

e sc

hoo

l da

nce

.

b.

If d

is

$165

0, w

hat

per

cen

t of

th

e bu

dget

did

th

e ba

nd

acco

un

t fo

r? 2

5%

20. P

HY

SIC

AL

SCIE

NC

E M

r. K

amin

ksi

plan

s to

dis

lodg

e a

tree

stu

mp

in h

is y

ard

by u

sin

g a

6-fo

ot b

ar a

s a

leve

r. H

e pl

aces

th

e ba

r so

th

at 0

.5 f

oot

exte

nds

fro

m t

he

fulc

rum

to

the

end

of t

he

bar

un

der

the

tree

stu

mp.

In

th

e di

agra

m, b

rep

rese

nts

th

e to

tal

len

gth

of

the

bar

and

t re

pres

ents

th

e po

rtio

n o

f th

e ba

r be

yon

d th

e fu

lcru

m.

a. W

rite

an

equ

atio

n t

hat

can

be

use

d to

cal

cula

te t

he

m

ech

anic

al a

dvan

tage

.

b.

Wh

at i

s th

e m

ech

anic

al a

dvan

tage

? 11

c. I

f a

forc

e of

200

pou

nds

is

appl

ied

to t

he

end

of t

he

leve

r, w

hat

is

the

forc

e pl

aced

on

th

e tr

ee s

tum

p? 2

200

lb

b

fulc

rum

tree

stum

pt

11-3

MA

= b

- t

−

t

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2012

/20/

10

7:40

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-3

Cha

pte

r 11

21

G

lenc

oe A

lgeb

ra 1

Wor

d Pr

oble

m P

ract

ice

Sim

plify

ing

Rati

on

al

Exp

ressio

ns

1.PH

YSI

CA

L SC

IEN

CE

Pre

ssu

re i

s eq

ual

to

the

mag

nit

ude

of

a fo

rce

divi

ded

by t

he

area

ove

r w

hic

h t

he

forc

e ac

ts.

P=

F−

A

Gab

e an

d S

hel

by e

ach

pu

sh o

pen

a d

oor

wit

h o

ne

han

d. I

n o

rder

to

open

, th

e do

or

requ

ires

20

pou

nds

of

forc

e. T

he

surf

ace

area

of

Gab

e’s

han

d is

10

squ

are

inch

es,

and

the

surf

ace

area

of

Sh

elby

’s h

and

is

8 sq

uar

e in

ches

. Wh

ose

han

d fe

els

the

grea

ter

pres

sure

?

Sh

elby

’s:

2.5

lb−

in 2

(vs

Gab

e’s

2 lb

−

i n 2

)

2. G

RA

PHIN

G R

ecal

l th

at t

he

slop

e of

a

lin

e is

a r

atio

of

the

vert

ical

ch

ange

to

the

hor

izon

tal

chan

ge i

n c

oord

inat

es f

or

two

give

n p

oin

ts. W

rite

a r

atio

nal

ex

pres

sion

th

at r

epre

sen

ts t

he

slop

e of

th

e li

ne

con

tain

ing

the

poin

ts a

t (p

, r)

and

(7, -

3).

r +

3

−

p -

7

or

-3

- r

−

7 -

p

3. A

UTO

MO

BIL

ES T

he

forc

e n

eede

d to

ke

ep a

car

fro

m s

kidd

ing

out

of a

tu

rn o

n

a pa

rtic

ula

r ro

ad i

s gi

ven

by

the

form

ula

be

low

. Wh

at f

orce

is

requ

ired

to

keep

a

2000

-pou

nd

car

trav

elin

g at

50

mil

es p

er

hou

r on

a c

urv

e w

ith

rad

ius

of 7

50 f

eet

on t

he

road

? W

hat

val

ue

of r

is

excl

ude

d?

f

=

0.06

72w

s 2 −

r

f

= f

orce

in

pou

nds

w

= w

eigh

t in

pou

nds

s

= s

peed

in

mph

r

= r

adiu

s in

fee

t

4

48 lb

, r ≠

0

4.PA

CK

AG

ING

In

ord

er t

o sa

fely

sh

ip a

n

ew e

lect

ron

ic d

evic

e, t

he

dist

ribu

tion

m

anag

er a

t D

ata

Pro

duct

s C

ompa

ny

dete

rmin

es t

hat

th

e pa

ckag

e m

ust

co

nta

in a

cer

tain

am

oun

t of

cu

shio

nin

g on

eac

h s

ide

of t

he

devi

ce. T

he

devi

ce i

s sh

aped

lik

e a

cube

wit

h s

ide

len

gth

x,

and

som

e si

des

nee

d m

ore

cush

ion

ing

than

oth

ers

beca

use

of

the

devi

ce’s

de

sign

. Th

e vo

lum

e of

a s

hip

pin

g co

nta

iner

is

repr

esen

ted

by t

he

expr

essi

on (

x2 + 6

x +

8)(

x +

6).

Fin

d th

e po

lyn

omia

l th

at r

epre

sen

ts t

he

area

of

the

top

of t

he

box

if t

he

hei

ght

of t

he

box

is x

+ 2

.

5. S

CH

OO

L C

HO

ICE

Du

rin

g a

rece

nt

sch

ool

year

, th

e ra

tio

of p

ubl

ic s

choo

l st

ude

nts

to

priv

ate

sch

ool

stu

den

ts i

n

the

Un

ited

Sta

tes

was

app

roxi

mat

ely

7.6

to 1

. Th

e st

ude

nts

att

endi

ng

publ

ic

sch

ool

outn

um

bere

d th

ose

atte

ndi

ng

priv

ate

sch

ools

by

42,2

40,0

00.

a. W

rite

a r

atio

nal

exp

ress

ion

to

expr

ess

the

rati

o of

pu

blic

sch

ool

stu

den

ts t

o x

priv

ate

sch

ool

stu

den

ts.

x +

42,

240,

00

0

−

x

b.

How

man

y st

ude

nts

att

ende

d pr

ivat

e sc

hoo

l?

6,40

0,0

00

11-3

x +

2

x 2

+ 1

0x +

24

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2112

/20/

10

7:41

PM


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

22

G

lenc

oe A

lgeb

ra 1

Sh

an

no

n’s

Ju

gg

lin

g T

heo

rem

Mat

hem

atic

ian

s lo

ok a

t va

riou

s m

ath

emat

ical

way

s to

rep

rese

nt

jugg

lin

g. O

ne

way

th

ey

hav

e fo

un

d to

rep

rese

nt

jugg

lin

g is

Sh

ann

on’s

Ju

ggli

ng

Th

eore

m. S

han

non

’s J

ugg

lin

g T

heo

rem

use

s th

e ra

tion

al e

quat

ion

f +

d

−

v +

d =

b −

h

wh

ere

f is

th

e fl

igh

t ti

me,

or

how

lon

g a

ball

is

in t

he

air,

d i

s th

e dw

ell

tim

e, o

r h

ow l

ong

a ba

ll i

s in

a h

and,

v i

s th

e va

can

t ti

me,

or

how

lon

g a

han

d is

em

pty,

b i

s th

e n

um

ber

of b

alls

, an

d h

is

the

nu

mbe

r of

han

ds (

eith

er 1

or

2 fo

r a

real

-lif

e si

tuat

ion

, pos

sibl

y m

ore

for

a co

mpu

ter

sim

ula

tion

).S

o, g

iven

th

e va

lues

for

f, d

, v, a

nd

h, i

t is

pos

sibl

e to

det

erm

ine

the

nu

mbe

r of

bal

ls

bein

g ju

ggle

d. I

f th

e fl

igh

t ti

me

is 9

sec

onds

, th

e dw

ell

tim

e is

3 s

econ

ds, t

he

vaca

nt

tim

e is

1 s

econ

d, a

nd

the

nu

mbe

r of

han

ds i

s 2,

how

man

y ba

lls

are

bein

g ju

ggle

d?

f +

d

−

v +

d

= b −

h

Ori

gin

al equation

9 +

3

−

1 +

3

= b −

2 f =

9,

d =

3,

v =

1,

and h

= 2

12

−

4 =

b −

2 S

implif

y.

24

= 4

b C

ross m

ultip

ly.

6

= b

D

ivid

e.

So,

th

e n

um

ber

of b

alls

bei

ng

jugg

led

is 6

.

Giv

en t

he

foll

owin

g in

form

atio

n, d

eter

min

e th

e n

um

ber

of

bal

ls b

ein

g ju

ggle

d.

1. fl

igh

t ti

me

= 6

sec

onds

, vac

ant

tim

e =

1 s

econ

d,

dwel

l ti

me

= 1

sec

ond,

nu

mbe

r of

han

ds =

27

2. f

ligh

t ti

me

= 1

3 se

con

ds, v

acan

t ti

me

= 1

sec

ond,

dw

ell

tim

e =

5 s

econ

ds, n

um

ber

of h

ands

= 1

3

3. fl

igh

t ti

me

= 4

sec

onds

, vac

ant

tim

e =

1 s

econ

d,

dwel

l ti

me

= 1

sec

ond,

nu

mbe

r of

han

ds =

25

4. f

ligh

t ti

me

= 1

6 se

con

ds, v

acan

t ti

me

= 1

sec

ond,

dw

ell

tim

e =

2 s

econ

ds, n

um

ber

of h

ands

= 2

12

5. fl

igh

t ti

me

= 1

8 se

con

ds, v

acan

t ti

me

= 3

sec

onds

, dw

ell

tim

e =

2 s

econ

ds, n

um

ber

of h

ands

= 1

4

Enri

chm

ent

11-3

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2212

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-4

Cha

pte

r 11

23

G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

M

ult

iply

ing

an

d D

ivid

ing

Rati

on

al

Exp

ressio

ns

Mu

ltip

ly R

atio

nal

Exp

ress

ion

s T

o m

ult

iply

rat

ion

al e

xpre

ssio

ns,

you

mu

ltip

ly t

he

nu

mer

ator

s an

d m

ult

iply

th

e de

nom

inat

ors.

Th

en s

impl

ify.

F

ind

2 c 2 f

−

5a b

2 · a

2 b

−

3cf .

2 c 2 f

−

5a b

2 · a

2 b

−

3cf

= 2 a

2 b c 2 f

−

15a b

2 cf

M

ultip

ly.

=

1 (

abcf

)(2a

c)

−

1 (ab

cf)(

15b)

S

implif

y.

=

2a

c −

15b

Sim

plif

y.

F

ind

x 2

- 1

6 −

2x +

8

·

x +

4

−

x 2

+ 8

x +

16 .

x 2 -

16

−

2x +

8 ·

x +

4

−

x 2 +

8x

+ 1

6 =

(x

- 4

)(x

+ 4

)

−

2(x

+ 4

)

·

x +

4

−

(x +

4)(

x +

4)

Facto

r.

= (x

- 4

)(x

+ 4

)1

−

2(x

+ 4

) 1 ·

x

+ 4

1 −

(x +

4)(

x +

4) 1

Sim

plif

y.

=

x -

4

−

2x +

8

M

ultip

ly.

Exer

cise

sF

ind

eac

h p

rod

uct

.

1.

6ab

−

a 2 b

2 · a

2 −

b 2 6a

−

b 3

2. m

p 2

−

3 ·

4

−

mp

4p

−

3

3.

x +

2

−

x -

4 ·

x -

4

−

x -

1

x +

2

−

x -

1

4. m

- 5

−

8 ·

16

−

m -

5 2

5.

2n -

8

−

n +

2

· 2n

+ 4

−

n -

4

4 6.

x 2 -

64

−

2x +

16

·

x +

8

−

x 2 +

16x

+ 6

4

x -

8

−

2x +

16

7.

8x

+ 8

−

x 2 -

2x

+ 1

·

x -

1

−

2x +

2

4

−

x -

1

8. a

2 -

25

−

a +

2

· a

2 -

4

−

a -

5

(a +

5)(

a -

2)

9.

x 2 +

6x

+ 8

−

2 x 2

+ 9

x +

4 ·

2 x 2

- x

- 1

−

x 2 -

3x

+ 2

x

+ 2

−

x -

2

10.

m

2 -

1

−

2 m 2

- m

- 1

·

2m +

1

−

m 2

- 2

m +

1

m

+ 1

−

(m -

1 ) 2

11.

n

2 -

1

−

n 2

- 7

n +

10 ·

n

2 -

25

−

n2

+ 6

n +

5

n -

1

−

n -

2

12.

3p -

3r

−

10pr

· 20

p 2 r

2 −

p 2 -

r 2

6p

r −

p +

r

13.

a 2

+ 7

a +

12

−

a 2

+ 2

a -

8

· a

2 +

3a

- 1

0

−

a 2

+ 2

a -

8

14

. v 2

- 4

v -

21

−

3 v 2

+ 6

v

·

v 2 +

8v

−

v 2 +

11v

+ 2

4

v -

7

−

3v +

6

(a +

3)(

a +

5)

−

(a -

2)(

a +

4)

11-4

Exam

ple

1

Exam

ple

2

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2312

/20/

10

7:41

PM


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

24

G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Mu

ltip

lyin

g a

nd

Div

idin

g R

ati

on

al

Exp

ressio

ns

Div

ide

Rat

ion

al E

xpre

ssio

ns

To

divi

de r

atio

nal

exp

ress

ion

s, m

ult

iply

by

the

reci

proc

al o

f th

e di

viso

r. T

hen

sim

plif

y.

F

ind

12 c 2 f

−

5 a 2 b

2 ÷

c 2 f

2

−

10a

b .

12 c 2 f

−

5 a 2 b

2 ÷

c 2 f 2

−

10ab

= 12

c 2 f

−

5 a 2 b

2 × 10

ab

−

c 2 f 2

=

121 c2 f 1

−

15aa

2 b2 b ×

2 10 1 a

b 1

−

1 c 2 f

2 f

=

24

−

abf

F

ind

x

2 +

6x

- 2

7

−

x 2

+ 1

1x +

18

÷

x -

3

−

x 2

+ x

- 2

.

x 2 +

6x

- 2

7

−

x 2 +

11x

+ 1

8 ÷

x -

3

−

x 2 +

x -

2

=

x 2 +

6x

- 2

7

−

x 2 +

11x

+ 1

8 ×

x 2

+ x

- 2

−

x -

3

= (x

+ 9

)(x

- 3

)

−

(x +

9)(

x +

2)

×

(x +

2)(

x -

1)

−

x -

3

= 1 (x

+ 9

)(x

- 3

)1

−

1(x

+ 9

)(x

+ 2

) 1 × 1 (x

+ 2

)(x

- 1

)

−

x -

31

= x

- 1

Exer

cise

sF

ind

eac

h q

uot

ien

t.

1.

12ab

−

a 2 b

2 ÷ b −

a

12

−

b 2

2. n

−

4 ÷ n

−

p p

−

4

3.

3 xy 2

−

8 ÷

6xy

y

−

16

4. m

- 5

−

8 ÷

m -

5

−

16

2

5.

2n -

4

−

2n

÷

n 2

- 4

−

n

1 −

n +

2

6.

y 2 -

36

−

y 2 -

49 ÷

y +

6

−

y +

7

y -

6

−

y -

7

7.

x 2 -

5x

+ 6

−

5

÷ x

- 3

−

15

3(x

- 2

) 8.

a

2 b 3 c

−

3 r 2 t

÷

6 a 2 b

c −

8r t 2 u

4 b

2 tu

−

9r

9.

x 2 +

6x

+ 8

−

x 2 +

4x

+ 4

÷

x

+ 4

−

x +

2 1

10

. m

2 -

49

−

m

÷

m

2 -

13m

+ 4

2

−

3 m 2

3m

(m +

7)

−

m -

6

11.

n 2

- 5

n +

6

−

n 2

+ 3

n

÷

3 -

n

−

4n

+ 1

2 -

4(n

- 2

) −

n

12

. p

2 -

2pr

+ r

2

−

p +

r

÷

p

2 -

r 2

−

p +

r

p -

r

−

p +

r

13.

a 2

+ 7

a +

12

−

a 2

+ 3

a -

10

÷

a 2

- 9

−

a 2

- 2

5 (a

+ 4

)(a

- 5

)

−

(a -

2)(

a -

3)

14.

a

2 -

9

−

2 a 2

+ 1

3a -

7

÷

a +

3

−

4 a 2

- 1

(a

- 3

)(2a

+ 1

)

−

a +

7

11-4

Exam

ple

1

Exam

ple

2

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2412

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-4

Cha

pte

r 11

25

G

lenc

oe A

lgeb

ra 1

Skill

s Pr

acti

ceM

ult

iply

ing

an

d D

ivid

ing

Rati

on

al

Exp

ressio

ns

Fin

d e

ach

pro

du

ct.

1.

14

−

c 2 · c 5 −

2c 7

c2 2.

3 m 2

−

2t

· t 2 −

12

m 2 t

−

8

3.

2 a 2 b

−

b 2 c

· b −

a

2a

−

c

4. 2 x

2 y

−

3 x 2 y

· 3x

y −

4y

x

−

2

5.

3(4m

- 6

) −

18r

·

9 r

2 −

2(4m

- 6

) 3r

−

4 6.

4(

n +

2)

−

n(n

- 2

) · n

- 2

−

n +

2

4 −

n

7.

(y -

3)(

y +

3)

−

4

·

8 −

y +

3 2

y-

6 8.

(x

- 2

)(x

+ 2

)

−

x(8x

+ 3

)

·

2(8x

+ 3

) −

x -

2

2(x +

2)

−

x

9.

(a -

7)(

a +

7)

−

a(a

+ 5

)

· a

+ 5

−

a +

7

a -

7

−

a

10

.

4(b

+ 4

) −

(b -

4)(

b -

3)

· b

- 3

−

b +

4

4 −

b -

4

Fin

d e

ach

qu

otie

nt.

11.

c3 −

d3

÷ d

3 −

c3

c6

−

d6

12.

x3 −

y2 ÷

x3 −

y 1

−

y

13.

6a3

−

4f 2 ÷

2a

2 −

12f 2

9a

14.

4m3

−

rp2

÷ 2m

−

rp

2m2

−

p

15.

3b +

3

−

b +

2

÷ (

b +

1)

3

−

b +

2

16.

x -

5

−

x +

3 ÷

(x

- 5

)

1 −

x +

3

17.

x2 -

x -

12

−

6

÷ x

+ 3

−

x -

4

(x -

4)2

−

6

18.

a2 -

5a

- 6

−

3

÷ a

- 6

−

a +

1

(a +

1)2

−

3

19.

m2

+ 2

m +

1

−

10m

- 1

0

÷ m

+ 1

−

20

2(

m +

1)

−

m -

1

20

. y2

+ 1

0y +

25

−

3y -

9

÷

y

+ 5

−

y -

3

y +

5

−

3

21.

b

+ 4

−

b2 -

8b

+ 1

6 ÷ 2b

+ 8

−

b -

8

b

- 8

−

2(b

- 4

)2 22

. 6x

+ 6

−

x -

1

÷ x2

+ 3

x +

2

−

2x -

2

12

−

x +

2

11-4

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2512

/20/

10

7:41

PM


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

26

G

lenc

oe A

lgeb

ra 1

Prac

tice

M

ult

iply

ing

an

d D

ivid

ing

Rati

on

al

Exp

ressio

ns

Fin

d e

ach

pro

du

ct o

r q

uot

ien

t.

1.

18 x 2

−

10 y 2 ·

15 y 3

−

24x

9xy

−

8

2. 24

rt 2

−

8 r 4 t

3 · 12

r 3 t 2

−

36 r 2 t

1 −

r 2

3.

(x +

2)(

x +

2)

−

8

·

72

−

(x +

2)(

x -

2)

9(x +

2)

−

x -

2

4.

m

+ 7

−

(m -

6)(

m +

2)

· (m

- 6

)(m

+ 4

)

−

(m +

7)

m

+ 4

−

m +

2

5.

a

- 4

−

a 2

- a

- 1

2 · a

+ 3

−

a -

6

1 −

a -

6

6. 4x

+ 8

−

x 2 ·

x −

x 2 -

5x

- 1

4

4 −

x(x

- 7

)

7.

n 2

+ 1

0n +

16

−

5n -

10

·

n -

2

−

n 2

+ 9

n +

8

n

+ 2

−

5(n

+ 1

) 8.

3y

- 9

−

y 2 -

9y

+ 2

0 · y 2

- 8

y +

16

−

y -

3

3(

y -

4)

−

y -

5

9.

b 2 +

5b

+ 4

−

b 2 -

36

· b 2

+ 5

b -

6

−

b 2 +

2b

- 8

(b

+ 1

)(b

- 1

)

−

(b -

6)(

b -

2)

10.

t 2 +

6t

+ 9

−

t 2 -

10t

+ 2

5 ·

t 2 -

t -

20

−

t 2 +

7t

+ 1

2 t

+ 3

−

t -

5

11.

28a2

−

7b2

÷ 21

a3 −

35b

20

−

3ab

12.

mn

2 p3

−

x4 y2

÷ m

np2

−

x3 y

n

p

−

xy

13.

2a

−

a -

1 ÷

(a

+ 1

)

2a

−

(a +

1)(

a -

1)

14.

z2 -

16

−

3z

÷ (

z -

4)

z +

4

−

3z

15.

4y +

20

−

y -

3

÷

y +

5

−

2y -

6 8

16

. 4x

+ 1

2 −

6x -

24 ÷

2x +

6

−

x +

3

x +

3

−

3(x -

4)

17.

b2 +

2b

- 8

−

b2 -

11b

+ 1

8 ÷

2b -

8

−

2b -

18

b +

4

−

b -

4

18.

3x -

3

−

x2 -

6x

+ 9

÷

6x -

6

−

x2 -

5x

+ 6

x -

2

−

2(x -

3)

19.

a2 +

8a

+ 1

2

−

a2 -

7a

+ 1

0 ÷ a2

- 4

a -

12

−

a2 +

3a

- 1

0 2

0. y

2 +

6y

- 7

−

y2 +

8y

- 9

÷

y2

+ 9

y +

14

−

y2 +

7y

- 1

8

(a +

6)(

a +

5)

−

(a -

6)(

a -

5)

y -

2

−

y +

2

21. B

IOLO

GY

Th

e h

eart

of

an a

vera

ge p

erso

n p

um

ps a

bou

t 90

00 l

iter

s of

blo

od p

er d

ay.

How

man

y qu

arts

of

bloo

d do

es t

he

hea

rt p

um

p pe

r h

our?

(H

int:

On

e qu

art

is e

qual

to

0.94

6 li

ter.

) R

oun

d to

th

e n

eare

st w

hol

e n

um

ber.

396

qt/

h

22. T

RA

FFIC

On

Sat

urd

ay, i

t to

ok M

s. T

orre

s 24

min

ute

s to

dri

ve 2

0 m

iles

fro

m h

er h

ome

to h

er o

ffic

e. D

uri

ng

Fri

day’

s ru

sh h

our,

it t

ook

75 m

inu

tes

to d

rive

th

e sa

me

dist

ance

.

a. W

hat

was

Ms.

Tor

res’

s av

erag

e sp

eed

in m

iles

per

hou

r on

Sat

urd

ay?

50 m

ph

b.

Wh

at w

as h

er a

vera

ge s

peed

in

mil

es p

er h

our

on F

rida

y? 1

6 m

ph

11-4

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2612

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-4

Cha

pte

r 11

27

G

lenc

oe A

lgeb

ra 1

Wor

d Pr

oble

m P

ract

ice

Mu

ltip

lyin

g a

nd

Div

idin

g R

ati

on

al

Exp

ressio

ns

1.JO

BS

Ros

a ea

rned

$26

.25

for

baby

sitt

ing

for

3 1−

2 hou

rs. A

t th

is r

ate,

how

mu

ch w

ill

she

earn

bab

ysit

tin

g fo

r 5

hou

rs?

2. H

OM

EWO

RK

Ale

jan

dro

and

An

der

wer

e w

orki

ng

on t

he

foll

owin

g h

omew

ork

prob

lem

.

Fin

d n

- 1

0 −

n +

3

· 2n

+ 6

−

n +

3 .

Ale

jan

dro’

s S

olu

tion

n -

10

−

n +

3

· 2n

+ 6

−

n +

3

= 2(

n -

10)

(n +

3 ) 1

−

−

(n +

3)(

n +

3 ) 1

= 2n

- 2

0 −

n +

3

A

nde

r’s

Sol

uti

on

n -

10

−

n +

3

· 2n

+ 6

−

n +

3

= 2(

n -

10)

(n +

3 ) 1

−

−

n +

3 1

= 2

n -

20

Is e

ith

er o

f th

em c

orre

ct?

Exp

lain

.

Ale

jan

dro

is c

orr

ect

bec

ause

he

mu

ltip

lied

den

om

inat

ors

. An

der

tr

eate

d t

he

den

om

inat

or

like

it w

as

an a

dd

itio

n p

rob

lem

.

3. G

EOM

ETRY

Su

ppos

e th

e ra

tion

al

expr

essi

on 5k

2 m3

−

2ab

repr

esen

ts t

he

area

of a

sec

tion

in

a t

iled

flo

or a

nd

2km

−

a

r

epre

sen

ts t

he

sect

ion

’s l

engt

h. W

rite

a

rati

onal

exp

ress

ion

to

repr

esen

t th

e se

ctio

n’s

wid

th.

4. T

RA

VEL

Hel

ene

trav

els

800

mil

es f

rom

A

mar

illo

to

Bro

wn

svil

le a

t an

ave

rage

sp

eed

of 4

0 m

iles

per

hou

r. S

he

mak

es

the

retu

rn t

rip

driv

ing

an a

vera

ge o

f 60

mil

es p

er h

our.

Wh

at i

s th

e av

erag

e ra

te f

or t

he

enti

re t

rip?

(H

int:

Rec

all

that

t =

d ÷

r.)

48

mp

h

5. M

AN

UFA

CTU

RIN

G I

ndi

a w

orks

in

a

met

al s

hop

an

d n

eeds

to

dril

l eq

ual

ly

spac

ed h

oles

alo

ng

a st

rip

of m

etal

. Th

e ce

nte

rs o

f th

e h

oles

on

th

e en

ds o

f th

e st

rip

mu

st b

e ex

actl

y 1

inch

fro

m e

ach

en

d. T

he

rem

ain

ing

hol

es w

ill

be e

qual

ly

spac

ed.

a. I

f th

ere

are

x eq

ual

ly s

pace

d h

oles

, w

rite

an

exp

ress

ion

for

th

e n

um

ber

of

equ

al s

pace

s th

ere

are

betw

een

hol

es.

x -

1

b.

Wri

te a

n e

xpre

ssio

n f

or t

he

dist

ance

be

twee

n t

he

end

scre

ws

if t

he

len

gth

is

ℓ.

c. W

rite

a r

atio

nal

equ

atio

n t

hat

re

pres

ents

th

e di

stan

ce b

etw

een

th

e h

oles

on

a p

iece

of

met

al t

hat

is

� i

nch

es l

ong

and

mu

st h

ave

x eq

ual

ly

spac

ed h

oles

.

d.

How

man

y h

oles

wil

l be

dri

lled

in

a

met

al s

trip

th

at i

s 6

feet

lon

g w

ith

a

dist

ance

of

7 in

ches

bet

wee

n t

he

cen

ters

of

each

scr

ew?

11d

dd

dd

1 inch

1 inch

�

11-4

$37

.50

5km

2 −

4b

� -

2

d =

� -

2

−

x -

1

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2712

/20/

10

7:41

PM


An

swer

s

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

28

G

lenc

oe A

lgeb

ra 1

Geo

metr

ic S

eri

es

A g

eom

etri

c se

ries

is

a su

m o

f th

e te

rms

in a

geo

met

ric

sequ

ence

. Eac

h t

erm

of

a ge

omet

ric

sequ

ence

is

form

ed b

y m

ult

iply

ing

the

prev

iou

s te

rm b

y a

con

stan

t te

rm c

alle

d th

e co

mm

on

rati

o.

1 +

1 −

2 + 1 −

4 +

1

−

8 ← g

eom

etri

c se

quen

ce w

her

e th

e co

mm

on r

atio

is

1 −

2

Th

e su

m o

f a

geom

etri

c se

ries

can

be

repr

esen

ted

by t

he

rati

onal

exp

ress

ion

x 0 (r

)n -

1

−

r -

1 ,

wh

ere

x 0 is

th

e fi

rst

term

of

the

seri

es, r

is

the

com

mon

rat

io, a

nd

n i

s th

e

nu

mbe

r of t

erm

s.

In t

he

exam

ple

abov

e, 1

+ 1 −

2 + 1 −

4 + 1 −

8 = 1

․

( 1 −

2 ) 4 - 1

−

1 −

2 - 1

o

r 15

−

8 .

You

can

ch

eck

this

by

ente

rin

g 1

+ 1 −

2 + 1 −

4 + 1 −

8 a c

alcu

lato

r. T

he

resu

lt i

s th

e sa

me.

Rew

rite

eac

h s

um

as

a ra

tion

al e

xpre

ssio

n a

nd

sim

pli

fy.

1. 9

+ 3

+ 1

+ 1 −

3 + 1 −

9 12

1 −

9

2. 5

00 +

250

+ 1

25 +

62

1 −

2 18

75

−

2

3. 6

+ 1

+ 1 −

6 +

1 −

36

259

−

36

4. 1

00 +

20

+ 4

+ 4 −

5 62

4 −

5

5. 1

000

+ 1

00 +

10

+ 1

+

1 −

10 +

1

−

100 +

1

−

1000

1,

111,

111

−

100

0

6. 5

5 +

5 +

5 −

11 +

5

−

121

7320

−

121

Enri

chm

ent

11-4

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2812

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-4

Cha

pte

r 11

29

G

lenc

oe A

lgeb

ra 1

Spre

adsh

eet

Act

ivit

yR

evo

luti

on

s p

er

Min

ute

On

e of

th

e ch

arac

teri

stic

s th

at m

akes

a s

prea

dsh

eet

pow

erfu

l is

th

e ab

ilit

y to

rec

alcu

late

va

lues

in

for

mu

las

auto

mat

ical

ly. Y

ou c

an u

se t

his

abi

lity

to

inve

stig

ate

real

-wor

ld

situ

atio

ns.

Exer

cise

s

Use

th

e sp

read

shee

t of

rev

olu

tion

s p

er m

inu

te.

1. H

ow i

s th

e n

um

ber

of r

evol

uti

ons

affe

cted

if

the

spee

d of

a w

hee

l of

a g

iven

dia

met

er i

s do

ubl

ed?

RP

M is

cu

t in

hal

f.

2. N

ame

two

way

s th

at y

ou c

an d

oubl

e th

e R

PM

of

a bi

cycl

e w

hee

l. D

ou

ble

th

e sp

eed

o

r h

alve

th

e d

iam

eter

of

the

wh

eel,

keep

ing

th

e sa

me

spee

d.

U

se a

sp

read

shee

t to

in

vest

igat

e th

e ef

fect

of

dou

bli

ng

the

dia

met

er o

f a

tire

on

th

e n

um

ber

of

revo

luti

ons

the

tire

mak

es a

t a

give

n s

pee

d.

Use

dim

ensi

onal

an

alys

is t

o fi

nd

the

form

ula

for

th

e re

volu

tion

s pe

r m

inu

te

of a

tir

e w

ith

dia

met

er o

f x

inch

es t

rave

lin

g at

y m

iles

per

hou

r.

1 re

volu

tion

−

π ·

x

× y

−

1 ×

1

−

60 m

inu

tes ×

63

,360

−

1

=

1056

y re

volu

tion

s

−−

π ·

x

min

ute

s

Ste

p 1

U

se C

olu

mn

A o

f th

e sp

read

shee

t fo

r di

amet

er

of t

he

tire

in

in

ches

. Use

C

olu

mn

B f

or t

he

spee

d in

m

iles

per

hou

r.

Ste

p 2

C

olu

mn

C c

onta

ins

the

form

ula

for

th

e n

um

ber

of

rota

tion

s pe

r m

inu

te.

Not

ice

that

in

Exc

el, π

is

ente

red

as P

I( )

.

Ste

p 3

C

hoo

se v

alu

es f

or t

he

diam

eter

an

d sp

eed

and

stu

dy t

he

resu

lts

show

n i

n t

he

spre

adsh

eet.

To

com

pare

th

e re

volu

tion

s pe

r m

inu

te f

or d

oubl

ed d

iam

eter

s,

keep

th

e sp

eed

con

stan

t an

d ch

ange

th

e di

amet

ers.

It

app

ears

th

at w

hen

th

e di

amet

er i

s do

ubl

ed, t

he

nu

mbe

r of

rev

olu

tion

s pe

r m

inu

te i

s h

alve

d.

A1 3 4 5 62

BC

Dia

met

er (

in.)

Sp

eed

(m

ph

)R

PM

=(1

056*

B2)

/(A

2*P

I())

=(1

056*

B3)

/(A

3*P

I())

=(1

056*

B4)

/(A

4*P

I())

=(1

056*

B5)

/(A

5*P

I())

Sh

eet

1S

hee

t 2

Sh

eet

3

A1 3 4 5 62

BC

Dia

met

er (

in.)

Sp

eed

(m

ph

)R

PM

1848

7.44

9243

.72

4621

.86

2310

.93

1 2 4 8

55 55 55 55

Sh

eet

1S

hee

t 2

Sh

eet

3

11-4

Exam

ple

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

2912

/20/

10

7:41

PM


Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass




NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

30

G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

Div

idin

g P

oly

no

mia

ls

Div

ide

Poly

no

mia

ls b

y M

on

om

ials

To

divi

de a

pol

ynom

ial

by a

mon

omia

l, di

vide

ea

ch t

erm

of

the

poly

nom

ial

by t

he

mon

omia

l.

F

ind

(4r

2 -

12r

) ÷

(2r

).

(4r2

- 1

2r)

÷ 2

r =

4r

2 -

12r

−

2r

= 4r

2 −

2r -

12r

−

2r

Div

ide e

ach t

erm

.

=

2r

4r

2

−

1

2r -

12r

6

−

2r

1

Sim

plif

y.

= 2

r -

6

Sim

plif

y.

F

ind

(3x

2 -

8x

+ 4

) ÷

(4x

).

(3x2

- 8

x +

4)

÷ 4

x =

3x2

- 8

x +

4

−

4x

= 3x

2 −

4x -

8x

−

4x +

4 −

4x

= 3x

4 3x2

−

4x -

8x

−

4x +

4 −

4x

= 3x

−

4 -

2 +

1 −

x

Exer

cise

sF

ind

eac

h q

uot

ien

t.

1. (

x3 +

2x2

- x

) ÷

x x

2 +

2x -

1

2. (

2x3

+ 1

2x2

- 8

x) ÷

(2x

) x

2 +

6x -

4

3. (

x2 +

3x

- 4

) ÷

x x

+ 3

- 4 −

x

4. (

4m2

+ 6

m -

8)

÷ (

2m2 )

2 +

3 −

m -

4

−

m2

5. (

3x3

+ 1

5x2

- 2

1x)

÷ (

3x)

x2

+ 5

x -

7

6. (

8m2 p

2 +

4m

p -

8p)

÷ p

8m

2 p +

4m

- 8

7. (

8y4

+ 1

6y2

- 4

) ÷

(4y

2 ) 2

y2

+ 4

- 1 −

y2

8. (

16x4 y

2 + 2

4xy

+ 5

) ÷

(xy

) 16

x3 y

+ 2

4 +

5

−

xy

9.

15x2

- 2

5x +

30

−

5 3

x2

- 5

x +

6

10.

10a2 b

+ 1

2ab

- 8

b

−−

2a

5ab

+ 6

b -

4b

−

a

11.

6x3

+ 9

x2 +

9

−

3x

2x

2 +

3x +

3 −

x

12.

m2

- 1

2m +

42

−

3m2

1 −

3 - 4 −

m +

14

−

m2

13.

m2 p

2 -

5m

p +

6

−

m2 p

2 1

-

5 −

mp +

6

−

m2 p

2

14.

p2 -

4pr

+ 6

r2

−

pr

p

−

r -

4 +

6r

−

p

15.

6a2 b

2 -

8ab

+ 1

2

−−

2a2

3b

2 -

4b

−

a

+ 6 −

a2

16.

2x2 y

3 -

4x2 y

2 -

8xy

−

−

2xy

xy

2 -

2xy -

4

17.

9x2 y

2 z -

2xy

z +

12x

−

−

xy

18

. 2a

3 b3

+ 8

a2 b2

- 1

0ab

+ 1

2

−

−

2a2 b

2

9

xyz -

2z +

12

−

y

a

b +

4 -

5

−

ab +

6

−

a2 b

2

11-5

Exam

ple

1Ex

amp

le 2

2 1

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3012

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-5

Cha

pte

r 11

31

G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Div

idin

g P

oly

no

mia

ls

Div

ide

Poly

no

mia

ls b

y B

ino

mia

ls T

o di

vide

a p

olyn

omia

l by

a b

inom

ial,

fact

or t

he

divi

den

d if

pos

sibl

e an

d di

vide

bot

h d

ivid

end

and

divi

sor

by t

he

GC

F. I

f th

e po

lyn

omia

l ca

nn

ot b

e fa

ctor

ed, u

se l

ong

divi

sion

.

F

ind

(x2

+ 7

x +

10)

÷ (

x +

3).

Ste

p 1

D

ivid

e th

e fi

rst

term

of

the

divi

den

d, x

2 by

th

e fi

rst

term

of

the

divi

sor,

x.

x

x +

3 � �

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

x2 +

7x

+ 1

0

(-)

x2 +

3x

M

ultip

ly x

and x

+ 3

.

4x

Subtr

act.

Ste

p 2

B

rin

g do

wn

th

e n

ext

term

, 10.

Div

ide

the

firs

t te

rm o

f 4x

+ 1

0 by

x.

x +

4x

+ 3

� ��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

�

x2 +

7x

+ 1

0

x2 +

3x

4x

+ 1

0

(-

) 4x

+ 1

2 Mu

ltip

ly 4

and x

+ 3

.

-2

Subtr

act.

Th

e qu

otie

nt

is x

+ 4

wit

h r

emai

nde

r -

2. T

he

quot

ien

t ca

n b

e w

ritt

en a

s x

+ 4

+

-2

−

x +

3 .

Exer

cise

sF

ind

eac

h q

uot

ien

t.

1. (

b2 -

5b

+ 6

) ÷

(b

- 2

) b

- 3

2.

(x2

- x

- 6

) ÷

(x

- 3

) x +

2

3. (

x2 +

3x

- 4

) ÷

(x

- 1

) x +

4

4. (

m2

+ 2

m -

8)

÷ (

m +

4)

m -

2

5. (

x2 +

5x

+ 6

) ÷

(x

+ 2

) x +

3

6. (

m2

+ 4

m +

4)

÷ (

m +

2)

m +

2

7. (

2y2

+ 5

y +

2)

÷ (

y +

2)

2y +

1

8. (

8y2

- 1

5y -

2)

÷ (

y -

2)

8y +

1

9.

8x2

- 6

x -

9

−

4x +

3

2x

- 3

10

. m

2 -

5m

- 6

−

m -

6

m

+ 1

11.

x3 +

1

−

x -

2

x2 +

2x +

4 +

9

−

x -

2

12.

6m3

+ 1

1m2

+ 4

m +

35

−−

2m +

5

3m

2 -

2m

+ 7

13.

6a2

+ 7

a +

5

−

2a +

5

3a

- 4

+

25

−

2a +

5

14.

8p3

+ 2

7 −

2p +

3

4p

2 -

6p

+ 9

11-5

Exam

ple

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3112

/20/

10

7:41

PM


Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass

Chapter 11 A15 Glencoe Algebra 1


NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

32

G

lenc

oe A

lgeb

ra 1

Skill

s Pr

acti

ceD

ivid

ing

Po

lyn

om

ials

Fin

d e

ach

qu

otie

nt.

1. (

20x2

+ 1

2x)

÷ 4

x 5

x +

3

2. (

18n

2 +

6n

) ÷

3n

6n

+ 2

3. (

b2 -

12b

+ 5

) ÷

2b

b

−

2 - 6

+

5 −

2b

4. (

8r2

+ 5

r -

20)

÷ 4

r 2

r +

5 −

4 - 5 −

r

5.

12p3 r

2 +

18p

2 r -

6pr

−

−

6p2 r

2pr

+ 3

- 1 −

p

6. 15

k2 u -

10k

u +

25u

2

−−

5ku

3k

- 2

+ 5u

−

k

7. (

x2 -

5x

- 6

) ÷

(x

- 6

) x +

1

8. (

a2 -

10a

+ 1

6) ÷

(a

- 2

) a

- 8

9. (

n2

- n

- 2

0) ÷

(n

+ 4

) n

- 5

10

. ( y

2 +

4y

- 2

1) ÷

( y

- 3

) y +

7

11. (

h2

- 6

h +

9)

÷ (

h -

2)

h -

4 +

1

−

h -

2

12. (

b2 +

5b

- 2

) ÷

(b

+ 6

) b

- 1

+

4 −

b +

6

13. (

y2

+ 6

y +

1)

÷ (

y +

2)

y +

4 -

7

−

y +

2

14. (

m2

- 2

m -

5)

÷ (

m -

3)

m +

1 -

2

−

m -

3

15.

2c2

- 5

c -

3

−

2c +

1

c -

3

16.

2r2

+ 6

r -

20

−

2r -

4

r +

5

17.

x3 -

3x2

- 6

x -

20

−

−

x -

5

x

2 +

2x +

4

18.

p3 -

4p2

+ p

+ 6

−

p -

2

p

2 -

2p

- 3

19.

n3

- 6

n -

2

−

n +

1

n

2 -

n -

5 +

3

−

n +

1

20.

y3 -

y2

- 4

0

−

y -

4

y2

+ 3

y +

12

+

8 −

y -

4

11-5

017_

035_

ALG

1_A

_CR

M_C

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R_6

6028

5.in

dd

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/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-5

Cha

pte

r 11

33

G

lenc

oe A

lgeb

ra 1

Prac

tice

D

ivid

ing

Po

lyn

om

ials

Fin

d e

ach

qu

otie

nt.

1. (

6q2

- 1

8q -

9)

÷ 9

q 2.

(y2

+ 6

y +

2)

÷ 3

y 3.

12a2 b

- 3

ab2

+ 4

2ab

−

−

6a2 b

2q

−

3 -

2 -

1 −

q

y

−

3 + 2

+

2 −

3y

2 -

b

−

2a +

7 −

a

4.

2m3 p

2 +

56m

p -

4m

2 p3

−

−

8m3 p

5. (

x2 -

3x

- 4

0) ÷

(x

+ 5

)

6. (3

m2 -

20m

+ 1

2) ÷

(m -

6)

p

−

4 +

7 −

m2

-

p2

−

2m

x -

8

3m

- 2

7. (

a2 +

5a

+ 2

0) ÷

(a

- 3

)

8. (

x2 -

3x

- 2

) ÷

(x

+ 7

) 9.

(t2

+ 9

t +

28)

÷ (

t +

3)

a

+ 8

+

44

−

a -

3

x -

10

+

68

−

x +

7

t +

6 +

10

−

t +

3

10. (

n2

- 9

n +

25)

÷ (

n -

4)

11

. 6r

2 -

5r

- 5

6

−

3r +

8

12

. 20

w2

+ 3

9w +

18

−

−

5w +

6

n

- 5

+

5 −

n -

4

2r

- 7

4

w +

3

13. (

x3 +

2x2

- 1

6) ÷

(x

- 2

)

14. (

t3 -

11t

- 6

) ÷

(t

+ 3

)

x

2 +

4x +

8

t2

- 3

t -

2

15. x3

+ 6

x2 +

3x

+ 1

−

−

x -

2

16

. 6d3

+ d

2 -

2d

+ 1

7

−−

2d +

3

x

2 +

8x +

19

+

39

−

x -

2

3d

2 -

4d

+ 5

+

2 −

2d +

3

17. 2k

3 +

k2

- 1

2k +

11

−

−

2k -

3

18

. 9y

3 -

y -

1

−

3y +

2

k

2 +

2k -

3 +

2

−

2k +

3

3y

2 -

2y +

1 -

3

−

3y +

2

19. L

AN

DSC

API

NG

Joc

elyn

is

desi

gnin

g a

bed

for

cact

us

spec

imen

s at

a b

otan

ical

gar

den

. T

he

tota

l ar

ea c

an b

e m

odel

ed b

y th

e ex

pres

sion

2x2

+ 7

x +

3, w

her

e x

is i

n f

eet.

a. S

upp

ose

in o

ne

desi

gn t

he

len

gth

of

the

cact

us

bed

is 4

x, a

nd

in a

not

her

, th

e le

ngt

h i

s

2x +

1. W

hat

are

th

e w

idth

s of

th

e tw

o de

sign

s?

b.

If x

= 3

fee

t, w

hat

wil

l be

th

e di

men

sion

s of

th

e ca

ctu

s be

d in

eac

h o

f th

e de

sign

s?12

ft

by 3

.5 f

t; 7

ft

by 6

ft

20. F

UR

NIT

UR

E T

eri

is u

phol

ster

ing

the

seat

s of

fou

r ch

airs

an

d a

ben

ch. S

he

nee

ds 1 −

4 squ

are

yard

of

fabr

ic f

or e

ach

ch

air,

and

1 −

2 squ

are

yard

for

th

e be

nch

. If

the

fabr

ic a

t th

e st

ore

is 4

5 in

ches

wid

e, h

ow m

any

yard

s of

fab

ric

wil

l Ter

i n

eed

to c

over

th

e ch

airs

an

d th

e be

nch

if

ther

e is

no

was

te?

11-5

x

−

2 + 7 −

4 +

3 −

4x ;

x +

3

1 1 −

5 yd

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3312

/20/

10

7:41

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF 2nd



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

34

G

lenc

oe A

lgeb

ra 1

Wor

d Pr

oble

m P

ract

ice

Div

idin

g P

oly

no

mia

ls

1.TE

CH

NO

LOG

Y T

he

surf

ace

area

in

sq

uar

e m

illi

met

ers

of a

rec

tan

gula

r co

mpu

ter

mic

roch

ip i

s re

pres

ente

d by

th

e ex

pres

sion

x2 -

12x

+ 3

5, w

her

e x

is t

he

nu

mbe

r of

cir

cuit

s. I

f th

e w

idth

of

the

chip

is

x -

5 m

illi

met

ers,

wri

te a

po

lyn

omia

l th

at r

epre

sen

ts t

he

len

gth

.

2. H

OM

EWO

RK

You

r cl

assm

ate

Ava

w

rite

s h

er a

nsw

er t

o a

hom

ewor

k pr

oble

m o

n t

he

chal

kboa

rd. S

he

has

sim

plif

ied

6x2

- 1

2x

−

6

as x

2 - 1

2x. I

s th

is

cor

rect

? If

not

, wh

at i

s th

e co

rrec

t si

mpl

ific

atio

n?

Th

is is

no

t co

rrec

t. S

he

forg

ot

to f

acto

r 6

fro

m t

he

–12 x

ter

m. T

he

corr

ect

answ

er

s

ho

uld

be

6(x

2 -

2x)

−

6

= x

2 -

2x.

3. C

IVIL

EN

GIN

EER

ING

Su

ppos

e 54

00 t

ons

of c

oncr

ete

cost

s (5

00 +

d)

doll

ars.

Wri

te

a fo

rmu

la t

hat

giv

es t

he

cost

C o

f t

ton

s

of

con

cret

e.

4.SH

IPPI

NG

Th

e O

vers

eas

Sh

ippi

ng

Com

pan

y lo

ads

carg

o in

to a

con

tain

er t

o be

sh

ippe

d ar

oun

d th

e w

orld

. Th

e vo

lum

e of

th

eir

ship

pin

g co

nta

iner

s is

de

term

ined

by

the

foll

owin

g eq

uat

ion

.

x3+

21x

2+

99x

+ 1

35

Th

e co

nta

iner

’s h

eigh

t is

x +

3. W

rite

an

ex

pres

sion

th

at r

epre

sen

ts t

he

area

of

the

base

of

the

ship

pin

g co

nta

iner

. x

2 + 1

8x +

45

5.C

IVIL

EN

GIN

EER

ING

Gre

ensh

ield

’s

For

mu

la c

an b

e u

sed

to d

eter

min

e th

e am

oun

t of

tim

e a

traf

fic

ligh

t at

an

in

ters

ecti

on s

hou

ld r

emai

n g

reen

.

G=

2.1

n+

3.7

G

= g

reen

tim

e in

sec

onds

n

= a

vera

ge n

um

ber

of v

ehic

les

trav

elin

g in

eac

h l

ane

per

ligh

t cy

cle

W

rite

a s

impl

ifie

d ex

pres

sion

to

repr

esen

t th

e av

erag

e gr

een

lig

ht

tim

e

per

veh

icle

.

6. S

OLI

D G

EOM

ETRY

Th

e su

rfac

e ar

ea o

f a

righ

t cy

lin

der

is g

iven

by

the

form

ula

S

= 2

πr2 +

2π

rh.

a. W

rite

a s

impl

ifie

d ra

tion

al e

xpre

ssio

n

that

rep

rese

nts

th

e ra

tio

of t

he

surf

ace

area

to

the

circ

um

fere

nce

of

the

cyli

nde

r.

r +

h

−

1

b.

Wri

te a

sim

plif

ied

rati

onal

exp

ress

ion

th

at r

epre

sen

ts t

he

rati

o of

th

e su

rfac

e ar

ea t

o th

e ar

ea o

f th

e ba

se.

2 +

2h

−

r

11-5 x –

7 m

m

500t

+ d

t −

540

0

2.1

+ 3.

7 −

n

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3412

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-5

Cha

pte

r 11

35

G

lenc

oe A

lgeb

ra 1

Enri

chm

ent

Syn

theti

c D

ivis

ion

You

can

div

ide

a po

lyn

omia

l su

ch a

s 3x

3 -

4x2

- 3

x -

2 b

y a

bin

omia

l su

ch a

s x

- 3

by a

pr

oces

s ca

lled

syn

thet

ic d

ivis

ion

. Com

pare

th

e pr

oces

s w

ith

lon

g di

visi

on i

n t

he

foll

owin

g ex

plan

atio

n.

D

ivid

e (3

x 3 -

4x 2

- 3

x -

2)

by

(x -

3)

usi

ng

syn

thet

ic d

ivis

ion

.

1. S

how

th

e co

effi

cien

ts o

f th

e te

rms

in

desc

endi

ng

orde

r.

2. T

he

divi

sor

is x

- 3

. Sin

ce 3

is

to b

esu

btra

cted

, wri

te 3

in

th

e co

rner

.

3. B

rin

g do

wn

th

e fi

rst

coef

fici

ent,

3.

4. M

ult

iply

. 3

.

3 =

9

5. A

dd.

-4

+ 9

= 5

6. M

ult

iply

. 3

.

5 =

15

7. A

dd.

-3

+ 1

5 =

12

8. M

ult

iply

. 3

.

12

= 3

6

9. A

dd.

-2

+ 3

6 =

34

Ch

eck

U

se l

ong

divi

sion

.3x

2 +

5x

+ 1

2x

- 3

� ��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

3x3

- 4

x2 -

3x

- 2

3x

3 -

9x2

5x2

-

3x

5x

2 -

15x 12

x -

2

12x

- 3

6 34

Th

e re

sult

is

3x2

+ 5

x +

12

+

34

−

x -

3 .

Div

ide

by

usi

ng

syn

thet

ic d

ivis

ion

. Ch

eck

you

r re

sult

usi

ng

lon

g d

ivis

ion

.

1. (

x3 +

6x2

+ 3

x +

1)

÷ (

x -

2)

2. (x

3 -

3x2

- 6

x -

20)

÷ (

x -

5)

x

2 +

8x +

19

+

39

−

x -

2

x

2 +

2x +

4

3. (

2x3

- 5

x +

1)

÷ (

x +

1)

4. (3

x3 -

7x2

+ 4

) ÷

(x

- 2

)

2x

2 -

2x -

3 +

4

−

x +

1

3x

2 +

x -

2

5. (

x3 +

2x2

- x

+ 4

) ÷

(x

+ 3

) 6.

(x3

+ 4

x2 -

3x

- 1

1) ÷

(x

- 4

)

x

2 -

x +

2 -

2

−

x +

3

x2

+ 8

x +

29

+

105

−

x -

4

3 3

-4

-3

-2

9 15

36

3

5 12

34

3x2

+ 5

x +

12,

rem

ain

der

34

11-5

Exam

ple

017_

035_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3512

/23/

10

11:3

0 A

M


A15_A27_ALG1_A_CRM_C11_AN_660285.indd A16A15_A27_ALG1_A_CRM_C11_AN_660285.indd A16 12/23/10 11:47 AM12/23/10 11:47 AM

Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

36

G

lenc

oe A

lgeb

ra 1

11-5

TI-N

spir

e® A

ctiv

ity

Div

idin

g P

oly

no

mia

ls

You

can

use

a g

raph

ing

calc

ula

tor

wit

h a

com

pute

r al

gebr

a sy

stem

(C

AS

) to

div

ide

poly

nom

ials

wit

h a

ny

divi

sor.

On

e fu

nct

ion

is

use

d to

fin

d th

e qu

otie

nt,

wh

ile

anot

her

fu

nct

ion

is

use

d to

fin

d th

e re

mai

nde

r.

U

se C

AS

to

fin

d (

x 4

+ x

3 – 3

x 2

+ x

) ÷

( x

2 +

4x

+ 5

).

Ste

p 1

A

dd a

new

Cal

cula

tor

page

on

th

e T

I-N

spir

e.

Ste

p 2

F

rom

th

e m

enu

, sel

ect

Alg

ebra

, Pol

ynom

ial

Too

ls a

nd

Qu

otie

nt

of P

olyn

omia

l.

Ste

p 3

T

ype

the

divi

den

d, a

com

ma,

an

d th

e di

viso

r.

Th

e C

AS

in

dica

tes

that

( x 4

+ x

3 – 3

x 2 +

x)

÷ ( x

2 +

4x

+ 5

) is

x 2

− 3

x +

4.

We

nee

d to

det

erm

ine

wh

eth

er t

her

e is

a r

emai

nde

r.

Ste

p 4

U

se t

he

Rem

ain

der

of

a P

olyn

omia

l op

tion

fr

om t

he

Alg

ebra

, Pol

ynom

ial

Too

ls m

enu

to

det

erm

ine

the

rem

ain

der.

Th

en t

ype

the

divi

den

d, a

com

ma,

an

d th

e di

viso

r.

T

he

rem

ain

der

is −

20. T

her

efor

e, (

x 4 +

x 3

− 3

x 2 +

x)

÷ (

x 2 +

4x

+ 5

) is

x 2 −

3x

+ 4

−

20

−

x 2 +

4x

+ 5

.

Ch

eck

U

se t

he

Exp

and

opt

ion

fro

m t

he

Alg

ebra

men

u t

o co

nfi

rm y

our

answ

er. T

ype

the

quot

ien

t m

ult

ipli

ed b

y th

e su

m o

f th

e di

viso

r an

d th

e re

mai

nde

r ov

er t

he

divi

sor.

Th

e C

AS

con

firm

s th

at t

he

divi

sion

w

as c

orre

ct.

Exer

cise

s

Fin

d e

ach

qu

otie

nt.

1. (

x 4 +

2 x 3

– 1

9 x 2

+ 28

x – 1

2) ÷

( x 2

+ 5

x – 6

) 2.

(2 x

4 +

3 x 2

+ x

+ 9

) ÷

( x 2

− 2

x +

3)

x 2

– 3

x +

2

2 x

2 +

4x +

5 −

x +

6

−

x 2 – 2

x +

3

3. (

3 x 4

− 2

0 x 3

+ 5

8 x 2

− 3

0x +

8)

÷ (

3 x 2

− 2x

+ 1

) 4.

(4 x

4 −

7 x 3

+ 2

9 x 2

− 3

x −

40)

÷

(4

x 2 +

x +

5)

x 2

+ 6

x +

15

+

6x −

7

−

3 x 2

− 2

x +

1

x 2 −

2x −

8

5. (

2 x 5

+ 1

5 x 4 +

10 x

3 +

4 x 2

+ 1

5x +

16)

÷ ( x

2 +

3x

+ 5

)

2

x 3 +

9 x 2

− 2

7x +

40

+

30x −

184

−

x 2 +

3x +

5

6. (

x 5 +

4 x 4

− 5

x 3 +

12 x

2 +

40x

+ 6

0) ÷

( x 3

+ x

+ 8

) x 2

+ 4

x +

6 +

14x +

108

−

x 3 +

x +

8

7. T

hink

Abo

ut It

Whe

n a

poly

nom

ial i

s di

vide

d by

( x 2 +

3x

+ 5

), th

e qu

otie

nt is

( x 2 +

6x

− 8

) w

ith

a re

mai

nder

of 4

x +

10.

Wha

t is

the

pol

ynom

ial?

x 4

+ 9

x 3 +

15 x

2 +

10x

− 3

0

Exam

ple

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3612

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-6

Cha

pte

r 11

37

G

lenc

oe A

lgeb

ra 1

11-6

Stud

y G

uide

and

Inte

rven

tion

Ad

din

g a

nd

Su

btr

acti

ng

Rati

on

al

Exp

ressio

ns

Ad

d a

nd

Su

btr

act

Rat

ion

al E

xpre

ssio

ns

wit

h L

ike

Den

om

inat

ors

To

add

rati

onal

exp

ress

ion

s w

ith

lik

e de

nom

inat

ors,

add

th

e n

um

erat

ors

and

then

wri

te t

he

sum

ov

er t

he

com

mon

den

omin

ator

. To

subt

ract

fra

ctio

ns

wit

h l

ike

den

omin

ator

s, s

ubt

ract

th

e n

um

erat

ors.

If

poss

ible

, sim

plif

y th

e re

sult

ing

rati

onal

exp

ress

ion

.

F

ind

5n

−

15 +

7n

−

15 .

5n

−

15 +

7n

−

15 =

5n +

7n

−

15

A

dd t

he n

um

era

tors

.

=

12n

−

15

Sim

plif

y.

=

12n

−

15 5 4n

D

ivid

e b

y 3

.

=

4n

−

5 S

implif

y.

F

ind

3x +

2

−

x -

2

-

4x

−

x -

2 .

3x +

2

−

x -

2

-

4x

−

x -

2 =

3x +

2 -

4x

−

x -

2

The c

om

mon

denom

inato

r is

x -

2.

=

2 -

x

−

x -

2

Subtr

act.

=

-1(

x -

2)

−

x -

2

2 -

x =

-1(x

- 2

)

=

-1(

x -

2)

−

x -

2

=

-1

−

1 S

implif

y.

=

-

1

Exer

cise

sF

ind

eac

h s

um

or

dif

fere

nce

.

1.

3 −

a +

4 −

a

7 −

a

2.

x2 −

8 + x −

8 x

2 +

x

−

8

3.

5x

−

9 -

x −

9 4x

−

9 4.

11

x −

15y -

x

−

15y

2x

−

3y

5.

2a -

4

−

a -

4

+

-a

−

a -

4 1

6.

m

+ 1

−

2m -

1 +

3m -

3

−

2m -

1 2

7.

y +

7

−

y +

6 -

1

−

y +

6 1

8.

3y

+ 5

−

5 -

2y

−

5 y

+ 5

−

5

9.

x +

1

−

x -

2 +

x -

5

−

x -

2 2

10

. 5a

−

3b2

+ 10

a −

3b2

5a

−

b2

11.

x2 +

x

−

x -

x2 +

5x

−

x -

4 12

. 5a

+ 2

−

a2 -

4a +

2

−

a2 1

−

a

13.

3x +

2

−

x +

2

+ x

+ 6

−

x +

2 4

14

. a

- 4

−

a +

1 +

a +

6

−

a +

1 2

Exam

ple

1Ex

amp

le 2

1 1

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3712

/20/

10

7:41

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

38

G

lenc

oe A

lgeb

ra 1

11-6

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Ad

din

g a

nd

Su

btr

acti

ng

Rati

on

al

Exp

ressio

ns

Ad

d a

nd

Su

btr

act

Rat

ion

al E

xpre

ssio

ns

wit

h U

nlik

e D

eno

min

ato

rs A

ddin

g or

su

btra

ctin

g ra

tion

al e

xpre

ssio

ns

wit

h u

nli

ke d

enom

inat

ors

is s

imil

ar t

o ad

din

g an

d su

btra

ctin

g fr

acti

ons

wit

h u

nli

ke d

enom

inat

ors.

Ad

din

g a

nd

Su

btr

acti

ng

Rat

ion

al E

xpre

ssio

ns

Ste

p 1

F

ind

th

e L

CD

of

the

exp

ressio

ns.

Ste

p 2

C

ha

ng

e e

ach

exp

ressio

n in

to a

n e

qu

iva

len

t exp

ressio

n w

ith

th

e L

CD

as

the

de

no

min

ato

r.

Ste

p 3

A

dd

or

su

btr

act

just

as w

ith

exp

ressio

ns w

ith

lik

e d

en

om

ina

tors

.

Ste

p 4

S

imp

lify if

ne

ce

ssa

ry.

F

ind

n +

3

−

n

+ 8n

- 4

−

4n

.

Fact

or e

ach

den

omin

ator

.

n =

n

4n =

4 .

n L

CD

= 4

nS

ince

th

e de

nom

inat

or o

f 8n

- 4

−

4n

is

alre

ady

4n, o

nly

n +

3

−

n

nee

ds t

o be

ren

amed

.

n +

3

−

n

+ 8n

- 4

−

4n

= 4(

n +

3)

−

4n

+ 8n

- 4

−

4n

=

4n +

12

−

4n

+ 8n

- 4

−

4n

=

12n

+ 8

−

4n

=

3n +

2

−

n

F

ind

3x

−

x2 -

4x -

1

−

x -

4 .

3x

−

x2 -

4x -

1

−

x -

4 =

3x

−

x(x

- 4

) -

1 −

x -

4

=

3x

−

x(x

- 4

) -

1 −

x -

4 ·

x −

x

=

3x

−

x(x

- 4

) -

x −

x(x

- 4

)

=

2x

−

x(x

- 4

)

=

2

−

x -

4

Exer

cise

sF

ind

eac

h s

um

or

dif

fere

nce

.

1.

1 −

a +

7 −

3a

10

−

3a

2.

1 −

6x +

3 −

8 4

+ 9

x

−

24x

3.

5 −

9x -

1 −

x2 5x

- 9

−

9x2

4.

6 −

x2 -

3 −

x3 6x

- 3

−

x3

5.

8 −

4a2

+

6 −

3a

2 +

2a

−

a2

6.

4

−

h +

1 +

2

−

h +

2

6h

+ 1

0 −

(h +

1)(

h +

2)

7.

y

−

y -

3 -

3

−

y +

3

y

2 +

9

−

(y -

3)(

y +

3)

8.

y −

y -

7 -

y

+ 3

−

y2 -

4y

- 2

1 y -

1

−

y -

7

9.

a

−

a +

4 +

4

−

a -

4

a

2 +

16

−

(a +

4)(

a -

4)

10.

6

−

3(m

+ 1

) +

2 −

3(m

- 1

)

8m -

4

−

−

3(m

+ 1

)(m

- 1

)

11.

4

−

x -

2y -

2

−

x +

2y

2x

+ 1

2y

−

(x +

2y)(

x -

2y)

12.

a

- 6

b −

2a2

- 5

ab +

2b2 -

7

−

a -

2b

-

13a

+ b

−

−

(2a

- b

)(a

- 2

b)

13.

y

+ 2

−

y2 +

5y

+ 6

+

2 -

y

−

y2 +

y -

6 0

14

.

q −

q2 -

16 +

q

+ 1

−

q2 +

5q

+ 4

2q -

4

−

(q -

4)(

q +

4)

Exam

ple

1Ex

amp

le 2

Facto

r th

e

denom

inato

r.

The L

CD

is

x(x

- 4

).

1 ·

x =

x

Subtr

act

num

era

tors

.

Sim

plif

y.

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3812

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-6 Lesson 11-6

Cha

pte

r 11

39

G

lenc

oe A

lgeb

ra 1

11-6

Skill

s Pr

acti

ceA

dd

ing

an

d S

ub

tracti

ng

Rati

on

al

Exp

ressio

ns

Fin

d e

ach

su

m o

r d

iffe

ren

ce.

1.

2y

−

5 +

y −

5 3y

−

5 2.

4r

−

9 + 5r

−

9 r

3.

t +

3

−

7 -

t −

7 3

−

7 4.

c

+ 8

−

4 -

c +

6

−

4

1 −

2

5.

x +

2

−

3 +

x +

5

−

3

2x +

7

−

3

6.

g +

2

−

4 +

g -

8

−

4

g -

3

−

2

7.

x

−

x -

1 -

1

−

x -

1

1 8.

3r

−

r +

3 -

r

−

r +

3

2r

−

r +

3

Fin

d t

he

LC

M o

f ea

ch p

air

of p

olyn

omia

ls.

9.

4x2 y

, 12x

y2 1

2x2 y

2 10

. n

+ 2

, n -

3

(n +

2)(

n -

3)

11.

2r -

1, r

+ 4

(2

r -

1)(

r +

4)

12.

t +

4, 4

t +

16

4(t

+ 4

)

Fin

d e

ach

su

m o

r d

iffe

ren

ce.

13.

5 −

4r -

2 −

r2 5r

- 8

−

4r2

14

. 5x

−

3y2

- 2x

−

9y

15x -

2xy

−

9y2

15.

x

−

x +

2 -

4

−

x -

1

x

2 -

5x -

8

−

(x +

2)(

x -

1)

16.

d -

1

−

d -

2 -

3

−

d +

5

d

2 +

d +

1

−

(d -

2)(

d +

5)

17.

b

−

b -

1 +

2

−

b -

4

b

2 -

2b

- 2

−

(b -

1)(

b -

4)

18.

k

−

k -

5 +

k -

1

−

k +

5

2k

2 -

k +

5

−

(k -

5)(

k +

5)

19.

3x +

15

−

x2 -

25 +

x

−

x +

5

x2

- 2

x +

15

−

(x +

5)(

x -

5)

20.

x

- 3

−

x2 -

4x

+ 4

+ x

+ 2

−

x -

2

x2

+ x

- 7

−

(x -

2)2

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

3912

/20/

10

7:41

PM



Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

40

G

lenc

oe A

lgeb

ra 1

11-6

Prac

tice

A

dd

ing

an

d S

ub

tracti

ng

Rati

on

al

Exp

ressio

ns

Fin

d e

ach

su

m o

r d

iffe

ren

ce.

1.

n

−

8 + 3n

−

8 2.

7u

−

16 +

5u

−

16

3.

w

+ 9

−

9 +

w +

4

−

9

n

−

2 3u

−

4 2w

+ 1

3 −

9

4.

x -

6

−

2 -

x -

7

−

2

5.

n +

14

−

5 -

n -

14

−

5

6.

6 −

c -

1 -

-

2 −

c -

1

1 −

2 28

−

5

8 −

c -

1

7.

x -

5

−

x +

2 +

-

2 −

x +

2

8.

r +

5

−

r -

5 +

2r -

1

−

r -

5

9.

4p +

14

−

p +

4

+ 2p

+ 1

0 −

p +

4

x

- 7

−

x +

2

3r +

4

−

r -

5

6

Fin

d t

he

LC

M o

f ea

ch p

air

of p

olyn

omia

ls.

10. 3

a3 b2 ,

18ab

3 11

. w -

4, w

+ 2

12

. 5d

- 2

0, d

- 4

1

8a3 b

3 (

w -

4)(

w +

2)

5(d

- 4

)

13. 6

p +

1, p

- 1

14

. x2

+ 5

x +

4, (

x +

1)2

15. m

2 +

3m

- 1

0, m

2 -

4

(

6p +

1)(

p -

1)

(x +

4)(

x +

1)2

(m

+ 5

)(m

- 2

)(m

+ 2

)

Fin

d e

ach

su

m o

r d

iffe

ren

ce.

16.

6p

−

5x2

- 2p

−

3x

18p

- 1

0xp

−

15x

2

17.

m +

4

−

m -

3 -

2 −

m -

6

m2

- 4

m -

18

−

(m -

3)(

m -

6)

18.

y +

3

−

y2 -

16

+

3y -

2

−

y2 +

8y

+ 1

6 4y

2 -

7y +

20

−

(y +

4)2 (

y -

4)

19.

p

+ 1

−

p2 +

3p

- 4

+

p −

p +

4

p

2 +

1

−

(p +

4)(

p -

1)

20.

t

+ 3

−

t2 -

3t

- 1

0 -

4t -

8

−

t2 -

10t

+ 2

5 21

.

4y

−

y2 -

y -

6 -

3y +

3

−

y2 -

4

-3t

2 -

2t

+ 1

−

(t +

2)(

t -

5)2

y

2 -

2y +

9

−

−

(y -

3)(

y +

2)(

y -

2)

22. S

ERV

ICE

Mem

bers

of

the

nin

th g

rade

cla

ss a

t P

ine

Rid

ge H

igh

Sch

ool

are

orga

niz

ing

into

ser

vice

gro

ups

. Wh

at i

s th

e m

inim

um

nu

mbe

r of

stu

den

ts w

ho

mu

st p

arti

cipa

te f

or

all

stu

den

ts t

o be

div

ided

in

to g

rou

ps o

f 4,

6, o

r 9

stu

den

ts w

ith

no

one

left

ou

t? 3

6

23. G

EOM

ETRY

Fin

d an

exp

ress

ion

for

th

e pe

rim

eter

of

rect

angl

e

5a +

4b

2a +

bB

A

CD

3a +

2b

2a +

b

AB

CD

. Use

th

e fo

rmu

la P

= 2

� +

2w

.

4(

4a +

3b

) −

2a +

b

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4012

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-6 Lesson 11-6

Cha

pte

r 11

41

G

lenc

oe A

lgeb

ra 1

11-6

Wor

d Pr

oble

m P

ract

ice

Rati

on

al

Exp

ressio

ns w

ith

Un

like D

en

om

inato

rs

1.TE

XA

S O

f th

e 25

4 co

un

ties

in

Tex

as,

4 ar

e la

rger

th

an 6

000

squ

are

mil

es.

An

oth

er 2

1 co

un

ties

are

sm

alle

r th

an

300

squ

are

mil

es. W

hat

fra

ctio

n o

f th

e co

un

ties

are

300

to

6000

squ

are

mil

es

in s

ize?

2. S

WIM

MIN

G P

ower

Poo

ls i

nst

alls

sw

imm

ing

pool

s. T

o de

term

ine

the

appr

opri

ate

size

of

pool

for

a y

ard,

th

ey

mea

sure

th

e le

ngt

h o

f th

e ya

rd i

n m

eter

s an

d ca

ll t

hat

val

ue

x. T

he

len

gth

an

d w

idth

of

the

pool

are

cal

cula

ted

wit

h t

he

diag

ram

bel

ow. W

rite

an

exp

ress

ion

in

si

mpl

est

form

for

th

e pe

rim

eter

of

a re

ctan

gula

r po

ol f

or t

he

give

n v

aria

ble

dim

ensi

ons.

m x

4

m 2

x 5

3. E

GY

PTIA

N F

RA

CTI

ON

S A

nci

ent

Egy

ptia

ns

use

d on

ly u

nit

fra

ctio

ns,

wh

ich

ar

e fr

acti

ons

in t

he

form

1 −

n . T

hei

r m

ath

emat

ical

not

atio

n o

nly

all

owed

for

a

nu

mer

ator

of

1. W

hen

th

ey n

eede

d to

ex

pres

s a

frac

tion

wit

h a

nu

mer

ator

ot

her

th

an 1

, th

ey w

rote

it

as a

su

m o

f u

nit

fra

ctio

ns.

An

exa

mpl

e is

sh

own

be

low

.

5 −

6 = 1 −

3 + 1 −

2

S

impl

ify

the

foll

owin

g ex

pres

sion

so

it i

s a

sum

of

un

it f

ract

ion

s.

5x +

6

−

10x2

+ 1

2x +

2x

−

8x2

1 −

2x +

1

−

4x

4.IN

SUR

AN

CE

For

a h

ospi

tal

stay

, Pau

l’s

hea

lth

in

sura

nce

pla

n r

equ

ires

him

to

pay

2−

5 t

he

cost

of

the

firs

t da

y in

th

e

hos

pita

l an

d 1−

5 t

he

cost

of

the

seco

nd

and

thir

d da

ys. I

f P

aul’s

hos

pita

l st

ay i

s 3

days

an

d co

st h

im $

420,

wh

at w

as t

he

full

dai

ly c

ost?

5. P

AC

KA

GE

DEL

IVER

Y F

redr

icks

burg

P

arce

l E

xpre

ss d

eliv

ered

a t

otal

of

498

pack

ages

on

Mon

day,

Tu

esda

y, a

nd

Wed

nes

day.

On

Tu

esda

y, t

hey

del

iver

ed

7 le

ss t

han

2 t

imes

th

e n

um

ber

of

pack

ages

del

iver

ed o

n M

onda

y. O

n

Wed

nes

day,

th

ey d

eliv

ered

th

e av

erag

e n

um

ber

deli

vere

d on

Mon

day

and

Tu

esda

y.

a. W

rite

a r

atio

nal

equ

atio

n t

hat

re

pres

ents

th

e su

m o

f th

e n

um

bers

of

pac

kage

s de

live

red

on M

onda

y,

Tu

esda

y, a

nd

Wed

nes

day.

b.

How

man

y pa

ckag

es w

ere

deli

vere

d on

Mon

day?

11

3

13x

−

10

229

−

254

answ

er:

x +

2x -

7 +

3x -

7

−

2

or

498

Sam

ple

$525

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4112

/20/

10

7:41

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

42

G

lenc

oe A

lgeb

ra 1

11-6

Enri

chm

ent

Su

m a

nd

Dif

fere

nce o

f A

ny T

wo

Lik

e P

ow

ers

Th

e su

m o

f an

y tw

o li

ke p

ower

s ca

n b

e w

ritt

en a

n +

bn, w

her

e n

is

a po

siti

ve i

nte

ger.

Th

e di

ffer

ence

of

like

pow

ers

is a

n -

bn. U

nde

r w

hat

con

diti

ons

are

thes

e ex

pres

sion

s ex

actl

y di

visi

ble

by (

a +

b)

or (

a -

b)?

Th

e an

swer

dep

ends

on

wh

eth

er n

is

an o

dd o

r ev

en n

um

ber.

Use

lon

g d

ivis

ion

to

fin

d t

he

foll

owin

g q

uot

ien

ts. (

Hin

t: W

rite

a

3 +

b3

as a

3 +

0a

2 +

0a

+ b

3 .) I

s th

e n

um

erat

or e

xact

ly

div

isib

le b

y th

e d

enom

inat

or?

Wri

te y

es o

r n

o.

1.

a3 +

b3

−

a +

b

2.

a3 +

b3

−

a -

b

3.

a3

- b

3 −

a +

b

4. a3

- b

3 −

a -

b

ye

s n

o

no

ye

s

5.

a4 +

b4

−

a +

b

6. a4

+ b

4 −

a -

b

7. a4

- b

4 −

a +

b

8. a4

- b

4 −

a -

b

n

o

no

ye

s ye

s

9.

a5 +

b5

−

a +

b

10.

a5 +

b5

−

a -

b

11.

a5 -

b5

−

a +

b

12.

a5 -

b5

−

a -

b

ye

s n

o

no

ye

s

13. U

se t

he

wor

ds o

dd

an

d ev

en t

o co

mpl

ete

thes

e tw

o st

atem

ents

.

a

. an

+ b

n i

s di

visi

ble

by a

+ b

if

n i

s , a

nd

by n

eith

er

a +

b n

or a

- b

if

n i

s .

b

. an

- b

n i

s di

visi

ble

by a

- b

if

n i

s , a

nd

by b

oth

a +

b a

nd

a -

b i

f n

is

.

14. D

escr

ibe

the

sign

s of

th

e te

rms

of t

he

quot

ien

ts w

hen

th

e di

viso

r is

a -

b.

T

he

term

s ar

e al

l po

siti

ve.

15. D

escr

ibe

the

sign

s of

th

e te

rms

of t

he

quot

ien

t w

hen

th

e di

viso

r is

a +

b.

T

he

term

s ar

e al

tern

atel

y p

osi

tive

an

d n

egat

ive.

od

dev

en

od

dev

en

036_

054_

ALG

1_A

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M_C

11_C

R_6

6028

5.in

dd

4212

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-7

Cha

pte

r 11

43

G

lenc

oe A

lgeb

ra 1

11-7

Stud

y G

uide

and

Inte

rven

tion

Mix

ed

Exp

ressio

ns a

nd

Co

mp

lex F

racti

on

s

Sim

plif

y M

ixed

Exp

ress

ion

s A

lgeb

raic

exp

ress

ion

s su

ch a

s a

+ b −

c an

d 5

+ x

+ y

−

x +

3 a

re

call

ed m

ixed

exp

ress

ion

s. C

han

gin

g m

ixed

exp

ress

ion

s to

rat

ion

al e

xpre

ssio

ns

is s

imil

ar

to c

han

gin

g m

ixed

nu

mbe

rs t

o im

prop

er f

ract

ion

s.

S

imp

lify

5 +

2 −

n .

5 +

2 −

n =

5 · n

−

n

+ 2 −

n

LC

D is n

.

=

5n +

2

−

n

A

dd t

he n

um

era

tors

.

Th

eref

ore,

5 +

2 −

n =

5n +

2

−

n

.

S

imp

lify

2 +

3

−

n +

3 .

2 +

3

−

n +

3 =

2(n

+ 3

) −

n +

3

+

3 −

n +

3

=

2n +

6

−

n +

3

+

3 −

n +

3

=

2n +

6 +

3

−

n +

3

=

2n +

9

−

n +

3

Th

eref

ore,

2 +

3

−

n +

3 =

2n +

9

−

n +

3 .

Exer

cise

sW

rite

eac

h m

ixed

exp

ress

ion

as

a ra

tion

al e

xpre

ssio

n.

1. 4

+ 6 −

a 4a

+ 6

−

a

2.

1 −

9x -

3 1

- 2

7x

−

9x

3. 3

x -

1 −

x2 3x

3 -

1

−

x2

4.

4 −

x2 -

2 4

- 2

x2

−

x2

5. 1

0 +

60

−

x +

5

10x +

110

−

x +

5

6.

h

−

h +

4 +

2

3h +

8

−

h +

4

7.

y

−

y -

2 +

y2

y3

- 2

y2

+ y

−

y -

2

8.

4 -

4

−

2x +

1

8x

−

2x +

1

9. 1

+ 1 −

x

x +

1

−

x

10

.

4 −

m -

2 -

2m

4

- 2

m2

+ 4

m

−

m -

2

11. x

2 +

x +

2

−

x -

3

x3

- 3

x2

+ x

+ 2

−

−

x -

3

12

. a -

3 +

a -

2

−

a +

3

a2

+ a

- 1

1 −

a +

3

13. 4

m +

3p

−

2t

8mt

+ 3

p

−

2t

14

. 2q2

+

q −

p +

q

q +

2rq

2 +

2q

3

−

r +

q

15.

2

−

y2 -

1 -

4y2

2 -

4y

4 +

4y

2

−

y2

- 1

16. t

2 +

p +

t

−

p -

t

pt2

- t

3 +

p +

t

−

p -

t

Exam

ple

1Ex

amp

le 2

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4312

/20/

10

7:41

PM



Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

44

G

lenc

oe A

lgeb

ra 1

11-7

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Mix

ed

Exp

ressio

ns a

nd

Co

mp

lex F

racti

on

s

Sim

plif

y C

om

ple

x Fr

acti

on

s If

a f

ract

ion

has

on

e or

mor

e fr

acti

ons

in t

he

nu

mer

ator

or

den

omin

ator

, it

is c

alle

d a

com

ple

x fr

acti

on.

Sim

plif

yin

g a

C

om

ple

x Fr

acti

on

Any c

om

ple

x f

ractio

n a

−

b

−

c −

d

wh

ere

b ≠

0,

c ≠

0,

an

d d

≠ 0

, ca

n b

e e

xp

resse

d a

s ad

−

bc .

S

imp

lify

2 +

4 −

a

−

a +

2

−

3 .

2 +

4 −

a

−

a +

2

−

3 =

2a

−

a +

4 −

a

−

a +

2

−

3

F

ind t

he L

CD

for

the n

um

era

tor

and r

ew

rite

as lik

e f

ractions.

=

2a

+ 4

−

a

−

a +

2

−

3

S

implif

y t

he n

um

era

tor.

=

2a +

4

−

a ·

3

−

a +

2

Rew

rite

as t

he p

roduct

of

the n

um

era

tor

and t

he r

ecip

rocal of

the d

enom

inato

r.

=

2(a

+ 2

) −

a ·

3

−

a +

2

Facto

r.

=

6 −

a

Div

ide a

nd s

implif

y.

Exer

cise

s

Sim

pli

fy e

ach

exp

ress

ion

.

1.

2 2 −

5 −

3 3 −

4

16

−

25

2. 3

−

x

−

4 −

y

3y

−

4x

3. x

−

y3

−

x3 −

y2

1 −

x2 y

4.

1 -

1 −

x

−

1 +

1 −

x

x -

1

−

x +

1

5. 1

- 1 −

x

−

1 -

1 −

x2

x

−

x +

1

6.

1 −

x -

3

−

2

−

x2 -

9

x +

3

−

2

7.

x2 -

25

−

y

−

x3 -

5x2

x +

5

−

x2 y

8. x

-

12

−

x -

1

−

x -

8

−

x -

2

(x +

3)(

x -

2)

−

(x -

1)(

x +

2)

9.

3

−

y +

2 -

2

−

y -

2

−

1

−

y +

2 -

2

−

y -

2

y -

10

−

-y -

6

Exam

ple

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4412

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-7

Cha

pte

r 11

45

G

lenc

oe A

lgeb

ra 1

11-7

Skill

s Pr

acti

ceM

ixed

Exp

ressio

ns a

nd

Co

mp

lex F

racti

on

s

Wri

te e

ach

mix

ed e

xpre

ssio

n a

s a

rati

onal

exp

ress

ion

.

1. 6

+ 4 −

h

6h +

4

−

h

2.

7 +

6 −

p 7p

+ 6

−

p

3. 4

b +

b −

c 4b

c +

b

−

c

4.

8q

- 2q

−

r 8q

r -

2q

−

r

5. 2

+

4 −

d -

5

2d -

6

−

d -

5

6. 5

-

6 −

f +

2

5f +

4

−

f +

2

7. b

2 +

12

−

b +

3

b3

+ 3

b2

+ 1

2

−

b +

3

8.

m -

6

−

m -

7

m2

- 7

m -

6

−

m -

7

9. 2

a +

a -

2

−

a 2a

2 +

a -

2

−

a

10

. 4r

- r

+ 9

−

2r

8r2

- r

- 9

−

2r

Sim

pli

fy e

ach

exp

ress

ion

.

11.

2 1 −

2 −

4 3 −

4 10

−

19

12.

3 2 −

3 −

5 2 −

5 55

−

81

13.

r −

n2

−

r2 −

n

1 −

rn

14.

a2 −

b3 −

a −

b a

−

b2

15.

x2 y

−

c

−

xy3

−

c2

xc

−

y2

16.

r -

2

−

r +

3

−

r -

2

−

3

3 −

r +

3

17.

w +

4

−

w

−

w2

- 1

6 −

w

1 −

w -

4

18.

x2 -

1

−

x

−

x -

1

−

x2

x(x

+ 1

) 19

.

b2 -

4

−

b2

+ 7

b +

10

−

b -

2

1

−

b +

5

20.

k2 +

5k

+ 6

−

k2 -

9

−

k +

2

1

−

k -

3

21.

g +

12

−

g +

8

−

g +

6

g +

2

−

g +

8

22.

p +

9

−

p -

6

−

p -

3

p -

3

−

p -

6

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4512

/20/

10

7:41

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

46

G

lenc

oe A

lgeb

ra 1

11-7

Prac

tice

Mix

ed

Exp

ressio

ns a

nd

Co

mp

lex F

racti

on

s

Wri

te e

ach

mix

ed e

xpre

ssio

n a

s a

rati

onal

exp

ress

ion

.

1. 1

4 -

9 −

u

2. 7

d +

4d

−

c

3. 3

n +

6 -

n

−

n

14

u -

9

−

u

7d

c +

4d

−

c

3n

2 -

n +

6

−

n

4. 5

b -

b +

3

−

2b

5.

3 +

t +

5

−

t2 -

1

6. 2

a +

a -

1

−

a +

1

10

b2

- b

- 3

−

2b

3t

2 +

t +

2

−

t2 -

1

2a

2 +

3a

- 1

−

a +

1

7. 2

p +

p +

1

−

p -

3

8. 4

n2

+ n

- 1

−

n2

- 1

9.

(t

+ 1

) +

4

−

t +

5

2p

2 -

5p

+ 1

−

p -

3

4n

3 +

4n

2 +

1

−

n +

1

t2

+ 6

t +

9

−

t +

5

Sim

pli

fy e

ach

exp

ress

ion

.

10.

3 2 −

5 −

2 5 −

6 6 −

5 11

. m

2 −

6p

−

3m

−

p2

mp

−

18

12.

x2 -

y2

−

x2

−

x +

y

−

3x

3(x -

y)

−

x

13.

a -

4

−

a2

−

a2 -

16

−

a

1 −

a(a

+ 4

) 14

. q2

- 7

q +

12

−

q2 -

16

−

q -

3

1 −

q +

4

15.

k2

+ 6

k −

k2 +

4k

- 5

−

k

- 8

−

k2 -

9k

+ 8

k

(k +

6)

−

k +

5

16.

b2 +

b -

12

−

b2 +

3b

- 4

−

b -

3

−

b2 -

b

b

17.

g -

10

−

g +

9

−

g -

5

−

g +

4

(g +

10)

(g +

4)

−

(g +

9)(

g +

5)

18.

y +

6

−

y -

7

−

y -

7

−

y +

6

(y -

6)(

y +

6)

−

(y -

7)(

y +

7)

19. T

RA

VEL

Ray

an

d Ja

n a

re o

n a

12

1 −

2 -hou

r dr

ive

from

Spr

ingf

ield

, Mis

sou

ri, t

o C

hic

ago,

Illi

noi

s. T

hey

sto

p fo

r a

brea

k ev

ery

3 1 −

4 hou

rs.

a. W

rite

an

exp

ress

ion

to

mod

el t

his

sit

uat

ion

. 12

1 −

2 −

3 1 −

4

b.

How

man

y st

ops

wil

l R

ay a

nd

Jan

mak

e be

fore

arr

ivin

g in

Ch

icag

o? 3

20. C

AR

PEN

TRY

Tai

nee

ds s

ever

al 2

1 −

4 -in

ch w

oode

n r

ods

to r

ein

forc

e th

e fr

ame

on a

fu

ton

.

Sh

e ca

n c

ut

the

rods

fro

m a

24

1 −

2 -in

ch d

owel

pu

rch

ased

fro

m a

har

dwar

e st

ore.

How

man

y

woo

den

rod

s ca

n s

he

cut

from

th

e do

wel

? 10

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4612

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-7

Cha

pte

r 11

47

G

lenc

oe A

lgeb

ra 1

11-7

1.C

YC

LIN

G N

atal

ie r

ode

in a

bic

ycle

eve

nt

for

char

ity

on S

atu

rday

. It

took

her

2−

3 o

f an

hou

r to

com

plet

e th

e 18

-mil

e ra

ce.

Wh

at w

as h

er a

vera

ge s

peed

in

mil

es

per

hou

r?

27 m

ph

2. Q

UIL

TIN

G M

rs. T

anto

ra s

ews

and

sell

s

Am

ish

bab

y qu

ilts

. Sh

e bo

ugh

t 42

3 −

4 yar

ds

of b

acki

ng

fabr

ic, a

nd

2 1 −

4 yar

ds a

re

nee

ded

for

each

qu

ilt

she

sew

s. H

ow

man

y qu

ilts

can

sh

e m

ake

wit

h t

he

back

ing

fabr

ic s

he

bou

ght?

3. T

RA

VEL

Th

e F

ran

z fa

mil

y tr

avel

ed f

rom

G

alve

ston

to

Wac

o fo

r a

fam

ily

reu

nio

n.

Dri

vin

g th

eir

min

ivan

, th

ey a

vera

ged

30 m

iles

per

hou

r on

th

e w

ay t

o W

aco

and

45 m

iles

per

hou

r on

th

e re

turn

tri

p h

ome

to G

alve

ston

. Wh

at i

s th

eir

aver

age

rate

for

th

e en

tire

tri

p? (

Hin

t: R

emem

ber

that

ave

rage

rat

e eq

ual

s to

tal

dist

ance

di

vide

d by

tot

al t

ime

and

that

tim

e ca

n

be r

epre

sen

ted

as a

rat

io o

f di

stan

ce

x to

rat

e.)

36

mp

h

4. P

HY

SIC

AL

SCIE

NC

E T

he

volu

me

of a

gas

va

ries

dir

ectl

y as

th

e K

elvi

n t

empe

ratu

re

T a

nd

inve

rsel

y as

th

e pr

essu

re P

, wh

ere

k is

th

e co

nst

ant

of v

aria

tion

.

V =

k ( T

−

P )

I

f k

=

13

−

157 ,

fin

d th

e vo

lum

e in

lit

ers

of

hel

ium

gas

at

273

degr

ees

Kel

vin

an

d13

−

3 at

mos

pher

es o

f pr

essu

re. R

oun

d yo

ur

answ

er t

o th

e n

eare

st h

un

dred

th.

5.

22 L

5. S

AFE

TY T

he

Occ

upa

tion

al S

afet

y an

d H

ealt

h A

dmin

istr

atio

n p

rovi

des

safe

ty

stan

dard

s in

th

e w

orkp

lace

to

keep

w

orke

rs f

ree

from

dan

gero

us

wor

kin

g co

ndi

tion

s. O

SH

A r

ecom

men

ds t

hat

for

ge

ner

al c

onst

ruct

ion

th

ere

be 5

foo

t-ca

ndl

es o

f il

lum

inat

ion

in

wh

ich

to

wor

k.

A f

orem

an u

sin

g a

ligh

t m

eter

fin

ds t

hat

th

e il

lum

inat

ion

of

a co

nst

ruct

ion

lig

ht

on a

su

rfac

e 8

feet

fro

m t

he

sou

rce

is

11 f

oot-

can

dles

. Th

e il

lum

inat

ion

pr

odu

ced

by a

lig

ht

sou

rce

vari

es

inve

rsel

y as

th

e sq

uar

e of

th

e di

stan

ce

from

th

e so

urc

e.

I is

ill

um

inat

ion

(in

foo

t-ca

ndl

es).

I

= k −

d2

d i

s th

e di

stan

ce f

rom

th

e so

urc

e (i

n f

eet)

.k

is a

con

stan

t.

a. F

ind

the

illu

min

atio

n o

f th

e sa

me

ligh

t at

a d

ista

nce

of

15 3 −

4 fee

t. R

oun

d

you

r an

swer

to

the

nea

rest

hu

ndr

edth

. 2.

84 f

oo

t-ca

nd

les

b.

Is t

her

e en

ough

ill

um

inat

ion

at

this

di

stan

ce t

o m

eet

OS

HA

req

uir

emen

ts

for

ligh

tin

g?

no

c. I

n o

rder

to

com

ply

wit

h O

SH

A, w

hat

is

th

e m

axim

um

all

owab

le w

orki

ng

dist

ance

fro

m t

his

lig

ht

sou

rce?

Rou

nd

you

r de

cim

al a

nsw

er t

o n

eare

st t

enth

.11

.9 f

t

Wor

d Pr

oble

m P

ract

ice

Mix

ed

Exp

ressio

ns a

nd

Co

mp

lex F

racti

on

s

19

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4712

/20/

10

7:41

PM



Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

48

G

lenc

oe A

lgeb

ra 1

11-7

Enri

chm

ent

Co

nti

nu

ed

Fra

cti

on

s

Con

tin

ued

fra

ctio

ns

are

a sp

ecia

l ty

pe o

f co

mpl

ex f

ract

ion

. Eac

h f

ract

ion

in

a c

onti

nu

ed

frac

tion

has

a n

um

erat

or o

f 1.

1 +

1

−

2 +

1

−

3 +

1 −

4

E

valu

ate

the

con

tin

ued

fr

acti

on a

bov

e. S

tart

at

the

bot

tom

an

d

wor

k y

our

way

up

.

Ste

p 1

3

+ 1 −

4 =

12

−

4 +

1 −

4 = 13

−

4

Ste

p 2

1

−

3 +

1 −

4 =

4 −

13

Ste

p 3

2

+

4 −

13 =

26

−

13 +

4 −

13 =

30

−

13

Ste

p 4

1

−

30

−

13 =

13

−

30

Ste

p 5

1

+ 13

−

30 =

1 13

−

30

C

han

ge 13

−

7 in

to a

co

nti

nu

ed f

ract

ion

.

Ste

p 1

13

−

7 =

7 −

7 + 6 −

7 = 1

+ 6 −

7

Ste

p 2

6 −

7 = 1 −

7 −

6

Ste

p 3

7 −

6 = 1

+ 1 −

6

Sto

p,

because t

he n

um

era

tor

is 1

.

Th

us,

13

−

7 ca

n b

e w

ritt

en a

s 1

+

1 −

1 +

1 −

6 .

Exam

ple

1Ex

amp

le 2

Eva

luat

e ea

ch c

onti

nu

ed f

ract

ion

.

1. 0

+

1

−

2 +

1

−

4 +

1

−

6 +

1 −

8 2.

0 +

1

−

1 +

1

−

2 +

1 −

6

3. 3

+

1

−

1 +

1

−

1 +

1

−

1 +

1 −

2 4.

1 +

1

−

3 +

1

−

5 +

1 −

7

Ch

ange

eac

h f

ract

ion

in

to a

con

tin

ued

fra

ctio

n.

5.

71

−

26

2 +

1

−−

1 +

1

−

2 +

1

−

1 +

1

−

2 +

1 −

2 6

. 9 −

56 0

+

1

−

6 +

1

−

4 +

1 −

2 7.

4626

−

2065

2

+

1

−

4 +

1

−

6 +

1 −

8 +

1 −

10

20

4 −

457

13

−

19

3

5 −

8 1

36

−

115

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4812

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-8

Cha

pte

r 11

49

G

lenc

oe A

lgeb

ra 1

11-8

Stud

y G

uide

and

Inte

rven

tion

Rati

on

al

Eq

uati

on

s

Solv

e R

atio

nal

Eq

uat

ion

s R

atio

nal

eq

uat

ion

s ar

e eq

uat

ion

s th

at c

onta

in r

atio

nal

ex

pres

sion

s. T

o so

lve

equ

atio

ns

con

tain

ing

rati

onal

exp

ress

ion

s, m

ult

iply

eac

h s

ide

of t

he

equ

atio

n b

y th

e le

ast

com

mon

den

omin

ator

.R

atio

nal

equ

atio

ns

can

be

use

d to

sol

ve w

ork

pro

ble

ms

and

rate

pro

ble

ms.

Exer

cise

sS

olve

eac

h e

qu

atio

n. S

tate

an

y ex

tran

eou

s so

luti

ons.

1.

x -

5

−

5 +

x −

4 = 8

20

2.

3 −

x =

6

−

x +

1 1

3.

x -

1

−

5 =

2x -

2

−

15

1

4.

8

−

n -

1 =

10

−

n +

1 9

5.

t -

4

−

t +

3 =

t +

3

- 4

1 −

3 6.

m +

4

−

m

+ m

−

3 = m

−

3 -4;

7.

q +

4

−

q -

1 +

q

−

q +

1 =

2 -

3 −

2 8.

5 -

2x

−

2 -

4x +

3

−

6 =

7x +

2

−

6

10

−

17

9.

m +

1

−

m -

1 -

m

−

1 -

m =

1 -

2 10

. x 2

- 9

−

x -

3 +

x2

= 9

-3

or

2

11.

2

−

x 2 -

36 -

1

−

x -

6 =

0 -

4; 6

is

12.

4z

−

z 2 +

4z

+ 3

=

6 −

z +

3 +

4

−

z +

1 -

3 is

extr

aneo

us

extr

aneo

us

13.

4 −

4 -

p -

p2

−

p -

4 =

4 -

6, 2

14

. x2

- 1

6 −

x -

4

+ x

2 =

16

-4,

3;

4 is

ex

tran

eou

s

S

olve

x -

3

−

3 +

x −

2 = 4

.

x

- 3

−

3 +

x −

2 = 4

6 ( x

- 3

−

3 +

x −

2 ) =

6(4

) T

he L

CD

is 6

.

2(x

- 3

) +

3x

= 2

4 D

istr

ibutive

Pro

pert

y

2x

- 6

+ 3

x =

24

Dis

trib

utive

Pro

pert

y

5x

= 3

0 S

implif

y.

x

= 6

D

ivid

e e

ach s

ide b

y 5

.

Th

e so

luti

on i

s 6.

S

olve

15

−

x 2

- 1

=

5 −

2(x

- 1

) . S

tate

any

extr

aneo

us

solu

tion

s.

15

−

x 2 -

1 =

5

−

2(x

- 1

) O

rigin

al equation

30(x

- 1

) = 5

(x2

- 1

) C

ross m

ultip

ly.

30x

- 3

0 =

5x2

- 5

D

istr

ibutive

Pro

pert

y

0

= 5

x2 -

30x

+ 3

0 -

5

Add -

30x

+ 3

0 t

o

each s

ide.

0

= 5

x2 -

30x

+ 2

5 S

implif

y.

0

= 5

(x2

- 6

x +

5)

Facto

r.

0

= 5

(x -

1)(

x -

5)

Facto

r.

x =

1

or

x =

5

Zero

Pro

duct

Pro

pert

y

Th

e n

um

ber

1 is

an

ext

ran

eou

s so

luti

on, s

ince

1

is a

n e

xclu

ded

valu

e fo

r x.

So,

5 i

s th

e so

luti

on o

f th

e eq

uat

ion

.

Exam

ple

1Ex

amp

le 2

extr

aneo

us

0 is

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

4912

/20/

10

7:41

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

50

G

lenc

oe A

lgeb

ra 1

11-8

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Rati

on

al

Eq

uati

on

s

Use

Rat

ion

al E

qu

atio

ns

to S

olv

e Pr

ob

lem

s R

atio

nal

equ

atio

n c

an b

e u

sed

to

solv

e w

ork

pro

ble

ms

and

rate

pro

ble

ms.

W

OR

K P

RO

BLE

M M

arla

can

pai

nt

Per

cy’s

kit

chen

in

3 h

ours

. Per

cy

can

pai

nt

it i

n 2

hou

rs. W

ork

ing

toge

ther

, how

lon

g w

ill

it t

ake

Mar

la a

nd

Per

cy t

o p

ain

t th

e k

itch

en?

In t

hou

rs, M

arla

com

plet

es t

· 1 −

3 of

the

job

and

Per

cy c

ompl

etes

t ·

1 −

2 of

the

job.

So

an

equ

atio

n f

or c

ompl

etin

g th

e w

hol

e jo

b is

t −

3 + t −

2 = 1

.

t −

3 + t −

2 = 1

2t

+ 3

t =

6

Multip

ly e

ach t

erm

by 6

.

5t

= 6

A

dd lik

e t

erm

s.

t

= 6 −

5 S

olv

e.

So

it w

ill

take

Mar

la a

nd

Per

cy 1

1 −

5 hou

rs t

o pa

int

the

room

if

they

wor

k to

geth

er.

Exam

ple

Exer

cise

s1.

GR

EETI

NG

CA

RD

S It

tak

es K

enes

ha

45 m

inu

tes

to p

repa

re 2

0 gr

eeti

ng

card

s. I

t ta

kes

Pau

la 3

0 m

inu

tes

to p

repa

re t

he

sam

e n

um

ber

of c

ards

. Wor

kin

g to

geth

er a

t th

is r

ate,

h

ow l

ong

wil

l it

tak

e th

em t

o pr

epar

e th

e ca

rds?

2. B

OA

TIN

G A

mot

orbo

at w

ent

upst

ream

at

15 m

iles

per

hou

r an

d re

turn

ed d

owns

trea

m

at 2

0 m

iles

per

hou

r. H

ow f

ar d

id t

he b

oat

trav

el o

ne w

ay i

f th

e ro

und

trip

too

k 3.

5 ho

urs?

30

mi

3. F

LOO

RIN

G M

aya

and

Reg

inal

d ar

e in

stal

lin

g h

ardw

ood

floo

rin

g. M

aya

can

in

stal

l fl

oori

ng

in a

roo

m i

n 4

hou

rs. R

egin

ald

can

in

stal

l fl

oori

ng

in a

roo

m i

n 3

hou

rs. H

ow

lon

g w

ould

it

take

th

em i

f th

ey w

orke

d to

geth

er?

4. B

ICY

CLI

NG

Ste

fan

is

bicy

lin

g on

a b

ike

trai

l at

an

ave

rage

of

10 m

iles

per

hou

r. E

rik

star

ts b

icyc

lin

g on

th

e sa

me

trai

l 30

min

ute

s la

ter.

If E

rik

aver

ages

16

mil

es p

er h

our,

how

lon

g w

ill

it t

ake

him

to

pass

Ste

fan

? 50

min

12

−

7 h

or

abo

ut

1.71

ho

urs

18 m

in

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

5012

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-8

Cha

pte

r 11

51

G

lenc

oe A

lgeb

ra 1

11-8

Skill

s Pr

acti

ceR

ati

on

al

Eq

uati

on

sS

olve

eac

h e

qu

atio

n. S

tate

an

y ex

tran

eou

s so

luti

ons.

1.

5

−

c

=

2

−

c +

3 -

5 2.

3 −

q =

5 −

q +

4 6

3.

7

−

m +

1 =

12

−

m +

2

2 −

5 4.

3

−

x +

2 =

5

−

x +

8 7

5.

y

−

y -

2 =

y +

1

−

y -

5

1 −

2 6.

b -

2

−

b =

b +

4

−

b +

2 -

1

7.

3m

−

2 -

1 −

4 = 10

m

−

8 1

8.

7g

−

9 +

1 −

3 = 5g

−

6 6

9.

2a +

5

−

6 -

2a

−

3 =

-

1 −

2 4

10.

n -

3

−

10

+ n

- 5

−

5 =

1 −

2 6

11.

c +

2

−

c +

c +

3

−

c =

7 1

12

. 3b

- 4

−

b -

b -

7

−

b =

1 -

3

13.

m -

4

−

m

-

m –

11

−

m +

4

= 1 −

m 2

14

. f

+ 2

−

f -

f +

1

−

f +

5 =

1 −

f -

1

15.

r +

3

−

r -

1 -

r

−

r -

3 =

0 9

16

. u

+ 1

−

u -

2 -

u

−

u +

1 =

0 -

1 −

4

17.

- 2

−

x +

1 +

2 −

x =

1 -

2, 1

18

. 5

−

m –

4 -

m

−

2m

– 8

= 1

6

19. A

CTI

VIS

M M

aury

an

d T

yra

are

mak

ing

phon

e ca

lls

to s

tate

rep

rese

nta

tive

s’ o

ffic

es t

o lo

bby

for

an i

ssu

e. M

aury

can

cal

l al

l 12

0 st

ate

repr

esen

tati

ves

in 1

0 h

ours

. Tyr

a ca

n c

all

all

120

stat

e re

pres

enta

tive

s in

8 h

ours

. How

lon

g w

ould

it

take

th

em t

o ca

ll a

ll

120

stat

e re

pres

enta

tive

s to

geth

er?

4 4 −

9 hr

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

5112

/20/

10

7:41

PM



Co

pyr

ight

© G

lenc

oe/

McG

raw

-Hill

, a d

ivis

ion

of

The

McG

raw

-Hill

Co

mp

anie

s, In

c.

An

swer

s

PDF Pass



NA

ME

DAT

E

P

ER

IOD

N

AM

E

D

ATE

PE

RIO

D

Cha

pte

r 11

52

G

lenc

oe A

lgeb

ra 1

11-8

Prac

tice

R

ati

on

al

Eq

uati

on

sS

olve

eac

h e

qu

atio

n. S

tate

an

y ex

tran

eou

s so

luti

ons.

1.

5 −

n +

2 =

7

−

n +

6

2.

x −

x -

5 =

x +

4

−

x -

6

3. k

+ 5

−

k =

k -

1

−

k +

9

8

4

-

3

4.

2h

−

h -

1 =

2h +

1

−

h +

2

5. 4y

−

3 +

1 −

2 = 5y

−

6 6.

y -

2

−

4 -

y +

2

−

5 =

-1

-

1 −

5 -

1 -

2

7.

2q -

1

−

6 -

q −

3 = q

+ 4

−

18

8.

5

−

p -

1 -

3

−

p +

2 =

0

9.

3t

−

3t -

3 -

1

−

9t +

3 =

1

-

7 -

13

−

2 -

1 −

2

10.

4x

−

2x +

1 -

2x

−

2x +

3 =

1

11.

d -

3

−

d

- d

- 4

−

d -

2 =

1 −

d

12.

3y -

2

−

y -

2

+

y 2 −

2 -

y =

-3

3

−

2 4

4

; ex

tran

eou

s: 2

13.

2 −

m +

2 -

m +

2

−

m -

2 =

7 −

3 14

. n

+ 2

−

n

+ n

+ 5

−

n +

3 =

-

1 −

n

15.

1 −

z +

1 -

6 -

z

−

6z

= 0

-

1, 2 −

5 -

9 −

2 , -

1 -

3, 2

16.

2p

−

p -

2 +

p +

2

−

p 2

- 4

= 1

17

. x

+ 7

−

x 2 -

9 -

x

−

x +

3 =

1

18.

2n

−

n -

4 -

n

+ 6

−

n 2

- 1

6 = 1

-

3; e

xtra

neo

us:

-2

-2,

4

-5,

-2

19. P

UB

LISH

ING

Tra

cey

and

Ala

n p

ubl

ish

a 1

0-pa

ge i

nde

pen

den

t n

ewsp

aper

on

ce a

mon

th.

At

prod

uct

ion

, Ala

n u

sual

ly s

pen

ds 6

hou

rs o

n t

he

layo

ut

of t

he

pape

r. W

hen

Tra

cey

hel

ps, l

ayou

t ta

kes

3 h

ours

an

d 20

min

ute

s.

a. W

rite

an

equ

atio

n t

hat

cou

ld b

e u

sed

to d

eter

min

e h

ow l

ong

it w

ould

tak

e T

race

y to

do

th

e la

you

t by

her

self

.

b.

How

lon

g w

ould

it

take

Tra

cey

to d

o th

e jo

b al

one?

7 h

30

min

20. T

RA

VEL

Em

ilio

mad

e ar

ran

gem

ents

to

hav

e L

ynda

pic

k h

im u

p fr

om a

n a

uto

rep

air

shop

aft

er h

e dr

oppe

d hi

s ca

r of

f. H

e ca

lled

Lyn

da t

o te

ll h

er h

e w

ould

sta

rt w

alki

ng a

nd

to l

ook

for

him

on

the

way

. Em

ilio

and

Lyn

da l

ive

10 m

iles

fro

m t

he a

uto

shop

. It

take

s E

mil

io 2

1 −

4 hou

rs t

o w

alk

the

dist

ance

an

d L

ynda

15

min

ute

s to

dri

ve t

he

dist

ance

.

a. I

f E

mil

io a

nd

Lyn

da l

eave

at

the

sam

e ti

me,

wh

en s

hou

ld L

ynda

exp

ect

to s

pot

Em

ilio

on

th

e ro

ad?

b.

How

far

wil

l E

mil

io h

ave

wal

ked

wh

en L

ynda

pic

ks h

im u

p? 1

mi

Sam

ple

an

swer

: 1 −

6 ( 10

−

3 ) +

1 −

t ( 10

−

3 ) =

1

in 1

3 1 −

2 min

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

5212

/20/

10

7:41

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 11-8

Cha

pte

r 11

53

G

lenc

oe A

lgeb

ra 1

11-8

Wor

d Pr

oble

m P

ract

ice

Rati

on

al

Eq

uati

on

s

1. E

LEC

TRIC

ITY

Th

e cu

rren

t in

a s

impl

e el

ectr

ic c

ircu

it v

arie

s in

vers

ely

as t

he

resi

stan

ce. I

f th

e cu

rren

t is

20

amps

w

hen

th

e re

sist

ance

is

5 oh

ms,

fin

d th

e cu

rren

t w

hen

th

e re

sist

ance

is

8 oh

ms.

12.5

am

ps

2. M

ASO

NRY

Sam

an

d B

elai

are

mas

ons

wh

o ar

e w

orki

ng

to b

uil

d a

ston

e w

all

that

wil

l be

120

fee

t lo

ng.

Sam

wor

ks

from

on

e en

d an

d is

abl

e to

bu

ild

one

ten

-foo

t se

ctio

n i

n 5

hou

rs. B

elai

wor

ks

from

th

e ot

her

en

d an

d is

abl

e to

fin

ish

a

ten

-foo

t se

ctio

n i

n 4

hou

rs. H

ow l

ong

wil

l it

tak

e S

am a

nd

Bel

ai t

o fi

nis

h

buil

din

g th

e w

all?

26

ho

urs

an

d 4

0 m

inu

tes

3. N

UM

BER

S T

he

form

ula

to

fin

d th

e su

m

of t

he

firs

t n

wh

ole

nu

mbe

rs i

s

sum

= n

2 +

n

−

2 . I

n o

rder

to

enco

ura

ge

stu

den

ts t

o sh

ow u

p ea

rly

to a

sch

ool

dan

ce, t

he

dan

ce c

omm

itte

e de

cide

s to

ch

arge

les

s fo

r th

ose

wh

o co

me

to t

he

dan

ce e

arly

. Th

eir

plan

is

to c

har

ge t

he

firs

t st

ude

nt

to a

rriv

e 1

pen

ny.

Th

e se

con

d st

ude

nt

thro

ugh

th

e do

or i

s ch

arge

d 2

pen

nie

s; t

he

thir

d st

ude

nt

thro

ugh

th

e do

or i

s ch

arge

d 3

pen

nie

s,

and

so o

n. H

ow m

uch

mon

ey, i

n t

otal

, w

ould

pai

d by

th

e fi

rst

150

stu

den

ts?

11,3

25 p

enn

ies

or

$113

.25

4. N

AU

TIC

AL

A f

erry

cap

tain

kee

ps t

rack

of

th

e pr

ogre

ss o

f h

is s

hip

in

th

e sh

ip’s

lo

g. O

ne

day,

he

reco

rds

the

foll

owin

g en

try.

Wit

h t

he

rece

nt

spri

ng

snow

mel

t, th

e cu

rren

t is

ru

nn

ing

stro

ng

tod

ay. T

he

six-

mil

e tr

ip d

own

stre

am t

o W

hyt

e’s

lan

din

g w

as v

ery

quic

k. H

owev

er, w

e on

ly c

over

ed t

wo

mil

es i

n t

he

sam

e am

oun

t of

tim

e w

hen

we

hea

ded

ba

ck u

pstr

eam

.

W

rite

a r

atio

nal

equ

atio

n u

sin

g b

for

the

spee

d of

th

e bo

at a

nd

c fo

r th

e sp

eed

of

the

stre

am a

nd

solv

e fo

r b

in t

erm

s of

c.

6 −

b +

c =

2

−

b -

c ;

b =

2c

5. H

EALT

H C

AR

E T

he

tota

l n

um

ber

of

Am

eric

ans

wai

tin

g fo

r ki

dney

an

d h

eart

tr

ansp

lan

ts i

s ap

prox

imat

ely

66,5

00. T

he

rati

o of

th

ose

awai

tin

g a

kidn

ey

tran

spla

nt

to t

hos

e aw

aiti

ng

a h

eart

tr

ansp

lan

t is

abo

ut

20 t

o 1.

a. H

ow m

any

peop

le a

re o

n e

ach

of

the

wai

tin

g li

sts?

Rou

nd

you

r an

swer

s to

th

e n

eare

st h

un

dred

.

kid

ney

: 63

,30

0; h

eart

: 32

00

b. T

hes

e tw

o gr

oups

mak

e u

p ab

out

3 −

4 of

the

tran

spla

nt

can

dida

tes

for

all

orga

ns.

Abo

ut

how

man

y or

gan

tr

ansp

lan

t ca

ndi

date

s ar

e th

ere

alto

geth

er?

Rou

nd

you

r an

swer

to

the

nea

rest

th

ousa

nd.

89,0

00

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

5312

/20/

10

7:41

PM



Co

pyrig

ht © G

lencoe/M

cGraw

-Hill, a d

ivision o

f The M

cGraw

-Hill C

om

panies, Inc.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 11

54

G

lenc

oe A

lgeb

ra 1

11-8

Enri

chm

ent

Win

nin

g D

ista

nces

In 1

999,

Hic

ham

El

Gu

erro

uj s

et a

wor

ld r

ecor

d fo

r th

e m

ile

run

wit

h

a ti

me

of 3

:43.

13 (

3 m

in 4

3.13

s).

In 1

954,

Rog

er B

ann

iste

r ra

n t

he

firs

t m

ile

un

der

4 m

inu

tes

at 3

:59.

4. H

ad t

hey

ru

n t

hos

e ti

mes

in

th

e sa

me

race

, how

far

in

fro

nt

of B

ann

iste

r w

ould

El

Gu

erro

uj h

ave

been

at

th

e fi

nis

h?

Use

d

− t =

r. S

ince

3 m

in 4

3.13

s =

223

.13

s, a

nd

3 m

in 5

9.4

s =

239

.4 s

,

El

Gu

erro

uj’s

rat

e w

as

5280

ft

−

223.

13 s

an

d B

ann

iste

r’s

rate

was

5280

ft

−

239.

4 s .

rt

d

El G

ue

rro

uj

52

80

−

22

3.1

3

22

3.1

35

28

0 fe

et

Ba

nn

iste

r 5

28

0

−

23

9.4

2

23

.13

52

80

−

23

9.4

. 2

23

.13

or

49

21.

2 fe

et

Th

eref

ore,

wh

en E

l G

uer

rou

j hit

th

e ta

pe, h

e w

ould

be

5280

- 4

921.

2,

or 3

58.8

fee

t, a

hea

d of

Ban

nis

ter.

Let

’s s

ee w

het

her

we

can

dev

elop

a

form

ula

for

th

is t

ype

of p

robl

em.

L

et D

= t

he

dist

ance

rac

ed,

W

= t

he

win

ner

’s t

ime,

an

d L

= t

he

lose

r’s

tim

e.F

ollo

win

g th

e sa

me

patt

ern

, you

obt

ain

th

e re

sult

s sh

own

in

th

e ta

ble

at t

he

righ

t.T

he

win

nin

g di

stan

ce w

ill

be D

- D

W

−

L .

1. S

how

th

at t

he

expr

essi

on f

or t

he

win

nin

g di

stan

ce i

s eq

uiv

alen

t to

D(L

- W

) −

L

.

D

- D

W

−

L =

DL

−

L

- D

W

−

L =

DL -

DW

−

L

= D

(L -

W)

−

L

Use

th

e fo

rmu

la w

inn

ing

dis

tan

ce =

D(L

-

W)

−

L

to

fin

d t

he

win

nin

gd

ista

nce

to

the

nea

rest

ten

th f

or e

ach

of

the

foll

owin

g O

lym

pic

rac

es.

2. w

omen

’s 4

00 m

eter

rel

ay: E

ast

Ger

man

y 41

.6 s

(19

80);

Can

ada

48.4

s (

1928

) 56

.2 m

eter

s

3. m

en’s

200

met

er f

rees

tyle

sw

imm

ing:

Mic

hae

l P

hel

ps 1

min

42.

96 s

(20

08);

Yevg

eny

Sad

ovyi

1 m

in 4

6.70

s (

1992

) 7.

1 m

eter

s

4. m

en’s

50,

000

met

er w

alk:

Ale

x S

chw

azer

3 h

37

min

9 s

(20

08);

Har

twig

Gau

ter

3 h

49

min

24

s (1

980)

267

0.0

met

ers

5. w

omen

’s 4

00 m

eter

fre

esty

le s

wim

min

g re

lay:

Net

her

lan

ds 3

min

33.

76 s

(20

08);

Un

ited

Sta

tes

3 m

in 3

9.29

s (

1996

) 10

.1 m

eter

s

rt

d

Win

ne

r D

−

W

W D

−

W ·

W =

D

Lo

se

r D

−

L W

D

−

L · W

= D

W

−

L

036_

054_

ALG

1_A

_CR

M_C

11_C

R_6

6028

5.in

dd

5412

/20/

10

7:41

PM



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

PDF Pass


Chapter 11 Assessment Answer KeyQuiz 1 (Lessons 11-1 through 11-2) Quiz 3 (Lessons 11-5 and 11-6) Mid-Chapter TestPage 57 Page 58 Page 59

Quiz 2 (Lessons 11-3 and 11-4)

Page 57Quiz 4 (Lessons 11-7 and 11-8)

Page 58

1.

2. y

x

O

10

10 20-10-20

20

-10

-20

3.

4.

5.

D

xy = -18; y = 2

-2

x = -1, y = 2

1.

2.

3.

4.

5.

6.

7.

8.

9. 10.

A

16 −

5c

1

w 2 y

−

x 3 z 3

20j

m + 1 −

m(m - 1)

b + 1

−

b + 2

-4

7

3q

−

2 r 2

1. x −

2 + 3 - 5 −

2x

2. 5x - 2

3. 3x + 4

4. 24r2t

5. (x + 7)(x + 3)2

6. -

3n −

n 2 - 16

7.

4x + 14 −

x - 5

8. 2a

−

3

9. 4x + 7

−

x - 3

10.

D

1.

3 m 2 + m + 1 −

m

2. 3

−

xy 4 n 2

3.

x 2 + 3x - 1 −

x 2 + 3x + 2

4. x = -1

5. B

H

A

G

A

G

D1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. x + 2

−

2(x + 3)

x + 1

−

x - 5

xy = 21

9

x = 0, y = 2

-3


Copyri

ght

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

An

swer

s

PDF Pass


Chapter 11 Assessment Answer KeyVocabulary Test Form 1Page 60 Page 61 Page 62

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. F

A

J

C

F

A

G

D

H

B

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

D

H

C

F

A

G

D

H

A

G

B: 4 −

7

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

least common denominator

rational expression

work problem

extraneous solutions

inverse variation

complex fraction

excluded values

mixed expression

least common

multiple

rate problem

Sample answer: The

equation x1y

1 = x

2y

2 is the

product rule for inverse

variations. Use it to solve

problems involving inverse

variations.

Sample answer: A rational

equation is an equation that

is made up of rational

expressions.


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Chapter 11 Assessment Answer KeyForm 2A Form 2BPage 63 Page 64 Page 65 Page 66

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

B

G

A

F

C

H

D

G

B

G

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B

J

C

F

B

F

D

G

D

H

B: 6

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

D

H

A

H

A

J

D

G

A

J

11.

12.

13.

14.

15.

16.

17.

18.

19.

20. G

D

H

B

F

C

H

C

F

B

B: x + 2, x + 3


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Chapter 11 Assessment Answer KeyForm 2CPage 67 Page 68

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. 2n + 25

−

n 2 - 36

3 −

a + 3

8r + 4

−

r - 5

3b - 5 + 2 −

b - 4

2t + 4rt

−

3 - 3r

6(x + 7)

−

7(x - 5)

m + 5

−

5m

1 −

r - 4

2(y - 3)

a - 7

−

a - 1 ; -4, 1

-3, 5

xy = 126.4; 4

y

xO

10

-10

-10

-20

-2010 20

20

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B: 66 ft/s

1 ohm

2 8 −

11 hr

52 min

4.06 yd/yr

-7; extraneous 4

2

(x - 3)(x + 6)

−

(x + 2)(x + 1)

r 2 - 4

−

3r

a 2 + 3a + 7

−

a + 2

-9y + 20

−

5y(y - 7)

4 n 2 + 9

−

2n - 3

-3


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Chapter 11 Assessment Answer KeyForm 2DPage 69 Page 70

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. 3p + 32

−

p 2 - 49

4 −

v - 2

9t - 2

−

t + 6

5r + 2 + 8 −

r - 3

4 a 2 b - 5 b 2

−

3 - 5ab

5(x + 3)

−

2(x - 9)

x + 1

−

4x

1 −

r + 1

1 −

2(m - 2)

k + 5

−

k - 1 ; 1, 3

3, 4

xy = 110.4; 8

y

xO

10

10 20

20

-10-20

-10

-20

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B: 44 ft /s

9 ohms

1 1 −

5 hr

67.5 min

108 qt/day

-5; extraneous 3

-4

(x - 3)(x + 2)

−

(x + 1)(x - 2)

9 −

(x - 2 ) 2

n 2 + 4n + 7

−

n + 3

-8x + 3

−

3x(x - 5)

25 n 2 + 16

−

5n - 4

-2


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Chapter 11 Assessment Answer KeyForm 3Page 71 Page 72 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

6x - 7 −

(2x + 7 ) 2 (2x - 7)

2w + 4

−

5 w 2 - 1

(x + 4)(x - 4)

x 3 −

4 y 2 (x - 3)

1 −

w + 2

2b + 5

−

2b + 3

5a(a + 2)

−

4(a + 3)

2x - 3

−

2x - 1 ; -

1 −

3 , 1 −

2

-3, -1, 0

xy = 12 −

5 ;

3 −

5

y

xO

2

2 4

4

-2-4

-2

-4

x = 0.5, y = 0

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B: c = 5; k = 2

2x + 3, x + 3

29 1 −

3 ft/s

3 3 −

4 hr

2 1 −

7 hr

3; extraneous 2

1

10 −

27 r 2 u 2

1 −

x + 4

p 2 - p - 7

−

p - 4

2y + 2

−

y - 7

7w + 12

−

w + 5

5x - 2 −

4x - 1

-2 y 2 + 12y + 30

−

y 2 - 36


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Chapter 11 Assessment Answer KeyPage 73, Extended-Response Test

Scoring Rubric

Score General Description Specifi c Criteria

4 Superior

A correct solution that

is supported by well-

developed, accurate

explanations

• Shows thorough understanding of the concepts of inverse variations, rational expressions, division of a polynomial, simplifying complex fractions, and solving rational equations.

• Uses appropriate strategies to solve problems.

• Computations are correct.

• Written explanations are exemplary.

• Graphs are accurate and appropriate.

• Goes beyond requirements of some or all problems.

3 Satisfactory

A generally correct solution,

but may contain minor fl aws

in reasoning or computation

• Shows an understanding of the concepts of inverse variations, rational expressions, division of a polynomial, simplifying complex fractions, and solving rational equations.

• Uses appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are effective.

• Graphs are mostly accurate and appropriate.

• Satisfi es all requirements of problems.

2 Nearly Satisfactory

A partially correct

interpretation and/or

solution to the problem

• Shows an understanding of most of the concepts of inverse variations, rational expressions, division of a polynomial, simplifying complex fractions, and solving rational equations.

• May not use appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are satisfactory.

• Graphs are mostly accurate.

• Satisfi es the requirements of most of the problems.

1 Nearly Unsatisfactory

A correct solution with no

supporting evidence or

explanation

• Final computation is correct.

• No written explanations or work is shown to substantiate the fi nal

computation.

• Graphs may be accurate but lack detail or explanation.

• Satisfi es minimal requirements of some of the problems.

0 Unsatisfactory

An incorrect solution

indicating no mathematical

understanding of the

concept or task, or no

solution is given

• Shows little or no understanding of most of the concepts of inverse variations, rational expressions, division of a polynomial, simplifying complex fractions, and solving rational equations.

• Does not use appropriate strategies to solve problems.

• Computations are incorrect.

• Written explanations are unsatisfactory.

• Graphs are inaccurate or inappropriate.

• Does not satisfy requirements of problems.

• No answer may be given.


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Chapter 11 Assessment Answer Key Page 73, Extended-Response Test

Sample Answers

1. Sample answer: rt = 300; r = 100, t = 3; r = 50, t = 6; r = 30, t = 10; r = 10, t = 30; r = 4, t = 75;

t

rO

20

40

60

80

100

20 40 60 80 100

2a. Sample answer: (x - 9)(2x + 3) −

x - 9 ;

2x + 3; 9

2b. Sample answer: x - 2 − x , x −

x + 3 ;

(x - 2)(x + 3) −

x2

2c. Sample answer: 8x + 5 −

4x - 7 , 5x + 4 −

4x - 7 ;

13x + 9 −

4x - 7

3a. Sample answer: (x - 3)(2 x + 5) −

x - 3 ;

2x2 - x - 15 −

x - 3 ; 2x + 5

(Students should have used long division to simplify the expression.)

3b. The student should recognize that the two methods will sometimes yield the same result, and will sometimes yield different results. If the denominator is a factor of the numerator, then simplifying a rational expression by dividing out the greatest common factor will yield a polynomial. In this case using long division will also yield the same polynomial. If the denominator is not a factor of the numerator, then simplifying a rational expression by dividing out the greatest common factor will yield a rational expression. However, in this case using long division will yield the sum of a polynomial and a rational expression.

4. The student should explain that the number of flies Roshanda can make is

equivalent to 18 1 −

2 yards −

14 1 −

5 inches

per fly. The

answer can be found by converting yards to inches in the numerator of this complex fraction (666 in.), then performing the division that is implied. Roshanda can make 46 flies

with 12 4 −

5 inches of thread left over.

5a. Working together Pat and Harold will irrigate the fields faster than Pat will irrigate them working alone. The students should recognize that y will never equal x in this situation.

5b. Sample answer: 6; 2.4; 2.4 h

5c. Sample answer: 4; 5 + √ �� 17 ; 9.1 h

In addition to the scoring rubric found on page A34, the following sample answers may be used as guidance in evaluating extended-response assessment items.


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Chapter 11 Assessment Answer KeyStandardized Test PracticePage 74 Page 75

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. F G H J

8. F G H J

9. A B C D

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. F G H J

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

3 6

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

8 4 6. 17. 18.


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Chapter 11 Assessment Answer KeyStandardized Test PracticePage 76

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30a.

30b.

no, 192 ≠ 52 + 172

22.8 cm

15.6 in.

11 in.

y

xO 8 16-8-16

-8

-16

8

16

y

xO 2-2

4

8

12

16

y = 4(2x)

4

x = 1; (1, -3)

(x - 3)(x + 8)

0 or 6

-2

-1-2-3-5-4 0 1 2 63 4 5

∅

Sample answer:

y - 1500 =

-

670 −

59 (x - 0)

yes; 4x - 9y = 60


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