Chapter 7 Resource Masters - Commack Schools

Chapter 7 Resource Masters

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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 7 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 7 Test Form 2A and Form 2C.

Copyright © by The McGraw-Hill Companies, Inc.

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.

Send all inquiries to:McGraw-Hill Education8787 Orion PlaceColumbus, OH 43240

ISBN: 978-0-07-660281-0MHID: 0-07-660281-8

Printed in the United States of America.

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

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ContentsTeacher’s Guide to Using the Chapter 7

Resource Masters .........................................iv

Chapter Resources Chapter 7 Student-Built Glossary ...................... 1Chapter 7 Anticipation Guide (English) ............. 3Chapter 7 Anticipation Guide (Spanish) ............ 4

Lesson 7-1Multiplication Properties of ExponentsStudy Guide and Intervention ............................ 5Skills Practice .................................................... 7Practice ............................................................. 8Word Problem Practice ...................................... 9Enrichment ...................................................... 10

Lesson 7-2Division Properties of ExponentsStudy Guide and Intervention ...........................11Skills Practice .................................................. 13Practice ........................................................... 14Word Problem Practice .................................... 15Enrichment ...................................................... 16

Lesson 7-3Rational ExponentsStudy Guide and Intervention .......................... 17Skills Practice .................................................. 19Practice ........................................................... 20Word Problem Practice .................................... 21Enrichment ...................................................... 22

Lesson 7-4Scientifi c NotationStudy Guide and Intervention .......................... 23Skills Practice .................................................. 25Practice ........................................................... 26Word Problem Practice .................................... 27Enrichment ...................................................... 28

Lesson 7-5Exponential FunctionsStudy Guide and Intervention .......................... 29Skills Practice .................................................. 31Practice ........................................................... 32Word Problem Practice .................................... 33Enrichment ...................................................... 34

Lesson 7-6Growth and DecayStudy Guide and Intervention .......................... 35Skills Practice .................................................. 37Practice ........................................................... 38Word Problem Practice .................................... 39Enrichment ...................................................... 40Spreadsheet Activity ........................................ 41

Lesson 7-7Geometric Sequences as Exponential FunctionsStudy Guide and Intervention .......................... 42Skills Practice .................................................. 44Practice ........................................................... 45Word Problem Practice .................................... 46Enrichment ...................................................... 47

Lesson 7-8Recursive FormulasStudy Guide and Intervention .......................... 48Skills Practice .................................................. 50Practice ........................................................... 51Word Problem Practice .................................... 52Enrichment ...................................................... 53

AssessmentStudent Recording Sheet ............................... 55Rubric for Scoring Extended Response .......... 56Chapter 7 Quizzes 1 and 2 ............................. 57Chapter 7 Quizzes 3 and 4 ............................. 58Chapter 7 Mid-Chapter Test ............................ 59Chapter 7 Vocabulary Test ............................... 60Chapter 7 Test, Form 1 .................................... 61Chapter 7 Test, Form 2A ................................. 63Chapter 7 Test, Form 2B ................................. 65Chapter 7 Test, Form 2C ................................. 67Chapter 7 Test, Form 2D ................................. 69Chapter 7 Test, Form 3 .................................... 71Chapter 7 Extended Response Test ................ 73Standardized Test Practice ...............................74

Answers ........................................... A1–A36

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Teacher’s Guide to Using the Chapter 7 Resource Masters

Chapter ResourcesStudent-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 7-1. 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 ResourcesStudy 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.

The Chapter 7 Resource Masters includes the core materials needed for Chapter 7. 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.

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Assessment OptionsThe assessment masters in the Chapter 7 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master corresponds 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 11 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 ques-

tions 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 stu-dents. 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 7 1 Glencoe Algebra 1

7 Student-Built Glossary

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 7. 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.

Vocabulary TermFound

on PageDefi nition/Description/Example

binomialby·NOH·mee·uhl

constant

common ratio

compound interest

cube root

exponential decay function

exponential equation

exponential function

exponential growth function

(continued on the next page)

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Student-Built Glossary (continued)7

Vocabulary TermFound

on PageDefi nition/Description/Example

geometric sequence

monomial mah·NOH·mee·uhl

negative exponent

nth root

order of magnitude

rational exponent

recursive formula

scientific notation

zero exponent


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Anticipation GuideExponents and Exponential Functions

Before you begin Chapter 7

• 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).

STEP 1A, D, or NS

StatementSTEP 2A or D

1. When multiplying two powers that have the same base, multiply the exponents.

2. ( k 3 ) 4 is equivalent to k 12 .

3. To divide two powers that have the same base, subtract the exponents.

4. ( 2 − 5 )

3 is the same as 2 −

5 3 .

5. A polynomial may contain one or more monomials.

6. The degree of the polynomial 3 x 2 y 3 - 5y 2 + 8 x 3 is 3 because the greatest exponent is 3.

7. A function containing powers is called an exponential function.

8. Receiving compound interest on a bank account is one example of exponential growth.

After you complete Chapter 7

• 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.

7

Step 1

Step 2


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Capítulo 7 4 Álgebra 1 de Glencoe

Ejercicios preparatoriosExponentes y Funciones Exponenciales

7

Antes de comenzar el Capítulo 7

• 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)).

PASO 1A, D o NS

EnunciadoPASO 2A o D

1. Al multiplicar dos potencias con la misma base, multiplica los exponentes.

2. ( k 3 ) 4 es equivalente a k 12 .

3. Para dividir dos potencias que tienen la misma base, resta los exponentes.

4. ( 2 − 5 )

3 es lo mismo que 2 −

5 3 .

5. Un polinomio puede contener uno o más monomios.

6. El grado del polinomio 3x 2 y 3 - 5y 2 + 8x 3 es 3 porque el exponente más grande es 3.

7. Una función que contiene potencias se denomina función exponencial.

8. El recibir interés compuesto en una cuenta bancaria es un ejemplo de crecimiento exponencial.

Después de completar el Capítulo 7

• 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.

Paso 1

Paso 2

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Study Guide and InterventionMultiplication Properties of Exponents

Multiply Monomials A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. An expression of the form xn is called a power and represents the product you obtain when x is used as a factor n times. To multiply two powers that have the same base, add the exponents.

Product of Powers For any number a and all integers m and n, am ․ an = am + n.

Simplify (3x6)(5x2).(3x6)(5x2) = (3)(5)(x6 ․ x2) Group the coefficients

and the variables

= (3 ․ 5)(x6 + 2) Product of Powers

= 15x8 Simplify.

The product is 15x8.

Simplify (-4a3b)(3a2b5).(-4a3b)(3a2b5) = (-4)(3)(a3 ․ a2)(b ․ b5)

= -12(a3 + 2)(b1 + 5) = -12a5b6

The product is -12a5b6.

ExercisesSimplify each expression.

1. y(y5) 2. n2 ․ n7 3. (-7x2)(x4)

4. x(x2)(x4) 5. m ․ m5 6. (-x3)(-x4)

7. (2a2)(8a) 8. (rn)(rn3)(n2) 9. (x2y)(4xy3)

10. 1 − 3 ( 2a 3 b)( 6b 3 ) 11. (-4x3)(-5x7) 12. (-3j2k4)(2jk6)

13. ( 5a 2 bc 3 ) ( 1 − 5 abc 4 ) 14. (-5xy)(4x2)(y4) 15. (10x3yz2)(-2xy5z)

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Example 1 Example 2


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Simplify Expressions An expression of the form (xm)n is called a power of a power and represents the product you obtain when xm is used as a factor n times. To find the power of a power, multiply exponents.

Power of a Power For any number a and any integers m and p, (am)p = amp.

Power of a Product For any numbers a and b and any integer m, (ab)m = ambm.

We can combine and use these properties to simplify expressions involving monomials.

Simplify (-2ab2)3(a2)4.

(-2ab2)3(a2)4 = (-2ab2)3(a8) Power of a Power

= (-2)3(a3)(b2)3(a8) Power of a Product

= (-2)3(a3)(a8)(b2)3 Group the coefficients and the variables

= (-2)3(a11)(b2)3 Product of Powers

= -8a11b6 Power of a Power

The product is -8a11b6.

ExercisesSimplify each expression.

1. (y5)2 2. (n7)4 3. (x2)5(x3)

4. -3(ab4)3 5. (-3ab4)3 6. (4x2b)3

7. (4a2)2(b3) 8. (4x)2(b3) 9. (x2y4)5

10. (2a3b2)(b3)2 11. (-4xy)3(-2x2)3 12. (-3j2k3)2(2j2k)3

13. (25 a 2 b) 3 ( 1 − 5 abf)

2 14. (2xy)2(-3x2)(4y4) 15. (2x3y2z2)3(x2z)4

16. (-2n6y5)(-6n3y2)(ny)3 17. (-3a3n4)(-3a3n)4 18. -3(2x)4(4x5y)2

Study Guide and Intervention (continued)

Multiplication Properties of Exponents

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Example


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Skills PracticeMultiplication Properties of Exponents

Determine whether each expression is a monomial. Write yes or no. Explain.

1. 11

2. a - b

3. p 2

− r 2

4. y

5. j3k

6. 2a + 3b

Simplify.

7. a2(a3)(a6) 8. x(x2)(x7)

9. (y2z)(yz2) 10. (ℓ2k2)(ℓ3k)

11. (a2b4)(a2b2) 12. (cd2)(c3d2)

13. (2x2)(3x5) 14. (5a7)(4a2)

15. (4xy3)(3x3y5) 16. (7a5b2)(a2b3)

17. (-5m3)(3m8) 18. (-2c4d)(-4cd)

19. (102)3 20. (p3)12

21. (-6p)2 22. (-3y)3

23. (3pr2)2 24. (2b3c4)2

GEOMETRY Express the area of each figure as a monomial.

25.

x2

x5

26.

cd

cd

27.

4p

9p3

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PracticeMultiplication Properties of Exponents

Determine whether each expression is a monomial. Write yes or no. Explain your reasoning.

1. 21 a 2 − 7b

2. b 3 c 2 − 2

Simplify each expression.

3. (-5x2y)(3x4) 4. (2ab2f 2)(4a3b2f 2)

5. (3ad4)(-2a2) 6. (4g3h)(-2g5)

7. (-15x y 4 ) (- 1 − 3 x y 3 ) 8. (-xy)3(xz)

9. (-18 m 2 n) 2 (- 1 − 6 m n 2 ) 10. (0.2a2b3)2

11. ( 2 − 3 p)

2 12. ( 1 −

4 a d 3 )

2

13. (0.4k3)3 14. [(42)2]2

GEOMETRY Express the area of each figure as a monomial.

15.

6a2b4

3ab2

16. 5x3

17.

6ab3

4a2b

GEOMETRY Express the volume of each solid as a monomial.

18.

3h23h2

3h2

19.

m3nmn3

n 20.

7g2

3g

21. COUNTING A panel of four light switches can be set in 24 ways. A panel of five light switches can set in twice this many ways. In how many ways can five light switches be set?

22. HOBBIES Tawa wants to increase her rock collection by a power of three this year and then increase it again by a power of two next year. If she has 2 rocks now, how many rocks will she have after the second year?

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1. GRAVITY An egg that has been falling for x seconds has dropped at an average speed of 16x feet per second. If the egg is dropped from the top of a building, its total distance traveled is the product of the average rate times the time. Write a simplified expression to show the distance the egg has traveled after x seconds.

2. CIVIL ENGINEERING A developer is planning a sidewalk for a new development. The sidewalk can be installed in rectangular sections that have a fixed width of 3 feet and a length that can vary. Assuming that each section is the same length, express the area of a 4-section sidewalk as a monomial.

x

3 ft

3. PROBABILITY If you flip a coin 3 times in a row, there are 23 outcomes that can occur.

Outcomes

HHH TTT

HTT THH

HTH TTH

HHT THT

If you then flip the coin two more times, there are 23 × 22 outcomes that can occur. How many outcomes can occur if you flip the coin as mentioned above four more times? Write your answer in the form 2x.

4. SPORTS The volume of a sphere is given by the formula V = 4−

3 π r 3 , where r is the

radius of the sphere. Find the volume of air in three different basketballs. Use π = 3.14. Round your answers to the nearest whole number.

Ball Radius (in.) Volume (in3)

Child’s 4

Women’s 4.5

Men’s 4.8

5. ELECTRICITY An electrician uses the formula W = I2R , where W is the power in watts, I is the current in amperes, and R is the resistance in ohms.

a. Find the power in a household circuit that has 20 amperes of current and 5 ohms of resistance.

b. If the current is reduced by one half, what happens to the power?

Word Problem PracticeMultiplication Properties of Exponents


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Enrichment

An Wang An Wang (1920–1990) was an Asian-American who became one of the pioneers of the computer industry in the United States. He grew up in Shanghai, China, but came to the United States to further his studies in science. In 1948, he invented a magnetic pulse controlling device that vastly increased the storage capacity of computers. He later founded his own company, Wang Laboratories, and became a leader in the development of desktop calculators and word processing systems. In 1988, Wang was elected to the National Inventors Hall of Fame.

Digital computers store information as numbers. Because the electronic circuits of a computer can exist in only one of two states, open or closed, the numbers that are stored can consist of only two digits, 0 or 1. Numbers written using only these two digits are called binary numbers. To find the decimal value of a binary number, you use the digits to write a polynomial in 2. For instance, this is how to find the decimal value of the number 10011012. (The subscript 2 indicates that this is a binary number.)

10011012 = 1 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20

= 1 × 64 + 0 × 32 + 0 × 16 + 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1= 64 + 0 + 0 + 8 + 4 + 0 + 1= 77

Find the decimal value of each binary number.

1. 11112 2. 100002 3. 110000112 4. 101110012

Write each decimal number as a binary number.

5. 8 6. 11 7. 29 8. 117

9. The chart at the right shows a set of decimal code numbers that is used widely in storing letters of the alphabet in a computer’s memory. Find the code numbers for the letters of your name. Then write the code for your name using binary numbers.

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The American Standard Guide for

Information Interchange (ASCII)

A 65 N 78 a 97 n 110

B 66 O 79 b 98 o 111

C 67 P 80 c 99 p 112

D 68 Q 81 d 100 q 113

E 69 R 82 e 101 r 114

F 70 S 83 f 102 s 115

G 71 T 84 g 103 t 116

H 72 U 85 h 104 u 117

I 73 V 86 i 105 v 118

J 74 W 87 j 106 w 119

K 75 X 88 k 107 x 120

L 76 Y 89 l 108 y 121

M 77 Z 90 m 109 z 122


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Study Guide and InterventionDivision Properties of Exponents

Divide Monomials To divide two powers with the same base, subtract the exponents.

Quotient of Powers For all integers m and n and any nonzero number a, a m − a n

= a m - n .

Power of a Quotient For any integer m and any real numbers a and b, b ≠ 0, ( a −

b )

m = a m −

b m .

Simplify a 4 b 7 −

ab 2 . Assume

that no denominator equals zero.

a 4 b 7 − ab 2

= ( a 4 − a ) ( b 7 − b 2

) Group powers with the same base.

= (a4 - 1)(b7 - 2) Quotient of Powers

= a3b5 Simplify.

The quotient is a3b5.

Simplify (

2 a 3 b 5 −

3 b 2

)

3

. Assume

that no denominator equals zero.

( 2 a 3 b 5 − 3 b 2

) 3

= (2 a 3 b 5 ) 3 −

(3 b 2 ) 3 Power of a Quotient

= 2 3 ( a 3 ) 3 ( b 5 ) 3 −

(3 ) 3 ( b 2 ) 3 Power of a Product

= 8 a 9 b 15 − 27 b 6

Power of a Power

= 8 a 9 b 9 − 27

Quotient of Powers

The quotient is 8 a 9 b 9 − 27

.

ExercisesSimplify each expression. Assume that no denominator equals zero.

1. 5 5 − 5 2

2. m 6 − m 4

3. p 5 n 4

− p 2 n

4. a 2 − a 5. x 5 y 3

− x 5 y 2

6. -2 y 7

− 14 y 5

7. x y 6

− y 4 x

8. ( 2 a 2 b − a ) 3

9. ( 4 p 4 r 4

− 3 p 2 r 2

) 3

10. ( 2 r 5 w 3 − r 4 w 3

) 4

11. ( 3 r 6 n 3 − 2 r 5 n

) 4

12. r 7 n 7 t 2 − n 3 r 3 t 2

7-2

Example 1 Example 2


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Division Properties of Exponents

Negative Exponents Any nonzero number raised to the zero power is 1; for example, (-0.5)0 = 1. Any nonzero number raised to a negative power is equal to the reciprocal of the number raised to the opposite power; for example, 6 -3 = 1 −

6 3 . These definitions can be used to

simplify expressions that have negative exponents.

Zero Exponent For any nonzero number a, a0 = 1.

Negative Exponent Property For any nonzero number a and any integer n, a -n = 1 − a n

and 1 −

a -n = a n .

The simplified form of an expression containing negative exponents must contain only positive exponents.

Simplify 4 a -3 b 6 −

16 a 2 b 6 c -5 . Assume that no denominator equals zero.

4 a -3 b 6 − 16 a 2 b 6 c -5

= ( 4 − 16

) ( a -3 − a 2

) ( b 6 − b 6

) ( 1 − c -5

) Group powers with the same base.

= 1 − 4 ( a -3-2 )( b 6-6 )( c 5 ) Quotient of Powers and Negative Exponent Properties

= 1 − 4 a -5 b 0 c 5 Simplify.

= 1 − 4 ( 1 −

a 5 ) (1) c 5 Negative Exponent and Zero Exponent Properties

= c 5 − 4 a 5

Simplify.

The solution is c 5 − 4 a 5

.

ExercisesSimplify each expression. Assume that no denominator equals zero.

1. 2 2 − 2 -3

2. m − m -4

3. p -8

− p 3

4. b -4 − b -5

5. (- x -1 y) 0

− 4 w -1 y 2

6. ( a 2 b 3 ) 2 −

(a b) -2

7. x 4 y 0

− x -2

8. (6 a -1 b) 2 −

( b 2 ) 4 9. (3rt ) 2 u -4

− r -1 t 2 u 7

10. m -3 t -5 − ( m 2 t 3 ) -1

11. ( 4 m 2 n 2 − 8 m -1 �

) 0

12. (-2m n 2 ) -3

− 4 m -6 n 4

7-2

Example


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Skills PracticeDivision Properties of Exponents

Simplify each expression. Assume that no denominator equals zero.

1. 6 5 − 6 4

2. 9 12 − 9 8

3. x 4 − x 2

4. r 3 t 2 − r 3 t 4

5. m − m 3

6. 9 d 7 − 3 d 6

7. 12 n 5 − 36n

8. w 4 x 3 − w 4 x

9. a 3 b 5 − ab 2

10. m 7 p 2

− m 3 p 2

11. -21 w 5 x 2 − 7 w 4 x 5

12. 32 x 3 y 2 z 5

− -8xy z 2

13. ( 4 p 7

− 7 r 2

) 2

14. 4-4

15. 8-2 16. ( 5 − 3 )

-2

17. ( 9 − 11

) -1

18. h 3 − h -6

19. k0(k4)(k-6) 20. k-1(ℓ-6)(m3)

21. f -7

− f 4

22. ( 16 p 5 w 2

− 2 p 3 w 3

) 0

23. f -5 g 4

− h -2

24. 15 x 6 y -9

− 5xy -11

25. -15 t 0 u -1 − 5 u 3

26. 48 x 6 y 7 z 5

− -6x y 5 z 6

7-2


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PracticeDivision Properties of Exponents

Simplify each expression. Assume that no denominator equals zero.

1. 8 8 − 8 4

2. a 4 b 6 − ab 3

3. xy 2

− xy

4. m 5 np

− m 4 p

5. 5 c 2 d 3 − -4 c 2 d

6. 8 y 7 z 6

− 4 y 6 z 5

7. ( 4f 3g −

3 h 6 )

3

8. ( 6 w 5 − 7 p 6 r 3

) 2

9. -4 x 2 − 24 x 5

10. x3(y-5)(x-8) 11. p(q-2)(r-3) 12. 12-2

13. ( 3 − 7 )

-2 14. ( 4 −

3 )

-4 15. 22 r 3 s 2 −

11 r 2 s -3

16. -15 w 0 u -1 − 5 u 3

17. 8 c 3 d 2 f 4

− 4 c -1 d 2 f -3

18. ( x -3 y 5

− 4 -3

) 0

19. 6 f -2 g 3 h 5

− 54 f -2 g -5 h 3

20. -12 t -1 u 5 x -4 − 2 t -3 u x 5

21. r 4 − (3r ) 3

22. m -2 n -5 − ( m 4 n 3 ) -1

23. ( j -1 k 3 ) -4

− j 3 k 3

24. (2 a -2 b ) -3 −

5 a 2 b 4

25. ( q -1 r 3 −

q r -2 )

-5

26. ( 7 c -3 d 3 − c 5 d h -4

) -1

27. ( 2 x 3 y 2 z −

3 x 4 y z -2 )

-2

28. BIOLOGY A lab technician draws a sample of blood. A cubic millimeter of the blood contains 223 white blood cells and 225 red blood cells. What is the ratio of white blood cells to red blood cells?

29. COUNTING The number of three-letter “words” that can be formed with the English alphabet is 263. The number of five-letter “words” that can be formed is 265. How many times more five-letter “words” can be formed than three-letter “words”?

7-2


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Word Problem PracticeDivision Properties of Exponents

1. CHEMISTRY The nucleus of a certain atom is 10-13 centimeters across. If the nucleus of a different atom is 10-11 centimeters across, how many times as large is it as the first atom?

2. SPACE The Moon is approximately 254 kilometers away from Earth on average. The Olympus Mons volcano on Mars stands 25 kilometers high. How many Olympus Mons volcanoes, stacked on top of one another, would fit between the surface of the Earth and the Moon?

3. E-MAIL Spam (also known as junk e-mail) consists of identical messages sent to thousands of e-mail users. People often obtain anti-spam software to filter out the junk e-mail messages they receive. Suppose Yvonne’s anti-spam software filtered out 102 e-mails, and she received 104 e-mails last year. What fraction of her e-mails were filtered out? Write your answer as a monomial.

4. METRIC MEASUREMENT Consider a dust mite that measures 10-3 millimeters in length and a caterpillar that measures 10 centimeters long. How many times as long as the mite is the caterpillar?

5. COMPUTERS In 1995, standard capacity for a personal computer hard drive was 40 megabytes (MB). In 2010, a standard hard drive capacity was 500 gigabytes (GB or Gig). Refer to the table below.

Memory Capacity Approximate Conversions

8 bits = 1 byte

103 bytes = 1 kilobyte

103 kilobytes = 1 megabyte (meg)

103 megabytes = 1 gigabyte (gig)

103 gigabytes = 1 terabyte

103 terabytes = 1 petabyte

a. The newer hard drives have about how many times the capacity of the 1995 drives?

b. Predict the hard drive capacity in the year 2025 if this rate of growth continues.

c. One kilobyte of memory is what fraction of one terabyte?

7-2


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Enrichment

Patterns with PowersUse your calculator, if necessary, to complete each pattern.

a. 210 = b. 510 = c. 410 =

29 = 59 = 49 =

28 = 58 = 48 =

27 = 57 = 47 =

26 = 56 = 46 =

25 = 55 = 45 =

24 = 54 = 44 =

23 = 53 = 43 =

22 = 52 = 42 =

21 = 51 = 41 =

Study the patterns for a, b, and c above. Then answer the questions.

1. Describe the pattern of the exponents from the top of each column to the bottom.

2. Describe the patte rn of the powers from the top of the column to the bottom.

3. What would you expect the following powers to be?

20 50 40

4. Refer to Exercise 3. Write a rule. Test it on patterns that you obtain using 22, 25, and 24 as bases.

Study the pattern below. Then answer the questions.

03 = 0 02 = 0 01 = 0 00 = ? 0-1 does not exist. 0-2 does not exist. 0-3 does not exist.

5. Why do 0-1, 0-2, and 0-3 not exist?

6. Based upon the pattern, can you determine whether 00 exists?

7. The symbol 00 is called an indeterminate, which means that it has no unique value. Thus it does not exist as a unique real number. Why do you think that 00 cannot equal 1?

7-2


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Study Guide and InterventionRational Exponents

Rational Exponents For any real numbers a and b and any positive integer n, if an = b, then a is an nth root of b. Rational exponents can be used to represent nth roots.

Square Root b 1 − 2 = √ � b

Cube Root b 1 − 3 = 3 √ � b

nth Root b 1 − n =

n √ � b

7-3

Write (6xy ) 1 −

2 in radical

form.(6xy )

1 − 2 = √ �� 6xy Defi nition of b

1 −

2

Simplify 625 1 −

4 .

625 1 − 4 = 4 √ �� 625 b

1 − n =

n √ � b

= 4 √ �� 5 � 5 � 5 � 5 625 = 5 4

= 5 Simplify.

Example 1 Example 2

ExercisesWrite each expression in radical form, or write each radical in exponential form.

1. 14 1 − 2 2. 5 x

1 − 2 3. 17 y

1 − 2

4. 1 2 1 − 2 5. 19 ab

1 − 2 6. √ �� 17

7. √ �� 12n 8. √

��

18b 9. √ �� 37

Simplify.

10. 3 √ �� 343 11. 5 √ �� 1024 12. 51 2 1 − 3

13. 4 √ �� 2401 14. 6 √ ��

64 15. 24 3 1 − 5

16. 3 √ �� 1331 17. 4 √ ��

6561 18. 409 6 1 − 4


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Rational Exponents

7-3

Solve Exponential Equations In an exponential equation, variables occur as exponents. Use the Power Property of Equality and the other properties of exponents to solve exponential equations.

Solve 102 4 x − 1 = 4.

102 4 x − 1 = 4 Original equation

( 4 5 ) x − 1 = 4 Rewrite 1024 as 4 5 .

4 5x − 5 = 4 1 Power of a Power, Distributive Property

5x − 5 = 1 Power Property of Equality

5x = 6 Add 5 to each side.

x = 6 − 5 Divide each side by 5.

ExercisesSolve each equation.

1. 2 x = 128 2. 3 3x + 1 = 81 3. 4 x - 3 = 32

4. 5 x = 15,625 5. 6 3x + 2 = 216 6. 4 5x - 3 = 16

7. 8 x = 4096 8. 9 3x + 3 = 6561 9. 1 1 x - 1 = 1331

10. 3 x = 6561 11. 2 5x + 4 = 512 12. 7 x - 2 = 343

13. 8 x = 262,144 14. 5 5x = 3125 15. 9 2x - 6 = 6561

16. 7 x = 2401 17. 7 3x = 117,649 18. 6 2x - 7 = 7776

19. 9 x = 729 20. 8 3x + 1 = 4096 21. 1 3 3x - 8 = 28,561

Example


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7-3 Skills PracticeRational Exponents

Write each expression in radical form, or write each radical in exponential form.

1. (8 x) 1 − 2 2. 6z

1 − 2 3. √ �� 19

4. √ �� 11 5. 19 x 1 − 2 6. √ �� 34

7. √ �� 27g 8. 33 gh 1 − 2 9. √ �� 13abc

Simplify.

10. ( 1 − 16

) 1 − 4 11. 5 √ �� 3125 12. 72 9

1 −

3

13. ( 1 − 32

) 1 − 5 14. 6 √ �� 4096 15. 102 4

1 − 5

16. ( 16 − 625

) 1 − 4 17. 6 √ �� 15,625 18. 117,6 49

1 − 6

Solve each equation.

19. 2 x = 512 20. 3 x = 6561 21. 6 x = 46,656

22. 5 x = 125 23. 3 x - 3 = 243 24. 4 x - 1 = 1024

25. 6 x - 1 = 1296 26. 2 4x + 3 = 2048 27. 3 3x + 3 = 6561


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7-3 PracticeRational Exponents

Write each expression in radical form, or write each radical in exponential form.

1. √ �� 13 2. √ �� 37 3. √ �� 17x

4. (7ab ) 1 −

2 5. 2 1z

1 − 2 6. 13(a b)

1 − 2

Simplify.

7. ( 1 − 81

) 1 − 4 8. 5

√ �� 1024 9. 51 2 1 − 3

10. ( 32 − 1024

) 1 − 5 11. 4

√ �� 1296 12. 312 5 1 − 5


13. 3 x = 729 14. 4 x = 4096 15. 5 x = 15,625

16. 6 x + 3 = 7776 17. 3 x - 3 = 2187 18. 4 3x + 4 = 16,384

19. WATER The flow of water F in cubic feet per second over a wier, a small overflow dam,

can be represented by F = 1.26 H 3 − 2 , where H is the height of the water in meters above

the crest of the wier. Find the height of the water if the flow of the water is 10.08 cubic feet per second.


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7-3 Word Problem PracticeRational Exponents

1. VELOCITY The velocity v in feet per second of a freely falling object that has fallen h feet can be represented by

v = 8 h 1 − 2 . Find the distance that an object

has fallen if its velocity is 96 feet per second.

2. ELECTRICITY The relationship of the current, the power, and the resistance in an appliance can be modeled by

I R 1 − 2 = √ � P , where I is the current in

amperes, P is the power in watts, and R is the resistance in ohms. Find the power that an appliance is using if the current is 2.5 amps and the resistance is 16 ohms.

3. GEOMETRY The surface area T of a cube in square inches can be determined

by T = 6 V 2 − 3 , where V is the volume of

the cube in cubic inches.

a. Find the surface area of a cube that has a volume of 4096 cubic inches.

b. Find the volume of a cube that has a surface area of 96 square inches.

4. PLANETS The average distance d in astronomical units that a planet is from

the Sun can be modeled by d = t 2 − 3 , where

t is the number of Earth years that it takes for the planet to orbit the Sun.

a. Find the average distance a planet is from the Sun if the planet has an orbit of 27 Earth years.

b. Find the time that it would take a planet to orbit the Sun if its average distance from the Sun is 4 astronomical units.

5. BIOLOGY The relationship between the mass m in kilograms of an organism and its metabolism P in Calories per day can

be represented by P = 73.3 4 √ �� m 3 . Find

the mass of an organism that has a metabolism of 586.4 Calories per day.

6. MANUFACTURING The profit P of a company in thousands of dollars can be

modeled by P = 12.75 5 √ � c 2 , where c is the

number of customers in hundreds. If the profit of the company is $51,000, how many customers do they have?


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Enrichment7-3

CounterexamplesSome statements in mathematics can be proven false by counterexamples. Consider the following statement.For any numbers a and b, a - b = b - a.You can prove that this statement is false in general if you can find one example for which the statement is false.Let a = 7 and b = 3. Substitute these values in the equation above.

7 - 3 � 3 - 7 4 ≠ -4

In general, for any numbers a and b, the statement a - b = b - a is false. You can make the equivalent verbal statement: subtraction is not a commutative operation.

In each of the following exercises a, b, and c are any numbers. Prove that the statement is false by counterexample.

1. a - (b - c) � (a - b) - c 2. a ÷ (b ÷ c) � (a ÷ b) ÷ c

3. a ÷ b � b ÷ a 4. a ÷ (b + c) � (a ÷ b) + (a ÷ c)

5. a 1 − 2 + a

1 −

2 � a 1 6. a c � b d � (ab ) c + d

7. Write the verbal equivalents for Exercises 1, 2, and 3.

8. For the Distributive Property a(b + c) = ab + ac, it is said that multiplication distributes over addition. Exercise 4 proves that some operations do not distribute. Write a statement for the exercise that indicates this.


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Study Guide and InterventionScientifi c Notation

7-4

Scientific Notation Very large and very small numbers are often best represented using a method known as scientific notation. Numbers written in scientific notation take the form a × 10n, where 1 ≤ a < 10 and n is an integer. Any number can be written in scientific notation.

Express 34,020,000,000 in scientific notation.Step 1 Move the decimal point until it is to the right of the first nonzero digit. The result is a real number a. Here, a = 3.402.

Step 2 Note the number of places n and the direction that you moved the decimal point. The decimal point moved 10 places to the left, so n = 10.

Step 3 Because the decimal moved to the left, write the number as a × 10n.34,020,000,000 = 3.4020000000 × 1010

Step 4 Remove the extra zeros. 3.402 × 1010

Express 4.11 × 10-6 in standard notation.Step 1 The exponent is –6, so n = –6.

Step 2 Because n < 0, move the decimal point 6 places to the left. 4.11 × 10-6 ⇒ .00000411

Step 3 4.11 × 10-6 ⇒ 0.00000411Rewrite; insert a 0 before the decimal point.

Example 1 Example 2

ExercisesExpress each number in scientific notation.

1. 5,100,000 2. 80,300,000,000 3. 14,250,000

4. 68,070,000,000,000 5. 14,000 6. 901,050,000,000

7. 0.0049 8. 0.000301 9. 0.0000000519

10. 0.000000185 11. 0.002002 12. 0.00000771

Express each number in standard form.

13. 4.91 × 104 14. 3.2 × 10-5 15. 6.03 × 108

16. 2.001 × 10-6 17. 1.00024 × 1010 18. 5 × 105

19. 9.09 × 10-5 20. 3.5 × 10-2 21. 1.7087 × 107


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Scientifi c Notation

7-4

Products and Quotients in Scientific Notation You can use scientific notation to simplify multiplying and dividing very large and very small numbers.

Evaluate (9.2 × 10-3) ×(4 × 108). Express the result in both scientific notation and standard form.

(9.2 × 10-3)(4 × 108) Original expression

= (9.2 × 4)(10-3 × 108) Commutative andAssociative Properties

= 36.8 × 105 Product of Powers

= (3.68 × 101) × 105 36.8 = 3.68 × 10


= 3,680,000 Standard Form

Evaluate (2.76 × 10

7

)

−

(6.9 × 10

5

)

.

Express the result in both scientific notation and standard form. (2.76 × 10 7 )

− (6.9 × 10 5 )

= ( 2.76 − 6.9

) ( 10 7 −

10 5 ) Product rule for

fractions

= 0.4 × 102 Quotient of Powers

= 4.0 × 10-1 × 102 0.4 = 4.0 × 10-1


= 40 Standard form

Example 1 Example 2

ExercisesEvaluate each product. Express the results in both scientific notation and standard form.

1. (3.4 × 103)(5 × 104) 2. (2.8 × 10-4)(1.9 × 107)

3. (6.7 × 10-7)(3 × 103) 4. (8.1 × 105)(2.3 × 10-3)

5. (1.2 × 1 0 -4 ) 2 6. (5.9 × 1 0 5 ) 2

Evaluate each quotient. Express the results in both scientific notation and standard form.

7. (4.9 × 1 0 -3 )

− (2.5 × 1 0 -4 )

8. 5.8 × 10 4 − 5 × 10 -2

9. (1.6 × 1 0 5 )

− (4 × 1 0 -4 )

10. 8.6 × 10 6 − 1.6 × 10 -3

11. (4.2 × 1 0 -2 )

− (6 × 1 0 -7 )

12. 8.1 × 10 5 − 2.7 × 10 4


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Skills PracticeScientifi c Notation

7-4

Express each number in scientific notation.

1. 3,400,000,000 2. 0.000000312

3. 2,091,000 4. 980,200,000,000,000

5. 0.00000000008 6. 0.00142


7. 2.1 × 105 8. 8.023 × 10-7

9. 3.63 × 10-6 10. 7.15 × 108

11. 1.86 × 10-4 12. 4.9 × 105

Evaluate each product. Express the results in both scientific notation and standard form.

13. (6.1 × 105)(2 × 105) 14. (4.4 × 106)(1.6 × 10-9)

15. (8.8 × 108)(3.5 × 10-13) 16. (1.35 × 108)(7.2 × 10-4)

17. (2.2 × 1 0 -3 ) 2 18. (3.4 × 1 0 2 ) 2


19. (9.2 × 10-8) −

(2 × 10-6) 20. (4.8 × 104)

− (3 × 10-5)

21. (1.161 × 10-9) −

(4.3 × 10-6) 22.

(4.625 × 1 0 10 ) −

(1.25 × 10 4 )

23. (2.376 × 10-4) −

(7.2 × 10-8) 24. (8.74 × 10-3)

− (1.9 × 105)


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PracticeScientifi c Notation

7-4


1. 1,900,000 2. 0.000704

3. 50,040,000,000 4. 0.0000000661


5. 5.3 × 107 6. 1.09 × 10-4

7. 9.13 × 103 8. 7.902 × 10-6

Evaluate each product. Express the results in both scientific notation and standard form.

9. (4.8 × 104)(6 × 106) 10. (7.5 × 10-5)(3.2 × 107)

11. (2.06 × 104)(5.5 × 10-9) 12. (8.1 × 10-6)(1.96 × 1011)

13. (7.2 × 1 0 -5 ) 2 14. (5.29 × 1 0 6 ) 2


15. (4.2 × 1 0 5 ) −

(3 × 1 0 -3 ) 16. (1.76 × 1 0 -11 )

− (2.2 × 1 0 -5 )

17. (7.05 × 1 0 12 ) −

(9.4 × 1 0 7 ) 18. (2.04 × 1 0 -4 )

− (3.4 × 1 0 5 )

19. GRAVITATION Issac Newton’s theory of universal gravitation states that the equation F = G

m1m2 − r2

can be used to calculate the amount of gravitational force in newtons

between two point masses m1 and m2 separated by a distance r. G is a constant equal to 6.67 × 10-11 Nm2kg–2. The mass of Earth m1 is equal to 5.97 × 1024 kg, the mass of the Moon m2 is equal to 7.36 × 1022 kg, and the distance r between the two is 384,000,000 m.

a. Express the distance r in scientific notation.

b. Compute the amount of gravitational force between Earth and the Moon. Express your answer in scientific notation.


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Word Problem PracticeScientifi c Notation

7-4

1. PLANETS Neptune’s mean distance from the Sun is 4,500,000,000 kilometers. Uranus’ mean distance from the Sun is 2,870,000,000 kilometers. Express these distances in scientific notation.

2. PATHOLOGY The common cold is caused by the rhinovirus, which commonly measures 2 × 10-8 m in diameter. The E. coli bacterium, which causes food poisoning, commonly measures 3 × 10-6 m in length. Express these measurements in standard form.

3. COMMERCIALS A 30-second commercial aired during the 2007 Super Bowl cost $2,600,000. A 30-second commercial aired during the 1967 Super Bowl cost $40,000. Express these values in scientific notation. How many times more expensive was it to air an advertisement during the 2007 Super Bowl than the 1967 Super Bowl?

4. AVOGADRO’S NUMBER Avogadro’s number is an important concept in chemistry. It states that the number 6.022 × 1023 is approximately equal to the number of molecules in 12 grams of carbon 12. Use Avogadro’s number to determine the number of molecules in 5 × 10-7 grams of carbon 12.

5. COAL RESERVES The table below shows the number of kilograms of coal select countries had in proven reserve at the end of a recent year.

Coal Reserves

Country Coal (kg)

United States 2.46 × 1014

Russia 1.57 × 1014

India 9.24 × 1013

Romania 4.94 × 1011

Source: British Petroleum

a. Express each country’s coal reserves in standard form.

b. How many times more coal does the United States have than Romania?

c. One kilogram of coal has an energy density of 2.4 × 107 joules. What is the total energy density of the United States’ coal reserve? Express your answer in scientific notation.


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7-4 EnrichmentEngineering Notation

Engineering notation is a variation on scientific notation where numbers are expressed as powers of 1000 rather than as powers of 10. Engineering notation takes the familiar form of a × 10n, but n is restricted to multiples of three and 1 ≤ |a|< 1000.One advantage to engineering notation is that numbers can be neatly expressed using SI prefixes. These prefixes are typically used for scientific measurements.

1000n 10n SI Prefi x Symbol

10005 1015 peta P

10004 1012 tera T

10003 109 giga G

10002 106 mega M

10001 103 kilo k

1000-1 10-3 milli m

1000-2 10-6 micro μ

1000-3 10-9 nano n

1000-4 10-12 pico p

1000-5 10-15 femto f

NUCLEAR POWER The output of a nuclear power plant is measured to be 620,000,000 watts. Express this number in engineering notation and using SI prefixes.

To express a number in engineering notation, first convert the number to scientific notation.

Step 1 620,000,000 ⇒ 6.20000000 a = 6.20000000

Step 2 The decimal point moved 8 places to the left, so n = 8.

Step 3 620,000,000 = 6.20000000 × 108 = 6.2 × 108

Because 8 is not a multiple of 3, we need to round down n to the next multiple of 3.

Step 4 6.2 × 108 = (6.2 × 102) × 106 620 = 6.2 × 102

Step 5 = 620 × 106 Product of Powers

The output of the power plant is 620 × 106 watts. Using the chart above, the prefix for 106 is found to be mega, or M. The output of the power plant is 620 megawatts, or 620MW.

ExercisesExpress each number in engineering notation.

1. 40,000,000,000 2. 180,000,000,000,000

3. 0.00006 4. 0.00000000000039

Express each measurement using SI prefixes.

5. 0.00000000014 gram 6. 40,000,000,000 watts

7. 63,100,000,000,000 bytes 8. 0.0000002 meter

Example


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7-5

Graph Exponential FunctionsExponential Function a function defi ned by an equation of the form y = a bx, where a ≠ 0, b > 0, and b ≠ 1

You can use values of x to find ordered pairs that satisfy an exponential function. Then you can use the ordered pairs to graph the function.

Graph y = 3x. Find the y-intercept and state the domain and range.

x y

-2 1 −

9

-1 1 −

3

0 1

1 3

2 9

The y-intercept is 1. The domain is all real numbers, and the range is all positive numbers.

x

y

O

Graph y = (

1 −

4 )

x

. Find the y-intercept and state the domain and range.

The y-intercept is 1. The domain is all real numbers, and the range is all positive numbers.

x

y

O 2

8

ExercisesGraph each function. Find the y-intercept and state the domain and range.

1. y = 0.3x 2. y = 3x + 1 3. y = ( 1 − 3 )

x

+ 1

x

y

O 1

2

x

y

O 2

16

4

8

12

x

y

O 1

2

Example 1 Example 2

x y

-2 16

-1 4

0 1

1 1 −

4

2 1 −

16

Study Guide and InterventionExponential Functions


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7-5 Study Guide and Intervention (continued)

Exponential Functions

Identify Exponential Behavior It is sometimes useful to know if a set of data is exponential. One way to tell is to observe the shape of the graph. Another way is to observe the pattern in the set of data.

Determine whether the set of data shown below displays exponential behavior. Write yes or no. Explain why or why not.

Method 1: Look for a PatternThe domain values increase by regular intervals of 2, while the range values have a common factor of 1 −

2 . Since the domain

values increase by regular intervals and the range values have a common factor, the data are probably exponential.

Method 2: Graph the Data The graph shows

rapidly decreasing values of y as x increases. This is characteristic of exponential behavior.

Exercises


1. x 0 1 2 3

y 5 10 15 20

2. x 0 1 2 3

y 3 9 27 81

3. x -1 1 3 5

y 32 16 8 4

4. x -1 0 1 2 3

y 3 3 3 3 3

5. x -5 0 5 10

y 1 0.5 0.25 0.125

6. x 0 1 2 3 4

y 1 − 3 1 −

9 1 −

27 1 −

81 1

− 243

Example

x 0 2 4 6 8 10

y 64 32 16 8 4 2

x

y

O 4 8

16

32

48

72


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7-5 Skills PracticeExponential Functions

Graph each function. Find the y-intercept, and state the domain and range.

1. y = 2x 2. y = ( 1 − 3 )

x

3. y = 3(2x) 4. y = 3x + 2


5. x -3 -2 -1 0

y 9 12 15 18

6. x 0 5 10 15

y 20 10 5 2.5

7. x 4 8 12 16

y 20 40 80 160

8. x 50 30 10 -10

y 90 70 50 30

x

y

Ox

y

O

x

y

Ox

y

O


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7-5 Practice Exponential Functions

Graph each function. Find the y-intercept and state the domain and range.

1. y = ( 1 − 10

) x

2. y = 3x 3. y = ( 1 − 4 )

x

4. y = 4(2x) + 1 5. y = 2(2x - 1) 6. y = 0.5(3x - 3)


7. x 2 5 8 11

y 480 120 30 7.5

8. x 21 18 15 12

y 30 23 16 9

9. LEARNING Ms. Klemperer told her English class that each week students tend to forget one sixth of the vocabulary words they learned the previous week. Suppose a student learns 60 words. The number of words remembered can be described by the function

W(x) = 60 ( 5 − 6 )

x

, where x is the number of weeks that pass. How many words will the student remember after 3 weeks?

10. BIOLOGY Suppose a certain cell reproduces itself in four hours. If a lab researcher begins with 50 cells, how many cells will there be after one day, two days, and three days? (Hint: Use the exponential function y = 50(2x).)

x

y

Ox

y

O

x

y

O

x

y

Ox

y

Ox

y

O


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7-5

1. WASTE Suppose the waste generated by nonrecycled paper and cardboard products is approximated by the following function.

y = 1000(2)0.3x

Sketch the exponential function on the coordinate grid below.

y

xO

950650350

18501550

24502150

1250

4321-2-1-3-4

2. MONEY Tatyana’s grandfather gave her one penny on the day she was born. He plans to double the amount he gives her every day. Estimate how much she will receive from her grandfather on the 12th day of her life.

3. PICTURE FRAMESSince a picture frame includes a border, the picture must be smaller in area than the entire frame. The table shows the relationship between picture area and frame length for a particular line of frames. Is this an exponential relationship? Explain.

4. DEPRECIATION The value of Royce Company’s computer equipment is decreasing in value according to the following function.

y = 4000(0.87)x

In the equation, x is the number of years that have elapsed since the equipment was purchased and y is the value in dollars. What was the value 5 years after it was purchased? Round your answer to the nearest dollar.

5. METEOROLOGY The atmospheric pressure in millibars at altitude x meters can be approximated by the following function. The function is valid for values of x between 0 and 10,000.

f (x) = 1038(1.000134)-x

a. What is the pressure at sea level?

b. The McDonald Observatory in Texas is at an altitude of 2000 meters. What is the approximate atmospheric pressure there?

c. As altitude increases, what happens to atmospheric pressure?

Word Problem PracticeExponential Functions

Side Picture

Length Area

(in.) (in2)

5 6

6 12

7 20

8 30

9 42


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EnrichmentLogarithmic Functions

7-5

You have found the inverse of a linear function by interchanging x and y in the equation of the function. The inverse of an exponential function is called a logarithmic function.

Find the inverse function of the exponential function y = 2x. Make a table of values for each function. Then graph both functions.

To find the inverse function, interchange x and y. The inverse of y = 2 x is x = 2 y .

y = 2 x x = 2 y

x y x y

-3 1 − 8 1 −

8 -3

-2 1 − 4 1 −

4 -2

-1 1 − 2 1 −

2 -1

0 1 1 0

1 2 2 1

2 4 4 2

3 8 8 3

In general the inverse of y = b x is x = b y . In x = b y , the variable y is called the logarithm of x. This is usually written as y = lo g b x.

Exercises

1. What are the domain and range of y = lo g 2 x? How are the domain and range related to the domain and range of y = 2 x ?

2. How are the graphs of y = lo g 2 x and y = 2 x related?

3. Through which quadrants do the graphs of y = lo g 2 x and y = 2 x pass?

4. If an exponential function finds population as a function of time, what can you conclude about its inverse logarithmic function?

5. RESEARCH Investigate real-world uses of logarithmic functions. How are logarithmic functions used to model real-world situations?

Example

y

xO

12

8 12

8

4

4

y = x

y = 2x

x = 2y


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7-6 Study Guide and InterventionGrowth and Decay

Exponential Growth Population increases and growth of monetary investments are examples of exponential growth. This means that an initial amount increases at a steady rate over time.

The general equation for exponential growth is y = a(1 + r )t.

• y represents the fi nal amount.

Exponential Growth • a represents the initial amount.

• r represents the rate of change expressed as a decimal.

• t represents time.

POPULATION The population of Johnson City in 2005 was 25,000. Since then, the population has grown at an average rate of 3.2% each year.

a. Write an equation to represent the population of Johnson City since 2005.

The rate 3.2% can be written as 0.032.y = a(1 + r)t

y = 25,000(1 + 0.032)t

y = 25,000(1.032)t

b. According to the equation, what will the population of Johnson City be in 2015?

In 2015 t will equal 2015 - 2005 or 10.Substitute 10 for t in the equation from part a.y = 25,000(1.032)10 t = 10

≈ 34,256In 2015 the population of Johnson City will be about 34,256.

INVESTMENT The Garcias have $12,000 in a savings account. The bank pays 3.5% interest on savings accounts, compounded monthly. Find the balance in 3 years.

The rate 3.5% can be written as 0.035. The special equation for compound interest is A = P (1 + r−n)

nt, where A

represents the balance, P is the initial amount, r represents the annual rate expressed as a decimal, n represents the number of times the interest is compounded each year, and t represents the number of years the money is invested.

A = P (1 + r−n)nt

= 12,000 (1 + 0.035−12 )

3(12)

≈ 13,326.49In three years, the balance of the account will be $13,326.49.

Exercises 1. POPULATION The population of the United 2. INVESTMENT Determine the

States has been increasing at an average value of an investment of $2500 annual rate of 0.91%. If the population was if it is invested at an interest rate of about 303,146,000 in 2008, predict the 5.25% compounded monthly for population in 2012. 4 years.

3. POPULATION It is estimated that the 4. INVESTMENT Determine the population of the world is increasing at value of an investment of $100,000 an average annual rate of 1.3%. If the if it is invested at an interest rate of 2008 population was about 6,641,000,000, 5.2% compounded quarterly for predict the 2015 population. 12 years.

Example 1 Example 2


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7-6 Study Guide and Intervention (continued)

Growth and Decay

Exponential Decay Radioactive decay and depreciation are examples of exponential decay. This means that an initial amount decreases at a steady rate over a period of time.

The general equation for exponential decay is y = a(1 - r)t .

• y represents the fi nal amount.

Exponential Decay • a represents the initial amount.

• r represents the rate of decay expressed as a decimal.

• t represents time.

DEPRECIATION The original price of a tractor was $45,000. The value of the tractor decreases at a steady rate of 12% per year.

a. Write an equation to represent the value of the tractor since it was purchased.

The rate 12% can be written as 0.12.y = a(1 - r)t General equation for exponential decay

y = 45,000(1 - 0.12)t a = 45,000 and r = 0.12

y = 45,000(0.88)t Simplify.

b. What is the value of the tractor in 5 years?

y = 45,000(0.88)t Equation for decay from part a

y = 45,000(0.88)5 t = 5

y ≈ 23,747.94 Use a calculator.

In 5 years, the tractor will be worth about $23,747.94.

Exercises 1. POPULATION The population of Bulgaria has been decreasing at an annual

rate of 0.89%. If the population of Bulgaria was about 7,450,349 in the year 2005, predict its population in the year 2015.

2. DEPRECIATION Mr. Gossell is a machinist. He bought some new machinery for about $125,000. He wants to calculate the value of the machinery over the next 10 years for tax purposes. If the machinery depreciates at the rate of 15% per year, what is the value of the machinery (to the nearest $100) at the end of 10 years?

3. ARCHAEOLOGY The half-life of a radioactive element is defined as the time that it takes for one-half a quantity of the element to decay. Radioactive carbon-14 is found in all living organisms and has a half-life of 5730 years. Consider a living organism with an original concentration of carbon-14 of 100 grams.

a. If the organism lived 5730 years ago, what is the concentration of carbon-14 today?

b. If the organism lived 11,460 years ago, determine the concentration of carbon-14 today.

4. DEPRECIATION A new car costs $32,000. It is expected to depreciate 12% each year for 4 years and then depreciate 8% each year thereafter. Find the value of the car in 6 years.

Example


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7-6 Skills PracticeGrowth and Decay

1. POPULATION The population of New York City increased from 8,008,278 in 2000 to 8,168,388 in 2005. The annual rate of population increase for the period was about 0.4%.

a. Write an equation for the population t years after 2000.

b. Use the equation to predict the population of New York City in 2015.

2. SAVINGS The Fresh and Green Company has a savings plan for its employees. If an employee makes an initial contribution of $1000, the company pays 8% interest compounded quarterly.

a. If an employee participating in the plan withdraws the balance of the account after 5 years, how much will be in the account?

b. If an employee participating in the plan withdraws the balance of the account after 35 years, how much will be in the account?

3. HOUSING Mr. and Mrs. Boyce bought a house for $96,000 in 1995. The real estate broker indicated that houses in their area were appreciating at an average annual rate of 7%. If the appreciation remained steady at this rate, what was the value of the Boyce’s home in 2009?

4. MANUFACTURING Zeller Industries bought a piece of weaving equipment for $60,000. It is expected to depreciate at an average rate of 10% per year.

a. Write an equation for the value of the piece of equipment after t years.

b. Find the value of the piece of equipment after 6 years.

5. FINANCES Kyle saved $500 from a summer job. He plans to spend 10% of his savings each week on various forms of entertainment. At this rate, how much will Kyle have left after 15 weeks?

6. TRANSPORTATION Tiffany’s mother bought a car for $9000 five years ago. She wants to sell it to Tiffany based on a 15% annual rate of depreciation. At this rate, how much will Tiffany pay for the car?


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7-6 Practice Growth and Decay

1. COMMUNICATIONS Sports radio stations numbered 220 in 1996. The number of sports radio stations has since increased by approximately 14.3% per year.

a. Write an equation for the number of sports radio stations for t years after 1996.

b. If the trend continues, predict the number of sports radio stations in 2015.

2. INVESTMENTS Determine the amount of an investment if $500 is invested at an interest rate of 4.25% compounded quarterly for 12 years.

3. INVESTMENTS Determine the amount of an investment if $300 is invested at an interest rate of 6.75% compounded semiannually for 20 years.

4. HOUSING The Greens bought a condominium for $110,000 in 2010. If its value appreciates at an average rate of 6% per year, what will the value be in 2015?

5. DEFORESTATION During the 1990s, the forested area of Guatemala decreased at an average rate of 1.7%.

a. If the forested area in Guatemala in 1990 was about 34,400 square kilometers, write an equation for the forested area for t years after 1990.

b. If this trend continues, predict the forested area in 2015.

6. BUSINESS A piece of machinery valued at $25,000 depreciates at a steady rate of 10% yearly. What will the value of the piece of machinery be after 7 years?

7. TRANSPORTATION A new car costs $18,000. It is expected to depreciate at an average rate of 12% per year. Find the value of the car in 8 years.

8. POPULATION The population of Osaka, Japan, declined at an average annual rate of 0.05% for the five years between 1995 and 2000. If the population of Osaka was 11,013,000 in 2000 and it continues to decline at the same rate, predict the population in 2050.


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7-6

1. DEPRECIATION The value of a new plasma television depreciates by about 7% each year. Aeryn purchases a 50-inch plasma television for $3000. What is its value after 4 years? Round your answer to the nearest hundred.

2. MONEY Hans opens a savings account by depositing $1200 in an account that earns 3 percent interest compounded weekly. How much will his investment be worth in 10 years? Assume that there are exactly 52 weeks in a year and round your answer to the nearest cent.

3. HIGHER EDUCATION The table lists the average costs of attending a four-year college in the United States during a recent year.

Source: College Board

Russ’s parents invested money in a savings account earning an average of 4.5 percent interest, compounded monthly. After 15 years, they have exactly the right amount to cover the tuition, fees, room and board for Russ’s first year at a public college. What was their initial investment? Round your answer to the nearest dollar.

4. POPULATION In 2007 the U.S. Census Bureau estimated the population of the United States estimated at 301 million. The annual rate of growth is about 0.89%. At this rate, what is the expected population at the time of the 2020 census? Round your answer to the nearest ten million.

5. MEDICINE When doctors prescribe medication, they have to consider the rate at which the body filters a drug from the bloodstream. Suppose it takes the human body 6 days to filter out half of the Flu-B-Gone vaccine. The amount of Flu-B-Gone vaccine remaining in the bloodstream x days after an injection is

given by the equation y = y0 (0.5) x − 6 , where

y0 is the initial amount. Suppose a doctor injects a patient with 20 μg (micrograms) of Flu-B-Gone.

a. How much of the vaccine will remain after 1 day? Round your answer to the nearest tenth.

b. How much of the vaccine will remain after 12 days? Round your answer to the nearest tenth.

c. After how many days will the amount of vaccine be less than 1 μg?

Word Problem PracticeGrowth and Decay

College

Sector

Tuition and

Fees

Room and

Board

Four-year

Public$5941 $6636

Four-year

Private $21,235 $7,791


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7-6

You can use the formula for compound interest A = P (1 + r − n ) nt when n, the number of times

a year interest is compounded, is known. A special type of compound interest calculation regularly used in finance is continuously compounded interest, where n approaches infinity.

INVESTING Mr. Rivera placed $5000 in an investment account that has an interest rate of 9.5% per year.

Exercises

1. SAVINGS Mr. Harris saves $20,000 in a money-market account at a rate of 5.2%.

a. Determine the value of his investment after 10 years if interest iscompounded quarterly.

b. Determine the value of his investment after 10 years if interest iscompounded continuously.

2. COLLEGE SAVINGS Shannon is choosing between two different savings accounts for her college fund. The first account compounds interest semiannually at a rate of 11.0%. The second account compounds interest continuously at a rate of 10.8%. If Shannon plans to keep her money in the account for 5 years, which account should she choose? Explain.

EnrichmentContinuously Compounding Interest

General formula: A = Pert

A = current amount of investment

P = initial amount of investment

e = natural logarithm, a constant approximately equal to 2.71828

r = annual rate of interest, expressed as a decimal

t = number of years the money is invested

Example

a. How much money will be in the account after 5 years if interest is compounded monthly?

A = P (1 + r − n ) nt

Compound interest

equation

= 5000 (1 + 0.095 − 12

) 12(5)

P = 5000, r = 0.095,

n = 12, and t = 5

= 5000(1.0079)60 Simplify.

= 8025.05 Use a calculator.

There will be about $8025.05 in the account if interest is compounded monthly.

b. How much money will be in the account after 5 years if interest is compounded continuously?

A = Pert Continuous interest

equation

= 5000(2.71828)0.095(5) P = 5000, e = 2.71828,

r = 0.095, and t = 5

= 5000(2.71828)0.475 Simplify.

= 8040.07 Use a calculator.

There will be about $8040.07 in the account if interest is compounded continuously.


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7-6 Spreadsheet ActivityCompound Interest

Exercises

Use the spreadsheet of accounts involving compound interest.

1. How are the doubling times affected if the accounts compound interest quarterly instead of monthly?

2. How long will it take each account to reach $4000 if the interest is compounded monthly? quarterly?

3. How do the interest rate and the number of times the interest is compounded affect the growth of an investment?

Banks often use spreadsheets to calculate and store financial data. One application is to calculate compound interest on an account.

Use a spreadsheet to find the time it will take an investment of $1000 to double. Suppose you can choose from investments that have annual interest rates of 5%, 8%, or 10% compounded monthly.The compound interest equation is A = P (1 + r − n )

nt , where P is the principal or initial

investment, A is the final amount of the investment, r is the annual interest rate, n is the number of times interest is paid, or compounded, each year, and t is the number of years. In this case, P = 1000 and n = 12.

Step 1 Use Column A of the spreadsheet for the years.

Step 2 Columns B, C, and D contain the formulas for the final amounts of the investments. Format the cells in these columns as currency so that the amounts are shown in dollars and cents. Each formula will use the values in Column A as the value of t. For example, the formula that is in cell C3 is =1000*(1 + (0.08/12))^(12 * A3).

Study the spreadsheet for the times when each investment exceeds $2000. At 5%, the $1000 will double in 14 years, at 8% it will double in 9 years, and at 10% it will double in 7 years.

A1

3456789

1011

2

B C D

1213141516

Years$1,051.16

$1,104.94

$1,161.47

$1,220.90

$1,283.36

$1,349.02

$1,418.04

$1,490.59

$1,566.85

$1,647.01

$1,731.27

$1,819.85

$1,912.96

$2,010.83

5% 8% 10%$1,083.00

$1,172.89

$1,270.24

$1,375.67

$1,489.85

$1,613.50

$1,747.42

$1,892.46

$2,049.53

$2,219.64

$2,403.87

$2,603.39

$2,819.47

$3,053.48

$1,104.71

$1,220.39

$1,348.18

$1,489.35

$1,645.30

$1,817.59

$2,007.92

$2,218.18

$2,450.45

$2,707.04

$2,990.50

$3,303.65

$3,649.58

$4,031.74

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7-7

Recognize Geometric Sequences A geometric sequence is a sequence in which each term after the first is found by multiplying the previous term by a nonzero constant r called the common ratio. The common ratio can be found by dividing any term by its previous term.

Exercises

Determine whether each sequence is arithmetic, geometric, or neither. Explain.

1. 1, 2, 4, 8, . . . 2. 9, 14, 6, 11, . . .

3. 2 − 3 , 1 −

3 , 1 −

6 , 1 −

12 , . . . 4. –2, 5, 12, 19, . . .

Find the next three terms in each geometric sequence.

5. 648, –216, 72, . . . 6. 25, –5, 1, . . .

7. 1 − 16

, 1 − 2 , 4, . . . 8. 72, 36, 18, . . .

Study Guide and InterventionGeometric Sequences as Exponential Functions

Example 1 Determine whether the sequence is arithmetic, geometric, or neither: 21, 63, 189, 567, . . .

Find the ratios of the consecutive terms. If the ratios are constant, the sequence is geometric.21 63 189 567

63 − 21

= 189 − 63

= 567 − 189

= 3

Because the ratios are constant, the sequence is geometric. The common ratio is 3.

Find the next three terms in this geometric sequence:-1215, 405, -135, 45, . . .

Step 1 Find the common ratio.–1215 405 –135 45

405 − -1215

= -135 − 405

= 45 − -135

= - 1 − 3

The value of r is - 1 − 3 .

Step 2 Multiply each term by the common ratio to find the next three terms.

45 –15 5 - 5 − 3

× (-

1 − 3 ) × (-

1 −

3 ) × (-

1 − 3 )

The next three terms of the sequence are –15, 5, and - 5 −

3 .

Example 2


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7-7

Geometric Sequences and Functions The nth term an of a geometric sequence with first term a1 and common ratio r is given by the following formula, where n is any positive integer: an = a1 · r

n – 1.

Exercises

1. Write an equation for the nth term of the geometric sequence –2, 10, –50, . . . . Find the eleventh term of this sequence.

2. Write an equation for the nth term of the geometric sequence 512, 128, 32, . . . . Find the sixth term of this sequence.

3. Write an equation for the nth term of the geometric sequence 4 − 9 , 4, 36, . . . .

Find the eighth term of this sequence.

4. Write an equation for the nth term of the geometric sequence 6, –54, 486, . . . . Find the ninth term of this sequence.

5. Write an equation for the nth term of the geometric sequence 100, 80, 64, . . . . Find the seventh term of this sequence.

6. Write an equation for the nth term of the geometric sequence 2 − 5 , 1 −

10 , 1 −

40 , . . . .

Find the sixth term of this sequence.

7. Write an equation for the nth term of the geometric sequence 3 − 8 , - 3 −

2 , 6, . . . .

Find the tenth term of this sequence.

8. Write an equation for the nth term of the geometric sequence –3, –21, –147, . . . . Find the fifth term of this sequence.


Geometric Sequences as Exponential Functions

Example a. Write an equation for the nth term of the geometric sequence 5, 20, 80, 320, . . .

The first term of the sequence is 320. So, a1 = 320. Now find the common ratio.5 20 80 320

20 − 5 = 80 −

20 = 320 −

80 = 4

The common ratio is 4. So, r = 4.an = a1 · r

n – 1 Formula for nth term

an = 5 · 4n – 1 a1 = 5 and r = 4

b. Find the seventh term of this sequence.

Because we are looking for the seventh term, n = 7.an = a1 · r

n – 1 Formula for nth term

a7 = 5 · 47 – 1 n = 7

= 5 · 46 Simplify.

= 5 · 4096 46 = 4096

= 20,480 Multiply.

The seventh term of the sequence is 20,480.


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7-7


1. 7, 13, 19, 25, … 2. –96, –48, –24, –12, …

3. 108, 66, 141, 99, … 4. 3, 9, 81, 6561, …

5. 7 − 3 , 14, 84, 504, … 6. 3 −

8 , - 1−

8, - 5 −

8 , - 9 −

8 , …


7. 2500, 500, 100, … 8. 2, 6, 18, …

9. –4, 24, –144, … 10. 4 − 5 , 2 −

5 , 1 −

5 , …

11. –3, –12, –48, … 12. 72, 12, 2, …

13. Write an equation for the nth term of the geometric sequence 3, – 24, 192, …. Find the ninth term of this sequence.

14. Write an equation for the nth term of the geometric sequence 9 − 16

, 3 − 8 , 1 −

4 , ….

Find the seventh term of this sequence.

15. Write an equation for the nth term of the geometric sequence 1000, 200, 40, …. Find the fifth term of this sequence.

16. Write an equation for the nth term of the geometric sequence – 8, – 2, - 1 − 2 , ….

Find the eighth term of this sequence.

17. Write an equation for the nth term of the geometric sequence 32, 48, 72, …. Find the sixth term of this sequence.

18. Write an equation for the nth term of the geometric sequence 3 − 100

, 3 − 10

, 3, …. Find the ninth term of this sequence.

Skills PracticeGeometric Sequences as Exponential Functions


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7-7


1. 1, -5, -11, -17, … 2. 3, 3 − 2 , 1, 3 −

4 , …

3. 108, 36, 12, 4, … 4. -2, 4, -6, 8, …


5. 64, 16, 4, … 6. 2, -12, 72, …

7. 3750, 750, 150, … 8. 4, 28, 196, …

9. Write an equation for the nth term of the geometric sequence 896, -448, 224, … . Find the eighth term of this sequence.

10. Write an equation for the nth term of the geometric sequence 3584, 896, 224, … . Find the sixth term of this sequence.

11. Find the sixth term of a geometric sequence for which a2 = 288 and r = 1 − 4

.

12. Find the eighth term of a geometric sequence for which a3 = 35 and r = 7.

13. PENNIES Thomas is saving pennies in a jar. The first day he saves 3 pennies, the second day 12 pennies, the third day 48 pennies, and so on. How many pennies does Thomas save on the eighth day?

PracticeGeometric Sequences as Exponential Functions


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7-7

1. WORLD POPULATION The CIA estimates the world population is growing at a rate of 1.167% each year. The world population for 2007 was about 6.6 billion.

a. Write an equation for the world population after n years. (Hint: The common ratio is not 0.01167.)

b. What will the estimated world population be in 2017?

2. MUSEUMS The table shows the annual visitors to a museum in millions. Write an equation for the projected number of visitors after n years.

3. BANKING Arnold has a bank account with a beginning balance of $5000. He spends one-fifth of the balance each month. How much money will be in the account after 6 months?

4. POPULATION The table shows the projected population of the United States through 2050. Does this table show an arithmetic sequence, a geometric sequenceor neither? Explain.

Source: U.S. Census Bureau

5. SAVINGS ACCOUNTS A bank offers a savings account with a 0.5% return each month.

a. Write an equation for the balance of a savings account after n months. (Hint: The common ratio is not 0.005.)

b. Given an initial deposit of $500, what will the account balance be after 15 months?

Word Problem PracticeGeometric Sequences as Exponential Functions

YearVisitors

(millions)

1 4

2 6

3 9

4 13 1 − 2

n ?

YearProjected

Population

2000 282,125,000

2010 308,936,000

2020 335,805,000

2030 363,584,000

2040 391,946,000

2050 419,854,000


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7-7

Pay it ForwardThe idea behind “pay it forward” is that on the first day, one person does a good deed for three different people. Then, on the second day, those three people will each perform good deeds for 3 more people, so that on Day 2, there are 3 × 3 or 9 good deeds being done. Continue this process to fill in the chart. A tree diagram will help you fill in the chart.

1. Graph the data you found in the chart as ordered pairs and connect with a smooth curve.

2. What type of function is your graph from Exercise 1? Write an equation that can be used to determine the number of good deeds on any given day, x.

3. How many good deeds will be performed on Day 21?

The formula T = 3 n + 1 - 3 − 2 can be used to determine the total number of good deeds that

have been performed, where n represents the day. For example, on Day 2, there have been 3 + 9 or 12 good deeds performed. Using the formula, you get, 3 2 + 1 - 3 −

2 or 3 3 - 3 −

2 + 27 - 3 −

2 or

a total of 12 good deeds performed.

4. Use the formula to determine the approximate number of good deeds that have been performed through Day 21.

5. Look up the world population. How does your number from Exercise 4 compare to the world population?

Person 1

Day 2Day 1

New Person 2

New Person 1

New Person 3

y

xO

240210180

120150

906030

432 7 8651

Day# of

Deeds

0 1

1 3

2 9

3

4

5

Enrichment


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Study Guide and Intervention Recursive Formulas

7-8

Using Recursive Formulas A recursive formula allows you to find the nth term of a sequence by performing operations on one or more of the terms that precede it.

Find the first five terms of the sequence in which a 1 = 5 and a n = -2 a n - 1 + 14, if n ≥ 2.

The given first term is a 1 = 5. Use this term and the recursive formula to find the next four terms. a 2 = -2 a 2 - 1 + 14 n = 2 a 4 = -2 a 4 - 1 + 14 n = 4

= -2 a 1 + 14 Simplify. = -2 a 3 + 14 Simplify.

= -2(5) + 14 or 4 a 1 = 5 = -2(6) + 14 or 2 a 3 = 6

a 3 = -2 a 3 - 1 + 14 n = 3 a 5 = -2 a 5 - 1 + 14 n = 5

= -2 a 2 + 14 Simplify. = -2 a 4 + 14 Simplify.

= -2(4) + 14 or 6 a 2 = 4 = -2(2) + 14 or 10 a 4 = 2

The first five terms are 5, 4, 6, 2, and 10.

ExercisesFind the first five terms of each sequence.

1. a 1 = -4, a n = 3 a n - 1 , n ≥ 2 2. a 1 = 5, a n = 2a n - 1 , n ≥ 2

3. a 1 = 8, a n = a n - 1 - 6, n ≥ 2 4. a 1 = -32, a n = a n - 1 + 13, n ≥ 2

5. a 1 = 6, a n = -3 a n - 1 + 20, n ≥ 2 6. a 1 = -9, a n = 2 a n - 1 + 11, n ≥ 2

7. a 1 = 12, a n = 2 a n - 1 - 10, n ≥ 2 8. a 1 = -1, a n = 4 a n - 1 + 3, n ≥ 2

9. a 1 = 64, a n = 0.5 a n - 1 + 8, n ≥ 2 10. a 1 = 8, a n = 1.5 a n - 1 , n ≥ 2

11. a 1 = 400, a n = 1 − 2 a n - 1 , n ≥ 2 12. a 1 = 1 −

4 , a n = a n - 1 + 3 −

4 , n ≥ 2

Example


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Recursive Formulas

Writing Recursive Formulas Complete the following steps to write a recursive formula for an arithmetic or geometric sequence.

Step 1 Determine if the sequence is arithmetic or geometric by fi nding a common

difference or a common ratio.

Step 2 Write a recursive formula.

Arithmetic Sequences a n = a n - 1 + d, where d is the common difference

Geometric Sequences an = r · an - 1, where r is the common ratio

Step 3 State the fi rst term and the domain for n.

Write a recursive formula for 216, 36, 6, 1, … .

Step 1 First subtract each term from the term that follows it.216 - 36 = 180 36 − 6 = 30 6 - 1 = 5There is no common difference. Check for a common ratio by dividing each term by the term that precedes it.

36 − 216

= 1 − 6 6 −

36 = 1 −

6 1 −

6 = 1 −

6

There is a common ratio of 1 − 6 . The sequence is geometric.

Step 2 Use the formula for a geometric sequence. a n = r · a n - 1 Recursive formula for geometric sequence

a n = 1 − 6 a n - 1 r = 1 −

6

Step 3 The first term a 1 is 216 and n ≥ 2.

A recursive formula for the sequence is a 1 = 216, a n = 1 − 6 a n - 1 , n ≥ 2.

ExercisesWrite a recursive formula for each sequence.

1. 22, 16, 10, 4, … 2. -8, -3, 2, 7, …

3. 5, 15, 45, 135, … 4. 243, 81, 27, 9, …

5. −3, 14, 31, 48, … 6. 8, -20, 50, -125, …

7-8

Example


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Skills PracticeRecursive Formulas

7-8

Find the first five terms of each sequence.

1. a 1 = -11, a n = a n − 1 − 7, n ≥ 2 2. a 1 = 14, a n = a n − 1 + 6.5, n ≥ 2

3. a 1 = -3, a n = 3a n − 1 + 10, n ≥ 2 4. a 1 = 187.5, a n = −0.8 a n - 1 , n ≥ 2

5. a 1 = 1 − 2 , a n = 3 −

2 a n - 1 + 1 −

4 , n ≥ 2 6. a 1 = 1 −

5 , a n = 5 −

2 a n - 1 - 1 −

2 , n ≥ 2

Write a recursive formula for each sequence.

7. 7, 28, 112, 448, … 8. 73, 52, 31, 10, …

9. 4, 1, 1 − 4 , 1 −

16 , … 10. -37, -15, 7, 29, …

11. 0.25, 0.37, 0.49, 0.61, … 12. 64, -80, 100, -125, 156.25, …

For each recursive formula, write an explicit formula. For each explicit formula, write a recursive formula.

13. a 1 = 38, a n = a n - 1 - 17, n ≥ 2 14. a n = 5n - 16

15. a n = 50(0.75 ) n - 1 16. a 1 = 16, a n = 4 a n − 1 , n ≥ 2


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PracticeRecursive Formulas

Find the first five terms of each sequence.

1. a 1 = 25, a n = a n - 1 - 12, n ≥ 2 2. a 1 = -101, a n = a n - 1 + 38, n ≥ 2

3. a 1 = 3.3, a n = a n - 1 + 2.7, n ≥ 2 4. a 1 = 7, a n = -3 a n - 1 + 20, n ≥ 2

5. a 1 = 20, a n = 1 − 5 a n - 1 , n ≥ 2 6. a 1 = 2 −

3 , a n = 1 −

3 a n - 1 - 2 −

9 , n ≥ 2


7. 80, -40, 20, -10, … 8. 87, 52, 17, -18, …

9. 1 − 3 , 4 −

15 , 16 −

75 , 64 −

375 , … 10. 4 −

5 , 3 −

10 , - 1 −

5 , - 7 −

10 , …

11. 2.6, 5.2, 7.8, 10.4, … 12. 100, 120, 144, 172.8, …

13. PIZZA The total costs for ordering one to five cheese pizzas from Luigi’s Pizza Palace are shown.

a. Write a recursive formula for the sequence.

b. Write an explicit formula for the sequence.

7-8

Total Number of

Pizzas OrderedCost

1 $7.00

2 $12.50

3 $18.00

4 $23.50

5 $29.00


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Word Problem PracticeRecursive Formulas

1. ASSEMBLY An assembly line can create 175 widgets in one hour, 350 widgets in two hours, 525 widgets in three hours, 700 widgets in four hours, and so on.

a. Find the next 5 terms in the sequence.

b. Write a recursive formula for the sequence.

c. Write an explicit formula for the sequence.

2. PAPER A piece of paper is folded several times. The number of sections into which the piece of paper is divided after each fold is shown below.

Number of Folds Sections

1 2

2 4

3 8

4 16

5 32

a. Find the next 5 terms in the sequence.

b. Write a recursive formula for the sequence.

c. Write an explicit formula for the sequence.

3. CLEANING An equation for the cost a n in dollars that a carpet cleaning company charges for cleaning n rooms is a n = 50 + 25(n - 1). Write a recursive formula to represent the cost a n.

4. SAVINGS A recursive formula for the balance of a savings account a n in dollars at the beginning of year n is a 1 = 500, an = 1.05 a n - 1 , n ≥ 2. Write an explicit formula to represent the balance of the savings account a n .

5. SNOW A snowman begins to melt as the temperature rises. The height of the snowman in feet after each hour is shown.

Hour Height (ft)

1 6.0

2 5.4

3 4.86

4 4.374

a. Write a recursive formula for the sequence.

b. Write an explicit formula for the sequence.

7-8


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EnrichmentArithmetic Series

7-8

Consider a situation in which an uncle gives his niece a dollar amount equal to her age each year on her birthday. An arithmetic sequence that represents this situation is 1, 2, 3, … . How much total money will the niece have received by her 30th birthday?

When the terms of an arithmetic sequence are added, an arithmetic series is formed. The solution of the problem above is the sum of the arithmetic series 1 + 2 + 3 + … + 30.

If we write the series twice, as shown below, and add the terms in each column, we can see a pattern. 1 + 2 + 3 + 4 + ... + 27 + 28 + 29 + 30 30 + 29 + 28 + 27 + ... + 4 + 3 + 2 + 1

31 + 31 + 31 + 31 + ... + 31 + 31 + 31 + 31

The sum of each pair of terms is 31. Multiplying 31 by the number of terms in the bottom row, 30, gives us the sum of the bottom row, or 930. Since the series was written twice, and resultantly, each term in the series was accounted for twice, we can divide 930 by 2 and find that the sum of the series 1 + 2 + 3 + … + 30 is 465.

1. Write an expression to represent the sum of the series above using the first term 1, the last term 30, and the number of terms in the series 30.

2. Write an expression to represent the sum of an arithmetic series with n terms, first term a 1 , and last term a n .

Use the expression that you found in Exercise 2 to find the sum of each arithmetic series.

3. 1 + 2 + 3 + … + 100 4. 5 + 10 + 15 + … + 100

5. 1 + 3 + 5 + … + 29 6. 2 + 4 + 6 + … + 60

7. 6 + 13 + 20 + … + 125 8. 8 + 12 + 16 + … + 84


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7 SCORE Student Recording SheetUse this recording sheet with pages 458–459 of the Student Edition.

7a.

7b.

8.

9.

10. (grid-in)

10.

Record your answers for Question 11 on the back of this paper.

Extended Response

Read each question. Then fill in the correct answer.

Multiple Choice

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.

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

Short Response/Gridded Response

9

8

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8

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5

4

3

2

1

0

. . . . .

9

8

7

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9

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055_066_ALG1_A_CRM_C07_AS_660281.indd 55055_066_ALG1_A_CRM_C07_AS_660281.indd 55 12/20/10 6:40 PM12/20/10 6:40 PM

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7 Rubric for Scoring Extended Response

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 11 Rubric

Score Specifi c Criteria

4 Part a identifies Mercury as the planet closest to the Sun, because it has the smallest exponent. In part b, students set up and solve the expression 2.28 × 108

− 1.50 × 108

to find that Mars is about 1.52 times further from the Sun than Earth. In part c, students set up and solve the expression 1.43 × 109

− 3 × 105

to find that it takes

approximately 4767 seconds for light from the Sun to reach Saturn.

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.


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Chapter 7 Quiz 2 (Lessons 7-3 and 7-4)

7

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7 SCORE

Simplify.

1. (r3)(2r5) 2. (x5)4

3. (-4t2n3)(3tn4) 4. (-5x4y2)3

5. (2cd)2(-4c3)2 6. (5y2w4)2 + 2( yw2)4

Simplify. Assume that no denominator is equal to zero.

7. 6 15 − 6 9

8. y 8

− y 3

9. r 6 n -7 − r 4 n 2

10. MULTIPLE CHOICE Write the ratio of the area of a circle with radius r to the circumference of the same circle.

A 2 − r B 2 C r − 2 D 1 −

2r

Simplify.

1. 4 √ �� 81 2. 4

3 − 2


3. 5 x = 125 4. 2 5x - 4 = 64

5. Write 2 √ �� 7xy in exponential form.


6. 14,000,000 7. 0.0000308

8. MULTIPLE CHOICE Evaluate (4 × 107)(3.6 × 10-4).

A 1.11 × 103 C 1.44 × 102

B 1.44 × 104 D 1.44 × 10-27

Evalute each quotient. Express the results in both scientific notation and standard form.

9. 7.2 × 10 4 − 8 × 1 0 -3

10. -3.6 × 1 0 -5 − 6 × 1 0 3


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

SCORE


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7 SCORE Chapter 7 Quiz 3(Lessons 7-5 and 7-6)

SCORE

Graph each function. Find the y-intercept and state the domain and range.

1. y = ( 1 − 3 )

x

2. y = 4(2)x - 1

Robin invests $1500 at 4.85% compounded quarterly.

3. Write an equation to represent the amount of money he will have in n years.

4. How much money will Robin have in 12 years?

5. MULTIPLE CHOICE The population of Camden, New Jersey, has been decreasing by 0.12% a year on average. If this trend continues, and the population was79,318 in 2006, estimate Camden’s population in 2015.

A 71,152 C 78,465 B 78,461 D 80,179

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5.

1. Determine whether the sequence 32, 16, 8, 4, . . . is arithmetic, geometric, or neither. Explain.

2. Find the next three terms of the geometric sequence -3, -12, -48, ... .

3. MULTIPLE CHOICE What is the eighth term of the geometric series -2, 10, -50, ...?

A -781,250 C 156,250 B -156,250 D 781,250

Write a recursive formula for each sequence. 4. 3, 20, 37, 54, …

5. 125, -25, 5, - 1 − 5

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5.


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7 SCORE

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

1. Simplify (n5)(n2)(r3)(r4).

A n10r12 B n7r7 C nr7 D nr14

2. Simplify (3w2r)2( - 2w5r2)3.

F -72w19r8 G 72w12r7 H -36w32r10 J 36w19r6

Simplify. Assume the denominator is not equal to zero.

3. m 6 n 3 − m 2 n 6

A m 4 − n 3

B - m 4 − n 3

C - m 8 − n 3

D m 8 − n 3

4. ( z 2 w -1 ) 3 −

( z 3 w 2 ) 2

F 1 − w 7

G z 12 − w 7

H w J 1 − w

5. Solve 2 3x + 10 = 128.

A 3 B 2 C 1 D -1 6. Express 0.000702 in scientific notation.

F 7.02 × 10-3 H 7.02 × 10-4 G 7.02-4 J 7.02 × 10-5

7. Evaluate 6.08 × 10 7 − 3.8 × 10 -2

.

A 1.6 × 105 B 1.6 × 109 C 2.3104 × 106 D 2.3104 × 1010

Part II Evalute each product. Express the results in both scientific notation and standard form.

8. (4.2 × 109)(1.6 × 10-5)

9. (7.8 × 10-3)(4 × 10-4)


10. 2 x 3 y 5 −

5( x 4 y 2 ) 3 11.

-10 m -1 y 0 r −

-14 m -7 y -3 r -4


12. 8 x = 16

13. 21 6 x + 1 = 6

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Chapter 7 Mid-Chapter Test (Lessons 7-1 through 7-4)


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7 SCORE Chapter 7 Vocabulary Test

State whether each sentence is true or false. If false, replace the underlinedword or phrase to make a true sentence.

1. The zero exponent property tells us that any nonzero number raised to the zero power is 1.

2. A very large or very small number written in the form a × 10n, where 1 ≤ a < 10 and n is an integer, is said to be in scientific notation.

3. The scientific notation of a quantity is the number rounded to the nearest power of 10.

4. The negative exponent of a geometric sequence can be found by dividing any term by its previous term.

5. Constants are monomials that are real numbers.

6. The equation A = P (1 + r − n ) nt

is the general equation for compound interest.

7. A recursive formula allows you to find the nth term of a sequence by performing operations to one or more of the terms that precede it.

8. Exponential decay is when an initial amount decreases by the same percent over a given period of time.

9. A number, a variable, or a product of a number and one or more variables is a geometric sequence.

Define each term in your own words.

10. exponential function

11. geometric sequence

common ratio

compound interest

constant

exponential decay

exponential function

exponential growth

geometric sequence

monomial

negative exponent

nth root

order of magnitude

rational exponent

recursive formula

scientifi c notation

zero exponent

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7 SCORE Chapter 7 Test, Form 1

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

1. Simplify y5 � y3.

A y2 B y8 C y15 D 2y8

2. Simplify (b4)3.

F b7 G 3b4 H b12 J 3b7

3. Simplify a7 −

a4 . Assume the denominator is not equal to zero.

A a11 B a28 C a3 D 1

4. A rectangle has a length of 25x3 and a width of 5x2. Find the area in square units.

F 25x6 G 25x5 H 125x6 J 125x5

5. Simplify m5r2 −

m2r3 . Assume the denominator is not equal to zero.

A m7r5 B m3 − r C m3r D r −

m3

6. Express 0.000024 in scientific notation.

F 2.4 × 105 G 0.24 × 10-4 H 2.4 × 10-4 J 2.4 × 10-5

7. Evaluate (7 × 108)(2.4 × 10-4).

A 1.68 × 104 B 2.92 × 104 C 1.68 × 105 D 1.68 × 1013

8. Evaluate 16 3 − 4 .

F 2 G 4 H 8 J 32

9. Solve 3 x + 2 = 81.

A 0 B 1 C 2 D 3

10. Which equation corresponds to the graph shown? F y = 2x + 2 H y = 2x - 2

G y = ( 1 − 2 )

x

- 2 J y = ( 1 − 2 )

x

+ 2

11. What is the y-intercept on the graph shown?

A 1 − 2 B 1 C 0 D -1

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7 Chapter 7 Test, Form 1 (continued)

12. TOURNAMENTS A chess tournament starts with 16 people competing.

The exponential function y = 16 ( 1 − 2 )

x

describes how many people are

remaining in the tournament after x rounds. How many people are left in the tournament after 2 rounds?

F 4 G 2 H 8 J 1

13. INVESTMENTS Determine the amount of an investment if $1000 is invested at an interest rate of 8% compounded quarterly for 2 years.

A $1160.00 B $1171.66 C $1040.40 D $1166.40

14. BIOLOGY If y = 10(2.5) t represents the number of bacteria in a culture at time t, how many will there be at time t = 6?

F 2441 G 244 H 24 J none

15. DEPRECIATION A $60,000 piece of machinery depreciates in value at a rate of 11% per year. About what will its value be in 5 years?

A $47,526 B $42,298 C $33,504 D $37,645

16. Which is the equation for the nth term of the geometric sequence -2, 8, -32, . . . ?

F a n = -2 � 4 n H a n = - 2 · 4 n-1 G a n = 4 � (-2 ) n J a n = - 2 · (-4)n-1

17. What is the ninth term of the geometric sequence 3, 9, 27, . . .? A 2187 B 6561 C 19,683 D 59,049

18. Find the third term of the sequence in which a 1 = 12 and a n = 5 a n - 1 - 14, if n ≥ 2.

F 1 G 46 H 216 J 1066

19. Find an explicit formula for a 1 = 17, a n = a n - 1 + 4, n ≥ 2. A a n = 4n + 13 C a n = 4n + 17 B a n = n + 4 D a n = 17n + 4

20. Find a recursive formula for the arithmetic sequence 18, 12, 6, 2, ... .

F a 1 = 18, a n = -6 a n - 1 , n ≥ 2 H a 1 = 18, a n = 2 − 3 a n - 1 , n ≥ 2

G a 1 = 18, a n = a n - 1 - 6, n ≥ 2 J a 1 = 18, a n = 1 − 2 a n - 1 + 9, n ≥ 2

Bonus Simplify (3n+1)(32n)4.

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7 SCORE


1. Simplify (x3)8.

A x24 B x11 C 8x24 D 8x11

2. Simplify (-2hk)4(4h3k5)2.

F 2h24k40 G -64h9k11 H - 256h10k14 J 256h10k14

3. Simplify 36 b 4 c 2 − 9 b -1 c 5

. Assume the denominator is not equal to zero.

A 27 b 4 − c 3

B 4 b 4 − c 3

C 27 b 3 − c 3

D 4 b 5 − c 3

4. Simplify (3 y 4 n 6 ) 2

− ( y 2 n -3 ) 4


F 9 − y 16

G 9 − n 24

H 9y16 J 9n24

5. Which monomial represents the number of square units in the area of a circle with radius 4x3 units?

A 16πx6 B 8πx6 C 16πx9 D 8πx5

6. Express 46,100,000 in scientific notation.

F 4.61 × 107 G 4.61 × 106 H 4.61 × 105 J 4.61 × 108

7. Evaluate 7 × 1 0 4 − 1.4 × 1 0 -5

.

A 5 × 109 B 5 × 10-20 C 5 × 10-1 D 5 × 101

8. ATTENDANCE The total attendance for a professional baseball team this season was 3.24 × 1 0 6 and two years ago was 2.43 × 1 0 6 . About how many times as large was this season’s attendance as attendance two years ago?

F 0.8 G 0.9 H 1.1 J 1.3

9. Write 10 y 1 −

2 in radical form.

A √ �� 10y B 10 √ � y C 10 √ �� 10y D y √ �� 10

10. Evaluate 8 1 3 − 4 .

F 3 G 9 H 27 J 243

11. Solve 5 x - 2 = 125.

A 2 B 3 C 4 D 5

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2.

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6.

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8.

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10.

11.

Chapter 7 Test, Form 2A


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7

12. Which is the equation for the nth term of the geometric sequence -4, 8, -16, ... ?

F a n = - 4 · 2 n-1 H a n = -4 . (-2) n-1 G a n = -2 . (-4) n-1 J a n = - 2 · 4 n

13. Which equation represents exponential growth? A y = 5( 0.84) x B y = 5x C y = 0.3x3 D y = 5 (1.06) x

14. Which equation corresponds to the graph shown? F y = ( 3) x + 1 H y = 2( 3 x ) G y = 2( 3 x + 1) J y = (2 . 3) x + 1

15. A weightlifter can increase the weight W(x) that she can lift according to W(x) = 315(1.0 5) x , where x represents the number of training cycles completed. How much will she lift after 4 training cycles?

A 365 lb B 383 lb C 378 lb D 402 lb

16. BIOLOGY A certain fast-growing bacteria increases 6% per minute. If there are 100 bacteria now, about how many will there be 12 minutes later?

F 172 G 201 H 48 J 190

17. POPULATION A city’s population is about 954,000 and is decreasing at an annual rate of 0.1%. Predict the population in 50 years.

A 577,176 B 906,300 C 1,002,888 D 907,450

18. Find the third term of the sequence in which a 1 = 7 and a n = -2 a n - 1 + 11, if n ≥ 2.

F -23 G -3 H 5 J 17

19. Find an explicit formula for a 1 = -4, a n = a n - 1 + 9, n ≥ 2. A a n = 9n - 13 C a n = 9n - 4 B a n = n + 9 D a n = -4n + 9

20. Find a recursive formula for the arithmetic sequence 24, 32, 40, 48, ... . F a 1 = 24, a n = 8 a n - 1 , n ≥ 2 H a 1 = 24, a n = 4 −

3 a n - 1 , n ≥ 2

G a 1 = 24, a n = 1 − 2 a n - 1 + 20, n ≥ 2 J a 1 = 24, a n = a n - 1 + 8, n ≥ 2

Bonus Simplify 7 x - 3 − 7 3x - 1

.

Chapter 7 Test, Form 2A (continued)

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7 SCORE


1. Simplify (m4)2.

A 6m B m8 C m6 D 2m4

2. Simplify (-2xy2)4(2x3y4)2.

F 4x24y32 G -8x9y6 H 64x10y16 J -4x10y16

3. Simplify 6 n -3 y

− 2 n -1 y -3


A 4 y 3

− n 2

B 3 y 4

− n 2

C 3 − n 4 y 2

D 3n2 −

y4

4. Simplify ( a -2 b 4 ) -6 −

( a 4 b -8 ) 3 . Assume the denominator is not equal to zero.

F ab3 G 1 H a 24 − b 48

J b 48 − a 24

5. Which monomial represents the number of square units in the area of a circle with radius 3x3 units?

A 9πx6 B 6πx6 C. 9πx9 D 6πx5

6. Express 8,450,000 in scientific notation.

F 8.45 × 104 G 8.45 × 107 H 8.45 × 105 J 8.45 × 106

7. Evaluate 4.65 × 1 0 -4 − 5 × 1 0 -6

.

A 9.3 × 1011 B 9.3 × 101 C 9.3 × 102 D 9.3 × 100

8. SOLAR SYSTEM The average distance Earth is from the Sun is about 9.296 × 1 0 7 miles, and the average distance Mars is from the Sun is about 1.4162 × 1 0 8 . About how many times as far is Mars from the Sun as Earth is from the Sun?

F 0.7 G 0.9 H 1.3 J 1.5

9. Write (8x ) 1 − 2 in radical form.

A 8 √ � x B √ � 8x C 8 √ � 8x D x √ � 8

10. Evaluate 12 5 2 − 3 .

F 5 G 25 H 625 J 3125

11. Solve 6 x + 1 = 1296.

A 1 B 2 C 3 D 4

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2.

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7.

8.

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10.

11.

Chapter 7 Test, Form 2B


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7

12. Which is the equation for the nth term of the geometric sequence 6, 12, 24, ... ? F a n = 6 . 2 n G a n = 2 . 3 n H a n = 3 . 2 n J a n = 3 . 2 n-1

13. Which equation represents exponential decay? A y = 0.5x3 B y = 0.5x2 - x C y = 0.5(1 .07) x D y = 0.5(.8 7) x

14. Which equation corresponds to the graph shown? F y = 3 x + 2 H y = 2( 3 x ) G y = 2( 3 x + 1) J y = (2 . 3 ) x + 1

15. A weight lifter can deadlift 275 pounds. She can increase the weight W(x) that she can lift according to the function W(x) = 275(1.05)x, where x represents the number of training cycles completed. How much will she deadlift after 5 training cycles?

A 334 lb B 369 lb C 344 lb D 351 lb

16. POPULATION A city’s population is about 763,000 and is increasing at an annual rate of 1.5%. Predict the population of the city in 50 years.

F 1,335,250 G 826,830,628 H 358,374 J 1,606,300

17. BUSINESS A printing press valued at $120,000 depreciates 12% per year. What will be the approximate value of the printing press in 7 years?

A $19,200 B $265,282 C $49,041 D $55,728

18. Find the third term of the sequence in which a 1 = -1 and a n = 5 a n - 1 - 3, if n ≥ 2.

F -218 G -43 H -8 J 12

19. Find an explicit formula for a 1 = 10, a n = a n - 1 - 3, n ≥ 2. A a n = n - 3 C a n = -3n + 10 B a n = 10n - 3 D a n = -3n + 13

20. Find a recursive formula for the arithmetic sequence 8, -2, -12, -22, ... . F a 1 = 8, a n = -10 a n - 1 , n ≥ 2 H a 1 = 8, a n = - 1 −

2 a n - 1 + 2, n ≥ 2

G a 1 = 8, a n = a n - 1 - 10, n ≥ 2 J a 1 = 8, a n = 1 − 2 a n - 1 - 6, n ≥ 2

Bonus Simplify 32n - 1 · ( 3 5n ) .

Chapter 7 Test, Form 2B (continued)

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7 SCORE

Simplify.

1. (9t3n5)(-2tn2)

2. (w5y4)3

3. 4a3n6 + 4(a3n)6 + 4(an2)3


4. 16 r 3 t -5 − 4 r -1 t 2

5. (-8 x 2 y 2 ) 2

− (4 x 3 y ) 3


Evalute each product or quotient. Express the results inboth scientific notation and standard form.

7. (2.8 × 10-4)(7.7 × 109)

8. 8.1 × 1 0 4 − 1.8 × 1 0 -4

9. ATTENDANCE The total home attendance for a professional basketball team in 2010 was about 8.2 × 1 0 5 , and in 2008 was about 7.175 × 1 0 5 . About how many times as large was the attendance in 2010 as the attendance in 2008?


10. 3 6 x - 2 = 6

11. 4 3x - 2 = 256

12. 3 2 x + 3 = 2

Chapter 7 Test, Form 2C

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7

Simplify.

13. 6 4 5 − 6

14. 8 1 5 − 4

Write each expression in radical form.

15. 2 2 1 − 3

16. 3 x 1 − 4

17. A weight lifter can bench-press 165 pounds. She plans to increase the weight W(x) in pounds that she is lifting according to the function W(x) = 165(1.05)x, where x represents the number of training cycles she completes. How much will she bench-press after 4 training cycles?

18. Graph y = 7x. Find the y-intercept and state the domain and range.

19. INVESTMENTS Determine the amount of an investment if $700 is invested at an interest rate of 8% compounded monthly for 9 years.

20. BUSINESS A new welding machine valued at $38,000 depreciates at a steady rate of 9% per year. What is the value of the welding machine in 10 years?

21. Write an equation for the nth term of the geometric sequence -3, 9, -27,. . . .

Find the first three terms of each sequence.

22. a 1 = 5, a n = a n - 1 - 3, n ≥ 2

23. a 1 = 3, a n = 2 a n - 1 + 4, n ≥ 2


24. 18, 29, 40, 51, …

25. 2, −12, 72, −432, …

Bonus Find the first term of the geometric sequence with a 6 = 256 and a 7 = 1024.

Chapter 7 Test, Form 2C (continued)

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7 SCORE

Simplify.

1. (3a2b5)(-2ab3)

2. (w3z7)3

3. 4a4b8 + 2(ab2)4 + 4(a2b4)2


4. 4 a -3 d 2 − 8 a 2 d -5

5. (3 r 3 t 5 ) 3 −

(-3 r 2 t 7 ) 2


Evaluate each product or quotient. Express the results in both scientific notation and standard form.

7. (4.6 × 103)(9.12 × 10-7)

8. 1.6 × 1 0 -3 − 8 × 1 0 -7

9. ATTENDANCE The total home attendance for a professional football team in 2010 was about 5.44 × 1 0 5 , and in 2008 was about 4.32 × 1 0 5 . About how many times as large was the attendance in 2010 as the attendance in 2008?


10. 12 5 x - 1 = 5

11. 3 3x + 1 = 81

12. 6 4 2x + 3 = 2

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5.

6.

7.

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9.

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11.

12.

Chapter 7 Test, Form 2D


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7

Simplify.

13. 100 0 2 − 3

14. 4 5 − 2


15. 5 7 1 − 4

16. 10 x 1 − 3

17. A weight lifter can bench-press 145 pounds. She plans to increase the weight W(x) in pounds that she is lifting according to the function W(x) = 145(1.05)x, where x represents the number of training cycles she completes. How much will she bench-press after 5 training cycles?

18. Graph y = ( 1 − 6 )

x

. Find the y-intercept and state the

domain and range.

19. ART An oil painting originally cost $2500 and increases in value at a rate of 6% per year. Find the value of the painting after 12 years.

20. CARS A new car valued at $16,500 depreciates at a steady rate of 12% per year. What is the value of the car in 10 years?

21. Write an equation for the nth term of the geometric sequence 4, 8, 16,… .


22. a 1 = -4, a n = a n − 1 + 7, n ≥ 2

23. a 1 = 1, a n = 4 a n - 1 − 2, n ≥ 2


24. 27, 19, 11, 3, …

25. 1296, 216, 36, 6, …

Bonus Find the first term of the geometric sequence with a 5 = 625 and a 6 = 3125.

Chapter 7 Test, Form 2D (continued)

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7 SCORE

Simplify.

1. ( 2 − 3 h 3 )

4

2. (43x+7)(42x-9)


3. (-2m x -3 ) -4 −

8 m -5 x 0

4. ( -3 a 2 b -3 − 6 a 3 b -4

) 2

( -5b − 4 a -3

) -3

5. Express 473 in scientific notation.

Evaluate each product or quotient. Express the results in both scientific notation and standard form.

6. (5.2 × 104)(5.9 × 106)

7. 2.485 × 1 0 3 − 7.1 × 1 0 9

8. ATTENDANCE The total home attendance for a professional football team was about 5.2 × 1 0 5 . In the same year, the total home attendance for a professional baseball team was about 2.673 × 1 0 6 . About how many times as large was the attendance for the baseball team as the attendance for the football team?


9. 4 3x - 2 = 256

10. 62 5 x - 1 = 5

11. 3 3x + 2 = 729

12. 25 6 4x - 5 = 2

Chapter 7 Test, Form 3

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7

Simplify.

13. 129 6 3 − 4

14. 8 7 − 3


15. 32 x 1 − 6

16. (4yz ) 1 − 8

17. WEIGHT LIFTING A weight lifter can squat 475 pounds. She plans to increase the weight W(x) in pounds that she is lifting according to the function W(x) = 475(1.05)x, where x represents the number of training cycles she completes. How much will she squat after 4 training sessions?

18. Graph y = 4(2x - 1). State the y-intercept.

19. INVESTMENTS Determine the amount of an investment if $96,000 is invested at an interest rate of 5.2% compounded daily for 3 years.

20. Suppose a car that sells for $40,000 depreciates 10% per year. How many years would it take for the car to have a value less than $25,000?

21. Write an equation for the nth term of the geometric sequence -7, 1.75, -0.4375,… .


22. a 1 = 2.25, a n = a n − 1 + 0.5, n ≥ 2

23. a 1 = 3.5, a n = −0.5 a n - 1 + 0.75, n ≥ 2


24. 50, 3, −44, −91 …

25. 1 − 8 , 1 −

4 , 1 −

2 , 1, …

Bonus Find the first term of the geometric sequence with a 6 = 1 −

4 and a 7 = 1 −

8 .

Chapter 7 Test, Form 3 (continued)

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B:

y

xO


Ass

essm

ent

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7

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. a. Simplify ( 4 a 3 − 2 a -2

) 4

by using the Quotient of Powers property first.

Then use the Power of a Power property.

b. Simplify ( 4 a 3 − 2 a -2

) 4

by using the Power of a Quotient property first.

Then use the Quotient of Powers property.

c. Write a statement that generalizes the results of part a and part b.

2. Give a counterexample for each statement.

a. As n increases in a geometric sequence, the value of a n will move farther away from zero.

b. In a recursive sequence, if a 1 = a 2, then a 2 = a 3 , and so on.

3. Honovi purchased a new car for $25,000 and has $5000 left to invest.

a. Choose an interest rate between 4% and 7% for Honovi’s investment, and find the length of time it would take for the investment to double.

b. Choose an annual depreciation rate from 8% to 10% for the new car that Honovi purchased, and find the length of time it would take for the car’s value to be equal to one-half of the purchase price.

c. Using the rates from part a and part b, find the length of time it would take for the investment to be equal to the value of the car. What is the value at that time?

4. The mass of Jupiter is 1.8987 × 10 27 kilograms and is roughly 317.8 times the mass of Earth. The mass of the solar system is about 1.992 × 10 30 kilograms. The Sun accounts for about 99.8% of this mass.

a. Find the mass of Earth in kilograms.

b. What is the approximate mass of all of the other objects in the solar system, not counting the Sun?

c. What percentage of the mass obtained in part b is represented by Jupiter?

Chapter 7 Extended-Response Test


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7 SCORE Standardized Test Practice(Chapters 1–7)

1. Write y - 11 = 1 − 2 (x - 12) in standard form. (Lesson 4-2)

A y = 1 − 2 x + 5 B y = 1 −

2 x - 17 C x - 2y = -10 D x - 2y = 17

2. Ashanti wants to collect more than 50 food items for the local homeless shelter. If he has already collected 13, how many more items must he collect? (Lesson 5-1)

F 37 G at least 38 H at least 37 J more than 36

3. Solve -5n + 22 ≥ -73. (Lesson 5-3)

A {n ⎥ n ≤ 19} B {n ⎥ n ≥ 19} C {n | n ≤ - 51 − 5 } D {n ⎥ n ≥ 90}

4. Solve ⎥ 5k + 2 ⎥ > 1. (Lesson 5-5)

F {k | k < - 3 − 5 or k > - 1 −

5 } H {k | k - 3 −

5 < k < 1 −

5 }

G ∅ J {k ⎥ k is a real number.}

5. Use substitution to solve the system y = -2xof equations. (Lesson 6-2) 5x + 3y = 4

A (-4, 8) B (8, -4) C (-8, 4) D (4, -8)

6. Use elimination to solve the system x + 3y = -6of equations. (Lesson 6-3) 2x + 3y = -9

F (-2, -4) G (-6, 1) H (-3, -1) J (3, -5)

7. Simplify ( 3 2 )( 3 3 ) −

( 3 -2 )( 3 -3 ) . (Lesson 7-2)

A 310 B 312 C -1 D 1 − 3

8. Simplify 4 √ 1296 . (Lesson 7-3)

F 36 G 9 H 6 J 3

9. Evaluate (3.2 × 1 0 6 )(4.6 × 1 0 -2 ) (Lesson 7-4)

A 1.472 × 1 0 4 C 1.472 × 1 0 3

B 1.472 × 10 5 D 14.72 × 1 0 4

10. Find the sixth term of the geometric sequence -16, 40, -100, … . (Lesson 7-7)

F −15,625 H 6250

G −6250 J 1562.5

11. Simplify 24 + 6 - {23 + 7(5 - 2)}. (Lesson 1-2)

A 43 B 3 C 1 D 45

1 .

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

A B C D

A B C D

F G H J

A B C D

A B C D

F G H J

Part 1: Multiple Choice

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

F G H J

F G H J

F G H J

A B C D

A B C D


Ass

essm

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7 Standardized Test Practice (continued)

12. Find the percent of change. (Lesson 2-7)

Original: $28 New: $21

F 25% G 7% H 33% J 14%

13. Solve a(b - c) = a − b for c. (Lesson 2-8)

A c = b 2 - 1 − b B c = -1 −

b - b C c = 1 −

b + ab D c = a −

b - ab

14. Write an equation in function notation for the relation. (Lesson 3-6)

F f(x) = -2x - 2 H f(x) = -2x + 2

G f(x) = 2x - 2 J f(x) = 1 − 2 x - 2

15. Find the slope of the line that passes through the points (-4, 4) and (6, -8). (Lesson 3-3)

A 6 − 5 B - 6 −

5 C - 5 −

6 D 5 −

6

16. Write y + 1 = 4(x - 2) in slope-intercept form. (Lesson 4-3)

F y = 4x - 3 G y = 4x + 7 H y = 4x - 9 J y = x - 9

17. Solve 14u + 9 - 10u > 21. (Lesson 5-3)

A u > 3 B u > -3 C u < 3 D u < -3

18. Simplify (4x 2) 2(-2x4) 3. (Lesson 7-1)

F -48x16 G -48x11 H 128x11 J -128x16

12.

13.

14.

15.

16.

17.

18. F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

xO

f (x)

19. Evaluate 4x(y2 - z2) if x = 3, 20. After receiving his allowance, y = 5 and z = 4. (Lesson 1-2) Spencer spent half of it on a

Mother’s Day card. He bought 2 toy cars for $0.49 each to give to his brothers, and a pack of gum for $0.35. How much did Spencer receive for his allowance if $0.42 is left over? (Lesson 2-3)

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

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 oval that corresponds to that entry.


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7 Standardized Test Practice (continued)

21. Translate the sentence into an equation. The quotient of 35 and 6 plus a number is equal to twice the sum of that number and 3. (Lesson 2-1)

22. TUITION The average cost of tuition and fees at Trenton College was $33,560 for a recent school year. Two school years before, the average cost was $31,110. Find the rate of change between the two school years. (Lesson 3-3)

23. Solve 3y - 4 ≤ 7. (Lesson 5-3)

24. RECREATION Barrington needs to buy snorkels and fins for his family for their annual beach vacation. Snorkels cost $8 a set and fins cost $12 a pair. Write an inequality that represents this situation if Barrington has $62 to spend. (Lesson 5-6)

25. Graph the system of equations and determine the number of solutions it has. If it has one solution, name it. (Lesson 6-1)

x + y = 4y = x

26. Determine the best method to solve the system of equations. Then solve the system. (Lesson 6-5)

5x - 2y = 72x + 5y = -3

27. Simplify 51x-1y3

− 17x2y

Assume the denominator is not equal

to zero. (Lesson 7-2)

28. Solve 9 4n - 3 = 3 6 . (Lesson 7-3)

29. Write a recursive formula for 39, 32, 25, 18, … . (Lesson 7-8)

30. As of the end of the 2010 football season, the Dallas Cowboys and the Denver Broncos had won a total of 7 Super Bowls. The Cowboys had won 2.5 times as many Super Bowls as the Broncos. (Lesson 6-2)

a. Define variables and write a system of linear equations to find each team’s wins.

b. How many Super Bowls had the Dallas Cowboys won?

Part 3: Short Response

Instructions: Write your answers in the space provided.

21.

22.

23.

24.

25. y

xO

26.

27.

28.

29.

30a.

30b.


An

swer

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PDF Pass

Chapter 7 A1 Glencoe Algebra 1

Chapter Resources

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

NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

3 G

lenc

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Ant

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Exp

on

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ts a

nd

Exp

on

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tial

Fu

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s

B

efor

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u b

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Ch

ap

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7

•

Rea

d ea

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tate

men

t.

•

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wh

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ou A

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(A

) or

Dis

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) w

ith

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•

Wri

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or

D i

n t

he

firs

t co

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R i

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re n

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wh

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disa

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Not

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EP

1A

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r N

SS

tate

men

tS

TE

P 2

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ult

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two

pow

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that

hav

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me

base

, m

ult

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th

e ex

pon

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.D

2.

( k 3 )

4 is

equ

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to k

12 .

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. T

o di

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o po

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s th

at h

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th

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4.

( 2 −

5 ) 3 i

s th

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D

5.

A p

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. T

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l 3 x

2 y 3 -

5y 2 +

8 x 3

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bec

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th

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xpon

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is 3

.D

7.

A f

un

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•

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each

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7 Step

1

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2

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NA

ME

DAT

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P

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IOD

Lesson 7-1

Cha

pte

r 7

5 G

lenc

oe A

lgeb

ra 1

Stud

y G

uide

and

Inte

rven

tion

Mu

ltip

licati

on

Pro

pert

ies o

f E

xp

on

en

ts

Mu

ltip

ly M

on

om

ials

A m

onom

ial

is a

num

ber,

a va

riab

le, o

r th

e pr

oduc

t of

a n

umbe

r an

d on

e or

mor

e va

riab

les

wit

h no

nneg

ativ

e in

tege

r ex

pone

nts.

An

expr

essi

on o

f th

e fo

rm x

n

is c

alle

d a

pow

er a

nd r

epre

sent

s th

e pr

oduc

t yo

u ob

tain

whe

n x

is u

sed

as a

fac

tor

n t

imes

. T

o m

ulti

ply

two

pow

ers

that

hav

e th

e sa

me

base

, add

the

exp

onen

ts.

Pro

du

ct o

f P

ow

ers

Fo

r a

ny n

um

be

r a

an

d a

ll in

teg

ers

m a

nd

n,

am ․

an

= a

m +

n.

S

imp

lify

(3x

6 )(5

x2 ).

(3x6 )

(5x2 )

= (

3)(5

)(x6

․ x2 )

Gr

oup t

he c

oeffic

ients

and t

he v

ari

able

s

=

(3

․ 5)

(x6

+ 2)

Pro

duct

of

Pow

ers

=

15x

8 S

implif

y.

Th

e pr

odu

ct i

s 15

x8 .

S

imp

lify

(-

4a3 b

)(3a

2 b5 )

.(-

4a3 b

)(3a

2 b5 )

= (

-4)

(3)(

a3 ․

a2 )(b

․ b

5 )

= -

12(a

3 +

2)(

b1 +

5)

=

-12

a5 b6

Th

e pr

odu

ct i

s -

12a5 b

6 .

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n.

1. y

(y5 )

2.

n2

․ n

7 3.

(-7x

2 )(x

4 )

y

6 n

9 -

7x6

4. x

(x2 )

(x4 )

5.

m ․

m5

6. (-

x3 )(-

x4 )

x

7 m

6 x

7

7. (

2a2 )

(8a)

8.

(rn

)(rn

3 )(n

2 )

9. (x

2 y)(

4xy3 )

1

6a3

r2 n

6 4

x3 y

4

10. 1

−

3 ( 2a

3 b)(

6b 3 )

11

. (-

4x3 )

(-5x

7 )

12. (

-3j

2 k4 )

(2jk

6 )

4

a3 b4

20x

10

-6j

3 k10

13. (

5a 2 b

c 3 ) ( 1 −

5 a

bc 4 )

14. (

-5x

y)(4

x2 )(y

4 )

15. (

10x3 y

z2 )(-

2xy5 z

)

a

3 b2 c

7 -

20x

3 y5

-20

x4 y

6 z3

7-1

Exam

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1Ex

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Answers (Anticipation Guide and Lesson 7-1)

A01_A14_ALG1_A_CRM_C07_AN_660281.indd A1A01_A14_ALG1_A_CRM_C07_AN_660281.indd A1 12/20/10 7:11 PM12/20/10 7:11 PM

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PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

6 G

lenc

oe A

lgeb

ra 1

Sim

plif

y Ex

pre

ssio

ns

An

exp

ress

ion

of

the

form

(xm

)n i

s ca

lled

a p

ower

of

a p

ower

an

d re

pres

ents

th

e pr

odu

ct y

ou o

btai

n w

hen

xm i

s u

sed

as a

fac

tor

n t

imes

. To

fin

d th

e po

wer

of

a po

wer

, mu

ltip

ly e

xpon

ents

.

Po

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of

a P

ow

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or

any n

um

be

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an

d a

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teg

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m a

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p,

(am)p

= a

mp.

Po

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of

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rod

uct

Fo

r a

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um

be

rs a

an

d b

an

d a

ny in

teg

er

m,

(ab

)m =

amb

m.

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can

com

bin

e an

d u

se t

hes

e pr

oper

ties

to

sim

plif

y ex

pres

sion

s in

volv

ing

mon

omia

ls.

S

imp

lify

(-

2ab2 )

3 (a

2 )4 .

(-2a

b2 )3 (

a2 )4

= (

-2a

b2 )3 (

a8 )

Pow

er

of

a P

ow

er

=

(-

2)3 (

a3 )(b

2 )3 (

a8 )

Pow

er

of

a P

roduct

=

(-

2)3 (

a3 )(a

8 )(b

2 )3

Gro

up t

he c

oeffic

ients

and t

he v

ari

able

s

=

(-

2)3 (

a11)(

b2 )3

Pro

duct

of

Pow

ers

=

-8a

11b6

Pow

er

of

a P

ow

er

Th

e pr

odu

ct i

s -

8a11

b6 .

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n.

1. (

y5 )2

2. (n

7 )4

3. (x

2 )5 (

x3 )

y

10

n28

x

13

4. -

3(ab

4 )3

5. (-

3ab4 )

3 6.

(4x2 b

)3

-

3a3 b

12

-27

a3 b

12

64x

6 b3

7. (

4a2 )

2 (b3 )

8.

(4x)

2 (b3 )

9.

(x2 y

4 )5

1

6a4 b

3 1

6x2 b

3 x

10y

20

10. (

2a3 b

2 )(b

3 )2

11. (

-4x

y)3 (

-2x

2 )3

12. (

-3j

2 k3 )

2 (2j

2 k)3

2

a3 b

8 5

12x

9 y3

72j

10k

9

13. (

25 a

2 b) 3 ( 1

−

5 abf

) 2 14

. (2x

y)2 (

-3x

2 )(4

y4 )

15. (

2x3 y

2 z2 )

3 (x2 z

)4

6

25a

8 b5 f

2 -

48x

4 y6

8x

17y

6 z10

16. (

-2n

6 y5 )

(-6n

3 y2 )

(ny)

3 17

. (-

3a3 n

4 )(-

3a3 n

)4 18

. -3(

2x)4 (

4x5 y

)2

1

2n12

y10

-

243a

15n

8 -

768x

14y

2

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Mu

ltip

licati

on

Pro

pert

ies o

f E

xp

on

en

ts

7-1

Exam

ple

001_

020_

ALG

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M_C

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1.in

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/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-1

Cha

pte

r 7

7 G

lenc

oe A

lgeb

ra 1

Skill

s Pr

acti

ceM

ult

iplicati

on

Pro

pert

ies o

f E

xp

on

en

ts

Det

erm

ine

wh

eth

er e

ach

exp

ress

ion

is

a m

onom

ial.

Wri

te y

es o

r n

o. E

xpla

in.

1. 1

1 Ye

s; 1

1 is

a r

eal n

um

ber

an

d a

n e

xam

ple

of

a co

nst

ant.

2. a

- b

No

; th

is is

th

e d

iffer

ence

, no

t th

e p

rod

uct

, of

two

var

iab

les.

3.

p 2

−

r 2 N

o;

this

is t

he

qu

oti

ent,

no

t th

e p

rod

uct

, of

two

var

iab

les.

4. y

Yes

; si

ng

le v

aria

ble

s ar

e m

on

om

ials

.

5. j

3 k Y

es;

this

is t

he

pro

du

ct o

f tw

o v

aria

ble

s.

6. 2

a +

3b

No

; th

is is

th

e su

m o

f tw

o m

on

om

ials

.

Sim

pli

fy.

7. a

2 (a3 )

(a6 )

a11

8.

x(x

2 )(x

7 ) x

10

9. (

y2 z)(

yz2 )

y3 z

3 10

. (ℓ2 k

2 )(ℓ

3 k)

�5 k

3

11. (

a2 b4 )

(a2 b

2 ) a

4 b 6

12. (

cd2 )

(c3 d

2 ) c

4 d4

13. (

2x2 )

(3x5 )

6x

7 14

. (5a

7 )(4

a2 ) 2

0a9

15. (

4xy3 )

(3x3 y

5 ) 1

2x4 y

8 16

. (7a

5 b2 )

(a2 b

3 ) 7

a7 b

5

17. (

-5m

3 )(3

m8 )

-15

m11

18

. (-

2c4 d

)(-

4cd

) 8c

5 d2

19. (

102 )

3 10

6 o

r 1,

00

0,0

00

20. (

p3 )12

p36

21. (

-6p

)2 36

p2

22. (

-3y

)3 -

27y

3

23. (

3pr2 )

2 9p

2 r4

24. (

2b3 c

4 )2

4b6 c

8

GEO

MET

RY E

xpre

ss t

he

area

of

each

fig

ure

as

a m

onom

ial.

25.

x2

x5

26

.

cd

cd

27

.

4p 9p3

x

7

c2 d

2 1

8p4

7-1

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

712

/20/

10

6:38

PM

Answers (Lesson 7-1)


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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

8 G

lenc

oe A

lgeb

ra 1

Prac

tice

Mu

ltip

licati

on

Pro

pert

ies o

f E

xp

on

en

ts

Det

erm

ine

wh

eth

er e

ach

exp

ress

ion

is

a m

onom

ial.

Wri

te y

es o

r n

o. E

xpla

in y

our

reas

onin

g.

1.

21 a

2 −

7b

N

o;

this

invo

lves

th

e q

uo

tien

t, n

ot

the

pro

du

ct, o

f va

riab

les.

2.

b 3 c 2

−

2 Y

es;

this

is t

he

pro

du

ct o

f a

nu

mb

er,

1 −

2 , an

d t

wo

var

iab

les.

Sim

pli

fy e

ach

exp

ress

ion

.

3. (

-5x

2 y)(

3x4 )

-15

x6 y

4.

(2ab

2 f 2 )

(4a3 b

2 f 2 )

8a

4 b4 f

4

5. (

3ad

4 )(-

2a2 )

-6a

3 d4

6. (4

g3 h)(

-2g

5 ) -

8g8 h

7. (

-15

x y 4 )

(- 1 −

3 x y

3 ) 5x

2 y7

8. (-

xy)3 (

xz)

-x

4 y3 z

9. (

-18

m 2 n

) 2 (- 1 −

6 m

n 2 )

-54

m5 n

4 10

. (0.

2a2 b

3 )2

0.04

a4 b

6

11. (

2 −

3 p

) 2 4 −

9 p

2 12

. ( 1 −

4 a

d 3 ) 2

1 −

16 a

2 d 6

13. (

0.4k

3 )3

0.06

4k9

14. [

(42 )

2 ]2

48 or

65,5

36

GEO

MET

RY E

xpre

ss t

he

area

of

each

fig

ure

as

a m

onom

ial.

15.

6a2 b

4

3ab2

16

. 5x

3

17

.

6ab3

4a2 b

1

8a3 b

6 (

25x

6 )π

1

2a3 b

4

GEO

MET

RY E

xpre

ss t

he

volu

me

of e

ach

sol

id a

s a

mon

omia

l.

18.

3h2

3h2

3h2

19

.

m3 n

mn3

n

20.

7g2

3g

2

7h6

m4 n

5 (

63g

4 )π

21. C

OU

NTI

NG

A p

anel

of

fou

r li

ght

swit

ches

can

be

set

in 2

4 w

ays.

A p

anel

of

five

lig

ht

swit

ches

can

set

in

tw

ice

this

man

y w

ays.

In

how

man

y w

ays

can

fiv

e li

ght

swit

ches

be

set

? 25

or

32

22. H

OB

BIE

S T

awa

wan

ts t

o in

crea

se h

er r

ock

coll

ecti

on b

y a

pow

er o

f th

ree

this

yea

r an

d th

en i

ncr

ease

it

agai

n b

y a

pow

er o

f tw

o n

ext

year

. If

she

has

2 r

ocks

now

, how

man

y ro

cks

wil

l sh

e h

ave

afte

r th

e se

con

d ye

ar?

26 o

r 64

7-1

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

812

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-1

7-1

Cha

pte

r 7

9 G

lenc

oe A

lgeb

ra 1

1. G

RA

VIT

Y A

n e

gg t

hat

has

bee

n f

alli

ng

for

x se

con

ds h

as d

ropp

ed a

t an

ave

rage

sp

eed

of 1

6x f

eet

per

seco

nd.

If

the

egg

is

drop

ped

from

th

e to

p of

a b

uil

din

g, i

ts

tota

l di

stan

ce t

rave

led

is t

he

prod

uct

of

the

aver

age

rate

tim

es t

he

tim

e. W

rite

a

sim

plif

ied

expr

essi

on t

o sh

ow t

he

dist

ance

th

e eg

g h

as t

rave

led

afte

r x

seco

nds

.

2. C

IVIL

EN

GIN

EER

ING

A d

evel

oper

is

plan

nin

g a

side

wal

k fo

r a

new

de

velo

pmen

t. T

he

side

wal

k ca

n b

e in

stal

led

in r

ecta

ngu

lar

sect

ion

s th

at

hav

e a

fixe

d w

idth

of

3 fe

et a

nd

a le

ngt

h

that

can

var

y. A

ssu

min

g th

at e

ach

se

ctio

n i

s th

e sa

me

len

gth

, exp

ress

th

e ar

ea o

f a

4-se

ctio

n s

idew

alk

as a

m

onom

ial.

x

3 ft

3. P

RO

BA

BIL

ITY

If

you

fli

p a

coin

3 t

imes

in

a r

ow, t

her

e ar

e 23

outc

omes

th

at c

an

occu

r.

Ou

tco

mes

HH

HT

TT

HT

TT

HH

HT

HT

TH

HH

TT

HT

If

you

th

en f

lip

the

coin

tw

o m

ore

tim

es,

ther

e ar

e 23

× 2

2 ou

tcom

es t

hat

can

oc

cur.

How

man

y ou

tcom

es c

an o

ccu

r if

yo

u f

lip

the

coin

as

men

tion

ed a

bove

fou

r m

ore

tim

es?

Wri

te y

our

answ

er i

n t

he

form

2x .

4. S

POR

TS T

he

volu

me

of a

sph

ere

is g

iven

by

the

form

ula

V=

4 − 3 π r 3 ,

wh

ere

r is

th

e r

adiu

s of

th

e sp

her

e. F

ind

the

volu

me

of

air

in t

hre

e di

ffer

ent

bask

etba

lls.

Use

π

= 3

.14.

Rou

nd

you

r an

swer

s to

th

e n

eare

st w

hol

e n

um

ber.

Bal

lR

adiu

s (i

n.)

Volu

me

(in

3 )

Ch

ild’s

426

8W

om

en’s

4.5

382

Me

n’s

4.8

463

5. E

LEC

TRIC

ITY

An

ele

ctri

cian

use

s th

e fo

rmu

la W

= I

2 R ,

wh

ere

W i

s th

e po

wer

in

wat

ts, I

is

the

curr

ent

in a

mpe

res,

an

d R

is

the

resi

stan

ce i

n o

hm

s.

a.

Fin

d th

e po

wer

in

a h

ouse

hol

d ci

rcu

it

that

has

20

ampe

res

of c

urr

ent

and

5 oh

ms

of r

esis

tan

ce.

b

. If

th

e cu

rren

t is

red

uce

d by

on

e h

alf,

wh

at h

appe

ns

to t

he

pow

er?

Wor

d Pr

oble

m P

ract

ice

Mu

ltip

licati

on

Pro

pert

ies o

f E

xp

on

en

ts

16x

2 12x

29

200

0 w

atts

Th

e p

ow

er is

on

e-fo

urt

h t

he

pre

vio

us

amo

un

t.

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

912

/20/

10

6:38

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 7

10

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

An

Wan

g

An

Wan

g (1

920–

1990

) w

as a

n A

sian

-Am

eric

an w

ho

beca

me

one

of t

he

pion

eers

of

the

com

pute

r in

dust

ry i

n t

he

Un

ited

Sta

tes.

He

grew

up

in S

han

ghai

, Ch

ina,

bu

t ca

me

to t

he

Un

ited

Sta

tes

to f

urt

her

his

stu

dies

in

sci

ence

. In

194

8, h

e in

ven

ted

a m

agn

etic

pu

lse

con

trol

lin

g de

vice

th

at v

astl

y in

crea

sed

the

stor

age

capa

city

of

com

pute

rs. H

e la

ter

fou

nde

d h

is o

wn

com

pan

y, W

ang

Lab

orat

orie

s, a

nd

beca

me

a le

ader

in

th

e de

velo

pmen

t of

des

ktop

ca

lcu

lato

rs a

nd

wor

d pr

oces

sin

g sy

stem

s. I

n 1

988,

Wan

g w

as e

lect

ed t

o th

e N

atio

nal

In

ven

tors

Hal

l of

Fam

e.

Dig

ital

com

pute

rs s

tore

in

form

atio

n a

s n

um

bers

. Bec

ause

th

e el

ectr

onic

cir

cuit

s of

a

com

pute

r ca

n e

xist

in

on

ly o

ne

of t

wo

stat

es, o

pen

or

clos

ed, t

he

nu

mbe

rs t

hat

are

sto

red

can

co

nsi

st o

f on

ly t

wo

digi

ts, 0

or

1. N

um

bers

wri

tten

usi

ng

only

th

ese

two

digi

ts a

re c

alle

d b

inar

y n

um

ber

s. T

o fi

nd

the

deci

mal

val

ue

of a

bin

ary

nu

mbe

r, yo

u u

se t

he

digi

ts t

o w

rite

a

poly

nom

ial

in 2

. For

in

stan

ce, t

his

is

how

to

fin

d th

e de

cim

al v

alu

e of

th

e n

um

ber

1001

101 2.

(Th

e su

bscr

ipt

2 in

dica

tes

that

th

is i

s a

bin

ary

nu

mbe

r.)

1001

101 2

= 1

× 2

6 +

0 ×

25

+ 0

× 2

4 + 1

× 2

3 +

1 ×

22

+ 0

× 2

1 +

1 ×

20

= 1

× 6

4 +

0 ×

32

+ 0

× 1

6 +

1 ×

8 +

1 ×

4 +

0 ×

2 +

1 ×

1=

6

4

+

0

+

0

+

8

+

4

+

0

+

1=

77

Fin

d t

he

dec

imal

val

ue

of e

ach

bin

ary

nu

mb

er.

1. 1

111 2

15

2. 1

0000

2 16

3.

110

0001

1 2 19

5 4.

101

1100

1 2 18

5

Wri

te e

ach

dec

imal

nu

mb

er a

s a

bin

ary

nu

mb

er.

5. 8

10

00 2

6. 1

1 10

112

7. 2

9 11

101 2

8. 1

17 1

1101

012

9.

Th

e ch

art

at t

he

righ

t sh

ows

a se

t of

dec

imal

co

de n

um

bers

th

at i

s u

sed

wid

ely

in s

tori

ng

lett

ers

of t

he

alph

abet

in

a c

ompu

ter’

s m

emor

y.

Fin

d th

e co

de n

um

bers

for

th

e le

tter

s of

you

r n

ame.

Th

en w

rite

th

e co

de f

or y

our

nam

e u

sin

g bi

nar

y n

um

bers

. An

swer

s w

ill v

ary.

7-1

Th

e A

mer

ican

Sta

nd

ard

Gu

ide

for

Info

rmat

ion

Inte

rch

ang

e (A

SC

II)

A

65

N

78

a 9

7n

11

0

B

66

O

79

b

98

o

111

C

67

P

80

c 9

9p

11

2

D

68

Q

81

d

10

0q

11

3

E

69

R

82

e 101

r 11

4

F

70

S

83

f 10

2s

115

G

71

T

84

g

10

3t

116

H

72

U

85

h

10

4u

11

7

I 7

3V

8

6i

10

5v

118

J 74

W

87

j 10

6w

11

9

K

75

X

88

k 10

7x

12

0

L

76

Y

89

l 10

8y

12

1

M

77

Z

90

m

10

9z

12

2

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1012

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-2

Cha

pte

r 7

11

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Div

isio

n P

rop

ert

ies o

f E

xp

on

en

ts

Div

ide

Mo

no

mia

ls T

o di

vide

tw

o po

wer

s w

ith

th

e sa

me

base

, su

btra

ct t

he

expo

nen

ts.

Qu

oti

ent

of

Po

wer

sF

or

all

inte

ge

rs m

an

d n

an

d a

ny n

on

ze

ro n

um

be

r a,

a m

−

a n =

a m

- n .

Po

wer

of

a Q

uo

tien

tF

or

any in

teg

er

m a

nd

any r

ea

l nu

mb

ers

a a

nd

b,

b ≠

0,

( a −

b ) m

= a

m

−

b m .

S

imp

lify

a 4 b

7 −

ab

2 . A

ssu

me

that

no

den

omin

ator

eq

ual

s ze

ro.

a 4 b

7 −

ab

2 = ( a

4 −

a

) ( b 7

−

b 2 ) G

roup p

ow

ers

with t

he s

am

e b

ase.

=

(a4

- 1)(

b7 -

2)

Quotient

of

Pow

ers

=

a3 b

5 S

implif

y.

Th

e qu

otie

nt

is a

3 b5 .

S

imp

lify

( 2 a

3 b 5

−

3 b 2

) 3 . Ass

um

e

that

no

den

omin

ator

eq

ual

s ze

ro.

( 2 a 3 b

5 −

3 b

2 ) 3 = (2

a 3 b

5 ) 3

−

(3 b 2 )

3 P

ow

er

of

a Q

uotient

=

2 3 ( a

3 ) 3 (

b 5 ) 3

−

(3 ) 3 (

b 2 ) 3

P

ow

er

of

a P

roduct

=

8 a 9 b

15

−

27 b 6

Pow

er

of

a P

ow

er

=

8 a 9 b

9 −

27

Quotient

of

Pow

ers

Th

e qu

otie

nt

is 8 a

9 b 9

−

27

.

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n. A

ssu

me

that

no

den

omin

ator

eq

ual

s ze

ro.

1.

5 5 −

5 2 5

3 o

r 12

5 2.

m

6 −

m

4 m2

3.

p 5 n

4 −

p

2 n

p3 n

3

4.

a 2

−

a a

5.

x 5 y

3 −

x 5 y

2 y

6.

-

2 y 7

−

14 y 5

-

1 −

7 y 2

7.

x y 6

−

y 4 x y

2 8.

( 2 a 2 b

−

a ) 3 8

a3 b

3 9.

( 4 p 4 r

4 −

3 p

2 r 2 ) 3 64

−

27 p

6 r 6

10. ( 2 r

5 w 3

−

r 4 w 3 ) 4 1

6r 4

11. (

3 r 6 n

3 −

2 r

5 n ) 4 81

−

16 r

4 n 8

12.

r 7 n 7 t

2 −

n

3 r 3 t

2 r 4 n

4

7-2

Exam

ple

1Ex

amp

le 2

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1112

/21/

10

4:49

PM

Answers (Lesson 7-1 and Lesson 7-2)


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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

12

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Div

isio

n P

rop

ert

ies o

f E

xp

on

en

ts

Neg

ativ

e Ex

po

nen

ts A

ny

non

zero

nu

mbe

r ra

ised

to

the

zero

pow

er i

s 1;

for

exa

mpl

e,

(-0.

5)0

= 1

. An

y n

onze

ro n

um

ber

rais

ed t

o a

neg

ativ

e po

wer

is

equ

al t

o th

e re

cipr

ocal

of

the

nu

mbe

r ra

ised

to

the

oppo

site

pow

er; f

or e

xam

ple,

6 -

3 =

1 −

6 3 . T

hes

e de

fin

itio

ns

can

be

use

d to

si

mpl

ify

expr

essi

ons

that

hav

e n

egat

ive

expo

nen

ts.

Zer

o E

xpo

nen

tF

or

any n

on

ze

ro n

um

be

r a,

a0 =

1.

Neg

ativ

e E

xpo

nen

t P

rop

erty

Fo

r a

ny n

on

ze

ro n

um

be

r a

an

d a

ny in

teg

er

n, a

-n

=

1

−

a n a

nd

1

−

a -

n = a

n .

Th

e si

mpl

ifie

d fo

rm o

f an

exp

ress

ion

con

tain

ing

neg

ativ

e ex

pon

ents

mu

st c

onta

in o

nly

po

siti

ve e

xpon

ents

.

S

imp

lify

4 a

-3 b

6 −

16 a

2 b 6 c

-5 .

Ass

um

e th

at n

o d

enom

inat

or e

qu

als

zero

.

4 a

-3 b

6 −

16

a 2 b

6 c -

5 =

( 4 −

16

) ( a -

3 −

a

2 ) ( b 6 −

b 6 ) (

1 −

c -5 )

Gro

up p

ow

ers

with t

he s

am

e b

ase.

=

1 −

4 (

a -

3-2 )

( b 6-

6 )( c

5 )

Quotient

of

Pow

ers

and N

egative

Exponent

Pro

pert

ies

=

1 −

4 a

-5 b

0 c 5

Sim

plif

y.

=

1 −

4 ( 1

−

a 5 ) (1

) c 5

Negative

Exponent

and Z

ero

Exponent

Pro

pert

ies

=

c 5 −

4 a

5 S

implif

y.

Th

e so

luti

on i

s c 5 −

4 a

5 .

Exer

cise

sS

imp

lify

eac

h e

xpre

ssio

n. A

ssu

me

that

no

den

omin

ator

eq

ual

s ze

ro.

1.

2 2 −

2 -

3 25

or

32

2.

m

−

m -

4 m5

3.

p -

8 −

p

3 1 −

p 11

4.

b -4

−

b -5 b

5.

(-

x -1 y

) 0 −

4 w

-1 y

2 w

−

4 y 2

6.

( a 2 b

3 ) 2

−

(a b)

-2

a6 b

8

7.

x 4 y 0

−

x -2

x6

8.

(6 a

-1 b

) 2 −

( b

2 ) 4

36

−

a 2 b

6 9.

(3

rt ) 2 u

-4

−

r -1 t

2 u 7

9 r 3

−

u 11

10.

m -

3 t -

5 −

( m

2 t 3 )

-1

1 −

mt 2

11. (

4 m 2 n

2 −

8 m

-1 �

) 0 1

12.

( -2m

n 2 )

-3

−

4 m -

6 n 4

-

m 3

−

32 n

10

7-2

Exam

ple

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1212

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-2

Cha

pte

r 7

13

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceD

ivis

ion

Pro

pert

ies o

f E

xp

on

en

ts

Sim

pli

fy e

ach

exp

ress

ion

. Ass

um

e th

at n

o d

enom

inat

or e

qu

als

zero

.

1.

6 5 −

6 4 6

1 o

r 6

2.

9 12

−

9 8 94

or

6561

3.

x 4 −

x 2 x

2 4.

r 3 t

2 −

r 3 t

4 1 −

t 2

5.

m

−

m 3

1 −

m 2

6.

9 d 7

−

3 d 6 3

d

7.

12 n

5 −

36

n

n 4

−

3 8.

w

4 x 3

−

w 4 x

x

2

9.

a 3 b

5 −

ab

2 a

2 b3

10.

m 7 p

2 −

m

3 p 2 m

4

11.

-21

w 5 x

2 −

7 w

4 x 5

-

3w

−

x 3

12.

32 x 3 y

2 z 5

−

-8x

y z 2

-4x

2 yz

3

13.

( 4 p 7

−

7 r 2 ) 2

16 p

14

−

49 r 4

14. 4

-4

1 −

44 o

r

1 −

256

15. 8

-2

1 −

82 o

r 1 −

64

16.

( 5 −

3 ) -

2 9

−

25

17.

( 9 −

11 ) -

1 11

−

9 18

. h

3 −

h

-6 h

9

19. k

0 (k4 )

(k-

6 )

1 −

k 2

20. k

-1 (

ℓ-6 )

(m3 )

m

3 −

k �

6

21.

f -7

−

f 4

1 −

f 11

22.

( 16 p

5 w 2

−

2 p 3 w

3 ) 0 1

23.

f -5 g

4 −

h

-2

g 4 h

2 −

f 5

24.

15 x 6 y

-9

−

5xy -

11

3x

5 y2

25.

-15

t 0 u -

1 −

5 u

3 -

3 −

u 4

26.

48 x 6 y

7 z 5

−

-6x

y 5 z 6

-

8 x 5 y

2 −

z

7-2

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1312

/20/

10

6:38

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 7

14

Gle

ncoe

Alg

ebra

1

Prac

tice

Div

isio

n P

rop

ert

ies o

f E

xp

on

en

ts

Sim

pli

fy e

ach

exp

ress

ion

. Ass

um

e th

at n

o d

enom

inat

or e

qu

als

zero

.

1.

8 8 −

8 4 8

4 o

r 40

96

2. a

4 b 6

−

ab 3

a3 b

3 3.

xy

2 −

xy

y

4.

m 5 n

p −

m

4 p

mn

5.

5 c

2 d 3

−

-4 c

2 d -

5 d 2

−

4 6.

8 y

7 z 6

−

4 y 6 z

5 2yz

7. ( 4f

3 g

−

3 h 6 ) 3

64 f 9 g

3 −

27 h

18

8. ( 6 w

5 −

7 p

6 r 3 ) 2

36 w

10

−

49 p

12 r 6

9.

-4 x

2 −

24

x 5 -

1 −

6 x 3

10. x

3 (y-

5 )(x

-8 )

1 −

x 5 y

5 11

. p(q

-2 )

(r-

3 )

p

−

q 2 r 3

12. 1

2-2

1

−

144

13. (

3 −

7 ) -

2 49

−

9 14

. ( 4 −

3 ) -

4 81

−

256

15.

22 r 3 s

2 −

11

r 2 s -

3 2rs

5

16.

-15

w 0 u

-1

−

5 u 3

-

3 −

u 4

17.

8 c 3 d

2 f 4

−

4 c -

1 d 2 f

-3 2

c4 f

7 18

. ( x -

3 y 5

−

4 -3 ) 0 1

19.

6 f -

2 g 3 h

5 −

54

f -2 g

-5 h

3 g

8 h 2

−

9

20.

-12

t -1 u

5 x -

4

−

2 t -

3 u x 5

-

6 t 2 u

4 −

x 9

21

.

r 4 −

(3

r ) 3

r −

27

22.

m -

2 n -

5 −

( m

4 n 3 )

-1

m 2

−

n 2

23.

( j -

1 k 3 )

-4

−

j 3 k 3

j −

k 15

24

. (2

a -

2 b ) -

3 −

5 a

2 b 4

a 4

−

40 b

7

25. ( q -

1 r 3

−

q r -

2 ) -5 q

10

−

r 25

26. (

7 c -

3 d 3

−

c 5 d h

-4 ) -

1 c

8 −

7 d 2 h

4 27

. ( 2 x 3 y

2 z

−

3 x 4 y

z -2 ) -

2 9 x

2 −

4 y 2 z

6

28. B

IOLO

GY

A l

ab t

ech

nic

ian

dra

ws

a sa

mpl

e of

blo

od. A

cu

bic

mil

lim

eter

of

the

bloo

d co

nta

ins

223

wh

ite

bloo

d ce

lls

and

225

red

bloo

d ce

lls.

Wh

at i

s th

e ra

tio

of w

hit

e bl

ood

cell

s to

red

blo

od c

ells

?

1 −

484

29. C

OU

NTI

NG

Th

e n

um

ber

of t

hre

e-le

tter

“w

ords

” th

at c

an b

e fo

rmed

wit

h t

he

En

glis

h

alph

abet

is

263 .

Th

e n

um

ber

of f

ive-

lett

er “

wor

ds”

that

can

be

form

ed i

s 26

5 . H

ow m

any

tim

es m

ore

five

-let

ter

“wor

ds”

can

be

form

ed t

han

th

ree-

lett

er “

wor

ds”?

676

7-2

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1412

/24/

10

8:19

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-2

Cha

pte

r 7

15

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Div

isio

n P

rop

ert

ies o

f E

xp

on

en

ts

1. C

HEM

ISTR

Y T

he

nu

cleu

s of

a c

erta

in

atom

is

10-

13 c

enti

met

ers

acro

ss. I

f th

e n

ucl

eus

of a

dif

fere

nt

atom

is

10-

11 c

enti

met

ers

acro

ss, h

ow m

any

tim

es a

s la

rge

is i

t as

th

e fi

rst

atom

?

2. S

PAC

E T

he

Moo

n i

s ap

prox

imat

ely

254

kilo

met

ers

away

fro

m E

arth

on

av

erag

e. T

he

Oly

mpu

s M

ons

volc

ano

on

Mar

s st

ands

25

kilo

met

ers

hig

h. H

ow

man

y O

lym

pus

Mon

s vo

lcan

oes,

sta

cked

on

top

of

one

anot

her

, wou

ld f

it b

etw

een

th

e su

rfac

e of

th

e E

arth

an

d th

e M

oon

?

3. E

-MA

IL S

pam

(al

so k

now

n a

s ju

nk

e-m

ail)

con

sist

s of

ide

nti

cal

mes

sage

s se

nt

to t

hou

san

ds o

f e-

mai

l u

sers

. Peo

ple

ofte

n o

btai

n a

nti

-spa

m s

oftw

are

to f

ilte

r ou

t th

e ju

nk

e-m

ail

mes

sage

s th

ey

rece

ive.

Su

ppos

e Y

von

ne’

s an

ti-s

pam

so

ftw

are

filt

ered

ou

t 10

2 e-

mai

ls, a

nd

she

rece

ived

104

e-m

ails

las

t ye

ar. W

hat

fr

acti

on o

f h

er e

-mai

ls w

ere

filt

ered

ou

t?

Wri

te y

our

answ

er a

s a

mon

omia

l.

4. M

ETR

IC M

EASU

REM

ENT

Con

side

r a

dust

mit

e th

at m

easu

res

10-

3 m

illi

met

ers

in l

engt

h a

nd

a ca

terp

illa

r th

at m

easu

res

10 c

enti

met

ers

lon

g. H

ow m

any

tim

es a

s lo

ng

as t

he

mit

e is

th

e ca

terp

illa

r?

5. C

OM

PUTE

RS

In 1

995,

sta

nda

rd c

apac

ity

for

a pe

rson

al c

ompu

ter

har

d dr

ive

was

40

meg

abyt

es (

MB

). In

201

0, a

sta

nda

rd

har

d dr

ive

capa

city

was

500

gig

abyt

es

(GB

or

Gig

). R

efer

to

the

tabl

e be

low

.

Mem

ory

Cap

acit

y A

pp

roxi

mat

e C

onv

ersi

on

s

8 b

its =

1 b

yte

10

3 b

yte

s =

1 k

ilobyte

10

3 k

ilobyte

s =

1 m

ega

byte

(m

eg

)

10

3 m

ega

byte

s =

1 g

iga

byte

(g

ig)

10

3 g

iga

byte

s =

1 t

era

byte

10

3 t

era

byte

s =

1 p

eta

byte

a. T

he

new

er h

ard

driv

es h

ave

abou

t h

ow m

any

tim

es t

he

capa

city

of

the

1995

dri

ves?

b.

Pre

dict

th

e h

ard

driv

e ca

paci

ty i

n t

he

year

202

5 if

th

is r

ate

of g

row

th

con

tin

ues

.

c. O

ne

kilo

byte

of

mem

ory

is w

hat

fr

acti

on o

f on

e te

raby

te?

7-2

100

12,5

00

253 =

15,

625

10-

2

6.25

pet

abyt

es

1 −

109

= 1

0 -

9

104 =

10,

00

0

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1512

/20/

10

6:38

PM


A01_A14_ALG1_A_CRM_C07_AN_660281.indd A6A01_A14_ALG1_A_CRM_C07_AN_660281.indd A6 12/24/10 9:29 AM12/24/10 9:29 AM

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

16

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Patt

ern

s w

ith

Po

wers

Use

you

r ca

lcu

lato

r, if

nec

essa

ry, t

o co

mpl

ete

each

pat

tern

.

a. 2

10 =

b.

510 =

c.

410 =

2

9 =

59

=

49 =

2

8 =

58

=

48 =

2

7 =

57

=

47 =

2

6 =

56

=

46 =

2

5 =

55

=

45 =

2

4 =

54

=

44 =

2

3 =

53

=

43 =

2

2 =

52

=

42 =

2

1 =

51

=

41 =

Stu

dy

the

pat

tern

s fo

r a,

b, a

nd

c a

bov

e. T

hen

an

swer

th

e q

ues

tion

s.

1. D

escr

ibe

the

patt

ern

of

the

expo

nen

ts f

rom

th

e to

p of

eac

h c

olu

mn

to

the

bott

om.

Th

e ex

po

nen

ts d

ecre

ase

by o

ne

fro

m e

ach

ro

w t

o t

he

on

e b

elo

w.

2. D

escr

ibe

the

patt

e rn

of

the

pow

ers

from

th

e to

p of

th

e co

lum

n t

o th

e bo

ttom

. To

get

ea

ch p

ow

er, d

ivid

e th

e p

ow

er o

n t

he

row

ab

ove

by t

he

bas

e (2

, 5, o

r 4)

.

3. W

hat

wou

ld y

ou e

xpec

t th

e fo

llow

ing

pow

ers

to b

e?

20

1 50

1 40

1

4. R

efer

to

Exe

rcis

e 3.

Wri

te a

ru

le. T

est

it o

n p

atte

rns

that

you

obt

ain

usi

ng

22, 2

5, a

nd

24

as b

ases

.

Stu

dy

the

pat

tern

bel

ow. T

hen

an

swer

th

e q

ues

tion

s.

03 =

0

02 =

0

01 =

0

00 =

?

0-

1 do

es n

ot e

xist

. 0-

2 do

es n

ot e

xist

. 0-

3 do

es n

ot e

xist

.

5. W

hy

do 0

-1 ,

0-2 ,

and

0-3

not

exi

st?

Neg

ativ

e ex

po

nen

ts a

re n

ot

defi

ned

un

less

th

e b

ase

is n

on

zero

.

6. B

ased

upo

n t

he

patt

ern

, can

you

det

erm

ine

wh

eth

er 0

0 ex

ists

?

No

, sin

ce t

he

pat

tern

0n =

0 b

reak

s d

ow

n f

or

n <

1.

7. T

he

sym

bol

00 is

cal

led

an i

nd

eter

min

ate,

wh

ich

mea

ns

that

it

has

no

un

iqu

e va

lue.

T

hu

s it

doe

s n

ot e

xist

as

a u

niq

ue

real

nu

mbe

r. W

hy

do y

ou t

hin

k th

at 0

0 ca

nn

ot e

qual

1?

A

nsw

ers

will

var

y. O

ne

answ

er is

th

at if

00

= 1

, th

en 1

= 1 −

1 = 1

0 −

0 0 =

( 1 −

0 ) 0 ,

w

hic

h is

a f

alse

res

ult

, sin

ce d

ivis

ion

by

zero

is n

ot

allo

wed

. Th

us,

00

can

no

t eq

ual

1.

45

216

254

6412

58

256

625

1610

2431

2532

4096

15,6

2564

16,3

8478

,125

128

65,5

3639

0,62

525

626

2,14

41,

953,

125

512

1,04

8,57

69,

765,

625

7-2

1024

Any

no

nze

ro n

um

ber

to

th

e ze

ro p

ow

er e

qu

als

on

e.

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1612

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-3

Cha

pte

r 7

17

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Rati

on

al

Exp

on

en

ts

Rat

ion

al E

xpo

nen

ts F

or a

ny

real

nu

mbe

rs a

an

d b

and

any

posi

tive

in

tege

r n

, if

an =

b, t

hen

a i

s an

nth

roo

t of

b. R

atio

nal

exp

onen

ts c

an b

e u

sed

to r

epre

sen

t n

th r

oots

.

Squ

are

Roo

t b

1 −

2 =

√ �

b

Cu

be R

oot

b 1 −

3 =

3 √ �

b

nth

Roo

t b

1 −

n =

n √ �

b

7-3

W

rite

(6x

y ) 1 −

2 in

rad

ical

fo

rm.

(6xy

) 1 −

2 =

√ �

�

6xy

Defi nitio

n o

f b

1

−

2

S

imp

lify

625

1 −

4 .

625 1

−

4 = 4 √

��

625

b 1

−

n =

n √

�

b

=

4 √ �

��

��

5 �

5 �

5 �

5

625 =

5 4

=

5

Sim

plif

y.

Exam

ple

1Ex

amp

le 2

Exer

cise

sW

rite

eac

h e

xpre

ssio

n i

n r

adic

al f

orm

, or

wri

te e

ach

rad

ical

in

exp

onen

tial

for

m.

1. 1

4 1 −

2 √

��

14

2.

5 x 1

−

2 5 √

�

x

3. 1

7 y 1 −

2 1

7 √

�

y

4. 1

2 1 −

2 √

��

12

5.

19 a

b 1 −

2 1

9a √

�

b

6.

√ �

�

17 1

7 1 −

2

7.

√ �

�

12n

(12

n ) 1

−

2 8.

√ �

�

18

b (1

8b ) 1

−

2 9.

√ �

�

37 3

7 1 −

2

Sim

pli

fy.

10.

3 √ �

�

343

7

11.

5 √ �

�

1024

4

12. 5

1 2 1 −

3 8

13.

4 √ �

�

2401

7

14.

6 √

��

64

2

15. 2

4 3 1 −

5 3

16.

3 √ �

�

1331

11

17.

4 √

��

�

6561

9

18. 4

09 6 1

−

4 8

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1712

/20/

10

6:38

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 7

18

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Rati

on

al

Exp

on

en

ts

7-3

Solv

e Ex

po

nen

tial

Eq

uat

ion

s In

an

exp

onen

tial

eq

uat

ion

, var

iabl

es o

ccu

r as

ex

pon

ents

. Use

th

e P

ower

Pro

pert

y of

Equ

alit

y an

d th

e ot

her

pro

pert

ies

of e

xpon

ents

to

solv

e ex

pon

enti

al e

quat

ion

s.

S

olve

102

4 x

− 1 =

4.

10

2 4 x

− 1 =

4

Ori

gin

al equation

( 4

5 ) x

− 1 =

4

Rew

rite

1024 a

s 4

5 .

4 5x

− 5 =

4 1

Pow

er

of

a P

ow

er,

Dis

trib

utive

Pro

pert

y

5x

− 5

= 1

P

ow

er

Pro

pert

y o

f E

qualit

y

5x

= 6

A

dd 5

to e

ach s

ide.

x =

6 −

5

Div

ide e

ach s

ide b

y 5

.

Exer

cise

sS

olve

eac

h e

qu

atio

n.

1. 2

x = 1

28 7

2.

3 3x

+ 1 =

81

1 3.

4 x

- 3 =

32

11

−

2

4. 5

x = 1

5,62

5 6

5. 6

3x +

2 =

216

1 −

3 6.

4 5x

- 3 =

16

1

7. 8

x = 4

096

4 8.

9 3x

+ 3 =

656

1 1 −

3 9.

1 1 x

- 1 =

133

1 4

10. 3

x = 6

561

8 11

. 2 5x

+ 4 =

512

1

12. 7

x -

2 =

343

5

13. 8

x = 2

62,1

44 6

14

. 5 5x

= 3

125

1 15

. 9 2x

- 6 =

656

1 5

16. 7

x = 2

401

4 17

. 7 3x

= 1

17,6

49 2

18

. 6 2x

- 7 =

777

6 6

19. 9

x = 7

29 3

20

. 8 3x

+ 1 =

409

6 1

21. 1

3 3x -

8 =

28,

561

4

Exam

ple

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1812

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-3

Cha

pte

r 7

19

Gle

ncoe

Alg

ebra

1

7-3

Skill

s Pr

acti

ceR

ati

on

al

Exp

on

en

tsW

rite

eac

h e

xpre

ssio

n i

n r

adic

al f

orm

, or

wri

te e

ach

rad

ical

in

exp

onen

tial

for

m.

1. (

8 x) 1

−

2 √ �

�

8x

2.

6z 1

−

2 6 √

�

z

3. √

��

19 1

9 1 −

2

4.

√ �

�

11 1

1 1 −

2 5.

19 x

1 −

2 1

9 √

�

x

6. √

��

34 3

4 1 −

2

7.

√ �

�

27g

(27g

) 1 −

2 8.

33 g

h 1 −

2 3

3g √

�

h

9.

√ �

��

13ab

c (1

3ab

c ) 1

−

2

Sim

pli

fy.

10. (

1 −

16 ) 1

−

4 1 −

2 11

. 5 √

��

3125

5

12. 7

2 9 1 −

3 9

13. (

1 −

32 ) 1

−

5 1 −

2 14

. 6 √

��

4096

4

15. 1

02 4 1

−

5 4

16. (

16

−

625 ) 1

−

4 2 −

5 17

. 6 √

��

�

15,6

25 5

18

. 117

,6 49

1 −

6 7

Sol

ve e

ach

eq

uat

ion

.

19. 2

x = 5

12 9

20

. 3 x =

656

1 8

21. 6

x = 4

6,65

6 6

22. 5

x = 1

25 3

23

. 3 x

- 3 =

243

8

24. 4

x -

1 =

102

4 6

25. 6

x -

1 =

129

6 5

26. 2

4x +

3 =

204

8 2

27. 3

3x +

3 =

656

1 5 −

3

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

1912

/20/

10

6:38

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

20

Gle

ncoe

Alg

ebra

1

7-3

Prac

tice

Rati

on

al

Exp

on

en

tsW

rite

eac

h e

xpre

ssio

n i

n r

adic

al f

orm

, or

wri

te e

ach

rad

ical

in

exp

onen

tial

for

m.

1.

√ �

�

13 1

3 1 −

2 2.

√ �

�

37 3

7 1 −

2 3.

√ �

�

17x

(17 x

) 1 −

2

4. (

7ab )

1 −

2 √ �

�

7a

b

5. 2

1z 1 −

2 2

1 √

�

z

6. 1

3(a b

) 1 −

2 1

3 √

��

ab

Sim

pli

fy.

7. (

1 −

81 ) 1

−

4 1 −

3 8.

5 √ �

�

1024

4

9. 5

1 2 1 −

3 8

10. (

32

−

1024

) 1 −

5 1

−

2 11

. 4 √

��

1296

6

12. 3

12 5 1

−

5 5

Sol

ve e

ach

eq

uat

ion

.

13. 3

x = 7

29 6

14

. 4 x =

409

6 6

15. 5

x = 1

5,62

5 6

16. 6

x +

3 =

777

6 2

17. 3

x -

3 =

218

7 10

18

. 4 3x

+ 4 =

16,

384

1

19. W

ATE

R T

he

flow

of

wat

er F

in

cu

bic

feet

per

sec

ond

over

a w

ier,

a sm

all

over

flow

dam

,

can

be

repr

esen

ted

by F

= 1

.26 H

3 −

2 ,

wh

ere

H i

s th

e h

eigh

t of

th

e w

ater

in

met

ers

abov

e th

e cr

est

of t

he

wie

r. F

ind

the

hei

ght

of t

he

wat

er i

f th

e fl

ow o

f th

e w

ater

is

10.0

8 cu

bic

feet

per

sec

ond.

4 f

t

001_

020_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2012

/20/

10

6:38

PM

Lesson X-3


NA

ME

DAT

E

P

ER

IOD

Lesson 7-3

Cha

pte

r 7

21

Gle

ncoe

Alg

ebra

1

7-3

Wor

d Pr

oble

m P

ract

ice

Rati

on

al

Exp

on

en

ts

1. V

ELO

CIT

Y T

he

velo

city

v i

n f

eet

per

seco

nd

of a

fre

ely

fall

ing

obje

ct t

hat

has

fa

llen

h f

eet

can

be

repr

esen

ted

by

v=

8 h

1 −

2 . F

ind

the

dist

ance

th

at a

n o

bjec

t h

as f

alle

n i

f it

s ve

loci

ty i

s 96

fee

t pe

r se

con

d. 1

44 f

t

2. E

LEC

TRIC

ITY

Th

e re

lati

onsh

ip o

f th

e cu

rren

t, t

he

pow

er, a

nd

the

resi

stan

ce

in a

n a

ppli

ance

can

be

mod

eled

by

I R 1 −

2 =

√ �

P

, w

her

e I

is t

he

curr

ent

in

ampe

res,

P i

s th

e po

wer

in

wat

ts, a

nd

R

is t

he

resi

stan

ce i

n o

hm

s. F

ind

the

pow

er

that

an

app

lian

ce i

s u

sin

g if

th

e cu

rren

t is

2.5

am

ps a

nd

the

resi

stan

ce i

s 16

oh

ms.

10

0 w

atts

3. G

EOM

ETRY

Th

e su

rfac

e ar

ea T

of

a cu

be i

n s

quar

e in

ches

can

be

dete

rmin

ed

by T

= 6

V 2 −

3 ,

wh

ere

V i

s th

e vo

lum

e of

the

cube

in

cu

bic

inch

es.

a. F

ind

the

surf

ace

area

of

a cu

be t

hat

h

as a

vol

um

e of

409

6 cu

bic

inch

es.

15

36 i n

2

b.

Fin

d th

e vo

lum

e of

a c

ube

th

at h

as a

su

rfac

e ar

ea o

f 96

squ

are

inch

es.

64

i n 3

4. P

LAN

ETS

Th

e av

erag

e di

stan

ce d

in

as

tron

omic

al u

nit

s th

at a

pla

net

is

from

the

Su

n c

an b

e m

odel

ed b

y d

=t 2

−

3 , w

her

e t

is t

he

nu

mbe

r of

Ear

th y

ears

th

at i

t ta

kes

for

the

plan

et t

o or

bit

the

Su

n.

a. F

ind

the

aver

age

dist

ance

a p

lan

et i

s fr

om t

he

Su

n i

f th

e pl

anet

has

an

or

bit

of 2

7 E

arth

yea

rs.

9 as

tro

no

mic

al u

nit

s

b.

Fin

d th

e ti

me

that

it

wou

ld t

ake

a pl

anet

to

orbi

t th

e S

un

if

its

aver

age

dist

ance

fro

m t

he

Su

n i

s 4

astr

onom

ical

un

its.

8

Ear

th y

ears

5. B

IOLO

GY

Th

e re

lati

onsh

ip b

etw

een

th

e m

ass

m i

n k

ilog

ram

s of

an

org

anis

m a

nd

its

met

abol

ism

P i

n C

alor

ies

per

day

can

be r

epre

sen

ted

by P

= 7

3.3

4 √ �

�

m 3

. Fin

d th

e m

ass

of a

n o

rgan

ism

th

at h

as a

m

etab

olis

m o

f 58

6.4

Cal

orie

s pe

r da

y.

16 k

g

6. M

AN

UFA

CTU

RIN

G T

he

prof

it P

of

a co

mpa

ny

in t

hou

san

ds o

f do

llar

s ca

n b

e

mod

eled

by

P =

12.

75 5 √

�

c 2 , w

her

e c

is t

he

nu

mbe

r of

cu

stom

ers

in h

un

dred

s. I

f th

e pr

ofit

of

the

com

pan

y is

$51

,000

, how

m

any

cust

omer

s do

th

ey h

ave?

320

0 cu

sto

mer

s

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2112

/20/

10

6:38

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 7

22

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

7-3

Co

un

tere

xam

ple

sS

ome

stat

emen

ts i

n m

ath

emat

ics

can

be

prov

en f

alse

by

cou

nte

rexa

mp

les.

C

onsi

der

the

foll

owin

g st

atem

ent.

For

an

y n

um

bers

a a

nd

b, a

- b

= b

- a

.Yo

u c

an p

rove

th

at t

his

sta

tem

ent

is f

alse

in

gen

eral

if

you

can

fin

d on

e ex

ampl

e fo

r w

hic

h

the

stat

emen

t is

fal

se.

Let

a =

7 a

nd

b =

3. S

ubs

titu

te t

hes

e va

lues

in

th

e eq

uat

ion

abo

ve.

7 -

3 �

3 -

7

4 ≠

-4

In g

ener

al, f

or a

ny

nu

mbe

rs a

an

d b,

th

e st

atem

ent

a -

b =

b -

a i

s fa

lse.

You

can

mak

e th

e eq

uiv

alen

t ve

rbal

sta

tem

ent:

su

btra

ctio

n i

s n

ot a

com

mu

tati

ve o

pera

tion

.

In e

ach

of

the

foll

owin

g ex

erci

ses

a, b

, an

d c

are

an

y n

um

ber

s. P

rove

th

at t

he

stat

emen

t is

fal

se b

y co

un

tere

xam

ple

.

1. a

- (

b -

c)

� (

a -

b)

- c

2.

a ÷

(b

÷ c

) �

(a

÷ b

) ÷

c

3. a

÷ b

� b

÷ a

4.

a ÷

(b

+ c

) �

(a

÷ b

) +

(a

÷ c

)

5. a

1 −

2 +

a 1 −

2 � a

1 6.

a c �

b d �

(ab

) c +

d

7. W

rite

th

e ve

rbal

equ

ival

ents

for

Exe

rcis

es 1

, 2, a

nd

3.

8. F

or t

he

Dis

trib

uti

ve P

rope

rty

a(b

+ c

) =

ab

+ a

c, i

t is

sai

d th

at m

ult

ipli

cati

on

dist

ribu

tes

over

add

itio

n. E

xerc

ise

4 pr

oves

th

at s

ome

oper

atio

ns

do n

ot d

istr

ibu

te. W

rite

a

stat

emen

t fo

r th

e ex

erci

se t

hat

in

dica

tes

this

.

Sam

ple

an

swer

s ar

e g

iven

.

6

- (

4 -

2)

� (

6 -

4)

- 2

6 ÷

(4

÷ 2

) �

(6

÷ 4

) ÷

2

6

- 2

� 2

- 2

6 −

2 �

1.5

−

2

4

≠ 0

3

≠ 0

.75

6

÷ 4

� 4

÷ 6

6 ÷

(4

+ 2

) �

(6

÷ 4

) +

(6

÷ 2

)

3 −

2 ≠

2 −

3

6 ÷

6 �

1.5

+ 3

1

≠ 4

.5

9

1 −

2 + 9

1 −

2 � 9

1

2 3

� 5

2 �

[(2

)(5)

] (3 +

2)

3

+ 3

� 9

8 � 2

5 �

10

5

6

≠ 9

200

≠ 1

00,

00

0

1

. Su

btr

acti

on

is n

ot

an a

sso

ciat

ive

op

erat

ion

.

2. D

ivis

ion

is n

ot

an a

sso

ciat

ive

op

erat

ion

.

3. D

ivis

ion

is n

ot

a co

mm

uta

tive

op

erat

ion

.

4

. Div

isio

n d

oes

no

t d

istr

ibu

te o

ver

add

itio

n.

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2212

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-4

Cha

pte

r 7

23

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

Scie

nti

fi c N

ota

tio

n

7-4

Scie

nti

fic

No

tati

on

Ver

y la

rge

and

very

sm

all

nu

mbe

rs a

re o

ften

bes

t re

pres

ente

d u

sin

g a

met

hod

kn

own

as

scie

nti

fic

not

atio

n. N

um

bers

wri

tten

in

sci

enti

fic

not

atio

n t

ake

the

form

a ×

10

n, w

her

e 1

≤ a

< 1

0 a

nd

n i

s an

in

tege

r. A

ny

nu

mbe

r ca

n b

e w

ritt

en i

n

scie

nti

fic

not

atio

n.

E

xpre

ss 3

4,02

0,00

0,00

0 in

sc

ien

tifi

c n

otat

ion

.S

tep

1

Mov

e th

e de

cim

al p

oin

t u

nti

l it

is

to t

he

righ

t of

th

e fi

rst

non

zero

dig

it. T

he

resu

lt i

s a

real

nu

mbe

r a.

Her

e, a

= 3

.402

.

Ste

p 2

N

ote

the

nu

mbe

r of

pla

ces

n a

nd

the

dire

ctio

n t

hat

you

mov

ed t

he

deci

mal

po

int.

Th

e de

cim

al p

oin

t m

oved

10

plac

es t

o th

e le

ft, s

o n

= 1

0.

Ste

p 3

B

ecau

se t

he

deci

mal

mov

ed t

o th

e le

ft, w

rite

th

e n

um

ber

as a

× 1

0n.

34,0

20,0

00,0

00 =

3.4

0200

0000

0 ×

1010

Ste

p 4

Rem

ove

the

extr

a ze

ros.

3.4

02 ×

1010

E

xpre

ss 4

.11

× 1

0-6 in

st

and

ard

not

atio

n.

Ste

p 1

T

he

expo

nen

t is

–6,

so

n =

–6.

Ste

p 2

B

ecau

se n

< 0

, mov

e th

e de

cim

al

poin

t 6

plac

es t

o th

e le

ft.

4.

11 ×

10

-6 ⇒

.0

00

00

41

1

Ste

p 3

4.

11 ×

10

-6 ⇒

0.0

00

00

41

1R

ew

rite

; in

sert

a 0

befo

re t

he d

ecim

al poin

t.

Exam

ple

1Ex

amp

le 2

Exer

cise

sE

xpre

ss e

ach

nu

mb

er i

n s

cien

tifi

c n

otat

ion

.

1. 5

,100

,000

2.

80,

300,

000,

000

3. 1

4,25

0,00

0

5.1

× 1

06 8

.03

× 1

010

1.4

25 ×

107

4. 6

8,07

0,00

0,00

0,00

0 5.

14,

000

6. 9

01,0

50,0

00,0

00

6.8

07 ×

1013

1

.4 ×

104

9.0

105

× 1

011

7. 0

.004

9 8.

0.0

0030

1 9.

0.0

0000

0051

9

4.9

× 1

0-3

3.0

1 ×

10-

4 5

.19

× 1

0-8

10. 0

.000

0001

85

11. 0

.002

002

12. 0

.000

0077

1

1.8

5 ×

10-

7 2

.002

× 1

0-3

7.7

1 ×

10-

6

Exp

ress

eac

h n

um

ber

in

sta

nd

ard

for

m.

13. 4

.91

× 1

04 14

. 3.2

× 1

0-5

15. 6

.03

× 1

08

4

9,10

0 0

.00

003

2 6

03,0

00,

00

0

16. 2

.001

× 1

0-6

17. 1

.000

24 ×

1010

18

. 5 ×

105

0

.00

00

020

01

10,

002

,40

0,0

00

50

0,0

00

19. 9

.09

× 1

0-5

20. 3

.5 ×

10-

2 21

. 1.7

087

× 1

07

0

.00

009

09

0.0

35

17,

087,

00

0

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2312

/20/

10

6:38

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

24

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Scie

nti

fi c N

ota

tio

n

7-4

Pro

du

cts

and

Qu

oti

ents

in S

cien

tifi

c N

ota

tio

n Y

ou c

an u

se s

cien

tifi

c n

otat

ion

to

sim

plif

y m

ult

iply

ing

and

divi

din

g ve

ry l

arge

an

d ve

ry s

mal

l n

um

bers

.

E

valu

ate

(9.2

× 1

0-3 )

×(4

× 1

08 ). E

xpre

ss t

he

resu

lt i

n b

oth

sc

ien

tifi

c n

otat

ion

an

d s

tan

dar

d f

orm

.

(9.2

× 1

0-3 )

(4 ×

108 )

O

rigin

al expre

ssio

n

= (

9.2

× 4

)(10

-3

× 1

08 )

Com

muta

tive

and

Associa

tive

Pro

pert

ies

=

36.

8 ×

105

Pro

duct

of

Pow

ers

=

(3.

68 ×

101 )

× 1

05 36.8

= 3

.68 ×

10

=

3.6

8 ×

106

Pro

duct

of

Pow

ers

=

3,6

80,0

00

Sta

ndard

Form

E

valu

ate

( 2.7

6 ×

1

0 7

) −

( 6.9

× 1

0 5

) .

Exp

ress

th

e re

sult

in

bot

h s

cien

tifi

c n

otat

ion

an

d s

tan

dar

d f

orm

. ( 2

.76

× 1

0 7 )

−

( 6.9

× 1

0 5 )

=

( 2.76

−

6.9 ) (

10 7

−

10 5 )

Product

rule

for

fractions

= 0

.4 ×

102

Quotient

of

Pow

ers

= 4

.0 ×

10-

1 ×

102

0.4

= 4

.0 ×

10

-1

= 4

.0 ×

101

Pro

duct

of

Pow

ers

= 4

0 S

tandard

form

Exam

ple

1Ex

amp

le 2

Exer

cise

sE

valu

ate

each

pro

du

ct. E

xpre

ss t

he

resu

lts

in b

oth

sci

enti

fic

not

atio

n a

nd

st

and

ard

for

m.

1. (

3.4

× 1

03 )(5

× 1

04 )

2. (2

.8 ×

10-

4 )(1

.9 ×

107 )

1

.7 ×

108 ;

170

,00

0,0

00

5.3

2 ×

103 ;

532

0

3. (

6.7

× 1

0-7 )

(3 ×

103 )

4.

(8.1

× 1

05 )(2

.3 ×

10-

3 )

2.0

1 ×

10-

3 ; 0

.002

01

1.8

63 ×

103 ;

186

3

5. (

1.2

× 1

0 -4 )

2 6.

(5.9

× 1

0 5 ) 2

1

.44

× 1

0 -

8 ; 0

.00

00

00

0144

3

.481

× 1

0 11

; 34

8,10

0,0

00,

00

0

Eva

luat

e ea

ch q

uot

ien

t. E

xpre

ss t

he

resu

lts

in b

oth

sci

enti

fic

not

atio

n a

nd

st

and

ard

for

m.

7.

( 4.9

× 1

0 -

3 )

−

( 2.5

× 1

0 -

4 )

8.

5.8

× 1

0 4

−

5 ×

10

-2

1

.96

× 1

01 ; 1

9.6

1.1

6 ×

106 ;

1,1

60,0

00

9.

( 1.6

× 1

0 5 )

−

( 4 ×

1 0

-4 )

10

. 8.

6 ×

10 6

−

1.6

× 1

0 -3

4

.0 ×

108 ;

40

0,0

00,

00

0 5

.375

× 1

09 ; 5

,375

,00

0,0

00

11.

( 4.2

× 1

0 -

2 )

−

( 6 ×

1 0

-7 )

12

. 8.

1 ×

10 5

−

2.7

× 1

0 4

7

× 1

04 ; 7

0,0

00

3 ×

101 ;

30

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2412

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-4

Cha

pte

r 7

25

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceS

cie

nti

fi c N

ota

tio

n

7-4

Exp

ress

eac

h n

um

ber

in

sci

enti

fic

not

atio

n.

1. 3

,400

,000

,000

2.

0.0

0000

0312

3

.4 ×

109

3.1

2 ×

10-

7

3. 2

,091

,000

4.

980

,200

,000

,000

,000

2

.091

× 1

06 9

.802

× 1

014

5. 0

.000

0000

0008

6.

0.0

0142

8

× 1

0-11

1

.42

× 1

0-3

Exp

ress

eac

h n

um

ber

in

sta

nd

ard

for

m.

7. 2

.1 ×

105

8. 8

.023

× 1

0-7

2

10,0

00

0.0

00

00

0802

3

9. 3

.63

× 1

0-6

10. 7

.15

× 1

08

0

.00

00

0363

7

15,0

00,

00

0

11. 1

.86

× 1

0-4

12. 4

.9 ×

105

0

.00

0186

4

90,0

00

Eva

luat

e ea

ch p

rod

uct

. Exp

ress

th

e re

sult

s in

bot

h s

cien

tifi

c n

otat

ion

an

d

stan

dar

d f

orm

.

13. (

6.1

× 1

05 )(2

× 1

05 )

14. (

4.4

× 1

06 )(1

.6 ×

10-

9 )

1.2

2 ×

1011

; 12

2,0

00,

00

0,0

00

7.0

4 ×

10-

3 ; 0

.007

04

15. (

8.8

× 1

08 )(3

.5 ×

10-

13)

16. (

1.35

× 1

08 )(7

.2 ×

10-

4 )

3.0

8 ×

10-

4 ; 0

.00

0308

9

.72

× 1

04 ; 9

7,20

0

17. (

2.2

× 1

0 -3 ) 2

18. (

3.4

× 1

0 2 ) 2

4

.84

× 1

0 -

6 ; 0

.00

00

0484

1

.156

× 1

0 5 ;

115

,60

0

Eva

luat

e ea

ch q

uot

ien

t. E

xpre

ss t

he

resu

lts

in b

oth

sci

enti

fic

not

atio

n a

nd

st

and

ard

for

m.

19. (9

.2 ×

10-

8 )

−

(2

× 1

0-6 )

20

. (4

.8 ×

104 )

−

(3 ×

10-

5 )

4

.6 ×

10-

2 ; 0

.046

1

.6 ×

109 ;

1,6

00,

00

0,0

00

21.

(1.1

61 ×

10-

9 )

−

(4.3

× 1

0-6 )

22.

( 4.6

25 ×

1 0 1

0 )

−

( 1.2

5 ×

10 4

)

2

.7 ×

10-

4 ; 0

.00

027

3.7

× 1

06 ; 3

,70

0,0

00

23.

(2.3

76 ×

10-

4 )

−

(7.2

× 1

0-8 )

24.

(8.7

4 ×

10-

3 )

−

(1

.9 ×

105 )

3

.3 ×

103 ;

3,3

00

4.6

× 1

0-8 ;

0.0

00

00

004

6

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2512

/20/

10

6:38

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 7

26

Gle

ncoe

Alg

ebra

1

Prac

tice

Scie

nti

fi c N

ota

tio

n

7-4

Exp

ress

eac

h n

um

ber

in

sci

enti

fic

not

atio

n.

1. 1

,900

,000

2.

0.0

0070

4

1.9

× 10

6

7.0

4 ×

10-

4

3. 5

0,04

0,00

0,00

0 4.

0.0

0000

0066

1

5.0

04 ×

1010

6

.61

× 1

0-8

Exp

ress

eac

h n

um

ber

in

sta

nd

ard

for

m.

5. 5

.3 ×

107

6. 1

.09

× 1

0-4

5

3,0

00,

00

0 0

.00

0109

7. 9

.13

× 1

03 8.

7.9

02 ×

10-

6

9

130

0.0

00

007

902

Eva

luat

e ea

ch p

rod

uct

. Exp

ress

th

e re

sult

s in

bot

h s

cien

tifi

c n

otat

ion

an

d

stan

dar

d f

orm

.

9. (

4.8

× 1

04 )(6

× 1

06 )

10. (

7.5

× 1

0-5 )

(3.2

× 1

07 )

2.8

8 ×

1011

; 28

8,0

00,

00

0,0

00

2.4

× 1

03 ; 2

400

11. (

2.06

× 1

04 )(5

.5 ×

10-

9 )

12. (

8.1

× 1

0-6 )

(1.9

6 ×

1011

)

1.1

33 ×

10-

4 ; 0

.00

0113

3 1

.587

6 ×

106 ;

1,5

87,6

00

13. (

7.2

× 1

0 -5 )

2 14

. (5.

29 ×

1 0 6 )

2

5.1

84 ×

1 0

-9 ;

0.0

00

00

00

0518

4 2.

7984

1 ×

10

13 ;

27,9

84,1

00,

00

0,0

00

Eva

luat

e ea

ch q

uot

ien

t. E

xpre

ss t

he

resu

lts

in b

oth

sci

enti

fic

not

atio

n a

nd

st

and

ard

for

m.

15.

(4.2

× 1

0 5 )

−

(3 ×

1 0 -

3 )

16.

(1.7

6 ×

1 0 -

11 )

−

(2.2

× 1

0 -5 )

1

.4 ×

108 ;

140

,00

0,0

00

8 ×

10-

7 ; 0

.00

00

008

17.

(7.0

5 ×

1 0 12

)

−

(9.4

× 1

0 7 )

18.

(2.0

4 ×

1 0 -

4 )

−

(3

.4 ×

1 0 5 )

7

.5 ×

104 ;

75,

00

0 6

× 1

0-10

; 0.

00

00

00

00

06

19. G

RA

VIT

ATI

ON

Iss

ac N

ewto

n’s

th

eory

of

un

iver

sal

grav

itat

ion

sta

tes

that

th

e eq

uat

ion

F

= G

m1m

2 −

r2

can

be

use

d to

cal

cula

te t

he

amou

nt

of g

ravi

tati

onal

for

ce i

n n

ewto

ns

betw

een

tw

o po

int

mas

ses

m1

and

m2

sepa

rate

d by

a d

ista

nce

r. G

is

a co

nst

ant

equ

al

to 6

.67

× 1

0-

11 N

m2 k

g–2 .

Th

e m

ass

of E

arth

m1

is e

qual

to

5.97

× 1

02

4 k

g, t

he

mas

s of

th

e M

oon

m2

is e

qual

to

7.36

× 1

022 k

g, a

nd

the

dist

ance

r b

etw

een

th

e tw

o is

384

,000

,000

m.

a. E

xpre

ss t

he

dist

ance

r i

n s

cien

tifi

c n

otat

ion

. 3.

84 ×

108

m

b.

Com

pute

th

e am

oun

t of

gra

vita

tion

al f

orce

bet

wee

n E

arth

an

d th

e M

oon

. Exp

ress

yo

ur

answ

er i

n s

cien

tifi

c n

otat

ion

. 1.

99 ×

1020

New

ton

s

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2612

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-4

Cha

pte

r 7

27

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Scie

nti

fi c N

ota

tio

n

7-4

1. P

LAN

ETS

Nep

tun

e’s

mea

n d

ista

nce

fro

m

the

Su

n i

s 4,

500,

000,

000

kilo

met

ers.

U

ran

us’

mea

n d

ista

nce

fro

m t

he

Su

n i

s 2,

870,

000,

000

kilo

met

ers.

Exp

ress

th

ese

dist

ance

s in

sci

enti

fic

not

atio

n.

N

eptu

ne:

4.5

× 1

09km

; U

ran

us:

2.

87 ×

109

km

2. P

ATH

OLO

GY

Th

e co

mm

on c

old

is

cau

sed

by t

he

rhin

ovir

us,

wh

ich

co

mm

only

mea

sure

s 2

× 1

0-

8 m

in

di

amet

er. T

he

E. c

oli

bact

eriu

m, w

hic

h

cau

ses

food

poi

son

ing,

com

mon

ly

mea

sure

s 3

× 1

0-

6 m

in

len

gth

. Exp

ress

th

ese

mea

sure

men

ts i

n s

tan

dard

for

m.

R

hin

ovir

us:

0.0

00

00

002

m;

E. c

oli:

0.

00

00

03 m

3. C

OM

MER

CIA

LS A

30-

seco

nd

com

mer

cial

air

ed d

uri

ng

the

2007

Su

per

Bow

l co

st $

2,60

0,00

0. A

30-

seco

nd

com

mer

cial

air

ed d

uri

ng

the

1967

Su

per

Bow

l co

st $

40,0

00. E

xpre

ss t

hes

e va

lues

in

sci

enti

fic

not

atio

n. H

ow m

any

tim

es

mor

e ex

pen

sive

was

it

to a

ir a

n

adve

rtis

emen

t du

rin

g th

e 20

07 S

upe

r B

owl

than

th

e 19

67 S

upe

r B

owl?

20

07 S

up

er B

ow

l: $

2.6

× 10

6 ; 1

967

Su

per

Bo

wl:

$ 4

× 1

04 ;

6.5

× 1

01 o

r 65

tim

es m

ore

ex

pen

sive

4. A

VO

GA

DR

O’S

NU

MB

ER A

voga

dro’

s n

um

ber

is a

n i

mpo

rtan

t co

nce

pt i

n

chem

istr

y. I

t st

ates

th

at t

he

nu

mbe

r 6.

022

× 1

02

3 i

s ap

prox

imat

ely

equ

al t

o th

e n

um

ber

of m

olec

ule

s in

12

gram

s of

ca

rbon

12.

Use

Avo

gadr

o’s

nu

mbe

r to

de

term

ine

the

nu

mbe

r of

mol

ecu

les

in

5 ×

10

-7 g

ram

s of

car

bon

12.

2.

509

× 1

016 m

ole

cule

s

5. C

OA

L R

ESER

VES

Th

e ta

ble

belo

w s

how

s th

e n

um

ber

of k

ilog

ram

s of

coa

l se

lect

co

un

trie

s h

ad i

n p

rove

n r

eser

ve a

t th

e en

d of

a r

ecen

t ye

ar.

Co

al R

eser

ves

Co

un

try

Co

al (

kg)

Un

ite

d S

tate

s2

.46

× 1

014

Ru

ssia

1.5

7 ×

10

14

Ind

ia9

.24

× 1

013

Ro

ma

nia

4.9

4 ×

10

11

Sour

ce: B

ritis

h Pe

trol

eum

a. E

xpre

ss e

ach

cou

ntr

y’s

coal

res

erve

s in

sta

nda

rd f

orm

. U

nit

ed S

tate

s:

246,

00

0,0

00,

00

0,0

00

kg;

Ru

ssia

: 15

7,0

00,

00

0,0

00,

00

0 kg

; In

dia

: 92

,40

0,0

00,

00

0,0

00

kg;

Ro

man

ia:

494,

00

0,0

00,

00

0 kg

b.

How

man

y ti

mes

mor

e co

al d

oes

the

Un

ited

Sta

tes

hav

e th

an R

oman

ia?

4.98

× 1

02 o

r 49

8 ti

mes

c. O

ne

kilo

gram

of

coal

has

an

en

ergy

de

nsi

ty o

f 2.

4 ×

10

7 jo

ule

s. W

hat

is

the

tota

l en

ergy

den

sity

of

the

Un

ited

S

tate

s’ c

oal

rese

rve?

Exp

ress

you

r an

swer

in

sci

enti

fic

not

atio

n.

5.

90 ×

1021

jou

les

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2712

/20/

10

6:38

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

28

Gle

ncoe

Alg

ebra

1

7-4

Enri

chm

ent

En

gin

eeri

ng

No

tati

on

En

gin

eeri

ng

not

atio

n i

s a

vari

atio

n o

n s

cien

tifi

c n

otat

ion

wh

ere

nu

mbe

rs a

re e

xpre

ssed

as

pow

ers

of 1

000

rath

er t

han

as

pow

ers

of 1

0. E

ngi

nee

rin

g n

otat

ion

tak

es t

he

fam

ilia

r fo

rm

of a

× 1

0n, b

ut

n i

s re

stri

cted

to

mu

ltip

les

of t

hre

e an

d 1

≤ |

a|<

1000

.O

ne

adva

nta

ge t

o en

gin

eeri

ng

not

atio

n i

s th

at n

um

bers

can

be

nea

tly

expr

esse

d u

sin

g S

I pr

efix

es. T

hes

e pr

efix

es a

re t

ypic

ally

use

d fo

r sc

ien

tifi

c m

easu

rem

ents

.

100

0n10

nS

I Pre

fi x

Sym

bo

l10

00

510

15

pe

taP

10

00

410

12

tera

T

10

00

310

9g

iga

G

10

00

210

6m

ega

M

10

00

110

3kilo

k

10

00

-1

10

-3

mill

im

10

00

-2

10

-6

mic

roμ

10

00

-3

10

-9

na

no

n

10

00

-4

10

-12

pic

op

10

00

-5

10

-15

fem

tof

N

UC

LEA

R P

OW

ER T

he

outp

ut

of a

nu

clea

r p

ower

pla

nt

is m

easu

red

to

be

620,

000,

000

wat

ts. E

xpre

ss t

his

nu

mb

er i

n e

ngi

nee

rin

g n

otat

ion

an

d u

sin

g S

I p

refi

xes.

To

expr

ess

a n

um

ber

in e

ngi

nee

rin

g n

otat

ion

, fir

st c

onve

rt t

he

nu

mbe

r to

sci

enti

fic

not

atio

n.

Ste

p 1

62

0,00

0,00

0 ⇒

6.2

0000

000

a =

6.2

0000000

Ste

p 2

T

he

deci

mal

poi

nt

mov

ed 8

pla

ces

to t

he

left

, so

n =

8.

Ste

p 3

62

0,00

0,00

0 =

6.2

0000

000

× 1

08 =

6.2

× 1

08

Bec

ause

8 i

s n

ot a

mu

ltip

le o

f 3,

we

nee

d to

rou

nd

dow

n n

to

the

nex

t m

ult

iple

of

3.

Ste

p 4

6.

2 ×

108

= (

6.2

× 1

02 ) ×

106

620 =

6.2

× 1

02

Ste

p 5

=

620

× 1

06 P

roduct

of

Pow

ers

Th

e ou

tpu

t of

th

e po

wer

pla

nt

is 6

20 ×

106

wat

ts. U

sin

g th

e ch

art

abov

e, t

he

pref

ix f

or 1

06 is

fou

nd

to b

e m

ega,

or

M. T

he

outp

ut

of t

he

pow

er p

lan

t is

620

meg

awat

ts, o

r 62

0MW

.

Exer

cise

sE

xpre

ss e

ach

nu

mb

er i

n e

ngi

nee

rin

g n

otat

ion

.

1. 4

0,00

0,00

0,00

0 40

× 1

09 2.

180

,000

,000

,000

,000

180

× 1

012

3. 0

.000

06 6

0 ×

10-

6 4.

0.0

0000

0000

0003

9 39

0 ×

10-

15

Exp

ress

eac

h m

easu

rem

ent

usi

ng

SI

pre

fixe

s.

5. 0

.000

0000

0014

gra

m

6. 4

0,00

0,00

0,00

0 w

atts

1

40 p

ico

gra

ms

(pg

) 4

0 g

igaw

atts

(G

W)

7. 6

3,10

0,00

0,00

0,00

0 by

tes

8. 0

.000

0002

met

er

63.

1 te

raby

tes

(TB

) 2

00

nan

om

eter

s (n

m)

Exam

ple

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2812

/20/

10

6:38

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-5

Cha

pte

r 7

29

Gle

ncoe

Alg

ebra

1

7-5

Gra

ph

Exp

on

enti

al F

un

ctio

ns

Exp

on

enti

al F

un

ctio

na

fu

nctio

n d

efi n

ed

by a

n e

qu

atio

n o

f th

e fo

rm y

= a

bx ,

wh

ere

a ≠

0,

b >

0,

an

d b

≠ 1

You

can

use

val

ues

of

x to

fin

d or

dere

d pa

irs

that

sat

isfy

an

exp

onen

tial

fu

nct

ion

. Th

en y

ou

can

use

th

e or

dere

d pa

irs

to g

raph

th

e fu

nct

ion

.

G

rap

h y

= 3

x . F

ind

th

e y-

inte

rcep

t an

d s

tate

th

e d

omai

n a

nd

ra

nge

.

xy

-2

1

−

9

-1

1

−

3

01

13

29

Th

e y-

inte

rcep

t is

1.

Th

e do

mai

n i

s al

l re

al n

um

bers

, an

d th

e ra

nge

is

all

posi

tive

nu

mbe

rs.x

y

O

G

rap

h y

= (

1 −

4 ) x . Fin

d t

he

y-in

terc

ept

and

sta

te t

he

dom

ain

an

d

ran

ge.

Th

e y-

inte

rcep

t is

1.

Th

e do

mai

n i

s al

l re

al n

um

bers

, an

d th

e ra

nge

is

all

pos

itiv

e n

um

bers

.

x

y

O2

8

Exer

cise

sG

rap

h e

ach

fu

nct

ion

. Fin

d t

he

y-in

terc

ept

and

sta

te t

he

dom

ain

an

d r

ange

.

1. y

= 0

.3x

2. y

= 3

x +

1

3. y

= ( 1

−

3 ) x + 1

x

y

O1

2

x

y

O2

16 4812

x

y

O1

2

1

; D

= {

all r

eal

2; D

= {

all r

eal

2;

D =

{al

l rea

ln

um

ber

s},

nu

mb

ers}

,

nu

mb

ers}

,R

= {

y|y >

0}

R =

{y

|y >

1}

R

= {

y|y >

1}

Exam

ple

1Ex

amp

le 2

xy

-2

16

-1

4

01

1 1

−

4

2 1

−

16

Stud

y G

uide

and

Inte

rven

tion

Exp

on

en

tial

Fu

ncti

on

s

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

2912

/20/

10

6:38

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 7

30

Gle

ncoe

Alg

ebra

1

7-5

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Exp

on

en

tial

Fu

ncti

on

s

Iden

tify

Exp

on

enti

al B

ehav

ior

It i

s so

met

imes

use

ful

to k

now

if

a se

t of

dat

a is

ex

pon

enti

al. O

ne

way

to

tell

is

to o

bser

ve t

he

shap

e of

th

e gr

aph

. An

oth

er w

ay i

s to

obs

erve

th

e pa

tter

n i

n t

he

set

of d

ata.

D

eter

min

e w

het

her

th

e se

t of

dat

a sh

own

bel

ow d

isp

lays

ex

pon

enti

al b

ehav

ior.

Wri

te y

es o

r n

o. E

xpla

in w

hy

or w

hy

not

.

Met

hod

1: L

ook

for

a P

atte

rnT

he

dom

ain

val

ues

in

crea

se b

y re

gula

r in

terv

als

of 2

, wh

ile

the

ran

ge v

alu

es h

ave

a co

mm

on f

acto

r of

1 −

2 .

Sin

ce t

he

dom

ain

va

lues

in

crea

se b

y re

gula

r in

terv

als

and

the

ran

ge v

alu

es h

ave

a co

mm

on f

acto

r, th

e da

ta a

re p

roba

bly

expo

nen

tial

.

Met

hod

2: G

raph

th

e D

ata

T

he

grap

h s

how

s ra

pidl

y de

crea

sin

g va

lues

of

y as

x

incr

ease

s. T

his

is

char

acte

rist

ic o

f ex

pon

enti

al

beh

avio

r.

Exer

cise

s

Det

erm

ine

wh

eth

er t

he

set

of d

ata

show

n b

elow

dis

pla

ys e

xpon

enti

al b

ehav

ior.

W

rite

yes

or

no.

Exp

lain

wh

y or

wh

y n

ot.

1.

x0

12

3

y5

10

15

20

2.

x

01

23

y3

92

78

1

N

o;

the

do

mai

n v

alu

es a

re a

t re

gu

lar

Yes

; th

e d

om

ain

val

ues

are

at

inte

rval

s, a

nd

th

e ra

ng

e va

lues

hav

e r

egu

lar

inte

rval

s, a

nd

th

e ra

ng

e a

com

mo

n d

iffer

ence

5.

val

ues

hav

e a

com

mo

n f

acto

r 3.

3.

x-

11

35

y3

216

84

4.

x

-1

01

23

y3

33

33

Y

es;

the

do

mai

n v

alu

es a

re a

t

No

; th

e d

om

ain

val

ues

are

at

reg

ula

r in

terv

als,

an

d t

he

ran

ge

r

egu

lar

inte

rval

s, b

ut

the

ran

ge

valu

es h

ave

a co

mm

on

fac

tor

1 −

2 . v

alu

es d

o n

ot

chan

ge.

5.

x-

50

510

y1

0.5

0.2

50

.12

5

6.

x

01

23

4

y 1

−

3

1

−

9

1

−

27

1

−

81

1

−

24

3

Y

es;

the

do

mai

n v

alu

es a

re a

t

Yes

; th

e d

om

ain

val

ues

are

at

reg

ula

r in

terv

als,

an

d t

he

ran

ge

r

egu

lar

inte

rval

s, a

nd

th

e ra

ng

e va

lues

hav

e a

com

mo

n f

acto

r 0.

5.

val

ues

hav

e a

com

mo

n f

acto

r 1 −

3 .

Exam

ple

x0

24

68

10

y6

43

216

84

2

x

y

O4

8

16324872

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3012

/24/

10

8:22

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-5

Cha

pte

r 7

31

Gle

ncoe

Alg

ebra

1

7-5

Skill

s Pr

acti

ceE

xp

on

en

tial

Fu

ncti

on

s

Gra

ph

eac

h f

un

ctio

n. F

ind

th

e y-

inte

rcep

t, a

nd

sta

te t

he

dom

ain

an

d r

ange

.

1. y

= 2

x 2.

y =

( 1 −

3 ) x

1

; D

= {

all r

eal n

um

ber

s},

1

; D

= {

all r

eal n

um

ber

s},

R =

{y |

y >

0}

R

= {

y |

y >

0}

3. y

= 3

(2x )

4.

y =

3x +

2

3;

D =

{al

l rea

l nu

mb

ers}

,

3; D

= {

all r

eal n

um

ber

s},

R

= {

y |

y >

0}

R

= {

y |

y >

2}

Det

erm

ine

wh

eth

er t

he

set

of d

ata

show

n b

elow

dis

pla

ys e

xpon

enti

al b

ehav

ior.

W

rite

yes

or

no.

Exp

lain

wh

y or

wh

y n

ot.

5.

x-

3-

2-

10

y9

12

15

18

6.

x

05

10

15

y2

010

52

.5

N

o;

the

do

mai

n v

alu

es a

re a

t

Yes

; th

e d

om

ain

val

ues

are

at

reg

ula

r in

terv

als

and

th

e ra

ng

e

reg

ula

r in

terv

als

and

th

e ra

ng

e va

lues

hav

e a

com

mo

n

val

ues

hav

e a

com

mo

n f

acto

r 0.

5.d

iffer

ence

3.

7.

x4

812

16

y2

04

08

016

0

8.

x

50

30

10

-10

y9

07

05

03

0

Y

es;

the

do

mai

n v

alu

es a

re a

t

No

; th

e d

om

ain

val

ues

are

at

reg

ula

r in

terv

als

and

th

e ra

ng

e

reg

ula

r in

terv

als

and

th

e ra

ng

e va

lues

hav

e a

com

mo

n f

acto

r 2.

v

alu

es h

ave

a co

mm

on

d

iffer

ence

20.x

y

Ox

y

O

x

y

Ox

y

O

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3112

/20/

10

6:39

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

32

Gle

ncoe

Alg

ebra

1

7-5

Prac

tice

E

xp

on

en

tial

Fu

ncti

on

sG

rap

h e

ach

fu

nct

ion

. Fin

d t

he

y-in

terc

ept

and

sta

te t

he

dom

ain

an

d r

ange

.

1. y

= (

1 −

10 ) x

2. y

= 3

x 3.

y =

( 1 −

4 ) x

4. y

= 4

(2x )

+ 1

5.

y =

2(2

x - 1

)

6. y

= 0

.5(3

x - 3

)

Det

erm

ine

wh

eth

er t

he

set

of d

ata

show

n b

elow

dis

pla

ys e

xpon

enti

al b

ehav

ior.

W

rite

yes

or

no.

Exp

lain

wh

y or

wh

y n

ot.

7.

x2

58

11

y4

80

12

03

07.

5

8.

x

21

18

15

12

y3

02

316

9

Y

es;

the

do

mai

n v

alu

es a

re a

t

No

; th

e d

om

ain

val

ues

are

at

reg

ula

r in

terv

als

and

th

e ra

ng

e

reg

ula

r in

terv

als

and

th

e ra

ng

eva

lues

hav

e a

com

mo

n

val

ues

hav

e a

com

mo

n

fact

or

0.25

. d

iffer

ence

7.

9. L

EAR

NIN

G M

s. K

lem

pere

r to

ld h

er E

ngl

ish

cla

ss t

hat

eac

h w

eek

stu

den

ts t

end

to f

orge

t on

e si

xth

of

the

voca

bula

ry w

ords

th

ey l

earn

ed t

he

prev

iou

s w

eek.

Su

ppos

e a

stu

den

t le

arn

s 60

wor

ds. T

he

nu

mbe

r of

wor

ds r

emem

bere

d ca

n b

e de

scri

bed

by t

he

fun

ctio

n

W

(x)

= 6

0 ( 5 −

6 ) x , w

her

e x

is t

he

nu

mbe

r of

wee

ks t

hat

pas

s. H

ow m

any

wor

ds w

ill

the

s

tude

nt

rem

embe

r af

ter

3 w

eeks

? ab

ou

t 35

10. B

IOLO

GY

Su

ppos

e a

cert

ain

cel

l re

prod

uce

s it

self

in

fou

r h

ours

. If

a la

b re

sear

cher

be

gin

s w

ith

50

cell

s, h

ow m

any

cell

s w

ill

ther

e be

aft

er o

ne

day,

tw

o da

ys, a

nd

thre

e da

ys?

(Hin

t: U

se t

he

expo

nen

tial

fu

nct

ion

y =

50(

2x ).)

32

00

cells

; 20

4,80

0 ce

lls;

13,1

07,2

00

cells

x

y

Ox

y

O

x

y

O

x

y

Ox

y

Ox

y

O

1; D

= {

all r

eal

nu

mb

ers}

;

R =

{ y

| y

> 0}

1; D

= {

all r

eal

nu

mb

ers}

;

R =

{ y

| y

> 0}

1; D

= {

all r

eal

num

bers

};

R =

{ y

| y

> 0}

5; D

= {

all

real

nu

mb

ers}

; R

= {

y |

y >

1}

0; D

= {

all

real

nu

mb

ers}

; R

= {

y |

y >

-2}

-1;

D =

{a

ll re

al

num

bers

};

R =

{ y

| y >

-1.

5}

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3212

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7 -5

Cha

pte

r 7

33

Gle

ncoe

Alg

ebra

1

7-5

1. W

AST

E S

upp

ose

the

was

te g

ener

ated

by

non

recy

cled

pap

er a

nd

card

boar

d pr

odu

cts

is a

ppro

xim

ated

by

the

foll

owin

g fu

nct

ion

.y

= 1

000(

2)0.

3x

Ske

tch

th

e ex

pon

enti

al f

un

ctio

n o

n t

he

coor

din

ate

grid

bel

ow. y

xO950

650

350

1850

1550

2450

2150

1250

43

21

-2-

1-

3-

4

2. M

ON

EY T

atya

na’

s gr

andf

ath

er g

ave

her

on

e pe

nn

y on

th

e da

y sh

e w

as b

orn

. He

plan

s to

dou

ble

the

amou

nt

he

give

s h

er

ever

y da

y. E

stim

ate

how

mu

ch s

he

wil

l re

ceiv

e fr

om h

er g

ran

dfat

her

on

th

e 12

th

day

of h

er l

ife.

abo

ut

$20

3.PI

CTU

RE

FRA

MES

Sin

ce a

pic

ture

fra

me

incl

ude

s a

bord

er, t

he

pict

ure

mu

st b

e sm

alle

r in

are

a th

an

the

enti

re f

ram

e. T

he

tabl

e sh

ows

the

rela

tion

ship

bet

wee

n

pict

ure

are

a an

d fr

ame

len

gth

for

a p

arti

cula

r li

ne

of f

ram

es. I

s th

is

an e

xpon

enti

al

rela

tion

ship

? E

xpla

in.

No

; th

ere

is n

o c

om

mo

n f

acto

r b

etw

een

th

e p

ictu

re a

reas

.

4.D

EPR

ECIA

TIO

N T

he

valu

e of

Roy

ce

Com

pan

y’s

com

pute

r eq

uip

men

t is

de

crea

sin

g in

val

ue

acco

rdin

g to

th

e fo

llow

ing

fun

ctio

n.

y=

400

0(0.

87)x

In t

he

equ

atio

n, x

is

the

nu

mbe

r of

ye

ars

that

hav

e el

apse

d si

nce

th

e eq

uip

men

t w

as p

urc

has

ed a

nd

y is

th

e va

lue

in d

olla

rs. W

hat

was

th

e va

lue

5 ye

ars

afte

r it

was

pu

rch

ased

? R

oun

d yo

ur

answ

er t

o th

e n

eare

st d

olla

r.$1

994

5. M

ETEO

RO

LOG

Y T

he

atm

osph

eric

pr

essu

re i

n m

illi

bars

at

alti

tude

x

met

ers

can

be

appr

oxim

ated

by

the

foll

owin

g fu

nct

ion

. Th

e fu

nct

ion

is

vali

d fo

r va

lues

of

x be

twee

n 0

an

d 10

,000

.

f (

x) =

103

8(1.

0001

34)-

x

a.

Wh

at i

s th

e pr

essu

re a

t se

a le

vel?

10

38 m

illib

ars

b

. T

he

McD

onal

d O

bser

vato

ry i

n T

exas

is

at

an a

ltit

ude

of

2000

met

ers.

Wh

at

is t

he

appr

oxim

ate

atm

osph

eric

pr

essu

re t

her

e?

794

mill

ibar

s

c.

As

alti

tude

in

crea

ses,

wh

at h

appe

ns

to

atm

osph

eric

pre

ssu

re?

It

dec

reas

es.

Wor

d Pr

oble

m P

ract

ice

Exp

on

en

tial

Fu

ncti

on

s

Sid

e P

ictu

re L

eng

th

Are

a (

in.)

(i

n2 )

5

6

6

12

7

2

0

8

3

0

9

4

2

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3312

/20/

10

6:39

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 7

34

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Lo

gari

thm

ic F

un

cti

on

s

7-5

You

hav

e fo

un

d th

e in

vers

e of

a l

inea

r fu

nct

ion

by

inte

rch

angi

ng

x an

d y

in t

he

equ

atio

n o

f th

e fu

nct

ion

. Th

e in

vers

e of

an

exp

onen

tial

fu

nct

ion

is

call

ed a

log

arit

hm

ic f

un

ctio

n.

F

ind

th

e in

vers

e fu

nct

ion

of

the

exp

onen

tial

fu

nct

ion

y =

2x .

Mak

e a

tab

le o

f va

lues

for

eac

h f

un

ctio

n. T

hen

gra

ph

bot

h f

un

ctio

ns.

To

fin

d th

e in

vers

e fu

nct

ion

, in

terc

han

ge x

an

d y.

Th

e in

vers

e of

y =

2 x i

s x

= 2

y .

y =

2 x

x =

2 y

xy

xy

-3

1

−

8

1

−

8

-3

-2

1

−

4

1

−

4

-2

-1

1

−

2

1

−

2

-1

0

11

0

1

22

1

2

44

2

3

88

3

In g

ener

al t

he

inve

rse

of y

= b

x is

x =

b y .

In x

= b

y , th

e va

riab

le y

is

call

ed t

he

loga

rith

m

of x

. Th

is i

s u

sual

ly w

ritt

en a

s y

= l

o g b

x.

Exer

cise

s

1. W

hat

are

th

e do

mai

n a

nd

ran

ge o

f y

= l

o g 2

x? H

ow a

re t

he

dom

ain

an

d ra

nge

rel

ated

to

the

dom

ain

an

d ra

nge

of

y =

2 x ?

D =

{x

|x >

0},

R =

{al

l rea

l nu

mb

ers}

; Th

e d

om

ain

of

y =

lo g

2 x is

th

e ra

ng

e o

f y =

2 x a

nd

th

e ra

ng

e o

f y =

lo g

2 x is

th

e d

om

ain

of

y =

2 x .

2. H

ow a

re t

he

grap

hs

of y

= l

o g 2

x an

d y

= 2

x rel

ated

?

Th

ey a

re r

efl e

ctio

ns

in t

he

line

y =

x.

3. T

hro

ugh

wh

ich

qu

adra

nts

do

the

grap

hs

of y

= l

o g 2

x an

d y

= 2

x pas

s?

y =

lo g

2 x p

asse

s th

rou

gh

I an

d II

; y =

2 x p

asse

s th

rou

gh

I an

d IV

.

4. I

f an

exp

onen

tial

fu

nct

ion

fin

ds p

opu

lati

on a

s a

fun

ctio

n o

f ti

me,

wh

at c

an y

ou c

oncl

ude

ab

out

its

inve

rse

loga

rith

mic

fu

nct

ion

? T

he

log

arit

hm

ic f

un

ctio

n fi

nd

s ti

me

as a

fu

nct

ion

of

po

pu

lati

on

.

5. R

ESEA

RC

H I

nve

stig

ate

real

-wor

ld u

ses

of l

ogar

ith

mic

fu

nct

ion

s. H

ow a

re l

ogar

ith

mic

fu

nct

ion

s u

sed

to m

odel

rea

l-w

orld

sit

uat

ion

s?S

amp

le a

nsw

er: T

he

Ric

hte

r sc

ale

is a

log

arit

hm

ic f

un

ctio

n u

sed

to

m

easu

re e

arth

qu

ake

inte

nsi

ty.

Exam

ple

y

xO12

812

8 4

4

y =

x

y =

2x

x =

2y

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3412

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-6

Cha

pte

r 7

35

Gle

ncoe

Alg

ebra

1

7-6

Stud

y G

uide

and

Inte

rven

tion

Gro

wth

an

d D

ecay

Exp

on

enti

al G

row

th P

opu

lati

on i

ncr

ease

s an

d gr

owth

of

mon

etar

y in

vest

men

ts a

re

exam

ples

of

exp

onen

tial

gro

wth

. Th

is m

ean

s th

at a

n i

nit

ial

amou

nt

incr

ease

s at

a s

tead

y ra

te o

ver

tim

e.

Th

e g

en

era

l e

qu

atio

n fo

r exp

on

en

tia

l gro

wth

is y

= a

(1 +

r)t .

• y

rep

rese

nts

th

e fi n

al a

mo

un

t.

Exp

on

enti

al G

row

th

•

a re

pre

se

nts

th

e in

itia

l a

mo

un

t.

• r

rep

rese

nts

th

e r

ate

of

ch

an

ge

exp

resse

d a

s a

de

cim

al.

• t

rep

rese

nts

tim

e.

POPU

LATI

ON

Th

e p

opu

lati

on o

f Jo

hn

son

Cit

y in

200

5 w

as

25,0

00. S

ince

th

en, t

he

pop

ula

tion

has

gr

own

at

an a

vera

ge r

ate

of 3

.2%

eac

h y

ear.

a. W

rite

an

eq

uat

ion

to

rep

rese

nt

the

pop

ula

tion

of

Joh

nso

n C

ity

sin

ce 2

005.

Th

e ra

te 3

.2%

can

be

wri

tten

as

0.03

2.y

= a

(1 +

r)t

y =

25,

000(

1 +

0.0

32)t

y =

25,

000(

1.03

2)t

b.

Acc

ord

ing

to t

he

equ

atio

n, w

hat

wil

l th

e p

opu

lati

on o

f J

ohn

son

Cit

y b

e in

201

5?

In 2

015

t w

ill

equ

al 2

015

- 2

005

or 1

0.S

ubs

titu

te 1

0 fo

r t

in t

he

equ

atio

n f

rom

pa

rta.

y =

25,

000(

1.03

2)10

t=

10

≈

34,

256

In 2

015

the

popu

lati

on o

f Jo

hn

son

Cit

y w

ill

be a

bou

t 34

,256

.

INV

ESTM

ENT

Th

e G

arci

as h

ave

$12,

000

in a

sav

ings

ac

cou

nt.

Th

e b

ank

pay

s 3.

5% i

nte

rest

on

sav

ings

acc

oun

ts, c

omp

oun

ded

m

onth

ly. F

ind

th

e b

alan

ce i

n 3

yea

rs.

Th

e ra

te 3

.5%

can

be

wri

tten

as

0.03

5.

Th

e sp

ecia

l eq

uat

ion

for

com

pou

nd

inte

rest

is

A=

P (1

+r − n

)nt , w

her

e A

repr

esen

ts t

he

bala

nce

, P i

s th

e in

itia

l am

oun

t, r

rep

rese

nts

th

e an

nu

al r

ate

expr

esse

d as

a d

ecim

al, n

rep

rese

nts

th

e n

um

ber

of t

imes

th

e in

tere

st i

s co

mpo

un

ded

each

yea

r, an

d t

repr

esen

ts

the

nu

mbe

r of

yea

rs t

he

mon

ey i

s in

vest

ed.

A=

P (1

+r − n

)nt

= 1

2,00

0 (1

+

0.03

5−

12)3(

12)

≈ 1

3,32

6.49

In t

hre

e ye

ars,

th

e ba

lan

ce o

f th

e ac

cou

nt

wil

l be

$13

,326

.49.

Exer

cise

s 1

. PO

PULA

TIO

N T

he

popu

lati

on o

f th

e U

nit

ed

2. IN

VES

TMEN

T D

eter

min

e th

e S

tate

s h

as b

een

in

crea

sin

g at

an

ave

rage

v

alu

e of

an

in

vest

men

t of

$25

00

ann

ual

rat

e of

0.9

1%. I

f th

e po

pula

tion

was

i

f it

is

inve

sted

at

an i

nte

rest

rat

e of

ab

out

303,

146,

000

in 2

008,

pre

dict

th

e

5.2

5% c

ompo

un

ded

mon

thly

for

po

pula

tion

in

201

2.

4

yea

rs.

$308

2.78

abo

ut

314,

332,

051

3. P

OPU

LATI

ON

It

is e

stim

ated

th

at t

he

4.

INV

ESTM

ENT

Det

erm

ine

the

popu

lati

on o

f th

e w

orld

is

incr

easi

ng

at

val

ue

of a

n i

nve

stm

ent

of $

100,

000

an

ave

rage

an

nu

al r

ate

of 1

.3%

. If

the

i

f it

is

inve

sted

at

an i

nte

rest

rat

e of

20

08 p

opu

lati

on w

as a

bou

t 6,

641,

000,

000,

5.2

% c

ompo

un

ded

quar

terl

y fo

r

pred

ict

the

2015

pop

ula

tion

.

12

year

s.

abou

t 7,

269,

417,

259

$

185,

888.

87

Exam

ple

1Ex

amp

le 2

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3512

/20/

10

6:39

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

36

Gle

ncoe

Alg

ebra

1

7-6

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Gro

wth

an

d D

ecay

Exp

on

enti

al D

ecay

Rad

ioac

tive

dec

ay a

nd

depr

ecia

tion

are

exa

mpl

es o

f ex

pon

enti

al

dec

ay. T

his

mea

ns

that

an

in

itia

l am

oun

t de

crea

ses

at a

ste

ady

rate

ove

r a

peri

od o

f ti

me.

Th

e g

en

era

l e

qu

atio

n fo

r exp

on

en

tia

l d

ecay is y

= a

(1 -

r)t .

• y

rep

rese

nts

th

e fi n

al a

mo

un

t.

Exp

on

enti

al D

ecay

•

a re

pre

se

nts

th

e in

itia

l a

mo

un

t.

• r

rep

rese

nts

th

e r

ate

of

de

cay e

xp

resse

d a

s a

de

cim

al.

• t

rep

rese

nts

tim

e.

D

EPR

ECIA

TIO

N T

he

orig

inal

pri

ce o

f a

trac

tor

was

$45

,000

. Th

e va

lue

of t

he

trac

tor

dec

reas

es a

t a

stea

dy

rate

of

12%

per

yea

r.

a. W

rite

an

eq

uat

ion

to

rep

rese

nt

the

valu

e of

th

e tr

acto

r si

nce

it

was

pu

rch

ased

.

Th

e ra

te 1

2% c

an b

e w

ritt

en a

s 0.

12.

y =

a(1

- r

)t G

enera

l equation for

exponential decay

y =

45,

000(

1 -

0.1

2)t

a =

45,0

00 a

nd r

= 0

.12

y =

45,

000(

0.88

)t S

implif

y.

b.

Wh

at i

s th

e va

lue

of t

he

trac

tor

in 5

yea

rs?

y =

45,

000(

0.88

)t E

quation for

decay f

rom

part

a

y =

45,

000(

0.88

)5 t

= 5

y ≈

23,

747.

94

Use a

calc

ula

tor.

In 5

yea

rs, t

he

trac

tor

wil

l be

wor

th a

bou

t $2

3,74

7.94

.

Exer

cise

s 1

. PO

PULA

TIO

N T

he

popu

lati

on o

f B

ulg

aria

has

bee

n d

ecre

asin

g at

an

an

nu

al

rate

of

0.89

%. I

f th

e po

pula

tion

of

Bu

lgar

ia w

as a

bou

t 7,

450,

349

in t

he

year

20

05, p

redi

ct i

ts p

opu

lati

on i

n t

he

year

201

5. a

bo

ut

6,81

3,20

4

2. D

EPR

ECIA

TIO

N M

r. G

osse

ll i

s a

mac

hin

ist.

He

bou

ght

som

e n

ew m

ach

iner

y fo

r ab

out

$125

,000

. He

wan

ts t

o ca

lcu

late

th

e va

lue

of t

he

mac

hin

ery

over

th

e n

ext

10 y

ears

for

ta

x pu

rpos

es. I

f th

e m

ach

iner

y de

prec

iate

s at

th

e ra

te o

f 15

% p

er y

ear,

wh

at i

s th

e va

lue

of t

he

mac

hin

ery

(to

the

nea

rest

$10

0) a

t th

e en

d of

10

year

s? a

bo

ut

$24,

600

3. A

RC

HA

EOLO

GY

Th

e h

alf-

life

of

a ra

dioa

ctiv

e el

emen

t is

def

ined

as

the

tim

e th

at i

t ta

kes

for

one-

hal

f a

quan

tity

of

the

elem

ent

to d

ecay

. Rad

ioac

tive

car

bon

-14

is f

oun

d in

al

l li

vin

g or

gan

ism

s an

d h

as a

hal

f-li

fe o

f 57

30 y

ears

. Con

side

r a

livi

ng

orga

nis

m w

ith

an

or

igin

al c

once

ntr

atio

n o

f ca

rbon

-14

of 1

00 g

ram

s.

a. I

f th

e or

gan

ism

liv

ed 5

730

year

s ag

o, w

hat

is

the

con

cen

trat

ion

of

carb

on-1

4 to

day?

50

gb

. If

th

e or

gan

ism

liv

ed 1

1,46

0 ye

ars

ago,

det

erm

ine

the

con

cen

trat

ion

of

carb

on-1

4 to

day.

25

g

4. D

EPR

ECIA

TIO

N A

new

car

cos

ts $

32,0

00. I

t is

exp

ecte

d to

dep

reci

ate

12%

eac

h y

ear

for

4 ye

ars

and

then

dep

reci

ate

8% e

ach

yea

r th

erea

fter

. Fin

d th

e va

lue

of t

he

car

in

6 ye

ars.

ab

ou

t $1

6,24

2.63

Exam

ple

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3612

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-6

Cha

pte

r 7

37

Gle

ncoe

Alg

ebra

1

7-6

Skill

s Pr

acti

ceG

row

th a

nd

Decay

1. P

OPU

LATI

ON

Th

e po

pula

tion

of

New

Yor

k C

ity

incr

ease

d fr

om 8

,008

,278

in

200

0 to

8,1

68,3

88 i

n 2

005.

Th

e an

nu

al r

ate

of p

opu

lati

on i

ncr

ease

for

th

e pe

riod

was

ab

out

0.4%

.

a. W

rite

an

equ

atio

n f

or t

he

popu

lati

on t

yea

rs a

fter

200

0. P

= 8

,008

,278

(1.0

04)t

b.

Use

th

e eq

uat

ion

to

pred

ict

the

popu

lati

on o

f N

ew Y

ork

Cit

y in

201

5.

abo

ut

8,50

2,46

5

2. S

AV

ING

S T

he

Fre

sh a

nd

Gre

en C

ompa

ny

has

a s

avin

gs p

lan

for

its

em

ploy

ees.

If

an

empl

oyee

mak

es a

n i

nit

ial

con

trib

uti

on o

f $1

000,

th

e co

mpa

ny

pays

8%

in

tere

st

com

pou

nde

d qu

arte

rly.

a. I

f an

em

ploy

ee p

arti

cipa

tin

g in

th

e pl

an w

ith

draw

s th

e ba

lan

ce o

f th

e ac

cou

nt

afte

r 5

year

s, h

ow m

uch

wil

l be

in

th

e ac

cou

nt?

$14

85.9

5

b.

If a

n e

mpl

oyee

par

tici

pati

ng

in t

he

plan

wit

hdr

aws

the

bala

nce

of

the

acco

un

t af

ter

35 y

ears

, how

mu

ch w

ill

be i

n t

he

acco

un

t? $

15,9

96.4

7

3. H

OU

SIN

G M

r. an

d M

rs. B

oyce

bou

ght

a h

ouse

for

$96

,000

in

199

5. T

he

real

est

ate

brok

er i

ndi

cate

d th

at h

ouse

s in

th

eir

area

wer

e ap

prec

iati

ng

at a

n a

vera

ge a

nn

ual

rat

e of

7%

. If

the

appr

ecia

tion

rem

ain

ed s

tead

y at

th

is r

ate,

wh

at w

as t

he

valu

e of

th

e B

oyce

’s h

ome

in 2

009?

ab

ou

t $2

47,5

39

4. M

AN

UFA

CTU

RIN

G Z

elle

r In

dust

ries

bou

ght

a pi

ece

of w

eavi

ng

equ

ipm

ent

for

$60,

000.

It

is

expe

cted

to

depr

ecia

te a

t an

ave

rage

rat

e of

10%

per

yea

r.

a. W

rite

an

equ

atio

n f

or t

he

valu

e of

th

e pi

ece

of e

quip

men

t af

ter

t ye

ars.

E

= 6

0,0

00(

0.90

)t

b.

Fin

d th

e va

lue

of t

he

piec

e of

equ

ipm

ent

afte

r 6

year

s. a

bo

ut

$31,

886

5. F

INA

NC

ES K

yle

save

d $5

00 f

rom

a s

um

mer

job.

He

plan

s to

spe

nd

10%

of

his

sav

ings

ea

ch w

eek

on v

ario

us

form

s of

en

tert

ain

men

t. A

t th

is r

ate,

how

mu

ch w

ill

Kyl

e h

ave

left

af

ter

15 w

eeks

? $1

02.9

5

6. T

RA

NSP

OR

TATI

ON

Tif

fan

y’s

mot

her

bou

ght

a ca

r fo

r $9

000

five

yea

rs a

go. S

he

wan

ts

to s

ell

it t

o T

iffa

ny

base

d on

a 1

5% a

nn

ual

rat

e of

dep

reci

atio

n. A

t th

is r

ate,

how

mu

ch

wil

l Tif

fan

y pa

y fo

r th

e ca

r? a

bo

ut

$399

3

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3712

/20/

10

6:39

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 7

38

Gle

ncoe

Alg

ebra

1

7-6

Prac

tice

G

row

th a

nd

Decay

1.C

OM

MU

NIC

ATI

ON

S S

port

s ra

dio

stat

ion

s n

um

bere

d 22

0 in

199

6. T

he

nu

mbe

r of

sp

orts

rad

io s

tati

ons

has

sin

ce i

ncr

ease

d by

app

roxi

mat

ely

14.3

% p

er y

ear.

a. W

rite

an

equ

atio

n f

or t

he

nu

mbe

r of

spo

rts

radi

o st

atio

ns

for

t ye

ars

afte

r 19

96.

R=

220

(1.1

43)t

b.

If t

he

tren

d co

nti

nu

es, p

redi

ct t

he

nu

mbe

r of

spo

rts

radi

o st

atio

ns

in 2

015.

ab

ou

t 27

88 s

tati

on

s

2. I

NV

ESTM

ENTS

Det

erm

ine

the

amou

nt

of a

n i

nve

stm

ent

if $

500

is i

nve

sted

at

an

inte

rest

rat

e of

4.2

5% c

ompo

un

ded

quar

terl

y fo

r 12

yea

rs.

$830

.41

3. I

NV

ESTM

ENTS

Det

erm

ine

the

amou

nt

of a

n i

nve

stm

ent

if $

300

is i

nve

sted

at

an

inte

rest

rat

e of

6.7

5% c

ompo

un

ded

sem

ian

nu

ally

for

20

year

s.

$113

1.73

4. H

OU

SIN

G T

he

Gre

ens

bou

ght

a co

ndo

min

ium

for

$11

0,00

0 in

201

0. I

f it

s va

lue

appr

ecia

tes

at a

n a

vera

ge r

ate

of 6

% p

er y

ear,

wh

at w

ill

the

valu

e be

in

201

5?

abo

ut

$147

,205

5. D

EFO

RES

TATI

ON

Du

rin

g th

e 19

90s,

th

e fo

rest

ed a

rea

of G

uat

emal

a de

crea

sed

at a

n

aver

age

rate

of

1.7%

.

a. I

f th

e fo

rest

ed a

rea

in G

uat

emal

a in

199

0 w

as a

bou

t 34

,400

squ

are

kilo

met

ers,

wri

te

an

equ

atio

n f

or t

he

fore

sted

are

a fo

r t

year

s af

ter

1990

. C

= 3

4,40

0(0.

983)

t

b.

If t

his

tre

nd

con

tin

ues

, pre

dict

th

e fo

rest

ed a

rea

in 2

015.

abo

ut

22,4

07.6

5 km

2

6. B

USI

NES

S A

pie

ce o

f m

ach

iner

y va

lued

at

$25,

000

depr

ecia

tes

at a

ste

ady

rate

of

10%

yea

rly.

Wh

at w

ill

the

valu

e of

th

e pi

ece

of m

ach

iner

y be

aft

er 7

yea

rs?

abo

ut

$11,

957

7. T

RA

NSP

OR

TATI

ON

A n

ew c

ar c

osts

$18

,000

. It

is e

xpec

ted

to d

epre

ciat

e at

an

ave

rage

ra

te o

f 12

% p

er y

ear.

Fin

d th

e va

lue

of t

he

car

in 8

yea

rs.

abo

ut

$647

3

8. P

OPU

LATI

ON

Th

e po

pula

tion

of

Osa

ka, J

apan

, dec

lin

ed a

t an

ave

rage

an

nu

al r

ate

of 0

.05%

for

th

e fi

ve y

ears

bet

wee

n 1

995

and

2000

. If

the

popu

lati

on o

f O

saka

was

11

,013

,000

in

200

0 an

d it

con

tin

ues

to

decl

ine

at t

he

sam

e ra

te, p

redi

ct t

he

popu

lati

on

in 2

050.

ab

ou

t 10

,741

,021

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3812

/24/

10

9:09

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-6

Cha

pte

r 7

39

Gle

ncoe

Alg

ebra

1

7-6

1. D

EPR

ECIA

TIO

N T

he

valu

e of

a n

ew

plas

ma

tele

visi

on d

epre

ciat

es b

y ab

out

7% e

ach

yea

r. A

eryn

pu

rch

ases

a 5

0-in

ch

plas

ma

tele

visi

on f

or $

3000

. Wh

at i

s it

s va

lue

afte

r 4

year

s? R

oun

d yo

ur

answ

er

to t

he

nea

rest

hu

ndr

ed.

abo

ut

$220

0

2. M

ON

EY H

ans

open

s a

savi

ngs

acc

oun

t by

dep

osit

ing

$120

0 in

an

acc

oun

t th

at

earn

s 3

perc

ent

inte

rest

com

pou

nde

d w

eekl

y. H

ow m

uch

wil

l h

is i

nve

stm

ent

be

wor

th i

n 1

0 ye

ars?

Ass

um

e th

at t

her

e ar

e ex

actl

y 52

wee

ks i

n a

yea

r an

d ro

un

d yo

ur

answ

er t

o th

e n

eare

st c

ent.

$1

619.

69

3. H

IGH

ER E

DU

CA

TIO

N T

he

tabl

e li

sts

the

aver

age

cost

s of

att

endi

ng

a fo

ur-

year

co

lleg

e in

th

e U

nit

ed S

tate

s du

rin

g a

rece

nt

year

.

S

ourc

e: C

olle

ge B

oard

Ru

ss’s

par

ents

in

vest

ed m

oney

in

a

savi

ngs

acc

oun

t ea

rnin

g an

ave

rage

of

4.5

perc

ent

inte

rest

, com

pou

nde

d m

onth

ly.

Aft

er 1

5 ye

ars,

th

ey h

ave

exac

tly

the

righ

t am

oun

t to

cov

er t

he

tuit

ion

, fee

s,

room

an

d bo

ard

for

Ru

ss’s

fir

st y

ear

at a

pu

blic

col

lege

. Wh

at w

as t

hei

r in

itia

l in

vest

men

t? R

oun

d yo

ur

answ

er t

o th

e n

eare

st d

olla

r. $6

182

4.PO

PULA

TIO

N I

n 2

007

the

U.S

. Cen

sus

Bu

reau

est

imat

ed t

he

popu

lati

on o

f th

e U

nit

ed S

tate

s es

tim

ated

at

301

mil

lion

. T

he

ann

ual

rat

e of

gro

wth

is

abou

t 0.

89%

. At

this

rat

e, w

hat

is

the

expe

cted

po

pula

tion

at

the

tim

e of

th

e 20

20

cen

sus?

Rou

nd

you

r an

swer

to

the

nea

rest

ten

mil

lion

.

340

mill

ion

5. M

EDIC

INE

Wh

en d

octo

rs p

resc

ribe

m

edic

atio

n, t

hey

hav

e to

con

side

r th

e ra

te a

t w

hic

h t

he

body

fil

ters

a d

rug

from

th

e bl

oods

trea

m. S

upp

ose

it t

akes

th

e h

um

an b

ody

6 da

ys t

o fi

lter

ou

t h

alf

of

the

Flu

-B-G

one

vacc

ine.

Th

e am

oun

t of

F

lu-B

-Gon

e va

ccin

e re

mai

nin

g in

th

e bl

oods

trea

m x

day

s af

ter

an i

nje

ctio

n i

s

giv

en b

y th

e eq

uat

ion

y =

y0 (

0.5)

x −

6 , w

her

e y 0

is t

he

init

ial

amou

nt.

Su

ppos

e a

doct

or

inje

cts

a pa

tien

t w

ith

20

μg

(mic

rogr

ams)

of

Flu

-B-G

one.

a. H

ow m

uch

of

the

vacc

ine

wil

l re

mai

n

afte

r 1

day?

Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

17

.8 μ

g

b.

How

mu

ch o

f th

e va

ccin

e w

ill

rem

ain

af

ter

12 d

ays?

Rou

nd

you

r an

swer

to

the

nea

rest

ten

th.

5

μg

c. A

fter

how

man

y da

ys w

ill

the

amou

nt

of v

acci

ne

be l

ess

than

1 μ

g?

afte

r 26

day

s

Wor

d Pr

oble

m P

ract

ice

Gro

wth

an

d D

ecay

Co

lleg

eS

ecto

rTu

itio

n a

nd

Fees

Ro

om

an

dB

oar

d

Fo

ur-

yea

r

Pu

blic

$5

94

1

$6

63

6

Fo

ur-

yea

r

Pri

vate

$2

1,2

35

$7,

79

1

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

3912

/20/

10

6:39

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

40

Gle

ncoe

Alg

ebra

1

7-6

You

can

use

the

form

ula

for

com

poun

d in

tere

st A

= P

(1 +

r −

n ) n

t whe

n n

, the

num

ber

of t

imes

a

year

int

eres

t is

com

poun

ded,

is

know

n. A

spe

cial

typ

e of

com

poun

d in

tere

st c

alcu

lati

on

regu

larl

y us

ed i

n fi

nanc

e is

con

tin

uou

sly

com

pou

nd

ed i

nte

rest

, whe

re n

app

roac

hes

infi

nity

.

INV

ESTI

NG

Mr.

Riv

era

pla

ced

$50

00 i

n a

n i

nve

stm

ent

acco

un

t th

at

has

an

in

tere

st r

ate

of 9

.5%

per

yea

r.

Exer

cise

s

1. S

AV

ING

S M

r. H

arri

s sa

ves

$20,

000

in a

mon

ey-m

arke

t ac

cou

nt

at a

rat

e of

5.2

%.

a. D

eter

min

e th

e va

lue

of h

is i

nve

stm

ent

afte

r 10

yea

rs i

f in

tere

st i

sco

mpo

un

ded

quar

terl

y. $

33,5

28.0

1

b.

Det

erm

ine

the

valu

e of

his

in

vest

men

t af

ter

10 y

ears

if

inte

rest

is

com

pou

nde

d co

nti

nu

ousl

y. $

33,6

40.5

4

2. C

OLL

EGE

SAV

ING

S S

han

non

is

choo

sin

g be

twee

n t

wo

diff

eren

t sa

vin

gs a

ccou

nts

for

her

co

lleg

e fu

nd.

Th

e fi

rst

acco

un

t co

mpo

un

ds i

nte

rest

sem

ian

nu

ally

at

a ra

te o

f 11

.0%

. Th

e se

con

d ac

cou

nt

com

pou

nds

in

tere

st c

onti

nu

ousl

y at

a r

ate

of 1

0.8%

. If

Sh

ann

on p

lan

s to

ke

ep h

er m

oney

in

th

e ac

cou

nt

for

5 ye

ars,

wh

ich

acc

oun

t sh

ould

sh

e ch

oose

? E

xpla

in.

Sh

e sh

ou

ld c

ho

ose

th

e co

nti

nu

ou

s co

mp

ou

nd

ing

acc

ou

nt

at

10.8

%. A

fter

5 y

ears

, th

e va

lue

of

the

con

tin

uo

usl

y co

mp

ou

nd

ing

acc

ou

nt

will

be

1.71

6P, w

her

eas

the

qu

arte

rly

com

po

un

din

g a

cco

un

t w

ill o

nly

h

ave

a va

lue

of

1.70

8P.

Enri

chm

ent

Co

nti

nu

ou

sly

Co

mp

ou

nd

ing

In

tere

st

Ge

ne

ral fo

rmu

la: A

= P

ert

A =

cu

rre

nt

am

ou

nt

of

inve

stm

en

t

P =

in

itia

l a

mo

un

t o

f in

vestm

en

t

e =

na

tura

l lo

ga

rith

m,

a c

on

sta

nt

ap

pro

xim

ate

ly e

qu

al to

2.7

18

28

r =

an

nu

al ra

te o

f in

tere

st,

exp

resse

d a

s a

de

cim

al

t =

nu

mb

er

of

yea

rs t

he

mo

ney is inve

ste

d

Exam

ple

a. H

ow m

uch

mon

ey w

ill

be

in t

he

acco

un

t af

ter

5 ye

ars

if i

nte

rest

is

com

pou

nd

ed m

onth

ly?

A

= P

(1 +

r −

n ) n

t Co

mpound inte

rest

equation

= 5

000

(1 +

0.09

5 −

12

) 12

(5) P

= 5

000,

r =

0.0

95,

n =

12,

and t

= 5

= 5

000(

1.00

79)60

S

implif

y.

= 8

025.

05

Use a

calc

ula

tor.

T

her

e w

ill

be a

bou

t $8

025.

05 i

n t

he

acco

un

t if

in

tere

st i

s co

mpo

un

ded

mon

thly

.

b.

How

mu

ch m

oney

wil

l b

e in

th

e ac

cou

nt

afte

r 5

year

s if

in

tere

st i

s co

mp

oun

ded

con

tin

uou

sly?

A

= P

ert

Continuous inte

rest

equation

= 5

000(

2.71

828)

0.09

5(5)

P =

5000,

e =

2.7

1828,

r =

0.0

95,

and t

= 5

= 5

000(

2.71

828)

0.47

5 S

implif

y.

= 8

04

0.0

7

Use a

calc

ula

tor.

T

her

e w

ill

be a

bou

t $8

040.

07 i

n t

he

acco

un

t if

in

tere

st i

s co

mpo

un

ded

con

tin

uou

sly.

021_

040_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4012

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-6

Cha

pte

r 7

41

Gle

ncoe

Alg

ebra

1

7-6

Spre

adsh

eet

Act

ivit

yC

om

po

un

d I

nte

rest

Exer

cise

s

Use

th

e sp

read

shee

t of

acc

oun

ts i

nvo

lvin

g co

mp

oun

d i

nte

rest

.

1. H

ow a

re t

he

dou

blin

g ti

mes

aff

ecte

d if

th

e ac

cou

nts

com

pou

nd

inte

rest

qu

arte

rly

inst

ead

of m

onth

ly?

Th

e 5%

an

d 8

% a

cco

un

ts d

ou

ble

in t

he

sam

e n

um

ber

of

year

s as

bef

ore

, bu

t th

e 10

% a

cco

un

t w

ill d

ou

ble

in 8

yea

rs.

2. H

ow l

ong

wil

l it

tak

e ea

ch a

ccou

nt

to r

each

$40

00 i

f th

e in

tere

st i

s co

mpo

un

ded

mon

thly

? qu

arte

rly?

mo

nth

ly:

5% in

28

yr, 8

% in

18

yr, 1

0% in

14

yr;

qu

arte

rly:

5%

in 2

8 yr

, 8%

in 1

8 yr

, 10%

in 1

5 yr

3. H

ow d

o th

e in

tere

st r

ate

and

the

nu

mbe

r of

tim

es t

he

inte

rest

is

com

pou

nde

d af

fect

th

e gr

owth

of

an i

nve

stm

ent?

Th

e ac

cou

nts

wit

h h

igh

er in

tere

st r

ate

and

inte

rest

co

mp

ou

nd

ed m

ore

oft

en g

row

mo

re q

uic

kly.

Ban

ks o

ften

use

spr

eads

hee

ts t

o ca

lcu

late

an

d st

ore

fin

anci

al d

ata.

On

e ap

plic

atio

n i

s to

ca

lcu

late

com

pou

nd

inte

rest

on

an

acc

oun

t.

U

se a

sp

read

shee

t to

fin

d t

he

tim

e it

wil

l ta

ke

an i

nve

stm

ent

of

$100

0 to

dou

ble

. Su

pp

ose

you

can

ch

oose

fro

m i

nve

stm

ents

th

at h

ave

ann

ual

in

tere

st r

ates

of

5%, 8

%, o

r 10

% c

omp

oun

ded

mon

thly

.T

he

com

pou

nd

inte

rest

equ

atio

n i

s A

= P

(1 +

r −

n ) n

t , wh

ere

P i

s th

e pr

inci

pal

or i

nit

ial

inve

stm

ent,

A i

s th

e fi

nal

am

oun

t of

th

e in

vest

men

t, r

is

the

ann

ual

in

tere

st r

ate,

n i

s th

e n

um

ber

of t

imes

in

tere

st i

s pa

id, o

r co

mpo

un

ded,

eac

h y

ear,

and

t is

th

e n

um

ber

of y

ears

. In

th

is c

ase,

P =

100

0 an

d n

= 1

2.

Ste

p 1

U

se C

olu

mn

A o

f th

e sp

read

shee

t fo

r th

e ye

ars.

Ste

p 2

C

olu

mn

s B

, C, a

nd

D c

onta

in t

he

form

ula

s fo

r th

e fi

nal

am

oun

ts o

f th

e in

vest

men

ts. F

orm

at t

he

cell

s in

th

ese

colu

mn

s as

cu

rren

cy s

o th

at t

he

amou

nts

are

sh

own

in

do

llar

s an

d ce

nts

. Eac

h f

orm

ula

w

ill

use

th

e va

lues

in

Col

um

n A

as

th

e va

lue

of t

. For

exa

mpl

e,

the

form

ula

th

at i

s in

cel

l C

3 is

=1

000*

(1 +

(0.0

8/12

))^(1

2 *

A3)

.S

tudy

th

e sp

read

shee

t fo

r th

e ti

mes

wh

en

each

in

vest

men

t ex

ceed

s $2

000.

At

5%, t

he

$100

0 w

ill

dou

ble

in 1

4 ye

ars,

at

8% i

t w

ill

dou

ble

in 9

yea

rs, a

nd

at 1

0% i

t w

ill

dou

ble

in 7

yea

rs.

A1 3 4 5 6 7 8 9 10 112

BC

D

12 13 14 15 16

Yea

rs$

1,0

51

.16

$1

,10

4.9

4

$1

,16

1.4

7

$1

,22

0.9

0

$1

,28

3.3

6

$1

,34

9.0

2

$1

,41

8.0

4

$1

,49

0.5

9

$1

,56

6.8

5

$1

,64

7.0

1

$1

,73

1.2

7

$1

,81

9.8

5

$1

,91

2.9

6

$2

,01

0.8

3

5%8%

10%

$1

,08

3.0

0

$1

,17

2.8

9

$1

,27

0.2

4

$1

,37

5.6

7

$1

,48

9.8

5

$1

,61

3.5

0

$1

,74

7.4

2

$1

,89

2.4

6

$2

,04

9.5

3

$2

,21

9.6

4

$2

,40

3.8

7

$2

,60

3.3

9

$2

,81

9.4

7

$3

,05

3.4

8

$1

,10

4.7

1

$1

,22

0.3

9

$1

,34

8.1

8

$1

,48

9.3

5

$1

,64

5.3

0

$1

,81

7.5

9

$2

,00

7.9

2

$2

,21

8.1

8

$2

,45

0.4

5

$2

,70

7.0

4

$2

,99

0.5

0

$3

,30

3.6

5

$3

,64

9.5

8

$4

,03

1.7

4

1 2 3 4 5 6 7 8 9

10

11

12

13

14

Sh

eet

1S

hee

t 2

Sh

eet

3

Exam

ple

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4112

/20/

10

6:39

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 7

42

Gle

ncoe

Alg

ebra

1

7-7

Rec

og

niz

e G

eom

etri

c Se

qu

ence

s A

geo

met

ric

sequ

ence

is

a se

quen

ce i

n w

hic

h

each

ter

m a

fter

th

e fi

rst

is f

oun

d by

mu

ltip

lyin

g th

e pr

evio

us

term

by

a n

onze

ro c

onst

ant

r ca

lled

th

e co

mm

on r

atio

. Th

e co

mm

on r

atio

can

be

fou

nd

by d

ivid

ing

any

term

by

its

prev

iou

s te

rm.

Exer

cise

s

Det

erm

ine

wh

eth

er e

ach

seq

uen

ce i

s a

rith

met

ic, g

eom

etri

c, o

r n

eith

er. E

xpla

in.

1. 1

, 2, 4

, 8, .

. .

2. 9

, 14,

6, 1

1, .

. .

3.

2 −

3 ,

1 −

3 ,

1 −

6 ,

1 −

12 ,

. . .

4. –

2, 5

, 12,

19,

. . .

Fin

d t

he

nex

t th

ree

term

s in

eac

h g

eom

etri

c se

qu

ence

.

5. 6

48, –

216,

72,

. . .

6.

25,

–5,

1, .

. .

7.

1 −

16 ,

1 −

2 ,

4, .

. .

8. 7

2, 3

6, 1

8, .

. .

Stud

y G

uide

and

Inte

rven

tion

Geo

metr

ic S

eq

uen

ces a

s E

xp

on

en

tial

Fu

ncti

on

s

Exam

ple

1

Det

erm

ine

wh

eth

er t

he

seq

uen

ce i

s a

rith

met

ic, g

eom

etri

c, o

r n

eith

er: 2

1, 6

3, 1

89, 5

67, .

. .

Fin

d th

e ra

tios

of

the

con

secu

tive

ter

ms.

If

th

e ra

tios

are

con

stan

t, t

he

sequ

ence

is

geom

etri

c.21

63

18

9 56

7

63

−

21

=

18

9 −

63

=

567

−

189

= 3

Bec

ause

the

rat

ios

are

cons

tant

, the

seq

uenc

e is

geo

met

ric.

The

com

mon

rat

io is

3.

F

ind

th

e n

ext

thre

e te

rms

in t

his

geo

met

ric

seq

uen

ce:

-12

15, 4

05, -

135,

45,

. . .

Ste

p 1

F

ind

the

com

mon

rat

io.

–121

5 40

5 –1

35

45

40

5 −

-

1215

=

-

135

−

405

=

45

−

-13

5 =

-

1

−

3

Th

e va

lue

of r

is

- 1 −

3 .

Ste

p 2

M

ult

iply

eac

h t

erm

by

the

com

mon

ra

tio

to f

ind

the

nex

t th

ree

term

s.

45

–1

5

5

- 5

−

3

× (-

1 −

3 )

×

(-

1

−

3 )

× (-

1 −

3 )

Th

e n

ext

thre

e te

rms

of t

he

sequ

ence

ar

e –15

, 5, a

nd

- 5

−

3

.

Exam

ple

2

Geo

met

ric;

co

mm

on

rat

io is

2.

Nei

ther

; th

ere

is n

o c

om

mo

n

diff

eren

ce o

r ra

tio

.

Geo

met

ric;

co

mm

on

rat

io is

1 −

2 .A

rith

met

ic; c

om

mo

n d

iffer

ence

is 7

.

-24

, 8, -

2 2 −

3

32, 2

56, 2

048

- 1 −

5 , 1 −

25 ,

-

1 −

125

9, 4

1 −

2 , 2 1 −

4

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4212

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-7

Cha

pte

r 7

43

Gle

ncoe

Alg

ebra

1

7-7

Geo

met

ric

Seq

uen

ces

and

Fu

nct

ion

s T

he

nth

ter

m a

n o

f a

geom

etri

c se

quen

ce

wit

h f

irst

ter

m a

1 an

d co

mm

on r

atio

r i

s gi

ven

by

the

foll

owin

g fo

rmu

la, w

her

e n

is

any

posi

tive

in

tege

r: a

n =

a1

· rn

– 1.

Exer

cise

s

1. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

–2,

10,

–50

, . .

. .

Fin

d th

e el

even

th t

erm

of

this

seq

uen

ce.

2. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

512

, 128

, 32,

. . .

. F

ind

the

sixt

h t

erm

of

this

seq

uen

ce.

3. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

4 −

9 ,

4, 3

6, .

. . .

Fin

d th

e ei

ghth

ter

m o

f th

is s

equ

ence

.

4. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

6, –

54, 4

86, .

. . .

F

ind

the

nin

th t

erm

of

this

seq

uen

ce.

5. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

100

, 80,

64,

. . .

. F

ind

the

seve

nth

ter

m o

f th

is s

equ

ence

.

6. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

2 −

5 ,

1 −

10 ,

1 −

40

, . .

. .

Fin

d th

e si

xth

ter

m o

f th

is s

equ

ence

.

7. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

3 −

8 ,

- 3 −

2 ,

6, .

. . .

Fin

d th

e te

nth

ter

m o

f th

is s

equ

ence

.

8. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

–3,

–21

, –14

7, .

. . .

Fin

d th

e fi

fth

ter

m o

f th

is s

equ

ence

.

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Geo

metr

ic S

eq

uen

ces a

s E

xp

on

en

tial

Fu

ncti

on

s

Exam

ple

a.

Wri

te a

n e

qu

atio

n f

or

the

nth

ter

m o

f th

e ge

omet

ric

seq

uen

ce

5, 2

0, 8

0, 3

20, .

. .

Th

e fi

rst

term

of

the

sequ

ence

is

320.

So,

a 1

= 3

20. N

ow f

ind

the

com

mon

rat

io.

5

20

80

320

20

−

5 =

80

−

20

=

32

0 −

80

=

4

Th

e co

mm

on r

atio

is

4. S

o, r

= 4

.a n

= a

1 ·

rn –

1

Form

ula

for

nth t

erm

a n =

5 ·

4n

– 1

a 1 =

5 a

nd r

= 4

b.

Fin

d t

he

seve

nth

ter

m o

f th

is

seq

uen

ce.

Bec

ause

we

are

look

ing

for

the

seve

nth

ter

m,

n =

7.

a n =

a1

· rn

– 1

Form

ula

for

nth t

erm

a 7 = 5

· 4

7 – 1

n =

7

=

5 ·

46

Sim

plif

y.

=

5 ·

409

6 4

6 =

4096

=

20,

480

Multip

ly.

Th

e se

ven

th t

erm

of

the

sequ

ence

is

20,4

80.

an =

-

3 · 7n

- 1; -

7203

an =

3 −

8 · (-

4)n

- 1; -

98,3

04

an =

2 −

5 · (

1 −

4 ) n -

1 ;

1 −

2560

an =

10

0 · (

4 −

5 ) n -

1 ; 2

6 13

4 −

625

an =

6

· (-

9)n

- 1; 2

58,2

80,3

26

an =

4 −

9 · 9

n -

1; 2

,125

,764

an =

51

2 · ( 1

−

4 ) n -

1 ;

1 −

2

an =

-

2 · (-

5)n

- 1; -

19,5

31,2

50

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4312

/20/

10

6:39

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.

PDF 2nd



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

44

Gle

ncoe

Alg

ebra

1

7-7

Det

erm

ine

wh

eth

er e

ach

seq

uen

ce i

s a

rith

met

ic, g

eom

etri

c, o

r n

eith

er. E

xpla

in.

1. 7

, 13,

19,

25,

…

2. –

96, –

48, –

24, –

12, …

3. 1

08, 6

6, 1

41, 9

9, …

4.

3, 9

, 81,

656

1, …

5.

7 −

3 ,

14,

84,

504

, …

6. 3 −

8 ,

- 1 −

8 , -

5 −

8 ,

- 9 −

8 ,

…

Fin

d t

he

nex

t th

ree

term

s in

eac

h g

eom

etri

c se

qu

ence

.

7. 2

500,

500

, 100

, …

8. 2

, 6, 1

8, …

9. –

4, 2

4, –

144,

…

10.

4 −

5 ,

2 −

5 ,

1 −

5 ,

…

11. –

3, –

12, –

48, …

12

. 72,

12,

2, …

13. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

3, –

24,

192

, ….

Fin

d th

e n

inth

ter

m o

f th

is s

equ

ence

.

14. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

9 −

16

, 3 −

8 ,

1 −

4 ,

….

Fin

d th

e se

ven

th t

erm

of

this

seq

uen

ce.

15. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

100

0, 2

00, 4

0, …

. F

ind

the

fift

h t

erm

of

this

seq

uen

ce.

16. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

– 8

, – 2

, - 1 −

2 ,

….

Fin

d th

e ei

ghth

ter

m o

f th

is s

equ

ence

.

17. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

32,

48,

72,

….

Fin

d th

e si

xth

ter

m o

f th

is s

equ

ence

.

18. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

3

−

100 ,

3 −

10

, 3,

….

Fin

d th

e n

inth

ter

m o

f th

is s

equ

ence

.

Skill

s Pr

acti

ceG

eo

metr

ic S

eq

uen

ces a

s E

xp

on

en

tial

Fu

ncti

on

s

Ari

thm

etic

; co

mm

on

d

iffer

ence

is 6

.G

eom

etri

c; t

he

com

mo

n

rati

o is

1 −

2 .

Nei

ther

; th

ere

is n

o c

om

mo

n

diff

eren

ce o

r ra

tio

.N

eith

er; t

her

e is

no

co

mm

on

d

iffer

ence

or

rati

o.

Geo

met

ric;

co

mm

on

ra

tio

is 6

.A

rith

met

ic; c

om

mo

n

diff

eren

ce is

-

1 −

2 .

20, 4

, 4 −

5 54

, 162

, 486

864;

-51

84; 3

1,10

4

-19

2, -

768,

-30

72

1 −

10 ,

1 −

20 ,

1 −

40

1 −

3 , 1 −

18 ,

1

−

108

3(-

8) n

- 1 ;

50,

331,

648

3

−

100 ·

10

n -

1 ;

3,0

00,

00

0

9 −

16 ( 2

−

3 ) n -

1 ;

4 −

81

100

0 ( 1 −

5 ) n

- 1 ; 8

−

5

32 ( 3

−

2 ) n -

1

; 243

-8 (

1 −

4 ) n -

1 ; -

1

−

2048

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4412

/24/

10

9:22

AM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-7

Cha

pte

r 7

45

Gle

ncoe

Alg

ebra

1

7-7

Det

erm

ine

wh

eth

er e

ach

seq

uen

ce i

s a

rith

met

ic, g

eom

etri

c, o

r n

eith

er. E

xpla

in.

1. 1

, -5,

-11

, -17

, …

2. 3

, 3 −

2 ,

1, 3 −

4 ,

…

3. 1

08, 3

6, 1

2, 4

, …

4. -

2, 4

, -6,

8, …

Fin

d t

he

nex

t th

ree

term

s in

eac

h g

eom

etri

c se

qu

ence

.

5. 6

4, 1

6, 4

, …

6. 2

, -12

, 72,

…

7. 3

750,

750

, 150

, …

8. 4

, 28,

196

, …

9. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

896

, -44

8, 2

24, …

. F

ind

the

eigh

th t

erm

of

this

seq

uen

ce.

10. W

rite

an

equ

atio

n f

or t

he

nth

ter

m o

f th

e ge

omet

ric

sequ

ence

358

4, 8

96, 2

24, …

. F

ind

the

sixt

h t

erm

of

this

seq

uen

ce.

11. F

ind

the

sixt

h t

erm

of

a ge

omet

ric

sequ

ence

for

wh

ich

a2 =

288

an

d r

= 1

−

4

.

12. F

ind

the

eigh

th t

erm

of

a ge

omet

ric

sequ

ence

for

wh

ich

a3 =

35

and

r =

7.

13. P

ENN

IES

Th

omas

is

savi

ng

pen

nie

s in

a ja

r. T

he

firs

t da

y h

e sa

ves

3 pe

nn

ies,

th

e se

con

d da

y 12

pen

nie

s, t

he

thir

d da

y 48

pen

nie

s, a

nd

so o

n. H

ow m

any

pen

nie

s do

es

Th

omas

sav

e on

th

e ei

ghth

day

?

Prac

tice

Geo

metr

ic S

eq

uen

ces a

s E

xp

on

en

tial

Fu

ncti

on

s

Ari

thm

etic

; th

e co

mm

on

d

iffer

ence

is -

6.N

eith

er; t

her

e is

no

co

mm

on

d

iffer

ence

or

rati

o.

Geo

met

ric;

th

e co

mm

on

ra

tio

is 1 −

3 .N

eith

er; t

her

e is

no

co

mm

on

d

iffer

ence

or

rati

o.

1; 1 −

4 ; 1 −

16

-43

2; 2

592;

-15

,552

30; 6

; 6 −

5 13

72; 9

604;

67,

228

an =

896

(- 1 −

2 ) n -

1 ;

-7

an =

358

4 ( 1 −

4 ) n -

1 ;

3.5

1.12

5

588,

245

49,1

52

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4512

/20/

10

6:39

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 7

46

Gle

ncoe

Alg

ebra

1

7-7

1. W

OR

LD P

OPU

LATI

ON

Th

e C

IA

esti

mat

es t

he

wor

ld p

opu

lati

on i

s gr

owin

g at

a r

ate

of 1

.167

% e

ach

yea

r. T

he

wor

ld p

opu

lati

on f

or 2

007

was

abo

ut

6.6

bill

ion

.

a. W

rite

an

equ

atio

n f

or t

he

wor

ld

popu

lati

on a

fter

n y

ears

. (H

int:

Th

e co

mm

on r

atio

is

not

0.0

1167

.)

b.

Wh

at w

ill

the

esti

mat

ed w

orld

po

pula

tion

be

in 2

017?

2. M

USE

UM

S T

he

tabl

e sh

ows

the

ann

ual

vi

sito

rs t

o a

mu

seu

m i

n m

illi

ons.

Wri

te

an e

quat

ion

for

th

e pr

ojec

ted

nu

mbe

r of

vi

sito

rs a

fter

n y

ears

.

3. B

AN

KIN

G A

rnol

d h

as a

ban

k ac

cou

nt

wit

h a

beg

inn

ing

bala

nce

of

$500

0. H

e sp

ends

on

e-fi

fth

of

the

bala

nce

eac

h

mon

th. H

ow m

uch

mon

ey w

ill

be i

n t

he

acco

un

t af

ter

6 m

onth

s?

4. P

OPU

LATI

ON

Th

e ta

ble

show

s th

e pr

ojec

ted

popu

lati

on o

f th

e U

nit

ed S

tate

s th

rou

gh 2

050.

Doe

s th

is t

able

sh

ow a

n

arit

hm

etic

seq

uen

ce, a

geo

met

ric

sequ

ence

or n

eith

er?

Exp

lain

.

Sour

ce: U

.S. C

ensu

s B

urea

u

5. S

AV

ING

S A

CC

OU

NTS

A b

ank

offe

rs a

sa

vin

gs a

ccou

nt

wit

h a

0.5

% r

etu

rn e

ach

m

onth

.

a. W

rite

an

equ

atio

n f

or t

he

bala

nce

of

a sa

vin

gs a

ccou

nt

afte

r n

mon

ths.

(H

int:

Th

e co

mm

on r

atio

is

not

0.0

05.)

b.

Giv

en a

n i

nit

ial

depo

sit

of $

500,

w

hat

wil

l th

e ac

cou

nt

bala

nce

be

afte

r 15

mon

ths?

Wor

d Pr

oble

m P

ract

ice

Geo

metr

ic S

eq

uen

ces a

s E

xp

on

en

tial

Fu

ncti

on

s

Year

Vis

itors

(mill

ion

s)

14

26

39

413

1

−

2

n?

Year

Pro

ject

edP

op

ula

tio

n

20

00

28

2,1

25

,00

0

2010

30

8,9

36

,00

0

20

20

33

5,8

05

,00

0

20

30

36

3,5

84

,00

0

20

40

39

1,9

46

,00

0

20

50

419

,85

4,0

00

an =

6,

600,

00

0,0

00

· 1.

0116

7 n

- 1

abo

ut

7.4

bill

ion

an =

4

· ( 3 −

2 ) n -

1

$131

0.72

Nei

ther

, th

ere

is n

o c

om

mo

n r

atio

or

diff

eren

ce.

an =

p

· 1.

005

n

$538

.84

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4612

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7 -7

Cha

pte

r 7

47

Gle

ncoe

Alg

ebra

1

7-7

Pay i

t Fo

rward

Th

e id

ea b

ehin

d “p

ay i

t fo

rwar

d” i

s th

at o

n t

he

firs

t da

y, o

ne

pers

on d

oes

a go

od d

eed

for

thre

e di

ffer

ent

peop

le. T

hen

, on

th

e se

con

d da

y, t

hos

e th

ree

peop

le w

ill

each

per

form

goo

d de

eds

for

3 m

ore

peop

le, s

o th

at o

n D

ay 2

, th

ere

are

3 ×

3 o

r 9

good

dee

ds b

ein

g do

ne.

C

onti

nu

e th

is p

roce

ss t

o fi

ll i

n t

he

char

t. A

tre

e di

agra

m w

ill

hel

p yo

u f

ill

in t

he

char

t.

1. G

raph

th

e da

ta y

ou f

oun

d in

th

e ch

art

as o

rder

ed p

airs

an

d co

nn

ect

wit

h a

sm

ooth

cu

rve.

2. W

hat

typ

e of

fu

nct

ion

is

you

r gr

aph

fro

m E

xerc

ise

1?

Wri

te a

n e

quat

ion

th

at c

an b

e u

sed

to d

eter

min

e th

e n

um

ber

of g

ood

deed

s on

an

y gi

ven

day

, x.

exp

on

enti

al;

y =

3x

3. H

ow m

any

good

dee

ds w

ill

be p

erfo

rmed

on

Day

21?

10

,460

,353

,203

Th

e fo

rmu

la T

= 3 n

+ 1 -

3

−

2 c

an b

e u

sed

to d

eter

min

e th

e to

tal

nu

mbe

r of

goo

d de

eds

that

hav

e be

en p

erfo

rmed

, wh

ere

n r

epre

sen

ts t

he

day.

For

exa

mpl

e, o

n D

ay 2

, th

ere

hav

e be

en

3 +

9 o

r 12

goo

d de

eds

perf

orm

ed. U

sin

g th

e fo

rmu

la, y

ou g

et,

3 2 +

1 -

3

−

2 o

r 3 3

- 3

−

2 +

27 -

3

−

2 o

r a

tota

l of

12

good

dee

ds p

erfo

rmed

.

4. U

se t

he

form

ula

to

dete

rmin

e th

e ap

prox

imat

e n

um

ber

of g

ood

deed

s th

at h

ave

been

pe

rfor

med

th

rou

gh D

ay 2

1.

15,6

90,5

29,8

00

5. L

ook

up

the

wor

ld p

opu

lati

on. H

ow d

oes

you

r n

um

ber

from

Exe

rcis

e 4

com

pare

to

the

wor

ld p

opu

lati

on?

T

he

wo

rld

po

pu

lati

on

is a

pp

roxi

mat

ely

6,56

0,52

6,04

6. T

he

nu

mb

er o

f g

oo

d d

eed

s p

erfo

rmed

is g

reat

er t

han

th

e en

tire

wo

rld

po

pu

lati

on

.

Per

son

1

Day

2D

ay 1

New

Per

son

2

New

Per

son

1

New

Per

son

3

y

xO240

210

180

120

150 90 60 30

43

27

86

51

Day

# o

f D

eed

s

01

13

29

3 27

481

524

3Enri

chm

ent

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4712

/20/

10

6:39

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

48

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

R

ecu

rsiv

e F

orm

ula

s

7-8

Usi

ng

Rec

urs

ive

Form

ula

s A

rec

urs

ive

form

ula

all

ows

you

to

fin

d th

e n

th t

erm

of

a se

quen

ce b

y pe

rfor

min

g op

erat

ion

s on

on

e or

mor

e of

th

e te

rms

that

pre

cede

it.

F

ind

th

e fi

rst

five

ter

ms

of t

he

seq

uen

ce i

n w

hic

h a

1 =

5 a

nd

a

n =

-2 a

n -

1 +

14,

if

n ≥

2.

Th

e gi

ven

fir

st t

erm

is

a 1

= 5

. Use

th

is t

erm

an

d th

e re

curs

ive

form

ula

to

fin

d th

e n

ext

fou

r te

rms.

a

2 =

-2 a

2 -

1 +

14

n =

2

a 4

= -

2 a 4

- 1 +

14

n =

4

=

-2 a

1 +

14

Sim

plif

y.

= -

2 a 3

+ 1

4 S

implif

y.

=

-2(

5) +

14

or 4

a 1

= 5

=

-2(

6) +

14

or 2

a 3

= 6

a

3 =

-2 a

3 -

1 +

14

n =

3

a 5

= -

2 a 5

- 1 +

14

n =

5

=

-2 a

2 +

14

Sim

plif

y.

= -

2 a 4

+ 1

4 S

implif

y.

=

-2(

4) +

14

or 6

a 2

= 4

=

-2(

2) +

14

or 1

0 a 4

= 2

Th

e fi

rst

five

ter

ms

are

5, 4

, 6, 2

, an

d 10

.

Exe

rcis

esF

ind

th

e fi

rst

five

ter

ms

of e

ach

seq

uen

ce.

1. a

1 =

-4,

a n =

3 a

n -

1 , n

≥ 2

2.

a 1

= 5

, a n =

2a

n -

1 , n

≥ 2

-

4, -

12, -

36, -

108,

-32

4 5

, 10,

20,

40,

80

3. a

1 =

8, a

n =

a n

- 1 -

6, n

≥ 2

4.

a 1

= -

32, a

n =

a n

- 1 +

13,

n ≥

2

8

, 2, -

4, -

10, -

16

-32

, -19

, -6,

7, 2

0

5. a

1 =

6, a

n =

-3 a

n -

1 +

20,

n ≥

2

6. a

1 =

-9,

a n =

2 a

n -

1 +

11,

n ≥

2

6

, 2, 1

4, -

22, 8

6 -

9, -

7, -

3, 5

, 21

7. a

1 =

12,

a n =

2 a

n -

1 -

10,

n ≥

2

8. a

1 =

-1,

a n =

4 a

n -

1 +

3, n

≥ 2

1

2, 1

4, 1

8, 2

6, 4

2 -

1, -

1, -

1, -

1, -

1

9. a

1 =

64,

a n =

0.5

a n

- 1 +

8, n

≥ 2

10

. a 1

= 8

, a n =

1.5

a n

- 1 , n

≥ 2

6

4, 4

0, 2

8, 2

2, 1

9 8

, 12,

18,

27,

40.

5

11. a

1 =

400

, a n =

1 −

2 a

n -

1 , n

≥ 2

12

. a 1

= 1 −

4 ,

a n =

a n

- 1 +

3 −

4 ,

n ≥

2

4

00,

20

0, 1

00,

50,

25

1 −

4 , 1,

7 −

4 , 5 −

2 , 13

−

4

Exam

ple

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4812

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-8

Cha

pte

r 7

49

Gle

ncoe

Alg

ebra

1

Stud

y G

uide

and

Inte

rven

tion

(co

nti

nu

ed)

Recu

rsiv

e F

orm

ula

s

Wri

tin

g R

ecu

rsiv

e Fo

rmu

las

Com

plet

e th

e fo

llow

ing

step

s to

wri

te a

rec

urs

ive

form

ula

for

an

ari

thm

etic

or

geom

etri

c se

quen

ce.

Ste

p 1

De

term

ine

if

the

se

qu

en

ce

is a

rith

me

tic o

r g

eo

me

tric

by fi n

din

g a

co

mm

on

diff

ere

nce

or

a c

om

mo

n r

atio.

Ste

p 2

Wri

te a

re

cu

rsiv

e fo

rmu

la.

Ari

thm

etic

Seq

uen

ces

a n

= a

n -

1 +

d,

wh

ere

d is t

he

co

mm

on

diff

ere

nce

Geo

met

ric

Seq

uen

ces

a n =

r ·

an

- 1,

wh

ere

r is t

he

co

mm

on

ra

tio

Ste

p 3

Sta

te t

he

fi r

st

term

an

d t

he

do

ma

in fo

r n.

W

rite

a r

ecu

rsiv

e fo

rmu

la f

or 2

16, 3

6, 6

, 1, …

.

Ste

p 1

F

irst

su

btra

ct e

ach

ter

m f

rom

th

e te

rm t

hat

fol

low

s it

.21

6 -

36

= 1

80

36 −

6 =

30

6 -

1 =

5T

her

e is

no

com

mon

dif

fere

nce

. Ch

eck

for

a co

mm

on r

atio

by

divi

din

g ea

ch t

erm

by

the

term

th

at p

rece

des

it.

36

−

216 =

1 −

6

6 −

36 =

1 −

6

1 −

6 =

1 −

6

Th

ere

is a

com

mon

rat

io o

f 1 −

6 . T

he

sequ

ence

is

geom

etri

c.

Ste

p 2

U

se t

he

form

ula

for

a g

eom

etri

c se

quen

ce.

a n =

r ·

a n

- 1

Recurs

ive form

ula

for

geom

etr

ic s

equence

a n =

1 −

6 a

n -

1

r = 1

−

6

Ste

p 3

T

he

firs

t te

rm a

1 is

216

an

d n

≥ 2

.

A r

ecu

rsiv

e fo

rmu

la f

or t

he

sequ

ence

is

a 1

= 2

16, a

n =

1 −

6 a

n -

1 , n

≥ 2

.

Exer

cise

sW

rite

a r

ecu

rsiv

e fo

rmu

la f

or e

ach

seq

uen

ce.

1. 2

2, 1

6, 1

0, 4

, …

2. -

8, -

3, 2

, 7, …

a

1 =

22,

a n =

a n

- 1 -

6, n

≥ 2

a

1 =

-8,

a n =

a n

- 1 +

5, n

≥ 2

3. 5

, 15,

45,

135

, …

4. 2

43, 8

1, 2

7, 9

, …

a

1 =

5, a

n =

3 a

n -

1 , n

≥ 2

a

1 =

243

, a n =

1 −

3 a n

- 1 , n

≥ 2

5. −

3, 1

4, 3

1, 4

8, …

6.

8, -

20, 5

0, -

125,

…

a

1 =

-3,

a n =

a n

- 1 +

17,

n ≥

2

a 1

= 8

, a n =

-2.

5 a n

− 1 , n

≥ 2

7-8

Exam

ple

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

4912

/20/

10

6:39

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 7

50

Gle

ncoe

Alg

ebra

1

Skill

s Pr

acti

ceR

ecu

rsiv

e F

orm

ula

s

7-8

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h s

equ

ence

.

1. a

1 =

-11

, a n =

a n

− 1 −

7, n

≥ 2

2.

a 1

= 1

4, a

n =

a n

− 1 +

6.5

, n ≥

2

-11

, -18

, -25

, -32

, -39

1

4, 2

0.5,

27,

33.

5, 4

0

3. a

1 =

-3,

a n =

3a

n −

1 +

10,

n ≥

2

4. a

1 =

187

.5, a

n =

−0.

8 a n

- 1 , n

≥ 2

-

3, 1

, 13,

49,

157

1

87.5

, -15

0, 1

20, -

96, 7

6.8

5. a

1 =

1 −

2 ,

a n =

3 −

2 a

n -

1 +

1 −

4 ,

n ≥

2

6. a

1 =

1 −

5 ,

a n =

5 −

2 a

n -

1 -

1 −

2 ,

n ≥

2

1 −

2 , 1,

7 −

4 , 23

−

8 , 73

−

16

1 −

5 , 0,

-

1 −

2 , -

7 −

4 , -

39

−

8

Wri

te a

rec

urs

ive

form

ula

for

eac

h s

equ

ence

.

7. 7

, 28,

112

, 448

, …

8. 7

3, 5

2, 3

1, 1

0, …

a

1 =

7, a

n =

4 a

n −

1 , n

≥ 2

a

1 =

73,

a n =

a n

- 1 -

21,

n ≥

2

9. 4

, 1,

1 −

4 ,

1 −

16 ,

…

10. -

37, -

15, 7

, 29,

…

a

1 =

4, a

n =

1 −

4 a n

− 1 , n

≥ 2

a

1 =

-37

, a n =

a n

- 1 +

22,

n ≥

2

11. 0

.25,

0.3

7, 0

.49,

0.6

1, …

12

. 64,

-80

, 100

, -12

5, 1

56.2

5, …

a

1 =

0.2

5, a

n =

a n

- 1 +

0.1

2, n

≥ 2

a

1 =

64,

a n =

-1.

25 a

n -

1 , n

≥ 2

For

eac

h r

ecu

rsiv

e fo

rmu

la, w

rite

an

exp

lici

t fo

rmu

la. F

or e

ach

exp

lici

t fo

rmu

la,

wri

te a

rec

urs

ive

form

ula

.

13. a

1 =

38,

a n =

a n

- 1 -

17,

n ≥

2

14. a

n =

5n

- 1

6

a n =

-17

n +

55

a 1

= -

11, a

n =

a n

- 1 +

5, n

≥ 2

15. a

n =

50(

0.75

) n -

1

16. a

1 =

16,

a n =

4 a

n −

1 , n

≥ 2

a

1 =

50,

a n =

0.7

5 a n

- 1 , n

≥ 2

a

n =

16(

4 ) n

- 1

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

5012

/20/

10

6:39

PM


NA

ME

DAT

E

P

ER

IOD

Lesson 7-8

Cha

pte

r 7

51

Gle

ncoe

Alg

ebra

1

Prac

tice

Recu

rsiv

e F

orm

ula

s

Fin

d t

he

firs

t fi

ve t

erm

s of

eac

h s

equ

ence

.

1. a

1 =

25,

a n =

a n

- 1 -

12,

n ≥

2

2. a

1 =

-10

1, a

n =

a n

- 1 +

38,

n ≥

2

25,

13,

1, -

11, -

23

-10

1, -

63, -

25, 1

3, 5

1

3. a

1 =

3.3

, a n =

a n

- 1 +

2.7

, n ≥

2

4. a

1 =

7, a

n =

-3 a

n -

1 +

20,

n ≥

2

3.3

, 6, 8

.7, 1

1.4,

14.

1 7

, -1,

23,

-49

, 167

5. a

1 =

20,

a n =

1 −

5 a

n -

1 , n

≥ 2

6.

a 1

= 2 −

3 ,

a n =

1 −

3 a

n -

1 -

2 −

9 , n

≥ 2

2

0, 4

, 4 −

5 , 4 −

25 ,

4 −

125

2 −

3 , 0,

-

2 −

9 , -

8 −

27 ,

-

26

−

81

Wri

te a

rec

urs

ive

form

ula

for

eac

h s

equ

ence

.

7. 8

0, -

40, 2

0, -

10, …

8.

87,

52,

17,

-18

, …

a

1 =

80,

a n =

-0.

5 a n

- 1 , n

≥ 2

a

1 =

87,

a n =

a n

- 1 -

35,

n ≥

2

9.

1 −

3 ,

4 −

15 ,

16

−

75 ,

64

−

375 ,

…

10.

4 −

5 ,

3 −

10 ,

- 1 −

5 ,

- 7 −

10

, …

a

1 =

1 −

3 , a

n =

4 −

5 a n

- 1 , n

≥ 2

a

1 =

4 −

5 , a

n =

a n

- 1 -

1 −

2 , n

≥ 2

11. 2

.6, 5

.2, 7

.8, 1

0.4,

…

12. 1

00, 1

20, 1

44, 1

72.8

, …

a 1

= 2

.6, a

n =

a n

- 1 +

2.6

, n ≥

2

a 1

= 1

00,

a n =

1.2

a n

- 1 , n

≥ 2

13. P

IZZA

Th

e to

tal

cost

s fo

r or

deri

ng

one

to f

ive

chee

se

pizz

as f

rom

Lu

igi’s

Piz

za P

alac

e ar

e sh

own

.

a. W

rite

a r

ecu

rsiv

e fo

rmu

la f

or t

he

sequ

ence

. a 1

= 7

, a n =

a n

- 1 +

5.5

, n ≥

2

b.

Wri

te a

n e

xpli

cit

form

ula

for

th

e se

quen

ce.

an =

5.5

n +

1.5

7-8

Tota

l Nu

mb

er o

f P

izza

s O

rder

edC

ost

1 $

7.0

0

2$

12

.50

3$

18

.00

4$

23

.50

5$

29

.00

041_

054_

ALG

1_A

_CR

M_C

07_C

R_6

6028

1.in

dd

5112

/20/

10

6:39

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.

PDF Pass



NA

ME

DAT

E

P

ER

IOD

Cha

pte

r 7

52

Gle

ncoe

Alg

ebra

1

Wor

d Pr

oble

m P

ract

ice

Recu

rsiv

e F

orm

ula

s

1. A

SSEM

BLY

An

ass

embl

y li

ne

can

cre

ate

175

wid

gets

in

on

e h

our,

350

wid

gets

in

tw

o h

ours

, 525

wid

gets

in

th

ree

hou

rs,

700

wid

gets

in

fou

r h

ours

, an

d so

on

.

a. F

ind

the

nex

t 5

term

s in

th

e se

quen

ce.

875,

105

0, 1

225,

140

0, 1

575

b.

Wri

te a

rec

urs

ive

form

ula

for

th

e se

quen

ce.

a 1

= 1

75, a

n=

a n

- 1

+ 1

75, n

≥ 2

c. W

rite

an

exp

lici

t fo

rmu

la f

or t

he

sequ

ence

. a

n=

175

n

2. P

APE

R A

pie

ce o

f pa

per

is f

olde

d se

vera

l ti

mes

. Th

e n

um

ber

of s

ecti

ons

into

wh

ich

th

e pi

ece

of p

aper

is

divi

ded

afte

r ea

ch

fold

is

show

n b

elow

.

Nu

mb

er o

f Fo

lds

Sec

tio

ns

12

24

38

416

53

2

a. F

ind

the

nex

t 5

term

s in

th

e se

quen

ce.

64, 1

28, 2

56, 5

12, 1

024

b.

Wri

te a

rec

urs

ive

form

ula

for

th

e se

quen

ce.

a 1

= 2

, a n =

2 a

n -

1 , n

≥ 2

c. W

rite

an

exp

lici

t fo

rmu

la f

or t

he

sequ

ence

. a

n =

2 n

3. C

LEA

NIN

G A

n e

quat

ion

for

th

e co

st

a n i

n d

olla

rs t

hat

a c

arpe

t cl

ean

ing

com

pan

y ch

arge

s fo

r cl

ean

ing

n r

oom

s is

a

n=

50

+ 2

5(n

- 1

). W

rite

a r

ecu

rsiv

e fo

rmu

la t

o re

pres

ent

the

cost

a n.

a 1

= 5

0, a

n=

a n

- 1

+ 2

5, n

≥ 2

4. S

AV

ING

S A

rec

urs

ive

form

ula

for

th

e ba

lan

ce o

f a

savi

ngs

acc

oun

t a

n i

n

doll

ars

at t

he

begi

nn

ing

of y

ear

n i

s a

1 =

500

, an =

1.0

5 a n

- 1 , n

≥ 2

. Wri

te a

n

expl

icit

for

mu

la t

o re

pres

ent

the

bala

nce

of

th

e sa

vin

gs a

ccou

nt

a n .

a n =

50

0(1.

0 5) n

- 1

5. S

NO

W A

sn

owm

an b

egin

s to

mel

t as

th

e te

mpe

ratu

re r

ises

. Th

e h

eigh

t of

th

e sn

owm

an i

n f

eet

afte

r ea

ch h

our

is

show

n.

Ho

ur

Hei

gh

t (f

t)

16

.0

25

.4

34

.86

44

.374

a. W

rite

a r

ecu

rsiv

e fo

rmu

la f

or t

he

sequ

ence

.

a 1

= 6

, a n =

0. 9

a n

- 1 , n

≥ 2

b.

Wri

te a

n e

xpli

cit

form

ula

for

th

e se

quen

ce.

a n =

6(0

.9 ) n

- 1

7-8

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NA

ME

DAT

E

P

ER

IOD

Lesson 7-8

Cha

pte

r 7

53

Gle

ncoe

Alg

ebra

1

Enri

chm

ent

Ari

thm

eti

c S

eri

es

7-8

Con

side

r a

situ

atio

n i

n w

hic

h a

n u

ncl

e gi

ves

his

nie

ce a

dol

lar

amou

nt

equ

al t

o h

er

age

each

yea

r on

her

bir

thda

y. A

n a

rith

met

ic s

equ

ence

th

at r

epre

sen

ts t

his

sit

uat

ion

is

1, 2

, 3, …

. H

ow m

uch

tot

al m

oney

wil

l th

e n

iece

hav

e re

ceiv

ed b

y h

er 3

0th

bir

thda

y?

Wh

en t

he

term

s of

an

ari

thm

etic

seq

uen

ce a

re a

dded

, an

ari

thm

etic

ser

ies

is f

orm

ed. T

he

solu

tion

of

the

prob

lem

abo

ve i

s th

e su

m o

f th

e ar

ith

met

ic s

erie

s 1

+ 2

+ 3

+ …

+ 3

0.

If w

e w

rite

th

e se

ries

tw

ice,

as

show

n b

elow

, an

d ad

d th

e te

rms

in e

ach

col

um

n, w

e ca

n

see

a pa

tter

n.

1

+

2 +

3

+

4 +

... +

27

+ 2

8 +

29

+ 3

0

30 +

29

+ 2

8 +

27

+ ..

. +

4 +

3

+

2 +

1

31

+ 3

1 +

31

+ 3

1 +

... +

31

+ 3

1 +

31

+ 3

1

Th

e su

m o

f ea

ch p

air

of t

erm

s is

31.

Mu

ltip

lyin

g 31

by

the

nu

mbe

r of

ter

ms

in t

he

bott

om

row

, 30,

giv

es u

s th

e su

m o

f th

e bo

ttom

row

, or

930.

Sin

ce t

he

seri

es w

as w

ritt

en t

wic

e, a

nd

resu

ltan

tly,

eac

h t

erm

in

th

e se

ries

was

acc

oun

ted

for

twic

e, w

e ca

n d

ivid

e 93

0 by

2 a

nd

fin

d th

at t

he

sum

of

the

seri

es 1

+ 2

+ 3

+ …

+ 3

0 is

465

.

1. W

rite

an

exp

ress

ion

to

repr

esen

t th

e su

m o

f th

e se

ries

abo

ve u

sin

g th

e fi

rst

term

1, t

he

last

ter

m 3

0, a

nd

the

nu

mbe

r of

ter

ms

in t

he

seri

es 3

0.

S

amp

le a

nsw

er:

30(1

+ 3

0)

−

2

2. W

rite

an

exp

ress

ion

to

repr

esen

t th

e su

m o

f an

ari

thm

etic

ser

ies

wit

h n

ter

ms,

fir

st

term

a 1 ,

and

last

ter

m a

n .

S

amp

le a

nsw

er:

n( a

1 +

a n )

−

2

Use

th

e ex

pre

ssio

n t

hat

you

fou

nd

in

Exe

rcis

e 2

to f

ind

th

e su

m o

f ea

ch

arit

hm

etic

ser

ies.

3. 1

+ 2

+ 3

+ …

+ 1

00

5050

4.

5 +

10

+ 1

5 +

… +

100

10

50

5. 1

+ 3

+ 5

+ …

+ 2

9 2

25

6. 2

+ 4

+ 6

+ …

+ 6

0 9

30

7. 6

+ 1

3 +

20

+ …

+ 1

25

1179

8.

8 +

12

+ 1

6 +

… +

84

920

041_

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PM



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Chapter 7 Assessment Answer Key

An

swer

s

Quiz 2 (Lessons 7-3 and 7-4)Page 57

Quiz 4 (Lessons 7-7 and 7-8)Page 58

Quiz 1 (Lessons 7-1 and 7-2) Quiz 3 (Lessons 7-5 and 7-6) Mid-Chapter TestPage 57 Page 58 Page 59

B

3.08 × 10-5

1.4 × 107

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

3

8

3

2

2(7xy ) 1 − 2

9.0 × 106;9,000,000

-6 × 10-9;-0.000000006

2r 8

x20

-12t3n 7

-125x 12y 6

64c 8d 2

27y 4w 8

66 or 46,656

y 5

r2

− n9

C

a 1 = 3, a n = a n - 1 + 17, n ≥ 2

a 1 = 125, a n = -

1 −

5 a n - 1 ,

n ≥ 2

1.

2.

3.

4.

5.

-192, -768, -3072

C

Geometric; the common ratio is 1 −

2 .

1; D = all real numbers, R = {y|y > 0}

1.

y

xO

y = 13( )x

2.

y

xO

y = 4(2) -1 x

3.

4.

5.

A = 1500 (1 + 0.0485

−

4

) 4

˙

n

$2675.04

C

3; D = all real numbers, R = {y|y > -1}

1.

2.

3.

4.

5.

6.

7. B

G

D

F

A

F

B

8.

9.

10.

11.

12.

13.

5 m 6 y 3 r 5

−

7

2 −

5 x 9 y

6.72 × 104; 67,200

3.12 × 10-6; 0.00000312

4 −

3

-

2 −

3

Part II

Part I


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Chapter 7 Assessment Answer KeyVocabulary Test Form 1Page 60 Page 61 Page 62

Sample answer: A sequence in which each term after the fi rst is found by multiplying the previous term by a constant r, called the common ratio.

Sample answer: A function that can be described by an equation of the form y = a b x .

false; monomial

true

true

true

true

false; common ratio

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

true

false; order of magnitude

true

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

C

H

B

J

C

H

B

J

C

H

D

F 12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 39n + 1

B

F

C

J

C

H

A

G


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An

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Chapter 7 Assessment Answer KeyForm 2A Form 2BPage 63 Page 64 Page 65 Page 66

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

A

J

D

J

A

F

A

J

B

H

D

H 12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 7-2x - 2

D

F

B

G

D

J

A

J

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

B

H

B

G

A

J

B

J

B

G

C

12.

13.

14.

15.

16.

17.

18.

19.

20.

B: 37n - 1

H

D

H

D

J

C

G

D

G


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Chapter 7 Assessment Answer KeyForm 2CPage 67 Page 68

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. -

14 −

5

2

5 −

2

1.1

4 r 4 −

t 7

y −

x 5

8a 3n 6 + 4a18n 6

w 15y 12

-18t4n7

6.07 × 10-7

2.156 × 106; 2,156,000

4.5 × 108; 450,000,000

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B:

about $14,797.81

5, 2, −1

3, 10, 24

1 −

4

a 1 = 18, a n = a n - 1 + 11, n ≥ 2

a 1 = 2, a n = -6 a n - 1 , n ≥ 2

$1434.67

an = -3 · (-3)n - 1


243

32

about 201 lb

3 √

��

22

3 4 √ �

x

y

xO

y = 7x


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An

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Chapter 7 Assessment Answer KeyForm 2DPage 69 Page 70

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

4.02 × 10-6

3r5t

d 7 −

2 a 5

10a 4b 8

w 9z 21

-6a 3b 8

1.3

1

-

17 −

12

4 −

3

4.1952 × 10-3;

0.0041952

2 × 103; 2000

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B:

a n = 4 · 2 n - 1

about $4595.27

about $5030.49

about 185 lb

10 3 √ �

x

1

a 1 = 27, a n = a n − 1 − 8, n ≥ 2

a 1 = 1296, a n =

1 −

6 a n − 1 , n ≥ 2

1, 2, 6

−4, 3, 10

4 √ ��

57

32

100

y

xO

y = 16( )x



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Chapter 7 Assessment Answer KeyForm 3Page 71 Page 72

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. 41

−

32

4 −

3

5 −

4

2

5.1

- 16 −

125a11b

mx 12 −

128

45x - 2

16

−

81 h 12

3.068 × 1011; 306,800,000,000

4.73 × 102

3.5 × 10-7;

0.00000035

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

B:

a n = -7 · (-

1 −

4 )

n - 1

5 years

$112,206.07

0

about 577 lbs

a 1 = 50,

a n = a n − 1 − 47, n ≥ 2

a 1 = 1 −

8 , a n = 2 a n − 1 ,

n ≥ 2

3.5, −1, 1.25

2.25, 2.75, 3.25

8

8

√ ��

4yz

32 6 √ �

x

128

216

y

xOy = 4(2x - 1)


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Chapter 7 Assessment Answer Key Page 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 multiplying and dividing monomials, the degree of a polynomial, and adding, subtracting, and multiplying polynomials.

• 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 multiplying and dividing monomials, the degree of a polynomial, and adding, subtracting, and multiplying polynomials.

• 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 multiplying and dividing monomials, the degree of a polynomial, and adding, subtracting, and multiplying polynomials.

• 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

multiplying and dividing monomials, the degree of a polynomial, and adding, subtracting, and multiplying polynomials.

• 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 7 Assessment Answer Key Page 73, Extended-Respo nse Test Sample Answers

1a. ( 4a3

−

2a-2 )

4 = (2a5)4

= 16a20

1b. ( 4a3

−

2a-2 )

4 =

(4a3) 4 −

(2a-2) 4 = 256a12

−

16a-8 = 16a20

1c. Sample answer: When simplifying monomials, the order of applying the Quotient of Powers property and Power of a Quotient property does not matter.

2a. Sample answer: In the geometric sequence 6, 3, 1.5, …, the value of r is 0.5 and the absolute value of a n + 1 will be closer to zero than the value of a n .

2b. Sample answer: In the recursive sequence, a 1 = 3, a 2 = 3, a n + 2 = a n + a n + 1 , the values of a 1 , a 2 , and a 3 are 3, 3, and 6, respectively. a 1 = a 2 , but a 2 ≠ a 3 .

3a. Sample answer: 5%; about 14.2 years

3b. Sample answer: 10%; about 6.6 years

3c. Sample answer: about 10.4 years; about $8320

4a. 5.97 × 1 0 24 kg

4b. 3.984 × 1 0 27 kg

4c. 47.66%

In addition to the scoring rubric found on page A33, the following sample answers

may be used as guidance in evaluating open-ended assessment items.


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Chapter 7 Assessment Answer KeyStandardized Test PracticePage 74 Page 75

19. 20.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

A B C D

A B C D

F G H J

A B C D

A B C D

F G H J

F G H J

F G H J

F G H J

12.

13.

14.

15.

16.

17.

18. F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

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

81 3 5

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

. 0

A B C D

A B C D


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Chapter 7 Assessment Answer KeyStandardized Test PracticePage 76

one solution; (2, 2)

35 −

6 + n = 2(n + 3)

3y2

−

x3

elimination (x); (1, –1)

Dallas Cowboys won

5 Super Bowls

Sample answer: c = Dallas Cowboys winsb = Denver Broncos wins; c + b = 7; c = 2.5b

a 1 = 39, a n =

a n - 1 - 7, n ≥ 2.

1.5

8x + 12y ≤ 62

$1225 per year

21.

22.

23.

24.

25. y

xO

x + y = 4

y = x

(2, 2)

26.

27.

28.

29.

30a.

30b.

y | y ≤ 3 2 −

3 { }


Chapter 7 Resource Masters - Commack Schools

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Transcript of Chapter 7 Resource Masters - Commack Schools