JUDITH A. PENNA - Pearson

1 STUDENT S solutions manual JUDITH A. PENNA Indiana University Purdue University Indianapolis COLLEGE ALGEBRA: GRAPHS AND MODELS FIFTH EDITION Marvin L. Bittinger Indiana University Purdue University Indianapolis JUDITH A. Beecher Indiana University Purdue University Indianapolis David J. Ellenbogen Community College of Vermont JUDITH A. PENNA Indiana University Purdue University Indianapolis Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo The author and publisher of this book have used their best efforts in preparing this book.

2 These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright 2013, 2009, 2005 Pearson Education, Inc. Publishing as Pearson , 75 Arlington Street, Boston, MA 02116.

3 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-79125-2 ISBN-10: 0-321-79125-8 1 2 3 4 5 6 BRR 15 14 13 12 11 Copyright 2013 Pearson Education, Inc. 05 50 3 10 RBasic Concepts of numbers:23,6, , , 11,3 27, 516, 87,0, numbers: 3,6 26, , 35,5 3(Although there is a pattern in , there isno repeating block of digits.) numbers: 6,3 27, 0, but not natural numbers: 11, numbers but not integers:23, , , 516, is a closed interval, so we use brackets.

4 Interval no-tation is [ 5,5]. is a half-open interval. We use a parenthesis onthe left and a bracket on the right. Interval notation is( 3, 1]. interval is of unlimited extent in the negative direc-tion, and the endpoint 2 is included. Interval notation is( , 2]. interval is of unlimited extent in the positive direc-tion, and the endpoint is not included. Interval nota-tion is ( , ).19.{x|7<x},or{x|x>7}.This interval is of unlimited extent in the positive directionand the endpoint 7 is not included. Interval notation is(7, ). endpoints 0 and 5 are not included in the interval, sowe use parentheses. Interval notation is (0,5).))

5 Endpoint 9 is included in the interval, so we use abracket before the 9. The endpoint 4 is not included,so we use a parenthesis after the 4. Interval notation is[ 9, 4). endpoints are included in the interval, so we usebrackets. Interval notation is [x, x+h]. endpointpis not included in the interval, so we use aparenthesis before thep. The interval is of unlimited ex-tent in the positive direction, so we use the infinity symbol . Interval notation is (p, ). 6 is an element of the set of natural numbers, thestatement is is not an element of the set of integers, the state-ment is 115is an element of the set of rational numbers,the statement is 11 is an element of the set of real numbers, thestatement is 24 is an element of the set of whole numbers, thestatement is is not an element of the set of irrational num-bers, the statement is every whole number is an integer, the statement every rational number is a real number, the state-ment is there are real numbers that are not integers.]

6 Thestatement is sentence 3 +y=y+ 3 illustrates the commutativeproperty of sentence 3 1= 3 illustrates the multiplicativeidentity sentence 5 x=x 5 illustrates the commutative prop-erty of sentence 2(a+b)=(a+b)2 illustrates the commutativeproperty of sentence 6(m+n)= 6(n+m) illustrates the com-mutative property of sentence 8 18= 1 illustrates the multiplicative inverseproperty. Copyright 2013 Pearson Education, Inc. 2 Chapter R: Basic Concepts of distance of from 0 is , so| |= distance 295 from 0 is 295, so|295|= distance of 97 from 0 is 97, so| 97|= distance of 0 from 0 is 0, so|0|= distance of54from 0 is54,so 54 = |14 ( 8)|=|14+8|=|22|= 22, or| 8 14|=| 22|=2271.

7 | 3 ( 9)|=| 3+9|=|6|=6,or| 9 ( 3)|=| 9+3|=| 6|=673.| |=| |= , or| |=| |= 34 158 = 68 158 = 218 =218,or 158 34 = 158+34 = 158+68 = 218 =21877.| 7 0|=| 7|=7,or|0 ( 7)|=|0+7|=|7|= may vary. One such number may vary. Since 1101= and 1100= , one such number is 12+32= 10, the hypotenuse of a right triangle withlegs of lengths 1 unit and 3 units has a length of 10 units. 31cc2=12+32c2=10c= 7=137 a m=1am,a =0 that each exponent is negative. We move eachfactor to the other side of the fraction bar and change thesign of each 5y 4= that each exponent is negative. We move eachfactor to the other side of the fraction bar and change thesign of each 1n 12t 6=t6m1n12, 1 (For any nonzero real number,a0= 1.)

8 Z7=z0+7=z7,orz0 z7=1 z7= 5 6=58+( 6)=52, 5 m5=m 5+5=m0= y 7=y3+( 7)=y 4,or1y417.(x+3)4(x+3) 2=(x+3)4+( 2)=(x+3) 3 38 3=3 3+8+1=36, 3x2=2 3 x3+2=6x523.( 3a 5)(5a 7)= 3 5 a 5+( 7)= 15a 12,or 15a1225.(6x 3y5)( 7x2y 9)=6( 7)x 3+2y5+( 9)= 42x 1y 4,or 42xy427.(2x)4(3x)3=24 x4 33 x3=16 27 x4+3= 432x729.( 2n)3(5n)2=( 2)3n3 52n2= 8 25 n3+2= 31= 7b12=b 7 12=b 19, 2x 1y=x2 ( 1)y 2 1=x3y 3, 4y34x 5y8=324x 4 ( 5)y3 8=8xy 5,or8xy539.(2x2y)4=24(x2)4y4=16x2 4y4=16x8y441.( 2x3)5=( 2)5(x3)5=( 2)5x3 5= 32x1543.( 5c 1d 2) 2=( 5) 2c 1( 2)d 2( 2)=c2d4( 5)2=c2d42545.(3m4)3(2m 5)4=33m12 24m 20=27 16m12+( 20)= 432m 8,or432m847. 2x 3y7z 1 3=(2x 3y7)3(z 1)3=23x 9y21z 3=8x 9y21z 3,or8y21z3x949.

9 24a10b 8c712a6b 3c5 5=(2a4b 5c2) 5=2 5a 20b25c 10,orb2532a20c10 Copyright 2013 Pearson Education, Inc. Exercise Set 16,500,000 to scientific want the decimal point to be positioned between the1 and the 6, so we move it 7 places to the left. Since16,500,000 is greater than 10, the exponent must be ,500,000= to scientific want the decimal point to be positioned between the4 and the 3, so we move it 7 places to the right. is a number between 0 and 1, the exponentmust be = 10 234,600,000,000 to scientific notation. We wantthe decimal point to be positioned between the 2 and the3, so we move it 11 places to the left.

10 Since 234,600,000,000is greater than 10, the exponent must be ,600,000,000 = to scientific notation. We want the deci-mal point to be positioned between the 1 and the last 0, sowe move it 3 places to the right. Since is a numberbetween 0 and 1, the exponent must be = 10 to want the decimal point to be positioned between the1 and the 6, so we move it 27 places to the right. is a number between 0and 1, the exponent must be = 10 105to decimal exponent is positive, so the number is greater than10. We move the decimal point 5 places to the 105= 760, 10 7to decimal exponent is negative, so the number is between 0 and1.