(Solution Manual) College Algebra Graphs And Models

Bittinger,Beecher,Ellenbogen,Penna, 6

Math & Statistics

Solution Manual

6th Edition

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Solution Manual for College Algebra Graphs And Models, The downloadable file consists of 565 pages, and the total size of the file is 8.97 MB. Purchase now only for highly discounted price.

Author(s): Bittinger, Beecher, Ellenbogen, Penna

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SOLUTIONS MANUAL COLLEGE ALGEBRA GRAPHS AND MODELS SIXTH EDITION Marvin L. Bittinger Judith A. Beecher David J. Ellenbogen Judith A. Penna Just-in-Time Review 1. Real Numbers 1.Rational numbers:2 3,6 ,−2.45, 18 .4,−11,3√ 27, 51 6,−8 7, 0,√ 16 2.Rational numbers but not integers:2 3,−2.45, 18 .4, 51 6, −8 7 3.Irrational numbers:√ 3,6√ 26, 7 .151551555 ...,−√ 35,5√ 3 (Although there is a pattern in 7 .151551555 ..., there is no repeating block of digits.) 4.Integers: 6, −11,3√ 27, 0,√ 16 5.Whole numbers: 6,3√ 27, 0,√ 16 6.Real numbers: All of them 2. Properties of Real Numbers 1.−24 + 24 = 0 illustrates the additive inverse property. 2.7(xy)=( 7 x)yillustrates the associative property of mul- tiplication. 3.9(r−s)=9 r−9sillustrates a distributive property. 4.11 + z=z+ 11 illustrates the commutative property of addition. 5.−20·1=−20 illustrates the multiplicative identity prop- erty. 6.5(x+y)=( x+y)5 illustrates the commutative property of multiplication. 7.q+0= qillustrates the additive identity property. 8.75·1 75= 1 illustrates the multiplicative inverse property. 9.(x+y)+w=x+(y+w) illustrates the associative property of addition. 10.8(a+b)=8 a+8billustrates a distributive property. 3. Order on the Number Line 1.9 is to the right of −9 on the number line, so it is false that 9 <−9. 2.−10 is to the left of −1 on the number line, so it is true that−10≤−1.3.−5=−√ 25, and −√ 26 is to the left of −√ 25, or −5, on the number line. Thus it is true that −√ 26<−5. 4.√ 6=√ 6, so it is true that√ 6≤√ 6. 5.−30 is to the left of −25 on the number line, so it is false that−30>−25. 6.−4 5=−16 20and−5 4=−25 20;−16 20is to the right of −25 20, so it is true that −4 5>−5 4. 4. Absolute Value 1.|−98|=9 8 ( |a|=−a,i fa<0.) 2.|0|=0 ( |a|=a,i fa≥0.) 3.|4.7|=4.7(|a|=a,ifa≥0.) 4./vextendsingle/vextendsingle/vextendsingle/vextendsingle−2 3/vextendsingle/vextendsingle/vextendsingle/vextendsingle=2 3(|a|=−a,ifa<0.) 5.|−7−13|=|−20|= 20, or |13−(−7)|=|1 3+7 |=|20|=2 0 6.|2−14.6|=|−12.6|=1 2 .6, or |14.6−2|=|12.6|=1 2 .6 7.|−39−(−28)|=|−3 9+2 8 |=|−11|= 11, or |−28−(−39)|=|−2 8+3 9 |=|11|=1 1 8./vextendsingle/vextendsingle/vextendsingle/vextendsingle−3 4−15 8/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle−6 8−15 8/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle−21 8/vextendsingle/vextendsingle/vextendsingle/vextendsingle=21 8,o r /vextendsingle/vextendsingle/vextendsingle/vextendsingle15 8−/parenleftbigg −3 4/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle15 8+6 8/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle21 8/vextendsingle/vextendsingle/vextendsingle/vextendsingle=21 8 5. Operations with Real Numbers 1.8−(−11) = 8 + 11 = 19 2.−3 10·/parenleftbigg −1 3/parenrightbigg =3·1 10·3=3 3·1 10=1·1 10=1 10 3.15÷(−3) =−5 4.−4−(−1) =−4+1= −3 5.7·(−50) = −350 6.−0.5−5=−0.5+(−5) =−5.5 7.−3+2 7=2 4 8.−400÷−40 = 10 9.4.2·(−3) =−12.6 Copyright c/circlecopyrt2017 Pearson Education, Inc. 2 Just-in-Time Review 10.−13−(−33) = −1 3+3 3=2 0 11.−60 + 45 = −15 12.1 2−2 3=1 2+/parenleftbigg −2 3/parenrightbigg =3 6+/parenleftbigg −4 6/parenrightbigg =−1 6 13.−24÷3=−8 14.−6+(−16) = −22 15.−1 2÷/parenleftbigg −5 8/parenrightbigg =−1 2·/parenleftbigg −8 5/parenrightbigg =1·8 2·5=1·2/·4 2/·5=4 5 6. Interval Notation 1.This is a closed interval, so we use brackets. Interval no- tation is [ −5,5]. 2.This is a half-open interval. We use a parenthesis onthe left and a bracket on the right. Interval notation is(−3,−1]. 3.This interval is of unlimited extent in the negative direc-tion, and the endpoint −2 is included. Interval notation is (−∞,−2]. 4.This interval is of unlimited extent in the positive direc-tion, and the endpoint 3 .8 is not included. Interval nota- tion is (3 .8,∞). 5.{x|7<x},o r{x|x>7}. This interval is of unlimited extent in the positive directionand the endpoint 7 is not included. Interval notation is(7,∞). 6.The endpoints −2 and 2 are not included in the interval, so we use parentheses. Interval notation is ( −2,2). 7.The endpoints −4 and 5 are not included in the interval, so we use parentheses. Interval notation is ( −4,5). 8.The interval is of unlimited extent in the positive direc-tion, and the endpoint 1.7 is included. Internal notationis [1.7,∞). 9.The endpoint −5 is not included in the interval, so we use a parenthesis before −5. The endpoint −2 is included in the interval, so we use a bracket after −2. Interval notation is (−5,−2]. 10.This interval is of unlimited extent in the negative direc-tion, and the endpoint√ 5 is not included. Interval nota- tion is ( −∞,√ 5). 7. Integers as Exponents 1.3−6=1 36Using a−m=1 am 2.1 (0.2)−5=( 0.2)5Using a−m=1 am3.w−4 z−9=z9 w4Usinga−m b−n=bn am 4./parenleftbiggz y/parenrightbigg2 =z2 y2Raising a quotient to a power 5.1000= 1 Using a0=1 , a/negationslash=0 6.a5 a−3=a5−(−3)=a5+3=a8Using the quotient rule 7.(2xy3)(−3x−5y)=2 ( −3)x·x−5·y3·y =−6x1+(−5)y3+1 =−6x−4y4,or−6y4 x4 8.x−4·x−7=x−4+(−7)=x−11,o r1 x11 9.(mn)−6=m−6n−6,o r1 m6n6 10.(t−5)4=t−5·4=t−20,o r1 t20 8. Scientific Notation 1.Convert 18,500,000 to scientific notation. We want the decimal point to be positioned between the 1 and the 8, so we move it 7 places to the left. Since18,500,000 is greater than 10, the exponent must be posi-tive. 18,500,000 = 1 .85×10 7 2.Convert 0.000786 to scientific notation. We want the decimal point to be positioned between the 7 and the 8, so we move it 4 places to the right. Since0.000786 is between 0 and 1, the exponent must be nega-tive. 0.000786 = 7 .86×10 −4 3.Convert 0.0000000023 to scientific notation. We want the decimal point to be positioned between the 2 and the 3, so we move it 9 places to the right. Since0.0000000023 is between 0 and 1, the exponent must benegative. 0.0000000023 = 2 .3×10 −9 4.Convert 8,927,000,000 to scientific notation. We want the decimal point to be positioned between the 8 and the 9, so we move it 9 places to the left. Since8,927,000,000 is greater than 10, the exponent must bepositive. 8,927,000,000 = 8 .927×10 9 5.Convert 4 .3×10−8to decimal notation. The exponent is negative, so the number is between 0 and1. We move the decimal point 8 places to the left. 4.3×10 −8=0.000000043 Copyright c/circlecopyrt2017 Pearson Education, Inc. Just-in-Time Review 3 6.Convert 5 .17×106to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 5.17×106=5,170,000 7.Convert 6 .203×1011to decimal notation. The exponent is positive, so the number is greater than10. We move the decimal point 11 places to the right. 6.203×10 11= 620 ,300,000,000 8.Convert 2 .94×10−5to scientific notation. The exponent is negative, so the number is between 0 and1. We move the decimal point 5 places to the left. 2.94×10 −5=0.0000294 9. Order of Operations 1.3+1 8 ÷6−3=3+3 −3 Dividing =6−3 = 3 Adding and subtracting 2.=5·3+8·32+ 4(6−2) =5·3+8·32+4·4 Working inside parentheses =5·3+8·9+4·4 Evaluating 32 = 15 + 72 + 16 Multiplying = 87 + 16 Adding in order = 103 from left to right 3. 5[3−8·32+4·6−2] = 5[3−8·9+4·6−2] = 5[3−7 2+2 4 −2] =5 [−6 9+2 4 −2] =5 [−45−2] =5 [−47] =−235 4. 16÷4·4÷2·256 =4·4÷2·256 Multiplying and dividing in order from left to right =1 6÷2·256 =8·256 = 2048 5. 26·2−3÷210÷2−8 =23÷210÷2−8 =2−7÷2−8 =26.4(8−6)2−4·3+2·8 31+1 90 =4·22−4·3+2·8 3+1Calculating in the numerator and in the denominator =4·4−4·3+2·8 4 =16−1 2+1 6 4 =4+1 6 4 =20 4 =5 7.64÷[(−4)÷(−2)] = 64 ÷2=3 2 8. 6[9−(3−2)] + 4(2 −3) = 6[9−1] + 4(2 −3) =6·8+4 (−1) =4 8−4 =4 4 10. Introduction to Polynomials 1.5−x6 The term of highest degree is −x6, so the degree of the polynomial is 6. 2.x2y5−x7y+4 The degree of x2y5is 2 + 5, or 7; the degree of −x7yis 7+1, or 8; the degree of 4 is 0 (4 = 4 x0). Thus the degree of the polynomial is 8. 3.2a4−3+a2 The term of highest degree is 2 a4, so the degree of the polynomial is 4. 4.−41 =−41x0, so the degree of the polynomial is 0. 5.4x−x3+0.1x8−2x5 The term of highest degree is 0 .1x8, so the degree of the polynomial is 8. 6.x−3 has two terms. It is a binomial. 7.14y5has one term. It is a monomial. 8.2y−1 4y2+ 8 has three terms. It is a trinomial. 11. Add and Subtract Polynomials 1. (8y−1)−(3−y) =( 8y−1 )+(−3+y) =( 8+1 ) y+(−1−3) =9y−4 Copyright c/circlecopyrt2017 Pearson Education, Inc. 4 Just-in-Time Review 2. (3x2−2x−x3+2 )−(5x2−8x−x3+4 ) =( 3x2−2x−x3+2 )+( −5x2+8x+x3−4) =( 3−5)x2+(−2+8 ) x+(−1+1 ) x3+( 2−4) =−2x2+6x−2 3. (2x+3y+z−7 )+( 4 x−2y−z+ 8)+ (−3x+y−2z−4) =( 2+4 −3)x+( 3−2+1 ) y+( 1−1−2)z+ (−7+8−4) =3x+2y−2z−3 4. (3ab2−4a2b−2ab+ 6)+ (−ab2−5a2b+8ab+4 ) =( 3−1)ab2+(−4−5)a2b+(−2+8 ) ab+( 6+4 ) =2ab2−9a2b+6ab+1 0 5. (5x2+4xy−3y2+2 )−(9x2−4xy+2y2−1) =( 5x2+4xy−3y2+2 )+( −9x2+4xy−2y2+1 ) =( 5−9)x2+( 4+4 ) xy+(−3−2)y2+( 2+1 ) =−4x2+8xy−5y2+3 12. Multiply Polynomials 1.(3a2)(−7a4) = [3( −7)](a2·a4) =−21a6 2. (y−3)(y+5 ) =y2+5y−3y−15 Using FOIL =y2+2y−15 Collecting like terms 3. (x+ 6)( x+3 ) =x2+3x+6x+ 18 Using FOIL =x2+9x+ 18 Collecting like terms 4. (2a+ 3)( a+5 ) =2a2+1 0a+3a+ 15 Using FOIL =2a2+1 3a+ 15 Collecting like terms 5. (2x+3y)(2x+y) =4x2+2xy+6xy+3y2Using FOIL =4x2+8xy+3y2 6. (11t−1)(3t+4 ) =3 3 t2+4 4t−3t−4 Using FOIL =3 3 t2+4 1t−4 13. Special Products of Binomials 1. (x+3 )2 =x2+2·x·3+32 [(A+B)2=A2+2AB+B2] =x2+6x+92. (5x−3)2 =( 5x)2−2·5x·3+32 [(A−B)2=A2−2AB+B2] =2 5 x2−30x+9 3. (2x+3y)2 =( 2x)2+ 2(2 x)(3y)+( 3 y)2 [(A+B)2=A2+2AB+B2] =4x2+1 2xy+9y2 4. (a−5b)2 =a2−2·a·5b+( 5b)2 [(A−B)2=A2−2AB+B2] =a2−10ab+2 5b2 5. (n+ 6)( n−6) =n2−62[(A+B)(A−B)=A2−B2] =n2−36 6. (3y+ 4)(3 y−4) =( 3y)2−42[(A+B)(A−B)=A2−B2] =9y2−16 14. Factor Polynomials; The FOIL Method 1.3x+1 8=3 ·x+3·6=3 ( x+6 ) 2.2z3−8z2=2z2·z−2z2·4=2 z2(z−4) 3. 3x3−x2+1 8x−6 =x2(3x−1) + 6(3 x−1) =( 3x−1)(x2+6 ) 4. t3+6t2−2t−12 =t2(t+6 )−2(t+6 ) =(t+ 6)( t2−2) 5.w2−7w+1 0 We look for two numbers with a product of 10 and a sum of−7. By trial, we determine that they are −5 and −2. w2−7w+1 0=( w−5)(w−2) 6.t2+8t+1 5 We look for two numbers with a product of 15 and a sumof 8. By trial, we determine that they are 3 and 5. t 2+8t+1 5=( t+ 3)( t+5 ) 7.2n2−20n−4 8=2 ( n2−10n−24) Now factor n2−10n−24. We look for two numbers with a product of −24 and a sum of −10. By trial, we determine that they are 2 and −12. Then n2−10n−24 = (n+ 2)( n−12). We must include the common factor, 2, to have a factorization of the original trinomial. 2n2−20n−4 8=2 ( n+ 2)( n−12) Copyright c/circlecopyrt2017 Pearson Education, Inc. Just-in-Time Review 5 8.y4−9y3+1 4y2=y2(y2−9y+ 14) Now factor y2−9y+ 14. Look for two numbers with a product of 14 and a sum of −9. The numbers are −2 and −7. Then y2−9y+1 4=( y−2)(y−7). We must include the common factor, y2, in order to have a factorization of the original trinomial. y4−9y3+1 4y2=y2(y−2)(y−7) 9.2n2+9n−56 1.There is no common factor other than 1 or −1. 2.The factorization must be of the form (2n+) ( n+) . 3.Factor the constant term, −56. The possibilities are−1·56, 1(−56),−2·28, 2(−28),−4·16, 4(−16), −7·8, and 7( −8). The factors can be written in the opposite order as well: 56( −1),−56·1, 28(−2), −28·2, 16(−4),−16·4, 8(−7), and −8·7. 4.Find a pair of factors for which the sum of the outerand the inner products is the middle term, 9 n.B y trial, we determine that the factorization is (2 n− 7)(n+ 8). 10.2y 2+y−6 1.There is no common factor other than 1 or −1. 2.The factorization must be of the form(2y+) ( y+) . 3.Factor the constant term, −6. The possibilities are −1·6, 1(−6),−2·3, and 2( −3). The factors can be written in the opposite order as well: 6( −1),−6·1, 3(−2) and −3·2. 4.Find a pair of factors for which the sum of the outerand the inner products is the middle term, y.B y trial, we determine that the factorization is(2y−3)(y+ 2). 11.b 2−6bt+5t2 We look for two numbers with a product of 5 and a sumof−6. By trial, we determine that they are −1 and −5. b 2−6bt+5t2=(b−t)(b−5t) 12.x4−7x2−30 = ( x2)2−7x2−30 We look for two numbers with a product of −30 and a sum of−7. By trial, we determine that they are 3 and −10. x4−7x2−30 = ( x2+ 3)( x2−10) 15. Factoring Polynomials; The ac-Method 1.8x2−6x−9 1.There is no common factor other than 1 or −1. 2.Multiply the leading coefficient and the constant:8(−9) =−72. 3.Try to factor −72 so that the sum of the factors is the coefficient of the middle term, −6. The factors we want are −12 and 6.4.Split the middle term using the numbers found instep (3): −6x=−12x+6x 5.Factor by grouping. 8x 2−6x−9=8 x2−12x+6x−9 =4x(2x−3) + 3(2 x−3) =( 2x−3)(4x+3 ) 2.10t2+4t−6 1.Factor out the largest common factor, 2. 10t2+4t−6 = 2(5 t2+2t−3) Now factor 5 t2+2t−3. 2.Multiply the leading coefficient and the constant:5(−3) =−15. 3.Try to factor −15 so that the sum of the factors is the coefficient of the middle term, 2. The factors wewant are 5 and −3. 4.Split the middle term using the numbers found instep (3): 2t=5t−3t. 5.Factor by grouping. 5t 2+2t−3=5 t2+5t−3t−3 =5t(t+1 )−3(t+1 ) =(t+ 1)(5 t−3) Include the largest common factor in the final factoriza-tion. 10t 2+4t−6=2 ( t+ 1)(5 t−3) 3.18a2−51a+1 5 1.Factor out the largest common factor, 3. 18a2−51a+ 15 = 3(6 a2−17a+5 ) Now factor 6 a2−17a+5 . 2.Multiply the leading coefficient and the constant:6(5) = 30. 3.Try to factor 30 so that the sum of the factors is thecoefficient of the middle term, −17. The factors we want are −2 and −15. 4.Split the middle term using the numbers found instep (3): −17a=−2a−15a. 5.Factor by grouping. 6a 2−17a+5=6 a2−2a−15a+5 =2a(3a−1)−5(3a−1) =( 3a−1)(2a−5) Include the largest common factor in the final factoriza-tion. 18a 2−51a+ 15 = 3(3 a−1)(2a−5) Copyright c/circlecopyrt2017 Pearson Education, Inc. 6 Just-in-Time Review 16. Special Factorizations 1.z2−81 = z2−92=(z+ 9)( z−9) 2.16x2−9=( 4 x)2−32=( 4x+ 3)(4 x−3) 3.7pq4−7py4=7p(q4−y4) =7p[(q2)2−(y2)2] =7p(q2+y2)(q2−y2) =7p(q2+y2)(q+y)(q−y) 4.x2+1 2x+3 6= x2+2·x·6+62 =(x+6 )2 5.9z2−12z+4=( 3 z)2−2·3z·2+22=( 3z−2)2 6. a3+2 4a2+ 144 a =a(a2+2 4a+ 144) =a(a2+2·a·1 2+1 22) =a(a+ 12)2 7.x3+6 4= x3+43 =(x+ 4)( x2−4x+ 16) 8.m3−216 = m3−63 =(m−6)(m2+6m+ 36) 9.3a5−24a2=3a2(a3−8) =3a2(a3−23) =3a2(a−2)(a2+2a+4 ) 10. t6+1=( t2)3+13 =(t2+ 1)( t4−t2+1 ) 17. Equation-Solving Principles 1.7t=7 0 t= 10 Dividing by 7 The solution is 10. 2.x−5=7 x= 12 Adding 5 The solution is 12. 3.3x+4= −8 3x=−12 Subtracting 4 x=−4 Dividing by 3 The solution is −4. 4.6x−1 5=4 5 6x= 60 Adding 15 x= 10 Dividing by 6 The solution is 10.5.7y−1=2 3 −5y 12y−1 = 23 Adding 5 y 12y= 24 Adding 1 y= 2 Dividing by 12 The solution is 2. 6.3m−7=−13 + m 2m−7=−13 Subtracting m 2m=−6 Adding 7 m=−3 Dividing by 2 The solution is −3. 7.2(x+7 )=5 x+1 4 2x+1 4=5 x+1 4 −3x+ 14 = 14 Subtracting 5 x −3x= 0 Subtracting 14 x=0 The solution is 0. 8.5y−(2y−1 0 )=2 5 5y−2y+1 0=2 5 3y+ 10 = 25 Collecting like terms 3y= 15 Subtracting 10 y= 5 Dividing by 3 The solution is 5. 18. Inequality-Solving Principles 1.p+2 5≥−100 p≥−125 Subtracting 25 The solution set is [ −125,∞). 2.−2 3x>6 x<−3 2·6 Multiplying by −3 2and reversing the inequality symbol x<−9 The solution set is ( −∞,−9). 3.9x−1<17 9x<18 Adding 1 x<2 Dividing by 9 The solution set is ( −∞,2). 4.−x−16≥40 −x≥56 Adding 6 x≤−56 Multiplying by −1 and reversing the inequality symbol The solution set is ( −∞,−56]. Copyright c/circlecopyrt2017 Pearson Education, Inc. Just-in-Time Review 7 5.1 3y−6<3 1 3y<9 Adding 6 y<27 Multiplying by 3 The solution set is ( −∞,27). 6.8−2w≤−14 −2w≤−22 Subtracting 8 w≥11 Dividing by −2 and reversing the inequality symbol The solution set is [11 ,∞). 19. The Principle of Zero Products 1.2y2+4 2y=0 2y(y+ 21) = 0 2y=0 or y+2 1=0 y=0 or y =−21 The solutions are 0 and −21. 2.(a+ 7)( a−1) = 0 a+7=0 or a−1=0 a=−7or a =1 The solutions are −7 and 1. 3.(5y+ 3)( y−4) = 0 5y+3=0 or y−4=0 5y=−3or y =4 y=−3 5or y =4 The solutions are −3 5and 4. 4.6x2+7x−5=0 (3x+ 5)(2 x−1) = 0 3x+5=0 or2x−1=0 3x=−5or 2x=1 x=−5 3or x =1 2 The solutions are −5 3and1 2. 5.t(t−8) = 0 t=0 or t−8=0 t=0 or t =8 The solutions are 0 and 8. 6.x2−8x−33 = 0 (x+ 3)( x−11) = 0 x+3=0 or x−1 1=0 x=−3or x =1 1 The solutions are −3 and 11.7.x2+1 3x=3 0 x2+1 3x−30 = 0 (x+ 15)( x−2) = 0 x+1 5=0 or x−2=0 x=−15or x =2 The solutions are −15 and 2. 8.12x2−7x−12 = 0 (4x+ 3)(3 x−4) = 0 4x+3=0 or3x−4=0 4x=−3or 3x=4 x=−3 4or x =4 3 The solutions are −3 4and4 3. 20. The Principle of Square Roots 1.x2−3 6=0 x2=3 6 x=√ 36or x =−√ 36 x=6 or x =−6 The solutions are 6 and −6, or±6. 2.2y2−20 = 0 2y2=2 0 y2=1 0 y=√ 10or y=−√ 10 The solutions are√ 10 and −√ 10, or ±√ 10. 3.6z2=1 8 z2=3 z=√ 3or z=−√ 3 The solutions are√ 3 and −√ 3, or±√ 3. 4.3t2−1 5=0 3t2=1 5 t2=5 t=√ 5or t=−√ 5 The solutions are√ 5 and −√ 5, or±√ 5. 5.z2−1=2 4 z2=2 5 z=√ 25or z=−√ 25 The solutions are 5 and −5, or±5. 6.5x2−7 5=0 5x2=7 5 x2=1 5 x=√ 15or x=−√ 15 The solutions are√ 15 and −√ 15, or ±√ 15. Copyright c/circlecopyrt2017 Pearson Education, Inc. 8 Just-in-Time Review 21. Simplify Rational Expressions 1.3x−3 x(x−1) The denominator is 0 when the factor x= 0 and also when x−1 = 0, or x= 1. The domain is the set of all real numbers except 0 and 1. 2.y+6 y2+4y−21=y+6 (y+ 7)( y−3) The denominator is 0 when y=−7o ry= 3. The domain is the set of all real numbers except −7 and 3. 3.x2−4 x2−4x+4=(x+ 2)( x−2)✏ (x−2)(x−2)✏=x+2 x−2 4.x2+2x−3 x2−9=(x−1)(x+3)✏ (x+3)✏(x−3)=x−1 x−3 5.x3−6x2+9x x3−3x2=x(x2−6x+9 ) x2(x−3) =x/(x−3)✏(x−3) x/·x(x−3)✏ =x−3 x 6.6y2+1 2y−48 3y2−9y+6=6(y2+2y−8) 3(y2−3y+2 ) =2·3/·(y+ 4)( y−2)✏ 3/(y−1)(y−2)✏ =2(y+4 ) y−1 22. Multiply and Divide Rational Expressions 1.r−s r+s·r2−s2 (r−s)2=(r−s)(r2−s2) (r+s)(r−s)2 =(r−s)✏(r−s)✏(r+s)✏·1 (r+s)✏(r−s)✏(r−s)✏ =1 2.m2−n2 r+s÷m−n r+s =m2−n2 r+s·r+s m−n =(m+n)(m−n)✏(r+s)✏ (r+s)✏(m−n)✏ =m+n 3.4x2+9x+2 x2+x−2·x2−1 3x2+x−2 =(4x+ 1)( x+2)✏(x+1)✏(x−1)✏ (x+2)✏(x−1)✏(3x−2)(x+1)✏ =4x+1 3x−24.3x+1 2 2x−8÷(x+4 )2 (x−4)2 =3x+1 2 2x−8·(x−4)2 (x+4 )2 =3(x+4)✏(x−4)✏(x−4) 2(x−4)✏(x+4)✏(x+4 ) =3(x−4) 2(x+4 ) 5.a2−a−2 a2−a−6÷a2−2a 2a+a2 =a2−a−2 a2−a−6·2a+a2 a2−2a =(a−2)✏(a+1 )( a)✧(2+a)✏ (a−3)(a+2)✏(a)✧(a−2)✏ =a+1 a−3 6.x2−y2 x3−y3·x2+xy+y2 x2+2xy+y2 =(x+y)(x−y)(x2+xy+y2) (x−y)(x2+xy+y2)(x+y)(x+y) =1 x+y·(x+y)(x−y)(x2+xy+y2) (x+y)(x−y)(x2+xy+y2) =1 x+y·1 Removing a factor of 1 =1 x+y 23. Add and Subtract Rational Expressions 1.a−3b a+b+a+5b a+b=2a+2b a+b =2(a+b)✏ 1·(a+b)✏ =2 2.x2−5 3x2−5x−2+x+1 3x−6 =x2−5 (3x+ 1)( x−2)+x+1 3(x−2) =x2−5 (3x+ 1)( x−2)·3 3+x+1 3(x−2)·3x+1 3x+1 =3(x2−5 )+( x+ 1)(3 x+1 ) 3(3x+ 1)( x−2) =3x2−1 5+3 x2+4x+1 3(3x+ 1)( x−2) =6x2+4x−14 3(3x+ 1)( x−2) Copyright c/circlecopyrt2017 Pearson Education, Inc. Just-in-Time Review 9 3.a2+1 a2−1−a−1 a+1 =a2+1 (a+ 1)( a−1)−a−1 a+1,LCD is ( a+ 1)( a−1) =a2+1−(a−1)(a−1) (a+ 1)( a−1) =a2+1−a2+2a−1 (a+ 1)( a−1) =2a (a+ 1)( a−1) 4.9x+2 3x2−2x−8+7 3x2+x−4 =9x+2 (3x+ 4)( x−2)+7 (3x+ 4)( x−1), LCD is (3 x+ 4)( x−2)(x−1) =9x+2 (3x+4)( x−2)·x−1 x−1+7 (3x+4)( x−1)·x−2 x−2 =9x2−7x−2 (3x+4)( x−2)(x−1)+7x−14 (3x+4)( x−1)(x−2) =9x2−16 (3x+ 4)( x−2)(x−1) =(3x+4)✘ (3x−4) (3x+4)✘ (x−2)(x−1) =3x−4 (x−2)(x−1) 5.y y2−y−20−2 y+4 =y (y+ 4)( y−5)−2 y+4,LCD is ( y+ 4)( y−5) =y (y+ 4)( y−5)−2 y+4·y−5 y−5 =y (y+ 4)( y−5)−2y−10 (y+ 4)( y−5) =y−(2y−10) (y+ 4)( y−5) =y−2y+1 0 (y+ 4)( y−5) =−y+1 0 (y+ 4)( y−5)6.3y y2−7y+1 0−2y y2−8y+1 5 =3y (y−2)(y−5)−2y (y−5)(y−3), LCD is ( y−2)(y−5)(y−3) =3y(y−3)−2y(y−2) (y−2)(y−5)(y−3) =3y2−9y−2y2+4y (y−2)(y−5)(y−3) =y2−5y (y−2)(y−5)(y−3) =y(y−5)✏ (y−2)(y−5)✏(y−3) =y (y−2)(y−3) 24. Simplify Complex Rational Expressions 1.x y−y x 1 y+1 x=x y−y x 1 y+1 x·xy xy,LCM is xy =/parenleftBigx y−y x/parenrightBig (xy) /parenleftBig1 y+1 x/parenrightBig (xy) =x2−y2 x+y =(x+y)✏(x−y) (x+y)✏·1 =x−y 2.a−b b a2−b2 ab=a−b b·ab a2−b2 =a−b b·ab (a+b)(a−b) =ab✧(a−b)✏ b/(a+b)(a−b)✏ =a a+b Copyright c/circlecopyrt2017 Pearson Education, Inc. 10 Just-in-Time Review 3.w+8 w2 1+2 w=w·w2 w2+8 w2 1·w w+2 w =w3+8 w2 w+2 w =w3+8 w2·w w+2 =(w+2)✏(w2−2w+4 )w/ w/·w(w+2)✏ =w2−2w+4 w 4.x2−y2 xy x−y y=x2−y2 xy·y x−y =(x+y)(x−y)✏ y✧ xy✧(x−y)✏ =x+y x 5.a b−b a 1 a−1 b=a2−b2 b−aMultiplying byab ab =(a+b)(a−b) b−a =(a+b)(a−b)✏ −1·(a−b)✏ =−a−b 25. Simplify Radical Expressions 1./radicalbig (−21)2=|−21|=2 1 2./radicalbig 9y2=/radicalbig (3y)2=|3y|=3y 3./radicalbig (a−2)2=a−2 4.3√ −27x3=3/radicalbig (−3x)3=−3x 5.4√ 81x8=4/radicalbig (3x2)4=3x2 6.5√ 32 =5√ 25=2 7.4/radicalbig 48x6y4=4/radicalbig 16x4y4·3x2=2xy4√ 3x2= 2xy4√ 3x2 8.√ 15√ 35 =√ 15·35 =√ 3·5·5·7=√ 52·3·7=√ 52·√ 3·7=5√ 21 9.√40xy√ 8x=/radicalbigg 40xy 8x=/radicalbig 5y 10.3√ 3x2 3√ 24x5=3/radicalbigg 3x2 24x5=3/radicalbigg 1 8x3=1 2x11.√ x2−4x+4=/radicalbig (x−2)2=x−2 12./radicalbig 2x3y√12xy=/radicalbig 24x4y2=/radicalbig 4x4y2·6=2 x2y√ 6 13.3/radicalbig 3x2y3√ 36x=3/radicalbig 108x3y=3/radicalbig 27x3·4y=3x3√4y 14.5√ 2+3√ 32 = 5√ 2+3√ 16·2 =5√ 2+3·4√ 2 =5√ 2+1 2√ 2 = (5 + 12)√ 2 =1 7√ 2 15.7√ 12−2√ 3=7·2√ 3−2√ 3=1 4√ 3−2√ 3=1 2√ 3 16.2√ 3 2+3√ 8−4√ 18 = 2 ·4√ 2+3·2√ 2−4·3√ 2= 8√ 2+6√ 2−12√ 2=2√ 2 17.6√ 20−4√ 45 +√ 8 0=6√ 4·5−4√ 9·5+√ 16·5 =6·2√ 5−4·3√ 5+4√ 5 =1 2√ 5−12√ 5+4√ 5 = (12 −1 2+4 )√ 5 =4√ 5 18. (2 +√ 3)(5 + 2√ 3) =2·5+2·2√ 3+√ 3·5+√ 3·2√ 3 =1 0+4√ 3+5√ 3+3·2 =1 0+9√ 3+6 =1 6+9√ 3 19./parenleftbig√ 8+2√ 5/parenrightbig/parenleftbig√ 8−2√ 5/parenrightbig =/parenleftbig√ 8/parenrightbig2−/parenleftbig 2√ 5/parenrightbig2 =8−4·5 =8−20 =−12 20.(1 +√ 3)2=12+2·1·√ 3+(√ 3)2 =1+2√ 3+3 =4+2√ 3 26. Rationalizing Denominators 1.4√ 11=4√ 11·√ 11√ 11=4√ 11 11 2./radicalbigg 3 7=/radicalbigg 3 7·7 7=/radicalbigg 21 49=√ 21√ 49=√ 21 7 3.3√ 7 3√ 2=3√ 7 3√ 2·3√ 4 3√ 4=3√ 28 3√ 8=3√ 28 2 4.3/radicalbigg 16 9=3/radicalbigg 16 9·3 3=3/radicalbigg 48 27=3√ 48 3√ 27= 3√ 8·6 3=23√ 6 3 Copyright c/circlecopyrt2017 Pearson Education, Inc. Just-in-Time Review 11 5.3√ 30−4=3√ 30−4·√ 3 0+4√ 3 0+4 =3√ 3 0+1 2 (√ 30)2−42 =3√ 3 0+1 2 30−16 =3√ 3 0+1 2 14 6.4√ 7−√ 3=4√ 7−√ 3·√ 7+√ 3√ 7+√ 3 =4√ 7+4√ 3 (√ 7)2−(√ 3)2 =4√ 7+4√ 3 7−3 =4√ 7+4√ 3 4=4(√ 7+√ 3) 4 =√ 7+√ 3 7.6√m−√n=6√m−√n·√m+√n√m+√n =6(√m+√n) (√m)2−(√n)2 =6√m+6√n m−n 8.1−√ 2√ 3−√ 6=1−√ 2√ 3−√ 6·√ 3+√ 6√ 3+√ 6 =√ 3+√ 6−√ 6−√ 12 3−6 =√ 3+√ 6−√ 6−2√ 3 3−6 =−√ 3 −3=√ 3 3 27. Rational Exponents 1.y5/6=6/radicalbig y5 2.x2/3=3√ x2 3.163/4= (161/4)3=/parenleftbig4√ 16/parenrightbig3=23=8 4.47/2=(√ 4)7=27= 128 5.125−1/3=1 1251/3=1 3√ 125=1 5 6.32−4/5=/parenleftbig5√ 32/parenrightbig−4=2−4=1 16 7.12/radicalbig y4=y4/12=y1/3 8.√ x5=x5/29.x1/2·x2/3=x1/2+2/3=x3/6+4/6=x7/6=6√ x7=x6√x 10.(a−2)9/4(a−2)−1/4=(a−2)9/4+(−1/4)= (a−2)8/4=(a−2)2 11.(m1/2n5/2)2/3=m1 2·2 3n5 2·2 3=m1/3n5/3= 3√m3√ n5=3√ mn5=n3√ mn2 28. The Pythagorean Theorem 1. a2+b2=c2 82+1 52=c2 64 + 225 = c2 289 = c2 17 = c 2.a2+b2=c2 42+42=c2 1 6+1 6= c2 32 = c2 √ 32 = c 5.657≈c 3.a2+b2=c2 52+b2=1 32 25 + b2= 169 b2= 144 b=1 2 4. a2+b2=c2 a2+1 22=1 32 a2+ 144 = 169 a2=2 5 a=5 5. a2+b2=c2 (√ 5)2+b2=62 5+b2=3 6 b2=3 1 b=√ 31≈5.568 Copyright c/circlecopyrt2017 Pearson Education, Inc.4 2 /H110022 /H1100244 2 /H110022 /H110024 xy (/H110021, 4) (/H110023, /H110025)(0, 2) (4, 0) (2, /H110022) 22 4 224 2224 24y x(25, 0) (2, 24)(1, 4)(0, 3) (24, 22)4 2 /H110022 /H1100244 /H110022 /H110024 xy (/H110025, 1)(2, 3) (0, 1) (5, 1) (2, /H110021) 22 4 224 2224 24y x(25, 2) (4, 23)(0, 21)(2, 0) (21, 25)Chapter 1 Graphs, Functions, and Models Exercise Set 1.1 1.Point A is located 5 units to the left of the y-axis and 4 units up from the x-axis, so its coordinates are ( −5,4). Point B is located 2 units to the right of the y-axis and 2 units down from the x-axis, so its coordinates are (2 ,−2). Point Cis located 0 units to the right or left of the y-axis and 5 units down from the x-axis, so its coordinates are (0,−5). Point D is located 3 units to the right of the y-axis and 5 units up from the x-axis, so its coordinates are (3 ,5). Point E is located 5 units to the left of the y-axis and 4 units down from the x-axis, so its coordinates are (−5,−4). Point F is located 3 units to the right of the y-axis and 0 units up or down from the x-axis, so its coordinates are (3,0). 2.G: (2 ,1); H: (0 ,0); I: (4 ,−3); J: ( −4,0); K: ( −2,3); L: (0 ,5) 3.To graph (4 ,0) we move from the origin 4 units to the right of the y-axis. Since the second coordinate is 0, we do not move up or down from the x-axis. To graph ( −3,−5) we move from the origin 3 units to the left of the y-axis. Then we move 5 units down from the x-axis. To graph ( −1,4) we move from the origin 1 unit to the left of the y-axis. Then we move 4 units up from the x-axis. To graph (0 ,2) we do not move to the right or the left of they-axis since the first coordinate is 0. From the origin we move 2 units up. To graph (2 ,−2) we move from the origin 2 units to the right of the y-axis. Then we move 2 units down from the x-axis. 4.5.To graph ( −5,1) we move from the origin 5 units to the left of the y-axis. Then we move 1 unit up from the x-axis. To graph (5 ,1) we move from the origin 5 units to the right of the y-axis. Then we move 1 unit up from the x-axis. To graph (2 ,3) we move from the origin 2 units to the right of the y-axis. Then we move 3 units up from the x-axis. To graph (2 ,−1) we move from the origin 2 units to the right of the y-axis. Then we move 1 unit down from the x-axis. To graph (0 ,1) we do not move to the right or the left of they-axis since the first coordinate is 0. From the origin we move 1 unit up. 6. 7.The first coordinate represents the year and the corre- sponding second coordinate represents the number of citiesserved by Southwest Airlines. The ordered pairs are(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),and (2014, 96). 8.The first coordinate represents the year and the secondcoordinate represents the percent of Marines who arewomen. The ordered pairs are (1960, 1%), (1970, 0.9%),(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), (2011, 6.8%),and (2014, 7.6%). 9.To determine whether ( −1,−9) is a solution, substitute −1 for xand−9 for y. y=7x−2 −9?7 ( −1)−2/vextendsingle/vextendsingle/vextendsingle−7−2 −9/vextendsingle/vextendsingle−9T R U E The equation −9=−9 is true, so ( −1,−9) is a solution. Copyright c/circlecopyrt2017 Pearson Education, Inc. 14 Chapter 1: Graphs, Functions, and Models To determine whether (0 ,2) is a solution, substitute 0 for xand 2 for y. y=7x−2 2?7 ·0−2/vextendsingle/vextendsingle/vextendsingle0−2 2/vextendsingle/vextendsingle−2 FALSE The equation 2 = −2 is false, so (0 ,2) is not a solution. 10.For/parenleftbigg1 2,8/parenrightbigg :y=−4x+1 0 8?−4·1 2+1 0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle−2+1 0 8/vextendsingle/vextendsingle8T R U E /parenleftbigg1 2,8/parenrightbigg is a solution. For (−1,6): y=−4x+1 0 6?−4(−1 )+1 0/vextendsingle/vextendsingle/vextendsingle4+1 0 6/vextendsingle/vextendsingle14 FALSE (−1,6) is not a solution. 11.To determine whether/parenleftBig2 3,3 4/parenrightBig is a solution, substitute2 3 forxand3 4fory. 6x−4y=1 6·2 3−4·3 4?1/vextendsingle/vextendsingle 4−3/vextendsingle/vextendsingle/vextendsingle 1/vextendsingle/vextendsingle1T R U E The equation 1 = 1 is true, so/parenleftBig2 3,3 4/parenrightBig is a solution. To determine whether/parenleftBig 1,3 2/parenrightBig is a solution, substitute 1 for xand3 2fory. 6x−4y=1 6·1−4·3 2?1/vextendsingle/vextendsingle 6−6/vextendsingle/vextendsingle/vextendsingle 0/vextendsingle/vextendsingle1 FALSE The equation 0 = 1 is false, so/parenleftBig 1,3 2/parenrightBig is not a solution. 12.For (1 .5,2.6): x2+y2=9 (1.5)2+( 2.6)2?9 2.2 5+6 .76/vextendsingle/vextendsingle/vextendsingle 9.01/vextendsingle/vextendsingle9 FALSE (1.5,2.6) is not a solution.For (−3,0): x 2+y2=9 (−3)2+02?9 9+0/vextendsingle/vextendsingle/vextendsingle 9/vextendsingle/vextendsingle9T R U E (−3,0) is a solution. 13.To determine whether/parenleftBig −1 2,−4 5/parenrightBig is a solution, substitute −1 2foraand−4 5forb. 2a+5b=3 2/parenleftBig −1 2/parenrightBig +5/parenleftBig −4 5/parenrightBig ?3/vextendsingle/vextendsingle −1−4/vextendsingle/vextendsingle/vextendsingle −5/vextendsingle/vextendsingle3 FALSE The equation −5 = 3 is false, so/parenleftBig −1 2,−4 5/parenrightBig is not a solu- tion. To determine whether/parenleftBig 0,3 5/parenrightBig is a solution, substitute 0 for aand3 5forb. 2a+5b=3 2·0+5·3 5?3/vextendsingle/vextendsingle 0+3/vextendsingle/vextendsingle/vextendsingle 3/vextendsingle/vextendsingle3T R U E The equation 3 = 3 is true, so/parenleftBig 0,3 5/parenrightBig is a solution. 14.For/parenleftBig 0,3 2/parenrightBig : 3m+4n=6 3·0+4·3 2?6/vextendsingle/vextendsingle 0+6/vextendsingle/vextendsingle/vextendsingle 6/vextendsingle/vextendsingle6T R U E /parenleftBig 0,3 2/parenrightBig is a solution. For/parenleftBig2 3,1/parenrightBig : 3m+4n=6 3·2 3+4·1?6/vextendsingle/vextendsingle 2+4/vextendsingle/vextendsingle/vextendsingle 6/vextendsingle/vextendsingle6T R U E The equation 6 = 6 is true, so/parenleftBig2 3,1/parenrightBig is a solution. 15.To determine whether ( −0.75,2.75) is a solution, substi- tute−0.75 for xand 2 .75 for y. x2−y2=3 (−0.75)2−(2.75)2?3 0.5625−7.5625/vextendsingle/vextendsingle/vextendsingle −7/vextendsingle/vextendsingle3 FALSE The equation −7 = 3 is false, so ( −0.75,2.75) is not a solution. Copyright c/circlecopyrt2017 Pearson Education, Inc.y x /H110024/H1100222 4 /H110024/H11002224 5x /H11002 3y /H11005 /H11002 15 (/H110023, 0)(0, 5)y x /H110024/H1100222 4 /H110024/H11002224 2x /H11002 4y /H11005 8(4, 0) (0, /H110022) (0, 4)y x /H110024/H1100222 4 /H110024/H110022242x /H11001 y /H11005 4 (2, 0) y x /H110024/H1100222 4 /H110022246(0, 6) 3x /H11001 y /H11005 6(2, 0)Exercise Set 1.1 15 To determine whether (2 ,−1) is a solution, substitute 2 forxand−1 for y. x2−y2=3 22−(−1)2?3 4−1/vextendsingle/vextendsingle/vextendsingle 3/vextendsingle/vextendsingle3T R U E The equation 3 = 3 is true, so (2 ,−1) is a solution. 16.For (2 ,−4): 5x+2y 2=7 0 5·2+2 ( −4)2?7 0 1 0+2 ·16/vextendsingle/vextendsingle/vextendsingle 1 0+3 2/vextendsingle/vextendsingle 42/vextendsingle/vextendsingle70 FALSE (2,−4) is not a solution. For (4 ,−5): 5x+2y 2=7 0 5·4+2 ( −5)2?7 0 2 0+2 ·25/vextendsingle/vextendsingle/vextendsingle 2 0+5 0/vextendsingle/vextendsingle 70/vextendsingle/vextendsingle70 TRUE (4,−5) is a solution. 17.Graph 5 x−3y=−15. To find the x-intercept we replace ywith 0 and solve for x. 5x−3·0=−15 5x=−15 x=−3 Thex-intercept is ( −3,0). To find the y-intercept we replace xwith 0 and solve for y. 5·0−3y=−15 −3y=−15 y=5 They-intercept is (0 ,5). We plot the intercepts and draw the line that contains them. We could find a third point as a check that theintercepts were found correctly.18. 19.Graph 2 x+y=4 . To find the x-intercept we replace ywith 0 and solve for x. 2x+0=4 2x=4 x=2 Thex-intercept is (2 ,0). To find the y-intercept we replace xwith 0 and solve for y. 2·0+y=4 y=4 They-intercept is (0 ,4). We plot the intercepts and draw the line that contains them. We could find a third point as a check that theintercepts were found correctly. 20. 21.Graph 4 y−3x= 12. To find the x-intercept we replace ywith 0 and solve for x. 4·0−3x=1 2 −3x=1 2 x=−4 Thex-intercept is ( −4,0). To find the y-intercept we replace xwith 0 and solve for y. Copyright c/circlecopyrt2017 Pearson Education, Inc.