Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition

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SOLUTIONSMANUALJUDITHA.PENNAALGEBRA&TRIGONOMETRYGRAPHS ANDMODELSSIXTHEDITIONPRECALCULUSGRAPHS ANDMODELS,ARIGHTTRIANGLEAPPROACHSIXTHEDITIONMarvin L. BittingerIndiana University Purdue University IndianapolisJudith A. BeecherDavid J. EllenbogenCommunity College of VermontJudith A. Penna

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ContentsJust-in-Time Review ....................... 1Chapter1..........................13Chapter2..........................73Chapter3.......................... 127Chapter4.......................... 185Chapter5.......................... 273Chapter6.......................... 333Chapter7.......................... 387Chapter8.......................... 429Chapter9.......................... 483Chapter10 . ......................... 583Chapter11 . ......................... 667ChapterR .......................... 709

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Just-in-Time Review1. Real Numbers1.Rational numbers: 23 , 6,−2.45, 18.4,−11,3√27, 5 16 ,−87 ,0,√162.Rational numbers but not integers:23 ,−2.45, 18.4, 5 16 ,−873.Irrational numbers:√3,6√26, 7.151551555. . .,−√35,5√3(Although there is a pattern in 7.151551555. . ., there isno repeating block of digits.)4.Integers: 6,−11,3√27, 0,√165.Whole numbers: 6,3√27, 0,√166.Real numbers: All of them2. Properties of Real Numbers1.−24 + 24 = 0 illustrates the additive inverse property.2.7(xy) = (7x)yillustrates the associative property of mul-tiplication.3.9(r−s) = 9r−9sillustrates a distributive property.4.11 +z=z+ 11 illustrates the commutative property ofaddition.5.−20·1 =−20 illustrates the multiplicative identity prop-erty.6.5(x+y) = (x+y)5 illustrates the commutative propertyof multiplication.7.q+ 0 =qillustrates the additive identity property.8.75·175 = 1 illustrates the multiplicative inverse property.9.(x+y)+w=x+(y+w) illustrates the associative propertyof addition.10.8(a+b) = 8a+ 8billustrates a distributive property.3. Order on the Number Line1.9 is to the right of−9 on the number line, so it is falsethat 9<−9.2.−10 is to the left of−1 on the number line, so it is truethat−10≤ −1.3.−5 =−√25, and−√26 is to the left of−√25, or−5, onthe 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 falsethat−30>−25.6.−45 =−1620 and−54 =−2520 ;−1620 is to the right of−2520 ,so it is true that−45>−54 .4. Absolute Value1.| −98|= 98(|a|=−a, ifa <0.)2.|0|= 0(|a|=a, ifa≥0.)3.|4.7|= 4.7(|a|=a,ifa≥0.)4.∣∣∣∣−23∣∣∣∣= 23(|a|=−a,ifa <0.)5.| −7−13|=| −20|= 20, or|13−(−7)|=|13 + 7|=|20|= 206.|2−14.6|=| −12.6|= 12.6, or|14.6−2|=|12.6|= 12.67.| −39−(−28)|=| −39 + 28|=| −11|= 11, or| −28−(−39)|=| −28 + 39|=|11|= 118.∣∣∣∣−34−158∣∣∣∣=∣∣∣∣−68−158∣∣∣∣=∣∣∣∣−218∣∣∣∣= 218 , or∣∣∣∣158−(−34)∣∣∣∣=∣∣∣∣158 + 68∣∣∣∣=∣∣∣∣218∣∣∣∣= 2185. Operations with Real Numbers1.8−(−11) = 8 + 11 = 192.−310·(−13)=3·110·3 = 33·110 = 1·110 =1103.15÷(−3) =−54.−4−(−1) =−4 + 1 =−35.7·(−50) =−3506.−0.5−5 =−0.5 + (−5) =−5.57.−3 + 27 = 248.−400÷ −40 = 109.4.2·(−3) =−12.6

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2Just-in-Time Review10.−13−(−33) =−13 + 33 = 2011.−60 + 45 =−1512.12−23 = 12 +(−23)= 36 +(−46)=−1613.−24÷3 =−814.−6 + (−16) =−2215.−12÷(−58)=−12·(−85)= 1·82·5 = 1·2/·42/·5= 456. Interval Notation1.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}, or{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 aparenthesis before−5. The endpoint−2 is included in theinterval, 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 Exponents1.3−6=136Usinga−m=1am2.1(0.2)−5= (0.2)5Usinga−m=1am3.w−4z−9=z9w4Usinga−mb−n=bnam4.(zy)2=z2y2Raising a quotient to a power5.1000= 1Usinga0= 1,a= 06.a5a−3=a5−(−3)=a5+3=a8Using the quotient rule7.(2xy3)(−3x−5y) = 2(−3)x·x−5·y3·y=−6x1+(−5)y3+1=−6x−4y4,or−6y4x48.x−4·x−7=x−4+(−7)=x−11, or1x119.(mn)−6=m−6n−6, or1m6n610.(t−5)4=t−5·4=t−20, or1t208. Scientific Notation1.Convert 18,500,000 to scientific notation.We want the decimal point to be positioned between the1 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 = 1.85×1072.Convert 0.000786 to scientific notation.We want the decimal point to be positioned between the7 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 = 7.86×10−43.Convert 0.0000000023 to scientific notation.We want the decimal point to be positioned between the2 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 = 2.3×10−94.Convert 8,927,000,000 to scientific notation.We want the decimal point to be positioned between the8 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 = 8.927×1095.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

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Just-in-Time Review36.Convert 5.17×106to decimal notation.The exponent is positive, so the number is greater than10. We move the decimal point 6 places to the right.5.17×106= 5,170,0007.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.6.203×1011= 620,300,000,0008.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= 0.00002949. Order of Operations1.3 + 18÷6−3 = 3 + 3−3Dividing= 6−3 = 3Adding and subtracting2.= 5·3 + 8·32+ 4(6−2)= 5·3 + 8·32+ 4·4Working inside parentheses= 5·3 + 8·9 + 4·4Evaluating 32= 15 + 72 + 16Multiplying= 87 + 16Adding in order= 103from left to right3.5[3−8·32+ 4·6−2]= 5[3−8·9 + 4·6−2]= 5[3−72 + 24−2]= 5[−69 + 24−2]= 5[−45−2]= 5[−47]=−2354.16÷4·4÷2·256= 4·4÷2·256Multiplying and dividingin order from left to right= 16÷2·256= 8·256= 20485.26·2−3÷210÷2−8= 23÷210÷2−8= 2−7÷2−8= 26.4(8−6)2−4·3 + 2·831+ 190= 4·22−4·3 + 2·83 + 1Calculating in thenumerator and inthe denominator= 4·4−4·3 + 2·84= 16−12 + 164= 4 + 164= 204= 57.64÷[(−4)÷(−2)] = 64÷2 = 328.6[9−(3−2)] + 4(2−3)= 6[9−1] + 4(2−3)= 6·8 + 4(−1)= 48−4= 4410. Introduction to Polynomials1.5−x6The term of highest degree is−x6, so the degree of thepolynomial is 6.2.x2y5−x7y+ 4The degree ofx2y5is 2 + 5, or 7; the degree of−x7yis7 + 1, or 8; the degree of 4 is 0 (4 = 4x0). Thus the degreeof the polynomial is 8.3.2a4−3 +a2The term of highest degree is 2a4, so the degree of thepolynomial is 4.4.−41 =−41x0, so the degree of the polynomial is 0.5.4x−x3+ 0.1x8−2x5The term of highest degree is 0.1x8, so the degree of thepolynomial is 8.6.x−3 has two terms. It is a binomial.7.14y5has one term. It is a monomial.8.2y−14y2+ 8 has three terms. It is a trinomial.11. Add and Subtract Polynomials1.(8y−1)−(3−y)= (8y−1) + (−3 +y)= (8 + 1)y+ (−1−3)= 9y−4

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4Just-in-Time Review2.(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−23.(2x+ 3y+z−7) + (4x−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−34.(3ab2−4a2b−2ab+ 6)+(−ab2−5a2b+ 8ab+ 4)= (3−1)ab2+ (−4−5)a2b+ (−2 + 8)ab+ (6 + 4)= 2ab2−9a2b+ 6ab+ 105.(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+ 312. Multiply Polynomials1.(3a2)(−7a4) = [3(−7)](a2·a4)=−21a62.(y−3)(y+ 5)=y2+ 5y−3y−15Using FOIL=y2+ 2y−15Collecting like terms3.(x+ 6)(x+ 3)=x2+ 3x+ 6x+ 18Using FOIL=x2+ 9x+ 18Collecting like terms4.(2a+ 3)(a+ 5)= 2a2+ 10a+ 3a+ 15Using FOIL= 2a2+ 13a+ 15Collecting like terms5.(2x+ 3y)(2x+y)= 4x2+ 2xy+ 6xy+ 3y2Using FOIL= 4x2+ 8xy+ 3y26.(11t−1)(3t+ 4)= 33t2+ 44t−3t−4Using FOIL= 33t2+ 41t−413. Special Products of Binomials1.(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]= 25x2−30x+ 93.(2x+ 3y)2= (2x)2+ 2(2x)(3y) + (3y)2[(A+B)2=A2+2AB+B2]= 4x2+ 12xy+ 9y24.(a−5b)2=a2−2·a·5b+ (5b)2[(A−B)2=A2−2AB+B2]=a2−10ab+ 25b25.(n+ 6)(n−6)=n2−62[(A+B)(A−B) =A2−B2]=n2−366.(3y+ 4)(3y−4)= (3y)2−42[(A+B)(A−B) =A2−B2]= 9y2−1614. Factor Polynomials; The FOIL Method1.3x+ 18 = 3·x+ 3·6 = 3(x+ 6)2.2z3−8z2= 2z2·z−2z2·4 = 2z2(z−4)3.3x3−x2+ 18x−6=x2(3x−1) + 6(3x−1)= (3x−1)(x2+ 6)4.t3+ 6t2−2t−12=t2(t+ 6)−2(t+ 6)= (t+ 6)(t2−2)5.w2−7w+ 10We look for two numbers with a product of 10 and a sumof−7. By trial, we determine that they are−5 and−2.w2−7w+ 10 = (w−5)(w−2)6.t2+ 8t+ 15We look for two numbers with a product of 15 and a sumof 8. By trial, we determine that they are 3 and 5.t2+ 8t+ 15 = (t+ 3)(t+ 5)7.2n2−20n−48 = 2(n2−10n−24)Now factorn2−10n−24. We look for two numbers with aproduct of−24 and a sum of−10. By trial, we determinethat they are 2 and−12. Thenn2−10n−24 =(n+ 2)(n−12). We must include the common factor, 2,to have a factorization of the original trinomial.2n2−20n−48 = 2(n+ 2)(n−12)

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Just-in-Time Review58.y4−9y3+ 14y2=y2(y2−9y+ 14)Now factory2−9y+ 14.Look for two numbers with aproduct of 14 and a sum of−9. The numbers are−2 and−7. Theny2−9y+ 14 = (y−2)(y−7). We must includethe common factor,y2, in order to have a factorization ofthe original trinomial.y4−9y3+ 14y2=y2(y−2)(y−7)9.2n2+ 9n−561. 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 possibilitiesare−1·56, 1(−56),−2·28, 2(−28),−4·16, 4(−16),−7·8, and 7(−8).The factors can be written inthe 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, 9n. Bytrial, we determine that the factorization is (2n−7)(n+ 8).10.2y2+y−61. There is no common factor other than 1 or−1.2. Thefactorizationmustbeoftheform(2y+)(y+).3. Factor the constant term,−6. The possibilities are−1·6, 1(−6),−2·3, and 2(−3). The factors can bewritten 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.Bytrial, we determine that the factorization is(2y−3)(y+ 2).11.b2−6bt+ 5t2We look for two numbers with a product of 5 and a sumof−6. By trial, we determine that they are−1 and−5.b2−6bt+ 5t2= (b−t)(b−5t)12.x4−7x2−30 = (x2)2−7x2−30We look for two numbers with a product of−30 and a sumof−7. By trial, we determine that they are 3 and−10.x4−7x2−30 = (x2+ 3)(x2−10)15. Factoring Polynomials; Theac-Method1.8x2−6x−91. 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 isthe coefficient of the middle term,−6. The factorswe want are−12 and 6.4. Split the middle term using the numbers found instep (3):−6x=−12x+ 6x5. Factor by grouping.8x2−6x−9 = 8x2−12x+ 6x−9= 4x(2x−3) + 3(2x−3)= (2x−3)(4x+ 3)2.10t2+ 4t−61. Factor out the largest common factor, 2.10t2+ 4t−6 = 2(5t2+ 2t−3)Now factor 5t2+ 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 isthe 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.5t2+ 2t−3 = 5t2+ 5t−3t−3= 5t(t+ 1)−3(t+ 1)= (t+ 1)(5t−3)Include the largest common factor in the final factoriza-tion.10t2+ 4t−6 = 2(t+ 1)(5t−3)3.18a2−51a+ 151. Factor out the largest common factor, 3.18a2−51a+ 15 = 3(6a2−17a+ 5)Now factor 6a2−17a+ 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 wewant are−2 and−15.4. Split the middle term using the numbers found instep (3):−17a=−2a−15a.5. Factor by grouping.6a2−17a+ 5 = 6a2−2a−15a+ 5= 2a(3a−1)−5(3a−1)= (3a−1)(2a−5)Include the largest common factor in the final factoriza-tion.18a2−51a+ 15 = 3(3a−1)(2a−5)

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6Just-in-Time Review16. Special Factorizations1.z2−81 =z2−92= (z+ 9)(z−9)2.16x2−9 = (4x)2−32= (4x+ 3)(4x−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+ 12x+ 36 =x2+ 2·x·6 + 62= (x+ 6)25.9z2−12z+ 4 = (3z)2−2·3z·2 + 22= (3z−2)26.a3+ 24a2+ 144a=a(a2+ 24a+ 144)=a(a2+ 2·a·12 + 122)=a(a+ 12)27.x3+ 64 =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 Principles1.7t= 70t= 10Dividing by 7The solution is 10.2.x−5 = 7x= 12Adding 5The solution is 12.3.3x+ 4 =−83x=−12Subtracting 4x=−4Dividing by 3The solution is−4.4.6x−15 = 456x= 60Adding 15x= 10Dividing by 6The solution is 10.5.7y−1 = 23−5y12y−1 = 23Adding 5y12y= 24Adding 1y= 2Dividing by 12The solution is 2.6.3m−7 =−13 +m2m−7 =−13Subtractingm2m=−6Adding 7m=−3Dividing by 2The solution is−3.7.2(x+ 7) = 5x+ 142x+ 14 = 5x+ 14−3x+ 14 = 14Subtracting 5x−3x= 0Subtracting 14x= 0The solution is 0.8.5y−(2y−10) = 255y−2y+ 10 = 253y+ 10 = 25Collecting like terms3y= 15Subtracting 10y= 5Dividing by 3The solution is 5.18. Inequality-Solving Principles1.p+ 25≥ −100p≥ −125Subtracting 25The solution set is [−125,∞).2.−23x >6x <−32·6Multiplying by−32 andreversing the inequality symbolx <−9The solution set is (−∞,−9).3.9x−1<179x <18Adding 1x <2Dividing by 9The solution set is (−∞,2).4.−x−16≥40−x≥56Adding 6x≤ −56Multiplying by−1 andreversing the inequality symbolThe solution set is (−∞,−56].

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Just-in-Time Review75.13y−6<313y <9Adding 6y <27Multiplying by 3The solution set is (−∞,27).6.8−2w≤ −14−2w≤ −22Subtracting 8w≥11Dividing by−2 andreversing the inequality symbolThe solution set is [11,∞).19. The Principle of Zero Products1.2y2+ 42y= 02y(y+ 21) = 02y= 0ory+ 21 = 0y= 0ory=−21The solutions are 0 and−21.2.(a+ 7)(a−1) = 0a+ 7 = 0ora−1 = 0a=−7ora= 1The solutions are−7 and 1.3.(5y+ 3)(y−4) = 05y+ 3 = 0ory−4 = 05y=−3ory= 4y=−35ory= 4The solutions are−35 and 4.4.6x2+ 7x−5 = 0(3x+ 5)(2x−1) = 03x+ 5 = 0or2x−1 = 03x=−5or2x= 1x=−53orx= 12The solutions are−53 and 12 .5.t(t−8) = 0t= 0ort−8 = 0t= 0ort= 8The solutions are 0 and 8.6.x2−8x−33 = 0(x+ 3)(x−11) = 0x+ 3 = 0orx−11 = 0x=−3orx= 11The solutions are−3 and 11.7.x2+ 13x= 30x2+ 13x−30 = 0(x+ 15)(x−2) = 0x+ 15 = 0orx−2 = 0x=−15orx= 2The solutions are−15 and 2.8.12x2−7x−12 = 0(4x+ 3)(3x−4) = 04x+ 3 = 0or3x−4 = 04x=−3or3x= 4x=−34orx= 43The solutions are−34 and 43 .20. The Principle of Square Roots1.x2−36 = 0x2= 36x=√36orx=−√36x= 6orx=−6The solutions are 6 and−6, or±6.2.2y2−20 = 02y2= 20y2= 10y=√10ory=−√10The solutions are√10 and−√10, or±√10.3.6z2= 18z2= 3z=√3orz=−√3The solutions are√3 and−√3, or±√3.4.3t2−15 = 03t2= 15t2= 5t=√5ort=−√5The solutions are√5 and−√5, or±√5.5.z2−1 = 24z2= 25z=√25orz=−√25The solutions are 5 and−5, or±5.6.5x2−75 = 05x2= 75x2= 15x=√15orx=−√15The solutions are√15 and−√15, or±√15.

Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition - Page 11 preview image

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8Just-in-Time Review21. Simplify Rational Expressions1.3x−3x(x−1)The denominator is 0 when the factorx= 0 and alsowhenx−1 = 0, orx= 1. The domain is the set of all realnumbers except 0 and 1.2.y+ 6y2+ 4y−21 =y+ 6(y+ 7)(y−3)The denominator is 0 wheny=−7 ory= 3. The domainis the set of all real numbers except−7 and 3.3.x2−4x2−4x+ 4 = (x+ 2)(x−2)✏(x−2)(x−2)✏=x+ 2x−24.x2+ 2x−3x2−9= (x−1)(x+3)✏(x+3)✏(x−3) =x−1x−35.x3−6x2+ 9xx3−3x2=x(x2−6x+ 9)x2(x−3)=x/(x−3)✏(x−3)x/·x(x−3)✏=x−3x6.6y2+ 12y−483y2−9y+ 6= 6(y2+ 2y−8)3(y2−3y+ 2)= 2·3/·(y+ 4)(y−2)✏3/(y−1)(y−2)✏= 2(y+ 4)y−122. Multiply and Divide Rational Expressions1.r−sr+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)✏= 12.m2−n2r+s÷m−nr+s=m2−n2r+s·r+sm−n= (m+n)(m−n)✏(r+s)✏(r+s)✏(m−n)✏=m+n3.4x2+ 9x+ 2x2+x−2·x2−13x2+x−2= (4x+ 1)(x+2)✏(x+1)✏(x−1)✏(x+2)✏(x−1)✏(3x−2)(x+1)✏= 4x+ 13x−24.3x+ 122x−8÷(x+ 4)2(x−4)2= 3x+ 122x−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−2a2−a−6÷a2−2a2a+a2=a2−a−2a2−a−6·2a+a2a2−2a= (a−2)✏(a+ 1) (a)✧(2+a)✏(a−3)(a+2)✏(a)✧(a−2)✏=a+ 1a−36.x2−y2x3−y3·x2+xy+y2x2+ 2xy+y2=(x+y)(x−y)(x2+xy+y2)(x−y)(x2+xy+y2)(x+y)(x+y)=1x+y·(x+y)(x−y)(x2+xy+y2)(x+y)(x−y)(x2+xy+y2)=1x+y·1Removing a factor of 1=1x+y23. Add and Subtract Rational Expressions1.a−3ba+b+a+ 5ba+b= 2a+ 2ba+b=2(a+b)✏1·(a+b)✏= 22.x2−53x2−5x−2 +x+ 13x−6=x2−5(3x+ 1)(x−2) +x+ 13(x−2)=x2−5(3x+ 1)(x−2)·33 +x+ 13(x−2)·3x+ 13x+ 1= 3(x2−5) + (x+ 1)(3x+ 1)3(3x+ 1)(x−2)= 3x2−15 + 3x2+ 4x+ 13(3x+ 1)(x−2)=6x2+ 4x−143(3x+ 1)(x−2)

Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition - Page 12 preview image

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Just-in-Time Review93.a2+ 1a2−1−a−1a+ 1=a2+ 1(a+ 1)(a−1)−a−1a+ 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+ 23x2−2x−8 +73x2+x−4=9x+ 2(3x+ 4)(x−2) +7(3x+ 4)(x−1),LCD is (3x+ 4)(x−2)(x−1)=9x+ 2(3x+4)(x−2)·x−1x−1 +7(3x+4)(x−1)·x−2x−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.yy2−y−20−2y+ 4=y(y+ 4)(y−5)−2y+ 4,LCD is (y+ 4)(y−5)=y(y+ 4)(y−5)−2y+ 4·y−5y−5=y(y+ 4)(y−5)−2y−10(y+ 4)(y−5)=y−(2y−10)(y+ 4)(y−5)=y−2y+ 10(y+ 4)(y−5)=−y+ 10(y+ 4)(y−5)6.3yy2−7y+ 10−2yy2−8y+ 15=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 Expressions1.xy−yx1y+ 1x=xy−yx1y+ 1x·xyxy ,LCM isxy=(xy−yx)(xy)(1y+ 1x)(xy)=x2−y2x+y= (x+y)✏(x−y)(x+y)✏·1=x−y2.a−bba2−b2ab=a−bb·aba2−b2=a−bb·ab(a+b)(a−b)=a b✧(a−b)✏b/ (a+b)(a−b)✏=aa+b

Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition - Page 13 preview image

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10Just-in-Time Review3.w+8w21 + 2w=w·w2w2+8w21·ww+ 2w=w3+ 8w2w+ 2w=w3+ 8w2·ww+ 2= (w+2)✏(w2−2w+ 4)w/w/·w(w+2)✏=w2−2w+ 4w4.x2−y2xyx−yy=x2−y2xy·yx−y= (x+y)(x−y)✏y✧x y✧(x−y)✏=x+yx5.ab−ba1a−1b=a2−b2b−aMultiplying byabab= (a+b)(a−b)b−a= (a+b)(a−b)✏−1·(a−b)✏=−a−b25. Simplify Radical Expressions1.√(−21)2=| −21|= 212.√9y2=√(3y)2=|3y|= 3y3.√(a−2)2=a−24.3√−27x3=3√(−3x)3=−3x5.4√81x8=4√(3x2)4= 3x26.5√32 =5√25= 27.4√48x6y4=4√16x4y4·3x2= 2xy4√3x2=2xy4√3x28.√15√35 =√15·35 =√3·5·5·7 =√52·3·7 =√52· √3·7 = 5√219.√40xy√8x=√40xy8x=√5y10.3√3x23√24x5=3√3x224x5=3√18x3=12x11.√x2−4x+ 4 =√(x−2)2=x−212.√2x3y√12xy=√24x4y2=√4x4y2·6 = 2x2y√613.3√3x2y3√36x=3√108x3y=3√27x3·4y= 3x3√4y14.5√2 + 3√32 = 5√2 + 3√16·2= 5√2 + 3·4√2= 5√2 + 12√2= (5 + 12)√2= 17√215.7√12−2√3 = 7·2√3−2√3 = 14√3−2√3 = 12√316.2√32 + 3√8−4√18 = 2·4√2 + 3·2√2−4·3√2 =8√2 + 6√2−12√2 = 2√217.6√20−4√45 +√80 = 6√4·5−4√9·5 +√16·5= 6·2√5−4·3√5 + 4√5= 12√5−12√5 + 4√5= (12−12 + 4)√5= 4√518.(2 +√3)(5 + 2√3)= 2·5 + 2·2√3 +√3·5 +√3·2√3= 10 + 4√3 + 5√3 + 3·2= 10 + 9√3 + 6= 16 + 9√319.(√8 + 2√5)(√8−2√5)=(√8)2−(2√5)2= 8−4·5= 8−20=−1220.(1 +√3)2= 12+ 2·1· √3 + (√3)2= 1 + 2√3 + 3= 4 + 2√326. Rationalizing Denominators1.4√11 =4√11·√11√11 = 4√11112.√37 =√37·77 =√2149 =√21√49 =√2173.3√73√2 =3√73√2·3√43√4 =3√283√8=3√2824.3√169=3√169·33 =3√4827 =3√483√27 =3√8·63= 23√63

Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition - Page 14 preview image

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Just-in-Time Review115.3√30−4 =3√30−4·√30 + 4√30 + 4=3√30 + 12(√30)2−42= 3√30 + 1230−16= 3√30 + 12146.4√7− √3 =4√7− √3·√7 +√3√7 +√3=4√7 + 4√3(√7)2−(√3)2= 4√7 + 4√37−3= 4√7 + 4√34= 4(√7 +√3)4=√7 +√37.6√m− √n=6√m− √n·√m+√n√m+√n=6(√m+√n)(√m)2−(√n)2= 6√m+ 6√nm−n8.1− √2√3− √6 =1− √2√3− √6·√3 +√6√3 +√6=√3 +√6− √6− √123−6=√3 +√6− √6−2√33−6=−√3−3=√3327. Rational Exponents1.y5/6=6√y52.x2/3=3√x23.163/4= (161/4)3=(4√16)3= 23= 84.47/2= (√4)7= 27= 1285.125−1/3=11251/3=13√125 = 156.32−4/5=(5√32)−4= 2−4=1167.12√y4=y4/12=y1/38.√x5=x5/29.x1/2·x2/3=x1/2+2/3=x3/6+4/6=x7/6=6√x7=x6√x10.(a−2)9/4(a−2)−1/4= (a−2)9/4+(−1/4)=(a−2)8/4= (a−2)211.(m1/2n5/2)2/3=m12·23n52·23=m1/3n5/3=3√m3√n5=3√mn5=n3√mn228. The Pythagorean Theorem1.a2+b2=c282+ 152=c264 + 225 =c2289 =c217 =c2.a2+b2=c242+ 42=c216 + 16 =c232 =c2√32 =c5.657≈c3.a2+b2=c252+b2= 13225 +b2= 169b2= 144b= 124.a2+b2=c2a2+ 122= 132a2+ 144 = 169a2= 25a= 55.a2+b2=c2(√5)2+b2= 625 +b2= 36b2= 31b=√31≈5.568

Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition - Page 15 preview image

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Solution Manual for Algebra and Trigonometry: Graphs and Models, 6th Edition - Page 16 preview image

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42244224xy(1, 4)(3,5)(0, 2)(4, 0)(2,2)224224222424yx(25, 0)(2,24)(1, 4)(0, 3)(24,22)4224424xy(5, 1)(2, 3)(0, 1)(5, 1)(2,1)224224222424yx(25, 2)(4,23)(0,21)(2, 0)(21,25)Chapter 1Graphs, Functions, and ModelsExercise Set 1.11.Point A is located 5 units to the left of they-axis and4 units up from thex-axis, so its coordinates are (−5,4).Point B is located 2 units to the right of they-axis and2 units down from thex-axis, so its coordinates are (2,−2).Point Cis located 0 units to the right or left of they-axisand 5 units down from thex-axis, so its coordinates are(0,−5).Point D is located 3 units to the right of they-axis and5 units up from thex-axis, so its coordinates are (3,5).Point E is located 5 units to the left of they-axis and4unitsdownfromthex-axis,soitscoordinatesare(−5,−4).Point F is located 3 units to the right of they-axis and0 units up or down from thex-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 rightof they-axis. Since the second coordinate is 0, we do notmove up or down from thex-axis.To graph (−3,−5) we move from the origin 3 units to theleft of they-axis.Then we move 5 units down from thex-axis.To graph (−1,4) we move from the origin 1 unit to the leftof they-axis. Then we move 4 units up from thex-axis.To graph (0,2) we do not move to the right or the left ofthey-axis since the first coordinate is 0. From the originwe move 2 units up.To graph (2,−2) we move from the origin 2 units to theright of they-axis. Then we move 2 units down from thex-axis.4.5.To graph (−5,1) we move from the origin 5 units to theleft of they-axis. Then we move 1 unit up from thex-axis.To graph (5,1) we move from the origin 5 units to the rightof they-axis. Then we move 1 unit up from thex-axis.To graph (2,3) we move from the origin 2 units to the rightof they-axis. Then we move 3 units up from thex-axis.To graph (2,−1) we move from the origin 2 units to theright of they-axis.Then we move 1 unit down from thex-axis.To graph (0,1) we do not move to the right or the left ofthey-axis since the first coordinate is 0. From the originwe move 1 unit up.6.7.The first coordinate represents the year and the corre-sponding second coordinate represents the number of citiesservedbySouthwestAirlines.Theorderedpairsare(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),and (2014, 96).8.The first coordinate represents the year and the secondcoordinaterepresentsthepercentofMarineswhoarewomen.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 forxand−9 fory.y= 7x−2−9 ? 7(−1)−2∣∣∣−7−2−9∣∣−9TRUEThe equation−9 =−9 is true, so (−1,−9) is a solution.

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