Solution Manual for Basic College Mathematics, 13th Edition

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SOLUTIONSMANUALBASICCOLLEGEMATHEMATICSTHIRTEENTHEDITIONMarvin L. BittingerIndiana University Purdue University IndianapolisJudith A. BeecherBarbara L. JohnsonIvy Tech Community College of Indiana

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ContentsChapter1........................... 1Chapter2..........................25Chapter3.........................43Chapter4.........................71Chapter5..........................95Chapter6......................... 115Chapter7......................... 139Chapter8......................... 157Chapter9......................... 179Chapter 10......................... 199Chapter 11......................... 213

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Chapter 1Whole NumbersExercise Set 1.1RC2.In 615,702, the number 615 is in the thousands period.RC4.The number 721 is written in standard notation.CC2.A word name for 42,000,000 is forty-two million.CC4.A word name for 18,000,000,000 is eighteen billion.CC6.A word name for 40,000,000,000,000 is forty trillion.2.5 ten thousands4.5 hundred thousands6.98.310.6 thousands + 6 hundreds + 8 tens + 8 ones12.1 thousand + 7 hundreds + 8 tens + 6 ones14.3 ten thousands + 8 thousands + 4 hundreds + 5 tens +3 ones16.1 hundred thousand + 3 ten thousands + 5 thousands +0 hundreds + 8 tens + 0 ones, or 1 hundred thousand +3 ten thousands + 5 thousands + 8 tens18.1 bil l ion + 2 hundred millions + 6 ten millions + 6millions + 8 hundred thousands + 8 ten thousands + 3thousands + 5 hundreds + 9 tens + 8 ones20.3 hundred millions + 2 ten millions + 3 millions + 9 hun-dred thousands + 9 ten thousands + 5 thousands + 5hundreds + 2 tens + 8 ones22.Forty-eight24.Forty-five thousand, nine hundred eighty-seven26.One hundred eleven thousand, thirteen28.Forty-three billion, five hundred fifty million, six hundredfifty-one thousand, eight hundred eight30.Ninety-nine thousand, eight hundred fifty-three32.Two hundred twenty-six million, one thousand, two hun-dred eighty-eight34.354,70236.17,11238.19,610,43940.700,000,00042.26,000,000,00044.200,01746.2,793,000,00048.All digits are 9’s. Answers may vary. For an 8-digit read-out, for example, it would be 99,999,999. This number hasthree periods.Exercise Set 1.2RC2.In the addition 5 + 2 = 7, the number 7 is the sum.RC4.The distance around an object is its perimeter.CC2.110CC4.1700CC6.10,0002.1 5 2 1+3 4 81 8 6 94.17 3+ 6 91 4 26.17 5 0 3+ 2 6 8 31 0,1 8 68.119 9 9+1 11 0 1 010.12 7 1+ 3 3 3 83 6 0 912.12 8 0+ 3 4,9 0 23 5,1 8 214.1111 0,1 2 01 2,9 8 9+5 7 3 82 8,8 4 716.13 6 5 4+ 2 7 0 06 3 5 418.1114 5,8 7 9+ 2 1,7 8 66 7,6 6 5

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2Chapter 1:Whole Numbers20.11119 9,9 9 9+1 1 21 0 0,1 1 122.23 82 73 21 4+ 7 61 8 724.11114 2,4 8 78 3,1 4 1+ 3 6,7 1 21 6 2,3 4 026.229 8 95 6 68 3 49 2 0+ 7 0 34 0 1 228.Perimeter = 14 mi + 13 mi + 8 mi + 10 mi + 47 mi +22 miWe carry out the addition.21 41 381 04 7+ 2 21 1 4The perimeter of the figure is 114 mi.30.Perimeter = 62 yd + 39 yd + 54 yd + 46 yd + 28 ydWe carry out the addition.26 23 95 44 6+ 2 82 2 9The perimeter of the figure is 229 yd.32.90 ft + 90 ft + 90 ft + 90 ft = PerimeterWe carry out the addition.9 09 09 0+ 9 03 6 0The batter travels 360 ft when a home run is hit.34.Nine billion, three hundred forty-six million, three hundredninety-nine thousand, four hundred sixty-eightExercise Set 1.3RC2.subtraction symbolRC4.differenceCC2.9 101 0 0/✥✥−79 3CC4.101/ 0/ 0 0−4 0 06 0 0CC6.99 101 0 0 0/✥✥✥−9 9 912.8 7−3 45 34.5 2 6−3 2 32 0 36.6 137/ 3/−2 84 58.5 146/ 4/−1 94 510.1868/157/ 9/5/−3 983 9712.1020/163/ 1/6/−2 476914.1777/178/ 8/7/−6 981 8916.3 123 4/ 2/−2 1 71 2 5

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Exercise Set 1.4318.13 1253/2/116/ 4/3/1/−2 8963 53520.12 1572/5/148/ 3/6/4/−5 3752 98922.11 1381/3/119/ 2/4/1/−5 6433 59824.6 157/ 5/ 8 3−3 6 4 13 9 4 226.11 1151/1/121 6/,2/2/2/−5,8881 0,33428.1121/11 8 143/ 2/,1/ 9/ 4/−2 9,2 3 629 5 830.8 109/ 0/−7 81 232.79 138 0 3✥✥−4 1 83 8 534.39 159 4 0 5/✥✥−2 5 89 1 4 736.1464/9 107/ 5 0 0/✥✥−3 6 0 43 8 9 638.69138 4,7 03/✥✥−2 988 4,4 0540.144/10 0 171/ 5/,0/ 1/ 7/−78 0 972 0 842.799 138 0 0 3/✥✥✥−5 9 97 4 0 444.699 101 7,0 0 0/✥✥✥−1 1,5 9 85 4 0 246.3999164 0,00 6/✭✭✭✭−14 73 9,85 948.2999143 0,00 4/✭✭✭✭−674 92 3,25 550.9 0 1+ 2 39 2 452.19 9 0 9+ 1 0 1 11 0,9 2 054.9 thousands + 1 hundred + 0 tens + 3 ones, or 9 thousands+ 1 hundred + 3 ones56.3,928,124−1,098,947Using a calculator to carry out the subtraction, we findthat the difference is 2,829,177.Exercise Set 1.4RC2.In the multiplication 4×3 = 12, 12 is the product.RC4.The product of 1 and any numberaisa.CC2.4800CC4.80,000CC6.400,0002.28 7×43 4 84.57 6×96 8 46.48 0 6×75 6 4 28.3227 8 6 7×43 1,4 6 8

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4Chapter 1:Whole Numbers10.47 8×6 04 6 8 012.228 7×3 43 4 82 6 1 02 9 5 814.2 3 4 0×1 0 0 02,3 4 0,0 0 016.8 0 0×7 0 05 6 0,0 0 018.54547 7 7×7 75 4 3 95 4 3 9 05 9,8 2 920.37375 4 9×8 84 3 9 24 3 9 2 04 8,3 1 222.21114 3 2×3 7 52 1 6 03 0 2 4 01 2 9 6 0 01 6 2,0 0 024.2323453 4 6×6 5 93 1 1 41 7 3 0 02 0 7 6 0 02 2 8,0 1 426.2261 5118 9 2 8×3 1 7 21 7 8 5 66 2 4 9 6 08 9 2 8 0 02 6 7 8 4 0 0 02 8,3 1 9,6 1 628.2424136 4 0 8×6 0 6 42 5 6 3 23 8 4 4 8 03 8 4 4 8 0 0 03 8,8 5 8,1 1 230.1 14 44 43 5 5×2 9 93 1 9 53 1 9 5 07 1 0 0 01 0 6,1 4 532.122416 5 2 1×3 4 4 95 8 6 8 92 6 0 8 4 02 6 0 8 4 0 01 9 5 6 3 0 0 02 2,4 9 0,9 2 934.34444 5 0 6×7 8 0 03 6 0 4 8 0 03 1 5 4 2 0 0 03 5,1 4 6,8 0 036.126 0 0 9×2 0 0 31 8 0 2 71 2 0 1 8 0 0 01 2,0 3 6,0 2 738.A=l×w= 129 yd×65 yd = 8385 sq yd40.A=l×w= 200 ft×85 ft = 17,000 sq ft42.1229 8 7 68 7 67 6+61 0,8 3 444.1323/108 183/ 4/ 0/,7 9/ 8/−8 6,6 7 92 5 4,1 1 946.0

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Exercise Set 1.5548.Seven million, four hundred thirty-two thousandExercise Set 1.5RC2.dividendRC4.divisorCC2.341 51 23The answer is 3 R 3.CC4.4 829 781 71 61The answer is 48 R 1.2.54÷9 = 6 because 54 = 9·6.4.3737 = 1 Any nonzero number divided by itself is 1.6.561= 56 Any number divided by 1 is that same number.8.032 = 0Zero divided by any nonzero number is 0.10.74÷0 is not defined, because division by 0 is not defined.12.204= 5 because 20 = 4·5.14.2 3 336 9 9699990The answer is 233.16.1 0 888 6 986 96 45The answer is 108 R 5.18.7 0 832 1 2 42 12 42 40The answer is 708.20.1 0 1 299 1 1 091 192 01 82The answer is 1012 R 2.22.1 9 423 8 921 81 8981The answer is 194 R 1.24.1 4 668 8 162 82 44 13 65The answer is 146 R 5.26.2 0 0 936 0 2 762 72 70The answer is 2009.28.5 1 784 1 3 94 01 385 95 63The answer is 517 R 3.30.1 2 7 01 0 01 2 7,0 0 01 0 02 7 02 0 07 0 07 0 0000The answer is 1270.

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6Chapter 1:Whole Numbers32.4 2 61 04 2 6 04 02 62 06 06 00The answer is 426.34.2 8 92 05 7 9 84 01 7 91 6 01 9 81 8 01 8The answer is 289 R 18.36.2 44 09 8 78 01 8 71 6 02 7The answer is 24 R 27.38.4 02 39 4 29 22 2The answer is 40 R 22.40.5 05 42 7 2 92 7 02 9The answer is 50 R 29.42.5 51 0 25 6 1 25 1 05 1 25 1 02The answer is 55 R 2.44.1 0 777 4 974 94 90The answer is 107.46.8 0 897 2 7 37 27 37 21The answer is 808 R 1.48.1 0 1 077 0 7 47 0774The answer is 1010 R 4.50.3 0 12 47 2 4 27 24 22 41 8The answer is 301 R 18.52.1 0 24 84 8 9 94 89 99 63The answer is 102 R 3.54.2 1 03 67 5 6 37 23 63 63The answer is 210 R 3.56.8 0 33 62 8,9 2 92 8 81 2 91 0 82 1The answer is 803 R 21.58.9 8 49 08 8,5 6 08 1 07 5 67 2 03 6 03 6 00The answer is 984.

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Chapter 1 Mid-Chapter Review760.2 9 0 43 0 68 8 8,8 8 86 1 22 7 6 82 7 5 41 4 8 81 2 2 42 6 4The answer is 2904 R 264.62.7 0 0 28 0 35,6 2 2,6 0 65 6 2 11 6 0 61 6 0 60The answer is 7002.64.8 8,7 7 7−2 2,3 3 36 6,4 4 466.2 6 8×3 51 3 4 08 0 4 09 3 8 068.A=l×w= 11 ft×9 ft =99 sq ft70.Pairs of factors whose product is 36 are:1 and 362 and 183 and 124 and 96 and 6a) The pair above whose sum is 13 is 4 and 9.b) The pair above whose difference is 0 is 6 and 6.c) The pair above whose sum is 20 is 2 and 18.d) The pair above whose difference is 9 is 3 and 12.72.34,584,132÷76= 4,386Consider the related multiplication sentence:4,386·76= 34,584,132Since the ones digit of the product is 2, the missing onesdigit must be either 2 or 7 (6·2 = 12 and 6·7 = 42).We try 2:34,584,132÷762 = 45,386We see that the missing ones digit is 2 and the missingthousands digit is 5.Chapter 1 Mid-Chapter Review1.The statement is false.For example, 8−5 = 3, but 5 isnot equalto 8 + 3.2.The statement is true. See page 19 in the text.3.The statement is true. See page 19 in the text.4.The statement is false. For example, 3·0 = 0 and 0 is notgreater than 3. Also, 1·1 = 1 and 1 is not greater than 1.5.It is true that zero divided by any nonzero number is 0.6.The statement is false.Any number divided by 1 is thenumber itself. For example, 271= 27.7.95︸︷︷︸, 406︸︷︷︸, 237︸︷︷︸↑↑↑Ninety-five million,four hundred six thousand,two hundred thirty-seven8.59 146 0 4/✥✥−4 9 71 0 79.269 8The digit 6 names the number of hundreds.10.61, 2 0 4The digit 6 names the number of ten thousands.11.1 46 , 2 3 7The digit 6 names the number of thousands.12.5 86The digit 6 names the number of ones.13.3 0 6, 4 5 8, 129The digit 2 names the number of tens.14.3 06 , 4 5 8, 1 2 9The digit 6 names the number of millions.15.3 0 6, 458, 1 2 9The digit 5 names the number of ten thousands.16.3 0 6, 4 5 8,12 9The digit 1 names the number of hundreds.17.5602 = 5 thousands + 6 hundreds + 0 tens + 2 ones, or 5thousands + 6 hundreds + 2 ones18.69,345 = 6 ten thousands + 9 thousands + 3 hundreds +4 tens + 5 ones19.A word name for 136 is one hundred thirty-six.20.A word name for 64,325 is sixty-four thousand, three hun-dred twenty-five.21.Standard notation for three hundred eight thousand, sevenhundred sixteen is 308,716.22.Standard notation for four million, five hundred sixty-seven thousand, two hundred ninety-one is 4,567,291.

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8Chapter 1:Whole Numbers23.3 1 6+ 4 8 27 9 824.115 9 3+ 4 3 71 0 3 025.112 6 3 8+ 5 2 8 47 9 2 226.1114 6 1 72 4 3 6+4 8 17 5 3 427.7 8 6−3 2 14 6 528.1151/146/ 2/4/−2 853 3929.1525/9 123/ 6 0 2/✥✥−1 7 4 81 8 5 430.499 145 0 0 4/✥✥✥−6 7 64 3 2 831.33 6×62 1 632.11555 6 7×2 84 5 3 61 1 3 4 01 5,8 7 633.2134 0 7×3 2 52 0 3 58 1 4 01 2 2 1 0 01 3 2,2 7 534.22 319 4 3 5×6 0 21 8 8 7 05 6 6 1 0 0 05,6 7 9,8 7 035.2 5 341 0 1 282 12 01 21 20The answer is 253.36.1 1 23 84 2 6 13 84 63 88 17 65The answer is 112 R 5.37.2 36 01 3 9 91 2 01 9 91 8 01 9The answer is 23 R 19.38.1 4 45 68 0 9 55 62 4 92 2 42 5 52 2 43 1The answer is 144 R 31.39.Perimeter = 10 m + 4 m + 8 m + 3 m = 25 m40.A= 4 in.×2 in. = 8 sq in.41.When numbers are being added, it does not matter howthey are grouped.42.Subtraction is not commutative. For example, 5−2 = 3,but 2−5= 3.43.Answers will vary. Suppose one coat costs $150. Then themultiplication 4·$150 gives the cost of four coats.Suppose one ream of copy paper costs $4. Then the mul-tiplication $4·150 gives the cost of 150 reams.44.Using the definition of division, 0÷0 =asuch thata·0 = 0.We see thatacould beanynumber sincea·0 = 0 for anynumbera.Thus, we cannot say that 0÷0 = 0.This iswhy we agree not to allow division by 0.

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Exercise Set 1.69Exercise Set 1.6RC2.Because the digit in the hundreds place, 5, is 5 orhigher, we round up. The statement is false.RC4.The statement is true. See page 40 in the text.CC2.10002.5304.89506.508.80010.90012.70014.460016.198,40018.500020.200022.736,00024.6,713,00026.6 26 09 71 0 04 65 0+ 8 1+8 02 9 028.6 7 36 7 0−2 8−3 06 4 030.4 14 02 12 05 56 0+6 0+6 01 7 71 8 032.8 3 68 4 03 7 43 7 07 9 47 9 0+ 9 3 8+ 9 4 03 9 4 72 9 4 03947 seems to be incorrect.34.5 6 86 0 04 7 25 0 09 3 89 0 0+ 4 0 2+ 4 0 02 4 0 036.9 4 3 89 4 0 0−2 7 8 7−2 8 0 06 6 0 038.4 8 15 0 07 0 27 0 06 2 36 0 0+ 1 0 4 3+ 1 0 0 01 8 4 92 8 0 01849 seems to be incorrect.40.3 2 63 0 02 7 53 0 07 5 88 0 0+ 9 4 3+ 9 0 02 3 0 22 3 0 042.7 6 4 88 0 0 09 3 4 89 0 0 07 8 4 28 0 0 0+ 2 2 2 2+ 2 0 0 02 7, 0 0 044.8 4, 8 9 08 5, 0 0 0−1 1, 1 1 0−1 1, 0 0 07 4, 0 0 046.5 15 0×7 8×8 04 0 0 048.6 36 0×5 4×5 03 0 0 050.3 5 54 0 0×2 9 9×3 0 01 2 0, 0 0 052.7 8 98 0 0×4 3 4×4 0 03 2 0, 0 0 054.454÷87≈450÷90 = 556.1263÷29≈1260÷30 = 4258.3641÷571≈3600÷600 = 660.32,854÷748≈32,900÷700 = 4762.$ 4 9 8$ 5 0 02 8 93 0 0+ 1 4 5+ 1 0 0$ 9 0 064.$ 4 9 8$ 5 0 02 8 93 0 06 21 0 01 5 92 0 0+ 3 2 0+3 0 0$ 1 4 0 0The budget covers the choices.66.Answers will vary depending on the options chosen.

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10Chapter 1:Whole Numbers68.a) Totalcost of attending:$250·490 = $122,500Totalcost of hotelrooms:$170·2·320 = $108,800Totalamount spent:$122,500 + $108,800 = $231,300b) Totalcost of attending:$200·500 = $100,000Totalcost of hotelrooms:$200·2·300 = $120,000Totalamount spent:$100,000 + $120,000 = $220,00070.$2211÷$11≈$2200÷$10 = 220 rolls72.32>074.28>1876.77<11778.999>99780.345<45682.12<3284.1,014,023>758,708, or 758,708<1,014,02386.843,393>842,583, or 842,583<843,39388.9 0 0 2+ 4 5 8 71 3,5 8 990.899 129 0 0 2/✥✥✥−4 5 8 74 4 1 592.343 0 6×5 82 4 4 81 5 3 0 01 7,7 4 894.2 0 82 34 7 8 44 61 8 41 8 40The answer is 208.96.27,060; this is close to the estimated sum found in Exercise42.98.73,780; this is close to the estimated difference found inExercise 44.Exercise Set 1.7RC2.(a)RC4.(b)CC2.672 = 6·n672 ? 6·112∣∣∣672TRUE112 is a solution.CC4.5462 = 3189 +t5462 ? 3189 + 2327∣∣∣5516FALSE2327 is not a solution.2.254.86.t= 5678 + 9034 = 14,7128.m= 9007−5667 = 334010.z= 34·15 = 51012.w= 256÷16 = 1614.t= 22−15 = 716.t= 16−16 = 018.x= 57−20 = 3720.w= 53−17 = 3622.x= 426= 724.m= 1629= 1826.y= 964= 2428.t= 7413= 24730.y= 9281−8322 = 95932.p= 92−56 = 3634.y= 23×78 = 179436.z= 133−67 = 6638.w= 34044= 85140.x= 807−438 = 36942.q= 10,534÷458 = 2344.x= 608019= 32046.x= 150020= 75

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Exercise Set 1.81148.t= 9281−8322 = 95950.n= 3004−1745 = 125952.n= 66012= 5554.x= 22,135233= 9556.z= 512−63 = 44958.1 4 291 2 7 893 73 61 81 80The answer is 142.60.3 3 41 75 6 8 95 15 85 17 96 81 1The answer is 334 R 11.62.342>33964.0<1166.6,375,60068.x= 14,332,38848,916= 293Exercise Set 1.8RC2.Translate.RC4.Check.CC2.Letx= the number of miles by which Wednesday’sdrive exceeded Thursday’s drive. Then we have 500 =x+ 125. The correct answer is (b).CC4.Letx= the totalamount spent on groceries in Mayand June. Then we have 500 + 125 =x. The correctanswer is (a).2.Letw= the number of pounds by which the waste gener-ated annually per capita in the United States exceeds thewaste generated in Denmark.Solve: 1473 +w= 1606w= 123 lb4.Letw= the number of pounds of waste generated annuallyper capita in Iceland.Solve:w+ 531 = 1713w= 1182 lb6.Letn= the number of entries in each row.Solve: 504÷36 =nn= 14 entries8.Letr= the number of active rotary oilrigs in 2007.Solve:r+ 687 = 984r= 297 rigs10.Letn= the number of miles by which the length of theNile exceeds the length of the Missouri-Mississippi.Solve: 3860 +n= 4135n= 275 mi12.Letp= the totalnumber of squares in the puzzle.Solve: 15·15 =pp= 225 squares14.Letc= the number of milligrams of caffeine in a 20-ozbottle of Coca Cola.Solve: 25 + 32 =cc= 57 milligrams16.Letm= the number of minutes in a day.Solve: 60·24 =mm= 1440 minutes18.Letr= the amount by which the average monthly rent inAtlanta exceeds the average monthly rent in Indianapolis.Solve: 905 +r= 1401r= $496 per month20.Letr= the amount of rent each sister pays.Solve: 2r= 936r= $468 per month22.Letr= the amount of rent a tenant would pay for a one-bedroom apartment, on average, in Seattle during a 6-month period.Solve: 6·2063 =rr= $12,37824.Lets= the speed limit for trucks.Solve:s+ 10 = 75s= 65 mph26.Letq= the number of quires in a ream.Solve: 25·q= 500q= 20 quires

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12Chapter 1:Whole Numbers28.Lets= the amount by which spending by visitors to theUnited States exceeded spending by Americans travelingabroad.Solve: 110,500,000,000 +s= 153,700,000,000s= $43,200,000,00030.Letc= the totalcost of the purchase.Solve: 96·88 =cc= $844832.Letw= the number of full weeks that will pass before thestation must begin re-airing episodes.Solve: 5·w= 208w= 41 R 3, so 41 full weeks will pass and 3 episodes will beshown the following week before previously aired episodesare rerun.34.Letl= the number of labels on each sheet.Solve: 25·l= 750l= 30 labels36.Letg= the number of gallons required for 3795 mi of citydriving.Solve: 3795÷23 =gg= 165 gal38.a) LetA= the area of the court, in square feet.Solve:A= 94·50A= 4700 square feetb) LetP= the perimeter of the court, in feet.Solve:P= 94 + 50 + 94 + 50P= 288 ftc) Leta= the amount by which the area of a col-lege court exceeds the area of a high school court,in square feet.Solve: 4200 +a= 4700a= 500 square feet40.Letc= the number of cartons needed.Solve: 528÷12 =cc= 44 cartons42.Letm= the distance on the map, in inches, between twocities that, in reality, are 2016 mi apart.Solve: 2016÷288 =mm= 7 in.Letr= the distance in miles, in reality, between two citiesthat are 8 in. apart on the map.Solve: 288·8 =rr= 2304 mi44.Letm= the number of months required to pay off theloan.Solve: 7824÷163 =mm= 48 months46.Letn= the number of 100’s in 3500.Solve: 3500÷100 =nn= 35Lett= the number of minutes you must golf, walking, inorder to lose one pound.Solve:t= 35·20t= 700 min; we could also express this as 11 hr, 40 min.48.Letn= the number of new jobs that will be created formarketing managers and accountants.n= 19,700 + 142,400 = 162,100Lets= the number of new jobs that will be created forsales managers.Solve:s+ 143,100 = 162,100s= 19,000 jobs50.LetF= the number of seats in first class,E= the numberof seats in economy class, andT= the totalnumber ofseats.Solve: 3·4 =F, 23·6 =E, andT=F+EF= 12,E= 138,T= 12 + 138 = 150 seats52.Letc= the totalcost of the 5 video games.Solve: 5·64 =cc= $320Then letn= the number of $20 bills required.Solve: 320÷20 =nn= 16 $20 bills54.Letb= the new balance.Solve: 749−34−65 + 123 =bb= $77356.Letl= the total length of the bookshelves, in feet.Solve: 6·3 =ll= 18 ftSince the total length of the bookshelves is greater than16 ft, the shelves cannot be put side by side on the 16-ftwall.58.1585/9 129/ 6 0 2/✥✥−1 8 4 37 7 5 960.1 4 73 24 7 0 83 21 5 01 2 82 2 82 2 44The answer is 147 R 4.62.A=l×w= 211 ft×46 ft = 9706 sq ft

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Exercise Set 1.91364.x= 81−15 = 6666.Consider one student taking one class a “student-classunit.” Then lets= the totalnumber of student-class unitsandp= the number of students taught by each instructor.Solve: 1200·5 =s, 4·30 =ps= 6000,p= 120Now letn= the number of instructors.Solve: 6000÷120 =nn= 50 instructorsExercise Set 1.9RC2.The expression 92can be read “nine squared.”RC4.To find the average of 7, 8, and 9, we add the numbersand divide the sum by 3.CC2.DivisionCC4.Multiplication2.254.1336.928.1410.12512.6414.100,00016.6418.(12 + 6) + 18 = 18 + 18= 3620.(52−40)−8 = 12−8= 422.(1000÷100)÷10 = 10÷10= 124.256÷(64÷4) = 256÷16= 1626.22+ 52= 4 + 25= 2928.(32−27)3+ (19 + 1)3= 53+ 203= 125 + 8000= 812530.23 + 18·20 = 23 + 360= 38332.10·7−4 = 70−4= 6634.90−5·5·2 = 90−50= 4036.82−8·2 = 64−8·2= 64−16= 4838.1000÷25−(15 + 5) = 1000÷25−20= 40−20= 2040.3·8 + 5·8 = 24 + 40= 6442.144÷4−2 = 36−2= 3444.7·(10−3)2−2·(3 + 1)2= 7·72−2·42= 7·49−2·16= 343−32= 31146.62−34÷33= 36−81÷27= 36−3= 3348.72+ 20·4−(28 + 9·2)= 72+ 20·4−(28 + 18)= 72+ 20·4−46= 49 + 20·4−46= 49 + 80−46= 8350.8×9−(12−8)÷4−(10−7)= 8×9−4÷4−3= 72−1−3= 6852.80−24·15÷(7·5−45÷3)= 80−24·15÷(35−15)= 80−24·15÷20= 80−16·15÷20= 80−240÷20= 80−12= 6854.27÷25·24÷22= 128÷32·16÷4= 4·16÷4= 64÷4= 1656.86 + 92 + 80 + 784= 3364= 8458.$1025 + $775 + $2062 + $942 + $37215= $85255= $170560.72÷6− {2×[9−(4×2)]}= 72÷6− {2×[9−8]}= 72÷6− {2×1}= 72÷6−2= 12−2= 1062.[92×(6−4)÷8] + [7×(8−3)]= [92×2÷8] + [7×5]= [184÷8] + 35= 23 + 35= 58

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