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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Document preview page 1

Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 1

Document preview content for Solution Manual for Basic College Mathematics with Early Integers, 3rd edition

Solution Manual for Basic College Mathematics with Early Integers, 3rd edition

Simplify your textbook learning with Solution Manual for Basic College Mathematics with Early Integers, 3rd edition, providing step-by-step solutions to every chapter.

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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 1 preview imageContentsChapter1........................... 1Chapter2..........................25Chapter3.........................35Chapter4.........................55Chapter5..........................83Chapter6......................... 107Chapter7......................... 127Chapter8......................... 153Chapter9......................... 167Chapter 10......................... 189Chapter 11......................... 209
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 2 preview image
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 3 preview imageChapter 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.2.5 ten thousands4.5 hundred thousands6.88.110.1 thousand + 7 hundreds + 7 tens + 6 ones12.1 thousand + 0 hundreds + 8 tens + 1 one, or 1 thousand +8 tens + 1 one14.3 thousands + 0 hundreds + 9 tens + 7 ones, or 3 thou-sands + 9 tens + 7 ones16.3 ten thousands + 8 thousands + 4hundreds + 5 tens +3 ones18.1 billion + 2 hundred millions + 0 ten millions + 5 millions+ 0 hundred thousands + 7 ten thousands + 3 thousands+ 6 hundreds + 1 ten + 2 ones, or 1 billion + 2 hundredmillions + 5 millions + 7 ten thousands + 3 thousands +6 hundreds + 1 ten + 2 ones20.3 hundred millions + 1 ten million + 3 millions + 8 hun-dred thousands + 4ten thousands + 7 thousands + 4hundreds + 6 tens + 5 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-one thousand, two hundred forty-five32.Eighty-nine million, three hundred thirty-one thousand,six hundred twenty-two34.354,70236.17,11238.19,610,43940.700,000,00042.26,000,000,00044.200,01746.2,793,000,00048.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.2.1 5 2 1+3 481 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 6 520.11119 9,9 9 9+1 1 21 0 0,1 1 1
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 4 preview image2Chapter 1:Whole Numbers22.23 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 = 14mi + 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 114mi.30.Perimeter = 62 yd + 39 yd + 54yd + 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.difference2.8 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 4 76914.1777/178/ 8/7/−6 981 8916.3 123 4/ 2/−2 1 71 2 518.13 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 64 33 598
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 5 preview imageExercise Set 1.4324.6 157/ 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 4,40540.144/10 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/✭✭✭✭−1473 9,85 948.2999143 0,00 4/✭✭✭✭−67492 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.2.28 7×43 4 84.57 6×96 8 46.48 0 6×75 6 4 28.3227 8 6 7×43 1,4 6 810.47 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 0
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 6 preview image4Chapter 1:Whole Numbers18.54547 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 1 946.048.Seven million, four hundred thirty-two thousandExercise Set 1.5RC2.dividendRC4.divisor2.54÷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.
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 7 preview imageExercise Set 1.5514.2 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 194R 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 1270.32.4 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 24R 27.
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 8 preview image6Chapter 1:Whole Numbers38.4 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 984.60.2 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 2904R 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 ft
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 9 preview imageChapter 1 Mid-Chapter Review770.Pairs of factors whose product is 36 are:1 and 362 and 183 and 124and 96 and 6a) The pair above whose sum is 13 is 4and 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 equal to 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︸︷︷︸, 4 06︸︷︷︸, 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 +4tens + 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 4,567,291.23.3 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+48 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 4
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 10 preview image8Chapter 1:Whole Numbers30.499 145 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 + 4m + 8 m + 3 m = 25 m40.A= 4in.×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 $4·150 gives the cost of 150 reams.44.Using the definition of division, 0÷0 =asuch thata·0 = 0.We see thatacould beanynumber sincea·0 = 0 for anynumbera.Thus, we cannot say that 0÷0 = 0.This iswhy we agree not to allow division by 0.Exercise Set 1.6RC2.Because the digit in the hundreds place, 5, is 5 orhigher, we round up. The statement is false.RC4.The statement is true. See page 40 in the text.2.5304.89506.508.80010.90012.70014.460016.198,40018.500020.200022.736,00024.6,713,00026.6 26 09 71 0 0465 0+ 8 1+8 02 9 0
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 11 preview imageExercise Set 1.6928.6 7 36 7 0−2 8−3 06 4 030.41402 12 05 56 0+6 0+6 01 7 71 8 032.8 3 68 4 03 7 4 3 7 07 9 4 7 9 0+ 9 3 8+ 9 4 03 9 4 72 9 4 03947 seems to be incorrect.34.5 6 86 0 04 7 25 0 09 3 89 0 0+ 4 0 2+ 4 0 02 4 0 036.9 4 3 89 4 0 0−2 7 8 7−2 8 0 06 6 0 038.4 8 15 0 07 0 27 0 06 2 36 0 0+ 1 0 4 3+ 1 0 0 01 8 4 92 8 0 01849 seems to be incorrect.40.3 2 63 0 02 7 53 0 07 5 88 0 0+ 9 4 3+ 9 0 02 3 0 22 3 0 042.7 6 4 88 0 0 09 3 4 89 0 0 07 8 4 28 0 0 0+ 2 2 2 2+ 2 0 0 02 7, 0 0 044.8 4 , 8 9 08 5, 0 0 0−1 1, 1 1 0−1 1, 0 0 07 4 , 0 0 046.5 15 0×7 8×8 04 0 0 048.6 36 0×5 4×5 03 0 0 050.3 5 54 0 0×2 9 9×3 0 01 2 0, 0 0 052.7 8 98 0 0×4 3 4×4 0 03 2 0, 0 0 054.454÷87≈450÷90 = 556.1263÷29≈1260÷30 = 4258.3641÷571≈3600÷600 = 660.32,854÷748≈32,900÷700 = 4762.$ 6 8 6$ 7 0 09 51 0 0+ 1 9 9+2 0 0$ 1 0 0 064.$ 5 3 6$ 5 0 02 8 93 0 09 51 0 01 3 01 0 0+ 1 9 9+2 0 0$ 1 2 0 0Alyssa’s budget covers her choices.66.Answers will vary depending on the options chosen.68.a) Total cost of attending:$250·490 = $122,500Total cost of hotel rooms:$170·2·320 = $108,800Total amount spent:$122,500 + $108,800 = $231,300b) Total cost of attending:$200·500 = $100,000Total cost of hotel rooms:$200·2·300 = $120,000Total amount spent:$100,000 + $120,000 = $220,00070.$2211÷$11≈$2200÷$10 = 220 boxes72.32>074.28>1876.77<11778.999>99780.345<45682.12<3284.4,134,519>1,532,623, or 1,532,623<4,134,51986.82>77, or 77<8288.9 0 0 2+ 4 5 8 71 3,5 8 990.899 129 0 0 2/✥✥✥−4 5 8 74 4 1 5
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 12 preview image10Chapter 1:Whole Numbers92.343 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)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= 24 730.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= 7548.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 334R 11.62.342>33964.0<1166.6,375,60068.x= 14,332,38848,916= 293Exercise Set 1.8RC2.Translate.RC4.Check.2.Leth= the number of feet by which the height of theGrollo Tower would have exceeded the height of the Miglin-Beitler Skyneedle.Solve: 2001 +h= 2224h= 223 ft4.Leth= the height of the Burj Khalifa, in feet.Solve:h+ 1883 = 4600h= 2717 ft
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 13 preview imageExercise Set 1.8116.Letc= the number of milligrams of caffeine in a 20-ozbottle of Coca Cola.Solve: 25 + 32 =cc= 57 milligrams8.Letn= the number of entries in each row.Solve: 504÷36 =nn= 14entries10.Letr= the number of active rotary oil rigs in 2007.Solve:r+ 687 = 984r= 297 rigs12.Letn= the number of miles by which the length of theNile exceeds the length of the Missouri-Mississippi.Solve: 3860 +n= 4135n= 275 mi14.Letp= the total number of squares in the puzzle.Solve: 15·15 =pp= 225 squares16.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 Phoenix.Solve: 625 +r= 84 6r= $221 per month20.Letr= the amount of rent each sister pays.Solve: 2r= 1030r= $515 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: 6·1303 =rr= $781824.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 quires28.Lets= the amount by which spending by visitors to theUnited States exceeded spending by Americans travelingabroad.Solve: 110,200,000,000 +s= 153,000,000,000s= $4 2,800,000,00030.Letc= the total cost 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 3465 mi of citydriving.Solve: 3465÷21 =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= 2304mi44.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= 35
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 14 preview image12Chapter 1:Whole NumbersLett= the number of minutes you must bicycle in orderto lose one pound.Solve:t= 35·15t= 525 min; we could also express this as 8 hr, 45 min.48.Letn= the number of new jobs that will be created forpostsecondary teachers and elementary teachers.n= 305,700 + 248,800 = 554,500Lets= the number of new jobs that will be created forsecondary teachers.Solve:s+ 4 84,600 = 554,500s= 69,900 jobs50.LetF= the number of seats in first class,E= the numberof seats in economy class, andT= the total number ofseats.Solve: 3·4 =F, 21·6 =E, andT=F+EF= 12,E= 126,T= 12 + 126 = 138 seats52.Letc= the total cost 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 ft64.x= 81−15 = 6666.Consider one student taking one class a “student-classunit.” Then lets= the total number 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 3.2.254.1336.928.1410.12512.6414.100,00016.6418.(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= 4 036.82−8·2 = 64−8·2= 64−16= 4 8
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 15 preview imageChapter 1 Summary and Review: Concept Reinforcement1338.1000÷25−(15 + 5) = 1000÷25−20= 4 0−20= 2040.3·8 + 5·8 = 24 + 4 0= 6442.144÷4−2 = 36−2= 3444.7·(10−3)2−2·(3 + 1)2= 7·72−2·42= 7·49−2·16= 34 3−32= 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= 4 9 + 20·4−46= 4 9 + 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= 5864.(18÷2)· {[(9·9−1)÷2]−[5·20−(7·9−2)]}= 9· {[(81−1)÷2]−[5·20−(63−2)]}= 9· {[80÷2]−[5·20−61]}= 9· {40−[100−61]}= 9· {40−39}= 9· {1}= 966.15(23−4·2)3÷(3·25)= 15(23−8)3÷75Multiplying inside parentheses= 15·153÷75Subtracting inside parentheses= 15·3375÷75Evaluating the exponentialexpression= 50,625÷75Doing all multiplication and= 675divisions in order from left to right68.(19−24)5−(141÷47)2= (19−16)5−32= 35−32= 24 3−9= 23470.x= 5032−4197 = 83572.y= 155442= 3774.t= 10,000100= 10076.Letg= the total number of gallons of gasoline purchased.Solve: 23 + 24+ 26 + 25 =gg= 98 gallons78.12÷4 + 2·3−2 = 3 + 6−2= 7Correct answer12÷(4+ 2)·(3−2) = 280.Answers may vary. One correct answer is9·8 + 7·6−5·4 + 3·2·1 = 100.Chapter 1 Vocabulary Reinforcement1.The distance around an object is its perimeter.2.The minuend is the number from which another numberis being subtracted.3.For large numbers, digits are separated by commas intogroups of three, called periods.4.In the sentence 28÷7 = 4, the dividend is 28.5.In the sentence 10×1000 = 10,000, 10 and 1000 are calledfactors and 10,000 is called the product.6.The number 0 is called the additive identity.7.The sentence 3×(6×2) = (3×6)×2 illustrates theassociative law of multiplication.8.We can use the following rule to check division:quotient·divisor + remainder = dividend.
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Solution Manual for Basic College Mathematics with Early Integers, 3rd edition - Page 16 preview image14Chapter 1:Whole NumbersChapter 1 Concept Reinforcement1.The statement is true. See page 42 in the text.2.a÷a=aa= 1,a= 0; the statement is true.3.a÷0 is not defined, so the statement is false.4.The statement is false.For example, 1 + 2 = 5 is not atrue equation.5.The statement is true. See page 71 in the text.6.The statement is false. See page 73 in the text.Chapter 1 Study Guide1.4 32 , 0 7 9The digit 2 names the number of thousands.2.1113 6,0 4 7+ 2 9,2 5 56 5,3 0 23.79 154 8 0 5/✥✥−1 5 6 83 2 3 74.211736 8 4×3 2 96 1 5 6Multiplying by 91 3 6 8 0Multiplying by 202 0 5 2 0 0Multiplying by 3002 2 5,0 3 65.3 1 52 78 5 1 98 14 12 71 4 91 3 51 4The answer is 315 R 14.6.Round 36,468 to the nearest thousand.3 6,46 8↑The digit 6 is in the thousands place.Consider the nextdigit to the right. Since the digit, 4, is 4 or lower, rounddown, meaning that 6 thousands stays as 6 thousands.Then change the digits to the right of the thousands digitto zeros.The answer is 36,000.7.Since 78 is to the left of 81 on the number line, 78<81.8.24·x= 86424·x24= 86424Dividing by 24x= 36Check:24·x= 86424·36 ? 864864∣∣∣TRUEThe solution is 36.9.63= 6·6·6 = 216Chapter 1 Review Exercises1.4 , 6 7 8, 9 5 2The digit 8 means 8 thousands.2.13, 7 6 8, 9 4 0The digit 3 names the number of millions.3.2793 = 2 thousands + 7 hundreds + 9 tens + 3 ones4.56,078 = 5 ten thousands + 6 thousands + 0 hundreds+ 7 tens + 8 ones, or 5 ten thousands + 6 thousands +7 tens + 8 ones5.4,007,101 = 4 millions + 0 hundred thousands + 0 tenthousands + 7 thousands + 1 hundred + 0 tens + 1 one,or 4millions + 7 thousands + 1 hundred + 1 one6.67︸︷︷︸, 819︸︷︷︸↑↑Sixty-seven thousand,eight hundred nineteen7.2︸︷︷︸, 781︸︷︷︸, 4 27︸︷︷︸↑↑↑Two million,seven hundred eighty-one thousand,four hundred twenty-seven8.Four hundred seventy-six thousand,five hundred eighty-eight↓↓Standard notation is︷︸︸︷476 ,︷︸︸︷588 .9.One billion,six hundred forty million,↓↓Standard notation is︷︸︸︷1,︷︸︸︷640 ,000,000.10.117 3 0 4+ 6 9 6 81 4,2 7 2
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Basic College Mathematics by Marvin...

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