(Solution Manual) Algebra Foundations Basic Math, Introductory and Intermediate Algebra

Bittinger,Beecher, 1

Math & Statistics

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Solution Manual for Algebra Foundations Basic Math, Introductory and Intermediate Algebra, The downloadable file consists of 635 pages, and the total size of the file is 9.28 MB. Purchase now only for highly discounted price.

Author(s): Bittinger, Beecher

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SOLUTIONS MANUAL ALGEBRA FOUNDATIONS BASIC MATHEMATICS, INTRODUCTORY ALGEBRA , AND INTERMEDIATE ALGEBRA Marvin Bittin ger Judith Beecher Barbara Johnso n Contents Chapter 1 ........................... 1 Chapter 2 ..........................2 5 Chapter 3 .........................4 3 Chapter 4 .........................6 7 Chapter 5 ..........................8 7 Chapter 6 ......................... 1 0 5 Chapter 7 ......................... 1 2 7 Chapter 8 ......................... 1 3 7 Chapter 9 ......................... 1 5 5 Chapter 10 ......................... 1 7 1 Chapter 11 ......................... 1 8 7 Chapter 12 ......................... 2 1 3 Chapter 13 ......................... 2 3 5 Chapter 14 ......................... 2 5 7 Chapter 15 ......................... 2 8 3 Chapter 16 ......................... 3 1 9 Chapter 17 ......................... 3 4 5 Chapter 18 ......................... 3 7 5 Chapter 19 ......................... 3 9 9 Chapter 20 ......................... 4 2 1 Chapter 21 ......................... 4 6 1 Chapter 1 Whole Numbers Exercise Set 1.1 RC2. In 615,702, the number 615 is in the thousands period . RC4. The number 721 is written in standard notation. 2.5 ten thousands 4.5 hundred thousands 6.8 8.1 10.1 thousand + 7 hundreds + 7 tens + 6 ones 12.1 thousand + 0 hundreds + 8 tens + 1 one, or 1 thousand + 8 tens + 1 one 14.3 thousands + 0 hundreds + 9 tens + 7 ones, or 3 thou-sands + 9 tens + 7 ones 16.3 ten thousands + 8 thousands + 4hundreds + 5 tens +3 ones 18.1 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 ones 20.3 hundred millions + 1 ten millio n + 3 millions + 8 hun- dred thousands + 4ten thousands + 7 thousands + 4hundred s + 6 tens + 5 ones 22.Forty-eight 24.Forty-five thousand, nine hundred eighty-seven 26.One hundred eleven thousand, thirteen 28.Forty-three billion, five hundred fifty million, six hundredfifty-one thousand, eight hundred eight 30.Ninety-one thousand, two hundred forty-five 32.Eighty-nine million, three hundred thirty-one thousand,six hundred twenty-two 34.354,702 36.17,112 38.19,610,439 40.700,000,000 42.26,000,000,000 44.200,01746.2,793,000,000 48.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 1.2 RC2. In the addition 5 + 2 = 7, the number 7 is the sum . RC4. The distance around an object is its perimeter . 2. 1521 +3 48 1869 4.1 73 +69 142 6.1 7503 +2683 10,186 8.11 999 +1 1 1010 10.1 271 +3338 3609 12.1 280 +34 ,902 35,182 14.11 1 10,120 12,989 +5 7 3 8 28,847 16.1 3654 +2700 6354 18.11 1 45,879 +21 ,786 67,665 20.111 1 99,999 +1 1 2 100 ,111 Copyright c/circlecopyrt2015 Pearson Education, Inc. 2 Chapter 1: Whole Numbers 22.2 38 273214 +76 187 24.111 1 42,487 83,141 +36 ,712 162 ,340 26.22 989566834920 +703 4012 28.Perimeter = 14mi + 13 mi + 8 mi + 10 mi + 4 7 mi +22 mi We carry out the addition. 2 14 13 8 1047 +22 114 The perimeter of the figure is 114mi. 30.Perimeter = 62 yd + 39 yd + 54yd + 4 6 yd + 28 yd We carry out the addition. 2 62 395446 +28 229 The perimeter of the figure is 229 yd. 32.90 ft + 90 ft + 90 ft + 90 ft = Perimeter We carry out the addition. 90 9090 +90 360 The 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 1.3 RC2. subtraction symbol RC4. difference 2. 87 −34 53 4. 526 −323 203 6.61 3 7/3/ −28 45 8.51 4 6/4/ −19 45 10.18 68/15 7/9/5/ −39 8 39 7 12.10 20/16 3/1/6/ −247 69 14.17 77/17 8/8/7/ −69 8 18 9 16.31 2 34/2/ −21 7 12 5 18.13 12 53/2/11 6/4/3/1/ −28 9 6 35 3 5 20.12 15 72/5/14 8/3/6/4/ −53 7 5 29 8 9 22.11 13 81/3/11 9/2/4/1/ −56 43 35 9 8 Copyright c/circlecopyrt2015 Pearson Education, Inc. Exercise Set 1.4 3 24.61 5 7/5/8 3 −3641 3942 26.11 11 51/1/12 16/,2/2/2/ −5,888 10,334 28.11 21/11 8 14 3/2/,1/9/4/ −29,236 29 5 8 30.81 0 9/0/ −78 12 32.79 1 3 803✥✥ −418 385 34.39 1 5 9405 /✥✥ −258 9147 36.14 64/91 0 7/500 /✥✥ −3604 3896 38.691 3 84,70 3/✥✥ − 29 8 84,405 40.14 4/10 0 17 1/5/,0/1/7/ −78 0 9 72 0 8 42.799 1 3 8003 /✥✥✥ −599 7404 44.699 1 0 17,000/✥✥✥ −11,598 5402 46.3999 1 6 40,00 6/ ✭✭✭✭ − 147 39,85 948.2999 1 4 30,00 4/ ✭✭✭✭ −6749 23,25 5 50. 901 +23 924 52.1 9909 +1011 10,920 54.9 thousands + 1 hundred + 0 tens + 3 ones, or 9 thousands + 1 hundred + 3 ones 56.3,928,124−1,098,947 Using a calculator to carry out the subtraction, we findthat the difference is 2,829,177. Exercise Set 1.4 RC2. In the multiplication 4 ×3 = 12, 12 is the product . RC4. The product of 1 and any number aisa. 2.2 87 ×4 348 4.5 76 ×9 684 6.4 806 × 7 5642 8.32 2 7867 × 4 31,468 10.4 78 ×60 4680 12.2 2 87 ×34 348 2610 2958 14. 2340 ×1000 2,340 ,000 16. 800 ×700 560 ,000 Copyright c/circlecopyrt2015 Pearson Education, Inc. 4 Chapter 1: Whole Numbers 18.54 54 777 ×77 5439 54390 59,829 20.3737 549 ×88 4392 43920 48,312 22.21 11 432 ×375 2160 30240 129600 162 ,000 24.2323 45 346 ×659 3114 17300 207600 228 ,014 26.2261 5 11 8928 ×3172 178 5 6 624960 892800 26784000 28,319 ,616 28.24 24 13 6408 ×6064 256 3 2 384480 38448000 38,858 ,112 30.11 4444 355 ×299 3195 31950 71000 106 ,14532.1 2 2 41 6521 ×3449 586 8 9 260840 2608400 19563000 22,490 ,929 34.34 44 4506 ×7800 3604800 31542000 35,146 ,800 36.1 2 6009 ×2003 180 2 7 12018000 12,036 ,027 38.A=l×w= 129 yd ×65 yd = 8385 sq yd 40.A=l×w= 200 ft ×85 ft = 17 ,000 sq ft 42.12 2 9876 876 76 +6 10,834 44.13 23/10 8 18 3/4/0/,79/8/ −86,67 9 254 ,11 9 46.0 48.Seven million, four hundred thirty-two thousand Exercise Set 1.5 RC2. dividend RC4. divisor 2.54÷9 = 6 because 54= 9 ·6. 4.37 37= 1 Any nonzero number divided by itself is 1. 6.56 1= 56 Any number divided by 1 is that same number. 8.0 32= 0 Zero divided by any nonzero number is 0. 10.74÷0 is not defined, because division by 0 is not defined. 12.20 4= 5 because 20 = 4 ·5. Copyright c/circlecopyrt2015 Pearson Education, Inc. Exercise Set 1.5 5 14.233 3699 6 99 99 0 The answer is 233. 16.108 88698 6964 5 The answer is 108 R 5. 18.708 3212421 2424 0 The answer is 708. 20.1012 991109 11 9 2018 2 The answer is 1012 R 2. 22.194 23892 1818 98 1 The answer is 194R 1. 24.146 68816 2824 4136 5 The answer is 146 R 5.26.2009 360276 2727 0 The answer is 2009. 28.517 8413940 13 8 5956 3 The answer is 517 R 3. 30.1270 100 127 ,000 100 270200 700700 00 0 The answer is 1270. 32.426 10426040 2620 6060 0 The answer is 426. 34.289 20579840 179160 198180 18 The answer is 289 R 18. 36.24 4098780 187160 27 The answer is 24R 27. Copyright c/circlecopyrt2015 Pearson Education, Inc. 6 Chapter 1: Whole Numbers 38.40 23942 92 22 The answer is 40 R 22. 40.50 542729270 29 The answer is 50 R 29. 42.55 102 5612510 512510 2 The answer is 55 R 2. 44.107 77497 4949 0 The answer is 107. 46.808 9727372 7372 1 The answer is 808 R 1. 48.1010 7707470 77 4 The answer is 1010 R 4. 50.301 24724272 4224 18 The answer is 301 R 18. 52.102 48489948 9996 3 The answer is 102 R 3.54.210 36756372 3636 3 The answer is 210 R 3. 56.803 3628,929 288 129108 21 The answer is 803 R 21. 58.984 9088,560 810 756720 360360 0 The answer is 984. 60.2904 306 888 ,888 612 27682754 14881224 264 The answer is 2904R 264 . 62.7002 803 5,622 ,606 5621 16061606 0 The answer is 7002. 64. 88,777 −22,333 66,444 66. 268 ×35 13408040 9380 68.A=l×w=1 1f t ×9 ft = 99 sq ft Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Mid-Chapter Review 7 70.Pairs of factors whose product is 36 are: 1 and 36 2 and 183 and 124and 96 and 6 a) 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 ,386 Consider the related multiplication sentence: 4,386·76 =3 4 ,584,132 Since the ones digit of the product is 2, the missing ones digit must be either 2 or 7 (6 ·2 = 12 and 6 ·7 = 42). We try 2: 34,584,132÷762 = 45 ,386 We see that the missing ones digit is 2 and the missing thousands digit is 5. Chapter 1 Mid-Chapter Review 1.The statement is false. For example, 8 −5 = 3, but 5 is not 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 not greater 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 the number itself. For example,27 1= 27. 7. 95/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright,40 6/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright, 237/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ↑↑↑ Ninety-five million, four hundred six thousand, two hundred thirty-seven 8.59 1 4 604/✥✥ −497 107 9.2698 The digit 6 names the number of hundreds. 10.61 ,204 The digit 6 names the number of ten thousands. 11.146,237 The digit 6 names the number of thousands.12.586 The digit 6 names the number of ones. 13.306 ,458 ,1 29 The digit 2 names the number of tens. 14.306,458 ,129 The digit 6 names the number of millions. 15.3 0 6, 4 58 ,129 The digit 5 names the number of ten thousands. 16.306 ,458 , 129 The digit 1 names the number of hundreds. 17.5602 = 5 thousands + 6 hundreds + 0 tens + 2 ones, or 5 thousand s+6h undreds + 2 ones 18.69,345 = 6 ten thousands + 9 thousands + 3 hundreds +4tens + 5 ones 19.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. 316 +482 798 24.11 593 +437 1030 25.11 2638 +5284 7922 26.111 46172436 +48 1 7534 27. 786 −321 465 28.11 51/14 6/2/4/ −28 5 33 9 29.15 25/91 2 3/602 /✥✥ −1748 1854 Copyright c/circlecopyrt2015 Pearson Education, Inc. 8 Chapter 1: Whole Numbers 30.499 1 4 5004 /✥✥✥ −676 4328 31.3 36 ×6 216 32.11 55 567 ×28 4536 11340 15,876 33.21 3 407 ×325 2035 8140 122100 132 ,275 34.22 3 1 9435 ×602 1 8870 566 1000 5,679 ,870 35.253 41012 8 2120 1212 0 The answer is 253. 36.112 38426138 4638 8176 5 The answer is 112 R 5. 37.23 601399120 199180 19 The answer is 23 R 19.38.144 56809556 249224 255224 31 The answer is 144 R 31. 39.Perimeter = 10 m + 4m + 8 m + 3 m = 25 m 40.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/negationslash=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 that a·0=0 . We see that acould be anynumber since a·0 = 0 for any number a. Thus, we cannot say that 0 ÷0 = 0. This is why we agree not to allow division by 0. Exercise Set 1.6 RC2. 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. 2.530 4.8950 6.50 8.800 10.900 12.700 14.4600 16.198,400 18.5000 20.2000 22.736,000 24.6,713,000 26. 62 6097 100465 0 +81 + 80 290 Copyright c/circlecopyrt2015 Pearson Education, Inc. Exercise Set 1.6 9 28. 673 670 −28 −30 640 30. 4140 21 2055 60 +6 0 +6 0 177 180 32. 836 840374370794790 +938 +940 3947 2940 3947 seems to be incorrect. 34. 568 600472 500938 900 +402 +400 2400 36. 9438 9400 −2787 −2800 6600 38. 481 500702 700623 600 +1043 +1000 1849 2800 1849 seems to be incorrect. 40. 326 300275 300758 800 +943 +900 2302 2300 42. 7648 80009348 90007842 8000 +2222 +2000 27 ,000 44. 84,890 85 ,000 −11 ,110 −11 ,000 74,000 46. 51 50 ×78 ×80 4000 48. 63 60 ×54 ×50 3000 50. 355 400 ×299 ×300 120 ,000 52. 789 800 ×434 ×400 320 ,00054.454÷87≈450÷90 = 5 56.1263÷29≈1260÷30 = 42 58.3641÷571≈3600÷600 = 6 60.32,854÷748≈32,900÷700 = 47 62. $686 $700 95 100 +199 + 200 $1000 64. $536 $500 289 300 95 100 130 100 +199 + 200 $1200 Alyssa’s budget covers her choices. 66.Answers will vary depending on the options chosen. 68.a)Total cost of attending: $250·490 = $122 ,500 Total cost of hotel rooms: $170·2·320 = $108 ,800 Total amount spent: $122,500 + $108 ,800 = $231 ,300 b)Total cost of attending: $200·500 = $100 ,000 Total cost of hotel rooms: $200·2·300 = $120 ,000 Total amount spent: $100,000 + $120 ,000 = $220 ,000 70.$2211 ÷$11≈$2200 ÷$10 = 220 boxes 72.32>0 74.28>18 76.77<117 78.999>997 80.345<456 82.12<32 84.4,134,519>1,532,623, or 1 ,532,623<4,134,519 86.82>77, or 77 <82 88. 9002 +4587 13,589 90.899 1 2 9002 /✥✥✥ −4587 4415 Copyright c/circlecopyrt2015 Pearson Education, Inc. 10 Chapter 1: Whole Numbers 92.3 4 306 ×58 2448 15300 17,748 94.208 234784 46 184184 0 The 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 1.7 RC2. (a) RC4. (b) 2.25 4.8 6.t= 5678 + 9034= 14 ,712 8.m= 9007 −5667 = 3340 10.z=3 4·15 = 510 12.w= 256 ÷16 = 16 14.t=2 2−15 = 7 16.t=1 6−16 = 0 18.x=5 7−20 = 37 20.w=5 3−17 = 36 22.x=42 6=7 24.m=162 9=1 8 26.y=96 4=2 4 28.t=741 3=2 47 30.y= 9281 −8322 = 959 32.p=9 2−56 = 36 34.y=2 3×78 = 1794 36.z= 133 −6 7=6 6 38.w=3404 4= 851 40.x= 807 −438 = 36942.q=1 0 ,534÷458 = 23 44.x=6080 19= 320 46.x=1500 20=7 5 48.t= 9281 −8322 = 959 50.n= 3004 −1745 = 1259 52.n=660 12=5 5 54.x=22,135 233=9 5 56.z= 512 −63 = 449 58.142 91278 9 3736 1818 0 The answer is 142. 60.334 17568951 5851 7968 11 The answer is 334R 11. 62.342>339 64.0<11 66.6,375,600 68.x=14,332,388 48,916= 293 Exercise Set 1.8 RC2. Translate. RC4. Check. 2.Leth= the number of feet by which the height of the Grollo Tower would have exceeded the height of the Miglin-Beitler Skyneedle. Solve: 2001 + h= 2224 h= 223 ft 4.Leth= the height of the Burj Khalifa, in feet. Solve: h+ 1883 = 4600 h= 2717 ft Copyright c/circlecopyrt2015 Pearson Education, Inc. Exercise Set 1.8 11 6.Letc= the number of milligrams of caffeine in a 20-oz bottle of Coca Cola. Solve: 25 + 32 = c c= 57 milligrams 8.Letn= the number of entries in each row. Solve: 504 ÷36 = n n= 14entries 10.Letr= the number of active rotary oil rigs in 2007. Solve: r+ 687 = 984 r= 297 rigs 12.Letn= the number of miles by which the length of the Nile exceeds the length of the Missouri-Mississippi.Solve: 3860 + n= 4135 n= 275 mi 14.Letp= the total number of squares in the puzzle. Solve: 15 ·15 = p p= 225 squares 16.Letm= the number of minutes in a day. Solve: 60 ·24= m m= 1440 minutes 18.Letr= the amount by which the average monthly rent in Atlanta exceeds the average monthly rent in Phoenix.Solve: 625 + r=8 46 r= $221 per month 20.Letr= the amount of rent each sister pays. Solve: 2 r= 1030 r= $515 per month 22.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 = r r= $7818 24.Lets= the speed limit for trucks. Solve: s+1 0=7 5 s= 65 mph 26.Letq= the number of quires in a ream. Solve: 25 ·q= 500 q= 20 quires 28.Lets= the amount by which spending by visitors to the United States exceeded spending by Americans traveling abroad. Solve: 110 ,200,000,000 + s= 153 ,000,000,000 s=$ 42 ,800,000,00030.Letc= the total cost of the purchase. Solve: 96 ·88 = c c= $8448 32.Letw= the number of full weeks that will pass before the station must begin re-airing episodes.Solve: 5 ·w= 208 w= 41 R 3, so 41 full weeks will pass and 3 episodes will be shown the following week before previously aired episodes are rerun. 34.Letl= the number of labels on each sheet. Solve: 25 ·l= 750 l= 30 labels 36.Letg= the number of gallons required for 3465 mi of city driving. Solve: 3465 ÷21 = g g= 165 gal 38.a)LetA= the area of the court, in square feet. Solve: A=9 4·50 A= 4700 square feet b)LetP= the perimeter of the court, in feet. Solve: P=9 4+5 0+9 4+5 0 P= 288 ft c)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= 4700 a= 500 square feet 40.Letc= the number of cartons needed. Solve: 528 ÷12 = c c= 44 cartons 42.Letm= the distance on the map, in inches, between two cities that, in reality, are 2016 mi apart.Solve: 2016 ÷288 = m m= 7 in. Letr= the distance in miles, in reality, between two cities that are 8 in. apart on the map.Solve: 288 ·8=r r= 2304mi 44.Letm= the number of months required to pay off the loan.Solve: 7824 ÷163 = m m= 48 months 46.Letn= the number of 100’s in 3500. Solve: 3500 ÷100 = n n=3 5 Copyright c/circlecopyrt2015 Pearson Education, Inc. 12 Chapter 1: Whole Numbers Lett= the number of minutes you must bicycle in order to lose one pound. Solve: t=3 5·15 t= 525 min; we could also express this as 8 hr, 45 min. 48.Letn= the number of new jobs that will be created for postsecondary teachers and elementary teachers.n= 305 ,700 + 248 ,800 = 554 ,500 Lets= the number of new jobs that will be created for secondary teachers.Solve: s+48 4 ,600 = 554 ,500 s=6 9 ,900 jobs 50.LetF= the number of seats in first class, E= the number of seats in economy class, and T= the total number of seats.Solve: 3 ·4=F,2 1·6=E, and T=F+E F= 12, E= 126, T= 12 + 126 = 138 seats 52.Letc= the total cost of the 5 video games. Solve: 5 ·64= c c= $320 Then let n= the number of $20 bills required. Solve: 320 ÷20 = n n= 16 $20 bills 54.Letb= the new balance. Solve: 749 −34−65 + 123 = b b= $773 56.Letl= the total length of the bookshelves, in feet. Solve: 6 ·3=l l=1 8f t Since the total length of the bookshelves is greater than 16 ft, the shelves cannot be put side by side on the 16-ftwall. 58. 15 85/91 2 9/602 /✥✥ −1843 7759 60.147 32470832 150128 228224 4 The answer is 147 R 4. 62.A=l×w= 211 ft ×46 ft = 9706 sq ft 64.x=8 1−15 = 6666.Consider one student taking one class a “student-classunit.” Then let s= the total number of student-class units andp= the number of students taught by each instructor. Solve: 1200 ·5=s,4·30 = p s= 6000, p= 120 Now let n= the number of instructors. Solve: 6000 ÷120 = n n= 50 instructors Exercise Set 1.9 RC2. 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 . 2.25 4.133 6.92 8.14 10.125 12.64 14.100,000 16.64 18.( 1 2+6 )+1 8=1 8+1 8 =3 6 20.(52−40)−8=1 2 −8 =4 22.(1000 ÷100)÷10 = 10 ÷10 =1 24.256÷(64÷4) = 256 ÷16 =1 6 26.22+52=4+2 5=2 9 28.(32−27) 3+ (19 + 1)3=53+2 03 = 125 + 8000 = 8125 30.2 3+1 8 ·20 = 23 + 360 = 383 32.10·7−4=7 0 −4 =6 6 34.90−5·5·2=9 0 −50 =40 36.82−8·2=6 4 −8·2 =6 4−16 =48 Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Summary and Review: Concept Reinforcement 13 38.1000÷25−(15 + 5) = 1000 ÷25−20 =40−20 =2 0 40.3·8+5·8=2 4+4 0 =6 4 42.144÷4−2=3 6 −2 =3 4 44.7·(10−3)2−2·( 3+1 )2=7·72−2·42 =7·49−2·16 =3 43 −32 = 311 46.62−34÷33=3 6−81÷27 =3 6−3 =3 3 48.72+2 0·4−( 2 8+9 ·2) =72+2 0·4−(28 + 18) =72+2 0·4−46 =49+2 0 ·4−46 =49+8 0 −46 =8 3 50.8×9−(12−8)÷4−(10−7) =8×9−4÷4−3 =7 2−1−3 =6 8 52. 80−24·15÷(7·5−45÷3) =8 0−24·15÷(35−15) =8 0−24·15÷20 =8 0−16·15÷20 =8 0−240÷20 =8 0−12 =6 8 54. 27÷25·24÷22 = 128 ÷32·16÷4 =4·16÷4 =6 4÷4 =1 6 56.8 6+9 2+8 0+7 8 4=336 4=8 4 58.$1025 + $775 + $2062 + $942 + $3721 5=$8525 5= $1705 60.72÷6−{2×[9−(4×2)]} =7 2÷6−{2×[9−8]} =7 2÷6−{2×1} =7 2÷6−2 =1 2−2 =1 0 62.[92×(6−4)÷8 ]+[ 7 ×(8−3)] = [92 ×2÷8 ]+[ 7 ×5] = [184 ÷8 ]+3 5 =2 3+3 5 =5 864.(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} =9 66.15(23 −4·2)3÷(3·25) = 15(23 −8)3÷75 Multiplying inside parentheses =1 5·153÷75 Subtracting inside parentheses =1 5·3375÷75 Evaluating the exponential expression =5 0 ,625÷75 Doing all multiplication and = 675 divisions in order from left to right 68. (19−24)5−(141÷47)2 = (19 −16)5−32 =35−32 =2 43 −9 = 234 70.x= 5032 −4197 = 835 72.y=1554 42=3 7 74.t=10,000 100= 100 76.Letg= the total number of gallons of gasoline purchased. Solve: 23 + 24+ 26 + 25 = g g= 98 gallons 78.12÷4+2·3−2=3+6 −2 = 7 Correct answer 12÷(4+ 2) ·(3−2) = 2 80.Answers may vary. One correct answer is 9·8+7·6−5·4+3·2·1 = 100 . Chapter 1 Vocabulary Reinforcement 1.The distance around an object is its perimeter . 2.The minuend is the number from which another number is being subtracted. 3.For large numbers, digits are separated by commas into groups 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 called factors 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 the associative law of multiplication. 8.We can use the following rule to check division: quotient ·divisor + remainder = dividend . Copyright c/circlecopyrt2015 Pearson Education, Inc. 14 Chapter 1: Whole Numbers Chapter 1 Concept Reinforcement 1.The statement is true. See page 42 in the text. 2.a÷a=a a=1 , a/negationslash= 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 a true 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 Guide 1.432,079 The digit 2 names the number of thousands. 2.11 1 36,047 +29 ,255 65,302 3.79 1 5 4805 /✥✥ −1568 3237 4.21 173 684 ×329 6 1 5 6 Multiplying by 9 1 3 6 8 0 Multiplying by 20 2 0 5 2 0 0 Multiplying by 300 225 ,036 5.315 278519 81 4127 149135 14 The answer is 315 R 14. 6.Round 36,468 to the nearest thousand. 36,468 ↑ 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= 864 24·x 24=864 24Dividing by 24 x=3 6 Check: 24 ·x= 864 24·36 ? 864 864/vextendsingle/vextendsingle/vextendsingle TRUE The solution is 36. 9.63=6·6·6 = 216 Chapter 1 Review Exercises 1.4,67 8,952 The digit 8 means 8 thousands. 2.13,768 ,940 The digit 3 names the number of millions. 3.2793 = 2 thousand s+7h undreds + 9 tens + 3 ones 4.56,078 = 5 ten thousand s + 6 thousands + 0 hundreds + 7 tens + 8 ones, or 5 ten thousands + 6 thousands + 7 tens + 8 ones 5.4,007,101 = 4 millions + 0 hundred thousands + 0 tenthousands + 7 thousands + 1 hundred + 0 tens + 1 one,or 4millions + 7 thousands + 1 hundred + 1 one 6. 67/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright,819/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ↑↑ Sixty-seven thousand, eight hundred nineteen 7. 2/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright, 781/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright,42 7/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ↑↑↑ Two million, seven hundred eighty-one thousand, four hundred twenty-seven 8.Four hundred seventy-six thousand, five hundred eighty-eight ↓↓ Standard notation is/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright 476,/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright 588. 9. One billion, six hundred forty million, ↓↓ Standard notation is/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright 1,/bracehtipdownleft/bracehtipupright/bracehtipupleft/bracehtipdownright 640,000,000. 10.11 7304 +6968 14,272 Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Summary and Review: Review Exercises 15 11.11 1 27,609 +38 ,415 66,024 12.11 2703 41256004 +8956 21,788 13.11 91,426 +7 ,495 98,921 14.13 793 /15 804/5/✥✥ −2897 5148 15.899 1 1 9001 /✥✥✥ −7312 1689 16.599 1 3 6003 /✥✥✥ −3729 2274 17.1613 26/3/91 5 3/7/,405/✥✥ −19,648 17,757 18.2 17,000 ×300 5,100 ,000Multiplying by 300 (Write 00 and then multiply 17,000 by 3.) 19.63 4 7846 ×800 6,276 ,800Multiplying by 800 (Write 00 and then multiply 7846 by 8.) 20.13 2524 726 ×698 5808 65340 435600 506 ,748Multiplying by 8 Multiplying by 90Multiplying by 600 21.32 64 587 ×47 41 0 9 Multiplying by 7 2 3 48 0 Multiplying by 4 0 27,58922. 8305 ×642 166 1 0 332200 4983000 5,331 ,810 23.12 563 5 1310 3 The answer is 12 R 3. 24.5 168080 0 The answer is 5. 25.913 7639463 97 2421 3 The answer is 913 R 3. 26.384 8307324 6764 3332 1 The answer is 384R 1. 27.4 60286240 46 The answer is 4R 4 6. 28.54 794266395 316316 0 The answer is 54. 29.452 3817,176 15 2 197190 7676 0 The answer is 452. Copyright c/circlecopyrt2015 Pearson Education, Inc. 16 Chapter 1: Whole Numbers 30.5008 1470,112 70 112 112 0 The answer is 5008. 31.4389 1252,668 48 4636 106 96 108108 0 The answer is 4389. 32.Round 345,759 to the nearest hundred. 345 ,759 ↑ The digit 7 is in the hundreds place. Consider the nextdigit to the right. Since the digit, 5, is 5 or higher, round7 hundreds up to 8 hundreds. Then change the digits tothe right of the hundreds digit to zero. The answer is 345,800. 33.Round 345,759 to the nearest ten. 345 ,75 9 ↑ The digit 5 is in the tens place. Consider the next digit to the right. Since the digit, 9, is 5 or higher, round 5 tens upto 6 tens. Then change the digit to the right of the tensdigit to zero. The answer is 345,760. 34.Round 345,759 to the nearest thousand. 345 , 759 ↑ The digit 5 is in the thousands place. Consider the next digit to the right. Since the digit, 7, is 5 or higher, round5 thousands up to 6 thousands. Then change the digits tothe right of the thousands digit to zero. The answer is 346,000. 35.Round 345,759 to the nearest hundred thousand. 3 45,759 ↑ The digit 3 is in the hundred thousands place. Consider the next digit to the right. Since the digit, 4, is 4 or lower,round down, meaning that 3 hundred thousands stays as3 hundred thousands. Then change the digits to the rightof the hundred thousands digit to zero. The answer is 300,000. 36.Since 67 is to the right of 56 on the number line, 67 >56.37.Since 1 is to the left of 23 on the number line, 1 <23. 38. Rounded to the nearest hundred 41 ,348 41 ,300 +19 ,749 +19 ,700 61 ,000 ←Estimated answer 39. Rounded to the nearest hundred 38 ,652 38 ,700 −24,54 9 −24,500 14,200 ←Estimated answer 40. Rounded to the nearest hundred 396 400 ×748 ×700 280 ,000 ←Estimated answer 41. 46·n= 368 46·n 46=368 46 n=8 Check: 46 ·n= 368 46·8 ? 368 368/vextendsingle/vextendsingle/vextendsingle TRUE The solution is 8. 42. 47+ x=9 2 47+ x−47=9 2 −47 x=45 Check: 47 + x=9 2 47 + 45?9 2 92/vextendsingle/vextendsingle/vextendsingle TRUE The solution is 45. 43.1·y=5 8 y=5 8 ( 1 ·y=y) The number 58 checks. It is the solution. 44. 24= x+2 4 24−24= x+2 4−24 0=x The number 0 checks. It is the solution. 45.Exponential notation for 4 ·4·4is 4 3. 46.104=1 0·10·10·10 = 10 ,000 47.62=6·6=3 6 48.8·6 + 17 = 48 + 17 Multiplying = 65 Adding Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Summary and Review: Review Exercises 17 49. 10·24−( 1 8+2 ) ÷4−(9−7) =1 0·24−20÷4−2 Doing the calculations inside the parentheses =2 40 −5−2 Multiplying and dividing = 235 −2 Subtracting from = 233 left to right 50. (80÷16)×[(20−56÷8 )+( 8 ·8−5·5)] =5×[(20−7) + (64 −25)] =5×[13 + 39] =5×52 = 260 51.We add the numbers and divide by the number of addends. 157 + 170 + 168 3=495 3= 165 52.Familiarize . Let x= the additional amount of money, in dollars, Natasha needs to buy the desk. Translate . Money available/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightplusAdditional amount/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisPrice of desk/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 196 + x = 698 Solve . We subtract 196 on both sides of the equation. 196 + x= 698 196 + x−196 = 698 −196 x= 502 Check . We can estimate. 196 + 502 ≈200 + 500 ≈700≈698 The answer checks. State . Natasha needs $502 dollars. 53.Familiarize . Let b= the balance in Toni’s account after the deposit.Translate . Original balance/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightplus Deposit/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightis New balance/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 406 + 78 = b Solve . We add on the left side. 40 6+7 8= b 484 = b Check . We can repeat the calculation. The answer checks. State . The new balance is $484. 54.Familiarize . Let y= the year in which the copper content of pennies was reduced. Original year/bracehtipupleft /bracehtipdownright/bracehtipdownleft/bracehtipuprightplus 73 yr/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightisYear of copper reduction/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 1909 + 73 = ySolve . We add on the left side. 1909 + 73 = y 1982 = y Check . We can estimate. 1909 + 73 ≈1910 + 70 ≈1980≈1982 The answer checks. State . The copper content of pennies was reduced in 1982. 55.Familiarize . We first make a drawing. Let c= the num- ber of cartons filled. ··· ··· ···... ...... ··· ···  12 in each row How many rows? Translate . Number of cans/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightdivided byNumber per carton/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisNumber of cartons/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 228 ÷ 12 = c Solve . We carry out the division. 19 12228 12 108108 0 Thus, 19 = c,o rc= 19. Check . We can check by multiplying: 12 ·19 = 228. Our answer checks. State . 19 cartons were filled. 56.Familiarize . This is a multistep problem. Let s= the cost of 13 stoves, r= the cost of 13 refrigerators, and t= the total cost of the stoves and refrigerators.Translate . Number of stoves/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprighttimesPrice per stove/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal cost of stoves/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 13 · 425 = s Number of refrigerators/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprighttimesPrice per refrigerator/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal cost of refrigerators/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 13 · 620 = r Cost of stoves/bracehtipupleft /bracehtipdownright/bracehtipdownleft/bracehtipuprightplusCost of refrigerators/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightis Total cost/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt s + r = t Solve . We first carry out the multiplications in the first two equations. 13·425 = s 13·620 = r 5525 = s 8060 = r Copyright c/circlecopyrt2015 Pearson Education, Inc. 18 Chapter 1: Whole Numbers Now we substitute 5525 for sand 8060 for rin the third equation and then add on the left side. s+r=t 5525 + 8060 = t 13,585 = t Check . We repeat the calculations. The answer checks. State . The total cost was $13,585. 57.Familiarize . Let b= the number of beehives the farmer needs. ··· ··· ···... ...... ··· ···  30 in each row How many rows? Translate . Number of trees/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightdivided byNumber of trees pollinated by each hive/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisNumber of hives needed/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 420 ÷ 30 = b Solve . We carry out the division. 14 30420 30 120120 0 Thus, 14= b,o rb=1 4. Check . We can check by multiplying: 30 ·14= 4 20. The answer checks. State . The farmer needs 14beehives. 58.A=l·w=1 4f t ·7 ft = 98 square ft Perimeter = 14f t+7f t+1 4f t+7f t=4 2f t 59.Familiarize . We make a drawing. Let b= the number of beakers that will be filled. 1m L 1m L ··· 1m L 1m L ··· ···... ...... 1m L ··· ··· 1m L  20 in each row How many rows? Translate . Amount of alcohol/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightdivided byAmount per beaker/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisNumber of beakers filled/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 2753 ÷ 20 = bSolve . We carry out the division. 137 202753 20 7560 153140 13 Thus, 137 R 13 = b. Check . We can check by multiplying the number of beakers by 137 and then adding the remainder, 13. 137·20 = 2740 and 2740 + 13 = 2753 The answer checks. State . 137 beakers can be filled; 13 mL will be left over. 60.Familiarize . This is a multistep problem. Let b= the to- tal amount budgeted for food, clothing, and entertainment and let r= the income remaining after these allotments. Translate . Food and clothing budget/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightplusEntertainment budget/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal of these allotments/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 7825 + 2860 = b Food, clothing, and entertainment allotments/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightplusRemaining income/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal income/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt b + r =3 8 ,283 Solve . We add on the left side to solve the first equation. 7825 + 2860 = b 10,685 = b Now we substitute 10,685 for bin the second equation and solve for r. b+r=3 8 ,283 10,685 + r=3 8 ,283 10,685 + r−10,685 = 38 ,283−10,685 r=2 7 ,598 Check . We repeat the calculations. The answer checks. State . After the allotments for food, clothing, and enter- tainment, $27,598 remains. 61.7+( 4+3 ) 2=7+72 =7+49 =5 6 Answer B is correct. 62.7+42+32=7+1 6+9=2 3+9=3 2 Answer A is correct. Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Test 19 63. [46−(4−2)·5]÷2+4 =[ 46 −2·5]÷2+4 =[ 46 −10]÷2+4 =3 6÷2+4 =1 8+4 =2 2 Answer D is correct. 64. 9d ×d2 8036 By using rough estimates, we see that the factor d2≈ 8100÷9 0=9 0o r d2≈8000÷100 = 80. Since 99 ×92 = 9108 and 98 ×82 = 8036, we have d=8 . 65. 9a1 2b1236 ,421 Since 250 ×1000 = 250 ,000≈236,421 we deduce that 2b1≈250 and 9 a1≈1000. By trial we find that a=8 andb=4. 66.At the beginning of each day the tunnel reaches 500 ft − 200 ft, or 300 ft, farther into the mountain than it did the day before. We calculate how far the tunnel reaches intothe mountain at the beginning of each day, starting withDay 2. Day 2: 300 ftDay 3: 300 ft + 300 ft = 600 ftDay 4: 600 ft + 300 ft = 900 ftDay 5: 900 ft + 300 ft = 1200 ftDay 6: 1200 ft + 300 ft = 1500 ftWe see that the tunnel reaches 1500 ft into the mountain at the beginning of Day 6. On Day 6 the crew tunnels anadditional 500 ft, so the tunnel reaches 1500 ft + 500 ft, or2000 ft, into the mountain. Thus, it takes 6 days to reachthe copper deposit. Chapter 1 Discussion and Writing Exercises 1.No; if subtraction were associative, then a−(b−c)= (a−b)−cfor any a,b, and c. But, for example, 12−(8−4) = 12 −4=8 , whereas (12−8)−4=4 −4=0 . Since 8 /negationslash= 0, this example shows that subtraction is not associative. 2.By rounding prices and estimating their sum a shoppercan estimate the total grocery bill while shopping. This isparticularly useful if the shopper wants to spend no morethan a certain amount. 3.Answers will vary. Anthony is driving from Kansas Cityto Minneapolis, a distance of 512 miles. He stops for gasafter driving 183 miles. How much farther must he drive?4.The parentheses are not necessary in the expression9−(4·2). Using the rules for order of operations, the multiplication would be performed before the subtractioneven if the parentheses were not present. The parentheses are necessary in the expression (3 ·4) 2; (3·4)2=1 22= 144, but 3 ·42=3·16 = 48. Chapter 1 Test 1.546 ,789 The digit 5 tells the number of hundred thousands. 2.8843 = 8 thousand s+8h undreds + 4tens + 3 ones 3. 38/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright,40 3/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright,277/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ↑↑ ↑ Thirty-eight million, four hundred three thousand, two hundred seventy-seven 4. 6811 +3178 9989Add ones, add tens, add hundreds, and then add thousands. 5.11 1 45,889 +17 ,902 63,791 6.211 1239 843301 + 782 3165 7. 6203 +43 1 2 10,515 8. 7983 −4353 3630Subtract ones, subtract tens, subtracthundreds, and then subtract thousands. 9. 61 4 297/4/ −193 5 103 9 10.891 7 8907 /✥✥ −2059 6848 11.12 12/91 6 2/3,06/7✥✥ −17,892 5175 Copyright c/circlecopyrt2015 Pearson Education, Inc. 20 Chapter 1: Whole Numbers 12.56 7 4568 × 9 41,112 13.54 3 8876 ×600 5,325 ,600Multiply by 6 hundreds (We write 00 and then multiply 8876 by 6.) 14. 65 ×37 455 1950 2405Multiplying by 7Multiplying by 30Adding 15. 678 ×788 5424 54240 474600 534 ,264 16.3 41512 3 The answer is 3 R 3. 17.70 642042 00 0 The answer is 70. 18.97 898633801 623623 0 The answer is 97. 19.805 4435,428 352 228220 8 The answer is 805 R 8. 20.Round 34,528 to the nearest thousand. 34,528 ↑ The digit 4is in the thousands place. Consider the nextdigit to the right, 5. Since 5 is 5 or higher, round 4thou-sands up to 5 thousands. Then change all digits to theright of thousands to zeros. The answer is 35,000.21.Round 34,528 to the nearest ten. 34,52 8 ↑ The digit 2 is in the tens place. Consider the next digit to the right, 8. Since 8 is 5 or higher, round 2 tens up to 3tens. Then change the digit to the right of tens to zero. The answer is 34,530. 22.Round 34,528 to the nearest hundred. 34,5 28 ↑ The digit 5 is in the hundreds place. Consider the next digit to the right, 2. Since 2 is 4or lower, round down,meaning that 5 hundreds stays as 5 hundreds. Then changeall digits to the right of hundreds to zero. The answer is 34,500. 23. Rounded to the nearest hundred 23 ,649 23 ,600 +54,74 6 + 54,700 78 ,300 ←Estimated answer 24. Rounded to the nearest hundred 54,751 54,800 −23 ,649 −23 ,600 31 ,200 ←Estimated answer 25. Rounded to the nearest hundred 824800 ×489 ×500 400 ,000 ←Estimated answer 26.Since 34is to the right of 17 on the number line, 34 >17. 27.Since 117 is to the left of 157 on the number line, 117<157. 28. 28 + x=7 4 28 + x−2 8=7 4 −28 Subtracting 28 on both sides x=46 Check: 28 + x=7 4 2 8 + 46?7 4 74/vextendsingle/vextendsingle/vextendsingle TRUE The solution is 46. 29.169÷13 = n We carry out the division. 13 13169 13 3939 0 The solution is 13. Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Test 21 30. 38·y= 532 38·y 38=532 38Dividing by 38 on both sides y=1 4 Check: 38 ·y= 532 38·14? 532 532/vextendsingle/vextendsingle/vextendsingle TRUE The solution is 14. 31.3 8 1=0+ a 381 = a Adding on the right side The solution is 381. 32.Familiarize . Let s= the number of calories in an 8-oz serving of skim milk. Translate . Number of calories in skim milk/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightplusHow many more calories/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisNumber of calories in whole milk/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt s + 63 = 146 Solve . We subtract 63 on both sides of the equation. s+6 3=1 46 s+6 3−6 3=1 46 −63 s=8 3 Check . Since 63 calories more than 83 calories is 83 + 63, or 146 calories, the answer checks. State . An 8-oz serving of skim milk contains 83 calories. 33.Familiarize . Let s= the number of staplers that can be filled. We can think of this as repeated subtraction, taking successive sets of 250 staples and putting them intosstaplers. Translate . Number of staples/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightdivided byNumber in each stapler/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisNumber of staplers filled/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 5000 ÷ 250 = s Solve . We carry out the division. 20 250 5000 500 00 0 Then 20 = s. Check . We can multiply the number of staplers filled by the number of staples in each one. 20·250 = 5000 The answer checks. State . 20 staplers can be filled from a box of 5000 staples.34.Familiarize . Let a= the total land area of the five largest states, in square meters. Since we are combining the areas of the states, we can add. Translate . 571,951 + 261 ,797 + 155 ,959 + 145 ,552 + 121 ,356 = a Solve . We carry out the addition. 2 133 2 571 ,951 261 ,797 155 ,959 145 ,552 +121 ,356 1,256 ,615 Then 1 ,256,615 = a. Check . We can repeat the calculation. We can also esti- mate the result by rounding. We will round to the nearest ten thousand. 571,951 + 261 ,797 + 155 ,959 + 145 ,552 + 121 ,356 ≈570,000 + 260 ,000 + 160 ,000 + 150 ,000 + 120 ,000 =1,260,000 Since 1 ,260,000≈1,256,615, we have a partial check. State . The total land area of Alaska, Texas, California, Montana, and New Mexico is 1,256,615 sq mi. 35.a)We will use the formula Perimeter = 2 ·length + 2·width to find the perimeter of each pool table in inches. We will use the formula Area = length ·width to find the area of each pool table, in sq in. For the 50 in. by 100 in. table: Perimeter = 2 ·100 in. + 2 ·50 in. = 200 in. + 100 in. = 300 in. Area = 100 in. ·50 in. = 5000 sq in. For the 44 in. by 88 in. table: Perimeter = 2 ·88 in. + 2 ·44 in. = 176 in. + 88 in.= 264in. Area = 88 in. ·44 in. = 3872 sq in. For the 38 in. by 76 in. table: Perimeter = 2 ·76 in. + 2 ·38 in. = 152 in. + 76 in.= 228 in. Area = 76 in. ·38 in. = 2888 sq in. b)Leta= the number of square inches by which the area of the largest table exceeds the area of the smallest table. We subtract to find a. a= 5000 sq in. −2888 sq in. = 2112 sq in. 36.Familiarize . Let n= the number of 12-packs that can be filled. We can think of this as repeated subtraction, takingsuccessive sets of 12 snack cakes and putting them into n packages. Copyright c/circlecopyrt2015 Pearson Education, Inc.12 in.8 in.6 in.22 Chapter 1: Whole Numbers Translate . Number of cakes/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightdivided byNumber in each package/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightisNumber of 12-packs/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 22,231 ÷ 12 = n Solve . We carry out the division. 1852 1222,231 12 102 96 63 60 3124 7 Then 1852 R 7 = n. Check . We multiply the number of packages by 12 and then add the remainder, 7. 12·1852 = 22 ,224 22,224+ 7 = 22 ,231 The answer checks. State . 1852 twelve-packs can be filled. There will be 7 cakes left over. 37.Familiarize . This a multistep problem. Let b= the total cost of the black cartridges, p= the total cost of the photo cartridges, and t= the total cost of the entire purchase. Translate . For the black ink cartridges: Number of black cartridges/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprighttimesPrice per cartridge/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal cost of black cartridges/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright /arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 3 · 15 = b For the photo cartridges: Number of photo cartridges/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprighttimesPrice per cartridge/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal cost of photo cartridges/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 2 · 25 = p For the total cost of the order: Cost of black cartridges/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightplusCost of photo cartridges/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal cost of purchase/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt b + p = t Solve . We solve the first two equations and then add the solutions. 3·15 = b 45= b2·25 = p 50 = p b+p=t 45+5 0= t 95 = t Check . We repeat the calculations. The answer checks. State . The total cost of the purchase was $95. 38.Exponential notation for 12 ·12·12·12 is 12 4. 39.73=7·7·7 = 343 40.105=1 0·10·10·10·10 = 100 ,000 41. 35−1·28÷4+3 =3 5−28÷4+ 3 Doing all multiplications and =3 5−7 + 3 divisions in order from left to right = 28 + 3 Doing all additions and subtractions = 31 in order from left to right 42. 102−22÷2 = 100 −4÷2 Evaluating the exponential expressions = 100 −2 Dividing = 98 Subtracting 43. (25−15)÷5 =1 0÷5 Doing the calculation inside the parentheses = 2 Dividing 44. 24+2 4÷12 =1 6+2 4 ÷12 Evaluating the exponential expression = 16 + 2 Dividing= 18 Adding 45. 8×{(20−11)·[(12 + 48) ÷6−(9−2)]} =8×{9·[60÷6−7]} =8×{9·[10−7]} =8×{9·3} =8×27 = 216 46.We add the numbers and then divide by the number of addends. 9 7+9 9+8 7+8 9 4=372 4=9 3 Answer A is correct. 47.Familiarize . We make a drawing. Copyright c/circlecopyrt2015 Pearson Education, Inc. Chapter 1 Test 23 Observe that the dimensions of two sides of the container are 8 in. by 6 in. The area of each is 8 in. ·6 in. and their total area is 2 ·8 in.·6 in. The dimensions of the other two sides are 12 in. by 6 in. The area of each is 12 in. ·6 in. and their total area is 2 ·12 in. ·6 in. The dimensions of the bottom of the box are 12 in. by 8 in. and its areais 12 in. ·8 in. Let c= the number of square inches of cardboard that are used for the container. Translate . We add the areas of the sides and the bottom of the container. 2·8 in.·6i n .+2 ·12 in. ·6 in. + 12 in. ·8 in. = c Solve . We carry out the calculation. 2·8 in.·6 in. + 2 ·12 in. ·6 in. + 12 in. ·8 in. = c 96 sq in. + 144 sq in. + 96 sq in. = c 336 sq in. = c Check . We can repeat the calculations. The answer checks.State . 336 sq in. of cardboard are used for the container. 48.We can reduce the number of trials required by simplifying the expression on the left side of the equation and thenusing the addition principle. 359−46+ a÷3×25−7 2= 339 359−46+ a÷3×25−49 = 339 359−46+a 3×25−49 = 339 359−46+25·a 3−49 = 339 313 +25·a 3−49 = 339 264+25·a 3= 339 264+25·a 3−264= 339 −264 25·a 3=7 5 We see that when we multiply aby 25 and divide by 3, the result is 75. By trial, we find that25·9 3=225 3= 75, soa= 9. We could also reason that since 75 = 25 ·3 and 9/3 = 3, we have a=9 . 49.Familiarize . First observe that a 10-yr loan with monthly payments has a total of 10 ·12, or 120, payments. Let m= the number of monthly payments represented by $9160 andletp= the number of payments remaining after $9160 has been repaid. Translate . First we will translate to an equation that can be used to find m. Then we will write an equation that can be used to find p. Payments to date/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightdivided byAmount of each payment/bracehtipupleft/bracehtipdownright/bracehtipdownleft /bracehtipuprightisNumber of payments made/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt 9160 ÷ 229 = mPayments already made/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightplusRemaining payments/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipuprightisTotal number of payments/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright/arrowbt/arrowbt/arrowbt/arrowbt/arrowbt m + p = 120 Solve . To solve the first equation we carry out the divi- sion. 40 229 9160 916 00 0 Thus, m=40 . Now we solve the second equation. m+p= 120 40+ p= 120 Substituting 40 for m 40+ p−40 = 120 −40 p=8 0 Check . We can approach the problem in a different way to check the answer. In 10 years, Cara’s loan paymentswill total 120 ·$229, or $27,480. If $9160 has already been paid, then $27 ,480−$9160, or $18,320, remains to be paid. Since 80 ·$229 = $18 ,320, the answer checks. State . 80 payments remain on the loan. Copyright c/circlecopyrt2015 Pearson Education, Inc.