311 Chapter 10 Modeling with Sequences and Series Homework 10.1 1.2 7 5, 5 2 7 = = The sequence does not have a common difference.It is not arithmetic 3.38 40 2, 36 38 2, 34 36 2,32 34 2... = = = = The sequence has a common difference2. It is arithmetic. 5.11 3 8,19 11 8, 27 19 8,35 27 8 = = = = The sequence has a common difference of 8.It is arithmetic. 7.44 4 40, 444 44 400 = =The sequence does not have a common difference.It is not arithmetic 9.The sequence has a common difference of 6. 5 ( 1) 65 6 66 1nna nna n= + = + = 11.The sequence has a common difference of11. 4 ( 1) 114 11 1111 7nna nna n= + = + += + 13.The sequence has a common difference of6. 100 ( 1) 6100 6 66 106nna nna n= + = += + 15.The sequence has a common difference of 2. 1 ( 1) 21 2 22 1nna nna n= + = + = 17.The sequence has a common difference of 3. 375 (37 1) 3 5 36 3 5 108 113 a = + = + = + =19.The sequence has a common difference of9. 45200 (45 1) 9 200 44 9200 396 196a = + = + = + = 21.The sequence has a common difference of 1.6. 964.1 (96 1) 1.6 4.1 95 1.64.1 152 156.1a = + = + = + = 23.The sequence has a common difference of7. 648 (64 1) 7 8 63 78 441 433a = + = + = + = 25.The sequence has a common difference of 5. 533 3 ( 1) 5533 3 5 5533 5 2535 5107nnnnn= + = + = == 533 is the 107th term. 27.The sequence has a common difference of 8. 695 7 ( 1) 8695 7 8 8695 8 1696 887nnnnn= + = + = == 695 is the 87th term. 29.The sequence has a common difference of 8. 2469 27 ( 1) 82469 27 8 82469 8 352504 8313nnnnn= + = + = == 2469 is the 313th term. 31.The sequence has a common difference of4. 14251 29 ( 1) 414251 29 4 414251 33 414284 43571nnnnn = + = + = = = -14251 is the 3571st term. Homework 10.1SSM: Intermediate Algebra 312 33. EQ 1EQ 2 1 11 11 124 (7 1) 66 (13 1)24 6 66 1224 6 66 12a d a da d a dd a d a= + = + = + = + = = Solve the system of linear equations by setting the left side of each equation equal to each other. 24 6 66 126 427d ddd = == Use EQ 1 and7 d = to find 1. a1124 6(7)18aa = = 4018 (40 1) 7 18 39 7 255 a = + = + =35.EQ 1EQ 2 1 11 11 1500 (41 1) 500 +(81 1)500 40500 80500 40500 80a d a da d a dd a d a= + = = + = + = = Solve the system of linear equations by setting the left side of each equation equal to each other. 500 40 500 8040 00d ddd = == Use EQ 1 and0 d = to find 1. a11500 40 0500aa == 990500 (990 1) 0 500 a = + =37.(1) 4(1) 2 2(2) 4(2) 2 6(3) 4(3) 2 10fff= == == = Yes, this sequence is arithmetic with a common difference equal to the slope, 4. 39. 222(1) 1 1(2) 2 4(3) 3 9fff= == == = No, this sequence is not arithmetic.There is no common difference. 41.The students work is not correct.The sequence is not arithmetic because there is no common difference. 43.Since9 d = and 18, a =( ) 8 ( 1) 9 8 9 9 9 1 f x x x x = + = + = 45. a.27, 500 ( 1) 80027500 800 80026700 800nna nna n= + = + = + b. 2226700 800 22 $44300 a = + =The salary for the 22nd year will be $44,300. c.50000 26700 80023300 80029.125nnn= +== The salary will first be above $50,000 in the 30th year. 47.a.1356na n = +b. 1234135 1 35.176135 2 35.336135 3 35.56135 4 35.676aaaa= + = + = + == + These values represent the number of hours the instructor would work if she had 1, 2, 3, or 4 students, respectively. c. 130135 130 56.676a = + hours per week d. 160 3561256150nnn= +== The greatest number is 150 students. 49.a.The band collects0.3(6) $1.80 = per cover charge, so the common difference is 1.8.50 0.3(6) 50 1.8na n n = + = +b.256 50 1.8306 1.8170nnn= +== 170 people paid the cover charge. SSM: Intermediate AlgebraHomework 10.2 313 c.Total number of people who pay is 200 18 11 6 165 = . 16550 1.8(165) $247 a = + =Their maximum profit is $247.00. d.Little Muddys profit sequence 84na28 30 32 34 360121620DollarsN u m be rofPe o pl e 51.a.0.37 ( 1) 0.230.37 0.23 0.230.14 0.23nna nna n= + = + = + b. 130.14 0.23(13) $3.13 a = + =c. 160.14 0.23(16) $3.82 a = + =No, $3.85 is not the better deal. d. 800.14 0.23(80) $18.54 a = + =53.Written response Homework 10.2 1. 28 196 1372 96047, 7, 7, 7...4 28 196 1372= = = =The sequence has a common ratio of 7.It is geometric.3.( )( )6 13 7, 1 6 7, 8 1 7,15 8 7... = = = = The sequence has a common difference of7. It is arithmetic. 5. 4 61.33, 1.53 4 =The sequence has no common ratio. 4 3 1, 6 4 2 = =The sequence has no common difference.The sequence is neither. 7. 8 85 258540 1 8 1 1 1, , ,200 5 40 5 8 5 5= = = =The sequence has a common ratio of 1.5It isgeometric. 9.The sequence has a common ratio of 2. ( )113 23(2) (2)3(2) 3(2)2 2nnnnnnaa=== = 11.The sequence has a common ratio of 1.4 ( )11180041 18004 41800 44132004nnnnnnaa = = = = 13.The sequence has a common ratio of 1.2 ( )11110021 11002 21100 2212002nnnnnnaa = = = = 15.The sequence has a common ratio of 4. ( )111 44 (4)1441(4)4nnnnnnaa== = = Homework 10.2SSM: Intermediate Algebra 314 17.The sequence has a common difference of 5. 14 ( 1) 5 14 5 5 5 9na n n n = + = + = +19.The sequence has a common ratio of 5. 34 1 33 23344(5) 4(5) 4.66 10 a= = 21.The sequence has a common ratio of 1.2 27 1 266271 180 80 1.19 102 2a = = 23.The sequence has a common ratio of 2. 23 1 22238(2) 8(2) 33, 554, 432 a= = =25.The sequence has a common difference of 3.4. 969.5 (96 1) 3.4 9.5 95 3.4 332.5 a = + = + =27.The sequence has a common ratio of 1.2 11110.46875 240210.00195312521log(0.001953125) log21log(0.001953125) ( 1) log2nnnn = = = = log(0.001953125)11log2n = log(0.001953125)1 101log2n = + = 29.Use the sequence 0.0024, 0.0112, 0.056, 0.28, 1.4,,109,375.This sequence has a common ratio of 5.111109, 375 0.00224(5)48828125 5log(48828125) log(5 )log(48828125) ( 1) log5log(48828125)1log5log(48828125)1 12log5nnnnnn==== = = + = 31.The sequence has a common difference of 3. 367 4 ( 1) 3367 4 3 3367 1 3366 3122nnnnn= + = + = +== 33.The sequence has a common ratio of 2. 1113, 407, 872 13(2 )262144 2log(262144) log(2 )log(262144) ( 1) log 2log(262144)1log 2log(262144)1 19log 2nnnnnn==== = = + = 35.The sequence has a common ratio of 3. 11128, 697, 814 2(3)14, 348, 907 3log(14, 348, 907) log(3 )log(14, 348, 907) ( 1) log3log(14, 348, 907)1log3log(14, 348, 907)1 16log3nnnnnn==== = = + = 37.The sequence is geometric.The common ratio is 5. 39.The sequence is arithmetic.The common difference is the slope which is 7. 41.The sequence has a common ratio of 3. 18(3)nna= OR ( ) ( )18 3xf x=43.The students work is not correct.The student is using the formula for a geometric sequence but the sequence is arithmetic. SSM: Intermediate AlgebraHomework 10.3 315 45.a. 127000(1.04)nna=b. 10 110927000(1.04)27000(1.04)$38429.42a== The persons salary will be $38,429.42 for the 10th year. c. ( )11150000 27000(1.04)501.042750log log 1.042750log ( 1) log1.0427nnnn== = = 50log271log1.0450log271 16.7log1.04nn = = + The salary will first be above $50,000 in the 17th year. 47.a.2, 4, 8,16, 32b. The sequence has a common ratio of 2. 1 1 1 1 12(2) 2 2 2 2n n n nna + = = = =c. 882 256 a = = ancestors d. 35 10352 3.44 10 34.4 billion ancestorsa = = Model breakdown has occurred.This is much higher than the worlds population now! 49.a. 15(3)nna=b. 5 1 455(3) 5(3) 405 a= = = students c.( ) ( )11 1 10115 3 5 3 295, 245 a= = = students Model breakdown has occurred.We are assuming there are this many students on campus. d.Answers may vary. One of the assumptions was that a student would tell the rumor to 3 other students who have not heard the rumor yet.This assumption is reasonable for the first several days.However, as the number of students who heard the rumor grows larger, it is unlikely that those students would know 3 other students who had not heard the rumor yet. 51.Answers may vary. Homework 10.3 1. ( )9090 2 447202052S+= =3. 108108(13 548)302942S+= =5. 7272(37 1099)382322S+ = = 7.The series is arithmetic with a common difference of8. 7474(5 589)219782S+= =9.The series is arithmetic with a common difference of -4. 101101(93 307)108072S+ = = 11.The series is arithmetic with a common difference of 0. 117117(4 4)4682S+= =13.The series is arithmetic with a common difference of 10. 1251253 (125 1) 10 3 124 10 1243125(3 1243)778752aS= + = + =+= = 15.The series is arithmetic with a common difference of 11. 81818 (81 1) 11 8 80 11 88881(8 888)362882aS= + = + =+= = Homework 10.3SSM: Intermediate Algebra 316 17.The series is arithmetic with a common difference of 13. 15215215 151 13 1978152(15 1978)1514682aS= + =+= = 19.The series is arithmetic with a common difference of3. 10010040 99 3 257100( 40 257)108502aS= + = += = 21.The series is arithmetic with a common difference of 6. 247 19 ( 1) 6247 19 6 6247 13 6234 639nnnnn= + = + = +== So 39247 a = 3939(19 247)51872S+= =23.The series is arithmetic with a common difference of8. 900 900 ( 1) 8900 900 8 8900 908 81808 8226nnnnn = + = + = = = 226226So900226(900 900)02aS =+ = = 25.The series is arithmetic with a common difference of 3.340 4 ( 1) 3340 4 3 3340 3 1339 3113nnnnn= + = + = +== So 113340 a = 113113(4 340)194362S+= = 27.The series is arithmetic with a common difference of 1. 10, 000 1 ( 1) 110, 000 1 110, 000nnn= + = + = So 1000010, 000 a = 1000010000(1 10000)50, 005, 0002S+= =29. nS is positive because your sum will be the sum ofn positive numbers. 31. nS is positive because your sum will be the sum of a couple of negative and lots of larger positive numbers. 33.Yes, the series is arithmetic.The common difference is the slope, 7. 35.a. 2828500 (28 1) 110028500 27 1100 $58200a = + = + = b. 2828(28500 58200)$1, 213, 8002S+= =37.Company A 202035000 (20 1) 700 4830020(35000 48300)$833, 0002aS= + =+= = Company B 202027000 (20 1) 1500 $55, 50020(27000 55500)$825, 0002aS= + =+= = Your total earnings for 20 years would be greater at Company A by $8000. 39.a. 5016 (50 1) 2 114 a = + =There are 114 seats in the 50th row. 5050(16 114)32502S+= = seats in the auditorium SSM: Intermediate AlgebraHomework 10.4 317 b. 1016 9 2 34 a = + = seats in the 10th row 1010(16 34)2502S+= = total seats in the first ten rows This leaves3250 250 3000 = seats in rows 11 through 50. So total revenue is 250($20) 3000($15) $50, 000. + =c.If they sit in the first ten rows first and then fill in the other rows, the revenue would be 250($20) 2650(15) $44, 750. + = This would give us the maximum revenue.If all 2900 people sit in rows 11 through 50 (i.e. none of them sit in the first ten rows), the revenue will2900($15) $43, 500. = This would give us the minimum revenue. 41.a. b.Using linear regression gives us( ) 12.62 174.4 f t x = +c.(0) 12.62(0) 174.4 174.4 f = + =In 1990 sales at food and drink places was about $174,400,000,000. d.(19) 12.62(19) 174.4 414.18 f = + =In 2009 sales at food and drink places will be about $414,180,000,000. e. 20(0) (1) (2) ... (19)20(174.4 414.18)5885.82f f f fS+ + + ++= = = This means that the total sales from 1990 through 2009 will be $5,885,800,000,000. f.60%of the total sales is $3531.48 trillion Assume restaurant owners report 9% while the wait staff receives 15%.The difference, 6%, is the unreported income.6% of $3531.48 trillion is $211.89 trillion. 43.Written response Homework 10.4 1. 27275(1 2 )671, 088, 6351 2S= = 3. 30306(1 1.3 )52, 379.911 1.3S= 5. 272713(1 0.8 )64.841 0.8S= 7. 50502.3(1 0.9 )22.881 0.9S= 9.The series is geometric with a common ratio of 5. 13132(1 5 )610, 351, 5621 5S= = 11.The series is geometric with a common ratio of 0.3. 1111600(1 0.3 )857.141 0.3S= 13.The series is geometric with a common ratio of 2.3 ( )( )102310233 18.841S= 15.The series is geometric with a common ratio of 4. 1167,108, 864 1(4)log67,108, 864 log 4log67,108, 864 ( 1) log 4log67,108, 8641log 4log67,108, 8641 14log 4nnnnn=== = = + = 14141(1 4 )89, 478, 4851 4S= = Homework 10.4SSM: Intermediate Algebra 318 17.The series is geometric with a common ratio of 1.2. 11121.4990848 5(1.2)4.29981696 1.2log 4.29981696 log1.2log 4.29981696 ( 1) log1.2log 4.299816961log1.2log 4.299816961 9log1.2nnnnnn==== = = + = 995(1 1.2 )103.991 1.2S= 19.The series is geometric with common ratio of 1.2 11114.8828125 10, 000210.0004882812521log0.00048828125 log21log0.00048828125 ( 1) log2log0.0004882812511log2log0.000488281251 121log2nnnnnn = = = = = = + = ( )( )1212121210000 1199995.121S= 21.The series is arithmetic with a common difference of 0. 100100(1 1)1002S+= = 23.The series is arithmetic with a common difference of 6. 351 3 ( 1) 6351 3 6 6351 6 3354 659nnnnn= + = + = == 5959(3 351)10, 4432S+= =25.The series is geometric with a common ratio of 1.6 1115 130362, 797, 056 61 12,176, 782, 336 61 1log log2,176, 782, 336 61 1log ( 1) log2,176, 782, 336 61log2,176, 782, 33611log61log2,176, 782, 3361log6nnnnnn = = = = = =1 13 + = ( )( )1316131630 1361S= = 27. nS must be positive because it is the sum of all positive values. 29.The series is arithmetic because( ) f x is linear.The common difference of the series is the slope of( ) f x which is1. 31. 202023500(1 1.04 )$699, 784.851 1.04S= The persons total earnings after 20 years of work will be $699,784.85. SSM: Intermediate AlgebraChapter 10 Review Exercises 319 33.Company A 303026000(1 1.05 )$1, 727, 410.041 1.05S= Company B 303031000(1 1.03 )$1, 474, 837.891 1.03S= The earnings at Company A after 30 years will be $252,572.15 more than Company B. 35.Recall, the number of ancestorsn generations back is a geometric series with a common ratio of 2. 102(1 2 )20461 2nS= =ancestors 37.a.The entrepreneurs name would be taken off the list in the 11th round.The amount of money sent to the entrepreneur each round is a geometric series with common ratio 8. 101040(1 8 )6,135, 667, 5601 8S= = The entrepreneur could receive as much as approximately $6.14 billion. b. 8(1 8 )6, 000, 000, 0001 886, 000, 000, 000 (1 8 )75, 250, 000, 000 8 18 5, 250, 000, 001log8 log5, 250, 000, 001log8 log5, 250, 000, 001log5, 250, 000, 001log810.76nnnnnnnn== = ==== There will be ten full rounds and part of an 11th round. c.The money from the first 10 rounds will go to the entrepreneur.The chain letter runs out of people (and money!) to complete the 11th round.All the money from the 11th round would go to the first 8 people besides the entrepreneur.This amount is (in billions): 30 6.1362.988So, 9 people will receive money from the chain letters.The entrepreneur will receive approximately $6.14 billion.The 8 people besides the entrepreneur will receive an average of $2.98 billion. 39.a.Use an exponential regression to find( ). f t( ) 85.89(0.47)tf t =b. 0(0) 85.89(0.47) 85.89 f = =In 1980 85.89 million eight-track cartridges were sold. c.There are 6 terms in the series son must be 6. 6685.89(1 0.47 )160.311 0.47S= There were a total of 160.31 million eight-track cartridges sold from 1980 through 1985. d.The actual total was 158 million eight-track cartridges. e. 292985.89(1 0.47 )162.061 0.47S= There will be a total of 162.06 million eight-track cartridges sold between 1980 and 2008. Chapter 10 Review Exercises 1. 40 1 10 1 2.5 1 0.625 1, , ,160 4 40 4 10 4 2.5 4= = = =The sequence is geometric with a common ratio of 1.4 2. 24 13 11, 35 24 11, 46 35 11,57 46 11 = = = = The series is arithmetic with a common difference of 11. 3.95 101 6, 89 95 6,83 89 6, 77 83 6 = = = = The sequence is arithmetic with a common difference of6. 4. 18 36 72 1442, 2, 2, 29 18 36 72= = = =The series is geometric with a common ratio of 2. Chapter 10 Review ExercisesSSM: Intermediate Algebra 320 5. 7 7 7 75 25 125 6257 7 75 25 1251 1 1 1, , ,7 5 5 5 5= = = =The series is geometric with a common ratio of 1.5 6.4 3 1, 6 4 2 = =The sequence is not arithmetic because it does not have a common difference. 4 61.33, 1.53 4 =The sequence is not geometric because it does not have a common ratio.It is neither. 7.The sequence is geometric with a common ratio of 3. 112(3)2(3) (3)12(3)32(3)3nnnnnnaa== = = 8.The sequence is arithmetic with a common difference of 3. 25 ( 1) 325 3 322 3nna nna n= + = + = + 9.The sequence is arithmetic with a common difference of5. 9 ( 1) 59 5 514 5nna nna n= + = += 10.The sequence is geometric with a common ratio of 1.2 ( )11120021 12002 21200 2214002nnnnnnaa = = = = 11.The sequence is arithmetic with a common difference of 2.7. 3.2 ( 1) 2.73.2 2.7 2.72.7 0.5nna nna n= + = + = + 12.The sequence is geometric with a common ratio of 0.7. 11800(0.7)800(0.7) (0.7)10800(0.7)78000(0.7)7nnnnnnaa== = = 13.The sequence is geometric with a common ratio 2.47 1 46 14476(2) 6(2) 4.22 10 a= = 14.The sequence is geometric with a common ratio 1.4 9 1 891 1768 768 0.01174 4a = = 15.The sequence is arithmetic with a common difference of3. 9887 (98 1) 3 204 a = + = 16.The sequence is arithmetic with a common difference of 2.6. 872.3 (87 1) 2.6 225.9 a = + = SSM: Intermediate AlgebraChapter 10 Review Exercises 321 17.The sequence is arithmetic with a common difference 4. 2023 7 ( 1) 42023 7 4 42023 3 42020 4505nnnnn= + = + = +== 2023 is the 505th term. 18.The sequence is arithmetic with a common difference8. 107 501 ( 1) 8107 501 8 8107 509 8616 877nnnnn = + = + = = = 107 is the 77th term. 19.The sequence is geometric with a common ratio 3.111470, 715, 894,135 5(3)94,143,178, 827 3log94,143,178, 827 log3log94,143,178, 827 ( 1) log3log94,143,178, 8271log3log94,143,178, 8271 24log3nnnnnn==== = = + = 470,715,894,135 is the 24th term. 20.EQ 1 EQ 2 1 11 11 152 (5 1) 36 +(9 1)52 436 852 436 8a d a da d a dd a d a= + = = + = + = = Solve the system of linear equations by setting the left side of each equation equal to each other. 52 4 36 84 164d ddd = = = Use EQ 1 and4 d = to find 1. a1152 4 468aa == 6968 (69 1) 4 204 a = + = 21. 4343(52 200)31822S+ = = 22. 22224(1 1.7 )671,173.071 1.7S= 23.The series is geometric with a common ratio of 2. 1111, 610, 612, 736 3(2)536, 870, 912 2log536, 870, 912 log 2log536, 870, 912 ( 1) log 2log536, 870, 9121log 2log536, 870, 9121 30log 2nnnnnn==== = = + = 30303(1 2 )3, 221, 225, 4691 2S= = 24.The series is arithmetic with a common difference of 6. 1200 30 ( 1) 61200 30 6 61200 24 61176 6196nnnnn= + = + = +== 196196(30 1200)120, 5402S+= =25.The series is arithmetic with a common difference of4. 1197 11 ( 1) 41197 11 4 41197 15 41212 4303nnnnn = + = + = = = 303303(11 1197)179, 6792S+ = = Chapter 10 TestSSM: Intermediate Algebra 322 26.The series is geometric with a common ratio of 1.3 11111 531, 44131 1531, 441 31 1log log531, 441 31 1log ( 1) log531, 441 31log531, 44111log31log531, 4411 131log3nnnnnn = = = = = = + = ( )( )13131313531, 441 1797,1611S= = 27.(1), (2), (3),..., (80) f f f f is a geometric sequence with a common ratio of 5. 28.(1), (2), (3),..., (80) f f f f is an arithmetic sequence with a common difference of9. 29.a. Company A 25 12528, 000(1.04) $71, 772.52 a= Company B 2534, 000 (24 1) 1500 $70, 000 a = + =b.Company A 252528, 000(1 1.04 )$1,166, 085.431 1.04S= Company B 2525(34000 70000)$1, 300, 0002S+= =c.Answers may vary. 30.a.Using linear regression, we get( ). f t( ) 0.32 5.42 f t t = +b.(19) 0.32(19) 5.42 11.5 f = + =In 2009, there will be about 11,500 deaths. c.Sincef is linear we can treat the sum of all the terms as an arithmetic series with a common difference of 0.32 and 15.42. a =2020(5.42 11.5)169.22S+= =From 1990 through 2009 there will be about 169,200 deaths. d.Answers may vary. Chapter 10 Test 1. 6 12 24 482, 2, 2, 23 6 12 24= = = = It is a geometric sequence with a common ratio of 2. 2.19 20 1,17 19 219 170.95, 0.8920 19 = = = The sequence has neither a common difference nor a common ratio, so it is none of these. 3. 35 175 875 43755, 5, 5, 57 35 175 875= = = =It is a geometric series with a common ratio of 5. 4.61 69 8, 53 61 8,45 53 9, 37 45 8 = = = = It is an arithmetic series with a common difference of8. 5.The sequence is arithmetic with a common difference of6. 31 ( 1) 631 6 637 6nna nna n= + = += SSM: Intermediate AlgebraChapter 10 Test 323 6.The sequence is geometric with a common ratio of 6. 116(4)6(4) (4)16(4)43(4)2nnnnnnaa== = = 7.The sequence is arithmetic with a common difference of 3. 874 (87 1) 3 262 a = + =8.The sequence is geometric with a common ratio of 1.2 31 1311100 0.0000000932a = 9.The sequence is arithmetic with a common difference of 4. 1789 27 ( 1) 41789 27 4 41789 31 41820 4455nnnnn= + = + = +== 1789 is the 455th term in the sequence. 10.The series is geometric with a common ratio of 1.1. 111428.717762 200(1.1)2.14358881 (1.1)log 2.14358881 log1.1log 2.14358881 ( 1) log1.1log 2.143588811log1.1log 2.143588811 9log1.1nnnnnn==== = = + = 428.717762 is the 9th term in the sequence. 11.The series is geometric with a common ratio of 1.3 ( )( )2013201327 140.501S= 12.The series is geometric with a common ratio of 2. 1112,147, 483, 648 4(2)536, 870, 912 2log536, 870, 912 log 2log536, 870, 912 ( 1) log 2log536, 870, 9121log 2log536, 870, 9121 30log 2nnnnnn==== = = + = 30304(1 2 )4, 294, 967, 2921 2S= = 13.The series is arithmetic with a common difference of4. 78 50 ( 1) 478 50 4 478 54 4132 433nnnnn = + = + = = = 3333(50 78)4622S+ = = 14.The series is arithmetic with a common difference of 14. 40019 (400 1) 14 5605 a = + =400400(19 5605)1,124, 8002S+= =15. 2 3 45 202 3 4 5 2020(7 2) (7 2 2 ) (7 3 2 ) (7 4 2 )(7 5 2 ) ... (7 20 2 )(7 7 2 7 3 7 4 7 5 ... 7 20)(2 2 2 2 2 ... 2 )20(7 140) 2(1 2 )2 1 21470 2, 097,150 2, 098, 260+ + + + + + ++ + + + += + + + + + + + + + + + + ++ = += + = 16. 222(1) 3(1) 1 4(2) 3(2) 1 13(3) 3(3) 1 28fff= + == + == + = This series is neither geometric nor arithmetic. 17. nS is negative because most of the terms of the series will be negative.The sum of negative numbers is negative.Cumulative Review Chapters 1-10SSM: Intermediate Algebra 324 18.a. Use a linear regression to find( ). f t( ) 6.69 41.67 f t t = b.(7) 6.69(7) 41.67 5.16(8) 6.69(8) 41.67 11.85(9) 6.69(9) 41.67 18.54(10) 6.69(10) 41.67 25.23(11) 6.69(11) 41.67 31.92fffff= == == == == = These values estimate the number of fatalities in 1987, 1988, 1989, 1990, and 1991, respectively c.( )2127 6.69(27) 41.67 138.96 a f = = =2121(5.16 138.96)1513.262S+= =From 1987 to 2007 the total number of wet-bike fatalities is about 1513. 19.a. 132000(1.03)nna=b.( )11140000 32000 1.031.25 1.03log1.25 log1.03log1.25 ( 1) log1.03log1.251log1.03log1.251 8.5log1.03nnnnnn==== = + The salary will first be above $40,000 in the 9th year. c. 242532000(1.03) $65049.41 a = d. 252532000(1 1.03 )$1,166, 696.461 1.03S= Cumulative Review Chapters 1-10 1. ( )( )24 49 02 7 2 7 0xx x = + = 2 7 0or2 7 07 7or2 2x xx x = + == = 2. ( )5 9898598ln ln598ln 2.975535xxxeeex== = = 3. 7771/ 772 3 512 54272727 1.6013bbbbb ===== 4.( )34log 4 7 44 7 34 7 814 8822xxxxx = = === 5. ( )( )226 13 56 13 5 03 1 2 5 0x xx xx x+ =+ = + = 3 1 0or2 5 01 5or3 2x xx x = + == = 6.( )( )223 4 53 4 12 517 0x xx x xx x+ =+ = = ( ) ( ) ( )( )( )21 1 4 1 172 11 692x == SSM: Intermediate AlgebraCumulative Review Chapters 1-10 325 7. ( ) ( )3 64ln 2 ln 8 9 x x =( ) ( )( ) ( )( )43 612 612666 99696ln 2 ln 8 9ln 16 ln 8 916ln 98ln 2 9223.99272x xx xxxxx eexex = = = ==== 8. ( )( )( )36 3 5 526 3 57193219log2ln 19/ 2ln 32.0492xxxx ===== 9.( ) ( )( )( )2223 2 5 4 36 15 4 6 912 20 010 2 0x xx x xx xx x + = + = + + = = 10 0or 2 010or 2x xx x = == = 10.( )44 4log 81 48133bbbb==== 11.( )( )21 22 3 61 23 2 2 3x xx x x xx xx x x x =+ = + + ( )( )( )( )( )( )( )( )( )( )( ) ( )( )( )( )( )( )( )( )2 22 2223 2 2 13 2 3 2 3 21 3 43 2 3 24 1 303 22 3 503 22 3 5 0 3, 22 5 1 0x x x xx x x x x xx x xx x x xx x xx xx xx xx x xx x + = + + + + = + + + = + = + = + =

2 5 0or 1 05or 12x xx x = + == = 12.( )( )( )2225 3 2 7 175 3 2 103 2 23 2 23 2 22 23xxxxxx + = = = = = = 13.( ) 20 4 7 2 920 4 14 6318 434318x xx xxx = + = + == Cumulative Review Chapters 1-10SSM: Intermediate Algebra 326 14. ( ) ( )2 21 2 5 11 1 2 51 1 2 51 1 2 2 5 2 5x xx xx xx x x+ + + + + + + + ( )( )( )( )( )222225 2 2 55 2 2 510 25 4 2 510 25 8 2018 45 015 3 0x xx xx x xx x xx xx x + + + + + 15 0or 3 015or 3x xx x Check15 x ( ) ( )??15 1 2 15 5 11 1false+ Check3 x ( ) ( )?3 1 2 3 5 11 1true+ The solution is 3. 15.( ) ( )( )( )( )( )6 661222log 3 log 1 1log 3 1 13 1 63 3 63 3 6 02 02 1 0x xx xx xx xx xx xx x+ 1 ] +

2 0or 1 02or 1x xx x + 1 x will not work in the original equation. The solution is 2. 16. 22223 5 1 03 5 153 135 25 253 1 33 36 36x xx xx xx x + _ , _ + + , 2225 753 16 365 3936 365 136 365 136 365 136 65 136xxxxxx _ + , _ , _ , t tt 17.2 4 05 3 7x yx y+ + Multiply the first equation by 3 and the second equation by 4. 6 12 020 12 28x yx y + Add the two equations and solve the result for x. 14 282xx Substitute this result for x in the first equation. ( ) 2 2 4 04 4 04 41yyyy+ + The solution to the system is( ) 2, 1 . 18.3 94 2 2y xx y ++ Substitute the first equation for y in the second equation. ( ) 4 2 3 9 24 6 18 210 202x xx xxx+ + + + SSM: Intermediate AlgebraCumulative Review Chapters 1-10 327 Substitute this value for x into the first equation. ( )3 93 2 96 93y x = += += += The solution to the system is( ) 2, 3 . 19.( ) 5 2 3 5 1 2 45 6 10 1 2 416 6 2 42 147x xx xx xxx + + + Interval:( ] , 7 0 7 20. ( ) ( )( )4 22 3 5 2 8 12 10 48 10 12 418 818 83 6 81 36291629162916b c b c b c b cb cb cb c + + = === 21. 4 71 31/ 2 4/ 33 32 43/ 4 7/ 34 7 2 33 3 4 41/ 4 3/ 31/ 48 45 10454545b cb cb cb cb ccb + ==== 22. ( ) ( )223 53 9 5 49 4212 924 1814 23 212349b cb cb cb cb c = = = 23. ( )( )1 12 2a bab a bb aaa bbab aba baba b a bab = = = += 24. 3 223 8 2 18 3 4 2 2 9 23 4 2 2 9 23 2 2 2 3 26 2 6 20x x x x x x xx x x xx x x xx x x x = = = = = 25. 7 66312 4 34 32 3x x xx xx x= == 26.( ) ( ) ( )( )( )21 204 4204 4545 7 5 7 5 75 7 5 75 7 5 7x x xx xx x = = = 27. 3333 2 333 23 23 23 24 4444x xxxxxx xxx== == 28. ( )( )225 4 5 4 3 23 2 3 2 3 25 3 4 3 5 2 4 23 215 12 10 89 415 22 89 4x x xx x xx x x xxx x xxx xx = + + + = += += Cumulative Review Chapters 1-10SSM: Intermediate Algebra 328 29. ( ) ( ) ( ) ( )( ) ( )( )( )2 34 9 4 98 278 27352ln 3 3ln 5 ln 3 ln 5ln 9 ln 125ln 9 125ln 1125x x x xx xx xx+ = += += = 30. ( ) ( )( ) ( )( ) ( )5 24 55 220 102010104log 2 5log 2log 2 log 2log 16 log 3216log32log2b bb bb bbbx xx xx xxxx= = = = 31.( ) ( ) ( )( ) ( )2 2 223 5 3 2 3 5 59 30 25x x xx x = += + 32. ( )( )3 4 2 73 2 4 2 3 7 4 76 8 21 286 13 28x xx x x xx x xx x += + = + = + 33. ( )( )( )( )2 33 33 3 3 36 4 32 1 52 2 52 2 2 5 2 52 2 10 10x x xx x xx x x x x xx x x x + += += = 34.( )( )22 23 2 23 24 1 2 34 1 4 2 1 2 4 3 1 34 8 2 12 34 9 10 3x x xx x x x x x xx x x x xx x x = + + = + += + 35. ( )22 22 229 5 159 20 4 59 4 55 15 9 203x xx x x xx x xx x xx + = +=( )( )35xx+ ( )( ) 54xx( )( )15 3xx+( )( )( )3 15 4x xx+ += 36. ( )( ) ( )( )2 23 210 25 7 103 25 5 5 2x xx x x xx xx x x x+ + ++= ( )( )( )( )( )( )( )( )( )( ) ( ) ( ) ( )( ) ( )( ) ( )2 22 22 22223 2 2 55 5 2 5 5 23 6 3 105 2 5 23 6 3 105 22 3 105 2x x x xx x x x x xx x x xx x x xx x x xx xx xx x + = = + += += 37. ( )( )( )( )( )( )( )22 24 7 216 10 4 8 164 7 3 12 3 1 2 4 44x x xx x x xx x xx x x xx x + + = + =( ) 2 3 1 x ( )( ) 7 3 12xx + ( ) ( ) 4 4 x x ( )( )72 2 4xx x=+ SSM: Intermediate AlgebraCumulative Review Chapters 1-10 329 38. ( )( ) ( ) ( )( )( )( )( )2 3 2221 212 27 9 91 29 3 1 9 11 29 31 9xx x x x xxx x x x xxx xx x+++ + + += ++ + + ++= ++ ++ ( )( ) ( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )2 223 1 2 99 3 3 1 1 9 3 91 29 3 1 3 33 3 2 9 189 3 3 12 9 159 3 3 1x x x xx x x x x x x xxx x x x xx x x x x xx x x xx xx x x x + + += ++ + + + + ++= ++ + + + + + + + +=+ + ++ +=+ + + 39. ( )22 2 222 3 24644 4 64 4 43 242xx xxx x x x xxx++ += + + + +++=( ) ( ) 8 8 x x +( ) 3 8 x +( ) ( ) 2 2 x x + +( )( )38 2 x x= + 40.( ) ( )( )22223 3 73 6 9 73 18 27 73 18 34f x xx xx xx x= + = + + = = ( )23 18 34 f x x x = 41. ( ) ( )( )( )( )( )( )3 2224 8 25 504 2 25 22 4 252 2 5 2 5x x xx x xx xx x x += = = + 42. ( )( )( )3 2 22 4 30 2 2 152 5 3x x x x x xx x x = = + 43. ( )( )( )2 26 2 20 2 3 102 3 5 2x x x xx x+ = + = + 44.( )( )2100 1 10 1 10 1 x x x = +45.( ) 2 3 f =46.When( ) 3 f x = ,0 x =or2 x = . 47.When( ) 4 f x = ,1 x = . 48.There are no values of x such that( ) 5 f x = . 49.The y-intercept is( ) 0, 3 . 50.The x-intercepts are( ) 1, 0 and( ) 3, 0 . 51.The graph is quadratic so the function is of the form( ) ( )2f x a x h k = + . The vertex is ( ) ( ) , 1, 4 h k =so we have( ) ( )21 4 f x a x = + . Choosing another point on the graph,( ) 0, 3 , we can find the value of a. ( )23 0 1 43 41aaa= += + = Thus, the function is( ) ( )221 42 3f x xx x= += + + 52.( ) 5 2xy =44yx84 Cumulative Review Chapters 1-10SSM: Intermediate Algebra 330 53.( )23 4 y x = 444 4yx 54. 345y x = +444 4yx 55.( ) 3 2 2 26 3 6 24 7 8724x y y xx y y xy xy x = + = + = += 444 4yx 56.2 5 4 y x = + 444 4yx 57. 1153xy = 42 0yx1 04 58. 22 5 1 y x x = + 444 4yx 59.( ) ( )222 3 5 12 6 5 52 11 5x x y xx x y xy x x + = + + = += + + 168yx4 8 60.( ) 3, 2 and( ) 2, 5 ( )5 2 72 3 5m = = Using the slope m and the point( ) 3, 2 , we get: ( )72 35115y mx bbb= += += The equation of the line is 7 115 5y x = or 7 5 11 x y + = . SSM: Intermediate AlgebraCumulative Review Chapters 1-10 331 61.( ) 3, 95and( ) 6,12We want to fit the model( )xy ab = . Plug both points into the equation. ( )( )369512abab== Divide the second equation by the first. 63331295129512950.502ababbbb=== Substitute the point( ) 3, 95into the equation ( ) 0.502xy a = . ( )3395 0.50295750.9520.502aa== The equation is( ) 750.92 0.502xy = . 62.( ) ( ) ( ) 2,1 , 3, 6 ,and4,15Substitute the points into the equation 2y ax bx c = + + . ( ) ( )( ) ( )( ) ( )2221 2 26 3 315 4 4a b ca b ca b c= + += + += + + Rewrite as: 4 2 19 3 616 4 15a b ca b ca b c+ + =+ + =+ + = Multiply the first equation by1 and add to both the second and third equations. 4 2 15 512 2 14a b ca ba b+ + =+ =+ = Multiply the second equation by2 and add to the third equation. 4 2 15 52 4a b ca ba+ + =+ == Solve the third equation for a. 2 42aa== Substitute this value into the second equation and solve for b. ( ) 5 2 510 55bbb+ =+ == Substitute the values for a and b into the first equation and solve for c. ( ) ( ) 4 2 2 5 18 10 13ccc+ + = + == The equation is 22 5 3 y x x = + . 63.( ) 2, 5and( ) 6,17Substitute the points into the equation y a x b = + . 5 217 6a ba b= += + Rewrite as: 1.4142 52.4495 17a ba b+ =+ = Multiply the first equation by1 and add to the second equation. 1.4142 51.0353 12a ba+ == Solve the second equation for a. 1.0353 1211.591aa= Substitute this value into the first equation and solve for b. ( ) 1.4142 11.591 511.392bb+ = The equation is roughly11.591 11.392 y x = . 64.a.Linear: ( ) f x mx b = +The y-intercept is( ) 0, 2so2 b = . 4 221 0m= = ( ) 2 2 f x x = + Exponential: ( )xg x ab = The y-intercept is( ) 0, 2so2 a = . Cumulative Review Chapters 1-10SSM: Intermediate Algebra 332 Now plug in the point( ) 1, 4 . ( )14 24 22bbb=== ( ) ( ) 2 2xg x = Quadratic: ( )2h x ax bx c = + +Answers may vary. One possibility: Let2 a =so we have( )22 h x x bx c = + + . Plug in the point( ) 0, 2 . ( ) ( )22 2 0 02b cc= + += ( )22 2 h x x bx = + +Plug in the point( ) 1, 4 . ( ) ( )24 2 1 1 24 2 24 40bbbb= + += + += += ( )22 2 h x x = +b.

fgh 65.( )( )43 33log 81 log 34log 34 14=== = 66. ( ) ( )1/ 2log log1log211212b bbb bb=== = 67.( )( )4log 0.0001 log 104log104 14== = = 68.( )3log8log 8 1.8928log3= 69.( )2log y g x x = =Switch x and y, and solve for y. 2log2xx yy== ( )12xg x=70.( ) 4 7 y f x x = = Switch x and y, and solve for y. 4 74 71 74 4x yy xy x= = = ( )11 74 4f x x= 71.( )( )( )23 37 5 2 35x xf xx x x x = = + The domain is all real numbers except 7 and 5 since these values make the denominator equal zero. Domain:, 5, 7 x 72.This is a geometric sequence with a common ratio of 12r = .Since 196 a = , the sequence formula is 11962nna = . SSM: Intermediate AlgebraCumulative Review Chapters 1-10 333 73.This is a geometric sequence with a common ratio of3 r = . Since 12 a = , the 10th term is ( )( )( )10 11092 32 32 19, 68339, 366a==== 74.This is an arithmetic sequence with a common difference of4 d = . Since 186 a = and 170na = , we have ( )( )( )11170 86 4 1256 4 164 165na a d nnnnn= + = + = = = The last term in the sequence is term 65. 75.This is a geometric series with 12r =and 198, 304 a = . ( )111113 9830423 198304 23 1ln 1 ln98304 23ln9830411ln215 116nnnna a rnnnn= = = = = = = There are 16 terms in the series. ( )161161611198304 12112196, 605a rSr= == The sum of the series is 196,605. 76.This is an arithmetic sequence with3 d =and 111 a = . ( )( )( )11182 11 1 3171 3 157 158na a n dnnnn= + = + = = = There are 58 terms in the series. ( )( )( )( )116258 11 182229 11 18229 1935597nnna aSS+=+== +== 77.a.Start by plotting the data. A linear model may be reasonable. Use a linear regression to find( ) Bt . The model is 1.24 6.82 y t = + . b.Use a quadratic regression to find( ) Rt .

The model is20.97 0.29 177.04 y t t = + + . c.( )2124 6820.97 0.29 177.04tPtt t+=+ + d.If( ) Rtis increasing at a faster rate than ( ) Bt , then the ratio( )( )( )BtPtRt=will decrease. Cumulative Review Chapters 1-10SSM: Intermediate Algebra 334 e. 222124 6823.80.97 0.29 177.043.69 1.10 672.75 124 6823.69 122.90 9.25 0tt tt t tt t+=+ ++ + = + = ( ) ( ) ( )( )( )2122.9 122.9 4 3.69 9.252 3.69122.9 15240.947.3833.38or 0.075tt t === = 3.8% of recreational expenditures will consist of book sales in 2013. 78.a.We are given two data points for Indias population:( ) 0, 690and( ) 20,1014The y-intercept is( ) 0, 690so690 b = . The slope of the line can be found by using the two given points. 1014 69016.220 0m= = The linear model is ( ) 16.2 690 Lt t = + . b.We now fit the model ( )xy ab = . Since the y-intercept is( ) 0, 690 , we have690 a = . Substitute the point( ) 20,1014into the equation( ) 690xy b =and solve for b. ( )2020201014 690101469010141.019690bbb=== The model is ( ) ( ) 690 1.019tEt = . c.( ) ( )( ) ( )7070 16.2 70 690 182470 690 1.019 2576.57LE= + == According to the linear model, Indias population will be 1.824 billion in 2050. According to the exponential model, Indias population will be 2.577 billion in 2050. d.( ) ( ) 70 70 2.577 1.824 0.753 E L = =The exponential models predicted population exceeds that of the linear model by 753 million. 79.a.Start by plotting the data. A square root model of the form seems reasonable. The model is y a x b = + . The y-intercept is( ) 0, 9.4so9.4 b = . Substitute the point( ) 21, 46 into the equation9.4 y a x = +and solve for a. 46 21 9.421 36.67.99aaa= += The model is( ) 7.99 9.4 f t t = + . b.( ) 23 7.99 23 9.4 47.7 f = + About 47.7% of passenger vehicles sold in 2002 were light trucks. c.49 7.99 9.47.99 39.639.67.9924.56tttt= +== In 2004, 49% of passenger vehicles sold will be light trucks. 80.a.Start by plotting the data. A quadratic model may be reasonable. Use a quadratic regression. The model is ( )20.0058 0.396 10.023 f t t t = + . b.( ) ( ) ( )249 0.0058 49 0.396 49 10.0234.54f = + In 2009, will be about 4.5 thousand men studying to be priests in graduate programs. SSM: Intermediate AlgebraCumulative Review Chapters 1-10 335 c. 225 0.0058 0.396 10.0230.0058 0.396 5.023 0t tt t= + + = ( ) ( ) ( )( )( )20.396 0.396 4 0.0058 5.0232 0.005851.44or 16.84tt t = There were 5 thousand men studying to be priests in a graduate program in 1977 and there will be that many again in 2011. d.The function f is decreasing for34.14 t . The number of men studying to be priests in a graduate program was declining up until 1994 and it has been increasing since 1994 (and will continue to do so).