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Mathematical Applications For The Management, Life, And Social Sciences, 11th Edition Solution Manual

Mathematical Applications For The Management, Life, And Social Sciences, 11th Edition Solution Manual makes textbook problem-solving easy with a comprehensive guide that explains every step.

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    Chapter 0: Algebraic Concepts 1 Exercises 0.1 __________________________________________________________________ 1. 12 {1, 2, 3, 4,...}  2. 5  { x : x is a natural number greater than 5} 3. 6 {1, 2, 3, 4, 5}  4. 3  5. {1, 2, 3, 4, 5, 6, 7} 6. {7, 8, 9} 7. { x: x is a natural number greater than 2 and less than 8} 8. { x : x is a natural number greater than 6} 9. A   since  is a subset of every set. A B  since every element of A is an element of B . B B  since a set is always a subset of itself. 10. A   since  is a subset of every set. A B  since every element of A is an element of B . B B  since a set is always a subset of itself. 11. No. c A  but . c B  12. No. 12 A  but 12 . B  13. D C  since every element of D is an element of C . 14. E F  since every element of E is an element of F . 15. A B  and B A  . (Also A B  .) 16. and . (Also .) D F F D D F    17. Yes. A B  and B A  . Thus, A B  . 18. A D  19. No. D E  because 4 E  and 4 . D  20. F = G 21. A and B are disjoint since they have no elements in common. B and D are disjoint since they have no elements in common. C and D are disjoint. 22.  23. {4, 6} A B   since 4 and 6 are elements of each set. 24. { , , } A B a d e   since a , d , and e are elements of each set. 25. A  B =  since they have no common elements. 26. {3} A B   27. {1, 2, 3, 4, 5} A B   28. { , , , , , , , } A B a b c d e i o u   29. {1, 2, 3, 4} A B   or . A B B   30. A B   { x : x is a natural number not equal to 5} For problems 31 - 42, we have U = {1, 2, 3, . . . , 9, 10}. 31. {4, 6, 9, 10} A   since these are the only elements in U that are not elements of A . 32. {1, 2, 5, 6, 7, 9} B   since these are the only elements in U that are not elements of B . 33. {1, 2, 5, 6, 7, 9} B   {1, 2, 5,7} A B    34. {4, 6, 9, 10} {1, 2, 5, 6, 7, 9} {6, 9} A B A B         35.   1, 2, 3, 4, 5, 7, 8, 10 A B   ( ) {6, 9} A B    36. {3, 8} ( ) {1, 2, 4, 5, 6, 7, 9, 10} A B A B      37. {4, 6, 9, 10} A   {1, 2, 5, 6, 7, 9} B   {1, 2, 4, 5, 6, 7, 9, 10} A B    

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    Chapter 0: Algebraic Concepts 2 38. {4, 6, 9, 10} {3, 4, 8, 10} {3, 4, 6, 8, 9, 10} ( ) {1, 2, 5, 7} A B A B A B           39. {1, 2, 5, 6, 7, 9} B   {1, 3, 5, 7, 9} C         1, 2, 3, 5, 7, 8 1, 2, 5, 6, 7, 9 1, 2, 5, 7 A B          1, 2, 3, 5, 7, 9 A B C      40. {1, 3, 5, 8, 7, 2} {1, 2, 5, 6, 7, 9} {1, 3, 5, 7, 9} {1, 2, 3, 5, 6, 7, 9} ( ) {1, 2, 3, 5, 7} A B C B C A B C               41. {1, 2, 5, 6, 7, 9} B         1, 2, 3, 5, 7, 8 1, 2, 5, 6, 7, 9 1, 2, 5, 7 A B              3, 4, 6, 8, 9, 10 2, 4, 6, 8, 10 4, 6, 8, 10 A B C        42. {2, 3, 4, 6, 8, 10} ( ) {2, 3, 8} B C A B C      For problems 43 - 46, we have U = {1, 2, 3, . . . , 8, 9}. 43. A – B = {1, 3, 7, 9} – {3, 5, 8, 9} = {1, 7} 44. A – B = {1, 2, 3, 6, 9} – {1, 4, 5, 6, 7} = {2, 3, 9} 45. A – B = {2, 1, 5} – {1, 2, 3, 4, 5, 6} =  or   46. A – B = {1, 2, 3, 4, 5} – {7, 8, 9} = {1, 2, 3, 4, 5} 47. a. L = {2000, 2001, 2004, 2005, 2006, 2007, 2010, 2011, 2012} H = {2000, 2001, 2006, 2007, 2008, 2010, 2011, 2012} C = {2001, 2002, 2003, 2008, 2009} b. no c. C  is the set of all years when the percentage change from low to high was 35% or less. d. H  = {2002, 2003, 2004, 2005, 2009} C  = {2000, 2004, 2005, 2006, 2007, 2010, 2011, 2012} H C    = {2000, 2002, 2003, 2004, 2005, 2006, 2007, 2009, 2010, 2011, 2012}. H C    is the set of years when the high was less than or equal to 11,000 or the percent change was less than or equal to 35%. e. L  = {2002, 2003, 2008, 2009} L   C = {2002, 2003, 2008, 2009}. L   C is the set of years when the low was less than or equal to 8,000 and the percent change was more than 35%. 48. a. A = {O, L, P} B = {L, P} C = {O, M, P} b. B A  c. {O, P}; A C   this is the set of cities with at least 2,000,000 jobs in 2000 or 2025 and projected annual growth rates of at least 2.5%. d. B  is the set of cities with fewer than 1,500,000 jobs in 2000. 49. a. From the table, there are 100 white Republicans and 30 non-white Republicans who favor national health care, for a total of 130. b. From the table, there are 350 + 40 Republicans, and 250 + 200 Democrats who favor national health care, for a total of 840. c. From the table, there are 350 white Republicans, and 150 white Democrats and 20 non-whites who oppose national health care, for a total of 520.

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    Chapter 0: Algebraic Concepts 3 50. a. From the table, 250 white Republicans and 150 white Democrats oppose national health care, for a total of 400. b. From the table, there are 750 whites and there are 20 non-whites who oppose national health care. The total of this group is 770. c. From the table, there are 200 non-white Democrats who favor national health care. 51. a. The key to solving this problem is to work from "the inside out". There are 40 aides in E  F . This leaves 65 – 40 = 25 aides who speak English but do not speak French. Also we have 60 – 40 = 20 aides who speak French but do not speak English. Thus there are 40 + 25 + 20 = 85 aides who speak English or French. This means there are 15 aides who do not speak English or French. b. From the Venn diagram E  F has 40 aides. c. From the Venn diagram E  F has 85 aides. d. From the Venn diagram E F   has 25 aides. 52. a. There are 14 advertisers in the intersection of the sets. Since 30 advertised in These Times and U.S. News and we already have 14 in the center, 16 advertised in These Times and U.S. News and not in World . Since 26 advertised in World and U.S. News and we already have 14 in the center, 12 advertised in World and U.S. News and not in These Times . Since 27 advertised in World and These Times and we already have 14 in the middle, 13 advertised in World and These Times and not in U.S. News . 60 advertised in These Times and we have already accounted for 43, so 17 advertised in These Times only. 52 advertised in U.S. News and we have already accounted for 42, so 10 advertised in U.S. News only. 50 advertised in World and we have already accounted for 39, so 11 advertised in World only. b. In the union of the 3 publications we have 10 + 16 + 17 + 14 + 12 + 13 + 11 = 93 advertisers. Thus, there are 100 – 93 = 7 who advertised in none of these publications. c. There are 17 advertisers in the These Times circle that are not in an intersection. d. In the union of U.S. News and These Times we have 10 + 12 + 16 + 14 + 17 + 13 = 82 advertisers. 53. Since 12 students take M and E but not FA , and 15 take M and E , 3 take all three classes. Since 9 students take M and FA and we have already counted 3, there are 6 taking M and FA which are not taking E . Since 4 students take E and FA and we have already counted 3, there is only 1 taking E and FA but not taking M also. Since 20 students take E and we already have 16 enrolled in E , this leaves 4 taking only E . Since 42 students take FA and we already have 10 enrolled in FA , this leaves 32 taking only FA . Since 38 students take M and we already have 21 enrolled in M , this leaves 17 taking only M . a. In the union of the 3 courses we have 17 + 12 + 3 + 6 + 32 + 1 + 4 = 75 students enrolled. Thus, there are 100 – 75 = 25 students who are not enrolled in any of these courses. b. In M  E we have 17 + 12 + 3 + 6 + 1 + 4 = 43 enrolled. c. We have 17 + 32 + 4 = 53 students enrolled in exactly one of the courses. U E F 25 40 20 15 U U.S. News 10 12 13 11 These Times World 16 17 14 7 U Math 17 12 1 4 Fine Arts Economics 6 32 3 25

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    Chapter 0: Algebraic Concepts 4 54. Start by filling in the parts of the diagram for AL, since we have more information about it. 21 liked AL only. Since 30 liked AL but not PT, 9 liked AL or PT exclusively. 25 liked PT or AL but not DH, and 63 liked AL. That leaves 63 – (21 + 25 + 9) = 8 in the intersection of all 3. Since 18 liked PT and DH, only 10 liked PT and DH but not AL. Since 27 liked DH, 27 – (9 + 8 + 10) = 0 liked DH only. And since 58 liked PT, 58 – (25 + 8 + 10) = 15 liked PT only. a. The number of students that liked PT or DH is 25 + 15 + 9 + 8 + 10 + 0 = 67. b. The number that liked all three is 8. c. The number that liked only DH is 0. 55. a. and b. c. : 34%; A  : 9% B  ; : 38% O  ; : 3% AB  ; : 7% O  ; : 6% A  ; : 2% B  ; : 1% AB  Exercises 0.2 __________________________________________________________________ 1. a. Note that 1 10 10             , where  is irrational and 1 10  is rational. The product of a rational number other than 0 and an irrational number is an irrational number. b. – 9 is rational and an integer. c. 9 3 3 3 1   . This is a natural number, an integer, and a rational number. d. Division by zero is meaningless. 2. a. 0 0 6  is rational and an integer. b. rational c. rational d. rational 3. a. Commutative b. Distributive c. Associative d. Additive Identity 4. a. Multiplicative Identity b. Additive Inverse c. Multiplicative Inverse d. Commutative 5. – 6 < 0 6. 2 > – 20 7. – 14 < – 3 8. 3.14   9. 1 1 0.333 0.3333 3 3          10. 1 1 5 3 2 6   11. | 3 | | 5 | | 3 5 |      12. | 9 3 | | 9 | | 3 |      (12 = 12) 13. 2 2 3 10 2 3 20 9 20 11           14. 2 ( 3) 10 2 9 20 29       U AL 21 9 10 0 PT DH 25 15 8 22 U AL 21 9 10 0 PT DH 25 15 8 22 U A  A  B  O  B  AB  AB  Rh  A B O 

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    Chapter 0: Algebraic Concepts 5 15. 2 4 2 4 4 8 4 2 2 2      16. 2 2 (4 2) 6 36 18 2 2 2     17. 16 ( 4) 16 4 20 2 8 ( 2) 8 2 10          18. ( 5)( 3) ( 2)(3) 15 ( 6) 15 6 9 2 7 7 21 3 7                  19. | 5 2 | | 7 | | 3 | | 7 | 3 7 4 | 5 2 | | 3 | 3 3            20. 2 2 | 3 | 4 11|| | 3 | 7 || | 25 9 | | 5 3 | | 3 7 | |16 | | 4 | 16 4 1 16 4                    21. 2 2 ( 3) 2 3 6 9 6 6 9 3 4 4 3 3 4 2 3              22. 2 2 6 4( 3)( 2) 36 ( 12)( 2) 6 36 4 6 6 4 36 24 6 9 12 3 4                   23. 2 2 4 5 2 3 16 5 6 17 17 5 16 11 11 5 4               24. 2 2 3 2(5 2) 3 2 3 3 1 4 4 3 3 ( 2) 2 3               25. The entire line 26. The interval notation corresponding to 0 x  is [0, )  . 27. (1, 3]; half-open interval 28. [ – 4, 3]; closed interval 29. (2, 10); open interval 30. [2, );  half-open interval 31. 3 5 x    32. x > – 2 33. x > 4 34. 0 5 x   35. ( , 4) ( 3, ) ( 3, 4)       36. [ – 4, 17)  [ – 20, 10] = [ – 4, 10] 37. x > 4 and 0 x  = (4, )  38. x < 10 and x < – 1 is x < – 1 or ( , 1)   . 39. [0, ) [ 1, 5] [ 1, )       40. ( , 4)   (0, 2) = ( , 4)  41. ( ,0) (7, )    42. x > 4 and x < 0 The intersection is the empty set. 43. – 0.000038585 44. 0.404787025 45. 9122.387471 1 2 3 4 5 3  1  0 2  4 6 8 10 12 6  2  0 4  1 2 3 4 5 6 1  0 2  1 2 3 4 5 6 1  0 2  1 2 3 4 5 6 1  0 2  1 2 3 4 5 6 1  0 2  1 2 3 4 5 6 1  0 7 1 2 3 4 5 6 1  0 2 

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    Chapter 0: Algebraic Concepts 6 46. 11.80591621 47. 6 2500 2500 3240.184509 0.771561 [(1.1 ) 1]    Equation (2) gives     2 0.00454 13 0.126 13 0.271 2.68 billion y     48. 1591.712652 49. a. $300.00 $788.91 $1088.91   b. Federal withholding = 0.25(1088.91 – 54.45) = $258.62 c. Retirement: 0.05(1088.91) = $54.45 State tax = Retirement = $54.45 Local tax = 0.01(1088.91) = $10.89 Federal tax = $258.62 (from b. above) Social Security and Medicare tax d = 0.0765(1088.91) = $83.30 Total Withholding = $461.71 Take-home = 1088.91 – 461.71 = $627.20 50. a. 2010 2000 10 t    b.     2 5.03 10 100 10 1380 E    $2883 billion  c.     2 2015 2000 15 50.3 15 100 15 1380 $4011.75 billion t E        51. a. Equation (1) is more accurate. d Equation (1) gives d   0.207 13 0.000370 2.69 billion y    b. For 2018, Equation (1) gives   0.207 18 0.000370 3.73 billion y    For 2018, Equation (2) gives     2 0.00454 18 0.126 18 0.271 4.01 billion y     52. a.       2.31 10.5 31.26 55.515 inches Upper: 1.05 55.515 58.29 inches Lower: 0.95 55.515 52.74 inches 52.74 58.29 H H        b.       2.31 5.75 31.26 44.5425 inches Upper: 1.05 44.5425 46.77 inches Lower: 0.95 44.5425 42.32 inches 42.32 46.77 H H        53. a. $82, 401 171,850; $171,851 $373, 650; $373, 650 I I I      b. $4681.25 for $34, 000 $16, 781.25 for $82, 400 T I T I     c.   4681.25, 16,781.25 Exercises 0.3 __________________________________________________________________ 1.        4 4 4 4 4 4 256        2. 3 5 1 5 5 5 125         3. 6 2 1 2 2 2 2 2 2 64            4.         5 2 2 2 2 2 2 32          5. 2 2 1 1 3 9 3    6. 1 1 6 6   7.   2 3 3 3 9 1 2 2 2 4                     8. 3 3 3 2 2 8 3 27 3         9. 1.2 4 2.0736 x y  10. 3.7 3 50.653 x y    11. 1.5 5 0.1316872428 x y   12. 0.8 9 7.450580597 x y    

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    Chapter 0: Algebraic Concepts 7 13. 5 3 5 3 8 6 6 6 6     14. 4 2 4 2 1 7 8 8 8 8 8       15. 8 8 9 1 9 10 1 10 10 10 10      16. 8 8 3 5 3 7 7 7 7    17. 4 7 4 ( 7) 3 3 ( 3) 0 3 3 3 9 9 9 9 9 9 1 9 9 9                 18.   4 4 4 4 1 3 2 3 1 2 3 5 5 5 5 5 5 5 5 5          19.   3 3 3 3 9 3 3 3    20.   2 3 ( 3) ( 2) 6 2 2 2        21. 2 2 2 3 9 3 2 4                22. 4 4 4 2 5 5 5 2 2                         23. 3 3 3 3 1 1 1 1 x x x x            24. 4 4 1 x x   25. 2 0 2 2 1 1 x xy z x y y      26. 1 0 2 2 2 1 1 1 4 1 4 4 x y y y       27. 3 4 3 4 7 x x x x     28. 5 5 1 6 a a a a     29. 5 3 5 3 2 2 1 x x x x x         30. 5 2 5 ( 2) 7 7 1 y y y y y           31. 8 8 4 4 4 x x x x    32. 5 5 ( 1) 5 1 6 1 a a a a a        33.   5 5 7 12 7 y y y y      34. 3 3 ( 4) 3 4 1 4 y y y y y y            35.   3 4 3 4 12 x x x    36. 3 2 3 ( 2) 6 6 1 ( ) y y y y        37. 2 2 2 ( ) xy x y  38. 3 3 3 3 (2 ) 2 8 m m m   39.   4 4 5 4 5 4 20 5 2 2 16 16 x x x x           40. 3 3 3 3 3 3 3 9 8 8 512 512 ( ) a a a a           41.   8 4 2 4 8 4 4 2 2 16 x x y x y y       42.     3 3 5 3 5 5( 3) 3 15 15 32 ( 32) 1 ( 32) 1 32768 1 32768 x x x x x               

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    Chapter 0: Algebraic Concepts 8 43.    3 2 5 4 3 5 2 4 2 2 2 2 8 2 16 16 16 a b a b a b a b a b              44.    2 1 3 1 2 ( 3) 1 ( 1) 1 2 2 2 3 2 6 6 1 1 6 6 m y m y m y m y m y my                                  45.     2 2 1 2 2 2 2 2 2 2 2 2 y x x x y x x x y xy         46.     3 2 3 2 5 4 5 4 3 2 5 4 8 2 8 8 2 2 8 2 4 a b c a b c a b a b a b c a b c a b               47. 3 3 9 2 6 9 6 9 6 1 1 1 x x y y x y x y                48. 3 2 2 3 ( 2)( 3) 6 3 6 3 3 3 6 3 ( ) 1 x x x x y x y y y y x y                        49. 3 3 3 2 1 4 2 6 4 18 12 4 3 0 6 4 2 6 a b c b a c a c a b c a c b b                                  50. 2 2 1 40 1 4 40 ( 10) 2 4 10 2 4 4 2 (1/ 2) x y x y x y                               5 30 2 2 5 2 30 2 ( 5)( 2) ( 30)( 2) 2 10 60 10 60 (16 ) (16) ( ) ( ) 1 (16) 1 256 256 x y x y x y x y x y                   51. a. 2 2 2 2 2 2 4 2 1 1 1 1 1 2 2 (2 ) (2 ) 4 2 x x x x x x x         b.         2 2 2 2 2 2 4 2 1 1 1 1 1 4 4 16 2 2 2 x x x x x x x       c. 2 2 2 2 4 2 1 1 1 2 2 2 x x x x x      d.     2 2 2 2 2 2 2 1 1 2 2 2 4 8 2 x x x x x x          52. a.   1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 x x x x x           3 4 4 1 2 8 x x     b. 1 2 1 1 2 2 2 4 2 4 2 1 2 2 2 4 x x x x x            c.        1 2 2 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 x x x x x x                  1 2 2 2 4 2 2 x x        d. 2 1 1 ( 2)( 1) 1 2 2 2 2 (2 ) 2 2 2 2 2 x x x x x x         1 1 2 2 1 2 4 x      53. 1 1 x x   54. 2 2 1 x x   55.   3 3 3 3 2 2 8 x x x   56. 2 2 2 2 (3 ) 3 9 x x x  

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    Chapter 0: Algebraic Concepts 9 57. 2 2 2 1 1 1 1 4 4 (4 ) x x x     58. 4 4 4 3 3 1 3 2 2 (2 ) x x x     59. 3 3 3 3 1 2 8 2 x x x            60. 2 2 2 2 2 ( ) 1 3 9 9 3 x x x x            61. 5 5 1200, 0.12, 5 (1 ) 1200(1 0.12) 1200(1.12) $2114.81 n P i n S P i          – 2114.81–1200 $914.81 I S P    62. 1800 P  , 0.10 i  , 7 n  7 7 (1 ) 1800(1 0.10) 1800(1.10) 1800(1.9487171) $3507.69 n S P i        3507.69 1800 $1707.69 I S P      63. 6 6 5000, 0.115, 6 (1 ) 5000(1 0.115) 5000(1.115) $9607.70 n P i n S P i          – 9607.70 – 5000 $4607.70 I S P    64. 800 P  , 0.105, i  20 n  20 20 (1 ) 800(1 0.105) 800(1.105) 5892.99 n S P i       5892.99 800 $5092.99 I S P      65. S = 15,000, n = 6, i = 0.115       6 6 1 15, 000 1 0.115 15, 000 1.115 $7806.24 n P S i          66. 80,000, 20, 0.105 S n i    20 20 (1 ) 80, 000(1 0.105) 80, 000(1.105) $10,860.37 n P S i          67.   492.4 1.070 t I    58 Year 1980 2000 2008 . -value 20 40 48 . Income $1905 $7373 $12, 669 (in billions) . 492.4 1.070 $24,922 billion t I   a b c 68. a. 2019 2010 9 t    b.   9 0.012 1.75 1.8 billion cubic feet y   c.   12 0.012 1.75 9.9 billion cubic feet y  

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    Chapter 0: Algebraic Concepts 10 69.   1095 1 10.12 1.212 t y    . Year 1990 2003 2012 value 10 23 32 Predicted number of 442 976 1072 endangered species t  a b. Year 2020: 40 t  ;   40 1095 1090 1 10.12 1.212 y     Increase between 2007 and 2020 is 1090 1037 53   species c. Two possibilities might be more environmental protections and the fact that there are only a limited number of species. d. There are only a limited number of species. Also, below some threshold level the ecological balance might be lost, perhaps resulting in an environmental catastrophe (which the equation could not predict). To find the upper limit, which is 1095, compute y for large t -values: Year 2040 2100 2200 value 60 120 220 Predicted number of 1094.9 1095 1095 endangered species t  70.   249.6 1 1.915 1.028 t p    . Year 1980 2000 2020 value 30 50 70 U.S. population (age 20 - 64) 135.92 168.49 195.44 in millions t  a b. Year 2025: 75 t  ;   75 249.6 201.07 million 1 1.915 1.028 p     Year 2045: 95 t  ;   95 249.6 219.15 million 1 1.915 1.028 p     The increase from 2025 and 2045 is predicted to be 219.15 201.07 18.08   million. This is less than the predicted 28.4 million increase from 2000 to 2020. c. It is reasonable for a formula such as this to have an upper limit that cannot be exceeded because there are limited resources and space. To find the upper limit, which is 249.6 million, compute p for large t -values: Year 2150 2350 2450 value 200 400 500 U.S. population (age 20 - 64) 247.7 249.6 249.6 in millions t  71.   738.1 1.065 t H  a. 10 t  corresponds to 2000.

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    Chapter 0: Algebraic Concepts 11 b.   10 738.1 1.065 $1385.5 billion H   c.   20 738.1 1.065 $2600.8 billion H   d.   28 738.1 1.065 $4304.3 billion H   Exercises 0.4 __________________________________________________________________ 1. a. Since 2 16 256 3 9        we have 256 16 3 9 3 5. 3   b. 1.44 1.2  2. a.   5 5 5 3 3 3 3/5 3 3 5 32 1 32 32 (32) 32 (2) 8               b. 4 5 4 16 1048576    The square root of a negative number is not real. 3. a.   3 3/4 3 4 16 16 2 8    b.   3 3/2 ( 16) 16      The square root of a negative number is not real. 4. a.   1/3 1/3 3 1 1 27 27 3 27          b.   3 3 5 3 5 32 32 2 8    5. 2 2/3 2/3 2 3 8 27 27 3 9 27 8 8 2 4                                6. 3 3/2 3 4 4 2 8 9 9 3 27                        7. a.   2 2/3 2 3 64 64 4 16    b.         2/3 2/3 2 3 2 1 1 64 64 64 1 1 16 4          8. a.   2/3 2/3 2 2 3 1 1 1 1 64 16 64 4 64      b.     2 2 2/3 3 64 64 4 16        9.   4/9 4 9 (6.12) 6.12 2.237   10.   1/12 12 4.96 4.96 1.1428   11. 3 3/2 m m  12. 3 5 5/3 x x  13.   1/4 4 2 5 2 5 2/4 5/4 1/2 5/4 m n m n m n m n    14. 5 3 3/5 x x  15. 1 2 2 2 x x  16. 1 4 4 12 12 x x  17. 6 7/6 7 x x  18. 11/5 11 5 y y  19. 5/4 5/4 4 5 1 1 1 1 4 4 4 x x x              20. 3 5/3 5 3 5 1 x x x        21. 1/4 1/2 (1/4) (1/2) 3/4 y y y y     22. 2/3 1/5 (2/3) (1/5) (10/15) (3/15) 13/15 x x x x x       23. 3/4 4 (3/4) (16/4) 19/4 z z z z    

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    Chapter 0: Algebraic Concepts 12 24. 2/3 2 ( 2/3) 2 ( 2/3) (6/3) 4/3 x x x x x          25. 3/2 1 ( 3/2) (2/2) 5/2 5/2 1 y y y y y          26. 2 5/3 2 (5/3) ( 6/3) (5/3) 1/3 1/3 1 z z z z z z            27. 1 1 2 3 3 3 3 3 2 3 x x x x x                   28. 1/2 ( 1/2) ( 3/2) ( 1/2) (3/2) 2/2 3/2 x x x x x x            29. 5/2 ( 5/2) ( 2/5) ( 25/10) (4/10) 2/5 y y y y          21/10 21/10 1 y y    30. 4/9 (4/9) (1/12) (16/36) (3/36) 13/36 1/12 x x x x x      31. 2/3 3/4 (2/3)(3/4) 2/4 1/2 ( ) x x x x    32. 4/5 3 (4/5)(3) 12/5 ( ) x x x   33. 1/2 2 1 1 ( ) x x x     34. 2/3 2/5 ( 2/3)( 2/5) 4/15 ( ) x x x       35. 4 2 64 8 x x  36. 3 6 3 6 3 2 3 3 3 64 64 4 x y x y x y        37. 4 5 4 4 4 4 2 2 128 64 2 64 2 8 2 x y x y y x y y x y y        38. 3 3 3 5 8 5 8 3 3 3 3 2 2 2 2 2 2 3 54 54 3 2 3 2 x z x z x x z z xz x z        39. 8 5 6 3 2 2 3 3 3 6 3 2 2 3 3 3 2 2 2 3 40 8 5 8 5 2 5 x y x y x y x y x y x y x y        40. 5 5 2 32 32 4 2 x y x y x x y       2 4 2 x xy  41. 3 2 5 2 5 2 2 12 3 36 36 6 x y x y x y x y x y x       42. 3 2 2 4 2 4 2 3 3 3 3 3 2 2 3 3 3 3 16 3 48 48 2 6 2 6 x y x y x y x y x x y x xy          43. 5 3 2 4 2 2 2 3 2 63 28 9 7 4 7 3 7 2 7 42 x y x y x y xy x y x y xy x y x y x         44. 10 17 18 11 18 11 9 5 9 5 10 30 300 300 10 3 10 3 xz x z x z x z x z z x z z          45. 3 12 2 10 5 2 12 4 2 9 3 27 x y x y xy xy   46. 7 4 7 4 5 4 17 2 16 17 2 5 4 16 2 2 8 2 2 8 250 250 125 18 9 18 125 9 5 5 3 5 5 3 xy z xy z y z x y x x y y z x y y z x y z y x        47. 4 9 5 4 4 4 8 2 4 17 32 16 2 1 81 3 162 a b b b b b a a a   

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    Chapter 0: Algebraic Concepts 13 48. 3 3 4 3 2 3 4 3 3 3 2 3 2 3 2 2 3 3 16 16 128 8 128 2 2 x y x y x y y y x y x y          49. 9 9 9 1 ( ) 9 1 1 9 x x x A A A A x x     50. 20 20 1 ( ) 20 1 1 20 x x B B B B x x     51.   /7 7 /7 1 1 7 7 x x x R R R R x x     52.   3 3 1/2 1 3/2 1 3 /2 1 (( ) ) ( ) 3 1 2 2 3 x x x x T T T T T T T T x x       53. 2 3 2 3 6 3 3 3 3 3      54. 5 8 5 8 40 2 10 10 8 8 4 8 8 8 64       55. 2 2 m x m m x mx x x x x mx      56. 3 3 3 2 2 2 2 5 5 5 5 2 2 4 4 x w x x w x x w x x x xw x xw x w     57. 3 3 3 2 2 2 3 3 3 3 3 3 3 5 4 3 2 3 3 m x m m x mx mx x x x x x x      3 2 2 mx x  58. 2 3 3 2 3 3 2 3 4 3 4 4 4 2 2 5 2 3 4 8 4 4 4 y z mx y z mx y z mx yz y z y z y z    59. 2/3 2/3 3 2 2 2 1 2 3 3 3 x x x        60. 3/4 3/4 3/4 4 3 2 2 2 2 3 3 3 3 x x x x          61. 1/2 3/2 3 3 3 x x x x x    62. 1/2 1/3 (1/2) (1/3) (3/6) (2/6) 3 x x x x x x       5/6 x  63. 1/2 3 3 2 2 x x  64. 3 1/3 3 4 4 4 3 3 3 x x x   65. 1/2 1/2 1 1 1 1 2 2 2 x x x     66. 3/2 3/2 3 1 1 1 2 2 2 x x x       67. a. 17/2 17 17 8.5 10 10 2 R I     b. 9.0 10 1,000,000,000 I   c. 9.0 2.1 2011 6.9 1989 10 10 125.9 10 I I    68. a. 10 /10 1/10 10 (10 ) 10 D D D   b. 10 32 1 10 1584.89 I  

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    Chapter 0: Algebraic Concepts 14 c. 10 140 10 10 140 32 108 2 10 32 1 (108)(1/10) 10.8 10 10 10 10 10 10 10 6.31 10 I I         69. a. 5 1000 1 100 r S         b. 5 6.6 1000 1 $1173.26 100 S          70. a. 100 21 21 0.21 , so 29 100 L x   b.   0.21 29 115 78.5 years L   71. a. 100 13/100 13 0.924 0.924 P t t   b. Year Population 2005 5 1.1390 2010 10 1.2464 2045 45 1.5156 2050 50 1.5365 t Change from 2005 to 2010 : 0.1074 billion Change from 2045 to 2050 : 0.0209 billion By 2045 and 2050 the population is much larger than earlier in the 21 st century, and there is a limited number of people that any land can support — in terms of both space and food. 72. a. 50 0.38 38/100 19/50 19 14.32 14.32 14.32 14.32 p t t t t     b. Year Percent of Roads Paved 1970 20 44.7 1980 30 52.1 2000 50 63.3 2010 60 67.9 t Change from 1970 to 1980 : 7.4% Change from 2000 to 2010 : 4.6% The equation estimates a greater percent change from 1970 to 1980 than from 2000 to 2010. There were fewer roads left to be paved from 2000 to 2010. c. When a t -value makes 100%, p  you can be certain that the equation is no longer valid since you cannot pave more than 100% of the roads. 73. 25, 10, k t   0 98 q  / 0 10/25 2/5 (2 ) 98(2 ) 98(2 ) 74 kg t k q q        74. 5600, 10,000, k t   0 40 q  / 0 10,000/5600 25/14 (2 ) 40(2 ) 40(2 ) 11.6 g t k q q        75.       0.03 10 0 2.5 30, 000 2.5 ht P P     0.3 30, 000 2.5 39, 491  

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    Chapter 0: Algebraic Concepts 15 76. 10 x          0.1 0.1 10 1 2000 2 2000 2 2000 2 1 2000 2 $1000 x S          77. a. (0.7) 500(0.02) t N  ; at 0 t  we have 0 (0.7) 1  . Thus, 1 500(0.02) 10. N   b.     5 0.7 500 0.02 N    0.16807 500 0.02 259   Exercises 0.5 __________________________________________________________________ 1. 2 10 3 x x   a. The largest exponent is 2. The degree of the polynomial is 2. b. The coefficient of 2 x is – 1. c. The constant term is 10. d. It is a polynomial of one variable x . 2. 4 9 5 2 7 x x   a. The largest exponent is 9. The degree of the polynomial is 9. b. The coefficient of 9 x is – 2. c. The constant term is 7. d. It is a polynomial of one variable x . 3. 2 3 7 14 x y xy z  a. The sum of the exponents in each term is 3 and 5, respectively. The degree of the polynomial is 5. b. The coefficient of 3 xy z is – 14. c. The constant term is zero. d. It is a polynomial of several (three) variables: x , y , and z . 4. 5 2 3 6 2 7 5 x x y y   a. The sum of the exponents of each term is 5, 5 and 6, respectively. The degree of the polynomial is 6. b. The coefficient of 6 y is – 5. c. The constant term is 0. d. It is a polynomial of two variables; x and y . 5. 5 2 2 3 5 x x   a. n n a x means 5 2 a  . b. 3 0 a  (Term is 3 0 x ) c. 2 3 a   d. 0 5 a   , the constant term. 6. 3 5 4 17 x x   a. 3 5 a  b. 1 4 a   (Term is – 4 x ) c. 2 0 a  d. 0 17 a   7. 2 4 x x  When 2, x   2 2 4 4( 2) ( 2) 8 4 12. x x           8.         2 2 2 10 6 4 When 1, 10 6 4 10 6 4 1 10 150 140. x x x              

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