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Date Created: 11/15/15
MATH 119 Written Assignment 1 2 26.  3w 8w4  (3w2)(w2) Section 1.2 3 2 18.    (2p 5p7)(4p 8p2)  3 2 2 2p 5p74p 8p2  32.  12z  z 1 (4z 1)(3z 1)           2p 4p 5p8p72  2 2p 4p 3p9 58.  81y 100  (9y10)(9y10) 4 2 2 2 3 2 2 72.  y 7y 10  (y 5)(y 2) 22.  (3y 9y 11y8)(4y 10y6)  3y 9y 11y84y 10y6          3y 9y 4y 11y10y86  Section 1.4 3y 13y 21y14 6(x2) 6 6 8.      24.  2a(4a 6a8)  (x4)(x2) x4 x4 (4a )(2a)(6a)(2a)(8)(2a)  2         r r 6 (r 3)(r 2) r 2 8a 12a 16a 12.   2   r r 12 (r 3)(r 4) r 4 32.  (.012x.17)(.3x.54)  2k 8 3k 12 (.012x)(.3x)(.012x)(.54) 22.     6 3               (.17)(.3x)(.17)(.54)          2 .0036x .00648x.051x.0918  2k 8 3k 12 2   .0036x .04452x.0918 6 3 2k 8 3 Section 1.3 6 3k 12  2.  y65xy  5y(1)5y(13x)  2(k 4)  3                6 3(k 4) 5y(113x) Since 113y  is prime, you are finished. k 4 3 3 3(k 4)  6. 5x 55x 10x  3 2 5x 55x 10x  1 3 1 1 1       2     5x(x 11x2) 3 3 3 1 3 2 Since  x 11x2  is prime so you are finished. 2 24.  r 16r 60  (r 6)(r 10) 4 3 Section 1.5 40.    3(k 1) k 1 (6) 14 146 8 2.   6  (6)  (6) 1,679,616 The common denominator is  3(k 1) (6) 4 3 3 3 3   6.   5  5 5 125 3(k 1) k 1    3  3 3 3 3  xy (xy) x y x y 4 3 3 4 1 1    14.  (x)  4  4 3(k 1) 3 k 1 (x) x 1 4 9 24.  8  x           1 3 1 3(k 1) 3(k 1)  3  x 3 8  x         3     or     8  x 8  x 49 x  2 2  x  3(k 1) 3 1 3 1 1 1 38.  12  412 4 12 4 4 12  2  1 13 12 2 3(k 1) (4x) 12 xy 58.    11 5 x y 2 42.    3(p4) 6(p4) 1 (4x) 2 xy The common denominator is  6(p4) 32 2  x y 11  5  12 12 3(p4) 6(p4) (4x) (xy) 2 2  x y 2 11 5    12 12 2 2 2 3(p4) 6(p4) 4 x x y  x y 2         22  5  6(p4) 6(p4) 2xy 12 3  x y 2 225 2x 12y122   6(p4) 12 32 2x y  2  2 17 x y 32 x y 2 6(p4) 8.  2 (k 2)3(k 1) 12k 80.   49  16  74  3 2 k 83k 3 142k 88.   5  2  5  2   5  2  5  2  2  5 42k 2 2           2 5  2          2k 10 142k 5 10  10 2 52 3 2k 2k 10 142k 2k Section 1.6      0 144k 2.  5y 19 1014 14144k 4(4)5y 19(4) 5y 15 24  4k 5y 15        5 5 24 4k  y  3 4 4 6  k 6.  (k 2)6  4k (3k 1) 3k 66  4k 3k 1 10.  4(x2) 1  2 3 x1 3 2 4  3k 12  k 1 4 1 3  (x2)  2  x1         2 4  3k (k)12  k (k)1 4x8 1 6x   1 3 2 4 2k 12 1 The least common denominator is 12 so you       multiply both sides by 12: 2k 1212 112 12 x8 1  1 6x        2k 13 1  3 2  4  2k 13     16x326 18x12  2 2 16x26 18x12 13 k  2 16x16x26 18x16x12 8 5 18.     4 26  2x12 3k 9 k 3 The least common denominator is   3k 9   so 2612  2x1212 multiply both sides by 3k 9 :          14  2x 3k 9  8  5   4(3k 9) 1 3k 9 k 3 14 2x  815 12k 36 2 2 x  7 7 12k 36 14.   3k  9k 5  11k 8 736 12k 3636 2 6 k         The least common denominator is   6k   so you 29 12k multiply both sides by 6k : 29 12k 6k  3k 9k 5 6 11k         12 12 1  2 6  1 k 29 2 2 k  9k 9k 5k  66k 48 12 24.  (ab)bx  a(x2) 5k  66k 48 3abbx  a(x2) 3abbx  ax2a          (66k)  66k 66k 48 3abbxbx  ax2abx 3ab  axbx2a 71k  48 3a 2ab  axbx2a  2a 5ab  axbx 71k  48         71 71 5ab  x(ab) 48 5ab  x(ab) k   71 ab ab x 5ab ab 2 16.  z(2z 7)  4 30.  S  S 0 gt k     Solve for g. S S k  S S  gt k k 2 2z 7z  4 0 0 0 2 S S 0  gt 2 2z 7z 4  0 (2z 1)(z 4)  0         2 2z 1 0  or  z 4  0 S S 0k  gt t2 t2 1 z  2   or  z  4 S S 0k g  2 22.  (b4)  27 t (b4)  27 b4  27  or  b4   27 Section 1.7      b4  3 3  or  b4  3 3 2.  (p16)(p5)  0 b  43 3  or  b  43 3       16  0  or p5  0        16  or p  5 b  43 3 36.   2 4.  x 2x  0 3k k  6 3k k 6  0 x(x2)  0      x  0  or  x2  0 a  3, b 1, and c  6 2 x  0  or  x  2 1 (1) 4(3)(6) k  2(3) 2 8.  k 4k 5 0 (k 5)(k 1)  0         1 172      k 5  0  or  k 1 0 k  6 k  5  or  k  1 10.  15z  z  2 1 73 2 k  6 15z  z 2  0 (5z 2)(3z 1)  0 k 1.257  or  k  1.591      5z 2  0  or  3z 1 0 2 1 z     or  z  4 1 5 3 38.  5  2  0 k k 2 14.  6a 17a12  0 The least common denominator is  k2 so multiply (3a4)(2a3)  0 2 both sides by  k : 3a4  0  or  2a3 0         4 3 2 a     or  a           5 4  1   0 3 2 1  k k2 5k 4k 1 0 a  5, b  4, and c  1 2 a  (4) (4) 4(5)(1) 2(5) 4 1620 a  10 4 36 a          10 46 a  10 a  46   or  a 6 10 10 10 2 a    or  x  10 10 a 1  or  a   5 2 46. 8.06x 25.8726x  25.047256 8.06x 25.8726x25.047256 0 a 8.06, b  25.8726, and c  25.047256 25.8726 (25.8726) 4(8.06)(25.047256) x  2(8.06)         25.8726 1476.9187 x  16.12 25.872638.4307 x  16.12 x 25.872638.4307 .779  or  x 25.872638.4307  3.989 16.12 16.12

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