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Date Created: 11/15/15
MATH 119 Written Assignment 2                                 Section 2.1 2 The x­intercepts are  2 and  3 . 2 2 4.  (2,−1)  is a solution to  x  y 6x8y  15 Let  x  0  and substitute: because substituting in the point gives: 2 (1) 6(2) 8(1)  15 . y  3(0) 4(0)4 y  4 2 2 The y­intercept is −4. x y 6.  (1,−1)  is a solution to    4  because 2 3 substituting in the point gives: 30.  x­intercept is 0 and y­intercept is 0. 2 2 1 (1) 5    not −4. 2 3 6 y (­2,8) 8 (2,8) 16.  The x­intercept is 1.  There is no y­intercept. 6 18.  For  x2y 5 Let  y  0  and substitute: 4 x2(0)  5 x  5 (­122) (1,2) The x­intercept is 5. x (0,0) Let  x  0  and substitute: ­4 ­3 ­2 ­1 1 2 3 4 02y  5   2y  5 5 y    40.  x­intercepts are  3  and y­intercept is 3 2 5 The y­intercept is   . y 2 4 (0,3) 26.  For  y  3x 4x4 (­1,2sq3t((1,2sqrt(2)) Let  y  0  and substitute: 2 3x 4x4  0 1 (3x2)(x2)  0 (­3,0) (3,0) x 3x2  0     x2 0 ­4 ­3 ­2 ­1 1 2 3 4 2 ­1 x             x  2 3 b  3 and m  4 48.  a.  About \$250,000 3 y  mxb        b.  About \$1,250,000 14.   4 y  x3        c.  About \$1,500,000 3 20.  Re­write the equation into slope­intercept form: 82.  The total assets were lowest at x  2.1 which corresponds to 2002.  There were about \$2652.38 in early 2002. 4x2y  0 2y  4x y y  2x 6000 m  2 and b  0 5000 24.  There are many correct answers. 4000 3000 (2.1,2652.41) 34.  x5y  7 and 15y5 6x 2000 Now solve both equations for y: 1000 2x5y  7                15y5  6x x 5y  2x7           15y  6x5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 2 7 6 5 y  x                  y x                              Section 2.2 5 5 15 15 4.  Through (−5,−2) and (−4, 11): 2 1                          5 x 3    y  m  211  13  13 13 5(4) 54 1 2 The slopes for both equations is m   so the lines 3 are parallel. 6.  Through (0, 0) and (8,−2): 20 2 1 36.  x3y  4 and y 13x m  80  8  4 Now solve both equations for y: x3y  4                        y 13x perpendicular line will have a slope of   m   1 3y  x4            2  1 because  (2)   1 . 1 4  2 y  x                   1 3 3 (x1, 1 )  (6, 8) and m                                       2 The product of the two slopes is −1 so these lines y y  m(x x ) 1 1 are perpendicular. 1 y8  2 x6  y8  x3 2 y   x11 2 (x , y )  (5,2) and m  4 1 1 5                               Section 2.3 y y 1 m(x x ) 1 4 6.  Let x , y )  (0, 1.9 and  (x , y )  (10, 1.2) 42.  y(2)  x(5)  1 1 2 2 5 First, find the slope:   4 y2  x4 5 m  1.21.9  .7  .07 4 100 10 y  5 x2 Since you have the point (0, 1.9), the y­intercept is 1.9 which gives the equation: (x , y )  (3,9) and m  0 1 1 y  .07x1.9 y y 1 m(x x ) 1 46.  y(9)  0 x(3)  The value of x comes from the returns examined in y9  0 2008.  This gives: y  9 x  20081996 12 54.     Through   the   origin   and   horizontal.     A y  .07(12)1.9 horizontal line has slope 0.  y  .841.9 y 1.06 (x1, y1)  (0, 0) and m  0 y y1 m(x x )1 There were about 1.06 returns examined in 2008. y0  0 0  y  0 18.  a.  You have the points  (x1, 1 )  (0, 62.3 and (x , y )  (15, 53.3).  This gives the slope: 2 2 58.  Through (6, 8) and perpendicular to the line y  2x3.  The slope of this line is  53.362.3 9 m  2  so a m    .6 150 15 Since the division is by a negative number, reverse Since you have the point (0, 62.3), the y­intercept is the direction of the inequality: 62.3 which gives the equation: 8k  32 y  .6x62.3 k  4 k  4 Using a graphing calculator or graphing software, the regression equation is: The solution is 4,   y  .54x61.7       b.  Using the two­point model,  x  20 for 2010           and  x  22 for 2012 :           −4 y  .6(20)62.3 5.   2b  0 y  50.3 million   y  .6(22)62.3       b  0 Since the division is by a negative number, reverse y  49.1 million the direction of the inequality: Using the regression­line model: 2b  0 b  0 y  .54(20)61.7 b  0 y  50.9 million y  .54(22)61.7 The solution is , 0  y  49.8 million                                                                                  24a.     Using  a graphing   calculator  or  graphing                                                                    0 software, the regression equation is: m(42m)3 2m2 y .718x9.9 13.  m42m3 2m2 m1 2m2      b.  Answers will vary. 1 m      c.  Men are gaining more in median income each year. The solution is 1,                                   Section 2.4                      −1 3.   8k  32       k  4 21.  83r 116 The   absolute   value   of   a   number   is   always       7 3r 15 nonnegative, therefore, this is always true for all 7      3  r  5 real numbers.  The graph is the entire real number line. 7 , 5 The solution is  (, ) The solution is  3            56.  Let x represent the number of miles traveled 7 and let y be the cost of the travel.  This gives:                                                        5 3 y  2.5.4(5x) 2k 1 8.5 2.5.4(5x) 12.5 23.   4  3  2 6  2x 10       12  2k 1 6 3 x  5        11 2k  7       11  k  7 This means you can travel between 3 and 5 miles. 2 2  11 7  The solution is   ,   58.     Let  C 100x 6000 and R  500x .     Since  2  R  C , you have: 500x 100x6000 400x  6000           x 15 11 7        2                                2            The number of units of squash must be someplace in the interval 15,  . 34.   p  7 62.  Let C  25000x21,700,000        p  7 or p  7 R 102,500x Let  .  Since  R C  you have: The solution is  (,7) or (7, ) 102,500x  25,000x21,700,000 77500x  21,700,000 x  280           The value of x must be in the interval  280,  .                       −7                            7 38.   b  5

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