By Nicholas A. Macri (Auth.)
Real looking and appropriate functions from a number of disciplines support inspire enterprise and social technology scholars taking a finite arithmetic path. a versatile company permits teachers to tailor the booklet to their path
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Extra resources for Study Guide for Applied Finite Mathematics
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1 |l 1 -2 -3| 2 3 1 10 I3 -1 2 111 Add -2 times the first row to the second row. This yields the new matrix |l 1 -2 0 1 5 16 -1 2 111 [3 -3| Add -3 times the first row to the third row. ["l 1 -2 0 1 5 16 0 -4 8 20 This yields the new matrix -3l Add -1 times the second row to the first row. This yields the new matrix I1 0 -7 0 1 5 16 -4 8 20 J [0 -191 Add 4 times the second row to the third row. J1 0 -7 0 1 5 16 I 0 0 28 84 Multiply the third row by -^p This yields the new matrix -19] This yields the new matrix |l 0 -7 -191 0 1 5 16 0 0 1 3 47 4: Matrices and Linear Systems At this point we can easily solve the new corresponding linear system of equations, which is x - 7z = -19 y+ 5z = 16 z = 3.
0 + 8 9 . 0 and ( 0 + 1 + 2 + 3 + 4 + 5 ) b + ( 0 + 1 + 4 + 9 + 1 6 + 25)m = 6 8 . 7 + 1 4 2 . 6 + 2 4 3 . 3 + 3 2 8 . 0 + 4 4 5 . 6. 18. Exercise 13 Let x = the number of meals prepared and sold in a week. 50x. To find the break-even point, we set R = C and solve for x. 25x = $1375, and x = $1100. 5 Results to remember: 1. When determining the region defined by an inequality, first draw the corresponding line, then choose a test point. a) Suppose the test point is above the line: (1) If the coordinates of the point satisfy the inequality, then the solution set is the half plane above the line.
2 Find the minimum of z = lOOOx + 600y subject to x + y < 30. CD 3 < x. (2) x < 10. (3) y ^ 18. (4) In the discussion that follows, refer to Figure 3. We first determine the feasible region using inequalities (1), (2), (3), and (4). From inequality (1), consider the equation x + y = 30. This equation is satisfied by points (0,30) and (30,0). These two points determine line £in Figure 3. (1). The point (0,0) lies below line l~ and satisfies inequality Therefore the feasible region lies below line JL.