kellyspeakslouder
kellyspeakslouder kellyspeakslouder
  • 07-03-2017
  • Mathematics
contestada

The length of a rectangle is 5yd less than twice the width, and the area of the rectangle is 52yd^2. Find the dimensions of the rectangle.

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Aliwohaish12
Aliwohaish12 Aliwohaish12
  • 07-03-2017
52 = (2*W - 5) * W
52 = 2*W^2 - 5*W
0 = 2*W^2 - 5*W - 52
0 = (2W - 13)*(W + 4)

2W - 13 = 0 or W + 4 = 0
W = 13/2 or W = -4

Width cannot be negative so discard W = -4

W = 13/2 L = 2*(13/2) - 5 = 13 - 5 = 8
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