The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle? If x = the width of the rectangle, which of the following equations is used in the process of solving this problem? a) 2x2 + 3x - 119 = 0 b) 3x2 + 3x - 119 = 0 c) 6x2 - 119 = 0

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The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle? If x = the width of the rectangle, which of the following equations is used in the process of solving this problem? a) 2x2 + 3x - 119 = 0 b) 3x2 + 3x - 119 = 0 c) 6x2 - 119 = 0

Mathematics
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119 = A = WL; L=2W+3 119= W(2W+3) 119=2W^2 +3W 0= 2W^2 +3W -119 (a)
A= L * W L=2x+3 W=x (2x+3) * x= 119 2x^2 +3x =119 2x^2 +3x -119 = 0 ---->(a)

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