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anonymous
 5 years ago
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
anonymous
 5 years ago
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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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0119 = A = WL; L=2W+3 119= W(2W+3) 119=2W^2 +3W 0= 2W^2 +3W 119 (a)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A= 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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