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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?

Respuesta :

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A = LW = 44
and
L = 3 + 2W
now substitute this into the first equation...
(3+2W)W=44
3W+2W^2=44

Rearranging
2W^2 + 3W - 44 = 0
Factoring we get
(2W + 11)(W - 4) = 0
so that
W = 4 inches
L = 11 inches
let width = x   then length  = 2x + 3

so area = 119 = x(2x + 3)

2x^2 + 3x - 119 = 0
(2x + 17)(x - 7)  
x = 7 , -8.5
so x = 7   ( -8.5 can be ignored)

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