The area of a rectangle is 45 ft^2 , and the length of the rectangle is 1 ft more than double the width. Find the dimensions of the rectangle

Let w be the width of the rectange, its length is: 2w + 1. The area of a rectangle is 45 ft^2, then w(2w + 1) = 45 2w^2 + w - 45 = 0 (w + 5)(2w - 9) = 0 w = -5 (not possible) or w = 9/2 ft. Length = 2(9/2) + 1 = 10 ft.

Let w be the width of the rectange, its length is: 2w + 1. The area of a rectangle is 45 ft^2, then w(2w + 1) = 45 2w^2 + w - 45 = 0 (w + 5)(2w - 9) = 0 w = -5 (not possible) or w = 9/2 ft. Length = 2(9/2) + 1 = 10 ft.

Weegy: Let w be the width of this rectangle then its length is 3w - 4.
Its area = w(3w - 4) = 55;
3w^2 - 4w - 55 = 0;
(w - 5)(3 + 11) = 0;
so w = 5 feet.
And the length is 3(5) - 4 = 15 - 4 = 11 feet.
The length of this rectangle is 11 feet and the [ width is 5 feet. ] (More)

Weegy: SOLUTION: The width of a rectangle is 7m more than the length. Find the dimensions if the ... x^2 + 7x - 60 = 0 {subtracted 60 from both sides} (x + 12)(x - 5) = 0 ... › Customizable Word Problem Solvers › Geometry (More)

Let w be the width of the rectange, its length is: 2w + 7. The area of a rectangle is 99 m^2, that is: w(2w + 7) = 99 2w^2 + 7w - 99 = 0 (w + 9)(2w - 11) = 0 w = -9 (impossible) or w = 11/2 m Length = 2(11/2) + 7 = 18 m.

Added 5/19/2014 12:35:20 AM

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