Asked by Liv
The length of a rectangle is
3yd less than twice the width, and the area of the rectangle is 65ydsquare. Find the dimensions of the rectangle.
3yd less than twice the width, and the area of the rectangle is 65ydsquare. Find the dimensions of the rectangle.
Answers
Answered by
Damon
L = 2 w - 3
and
L w = 65 so L = 65/w
65/w = 2 w - 3
65 = 2 w^2 - 3 w
2 w^2 - 3w - 65 = 0
(w+5))(2w-13)
positive w = 13/2 = 6.5
etc
and
L w = 65 so L = 65/w
65/w = 2 w - 3
65 = 2 w^2 - 3 w
2 w^2 - 3w - 65 = 0
(w+5))(2w-13)
positive w = 13/2 = 6.5
etc
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