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the length of a rectangle is 10 inches more than 1.5 its width. if the perimeter of the rectangle is 44, what are the dimensions
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Supposed to say 1/5
P = 44 = 2w + 2[(1/5)w + 10] You now have one equation in one variable that you can solve. The factor of 2 in front of the two terms is from 2w + 2L because its a rectangle.
(12/5)w = 24 -> w = 10
@mweisberg the problem is finished now.
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