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22 June, 02:50

A rectangle has a length that is nine feet less than four times its width. Its area is 90 ft. Algebraically determine the length of its width and length. Show work

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  1. 22 June, 03:16
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    Ok ... after much confusion, I've come to the conclusion that you must have written the question wrong and it's actually the perimeter that is 90 feet so that's how I'm going to work it ...

    Width = w

    Length = 4w - 9

    P = 2w x 2L

    90 = 2w + 2 (4w - 9)

    90 = 2w + 8w - 18

    90 = 10w - 18

    108 = 10w

    108/10 = w

    10 4/5 = w

    Width is 10 4/5 ft

    Length = 4w - 9

    L = 4 (10 4/5) - 9

    L = 43 1/5 - 9

    L = 34 1/5 ft

    Length 34 1/5 ft Width 10 4/5 ft or

    Length 34.2 ft and width 10.8 ft
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