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Today, 02:25

The length of a rectangle is 12 in. and the perimeter is 56 in. find the width of the rectangle

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  1. Today, 06:25
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    The perimeter p of the rectangle with length l and width w is:

    p = 2l + 2w

    that is, two times its length and two times its width, lets solve for the width w and substitute known dа ta:

    p = 2l + 2w

    2w = p - 2l

    w = (p - 2l) / 2

    w = (56 - (2*12)) / 2

    w = (56 - 24) / 2

    w = 32/2

    w = 16

    therefore the width of the rectangle is 16 in
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