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15 January, 13:11

A regulation tennis court for a double match is laid out so that it's length is 6 ft more than two times its width. the area of a doubles court is 2808 square feet what is the length and width of the singles Court

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  1. 15 January, 13:23
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    Area=LW

    L=6+2W

    given

    A=2808

    subsitute

    2808=LW

    subsitute 6+2W for L

    2808 = (6+2W) (W)

    2808=2W²+6W

    divide both sides by 2

    1404=W²+3W

    minus 1404 from both sides

    0=W+3W-1404

    farctor, what 2 numbers multiply to get - 1404 and add to get 3

    -36 and 39

    0 = (W-36) (W+39)

    set each to zero

    0=W-36

    36=W

    0=W+39

    -39=W

    false, can't have negative dimentions

    W=36

    L=6+2W

    L=6+2 (36)

    L=6+72

    L=78

    the length is 78ft

    the width is 36ft

    0=
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