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Posted by on Thursday, January 6, 2011 at 6:45pm.

the area of a certain rectangle is 288 yd squared. the perimeter is 68 yards. f you double the length and width what will be the area and perimeter of the new rectangle

  • math - , Thursday, January 6, 2011 at 7:31pm

    Area (A) = Lw = 288
    Perimeter (P) = 2L + 2w = 68
    L = length, w = width
    Solve 2L + 2w = 68 for either L or w
    I'll choose L
    2L + 2w = 68
    2L = 68 - 2w
    2L = 2(34 - w)
    L = 34 - w

    Substitute L = 34 - w, for L in
    A = Lw = 288
    (34 - w)(w) = 288

    34w - w^2 = 288
    w^2 - 34w + 288 = 0
    w = 18, 16
    So, this certain rectangle has dimensions of 18 by 16.

    you should be able to calculate the area and perimeter of the new rectangle now. (double the length and width and solve the equations)

  • math - , Monday, November 19, 2012 at 5:08pm

    r = 2/3t + v, for t

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