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alg 150 help

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the width of a rectangle is 9 less than twice the length. if the area os the rectangle is 41cm^2 what is the length of the diagonal?

  • alg 150 help -

    if x is the length, 2x-9 = width

    x(2x-9) = 41
    2x^2 - 9x - 41 = 0
    x = 1/4 (9±√409)
    since we want a positive length,

    length = 1/4 (9+√409)
    width = 1/2 (9+√409)-9 = (√409 - 9)/2

    diagonal d^2 = 1/16 ((9+√409)^2 + 4(√409-9)^2)
    = 1/8 (1225-27√409)
    d = √(that)

    that is an unusual answer. I suspect a typo somewhere. If so, make the adjustment and redo the calculations.

  • alg 150 help -

    The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 55 cm22, what is the length of the diagonal?

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