Algebra II

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Suppose that y varies directly as the 3/2 power of x, and that y=192 when x= 16. find y when x=9.

  • Algebra II -

    y = k * x^(3/2)

    192 = k * 16^(3/2)

    192 = k * 4^3

    192 = k * 64

    k = 3
    so
    y = 3 x^(3/2)
    if x = 9
    y = 3 * 3^3
    y = 3^4
    y = 9*9
    y = 81

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