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Math Help - This inverse variation is incorrect, please help me fix it.

  1. #1
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    This inverse variation is incorrect, please help me fix it.

    “b” varies inversely as the square root of “c”. If b = 1 when c = 16, find b when c = 6.

    b = k/x
    1= k/SQRT(16)
    k/SQRT(16) = 1
    k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
    k = 1 * SQRT(16)
    k = 4

    Now that I have the variation constant, I can plug in the x-value they gave me, and find the value of y:

    y = 4/6

    y = 2/3
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  2. #2
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    Quote Originally Posted by bobbyboy1111 View Post
    “b” varies inversely as the square root of “c”. If b = 1 when c = 16, find b when c = 6.

    b = k/x Mr F says: This makes no sense. The denominator is meant to be {\color{red}\sqrt{c}}.
    1= k/SQRT(16)
    k/SQRT(16) = 1
    k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
    k = 1 * SQRT(16)
    k = 4

    Now that I have the variation constant, I can plug in the x-value they gave me, and find the value of y: Mr F says: There are no x's or y's in the question. The pronumerals are b and c.

    y = 4/6

    y = 2/3
    Given: b = \frac{k}{\sqrt{c}}.

    k = 4.

    Therefore: b = \frac{4}{\sqrt{c}}. Now substitute c = 6.
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  3. #3
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    So....????

    b = k/SQRT(c)
    1= k/SQRT(16)
    k/SQRT(16) = 1
    k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
    k = 1 * SQRT(16)
    k = 4


    ....


    b = 4/SQRT(c)

    b = 4/SQRT(6)

    b = 4/2.44

    b = 1.63 or 163/100

    Thanks again, Mr. Fantastic
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  4. #4
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    Quote Originally Posted by bobbyboy1111 View Post
    So....????

    b = k/SQRT(c)
    1= k/SQRT(16)
    k/SQRT(16) = 1
    k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
    k = 1 * SQRT(16)
    k = 4


    ....


    b = 4/SQRT(c)

    b = 4/SQRT(6)

    b = 4/2.44

    b = 1.63 or 163/100


    Thanks again, Mr. Fantastic
    You should leave the final answer in exact form unless instructed otherwise by the question: b = \frac{4}{\sqrt{6}} = \frac{2 \sqrt{6}}{3}.
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