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Math Help - inverse variation

  1. #1
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    Red face inverse variation

    The illumination produced by a light source varies inversely as the square of the distance from the source. If the illumination produced 4m from a light source is 75 units, find the illumination produced 8 meters from the same source.

    I understand variations but inverse variations mess me up...
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  2. #2
    Moo
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    Hello,

    Just translate the text !

    If i is the illumination and d the distance, we have, with k a constant of proportionnality :

    i = \frac{k}{d^2}

    You've been given d=4 and i=75. So if you want, you can try to calculate k. But it's not necessary :

    You want the light i' produced at a distance of 8 = d'. d'=2d.

    So i' = \frac{k}{d'^2} = \frac{k}{(2d)^2} = \frac{1}{4} \frac{k}{d^2}= \frac{i}{4}
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  3. #3
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    I still don't understand how to work through the equation you gave me. In my math book, the answer is 18.75 units..how did they come to this answer?
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  4. #4
    Super Member wingless's Avatar
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    I think it's just a simple proportion.

    <br />
\begin{array}{lll}<br />
75&\longrightarrow&4^2\\<br />
x&\longrightarrow&8^2 \end{array}<br />

    75\cdot 16 = x \cdot 64

    x = \frac{75\cdot 16}{64} = \frac{75}{4} = 18.75
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