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y varies directly as the square of x. when x=3, y=8. Find y when x=2
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sorry what;s the answer ?
f(x) is \[y=\sqrt{x}\]
Sorry I lost it :(
no...it should be done in this way....
\[y=k{x^2}\] when x=3, y=8 8=k*3^2 k=8/9 y=8x^2/9 when x=2 y=8*2^2/9=32/9
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