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suppose z varies directly as the square of x and inversely as y. if z=8 when x=4 and y=6, find z when x=6 and y=12
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\[\Huge z = \frac{kx^2}{y}\]
find k in the first step!
\[\large 8 = \frac{4(k)}{6}\]
can u do it now?
yea, thanks
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welks!
\[Z=C*(x ^{2}/y)\] 8=C(16/6) FIND C C=3 then Z=3∗(x2/y) Z=3∗(36/12) Z=9
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