<PRE><B>Question 688182</B>
If y varies jointly as x and the square root of z, 
and inversely as the cube of w,
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y = k*x*sqrt(z)/w^3
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Solve for &quot;k&quot; using &quot;y = 8 when x=3, z=16 and w=2&quot;
8 = k*3*4/2^3
8 = k(3/2)
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k = 16/3
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Equation:
y = (16/3)*x*sqrt(z)/w^3
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Find y when x=5, z=25 and w=2.
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y = (16/3)*5*5/8
y = (2/3)25
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y = 16 2/3
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Cheers,
Stan H.

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