Question 1063818
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y varies directly as x and inversely as the square of z {{{cross(calculator)}}}. 
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k, variation constant

{{{y=k(x/z^2)}}}

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{{{yz^2=kx}}}

{{{k=(yz^2)/x}}}


You were then given some values to find what is k value.

{{{k=32(6^2)/114}}}

{{{k=32(36/114)}}}
or better to try
{{{k=(2*2*2*2*2*2^2*3^2)/(2*3*19)}}}

{{{k=(32*6)/19}}}

{{{k=192/19}}}--------which might be how you want it;


{{{highlight_green(y=(192/19)(x/z^2))}}}-------your variation model.