Question 1172670
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(for tutor @ikleyn)....<br>
... varies jointly as x and the CUBE of y..<br>
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Tutor @ikleyn has provided a response showing one way (of many!) to approach a problem like this.<br>
Another common method is to determine the constant of variation from the given data...
{{{z = kxy^3}}}
{{{96 = k(2)(2^3) = 16k}}}
{{{k = 96/16 = 6}}}<br>
... and then evaluate the formula with that constant and the new data.
{{{z = 6xy^3}}}
{{{z = 6(3)(3^3) = 6(81) = 486}}}<br>
And here is the way I myself find easiest to use....<br>
z is 96 when x is 2 and y is 2.
When x changes from 2 to 3, the value of z changes by a factor of 3/2.
When y changes from 2 to 3, the value of z changes by a factor of (3/2)^3.<br>
The "new" value of z is<br>
{{{(96)(3/2)(3/2)^3 = 96(81/16) = 81*6 = 486}}}<br>
Different students will find different methods more to their liking; try these different methods and find what "works" for you.<br>