Jointly and cube power

If y varies jointly as x and the cube of z and y=16 when x=4 and z=2, find an equation that represent this relationship

Result

y = (Correct answer is: x z^3) Wrong answer

Step-by-step explanation:

y1=16 x1=4 z1=2  y=k x z3  k=x1 z12y1=4 2216=1  y=k x z3  y=xz3



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