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 representing this relationship.
a. y=xz
b. y = 2xz³
c.y = 2kx/z³
d.y = xz³/2​

Correct answer:

y = x z^3

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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