I am also having trouble with this one, thanks!

If y varies jointly as x and the cube root of z, and y=120 when x=3 and z=8, find y when x=4 and z=27.

2 answers

y=k*x*cubroot z

120=k*3*2 find k, then put that k in to
y=k*x*cubroot z to solve the last.
Y = K * X * ∛Z
120= K (3)(2)...CUBE ROOT OF 8 IS 2
120= 6 K
120/6 =K
20= K

Y= (20)(4)(3)...CUBE ROOT OF 27 IS 3
Y= 240