SOLUTION: if y varies jointly as x and the cube of z, and y=378 when x=4 and z=3, find y when x=9 and z=2. A)y=283.5 B)y=84 C)y=567 D)y=252

Question 208781: if y varies jointly as x and the cube of z, and y=378 when x=4 and z=3, find y when x=9 and z=2.
A)y=283.5
B)y=84
C)y=567
D)y=252
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
if y varies jointly as x and the cube of z, and y=378 when x=4 and z=3, find y when x=9 and z=2.
A)y=283.5
B)y=84
C)y=567
D)y=252
-----------------------
y = kxz^3
----
Find "k":
378 = k*4*3^3
k = 3.5
-------------------
Equation:
y = (3.5)xz^3
-------------------
If x = 9 and z = 2 you get:
y = (3.5)*9*(2^3)
y = 252
===================================
Cheers,
Stan H.
PS: Reply to stanbon@comcast.net since the algebra.com
feedback function is not working.

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