Suppose z varies jointly as y and the cube of x. If z=168 when x=2 and y=7, what is z when x=10 and y=4

From the statement, z = kx^{3}y, where k is a constant. Set x = 2, y = 7, and z = 168. Then 168 = k·2^{3}·7 = 56k, and k = 3. Finally, if x = 10 and y = 4, then z = 3·10^{3}·4 = 12000.

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up