SOLUTION: Can you please help me? find the variation equation for the variation statement. z varies jointly as y and the cube of x; z=160 when x=2 and y=-5 Thanks again

Question 134911: Can you please help me?
find the variation equation for the variation statement.
z varies jointly as y and the cube of x; z=160 when x=2 and y=-5
Thanks again
Answer by nycsharkman(136) (Show Source): You can put this solution on YOUR website!
This question is asking for the EQUATION and thus it is NOT asking you to solve for any letters in particular.
If z varies jointly as y and the cube of x, then we have this equation:
z = k(x^3)(y), where k is the constant of variation.
Let plug the given numbers WITHOUT solving for k.
160 = k(2)^3(-5)
160 = k(8)(-5)
160 = -40k
==================================================
You said:
"I was looking for: y=-4xy^3, y=4x^3y, y=4xy^3, y=-4x^3y"
I said:

In the question, you did not indicate that the constant of variation, which is k needed to be found.

I found this equation:

160 = -40k

Divide both sides by -40 and k = -4.

Then your equation and answer is:

z = -4x^3y

It cannot be y = -4x^3y because z varies jointly as y and the cube of x.

In other words, if your question has a letter z, then z MUST appear or be part of your equation.

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