I am having trouble with this problem, thanks!
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 .
see above
To solve this problem, we need to use the concept of direct variation or joint variation. We know that y varies jointly as x and the cube of z, which can be represented as:
y = kx * z^3
Where k is the constant of variation.
To find the value of k, we can substitute the given values of y, x, and z into the equation and solve for k.
Given: y = 378, x = 4, and z = 3
378 = k * 4 * 3^3
Solving this equation will give us the value of k.
First, find the value of 3^3, which is 27.
378 = k * 4 * 27
Next, simplify the equation by multiplying 4 and 27.
378 = 108k
To find the value of k, divide both sides of the equation by 108.
378/108 = k
k ≈ 3.5
Now that we have found the value of k, we can use this value to find y when x = 9 and z = 2.
Plug the values of x, z, and k into the variation equation:
y = k * x * z^3
y = 3.5 * 9 * 2^3
y = 3.5 * 9 * 8
Simplify the equation by multiplying 9 and 8.
y = 252
Therefore, when x = 9 and z = 2, y is equal to 252.