suppose that y varies directly as the cube of x and that y=17 when x=4 what is y when x=3
To find the value of y when x=3, we need to use the direct variation equation. The equation would be in the form of y = kx^3, where k is the constant of variation.
We can find the value of k by substituting the given values of y and x into the equation:
17 = k(4^3)
17 = k(64)
k = 17/64
Now that we have the value of k, we can substitute it into the equation to find y when x=3:
y = (17/64)(3^3)
y = (17/64)(27)
y ≈ 7.21875
Therefore, when x=3, y is approximately equal to 7.21875.
To solve this problem, we'll use the direct variation equation:
y = k * x^3
where k is the constant of variation.
Step 1: Find the value of k.
Given that y = 17 when x = 4, we can substitute these values into the equation to solve for k:
17 = k * 4^3
17 = k * 64
k = 17 / 64
k = 0.265625
Step 2: Use the value of k to find y when x = 3.
Now that we have the value of k, we can substitute x = 3 into the equation:
y = 0.265625 * 3^3
y = 0.265625 * 27
y = 7.171875
So, when x = 3, y = 7.171875.