If x varies directly as \(y^3\) and x = 2 when y = 1, find x when y = 5
The correct answer is D. 250
Since x varies directly as \(y^3\), we can write the relationship between x and y as x = ky^3, where k is the constant of variation. We are given that x = 2 when y = 1, so we can use this information to find the value of k:
x = ky^3
=> 2 = k(1)^3
=> k = 2
Now that we know the value of k, we can use the relationship x = ky^3 to find the value of x when y = 5:
x = ky^3
=> x = 2(5)^3
=> x = 250
Hence, when y = 5, the value of x is 250.
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