If x varies directly as y3 and x = 2 when y = 1, find x when y = 5
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Correct Answer: Option D
Explanation:
x \(\alpha\) y3
x = ky3
k = \(\frac {x}{y^3}\)
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
x \(\alpha\) y3
x = ky3
k = \(\frac {x}{y^3}\)
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250