H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y.
A
H = \(\frac{p}{4y^2}\)
B
H = \(\frac{2p}{y^2}\)
C
H = \(\frac{p}{2y^2}\)
D
H = \(\frac{p}{y^2}\)
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
H \(\propto\) \(\frac{p}{y^2}\)
H = \(\frac{pk}{y^2}\)
1 = \(\frac{8k}{2^2}\)
k = \(\frac{4}{8}\)
= \(\frac{1}{2}\)
H = \(\frac{p}{2y^2}\)
Source: Myschool Past Questions Archive (Record #57165)