\(\frac{k^2r + p}{m^2}\)
\(\frac{k^2r + pm^2}{m^2}\)
\(\frac{k^2r - p}{m^2}\)
\(\frac{k^2r + p^2}{m^2}\)
Correct answer is B
square both sides to remove the square root
k\(^2\) = m\(^2\) \(\frac{t-p}{r}\)
\(\frac{k^2r}{m^2}\) = t - p
t = \(\frac{k^2r}{m^2}\) + p
t = \(\frac{k^2r + pm^2}{m^2}\)
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