Make t the subject of k = \(m \sqrt \frac{t-p}{r}\)

  • A \(\frac{k^2r + p}{m^2}\)
  • B \(\frac{k^2r + pm^2}{m^2}\)
  • C \(\frac{k^2r - p}{m^2}\)
  • D \(\frac{k^2r + p^2}{m^2}\)
waec 2022mathematics

The correct answer is B. \(\frac{k^2r + pm^2}{m^2}\)

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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