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

Assessment: WAEC SSCE - General Mathematics - 2022 (Objective) Subject: General Mathematics

Question 1 Report

\frac{k^{2} r + p}{m^{2}}

k2r+pm2m2

\frac{k^{2} r - p}{m^{2}}

\frac{k^{2} r + p^{2}}{m^{2}}

Answer Details

square both sides to remove the square root

k $^{2}$ = m $^{2}$ $\frac{t - p}{r}$

$\frac{k^{2} r}{m^{2}}$ = t - p

t = $\frac{k^{2} r}{m^{2}}$ + p

t = $\frac{k^{2} r + p m^{2}}{m^{2}}$

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