Make t the subject of k = m {t-p/r}

Correct Answer is : k2r+pm2m2{k²r + p m²}{m²}m2k2r+pm2

Correct Answer is : k2r+pm2m2{k²r + p m²}{m²}m2k2r+pm2

Test Index

WAEC Mathematics 2022 Paper

Question : 18

Total: 47

m \sqrt{\frac{t-p}{r}}

Solution:

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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