Question on: WAEC Mathematics - 2006
If p = [(\frac{Q(R - T)}{15})](^ \frac{1}{3}), make T the subject of the relation
A
T = R + \(\frac{P^3}{15Q}\)
B
T = R - \(\frac{15P^3}{Q}\)
C
T = R + \(\frac{P^3}{15Q}\)
D
T = 15R - \(\frac{Q}{P^3}\)
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Correct Option: B
Cubic both sides; P3 = (\frac{Q(R - T)}{15})
(cross multiplication) Q(R - T) = 15P3
(divide both sides by Q); R - T = 15(\frac{1}{Q})
(subtract r from both sides) - T = (\frac{15P^3}{Q - R})
T = R - (\frac{15P^3}{Q})
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