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 - \(\frac{15P^3}{Q}\)
(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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