Question on: WAEC Mathematics - 2015
Make K the subject of the relation T = \(\sqrt{\frac{TK - H}{K - H}}\)
A
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
B
K = \(\frac{HT}{(T - 1)^2}\)
C
K = \(\frac{H(T^2 + 1)}{T}\)
D
K = \(\frac{H(T - 1)}{T}\)
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Correct Option: A
T = \(\sqrt{\frac{TK - H}{K - H}}\)
Taking the square of both sides, give
T2 = \(\frac{TK - H}{K - H}\)
T2(K - H) = TK - H
T2K - T2H = TK - H
T2K - TK = T2H - H
K(T2 - T) = H(T2 - 1)
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
Taking the square of both sides, give
T2 = \(\frac{TK - H}{K - H}\)
T2(K - H) = TK - H
T2K - T2H = TK - H
T2K - TK = T2H - H
K(T2 - T) = H(T2 - 1)
K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
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