Make K the subject of the relation T = (sqrt{frac{TK – H}{K – H}})
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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})
Correct Answer: Option A
Explanation
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})