Make K the subject of the relation T = \(\sqrt{\frac{TK - H}{K - H}}\)...
MATHEMATICS
WAEC 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 = \(\frac{H(T - 1)}{T}\)
Correct Answer: A. K = \(\frac{H(T^2 - 1)}{T^2 - T}\)
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 = \(\frac{H(T^2 - 1)}{T^2 - T}\)
Post an Explanation Or Report an Error
If you see any wrong question or answer, please leave a comment below and we'll take a look. If you doubt why the selected answer is correct or need additional more details? Please drop a comment or Contact us directly. Your email address will not be published. Required fields are marked *