WAEC - Mathematics (2020 - No. 13)
Make m the subject of the relation k = \(\sqrt{\frac{m - y}{m + 1}}\)
m = \(\frac{y + k^2}{k^2 + 1}\)
m = \(\frac{y + k^2}{1 - k^2}\)
m = \(\frac{y - k^2}{k^2 + 1}\)
m = \(\frac{y - k^2}{1 - k^2}\)
Giải thích
k = \(\sqrt{\frac{m - y}{m + 1}}\)
k\(^2\) = \(\frac{m - y}{m + 1}\)
k\(^2\)m + k\(^2\) = m - y
k\(^2\) + y = m - k\(^2\)m
\(\frac{k^2 + y}{1 - k^2}\) = m\(\frac{(1 - k^2)}{1 - k^2}\)
m = \(\frac{y + k^2}{1 - k^2}\)