If y frac y 2 sqrt x 2 m 3N make x the subject ... - WAEC Mathematics 2010 Question
If y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular
A
\(\frac{\sqrt{9y^2 N^2 - 2m}}{3}\)
B
\(\frac{\sqrt{9y^2 N^2 - 4m}}{2}\)
C
\(\frac{\sqrt{9y^2 N^2 - 3m}}{2}\)
D
\(\frac{\sqrt{9y^2 N - 3m}}{2}\)
correct option: b
y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\)
3yN = 2(\(\sqrt{x^2 + m})\)
\(\frac{3yN}{2} = \sqrt{x^2 + m}\)
(\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\)
\(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\)
x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\)
x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)
3yN = 2(\(\sqrt{x^2 + m})\)
\(\frac{3yN}{2} = \sqrt{x^2 + m}\)
(\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\)
\(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\)
x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\)
x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)
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