Make F the subject of the formula t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)

  • A \(\frac{gv-t^2}{gt^2}\)
  • B \(\frac{gt^2}{gv-t^2}\)
  • C \(\frac{v}{\frac{1}{t^2} - \frac{1}{g}}\)
  • D \(\frac{gv}{t^2 - g}\)
jamb 1997mathematics

The correct answer is B. \(\frac{gt^2}{gv-t^2}\)

t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\)

t

= \(\frac{v}{\frac{1}{f} + \frac{1}{g}}\)

= \(\frac{vfg}{ftg}\)

\(\frac{1}{f} + \frac{1}{g}\) = \(\frac{v}{t^2}\)

= (g + f)t

= vfg

gt

= vfg - ft

gt

= f(vg - t

)

f = \(\frac{gt^2}{gv-t^2}\)

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