does anybodt know how to rearange the equation so n is the suject? any help would be great. thanks!!
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Originally Posted by rolfharris does anybodt know how to rearange the equation so n is the suject? any help would be great. thanks!! $\displaystyle a=1.33\sqrt{\frac{d}{r}}(\frac{n}{b^2})$ $\displaystyle \frac{a}{1.33}=\frac{\sqrt{d}}{\sqrt{r}}(\frac{n}{ b^2})$ $\displaystyle n=\frac{b^2a\sqrt{r}}{1.33\sqrt{d}} $
fantastic!!
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