Answer:
[tex]\bold{K=\frac{2r}m-c=\frac{2r-cm}m}[/tex]
Step-by-step explanation:
[tex]\bold{r= \frac m2 (c + K)}\\^{\div\frac m2}\qquad^{\div\frac m2}\\ \bold{r\cdot\frac2m=c+K}\\{}\ ^{-c}\qquad^{-c}\\\bold{\frac{2r}m-c=K}[/tex]
Or if you mean r= m/[2(c + K)]
[tex]\bold{\quad r\ =\ \frac m{ 2(c + K)}}\\^{\cdot(c+K)}\quad^{\cdot(c+K)}\\ \bold{r(c+K)=\frac m2}\\{}\qquad ^{\div r}\qquad^{\div r}\\\bold{c+K=\frac m2\cdot\frac1r}\\{}\quad^{-c}\qquad^{-c}\\\bold{K=\frac m{2r}-c}[/tex]
