$\frac{r}{10}+4=5$

$\left(x^2-7x-3\right)\left(2-3x\right)$

$\int\frac{1}{\left(x^2-5x+5\right)}dx$

$\lim_{x\to2}\left(\frac{x^3-8}{x^3+x^2+4}\right)$

$\frac{dy}{dx}\left(y=ln\left(sech\right)tanh^2x\right)$

$y^{\prime}=x\left(y^{2}+1\right)$

$100=\arctan\left(\frac{250}{x}\right)$