$\lim_{x\to\infty}\left(\frac{5x^2-3x+1}{x-3}\right)$

$y=e^{x^2-3x}$

$xy=-5$

$\int\frac{\left(sin\left(3x\right)\cdot cos\left(3x\right)\right)}{4}dx$

$12.8\cdot40+3\left(x\right)$

$\frac{sin\left(x\right)}{1-\cos\left(x\right)}+\frac{sin\left(x\right)}{1+cos\left(x\right)}=\tan\left(x\right)$

$\left(-10m^6p\right)\left(-5m^2p^3\right)$