$2700\left(1+\frac{0.0129}{12}\right)^{12\left(3\right)}$

$\lim_{x\to-\infty}\left(\frac{6x-4}{3x+1}\right)$

$x\cdot lx$

$\int\left(\left(sin\left(x\right)\right)^2+9\right)dx$

$\cos\left(x\right)-\tan\left(x\right)\cos\left(x\right)=0$

$\frac{\left(2x+2\right)}{\left(x+1\right)}>4$

$33-\left(-19\right)$