$\left(2x^3+4y^2\:\right)^3$
$\int_2^4\left(\sqrt{x^2-8}\right)dx$
$3x+17\ge68$
$x^2+10x+35=0$
$x^2-30x+196$
$\lim_{x\to\infty}\left(\frac{\left(x+1\right)}{\left(x-3\right)}\right)$
$\frac{dy}{dt}=-0.02y^2$
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