$3x+1\le x+4$
$\frac{1}{\tan x}+\tan x\left(\sin x\cdot\cos x\right)$
$\left(\:-\:5\:\right)\left(1\right)\left(\:1\:-\:2\:\right)$
$\left(\frac{4}{5}d+3\right)\left(\frac{8}{10}d+6\right)$
$4x^2+x-14$
$\frac{d}{dx}x+4y=1$
$\frac{\left(-6\right)\cdot8}{-12}$
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