$-28\:\cdot\:-13$
$\int\arccot\left(\sqrt{x}\right)dx$
$\sqrt[3]{56x^3}$
$\sqrt{\sqrt[3]{t^4}}$
$\frac{dy}{dx}=\frac{1}{2}\cdot y\left(100-y\right)$
$\:6b^2-23b+21$
$5\:-\:x\:=12$
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