$\frac{5}{6}=\:\:\:\frac{.1}{6}$
$\lim_{x\to-4}\left(\frac{x-4}{x^2-16}\right)$
$h^5h^2h^{10}$
$\left(6x-1\right)^3$
$8x+10y=-60$
$\lim_{x\to16}\left(\frac{4-\sqrt[2]{x}}{16-x^2}\right)$
$\int\frac{28x^3\sec^2\left(7x^4\right)}{\tan\left(7x^4\right)}dx$
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