$-10x=10$
$16a\:^4-\:c^6$
$8x\cdot2x^2\cdot2x^6$
$\int_1^{\infty}sin\left(2teta\right)dx$
$\frac{dy}{dx}=y\left(x\right)\left(4-y\left(x\right)\right)-3$
$\lim_{x\to2}\left(\frac{10-2x}{4x-2x^2}\right)$
$\lim_{x\to\infty}\left(\frac{4x+8}{3x^2-3x-5}\right)$
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