$-12=y-36$
$\lim_{x\to\infty}\left(\frac{9x^2-3x}{\sqrt{4x^2+x}}\right)$
$x-6\:+3x\:+4$
$\frac{d}{dx}x^2+y^2-3y=12$
$6\left(+x^3-3x^2-144x+440\right)$
$\frac{dy}{dx}=\frac{2x-1}{2y-2}$
$5x+3x-7+9$
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