$\lim_{x\to0}\left(\frac{-5}{x^2}\right)$
$12\left(x+y\right)$
$5 y + 2 > 3 y - 8$
$11-7x\le-12$
$h\left(-4\right)=\left(-4\right)+\frac{1}{2}\left(-4\right)^4-\frac{3}{4}\left(-4\right)^3+10$
$y'-3y=xe^{4x}$
$x\left|4-x^2\right|=\left|-x^2+4\right|x$
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