$28-2x=4$
$e^3=y$
$\lim_{x\to0}\left(\frac{1}{x^2}-\frac{1}{tan\left(x\right)}\right)$
$8x+4^2$
$-\frac{xe^{\frac{1}{2}x}\left(e^{\frac{1}{2}x}\left(1-x^2\right)^{\frac{1}{2\:}}\right)}{e^{\frac{1}{2}x}}$
$\left(8b^2+5c^3\right)^2$
$-2\left(-\infty\right)+1$
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