$8.4=\log\left(b\right)$
$x^2+18=11x$
$\lim_{x\to0}\left(\frac{\tan3\left(x+y\right)-\tan\left(3x\right)}{y}\right)$
$12-36+15.3-3-6\left(-9\right)$
$-2x\le-6$
$\frac{x^2+10x+12}{x+3}$
$-x^2+4x-6=0$
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