$x+8=18.48$
$\lim\:_{x\to\:0\:}\left(\frac{4xy}{3x^2+y^2}\right)$
$\frac{dy}{dx}=\frac{2x-9}{y}$
$2\cdot\left(3\right)-\left(-3\right)\cdot\left(-2\right)$
$\left(2x^2+z\right)^5$
$\left(x^{2}+x+1\sqrt{3x^{3}+4x^{2}+2x+7}$
$3x+18=48$
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