$16\left(\frac{-3}{8}\right)+16\left(\frac{1}{4}\right)$
$\left(3x^2+2y^2\right)^5$
$6x\left(6x-y\right)-4\left(-y-6x\right)$
$.08\left(11-x\right)$
$10x-6x+x$
$\frac{dy}{dx}=\left(y^2+3y+2\right)x^2$
$6\cdot m^5\cdot\frac{1}{m}$
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