$3\left(x-2\right)\ge4x-4$
$y'=cy$
$x^2y^3-6xy^3+9x$
$tan\:\left(x\right)=2.5$
$\frac{\sin\:^3\left(x\right)+\sin\left(x\right).\cos^2\left(x\right)}{\tan\left(x\right)}$
$7-19+15-8+6-13-6$
$\frac{x^3}{3}-\frac{3}{5}^3$
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