$8x\le40$
$-x+1>-\frac{10}{7}$
$\int\left(\frac{\sec^2\left(x\right)}{\left(1+\tan\left(x\right)\right)^3}\right)dx$
$-41+61\left(-10\right)+12$
$4-\left(-2\right)-5+1$
$24x^3-54x$
$9\left(6+2y\right)-5+2y$
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