$\left(-\infty\right)^6$
$\left(-9x^3+7y\right)^2=\:$
$\:\frac{1}{x^2-3x+2}$
$\frac{\sin\left(x\right)-\tan\left(x\right)}{\cos\left(x\right)-1}=\tan\left(x\right)$
$-94-\left(-84-10\right)$
$\left(3+2\cdot4\right)\cdot3-3\cdot7$
$3x+3;\:x=-8;\:y=-2$
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