$13+12x-x^2$
$1+7x=8$
$\frac{dy}{dx}\left(5y_{3x^4y^3}=2x^4+3\right)$
$\left(-8y^5\right)\left(-6y^4\right)$
$90-\left(-5-12\right)$
$\sec^4\left(x\right)=\tan^2\left(x\right)\sec^2\left(x\right)+\sec^2\left(x\right)$
$-7=1+\frac{x}{3}$
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