$0=-4$
$m=2x+6x+x^2+3$
$-2x^2-16x+40>0$
$\frac{\sec\left(1\right)+\tan\left(1\right)}{\csc\left(1\right)+1}=\tan\left(1\right)$
$4a^2-5a+3$
$49a^2-1$
$\frac{dy}{dx}=\frac{2x}{3y^2},\:y\left(0\right)=6$
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