$\left(2x^3\:+\:3y\right)\:dx\:+\:\left(3x\:+\:y\:-\:4\right)\:\:\:dy\:=\:0$
$3\cos^2x+cosx=0$
$-2x^2-4x-45$
$\frac{5}{4.83}$
$x^2-4x+2^2$
$\lim_{x\to4}\left(\frac{2x-8}{5\:-\:\sqrt{5x\:+\:5}}\right)$
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