$xy^{'\:}+y=\frac{4}{3}x^3+3x^2$
$\frac{-18b^7}{3b^4}$
$3x^2+6x>0$
$\left(-1260\right)\left(\left(4\right)\left(-5\right)\right)$
$2365.125+984.65+694.584$
$2x+5\cdot3x+4$
$n^2-12n+36$
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