$\frac{1}{4}x^2-7x+8+3x$
$y'-2x-7y=0$
$19+x=-30$
$\left(4x+\right)\left(x+3\right)$
$x+1\le5$
$\frac{\left(sin\left(x\right)+sin^2\left(x\right)\right)}{cos\left(x\right)+cos\left(x\right)\cdot sin\left(x\right)}=tan\left(x\right)$
$a\:-\:\left(-3\right)=\:9$
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