$-y^2+4y+16$
$\frac{40}{\frac{\:2}{5}}$
$y=\frac{\left(3x^2+1\right)^4\sqrt{3-2x}}{\left(x+4\right)^2}$
$x^2\:14x+49$
$3^3\cdot3^4\cdot3$
$\sin\left(x\right)\:x\:sec\:x\:=tg\:x$
$\left(3x+8y=15\right)\left(2x-8y=10\right)$
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