$a1=sin^{-1}\left(\frac{r\cdot sin40}{a}\right)$
$\frac{15}{3}+\left(6-4\right)^2$
$2x-3\ge x+2$
$2sin^2\theta\:-sin\theta\:=1$
$\left(\frac{1}{2}x^4-\frac{2}{3}y\right)^3$
$\left(5x^4-3x^2-6\right)\left(3x-4\right)$
$3b=48$
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