$-3\:+\:17\:+\:1\:$
$\left(\frac{1}{4}x\right)\left(12y^2\right)$
$-8x=40$
$2\cos^2\left(\theta\right)-\sqrt{3}\cos\left(\mathit{\theta}\right)=0$
$\:-5m-2m+3$
$4=-16t^2+148$
$\left(-7.3+13\right)$
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