$\frac{1}{2}t^5+\frac{3}{4}w^3\:^2$
$4+2\left(3+x\right)=4x-10$
$\frac{d^2u}{dx^2}=-5\frac{d^2}{dxdy}$
$+100+2$
$\:-132.51\:-1.39+.39+2.39+2.39+3.30+11.39+21.39+26.39\:+0.39$
$6\csc x+7=0$
$a-3a+2ac-3bc$
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