$\left(4u^3-3v^2\right)^2$
$x+4.18cm+5.89cm$
$x\left[\ln\left(\frac{x}{y}\right)+1\right]dy-ydx=0$
$\int_{x\left(x^2+1\right)}^1dx$
$sin^{-1}\frac{28.3}{53.1}$
$x^2-50=0$
$\lim_{x\to5}\left(\sin\left(x\pi\right)\right)$
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