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[math]\displaystyle{ \int_{0}^{2\pi}\!\! \int_{0}^{\infty} e^{-r^2}r\,dr\,d\theta = \pi }[/math]

$\mathbb{R}^n$

<math>\mathbb{R}^n</math>

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$

$c^2 = \sqrt{a^2+b^2}$

File:Chicken.pdf