$\left(3^{-1}\right)^2$
$0-5\:+\:650$
$a^2-4a-96=0$
$\left(tan^{\left(5\right)}a+tan^{\left(7\right)}a\right)sec^{\left(2\right)}a$
$\left(5n-2p\right)^2$
$7\:-4\:3\:28$
$\lim_{n\to\infty}\left(\frac{3^{n+1}+1}{3^n}\right)$
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