$-64-293+18-45-48$
$\int_{\infty}^0\left(4e^{5x}\right)dx$
$\left(3r^3+5a^4\right)^2$
$\lim_{x\to3}\left(\frac{x^2-19x+48}{x-3}\right)$
$\int25dy$
$\int_0^9\left(-396+72x-4x^2\right)\cdot\left(-9+x\right)dx$
$4-2t\:>t-5$
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