$\int\left(te^{5t^2}\right)dx$
$2\cdot5^{x-4}+15=265$
$y\cdot y\cdot y$
$\left(-16+34\right)-8+\sqrt{49}$
$-20+-1$
$4x\:-\:8\:-\:8$
$y'+4y=cosx$
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