$5x-3x+4y$
$x^5-3x^{3\:}$
$-3\cdot\left(-4+10\right)$
$3x\:+\:4x\:-\:2x$
$\sec\left(x\right)\cdot\cos\left(x\right)=\cot\left(x\right)$
$-\frac{7}{2}\sec\left(bx^2+7\right)$
$\frac{3^6-729}{x-3}$
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