$\frac{42}{7}+4$
$\frac{7-7\tan^4\left(x\right)}{sec^2x}=7\left(1-\tan^2\left(x\right)\right)$
$\lim_{x\to\infty}\:\left(3^x\right)\left(x-2^x\right)$
$\frac{dx}{9-\:e^{3x}}=\:\frac{dy}{ye^{3x}}$
$-8+4-15$
$xy+4y=x^3-x$
$3x^2-4x+7;x=-4$
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