$9x\:+\:4\:-\:3x$
$\lim_{x\to\infty}\left(\frac{\ln\left(2^n+1\right)}{2^x}\right)$
$\left(1.6\left(-7.4\right)\right)$
$a+b+c=3za$
$4-n=13$
$100w^2-10w+4$
$\frac{1}{\cos\left(x\right)}-\frac{\cos\left(x\right)}{1+\sin\left(x\right)}=sec\left(x\right)$
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