$\left(62.4\:\right)\int_{-4}^4\left(10-x\right)\:\left(\sqrt{16-x^2}\right)dx$
$x^{6\:}-y^6:\left(x+y\right)$
$\frac{x^4-3x^2+2x-5}{x+3}$
$9+y=13$
$a+8a^2+16a$
$\frac{a^{2}-2}{a-\sqrt{2}}$
$25+40x+16x^2$
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