$10y^2-x=0$
$\lim_{x\to-\infty}\left(\frac{\sqrt{1+81x^6}}{4-x^3}\right)$
$\text{cota}\left(\tan+\text{cota}\right)=\csc^2a$
$xy-x^{1.7}y^{2.2}$
$1+\frac{\csc\left(x\right)}{\sec\left(x\right)}-\cot\left(x\right)=\cos\left(x\right)$
$\int_4^{2y+8}dy$
$16x^2-16x+4$
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