$\frac{sin^2\left(x\right)}{cos^2\left(x\right)}=tan^2\left(x\right)$
$x^2+10x-6$
$\lim_{x\to\infty}\left(x^3\cdot\left(\frac{1}{x}-sin\left(\frac{1}{x}\right)\right)\right)$
$y=\left(10x\right)^{ln\:10x}$
$4x+9<5x-1$
$y+3.4=0.5$
$\frac{\left(-2x^2-6y^2\right)}{4xy^3}$
Get a preview of step-by-step solutions.
Earn solution credits, which you can redeem for complete step-by-step solutions.
Save your favorite problems.
Become premium to access unlimited solutions, download solutions, discounts and more!