$6x^2+9x+5x+3+x+2x^2$
$2\sin^2x=3\cos x$
$\frac{9\cdot sin\left(x\right)sec\left(x\right)}{tan\left(x\right)}$
$\left(-39\right)\::\:\left(-13\right)\:$
$5\cdot216-4\cdot6+6$
$sin\left(x\right)xcot\left(x\right)=cos\left(x\right)$
$-7x+6\le\frac{1}{2}x-13$
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