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\(e^2cos(x)\)
\(f(x)=x^2+4x-12\) \(y=x^2-2x+6x-12\) \(y=x(x-2)+6(x-2)\) \(y=(x-2)(x+6)\) thus: \(x_1=2\) and \(x_2=-6\)
\(\frac{2}{cos(x)}\)
\[\sqrt\frac{1}{x^2+2}-\frac{2x}{6y}\] \(\frac{33}{26}\)
\[sin(\theta)\times cos(\theta)\]
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\[360=2\pi\]
\[\sqrt{16}=\pm4\]
\[\huge{size}\] \huge{size}
\[\huge{\sqrt[n]{x}}\]
\[\large{ f(x)=\sqrt[3]{x} \\ y=\sqrt[3]{x} \\ f^{-1}(x)=x=\sqrt[3]{y} \\ f^{-1}(x)=x^3=y \\ f^{-1}(x)=x^3 }\]
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