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x^{2}-x-6=0 -x+3\gt 2x+1 ; line\:(1,\:2),\:(3,\:1) f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120)

Rewrite 1-t^ {2} as 1^ {2}-t^ {2}. The difference of squares can be factored using the rule: a^ {2}-b^ {2}=\left (a-b\right)\left (a+b\right). 81-t2 Final result : (t + 9) • (9 - t) Step by step solution : Step 1 :Trying to factor as a Difference of Squares : 1.1 Factoring: 81-t2 Theory : A difference of two perfect squares, ...

Jun 24, 2010 · In uniformly accelerated linear motion v(t)=a*t, so each point have coordinates (t, a*t). Then area under the curve it triangle. Area of the triangle equals 1/2*a*h where a is the length of the base of the triangle, and h is the height or altitude of the triangle. In this case s=1/2*t*(a*t) = 1/2*a*t^2 (see image) regards

What is the integral of 1/ (t^2+1) ?

Find the Fourier transform of $$f(t)=\frac1{1+t^2}$$ using contour integration that $$F\{f(t)\}=\int^\infty_{-\infty}\frac1{1+t^2}e^{2\pi ft}dt$$ How can I do this?