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Nadusha1986 [10]
3 years ago
7

Can someone check my answer? Is it B? Thanks!

Mathematics
2 answers:
zloy xaker [14]3 years ago
8 0
Solution is where the 2 lines intersect

I would say the answer is 3rd option, whre both lines pass through that point
koban [17]3 years ago
4 0

Answer:

(1, 4), because both lines pass through this point

Step-by-step explanation:

i just took the test

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_ 1. If U = {1,2,3,4,5) and A= {2,4} then A' should be?
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alina1380 [7]

Answer:

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a=2d+3

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

but the integrand is even, so this is really just

\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

Substitute x = 1/2 arccot(u/2), which transforms the integral to

\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du

There are lots of ways to compute this. What I did was to consider the complex contour integral

\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}

and it follows that

\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}

7 0
2 years ago
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