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Marina86 [1]
3 years ago
7

PLEASE HELP ASAP!!!!!!!!

Mathematics
2 answers:
-Dominant- [34]3 years ago
6 0

Answer:

It’s 2

Step-by-step explanation:

exis [7]3 years ago
3 0

Answer:

2

Step-by-step explanation:

Multiply that 3 by 2, obtaining:

 4       8

----- = -----

 3        6

These two ratios are equivalent.

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PLEASE HELP FAST!! I CANNOT GET THIS WRONG!!!!
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6

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Viefleur [7K]

Answer:

Start at the origin and plot the points.

Step-by-step explanation:

For A, start at the origin and then go 4 units to the left.

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6 0
2 years ago
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Can someone pls solve this
vredina [299]

Answer:

G=23/16

Step-by-step explanation:

8 0
2 years ago
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<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2
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
1 year ago
Will someone help me with this please
ANEK [815]

the first answer choice, associative property of addition

hope this helps. gl!

8 0
3 years ago
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