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Diano4ka-milaya [45]
3 years ago
5

Plesee help me im so confused

Mathematics
2 answers:
ikadub [295]3 years ago
8 0
D. because that is the description of an octagon
Tom [10]3 years ago
4 0
"Octo..." always means 8 of something ('octopus'), and if you're talking about polygons, then "regular" means that all sides have the same length and all angles have the same measure.
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HELP ASAP ILL GIVE YOU BRAINLEST
bezimeni [28]

Answer: C. f(x)=x^2+16x+63

 

Step-by-step explanation:

Vertex=(8,-1)

Focus=(8,-3/4)

Axis of Symmertry= x=8

Directrix= y=-5/4

I hope this helps you!

4 0
4 years ago
Subtrahend is when you subtract a number from another
aniked [119]

Didn't know if this was a question but if yes it is



4 0
4 years ago
HURRY!!!This table represents a proportional relationship. Use the table to identify the constant of proportionality between the
Shtirlitz [24]

Answer:

its 12

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Prove that dx/x^4 +4=π/8
insens350 [35]
\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}

Consider the complex-valued function

f(z)=\dfrac1{z^4+4}

which has simple poles at each of the fourth roots of -4. If \omega^4=-4, then

\omega^4=4e^{i\pi}\implies\omega=\sqrt2e^{i(\pi+2\pi k)/4} where k=0,1,2,3

Now consider a semicircular contour centered at the origin with radius R, where the diameter is affixed to the real axis. Let C denote the perimeter of the contour, with \gamma_R denoting the semicircular part of the contour and \gamma denoting the part of the contour that lies in the real axis.

\displaystyle\int_Cf(z)\,\mathrm dz=\left\{\int_{\gamma_R}+\int_\gamma\right\}f(z)\,\mathrm dz

and we'll be considering what happens as R\to\infty. Clearly, the latter integral will be correspond exactly to the integral of \dfrac1{x^4+4} over the entire real line. Meanwhile, taking z=Re^{it}, we have

\displaystyle\left|\int_{\gamma_R}\frac{\mathrm dz}{z^4+4}\right|=\left|\int_0^{2\pi}\frac{iRe^{it}}{R^4e^{4it}+4}\,\mathrm dt\right|\le\frac{2\pi R}{R^4+4}

and as R\to\infty, we see that the above integral must approach 0.

Now, by the residue theorem, the value of the contour integral over the entirety of C is given by 2\pi i times the sum of the residues at the poles within the region; in this case, there are only two simple poles to consider when k=0,1.

\mathrm{Res}\left(f(z),\sqrt2e^{i\pi/4}\right)=\displaystyle\lim_{z\to\sqrt2e^{i\pi/4}}f(z)(z-\sqrt2e^{i\pi/4})=-\frac1{16}(1+i)
\mathrm{Res}\left(f(z),\sqrt2e^{i3\pi/4}\right)=\displaystyle\lim_{z\to\sqrt2e^{i3\pi/4}}f(z)(z-\sqrt2e^{i3\pi/4})=\dfrac1{16}(1-i)

So we have

\displaystyle\int_Cf(z)\,\mathrm dz=\int_{\gamma_R}f(z)\,\mathrm dz+\int_\gamma f(z)\,\mathrm dz
\displaystyle=0+2\pi i\sum_{z=z_k}\mathrm{Res}(f(z),z_k) (where z_k are the poles surrounded by C)
=2\pi i\left(-\dfrac1{16}(1+i)+\dfrac1{16}(1-i)\right)
=\dfrac\pi4

Presumably, we wanted to show that

\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}=\frac\pi8

This integrand is even, so

\displaystyle\int_0^\infty\frac{\mathrm dx}{x^4+4}=\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^4+4}=\frac12\frac\pi4=\frac\pi8

as required.
6 0
4 years ago
Triangle ABC has vertices A(2,3), B(8,3), C(5,8). show that the triangle is isosceles​
Archy [21]
The coordinates match up, as B has 3 and A also has the coordinate 3. This means they are in the line and are isosceles
6 0
3 years ago
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