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german
3 years ago
5

What two expressions represent 3/12​?

Mathematics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

A and C

3x1/12= 3/12

3x1/2= 3/12

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NEED DONE FAST GIVING 40 POINTS
dlinn [17]

To solve this problem you must apply the proccedure shown below:

1. The perimeter of a rectangle can be calculated with the following formula:

Perimeter=2Width+2Length

2. The width is:

Tan\alpha=opposite/adjacent\\Tan(72)=12/W\\ W=(12)(Tan(72))\\ W=3.89 in

3. Now, you must substitute the values into the formula for calculate the perimeter:

Perimeter=2(3.89in)+2(12in)\\ Perimeter=31.78in

Therefore, the answer is: 31.78 in

7 0
3 years ago
Can someone help from least to greatest ?
Misha Larkins [42]

Answer:

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3 years ago
Find the volume of the figure below.
Aliun [14]

Answer:

28

Step-by-step explanation:

Volume = Area of cross section x length

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3 years ago
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(x+x+2)÷4=189.5 solve for x
Stells [14]
First multiply by 4 to clear the division.
(X+x+2)=750
Simplify the parenthesis
2x + 2 = 750
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8 0
3 years ago
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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
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