Answer:
90°
Step-by-step explanation:
This is only if it's a regular rectangle or a regular square, this doesn't work for something like a trapezium
Answer:
I think it is
Step-by-step explanation:
4,8,5,15,6,12,7,6
8,5,15,6,12,7
5,15,6,12
15,6
15+6=21
For the first integral, z = π/4 is a pole of order 3 and lies inside the contour |z| = 1. Compute the residue:
![\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = \lim_{z\to\frac\pi4}\frac1{(3-1)!} \frac{d^{3-1}}{dz^{3-1}}\left[e^z\cos(z)\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmathrm%7BRes%7D%5Cleft%28%5Cfrac%7Be%5Ez%5Ccos%28z%29%7D%7B%5Cleft%28z-%5Cfrac%5Cpi4%5Cright%29%5E3%7D%2C%20z%3D%5Cfrac%5Cpi4%5Cright%29%20%3D%20%5Clim_%7Bz%5Cto%5Cfrac%5Cpi4%7D%5Cfrac1%7B%283-1%29%21%7D%20%5Cfrac%7Bd%5E%7B3-1%7D%7D%7Bdz%5E%7B3-1%7D%7D%5Cleft%5Be%5Ez%5Ccos%28z%29%5Cright%5D)
We have
![\dfrac{d^2}{dz^2}[e^z\cos(z)] = -2e^z \sin(z)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2%7D%7Bdz%5E2%7D%5Be%5Ez%5Ccos%28z%29%5D%20%3D%20-2e%5Ez%20%5Csin%28z%29)
and so
Then by the residue theorem,
For the second integral, z = 2j and z = j/2 are both poles of order 2. The second poles lies inside the rectangle, so just compute the residue there as usual:
![\displaystyle \mathrm{Res}\left(\frac{\cosh(2z)}{(z-2j)^2\left(z-\frac j2\right)^2}, z=\frac j2\right) = \lim_{z\to\frac j2}\frac1{(2-1)!} \frac{d^{2-1}}{dz^{2-1}}\left[\frac{\cosh(2z)}{(z-2j)^2}\right] = \frac{16\cos(1)-24\sin(1)}{27}j](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmathrm%7BRes%7D%5Cleft%28%5Cfrac%7B%5Ccosh%282z%29%7D%7B%28z-2j%29%5E2%5Cleft%28z-%5Cfrac%20j2%5Cright%29%5E2%7D%2C%20z%3D%5Cfrac%20j2%5Cright%29%20%3D%20%5Clim_%7Bz%5Cto%5Cfrac%20j2%7D%5Cfrac1%7B%282-1%29%21%7D%20%5Cfrac%7Bd%5E%7B2-1%7D%7D%7Bdz%5E%7B2-1%7D%7D%5Cleft%5B%5Cfrac%7B%5Ccosh%282z%29%7D%7B%28z-2j%29%5E2%7D%5Cright%5D%20%3D%20%5Cfrac%7B16%5Ccos%281%29-24%5Csin%281%29%7D%7B27%7Dj)
The other pole lies on the rectangle itself, and I'm not so sure how to handle it... You may be able to deform the contour and consider a principal value integral around the pole at z = 2j. The details elude me at the moment, however.
If the answer is not in integer form, I put it in decimal
2. 5
3. 11
4. 13
5. 15
6. 2.82842712
7. 24
8. 8
9. 20
10. 100.01499887
11. 10
12. 7
13. 4i
14. i√83
15. i√2163
16. 15.58845726
17. i58.59180830
18.i3.60555127
19. i6
20. 0
hard work!
Answer:
<u>3 batches of biscuits.</u>
Step-by-step explanation:
For one batch of biscuits, Mito uses 1/4 cup of flour to coat the countertop and the rolling pin.He also uses 2 1/2 cups of flour for each batch of biscuits he bakes
When he uses 7 3/4 cups of flour
Let the number of batches = x
convert the given data to decimal:
2 1/2 = 2.5 and 1/4 = 0.25 and 7 3/4 = 7.75
So, the equation representing the problem is:
2.5 x + 0.25 = 7.75
Solve for x
2.5x = 7.75 - 0.25 = 7.5
x=7.5/2.5 = 75/25 = 3
<u>So, the number of batches of biscuits = 3 batches.</u>
