Answer:
it is
Step-by-step explanation:
Answer:183
Step-by-step explanation:18but with 3 183
Nice description. But you've omitted or overlooked the instructions. I'm assuming that you want to find the equation of this linear function.
13-(-2) 15 -5
The slope of this line is m = ----------- = --------- = ----
-4 -2 -6 2
Use the point-slope form of the equation of a straight line:
y-(-2) = (-5/2)(x-2), or y = -2 - (5/2)x + 5, or y = (-5/2)x + 3.
Answer:
40 bucks and i hope the help i not good englesh
Answer:
The answer is "
".
Step-by-step explanation:
Let add equation 1 and 2:
Using formula:
![x^2+y^2=1 \\\\](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D1%20%5C%5C%5C%5C)
convert to polar coordinates
![r=2](https://tex.z-dn.net/?f=r%3D2)
![=\int^{2\pi}_{0}\int^{1}_{0} (Z_2-z_1)r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r) \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0} \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4}) \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\](https://tex.z-dn.net/?f=%3D%5Cint%5E%7B2%5Cpi%7D_%7B0%7D%5Cint%5E%7B1%7D_%7B0%7D%20%20%28Z_2-z_1%29r%20%20%5C%20dr%20%5Cd%20%5Ctheta%5C%5C%5C%5C%3D%5Cint%5E%7B2%5Cpi%7D_%7B0%7D%5Cint%5E%7B1%7D_%7B0%7D%201-%20%28-1-x%5E2-y%5E2%29%20r%20%5C%20dr%20%5Cd%20%5Ctheta%5C%5C%5C%5C%3D%5Cint%5E%7B2%5Cpi%7D_%7B0%7D%5Cint%5E%7B1%7D_%7B0%7D%20%28%2B1%20%5Cpm%20r%5E2%29%20r%20%5C%20dr%20%5Cd%20%5Ctheta%5C%5C%5C%5C%3D%5Cint%5E%7B2%5Cpi%7D_%7B0%7D%5Cint%5E%7B1%7D_%7B0%7D%20%28-r%5E3%20%2B%20r%29%20%20%5C%20dr%20%5Cd%20%5Ctheta%5C%5C%5C%5C%3D%5Cint%5E%7B2%5Cpi%7D_%7B0%7D%20%28-%5Cfrac%7Br%5E4%7D%7B4%7D%2B%5Cfrac%7Br%5E2%7D%7B1%7D%29%5E%7B1%7D_%7B0%7D%20%20%5Cd%20%5Ctheta%5C%5C%5C%5C%3D%5Cint%5E%7B2%5Cpi%7D_%7B0%7D%20%28%5Cfrac%7B1%7D%7B4%7D%29%20%20%5Cd%20%5Ctheta%5C%5C%5C%5C%3D%28%5Cfrac%7B2%20%5Cpi%7D%7B4%7D%29%20%5C%5C%5C%5C%3D%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%20%5C%5C%5C%5C)