Answer:
it equals 89/90
Step-by-step explanation:
the gcf of 89, 90 is 1, so the fraction can not be simplified. The answer is 89/90.
We are given a diagram with different rays and lines.
Note: A line can be extend more and more upto infinitely length from both of the ends but a ray can be extend from from end only.
We need to find the rays for the line BE.
We can see BE line has a point A in between line.
So, we can name the rays formed with end point A as AB and AE.
<h3>Therefore, AB and AE rays are part of line BE.</h3><h3>So, the correct option is 2nd option :</h3><h3>AB and AE.</h3>
Answer:
20
Step-by-step explanation:
If you work this out you get 1:
5/100*20
5 divided by 100 is 0.05 then 0.05 times 20 is 1
Answer:
So, the volume is:
Step-by-step explanation:
We get the limits of integration:
We use the spherical coordinates and we calculate a triple integral:
![V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4 \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\](https://tex.z-dn.net/?f=V%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B4%7D%7D%5Cint_0%5E4%20%20%5Crho%5E2%20%5Csin%20%5Cvarphi%20%5C%2C%20d%5Crho%5C%2C%20d%5Cvarphi%5C%2C%20d%5Ctheta%5C%5C%5C%5CV%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B4%7D%7D%20%5Csin%20%5Cvarphi%20%5Cleft%5B%5Cfrac%7B%5Crho%5E3%7D%7B3%7D%5Cright%5D_0%5E4%5C%2C%20d%5Cvarphi%5C%2C%20d%5Ctheta%5C%5C%5C%5CV%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B4%7D%7D%20%5Csin%20%5Cvarphi%20%5Ccdot%20%5Cfrac%7B64%7D%7B3%7D%20%5C%2C%20d%5Cvarphi%5C%2C%20d%5Ctheta%5C%5C%5C%5CV%3D%5Cfrac%7B64%7D%7B3%7D%20%5Cint_0%5E%7B2%5Cpi%7D%20%5B-%5Ccos%20%5Cvarphi%5D_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B4%7D%7D%20%20%5C%2C%20d%5Ctheta%5C%5C%5C%5CV%3D%5Cfrac%7B64%7D%7B3%7D%20%5Cint_0%5E%7B2%5Cpi%7D%20%5Csqrt%7B2%7D%20%5C%2C%20d%5Ctheta%5C%5C%5C%5C)
we get:
![V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B64%7D%7B3%7D%20%5Cint_0%5E%7B2%5Cpi%7D%20%5Csqrt%7B2%7D%20%5C%2C%20d%5Ctheta%5C%5C%5C%5CV%3D%5Cfrac%7B64%5Csqrt%7B2%7D%7D%7B3%7D%5Ccdot%5B%5Ctheta%5D_0%5E%7B2%5Cpi%7D%5C%5C%5C%5CV%3D%5Cfrac%7B128%5Csqrt%7B2%7D%5Cpi%7D%7B3%7D)
So, the volume is:
For this case we have the following expression:
p ^ 0
By properties of exponents we have:
Any number raised to the exponent zero is equal to one.
We have then, that applying this property:
p ^ 0 = 1
Answer:
the simplified form of p ^ 0 is:
D. 1
