I would say no it’s not correct, the point is not in the correct place
800-20 is 780. A third of 780 is 780(1/3) = 260. 780 - 260 = 520. 25% of $520 is 0.25(520) = 130. 520 - 130 = 390. One sixth of 390 is 390(1/6) = 65. 390 - 65 = 325. So, Bob kept $325 for himself.
The perimeter is 32
I'm having a little difficulty finding the area because I forgot how the get the height... if I knew the height I would tell you.
I am so sorry I can not help all the way.
but if you find the height Area= base (11) times height (?)
Take the common denominator and divide by 3 then youll get 58
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:
