Answer:
627
Step-by-step explanation:
Answer:
x=12
Step-by-step explanation:
16x - (8x + 6) =90
The first step is to distribute the - sign
16x -8x -6 =90
Now we combine the like terms
8x-6 = 90
Add 6 to each side
8x-6+6 = 90+6
8x = 96
Divide each side by 8
8x/8 = 96/8
x =12
B. Add -12x to both sides. Then divide all terms by -2. The positive 12 on the right side will be a -6 and the -42 will be a positive 21. The inequality sign is always switched whenever the equation is divided by a negative number.
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:
