Answer: 1.6, but if your looking for rounding numbers, its 2. I suggest 1.6 though.
Step-by-step explanation: hope this helps :)
Could you be a little more specific.
The 12th term in the sequence is -26.
Given,
B (n) = -4 – 2(n – 1)
We have to find the 12th term in the sequence:
To find the nth term in an arithmetic sequence:
An arithmetic sequence's nth term is determined by the formula a = a + (n - 1)d. The common difference, or d, is the difference between any two consecutive terms in an arithmetic series; it can be calculated by deducting any pair of terms starting with a and an+1.
Here,
First term, a = -4
Common difference, d = -2
nth term = 12
Now, let’s find 12th term
B(n) = -4 -2(n – 1)
B(12) = -4 – 2(12 – 1)
B(12) = -4 -2(11)
B(12) = -4 -22
B(12) = -26
That is, the 12th term in the sequence is -26.
Learn more about arithmetic sequence here:
brainly.com/question/11570734
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Answer: 1/3
<u>Step-by-step explanation:</u>
When adding or subtracting fractions, they must have the same denominator.
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:
