Answer:
yes it is arithmetic because:
28 - 10= 18
18 - 10= 8
8 - 10= -2
it keeps decreasing by the same number (10). Therefore, it is arithmetic.
2 / -18 and 4/ a (1/2 ) I THINKKK
The two parabolas intersect for
![8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2](https://tex.z-dn.net/?f=8-x%5E2%20%3D%20x%5E2%20%5Cimplies%202x%5E2%20%3D%208%20%5Cimplies%20x%5E2%20%3D%204%20%5Cimplies%20x%3D%5Cpm2)
and so the base of each solid is the set
![B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}](https://tex.z-dn.net/?f=B%20%3D%20%5Cleft%5C%7B%28x%2Cy%29%20%5C%2C%3A%5C%2C%20-2%5Cle%20x%5Cle2%20%5Ctext%7B%20and%20%7D%20x%5E2%20%5Cle%20y%20%5Cle%208-x%5E2%5Cright%5C%7D)
The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas,
. But since -2 ≤ x ≤ 2, this reduces to
.
a. Square cross sections will contribute a volume of
![\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x](https://tex.z-dn.net/?f=%5Cleft%282%28x%5E2-4%29%5Cright%29%5E2%20%5C%2C%20%5CDelta%20x%20%3D%204%28x%5E2-4%29%5E2%20%5C%2C%20%5CDelta%20x)
where ∆x is the thickness of the section. Then the volume would be
![\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-2%7D%5E2%204%28x%5E2-4%29%5E2%20%5C%2C%20dx%20%3D%208%20%5Cint_0%5E2%20%28x%5E2-4%29%5E2%20%5C%2C%20dx%20%5C%5C%5C%5C%20%3D%208%20%5Cint_0%5E2%20%28x%5E4-8x%5E2%2B16%29%20%5C%2C%20dx%20%5C%5C%5C%5C%20%3D%208%20%5Cleft%28%5Cfrac%7B2%5E5%7D5%20-%20%5Cfrac%7B8%5Ctimes2%5E3%7D3%20%2B%2016%5Ctimes2%5Cright%29%20%3D%20%5Cboxed%7B%5Cfrac%7B2048%7D%7B15%7D%7D)
where we take advantage of symmetry in the first line.
b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of
![\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x](https://tex.z-dn.net/?f=%5Cdfrac%5Cpi8%20%5Cleft%282%28x%5E2-4%29%5Cright%29%5E2%20%5C%2C%20%5CDelta%20x%20%3D%20%5Cdfrac%5Cpi2%20%28x%5E2-4%29%5E2%20%5C%2C%20%5CDelta%20x)
We end up with the same integral as before except for the leading constant:
![\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-2%7D%5E2%20%5Cfrac%5Cpi2%20%28x%5E2-4%29%5E2%20%5C%2C%20dx%20%3D%20%5Cpi%20%5Cint_0%5E2%20%28x%5E2-4%29%5E2%20%5C%2C%20dx)
Using the result of part (a), the volume is
![\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%5Cpi8%20%5Ctimes%208%20%5Cint_0%5E2%20%28x%5E2-4%29%5E2%20%5C%2C%20dx%20%3D%20%5Cboxed%7B%5Cfrac%7B256%5Cpi%7D%7B15%7D%7D%7D)
c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is
![\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt3%7D4%20%5Cleft%282%28x%5E2-4%29%5Cright%29%5E2%20%5C%2C%20%5CDelta%20x%20%3D%20%5Csqrt3%20%28x%5E2-4%29%5E2%20%5C%2C%20%5CDelta%20x)
and using the result of part (a) again, the volume is
![\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-2%7D%5E2%20%5Csqrt%203%28x%5E2-4%29%5E2%20%5C%2C%20dx%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%20%5Ctimes%208%20%5Cint_0%5E2%20%28x%5E2-4%29%5E2%20%5C%2C%20dx%20%3D%20%5Cboxed%7B%5Cfrac%7B512%7D%7B5%5Csqrt3%7D%7D)
Answer:
3.17x10^-3
Step-by-step explanation:
Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the
10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.