Answer:
3.5 gallons a week
Step-by-step explanation:
Well 64 gl Oz is basically half a gallon. So half a gallon of water a day would result in 0.5*7 (7 days in a week) and you would get 3.5 gallons.
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:
X = 6
Step-by-step explanation:
2x + (-6x) = -24
2x - 6x = -24
-4x = -24
4x = 24
x = 6
Answer:
Estimation is just giving a general answer to the situation and can be specified on specifics depending on the numbers used and whichever question it's asking. But also remember that whenever you estimate your answer will be an approximate answers, not an exact.