Your answer is 76m.
a^2+b^2=c^2
57^2+b^2=95^2
3249+b^2=9025
b^2=5776
Take sqr on both sides and you get:
b=76m
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)
Positive correlation as they go to bottom left to top right
Answer:
The answer is c or x=25
Step-by-step explanation:
Point slope from would be 12x-y+26=0.
Black is the equation, 12x-y+26=0.
- Orange is <em>y-14=12(x+1)</em>
- Blue is <em>y-14=12(x-1)</em>
- Red is <em>y+14=12(x-1)</em>
- Green is <em>y-14=-12(x+1)</em>