<u>Solution-</u>
The two parabolas are,
By solving the above two equations we calculate where the two parabolas meet,
Given the symmetry, the area bounded by the two parabolas is twice the area bounded by either parabola with the x-axis.
![\therefore Area=2\int_{-c}^{c}y.dx= 2\int_{-c}^{c}(16x^2-c^2).dx\\=2[\frac{16}{3}x^3-c^2x]_{-c}^{ \ c}=2[(\frac{16}{3}c^3-c^3)-(-\frac{16}{3}c^3+c^3)]=2[\frac{32}{3}c^3-2c^3]=2(\frac{26c^3}{3})\\=\frac{52c^3}{3}](https://tex.z-dn.net/?f=%5Ctherefore%20Area%3D2%5Cint_%7B-c%7D%5E%7Bc%7Dy.dx%3D%202%5Cint_%7B-c%7D%5E%7Bc%7D%2816x%5E2-c%5E2%29.dx%5C%5C%3D2%5B%5Cfrac%7B16%7D%7B3%7Dx%5E3-c%5E2x%5D_%7B-c%7D%5E%7B%20%5C%20c%7D%3D2%5B%28%5Cfrac%7B16%7D%7B3%7Dc%5E3-c%5E3%29-%28-%5Cfrac%7B16%7D%7B3%7Dc%5E3%2Bc%5E3%29%5D%3D2%5B%5Cfrac%7B32%7D%7B3%7Dc%5E3-2c%5E3%5D%3D2%28%5Cfrac%7B26c%5E3%7D%7B3%7D%29%5C%5C%3D%5Cfrac%7B52c%5E3%7D%7B3%7D)
![So \frac{52c^3}{3}=\frac{250}{3}\Rightarrow c=\sqrt[3]{\frac{250}{52}}=1.68](https://tex.z-dn.net/?f=So%20%5Cfrac%7B52c%5E3%7D%7B3%7D%3D%5Cfrac%7B250%7D%7B3%7D%5CRightarrow%20c%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B250%7D%7B52%7D%7D%3D1.68)
Its data that is Qualitative Instead of Quantative
Answer:
x=15
Step-by-step explanation:
3(2x+5)=3x+4x
6x+15=3x+4x. *multiplied parentheses by 3*
15=3x+4x+-6. *moves all x to one side*
15=× * added
x=15
You flip the 3/4. So it's 4/3. Then you multiply. Thats basicly how all division works. Flipping the last fraction and multiplying.
So the square root of the quatity(x+7)=x-5
so you square both sides and get rid of the square root
x+7=(x-5)^2
(x-5)^2=x^2-10x+25
x+7=x^2-10x+25
subtract 7 from both sides
x=x^2-10x+18
subtract x from both sides
0=x^2-11x+18
so if xy=0 we can assume that x or/and y =0
factor out x^2-11+18
(find what two numbers multiply to get 18 and add to get -11)
-2 times -9=18
-2+(-9)=-11
(x-2)(x-9)=0
set them to zero
x-2=0
x=2
x-9=0
x=9
there are two answers -2 and -9
