<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)
Answer:
<em>-5</em>
Step-by-step explanation:
-31=9+8x
we need to isolate the variable by moving 8x to the left side and -31 to the right side....therefore 8x becomes a negative, -8x, and -31 a positive....so now it would look like...
-8x=9+31
then we work the right side out...
-8x=40
to get x we need to divide -8x by 8 so we do the same to the right side as well....
-8x=40
÷8 ÷8
so...
<em>x=-5</em>
hope this helped you- have a good day bro cya)
Hey there!
When you multiply across, you get the following product:
<span>
8 </span>× –<span>23 </span>× –<span>21 </span>× –<span>29
</span>–184 × –21 × –29
3864 × –29
–112,056
The sign will be negative.
Hope this helped you out! :-)<span />
200 I’m pretty sure Yea your gonna have to trust your gut with this one
