For there to be a region bounded by the two parabolas, you first need to find some conditions on
. The two parabolas must intersect each other twice, so you need two solutions to
You have
which means you only need to require that
. With that, the area of any such bounded region would be given by the integral
since
for all
. Now,
by symmetry across the y-axis. Integrating yields
![=4\left[c^2x-\dfrac{16}3x^3\right]_{x=0}^{x=|c|/4}](https://tex.z-dn.net/?f=%3D4%5Cleft%5Bc%5E2x-%5Cdfrac%7B16%7D3x%5E3%5Cright%5D_%7Bx%3D0%7D%5E%7Bx%3D%7Cc%7C%2F4%7D)
Since
, you have
.
Label the vertices of the quadrilateral A, B, C, D. Angle C is supplementary to the angle of measure 60 degrees, so its own measure is 120 degrees.
AD and BC are parallel, as are AB and CD, so ABCD is a parallelogram. This means angle A also has measure 120 degrees. The angle adjacent to the one with measure 
, part of angle A, then has measure 
.
The three angles above the line through vertex A are supplementary, so we have
Answer:
x = -21/11
Step-by-step explanation:
6(2x + 4)2 = (2x + 4) + 2
12(2x +4) = 2x + 4 + 2
24x + 48 = 2x +6
22x = -42
x = -21/11
First multiply 0.8 times m, then multiply 0.8 times -5.
This will give you 0.8m-4 = 10
Then add four to both sides to get you 0.8m =14.
Divide both sides by 0.8 to get m=17.5
Hope this helped. Good luck! :)
