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
.
Answer:
Step-by-step explanation:
From the figure attached,
Point B has been dilated to form point B'.
B(3, 1) → B'(6, 2)
→ B'[(2 × 3), (2 × 1)]
Since rule for the dilation of a point (x, y) by a factor of k is,
B(x, y) → B'(kx, ky)
By comparing the coordinates k = 2 is the scale factor by which the point B has been dilated about the origin.
Therefore, other vertices of the quadrilateral will be,
A(-2, 3) → A'(-4, 6)
C(1, -1) → C'(2, -2)
D(-3, -2) → D'(-6, -4)
Answer:
Lee abajo
Step-by-step explanation:
Te preguntan la fraccion de cada ejemplo verdad? la longitud es como se llama el ejemplo (1m) tu foto esta cortada y no veo el ejemplo A) o C) pero te puedo decir que el ejemplo B) es simplemente la fraccion 5/9 metros (el "/" quiere decir sobre)
It would be four 40×.1= four
