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:
x = (-7 ±√89)/4 . . . . . the inverse of selection 3.)
Step-by-step explanation:
You have a quadratic ax²+bx+c=0 with a=2, b=7, c=-5. The solution is given by the quadratic formula as ...
... x = (-b ±√(b² -4ac))/(2a)
Putting your values into the formula gives ...
... x = (-7 ±√(7²-4(2)(-5)))/(2(2))
... x = (-7 ±√89)/4
_____
<em>Comment on answer choices</em>
For whatever reason, the answer choices appear to be "upside down," with numerator and denominator interchanged. The notation a/b means "a divided by b", which is to say that "b" is the denominator.
Answer:
The surface area of the larger prism is 4 times the surface area of the smaller one.
(which agrees with answer B in your list)
Step-by-step explanation:
Notice that the lateral surface area of prisms always consists of parallelograms, whose areas are given by the product base times height. = B x H
Therefore if the dimensions of the prism double, then the base and height of the parallelograms will double as well, making such individual ares of the lateral faces go from:
B x H to --> 2 B x 2 H = 4 (B x H)
that is 4 times the original lateral face's area.
with the faces for the top and bottom bases of the prism, something similar happens, so we conclude that the surface area of the larger prism is 4 times the surface area of the smaller one.
