The net area of the region in relation to the x-axis, is just the integral at those bounds, thus
![\bf \displaystyle \int\limits_{-\frac{\pi }{2}}^{\pi }~8cos(\theta )\cdot d\theta \implies 8\int\limits_{-\frac{\pi }{2}}^{\pi }~cos(\theta )\cdot d\theta \\\\\\ \left. 8sin(\theta )\cfrac{}{} \right]_{-\frac{\pi }{2}}^{\pi }\implies [8sin(\pi )]~-~\left[8sin\left(-\frac{\pi }{2} \right) \right]\implies [0]-[-8]\implies 8](https://tex.z-dn.net/?f=%5Cbf%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5E%7B%5Cpi%20%7D~8cos%28%5Ctheta%20%29%5Ccdot%20d%5Ctheta%20%5Cimplies%208%5Cint%5Climits_%7B-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5E%7B%5Cpi%20%7D~cos%28%5Ctheta%20%29%5Ccdot%20d%5Ctheta%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%208sin%28%5Ctheta%20%29%5Ccfrac%7B%7D%7B%7D%20%20%5Cright%5D_%7B-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5E%7B%5Cpi%20%7D%5Cimplies%20%5B8sin%28%5Cpi%20%29%5D~-~%5Cleft%5B8sin%5Cleft%28-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%20%20%5Cright%5D%5Cimplies%20%5B0%5D-%5B-8%5D%5Cimplies%208)
Factor the polynomial. 3x^2+11x+6
1 answer:
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Answer:
0 < y <= 6
Step-by-step explanation:
Given:
√(x² - 6x + 9)
if -3 <= x < 3
you can write (x² - 6x + 9)
as (x-3) * (x- 3) which is (x-3)²
so now you have √{(x-3)²}, which is the same as the absolute value of x-3. In mathematics it is written as:
| x - 3 |
So √(x² - 6x + 9) = | x - 3 |
See the graph in the attachment.
for -3 <= x < 3
you get the y values bigger then ( but not equal to) 0 and smaller then or equal to 6.
This is written as 0 < y <= 6
Given:
A right triangle on a coordinate plane.
To find:
The area of a triangle.
Solution:
We know that, the area of a triangle is
From the given figure it is clear that the base of the triangle is 4 units and the height is 3 units. So, the area of the given triangle is
The area of the given triangle is 6 units². Therefore, the correct option is B.