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)
When a fraction is divided by the difference of 1/2 and 1/6 we get 2/3 what is the fraction
1 answer:
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Answer:
A. (-1, -4)
Step-by-step explanation:
The vertex can be found by converting the equation from standard form to vertex form.
<h3>Vertex</h3>Considering the x-terms, we have ...
y = (x^2 +2x) -3
where the coefficient of x is 2. Adding (and subtracting) the square of half that, we get ...
y = (x^2 +2x +(2/2)^2) -3 -(2/2)^2
y = (x +1)^2 -4
Compare this to the vertex form equation ...
y = a(x -h)^2 +k
which has vertex (h, k).
We see that h=-1 and k=-4. The vertex is (h, k) = (-1, -4).
On the attached graph, the vertex is the turning point, the minimum.
