Ms. Geronimo has a $10 gift certificate to her local bakery.
1 answer:
You might be interested in
First let's factor a 5 from every term.
5v^2-30v+40
=5(v^2-6v+8)
The v^2-6x+8 can be further factored using the x method(as I call it, its not its official name of course), You can get
5(v^2-6v+8)
=5(v-2)(v-4)
Done! any questions just ask in the comment section.
**Edit: You needed the area model so here is how it goes, I cannot draw lines but I will try my best.
we have original equation 5v^2-30v+40
first simplify to 5(v^2-6v+8)
Now factor the second part but using area model.
v -2
_____________
v | | -2v |
--------------------------
-4 | -4v | 8 |
_____________
This is basically the area model the thing in the middle is suppose to be a rectangle split into 4 parts and then the numbers on top and left are suppose to be the factored equations while they multiplies to the number in the box.
<u>Is this efficient?</u>
Definitely not, it took me longer to draw the box than write my whole first explanation.
<u>Is this easy?</u>
No, again it took me about 170 times longer to think of this answer this way.
<u>Should you use this method?</u>
Sure if your teacher wants you to, but just saying later in algebra these methods are way to slow and there are way better methods to factor, so if you arn't solid in the area model just don't worry about it.
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)