http://www.purplemath.com/modules/systprob2.htm
Solve decompostition problem
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Find the equation of the parabola that passes through the points (–1, 9), (1, 5), and (2, 12).
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About Me
- egrasso10
- I like to play hockey. Food that is not good for me I still eat all the time, doritos and goldfish. I think it is cool when you watch Monday Night Football and the players give respect to their high school instead of college. WOW, that is the type of high school Norton should be, so amazing you want to mention how great it is whenever you can.
33 comments:
you break up the problem and devide the numerator by the factors of the denominator
Factor the denominator and then im lost
you separate the equation and divide the number above the division sign by the factors of the number below the division sign.
In the hood math is frowned upon
i get lost at A B C when u need to distribute
you break the problem down as much as you can and just then solve it and yeahh..
to solve this i would break up the denominator by factoring out the equation. i would seperate the denominator into seperate fractions and put variables on top. then you do something that reminds me of cramers rule. this will give you three equations which you then solve by any means; substitution, graphing (not as acurate) or any other way. solving these equations will give you the numerations from the begining equation.
conorhotblog.blogspot.com or mybetterisbetterthenyourbetter.blogspot.com
the distributing seems easy now that u look at the answer
you gotta seperate the problem, then simplify or something. i dont really get it.
There was an x in the original problem then they just start adding more letters for some dumb reason, then they factor it, i get that, then they multiply it by something and "compare coefficients" (w/e that means) then they do something that gets them more numbers that they claim are the answers.
Mr. Grasso this problem would be better without the suck in it. They lost me at the step:
=(A+B+C)x^2+(B+3C)x+(-A-2B+2C)1. I dont feel like putting any effort into this and I'm sitting next to Tim who wrote a thesis on this thing making me look like a tool.
You seperate the problem and then divide the numerator by the factors of the denominator...
I have the slightest idea about how to solve this problem
you simplify the problem by factoring the denominator and labeling the numerator as a,b,c etc. Then you you make an equation using a numerator term that will be factored to the other denominators, but you don't use its that numerators own denominator. Then you solve, and its confusing....
you factor the denominator, separate it and then im lost with the rest.
I think I might understand the decomposition problem. You factor and break it up and somehow it all works out. The only problem I really had with it was that the trees in the first problem were wicked cheap. I wish I could get trees that cheap...
well first you need to know the factors of the denominator. you then seperate the equation and divide the numerator by the factors of the denominator.
i get it now after a little help from nigel
you do A/B*C B/A*C C/A*B
THEN YOU PULL OUT FACTORS NOT AS HARD ONCE YOU GET IT
P.S. CAPS ARE WAY COOLER
i can factor the denominator then i get lost and have no clue what to do next.
DAN I MADE A TRADE WITH JAKE SO THAT 20 BUCKS WILL SOON BE MINE
read some of the comments for this post, and that's how you do the problem
First, you have to factor the denominatior. After this you put A B and C over the factors. Then you have to take the denominators of what is not under the letter to foil it out. Then you take the denominators of not B to use for B. You do that for C but use the denominators of A and B. Now you foil what you have. I do not get what you do on the next step. I can't figure out where the numbers are coming from. You set A+B+C equal to O. Then you take the equation equal with X and set it equal to 5 since that has an X. After you set the quation with the 1 and set it equal to 7 since that is a constant. Then you can solve the systems of equations to get A,B, and C. You put these over the factors.
First, you have to factor the denominatior. After this you put A B and C over the factors. Then you have to take the denominators of what is not under the letter to foil it out. Then you take the denominators of not B to use for B. You do that for C but use the denominators of A and B. Now you foil what you have. I do not get what you do on the next step. I can't figure out where the numbers are coming from. You set A+B+C equal to O. Then you take the equation equal with X and set it equal to 5 since that has an X. After you set the quation with the 1 and set it equal to 7 since that is a constant. Then you can solve the systems of equations to get A,B, and C. You put these over the factors.
You factor the denominator and set the factors under A,B and C. Then you multiply by the common denominator and get the numerator equal to "A" times the two other factors and so on. Then you regroup them with their factors. Finally, you set them equal to the factors in the numerator and solve
OR You factor the denominator and set the factors under A,B and C. Then you multiply by the common denominator and get the numerator equal to "A" times the two other factors and so on. Then you regroup them with their factors. Finally, you set them equal to the factors in the numerator and solve
First, you have to factor the denominatior. After this you put A B and C over the factors. Then you have to take the denominators of what is not under the letter to foil it out. Then you take the denominators of not B to use for B. You do that for C but use the denominators of A and B. Now you foil what you have. I do not get what you do on the next step. I can't figure out where the numbers are coming from. You set A+B+C equal to O. Then you take the equation equal with X and set it equal to 5 since that has an X. After you set the quation with the 1 and set it equal to 7 since that is a constant. Then you can solve the systems of equations to get A,B, and C. You put these over the factors.
Finding the partial fraction is basically just separating one huge fraction into three smaller fractions. First you are to factor out the denominator as much as you can. Then you are to place variables on top of these factored denominators. After that you create an equation by taking your first numerator and setting it equal to the variables and the opposite denominators. Then you are to solve the equation for each variable.
From Sarah..
we should watch Miracle in class
It has hockey and that quote i always think since family size doritios. "Cause we're a family"
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