Sensei

claw836 wrote: yo, could ya just post the mat problem ya have ? cous numbers make more sens then letter here .. btw wat fruit peep said was like = bullcrap or almost right ... just post the math problem plox I don't understand what you are saying, and when finding the derivative you need to keep the variables in there, or the whole thing will fall apart. 
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Sensei

Sure but Algebra has a lot of topics could you be more specific? 
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Sensei

Let's see...factorisation,if it's possible? ^^ 
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bubblegumsamurai wrote: Let's see...factorisation,if it's possible? ^^ since Isaosan is not available I can help in his place for now. I assume you are speaking of factoring to solve a quadratic function, am I right? 

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Sensei

fruitschinposamurai wrote: hijitaka wrote: hmmmmm...i thought this is the math class so what are you guys talking about?? ok i have a question: can someone explain me why cos(x) is the derivative of sin(x) thanks Use the difference quotient f`(x)=(f(x+deltax)f(x))/deltax so it would be (sin(x+deltax)sin(x))/deltax then it would become (sin(x)cos(deltax)+sin(deltax)cos(x)sin(x))/deltax factor sin(x)((cos(deltax)1)/deltax)+cos(x)((sin(deltax)/deltax) the limit as deltax goes to zero of (cos(deltax)1)/deltax is zero and the limit as deltax goes to zero of sin(deltax)/deltax is 1 so the first term is eliminated and we are left with cos(x). thank you! 
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Sensei

I think I should take over the position of assistant teacher in this class because Okita doesn't really understand mathematics.....

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Sensei

fruitschinposamurai wrote: I think I should take over the position of assistant teacher in this class because Okita doesn't really understand mathematics..... you are right! i think you should take his position^^ 
*sleeping*


Sensei

Musashi_like_person wrote: bubblegumsamurai wrote: Let's see...factorisation,if it's possible? ^^ since Isaosan is not available I can help in his place for now. I assume you are speaking of factoring to solve a quadratic function, am I right? Yup.I don't really understand.And the make y as a subject thing too. 
Leaving CR.


Sensei

Musashidono isn't here so I will take his place I am assuming you talking about quadratic functions that look like this y=ax^2+bx+c Where a is any number other than zero, and b and c are any number including zero. So lets start with an easy example x^2+2x+1 This is a perfect square trinomial because when factored it looks like this (x+1)^2 The trick to this is looking at the first and last terms first. IF both the first term and last term are squares, then you move on to part 2, which is looking at the middle term. The middle term has to be twice the value of the square root of the first and last terms. Lets try it out x^2+4x+4 the first term x^2 is a square because its degree is 2 and it has an "invisible" coefficient of 1. The last term 4 is a square because it is 2 squared. Now we know the square root of the first term is x and the square root of the last term is 2. Now we have to multiply these two square roots together; and after doing so we get 2x. The last step is to take the new number we got, and to multiply it by 2 because the original factor is squared. After multiplying 2 by 2x we get 4x which is the middle term. This means we are looking at a perfect square trinomial. Once we now that the rest is easy. First we put the square root of the first term and the square root of the last term 4 in the parentheses. Then we take the sign of the middle term and put it between the x and the 2. So the end result would look like this (x+2)^2 One last example 4x^2+4x+1 1. The first and last term are squares (the square root of 4x^2 is 2x and the square root of 1 is 1) 2. The middle term is twice the product of the two square roots (twice the product of 2x and 1 which makes it 4x) 3. We put the square of the first term and last term and put it into parentheses, then we take the sign of the middle term and put it between them 4. Through this we get (2x+1)^2 
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Sensei

fruitschinposamurai wrote: Musashidono isn't here so I will take his place I am assuming you talking about quadratic functions that look like this y=ax^2+bx+c Where a is any number other than zero, and b and c are any number including zero. So lets start with an easy example x^2+2x+1 This is a perfect square trinomial because when factored it looks like this (x+1)^2 The trick to this is looking at the first and last terms first. IF both the first term and last term are squares, then you move on to part 2, which is looking at the middle term. The middle term has to be twice the value of the square root of the first and last terms. Lets try it out x^2+4x+4 the first term x^2 is a square because its degree is 2 and it has an "invisible" coefficient of 1. The last term 4 is a square because it is 2 squared. Now we know the square root of the first term is x and the square root of the last term is 2. Now we have to multiply these two square roots together; and after doing so we get 2x. The last step is to take the new number we got, and to multiply it by 2 because the original factor is squared. After multiplying 2 by 2x we get 4x which is the middle term. This means we are looking at a perfect square trinomial. Once we now that the rest is easy. First we put the square root of the first term and the square root of the last term 4 in the parentheses. Then we take the sign of the middle term and put it between the x and the 2. So the end result would look like this (x+2)^2 One last example 4x^2+4x+1 1. The first and last term are squares (the square root of 4x^2 is 2x and the square root of 1 is 1) 2. The middle term is twice the product of the two square roots (twice the product of 2x and 1 which makes it 4x) 3. We put the square of the first term and last term and put it into parentheses, then we take the sign of the middle term and put it between them 4. Through this we get (2x+1)^2 Thanks,Commander! 
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Sensei

The Potato Theorem
One of the fundamental theorems of Physics and Mathematics 
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Sensei

fruitschinposamurai wrote: The Potato Theorem One of the fundamental theorems of Physics and Mathematics hi Kondosan!! i have a question!! how can i show that all parabels with the formel f(x)=ax²+x+3 meet at one fix point? here it is the point (0/3)......i don't know how i can prove it...can you help me???? thanks 
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Sensei

hijitaka wrote: hi Kondosan!! i have a question!! how can i show that all parabels with the formel f(x)=ax²+x+3 meet at one fix point? here it is the point (0/3)......i don't know how i can prove it...can you help me???? thanks I finally receive a question and it is a relatively simple one, how boring. Anyways, the proof is simple. The yintercept of all equations in the form ax^2+bx+3 is always three. This occurs because the f(x) intercept is the value of f(x) when x=0. In the case of second degree polynomials the f(x) intercept is always the constant "c" that is usually at the end, therefore a(0)^2+b(0)+3 is always 3. Now that you understand(hopefully), you can formulate a rigorous proof by yourself. 
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Sensei

fruitschinposamurai wrote: hijitaka wrote: hi Kondosan!! i have a question!! how can i show that all parabels with the formel f(x)=ax²+x+3 meet at one fix point? here it is the point (0/3)......i don't know how i can prove it...can you help me???? thanks I finally receive a question and it is a relatively simple one, how boring. Anyways, the proof is simple. The yintercept of all equations in the form ax^2+bx+3 is always three. This occurs because the f(x) intercept is the value of f(x) when x=0. In the case of second degree polynomials the f(x) intercept is always the constant "c" that is usually at the end, therefore a(0)^2+b(0)+3 is always 3. Now that you understand(hopefully), you can formulate a rigorous proof by yourself. yeah thanks kondosan! btw can you explain me what approximation is?? 
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Sensei

Well that is a very broad subject. Approximation is just like how it sounds, it does not give the exact value, rather it approximates the value. 
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