Recent content by M55ikael

Yup. But back to the N(t) value question. Do you take into account all the values of N(t) when solving the inequality and getting rid of the modulus? I guess you solve the inequality for N(0) = 200 up to N(x)= 10 000

My book states the following: N(t) ≤ 10 000 => N(t) - 10 000 ≤ 0 => l N(t) - 10 000 l = -(N(t) - 10 000) EDIT: You are misunderstanding, please note that I have reversed the values. Like I said, this is the approach the book took, not me. They multiplied by -1 before integrating and got a...

Yes it does :-) I think you meantN(t)-10^{4}< 0\Rightarrow |-10^{4}+N(t)|=-10^{4}+N(t)

I wasn't accusing you of anything, I'm only trying to figure out how this works. How do I determine the value of N(t) if at all? I am still clueless. Why is this true: N(t)-10^{4}\geqslant 0\Rightarrow |-10^{4}+N(t)|=-(-10^{4}-N(t)) 10^{4}-N(t)/geqslant 0\Rightarrow...

Thanks so much! How do you determine N(t) is less than zero? Should I enter a value for t or something? Furthermore, my book has taken a slightly different approach ending up with: l N(t) - 10 000) l = Ce^(-kt) They then conclude that N(t) = 10 000 - Ce^(-kt) because N(t) is equal to...

Wow, that was quick! Could you explain what A is equal to? I raise - kt - C with e and get e^{-kt}*e^{-C}. How did you get A? Also, it seems I have missed something when it comes to absolute values, could you explain why N(t) becomes negative and 10^4 becomes positive? The initial...

Homework Statement I'm trying to solve this equation: dN/dt = k(10 000 - N(t)) The attempt at a solution dN/10 000 - N(t) = k dt I integrate, and I'm left with: - ln l10 000 - N(t)l = kt +C I raise both sides with e. I don't know what e^- ln(xxx) is so I multiply both...

Thanks, I allways knew it was pretty simple. But it's so easy to get your pluses and minuses mixed up, I couldn't get it reduced properly.

Homework Statement Hi! I have this frustrating equation I can't solve even though I have the answer. I'm trying to make two Eigenvectors for my matrix. I have found λ1 and λ2, but I can't solve the resulting equations. Homework Equations A = \begin{bmatrix} 2 & 1 \\ -\frac{1}{2} & 1...

Khanacademy teaches the same method as Vela described here: I was wondering though, if it's ok for the columns to have constants other than zero above the 1, meaning you would have a row-echelon matrix as opposed to a reduced row-echelon matrix. My logic tells me that the vectors would be...

Are there any proofs? It seems so simple I wonder why it's not mentioned in my book.

What's a pivot?

Yup, I just don't know how to write things properly in this forum yet. Is there a tutorial btw? Thanks for helping out everybody.

Yeah I followed HallsofIvy's suggestion and tossed one of the non-zero vectors, in this case v1. Then I did the elimination again with only three vectors and arrived at this system: a= 6/5c b= 7/5 c c= 5/6a Then I tossed v2, did the elimination and found that both constants had to be zero...

Thanks! English isn't my first language, so I didn't quite get the final sentence. Does that mean that the basis would be: \left[ \begin{array}{ccc}-5&7\\-2&4\\1&0\\0&1\end{array}\right]. ?