Thank you all for the help! It looks to me Ottawa University is the best option as of now. Although University of London is considerably cheaper, but I don't want to wait until 2014 to graduate when I can be done by early March with Ottawa. I've sent the information to my adviser to look over...
Hi billblack. Thanks for replying. I already tried that. Unfortunately, all I've found are either lower division classes which doesn't help, or the school only offers the degree as a whole and not individual classes. I have already taken enough hours at my school to graduate from there so that's...
Hi, I am currently looking for universities that offers Mathematics courses as an online or independent study course. I am only 2 courses short of graduation but unfortunately, so have my college funding. So if it's possible to take them online (which I believe is usually cheaper than on...
Oh you have no idea how glad I am to hear that I have finally at least gotten a few more questions on the right track and even two of them right.
Now, I think I know where you're heading with number 2. I just have to prove that x/(x-2) < x+1 then I can conclude that f(x)=O(h(x)).
With number...
*I failed horribly. First I had everything typed up and was about to post, but it failed because I accidentally typed in the wrong password. Then when I was retyping everything, I accidentally hit back and lost everything again. I am really frustrated now and I hope you won't mind my answers...
First off, thank you Tedjn for your reply. It really helped me understand the question better.
1. I didn't know about the Dijkstra's algorithm and that really helped me with question 1.
2. I read the definition of the big o notation and here's my second try:
10(x-1)(x-2)= O(x!)...
1. Explain how to find a shortest path between any given pair of vertices, given the length of edges connecting the vertices and the shortest distance between vertices (after doing the shortest-path calculation)
So I am given a two matrices, as said one with length of edges connecting and the...
Oh, I see where you're going...
(sinx)(secx)+1 <----since cosx*secx = cosx*1/cosx
(sinx)(1/cosx)+1
[[(cosx)(cosx)-(sinx)(-sinx)]/(cosx)^2]+1
[(cos2x+sin2x)/cos2x]+1
(1/cos2x)+1?
is the +1 supposed to stay?
1. Is there a value of b that will make...
x+b, x<0
g(x) = <
cosx, x=>0
continuous at x = 0?
differentiable at x = 0?
give reasons.
2. I'm not sure what are related equations for this. Limits?
3. So I try to find how to make it continuous at x = 0...
Ok, after reading DH's reply, I tried the question again and here's what I did.
y=(sinx+cosx)1/cosx
y'=(sinx+cosx)(-1/sinx)+(1/cosx)(-sinx+cosx)
=cosx/-sinx + sinx/-sinx + cosx/cosx - sinx/cosx
=cosx/-sinx - sinx/cosx
does that seem simple enough for you guys? I wish I were given the...
1. As the title states, I need to fine d/dx(sinx+cosx)secx.
2. I am given pretty much most of the derivatives and trig functions.
3. Here's my attempt to solve the question:
y=(sinx+cos)secx
y'=(sinx+cosx)(secxtanx)+(secx)(cosx-sinx)
=sinxsecxtanx+cosxsecxtanx+cosxsecx-sinxsecx...
So the question asks me to use the following information to graph the function f over the closed interval [-2, 5]
i) The graph of f is made of closed line segments joined end to end.
ii) The graph starts at the point (-2, 3)
ii) The derivative of f is the step function in the figure shown...