jordan

You didn't explain what to do if the x is in the denominator, so I simply took it out. In this step: [math]\frac{x^a}{ax^n}*\frac{{a^{-1}}{x^{-n}}}{{a^{-1}}{x^{-n}}}[/math] I multiplied the donomators to get 1 and then that put everything in the numerator. I pretty sure but not positive that it is correct. From there I combined the x's in to get x2(a-n). I'm thinking the 2 doesn't have to be there now, but I'll leave it for a second. Now, the rule says I subtract 1 from the exponent to find the new one: x2(a-n)-1. Then, multiplying the coefficient by the exponent yeilds a-1*[2(a-n)] or a-1*(2a-2n). Distributing the a-1 gives me 2-(2n/a). This is the new coefficient.

I assume you ae saing hypnotism works. If so, could you explain how?

Typo: #2 should be 26x12 Simple mistake: #7 should be -45x-10 For 8-10 the method is questionable, but all the same. I'll show number 10: [math]\frac{x^a}{ax^n}*\frac{{a^{-1}}{x^{-n}}}{{a^{-1}}{x^{-n}}}[/math] [math]\frac{{x^a}{a^{-1}}{x^{-n}}}{1}[/math] [math]a^{-1}x^{2a-2n}[/math] As I type I think I see what happened to my a. [math](2-\frac{2n}{a})x^{(2a-2n)-1}[/math] I'm getting myself a little confused now. Let me know if you see any errors.

How about if I made the last one [math]-nx^{2(-n-1)}[/math]? I'm really out of time and really late so I'll check back later.

But the others are? I was hoping to use the spoiler feature but I forgot it didn't work. Hope this didn't make anyone mad but posting the answers like that.

I was looking for the first answer. I'm trying to find something very cheap to get some obscure peices (like the one I mentioned) but haven't found (or looked) a whole lot yet.

It's seems as though I can't make the exponent more than one figure long: as in #2 and #10 which are supposed to be ^12 and ^(-n-1) respectively. How do I do that?

[math]3x^2[/math] [math]16x^{12}[/math] [math]-2x^{-3}[/math] [math]x^{-\frac{4}{5}}[/math] [math]\frac{3}{2}x^2[/math] [math]\frac{-5}{2}x^\frac{-3}{2}[/math] [math]-45x^{-8}[/math] [math]-x^{\frac{-3}{2}}[/math] [math]-8x^{\frac{-13}{6}}[/math] [math]-nx^{(-n-1)}[/math]

Where do you get this music from?

So you're taking the stance that [math].999\bar9[/math] doesn't equal 1?

One I just found today in a pile of music and started to play through was Liszt's Rhapsodie Hongroise #2. One of my altime favorites...so far. I've played through the first movement a few times and just skimmed through movement #2. Haven't touched #3 yet. But the first two are great. That is, if you like solo classical piano music. Just thought I'd add it to the list.

So then [math].00\bar01[/math] is equal to zero also? Assumably by all the same proofs above.

Hhhmmm... I guess the tabloids really are filled with lies because that picture contains neither a power point presentation nor a bar chart. I was so hoping for a bar chart when I saw much time you used up to make that picture. Oh well.

Ouch. Maybe you didn't realize that was an actual study (done in your own country I believe). Maybe you did an didn't care. Whatever you like.

Study shows little kids are more likely to be picked on by big kids rather than the other way around. Oh, wait. That one was a real headline.

Wouldn't cameras give away the security system a bit. If they saw them, they'd probably just destroy them, run, and come back another day.

But VB definitly has it's advantages over phpBB. For one, I don't remember a quick reply box like the one I'm typing in now on phpBB.

91 is the right answer. What I'm not sure about is what "r" is. Bloodhound said up there something about "take a rby r square where r<=n" and I didn't follow that. Also, I don't understand how you got from "n^2 + (n-1)^2.......+1^2" to "which is equal to n(n+1)(2n+1)/6" that.

For a square six units on a side, yes.

Here's the old thread How did you get your answer, bloodhond?

Yeah, I didn't do that quite right. I think after a quick edit it'll be right. Take the number of squares in each row and call that n. Then, starting with with X=0, do (x+1)^2. After that, add 1 to X and repeat the process. Continue doing so until X=n. So for a square with 2 in a row: (0+1)2=1 (1+1)2=4 Now, X=1. This is one less than n (as asked for by the n-1 part). Therefore you stop and add all the terms together for a total of 5.

I don't see what the acronym tag does.

[math]\sum_{X=0}^{n-1}{(X+1)^2}[/math] That's seems about right.

Interesting. I can understand a lot of what you're saying because that sounds a lot like me. I never ever used to study. Now I try to and I can't read more than a page before I need to remind myself to focus on what I'm reading. I know what you mean with a lot of those symptoms: most I can relate to and some I can't. I find a lot of times, if I'm doing something like cutting the grass, my mind will wander and think about everything imaginable and I don't even realize everything I've thought about until I'm done cutting he grass. I honestly don't remember anything that happened for that hour, just a few thoughts I had. It's weird.

That's exactly what I do, Cap'n. Back before I had the subscribed threads turned off, I had over 400 emails, about 300 from blike and 100 from drug companies. It seems they've been deleted by AOL, though, because I'm down to around 50.