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Messages - jnesselr
Pages: 1 ... 136 137 [138] 139 140 ... 165
2056
« on: October 02, 2010, 09:27:31 pm »
Actually, I never posted these, but here is a mock screenie: 
2057
« on: October 02, 2010, 09:02:25 pm »
That's pretty much the way to do it. You'd be surprised just how fast one massive For() loop that OR's the contents of both buffers is, though. I calculated that the following loop runs about 40 times per second at 6MHz:
Repeat getKey(15)
For(A,0,383)
.That ﹢ symbol is the 16-bit OR operator
{A*2+L₆}ʳ﹢{A*2+L₃}ʳ→{A*2+L₆}ʳ
End
C+1→C
End
Whoa, how'd you get the small r to work in a code tag?
"ʳ" There is a symbol for it. Just copy/paste.
2058
« on: October 02, 2010, 09:00:57 pm »
well, if you get lucky at the red lobster.. you might just order up netham45 by accident then you'd meet him 
See, now I want to try that... "Yes, I'd like one Netham45, please."
2059
« on: October 02, 2010, 08:59:28 pm »
I would never have solved it if I knew this fight were to happen. Okay, so x implying y aside, the tangent lines must be the same for it to be continuos. This is because of the fact that as you approach from the right, eg the Ax2+Bx equation, we say the slope of the line is heading toward N. From the left, eg the 3-x equation, we say the slope of the line is heading toward M. So, for continuity, I'm sure that you would all agree that the two lines had to be connected at the point (1,2), since it was at x=1 and 3-x=2.
Okay, so I lied about the whole x not implying y thing. If it is differentiable, then it implies it is continuous. And yeah, the quiz was worded wrong I do believe. It should be differentiable not continuos. Either way, It doesn't matter. You are correct if it does not need to be differentiable. But, to make sure it is absolutely continuous, and getting exact answers, you should see if it is differentiable.
Wow, I just check recent-posts. This is gonna be ninja'd.
2060
« on: October 02, 2010, 08:53:43 pm »
Now I want to meet a DJ!
Actually, I would rather meet Netham45. I've always wondered how he fit into that blue lobster costume.
2061
« on: October 02, 2010, 06:55:31 pm »
Ya, I feel like I've seen him here before but I guess that was at UTI that I've seen the name.
I don't even really get what that "Internets" thing is really O.o
Also, now we need to get back on topic 
Okay, I'll start. I am glad someone finally did this, and will test it once you get libs fixed. ;-)
2062
« on: October 02, 2010, 06:51:19 pm »
oh... Wow, totally was thinking he had been here forever. lol, okay. Well, floppus needs to get his own little topic to introduce himself. THEN, we throw the peanuts.
also, how many people have that internets thing in their sig now? At least 4. Possibly >9000
2063
« on: October 02, 2010, 06:48:43 pm »
Who are you welcoming, qazz42?
2064
« on: October 02, 2010, 06:38:58 pm »
well, not really.
The tangent lines for the two functions have to be the same. So, what you do is find the derivative of the functions. So:
f(x)={ -1 x<1
{ 2ax+b x≥1
which means that at 1, since the tangent lines need to be the same, and therefore the slope, and therefore the derivative, -1 = 2ax+b. So at x=1, -1=2a+b. This can be rearranged to get b=-1-2ax
So, at x=1, the two functions must equal each other to connect, as well as have their derivatives be the same. so 3-x=ax2+bx. See, now we're having fun. So, they must equal at x=1, so 3-1=a+b. or 2=a+b as you have pointed out.
Now we have two equations that equal b at x=1. So, 2=a-1-2a, or 2=-a-1, or -2=a+1, or a=-3. WHOA! So, with either of our equations, b=-1-2(-3), or b=-1+6=5. With the other one, 2=a+b, we have 2=-3+b, or b=5.
So, -3X2+5x is your second function.
f(x)={ 3-x x<1
{ -3x2+5x x≥1
The point at x=1 is (1,2) for both equations, and the tangent line is -1 for both equations. That wasn't too bad, was it?
2065
« on: October 02, 2010, 05:20:56 pm »
Have some special TAO: Unleashed ones. 
2066
« on: October 02, 2010, 05:03:41 pm »
Pascal [who does't know it]
*Raises hand*
Same here, doesn't even ring a bell.
Ditto. Although I do know about it, I've never used it.
2067
« on: October 02, 2010, 06:29:24 am »
What Type of computer is it?
2068
« on: October 01, 2010, 04:59:19 pm »
If you don't use a shell, you will need to find the program in ram and edit it.
2069
« on: September 30, 2010, 09:12:55 pm »
I'Ll try to gather some later, probably. Do you mind if some ideas come from my games? I'll go check in my old comic book-based RPGs I created around 2000-2001 to see if I might not get some more ideas from there.
That is fine. There are 7 attack types. Normal, Fire, Water, Earth, Wind, Light and Dark. There are many items. (I don't remember off the top of my head how many I am allowing for) Thank you!
2070
« on: September 30, 2010, 08:50:13 pm »
May I suggest, for other users who might have the same problem, posting what you did to fix it, and showing the code differences. Thank you.
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