BobbyJoeCool

Goooooooooogle

this is all I'll say (because I know no one on here can actually do anything about it, and most don't really care, because they don't really know me). friend(a)+suidide=bad grandparent(only living one I've ever known)+cancer=worse money<0=uh oh. friend(b)+attempted suicide=OMFG friend(c&d&e&f&g)=backstabbing sons (and daughers) of...=Keep it Coming... friend(h)+cutting=what next... and that's as far as I'll go... (There's more, but... I'm prefer to focus on making this month better than last, and step one is to put the past in the past...)

ok... I apologize to anyone born in september... being born in the month of EVIL (especially since I was reffering to this last month specifically, as in Sept. 2005) does not mean that I think you are evil or anything else of that sort... I'd also apologize to anyone who had a good month... Mine was filled with EVIL... and I'm just glad it's over.

When you dream, you go through 4 stages, and then REM sleep in about 90 minutes (which is why 90 minutes of sleep is better than 120). In stage one, you are only really half alseep (you can appear awake when you're in this stage). Your brain waves change as you go from stage to stage. Basically, from awake through stage four, your brain waves become less powerful. The alpha waves when you're awake indicate a quite active brain whereas the delta waves while in stage 4 are, while still active, much less active brain. As you hit different stages of sleep, different parts of your brain "turn off." The entire time you're asleep, you're still at least sub-consciously aware of the environment around you. This is why you don't often roll out of bed, or smother the person your sleeping next to (unless you're intoxicated). Also, the whole time you're asleep, the part of your brain that recognizes someone saying your name is still there. If someone says your name, you process this and your brain wakes you up. But for instance, I don't remember which stage it was, but I think it was in stage 4, where your non-vital motor functions shut down. In REM sleep, you get the same activity in your brain you do as you're awake, except your brain has essentially cut off the central and peripheral nervous systems (only keeping life support functions like heartbeat and breathing). The neurons in your brain are still firing just as much as when you're awake. It's just that it's firing with little input... So, random memories surface (some you may not even remember, but are still stored). and the "dreams" are your brain attempting to make sence of all the psudo-random firing in your brain. Which is why, many times, dreams seem disjointed and complelty random. Although, sometimes, your brain can actually put things together to seem non-random. And, the reason most people experience things in their dreams that have something to do with something that happened to them recently (ie: have a test comming up, so they dream about it), is that those thoughts/feelings are on the surface, and easier to get to (so the "random" firing is more likely to hit it...). As for daydreaming... when you start to daydream, think back... Think backwards on how you go to thinking about whatever it is you were thinking about. Usually, it's you see, hear, taste, feel, or especially smell something that reminds you of something else with leads you to think about something else and so on... kindof like a converstaion about "nothing" can go from one thing to something entirely different. If you mean the type of daydream where you're completely distracted (like when you're having problems) and you're trying to figure out how to get through them... It's more important to you at the time than whatever you're trying to pay attention to (like my Calculus teacher). So, you think about it instead of what you "should" be, because it's more pertinant to you. Anyway (I don't swear by this 100%, but probably 95%), I hope this helps. I "stole" most of this out of my various psychology classes I've taken.

yes, both to me, and this "great" nation of ours.

I would like to be the first to say that I'm happy that the month of EVIL is over, and that October is here... Yay! *Raises 8th liter bottle of Mt. Dew in toast of the new month...*

[math]\LaTeX[/math] Link to tutorial thread And I misunderstood the question... so my answer is wrong...

isn't [math]g^n (x)[/math] the same as saying [math](g(x))^n[/math]? In which case the derivative of the function for any n can be given using the simple diferentiaition rules... [math]g(x)^n=(xe^{-x})^n[/math] Genera power rule states... (or the power rule with the chain rule) [math]g(x)=(a)u^b, g'(x)=(ab)u^{b-1}u'[/math] so we use this on the function... with.. [math]u=xe^{-x}[/math] and u'= to the derivative of u... using the product rule... [math]-xe^{-x}+e^{-x}=(1-x)e^{-x}[/math] [math]g'(x)^n=n(xe^{-x})^{n-1}(1-x)e^{-x}[/math] or would that be way off and too simple? EDIT: all those funky f's I had were changed to g's... opps. EDIT mark II: I just noticed that KG said to take the deriviatves and not Dave... so this is kinda pointless isn't it...

the calculus lessons in the calculus forum? I think it'd be very useful for those to be right there for referance in the calculus forum...

except that I have no way of knowing what the final answers are... the book doesn't have them.

Ok... Our class (as our homework assignment) needs to find the following derivatives using implicit.. [math]sin^{-1}(x), cos^{-1}(x), tan^{-1}(x), cot^{-1}(x), csc^{-1}(x), sec^{-1}(x)[/math] and, since our book doesn't give these derivatives, I don't know if I got them right (because we have to then use them in the assignment...) [math]sin^{-1}(x)=y[/math] [math]sin(sin^{-1}(x))=sin(y)[/math] [math]x=sin(y)[/math] [math]1=cos(y)y'[/math] [math]y'=\frac{1}{cos(y)}=sec(y)[/math] using triangles and trig identities, since arcsin(x)=y, y is the angle with x the side opposite... so x/1 would be opposite over hypotinuse... so the last side is [math]\sqrt{1^2-x^2}[/math] so, sec is the reciprical of the cos, so it's hypotinuse over adjecent... [math]sec(y)=\frac{1}{\sqrt{1-x^2}}[/math] and back subsititution... [math]y'=sec(y)=\frac{1}{\sqrt{1-x^2}}[/math] Then, to generize it using the chain rule... [math]y'=sec(y)=\frac{1}{\sqrt{1-x^2}}=\frac{u'}{\sqrt{1-u^2}}[/math] Is this right? And does the same process work for the others? Let's see... using the process, this is what I get for the inverse trig derivatives... [math]y=sin^{-1}(x), y'=\frac{u'}{\sqrt{1-u^2}}[/math] [math]y=cos^{-1}(x), y'=-\frac{u'}{\sqrt{1-u^2}}[/math] [math]y=tan^{-1}(x), y'=\frac{u'}{1+u^2}[/math] [math]y=cot^{-1}(x), y'=-\frac{u'}{1+u^2}[/math] [math]y=sec^{-1}(x), y'=\frac{u'}{u\sqrt{u^2-1}}[/math] [math]y=csc^{-1}(x), y'=-\frac{u'}{u\sqrt{u^2-1}}[/math] anyone notice an error here? or did I do it right?

You of course mean we always have... Just because we haven't been wiped out by our own devices yet doesn't mean it won't happen...

I don't see how any of the second half of these work. It seems to me to not work. Is this something I'd learn after Calculus I?

The thing about Thermodynamics... I know it's for a closed system, but can you tell me if this is right for what a closed system is? A system that is completly void of any factors the originate from outside the system... (as in, nothing that isn't in the system can affect anything in the system). *also watched Mokele throw away some garbage.*

[math]\frac{x}{2(x-1)}-\frac{1}{(x-1)(x+1)}[/math] [math]\frac{x}{2(x-1)}-(\frac{1}{2(x-1)}-\frac{1}{2(x+1)})[/math] How do you figure that?

Ok... if you have a completely static black hole (as in not moving, not rotating, etc.), is the Swartzchild Radius the event horizon radius? Wikipedia seemed to imply that but didn't really say it.

Unfortunatly, that costs money...

you mean that the mass of light cancils out with itself and that if light has no mass, you're in essence dividing both sides by 0, which is illegal? but a photon has mass, it's just that it's mass is relative (because it's moving at the speed of light, it seems to have no mass to us because we're not moving at at speed of light.) But if that's true, how can you determine where the EH is on a black hole?

[math]v_{e}=\sqrt{\frac{2Gm}{d}}[/math] Source named Wikipedia (I love that site) The event horizon exists at the point where the escape velocity is the speed of light. [math]c=\sqrt{\frac{2Gm}{d}}[/math] [math]c^2=\frac{2Gm}{d}[/math] [math]dc^2=2Gm[/math] [math]d=\frac{2Gm}{c^2}[/math] as long as the object has mass=/=0, d=/=0, because G and c are constants... You don't need a mass for light to calculate the EH location...

Ok... I think I can answer this now... You need to accelerate 155 km/hr WEST (to cancil your eastwards moment). You also need to accelerate 118 km/hr south, so that you're moving that direction. the resultant vector is those two added, so that you are moveing 118 km/hr south. Try doing it that way... (and remember, you have a 90°*angle ). I see where it's going now... I get, 3.45 km/(h*s) you use pythagoreans and then devide by the time for average acceleration... the angle is just some trig... help much at all?

and my point was that you are a singlularity, and the other object is light (which it has to be in order to determine an event horizon). and say that this light is occupying only a single point in space. if the light gets close enough to the singularity of mass 1kg, the force on the light created by gravity will make the escape velocity greater than that of the speed of light. Now, this never happens in real space because 1kg of mass has electromagnetic force much greater than 1kg of mass can generate. So, in effect, you need the MASSIVE force of a black hole in order for any of it to work, at which point, the EH is so large that it's notciable with the naked eye...

what answer does the book give, and we'll see if we can help you get there... EDIT: also, in the above explanation, you only end up with the change in velocity... you still need to divide it by the time it takes to do this... [math]\frac{v_{c}^2}{s}=a[/math]

Those are all your choice. As everyone else has the same choice. This is the land of the "free." If someone wants to say "one nation, under God" go right ahead. If they don't want to say it, that also if fine. This nations main doctrine is that you may choose to do (or not do) things. Some of those things have consequences. Saying the pledge is, and always was OPTIONAL! No one is making anyone say anything. Does not pledging my alegance to my country mean I don't respect it, or the people who fight to protect it? NO. I love this country! But I refuse to pledge alegance to it. It's a personal choice. I pledge my alegance to no one except myself. As long as this country is protecting my interests, I'll "fight" to protect it. But, to me, pledging alegance is a bit much.

you do have an angle... from east to south=90°. You have a beginning vector, and the ending vector, and the angle between them (90°). Finding the other side is quite easy... [math]v_{a}^2+v_{b}^2=v_{c}^2[/math] (note: this is only true because it's a right angle (right triangle)) Normally it'd be [math]v_{c}^2=v_{a}^2+v_{b}^2-2v_{a}v_{b}cos©[/math] Since cos(90)=0, 2(v)(v)(0)=0 and you're left with the above. [math]v_{c}[/math] is the vector you're solving for. [math]v_{c}=\sqrt{v_{a}^2+v_{b}^2-2v_{a}v_{b}cos©}[/math] input numbers... any help?