Ducky Havok

I think the key here is that the car has a constant acceleration. From the info given, you can find the average velocity (140/3.6=38.89) and from that and the final velocity, you can get the initial velocity (53-38.89=14.11). Also, you can divide the average velocity by the time to get the acceleration (since it's a constant). Hope that helped and good luck on your test.

But that's moving a number and a sign, not just a number, and would be quite similar to John's post (#5).

The range of a natural log is all real numbers, (as x-2 approaches 0, ln(x-2) approaches negative infinity, as x-2 approaches infinity, ln(x-2) approaches infinity) so you should be able to get your domain from that.

I would love to post these here, but they are flash animations so I can only provide the link. It's called Jokes with Einstein... that should enough to get your interest . They're hilarious. http://www.spinnerdisc.com/jokeswitheinstein.html

And, something I only discovered recently, if you turn over those papers they put on your trays, they have all the nutritional information on them. I think they're obligated to give them to you, they just don't want you to read them or else you'll realize you might just die. Comparing those though would be a good way to start.

No, it was just numbers, and brabos got it right. It's supposed to be 101-10^2=1. I do like Dak's idea, though.

I completely stole this from iGoogle brainteasers. Move one number to make this equation correct: 101-102=1

Tli tate hlluk ht ealaee e eme lopa vav lvlris eyeat lwnt rnh. This isn't nearly as hard as the some of the others, and can very easily be solved with a paper and pencil, but I thought it was fun. Plus, this section doesn't have too many recent posts, so I'm hoping to spark some interest. The sentence doesn't really make sense (It is all real words in a grammatically correct sentence, I just couldn't think of anything serious to write so I had fun with it). I'm just going to go ahead and take a guess, and say it'll last 4 hours

http://www.physorg.com/news98468776.html Here's an article I came across this article and thought you all might enjoy. I'm curious to here the opinions of people who are much more knowledgeable than in physics, since I'm just a student. How much would we have to relearn if this was true?

I might be wrong, but I believe pressure and surface area also effect solubility of a solid.

In the first example they fail to give you the original balanced equation, or the fact that you start with 1.5 mol of benzene and 7 mol of oxygen. It seems to me like they just left out the question itself. And as for the 157.9 g/mol, that's the molecular weight of Iron (3) oxide (found from a periodic table). As for the first part, do you not understand what a limiting reagant is, or just how to find one?

I thought all gases did, but it had to do with their pressure since most gases are very insoluble due to entropy? Thanks, btw. I typed it out, and somehow I still failed to realize that it was a solid. I guess I should slow down just a little bit in the future.

I took my final about a week ago, and I was curious about the answer to one of the questions. I haven't been able to ask my professor about it (since I'm no longer in that class), so I figured I'd try here. Consider the following balanced equation. The reaction is endothermic as written. BaSO4(s) + H(+)(aq) in equilibrium with Ba(2+)(aq) + HSO4(-)(aq) (I was going to use the LaTeX, but when I looked at the guide all the appeared were errors, sorry). Which of the following effects will favor a shift to the right side? (A) Decrease temperature (B) Decrease pH © Add a soluble barium salt (D) Increase the pressure in the container (E) Add more BaSO4 (s) I put E right away and moved on, but the correct answer is B. I understand why B is correct (lower pH, more H+), but why isn't E correct as well?

Okay, thanks Dave, your approach makes perfect sense, and I kind of forgot about that at the time (this is all self taught so I may have just overlooked it). But, that was for the first one which I think was a little easier because the 2nd part simplified. Using Tom's method, I get [math]\frac{x^6\ln(x+3)}{6}-\frac{1}{6}\int{\frac{x^6}{x+3}dx}[/math] And thus comes a new dilemma for me. Maybe I should just give up for the night and look at it again after sleep... that usually seems to help. Thanks for the help though. (I think I've become discouraged because I used an online integrator to solve it to check for an answer and I got something extremely long)

Thanks, but that stills leaves [math]\int{x^5\ln(x)dx}[/math] which is pretty much the same problem. I hadn't thought about that or else I probably could have gotten partial credit, though. My thoughts were to turn it into a Maclaurin series and then integrate it (because that was one of the last things we went over), but I had some trouble with that. And after all that I just realized something. Since I never wrote the question down, I typed it wrong. I'm sorry. Here's the corrected version. [math]\int{x^5\ln(x+3)}[/math] Thanks for the help on the first one though, hopefully you can help on the real version now.

Well, on my final today for AP Calc. BC, I was able to solve all the integrals except this one... It's bugging me because I feel like I'm missing something obvious, and my teacher put them in order from easiest to hardest and this was somewhere toward the middle. The test is over so it doesn't matter now, I'm just curious as to how it should have been solved (without a calculator). [math]\int{x^{5}\ln{(3x)}dx}[/math] Edit: Corrected version of problem (I wrote it wrong, I'm a loser ) [math]\int x^{5}\ln(x+3)dx[/math]

You can use natural logs to work this out quickly. First take the natural log of both sides and get [math]\ln{2}=\ln{3^{0.01t}}[/math] This can be rewritten as [math]\ln{2}=0.01t\ln{3}[/math], so [math]t=100\frac{\ln{2}}{\ln{3}}[/math]

No, that is not the inverse. I believe all you did was say 1 over the function, right? The way to do inverses is to switch your x's and y's in the equation, and the solve for y. [math]x=\frac{1+e^y}{1-e^y} [/math] [math]x-xe^y=1+e^y[/math] [math]x-1=(x+1)e^y[/math] [math]\frac{x-1}{x+1}=e^y[/math] [math]\ln{\frac{x-1}{x+2}}=y[/math] or [math]\ln{(x-1)}-\ln{(x+2)}=y [/math]

Might I add, just so you don't mess up with it in the future, that [math]\frac{\sin{x}}{x}[/math] does not always equal 1. [math]\lim_{x\to0}\frac{\sin{x}}{x}=1[/math], but when its not that it's just a regular function and does not simplify.

Also, you could probably do a sign test around the point. If it's a minimum, it'll be negative to the left of the point and positive to the right, and if it's a maximum it'll be positive then negative.

Without wanting to download it, I'm just kinda curious, what is it a video of?

I'm sure there is an easier way to do this, but for now I'll resort to L'Hospital's Rule, which says that if the limit ends up being a indeterminate form then the limit is equal to the limit of the derivative of the top over the derivative of the bottom (Please don't quote me on this, I'm working from memory so that might not be right word for word.) After using the rule once, it ends up being [math]\lim_{x\to0}\frac{2\cos{2x}\cos{3x}\sin{4x}-3\sin{2x}\sin{3x}\sin{4x}+4\sin{2x}\cos{3x}\cos{4x}}{\cos{5x}\sin{6x}+6x\cos{5x}\cos{6x}-5x\sin{5x}\sin{6x}}[/math] This is still an indeterminate, so use the rule again. For the sake of my typing, if it ends up having a sin(nx) then I just left it out because it'd go to zero, and it'd just be adding and subtracting 0. [math]\lim_{x\to0}\frac{8\cos{2x}\cos{3x}\cos{4x}+8\cos{2x}\cos{3x}\cos{4x}}{6\cos{5x}\cos{6x}+6\cos{5x}\cos{6x}}[/math] Plug in 0 and get (8+8)/(6+6), 16/12, 4/3. I'm assuming that since it's toward the beginning of the school year you haven't learned L'Hospital's rule yet, so now you're a step ahead of the other once you get there . And if that's still confusing I'm sorry, I'm sure someone will come along later and show a much easier way of doing the problem that I simply overcomplicated.

Well I can't pin down my very first memory, and sadly I have a horrible memory, but your OP brought a memory of when I was about 4 to mind. My dad and I were visiting a park, and I wanted to go on the big slide like my dad was (He really is just a big kid at heart). Well he helped me climb the ladder, but when I got up there I was to terrified to go down, so I latched on to the little handlebars on the side. My dad, not realizing I was holding on and getting rather impatient, decided to push me. Me, being the stupid little child I was, didn't let go of the handle and ended up careening over the edge and smashing headfirst into the woodchucks below me. I blame this for my bad short term memory and everything that has ever gone wrong in my life. On the bright side, my dad felt so guilty he bought me ice cream afterwards (I don't remember this part but he swears he did).

I think you are quite right, because I just looked through and those were all the latest posts. 3 minutes to answer a question that's bugged me for a little while maybe I should of thought of the more obvious possible reasons

Okay, this is something that's bugged me for a little while but I never bothered asking about. What's with the big blue dots on the left side of the forums? What I mean is why are some blue and others not? And it's always the same ones that are blue because at first I just thought it was whichever forums had the most members. Any real reason or did someone just decide the site looked better with them?