The Thing

The square root law [math] \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}} [/math] only applies for real positive values of x and y. It should actually be [math] \sqrt{\frac{x}{y}}=\pm\frac{\sqrt{x}}{\sqrt{y}} [/math]

I don't get it. What exactly is it? My questions: 1. Why is it that the greater the vapor pressure the more volatile something is? 2. Why is it that the boiling point of something is when its vapor pressure equals that of the atmospheric pressure? Thanks!

Yes you can. A couple of things to keep in mind: 1. The central atom of the tetrahedral molecule is the geometric center of the tetrahedron. 2. Assume the bond lengths are the same. (Like in CH4) 3. The tetrahedron is a regular tetrahedron. (Again, like in CH4.) So each face is an equilateral triangle. Sorry. I couldn't draw a tetrahedron on paint (died horribly:D ). But visualize a tetrahedron ABCD, with its center at O. Then the "angles" you're trying to find is the angle between any 2 vertices and O (such as Angle AOB, or Angle BOC, etc...) All of these angles are the same. So let's find, say, angle AOB. Try the following: 1. Give the tetrahedron a side length, like 2 or 3 or 4, any number will do. 2. Now, the center of the tetrahedron will be directly above the center of any of the faces. So it will be directly above the center of triangle BCD (let's use this as the base). 3. So knowing that, find the distance between B and the center of BCD. Try it! 4. Now, let's name the center of BCD, err, P. Consider the triangle ABP. 5. You have just found the distance BP in step 3. You know AB, so use Pythagoras to find AP. 6. In ABP, remember that the bond lengths are the same. So AO=BO. Now, you have to find out the distance of AO (or BO). Try this. Hint: Set up an equation that uses Pythagoras! 7. After you have the distance of AO and BO, you can use the cosine law on Angle AOB. 8. And you're done. The result should be [math]cos^{-1}(-\frac{1}{3})[/math] Good luck and try it! Tell me if you need any steps clarified.

Hi I have a weird question that is from a biology lab but has to do with chemistry. We were testing the effect of different pH on bacteria, and placed tablets soaked in different concentration of HCl on an agar plate and couple of days later we measured the circle of inhibition around the tablet. My question is, how far can the HCl spread to? Like one circle of inhibition was about 2 cm in diameter (with 2.0M of HCl). Going away from the center, how does the concentration of the HCl drop? I vaguely remember reading somewhere that the concentration of HCl decreases logarithmically when moving away from the center of the circle. Is that right? Thank you!

Hi everyone. I want to do an experiment on the growth of a certain kind of bacteria in acidic and basic environments. I have 2 questions: 1. How do I obtain a pure culture of just one kind of bacteria, without any contamination (preferably a non-pathogenic bacteria)? Do I need a sterilized lab for this? 2. How do I go about setting up the acidic and basic environments? I have pre-prepared agar plates at school that I will use. Do I just dump acid or base all over it? Or do I use some other way (like using the little discs of filter paper)? Thanks and excuse my ignorance. I know NOTHING about this.

Indirect proofs are proofs by contradiction. So you list out the possibilities, which in this case there are 2: square of an odd number is odd, or it's even. Then, you assume that the square of an odd number is EVEN, bring about a contradiction, which means that this possibility is impossible, then the remaining one possibility, that the square of an odd number is odd, must be true. Judging by your "indirect proof", I'd say your direct proof looked like this probably: Let n=2k+1. Then n^2=4k^2+4k+1=2(2k^2+2k)+1. That's in the form of 2m+1 where m is an integer, which means the expression is odd. Turn it around. Assume that 4k^2+4k+1 is even. Then raise a contradiction.

Hi I have a couple of questions about cis-trans isomers. 1. When a molecule has one double bond and exhibits one instance of cis-trans isomerism, I put the cis or trans at the very beginning of the name right? 2. When a molecule has two or more double bonds and has 2 or more places of cis-trans isomerism, then where do I put the extra cis or trans? 3. When there are no identical groups around the double bond, do I write cis or trans depending on the orientation of the parent chain? Or do I not write cis or trans at all? Thank you so much.

Hey, that's quite nice! Put the series in pairs like this: [math]1+n-1+2+n-2+...+n-1+1+n[/math] [math]=n+n+n...+n[/math]. There are [math]n[/math] n's adding together. Which is the square that John drew above. So: [math]n^2[/math] is the series's sum.

Arithmetic Series. Get the value of [math]1+2+3+...+n[/math], add it to [math](n-1)+(n-2)+...+1[/math]: [math]\frac{n(n+1)}{2}+\frac{n(n-1)}{2}[/math] Simplify, gives the series as [math]n^2[/math].

Swansont just said I shouldn't fit something like a 23rd degree polynomial to it just because it fits well... Well, thanks. But can you give me an example of what method WOULD be appropriate in determining whether a relationship is linear?

Never mind. Thanks a lot though. By the way, the other question. Is comparing the coefficient of determination for different fits a valid method of deciding on whether the relationship is linear?

Really? Hooke's Law can be derived? Can you show me how? Also is comparing the coefficient of determination for different fits a valid method of deciding on whether the relationship is linear? Thanks.

Okay, let's make the scenario into deciding whether there is a linear relationship. However, after you plot your points there are usually some high-degree polynomial curves (with some extremely strange coefficients) that fit the points better than a line. Of course a line looks pretty close to the data, but then the higher-degree polynomials are even closer. Say we know that the relationship is indeed linear, and the higher-degree polynomials are caused by inevitable random errors in the sample. It looks linear, but a high-degree polynomial can fit more points than the line. In this case, how can we show that the relationship is indeed linear despite the fact that a high-degree polynomial curve fit the data points better?

Not sure if this belongs in the Physics or Math forum. I'm leaning towards Math more. Say, when scientists do experiments, record data, and come up wtih a formula, how do they prove that the formula is correct? I'm talking about those formulae that cannot be mathematically derived. One example might by Hooke's Law.There will inevitably be some error so how do they prove, besides just looking at the graph, that the formula should still stand in spite of the errors affecting the data? I know that they can't prove that the formulae is 100% true, but they can show, using statistical analysis that the formula is highly unlikely to be untrue. My question is, HOW? Say I've been experimenting on Hooke's Law F=-kx with a spring, a ruler, and some masses. There will be imprecision and random errors from the ruler. And if I collect the data, plot it, it will not assume the shape of a perfect line. So my question is: how do I know that I should fit a linear regression to it? How can I prove that it is very likely to be a linear relationship and not some other, like an exponential relationship? I look at the graph, yes, it looks linear, but how do I prove on paper or with a software mathematically that it is highly likely that this is a linear relationship? Thanks a lot.

Let y = a negative number with a huge absolute value.

Oh okay. Thanks dave.

Is there anyway to solve an equation (without logarithms) such as: [math] 2^x=x^2 [/math] and [math] 2^x=x^3 [/math] where x is in the exponent and the base of two different terms? If so, how?

5614. It's the other diagonal. With your diagonal Triangle ABC is obtuse, which means that AB does not equal AC, contrary to what's given. Proof of that is quite simple: Isosceles triangles have two equal base angles, namely angles ACB and ABC here. If ABC is greater than 90, ACB is greater than 90 and there's the whole 180 degree in the triangle theorem. So yeah, draw the shorter diagonal, you find Angle ABC is (180-30)/2=75. I think Rajdilawar just had an arithmetic error =]. Yeah then opposite angles of a parallelogram are same then supplementary angles. So it is 105 just to confirm. Now I'm REALLY sure Rajdilawar just had a calculation error.

Materazzi said Zidane said it in a very arrogant way, sizing Materazzi up. In the video too you can see Zidane looking up and down at Materazzi. Materazzi said that Zidane's arrogance and attitude towards him provoked him to provoke Zidane. I think Zidane deserved what he got, and more. They should do something to him (banning doesn't work though) besides the red card. Again, use the Euro 2004 Totti & Poulsen example: Totti got a 3-match ban. I REALLY don't understand people can blame Totti and at the same time sympathize with Zidane. He's no different than Totti in the eyes of the referee and in the rules. I really don't understand how people can sympathize with someone who used physical violence instead of with someone who just dissed Zidane. And Materazzi wasn't even the one who started the whole thing. BTW, Zidane has had 12 red cards, including one that was a similar case - a vicious head butt. And who can forget his malicious stepping on a Saudi Arabian player's back in 1998 World Cup?

That hasn't been confirmed yet. People have been making up all sorts of things that Materazzi "said". So many different things that there is no way that Materazzi could have said them all in that short period of time. First, some people claimed that Materazzi called Zidane a "dirty terrorist". Materazzi has openly denied that. Then there, of course, is the "son of terrorist" version which is derived from someone reading Materazzi's lips. But the same reading of lips also produced another version: that Materazzi called Zidane's sister a prostitute twice. None of these have been confirmed. And frankly, it really doesn't matter what Materazzi said. Remember Euro 2004, when Franscesco Totti spat at Christian Poulsen because he was provoked? The reaction after that was mostly anger against Totti and everybody blamed him for the incident, and he was banned for 3 matches afterwards. Now let's drag Totti and Poulsen out of the scenario, and put Zidane and Materazzi in. It's the exact same scenario, but now, for some reason, people are saying how Materazzi's a dirty player and Zidane's not too guilty. Is it just because he's "Zizou"?

There's this pepsi one with Roberto Carlos in it. This Japanese kid offers him a pepsi and bows to him...That one totally owns. See the link below. http://www.youtube.com/watch?v=Ztkd2SPtVjI I also like the Jose +10 commercials with the two kids choosing stars. Cisse, Kaka, Zidane, Beckham, Dufoe, Kahn, Messi, Beckenbauer. Beckenbauer? Hahahahaha. Platini!

I can type a word much faster than I can spell it out in my brain and I think many people can type faster than they can spell with their minds too. And it is so much easier to think in words than to think in letters, not to mention having random thoughts like "this is stupid" floating around. So the thoughts-recognition softwares would have to do similar tasks as voic recognition softwares because most people think in complete words and have other background thoughts.

Package and Deployment wizard has (or rather, lack) one thing I don't like (at least for VB6) - there's no option to change the user's registry. Sometimes I want to enter a key that associates a file extension with my program into the registry and I have no way of doing it. But anyways, there are lots of good setup softwares floating around, including InstallShield that I use.

I built a pretty small trebuchet that launched walnuts. Really cool, much better than the lame mousetrap catapult I also built at the same time just for fun. But seriously, I hope to build a ballista. Those things look awesome (and by awesome I mean deadly) and the projectiles they launch would look better than a round walnut (maybe nails or something sharp into the wall).