BigMoosie

I was assuming in the first scenario that I had read some kind of statistical information.

If your'e going to kill the person you might as well just chuck em in a giant blender.

Try Opera if you haven't already. I've been using Firefox for about a year but recently switched to Opera as it loads much faster and uses far less memory. The advantages that Firefox has due to extentions is minimal as Opera has alot of features when you dig into it.

The probability of an event occuring is based upon what information you have about it. For instance the probability of me dying this year is 1%, if I mentioned that I do not smoke or do drugs then the probability might drop to 0.9%. So it depends on what you know, if you know that 2 things can occur but have no idea in what ratio then the probability of either event is EXACTLY 50% until you attain more information.

That is the only method I can think of, how much volume is displaced. The density of the human body can vary quite a bit so a weight will not suffice. I guess you could use one of those setups that use multiple photographs to build a 3d model of you but who has one of those lying around? Is this question hypothetical or are you actually trying to solve your volume?

That sounds fun, what are you making it for?

They will either be parallel or skew (http://mathworld.wolfram.com/SkewLines.html). Edit: Url fixed, thanks swansont.

I think in total it took about 12 hours. That includes deriving the formulas which I am afraid are off by a very small degree (I have never studied projective geometry), when viewed at side angles there is a slight distortion. In fact I wonder if anybody could suggest the correct ones... Basically the part that has the flaw is a function which takes 5 variables: x,y,z (normalised from -1 to +1 of the cube), then "theta" which is the sideways rotation in radians and "psi" which is the vertical rotation in radians. With these values I need to resolve the x and y pixels to draw the given point at.

I have made a 3D function plotter as a widget for Opera 9 users, you can find it here: http://widgets.opera.com/widget/4408 Please let me know what you think of it, I have never tested other 3d function plotters so am not sure what features are standard. Hopefully your feedback will help me improve it. Regards, Moosie

I am after every word that exists as valid English. And the size of the list is no issue, I will not be sifting through the thousands of words, my computer will be. Google searches are just turning up online dictionaries and other things irrelivant to me.

Hi, I am wanting to write a computer program for personal use. One thing I need to achieve my goal is a list of every word in the English language (UK english that is). If somebody could point me to a source where I could get each word in a text file, perhaps separated by commas or new lines etc. that would be great. Regards, Moosie

Given 3 points in 3D space as cartesian coordiantes (a triangle) I need to calculate the normal for a 3D graphing program I am making. I haven't done any calculus that might help with this and I don't know about linear algebra so Wolfram's solution is of no help: Can anybody please translate this into an algebraic solution for me? Thankyou.

Hi, I'm thinking about purchasing some super powerful magnets, something about the size of a 9V battery but strong enough to dam near break your fingers. I was wondering if anybody knows a reliable online shop which ships to Australia or anything in general about buying them. Thanks!

I need to write a function to find the largest square factor of a number so it can be taken out of a radical. I hope there is a known solution to this. The most efficient aproach I can think of is: set total to 1 for (each prime less than square root of n) { while (n is divisble by prime^2) { divide n by prime^2 multiply total by prime } } "total" should then be a square factor of n. The problem with this is that it is very slow. Is there a faster method?

Just some thoughts I was having today. Is there an easy known method of filtering alcohol out of blood? If so would it be difficult to create a device to you attatch to your arm which diverts blood into a filter to help you sober up quickly if the moment asks? I was also thinking that perhaps a similar device could inject oxygen into the circulatory system, perhaps for scuba divers who want to free their face up a little from the mask. Would it have to be attatched to several major arteries or could it simply be on the arm or leg?

Thankyou both, it seems I was after arc length and this is the formula I was looking for: [math]L = \int_0^1 \sqrt{1 + [f'(x)]^2}[/math]

I am interested in knowing how to find the distance along a function between two values. I am surprised to see that this kind of problem is never encountered in the highest level mathematics in high schools in Australia and decided to research it myself. I would appreciate a term that might be useful to google or perhaps a link to a wiki article or something. Thankyou, BigMoosie

My proof. It has been shown that all pythagorean triples can be formed by this formula where a and b are any integers and the third value is the hypotenuse. [math]a^2 - b^2, 2 a b, a^2 + b^2[/math] So our arithmetic series could be defined by either of the following. (1) [math](a^2 + b^2) - 2ab = 2ab - (a^2-b^2)[/math] (2) [math](a^2 + b^2) - (a^2-b^2) = (a^2 - b^2) - 2ab[/math] Simplifying (1): [math]2a^2 = 4ab^2[/math] [math]a=2b[/math] Substituting a back into our original formula: [math]3b^2, 4b^2, 5b^2[/math] The values are in the 3,4,5 ratio, therefore true for any a or b in the first form. Simplifying (2): [math]3b^2 + 2ab - a^2 = 0[/math] [math]b = \frac{-2a \pm \sqrt{4a^2 + 12a^2}}{b} = \frac{-a \pm 2a}{3}[/math] Taking the first root is trivial as [math]b=-a[/math] and the first value will be zero (not technically a pythagorean triple). The second root: [math]a = 3b[/math] Substituting a into the original formula: [math]8b^2, 6b^2, 10b^2[/math] In the ratio of 3,4,5. Since it is true for any a or b it is true for any pythagorean triple.

It's easier to find out with 4 dice what are the odds of not getting any six: (5/6)^4, then subtract it from 1 to get the complimentary = 0.5177 With one roll of two die the odds of not getting double six are 35/36, using the same idea as before: 1-(35/36)^36 = 0.6373

For a moving object: Something twice as heavy is twice as hard to slow down. Something twice as fast is twice as hard to slow down. So momentum is directly proportional to both mass and velocity. This means that mass*velocity = momentum.

This is why your methods are invalid: To solve a trig value such as sine you use radians, your pocket calculator will take it in degress then convert to radians using pi then solve using: [math]\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}[/math] For small values of sine you can approximate with a straight line, for example: [math]\lim_{n \rightarrow 0 } sin(n) = n[/math] Your sine in your function is limiting the same way so you can effectively remove the sine and convert to radians and you get: [math]\pi = \lim_{n \rightarrow \infty } \frac{n \sin(\tfrac{360}{n}) } {2}= \lim_{n \rightarrow \infty } \frac{n \tfrac{2\pi}{n} } {2} = \frac{2\pi}{2} = \pi[/math] That being said your origional function when in radians can be any value, not just pi, so would you say this is an identity for e: ? [math]e = \lim_{n \rightarrow \infty } \frac{n \sin(\tfrac{e}{n}) } {2}[/math] To get something copyrighted you smack a © <NAME> <YEAR> or something similar on your paper.

Convert them to radians and you will see why your methods are invalid.

Although you are quite a credible man, I cannot take your advice on faith, it would be unscientific.

There are several, however they all involve infinite sums, continued fractions or other transcendal numbers etc, here are some examples: [math]\frac{\pi ^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2}[/math] [math]e^{i\pi}+1=0[/math]