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You're lucky that I'm in a linear algebra course right now. Note:If you see a semi-colon in any of the matrices below, it indicates a new row. The most general equation is [math]Ax=b[/math] where A is a square matrix, x is the unknown vector, and b is a vector that matches the inner dimension of the A square matrix. Note: If A is not square, then this equation is invalid. The augmented matrix [math][A|b][/math] can be written as [math]A \cdot x=b[/math] Identity matrix is [math]I_{mxn}[/math]. i.e. [math]I_{2x2}[/math] is [1 0; 0 1] [math]A_{mxn} \cdot I_{nxm} = A[/math] If A is a SQUARE matrix, i.e. size nxn, then [math]A \cdot I = A[/math]. DEFINITION: Matrix A of size nxn is invertible if there is a matrix b of size nxn such that [math]AB=BA=I_{nxn}[/math]. Therefore, B is called an INVERSE of A, denoted by [math]A^{-1}[/math]. Thereom: If A has an inverse, then it has exactly one inverse. For 2x2 matrices, you can determine whether they are invertible or not. If A=[a b; c d], you have to use: [math]A^{-1} = \frac{1}{ad-bc}[d -b; -c(space) a][/math] If the matrices are 3x3 or greater, you can use the Gaussian-Jordan Elimination. If you have any more questions, let me know.

Bettina, you must be a beautiful woman. It's not your fault. (Like some old women would glare at me!!)

What if I was a crackpot? Do I seem like a bad person? Just kidding. (At first, I thought crackpot is similiar to pothead - people who smoke weeds frequently).

Very funny. Imagine yourself walking by yourself in a long hallway, and there's another person walking toward you in the opposite direction. This person is someone you really like. Would you glare at the person, or try to ignore him/her? Or you simply have the urge to look at the person? You also would want to know if the person is looking at you or not. That would describe an awkward moment.

I think this author is explicitly talking about how she has to walk past other students on the university campus.

OHHH!!!!!!!!!!! I was actually thinking that my problems were too hard for them to solve, so that was why people didn't participate. I'll put up a question that is soo hard that you'll kick yourself all the night!!!

This is an newspaper article from my university papers, and when I read it, I thought it was funny, and it's very true too. [math]\rightarrow[/math]GRINDS MY GEARS By Courtenay Courier One thing that really grinds my gears is those awkward moments as you pass someone on the sidewalk. I don't know why they have to be so difficult and uncomfortable. You're just walking by someone, yet they seem to want to make it weird for you. I just don't understand it. You'll notice if you walk by someone that they either make a point of looking away from you or they glare straight at you, both of which are annoying. Even more so are those people who glare at you, but as you get closer, look away so you won't know they were staring at you. But, in order to see these reactions you have to glare at them, making them uncomfortable. So really, you find yourself doing the very same thing that weirds you right out when it happens to you. But it's unavoidable. When you're just walking along, there's really nothing going on and people going by may be the only thing to watch. But, for some reason, you're scared to let them see you looking at them. So, when they look over, you look away, making all the more obvious that you were staring at them to begin with. It always makes you feel better when the person you walk by either offers a head nod or a simple "hi". This is what ultimately saves them from coming off as too much of a douche. Now, to be honest, I'm a little bit hypocritical: when I see certain people, I just get the urge to glare right at them. Not that I think they're really something to look at, but because it creeps them out. Engough people do this to me, so I figure why not torment them a little too? So, when I pass someone who looks like they can be easily intimmidated, I give them 'the glare.' You know what I mean, that stare that looks like I'm looking straight through them, analysing them, making them really uneasy and then they stop looking at me, and that adds just a little bit of joy to my day. Basically, what I'm saying is that I don't mind messing with peoples heads, but I don't like it when they do it to me. Or, rather, they do it to me, which makes me want to give it back big time. I just think that this is such an unneccessary and ridiculous hassle. Why is walking by someone so damned uncomfortable? Why do other people have to make it that way? I mean, you're only going to see them for about five seconds, why should alter the mood of your day at all? But it can. I guess some people just have a natural urge to glare. Maybe they're just stupid. Maybe they only do this to me because they hate me from first glance. I don't know. But my roommate Robin seemed to think it was a stupid but annoying situation as well. So, I'm glad it's not just me who notices these things. Otherwise, I'd have no one to gripe to. The point is, if you see someone walking past you on the sidwalk, don't make it too obvious that you're trying not to look at them, and don't stare at them. I suppose a glance with a head nod is in order. Just stop being a jerk about it. ----------------------------------------------------------- So what is your opinions on this article? Do you feel the same way this journalist (and I) felt?

Good job 5614. See it was not too hard was it? I don't understand why many people won't participate in this. (Maybe they are not really that smart). But if you got a wrong answer, that doesn't matter, because you will learn from your mistakes. The most general formula for forces is obviously [math]F_{net}=ma[/math]. I drew a FBD (Free Body Diagram), and you can see where your vectors' directions. So the [math]F_{net}[/math] is changed into: [math]F_{app} - F_{fk} = ma[/math] [math]F_{app} = m(.722) + /mumg[/math] [math]335N = m(.722) + m(3.2373)[/math] [math]335N = m(3.9593)[/math] [math]m=84.6 kg[/math]. I'll post another physics question soon.

Anybody? I'm sure there is a lot of physicists out there!! Tip: The answer that you might get is outrageous. (The chair is really HEAVY).

A winner never quits, a quitter never wins.

Don't ask us. Try the experiment yourself and make sure there's no flammable materials around you. You can also check MSDS data sheets (Material Safety Data Sheets), for the potential harms that the chemicals will pose.

Here is another problem: An applied force [math]F_{app}[/math] of 335 Newtons is required to accelerate a chair across a level floor at [math]0.722 m/s^2[/math]. If the coefficient of kinetic friction is .330, find the mass of the chair.

Newspapers never tells the full truth.

This has been a hot debate about this relating with abortion, saying whether embyro is declared a "human" or not. But on a smaller scale such as a blastocyst, I wouldn't consider it a human, because it's really early in its stage. I don't have any evidences to support my opinion, but it sounds reasonable. It's like a baby chicken from an egg. We can control whether the egg would become into a baby chicken or not, as it needs nurturing envoirnment for it to grow. But the egg in the woman's womb is already in a nurturing envoirnment so that's another issue. I'm talking really randomly here.

Farsight, try to use LaTex next time, as it would be easier for us to read your calculations. I noticed that you did it in a little different way than I did. My calculations: As swansont stated that the distance equals both way, and that the rock have an acceleration due to gravity, you use [math]d=\frac{1}{2}at^2[/math]. Also, for the sound, you use [math]d=vt[/math]. Keep in mind that [math]T_r[/math] (time for the rock), combined with [math]T_s[/math] (time for the sound) equals to [math]1.89[/math] seconds. ([math]T_r+T_s = 1.89s[/math]). 1) [math]d=\frac{1}{2}(9.81 m/s^2)(1.89-T_s)^2[/math] [math]\rightarrow[/math]Notice how [math]1.89-T_s[/math] leaves the time for [math]T_r[/math] (the amount of time for flight of the rock). 2) [math]d=(332 m/s)(1.89-T_r)[/math] Since the distance equals both way you have: 1 = 2 [math]\frac{1}{2}(9.81 m/s^2)(1.89-T_s)^2 = (332 m/s)(1.89-T_r)[/math] Now you have a quadratic equation: [math]4.905(1.89-T_s)^2 + 332T_r - 627.48 = 0[/math] Using the quadratic formula you get: [math]1.839981808 s[/math] and another negative number which you don't need. [math]T_s = 1.89 - T_r[/math] [math]T_s = 1.89 - 1.839981808 = 0.0500181917 seconds[/math] (That's the amount of time for the sound to arrive the human's ear at the top. Using the distance formulas stated above, you input the 2 different times in their formula, and you will get the same distance for each. The well is 16.60603963 metres deep. I have another challenge physics problem for you guys and I'll post it tomorrow night.

swansont, your answer is right. After I posted the answer of 65.8 metres, I went to bed. While I was sleeping, I was thinking about this problem. I knew 65.8 metres was incorrect since the rock can't go down 65.8 metres in 1.89 seconds. Therefore, I woke up again, and redid my calculations. In the end, I got 16.6 metres too. It was too late (1:30 AM) for me to correct my post, so I'll do it tonight. I'll explain this in more details later when I get home from school. I'll also post my calculations for the other people to see.

I got the answer of 65.8 metres. Did you got it too?

I would assume that you need to use the principle of "rigid body equilibrium". I didn't learn it yet as I will probably learn it next semester. Sorry if I can't be any helpful than this.

Heh, it's not my homework. Just an interesting physics question for all of us that would requires abit of thinking. Of course, it does involve a simultaneous equation, because the motion is divided into 2 parts since it have different time of flight.

A man dropped a rock into a well. He heard the sound of the rock hitting the water after 1.89 seconds. Considering the speed of sound to be 332 metres per second. How deep is the well? This is a tricky question, since the rock has an acceleration when it's going down, and the sound have a constant velocity going back up. You guys try it.

I have a DELL XPS M1710, and it's a really good laptop. Very reliable and fast. A lot of potential for upgrades and it's Windows Vista Capable. I would recommend you that one because I got 3 years warranty with it. It includes McAfee Security which is a really good virus protection program. I bought the laptop for approximately 3,000 bucks. (taxes included)

Yeah, I have been wondering where the homepage is? The homepage that shows all the new posts.

Yup, it's real.

Download Flock from Photobucket. It worked for me.

Guess what? Something funny happened and it's strangely odd too. I just downloaded Flock, and went on SFN. Apparently the LaTeX scribbles showed up clearly on Flock, not on normal Internet Explorer. Wierd. Flock is a very cool Internet Browser that allows you to drag pictures and add comments to it, so you can put them on blogs, websites, etc. You can also upload hundreds of pictures to it very quickly. Very efficient program if you have a Photobucket account. Sorry if I posted this twice!! I don't know how to delete post #10.