psi20

The only way I know of sending it is through email. You might want to see the whole thing too. It's a Hoedown in Whose Line and its pretty funny.

Sure, but how?

Well, I wouldn't say it's an animal spirit, but I videotaped a t.v. show . Then I used a digital camera's videeotaping abilities to record the videotape playing. Then I put the video onto the computer. I play it on Windows Media Player. I made sure that all the sounds that could distract the me recording the videotape were blanked out. I turned off the radio, stopped the laundry machine, etc. I know the noise was neither me nor my sister. I've also ruled out neighbors playing some instrument next door. From 0:52 to 0:58, there's this intriguing music or some humming. I played around with the EQ, dropping everything from 125 to 16K. The music is pretty loud. It's creeping me out because it sounds scary, demonic even. Any explanations?

Speaking of which, where's the edge of the earth?

"have you got your EQ set correctly? -db drop in the 300hz to 1khz will have a similar effect (it`s used alot in Karoke machines, to drop the voice part on a tape, so you can "Sing" your own over it)." Ah! That's the solution to my problem! I want to record a song and take away the lyrics. Exactly what does your post mean? I'm not computer literate.

I videotaped someone playing the piano and I want to see what notes they played. However, the clip is a bit blurry. How can I make it clearer?

Anyone know where I can find the notes to play the Hoedown song on a piano?

Do all sound cones vibrate with the same intensity? I'm thinking of using a sound cone from an "Electronic Playground." Will it not vibrate as much as a radio's?

Ah, thanks, yeah that's the one. YT, can you describe the rotating mirror one?

Ok I think I have a proof for tangents of G-circles for equilateral triangles intersecting. Equilateral triangle means z and y are equal, A and C are 60, tan 60 = n/m. More importantly, using my method, acute angles have negative m's. (mz + zn sin C - mz cos C)/ n = (y(1+ cos(A+C))/ sin (A+C)) becomes (-my+ yn*sin60 + my cos 60)/n = y(1 + cos(120))/(sin (120)) ...basically going down the steps with algebra, cancel out the y's, do some more algebra, substitute n=m tan60, you prove that this is true. Therefore, tangents intersect.

Oh yeah, you're right . However, I'm still sure the one for right triangles is accurate. Proving that y= (ab)/(-a+b+c) will be able to show that the tangents intersect for right triangles. (mz + zn sin C - mz cos C)/ n = (y(1+ cos(A+C))/ sin (A+C)) m=0, C is 90 degrees The equation becomes (0z + zn*1 - 0zcos90)/n = y(1+ cos(A+90))/ sin(A+90) zn/n = y(1+ -sinA)/ cosA z=y(1-sinA)/cosA Remember that z+y=a, so z= a-y. Remember that sinA is a/c and cosA is b/c from trig. Therefore (a-y) = y(1-(a/c))/(b/c) (a-y)(b/c) = y((c-a)/c) (ab-yb)/c = (yc-ya)/c ab - yb = yc - ya ab = yc - ya + yb ab = y(-a + b + c) y = ab/(-a + b + c) The reason, bad judgement on my part, to say for all triangles that y = 2(area)/(perimeter - 2a) is because it works for all right triangles and an obtuse triangle I tried (*coulda sworn I did it for the equilateral, perhaps bad memory). Anyways, if we can prove for right triangles that y = ab/(-a + b + c), it means that for right triangles, (mz + zn sin C - mz cos C)/ n = (y(1+ cos(A+C))/ sin (A+C)). Working our way back up, that means the y-coordinates are all the same when the x-coordinates are the same, meaning they intersect.

If you're 10 feet away from one radio and 5 feet from another, will turning one on (through a remote control) affect the intensity of the other radio?

It's like a laser connected to a radio or something. The lasers oscillate at frequencies that you control using the radio tuner. The laser makes cool pictures, including stars, circles, swirlies, squares, zig zags, etc.

For all triangles I've gotten it down to y = 2(area)/(perimeter - 2a)

For right triangles, I simplified some equations down to y= (ab)/(-a+b+c) If someone can prove this to be true for right triangles, then the proof for right triangles will be done. It seems true for the values I put in.

Yeah, it turns out that I subconsciously and automatically estimated the results and plugged the estimations into the calculator. Although I estimated to the thousandths, the extra decimals caused the difference at the end to appear. It seems that the answer to your question is yes, the pairwise tangent lines have a point in common whether they are obtuse, right, or acute. *Err, now we need a general proof rather than guess and check.

It took me 3 hours and 3 sheets of paper to figure it out, so I can't put all the stuff down except for the important stuff. There was also my lack of foresight in realizing that using x and y as G-circle radii wasn't a good idea. Diagrams will help in reading this at any rate. There were some interesting results. Skip to the bottom after reading next 3 paragraphs for the the intersection of tangents. My method of problem solving was to put an obtuse triangle on the coordinate plane with an endpoint called C at (0,0). Point A of triangle is in Quadrant II, and point B is on the x-axis to the right of the origin. The G-circle of point C has a radius of z. The G-circle of A has radius of x. The G circle of B has radius of y. The sides of the triangle opposite of the points are lower cases of the point. Meaning side "a" is on the x-axis, which is opposite of point "A" in Quadrant II. The angles are labeled by the upper case. Angle A is at point A. The line containing side "a" is y=0. That means the a pairwise tangent is at x=z, perpendicular to "a" and containing the point of intersection of the 2 G-circles. The tangent will be called GTAN1 Define the slope of line "b" as -n/m. That means the pairwise tangent of these 2 G-circles will have a slope of m/n. Secondly, the pairwise tangent goes through the intersection of the two G-circles, which lies on the point (z cos C, z sin C) using trigonometry. Having the slope and the point, you figure that the pairwise tangent is y= (mx + zn sin C - zm cos C)/ n . That's GTAN2 Line "c" contains (a, 0). The x-coordinate is the distance from the origin, which is the side "a." Say that (a,0) was the center of the G-circle centered at point B. The intersection of the two circles would be ( a + y cos (A+C), y sin (A+C)). This is because angles A + B + C = 180. Therefore, 180 - B = A + C . Adding the "a" in the x-coordinate is because you shift the graph "a" to the right from the origin. From the 2 points (a,0) and (a + y cos (A+C), y sin (A+C)), we figure the slope of line "c" to be tan(A+C). That means the slope of the pairwise tangent is -cot (A+C) going through (a + y cos (A+C), y sin (A+C)). We figure the pairwise tangent line to be y = (-x cos(A+C) + a cos (A+C) + *y)/ sin (A+C) which is called GTAN3. Notice that I put *y. That's y the distance, or the radius of the G-circle centered at B. Bad judgement on my part. Now we have 3 GTAN lines. The intersection of GTAN1 and GTAN2 is (z, (mz + zn sin C - mz cos C)/ n) . The intersection of GTAN1 and GTAN3 is (z, (-z cos(A+C) + a cos (A+C) + *y)/ sin (A+C)) Oh, using the equation y-a=-z and substituting it into the y-coordinate of the point, we get (*y(1+ cos(A+C))/ sin (A+C)) If the three lines have a point in common, (mz + zn sin C - mz cos C)/ n = (*y(1+ cos(A+C))/ sin (A+C)) When you have acute angles, the m turns into -m, meaning the distance changes direction from the left of the origin to the right of the origin. When you have a right tirangle, m = 0. When you have an obtuse triangle, m is positive. Make sure you put the right numbers into the calculator. It turns out equilateral triangles are do intersect. The obtuse triangle and 3-4-5 triangle I tried have really close intersections, but not exactly. *Hold that thought, doing it manually might give more accurate results than calculators.

test [math] sin( 2) [/math]

So you have 3 circles with radii x, y, z x + y = 5 y + z = 12 x + z = 13 When you solve it, you get x = 3, y = 2, z = 10. x + y = 20 y + z = 21 x + z = 29 x = 14, y = 6, z = 15

I'm guessing there is something in psychology relating to a variety of aggressive behaviors that people show, some more than others. And I'm also guessing there's more than one kind of bully. I know some general reasons why people bully others, but I want to know about this kind of bully in particular. I try to grab a rubber band on the floor and this guy slides down and grabs my wrist. He vehemently tries to pry my hand open, but I put it in my other hand. Then he starts twisting my hand, or at least the sleeve of the wrist I was wearing. Then he aims a rubber band gun at my face and threatens to shoot me. He then threatens to slap me if I don't give the rubber band to him. I decide to give it back to him as to not cause trouble. He then starts to act nice, saying he didn't mean any of it. Then he tries to befriend me by flattering me. Some other things about him is that he tries to be the center of attention, pushes people when they're in his way, and tries to be nice to the teacher. According to the way he tried to befriend me, it also appears that he believes I can easily be taken advantage of. Now, I don't want to just acquiesce next time. But I'll need to know where to hit him hard, not physically but mentally. What are the psychological characteristics of this kind of bully? What kind of mental games can I play with him? What do I say to make him cry? I know it's ironic, but I want my revenge! >:}

I didn't use straws, but it'd be interesting to see what would happen.

This book I just read for my English class is called The Sound and the Fury by William Faulkner. He won a Nobel Prize in literature, yet the book I read broke every rule of English I ever learned. It was sloppy. It had no punctuation in some areas. There was supposed to be chaos, confusion, etc. The plot skipped around through time. And my English teacher thought it was a good book.

I've had a grudge against English since a long time. Not only is it a really subjective class (the teacher grades in theory based on a rubric, but technically grades without guidelines and focuses more on whether or not they like the person), my class focuses on memory and logical structure rather than creativity. I used to think creativity was better than essay structure, especially in writing. You don't go out and read essays, you go out and read books. Books are written in infinitely many ways, and I think English should be more open-minded in this sense.

Yeah

Basically, when you connect radiuses of the circles so that the radiuses are colinear, you get a leg of the triangle. Since the radii are equal (congruent circles), they form an equilateral triangle.