When people keep asking me useless questions. Kind of like what you're doing right now. I was thinking more along the lines of "genius".
HAHA! why, then i should be in good shape -lol i'm starting to like u now uhm maybe, maybe not oh, what was your score then, if you don't mind me asking? lmao yeah that! can't believe I didn't kno what it was called. Stupid me -noclue gosh, that's so jacked up! btw, nothing math related but does anyone know what 3sf stands for in science?
I'm not sure about 3sf but i know 3s is the name of an orbital of an atom. f is also another name of an orbital, but I'm not sure what 3sf together is.
Great Hope everything's going well for you. Nope. Far from it actually. 2160 And that was after taking it a second time See above post. You may rest assured. I mean, surely Airree was just being sarcastic with that comment. Geniuses are often crazy. I have another problem though, but this one 'should' be easier because it's on the parabola. Unfortunately, I'm not good with geometrical problems. P is a point on the parabola x^2=4y The normal at P meets the parabola again at Q. The tangents at P and Q meet at T. S is the focus and QS=2PS a.) Prove that angle PSQ = 90 degrees b.) Prove that PQ=PT It's a parametrics question, which is an easier topic, but it's one that I struggle with for some reason.
lol, only because I need help But yeah, I noticed that there has been an influx of old threads on the first few pages of several sections in the forum. What happened?
we had an invader who was boosting his post counts well me thinks it was to download something but i dunno so he ran-sacked the lounge
Oh, I see. Well, whatever it was he was attempting to download, I'm sure he has more than enough post counts to get it now
a. Prove that PSQ = 90 degrees. The fuction of the parabola: y = x^2/4. -> The co-ordinate of P: P(xP, yP) with yP > 0 and yP = xP^2/4 (1). -> The co-ordinate of Q: Q(xQ, yQ) with yQ > 0 and yQ = xQ^2/4 (2). The co-ordinate of S is S(0,1), so we have: PS^2 = xP^2 + (yP - 1)^2 = (yP + 1)^2 <=> PS = yP + 1. QS^2= xQ^2 + (yQ - 1)^2 = (yQ + 1)^2 <=> QS = yQ + 1. Since PS = 2QS, we have: yP = 2yQ+1 (3). The fuction of the normal: y = ax + b with a < 0. Since the line y = ax + b is the normal of the parabola y = x^2/4 at Q(xQ, yQ), we have: a(xQ^2)'/4 = -1 <=> axQ/2 = -1 <=> axQ = -2 (4). P and Q belong to the line y = ax + b, so we have: yP = axP + b (5). yQ = axQ + b (6). From 1 2 3 4 5 6 we have the 6 variable system of equations: yP = xP^2/4 yQ = xQ^2/4 yP = 2yQ + 1 axQ = -2 yP = axP + b yQ = axQ + b => yP = xP^2/4 yQ = xQ^2/4 yP = axP + b yP = 2b - 3 yQ = b - 2 xQ = -2/a => yP = xP^2/4 yP = 2b - 3 2b - 3 = axP + b yQ = 1/a^2 yQ = b - 2 => yP = xP^2/4 yP = 2b - 3 xP = (b - 3)/a b - 2 = 1/a^2 => 2b - 3 = {[(b - 3)/a]^2}/4 (A) b = 2 + 1/a^2 (B) Replacing b in the equation (A) by (2 + 1/a^2) in (B) will give us: a = -1/2 then b = 6 xP = -6; yP = 9 xQ = 4; yQ = 4 => PQ^2 = 125. PS^2 = 100 QS^2 = 25. Since PQ^2 = PS^2 + QS^2 = 125, according to the Pythagorean theorem, the angle PSQ = 90 degrees. b. Prove that PQ = PT. We already have the co-ordinates of P and Q. Defining the co-ordinate of T, and then calculating PT and comparing it to PQ will give you the desired result. God d@mn, that was a lot of work. I should be a part time math tutor. 25 bucks per hour unless you're a hot Asian chick, you can pay me other ways.
Anyway, I certainly hope I wasn't expected to be able to questions like that... just look at the amount of working I get the feeling it's not a standard parametrics question though, so hopefully that kind of stuff is just to "challenge" us as opposed to actually doing it in a 2 hour exam. Thank you for all your help, though, Geometry is really not my thing. Surely a few simple high school level math problems such as these should require very little effort from a "genius" such as yourself.
ROARR I HATE PHYSICSS.. my teacher is a jerk who can't even teach.. he doesn't even check if we did our homework right =_= anyway.. i need help in phyiscs too!! ashdgaksfsldghaldjrhl;' so close to that 4.0 and this stupid class is keeping me from it DD: A child sitting 2.20 m from the center of a merry-go-round moves with a speed of 1.25 m/s. Calculate [a] the centripetal acceleration of the child, and the net horizontal force eserted on the child (mass= 25.0 kg)
