THE ganondorflol Posted February 28, 2007 Posted February 28, 2007 ''Gasps''- you are fluent in Japanese? *envies*
Supergrunch Posted February 28, 2007 Posted February 28, 2007 ''Gasps''- you are fluent in Japanese? *envies* Nowhere near... I can sort of read it a bit.
DCK Posted February 28, 2007 Posted February 28, 2007 If you really hate remembering things like gender, then try Japanese- no gender, no plurals, only present and past, and verbs don't conjugate for persons. Yeah, Japanese is a piece of cake. EDIT: Can you say 'to pun' as a means to say "the name puns on ..." instead of "the name is a pun on ..."?
MoogleViper Posted February 28, 2007 Posted February 28, 2007 For all of you sexy mathematically minded people:
Zell Posted February 28, 2007 Posted February 28, 2007 2s = tu + tv 2s - tu = tv (2s - tu)/t = v a/(b-c) = x/b ab/(b-c) = x t-g = h/200 200(t-g) = h
Supergrunch Posted February 28, 2007 Posted February 28, 2007 Yeah, Japanese is a piece of cake. EDIT: Can you say 'to pun' as a means to say "the name puns on ..." instead of "the name is a pun on ..."? You can, but it sounds a little bit odd. 2s = tu + tv2s - tu = tv (2s - tu)/t = v a/(b-c) = x/b ab/(b-c) = x t-g = h/200 200(t-g) = h I think that last one is wrong: t = g - h/200 h/200 = g - t h = 200(g - t) The others look right though. Edit: Lol, "nasty equations"...
Ashley Posted February 28, 2007 Posted February 28, 2007 EDIT: Can you say 'to pun' as a means to say "the name puns on ..." instead of "the name is a pun on ..."? Can't say I've ever heard it used that way, and my instinct is telling me no. Its an abstract verb (or have I made that up, its been nearly two years since I finished studying English Language) but I'd say stick with "the name is a pun on..."
Supergrunch Posted February 28, 2007 Posted February 28, 2007 Can't say I've ever heard it used that way, and my instinct is telling me no. Its an abstract verb (or have I made that up, its been nearly two years since I finished studying English Language) but I'd say stick with "the name is a pun on..." Mm, OED says it's right, but like you say, it sounds wrong so it isn't advisable.
DCK Posted March 1, 2007 Posted March 1, 2007 Okay, I've handed in "... the name punning on... " for my bilingual exams, but no worries
MoogleViper Posted March 3, 2007 Posted March 3, 2007 1/f = 1/u + 1/v rearrange this formulae to express v in terms of f and u.
Shino Posted March 3, 2007 Posted March 3, 2007 1/f = 1/u + 1/v rearrange this formulae to express v in terms of f and u. 1/f = 1/u + 1/v <=> <=> f = u + v <=> <=> v = f - u I think thats want you wanted.
Supergrunch Posted March 3, 2007 Posted March 3, 2007 Shino is right. And you have to love double implication signs.
Ginger_Chris Posted March 3, 2007 Posted March 3, 2007 erm, 1/f = 1/u + 1/v does not, ever, imply f = u+v. how about: 1/v = 1/f - 1/u 1/v = (u-f)/(u*f) v = (u*f)/(u-f) if 1/f = 1/u + 1/v ==> f= u+ v it would make lens focusing mathematics much much easier. (but they are hard, so it doesnt). it would also suggest that parrelle light woudnt be focused at any point, and basically much up the whole of optics. Not thats entirely a bad thing, I hate optics.
Zell Posted March 3, 2007 Posted March 3, 2007 You can, but it sounds a little bit odd. I think that last one is wrong: t = g - h/200 h/200 = g - t h = 200(g - t) The others look right though. Edit: Lol, "nasty equations"... Bugger, got the signs wrong my C1 skills are poor. Gonna get my january module results this thursday, I think I got close to full marks on my C1
MoogleViper Posted March 3, 2007 Posted March 3, 2007 erm, 1/f = 1/u + 1/v does not, ever, imply f = u+v. how about: 1/v = 1/f - 1/u 1/v = (u-f)/(u*f) v = (u*f)/(u-f) if 1/f = 1/u + 1/v ==> f= u+ v it would make lens focusing mathematics much much easier. (but they are hard, so it doesnt). it would also suggest that parrelle light woudnt be focused at any point, and basically much up the whole of optics. Not thats entirely a bad thing, I hate optics. I thought it was too simple. I managed to get to 1/v = (u-f)/(u*f) but didn't know where to go after that.
Shino Posted March 3, 2007 Posted March 3, 2007 erm, 1/f = 1/u + 1/v does not, ever, imply f = u+v. how about: 1/v = 1/f - 1/u 1/v = (u-f)/(u*f) v = (u*f)/(u-f) if 1/f = 1/u + 1/v ==> f= u+ v it would make lens focusing mathematics much much easier. (but they are hard, so it doesnt). it would also suggest that parrelle light woudnt be focused at any point, and basically much up the whole of optics. Not thats entirely a bad thing, I hate optics. You are correct. I knew the only time I get to answer is the one I screw up.
Supergrunch Posted March 3, 2007 Posted March 3, 2007 erm, 1/f = 1/u + 1/v does not, ever, imply f = u+v. how about: 1/v = 1/f - 1/u 1/v = (u-f)/(u*f) v = (u*f)/(u-f) if 1/f = 1/u + 1/v ==> f= u+ v it would make lens focusing mathematics much much easier. (but they are hard, so it doesnt). it would also suggest that parrelle light woudnt be focused at any point, and basically much up the whole of optics. Not thats entirely a bad thing, I hate optics. Haha, good point. Boy, my maths teacher would kill me if he knew I had said that was right.
Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 Could anyone help with these questions on my integration homework? 1) The gradient of a curve is given by dy = 6x^2 - 3x dx The curve passes through the point (-1,2). Find the equation of the curve. (I wasn't in when she went through questions like this) 3) i) Express x^5/2+1 .....................x^2 in the form x^a + x^b, where a and b are contants. (^ = to the power of) Any help is appreciated.
Fresh Posted March 11, 2007 Posted March 11, 2007 Could anyone help with these questions on my integration homework?1) The gradient of a curve is given by dy = 6x^2 - 3x dx The curve passes through the point (-1,2). Find the equation of the curve. (I wasn't in when she went through questions like this) 3) i) Express x^5/2+1 .....................x^2 in the form x^a + x^b, where a and b are contants. (^ = to the power of) Any help is appreciated. Dude would you mind checking if you have typed in question 1 right? it may be me but would you mind?
Supergrunch Posted March 11, 2007 Posted March 11, 2007 Could anyone help with these questions on my integration homework?1) The gradient of a curve is given by dy = 6x^2 - 3x dx The curve passes through the point (-1,2). Find the equation of the curve. (I wasn't in when she went through questions like this) 3) i) Express x^5/2+1 .....................x^2 in the form x^a + x^b, where a and b are contants. (^ = to the power of) Any help is appreciated. (1) dy/dx = 6(x^2) - 3x y = 2(x^3) - (3/2)(x^2) + c (by integrating) and where x = -1, y = 2 so: 2 = -2 - (3/2) + c c = 5.5 so the equation is: 2y = 4(x^3) + 3(x^2) + 11 (3) the expression is equivalent to: (x^(5/2))/(x^2) + 1/(x^2) which is equal to: x^(1/2) + x^(-2)
Fresh Posted March 11, 2007 Posted March 11, 2007 (1) dy/dx = 6(x^2) - 3xy = 2(x^3) - (3/2)(x^2) + c (by integrating) and where x = -1, y = 2 so: 2 = -2 - (3/2) + c c = 5.5 so the equation is: 2y = 4(x^3) + 3(x^2) + 11 (3) the expression is equivalent to: (x^(5/2))/(x^2) + 1/(x^2) which is equal to: x^(1/2) + x^(-2) Damn, forgot the +c. *shakes fist at air*
Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 Supergrunch you are a God. Fresh > I would hvae totally forgot about it if my teacher didnt go through the basics with us, and if grunchy didnt mention it.
Supergrunch Posted March 11, 2007 Posted March 11, 2007 It's a basic form of differential equation, so if you're used to them then it makes it easier to remember the constant of integration.
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