Zanarkin Posted June 24, 2011 Share Posted June 24, 2011 I had my math exam today and there is a question that is bugging me. The question was... t10= -21 and t19= -48. Find a and d, then S4. This is what i did. tn = a + (n-1)(d) -21 = a + (10-1) (d) a = 9d + 21 I called this one 1 Doing the same for t19 I got a = 18d + 48 I called it 2 2 - 1 a = 18d + 48 a = 9d + 21 ------------ 0 = 9d + 27 -27 = 9d -3 = d Substitute 3 into either eqn #1 or #2 a = 9(-3) + 21 a = -6 Checking it just to make sure... -21 = -6 + 9(-3) -21 =/= -33 Where did i go wrong? Quote Link to comment Share on other sites More sharing options...
Original Alear Posted June 24, 2011 Share Posted June 24, 2011 Since this is Far from the Forest, You went wrong a long time before this math equation, my friend. Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 24, 2011 Share Posted June 24, 2011 (edited) t10= -21 and t19= -48. Find a and d, then S4. tn = a + (n-1)(d) -21 = a + (10-1) (d) -a = 9d + 21 I called this one 1 a=-21-9d Doing the same for t19 I got a = -18d - 48 I called it 2 - 21 - 9d=-18d - 48 9d=-27 d=-3 a=54-48=6 Lazy quoted the jist of what you did wrong was you forgot to minus the a You should be able to get t4 I make it -3 Edited June 24, 2011 by mikethfc Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted June 24, 2011 Author Share Posted June 24, 2011 (edited) fuck that - sign :( Thanks though :) Edited June 24, 2011 by Jhen Mohran Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 24, 2011 Share Posted June 24, 2011 (edited) You should've seen the chatroom earlier, that had AEA stuff (The integral of h(x))0.5= The integral of (h(x))0.5 Where h(x) is (dy/dx)2 Show that dy/dx = 2(y+c) Would look simpler if I could do that integral sign Edited June 24, 2011 by mikethfc Quote Link to comment Share on other sites More sharing options...
Zanarkin Posted June 24, 2011 Author Share Posted June 24, 2011 : s AEA? Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 24, 2011 Share Posted June 24, 2011 : s AEA? Its like STEP papers only some questions can be done without breaking down into tears and doing a Van Gogh impression on your wrists. Quote Link to comment Share on other sites More sharing options...
NTG Posted June 24, 2011 Share Posted June 24, 2011 STEP, STEP, STEP... a step forward into mind-boggling m-m-mathematics. Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 24, 2011 Share Posted June 24, 2011 STEP, STEP, STEP... a step forward into mind-boggling m-m-mathematics. Its O.K. fun maths exists too Quote Link to comment Share on other sites More sharing options...
Kriemhild Posted June 25, 2011 Share Posted June 25, 2011 You should've seen the chatroom earlier, that had AEA stuff (The integral of h(x))0.5= The integral of (h(x))0.5 Where h(x) is (dy/dx)2 Show that dy/dx = 2(y+c) Would look simpler if I could do that integral sign As I said, the right side would be the integral of dy/dx (which is not y, btw, since the dx isn't being canceled out here). While we both thought the left side would be equal to the square root of ydy/dx, we would be making the same mistake. We're only integrating (dy/dx)^2, not (dy/dx)^2 dx Also, if you try to work backwards from the end... dy/dx = 2y => dy = 2ydx => integral of dy = integral of 2y dx => y = 2 * integral of y dx The 2 is quite annoying, but y seems like a function involving e. Not sure if that's supposed to help. Quote Link to comment Share on other sites More sharing options...
Jaybee Posted June 25, 2011 Share Posted June 25, 2011 math guys look at this. a, b, c, d are all integers larger than 0. x = sqrt(a2 + b2), y = sqrt(c2 + d2). Prove that xy >= ac + bd GAAAAAAAAAAAARGH Quote Link to comment Share on other sites More sharing options...
Nightmare Posted June 25, 2011 Share Posted June 25, 2011 !bottle All the math you can get. Quote Link to comment Share on other sites More sharing options...
SquareRootOfPi Posted June 25, 2011 Share Posted June 25, 2011 math guys look at this. a, b, c, d are all integers larger than 0. x = sqrt(a2 + b2), y = sqrt(c2 + d2). Prove that xy >= ac + bd GAAAAAAAAAAAARGH Have you tried squaring both sides? Quote Link to comment Share on other sites More sharing options...
Jaybee Posted June 25, 2011 Share Posted June 25, 2011 I think that's what anyone would do. Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 25, 2011 Share Posted June 25, 2011 Have you tried getting the question paper the sheet was on and burning the fucker. It won't help with the maths but it will make you feel better. Quote Link to comment Share on other sites More sharing options...
Kriemhild Posted June 25, 2011 Share Posted June 25, 2011 math guys look at this. a, b, c, d are all integers larger than 0. x = sqrt(a2 + b2), y = sqrt(c2 + d2). Prove that xy >= ac + bd GAAAAAAAAAAAARGH xy = sqrt(a2 + b2) * (c2 + d2) = sqrt((a2 + b2)(c2 + d2)) = sqrt(a2c2 + b2c2 + a2d2 + b2d2) All terms inside the square root are positive, but I'm not sure how that would help... Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 25, 2011 Share Posted June 25, 2011 xy = sqrt(a2 + b2) * (c2 + d2) = sqrt((a2 + b2)(c2 + d2)) = sqrt(a2c2 + b2c2 + a2d2 + b2d2) All terms inside the square root are positive, but I'm not sure how that would help... What if you square it from there so it becomes x2y2= (a2c2 + b2c2 + a2d2 + b2d2) As the middle two terms have to be positive the LHS has to be bigger, as the LHS is bigger than the RHS, the root of the LHS must also be bigger than the RHS. Quote Link to comment Share on other sites More sharing options...
Kriemhild Posted June 25, 2011 Share Posted June 25, 2011 What if you square it from there so it becomes x2y2= (a2c2 + b2c2 + a2d2 + b2d2) As the middle two terms have to be positive the LHS has to be bigger, as the LHS is bigger than the RHS, the root of the LHS must also be bigger than the RHS. Problem is, x2y2 >= a2c2 + b2d2 doesn't necessarily imply that xy >= ab + cd Actually, wait, the possibilities I was thinking of involve one of the integers being negative. I just reread the original post and saw that they're all positive. nvm, I got it. Thanks! Quote Link to comment Share on other sites More sharing options...
SquareRootOfPi Posted June 25, 2011 Share Posted June 25, 2011 (edited) Um, (ac+bd)^2 = a^2c^2 + 2abcd + b^2d^2 You have to do a bit more work. Edited June 25, 2011 by SquareRootOfPi Quote Link to comment Share on other sites More sharing options...
Mikethfc Posted June 25, 2011 Share Posted June 25, 2011 Um, (ac+bd)^2 = a^2c^2 + 2abcd + b^2d^2 You have to do a bit more work. Rookie error But it narrows it down to b2c2 + a2d2 > 2abcd I'll probably come back to this in about an hour or so. Quote Link to comment Share on other sites More sharing options...
Ansem Posted June 25, 2011 Share Posted June 25, 2011 this thread made my head explode Quote Link to comment Share on other sites More sharing options...
Jaybee Posted June 26, 2011 Share Posted June 26, 2011 Rookie error But it narrows it down to b2c2 + a2d2 > 2abcd I'll probably come back to this in about an hour or so. This is what I got to before I gave up. Quote Link to comment Share on other sites More sharing options...
SquareRootOfPi Posted June 26, 2011 Share Posted June 26, 2011 (edited) Well, perhaps you should try showing that a^2d^2 - 2abcd + b^2c^2 is not negative. Edited June 26, 2011 by SquareRootOfPi Quote Link to comment Share on other sites More sharing options...
Jaybee Posted June 26, 2011 Share Posted June 26, 2011 Well, perhaps you should try showing that a^2d^2 - 2abcd + b^2c^2 is not negative. That is equal to (ad-bc)2 which... Oh god I'm an idiot. Quote Link to comment Share on other sites More sharing options...
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