Posted March 15, 2008 Oops... sorry. I put the wrong crocodile. Way to go, me... Also, I'm pretty sure it's posible. If I get a proof I'll post it here. It's not true. Draw an equilateral triangle with side 9 and all of those criteria will be met but BC won't be even. Share this post Link to post Share on other sites

Posted March 15, 2008 It's not true. Draw an equilateral triangle with side 9 and all of those criteria will be met but BC won't be even. They aren't all met - in that case AB and AC would be greater than d. Share this post Link to post Share on other sites

Posted March 15, 2008 This is from a set of problems we were given by an exchange school. The maths teachers can't do it, so any maths enthusiasts - knock yourselves out! You are given a triangle ABC. The point M is the mid-point of BC. The length of MA is d. Prove that, given d < AB and d < AC, BC is always an even integer. Have fun! They aren't all met - in that case AB and AC would be greater than d. Which is what they are supposed to be. Share this post Link to post Share on other sites

Posted March 16, 2008 Which is what they are supposed to be. Ah, the post got edited... the first time round, it said d > AB &c. Share this post Link to post Share on other sites

Posted March 16, 2008 Ah, the post got edited... the first time round, it said d > AB &c. That's not possible. If M is on BC and d is greater than AB and AC then ABC can't be a triangle. The furthest point from A has to be one of the corners. Unless I've read it wrong. Oops... sorry. I put the wrong crocodile. Way to go, me... Also, I'm pretty sure it's posible. If I get a proof I'll post it here. You should have read the thread Grunchie. Anyway my post still stands. It's not true. Share this post Link to post Share on other sites

Posted March 16, 2008 It's not true. Draw an equilateral triangle with side 9 and all of those criteria will be met but BC won't be even. Sorry. I'm an idiot. All the sides and d are natural numbers. I must have been drunk or something that night. Getting my inequalities mixed up and forgetting essential information. Oops... Share this post Link to post Share on other sites

Posted March 16, 2008 Sorry. I'm an idiot. All the sides and d are natural numbers. I must have been drunk or something that night. Getting my inequalities mixed up and forgetting essential information. Oops... GRRR. It still isn't true as you could draw and equilateral triangle with sides of 9.323454235345345345 or whatever. Share this post Link to post Share on other sites

Posted March 16, 2008 GRRR. It still isn't true as you could draw and equilateral triangle with sides of 9.323454235345345345 or whatever. But all the sides are natural numbers (+ve integers). So no. Share this post Link to post Share on other sites

Posted March 16, 2008 But all the sides are natural numbers (+ve integers). So no. They aren't integers. They could have side pi and it would fit your criteria. Share this post Link to post Share on other sites

Posted March 16, 2008 They aren't integers. They could have side pi and it would fit your criteria. Natural numbers are the integers from 1 upwards, so 'twouldn't. (sometimes 0 is included, but here that's irrelevant) Share this post Link to post Share on other sites

Posted March 17, 2008 Natural numbers are the integers from 1 upwards, so 'twouldn't. (sometimes 0 is included, but here that's irrelevant) Oh right I missed that bit. I thought he meant my equilateral triangle had natural numbers. Share this post Link to post Share on other sites

Posted March 18, 2008 Another maths one... i, j and k are unit vectors in the x-, y- and z-directions respectively. The 3-dimensional vector r has a magnitude of 5, and makes angles of 1/4 pi radians with each of i and k. Write r as a column vector, leaving any square roots in your answer. whaaaat theee fuuuuuck. Share this post Link to post Share on other sites

Posted March 18, 2008 Well, if it makes angles of pi/4 radians (aka 45 degrees) with both i and k, then it stands to reason that it lies in the i k plane. (draw a diagram if you can't see why this is). So the j component is 0, but what of the other two? Well they'll both be 5cos(pi/4) (or 5sin(pi/4), same thing), which is 5/sqrt(2). So the i component is 5/sqrt(2), the j component is 0 and the k component is 5/sqrt(2), which you can write as a column vector. Share this post Link to post Share on other sites

Posted March 18, 2008 Well, if it makes angles of pi/4 radians (aka 45 degrees) with both i and k, then it stands to reason that it lies in the i k plane. (draw a diagram if you can't see why this is). So the j component is 0, but what of the other two? Well they'll both be 5cos(pi/4) (or 5sin(pi/4), same thing), which is 5/sqrt(2). So the i component is 5/sqrt(2), the j component is 0 and the k component is 5/sqrt(2), which you can write as a column vector. Assuming you mean 5/2 sqrt(2) then awesome. That what the answer apparently is, and now i know how to get to it ^_^. Just check on my calculator, 5cos(pi/4) is 5/2 sqrt(2) Share this post Link to post Share on other sites

Posted March 18, 2008 Assuming you mean 5/2 sqrt(2) then awesome. That what the answer apparently is, and now i know how to get to it ^_^. Just check on my calculator, 5cos(pi/4) is 5/2 sqrt(2) Actually, our answers are one and the same, you've just rationalised the denominator by multiplying top and bottom by sqrt(2): 5/(sqrt(2)) = (5sqrt(2))/((sqrt(2))(sqrt(2))) = (5sqrt(2))/2 = (5/2)sqrt(2). Share this post Link to post Share on other sites

Posted April 4, 2008 Need someone who knows Excel as its been years since I've used it. Basically I want a bar graph that along the side has number of mins, along the bottom is days and along the other side is a different set of numbers (percent of day) which is shown as a line graph. Basically I am doing a project which looks at American tv shown on Britain. The days will remain constant but the number of mins and percent are so different (say 200-odd would make a 8% or whatever) so the two range of numbers have to be different on either side otherwise the line graph just floats around the base. Thanks. Real example (sorry about formatting): W/C 16/02/08 – BBC1 Day No. of Mins % of day Saturday 225 15.625 Sunday 170 11.805556 Monday 220 15.277778 Tuesday 135 9.375 Wednesday 175 12.152778 Thursday 45 3.125 Friday 100 6.944444 As you can see as it stands the left side would read 45-225 and the left should read 3.125-15.625 (or be rounded up) Share this post Link to post Share on other sites

Posted April 4, 2008 Couldn't you do several graphs, one for each day? It would mean more pages, but it might be easier to understand and then you can just note the total number of minutes devoted to US television programmes. Share this post Link to post Share on other sites

Posted April 5, 2008 I need to condense it into a five minute presentation though... Im going to go at it from the per week angle instead, and have all five channels on one graph. Thanks though Share this post Link to post Share on other sites

Posted April 22, 2008 x= t-1/t y=t+1/t show dy/dx= (t-1)(t+1)/t^2+1 This is really annoying me, because we've done the problem essentially in a maths exercise before (except y and x other way round) but I can't get the final answer, because I differentiate y to get dy/dt= -t^-2+1 which means I have a negative which I don't want. Can someone help me please thanks. Share this post Link to post Share on other sites

Posted April 22, 2008 x= t-1/t y=t+1/t show dy/dx= (t-1)(t+1)/t^2+1 This is really annoying me, because we've done the problem essentially in a maths exercise before (except y and x other way round) but I can't get the final answer, because I differentiate y to get dy/dt= -t^-2+1 which means I have a negative which I don't want. Can someone help me please thanks. x= t-1/t y=t+1/t dx/dt = 1+t^-2 dy/dt = 1-t^-2 dy/dx = dy/dt X dt/dx dy/dx = 1-t^-2/1+t^-2 dy/dx = 1+t^2/1-t^2 dy/dx = (t+1)(t-1)/(t+1)(1-t) dy/dx = (t-1)/(1-t) Multiply by (-1/-1)__________[-1/-1 = 1, so it's correct] (t-1)/(1-t) * (-1/-1) = -1(t-1)/-1(1-t) = (-t+1)/(-1+t) = (1-t)/(1-t) QED: (t-1)/(1-t) = (1-t)/(t-1), therefore (t-1) = (1-t) dy/dx = (t-1)(t+1)/(t+1)(1-t) = (t-1)(t+1)/(t+1)(t+1) QED: dy/dx = (t-1)(t+1)/t^2+1 Yes, it's not actually numerically correct, but it is algebraically, oddly... Share this post Link to post Share on other sites

Posted April 22, 2008 dy/dx = 1+t^2/1-t^2 dy/dx = (t+1)(t-1)/(t+1)(1-t) Well you've confused me, how did you go from that to that? Surely (t+1)(t-1)=t^2-1 Share this post Link to post Share on other sites

Posted April 22, 2008 Your both wrong, I believe the answer is 6! Though seriously that maths is WAAAAAAAAAAAAY beyond my level i look at it and get an instant headache... Share this post Link to post Share on other sites

Posted April 22, 2008 I need to condense it into a five minute presentation though... Im going to go at it from the per week angle instead, and have all five channels on one graph. Thanks though Here's what I did with the data that you gave: Excel File Download. Hope it helps. You could probably add extra day data quite easily to that graph. Share this post Link to post Share on other sites

Posted April 22, 2008 What core maths is that? C4. Its so annoying, I'm sure I've made a silly mistake. I was hoping to have it figured out for when I take it in tomorrow but I just can't see where I've gone wrong, nor understand the fish. Also fish, when you said (t+1)(t+1) at the bottom = T^2+1, thats wrong. Its t^2+ 2t +1. Share this post Link to post Share on other sites