Drake Glau Posted March 6, 2011 Report Share Posted March 6, 2011 2f) sin(-360+x) (I'm using x for theta)You can move the - out so -sin(360+x) and on the unit circle 360=0 in terms of degrees (it's the same thing on the graph whether it's 360 or 0) so you have -sin(0+x) or just -sin(x)3a) I'm having troubles with this for some reason. I ended up with something weird... Reply Link to post Share on other sites More sharing options...
BlackHawk Posted March 6, 2011 Report Share Posted March 6, 2011 (edited) Help getting the drawing. I don't need the solutionIt's done with paint, bare with it Could you show how do you count it? Edited March 6, 2011 by BlackHawk Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 7, 2011 Report Share Posted March 7, 2011 Help getting the drawing. I don't need the solutionIt's done with paint, bare with it Could you show how do you count it?Start at point A, assume you are facing straight up and that is 0 degrees. Turn counter clockwise however many degrees it says (200 in that case) and then make your line and label it with the length or whatever. When you get to point B do the same, assume you are facing straight up and that's zero degrees then turn counter clockwise however many degrees it says Reply Link to post Share on other sites More sharing options...
BlackHawk Posted March 7, 2011 Report Share Posted March 7, 2011 Help getting the drawing. I don't need the solutionIt's done with paint, bare with it Could you show how do you count it?Start at point A, assume you are facing straight up and that is 0 degrees. Turn counter clockwise however many degrees it says (200 in that case) and then make your line and label it with the length or whatever. When you get to point B do the same, assume you are facing straight up and that's zero degrees then turn counter clockwise however many degrees it says How to get the AC? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 7, 2011 Report Share Posted March 7, 2011 (edited) The question was to calculate that wasn't it? And it's a triangle so after you have AB and BC you just connect AC and you have a shape. Question says C is due south of A so I assumed it was directly under A and that's where it is. Edited March 7, 2011 by Drake Glau Reply Link to post Share on other sites More sharing options...
BlackHawk Posted March 7, 2011 Report Share Posted March 7, 2011 The question was to calculate that wasn't it? And it's a triangle so after you have AB and BC you just connect AC and you have a shape. Question says C is due south of A so I assumed it was directly under A and that's where it is.You can find angles and side lengths and use the rule of sines on this problem too Can you write the solution cuz I still don't get what the angles are? Please Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 7, 2011 Report Share Posted March 7, 2011 (edited) The question was to calculate that wasn't it? And it's a triangle so after you have AB and BC you just connect AC and you have a shape. Question says C is due south of A so I assumed it was directly under A and that's where it is.You can find angles and side lengths and use the rule of sines on this problem too Can you write the solution cuz I still don't get what the angles are? PleaseAbout 107.264km Edited March 7, 2011 by Drake Glau 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted March 7, 2011 Author Report Share Posted March 7, 2011 Help getting the drawing. I don't need the solutionBearings are always measured clockwise from the north, so your diagram should look like:and Drake's answer of AC is correct Reply Link to post Share on other sites More sharing options...
nametaken Posted March 7, 2011 Report Share Posted March 7, 2011 Another quick question:Is there a way of figuring out the turning points of a quadratic, without drawing a graph?Say, for example with x^2+2x+5.Thanks for the help...just going through revision for my maths exam. Reply Link to post Share on other sites More sharing options...
fan Posted March 7, 2011 Report Share Posted March 7, 2011 (edited) yes there is(x^2+2x+ 5)a= 1 b = 2 and c = 5axis of symmetry for a quadratic is -b/2athe axis of symmetry is where a turning point isso for that equation our x value is (-2/2)= -1substitute x into the equation and we get our y value as 4so co-ordinates of our turning point are (-1,4)Ateeq I had given you some stuff on that Edited March 7, 2011 by fan 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted March 7, 2011 Author Report Share Posted March 7, 2011 Another quick question:Is there a way of figuring out the turning points of a quadratic, without drawing a graph?Say, for example with x^2+2x+5.Thanks for the help...just going through revision for my maths exam.Turning points are the stationary points, right?So dy/dx=0So just differentiate y=x²+2x+5 with respect to xdy/dx=2x+2when 0=2x+2, x=-1when x=-1, y=(-1)²+2(-1)+5=4So the turning point is (-1, 4) Reply Link to post Share on other sites More sharing options...
Rigel Posted March 8, 2011 Report Share Posted March 8, 2011 I got an easier formula for finding the turning point: y = ax^2+bx+cCoordinates for turning point: x y(-b/2a; -D(Discriminant)/4a)So your quadratic function is:x^2+2x+5Then:Value of x: -2/2 => x=-1Value of y: Discriminant => D = (b^2)-4ac => (2^2)-4(1)(5) = -16; then => -16/4 => y=4Then the turning point is: (-1;4) Reply Link to post Share on other sites More sharing options...
nametaken Posted March 8, 2011 Report Share Posted March 8, 2011 Ok...another question:Is there a quick way of remembering the coefficients in the Pascal triangle, without actually drawing the triangle? It's just that I think that there is a way of remembering the coefficients, I just can seem to remember it.Any ideas?Thanks once again!!! Reply Link to post Share on other sites More sharing options...
Rigel Posted March 8, 2011 Report Share Posted March 8, 2011 (edited) Hey guys i got a question. This is about series:Let a1=2and an+1=n/(n+1)Find the first 5 terms of the series.I need help!!! Thank you a lot guys.And nametaken, i wish i could help you, but i just started IB yesterday Edited March 8, 2011 by Ipos Manger Reply Link to post Share on other sites More sharing options...
genepeer Posted March 9, 2011 Report Share Posted March 9, 2011 @nametaken haha, I can only go as far as the 4Cr, the rest... Calculate it with the n!/(r!(n-r)!) formula or draw the triangle.Speaking of Pascal's triangle; here's a cool look at the first 196 rows of Pascal's Triangle. The gap's represent numbers divisible by 7. Similar patterns form for any number, not just 7.Cool, right?@Ipos 1. Do you mean, an+1 = an/(an+1) or an+1=n/(n+1)?2. Do you really mean first terms of the series or sequence? Reply Link to post Share on other sites More sharing options...
Rigel Posted March 9, 2011 Report Share Posted March 9, 2011 1. Do you mean, an+1 = an/(an+1) or an+1=n/(n+1)?2. Do you really mean first terms of the series or sequence?1. The second one.2. I think that of the sequence. (I'm new to IB Maths).I got another exercise that i can't solve, Let:a1=1/2andan+1=(an+2)/nFind the first 5 terms of the sequence. Reply Link to post Share on other sites More sharing options...
Keel Posted March 9, 2011 Report Share Posted March 9, 2011 I am having trouble with linear combinations of normal distributions.Under what circumstances is the variance of the combined normal distribution 10 x Var(X) and when is it (10)2 x Var (X).Thank you in advance. Reply Link to post Share on other sites More sharing options...
genepeer Posted March 9, 2011 Report Share Posted March 9, 2011 @IposWhen just insert in the numbers, for an+1 = n/(n+1):a2 = a1+1 = 1/(1+1) = 1a3 = a2+1 = 2/(2+1) = 2/3etc.the same for the second question. Except now, a2 = a1+1 = (a1+2)/1 = (1/2+2)/1 = 5/2etc.@IposWhen just insert in the numbers, for an+1 = n/(n+1):a2 = a1+1 = 1/(1+1) = 1a3 = a2+1 = 2/(2+1) = 2/3etc.the same for the second question. Except now, a2 = a1+1 = (a1+2)/1 = (1/2+2)/1 = 5/2etc.@keel, I've forgotten statistics completely! 1 Reply Link to post Share on other sites More sharing options...
nametaken Posted March 9, 2011 Report Share Posted March 9, 2011 I've forgotten!! How would you work out the area between a curve and an axis?Thanks!! Maths test in 10 minutes.... Reply Link to post Share on other sites More sharing options...
Keel Posted March 9, 2011 Report Share Posted March 9, 2011 I've forgotten!! How would you work out the area between a curve and an axis?Thanks!! Maths test in 10 minutes....Intergration of the curve between the x values. READ QUICK! http://www.mathsrevision.net/alevel/pages.php?page=2 Reply Link to post Share on other sites More sharing options...
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