Ginger_Chris Posted November 26, 2006 Posted November 26, 2006 Circumference of circle= 2*pi*radius or pi*diameter
Letty Posted November 26, 2006 Posted November 26, 2006 Circumference of circle= 2*pi*radius or pi*diameter Thanks .
Guest Stefkov Posted November 26, 2006 Posted November 26, 2006 Could someone help with maths knowledge. Question: Show that the surface area, A cm2, of a carton is given by A = 4x2 + 216 ............------ ................x The carton is dimensions: 2x cm by x cm by h cm. i have no idea how to do it. Plus if anyone knows any Differentiation/Calculus then could you help me with another question i would pm you? Too much is asking in one post if i added it.
Ginger_Chris Posted November 26, 2006 Posted November 26, 2006 ok the carton has dimentions x by 2x by h? then: A= 2(x*2x)+2(x*h)+2(2x*h)=6xh+4(x^2) there has to be a h in the formula, is h a function of x at all? for that formula that you have to show the carton isnt a cuboid, and you need more information (like its shape, or a formula for its shape), h doesnt seem to be a constant. Anyway, i have a decent enough knowlege of calculus so send the pm.
Supergrunch Posted November 26, 2006 Posted November 26, 2006 Could someone help with maths knowledge.Question: Show that the surface area, A cm2, of a carton is given by A = 4x2 + 216 ............------ ................x The carton is dimensions: 2x cm by x cm by h cm. i have no idea how to do it. Plus if anyone knows any Differentiation/Calculus then could you help me with another question i would pm you? Too much is asking in one post if i added it. Is that C2? I think I remember doing that question.
Supergrunch Posted November 29, 2006 Posted November 29, 2006 I have a nasty question that I just can't work out: (by the way, I'm using j = sqrt(-1)) Solve: (j-z)^n = (jz-1)^n (j-z)/(jz-1) = a , where a = e^(jb) is an nth root of unity. z = (j+a)/(ja+1) But where do I go from here? Apparently it can be rearranged as z = (cosb)/(1-sinb) , but I don't see how.
Ginger_Chris Posted November 30, 2006 Posted November 30, 2006 I have a nasty question that I just can't work out:(by the way, I'm using j = sqrt(-1)) Solve: (j-z)^n = (jz-1)^n (j-z)/(jz-1) = a , where a = e^(jb) is an nth root of unity. z = (j+a)/(ja+1) But where do I go from here? Apparently it can be rearranged as z = (cosb)/(1-sinb) , but I don't see how. Right, I can get you to the z=cos(b)/(1-sin(b)) fairly long windedly so stick with it. I'm not sure what to do next, but it is 5am in the morning and im slightly tired. Right here it goes: (oh and i'm using i instead of j, i'm not an engineer) z=(i+a)/(ia+1) a=e^(ib) z=(i+e^(ib))/(ie^(ib)+1) dividing through by e^(ib/2) z=(ie^(-ib/2)+e^(ib/2))/(ie^(ib/2)+e^(-ib/2)) putting in terms of cos and sin z=(icos(b/2)+sin(b/2)+cos(b/2)+isin(b/2))/(icos(b/2)-sin(b/2)+cos(b/2)-isin(b/2)) this simplifies: z=((1+i)(cos(b/2)+sin(b/2))/((1+i)(cos(b/2)-sin(b/2)) z=(cos(b/2)+sin(b/2))/(cos(b/2)-sin(b/2)) multiplying top and bottom by the denomenator z=(cos2(b/2)-sin2(b/2))/(cos2(b/2)+sin2(b/2)-2cos(b/2)sin(b/2)) z=(cos2(b/2)-sin2(b/2))/(1-2cos(b/2)sin(b/2)) using double and half angle identities (whichever, i cant remember) z=cos(b)/(1-sin(b)) what you do then is up to you. from the roots of unity part you know that: b=2*pi*x/n where n is the n at the start and x=0,1,2,3.....(n-1) (giving you all the roots) if I were you I'd leave it at that unless you can simplify it anymore. Hope that helps, I'm sorry the forum isnt very clear for mathematics so if you dont get my notation just ask.
Supergrunch Posted November 30, 2006 Posted November 30, 2006 Right, I can get you to the z=cos(b)/(1-sin(b)) fairly long windedly so stick with it. I'm not sure what to do next, but it is 5am in the morning and im slightly tired. Right here it goes: (oh and i'm using i instead of j, i'm not an engineer) z=(i+a)/(ia+1) a=e^(ib) z=(i+e^(ib))/(ie^(ib)+1) dividing through by e^(ib/2) z=(ie^(-ib/2)+e^(ib/2))/(ie^(ib/2)+e^(-ib/2)) putting in terms of cos and sin z=(icos(b/2)+sin(b/2)+cos(b/2)+isin(b/2))/(icos(b/2)-sin(b/2)+cos(b/2)-isin(b/2)) this simplifies: z=((1+i)(cos(b/2)+sin(b/2))/((1+i)(cos(b/2)-sin(b/2)) z=(cos(b/2)+sin(b/2))/(cos(b/2)-sin(b/2)) multiplying top and bottom by the denomenator z=(cos2(b/2)-sin2(b/2))/(cos2(b/2)+sin2(b/2)-2cos(b/2)sin(b/2)) z=(cos2(b/2)-sin2(b/2))/(1-2cos(b/2)sin(b/2)) using double and half angle identities (whichever, i cant remember) z=cos(b)/(1-sin(b)) what you do then is up to you. from the roots of unity part you know that: b=2*pi*x/n where n is the n at the start and x=0,1,2,3.....(n-1) (giving you all the roots) if I were you I'd leave it at that unless you can simplify it anymore. Hope that helps, I'm sorry the forum isnt very clear for mathematics so if you dont get my notation just ask. Ah, thank you so much... none of my class could do that question. Don't worry, I'm fine once it gets to the simplified trig thing. I think it's strange that our syllabus uses j instead of i... but I use j all the time so I don't use i by mistake and get it marked wrong. But i is better, because it looks more like 1.
Charlie Posted November 30, 2006 Posted November 30, 2006 What are the differences between Marxism and Neo-Marxism? (4 marks) That question is going to come up in my NAB tomorrow that I need to pass otherwise I get chucked off the course... Diemetrix, I'm looking in your direction...
AshMat Posted December 2, 2006 Posted December 2, 2006 How do i work out Standard Deviation and Range from a Grouped Frequency table (For maths Coursework)
Supergrunch Posted December 2, 2006 Posted December 2, 2006 How do i work out Standard Deviation and Range from a Grouped Frequency table (For maths Coursework) You approximate, by making each datum at the centre of its category.
AshMat Posted December 2, 2006 Posted December 2, 2006 You approximate, by making each datum at the centre of its category. A full working out method would be really usefull, doesn't ahve to be majorly long, but i don't know how to start, they haven't told us how..
Supergrunch Posted December 2, 2006 Posted December 2, 2006 A full working out method would be really usefull, doesn't ahve to be majorly long, but i don't know how to start, they haven't told us how.. Right, well you presumably have different data categories, like 40<x=<50 etc. Then you assume that every data point is in the middle of its category- for instance, every data point in the 40 to 50 category is 45. So you now have a frequency table. Now, for standard deviation, you need to work out the sum of all squares. This is the sum of f(x - mean)^2, for every x. Then divide by the sum of f and square root. As for range, I'm not entirely sure, but I'd imagine that you take the top and bottom midpoints and subtract the smaller from the larger.
AshMat Posted December 2, 2006 Posted December 2, 2006 ok, the working out in my notebook is: x|---D---|D Squared 3|-2.754-| Answers 5|-0.75--| go 7|1.25---| Here 8|2.25---| Where does the D column come from?
Ginger_Chris Posted December 2, 2006 Posted December 2, 2006 eww stats, thats not a nice branch of maths
AshMat Posted December 2, 2006 Posted December 2, 2006 Right, i figured out where the D column came from, now as it is a grouped frequency table, i guess i use the mid value? EDIT: Done Just need to do my range. And also copy this whole apge up neat because it's so damn scruffy.
Supergrunch Posted December 2, 2006 Posted December 2, 2006 eww stats, thats not a nice branch of maths I know what you mean... apparently it gets interesting at degree level, but that's because by then it's basically pure...
Shorty Posted December 2, 2006 Posted December 2, 2006 Circumference of circle= 2*pi*radius or pi*diameterThanks I was always taught to memorise that as "2 Pi r" and remember the song "Kumbuya my lord"
Supergrunch Posted December 2, 2006 Posted December 2, 2006 I was always taught to memorise that as "2 Pi r" and remember the song "Kumbuya my lord" That formula annoys me. I once thought it was the area of the circle, and got a question wrong as a result.
MoogleViper Posted December 4, 2006 Posted December 4, 2006 Could somebody please get me a game sprite of a wooden ship. Sort of like a pirate ship or a merchant's ship. Also could somebody please find me some maps of what places (preferably Britain) would look like if the polar ice caps melted.
Atomic Boo Posted December 10, 2006 Author Posted December 10, 2006 what does 'es lebe deutschland' mean is it something like ' this is germany' ? thanks for the help
weeyellowbloke Posted December 10, 2006 Posted December 10, 2006 If someone could write a 10,000 word report on the geological history of the St Davids area by next Wednesday, that would be just dandy. I've kind of banned myself from the internet until I get it done, but I suppose I've cracked already. Damn my lack of willpower.
Sarka Posted December 10, 2006 Posted December 10, 2006 For Standard Grade History - in Enquiry Skills, for a How fully questions, do you use direct quotes form the source or not? eg. How fully does Source D explain the 'Turnip Winter' in Germany?
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