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Atomic Boo

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Guest Stefkov

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.

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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.

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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.

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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.

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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.

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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.

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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.

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