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Forum locked This topic is locked, you cannot edit posts or make further replies.  [ 17 posts ]  Go to page Previous  1, 2
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 Post subject: Re: Need help fo my math Exam
PostPosted: Thu May 16, 2013 5:56 pm 
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Excl1 wrote:
i have to find out how many blocks are in the layers http://prntscr.com/1558e9
like that to find that out i have to use this http://prntscr.com/1558gp

but if i do that with 2 layers i get this
(2x2^3+3x2^2+2)
---------------
6
I get 5 >.> it suppose to be 4! only if i use layer i get the correct answer ? what am i doing wrong?

oh god what am i looking a- *head explodes from math*

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 Post subject: Re: Need help fo my math Exam
PostPosted: Fri May 17, 2013 5:43 am 
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the top one has 1 block always since it is a pyramide so if you know they are for example 5 layers you can do this:



a=1(since the top layer has one block)+1
layer 2 :a X a(in this case 2X2=4)
layer 3 :(a+1) X (a+1);(2+1) X (2+1)=3 X 3= 9
layer 4 :(a+2) X (a+2);(2+2) X (2+2)=4 X 4=16
layer 5 :(a+3) X (a+3);(2+3) X (2+3)+5 X 5 =25


that works if your math teacher lets you use different ways to solve the problem, but many math teacher want a problem solved in a formula not in an improvised way.

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 Post subject: Re: Need help fo my math Exam
PostPosted: Fri May 17, 2013 8:31 am 
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aister wrote:
how the hell can u get 5 from that :|

(2x2^3+3x2^2+2)
---------------
6

= (2 * 8 + 3 * 4 + 2) / 6 = 78 / 6 = 13

did you learn in 3rd grade that dots go before lines so in fact it is

2 * 8 + 3*4+2
16 + 12 + 2

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 Post subject: Re: Need help fo my math Exam
PostPosted: Fri May 17, 2013 11:16 am 
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agonasani wrote:
the top one has 1 block always since it is a pyramide so if you know they are for example 5 layers you can do this:

a=1(since the top layer has one block)+1
layer 2 :a X a(in this case 2X2=4)
layer 3 :(a+1) X (a+1);(2+1) X (2+1)=3 X 3= 9
layer 4 :(a+2) X (a+2);(2+2) X (2+2)=4 X 4=16
layer 5 :(a+3) X (a+3);(2+3) X (2+3)+5 X 5 =25

that works if your math teacher lets you use different ways to solve the problem, but many math teacher want a problem solved in a formula not in an improvised way.

But think how long it takes if there are 500 layers.

The easiest way is to work out the volume of a pyramid: WidthxLengthxHeight/3, and as W=L=H, it can be reduced to (n^3)/3.

But that pyramid has all the side-blocks at a slope, so you need to add a sloping half block for each outer block: +(n^2)/2

and then you need to add a bit more for the corner pieces, all the way up to the top, the corner blocks have 1/3rd the volume of a full block, and you've already added half a block from the previous addition, so you only need to add 1/6th more (1/3 + 1/2 +1/6 = 1), so you add on; +n/6

That leaves you with:
(n^3)/3 + (n^2)/2 + n/6 (which can be reduced to (2n^3 + 3n^2 +n)/6


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 Post subject: Re: Need help fo my math Exam
PostPosted: Fri May 17, 2013 12:28 pm 
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i will just use my way i didnt get a thing from the formulas

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 Post subject: Re: Need help fo my math Exam
PostPosted: Fri May 17, 2013 1:19 pm 
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Okay then, what figure did you get for a 500 layer pyramid then?


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 Post subject: Re: Need help fo my math Exam
PostPosted: Sat May 18, 2013 1:00 am 
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i didnt do it yet

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