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Help with tricky math received! (Confirmed):)


Clown

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I can't get my head around this one "equation."

 

 

 

 

 

 

 

So, I figured out the pattern when it comes to cutting a circle up and finding the resulting pieces. I can't make an equation for it (to my understanding), so I think I need to use a sigma command.

 

 

 

 

 

 

 

Cutting a circle with one line will give you two pieces, for example. Two lines will make four pieces. The maximum number of pieces created from any given cut is what the equation will be all about.

 

 

 

received

 

 

 

Here, I found this information out by cutting circles.

 

 

 

 

 

 

 

Number of cuts in a circle - Pieces Created (Max)

 

 

 

0 - 1

 

 

 

1 - 2

 

 

 

2 - 4

 

 

 

3 - 7

 

 

 

4 - 11

 

 

 

5 - 16

 

 

 

6 - 22

 

 

 

7 - 29

 

 

 

8 - 37

 

 

 

9 - 46

 

 

 

10 - 56

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I think it would looks something like this:

 

 

 

 

 

 

 

 

 

 

 

sigmafr0.png

 

 

 

 

 

 

 

X = The Number of Lines you are talking about.

 

 

 

The sigma SHOULD give you the Y (max pieces resulting from cut)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[(n/2)*(n+1)]+1

~

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Thats ridicules. These forums have way, way to many rules.

 

 

 

 

 

 

 

PLEASE TELL ME AT LEAST A LITTLE INFORMATION IF YOU PLAN TO LOCK ME.

 

 

 

 

 

 

 

He's right though.. we AREN'T here to do your work for you.

 

 

 

 

 

 

 

~Defender~

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He never asked anyone to. Give him an example of how to make an equation, but don't give him the answer. Pretty easy.

 

 

 

 

 

 

 

Edit: I'm no math expert, but being that you are dividing them, this looks like simple division. X/Y=Z, X being the circle, Y being the number of cuts and Z being the number of peices you end up with...But I think that you're asking for an equation to figure out how many you'd have continuing on with the pattern, and I can't help ya' there.

The popularity of any given religion today depends on the victories of the wars they fought in the past.

- Me!

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I don't really want an exact equation for this pattern. I just need some help, maybe somebody to confirm my sigma I gave, ect...

 

 

 

 

 

 

 

 

 

 

 

If you look at the data, you can see the difference between the maximum number of pieces is consecutively going up by one.

 

 

 

 

 

 

 

Look..

 

 

 

 

 

 

 

0 - 1

 

 

 

1 - 2 Difference = 1

 

 

 

2 - 4 Difference = 2

 

 

 

3 - 7 Difference = 3

 

 

 

4 - 11

 

 

 

5 - 16

 

 

 

6 - 22

 

 

 

7 - 29

 

 

 

8 - 37

 

 

 

9 - 46

 

 

 

10 - 56

 

 

 

 

 

 

 

See the pattern? I have no experience with equations of this kind.

~

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It's also going up by the number of cuts made in the last one + 1! :P

 

 

 

 

 

 

 

Personally, I'm more interested in how you can cut a circle 4 times and get 11 equal peices. I would have guess 8. :-s

The popularity of any given religion today depends on the victories of the wars they fought in the past.

- Me!

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It's also going up by the number of cuts made in the last one + 1! :P

 

 

 

 

 

 

 

Personally, I'm more interested in how you can cut a circle 4 times and get 11 equal peices. I would have guess 8. :-s

 

 

 

 

 

 

 

Trust me, it CAN be done. I've experimented.

 

 

 

 

 

 

 

It's looking like my sigma is accurate?

~

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Yea, it's accurate according to what you have there.

 

 

 

 

 

 

 

Someone should draw a picture of how you can draw an even number of lines (4) and end up with an odd number of peices, though (11). :P I'm just not seeing how that can be done at the moment...

 

 

 

 

 

 

 

Someone will show me and I'll probably slap my forehead and go "oh..durr"

The popularity of any given religion today depends on the victories of the wars they fought in the past.

- Me!

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Oh, I thought they had to be equal, like, halving the circle. So I'm thinkin' "Okay...Unless this is some weird [wagon] circle, that...Isn't possible..." :lol:

The popularity of any given religion today depends on the victories of the wars they fought in the past.

- Me!

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Heh, now you can see what I have to deal with lol :lol:

 

 

 

 

 

 

 

The equation in relation to cuts - resulting pieces is..a challenge lol

 

 

 

I'm finding it a bit stressful.

~

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We did this for extra credit in class. This is what I got, that too I got it checked by my dad, a math proffessor, so hopefully its right. I'll give you a hint. Its in the quadratic equation form.

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i believe the answer will be

 

 

 

 

 

 

 

N

 

 

 

1+ (sigma) n

 

 

 

n=0

 

 

 

 

 

 

 

 

 

 

 

simple as that... 1 plus every number up to the one your "at"

 

 

 

 

 

 

 

theres a simpler way of working that out too i believe...

 

 

 

 

 

 

 

i think 1+(N+1)(N)/2 will give you the answer at any given number of lines (where N is the number of lines)

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Yeah you're right about the equation. You're using sigma cause it's a summation factor, and basically the formula for the given situation is something along the lines of..

 

 

 

 

 

 

 

1+ ̢̮Ãâ¹Ã¢â¬Â Ã¢ââ¬ÃÅ n

 

 

 

 

 

 

 

(with a n on top of the sigma and then x=0 below it)

turnip3zk1.png
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Yeah you're right about the equation. You're using sigma cause it's a summation factor, and basically the formula for the given situation is something along the lines of..

 

 

 

 

 

 

 

1+ ̢̮Ãâ¹Ã¢â¬Â Ã¢ââ¬ÃÅ n

 

 

 

 

 

 

 

(with a n on top of the sigma and then x=0 below it)

 

 

 

 

 

 

 

That is not correct. I'm talking about a sigma for the number of resulting pieces from a given number of lines, NOT a sigma for "+1."

~

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0 - 1

 

 

 

1 - 2

 

 

 

2 - 4

 

 

 

3 - 7

 

 

 

4 - 11

 

 

 

5 - 16

 

 

 

6 - 22

 

 

 

7 - 29

 

 

 

8 - 37

 

 

 

9 - 46

 

 

 

10 - 56

 

 

 

Dont know if you noticed, but the answer seems to be the result of the last term plus n,

 

 

 

i.e. for the 6th term 22 = 16 + 6 and for the 10th term 56 = 46 + 10

 

 

 

 

 

 

 

Dont know how you display that as a formula though :x

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Yeah you're right about the equation. You're using sigma cause it's a summation factor, and basically the formula for the given situation is something along the lines of..

 

 

 

 

 

 

 

1+ ̢̮Ãâ¹Ã¢â¬Â Ã¢ââ¬ÃÅ n

 

 

 

 

 

 

 

(with a n on top of the sigma and then x=0 below it)

 

 

 

 

 

 

 

That is not correct. I'm talking about a sigma for the number of resulting pieces from a given number of lines, NOT a sigma for "+1."

 

 

 

 

 

 

 

You need the +1 at the beginning for the number of pieces with 0 lines since summation factor of 0 would give you 0. The sigma part is for anything that's over 0 lines which you'd add to 1 (from 0 lines) to in order to get the total parts in the circle given the number of lines.

turnip3zk1.png
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Yeah you're right about the equation. You're using sigma cause it's a summation factor, and basically the formula for the given situation is something along the lines of..

 

 

 

 

 

 

 

1+ ̢̮Ãâ¹Ã¢â¬Â Ã¢ââ¬ÃÅ n

 

 

 

 

 

 

 

(with a n on top of the sigma and then x=0 below it)

 

 

 

 

 

 

 

That is not correct. I'm talking about a sigma for the number of resulting pieces from a given number of lines, NOT a sigma for "+1."

 

 

 

 

 

 

 

You need the +1 at the beginning for the number of pieces with 0 lines since summation factor of 0 would give you 0. The sigma part is for anything that's over 0 lines which you'd add to 1 (from 0 lines) to in order to get the total parts in the circle given the number of lines.

 

 

 

 

 

 

 

No matter. The sigma has been found (if you wish to see it, it's on my opening post.)

~

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