DatMusic
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| Joined: 22 Jun 2012 |
| Total Posts: 95 |
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DatMusic
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| Joined: 22 Jun 2012 |
| Total Posts: 95 |
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| 27 Jul 2012 11:18 PM |
Colipolie's first place. I hate posting with this account. Durr.
Vive la République, vive la France! |
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| 27 Jul 2012 11:18 PM |
| 3K does not in fact equal 5K. |
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Dillsion1
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| Joined: 05 Jun 2010 |
| Total Posts: 1971 |
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| 27 Jul 2012 11:18 PM |
How about you make me it for 1 TIX and I give it to everyone?
~Disco General~ |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:20 PM |
3x X y^4-3(x^4)=-131 3(x^4)-y^4=131
[3^1/2 (x^2)]^2 - [y^2]^2=131 {[left]-[right]}{[left]+[right]}=131 --------expression (A) the {[left]+[right]} =5k
:P |
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| 27 Jul 2012 11:21 PM |
i might make one
but i dont have bc right now. |
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richy8447
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| Joined: 28 Jan 2010 |
| Total Posts: 6222 |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:23 PM |
Oyus:
3x X y^4-3(x^4)=-131 3(x^4)-y^4=131
[3^1/2 (x^2)]^2 - [y^2]^2=131 {[left]-[right]}{[left]+[right]}=131 --------expression (A^8) the {[left]+[right]} =5k
Therefore:
IF: x and y were positive integers then x^2 and y^2 would also be integers, SQRT(3) is not an integer, therefore (x^2)(sqrt(3)) cannot be an integer
we can expand {[left]-[right]} again as difference of squares
{(3^1/4)(x)}^2-y^2}= [3^1/4(x)-y][3^1/4+y] so we have - using "int" for integer and "non" for non integer, [non-int]=non [non+int]=non
So, in (A) we have, (non)(non)(non)=131
The fractional part of the first and second (non) is the fractional part of the root of 3 but the integer part is different so, it will not produce an integer. The fractional part of the third (non) is the fractional part of sqrt of 3 so, it will not produce an integer when multiplied with the first 2 (non). Therefore, these (non)(non)(non) will not produce an integer. But 131 is an integer. It follows that our initial supposition about x and y was not valid. |
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| 27 Jul 2012 11:25 PM |
sigh , omg yus
i will make a palace for 3k |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:26 PM |
yu want mesh edited, and stuff cframed? whut kind of palace? victorian age? roman? plz tell me. |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:27 PM |
| BTW you can make me a palace too. Futuristic Roman plz 2k |
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| 27 Jul 2012 11:27 PM |
| im sure sigh used algebra. |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:28 PM |
| Algebra? What is this nonsense!!? I used copy and paste. |
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richy8447
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| Joined: 28 Jan 2010 |
| Total Posts: 6222 |
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| 27 Jul 2012 11:31 PM |
people who wanted the palaces. message me back! |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:31 PM |
The science of gnomes:
3x X y^4-3(x^4)=-131 3(x^4)-y^4=131
[3^1/2 (x^2)]^2 - [y^2]^2=131 {[left]-[right]}{[left]+[right]}=131 --------expression (A^8) the {[left]+[right]} =5k
Therefore:
IF: x and y were positive integers then x^2 and y^2 would also be integers, SQRT(3) is not an integer, therefore (x^2)(sqrt(3)) cannot be an integer
we can expand {[left]-[right]} again as difference of squares
{(3^1/4)(x)}^2-y^2}= [3^1/4(x)-y][3^1/4+y] so we have - using "int" for integer and "non" for non integer, [non-int]=non [non+int]=non
So, in (A) we have, (non)(non)(non)=131
The fractional part of the first and second (non) is the fractional part of the root of 3 but the integer part is different so, it will not produce an integer. The fractional part of the third (non) is the fractional part of sqrt of 3 so, it will not produce an integer when multiplied with the first 2 (non). Therefore, these (non)(non)(non) will not produce an integer. But 131 is an integer. It follows that our initial supposition about x and y was not valid.
So
5x^2 -45x +10 = 0
2x/(4x +2) -x +9 = 0
12x^2/(32 -x) +1 = 4/x
And one simultaneous quadratic:
y^2 = 10x +6
5x - 2y = x^2 +4 Then: 1. if 3+5+7+.........+n terms/ 5+8+11+............+10 terms =7
the value of n is a) 35 b) 36 c) 37 d) 40
2. if A,B,C are the points z1,z2,z3 ant the angles B and C are each pie- alpha/2 then (z2-z3)^2 = 4(z3-z1)(z1-z2) sin^2 alpha/2
So gnomes= The mass of a blackhole during puberty.
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