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Re: Buying a palace [3k=5k]

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DatMusic is not online. DatMusic
Joined: 22 Jun 2012
Total Posts: 95
27 Jul 2012 11:15 PM
Linkjs
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DatMusic is not online. DatMusic
Joined: 22 Jun 2012
Total Posts: 95
27 Jul 2012 11:17 PM
sdfds
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GamesEater is not online. GamesEater
Joined: 23 Jul 2012
Total Posts: 20901
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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OmegaPowerful is not online. OmegaPowerful
Joined: 11 Jun 2011
Total Posts: 75827
27 Jul 2012 11:18 PM
3K does not in fact equal 5K.
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Dillsion1 is not online. Dillsion1
Joined: 05 Jun 2010
Total Posts: 1971
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 is not online. sigh3192
Joined: 24 Oct 2009
Total Posts: 4245
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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Slayer9116 is not online. Slayer9116
Joined: 07 May 2011
Total Posts: 14904
27 Jul 2012 11:21 PM
i might make one

but i dont have bc right now.
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richy8447 is not online. richy8447
Joined: 28 Jan 2010
Total Posts: 6222
27 Jul 2012 11:21 PM
Sigh you forgot the 8
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sigh3192 is not online. sigh3192
Joined: 24 Oct 2009
Total Posts: 4245
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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darkprimal is not online. darkprimal
Joined: 17 Feb 2011
Total Posts: 3393
27 Jul 2012 11:25 PM
sigh , omg yus

i will make a palace for 3k
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sigh3192 is not online. sigh3192
Joined: 24 Oct 2009
Total Posts: 4245
27 Jul 2012 11:26 PM
See, Dark understands.
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darkprimal is not online. darkprimal
Joined: 17 Feb 2011
Total Posts: 3393
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 is not online. sigh3192
Joined: 24 Oct 2009
Total Posts: 4245
27 Jul 2012 11:27 PM
BTW you can make me a palace too. Futuristic Roman plz 2k
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darkprimal is not online. darkprimal
Joined: 17 Feb 2011
Total Posts: 3393
27 Jul 2012 11:27 PM
im sure sigh used algebra.
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sigh3192 is not online. sigh3192
Joined: 24 Oct 2009
Total Posts: 4245
27 Jul 2012 11:28 PM
Algebra? What is this nonsense!!? I used copy and paste.
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richy8447 is not online. richy8447
Joined: 28 Jan 2010
Total Posts: 6222
27 Jul 2012 11:28 PM
Sigh..do it again!!!
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darkprimal is not online. darkprimal
Joined: 17 Feb 2011
Total Posts: 3393
27 Jul 2012 11:29 PM
lol'd
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darkprimal is not online. darkprimal
Joined: 17 Feb 2011
Total Posts: 3393
27 Jul 2012 11:31 PM
people who wanted the palaces.
message me back!
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sigh3192 is not online. sigh3192
Joined: 24 Oct 2009
Total Posts: 4245
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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