richy8447
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| Joined: 28 Jan 2010 |
| Total Posts: 6222 |
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| 27 Jul 2012 11:38 PM |
x2 + y1 = m M - a1 = f F = 15
So therefore dwarf = 97
See my logic? |
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| 27 Jul 2012 11:39 PM |
you just want to be like sigh. you will never top his skill |
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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 11:39 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.
BUT the mass of a blackhole is not static therefore we must assume that gnomes can harness this power and change their shape as they choose. |
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| 27 Jul 2012 11:40 PM |
Where exactly does dwarf enter the equation here? All you showed is that the *F* is dwarf is equivalent to 15. What about *D*, *W*, *A*, and *R*?
- Cubyfan2 |
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richy8447
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| Joined: 28 Jan 2010 |
| Total Posts: 6222 |
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| 27 Jul 2012 11:41 PM |
You forgot to input the desisive factor of evil gnomes, dwarves and warrior monkies...
Therefor gnomes = the mass of TWO black holes during puberty |
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TDFall
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| Joined: 07 May 2009 |
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| 27 Jul 2012 11:41 PM |
WTF did I just read?
- Cubyfan2 |
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| 27 Jul 2012 11:42 PM |
| i should stay out of this junk |
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sigh3192
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| Joined: 24 Oct 2009 |
| Total Posts: 4245 |
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| 27 Jul 2012 11:42 PM |
| BUT The mass of a blackhole is not static and two black holes in a binary hole is even less static. Therefore warrior monkeys would turn into gnomes. Dwarves are extinct, and all gnomes are evil. So your argument is invalid. |
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| 28 Jul 2012 12:06 AM |
| I'm pretty, sure my mind has been violated..... |
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Notom
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| Joined: 18 Nov 2008 |
| Total Posts: 5108 |
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| 28 Jul 2012 12:08 AM |
did i just get mindrped?
◙WAT U GON DOO?◙ |
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Flammus
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| Joined: 26 Aug 2011 |
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Bcs54
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| Joined: 19 Sep 2009 |
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| 28 Jul 2012 12:09 AM |
me an sigh, bringing genious equations to you. |
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| 28 Jul 2012 12:09 AM |
| OMG MADONNA, CAN'T EVEN FIGURE THIS OUT HAX! |
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| 28 Jul 2012 12:10 AM |
| If only "Honeydew" was, here he would know what to do =\ |
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eletrowiz
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| Joined: 08 Dec 2008 |
| Total Posts: 12438 |
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| 28 Jul 2012 12:11 AM |
Summary; some people have a lot of imaginations making up equations. The chances of this can be represented as F^2 If F = AC^2/CD*GH We can quickly see that it is the same odds of a troll being under a bridge. |
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| 28 Jul 2012 12:13 AM |
| Ya.... what "eletrowiz" said. |
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| 28 Jul 2012 12:14 AM |
but trolls are real, this is proved many times on the internet shall i sit under a bridge? |
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