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rudester Tommy

How did you survive being so retarded

rudester Tommy

Warmer

you get thorns village idiot

Originally posted by Blakemore
I don't live in Texas.

We'll forgive you for that. For now.

Nuke Nixon Senior Member

Texas is flat, come to the Rocky Mountains, enjoy the snow... ha hua muhahahahaa!

rudester Tommy

I feel like going somewhere hot. But it must be modern, **** no more 3rd world countries with no decent meals.

rudester Tommy

Imagine being trapped in a vehicle with no food or water for 8 hrs travelling the sites on a tourist bus. Pure torture.

rudester Tommy

I've always wanted to go to Iceland or Greenland, although not hot it does interests me

Thaf Restricted

The historical codex, theory of everything, cracks QM, unlike relativity and newton's laws of universal motion it's a function f(t)=

Part 1; an 216 columb and 2 row [x y] coordinate matrix (sphere worlds) plus another 216 columb 2 row coordinate matrix (quantum gravity).

[n n]

Etc...

Part 2; You take every integer in the first two matrices to the power of the product of those two matrices. Like Russian matryoshka dolls you have to do this n number times, n=radius of observable universe over the planck length. & we must root every exponent in the -z direction to superimpose the smaller and smaller x and y values in the +z direction. && everytime we do this we need to take out the top exponent from power towers and redo it for counting down to the x y coordinates of the z vector which is the center of the codex, let's call this operation V for vector. Problem with operation V, you will lose entire matrices of coordinates when you remove an exponent from the power tower, so now we must rectify that by putting op V through the power tower and plugging those values back into the base matrix, and then redoing the full power-tower with these adjusted values, but this time rooting every other iteration of exponential matrices in the power tower so that the values remain the same._

Ex)

Vector +3

[3.14^[√4.13^[√3.14_√4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]_4.13^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]]
[3.14^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]_4.13^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]]

&

Vector -3

[3.14^[4.13^[3.14_4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]_4.13^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]]
[3.14^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]_4.13^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]]

&&
Vector -2 part 1

[3.14^[4.13_3.14]
[4.13_3.14]_4.13^[4.13_3.14]
[3.14_4.13]]
[3.14^[4.13_3.14]
[4.13_3.14]]_
4.13^[4.13_3.14]
[4.13_3.14]]
etc...

&&&
Vector -2 part 2

[36.34^[1^[1 1]
[1 1]
Etc...Etc..

(Note! + and negative vectors shown in example are reverse from what they should be for the Historical Codex because the values for the matrices in this example are >1 but the actual values you want will all be <1)

EDIT:_And there's still one final grueling layer to part 2 I'm not accounting for. Of course the original 216 coordinates of three z matrices containing 9 spheres is not I'm 3.14 4.13 it's less than one meters! So that means in the code where the -z side of the power tower are all rooted exponents, eventually it's going to yield a value greater than 1! Eventually up to 13.8/2 billion light years over one meter! So wherever that happens, every exponent above it is_not going to be rooted, whatever iteration exponential matrices that is, will say the nth exponent.

Part 3;_Then you put your answer into a fibonacci sequence. The beginning of the sequence is 1 (first matrix) + 1 (second matrix how the gravity moves xyz up down forward back) = another matrix of sphere worlds which is multiplied by a different quantum gravity matrix which is the second iteration of this special fibonacci sequence, which is a potential universe at time equals 2 planck seconds, with 4 matrices. Etc..(note: you must know where the vectors talk to one another is equal to the generation of the gravitational matrices & when incorporating one gravitational matrix into another vector's matrices one must add or subtract it's transformational values by the difference of values between the matrices in those vectors)
Ex)

[n1 n1]____[n2 n2]____[n3 n3]

________+__________=_____________

[n1 n1]____[n2 n2]____[n3 n3]

&

[n1 n1]___[n3 n3]____[x1 y1]

_______+_________=2*______

[n1 n1]___[n3 n3]____[x1 y1]

&&

[n2 n2]___[n3 n3]__[x2 y2]

________+________=2*______

[n2 n2]___[n3 n3]__[x2 y2]

&&&

___[x1 y1]____[x2 ny2]__[n4 n4]

+/-_______+/-________=__________

___[x1 y1]____[x2 y2]__[n4 n4]

&&&&

[n3 n3]___[n4/2_n4/2]__[nx ny]___[n5 n5]

________+_____________+________=________

[n3 n3]___[n4/2 n4/2]__[nx ny]___[n5 n5]

where, nx & ny = geometric equations applied to the [n3 n3] columns' gravitational tug and [n4 n4] matrix gets divided by 2 because of the inverse square law for [n2 n2]'s gravitational tug applied to [n3 n3]'s position from when it was [n1 n1]

The practical use of the codex is evolving the observable universe from the moment of the big bang to now._

You will acquire a printout of all of the atoms of the periodic table, and how their quantum particles behave, with a magnification of up to 26 orders of magnitude greater than that offered by an electron microscope, and how micro lasers will effect a quantum system every planck second after your instruments perturb a system.

Unlike normal quantum teleportation, which involves changes over lateral time, the codex can instead be used to teleport quantum objects_using_particle confusion, which is a result of analog quantum gravity, as gravity propagates in two directions, moving n-number of sphere worlds that collide to generate one higgs function to the exact quantum phonetic frequency of another higgs function, will duplicate that higgs function precisely where the other higgs function is located.

This works because all potential universes are so similar at the quantum scale.

Thaf Restricted

wikipedia.org/wiki/Beal_conjecture

√3=0.547722557505

(BTW -i=-0.316227766)

So a⁴+b⁴=c³ is 1.31607401295⁴+1.31607401295⁴=1.817120592832³

(Where a & b = 4throot3 & c is cuberoot 6)

And of course a if Beal wants to jip the 3 trillion hobos seeing this he will say "you can't add fractions", however, all you'd have to do in that case is move the decimal over past the final digit.

131,607,401,295⁴+131,607,401,295⁴=1,817,120,592,832³ have GCPF of 3

I'd like my payment in cash please.

Thaf Restricted

Nobody gives a **** ACTUALLY

I can buy heroine with the money from Bealz conjecture but I can't afford any quality poontang unless it's a great big fat retarded black woman maybe I can go to Somalian Pirates with the money and get them to bring me sum cause nobody in the Navy is doing sht for me

Thaf Restricted

Originally posted by Thaf
wikipedia.org/wiki/Beal_conjecture
√3=0.547722557505
(BTW -i=-0.316227766)
So a⁴+b⁴=c³ is 1.31607401295⁴+1.31607401295⁴=1.817120592832³
(Where a & b = 4throot3 & c is cuberoot 6)
And of course a if Beal wants to jip the 3 trillion hobos seeing this he will say "you can't add fractions", however, all you'd have to do in that case is move the decimal over past the final digit.
131,607,401,295⁴+131,607,401,295⁴=1,817,120,592,832³ have GCPF of 3
I'd like my payment in cash please.

The historical codex, theory of everything, cracks QM, unlike relativity and newton's laws of universal motion it's a function f(t)=

Part 1; an 216 columb and 2 row [x y] coordinate matrix (sphere worlds) plus another 216 columb 2 row coordinate matrix (quantum gravity).

[n n]

Etc...

Part 2; You take every integer in the first two matrices to the power of the product of those two matrices. Like Russian matryoshka dolls you have to do this n number times, n=radius of observable universe over the planck length. & we must root every exponent in the -z direction to superimpose the smaller and smaller x and y values in the +z direction. && everytime we do this we need to take out the top exponent from power towers and redo it for counting down to the x y coordinates of the z vector which is the center of the codex, let's call this operation V for vector. Problem with operation V, you will lose entire matrices of coordinates when you remove an exponent from the power tower, so now we must rectify that by putting op V through the power tower and plugging those values back into the base matrix, and then redoing the full power-tower with these adjusted values, but this time rooting every other iteration of exponential matrices in the power tower so that the values remain the same._

Ex)

Vector +3

[3.14^[√4.13^[√3.14_√4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]_4.13^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]]
[3.14^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]_4.13^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]]

&

Vector -3

[3.14^[4.13^[3.14_4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]_4.13^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]]
[3.14^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]_4.13^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]]

&&
Vector -2 part 1

[3.14^[4.13_3.14]
[4.13_3.14]_4.13^[4.13_3.14]
[3.14_4.13]]
[3.14^[4.13_3.14]
[4.13_3.14]]_
4.13^[4.13_3.14]
[4.13_3.14]]
etc...

&&&
Vector -2 part 2

[36.34^[1^[1 1]
[1 1]
Etc...Etc..

(Note! + and negative vectors shown in example are reverse from what they should be for the Historical Codex because the values for the matrices in this example are >1 but the actual values you want will all be <1)

EDIT:_And there's still one final grueling layer to part 2 I'm not accounting for. Of course the original 216 coordinates of three z matrices containing 9 spheres is not I'm 3.14 4.13 it's less than one meters! So that means in the code where the -z side of the power tower are all rooted exponents, eventually it's going to yield a value greater than 1! Eventually up to 13.8/2 billion light years over one meter! So wherever that happens, every exponent above it is_not going to be rooted, whatever iteration exponential matrices that is, will say the nth exponent.

Part 3;_Then you put your answer into a fibonacci sequence. The beginning of the sequence is 1 (first matrix) + 1 (second matrix how the gravity moves xyz up down forward back) = another matrix of sphere worlds which is multiplied by a different quantum gravity matrix which is the second iteration of this special fibonacci sequence, which is a potential universe at time equals 2 planck seconds, with 4 matrices. Etc..(note: you must know where the vectors talk to one another is equal to the generation of the gravitational matrices & when incorporating one gravitational matrix into another vector's matrices one must add or subtract it's transformational values by the difference of values between the matrices in those vectors)
Ex)

[n1 n1]____[n2 n2]____[n3 n3]

________+__________=_____________

[n1 n1]____[n2 n2]____[n3 n3]

&

[n1 n1]___[n3 n3]____[x1 y1]

_______+_________=2*______

[n1 n1]___[n3 n3]____[x1 y1]

&&

[n2 n2]___[n3 n3]__[x2 y2]

________+________=2*______

[n2 n2]___[n3 n3]__[x2 y2]

&&&

___[x1 y1]____[x2 ny2]__[n4 n4]

+/-_______+/-________=__________

___[x1 y1]____[x2 y2]__[n4 n4]

&&&&

[n3 n3]___[n4/2_n4/2]__[nx ny]___[n5 n5]

________+_____________+________=________

[n3 n3]___[n4/2 n4/2]__[nx ny]___[n5 n5]

where, nx & ny = geometric equations applied to the [n3 n3] columns' gravitational tug and [n4 n4] matrix gets divided by 2 because of the inverse square law for [n2 n2]'s gravitational tug applied to [n3 n3]'s position from when it was [n1 n1]

The practical use of the codex is evolving the observable universe from the moment of the big bang to now._

You will acquire a printout of all of the atoms of the periodic table, and how their quantum particles behave, with a magnification of up to 26 orders of magnitude greater than that offered by an electron microscope, and how micro lasers will effect a quantum system every planck second after your instruments perturb a system.

Unlike normal quantum teleportation, which involves changes over lateral time, the codex can instead be used to teleport quantum objects_using_particle confusion, which is a result of analog quantum gravity, as gravity propagates in two directions, moving n-number of sphere worlds that collide to generate one higgs function to the exact quantum phonetic frequency of another higgs function, will duplicate that higgs function precisely where the other higgs function is located.

This works because all potential universes are so similar at the quantum scale.

rudester Tommy

Originally posted by Thaf
Nobody gives a **** ACTUALLY
I can buy heroine with the money from Bealz conjecture but I can't afford any quality poontang unless it's a great big fat retarded black woman maybe I can go to Somalian Pirates with the money and get them to bring me sum cause nobody in the Navy is doing sht for me

😮‍💨

Nuke Nixon Senior Member

Hawkeye isn't that bad, but it's boring and cringe.

Kate Bishop wasn't such a gomer in the comics, why do they make her a quip machine in this?

Doesn't make any sense.

rudester Tommy

Do you think I would do well in prison?

rudester Tommy

I feel like I would shower shank someone

Nuke Nixon Senior Member

Never know how you're going to do in potentially life-threatening situations until you're in one.

rudester Tommy

I'm in one every day cus life is a jungle

rudester Tommy

Just getting up.

Nuke Nixon Senior Member

That sounds exhausting. Get out of the city and give your nervous system a rest and rejuvenation.

rudester Tommy

At the moment I can't, dealing with crap but hopefully after the holidays. I've been so busy and so stressed