I think you already have assisted me a billion times over everyone's expectation.
I think you already have assisted me a billion times over everyone's expectation.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
And the smallest one and only even prime is the number 2.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
10 is half the sum of the first two primes 2 and 3 which is the one and only pair of consecutive primes separated by 1.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
Howzitgoin .... You Turkish Trickster.....
Its possible that Antonio has detected Time Reversal.
After all, Prime 2 + Prime 3 = 5. But 10 is twice the sum of 5, not half the sum. Unless Antonio has detected time reversal, in which case 10 could be considered half the sum of 5 .... lol
cool bananas ... greg![]()
'Blondie says I must hate all Brunettes. I'll try, but if I can't ... I'll love them both'
... graffiti on Tavern wall, Pompeii, circa AD 70.
Thanks, Greg. Too bad this new TOEQUEST does not allow me to go back in space and time to make the correction 10 is twice of 5 and although all of my 10 typing fingers are operational.
Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
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