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Dose Mahi Drysdale have a rowing teamate?
umm Mahi Drysdale doesn't have a Rowing team mate for Rowing
Where is mahi drysdale from?
New Zealand
What has the author N S Mahi Mukandpuri written?
N. S. Mahi Mukandpuri has written: 'The teachings of Guru Ravidass'
When was Mahi-mahi created?
Mahi-mahi was created in 1758.
What are marine animals from a-z?
A Beluga Whale Catfish Dolphin Eel Flying fish G H I J Krill L Mahi-mahi Narwhal Octopus Porpoise Q Rockfish Swordfish Tuna U V Whale X Yellow fish Z
Where did the name mahi mahi come from?
In the Hawaiian language Mahi Mahi means 'very strong'. The Mahi Mahi fish is respected in the ancient Hawaiian culture.
Mahi mahi a hawaiian dish was once called what?
Mahi mahi, a popular fish, are found around the world in tropical and subtropical ocean waters. Another name for the mahi mahi is the dolphin fish.
Can you refreeze Mahi Mahi?
Yes.
Is mahi-mahi kosher?
Yes.
How do you pronounce mahi?
If you mean the fish Mahi Mahi, it's pronounced "MAH-hee".
Is mahi mahi a carnivore?
Yes. The dolphinfish, dorado, or mahi mahi feeds primarily on smal fish and squid.
What is the newest Proof about Fermat's last theorem?
PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE (X^+Y^2)^2=/=X^4+Y^4. SO (Z^2)^2=/=X^4+Y^4. SO Z^4=/=X^4+Y^4. CASE 2. IF Z^4=X^4+Y^4 BECAUSE X^4+Y^4.=/= (X^2+Y^2.)^2 SO Z^4=/=(X^2+Y^2.)^2 SO (Z^2)^2=/=(X^2+Y^2.)^2 SO Z^2=/=X^2+Y^2. (1) AND (2)=> Z^4+Z^2=/=X^4+Y^4+X^2+Y^2. SO 2Z^4+2Z^2=/=2X^4+2Y^4+2X^2+Y^2. SO (Z^4+Z^2+2Z^3+Z^4+Z^2-2Z^3)=/=(X^4+X^2+2X^3+X^4+X^2-2X^3)+)(Y^4+Y^2+2Y^3+Y^4+Y^2-2Y^3) SO IF (Z^4+Z^2+2Z^3)/4=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4 => (Z^4+Z^2-2Z^3)/4=/=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3/4) AND SO IF (Z^4+Z^2-2Z^3)/4=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3)./4 => (Z^4+Z^2+2Z^3)/4=/=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4 BECAUSE (Z^4+Z^2+2Z^3)/4 - (Z^4+Z^2-2Z^3)/4 =Z^3. SO Z^3=/=X^3+Y^3. GENERAL CASE N>2. Z^N=/=X^N+Y^N. WE HAVE [X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=X^(N+1)/2+Y^(N+1)/2+ H. BECAUSE X*Y>0=>H>0. SO [X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=/= X^(N+1)/2+Y^(N+1)/2 CASE 1. IF Z^(N-1)/2=X^(N-1)/2+Y^(N-1)/2 SO [Z^(N-1)/2]^(N+1)/(N-1)=[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1). BECAUSE [X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2. SO [Z^(N-1)/2]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2. SO Z^(N+1)/2=/=X^(N+1)/2+Y^(N+1)/2. CASE 2. IF Z^(N+1)/2=X^(N+1)/2+Y^(N+1)/2 SO [Z^(N+1)/2]^(N-1)/(N+1)=[X^(N+1)/2+Y^(N+1)/2 ]^(N-1)/(N+1) BECAUSE [X^(N+1)/2+Y^(N+1)/2](N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2. SO [Z^(N+1)/2]^(N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2. SO Z^(N-1)/2=/=X(N-1)/2+Y^(N-1)/2.. SO (1) AND (2)=> Z^(N+1)/2+Z^(N-1)/2=/=X^(N+1)/2+Y^(N+1)/2+X^(N-1)/2+Y^(N-1)/2. SO 2[Z^(N+1)/2+Z^(N-1)/2]=/=2[X^(N+1)/2+Y^(N+1)/2]+2[X^(N-1)/2+Y^(N-1)/2.] SO [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]+[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]+[X^(N+1)/2+X^(N-1)/2-2X^N ]+[Y^(N+1)/2+Y^(N-1)/2+2Y^N ]+[Y^(N+1)/2+Y^(N-1)/2-2Y^N ] SO IF [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2+2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4=> [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4 AND IF [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4 => [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]/4 + [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4 BECAUSE [Z^(N+1)/2+Z^(N-1)/2+2Z^N ] /4- [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=Z^N. SO Z^N=/=X^N+Y^N HAPPY&PEACE. Trantancuong.
