Mathematicians

Yes. Click on the 'Left handed basketball players' link on this page to see other basketball players that were/are left handed.

No - smaller.

It is: 76/2 = 38

Josef Friedrich Matthes was born in 1886.

The Greek mathematician Diophantus has traditionally been known as the "father of algebra", but in more recent times there is much debate over whetherMuhammad ibn Musa al-Khwarizmi deserves that title instead.

-Wikipedia

Hooke

Nope! Jojo is an only child.

Mathematicians usually work in comfortable offices

Yes, Bohr died on November 18, 1962

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.

sorry i didnt know how to type the matrix out

basically the 5x5 matrix is...

1 2 5 0 1

2 3 7 1 9

1 2 3 0 3

0 0 1 0 0

3 2 -4 0 1

i understand that the 4th row/column is best to delete.

but what do i do from here?

please show step by step if you can :)

She never had children and i think she never got a boyfriend that's hat i think

because when i had read her biography's they never mention if she had

any children that's what i think I'm not sure

14

sin(z)= (e^(i*z)-e^(-i*z))/(2*i) where i=(-1)^(1/2)

100 days-1 hour

hn'fe njw ncxie

They are similar. In many problems, both methods can be used. You can view Fourier transform is the Laplace transform on the circle, that is |z|=1. When you do Fourier transform, you don't need to worry about the convergence region. However, you need to find the convergence region for each Laplace transform.

The discrete version of Fourier transform is discrete Fourier transform, and the discrete version of Laplace transform is Z-transform.

No offense but I don't think a lot of 18 year olds should be driving. And in answer to your question, no. 16...ya got two more and please be responsible, it is a privilege not a right.

2.0833

The Cauchy-Riemann equations are fundamental in complex analysis, providing conditions for a function to be holomorphic, meaning it is complex differentiable. These equations are essential in various fields, including fluid dynamics, where they describe potential flow, and in electrical engineering for analyzing electromagnetic fields. Additionally, they are used in conformal mapping, which allows for the transformation of complex shapes in a way that preserves angles, facilitating the solution of physical problems in engineering and physics.

she had 22 brothers and sisters together. 15 brothers and 7 sisters

Alan Bakke (white man) had higher qualifications than the minorities (black people), but he was turned down. He took the confusion to court and the supreme court stated that "The University could use race as an admission requirement BUT count NOT use quotas (specific number)."

some Muslim dude

86,400 seconds.

No way to determine, they are present in different forms in many different civilizations and probably had prehistoric origins.

The abacus is simply a highly portable form of the much earlier counting board or counting table, that used pebbles or clay counters in grooves on a wood board or a table covered with a thin layer of sand.