Proofs

Assuming the teams had the same number of people on each, each team would have 27 people.

(24+30) = 54/2 = 27

They bisect each other at an angle of 90 degrees

They do in some parallelograms, not in others.

NO! Then it wouldn't be a square! It would be another shape, definetely not a square!

The set of all points in a plane that are equidistant from the two sides of a given angle

In Formal Logic proofs, the contradiction is represented with an inverted T (or upside-down T) as follows: ┴

The contradiction symbol can be introduced at any time a logical contradiction is encounterd, for example, all of the the following contradictory logical statements (using different symbols) can be replaced with the contradiction symbol:

The ball is completely blue and the ball not completely blue.

P ^ ¬P

P & ~P

P & !P

P AND NOT P

is 09354755654454

Surface area = 2*(L*B + B*H + H*L) cubic unitswhere

L = length

B = Breadth

H = Height.

None. Perfect squares, by definition, are the squares of counting numbers and these cannot be fractions.

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.

36 divided by two then add 6

A rhombus, square, kite or arrowhead.

Yes, as shown by this example:

S={*}

G=(S,*)

*:S X S-->S

*(*,*)=*

However, I could not find any nontrivial examples.

direct observation

SAS

67

With sides of length A, B and C units, the total surface area is

2*(AB + BC + CA) square units.

We know -1=-1, make these two into different but equivalent fractions -1/1=1/-1, take the square root of both sides sqrt(-1/1)=sqrt(1/-1), the rule sqrt(x/y)=sqrt(x)/sqrt(y) is true so we can write... sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1), simplify the fractions i/1=1/i, multiply both sides by i -1/1=i/i, i/i=1 and -1/1 is -1 so we can write... -1=1 There's your proof i=sqrt(-1) spoiler:i is imaginary

There cannot be such a postulate because it is not true.

Consider a line segment AB and let C be any point on the line between A and B. If the three points are A, B and C, there can be no circle that goes through them. It is so easy to show that the postulate is false that no mathematician would want his (they were mostly male) name associated with such nonsense.

Well, if one of the points approach the line that goes through the other two points, the radius of the circle diverges. The line is the limit of the ever-growing circles. In the ordinary plane, the limit itself does not exist as a circle, but mathematicians have supplemented the plane with infinity to "hold" the centres of such "straight" circles.

Spherical balls have no corners. The American football has 2 corners.

A ball (of any shape) may be sewn together of polygons each of which has several corners (vertexes).

Tiiangle

decimeter

Yes. Because the diagonals are perpendicular to each other and intersect at their midpoints, they bisect each other.

There is no AAA theorem since it is not true.

SSS is, in fact a theorem, not a postulate. It states that if the three sides of one triangle are equal in magnitude to the corresponding three sides of another triangle, then the two triangles are congruent.