Proofs

In a right triangle, the altitude with the hypotenuse as base divides the hypotenuse into two sections p and q. If we denote the length of the altitude by h, we have the relation h2 = p*q (Euklids altitutude theorem).

So, first draw the square root of 2 as the diagonal of a square with side length 1, then construct a right triangle with p=1 and q= sqrt(2) by using the Thales theorem and its altitude will be fourth root of 2 according to Euclids theorem.

ewan

bkt alm mou

BY: H!KAR! AND ke!

A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel

Let ABCD be a quadrilateral in which ABCD and AB=CD, where means parallel to.

Construct line AC and create triangles ABC and ADC.

Now, in triangles ABC and ADC,

AB=CD (given)

AC = AC (common side)

Angle BAC=Angle ACD

(corresponding parts of corresponding triangles or CPCTC)

Triangle ABC is congruent to triangle CDA by Side Angle Side

Angle BCA =Angle DAC by CPCTC

And since these are alternate angles, ADBC.

Thus in the quadrilateral ABCD, ABCD and ADBC.

We conclude ABCD is a parallelogram. var content_characters_counter = '1032';

Int sqrt(1+x2)/x = sqrt(1+x2) + LN [(sqrt(1+x2) - x -1) / (sqrt(1+x2) - x +1)]

A square. All squares are parallelograms, but not all parallelograms are squares.

The acceleration of an object that falls from a certain height does not depend on its mass, in an ideal condition with no air resistance.

The value of acceleration is the acceleration due to gravity, which is 9.81 m s-2.

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However, in this case, air resistance is going to matter. 12000 feet is high enough for the person to accelerate to what we call terminal velocity. Terminal velocity is the velocity where the force of acceleration due to gravity (9.81 m s-2) is matched by the air resistance. That velocity varies, depending on the outline shape of the person, and is typically around 200 km/h or 125 mph. That will be the velocity of the fall.

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100,200

composition

No

It means you have a piece of silverplate that has 12 pennyweight of silver plated onto a base metal.

2*(L*B + B*H + H*L) cubic units where

L = length

B = Breadth

H = Height.

There is nothing true about the AAA theorem and the SSS postulate because the AAA postulate is not true!

Prove that if it were true then there must be a contradiction.

360 degrees from pole to pole and 180 degrees in circles parallel to the equator.

3, 8, 13, 18, 23, 28, 33, and 38 are all solutions.

Wrong answer above. A limit is not the same thing as a limit point. A limit of a sequence is a limit point but not vice versa. Every bounded sequence does have at least one limit point. This is one of the versions of the Bolzano-Weierstrass theorem for sequences. The sequence {(-1)^n} actually has two limit points, -1 and 1, but no limit.

mikel jeksan

No.

Let M be the subset of 2x2 matrices A such that det(A)=0 and tr(A'A)=4 (I shall use ' to denote transpose).

Recall that one of the three definitions of a k-dimensional manifold is:

M c R^n is a k-dimensional manifold if for any p in M there is a neighborhood W c R^n of p and a smooth function F:W-->R^n-k so that F^-1(0) = M intersection W and rank(DF(x))=n-k for every x in M intersection W.

In short, what we need to do is find a function F so that the inverse image of the zero vector under F gives M and the rank of the derivative of F is equal to the dimension of the codomain of F.

Let an arbitrary 2x2 matrix be written as:

[a c]

[b d]

Then the two constraints that define M are

1) ad-bc=0

2) a^2+b^2+c^2+d^2=norm(a,b,c,d)^2=4

Define F:R^4-->R^2 by F(a, b, c, d)=(ad-bc, norm(a,b,c,d)^2-4). Then clearly F^-1(0,0)=M. Furthermore, F is smooth on M because the multiplication and addition of smooth functions (a, b, c, d) is also smooth. (Note that we have taken the neighborhood W to be some superset of M. This guaranteed to exist because if we interpret the set of 2x2 matrices as R^4, every point in M has norm 2, so any ball centered at the origin with length greater than 2 will contain M).

All that remains to be done is to check that rank(DF(x))=2 for every x in M. Observe that [DF]= [d -c -b a]

[2a 2b 2c 2d]

We shall now argue by contradiction. Suppose rank(DF) did not equal 2 for every x in M. Then we know that the two rows are linearly dependent i.e.

h[d -c -b a] + k[2a 2b 2c 2d] = 0 and h and k are not both 0.

Suppose h is 0. Then we have 2ka = 2kb = 2kc = 2kd = 0, and since k cannot also be 0, this implies that a=b=c=d=0, therefore norm(a,b,c,d)^2=0. But, (a,b,c,d) must be in M, so this is a contradiction. Hence h cannot be 0.

Now suppose h is nonzero. Then we can divide it out and there exists a, b, c, d and a constant k so that

[d -c -b a] + k[2a 2b 2c 2d] = 0 i.e. we have:

d + 2ka = -c + 2kb = -b + 2kc = a + 2kd = 0. From this we can substitute to obtain:

a(1 - 4k^2) = b(4k^2 - 1) = c(4k^2 - 1) = d(1 - 4k^2) = 0 and hence

a^2(1 - 4k^2)^2 = b^2(4k^2 - 1)^2 = c^2(4k^2 - 1)^2 = d^2(1 - 4k^2)^2 = 0.

Note that (1 - 4k^2)^2 = (4k^2 - 1)^2. Now, adding the four above expressions together we get:

(a^2 + b^2 + c^2 + d^2)(1 - 4k^2)^2 = norm(a,b,c,d)^2(1 - 4k^2)^2 = 0. But, since we require (a,b,c,d) to be in M, this reduces to 4(1 - 4k^2)^2 = 0. This implies that

1 - 4k^2 = 0, and hence k = +/-(1/2).

Now, if k=+1/2, then we have a + d = b - c = 0, therefore d = -a and b = c. Since we have det(A) = 0 as one of our constraints on M, this implies that ad - bc =

-(a^2) - (b^2) = -(a^2 + b^2) = 0, which implies a = b = 0, by the property of norms. But, if a = b = 0, then (a,b,c,d) = (0,0,0,0) and hence norm(a,b,c,d)^2 = 0, which is a contradiction.

We can argue analogously for the case where k = -1/2. Hence, assuming that rank(DF) is not 2 for some (a,b,c,d) in M leads to a contradiction, so we conclude that rank(DF)=2 for all x in M.

Finally, from this result, we conclude that since F:R^4-->R^2=R^(4-2), M must be a 2-dimensional manifold.

Zillion is a made-up word. There is no such number.

What? Pyrrho died more than 18 Centuries before Descartes!

No

It very much depends on f.

If f is one-to-one and onto (injective and surjective) then yes, else no.

One-to-one means that for each element in the domain there is a different image in the range. This is not true for g(x) = x2 for example, where -3 and +3 are both mapped to +9. So g(x) does not have an inverse UNLESS you restrict the domain of g to non-negative reals. Then -3 is no longer in the domain.

Onto means that every element in the range of the function has a corresponding element in the domain which is mapped onto it. Again, a suitable changes to the domain and range can transform a function without an inverse into an invertible one.