Proofs

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.

D = diagonal, S = side, P = perimeter

S = D/sqrt(2)

P = 4S = 4D/sqrt(2)

So, for this problem:

P = 48/sqrt(2)

To the nearest tenth, this is 33.9 inches

Half of its diameter

The only number to not have a reciprocal is zero. This is because it can't be divided by anything.

A number is divisible by 6 if it is divisible by 2 and 3.

Look at 333-3 which is 330 The sum of the digits is 6 and it is even so it is divisible by 6 Now consider 222-2 which I picked because unlike 333, 222 has even digits. 222-2=220, one again even number so divisible by 2 but NOT divisible by 3 so NOT divisible by 6

So it look like this is not true for all n

For any odd n, we have the following

1. nnn-n ends in 0 so it is even if we can show it is divisible by 3 we are done. but 777-7 is 770 which is NOT divisible by 3 so it is NOT true.

For some n it is true, but not for all n... Now when will nnn-n be divisible by 3.

only when n+n is a multiple of 3, ie n=33,66, 99 an that is it!

So we could easily prove that nnn-n is divisible by 6 if and only if n=3,6,or 9

-----------------------------

If by nnn, you mean n3, a proof is as follows:

n=0,1,2,3,4, or 5 (mod 6)

If n=0 (mod 6), we have (0 (mod 6))((0(mod 6))2-1)=0 (mod 6). [Since the first term is zero]

If n=1 (mod 6), we have (1 (mod 6))((1(mod 6))2-1)=0 (mod 6) [Since 1-1=0].

If n=2 (mod 6), we have (2 (mod 6))((2(mod 6))2-1)=(2*3) (mod 6) = 6 (mod 6)=0 (mod 6).

If n=3 (mod 6), we have (3 (mod 6))((3(mod 6))2-1)=(3*8) (mod 6) = 24 (mod 6) = 0 (mod 6).

If n=4 (mod 6), we have (4 (mod 6))((4(mod 6))2-1)=(4*15) (mod 6) = 60 (mod 6) = 0 (mod 6).

If n=5 (mod 6), we have (5 (mod 6))((5(mod 6))2-1)=(5*24) (mod 6) = 120 (mod 6) = 0 (mod 6).

If you're not comfortable with the modular arethmetic, you can substitue 6m+_, where the blank is each of the numbers 0 through 5 (since every number can be expressed either as a multiple of six, or as a multiple of six plus some number between 1 and 5 --the remainder when the number is divided by six). Taking our example with 5, you would get:

(n)(n2-1) can be written as (6m+5)((6m+5)2-1), where m is an integer.

Simplifying this, you get:

(6m+5)((6m+5)2-1)

(6m+5)((6m2+60m+25-1)

6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*25+5(-1)

6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*24

Since m is an integer and each term is divisible by 6, (n)(n2-1) is divisible by 6 for integers that can be expressed as 6m+5. You would then repeat the process for each of 0 through 4 to complete the proof. Clearly, if you are comfortable with it, modular arithmetic is the less cumbersome way to proceed.

Prove that (axb)n[(bxc)x(cxa)] = [a]n(bxc)]^2 where a,b,and c are all vectors.

First, multiply out the cross products. Since the cross product of two vectors is itself a vector, we'll give the cross products some names to make this a little easier to understand:

(bxc)=(b2c3-b3c2)i-(b1c3-b3c1)j+(b1c2-b2c1)k = vector d

(cxa)=(c2a3-c3a2)i-(c1a3-c3a1)j+(c1a2-c2a1)k = vector v

(axb)=(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k = vector u

=> (axb)n[(bxc)x(cxa)] = un[dxv]

(dxv)=(d2v3-d3v2)i-(d1v3-d3v1)j+(d1v2-d2v1)k = vector w

=> un[dxv] = unw = u1w1 + u2w2 + u3w3

Now replace u and w with their vector coordinates (notice that the negative sign is factored into the middle terms, so the variables are switched).

u1w1 + u2w2 + u3w3= (a2b3-a3b2)w1 + (a3b1-a1b3)w2 + (a1b2-a2b1)w3

= (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1)

Now we need to expand the v terms back out:

(d2v3-d3v2) = d2(c1a2-c2a1) - d3(c3a1-c1a3) = d2c1a2- d2 c2a1- d3c3a1 + d3c1a3

(d3v1-d1v3) = d3(c2a3-c3a2) - d1(c1a2-c2a1) = d3c2a3 - d3c3a2- d1c1a2+ d1c2a1

(d1v2-d2v1) = d1(c3a1-c1a3) - d2(c2a3-c3a2) = d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2

So: (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) = (a2b3-a3b2)(d2c1a2- d2 c2a1- d3c3a1 + d3c1a3) + (a3b1 - a1b3)(d3c2a3 - d3c3a2- d1c1a2+ d1c2a1)+ (a1b2-a2b1)(d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2)

= d2c1a2 a2b3- d2 c2a1 a2b3- d3c3a1 a2b3 + d3c1a3 a2b3- d2c1a2 a3b2+ d2 c2a1 a3b2+ d3c3a1 a3b2- d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1- d1c1a2 a3b1+ d1c2a1 a3b1- d3c2a3 a1b3+ d3c3a2 a1b3+ d1c1a2 a1b3- d1c2a1 a1b3+ d1c3a1 a1b2 - d1c1a3 a1b2 - d2c2a3 a1b2 + d2c3a2 a1b2- d1c3a1 a2b1+ d1c1a3 a2b1+ d2c2a3 a2b1- d2c3a2 a2b1

Some of the terms cancel out, leaving us with;

= d2c1a2 a2b3 - d2 c2a1 a2b3 + d3c1a3 a2b3 - d2c1a2 a3b2 + d3c3a1 a3b2 - d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1 + d1c2a1 a3b1 - d3c2a3 a1b3 + d1c1a2 a1b3 - d1c2a1 a1b3 + d1c3a1 a1b2 - d1c1a3 a1b2 + d2c3a2 a1b2 - d1c3a1 a2b1 + d2c2a3 a2b1 - d2c3a2 a2b1

Now factor out d1 , d2 , and d3

= d1(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + d2(c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + d3(c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)

Now we can factor out a dot product of ( d1 + d2 + d3):

= ( d1 + d2 + d3)n[(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + (c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + (c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)]

(Remember, to keep from changing the value of the equation we still need to keep the terms grouped together so that they multiply by the correct d components.)

Now factor out all the "a" components within the brackets:

= ( d1 + d2 + d3)n[(a1 a3{c2b1 - c1b2} + a1 a2{c1b3 - c3b1} + a1 a1{c3b2 - c2b3}) + (a1 a2{c3b2 - c2b3} + a2 a2{c1b3 - c3b1} + a2 a3{c2b1 - c1b2}) + (a1 a3{c3b2 - c2b3} + a2 a3{c1b3 - c3b1} + a3 a3{c2b1 - c1b2})]

= dn[( a1 a3+ a1 a2+ a1 a1)n({c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3}) + (a1 a2 + a2 a2 + a2 a3)n({c1b3- c3b1} + {c2b1- c1b2} + {c3b2- c2b3}) + ( a1 a3+ a2 a3 + a3 a3)n({c3b2- c2b3} + {c1b3- c3b1} + {c2b1- c1b2})]

And we know that {c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3} = (bxc), so we factor out (bxc):

= dn[(bxc)n[(a1 a3+ a1 a2+ a1 a1) + (a1 a2 + a2 a2 + a2 a3) + ( a1 a3+ a2 a3 + a3 a3)]

= dn[(bxc)n[a1(a3+ a2 + a1) + a2 (a1 + a2 +a3) + a3(a1+ a2 + a3)]]

= dn[(bxc)n([a1 + a2 +a3]n[a1 + a2 +a3]) = dn[(bxc)n(a n a)]

(from above, remember that d = (bxc) )

= (bxc)n(bxc)n a n a

= [an (bxc)]^2

Yes

It is 169.

cannot be determined

A transversal is simply a line that passes through two or more other lines . (the lines should be in the same plane) Transversals are important becasue if a transverals passes through or intersects two parallel lines we get some special angle congruences. The alternate interior angles are on opposite sides of the transversal line and inside the lines being transversed. Alternate exterior angles are on opposite sides of the transferal line and outside the lines being transversed. The alternate interior and alternate exterior angles are congruent if the lines being crossed by the transversal are parallel ( NOT to each other, the two alternate interiors are congruent and the two alternate exteriors are congruent.