Proofs

Half of its diameter

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

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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

Let p, p+2, and p+4 be three consecutive odd numbers. Then, if p ≡ 0 (mod 3), p is divisible by 3; if p ≡ 1 (mod 3), then p+2 is divisible by 3, and if p ≡ 2 (mod 3), then p+4 is divisible by 3.

That is, at least one of p, p+2, and p+4 must be divisible by 3. The only prime that can be divisible by 3 is 3 itself; for any other positive integer divisible by 3 must have another factor, making it composite. This gives the possibilities

-1, 1, 3

1, 3, 5

3, 5, 7

The first two are eliminated because -1 and 1 are not primes, leaving 3, 5, 7 as the only set of three consecutive integers that are all prime.

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.

None. A zillion is not a real number.

You'd be 1,000,000 minutes old.

However, in larger units some assumptions will have to be made:

As months do not all have the same number of days, when the 1,000,000 minutes started would affect the number of days, so an average number of mins per month will be used; then:

525,949 min ≈ 1 year

→ 1,000,000 mins ≈ 1,000,000 ÷ 525,949 years = 1 year 474,051 mins

525,949 mins ÷ 12 months ≈ 43,829.1 mins/month

→ 474,051 mins ≈ 474,051 ÷ 43,829.1 months = 10 months 35760 mins

1440 mins = 1 day

→ 35760 mins = 35760 ÷ 1440 days = 24 days 1200 mins

60 mins = 1 hour

→ 1200 mins = 1200 ÷ 60 hours = 20 hours

→ 1,000,000 mins ≈ 1 year, 10 months, 24 days, 20 hours.

Or approx 1 year, 10 months, 25 days old.

It is 1.7321 (approx).

A counter example occurs when somebody makes a claim that all members of some category of things have a particular property, and then someone else proves that the claim is not true by showing an example of a thing in the category that does not have the property claimed.

For example, if someone claimed that "All presidents of the United States are dead white men", then Barack Obama would be a counter-example because he is a president of the United States but isn't a dead white man.

For another example, if someone claimed that all mammals bear live young, the echidna and the platypus would be counter-examples because they are mammals that lay eggs.

For another example, if someone claimed that the United States is the only country that has never defaulted on its debts, Australia and Tuvalu would be counter-examples. The claim is logically equivalent to saying that all countries in the category of being not the USA have defaulted on a debt at least once,....

For another example, if someone claimed that all prime numbers are odd, "2" would be a counter example. Or if someone claimed that all odd numbers are prime "9" would be a counter-example.

In short, a counter-example to a proposition or claim is an example that proves that the proposition or claim is not true.

I think you mean e to the (- infinity) power. The proof would be a limit proof.

The limit (as n-->infinity) of [( en) ] = 0

You should have some other limits in class that you have proven.

Show that your limit is less than one of those given for all values of n then you have your proof.

For instance, if you already know that lim (as n-->infinity) of [(1/n) ] = 0

then for n = 1, 1/e1 < 1/1 true

for n = 2, 1/e2 < 1/2 true

Then assume that it is true for n = k

so for n = K, 1/eK < 1/K assumed true

therefore: eK > K

multiply both by e

e(k+1) > ke

but we know ke > k+1 because e>1

SO: e(k+1) > ke > k+1 now take the reciprocal (reverses the inequalities)

1/e(k+1) < 1/ke < 1/ (k+1)

by transitive prop of inequalities eliminate the middle term

so that 1/e(k+1) < 1/ (k+1) this proves the case for n=K+1 and therefore will be true for all values of n since k was never a specified value.

And if: 1/e(k+1) < 1/ (k+1) by one of the properties of limits, since the lim of 1/n is zero, then the lim of 1/en is also zero when n --> infinity.

associative?

single replacement

The midpoint theorem says the following:

In any triangle the segment joining the midpoints of the 2 sides of the triangle will be parallel to the third side and equal to half of it

To determine net words a minute:

Total words (5 characters = 1 word), divided by number of minutes of the timing, minus 2 for each error = nwpm

So, if you typed 75 words in 3 minutes with 2 errors on the timing, you would calculate net words per minute as follows:

75 words, divided by 3 (number of minutes) = 25 gwpm (gross words per minute) minus 4 (2 errors) = 21 nwpm - (net words per minute)

1.96 times

For a general Lp space: In the notation of Lp norms:

Let f and g be Lp functions, then:

f+gp <= fp+gp

Specifically for p=2, using integrals, we have (where "S" means integral):

(S(f+g)2)1/2 <= (S(f)2)1/2+(S(g)2)1/2

and again, replacing p with 2 will yield the definition is a general Lp space.

Formally, the difference between two numbers, x and y, is abs(x - y) which is never negative. If the two numbers are the same then the difference is 0 else it is positive.

If you define the difference as x - y then it is

positive if x > y,

0 if x = y and

negative if x < y.

50

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.