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## #1 Re: Ganesh's Puzzles » Brain Teasers » 2007-04-27 04:56:46

good job Landof, I don't know what I was thinkin, if she started with 3 and did all the stuff the problem says she did, she would still have half an orange left.

## #2 Re: Ganesh's Puzzles » *** Problems » 2007-04-24 05:09:28

ganesh wrote:

***32. A and B are two independent events. The probability that both A and B occur is 1/6 and that of neither of them occuring is 1/3. Find the probability of occurence of A.

Sorry, not enough information to find a unique solution.

I found that P(A) is not equal to P(B).  And, since A and B are interchanged "without loss of generality," a.k.a. WOLOG,
then it seems there are two possibilities for P(A).

A#5   q=2

a7   1 and -1

## #5 Re: Ganesh's Puzzles » Trigonometry » 2007-04-23 07:21:27

The foot of the pole and the midpoint of the pole are in the same horizontal level. ??

## #6 Re: Ganesh's Puzzles » Brain Teasers » 2007-04-23 06:55:34

she sold 3 to the first customer, 1.5 to the second customer and .5 to the 3rd customer for a total of 5.

## #7 Re: Ganesh's Puzzles » Brain Teasers » 2007-04-23 06:41:02

Maybe I just missed it, but I never did see an answer to #5.
I don't really see anything that would exclude the possibility that the two numbers are the same number, so... 0 and 0 would be the smallest.

If you insist that these numbers are not only arbitrary whole numbers, but also disctinct, then I guess 1 and 0 would be the smallest.

Am I understanding the question here??

## #8 Introductions » I know a little » 2007-04-23 06:18:40

Eeyore
Replies: 2

Hi,

I'm pleased to find this forum, I teach math at a local university, and am currently involved in a program for reform in math education.  The motto of this program is "teach less, learn more".

Some of my responses may be slightly non-traditional, but accurate nonetheless.  I believe that math is a set of concepts that is a playground for everyone to explore.  Rather than a salt mine with an infinitely large set of algorithms to memorize.

## #9 Re: Help Me ! » Need Abstract Math Help! » 2007-04-23 06:07:06

I'm sorry but this is not what we are trying to prove.
If I can prove the Lemma then the theorem that we are trying to prove can be easily proven.

That is what a Lemma is.

Of course we MUST prove the Lemma  before we have a full proof of the theorem.

What I was trying to do was use a pun to point out the meaning of the word Lemma.  But thanks for your help.

## #10 Re: Help Me ! » Need Abstract Math Help! » 2007-04-23 05:06:49

If you would just Lemma use the theorem that "in an injection, the cardinality of the domain must be less than or equal to the cardinality of the range, then this proof would be easy. ## #11 Re: Exercises » Differential Calculus Ex.1 » 2007-04-18 05:04:07

I think this depends on your definition of "at rest", I 've always thought of a particle being "at rest" when its velocity is equal to zero.  If you define "at rest" to be when the acceleration is zero, then this problem is trivial you don't need any formulas at all.
So, let's just make believe that "at rest" means when the velocity is zero.  Now now set ds/dt=o and find the time at which the paricle is "at rest", then find the acceleration at that moment.
ds/dt= 45 + 22t - 3t^2, and that equal zero when t=(11±16)/3= 9 or -5/3  , so i'm goin with the 9 cuz i'm not even sure our domain includes negative numbers, but more importantly I don't like fractions, especially fractions with denominators with factors other than 2 and 5... but that's a different story, lemma finish.  The acceleration in general is the derivative of the velocity, so a=ds²/dt²=22 - 6t, so if t = 9 then the acceleration is 22-54 = -32 and if you have a teacher that wants both of the solutions (both times when the velocity is zero), then you must include the acceleration when t = -5/3 is 22 - (-5/3)6=22 + 10 = 32 so the acceleration is the same, but in different directions.
Just my opinion, I might be right.
Oh dear, oh my, oh no....