being one myself (year 7 grade, only know what i've been taught, in the advanced class as well){- with a number after it is a negative numbers} game rules: make up an equasion, someone else answers it and then they make up there own. starter: 7+ -5 x 25 - 2 x10 + -7 - -5= ?

= -140

(999!) / (1001!) = ?

Well, no wonder I can't come up with any decent fractal equations....I'm embarrassed to own up to being so far wrong with my answer of 478. Now everyone will understand why I was an Art Major, despite my deficiencies in that area as well. :D

good, your question got me stumped, may have an answer in 3 hours of paperwork or earlier.

Caedes already gave my answer...

caedes is that 999 factorial devided by 1001 factorial?

hmm well i would have to say that it aprox = to: 4.027896473371708673172461363569e+2570


ok my problem is a 20X20 matrix times a 20X1 matrix what is the resulting matrix?

[25 35 24 26 26 52 36 98 54 68 65 12 78 45 87 84 84 89 10 2]
[46 87 84 24 35 21 78 45 12 45 12 87 68 21 78 95 65 12 23 2]
[12 48 68 95 25 61 32 45 12 18 15 65 97 58 79 19 46 15 45 3]
[15 78 15 38 95 64 45 12 74 18 18 78 45 68 98 5 54 68 9 4]
[4 4 6 48 98 35 68 46 85 74 68 94 2 78 49 2 74 68 99 1]
[79 81 35 64 97 57 15 23 63 32 64 79 43 21 7 76 48 22 41 9]
[25 35 24 26 26 52 36 98 54 68 65 12 78 45 87 84 84 89 10 2]
[15 78 15 38 95 64 45 12 74 18 18 78 45 68 98 5 54 68 9 4]
[46 87 84 24 35 21 78 45 12 45 12 87 68 21 78 95 65 12 23 2]
[12 48 68 95 25 61 32 45 12 18 15 65 97 58 79 19 46 15 45 3]
[25 35 24 26 26 52 36 98 54 68 65 12 78 45 87 84 84 89 10 2]
[25 35 24 26 26 52 36 98 54 69 65 12 78 45 87 84 84 89 10 2]
[47 44 6 48 98 85 68 46 85 74 68 94 2 78 49 2 74 68 99 1]
[79 81 35 64 97 57 15 23 63 32 64 79 43 21 7 76 48 22 41 9]
[15 78 15 38 95 64 45 12 74 18 18 78 45 68 98 5 54 68 9 4]
[99 4 5 48 98 35 68 46 85 74 68 94 2 78 49 2 74 68 99 1]
[46 87 84 24 35 21 78 45 12 45 12 87 68 21 78 95 65 12 23 2]
[54 98 25 64 24 52 99 74 34 65 12 10 20 35 68 48 69 41 47 8]
[19 87 65 48 95 25 64 87 95 21 36 18 74 68 95 41 46 5 48 7]
[20 22 65 33 4 15 12 78 99 15 68 15 23 6 4 84 5 5 54 1]

times:

[54 89 123 486 985 12 5 3221 4 78 95 63 48 74 15 13 456 87 987 1000]

nope sherpa. It's a much more realistic number than that.

My problem is still on the board =p

I thought that seemed much too high of a number. The answer is 1/(1001*1000) = 1/1001000 = 9.99000x10^-7

The matrix is far too much work for me to really want to do, so I'll leave that to someone else. :-D

Mary: There's an 'order of operations' that must be observed. First is 'parenthesis', then 'exponetials', 'multiplication', 'division', 'addition', and finally 'subtraction'. You may have heard the mnemonic "Please Excuse My Dear Aunt Sally" which helps you remember. :)

is it 0. something?

0.998001998? o well i tried

too much work to mess with that matrix :(

Sherpa: The way you have the second matric written is not quite right, but I'll assume that you meant it to be a single column matrix. image of the answer

My problem: arcsin( (e^(i)-e^(-i))/2 ) = ?

oops yep i ment to have it verticle.....

asin((e^(i)-e^(-i))/2)
0.76472515401120709635539754902472i
EDIT:

I Redid it and got exactly the same thing.... are you wanting it in degrees or radians?
asin((e^(i)-e^(-i))/2 )
0.76472515401120709635539754902472i

and ya, the answer that i gave for your first problem is not right it should be:

9.99000999000999000999000999001e-7

arcsin( (e^(i)-e^(-i))/2 ) = ? isn't correct yet. =)

In the edit above i re did the eq in radians this time i did it in degrees....

asin((e^(i)-e^(-i))/2 )
43.815523812334042168947610401512i

... ... ...

Nope, It shouldn't matter if you use radians or degrees as long as you are consistant.

Hm... let's see... since (e^x - e^(-x))/2 = sinh(x), then your equation is equivalent to

arcsin(sinh(i)),

which my eighty-nine tells me is approximately 0.7647i. Here's something else I picked up that you guys can ponder over:

Say x = y

Multiplying both sides by x, we obtain x^2 = xy

Subtracting y^2 from both sides, x^2 - y^2 = xy - y^2

By the difference of squares, (x + y)*(x - y) = y*(x - y)

Simplifying, x + y = y

Since x = y, we have y + y = y

Or, 2y = 1y

Canceling, we have 2 = 1.

So, I've proved that 2=1, or have I?... If not, what's wrong with the proof?

And I don't care for matrices, so I'll avoid that one. =)

Oh darnit, I screwed up copying the formula over. It should of been asin((e^(i)-e^(-i))/(2i) )

which is equal to arcsin(sin(1)) = 1

I guess the earlier answer was correct.

LOL I thought that seemed like a funny problem. =)

oh well>>>>>>> now the truth comes out......

So, what about my question? "Two is the same as one"? That's almost too tempting, isn't it? Sort of Big Brother-ish? :-D

It's tricky trying to hide the fact that you've devided by zero in step #4, huh? In step #3, (x - y)=0, so to simplify you'd have to divide by that on both sides. That's a no-no. =p

Nice job, caedes. Very good eye. Here's one that stumped me when I first saw it. If you know how to handle it, it's easy enough:

d/dx(x^x) = ?

By the way, that's for all x > 0.

Or, an easier one:

1 + 3 + 5 + 7 + 9 + ... + 95 + 97 + 99 = ?

By adding the first and last of all the numbers together.. you get 100, times half the total number of numbers, which is.... uh, 25, maybe? So, that would be 2500.
That's prolly wrong, it's too late to be thinking about math :(

P.S. : It's not that hard working with matrices... just plug 'em into Excel and let it do all the work. (:

P.P.S. : Isn't it great how division by zero can tell you anything you want it to? My calc 2 professor used it to prove that 0/0 = 22.4.....

d/dx(x^x) = x^x * (1+ln(x)) : start by taking nat. log of both sides, after that it's easy.

warlow: Isn't that called a telescopic series?

Hm... telescopic series? Don't know; I've never heard it called that. It's simply the sum of the first n odd numbers. Check this out:

1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
...

Anyway, the pattern is that the sum of the first n odd numbers is n^2. The formula for the nth odd number is (2n - 1), so that would mean that 99 is the fiftieth odd number, and 50^2 is 2500. Kudos to bob_smith for figuring out the other (quite clever) way and to caedes for nailing the derivative. BTW, I learned to do it by rearranging to e^(x*ln(x)); tomayto, tomahto.... Caedes, you're a physics guy, aren't you? Wow. I enjoy math, but I just took the standard undergrad E&M class and it drove me up the wall (although it was mostly the bad prof I had).

Okay, here's a kind of interesting problem that "might someday prove useful to you in a somewhat bizarre set of circumstances" (-Tom Lehrer).

In a game of five-card-draw Poker, how many different possible hands are there (from the 52 cards)?

Easy enough if you've had the right mathematics. Otherwise, very challenging.

I've had statistics already, so I'll leave that one for someone else. =)

Cool. Wonder if anyone else will give it a shot? If not, it's yours. :-)

yes caedes that is a telescopic series.

seems I'm in the same boat as you mrwarlow, E&M is some bad stuff.

the answer if I'm not too tired to do math correctly is 52*51*50*49*48 or 3.1188E8...I'm too tired to think of a new one right now...

So, what's E&M? I'm trying to relate the letters to some kind of math or statistics... but it just doesn't work. I'm trying to think of some cool stuff from Mathcounts in ages past, but just can't somehow.... oh well.
Wait: here's one: (integral sign) xsinx = ?
What are you guys doing posting at 1:44 AM??
Hm... any other questions... (:

bob_smith: E&M is electricity and magnetism.
integral(x*sin(x)) = -x*cos(x) + sin(x) + C
(Integration by parts... yeah, I worked it out by hand... wink wink)
Also, I post at 1:30 in the morning because I'm stupid and won't go to sleep. :-)

rustectrum03: Very close -- you're almost there. The only problem is that you counted, for instance, the two, three, four, five, and six of spades as a "different" hand than the six, five, four, three, and two of spades. :-) How would you account for that? Telescopic Series... I'll remember that....

Hm... shoot, maybe the problem I was really trying to ask was: d/dx xsinx... no, that's easy. I thought there was no algebraic solution ever found for that one, but I may be thinking of something else.
E&M! Of course! And I say I'm studying electrical engineering.... AND physics! Oh well :(

LOL No prob Bob (sorry). Perhaps you're thinking of integral(e^(-x^2))?...

Prove that (3 + 2*sqrt(2))^(2n-1) + (3 - 2*sqrt(2))^(2n-1) - 2 is a
perfect integral square for every positive integer n.

sqrt - means square root

On behalf of the members of the American Headache Society, I thank you all for your continuing support of pain.

I may have to go with Tracy on this one, Sherpa. =) You like the headache problems, don't you? Well, let's see, I'll give it a shot. Don't want to say I didn't try...

[Later...] Well, I tried and couldn't do it. That's quite a difficult math problem. Matter of fact, I could probably sit here three weeks and not get it... :-P ...Just messing with you. Thanks for giving me something to mull over (and get stumped with).

haha, google is such a great tool, lol!!!!!!!!!

I agree. =)

By the way, the answer to the Poker question is that there are "52-choose-5" number of different hands possible. The formula for this is 52!/(47!*5!) which is the same as (52*51*50*49*48)/120 or: 2598960.

The answer is....


Fourty-two.

For the integral series would you like a proof or logic table? I can do both if it so pleases you. There are pros and cons to each like with the heavy amount of parentheses. It'll probably last all nigh expanding these notations though. :-)