Mbahchiboy: bross biko just finish d no 1 to fulfil all righteousness.....
to d no two how did u get dat answer?EXPLAIN!

Cashio: i wondered really if jackpot is truly a girl/lady cos i have never seen any female with such IQ....thumbs up sis/bro.

Laplacian:
wat is there 2 finish in no.1..biko....
Just dat he made a HARMLESS mistake in d genral solution...
y=(Ax+B)e^-x
...for number 2, d numer of digits in any given number n, say, is given by Logn (approx.)
number of digits in
2004^2004=2004*Log2004 (approx.)

Swagalord18: I have some unsolved math problems ....but if i post them now ... U guys will start saying that i'm testing ur ability or this and that...... No problem sha ....i see i'm not welcome here ....4 reasons not clear 2 me ....i'l kindly unfollow ur thread now .....enjoy

Mbahchiboy: no qualms 4 d no 1 sir.....
4 d no 2 i got wat u said but i wud like to know how e got those digits,i dont mean d number of digits but d digits itself

smurfy: Four married couples bought eight seats in a row for a concert. In how many ways can they be seated if
(a) each couple is to sit together
(b) each couple, except a particular newly-wed couple, is not to sit together
(c) a particular couple decided to head home after having a minor argument?

Swagalord18: I have some unsolved math problems ....but if i post them now ... U guys will start saying that i'm testing ur ability or this and that...... No problem sha ....i see i'm not welcome here ....4 reasons not clear 2 me ....i'l kindly unfollow ur thread now .....enjoy

jackpot: take Log₁₀ of 2004 and multiply by 2004.

1 plus the integral part of (2004*Log₁₀2004) gives you the number of decimal places of 2004^2004.


Elementary reasoning.

Mbahchiboy: TO MY GENERALS:MAKE UNA DO DIS 1 4 ME::
if S=a(r^n-1)/r-1....
make r d subject of d formular

jackpot: to get the digits itself, take antilog to base 10 of the mantissa of (2004*Log₁₀2004).

Mbahchiboy: TO MY GENERALS:MAKE UNA DO DIS
1 4 ME::
if S=a(r^n-1)/r-1....
make r d subject of d formular

Swagalord18: Okay.thanks y'all .was expectin a"good riddance"kind of reaction.
Here'z breakfast .dont let it get cold,,
Show that ^a-b/_{a_|b - b_|a} = ¹/_{_|a} + ¹/_{_|b}
*a_|b means (a root b)

jackpot: @Mbahchiboy, find below what you sought for (i.e., the few digits of 2004²⁰⁰⁴)

should I throw more light?

Laplacian:
wat is there 2 finish in no.1..biko....
Just dat he made a HARMLESS mistake in d genral solution...
y=(Ax+B)e^-x
...for number 2, d numer of digits in any given number n, say, is given by Logn (approx.)
number of digits in
2004^2004=2004*Log2004 (approx.)

jackpot: @Mbahchiboy, find below what you sought for (i.e., the few digits of 2004²⁰⁰⁴)

should I throw more light?

rhydex 247:
Hmmmmmmmmmmm. For the no 1 solution.
recall that in the case of REAL AND EQUAL ROOTS: m=m1 (twice).
hence the general soln is y=e^m1x(A+Bx).
comparing to my solution y=Ae^-x+Bxe^-x or y=e^-x(A+Bx). hence my answer still dey kampe.

For the no 2 answer no be magic na ogbanje spiritual calculator I take solve am. lollll

statusnet:

a) If each couple is to sit together then it will be 4! and each individual couple can sit in 2 ways
Then 4! * 2(4) = 192 ways

b) since three of the couples can sit in any arrangement then they're treated as 6 ppl and the remaining couple can sit in two

Therefore, 6! * 2 = 1440

c) we weren't told a particular arrangement and with the exception of a couple

6!= 720

smurfy:

Good try statusnet. You got (c) right.

Here goes...
(a) Each couple as a unit can sit in 2! ways and each of the four two-unit couples can sit in 4! ways. That gives 48.

(b) 8! - (2! * 7!). Total number of ways of sitting without restriction is 8! (=40320). Total number of ways of sitting such that two newly-weds are always together is 2! * 7! (=10080). Subtracting gives 30240.
(c) Easy. Six people are left and they can sit in 6! = 720 ways.

@statusnet
Now you see that even newly-weds can spoil a good show. So mind the way you treat your sweetheart for I'm sure the cocky husband started that argument.

statusnet:

Please can you explain how you got this : 2! * 7!

rhydex 247: Here is my question.
1). The linear map H:R^3---->R^3 defined by H(x,y,z)=(x+y-2z, x+2y+z, 2x+2y-3z). Is H non singular?. Find the formula for H^-1 and hence evaluate H^-1(1,2,3).

d citizen:
The five star general approach to the problem: cos@+cos2@+ cos3@+

Let C = cos@ +cos2@ +cos3@+

Let S= sin@+sin2@+sin3@+ ( dat is the complex number by of the expression)

C+iS= (cos@ +isin@) (cos2@+isin2@)+ (cos3@ +isin3@)+

Thus, C+iS=e^i@+e^i2@ +ei3@+

C+iS=e^i@/(1 -ei@)(using sum to infinity)

C+is= e^i@ * (1-e^-i@)/(1-e^i@)(1-e^-i@)

C+is= e^i@-1/2-e^i@-e^-i@

C+iS=cos@+isin@-1/2(1-cos@)

Taking the real part of the expression

Cos@+cos2@+cos3@ = cos@-1/2(1-cos@)

jackpot:
Hi Sir d citizen, first of all, I must start by acknowledging your mathematical prowess and ability to think critically.

But I hope you know that the formula for the sum to infinity of a GP is valid if only absolute value of the common ratio is less than 1; i.e., |r|<1.

But it is easy to see that the GP
e^i@, e^2i@, e^3i@, . . .
has common ratio r= e^i@
and its absolute value is
|r|=|e^i@|=|cos@+i sin@|
=sqrt{cos²@+sin²@}=1
and as such, applying the sum to infinity of a GP there is extremely fallacious.

Consequently, I'm afraid that your solution is wrong.
Granted, your solution is already wrong. Now your rationalization(as shown in the bolded) here is also wrong since
^{e^i@}/_(1-e^i@)= -1+¹/_(1-e^i@)
=-1 + ¹/_{{(1-cos@)-i sin@}}
=-1+^{{(1-cos@)+i sin@}}/_{{(1-cos@)²+sin²@}}
=-1+^{{(1-cos@)+i sin@}}/_{2(1-cos@)}
and if you take the real part of this, you should be able to get
-1+ ^(1-cos@)/_2(1-cos@)
=-1+¹/₂
=-¹/₂


My Regards



I remain
your humble servant-ress

JACKPOT

d citizen:
@jackpot,

Look closely at ur rationalisation, u will see an error there

-1+1/1-(cos@+isin@)

It is wrongly equated and rationalised.

Again, in ur elementary sum to infinity,

The r is not a modulus. It is an ordinary r dat is less than 1 not its absolute value

Let me give u an example if u have r= -1 as ur common ratio, then d modulus, is equal to 1, which go contrary to the principle of geometric series to infinity.
. I will like the person dat pose the questions to confirm if my answer is wrong. Although, i am solving d problem with my fone. I am certain dat the answer is right

Mbahchiboy: TO MY GENERALS:MAKE UNA DO DIS 1 4 ME::
if S=a(r^n-1)/r-1....
make r d subject of d formular