₦airaland Forum

Humphrey77: SOLVE FOR h in the equation : 6(raise to power h) -31(2)raise to power h+32 is equal to zero ( HINT : USE LINEAR APPROX)
:* HAPPY!

f(h) = 6^h - 31(2^h) + 32
f^'(h) = 6^h * In 6 - 31(2^h)(In 2)
Using Newton's method of approximation which states that if h_n is an approximated solution to the function f(h), then the following equation is a better approximated solution
h_n+1 = h_n - f(h_n)/f^'(h_n)
Starting with h₀ = 0.1,
h₁ = 0.1 - (-0.02875) / -20.91 = 0.09863
h₂ = 0.09863 - (-0.000143) / -20.87 = 0.09862
h₂ = 0.09862 - 0.0000656/-20.87 = 0.09862
h = 0.09862
Actually h = 0.098623143
Well if I'm to score myself, I'll give myself 10% just for my effort to attempt it because I didn't use the method I was asked to use.
Lwkmd

Show last time edited. When a post gets edited/modified, the time the change was made should be shown.

Just like this musician said in this video,
https://www.youtube.com/watch?v=B9QqDQ0BGX8
They've tried but I'll give them 20% mainly because of their slow pace.

Humphrey77: Consider the digit from 1to 9 , select any three digit form (1,2,3,4,5,6,7,8,9) with out repetition the sum of the digit's will give 15 in the column , diagonal and row
HAPPY!

This is known as the magic square. I came across this in MATLAB. A special type of matrix known as magic square matrix http://en.m.wikipedia.org/wiki/Magic_square
I don't know how it is solved sha but this is the square for a 3 by 3
2 7 6
9 5 1
4 3 8

Humphrey77: @ akpos ; your sequence did not satisfy laplacian question! that is three non- zero integers are in AP, the product of any two when increase by one is a perfect square.
:*HAPPY

Let me use equations to make it clearer to you. Hope you know that I have two sets? The first one agrees with the question while the second one is the square roots of the perfect squares from the question.

According to the question, if (a, b, c) agree with the condition,
ab + 1 = X²
ac + 1 = Y²
bc + 1 = Z²
(a, b, c) & (X, Y, Z) are the two sets that I posted. Let M_n = (a_n, b_n, c_n) & P_n = (X_n, Y_n, Z_n)

We were able to get M₁ = (1, 8, 15) but I noticed that the subsequent ones that were gotten, had a relationship with M₁ as follows
M₁ = (a₁, b₁, c₁)
M₂ = (a₂, b₂, c₂)
M₃ = (a₃, b₃, c₃
Where a₂ = 0.5b₁, b₂ = 2c₁, c₂ = 2b₂ - a₂
Likewise a₃ = 0.5b₂, b₃ = 2c₂, c₃ = 2b₃ - a₃
Generally
M₁ = (a₁, b₁, c₁) = (1, 8, 15)
M_(n+1) = (0.5b_n, 2c_n, 2b_(n+1) - a_(n+1))
As for P_n
P_(n+1) = (Z_n, Z_n + Y_n, 2Y_(n+1) + X_(n+1)
P₁ = (X₁, Y₁, Z₁) = (3, 4, 11)


M₁ = 1, 8, 15
M₂ = 4, 30, 56
M₃ = 15, 112, 209
M₄ = 56, 418, 780
M₅ = 209, 1560, 2911
M₆ = 780, 5822, 10864
M₇ = 2911, 21728, 40545
M₈ = 10864, 81090, 151316

P₁ = 3, 4, 11
P₂ = 11, 15, 41
P₃ = 41, 56, 153
P₄ = 153, 209, 571
P₅ = 571, 780, 2131
P₆ = 2131, 2911, 7953
P₇ = 7953, 10864, 29681
P₈ = 29681, 40545, 110771

Laplacian: i luv ur observation even though it has flaws...nice wrk!
1.)first, ur second sequence should b; y, 2z, 4z-y....so all subsequent sequences are wrong.
2.) secondly, if u make d above corrections, and fit in real numbers, d first four sequences 'll b correct but all subsequent ones do not satisfy d condition of d question... Tanks

I noticed the mistake while I was fixing in the values.
As for the subsequent ones not satisfying the condition of the question, I guess you made a mistake along the line. With the help of a spread-sheet, I have the square roots of the perfect squares formed from the initial condition as shown below
3, 4, 11
11, 15, 41
41, 56, 153
153, 209, 571
571, 780, 2131
2131, 2911, 7953
7953, 10864, 29681
29681, 40545, 110771
There is still a trend here also.
The 3rd term is the same as the 1st term for the next triple, the 2nd term of this new triple is the same as the sum of the 2nd and 3rd term of the previous triple while the 3rd one is the same as multiplying the 2nd term by 2 then add the 1st term to it.
Chai! e no easy to use words explain wetin mathematical equation go describe for one line easily. I for just use variables make the picture de clearer.

Humphrey77: take a look at the third sequence it is not an AP OK.
it ( 15,112,209) ,
:*@ HAPPY/

Let me make it clearer;
From what I posted earlier,
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,
* I just found out that I made a mistake*
That changes everything. Substituting
X = 1, Y = 4, Z = 15
should give us the right values if not for my mistake

Here are the right values
1, 8, 15
4, 30, 56
15, 112, 209
56, 418, 780
209, 1560, 2911
780, 5822, 10864
2911, 21728, 40545
10864, 81090, 151316
It goes on and on

Like I posted earlier, half the 2nd term will give you the 1st term on the next triple while twice the 3rd term will give you the 2nd term on the next triple. Since you've gotten the 1st & 2nd term for another triple, the third term of the new triple can be gotten by multiplying 2 by the 2nd term then subtract the 1st term from it.

jackpot: that one na baby formula na. A "very obvious" one

Lemme share another Pythagorean formula which I have observed for you:

Proposition
Let k be any number at all. Then, the numbers:
k²-1, 2k and k²+1
are Pythagorean triples.

Proof
It follows from the application of the pythagorean theorem. QED.

Application
For k=2. . . . . . .3, 4, 5
k=3. . . . . . . 8, 6, 10
k=4. . . . . . . 15, 8, 17
k=5. . . . . . . 24, 10, 26
k=6. . . . . . . 35, 12, 37
k=7. . . . . . . 48, 14, 50.


Here is another one which I deduced from the above:

Corollary
Let k be an odd number. Then the numbers:
¹/₂(k²-1), k, ¹/₂(k²+1)
are Pythagorean triples.

Application
k=3. . . . . . . .4, 3, 5
k=5. . . . . . . .12, 5, 13
k=7. . . . . . . .24, 7, 25
k=9. . . . . . . .40, 9, 41

Propositions (jackpot 2013)
The following are pythagorean triples:
k^2-1, 2k, k^2+1
k^2-4, 4k, k^2+4
k^2-9, 6k, k^2+9
k^2-16, 8k, k^2+16

generally,
k^2-m^2, 2m, k^2+m^2

where 1<m<k and both k,m€ lN.


Other deductions are possible(think along the line the Corollary stated).


Gracias.

Nice and awesome. There is more to that. If you notice, with the given condition, the difference between 2 of the triples formed is usually 1. You can form much more as shown below;
From the theorem,
A² + B² = C²
A² = C² - B²
A² = (C + B)(C - B) ..... (1)

This can be interpreted as the product of the sum of two pythagorean triples and the difference between them is the same as the square of the third.

Let C + B = s ..... (2)
C - B = d ..... (3)
Substitute into equation (1) to get
A² = s*d ..... (4)
s = A² / d ...... (5)
Solving equations (2) & (3),
C = 0.5(s + d), B = 0.5(s - d)
substitute equation (5) into the above to get
C = 0.5 A² /d + d
C = (A² + d²) / 2d
similarly
B = (A² - d²) /2d
With this, we get
A, (A² - d²) /2d, (A² + d²) /2d

Using A = 12, several triples can be formed by changing the value of d
When d = 1, we'll get
12, (144 - 1) /2, (144 + 1) /2 = 12 , 71.5, 72.5
When d = 2, we'll get
12, (144 - 4) /4, (144 + 4) /4 = 12, 35, 37
When d = 3, we'll get
12, (144 - 9) /6, (144 + 9) /6 = 12, 22.5, 25.5
When d = 4, we'll get
12, (144 - 16)/8, (144 + 16)/8 = 12, 16, 20
When d = 6, we'll get
12, (144 - 36)/12, (144 + 36)/12 = 12, 9, 15
When d = 8, we'll get
12, (144 - 64)/ 16, (144 + 64)/16 = 12, 5, 13

benbuks: ^^i was equally unable to post 4d pass 72hrs ,,for unknown reasons.say x.
hm. oga richiez pls wot happend again?

Mine was because of what I posted. The message was repeated and one of the repeated was made all bold. I wanted to modify right away only to find out that I've been blocked from posting on the education section thanks to antispam bot. It was a lesson learnt the hard way.

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of ANY two when when increased by one is a perfect square...find the numbers......

Laplacian: ...integers which satisfy my question in increasing order of the magnitude of their first term include;
1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545
...infinitely more...

jackpot: You said infinitely more? Can you prove it?

Alpha Maximus: Is there any correlation between such a triple and its successive/previous triple? Is there a general formula for obtaining such triples? Is there any correlation between the first, second and third terms of consecutive triples of this sort?
For example I discovered an easy way of producing Pythagorean Triples

I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,

Humphrey77: -if x is equalto (9-0)(9-1)(9-2)(9-3)...(9-n) for all n is an integer x is ? @ jackpot please kindly give me the solution to the above question with in 10 min

Humphrey77: HERE IT
SOLUTION
(9-0)(9-1)(9-2)(9-3)(9-4)(9-5)(9-6)(9-7)(9-(9-9)(9-10)... AT THE POINT (9-9) WE HAVE ZERO . CLEARLY , (: THE PRODUCT EXPANSION AS IT VAULE EQUAL TO ZERO WITH WHICHTHE BEAKING POINTEXIST AT (9-9) )

Humphrey77: THE EXPRESSION (9-0)(9-1)(9-2)(9-3)(9-4)... IS EQAUL TO ZERO

HAPPY

Yes you are right but only for when n is greater than or equal to 9
What about when n < 9
For n = 0, X = 9 = 9 * 8! / 8!
n = 1, X = 9 * 8 = 9 * 8 * 7! / 7!
n = 2, X = 9 * 8 * 7 = 9 * 8 * 7 * 6! / 6!
This is the same as X = 9! / (8 - n)!

Hence X = (9-0)(9-1)(9-2)(9-3)...(9-n) can be written as
For 0 < n < 9, X = 9! / (8 - n)!
n greater than or equal to 9, X = 0
Well as for n < 0, that one pass me o.

Chai! I don't know where to start or continue from. I was banned for over 3 days till now. He de pain me well well for some reply whe I be want give but no fit give right away. He de paine me well wellu o ga ga.
I'm happy to be back.

mencade6: akpos, sure we would like you to join our league. Though am in nairaland championship league (coca cola league, division 1), while some are in the main league (nairaland champions league). Its only chamotex that can tell if there is space to add you in. He is the comptroller general and administrator for the leagues.

Chamo, abeg if space dey make you add akpos for our coca cola league. Akpos put your e-mail here so that it could easy or he may pm you. Tanx

Thank you jare. This na my email; akscandium@yahoo.com

doubleDx: Here it goes:

x + y + z = 3 .....equation1
x² + y² + z² = 3....equation 2
x³ + y³ + z³ = 3.... equation 3

Before we proceed. We have to create an expression for x³ + y³ in terms of (x + y)

Using addition and subtraction method to factorize the polynomial x³ + y³ yields:

x³ + y³ can be expressed as : (x + y)³ - 3xy (x + y)

Also x² + y² can be expressed as : (x + y)² - 2xy

Now, having created an expressing for the above.

From equation 1, (x + y) = (3 - z) ....equation 4

Substituting the expression for x² + y² in equation 2 yields :
(x + y)² - 2xy + z² = 3 .... equation 5

Substituting the expression for x³ + y³ in equation 3 yields:
(x + y)³ - 3xy (x + y) + z³ = 3 .... equation 6

Since we already know the expression for x + y = (3 - z) in terms of z. We can now find an expression for xy from equation 5 so we can substitute back in equation 6 to give us the values of z.

Computing xy in terms of z, subtitute x + y = (3 - z)

(3 - z)² - 2xy + z² = 3
2xy = (3 - z)( 3 - z) - 3 + z²
2xy = 9 - 6z + z² - 3 +z²

continuing from here

2xy = 2z² - 6z + 6
:. xy = z² - 3z + 3

Having known the expression for xy in terms of z as written above. We can now substitute both expression (i.e the xpression for (x + y) and xy) in terms of z in equation 6. x + y = (3 - z) and xy = z² - 3z + 3

(x + y)³ - 3xy (x + y) + z³ = 3
substituting yields :
(3 - z)³ - 3( z² - 3z + 3) (3 - z) + z³ = 3

27 - 27z + 9z² - z³ + 3z³ - 18z²
+ 36z - 27 + z³ = 3

3z³ - 9z² + 9z - 3 = 0
Simplifying further:

z³ - 3z² + 3z - 1 = 0
Factoring the polynomial yields:
(z - 1)³ = 0
z = 1 thrice

The 3 Q's and T can occur in the following ways;
TQQQ, QTQQ, QQTQ, QQQT
Hence all Q to the right of T will be
N/4= 3360/4
= 840

akpos4uall: Firstly, we need to get the amount of ways they can be arranged without any given/restricted condition
The general formula for this is given as;
N= n! / (x! * y! * z!) where x,y,z are the number of times we have repeated/identical arrays.
In this case, we have 3 Q's and 2 P's, hence
N = 8! / (3! * 2!)
N = 3360
* to be continued *

From the total arrangement, the position of R will be before or after P in the following three ways
RPP, PRP or PPR i.e before the 2 P's, after the 2 P's or in between the 2 P's.
This will also occur the same amount of times for each of those three
Hence, the number of times R will be to the left of P
=N/3
=3360/3
=1120

smurfy: 2. In how many different ways can the letters P, P, Q, Q, Q, R, S, T be arranged if

(a) the letter R must be to the left of the letter P

(b) the letter Q must be to the right of the letter T?

Firstly, we need to get the amount of ways they can be arranged without any given/restricted condition
The general formula for this is given as;
N= n! / (x! * y! * z!) where x,y,z are the number of times we have repeated/identical arrays.
In this case, we have 3 Q's and 2 P's, hence
N = 8! / (3! * 2!)
N = 3360
* to be continued *

smurfy: 2. In how many different ways can the letters P, P, Q, Q, Q, R, S, T be arranged if

(a) the letter R must be to the left of the letter P

(b) the letter Q must be to the right of the letter T?

Does it mean that R must be to the left of any of the P's or all of the P's?
E.g does QPRPT agree with the condition?

Laplacian: what u have reasoned is correct because the arrangement is symmetrical....it follows immediately that we have deceitfully proved that;

{summation i=1 to n-1}
(n-2)P(i-1)*(n-i)!=n!/2

Can somebody prove this for us?

Laplacian: case1: if R appears as first letter to d right,
number of ways of arranging d other letters, n=4!=24
case2: if R appears as second letter to d right,
n=4P3*3P1=24*3=72
case3: if R appears as third letter to d right,
n=4P2*3P2=12*6=72
case4: if R appears as fourth letter to d right,
n=4P1*3P3=4*6=24

so there are 192 possible ways of arrangement

Generally, if u are to arrange r objects so that one is always to d right of the other, then
case i: if object appears as i-th object to d right,
n=(r-1)P(r-i)*(r-2)P(i-1)

Using your logic, I think I have the solution
Case 1: S as first from left
n = 4! = 24

Case 2: S as 2nd from left
We are left with any 3 of P,Q,T for the 1st slot and the remaining 3 slots can be done in 3! ways
n = 3 * 3! = 18

Case 3: S as 3rd from left
We are left with 2 of the 3 P,Q,T for 1st & 2nd slot and the remaining 2 slots will be done in 2! ways
n = (3*2)*2! = 12

Case 4: S as 4th from left
Here P,Q,T can be arranged in any order in 3! ways for the first 3 slots and R in 1 way for the last slot
n = 3!*1 = 6

Total possible = 6 +12 +18 +24 = 60

smurfy: In how many different ways can the letters P, Q, R, S, T be arranged if the letter R must be to the right of the letter S?

**Goes out to buy gift for whoever thrashes the question**

This is the same as taking R and S as a single letter (lemme say letter V). Hence it is like arranging the letters P, Q, V & T. This is same as arranging 4 letters = 4! = 24
Answer is 24

Benitez is the right man for chelsea not Mourinho. You can't win the league with this rate of goals being conceded.

Man fit still finish second sha

Arsenal have conceded the fewest goals in the league.

The Gunners have had under 2.5 goals in 9 of their 10 previous home encounters with top-six sides

Everton have lost just once all season, fewer than any other side this season.

Everton have drawn the first half in 6 of their last 8 visits to top-six opposition but have conceded first in 10 of their last 14 trips to the top-six.

Arsenal have opened the scoring 12 times this season recording 11 wins in those matches.

Arsenal have won eight and lost none of their last 12 Barclays Premier League games against Everton.

Everton have only managed more than one goal in a league game against Arsenal on one occasion in the last 21 meetings.

The Gunners have won 14 and lost none of the last 17 Premier League home games against the Toffees (D3).

There have been just four goals scored in the last four Premier League meetings between Everton and Arsenal.

Howard Webb will officiate Arsenal v Everton on Sunday. The Gunners have won their last eight Premier League games with him as the man in the middle.

Arsenal are unbeaten at home in the Premier League with Howard Webb as referee (W10 D4 L0).

Arsenal have scored in 13 of their 14 games and have led in 12 of their 14 games this season, both Premier League highs.

Mesut Ozil has three goals and six assists in 11 Premier League appearances.

The Gunners have kept six clean sheets in their last seven games in all competitions, conceding just one goal in the other match (0-1 at Man Utd).

The Toffees have kept three away clean sheets in a row for the first time since May 2009; they have not recorded four or more in a row since August 1995 (had a run of five).

Read more at http://www.fourfourtwo.com/features/fourfourtwo-preview-arsenal-vs-everton#iIEpAvkAsLTzbQje.99

http://www.fourfourtwo.com/statszone/8-2013/matches/695041/pre-match#hHwQ7sit4tTKmsP5.99

ajasa4link: abeg make una dey post funny Arsenal memes for here make all d haters dey see jare....GUNNERS for life

Talk of fun, check out this blog for some funny arsenal faceswap http://www.news.arseblog.com/2013/12/arsenal-face-swaps/
My mind no quick reach here if not I for no start this thread https://www.nairaland.com/1544264/football-face-swap

More...

It seems I can't upload more than one pic because of their sizes.

I hope this is the right section to post this.
I feel I should share these funny football stars face swap thanks to @FootyFaceSwap on twitter. You can also view some Arsenal Face swaps on www.news.arseblog.com/2013/12/arsenal-face-swaps/

def lover: yes you cannot sell it until someone wants to buy.......if it takes forever to get a buyer thats when you can sell and at that forever prize....

Thanks a lot for this info. Does it also mean that the volume to be sold should be the same as volume to be bought?
As for the Transcorp stuff, you've been right all along.