₦airaland Forum

smurfy: Your move wasn't registered. Did you click Submit?

k, i just submitted it

Mr Calculus: PLZ SIRS NO BODY HAS PROVIDED A SOLUTION TO DIS.
U GUYS SHOULD PLZ HELP ME OUT

z=-16 is a real number & cannot hav imaginary component...so if z=a+jb then a=-16, b=0...
If z^4=-16, then z=+4j or -4j

smurfy: Now click on the piece u want to move, wait a moment, click on where u want to put it, wait a moment then click Submit.
Note: You can play multiple games at the same time.

tanks alooooot bro...now we can row...its ur turn 2 move

d citizen:
The math general in the house,

Nobody has done justice to the third question posed by mikebis on page 57

I believe dat he extracted the question from a textbook called bouday

u^2/v + v^2/u =12

1/v + 1/u = 1/3.

i guess nobody attemptd it 'cos its straightforwrd...multipy both eqns by uv to get, u^3+v^3=12uv,
u+v=uv/3, recall dat
u^3+v^3=(u+v)^3-3uv(u+v), d first eqn bcoms
(u+v)^3-3uv(u+v)=12uv, substitut 4 u+v frm eqn1
(uv)^3/27-(uv)^2=12uv or
(uv)^2-27uv-324=0 or
uv=0 or 36 or -9 so dat from
u+v=uv/3
u+v=0 or 12 or -3, solvin d quadratics givs, u=6,-4.85 and
v=6, 1.85

d citizen:
The math general in the house,

Nobody has done justice to the third question posed by mikebis on page 57

I believe dat he extracted the question from a textbook called bouday

u^2/v + v^2/u =12

1/v + 1/u = 1/3.

i guess nobody attemptd it 'cos its straightforwrd...multipy both eqns by uv to get, u^3+v^3=12uv,
u+v=uv/3, recall dat
u^3+v^3=(u+v)^3-3uv(u+v), d first eqn bcoms
(u+v)^3-3uv(u+v)=12uv, substitut 4 u+v frm eqn1
(uv)^3/27-(uv)^2=12uv or
(uv)^2-27uv-324=0 or
uv=0 or 36 or -9 so dat from
u+v=uv/3
u+v=0 or 12 or -3, solvin d quadratics givs, u=6,-4.85 and
v=6, 1.85

smurfy: Change ur phone's image quality settings to high or something like that. That's what I did to my operamini.

dat's been done succefully...how do i make a move?

benbuks: M47h3m471c5.1ncr353 y0ur 1Q.17.3nh4nc35.your.cr3471v17y..57u6y m0r3 0f 17.bu7.d0n7.b3.20b355355.w17h.17

Mathematics increases your IQ...it enhances your creativity...study more of it but dont be obsessed with it...

...@Smurfy...all d pieces are alignd vertically on a straight line...how do i make it rectangular?...

rhydex 247: Soln.
Case 1. Suppose G has an element g of order 3. Then d cyclic subgroup generated by g contains three elements.{g,g^2,g^3=e}, where e is d identity. But the order of every subgroup must divide the order of G, nd dis is a contradiction. So G has no element of order 4.
Case 2. G is nt cyclic. Thus G={e,a,b,c} where e is d identity, nd each of a,b nd c has order 2. Nw let us take a look at d multiplicatn table.
i dnt knw hw to construct a table here. Buh here is a way 4ward. Draw a multiplicatn table.The ist row is (e,a,b,c), ist column (e,a,b,c) , 2nd row is (e,a,b,c), 3rd row is (a,e,c,b), 4th row is (b,c,e,a) nd d 5th row is (c,b,a,e). Nd nw by inspection, we see dat G is abelian. All is well.

precise...thumbs up...

Alpha Maximus: ........for question 3, there is an infinite number of values satisfying 3x + 4y=5, the first being x=-1 and y=2.....when obtaining the other values of x and y, it is observed that x increases by a constant amount of -4 while y increases simultaneously by an amount of 3 i.e: -5 and 5, -9 and 8, -13 and 11, etc

...very correct...

rhydex 247: solution.
Suppose we have 2 subgroups of G defined as H1 and H2. Nw the questn implies H1 intersectn H2 must also be a subgroup of G.
For any element a in H1 there exists a^-1 and H2 is closed. The same holds for H2. So the intersectn will only contain an element c in H1 intersectn H2. If c and c^-1 are in H1 and H2 they must contain e the identity of G thus H1 intersectn H2 cannot be empty. Hence the intersectn of any subgroups of a group G is a subgroup of G.

kudos 2 u...u did a nice wrk...

@ smurfy, wat devic do u use on chess.com?...cmp or mobil phon?

smurfy: pawn advances to c8, gets promoted to Queen (or Rook) ...checkmate!

simple enough i guess

smurfy: @Laplacian
Answer to que1 is 0.
So, what's ur username on chess.com?

answers mean nothin 2 me without steps...username is adokwu-ondoma

Alpha Maximus: I'm calculatorless, scientifically ..

but electronically u 're no...lol!!

...questn1; find a number which wen multiplied by 7 and d result added to 1 gives a perfect square
....questn2; List ALL d subgroups of a group of order 7
...questn3; find All integers satisfying; 3x+4y=5
....question4; black king on a8, white king on a6, white pawn on c7...white to move and mate in 1

...G is a group of order 4, show dat G is abelian...

...show that d intersection of any subgroups of a group G is a subgroup of G...

...sum d series;
cos@+cos2@+cos3@+...

...@Maximus....for 55,130,250 handshakes, there are 10,501 people in d conferenc room, while for 31,988,001 handshakes, there are 7,999 people...10,501-7,999=2,502people

@smurfy...i got dat...
@Maximus..ur solution comes in 5min's time

smurfy: Nice time? Did you just say nice time? Well, maybe getting trashed by ME is nice time to YOU...lol (I get mouth sha...)

am set...we 'll c abt dat

...@Maximux...suppose there are n people lockd in d conference room...d first man shakes n-1 ppl & he is then shut outside...second man shakes n-2...and d we have d follwn progresn afta each hanshak...n-1, n-2, n-3,....,2, 1 so dat d total handshakes is d sum of d first n-1 positv integers
=(n-1)n/2...equat and find n....
pls dont ask me 4 d value 'cos i've no calc on me...

Alpha Maximus: ....I can't believe no-one has attempted this question, guyz in the house, you done fall my hand, sorry, calculator

...@Maximus, d rational mathematician...till dusk...u'll c d solution...

smurfy: Not bad. Here goes...
Each side has eight pawns, and each pawn can move in 2! ways. So white's first pawn move can be done in 2! * 8 ways (= 16). Each side has two knights and each knight can move in 2! ways, giving 2! * 2 (=4). So white's first two moves can be done in 16+4=20 ways. Since each side has exactly the same number of ways of moving, then, using product principle, the first two moves can be made in 20*20=400 ways. By the way, are u on chess.com? ...feel like trashing someone right now lol

@Smurfy, chess.com?...i've not registerd yet...i'll do dat @ dusk so we can have a nice time...

smurfy: In how many ways can the first two moves (one by each player) be made in a game of chess?


What will the answer be for the first four moves (two by each player)?

...a single pawn move from white attracts 16 possible pawn moves from black...so d 16 possibl pawn moves for whit attracts a total of 256 pawn moves from black...a singl pawn move from white attracts 4 Knight moves from black, so 16 pawn moves from white attracts 64 Knight moves...a single Knight move from white attracts 16 pawn moves from black, so 4 knight moves attracts 64black pawn moves...finally, a single white's knight move attracts 4 black's knight moves, so that 4 attracts 16....so we have: 256+64+64+16=400...i guess ur second question can confidentently b answered using computer analysis...

...@Jackpot...i'll advic u 2 reserv all comments concernin dat gentleman, i belive by now it should b obvious 2 everyone dat he abhors criticism, more developd in rainin insults dan coherently providin answers 2 questions...Jackpot i want 2 ask u a question, here's a man who claims 2 promot knowledge...publicly postin insults, what kind of knowledg is he promotin?...check his record from d tim he had issue wit Doubledx, about 75% of his issues seems 2 lie in d title dat some ppl just chose 4 demselves (just 4 fun)...i sugest we let him chose a name 4 himself 2 stop all dis cat & rat...my concern is mostly 4 our youngstars...by d way wat's 'SMH'?...

jackpot: The bolded is just cock-and-bull.

First of all, what does the remainder theorem say?

Once a polynomial f(x) is divided by x-a, the remainder is another polynomial of 1 degree less than that of f(x) together with a remainder R still to be divided by x-a.

i.e., f(x)/(x-a)=g(x)+R/(x-a)
therefore f(x)=(x-a)g(x)+R
substituting x=a yields that
the remainder R=f(a)

back to the question. What is a polynomial? I would leave that for you to find out.

The equation x^x+y^y=13 is not a polynomial.

Why then are you applying the remainder theorem which specifically talks about polynomials?

And the funny thing is that you said the application yields y=3. SMH

Rubbish. Bros, no just dey form anyhow, you are still a BIG learner.

Converts his answer sheet to toilet paper

...@Jackpot...fear no allow me talk ooo...

jackpot: Given that x,y,z depends on three variables, viz,
x=x(u,v,w), y=y(u,v,w), z=z(u,v,w),
we know that the three transformed equations
u=u(x,y,z), v=v(x,y,z), w=w(x,y,z) is non-degenerate if the Jacobian of the transformation doesn't vanish.

Now, consider the equation
x=x(u,v,w).
Differentiating both sides partially w.r.t. x, we have
1=Xu Ux+ Xv Vx+ Xw Wx

so, the relationship is

Xu*Ux=1-(Xv Vx + Xw Wx)

tanks...our only 'female Fermat'

...@Math*Generals...any solution 2 Rhydex247's questns?....@Jackpot...ur solution is pretty...our differencies came essentially from our methods of obtainin Ux,Vx..etc...evri other thing are same...

Alpha Maximus: ....*brings out intecontinental ballistic missile of WRONG and blasts lanrexlan!!!*.....the no he can make is the same no he can smoke!!! Its not 12 though!! Appreciated attempt!! Any of our self-acclaimed generals wanna try out the 'simple' question before I finally post the solution? *alantas spiritually back to amala*

....wat's ur motive 4 bringin dis questn up?...promote knowledge or demote ppl?...anybody can post a hundred such questions dat u too cannot answer?...

jackpot: The number is 10x+y

The problem is
maximize F(x,y)= 10x+y
subject to x+y=9
10>x,y > 0 integers
you may use Lagrange multiplier method.

solve and you'll get x=5 or 4, y=4 or 5. Thus, the number is 54 or 45. Sorry, I am quite busy.

...workin plz...