Alpha Maximus: ..... Smh, fucktards everywhere and a mega person(jackpot) here!! Can you now see why there's always an issue of 'half-baked' grraduates? I'm sure you never even made it to the oven!!! You're asking me if x^x+y^x=31 is a polynomial!! You're are an illiterate of no comparison!!*salutes* Wasn't it (5-y)^2+y^2-31=0 that I used as a polynomial!! For your info, a polynomial is an expression constructed of variables and constants using operations of addition, subtraction and multiplication with non-negative integer exponents(culled from Wikipedia)....do you see any mention of a polynomial being stictly of the 3rd or 4th degree, you intellectually devoid slowpoke!!! And about why I used the remainder theorem, isn't the first step of using it trying out integer values of 'x' or'y' first? Calling me a learner, you must be truly insane!! You think complex looking questions are always the hardest to solve, irredeemable mugu!! Who is forming,online again? Weren't you the one saying your solution was sexy? Pride is one of the deadly sins oh, and your seat is being reserved in the VIP section of Hell!!! And about that 'turns my solution into tissue paper' comment, the last time I met something like you , I flushed it down the toilet!!! Pompous piece of shit

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

rhydex 247: lol @ jackpot u said d questn is nt correct. I dnt expect dat 4rm u. Anyway. All is well.
THE BEST DNT ALWAYS DO THE BEST.

Laplacian: ...@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'?...

Boladearo: Find x and y
x^1/3 + y^1/3 = 6..........(i)
X + y = 72...................(ii)

jackpot: ^you cajoled the other person saying his solution is trial and error and that you are going to do it direct. Here you are saying that the Remainder Theorem you applied is trial and error method.

Okay ooo

Now, why didnt you get y=2 when applying your theorem? I suspect super-film trick here

If you get y=3, why not substitute into the more simpler x+y=5 to get your value of x?

I repeat, the remainder theorem is not about trial and error. You may divide a polynomial with x-2 and get a remainder. You may choose to divide the same polynomial with x-5 and get another remainder.

Maybe you wanted to say Factor Theorem.

But still before you apply both, f(x) must be a polynomial and what you used is not a polynomial.

The sky is your limit though, only if you will accept your mistakes, be less-abusive and learn. You can still be the future Udeme guiness man of Mathematics

rhydex 247: here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

rhydex 247:

soln.
Suppose u=(3,4). If r & s € k. r=1 and s=2.
(r+s)u=ru+su.
(r+s)u=(1+2)(3,4)=(9,4).
ru+su=1(3,4)+2(3,4)
ru+su=(3,4)+(6,4)= (9,. Since (r+s)u is not equal to ru+su. Hence is not a vector space.

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)?

Alpha Maximus:
For your first question(first two, one by each player), there are 400 possible ways
For your second question, (first four, 2 by each), there are 318,979,564,000 possible ways...... Chacha!!

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)?

rhydex 247:

soln.
Suppose u=(3,4). If r & s € k. r=1 and s=2.
(r+s)u=ru+su.
(r+s)u=(1+2)(3,4)=(9,4).
ru+su=1(3,4)+2(3,4)
ru+su=(3,4)+(6,4)= (9, . Since (r+s)u is not equal to ru+su. Hence is not a vector space.

rhydex 247: here is the question again
Show that V=IR^2 is not a vector space over IR with respect to operation (a,b)+(c,d)=(a+c, b+d) and k(a,b)=(ka,kb)

2nioshine: @Jackpot u seems like my one tym gal who actually gave me a tough time and made me study like c.obi (hope u r nt d unizik gal) ...alpha i luv ur math spirit no doubt u r no baby mathematician so lets b a little r***tional here(seun no dy pay or give us grades abeg),...double dx i was kinda di***pointed,but now u apologised we r kul...u remainent my most respectd boss...# Still in my observation mood#....nice wuk @ all 4 ur tym nd contributn(wish i could list al ur names)..sti following...wait o! Where is Richez

Laplacian:
...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...