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pyost
post Apr 18 2006, 08:44 PM
Post #21


Deus Ex Makina
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Jos jedan problem!

x=0,9999^.

100x=99,9999^.

99x=100x-x=99,9999^.-0,9999^.

99x=99

x=\frac{99}{99}

x=1

Dakle, imamo x=0,9999^. i imamo x=1! Probajte da nadjete problem, ja nisam uspeo.


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Cek@
post Apr 18 2006, 09:56 PM
Post #22





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Nemam pojma brate... Ocigledno 0,999999999999999999 tezi kecu, pa je mozda to resenje...
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pyost
post Apr 19 2006, 08:23 AM
Post #23


Deus Ex Makina
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Ne, ovo se sve radi bez zaokruzivanja, u tome i jeste problem.


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Sid
post Apr 19 2006, 12:40 PM
Post #24





Group: Članovi
Joined: 19-April 06
Member No.: 54
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Ne postoji greska u tome. To nam je nastvnik analize pokazao na casu.
Broj 0,9 gde se 9 beskonacno puno puta ponavlja jeste u stvari 1.


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pyost
post Apr 19 2006, 12:49 PM
Post #25


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Ali kako to objasniti ako se nigde ne pojavljuje zaokruzivanje? Cim je 0,nesto to ne moze da bude jedan. Ja bih rado prihvatio to objasnjenje, ali nije dovoljno matematicki biggrin.gif


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Sid
post Apr 19 2006, 09:51 PM
Post #26





Group: Članovi
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ovo jeste tacno. Ako ti to nije dovoljno matematicki kako objasnjavas da je jedan od najvaznijih brojeva u matematici "i" gde je i^2=-1.
Matematika je jedna "cudna" nauka biggrin.gif biggrin.gif


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pyost
post Apr 19 2006, 10:28 PM
Post #27


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Da znas biggrin.gif Drzacu se ja od sada na malo vecoj udaljenosti od iste.........


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Sid
post Apr 19 2006, 10:41 PM
Post #28





Group: Članovi
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Kada bi to tako lako moglo da se izvede. . .
Sta mislite o godisnjem iz analize?


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^_NiN0_^
post Apr 20 2006, 08:38 PM
Post #29


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Ovako!!!!
0.99999....=1-0.0000000000...
Kad ce se bre pojaviti kec u ovom zapisu 0.00000... kad ima beskonacno devetki!!!Tj 0.0000.....=10^-beskonacno (ne znam da koristim Latex dry.gif ) itd.
da svi ukapiraju!!!


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pyost
post Apr 20 2006, 09:03 PM
Post #30


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Trebalo bi da bude ovako nesto

CODE
[for]10^{-\infty}[/for]


10^{-\infty}


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^_NiN0_^
post Apr 26 2006, 11:34 AM
Post #31


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Pogledajte sliku!!!! laugh.gif laugh.gif laugh.gif laugh.gif

This post has been edited by ser_fanky Lj@B: Apr 26 2006, 11:38 AM


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Yo!hambin,
Yo!hambina, Yo!hambin
Yo-yo, yo-yo, yo
Yo!hambin,
Yo!hambina, Yo!hambiiina
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Velika matura deca Yo!hambina)
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pyost
post Apr 27 2006, 06:54 PM
Post #32


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Hehhe, dobra fora, samo nemam pojma kakve veze ima sa ovom temom biggrin.gif


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^_NiN0_^
post Apr 27 2006, 10:16 PM
Post #33


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Prebaci je u drugi topic!!!Ma mrzelo me da pisem u drugom...


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Yo!hambin,
Yo!hambina, Yo!hambin
Yo-yo, yo-yo, yo
Yo!hambin,
Yo!hambina, Yo!hambiiina
(Mala matura deca kokaina®
Velika matura deca Yo!hambina)
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pyost
post Apr 30 2006, 09:17 PM
Post #34


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E pa mrzi i mene tongue.gif


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ViVitas
post Jun 8 2006, 04:55 PM
Post #35





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Ja od svih zackoljica najvise volim problem Bagdadskog trgovca dinjama.


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pyost
post Jun 8 2006, 07:17 PM
Post #36


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Detalji?


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Puzzler
post Jul 19 2006, 03:12 PM
Post #37





Group: Članovi
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Jeste li čuli za "Šahovski zadatak"? Nama iz MG-a je veoma jednostavan...
Priča kaže da je izumitelj šaha, kada mu je indijski vladar ponudio nagradu za genijalni izum, tražio sledeće: da nagrada budu samo zrna žita, ali da ih dobije na sledeći način.
Na prvo polje ide jedno zrno, na drugo dva, na treće četiri, na četvrto 8 itd. Kada je to čuo, vladar se samo nasmejao i rekao: "Nema problema!"... A problema je bilo itekako!
Može li neko da mi kaže tačan broj zrna žita koja je izumitelj šaha zatražio?


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pyost
post Jul 19 2006, 07:33 PM
Post #38


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Mislim da bi odgovor bio 11111111..11111111 (64 jedinice) u binarnom kodu, pa da se pretvori u dekadni. A mozda i ne biggrin.gif


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maxydelanoche
post Jul 20 2006, 12:31 PM
Post #39





Group: Članovi
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Da, i ja mislim da je to odgovor. Ja sam cula za taj problem (citala negde, nekad)... Pa je kralj bio nesrecan sto ne moze da ispuni taj zahtev, matematicari su mu rekli da toliko zita nema ni u svim kraljevim zitnicama, da ne bi bilo ni kada bi svo kopno bilo prekriveno njivama...


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Puzzler
post Jul 20 2006, 03:14 PM
Post #40





Group: Članovi
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Da, to je tačno rešenje! Evo i rešenja u dekadnom sistemu:

S=1+2+4+...+2^{\63} /*(2-1)
S=(2+4+8+...+2^{\64})-(1+2+4+...+2^{\63})
S=2^{\i64}-1
S=18446744073709551615 (wow!)

P.S. Pogledajte u temi "Logika" još jedan zadatak!


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