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> Resenje Please!

jeele
post Nov 25 2006, 02:43 PM
Post #61





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a tko je ugljesa?


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RZA
post Nov 25 2006, 03:15 PM
Post #62


Njeno Ljubičanstvo


Group: Članovi
Joined: 7-September 06
Member No.: 92
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smile.gif ..
ugljesa je moj drug.. smile.gif.. smile.gif.
kako ne znate ugljesu?.. unsure.gif ..
pa.. on je isto isao 7 i 8 u MG.. sa mnom u odeljenje.. smile.gif..
sad ide u RG..
ugljesa.. ne znaju te.. aj ih tuci.. smile.gif

This post has been edited by Spužva: Nov 25 2006, 03:26 PM


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maxydelanoche
post Nov 26 2006, 07:35 AM
Post #63





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Joined: 3-May 06
From: Zion
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Blago njima...


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SerGrimReaper
post Nov 26 2006, 03:11 PM
Post #64





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cime da ih tucem hristina? laugh.gif
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RZA
post Nov 26 2006, 03:41 PM
Post #65


Njeno Ljubičanstvo


Group: Članovi
Joined: 7-September 06
Member No.: 92
Status: Bivši učenik MGa



smile.gif
necim dugackim, tvrdim... smile.gif








palicom.. XD.gif


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Eli0t
post Nov 26 2006, 03:56 PM
Post #66


Voodoo People
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Dosta..
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SerGrimReaper
post Nov 26 2006, 04:34 PM
Post #67





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Dobro de ne mora za kriket... jees
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Ana
post Dec 30 2006, 10:03 PM
Post #68





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Matematika... Samo treba da se vezba.

Neko je ovde dokon. Jel tako bese Ugi!?

wacko.gif ph34r.gif laugh.gif


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jeele
post Mar 23 2008, 06:51 PM
Post #69





Group: Članovi
Joined: 6-September 06
From: van sebe
Member No.: 89
Status: Van MGa
Ime i prezime: Jelena Dimov
Škola/Razred: MG, IVe



nadovezacu se na NASLOV TEME ...

2^logx9=4/3^x

nije (4/3)x ... bas je 4/3^x ...
nesto mi ne ide da ga resim ... pa reko' da ga podelim sa vama ...


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Nostradamus
post Mar 23 2008, 07:34 PM
Post #70





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Jel ovo u eksponentu sa leve strane log sa osnovom x od 9?


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 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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jeele
post Mar 23 2008, 08:00 PM
Post #71





Group: Članovi
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From: van sebe
Member No.: 89
Status: Van MGa
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da
logx(9) ...
ne umem da koristim latex-kako-god ...


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pyost
post Mar 23 2008, 08:03 PM
Post #72


Deus Ex Makina
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Pa nauci...

2^{log_9x} = \frac{4}{3^x}


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jeele
post Mar 23 2008, 08:40 PM
Post #73





Group: Članovi
Joined: 6-September 06
From: van sebe
Member No.: 89
Status: Van MGa
Ime i prezime: Jelena Dimov
Škola/Razred: MG, IVe



pokusah ali mi izgleda crtice i zagrade nikad nisu dobre ...

no, hajd' iz petog puta XD.gif izgleda da sam gledala uputstvo za neku zastarelu verziju ...

2^{log_x9} = \frac{4}{3^x}



This post has been edited by jeele: Mar 23 2008, 08:43 PM


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The snorks
post Mar 23 2008, 09:36 PM
Post #74





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Za kad ti treba ovo?
Posto nemam vremena da ga radim sad,moram da ucim bios za sutra...
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Nostradamus
post Mar 24 2008, 09:15 AM
Post #75





Group: Članovi
Joined: 5-November 07
From: Bg
Member No.: 747
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Ovaj zadatak nema resenje


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 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Puzzler
post Mar 24 2008, 02:03 PM
Post #76





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Evo i zašto...

2^{log_x{9}}={\frac{4}{3^x}}
4^{log_x{3}}={\frac{4}{3^x}}
3^x=4^{1-log_x{3}}
4^{xlog_43}=4^{1-log_x{3}}
xlog_43=1-log_x3    / {\cdot}log_34
x=log_34-log_x3 {\cdot}log_34
x=log_34-log_x4
log_34-x=log_x4
log_34-x={\frac{1}{log_4x}}

Sada nacrtajmo funkcije sa leve (zelena) i desne (crvena) strane, i vidimo da nemaju preseka... To se može u detalje objasniti, ali da ne počinjem.

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I da, koristim priliku da se izvinim Jeleni pošto u petak nisam imao kredita.

SORRYYYYY!!!


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jeele
post Mar 24 2008, 04:46 PM
Post #77





Group: Članovi
Joined: 6-September 06
From: van sebe
Member No.: 89
Status: Van MGa
Ime i prezime: Jelena Dimov
Škola/Razred: MG, IVe



jok, ovo sam nasla ovako ... dok sam radila ...
hvala! hug.gif


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