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Stefan_chemist
post Sep 11 2007, 04:56 PM
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Group: Članovi
Joined: 3-September 07
From: Pančevo
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Ovako: gledao sam zadatke sa prijemnih u MG i naisao sam na jedan u kojem je zadat D_n (ukupan broj dijagonala u mnogoulglu) pravilnog mnogougla (iznosio je 135). I fora zadatka je bila da se izracuna zbir unutrasnjih uglova tog n-tougla (ono D_n je veliko D sa indeksom n).

Dakle, posto se D_n izracunava po obrascu D_n=(n(n-3))/2 (malo je nezgodno za pisanje, ali se nadam da se razume), logicno je da cemo prvo izracunati koliko je n, pa nakon toga po formuli S_n = (n-2) x 180 izracunati zbir unutrasnjih uglova. (ono 180 je 180 stepeni.)

E, tu se javlja problem. Da bih iz formule 135 = (n(n-3))/2 dobio n, potrebna mi je kvadratna jednacina. Znam da je to jednacina oblika ax^2+bx+c=0, i da se resava po obrascu x 1,2 = (-b+/- koren b^2-4ac) / 2a. Kako ja sad to mogu da primenim na zadatak?

Help please.... rolleyes.gif

This post has been edited by Stefan_chemist: Sep 11 2007, 04:59 PM


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Iva
post Sep 11 2007, 08:53 PM
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Joined: 30-June 06
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Dugo si pisala post, pocela si pre mene, a zavrsila posle

Ma zato kaj sam radila još neke druge stvari usput.. počela da pišem pa promenila tab i zaboravila malo.. blush.gif..

Nego Stefane.. ma ja sam se krivo izrazila wink.gif.


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Posts in this topic
Stefan_chemist   Kj   Sep 11 2007, 04:56 PM
NeverMore21   RE: Kj   Sep 11 2007, 05:20 PM
Iva   RE: Kj   Sep 11 2007, 05:21 PM
NeverMore21   RE: Kj   Sep 11 2007, 05:24 PM
pyost   RE: Kj   Sep 11 2007, 05:55 PM
NeverMore21   RE: Kj   Sep 11 2007, 05:58 PM
Stefan_chemist   RE: Kj   Sep 11 2007, 06:22 PM
Iva   RE: Kj   Sep 11 2007, 08:53 PM


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