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Stefan_chemist
post Sep 11 2007, 04:56 PM
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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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NeverMore21
post Sep 11 2007, 05:20 PM
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Ne treba ti kvadratna jednacina! Iz tvog izraza se dobija sledece:

CODE


n*(n-3)=2*135

n*(n-3)=270

n*(n-3)=18*15

n=18


Razumeo? smile.gif


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Iva
post Sep 11 2007, 05:21 PM
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Imaš 135 = (n(n-3))/2. To "sređuješ" (ili već kako da rečem), pa dobiješ 270 = n^2 - 3n; zatim n^2 - 3n - 270 = 0.

Ovo je kvadratna jednačina, gde je

x = n
a = 1
b = -3
c = - 270


Onda samo uvrstiš šablon i dobiješ rešenje (ja dobih [B]18[/B] pomoću "+ jednačine").



Edit: sad tek videh Nevermore-ov post.. razlika je 1minut.. XD.gif

This post has been edited by Iva: Sep 11 2007, 05:22 PM


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NeverMore21
post Sep 11 2007, 05:24 PM
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QUOTE(Iva @ Sep 11 2007, 06:21 PM)
Edit: sad tek videh Nevermore-ov post.. razlika je 1minut.. XD.gif
*



Dugo si pisala post, pocela si pre mene, a zavrsila posle tongue.gif!


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pyost
post Sep 11 2007, 05:55 PM
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Nikada se ne zna kada ce nekome zatrebati resenje odredjene kvadratne jednacine, pa zato...

ax^2+bx+c=0

x_{1,2} = \frac{-b^2\pm\sqrt{b^2-4ac}}{2a}


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NeverMore21
post Sep 11 2007, 05:58 PM
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QUOTE(pyost @ Sep 11 2007, 06:55 PM)
Nikada se ne zna kada ce nekome zatrebati resenje odredjene kvadratne jednacine, pa zato...

ax^2+bx+c=0

x_{1,2} = \frac{-b^2\pm\sqrt{b^2-4ac}}{2a}
*



Ovo moram da uramim, uvek sam gresio pri pisanju kvadratne jednacine rolleyes.gif !


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Stefan_chemist
post Sep 11 2007, 06:22 PM
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tongue.gif+-->
QUOTE(NeverMore....tongue.gif)
n*(n-3)=2*135

n*(n-3)=270

n*(n-3)=18*15

n=18


Aha, pretpostavio sam da je tako nesto.. XD.gif

QUOTE(Iva)
Imaš 135 = (n(n-3))/2. To "sređuješ" (ili već kako da rečem), pa dobiješ 270 = n^2 - 3n; zatim n^2 - 3n - 270 = 0.

Ovo je kvadratna jednačina, gde je

x = n
a = 1
b = -3
c = - 270

Onda samo uvrstiš šablon i dobiješ rešenje (ja dobih 18 pomoću "+ jednačine").


Pa da! Tako jednostavno! Ali ja jos nisam ucio sisteme... unsure.gif unsure.gif unsure.gif

Edit: izvinite, upravo sam lupio glupost... sad.gif sad.gif sad.gif sad.gif Mislio sam da "uvrstiti sablon" znaci nesto drugo... :smajlikojisedaviumorusuzakojejesamnapravio:

This post has been edited by Stefan_chemist: Sep 11 2007, 06:35 PM


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Iva
post Sep 11 2007, 08:53 PM
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QUOTE
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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