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Joshuatig
ID: 492
Medical Thesis
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9. Nov 2017 04:33
Betka
ID: 491
Ulohy 1 a 2
Ahoj :-) Ano nula je parne cislo, a v dvojke sa poklada za prirodzene. *THUMBS UP*
2. Nov 2017 09:35
Martina
ID: 490
Ulohy 1 a 2
Berie sa nula v ulohe 1 ako parne cislo?
A v ulohe 2 sa poklada nula za prirodzene cislo?
Vopred dakujem za odpoved.:-)
29. Oct 2017 17:51
hynek
ID: 486
1. uloha sezam
dakujeme za otazku,
odpoved je nie, nemozu. v jednej casti ohranicenej ciarami moze byt len jedno cislo. alebo inak, tam kde uz cisla su, je cela cast vyplnena, doplna sa len do prazdnych policok.
1. Oct 2017 14:53
Mato
ID: 484
prva uloha
Mal by som otazku k prvej ulohe, a to, ci mozu byt dve cisla v jednej casti vyznacenej ciarami, napr ci mozem k 11 dat este cislo 5. Vopred dakujem za odpoved:-D
1. Oct 2017 14:18
Lukáš Wiesenganger
ID: 483
Matematika
:-D
26. Sep 2017 15:19
Betka
ID: 470
druha uloha
Ahoj Maťo :-)
ďakujeme za postreh, už sme príklady aj vzoráky pridali. A teraz k príkladu. Na slávnosť prichádzajú po jednom. V zadaní sa píše nikdy neprišli dvaja naraz. A otázka sa pýta na to s koľkými ľuďmi si potriasli pri príchode. A myslí sa tým, že keď ten človek A prišiel, tak s koľkými ľuďmi si potriasol. Do tohto počtu sa už nezarátava ak si potriasol s niekým kto prišiel pol hodinu po ňom. Ak by si mal ešte nejaké otázky, neváhaj sa opýtať. Veľa zdaru pri riešení *BYE*
8. May 2017 11:43
Mato
ID: 467
druha uloha
strucne... v druhej ulohe prichadzaju po paroch, alebo po jednotlivcoch? A ked tam uz typek A je, po nom pridu dalsi, pre typka A sa rata podanie si ruk aj s tymi co prisli?
A okrem toho, dalo by sa dat priklady aj na stranku?
3. May 2017 16:55
Nobody cares.
ID: 465
Poradie série
na papieri od druhej serie je napísaná prvá séria (v tej hlavičke zadania)*ROFL**ROFL*
11. Apr 2017 18:09
hynek
ID: 464
ziadna chyba...
spytat sa ked nieco nie je jasne nie je nikdy chyba, vzdy radi odpovieme. a kazde zadanie sa da napisat este kusok lepsie, vaseotazku nam vtom pomahaju...
9. Apr 2017 21:33
matus
ID: 463
uloha 1
Dakujem za odpved. je to moja chyba, je to napísané jednoznačne akurát som si to čítal večer takže som si to asi nevšimol. Aj tak dakujem za odpoved.:-X
9. Apr 2017 20:03
hynek
ID: 462
zopakujem odpoved...
pre istotu zopakujem, co uz je nizsie

ide o typy trojuholnikov, ale aj o ich absolutny pocet. takze ak maju ine vrcholy, aj ked su rovnako velke (zhodne), pocitaju sa kazdy zvlast.
je to v tejto vete zadania:
Potom si postupne z vrcholov tohto osemuholníka vyberiete každú možnú trojicu napríklad A, B, C). ...

Teda su tam aj trojuholniky rovnakeho tvaru, ak su vrcholy ina trojica bodov.

A aj trohjuholniky so stranou napr. AB (ale k tej strane este nejake dalsie dve strany k nejakemu tretiemu vrcholu...)
8. Apr 2017 17:23
uloha 1
ID: 461
Odpoved
to uz bolo odpovedane hynkom
8. Apr 2017 15:19
matua
ID: 460
úloha 1
sú trojuholníky ABC a BCD rovnake amulety a počítajú sa ako jeden?
8. Apr 2017 12:48
Na mene nezalezi Hlavne ze ma otazku :-)
ID: 459
1. uloha
KAŽDÁ mozna kombinacia trojuholnikov znamena, ze mame spojit aj strany, ktore su na kraji osemuholnika, teda ak mame osemuholnik s poradim bodov ABCDEFGH, mame zaratat aj trojuholniky so stranou AB, teda susediace body? Dufam ze tuto dlhu vetu pochopite*SING*, som na nu hrdy :-D
8. Apr 2017 11:13
Lenka
ID: 455
Re: premia
premia sa vypocitava automaticky z dosiahnutych bodov a rocnika riesitela, takze bud mame chybu vo vypocitavani alebo v rocniku riesitela v databaze. Pozrieme sa na to a opravime. Mrzi nas to :-( a dakujeme za upozornenie!
3. Apr 2017 19:06
-
ID: 454
premia
niektorým súťažiacim ste udelili vyššie prémie ako iným napríklad blažekovej stele, za úlohy dostala 16,5 a je siedmačka, teda by mala dostať 3 body za prémie, ale vy ste jej udelili 5 bodov
voči niektorým súťažiacim je to veľmi nespravodlivé
3. Apr 2017 19:00
hynek
ID: 453
trochu inac...
ide o typy trojuholnikov, ale aj o ich absolutny pocet. takze ak maju ine vrcholy, aj ked su rovnako velke (zhodne), pocitaju sa kazdy zvlast.
je to v tejto vete zadania:
Potom si postupne z vrcholov tohto osemuholníka vyberiete každú možnú trojicu napríklad A, B, C). ...
27. Mar 2017 19:20
marián
ID: 451
1. úloha
chcel by som sa spýtať, ak je viac rovnakých trojuholníkov (napr. ABC a BCD) mám počítať všetky alebo len jeden.
27. Mar 2017 17:29
hynek
ID: 448
stvorka
vysledky sme si museli (ako pri vsetkych ulohach) zratat sami, ten priklad a jeho riesenie nikde nie je. mas pravdu, ze pre urcite skupiny cisel riesenia suviseli, pre niektore nie. (suvislosti by sa ukazali, ak by sme to riesli pre cisla od 0 po 50 alebo po 100). ale vzdy vedela pomoct nejaka logika zlomkov, a sikovny matematik si tiez casto vie pomoct nejakym programom alebo aspon excelom, potom mu to 2 hodiny nebude trvat...*BYE*
18. Mar 2017 10:14

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