Fórum: Érdekes matekfeladatok
Játékszabályok Technikai információ TeX tanfolyam Elfelejtettem a jelszavam Témák
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| [803] Suzy | 2005-02-27 15:59:51 |
 Köszönöm.Sikerült. 3.Kétjegyű szám tízeseinek jegye 5-tel több mint az egyeseinek a száma.Ha a felcserélt számjegyű számot hozzáadjuk akkor 143-at kapunk.Mekkora az eredeti kétjegyű szám? Ez is egyenletben kéne.De a megoldást tudom.
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| [802] lorantfy | 2005-02-27 15:28:39 |
 Legyen a gyerek most x éves, akkor az apa 3x. 50 év múlva mindketten 50 évvel lesznek idősebbek, tehát x+50 és 3x+50 évesek lesznek. Mivel ekkor az apa éveinek 1/4 részével idősebb a gyermekénél, így az apa életkorának 3/4-e pont a gyerek éveinek számát adja. Megvan az egyenlet!
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| Előzmény: [801] Suzy, 2005-02-27 14:32:44 |
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| [801] Suzy | 2005-02-27 14:32:44 |
 Kösz rájöttem a híbára.De még van 3 feladat. 2.Egy jó Betűszámvető gyermektől mivel nagyon gyenge volt azt kérdik az egsámenben hány esztendős? azt felei 1/3 rész annyi mint atyám :hát az atyád hány esztendős?-kérdik tőle:erre meg ezt feleli hogy 50 esztendő múlva ha élünk atyám éveinek 1/4részével lessz idősebb nálam.Melyik hány esztendős most??Ezt is egyenletbe kéne felírni és megoldani.Köszi:suzy
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| [800] Suzy | 2005-02-27 14:19:57 |
 x+7x/12+1200=x jól írtam fel? Segíts levezetni.Kösz.
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| [799] Suzy | 2005-02-27 14:09:51 |
 Köszi.Megpróbálom megoldani.
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| [798] lorantfy | 2005-02-27 14:03:51 |
 Kedves Suzy!
A torony teljes magasságát vegyed x-nek. Ekkor a földben lévő rész , a vízben lévő rész és a víz felett lévő rész 100 láb. Ha ezeket összeadod a torony teljes magasságát kapod, vagyis x-et. Meg is van az egyenlet! Aztán mindkét oldalt beszorzod 12-vel, rendezed és kész.
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| Előzmény: [797] Suzy, 2005-02-27 13:42:05 |
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| [797] Suzy | 2005-02-27 13:42:05 |
 Van egy torony.Ennek a harmadrésze a földben negyede pedig a vízben van.És száz láb magasan emelkedik a víz fölé.Hány láb az egész?? Egyenletbe kéne felírni és megoldani. Légyszi ha tudja valaki írja le.De még ma.Köszi:suzy
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| [794] lorantfy | 2005-02-21 23:09:21 |
 Hello Zoli!
Nem tudom, hogy ezt nevezik-e a Newton-féle módszernek, de itt van egy egyszerű algoritmus:
A p számot elosztjuk egy olyan q számmal, melynek négyzete közel áll p-hez. A hányados legyen r. Vesszük a q és r számtani közepét, ez lesz az új q. És kezdjük előlről..
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| Előzmény: [793] SchZol, 2005-02-21 22:52:44 |
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Játékszabályok Technikai információ TeX tanfolyam Elfelejtettem a jelszavam Témák