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Fórum ExoSpace.cz Fórum čtenářů ExoSpace.cz 2015-05-30T01:59:05+01:00 https://forum.exospace.cz/feed.php?f=115 2015-05-30T01:59:05+01:00 2015-05-30T01:59:05+01:00 https://forum.exospace.cz/viewtopic.php?t=227&p=805#p805 <![CDATA[Koutek na hraní • Hrátky s magnetickou plastelínou]]>

Statistiky: Napsal od Petr Valach — 30. kvě 2015, 01:59


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2015-05-05T07:55:28+01:00 2015-05-05T07:55:28+01:00 https://forum.exospace.cz/viewtopic.php?t=315&p=743#p743 <![CDATA[Koutek na hraní • Jsou všechna tělesa magnetická?]]>

Statistiky: Napsal od Petr Valach — 05. kvě 2015, 07:55


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2015-03-07T00:00:24+01:00 2015-03-07T00:00:24+01:00 https://forum.exospace.cz/viewtopic.php?t=255&p=564#p564 <![CDATA[Koutek na hraní • Ručičky na ciferníku]]> Statistiky: Napsal od Peter Kvačkay — 07. bře 2015, 00:00


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2015-03-06T18:37:33+01:00 2015-03-06T18:37:33+01:00 https://forum.exospace.cz/viewtopic.php?t=250&p=559#p559 <![CDATA[Koutek na hraní • Re: Koláčiky v zime]]> https://mega.co.nz/#!ZIFRFbQa!IEm83mXAH ... IsgwLYFEX4

Statistiky: Napsal od Petr Valach — 06. bře 2015, 18:37


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2015-02-25T21:56:40+01:00 2015-02-25T21:56:40+01:00 https://forum.exospace.cz/viewtopic.php?t=250&p=524#p524 <![CDATA[Koutek na hraní • Koláčiky v zime]]> Adam robí koláčiky v tvare kocky a Jozef v tvare guľôčky.
S koláčmi idú von, kde je 5 °C. Jeden centimeter štvorcový povrchu koláča odovzdá okoliu 0,5°C za pol hodinu.
Z trúby ich obaja vyberú naraz rozohriate na 100°C. Po hodine pobytu v zime sa jeden z nich sťažuje, že jeho koláčiky majú
už len 33°C. Kto to je, a koľko stupňov majú koláčiky toho druhého? (zaokrúhlené na celé °C)

Statistiky: Napsal od Peter Kvačkay — 25. úno 2015, 21:56


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2015-02-24T21:19:59+01:00 2015-02-24T21:19:59+01:00 https://forum.exospace.cz/viewtopic.php?t=244&p=520#p520 <![CDATA[Koutek na hraní • Re: Zobrazenie bodu v rovine]]> p. Jeho vzdialenosť od priamky p bude |M'p| = |Mp|, pričom M M'. Zo zadania si spravíme smernicový tvar priamky (y = k × x + q): y = x + 4. Vidíme, že k = 1. Smerový vektor tejto priamky je [a;b] = [1;1], z čoho vieme získať normálový vektor priamky (vektor kolmý na našu priamku) vymenením koeficientov a a b a zmenením znamienka na jednom z nich (nezáleží, na ktorom), takže
[a;b] => [b;-a],
[1;1] => [1;-1].
Vydelením y-ovej súradnice dostaneme smernicu, čiže v tomto prípade -1 : 1 = -1.
Tento postup by sme vedeli použiť na akúkoľvek priamku, ale keďže táto priamka je veľmi jednoduchá, stačí nám zmeniť znamienko pred x, teda dostaneme priamku y = -x + 4. Kolmú priamku máme, teraz potrebujeme, aby prechádzala cez bod M[1;1]. Smernicu k, x a y už poznáme, potrebujeme zistiť koeficient q. Keďže táto priamka je rovnobežná s priamkou ktorú hľadáme, musíme ju posunúť nižšie zmenou koeficientu q. Po nahradení y = k × x + q známymi členmi (y = 1, x = 1, k = -1) dostaneme:
1 = -1 + q
q = 2.
Máme rovnicu priamky (nazvime si ju n), ktorá prechádza bodom M: y = -x + 2. Potrebujeme teraz tento bod presunúť po tejto priamke o vzdialenosť 2 × |Mp|. Označme si priesečník našich priamok p a n Z. Jeho súradnice dostaneme vypočítaním ako bod, kde sa hodnota jednej funkcie rovná hodnote druhej funkcie:
x + 4 = -x + 2
x = -1.
Hodnotu x dosadíme do jednej z rovníc (ľubovoľne, ktoré) a máme y-ovú súradnicu:
y = x + 4
y = 3
Z[-1;3]
Bod M posunieme o vektor MZ 2krát. Súradnice vektora vypočítame ako rozdiel súradníc bodu Z a bodu M:
MZ[-1-1;3-1],
MZ[-2;2].
Teraz k bodu M pripočítame 2krát tento vektor a dostaneme sa k nášmu obrazu M'.
M'[1 + 2× (-2); 1 + (2 × 2)].

M'[-3;5]

Statistiky: Napsal od Peter Kvačkay — 24. úno 2015, 21:19


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2015-02-19T20:26:32+01:00 2015-02-19T20:26:32+01:00 https://forum.exospace.cz/viewtopic.php?t=244&p=500#p500 <![CDATA[Koutek na hraní • Zobrazenie bodu v rovine]]> p danú rovnicou x – y + 4 = 0 a bod M[1;1]. Zistite súradnice bodu M' zobrazeného osou súmernosti p.

Statistiky: Napsal od Peter Kvačkay — 19. úno 2015, 20:26


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2015-01-26T18:03:01+01:00 2015-01-26T18:03:01+01:00 https://forum.exospace.cz/viewtopic.php?t=227&p=378#p378 <![CDATA[Koutek na hraní • Pokusy s magnety]]>

magnet.jpg
magnets_cents.jpg
magnet_cents.jpg

Statistiky: Napsal od Petr Valach — 26. led 2015, 18:03


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2015-01-26T14:45:22+01:00 2015-01-26T14:45:22+01:00 https://forum.exospace.cz/viewtopic.php?t=220&p=377#p377 <![CDATA[Koutek na hraní • Re: Matematická úloha - obsah trojúhelníku]]> x=4-y
x^2=(2^2)+y^2.

y nám vyjde 1,5, takže S=1,5*2=3.
reseni.jpg

Statistiky: Napsal od Peter Kvačkay — 26. led 2015, 14:45


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