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Scholieren.be • View topic - wiskunde: ontbinden in factoren, merkwaardige producten

wiskunde: ontbinden in factoren, merkwaardige producten

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wiskunde: ontbinden in factoren, merkwaardige producten

Postby jelle8020 on Sun 17 Jun, 2007 18:21

ontbinden in factoren:

zou iemand mij alle soorten kunnen geven + uitleg per soort oif en stap voor stap. hetzelfde voor merkwaardige producten
ik zit in het 3 de middelbaar in de richting wetenschap 5 uur wiskunde.

alvast bedankt en l
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Postby Buys on Sun 17 Jun, 2007 18:37

Moet je op s.com vragen als je vandaag nog antwoord wilt :wink:
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Postby jelle8020 on Sun 17 Jun, 2007 18:48

:-? :-? :-?
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Postby Buys on Sun 17 Jun, 2007 18:49

scholieren.com :wink:
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Postby jelle8020 on Sun 17 Jun, 2007 19:32

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Postby vuurrobin on Sun 17 Jun, 2007 19:42

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Postby Buys on Sun 17 Jun, 2007 19:49

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Postby jelle8020 on Sun 17 Jun, 2007 20:06

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Postby Pierewiet on Sun 17 Jun, 2007 23:49

5x^2-19x+12 Zoek twee getallen die bij elkaar opgeteld -19 opleveren, dit is -5 en -4.
= 5x^2-15x-4x+12= (5x^2-15x)-(4x-12)
= 5x(x-3)-4(x-3)
= (5x-4)(x-3)

Andere methode:
5x^2-19x+12 vermenigvuldig 5x^2 met +12 => 5x^2*12=60x^2
De factoren van 60x^2 zijn
2x,30x,
3x,20x
4x,15x
-15x+-4x= -19x etc. etc.

Zie je het niet helemaal zitten?
gebruik de formule x = {-b +/- sqrt ((-b)^2-4ac)} / 2a.
sqrt=wortel a=5; b=-19 en c=12

1. x = [19 +/- sqrt(121)] / 10 = [19 +/- 11] / 10 = 3 of 4/5.
x= 3 of x = 4 / 5 geeft
(x - 3)(x - 4/5) = 0
(x - 3)*5(*(x - 4/5) = 5*0
(x - 3)(5x - 4) = 0

Suc6!
He who asks is a fool for 5 minutes, but he who doesn't ask remains a fool forever! ##Chinese Proverb##
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Postby jelle8020 on Mon 18 Jun, 2007 06:04

dank u voor de hulp allemaal
ik heb alles gevonden nu.
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