moderne wiskunde 11e ed havo 4 a uitwerkingen vaardigheden 1
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HAVO
Wiskunde A
4
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Vaardigheden 1
Pagina ó0
1a 3(x+2)=3x+6 d A(A-6)=A2-6A
b p(1,5-p)=1,5p-p' e 7,2(2x + 1,24) = 14,4x + 8,928
6(10 - ïà=60-so f x+2(x + 1) - y*2x+2- 3x+2
7x=119 9-x=ll
x=lJ (1inks en rechts : 7) -x-z (links en rechts -9)
x+10=27 ..- 1 (links en rechts x -1)
x=11 (links en rechts -10) f 2(t+51=24
1,2-p-0,3 2t+70=24 (haakjes weggewerkt)
-p = -0,9 (links en rechts -1,2) 2t=74 (links en rechts -10)
p = 0,9 (links en rechts x -1) l- I (links en rechts : 2)
Itu=l
k = 14 (links en rechts x 2)
3a Op de plaats van het bordje moet 6 staan, want f; = 3.
b l+2x-6
2x = 5 (links en rechts -1)
x = 2ï (links en rechts : 2)
18 18
,
't +'t."t!
=-=3.dusklopt.
6
In deze vergelijking komt de variabele x twee keer voor. Dan kun je de
methode van de bordjes niet gebruiken.
b Alex krijgt nu de vergelijking 2x - 2 = 8.
G 2x-2=8
2x = 70 (links en rechts + 2)
x=5 (links en rechts : 2)
Rechts van het = teken geeft 6, 5 - 2 = 30 - 2 = 28 en links van het = teken
geeft4.5+8=20+8=28
De uitkomstenzijngelijk, dus x = 5 is de oplossing van de vergelijking.
Pagina ó1
s(2\x + 6) = JJ
2l x + 6 = 11 (bordje gelegd op 2+-rr + 6)
2, x = 5 (links en rechts -6)
x=2 (links en rechts :2\)
Controle: sQr'2 + 6) = 5 ' 11 = 55, dus klopt.
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