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Samenvatting Antwoordenboek Getal & Ruimte 2 Havo/Vwo Deel 1 - Wiskunde
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Antwoorden van Wiskunde havo-vwo 2 getal in ruimte 13de editie a en / of 1
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Getal & RuimteUitwerkingen
2 havo/vwo deel 1
Dertiende editie, 2022 Auteurs
Noordhoff Groningen J. H. Dijkhuis
G. de Jong
H. J. Houwing
J. D. Kuis
F. ten Klooster
S. K. A. de Waal
J. van Braak
J. H. M. Liesting-Maas
M. Wieringa
R. D. Hiele
J. E. Romkes
M. Haneveld
S. Voets
M. Vos
J. M. M. van Haren
B. W. van Laarhoven
R. Meijerink
E. Terlaak
© Noordhoff Uitgevers bv
,
,Inhoud
1 Rekenen met letters 4
2 Afstand en oppervlakte 22
3 Lineaire formules en vergelijkingen 77
4 Procenten en diagrammen 106
Algemene vaardigheden 131
© Noordhoff Uitgevers bv
, 1 Rekenen met letters
Voorkennis Letterrekenen
Bladz�de 10
1 a 6a · 7b = 42ab e r · 2p · q = 2pqr
b −3b · 8a = −24ab f 4p · 2p · 3q = 24p2q
c 2y · −5y = −10y2 g −m · −1 · −n = −mn
d 7 · −3pq = −21pq h −5ab · −3a = 15a2b
2 a 3x + 4x = 7x d 12pq − 5pq = 7pq
b 8y − 6y = 2y e −3x + 7x − 3x = x
c 3p + 5p + 9p = 17p f −6p + 7p − p = 0
3 a 5xy + 4xy = 9xy d −2abc + 18abc = 16abc
b 15ab + 40ac kan niet e 3def − 8de kan niet
c 4m + 12 kan niet f pq + pq = 2pq
Bladz�de 11
4 a 6a − 5a = a d 10x + 2x = 12x g 8p · 7q = 56pq
b 6a · 5b = 30ab e 10x · 2x = 20x2 h 8p − 7q kan niet
c 6a + 5b kan niet f 10x − 2x = 8x i −8p · 7p = −56p2
5 a 4x − 5y + 7x + 3y = c −4x − 5y + 7x + 3y = e 4ab − 3ab + a + 3ab =
11x − 2y 3x − 2y 4ab + a
b 4x − 5y − 7x − 3y = d 4a − 7 + 5a = f −p + 8p − 2qr − 9p =
−3x − 8y 9a − 7 −2p − 2qr
6 a 2x · 8y + 7x · 3y = d −2p · 3q − 3q · 8p = g 2x · 3x − 3x · −5x =
16xy + 21xy = 37xy −6pq − 24pq = −30pq 6x2 + 15x2 = 21x2
b 3a · 5b − 9b · a = e 2p · 3q − 3q · 8p = h −4a · 3b · 2c + 25abc =
15ab − 9ab = 6ab 6pq − 24pq = −18pq −24abc + 25abc = abc
c 4x · 6 − 5 · 2x = f 3a · 8a + 12a · 2a = i 5 · −2x − 3x · −6x =
24x − 10x = 14x 24a2 + 24a2 = 48a2 −10x + 18x2
7 a 2a · 7b + 3a · 4b = d 2a · 7b − 3a · 4b = g 2a · 8 + 4 · 3a =
14ab + 12ab = 26ab 14ab − 12ab = 2ab 16a + 12a = 28a
b 2a + 7b + 3a + 4b = e 2a · −7b − 3a · −4b = h 2a + 8 + 4 + 3a =
5a + 11b −14ab + 12ab = −2ab 5a + 12
c 2+a+7+b= f 2a · 7a + 3a · 4a = i 2a + 8 + 4 · 3a =
9+a+b 14a2 + 12a2 = 26a2 2a + 8 + 12a = 14a + 8
1.1 Haakjes wegwerken
Bladz�de 12
1 a opp. rechthoek I = a · b = ab
opp. rechthoek II = a · c = bc
b opp. rechthoek ABCD = opp. rechthoek I + opp. rechthoek II, dus
a(b + c) = ab + ac.
2 Lilly en Claire hebben beiden gelijk, want 5 − 2(4x − 6) = 5 − 8x + 12 = 17 − 8x
en dat is gelijk aan −8x + 17.
Bladz�de 13
3 a 3(a + 2b) = 3a + 6b c x(5x + 3y) = 5x2 + 3xy
b 4(6a − 9) = 24a − 36 d 2p(5p − 1) = 10p2 − 2p
4 Hoofdstuk 1 © Noordhoff Uitgevers bv
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