Getal & Ruimte (13e editie) - 1 havo/vwo
'Rekenvolgorde'.
| 1 havo/vwo | 2.1 Bewerkingen |
opgave 1Bereken. 1p a \(24 : 4 : 3\) PositiefDrieDelen (1) 00ah - Rekenvolgorde - basis - 0ms a \(24 : 4 : 3 = 6 : 3 = 2 \text{.}\) 1p 1p b \(3 + 5 ⋅ 4\) PositiefDrieDelen (2) 00ai - Rekenvolgorde - basis - 0ms b \(3 + 5 ⋅ 4 = 3 + 20 = 23 \text{.}\) 1p 1p c \((3 + 9) ⋅ 2\) PositiefDrieDelen (3) 00aj - Rekenvolgorde - basis - 0ms c \((3 + 9) ⋅ 2 = 12 ⋅ 2 = 24 \text{.}\) 1p 1p d \(24 : 6 ⋅ 3\) PositiefDrieDelen (4) 00ak - Rekenvolgorde - basis - 0ms d \(24 : 6 ⋅ 3 = 4 ⋅ 3 = 12 \text{.}\) 1p opgave 2Bereken. 1p a \(9 - 16 : 2\) PositiefDrieDelen (5) 00al - Rekenvolgorde - basis - 0ms a \(9 - 16 : 2 = 9 - 8 = 1 \text{.}\) 1p 1p b \(6 + 2 ⋅ (9 + 3)\) PositiefVierDelen (1) 00am - Rekenvolgorde - basis - 0ms b \(6 + 2 ⋅ (9 + 3) = 6 + 2 ⋅ 12 = 6 + 24 = 30 \text{.}\) 1p 1p c \((7 + 6) ⋅ (5 + 3)\) PositiefVierDelen (2) 00an - Rekenvolgorde - basis - 0ms c \((7 + 6) ⋅ (5 + 3) = 13 ⋅ 8 = 104 \text{.}\) 1p 1p d \(10 - 2 ⋅ 3 + 8\) PositiefVierDelen (3) 00ao - Rekenvolgorde - basis - 0ms d \(10 - 2 ⋅ 3 + 8 = 10 - 6 + 8 = 4 + 8 = 12 \text{.}\) 1p opgave 3Bereken. 1p \((16 - (9 + 5)) ⋅ 3\) PositiefVierDelen (4) 00ap - Rekenvolgorde - basis - 0ms ○ \((16 - (9 + 5)) ⋅ 3 = (16 - 14) ⋅ 3 = 2 ⋅ 3 = 6 \text{.}\) 1p |