Getal & Ruimte (13e editie) - 1 vwo
'Rekenvolgorde'.
| 1 vwo | 2.1 Bewerkingen |
opgave 1Bereken. 1p a \(40 : 5 : 4\) PositiefDrieDelen (1) 00ah - Rekenvolgorde - basis - 0ms a \(40 : 5 : 4 = 8 : 4 = 2 \text{.}\) 1p 1p b \(7 + 2 ⋅ 5\) PositiefDrieDelen (2) 00ai - Rekenvolgorde - basis - 0ms b \(7 + 2 ⋅ 5 = 7 + 10 = 17 \text{.}\) 1p 1p c \((3 + 7) ⋅ 4\) PositiefDrieDelen (3) 00aj - Rekenvolgorde - basis - 0ms c \((3 + 7) ⋅ 4 = 10 ⋅ 4 = 40 \text{.}\) 1p 1p d \(48 : 6 ⋅ 3\) PositiefDrieDelen (4) 00ak - Rekenvolgorde - basis - 0ms d \(48 : 6 ⋅ 3 = 8 ⋅ 3 = 24 \text{.}\) 1p opgave 2Bereken. 1p a \(8 - 12 : 3\) PositiefDrieDelen (5) 00al - Rekenvolgorde - basis - 0ms a \(8 - 12 : 3 = 8 - 4 = 4 \text{.}\) 1p 1p b \(4 + 5 ⋅ (7 + 2)\) PositiefVierDelen (1) 00am - Rekenvolgorde - basis - 0ms b \(4 + 5 ⋅ (7 + 2) = 4 + 5 ⋅ 9 = 4 + 45 = 49 \text{.}\) 1p 1p c \((5 + 8) ⋅ (2 + 4)\) PositiefVierDelen (2) 00an - Rekenvolgorde - basis - 0ms c \((5 + 8) ⋅ (2 + 4) = 13 ⋅ 6 = 78 \text{.}\) 1p 1p d \(22 - 7 ⋅ 2 + 4\) PositiefVierDelen (3) 00ao - Rekenvolgorde - basis - 0ms d \(22 - 7 ⋅ 2 + 4 = 22 - 14 + 4 = 8 + 4 = 12 \text{.}\) 1p opgave 3Bereken. 1p \((16 - (5 + 9)) ⋅ 6\) PositiefVierDelen (4) 00ap - Rekenvolgorde - basis - 0ms ○ \((16 - (5 + 9)) ⋅ 6 = (16 - 14) ⋅ 6 = 2 ⋅ 6 = 12 \text{.}\) 1p |