Getal & Ruimte (13e editie) - 1 vwo
'Rekenvolgorde'.
| 1 vwo | 2.1 Bewerkingen |
opgave 1Bereken. 1p a \(40 : 4 : 5\) PositiefDrieDelen (1) 00ah - Rekenvolgorde - basis - 0ms a \(40 : 4 : 5 = 10 : 5 = 2 \text{.}\) 1p 1p b \(4 + 5 ⋅ 2\) PositiefDrieDelen (2) 00ai - Rekenvolgorde - basis - 0ms b \(4 + 5 ⋅ 2 = 4 + 10 = 14 \text{.}\) 1p 1p c \((6 + 9) ⋅ 5\) PositiefDrieDelen (3) 00aj - Rekenvolgorde - basis - 0ms c \((6 + 9) ⋅ 5 = 15 ⋅ 5 = 75 \text{.}\) 1p 1p d \(36 : 9 ⋅ 3\) PositiefDrieDelen (4) 00ak - Rekenvolgorde - basis - 0ms d \(36 : 9 ⋅ 3 = 4 ⋅ 3 = 12 \text{.}\) 1p opgave 2Bereken. 1p a \(9 - 12 : 2\) PositiefDrieDelen (5) 00al - Rekenvolgorde - basis - 0ms a \(9 - 12 : 2 = 9 - 6 = 3 \text{.}\) 1p 1p b \(6 + 5 ⋅ (9 + 2)\) PositiefVierDelen (1) 00am - Rekenvolgorde - basis - 0ms b \(6 + 5 ⋅ (9 + 2) = 6 + 5 ⋅ 11 = 6 + 55 = 61 \text{.}\) 1p 1p c \((9 + 3) ⋅ (6 + 8)\) PositiefVierDelen (2) 00an - Rekenvolgorde - basis - 0ms c \((9 + 3) ⋅ (6 + 8) = 12 ⋅ 14 = 168 \text{.}\) 1p 1p d \(17 - 5 ⋅ 2 + 3\) PositiefVierDelen (3) 00ao - Rekenvolgorde - basis - 0ms d \(17 - 5 ⋅ 2 + 3 = 17 - 10 + 3 = 7 + 3 = 10 \text{.}\) 1p opgave 3Bereken. 1p \((19 - (4 + 7)) ⋅ 9\) PositiefVierDelen (4) 00ap - Rekenvolgorde - basis - 0ms ○ \((19 - (4 + 7)) ⋅ 9 = (19 - 11) ⋅ 9 = 8 ⋅ 9 = 72 \text{.}\) 1p |