Moderne Wiskunde (13e editie) - 2 vmbo k(gt)
'Rekenvolgorde'.
| 2 vmbo k(gt) | 1.5 Rekenvolgorde |
opgave 1Bereken. 1p a \(270 : 6 : 5\) PositiefDrieDelen (1) 00ah - Rekenvolgorde - basis - 1ms a \(270 : 6 : 5 = 45 : 5 = 9 \text{.}\) 1p 1p b \(4 + 6 ⋅ 2\) PositiefDrieDelen (2) 00ai - Rekenvolgorde - basis - 1ms b \(4 + 6 ⋅ 2 = 4 + 12 = 16 \text{.}\) 1p 1p c \((6 + 2) ⋅ 8\) PositiefDrieDelen (3) 00aj - Rekenvolgorde - basis - 0ms c \((6 + 2) ⋅ 8 = 8 ⋅ 8 = 64 \text{.}\) 1p 1p d \(20 : 5 ⋅ 3\) PositiefDrieDelen (4) 00ak - Rekenvolgorde - basis - 0ms d \(20 : 5 ⋅ 3 = 4 ⋅ 3 = 12 \text{.}\) 1p opgave 2Bereken. 1p a \(7 - 30 : 5\) PositiefDrieDelen (5) 00al - Rekenvolgorde - basis - 0ms a \(7 - 30 : 5 = 7 - 6 = 1 \text{.}\) 1p 1p b \((6 + 8) ⋅ (5 + 2)\) PositiefVierDelen (2) 00an - Rekenvolgorde - basis - 0ms b \((6 + 8) ⋅ (5 + 2) = 14 ⋅ 7 = 98 \text{.}\) 1p 1p c \(79 - 9 ⋅ 8 + 5\) PositiefVierDelen (3) 00ao - Rekenvolgorde - basis - 0ms c \(79 - 9 ⋅ 8 + 5 = 79 - 72 + 5 = 7 + 5 = 12 \text{.}\) 1p |
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| 2 vmbo k(gt) | 8.vk Voorkennis |
opgave 1Bereken. 1p \(5 + 3 ⋅ (7 + 6)\) PositiefVierDelen (1) 00am - Rekenvolgorde - basis - 0ms ○ \(5 + 3 ⋅ (7 + 6) = 5 + 3 ⋅ 13 = 5 + 39 = 44 \text{.}\) 1p |