Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №1835518
А
Аноним
13 октября 2024 16:44 · публичный вопрос
А
Аноним

2,690*5/7-9,500:3+(56,3:8/7)

Нейросеть13 октября 2024 16:44

Let's solve the expression step by step:

  1. Calculate \(2,690 \times 5\): \[ 2,690 \times 5 = 13,450 \]

  2. Divide by 7: \[ \frac{13,450}{7} = 1,921.42857 \quad (\text{approximately}) \]

  3. Calculate \(9,500 \div 3\): \[ 9,500 \div 3 = 3,166.66667 \quad (\text{approximately}) \]

  4. Calculate \(56.3 \div 8\): \[ 56.3 \div 8 = 7.0375 \]

  5. Now divide by 7: \[ \frac{7.0375}{7} = 1.004285714 \quad (\text{approximately}) \]

  6. Combine all parts: Now substitute back into the main expression: \[ 1,921.42857 - 3,166.66667 + 1.004285714 \]

    • First perform \(1,921.42857 - 3,166.66667\): \[ 1,921.42857 - 3,166.66667 = -1,245.2381 \quad (\text{approximately}) \]

    • Now add \(1.004285714\): \[ -1,245.2381 + 1.004285714 = -1,244.233814286 \quad (\text{approximately}) \]

So the final result of the expression \(2,690 \times 5 \div 7 - 9,500 \div 3 + (56.3 \div 8 \div 7)\) is approximately: \[ \boxed{-1,244.23} \]

2,690*5/7-9,500:3+(56,3:8/7)
Let's solve the expression step by step: 1. **Calculate \(2,690 \times 5\)**: \[ 2,690 \times 5 = 13,450 \] 2. **Divide by 7**: \[ \frac{13,450}{7} = 1,921.42857 \quad (\text{approximately}) \] 3. **Calculate \(9,500 \div 3\)**: \[ 9,500 \div 3 = 3,166.66667 \quad (\text{approximately}) \] 4. **Calculate \(56.3 \div 8\)**: \[ 56.3 \div 8 = 7.0375 \] 5. **Now divide by 7**: \[ \frac{7.0375}{7} = 1.004285714 \quad (\text{approximately}) \] 6. **Combine all parts**: Now substitute back into the main expression: \[ 1,921.42857 - 3,166.66667 + 1.004285714 \] - First perform \(1,921.42857 - 3,166.66667\): \[ 1,921.42857 - 3,166.66667 = -1,245.2381 \quad (\text{approximately}) \] - Now add \(1.004285714\): \[ -1,245.2381 + 1.004285714 = -1,244.233814286 \quad (\text{approximately}) \] So the final result of the expression \(2,690 \times 5 \div 7 - 9,500 \div 3 + (56.3 \div 8 \div 7)\) is approximately: \[ \boxed{-1,244.23} \]