\[ \begin{array}{l} =(6+2)+4= \\ (16+2)+4=6+(2+4) \quad 12=12 \\ 8 \times(10 \times 4)= \end{array} \] Solve it: b. \( (7+3)+1= \) d. \( 4 \times(5 \times 2)= \) \[ 11 \times 31 \times 2= \]

Sure, let's solve each part step by step: --- **b. \( (7 + 3) + 1 \)** 1. **Add the numbers inside the first parentheses:** \[ 7 + 3 = 10 \] 2. **Add the result to 1:** \[ 10 + 1 = 11 \] **Answer:** \( 11 \) --- **d. \( 4 \times (5 \times 2) \)** 1. **Multiply the numbers inside the parentheses first:** \[ 5 \times 2 = 10 \] 2. **Multiply the result by 4:** \[ 4 \times 10 = 40 \] **Answer:** \( 40 \) --- **\( 11 \times 31 \times 2 \)** 1. **First, multiply 11 by 31:** \[ 11 \times 31 = 341 \] 2. **Then, multiply the result by 2:** \[ 341 \times 2 = 682 \] **Answer:** \( 682 \) ---