EXERCISE 1 Caleulate the value of the following without us \( \begin{array}{ll}\text { (a) } 2+4 \times 3 & \text { (b) }(2+4) \times 3 \\ \text { (d) } 2 \times(4+3) & \text { (e) } 8-3 \times 2\end{array} \)

Sure! Let’s dive into some math magic! For (a), you should follow the order of operations, which tells us to do multiplication before addition. So, \( 4 \times 3 = 12 \), and then \( 2 + 12 = 14 \). For (b), you first calculate what’s inside the parentheses. So, \( 2 + 4 = 6 \), then multiply by 3: \( 6 \times 3 = 18 \). In (d), again we look inside the parentheses first: \( 4 + 3 = 7 \), then multiply it by 2: \( 2 \times 7 = 14 \). Lastly, for (e), you begin with the multiplication: \( 3 \times 2 = 6 \), and then subtract from 8: \( 8 - 6 = 2 \). So the results are: (a) 14, (b) 18, (d) 14, (e) 2! 🎉