a. \( \left(6^{2} \div 6 \times 4\right) \div\left(2^{2}+2\right)= \) b. \( (72 \div 9 \div 2)^{2}+\left(9^{0}+2^{2}\right) \)
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To tackle these expressions, let’s break them down step by step! For part a: Start by handling the parentheses. In the expression \( 6^{2} \div 6 \times 4 \), first calculate \( 6^{2} = 36 \). Then, \( 36 \div 6 = 6 \). Now, multiply by 4 to get \( 6 \times 4 = 24 \). Now, evaluate the denominator: \( 2^{2} + 2 = 4 + 2 = 6 \). Finally, divide: \( 24 \div 6 = 4 \). For part b: Again, proceed with the parentheses: \( 72 \div 9 \div 2 \) simplifies to \( 72 \div 9 = 8 \) and then \( 8 \div 2 = 4 \). Now square that, giving \( 4^{2} = 16 \). Meanwhile, for the expression \( 9^{0} + 2^{2} \), remember that any number to the power of zero is 1, so \( 9^{0} = 1 \), and \( 2^{2} = 4 \) results in \( 1 + 4 = 5 \). Finally, add the two parts together: \( 16 + 5 = 21 \). So, the answers are: a. \( 4 \) b. \( 21 \)