\( \left(3^{4} \div 3^{2}+12\right)+\left(2^{4} \div 2^{3}+3 \times 10\right) \) \( \begin{array}{lll}\text { A) } 53 & \text { B) } 52 & \text { C) } 44 \\ \text { D) } 43 & & \text { E) } 63\end{array} \)
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To solve the expression \( \left(3^{4} \div 3^{2}+12\right)+\left(2^{4} \div 2^{3}+3 \times 10\right) \), we first break it down step by step. 1. Calculate \( 3^{4} \div 3^{2} \): \[ 3^{4} \div 3^{2} = 3^{4-2} = 3^{2} = 9 \] Then add 12: \[ 9 + 12 = 21 \] 2. Now calculate \( 2^{4} \div 2^{3} \): \[ 2^{4} \div 2^{3} = 2^{4-3} = 2^{1} = 2 \] Then calculate \( 3 \times 10 \): \[ 3 \times 10 = 30 \] Now add these two results: \[ 2 + 30 = 32 \] 3. Finally, we add the results from both parts: \[ 21 + 32 = 53 \] So, the answer is \( \boxed{53} \).