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Certainly! Let's solve the given problem step by step.
### Problem
We are tasked with solving the expressions [tex]\[2^{2+1}\][/tex] and [tex]\[e^{1-e}\][/tex]
### Step-by-Step Solution
1. Evaluating [tex]\(2^{2+1}\)[/tex]: - First, let's simplify the exponent [tex]\(2 + 1\)[/tex]: [tex]\[2 + 1 = 3\][/tex] - Next, we raise 2 to the power of 3: [tex]\[2^3 = 2 \times 2 \times 2\][/tex] - Performing the multiplication: [tex]\[2 \times 2 = 4\][/tex] [tex]\[4 \times 2 = 8\][/tex] - Thus, [tex]\[2^{2+1} = 8\][/tex]
2. Evaluating [tex]\(e^{1-e}\)[/tex]: - Here, [tex]\(e\)[/tex] is the base of the natural logarithm, approximately equal to [tex]\(2.71828\)[/tex]. - The expression we need to evaluate is [tex]\(e\)[/tex] raised to the power of [tex]\((1 - e)\)[/tex]. - Numerically, this evaluates to: [tex]\[e^{1-e} \approx 0.17937407873401723\][/tex]
### Final Answer - The value of [tex]\(2^{2+1}\)[/tex] is [tex]\(8\)[/tex]. - The value of [tex]\(e^{1-e}\)[/tex] is approximately [tex]\(0.17937407873401723\)[/tex].
Therefore, the results are: [tex]\[2^{2+1} = 8\][/tex] [tex]\[e^{1-e} \approx 0.17937407873401723\][/tex]
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