Question
1. \( \frac{2^{n} \sqrt{32}+2^{n} \sqrt{2}}{2^{n} \sqrt{50}} \)
Ask by John Cole. in South Africa
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The expression simplifies to 1.
Solution
1. Factor out \(2^n\) from the numerator:
\[
\frac{2^n \sqrt{32} + 2^n \sqrt{2}}{2^n \sqrt{50}} = \frac{2^n (\sqrt{32} + \sqrt{2})}{2^n \sqrt{50}}.
\]
2. Cancel the common term \(2^n\) in the numerator and denominator:
\[
\frac{\sqrt{32} + \sqrt{2}}{\sqrt{50}}.
\]
3. Simplify the square roots:
\[
\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}, \quad \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}.
\]
4. Substitute the simplified forms into the expression:
\[
\frac{4\sqrt{2} + \sqrt{2}}{5\sqrt{2}} = \frac{(4 + 1)\sqrt{2}}{5\sqrt{2}} = \frac{5\sqrt{2}}{5\sqrt{2}}.
\]
5. Since the numerator and denominator are equal, the expression simplifies to:
\[
1.
\]
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Simplify this solution Bonus Knowledge
To simplify the expression \( \frac{2^{n} \sqrt{32}+2^{n} \sqrt{2}}{2^{n} \sqrt{50}} \), we can factor out \(2^{n}\) from the numerator. This gives us: \[ \frac{2^{n} (\sqrt{32} + \sqrt{2})}{2^{n} \sqrt{50}} = \frac{\sqrt{32} + \sqrt{2}}{\sqrt{50}}. \] Now, calculate the square roots: \[ \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \quad \text{and} \quad \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}. \] Substituting these back into the expression gives: \[ \frac{4\sqrt{2} + \sqrt{2}}{5\sqrt{2}} = \frac{(4 + 1)\sqrt{2}}{5\sqrt{2}} = \frac{5\sqrt{2}}{5\sqrt{2}} = 1. \] The final answer is \(1\).
