$\sqrt{\32}$+$\sqrt{\2}$/$\sqrt{\48}$+$\sqrt{\12}$

asked by guest
on Dec 27, 2024 at 1:00 am



You asked:

Evaluate the expression: \(\sqrt{32} + \frac{\sqrt{2}}{\sqrt{48}} + \sqrt{12}\)

MathBot Answer:

\[\sqrt{32} + \frac{\sqrt{2}}{\sqrt{48}} + \sqrt{12} = \frac{\sqrt{6}}{12} + 2 \sqrt{3} + 4 \sqrt{2} \approx 9.32508000986206629044475458607603\]

\[\begin{aligned}\sqrt{32} + \frac{\sqrt{2}}{\sqrt{48}} + \sqrt{12}& = \sqrt{32} + \sqrt{2} \cdot \frac{1}{\sqrt{48}} + \sqrt{12}\\& = 4 \cdot \sqrt{2} + \sqrt{2} \cdot \frac{1}{\sqrt{48}} + \sqrt{12}\\& = 4 \cdot \sqrt{2} + \sqrt{2} \cdot \frac{1}{4 \cdot \sqrt{3}} + \sqrt{12}\\& = 4 \cdot \sqrt{2} + \frac{1}{12} \cdot \sqrt{6} + \sqrt{12}\\& = 4 \cdot \sqrt{2} + \frac{1}{12} \cdot \sqrt{6} + 2 \cdot \sqrt{3}\\& = \left(4 \cdot \sqrt{2} + \frac{1}{12} \cdot \sqrt{6}\right) + 2 \cdot \sqrt{3}\\& = 2 \cdot \sqrt{3} + 4 \cdot \sqrt{2} + \frac{1}{12} \cdot \sqrt{6}\end{aligned}\]

+ Submit Question