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

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



You asked:

Evaluate the expression: 32+248+12\sqrt{32} + \frac{\sqrt{2}}{\sqrt{48}} + \sqrt{12}

MathBot Answer:

32+248+12=612+23+42≈9.32508000986206629044475458607603\sqrt{32} + \frac{\sqrt{2}}{\sqrt{48}} + \sqrt{12} = \frac{\sqrt{6}}{12} + 2 \sqrt{3} + 4 \sqrt{2} \approx 9.32508000986206629044475458607603

32+248+12=32+2⋅148+12=4⋅2+2⋅148+12=4⋅2+2⋅14⋅3+12=4⋅2+112⋅6+12=4⋅2+112⋅6+2⋅3=(4⋅2+112⋅6)+2⋅3=2⋅3+4⋅2+112⋅6\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