The expression you’ve provided involves a square root of a negative number, which will yield an imaginary number. Let’s work through it step-by-step:

Given: $\frac{\sqrt{6}}{3\sqrt{-9}}$

1. Simplify the square root of -9: $\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i$ where $i$ is the imaginary unit, defined as $i = \sqrt{-1}$.

2. Substitute back into the expression: $\frac{\sqrt{6}}{3 \cdot 3i} = \frac{\sqrt{6}}{9i}$

3. Simplify by rationalizing the denominator (multiply numerator and denominator by $i$): $\frac{\sqrt{6}}{9i} \cdot \frac{i}{i} = \frac{\sqrt{6} \cdot i}{9i^2}$

4. Since $i^2 = -1$, this becomes: $= \frac{\sqrt{6} \cdot i}{9 \cdot -1} = -\frac{\sqrt{6} \cdot i}{9} = -\frac{\sqrt{6}}{9}i$

So, the simplified result is: $-\frac{\sqrt{6}}{9}i$

Would you like further details or have any questions?

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Related Questions

1. How do we handle imaginary numbers in division?
2. What is the significance of rationalizing the denominator?
3. How is $i$ defined and why does $i^2 = -1$?
4. Can you simplify other expressions with square roots of negative numbers?
5. What is the practical use of imaginary numbers in real-world applications?

Tip: Always remember to rationalize the denominator in complex expressions for cleaner, standardized results.