To simplify the expression:

$\frac{\sqrt{-8}}{\sqrt{-2} \cdot \sqrt{-9}}$

We first use the property of square roots with negative numbers, which involves imaginary numbers ($i = \sqrt{-1}$).

1. Simplify $\sqrt{-8}$: $\sqrt{-8} = \sqrt{-1} \cdot \sqrt{8} = i \cdot \sqrt{8} = i \cdot 2\sqrt{2}$

2. Simplify $\sqrt{-2}$: $\sqrt{-2} = \sqrt{-1} \cdot \sqrt{2} = i \cdot \sqrt{2}$

3. Simplify $\sqrt{-9}$: $\sqrt{-9} = \sqrt{-1} \cdot \sqrt{9} = i \cdot 3 = 3i$

Now, substitute these values back into the original expression:

$\frac{i \cdot 2\sqrt{2}}{(i \cdot \sqrt{2}) \cdot (3i)}$

4. Simplify the denominator: $(i \cdot \sqrt{2}) \cdot (3i) = 3i^2 \cdot \sqrt{2} = 3(-1) \cdot \sqrt{2} = -3\sqrt{2}$

5. Simplify the entire expression: $\frac{i \cdot 2\sqrt{2}}{-3\sqrt{2}} = \frac{2i}{-3} = -\frac{2i}{3}$

So, $B = -\frac{2i}{3}$.

Would you like a more detailed explanation on any step or have further questions?

Here are 5 related questions to explore:

1. How do we handle square roots of negative numbers in general?
2. What is the imaginary unit $i$ and its significance in mathematics?
3. Can all negative square roots be simplified in this way?
4. How do we multiply complex numbers involving $i$?
5. What are some practical applications of complex numbers?

Tip: Remember, $i^2 = -1$ is crucial when simplifying expressions involving complex numbers.