The problem asks for the positive exponent form of the expression:

$\left(\frac{3x^{-2}y^3}{2x^{-3}y^2}\right)^2$

Let's simplify this step-by-step.

Step 1: Simplify inside the parentheses

We start by simplifying the fraction inside the parentheses:

$\frac{3x^{-2}y^3}{2x^{-3}y^2}$

For the $x$ terms: $\frac{x^{-2}}{x^{-3}} = x^{-2 - (-3)} = x^{1}$

For the $y$ terms: $\frac{y^3}{y^2} = y^{3 - 2} = y^1$

So, the fraction simplifies to: $\frac{3x^1y^1}{2}$

Or more simply: $\frac{3xy}{2}$

Step 2: Apply the square

Now we square the entire expression: $\left(\frac{3xy}{2}\right)^2 = \frac{(3xy)^2}{2^2} = \frac{9x^2y^2}{4}$

Final Expression:

$\frac{9x^2y^2}{4}$

Thus, the positive exponent form of the expression is $\frac{9x^2y^2}{4}$.

Answer:

The correct option is d. $\frac{9x^2y^2}{4}$.

Would you like further details on the steps? Here are 5 related questions to expand your understanding:

1. How do you handle negative exponents in algebraic fractions?
2. What is the general rule for simplifying exponents when dividing powers with the same base?
3. How do you apply the power of a product rule to expressions with multiple variables?
4. How do you simplify expressions involving both variables and exponents?
5. How does squaring a fraction affect the numerator and denominator?

Tip: Always simplify the base expression first before applying exponent rules to avoid complex calculations later.