Simplify $\frac{4}{3x^{-3}} \cdot \frac{3x^{2}}{2}$.
First, we recognize that $\frac{4}{3x^{-3}}$ can be rewritten as $\frac{4x^3}{3}$. Thus, we have  \begin{align*}
\frac{4}{3x^{-3}} \cdot \frac{3x^{2}}{2} & = \frac{4x^3}{3} \cdot \frac{3x^2}{2} \\
& = \frac{(4 \cdot 3)(x^3 \cdot x^2)}{3 \cdot 2} \\
& = 2x^{3+2} \\
& = \boxed{2x^5}.
\end{align*}
2x^5
Level 3
Algebra
Problem: Simplify $\frac{4}{3x^{-3}} \cdot \frac{3x^{2}}{2}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first, rewrite the term with the negative exponent. The term $x^{-3}$ can be expressed as $\frac{1}{x^3}$. Thus, $\frac{4}{3x^{-3}}$ becomes $\frac{4}{3} \cdot x^3$. 

Next, multiply this by the second fraction $\frac{3x^{2}}{2}$. The multiplication of fractions involves multiplying the numerators and the denominators separately. 

The numerator becomes $4 \cdot 3x^{2} = 12x^{2}$. The denominator becomes $3 \cdot 2 = 6$. Therefore, the expression simplifies to $\frac{12x^{2}}{6x^{3}}$.

Now, simplify the fraction $\frac{12}{6}$, which equals $2$. The variable part simplifies as follows: $x^{2} \div x^{3} = x^{2-3} = x^{-1}$. 

Combining these results, we have $2x^{-1}$. To express this in a more standard form, rewrite $x^{-1}$ as $\frac{1}{x}$. Thus, $2x^{-1}$ can be rewritten as $\frac{2}{x}$.

However, we need to consider the multiplication of the original terms more carefully. The correct approach is to combine the fractions before simplifying. 

Revisiting the multiplication, we have $\frac{4 \cdot 3x^{2}}{3 \cdot 2x^{-3}}$. The $3$ in the numerator and denominator cancels out, leaving $\frac{4x^{2}}{2x^{-3}}$. 

Now, simplify $\frac{4}{2}$ to get $2$. The variable part becomes $x^{2} \cdot x^{3} = x^{2+3} = x^{5}$. 

Thus, the final simplified expression is $2x^{5}$.<|END_OF_LATENT|> The final answer is $2x^5$.