1. **State the problem:** Determine the sign of the expression $$\frac{x}{y} \cdot 7y^3$$ given that $$x < 0$$ and $$y > 0$$. 2. **Rewrite the expression:** $$\frac{x}{y} \cdot 7y^3 = 7x \cdot \frac{y^3}{y} = 7x y^{3-1} = 7x y^2$$ 3. **Analyze the signs:** - Since $$x < 0$$, $$x$$ is negative. - Since $$y > 0$$, $$y^2 = y \cdot y$$ is positive because the square of a positive number is positive. - The constant 7 is positive. 4. **Combine the signs:** - The product $$7x y^2$$ is positive $7$ times negative $x$ times positive $y^2$. - Multiplying positive $7$ and positive $y^2$ gives positive. - Multiplying positive result by negative $x$ gives negative. 5. **Conclusion:** The expression $$\frac{x}{y} \cdot 7y^3$$ is negative under the given conditions. **Final answer:** (B) Negative