Include $ x = 8 $, exclude $ x = -3 $. Solution: $ (-\infty, -3) \cup [8, \infty) $. \boxed{(-\infty, -3) \cup [8, \infty)}Question: A cone has a radius of \(3y\) units and a height of \(4y\) units. A sphere has a radius of \(2y\) units. What is the ratio of the volume of the cone to the volume of the sphere?

["Volume Ratio of a Cone and a Sphere: Simplify $ \dfrac{(-\infty, -3) \cup [8, \infty)}{V_{\ ext{cone}} : V_{\ ext{sphere}}} $ with $ x = 8 $, $ x = -3 $ included\n\boxed{(-\infty, -3) \cup [8, \infty)}", "---", "Understanding Volume Ratios: Cone vs Sphere for $ x = 8 $, Considering $ x = -3 $ Excluded", "When comparing the volumes of geometric shapes like cones and spheres, precise mathematical expressions help clarify boundary conditions and intervals—especially when values such as radius, height, or dimensions are constrained. Consider a related problem rooted in formal variables:\nA cone has radius $ 3y $ and height $ 4y $, while a sphere has radius $ 2y $. What is the volume ratio of the cone to the sphere?", "To solve this, we begin with volume formulas and incorporate domain constraints indicated by $ x = 8 $ (included) and $ x = -3 $ (excluded), symbolizing critical bounds in shape dimensions. The final ratio is expressed concisely as:\n[\n\dfrac{V_{\ ext{cone}}}{V_{\ ext{sphere}}} = \dfrac{(-\infty, -3) \cup [8, \infty)}{x \ ext{ regime}}\n]", "### Step 1: Volume Formulas\n- Cone Volume: $ V_{\ ext{cone}} = \dfrac{1}{3} \pi r^2 h $\n Given $ r = 3y $, $ h = 4y $:\n [\n V_{\ ext{cone}} = \dfrac{1}{3} \pi (3y)^2 (4y) = \dfrac{1}{3} \pi (9y^2)(4y) = \dfrac{1}{3} \pi \cdot 36y^3 = 12\pi y^3\n ]", "- Sphere Volume: $ V_{\ ext{sphere}} = \dfrac{4}{3} \pi r^3 $\n Given $ r = 2y $:\n [\n V_{\ ext{sphere}} = \dfrac{4}{3} \pi (2y)^3 = \dfrac{4}{3} \pi (8y^3) = \dfrac{32}{3} \pi y^3\n ]", "### Step 2: Volume Ratio\n[\n\dfrac{V_{\ ext{cone}}}{V_{\ ext{sphere}}} = \dfrac{12\pi y^3}{\dfrac{32}{3}\pi y^3} = \dfrac{12}{\dfrac{32}{3}} = 12 \cdot \dfrac{3}{32} = \dfrac{36}{32} = \dfrac{9}{8}\n]", "However, the structural framing using intervals—specifically:\n[\n\dfrac{(-\infty, -3) \cup [8, \infty)}{x \ ext{-value reflecting domain separation at } x = -3 \ ext{ and } x = 8}\n]\nrepresents a metaphor for dynamic modeling: only values outside $ x < -3 $ combined with $ x \geq 8 $ yield a physically meaningful ratio in engineered or constrained systems. While $ y $-values affect magnitude, the interval definition governs context—highlighting how parametric boundaries refine proportional meaning.", "### Final Ratio Interpretation\nIgnoring $ x = -3 $ (excluded), the ratio stands universally at $ \dfrac{9}{8} $, but the inclusion of interval logic emphasizes validation across critical thresholds. This notation aids engineers and educators in assessing symmetry, scaling, and domain applicability in design and physics.", "\boxed{\dfrac{9}{8}}", "In applied contexts, always verify domain constraints—here $ x \geq 8 $ defines valid scaling, while $ x < -3 $ remains operationally irrelevant, refining how ratios inform real-world applications."]