Compare the following quantities and choose the correct relationship code: **Quantity I:** $$\frac{360 m^{7} n^{9}}{120 m^{-2} n^{3}} imes 24 m^{-4} n^{4}, \quad m 0, n

Question

Compare the following quantities and choose the correct relationship code: **Quantity I:** $$\frac{360 m^{7} n^{9}}{120 m^{-2} n^{3}} imes 24 m^{-4} n^{4}, \quad m > 0, n < 0$$ **Quantity II:** $$\frac{240 x^{9} y^{7}}{60 x^{4} y^{3}} \div 3 x^{-2} y^{3}, \quad x < 0, y < 0$$ **Quantity III:** $$\frac{48 a^{8} b^{12} imes 5 a^{3} b^{-4}}{6 a^{6} b}, \quad a > 0, b < 0$$ Use the following codes to express the relationships between Quantity I and Quantity II, and Quantity II and Quantity III respectively: - @ → > - & → < - * → ≥ - $ → ≤ - # → = (or relationship cannot be established) Select the correct option based on the relationships.

Accepted Answer

**Solution:** 1. **Quantity I:** Calculate stepwise: $$\frac{360}{120} \times 24 \times m^{7 - (-2) - 4} \times n^{9 - 3 + 4} = 3 \times 24 \times m^{7 + 2 - 4} \times n^{10} = 72 m^{5} n^{10}$$ Since $m > 0$ and $n < 0$, and $n^{10}$ is an even power, $n^{10} > 0$. Therefore, Quantity I is positive. 2. **Quantity II:** Simplify the expression: $$\frac{240}{60 \times 3} \times x^{9 - 4 + 2} \times y^{7 - 3 - 3} = \frac{240}{180} \times x^{7} \times y^{1} = \frac{4}{3} x^{7} y$$ Given $x < 0$ and $y < 0$, $x^{7}$ is negative (odd power), $y$ is negative, so their product $x^{7} y$ is positive (negative × negative = positive). Hence, Quantity II is positive. 3. **Quantity III:** Simplify: $$\frac{48 \times 5}{6} \times a^{8 + 3 - 6} \times b^{12 - 4 - 1} = 40 a^{5} b^{7}$$ Since $a > 0$ and $b < 0$, $b^{7}$ is negative (odd power), so Quantity III is negative. 4. **Comparisons:** - Quantity I and Quantity II are both positive, but their exact relationship cannot be definitively established without numerical values. - Quantity II is positive and Quantity III is negative, so Quantity II > Quantity III. Therefore, the relationship between Quantity I and Quantity II is '#', and between Quantity II and Quantity III is '@'. **Answer:** (#, @)

Answer

Answer

Answer