1. The problem asks to find which card matches Card C's simplified expression $x^{-15}$. 2. Card C is $\frac{(x^6)^{-2}}{x^3}$. 3. Simplify Card C step-by-step: - Use the power of a power rule: $$(x^6)^{-2} = x^{6 \times (-2)} = x^{-12}$$ - So Card C becomes $$\frac{x^{-12}}{x^3}$$ 4. Use the quotient rule for exponents: $$\frac{x^a}{x^b} = x^{a-b}$$ - Apply it: $$x^{-12 - 3} = x^{-15}$$ 5. Now find which card on the right also simplifies to $x^{-15}$. 6. Check Card E: $\frac{x^{-7}}{8y^2} \times \frac{8}{x^8 y^{-2}} = \frac{x^{-7} \times 8}{8 y^2 \times x^8 y^{-2}} = \frac{x^{-7} \times \cancel{8}}{\cancel{8} y^2 x^8 y^{-2}} = \frac{x^{-7}}{x^8 y^{2 - (-2)}} = \frac{x^{-7}}{x^8 y^{4}} = x^{-7 - 8} y^{-4} = x^{-15} y^{-4}$ which is not $x^{-15}$ alone. 7. Check Card F: $\frac{80xy}{10(xy)^{-2}} = \frac{80xy}{10 x^{-2} y^{-2}} = \frac{80xy}{10} \times x^{2} y^{2} = 8xy \times x^{2} y^{2} = 8 x^{1+2} y^{1+2} = 8 x^{3} y^{3}$ not $x^{-15}$. 8. Check Card G: $\frac{x^{2} y^{-2}}{x^{-4} y^{-2}} = x^{2 - (-4)} y^{-2 - (-2)} = x^{6} y^{0} = x^{6}$ not $x^{-15}$. 9. Check Card H: $\frac{8 y^{-8}}{x^{3} y^{-5}} = 8 x^{-3} y^{-8 - (-5)} = 8 x^{-3} y^{-3}$ not $x^{-15}$. 10. None of the cards on the right simplify exactly to $x^{-15}$. Final conclusion: Card C simplifies to $x^{-15}$ and does not have an equivalent card on the right with the same expression. Therefore, the blank in the sentence "Card C and Card ______ simplify to ______" should be filled as: "Card C and Card (none) simplify to $x^{-15}$". "q_count" is 8 because there are 8 cards/problems in total.