1. Let's solve the first problem: $\frac{4}{10} = \frac{\_}{60}$. We want to find the missing number in the numerator. 2. To find the missing number, we use the rule that if two fractions are equal, their cross products are equal: $4 \times 60 = 10 \times \_$. 3. Calculate the left side: $4 \times 60 = 240$. 4. So, $240 = 10 \times \_$. To find the missing number, divide both sides by 10: $$\cancel{10} \times \_ = \frac{240}{\cancel{10}}$$ $$\_ = 24$$ 5. The missing number is 24, so $\frac{4}{10} = \frac{24}{60}$. 6. Now, the second problem: $\frac{3}{5} = \frac{12}{\_}$. Find the missing denominator. 7. Cross multiply: $3 \times \_ = 5 \times 12$. 8. Calculate the right side: $5 \times 12 = 60$. 9. So, $3 \times \_ = 60$. Divide both sides by 3: $$\cancel{3} \times \_ = \frac{60}{\cancel{3}}$$ $$\_ = 20$$ 10. The missing denominator is 20, so $\frac{3}{5} = \frac{12}{20}$. 11. Third problem: $\_ = \frac{42}{77} \div 6$. We want to find the missing numerator. 12. Dividing by 6 is the same as multiplying by $\frac{1}{6}$: $$\_ = \frac{42}{77} \times \frac{1}{6} = \frac{42 \times 1}{77 \times 6} = \frac{42}{462}$$ 13. Simplify $\frac{42}{462}$ by dividing numerator and denominator by 42: $$\frac{\cancel{42}}{\cancel{462}} = \frac{1}{11}$$ 14. So, the missing numerator is $\frac{1}{11}$. 15. Fourth problem: $\frac{\_}{9} = \frac{36}{81}$. Find the missing numerator. 16. Cross multiply: $\_ \times 81 = 36 \times 9$. 17. Calculate right side: $36 \times 9 = 324$. 18. So, $81 \times \_ = 324$. Divide both sides by 81: $$\cancel{81} \times \_ = \frac{324}{\cancel{81}}$$ $$\_ = 4$$ 19. The missing numerator is 4, so $\frac{4}{9} = \frac{36}{81}$. Final answers: - $\frac{4}{10} = \frac{24}{60}$ - $\frac{3}{5} = \frac{12}{20}$ - $\_ = \frac{1}{11}$ for $\frac{6\_}{42/77}$ (interpreted as $\frac{6\_}{1} = \frac{42}{77}$ divided by 6) - $\frac{4}{9} = \frac{36}{81}$