Simplify $15 \cdot \frac{7}{10} \cdot \frac{1}{9}$.
Both 15 and 10 share a factor of 5, so the expression simplifies to $\frac{\cancelto{3}{15}}{1} \cdot \frac{7}{\cancelto{2}{10}} \cdot \frac{1}{9}.$ In addition, the 3 and the 9 share a factor of 3, so the expression simplifies to $\frac{\cancel{3}}{1} \cdot \frac{7}{2} \cdot \frac{1}{\cancelto{3}{9}}.$ Thus, the expression simplifies to $\frac{7}{2} \cdot \frac{1}{3} = \frac{(7)(1)}{(2)(3)} = \boxed{\frac{7}{6}}.$
\frac{7}{6}
Level 3
Prealgebra
Problem: Simplify $15 \cdot \frac{7}{10} \cdot \frac{1}{9}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first multiply the constants and fractions together. Start with the multiplication of the whole number and the first fraction: 

15 multiplied by 7 gives 105. 

Next, consider the denominators. The denominator of the first fraction is 10, and the second fraction is 9. Multiply these denominators together: 

10 multiplied by 9 gives 90. 

Now, we have the fraction $\frac{105}{90}$. 

To simplify this fraction, find the greatest common divisor (GCD) of 105 and 90. The prime factorization of 105 is 3 × 5 × 7, and for 90, it is 2 × 3 × 3 × 5. The common factors are 3 and 5, so the GCD is 15. 

Now divide both the numerator and the denominator by their GCD: 

105 divided by 15 gives 7, and 90 divided by 15 gives 6. 

Thus, the simplified fraction is $\frac{7}{6}$. 

This process confirms that the final answer is $\frac{7}{6}$.<|END_OF_LATENT|> The final answer is $\frac{7}{6}$.