Solve.
-$\frac{5 \frac{4}{7}}{10 \frac{2}{7}}=\frac{8 \frac{1}{2}}{n}$
A) $\frac{48}{221}$
B) $\frac{13}{204}$
C) $15 \frac{9}{13}$
D) $4 \frac{29}{48}$

Question

Quizplus · Accepted Answer

First, convert the mixed numbers to improper fractions: $5 \frac{4}{7} = \frac{39}{7}$ and $10 \frac{2}{7} = \frac{72}{7}$. The left side of the equation becomes $\frac{39/7}{72/7} = \frac{39}{72}$. Convert $8 \frac{1}{2}$ to an improper fraction: $8 \frac{1}{2} = \frac{17}{2}$. The equation is now $\frac{39}{72} = \frac{17}{2n}$. Cross-multiply to solve for $n$: $39 \cdot 2n = 72 \cdot 17$, which simplifies to $78n = 1224$. Solving for $n$ gives $n = 15 \frac{9}{13}$.

Solve - 5471027=812n\frac{5 \frac{4}{7}}{10 \frac{2}{7}}=\frac{8 \frac{1}{2}}{n}1072​574​​=n821​​ A) 48221\frac{48}{221}22148​ B) 13204\frac{13}{204}20413​ C) 1591315 \frac{9}{13}15139​ D) 429484 \frac{29}{48}44829​

Solve
- $\frac{5 \frac{4}{7}}{10 \frac{2}{7}}=\frac{8 \frac{1}{2}}{n}$ A) $\frac{48}{221}$ B) $\frac{13}{204}$ C) $15 \frac{9}{13}$ D) $4 \frac{29}{48}$