Question

each of the missing digits is included below. drag a digit to a white rectangle to make the solution correct. 1. $4\frac{1}{2}+1\frac{3}{4}$ $\frac{square}{2}div\frac{square}{4}$ $\frac{9}{2}cdot\frac{square}{7}=\frac{18}{square}$ $square\frac{4}{7}$ 2. $3\frac{1}{3}+1\frac{5}{6}$ $\frac{square}{3}div\frac{square}{6}$ $\frac{10}{3}cdot\frac{square}{11}=\frac{square}{11}$ $1\frac{square}{11}$ dividing mixed numbers 2 11 10 7 4 9 7 6 9 20

Explanation:

Step1: Solve the first mixed - number addition

First, convert the mixed numbers to improper fractions: $4\frac{1}{2}=\frac{4\times2 + 1}{2}=\frac{9}{2}$ and $1\frac{3}{4}=\frac{1\times4+3}{4}=\frac{7}{4}$. Then $\frac{9}{2}\div\frac{7}{4}=\frac{9}{2}\times\frac{4}{7}=\frac{9\times4}{2\times7}=\frac{18}{7}$. So the missing digits for the first part are 9 and 7.

Step2: Solve the second mixed - number addition

Convert the mixed numbers to improper fractions: $3\frac{1}{3}=\frac{3\times3 + 1}{3}=\frac{10}{3}$ and $1\frac{5}{6}=\frac{1\times6 + 5}{6}=\frac{11}{6}$. Then $\frac{10}{3}\div\frac{11}{6}=\frac{10}{3}\times\frac{6}{11}=\frac{10\times6}{3\times11}=\frac{20}{11}=1\frac{9}{11}$. So the missing digits for the second part are 20, 9.

Answer:

For the first part: 9, 7
For the second part: 20, 9