First, we write $2 \, \dfrac{1}{4}$ as an improper fraction:

\[ 2 \,\dfrac{1}{4} = \dfrac{(2 \times 4) + 1}{4}= \dfrac{9}{4} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[ \dfrac{9}{4} \times \dfrac{3}{7} = \dfrac {9 \times 3}{4 \times 7} = \dfrac{\color{blue}27}{28} \]

Therefore, the missing number is ${\color{blue}{27}}.$

First, we write $2\,\dfrac 1 {6} $ as an improper fraction:

\[ 2\,\dfrac 1 {6}=\dfrac{(2\times 6)+1}{6}= \dfrac {13} {6}\]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[\dfrac{1}{5}\times\dfrac {13} {6}= \dfrac {1 \times 13} {5 \times 6} = \dfrac{\color{blue}13}{30} \]

Therefore, the missing number is ${\color{blue}{13}}.$

First, we write $1\,\dfrac 2{5} $ as an improper fraction: \[ 1\,\dfrac 2 {5}=\dfrac{(1\times 5)+2}{5}= \dfrac {7} {5} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac {7} {5} \times \dfrac{3}{8} = \dfrac {7\times 3} {5 \times 8} = \bbox[4pt,border: 1px solid lightgray]{\dfrac{21}{40}} \]

First, we write $1\,\dfrac 2 {3} $ as an improper fraction:

\[ 1\,\dfrac 2 {3}=\dfrac{(1\times 3)+2}{3}= \dfrac 5 {3}\]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[ \dfrac 5 {3} \times \dfrac{1}{4} = \dfrac {5\times 1} {3\times 4} = \dfrac{5}{12} \]

First, we write $1 \,\dfrac{2}{5}$ as an improper fraction: \[ 1 \,\dfrac{2}{5}=\dfrac{(1 \times 5) + 2}{5}= \dfrac{7}{5} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{2}{3} \times \dfrac{7}{5} = \dfrac{2 \times 7}{3 \times 5} = \bbox[4pt,border: 1px solid lightgray]{\dfrac{14}{15}} \]

First, we write $2 \, \dfrac{1}{2}$ as an improper fraction:

\[ 2 \,\dfrac{1}{2} = \dfrac{(2 \times 2) + 1}{2}= \dfrac{5}{2} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[ \dfrac{5}{2} \times \dfrac{3}{4} = \dfrac {5 \times 3}{2 \times 4} = \dfrac{15}{8} \]

Finally, we write the resulting improper fraction as a mixed number:

\[ \dfrac{15}{8} = 1 \, \textrm{R} 7 = 1 \, \dfrac{\color{blue}7}{8} \]

Therefore, the missing number is ${\color{blue}{7}}.$

First, we write $3 \, \dfrac{1}{2}$ as an improper fraction:

\[ 3 \,\dfrac{1}{2} = \dfrac{(3 \times 2) + 1}{2}= \dfrac{7}{2} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[ \dfrac{5}{9} \times \dfrac{7}{2} = \dfrac{5 \times 7}{9 \times 2} = \dfrac{35}{18} \]

Finally, we write the resulting improper fraction as a mixed number:

\[ \dfrac{35}{18} = 1 \, \textrm{R} 17 = 1 \, \dfrac{\color{blue}17}{18} \]

Therefore, the missing number is ${\color{blue}{17}}.$

First, we write $1 \,\dfrac{1}{2}$ as an improper fraction: \[ 1 \,\dfrac{4}{7} = \dfrac{(1 \times 7) + 4}{7} = \dfrac{11}{7} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{11}{7} \times \dfrac{2}{3} = \dfrac{11 \times 2}{7 \times 3} = \dfrac{22}{21} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{22}{21} = 1 \, \textrm{R} 1 = \bbox[4pt,border: 1px solid lightgray]{1 \, \dfrac{1}{21}} \]

First, we write $1 \,\dfrac{5}{7}$ as an improper fraction: \[ 1 \,\dfrac{5}{7} = \dfrac{(1 \times 7) + 5}{7} = \dfrac{12}{7} \]

Now, we can multiply the two numbers. \begin{align*} \dfrac{3}{\color{red}4}\times\dfrac {12} {\color{blue}7} \end{align*}

We can simplify this product by swapping the denominators: \begin{align*} \dfrac{3}{\color{red}4}\times\dfrac {12} {\color{blue}7} &= \dfrac{3}{\color{blue}7}\times\dfrac {12} {\color{red}4} \\[5pt] &= \dfrac37 \times 3 \\[5pt] &=\dfrac{9}{7} \end{align*}

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{9}{7} =1\,\textrm{R} 2 = \bbox[4pt,border: 1px solid lightgray]{1\,\dfrac 27} \]