First, we write $3$ as an improper fraction:

\[ 3=\dfrac{3}{1} \]

We also 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 numbers. We multiply the numerators, and we multiply the denominators:

\[\dfrac{3}{1}\times\dfrac {9} {4}= \dfrac {3\times 9} {1\times 4} = \dfrac{\color{blue}27}{4} \]

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

First, we write $2\,\dfrac{2}{3}$ as an improper fraction: \[ 2\,\dfrac{2}{3} = \dfrac{(2 \times 3) + 2}{3} = \dfrac{8}{3} \]

We also write $4$ as an improper fraction: \[ 4 = \dfrac{4}{1} \]

Now, we can multiply the numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{8}{3} \times \dfrac{4}{1} = \dfrac{8 \times 4}{3 \times 1} = \bbox[4pt,border: 1px solid lightgray]{\dfrac{32}{3}} \]

We can also write this improper fraction as a mixed number: \[ \dfrac{32}{3} = 10\,\textrm{R}\,2 = \bbox[4pt,border: 1px solid lightgray]{10\,\dfrac{2}{3}} \]

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

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

Now, we can multiply the numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{3}{1} \times \dfrac{11}{7} = \dfrac {3 \times 11}{1 \times 7} = \dfrac{33}{7} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{33}{7} = 4\,\textrm{R} \,5 = \bbox[4pt,border: 1px solid lightgray]{4\,\dfrac{5}{7}} \]

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

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

We also write $2$ as an improper fraction:

\[ 2=\dfrac{2}{1} \]

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

\[ \dfrac 9 {4} \times \dfrac{2}{1} = \dfrac {9\times 2} {4\times 1} = \dfrac{18}{4} \]

Next, we reduce the fraction to its lowest terms by dividing the numerator and denominator by $2{:}$

\[ \dfrac{18}{4}=\dfrac{18\div 2}{4\div 2}=\dfrac{9}{2} \]

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

\[ \dfrac{9}{2} =4\,\textrm{R} 1 = 4\,\dfrac 1 {2}\]

First, we write $1\,\dfrac{1}{6}$ as an improper fraction: \[ 1\,\dfrac{1}{6} = \dfrac{(1 \times 6) + 1}{6} = \dfrac{7}{6} \]

We also write $3$ as an improper fraction: \[ 3 = \dfrac{3}{1} \]

Now, we can multiply the numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{7}{6} \times \dfrac{3}{1} = \dfrac{7 \times 3}{6 \times 1} = \dfrac{21}{6} \]

Next, we reduce the fraction to its lowest terms by dividing the numerator and denominator by $3{:}$ \[ \dfrac{21}{6} = \dfrac{21 \div 3}{6 \div 3} = \dfrac{7}{2} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{7}{2} = 3\,\textrm{R}\,1 = \bbox[4pt,border: 1px solid lightgray]{3\,\dfrac{1}{2}} \]

First, we write $1\,\dfrac{1}{8}$ as an improper fraction: \[ 1\,\dfrac{1}{8} = \dfrac{(1 \times 8) + 1}{8} = \dfrac{9}{8} \]

We also write $8$ as an improper fraction: \[ 8 = \dfrac{8}{1} \]

Now, we can multiply the numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{9}{8} \times \dfrac{8}{1} = \dfrac{9 \times 8}{8 \times 1} = \dfrac{72}{8} \]

Finally, we reduce the fraction to its lowest terms by dividing the numerator and denominator by $8{:}$ \[ \dfrac{72}{8} = \dfrac{72 \div 8}{8 \div 8} = \dfrac{9}{1} = 9 \]

First, we write $3$ as an improper fraction:

\[ 3=\dfrac{3}{1} \]

We also write $1\,\dfrac 2 {9} $ as an improper fraction:

\[ 1\,\dfrac 2 {9}=\dfrac{(1\times 9)+2}{9}= \dfrac {11} {9}\]

So now, we need to solve the following multiplication problem:

\[ \dfrac{\color{red}3}{1} \times \dfrac{11}{\color{blue}9} \]

Notice that the first fraction's numerator ($\color{red}3$) and the second fraction's denominator ($\color{blue}9$) have a common factor of $3.$

Therefore, we can simplify our problem by swapping the denominators first and then multiplying the resulting fractions:

\begin{align*} \dfrac{\color{red}3}{1} \times \dfrac{11}{\color{blue}9} &= \dfrac{\color{red}3}{\color{blue}9} \times \dfrac{11}{1} \\[5pt] &=\dfrac13\times 11 \\[5pt] &=\dfrac{1\times 11}{3} \\[5pt] &=\dfrac{11}{3} \end{align*}

Therefore, the missing number is ${\color{blue}{11}}.$

First, we write $2$ as an improper fraction:

\[ 2=\dfrac{2}{1} \]

We also write $2\,\dfrac 5 {6} $ as an improper fraction:

\[ 2\,\dfrac 5 {6}=\dfrac{(2\times 6)+5}{6}= \dfrac{17}{6}\]

So now, we need to solve the following multiplication problem:

\[ \dfrac{\color{red}2}{1} \times \dfrac{17}{\color{blue}6} \]

Notice that the first fraction's numerator ($\color{red}2$) and the second fraction's denominator ($\color{blue}6$) have a common factor of $2.$

Therefore, we can simplify our problem by swapping the denominators first and then multiplying the resulting fractions:

\begin{align*} \dfrac{\color{red}2}{1} \times \dfrac{17}{\color{blue}6} &= \dfrac{\color{red}2}{\color{blue}6} \times \dfrac{17}{1} \\[5pt] &=\dfrac13\times 17 \\[5pt] &=\dfrac{1\times 17}{3} \\[5pt] &=\dfrac{17}{3} \end{align*}

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

\[ \dfrac{17}{3} =5\,\textrm{R} 2 = 5\,\dfrac 2 {3} \]

First, we write $8$ as an improper fraction: \[ 8 = \dfrac{8}{1} \]

We also write $2 \, \dfrac{1}{4}$ as an improper fraction: \[ 2 \, \dfrac{1}{4} = \dfrac{(2 \times 4) + 1}{4} = \dfrac{9}{4} \]

So now, we need to solve the following multiplication problem: \[ \dfrac{\color{red}8}{1} \times \dfrac{9}{\color{blue}4} \]

Notice that the first fraction's numerator ($\color{red}8$) and the second fraction's denominator ($\color{blue}4$) have a common factor of $4.$

Therefore, we can simplify our problem by swapping the denominators first and then multiplying the resulting fractions: \begin{align} \dfrac{\color{red}8}{1} \times \dfrac{9}{\color{blue}4} & = \dfrac{\color{red}8}{\color{blue}4} \times \dfrac{9}{1} \\[5pt] & = 2 \times 9\\[5pt] & = \bbox[4pt,border: 1px solid lightgray]{18} \end{align}