First, we ignore the decimal point and multiply as if both numbers were whole numbers. We ignore any leading ${\color{red}{0}}$'s during the multiplication:

\[ 2 \times 7 = 14 \]

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $0.02$ and there is $\color{blue}1$ decimal place in $0.7,$ so their product will have ${\color{blue}2} + {\color{blue}1} = 3$ decimal places.

We take our value of $14$ and insert a decimal point to make a number with $3$ decimal places:

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{014}_{\large\text{$3$ digits}}\!\!\! \]

Therefore, $0.02 \times 0.7 = 0.014\,.$

We proceed by multiplying the two numbers just as we would with whole numbers:

Notice that both numbers have leading zeros, which we ignore during the multiplication:

\begin{align*} & \begin{array}{ccccc} & & \!\!\!\!\! \substack{ \\ \color{blue}1}{} \!\!\!\! & \\ & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 1 \!\!\!\! & \!\!\!\! 3 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . {5} \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 5 \!\!\!\! \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $0.13$ and there is $\color{blue}1$ decimal place in $0.5.$ Therefore, their product will have ${\color{blue}{2}} + {\color{blue}{1}} = 3$ decimal places.

So, we take our value of $65$ and add a decimal point to make a number with $\color{blue}3$ decimal places.

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{065}_{\large\text{$3$ digits}}\!\!\! \]

Therefore, $0.13 \times 0.5 = 0.065.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers. We ignore any leading ${\color{red}{0}}$'s during the multiplication:

\begin{align*} & \begin{array}{ccccc} & & \!\!\!\!\! \substack{ \\ \color{blue}2}{} \!\!\!\! & \!\!\!\!\! \substack{ \\ \color{blue}1}{} \!\!\!\! & \\ & & \!\!\!\! 3 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \hline & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $3.42$ and there is $\color{blue}1$ decimal place in $0.5,$ so their product will have ${\color{blue}2} + {\color{blue}1} = 3$ decimal places.

We take our value of $1710$ and insert a decimal point to make a number with $3$ decimal places:

\[ 1\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{710}_{\large\text{$3$ digits}} \]

Therefore, $3.42 \times 0.5 = 1.710 = 1.71\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers. We ignore any leading ${\color{red}{0}}$'s during the multiplication:

\begin{align*} & \begin{array}{ccccc} & & \!\!\!\!\! \substack{ \\ \color{blue}1}{} \!\!\!\! & \!\!\!\!\! \substack{ \\ \color{blue}1}{} \!\!\!\! & \\ & & \!\!\!\! 4 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 3 \!\!\!\! \\ \hline & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $4.56$ and there is $\color{blue}1$ decimal place in $0.3,$ so their product will have ${\color{blue}2} + {\color{blue}1} = 3$ decimal places.

We take our value of $1368$ and insert a decimal point to make a number with $3$ decimal places:

\[ 1\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{368}_{\large\text{$3$ digits}} \]

Therefore, $4.56 \times 0.3 = 1.368\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\! & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\! & \\ & & & \!\!\!\! 2 \!\!\!\! & \!\!\!\!\!\!\! . 6 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\!\!\!\! . 1 \!\!\!\! \\ \hline & & \!\!\!\!\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!8\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!8\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\!\!\!\!\! & \!\!\!\!8\!\!\!\! & \!\!\!\!3\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!8\!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $2.68$ and there is $\color{blue}1$ decimal place in $3.1,$ so their product will have ${\color{blue}{2}} + {\color{blue}{1}} = 3$ decimal places.

We take our value of $8308$ and insert a decimal point to make a number with $3$ decimal places:

\[ 8\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{308}_{\large\text{$3$ digits}} \]

Therefore, $2.68 \times 3.1 = 8.308\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\! & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}3$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & \!\!\!\! 3 \!\!\!\! & \!\!\!\!\!\!\! . 0 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! \\ \hline & & \!\!\!\!1\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!4\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!5\!\!\!\! & \!\!\!\!3\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\!1\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!5\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!4\!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $3.06$ and there is $\color{blue}1$ decimal place in $5.4,$ so their product will have ${\color{blue}{2}} + {\color{blue}{1}} = 3$ decimal places.

We take our value of $16524$ and insert a decimal point to make a number with $3$ decimal places:

\[ 16\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{524}_{\large\text{$3$ digits}} \]

Therefore, $3.06 \times 5.4 = 16.524\,.$

To determine how far he can travel $7.5$ minutes, we multiply $7.5$ by $0.04.$

We proceed by multiplying the two numbers just as we would with whole numbers:

Notice that the second number has $2$ leading $\color{red}0$s, which we ignore during the multiplication:

\begin{align*} & \begin{array}{ccccc} & & \!\!\!\!\! \substack{ \\ \color{blue}2}{} \!\!\!\! & \\ & & \!\!\!\! 7 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . \color{red}0 \!\!\!\! & \!\!\!\! {4} \!\!\!\! \\ \hline & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $7.5$ and there are $\color{blue}2$ decimal places in $0.04.$ Therefore, their product will have ${\color{blue}{1}} + {\color{blue}{2}} = 3$ decimal places.

So, we take our value of $300$ and add a decimal point to make a number with $\color{blue}3$ decimal place.

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{300}_{\large\text{$3$ digits}} \]

Therefore, $7.5 \times 0.04 = 0.300.$

Therefore, little Alan could cover a distance of $0.3$ kilometers in $7.5$ minutes.

To determine how much Vanessa paid for her carrots, we multiply $2.1$ by $0.72.$

We proceed by multiplying the two numbers just as we would with whole numbers:

Notice that the second number has a leading $\color{red}0$, which we ignore during the multiplication:

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\! 2 \!\!\!\! & \!\!\!\!\!\!\! . 1 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\color{red}0\!\!\!\! & \!\!\!\!\!\!\! . 7 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \hline & & \!\!\!\! \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $2.1$ and there are $\color{blue}2$ decimal places in $0.72.$ Therefore, their product will have ${\color{blue}{1}} + {\color{blue}{2}} = 3$ decimal places.

So, we take our value of $1512$ and add a decimal point to make a number with $3$ decimal places.

\[ 1\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{512}_{\large\text{$3$ digits}} \]

So, $2.1 \times 0.72 = 1.512.$

Therefore, Vanessa paid a total of $\$1.51$ for her carrots (rounding the nearest cent).

To determine how many miles the racer will drive in total after $4.5$ laps, we multiply $4.5$ by $5.04.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers. We ignore any leading $\color{red}0$'s during the multiplication:

\begin{align*} & \begin{array}{ccccc} & & & & & \!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & & & \!\!\!\! 4 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\!\!\!\! . 0 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \hline & & & \!\!\!\!\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!8\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \!\!\!\!+\!\!\!\! & & \!\!\!\!\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!5\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $4.5$ and there are $\color{blue}2$ decimal places in $5.04,$ so their product will have ${\color{blue}{1}} + {\color{blue}{2}} = 3$ decimal places.

We take our value of $22680$ and insert a decimal point to make a number with $3$ decimal places:

\[ 22\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{680}_{\large\text{$3$ digits}} \]

Hence, $4.5 \times 5.04 = 22.680 = 22.68\,.$

We conclude that the racer will drive a total of $22.68.$