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

We proceed by multiplying the two numbers just as we would with whole numbers. So, \[ 3 \times 7 = 21. \]

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

There is $\color{blue}1$ decimal place in $0.3$ and there is $\color{blue}1$ decimal place in $0.7.$ Therefore, their product will have ${\color{blue}1} + {\color{blue}1} = 2$ decimal places.

So, we take our value of $21$ and add a decimal point to make a number with $2$ decimal places.

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{21}_{\large\text{$2$ digits}} \]

Therefore, $0.3 \times 0.7 = 0.21.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers: \[ 8 \times 7 = 56. \]

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

There is $\color{blue}1$ decimal place in $0.8$ and there is $\color{blue}1$ decimal place in $0.7,$ so their product will have ${\color{blue}1} + {\color{blue}1} = 2$ decimal places.

We take our value of $56$ and insert a decimal point to make a number with $2$ decimal places:

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{56}_{\large\text{$2$ digits}} \]

Therefore, $0.8 \times 0.7 = 0.56\,.$

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

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

\begin{align*} & \begin{array}{ccccc} & & \!\!\!\!\! \substack{ \\ \color{blue}3}{} \!\!\!\! & \\ & & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . {9} \!\!\!\! \\ \hline & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \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 $1.4$ and there is $\color{blue}1$ decimal place in $0.9.$ Therefore, their product will have ${\color{blue}1} + {\color{blue}1} = 2$ decimal places.

So, we take our value of $126$ and add a decimal point to make a number with $2$ decimal places.

\[ 1\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{26}_{\large\text{$2$ digits}} \]

Therefore, $1.4 \times 0.9 = 1.26.$

First, we ignore the decimal points and multiply as if both numbers were whole numbers: \[ \begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}3$} } \!\!\!\! & \\ & & & \!\!\!\! 0 \!\!\!\! & \!\!\!\!\!\!\! . 7 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \hline & & \!\!\!\! \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 5 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 5 \!\!\!\! \\ \end{array} \end{align*} \] We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $0.7$ and there is $\color{blue}1$ decimal place in $3.5,$ so their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

We take our value of $245$ and insert a decimal point to make a number with $2$ decimal places.

\[ 2\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{45}_{\large\text{$2$ digits}} \]

Therefore, $0.7 \times 3.5 = 2.45\,.$

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

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \\ & & & \!\!\!\! 5 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 3 \!\!\!\! \\ \hline & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 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 $5.4$ and there is $\color{blue}1$ decimal place in $1.3.$ Therefore, their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

So, we take our value of $702$ and add a decimal point to make a number with $2$ decimal places.

\[ 7\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{02}_{\large\text{$2$ digits}} \]

Therefore, $5.4 \times 1.3 = 7.02.$

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

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & \!\!\!\! 4 \!\!\!\! & \!\!\!\!\!\!\! . 3 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\!\!\!\! . 7 \!\!\!\! \\ \hline & & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 1 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 1 \!\!\!\! \\ \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 $4.3$ and there is $\color{blue}1$ decimal place in $8.7,$ so their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

We take our value of $3741$ and insert a decimal point to make a number with $2$ decimal places:

\[ 37\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{41}_{\large\text{$2$ digits}} \]

Therefore, $4.3 \times 8.7 = 37.41\,.$

Notice that $1.60=1.6.$

To determine how much she needs to pay for the flaked almonds, we have to multiply $1.4$ by $1.6.$

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

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 6 \!\!\!\! \\ \hline & & \!\!\!\! \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \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 $1.4$ and there is $\color{blue}1$ decimal place in $1.6,$ so their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

We take our value of $224$ and insert a decimal point to make a number with $2$ decimal places:

\[ 2\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{24}_{\large\text{$2$ digits}} \]

So, $1.4 \times 1.6 = 2.24\,.$

We conclude that Tara will have to pay $\$2.24.$

To find out how meters the horse will run in $8.5$ seconds, we have to multiply $7.2$ by $8.5.$

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

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}1$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \\ & & & \!\!\!\! 7 \!\!\!\! & \!\!\!\!\!\!\! . 2 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \hline & & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 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 $7.2$ and there is $\color{blue}1$ decimal place in $8.5,$ so their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

We take our value of $6120$ and insert a decimal point to make a number with $2$ decimal places:

\[ 61\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{20}_{\large\text{$2$ digits}} \]

Therefore, $7.2 \times 8.5 = 61.20\,.$

We conclude that Lenny's horse will run a distance of $61.2$ meters.