First, we ignore the decimal point and multiply as if both numbers were whole numbers. We ignore any leading zeros during the multiplication: \[ 4 \times 8 = 32 \]

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

There are $\color{blue}2$ decimal place in $0.02$ and there are $\color{blue}2$ decimal places in $0.08,$ so their product will have ${\color{blue}2} + {\color{blue}2} = 4$ decimal places.

We take our value of $32$ and insert a decimal point to make a number with $4$ decimal places:

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\underbrace{0032}_{\large\text{$4$ digits}} \]

Therefore, $0.04 \times 0.08 = 0.0032\,.$

To find how much water the tank loses in $0.05$ hours, we need to calculate $0.08 \times 0.05.$

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

We proceed by multiplying the two numbers just as we would with whole numbers. So, \[ 8 \times 5 = 40. \]

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

There are $\color{blue}2$ decimal place in $0.08$ and there are $\color{blue}2$ decimal places in $0.05.$ Therefore, their product will have ${\color{blue}2} + {\color{blue}2} = 4$ decimal places.

So, we take our value of $40$ and add a decimal point to make a number with $4$ decimal places.

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\underbrace{0040}_{\large\text{$4$ digits}} \]

Therefore, the tank will lose $0.004$ gallons of water in $0.05$ hours.

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}2}{} \!\!\!\! & \\ & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 1 \!\!\!\! & \!\!\!\! 7 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . \color{red}0 \!\!\!\! & \!\!\!\! {4} \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 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 $0.17$ and there are $\color{blue}2$ decimal places in $0.04.$ Therefore, their product will have ${\color{blue}{2}} + {\color{blue}{2}} = 4$ decimal places.

So, we take our value of $68$ and add a decimal point to make a number with $4$ decimal place.

\[ 0.\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\underbrace{0068}_{\large\text{$4$ digits}} \]

Therefore, $0.17 \times 0.04 = 0.0068.$

To find the amount of copper in $0.52$ ounces of the alloy, we need to calculate $0.07 \times 0.52.$

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*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . \color{red}0 \!\!\!\! & \!\!\!\! 7 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \hline & & \!\!\!\! \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 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 $0.07$ and there are $\color{blue}2$ decimal places in $0.52.$ Therefore, their product will have ${\color{blue}{2}} + {\color{blue}{2}} = 4$ decimal places.

So, we take our value of $364$ and add a decimal point to make a number with $4$ decimal places.

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\underbrace{0364}_{\large\text{$4$ digits}} \]

Hence, there are $0.0364$ grams of copper in $0.52$ ounces of the alloy.

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*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \\ & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\color{red}0\!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \hline & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \\ \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 $0.44$ and there are $\color{blue}2$ decimal places in $0.54,$ so their product will have ${\color{blue}{2}} + {\color{blue}{2}} = 4$ decimal places.

We take our value of $2376$ and insert a decimal point to make a number with $4$ decimal places:

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\underbrace{2376}_{\large\text{$4$ digits}} \]

Therefore, $0.44 \times 0.54 = 0.2376\,.$

To find the time taken the rocket to travel $0.72$ kilometers, we need to calculate $0.12 \times 0.72.$

So, 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*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}1$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\! & \\ & & \!\!\!\! \color{red}0 \!\!\!\! & \!\!\!\!\!\!\! . 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\color{red}0\!\!\!\! & \!\!\!\!\!\!\! . 7 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \hline & & \!\!\!\! \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \end{array} \end{align*}

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

There is $\color{blue}2$ decimal place in $0.12$ and there are $\color{blue}2$ decimal places in $0.72.$ Therefore, their product will have ${\color{blue}{2}} + {\color{blue}{2}} = 4$ decimal places.

So, we take our value of $864$ and add a decimal point to make a number with $4$ decimal places.

\[ 0\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\underbrace{0864}_{\large\text{$4$ digits}} \]

Therefore, $0.12 \times 0.72 = 0.0864.$

So, the rocket traveled $0.72$ kilometers in $0.0864$ minutes

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

\begin{align*} & \begin{array}{ccccc} & & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\!\! & \!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \\ & & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 6 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\!\!\!\! . 2 \!\!\!\! \\ \hline & & & \!\!\!\!\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!3\!\!\!\! & \!\!\!\!2\!\!\!\! \\ \!\!\!\!+\!\!\!\! & & \!\!\!\!\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 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 $11.6$ and there is $\color{blue}1$ decimal place in $10.2.$ Therefore, their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

So, we take our value of $11832$ and add a decimal point to make a number with $2$ decimal places.

\[ 118\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{32}_{\large\text{$2$ digits}}\!\!\! \]

Therefore, $11.6 \times 10.2 = 118.32.$

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

\begin{align*} & \begin{array}{ccccc} & & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\ \fbox{$\color{blue}1$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\!\! & \!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\ \fbox{$\color{blue}3$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\!\!\!\! . 6 \!\!\!\! \\ \hline & & & \!\!\!\!\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!4\!\!\!\! \\ \!\!\!\!+\!\!\!\! & & \!\!\!\!\!\!\!\! & \!\!\!\!9\!\!\!\! & \!\!\!\!9\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 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 $12.4$ and there is $\color{blue}1$ decimal place in $18.6,$ so their product will have ${\color{blue}{1}} + {\color{blue}{1}} = 2$ decimal places.

We take our value of $23064$ and insert a decimal point to make a number with $2$ decimal places:

\[ 230\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{64}_{\large\text{$2$ digits}} \]

Therefore, $12.4 \times 18.6 = 230.64\,.$