We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 10\times 2\); therefore, the rectangular array should have \(\displaystyle 10\) squares across the top row and \(\displaystyle 2\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 20\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 2\) is the multiplier and \(\displaystyle 10\) is the multiplicand. 

First, we multiply \(\displaystyle 2\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 0}\\ \times\ {\color{Red} 2}\end{array}}{\ \ \ \ 0}\)

Next, we multiply\(\displaystyle 2\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}0\\ \times\ {\color{Red} 2}\end{array}}{\ \ \ 20}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 10\times 3\); therefore, the rectangular array should have \(\displaystyle 10\) squares across the top row and \(\displaystyle 3\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 30\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 3\) is the multiplier and \(\displaystyle 10\) is the multiplicand. 

First, we multiply \(\displaystyle 3\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 0}\\ \times\ {\color{Red} 3}\end{array}}{\ \ \ \ 0}\)

Next, we multiply \(\displaystyle 3\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}0\\ \times\ {\color{Red} 3}\end{array}}{\ \ \ 30}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 10\times 4\); therefore, the rectangular array should have \(\displaystyle 10\) squares across the top row and \(\displaystyle 4\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 40\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 4\) is the multiplier and \(\displaystyle 10\) is the multiplicand. 

First, we multiply \(\displaystyle 4\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 0}\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ \ 0}\)

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}0\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ 40}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 11\times 5\); therefore, the rectangular array should have \(\displaystyle 11\) squares across the top row and \(\displaystyle 5\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 55\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 5\) is the multiplier and \(\displaystyle 11\) is the multiplicand. 

First, we multiply \(\displaystyle 5\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 1}\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ \ 5}\)

Next, we multiply \(\displaystyle 5\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}1\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ 55}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 11\times 6\); therefore, the rectangular array should have \(\displaystyle 11\) squares across the top row and \(\displaystyle 6\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 66\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 6\) is the multiplier and \(\displaystyle 11\) is the multiplicand. 

First, we multiply \(\displaystyle 6\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 1}\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ \ 6}\)

Next, we multiply \(\displaystyle 6\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}1\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ 66}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 11\times 4\); therefore, the rectangular array should have \(\displaystyle 11\) squares across the top row and \(\displaystyle 4\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 44\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 4\) is the multiplier and \(\displaystyle 11\) is the multiplicand. 

First, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 1}\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ \ 4}\)

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}1\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ 44}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 12\times 4\); therefore, the rectangular array should have \(\displaystyle 12\) squares across the top row and \(\displaystyle 4\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 48\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 4\) is the multiplier and \(\displaystyle 12\) is the multiplicand. 

First, we multiply \(\displaystyle 4\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 2}\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ \ 8}\)

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}2\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ 48}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 12\times 5\); therefore, the rectangular array should have \(\displaystyle 12\) squares across the top row and \(\displaystyle 5\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 60\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 5\) is the multiplier and \(\displaystyle 12\) is the multiplicand. 

First, we multiply \(\displaystyle 5\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}^11{\color{Red} 2}\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ \ 0}\)

Because the product is greater than \(\displaystyle 9\), we carry the \(\displaystyle 1\) and add it to the next product. 

Next, we multiply \(\displaystyle 5\) and \(\displaystyle 1\) and then add the \(\displaystyle 1\) we carried. 

 \(\displaystyle \frac{\begin{array}[b]{r}^1{\color{Red} 1}2\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ 60}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 12\times 6\); therefore, the rectangular array should have \(\displaystyle 12\) squares across the top row and \(\displaystyle 6\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 72\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 6\) is the multiplier and \(\displaystyle 12\) is the multiplicand. 

First, we multiply \(\displaystyle 6\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}^11{\color{Red} 2}\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ \ 2}\)

Because the product is greater than \(\displaystyle 9\), we carry the \(\displaystyle 1\) and add it to the next product. 

Next, we multiply \(\displaystyle 6\) and \(\displaystyle 1\) and then add the \(\displaystyle 1\) we carried. 

 \(\displaystyle \frac{\begin{array}[b]{r}^1{\color{Red} 1}2\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ 72}\)

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 13\times 6\); therefore, the rectangular array should have \(\displaystyle 13\) squares across the top row and \(\displaystyle 6\) squares down the first column. The rectangular array should look like the image below:

We can count the squares to find our answer. The correct answer is is \(\displaystyle 78\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 6\) is the multiplier and \(\displaystyle 13\) is the multiplicand. 

First, we multiply \(\displaystyle 6\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}^11{\color{Red} 3}\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ \ 8}\)

Because the product is greater than \(\displaystyle 9\), we carry the \(\displaystyle 1\) and add it to the next product. 

Next, we multiply \(\displaystyle 6\) and \(\displaystyle 1\) and then add the \(\displaystyle 1\) we carried. 

 \(\displaystyle \frac{\begin{array}[b]{r}^1{\color{Red} 1}3\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ 78}\)