This is an example of commutative property. Commutative property of addition means that you can change the order of the smallest two numbers when adding, and it will always equal the biggest number.

When counting by \(\displaystyle 100\)s, we add \(\displaystyle 100\) to our previous number. 

\(\displaystyle \frac{\begin{array}[b]{r}500\\ +\ 100\end{array}}{ \ \ \ \space 600}\)

When adding any number by \(\displaystyle 100\) the only number that will change in your answer is the hundreds place. 

\(\displaystyle 2+0=2\)

\(\displaystyle 3+0=3\)

\(\displaystyle 4+1=5\)

\(\displaystyle \frac{\begin{array}[b]{r}432\\ +\ 100\end{array}}{\ \ \ \space 532}\)

Each dollar bill is worth \(\displaystyle \$1\), each penny is worth \(\displaystyle 1\cent\) and each quarter is worth \(\displaystyle 25\cent\).

We have \(\displaystyle 3\) dollar bills, \(\displaystyle 2\) pennies, and \(\displaystyle 3\) quarters.

\(\displaystyle \frac{\begin{array}[b]{r}\$1\\ \ \$1\\ +\$1\end{array}}{ \ \ \space\$3}\)   \(\displaystyle \frac{\begin{array}[b]{r}1\cent\\ +\ 1\cent\end{array}}{ \ \ \ \space 2\cent}\)   \(\displaystyle \frac{\begin{array}[b]{r}25\cent\\25\cent\\ +\ 25\cent\end{array}}{ \ \ \ \space 75\cent}\)

When we add dollars and cents, we add dollars to dollars and cents to cents. 

\(\displaystyle \frac{\begin{array}[b]{r}2\cent\\ +\ 75\cent\end{array}}{ \ \ \ \space 77\cent}\) 

We have \(\displaystyle \$3\) and \(\displaystyle 77\cent\) which is written as \(\displaystyle \$3.77\)

The only sequence from above that adds \(\displaystyle 10\) each time is \(\displaystyle 34,44,54\)

\(\displaystyle 34+10=44\)

\(\displaystyle 44+10=54\)

When we count the number of squares in this picture we get \(\displaystyle 10\). We have \(\displaystyle 5\) rows with \(\displaystyle 2\) squares in each row. Our math equation for this picture could be written as \(\displaystyle 2+2+2+2+2=10.\) 

The only sequence from above that adds \(\displaystyle 5\) each time is \(\displaystyle 23, 28, 33\)

\(\displaystyle 23+5=28\)

\(\displaystyle 28+5=33\)

This is an addition problem because we want to know how many total points were scored when we put all the points together. When we put things together we are adding. 

We can start at \(\displaystyle 2\) and count up \(\displaystyle 6\).

\(\displaystyle 2,3,4,5,6,7,8\)

\(\displaystyle \frac{\begin{array}[b]{r}2\\ 0\\+\ 6\end{array}}{ \ \ \ \space 8}\)

The associative property says that, when you are only adding or only multiplying, the way you group the items in a series does not matter.  Given 6 + 3 + 2 = 11, whether you group (6 + 3) + 2 or 6 + (3 + 2), the outcome does not change as the order in which you sum the numbers does not matter.

Write 12 down. Then write the 3 below the 2 in 12.

Multiply the bottom number with the top numbers, starting on the right:

\(\displaystyle \dpi{100} 3\times 2=6\)

\(\displaystyle \dpi{100} 3\times 1=3\)

The answer is 36.