Certainly! Let's work through each part of the problem step by step.
### Part 1: [tex]\(0.6 + 0.13\)[/tex]
1. Align the decimal points:
```
0.60
+ 0.13
```
2. Add each column, starting from the rightmost side:
- The hundredths place: [tex]\(0 + 3 = 3\)[/tex]
- The tenths place: [tex]\(6 + 1 = 7\)[/tex]
- The units place: [tex]\(0 + 0 = 0\)[/tex]
3. Combine the results:
```
0.60
+ 0.13
______
0.73
```
So, [tex]\(0.6 + 0.13 = 0.73\)[/tex].
### Part 2: [tex]\(0.22 - 0.17\)[/tex]
1. Align the decimal points:
```
0.22
- 0.17
```
2. Subtract each column starting from the rightmost side:
- The hundredths place: [tex]\(2 - 7\)[/tex]. Since 2 is less than 7, we need to borrow 1 from the tenths place.
- After borrowing: [tex]\(12 - 7 = 5\)[/tex]
- The tenths place: Since we borrowed 1, now it's [tex]\(1 - 1 = 0\)[/tex]
- The units place: [tex]\(0 - 0 = 0\)[/tex]
3. Combine the results:
```
0.22
- 0.17
______
0.05
```
So, [tex]\(0.22 - 0.17 = 0.05\)[/tex].
In conclusion:
[tex]\[
\begin{tabular}{|c|c|}
\hline $0.6 + 0.13 =$ & $0.22 - 0.17 =$ \\
\hline 0.73 & 0.05 \\
\hline
\end{tabular}
\][/tex]
