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Answer :
Sure, let's work through the problem step-by-step to understand what the code segment does and determine the final output.
1. Initialization of variables:
```java
int j = 10;
int k = 8;
```
Here, the variable `j` is initialized to 10 and the variable `k` is initialized to 8.
2. First update:
```java
j += 2;
```
This line increases the value of `j` by 2. So, `j` which was 10, will now become 12.
3. Second update:
```java
k += j;
```
This line increases the value of `k` by the current value of `j`. At this point, `j` is 12. So, the value of `k` will become:
[tex]\[
k = k + j = 8 + 12 = 20
\][/tex]
4. Printing the values:
```java
System.out.print(j);
System.out.print(" ");
System.out.println(k);
```
These three lines will print the values of `j` and `k` followed by a space. So, it first prints the value of `j`, which is 12, then a space, and finally the value of `k`, which is 20.
Therefore, the output of the code segment will be:
```
12 20
```
So, the correct answer is:
(D) [tex]\(12 \quad 12\)[/tex]
1. Initialization of variables:
```java
int j = 10;
int k = 8;
```
Here, the variable `j` is initialized to 10 and the variable `k` is initialized to 8.
2. First update:
```java
j += 2;
```
This line increases the value of `j` by 2. So, `j` which was 10, will now become 12.
3. Second update:
```java
k += j;
```
This line increases the value of `k` by the current value of `j`. At this point, `j` is 12. So, the value of `k` will become:
[tex]\[
k = k + j = 8 + 12 = 20
\][/tex]
4. Printing the values:
```java
System.out.print(j);
System.out.print(" ");
System.out.println(k);
```
These three lines will print the values of `j` and `k` followed by a space. So, it first prints the value of `j`, which is 12, then a space, and finally the value of `k`, which is 20.
Therefore, the output of the code segment will be:
```
12 20
```
So, the correct answer is:
(D) [tex]\(12 \quad 12\)[/tex]
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