In this code's loop, notice that it takes the absolute value of each element.  This is done on the line:

int y = Math.abs(x[i]);

This value is then used for the second loop, which goes for y times, each time outputting the value of the given member of the original array—but now with its particular sign value.  Thus, even numbers like  will be output multiple times (i.e. 8).  This is done line by line for each member of the parameter array.

The graphics method takes the 2D array matrix and then runs through each element.  This is the point of the double for loop in the method itself.  Notice that for each element it does several things.  First of all, it is clearly accumulating a value into the variable r.  Now, for each element, you are adding:

x[i][j] (the current value in the 2D array iteration)

TIMES

(i + 1), or, the current row number.  Thus, for the data given you are doing the following:

1* 1 + 1 * 6 + 1 * 7 + 2 * 1 + 2 * 4 + 2 * 5 = 34

First of all, you can eliminate the following two options very quickly, as the fun method returns a double value:

[D@4b71bbc9]

[4,9,11,19]

These are trying to trick you into thinking that it returns an array.  (The first value is really what would be the output, since there is no toString defined for the double array by default.)

Now, in fun, the loop simply runs through the values in the array, summing those up.  Notice, it runs "backwards" through the array, from the end to the beginning.  It then divides by the length of the array, giving you the average of the values in the array.

This code is using what are called parallel arrays.  The boolean array has one value for each character in the String s.  Notice the logic in the loop.  There is a condition:

 if(!b[i]) {

         c = Character.toUpperCase(c);

 } else {

         c = Character.toLowerCase(c);  

}

This means that if the boolean is false, then you will make the given letter upper case.  (This is because of the ! in the condition. Be careful!)  Otherwise, you make it lower case.

Carefully walking through the code, you will get:

i rEAd LoGiC fOr FuN!

Let's look at this line of code:

int * data = new int[12]

An int pointer is created and an array of size 12 is assigned to it. 

When data[5] is called, the "data" is dereferenced and the value of the 6th position is returned. 

The only one of the choices that does this is 

*(data + 5)

In line above, the pointer is incremented ot the 6th position and is then dereferenced to get the value.

If the code was 

*data

The first item at the first position will be returned.

*(data+1)

This will return the item and the second position and so on.

When a two dimensional array is created and initialized, the way to access the items inside the matrix is by calling the array with the row and column (i.e. myArray[ROW][COLUMN]). Keeping in mind that arrays start at 0, the number four would be in row 1, column 0. Therefore to save that number into the variable "fourth" we'll do the following:

fourth = myArray[1][0];

*Note: myArray[1][0] is not the same as myArray[0][1].

myArray[1][0] =4 because it is the item located at row=1 and column = 0. 

myArray[0][1] =12 because it is the item located at row=0 and column = 1. 

Arrays can be two-dimensional. However, when trying to keep track of keys and values it can become complicated when using an array. HashMaps are the best way to represent data containing keys and values. 

In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: [Int] and then initialize. Therefore, we have var arr: [Int] = []. var arr = [] is technically correct, but the prompt asks you to unwrap the variable. 

In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: [String] and then initialize. Therefore, we have var arr: [String] = []. var arr = [] is technically correct, but the prompt asks you to unwrap the variable. 

The code segment will work only when the array has at least one integer. If the array has no negative integers, then the while loop will continue running, incrementing i past the length of the array, in which case an Out of Bounds Exception will occur.