To solve this problem. First set up a mathematical equation that represents this scenario.

"The sum of three consecutive even numbers is 42."

Let 

and recalling that each consecutive even number is two values greater that the previous even number, the mathematical equation becomes

From here isolate the variable on one side of the equations with all other constants on the other side.

Subtract six from both sides. 

Now divide by three.

Since  represents the beginning number in this sequence it also represents the smallest even number in this sequence. Therefore the smallest even number is 12.

After the first term, the next four terms are obtained by adding, multiplying, subtracting, and dividing by 2, in order. The four terms after that are obtained by carrying out the same operations with 4. The next term is obtained by adding 6, so the operations can be expected to be repeated with 6, the next even number.

Observe the operations as carried out with 2:

Carrying out the same steps, in order, with 4:

Carrying out the same steps, in order, with 6:

, the number that replaces the circle;

, the number that replaces the square.

Terms are obtained by alternately adding 5 and multiplying by 3:

The missing two elements are found by multiplying by 3, then by adding 5:

Multiplying by 3 confirms that this is correct:

When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.

If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values. 

Let's look at f(x) = 2x + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2x + 3 isn't the correct answer.

Next, let's look at f(x) = x² – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.

Now, let's examine f(x) = x² – 2x. We know that x² by itself can never be negative. However, if x² is really small, then adding –2x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.

Next, let's examine f(x) = x⁴ + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, x⁴ cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then x⁴ would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)⁴ + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x). 

By process of elimination, the answer is f(x) = x⁴ + (1 – x)² . This makes sense because x⁴ can't be negative, and because (1 – x)² can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add x⁴ and (1 – x)², the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = x⁴ + (1 – x)² will not have any negative values, and |f(x)| will be the same as f(x) for all values of x. 

The answer is f(x) = x⁴ + (1 – x)² .

We need to solve the two equations |2a – 3| = 5 and |3 – 4b| = 11, in order to determine the possible values of a and b. When solving equations involving absolute values, we must remember to consider both the positive and negative cases. For example, if |x| = 4, then x can be either 4 or –4.

Let's look at |2a – 3| = 5. The two equations we need to solve are 2a – 3 = 5 and 2a – 3 = –5.

2a – 3 = 5 or 2a – 3 = –5

Add 3 to both sides.

2a = 8 or 2a = –2

Divide by 2.

a = 4 or a = –1

Therefore, the two possible values for a are 4 and –1. However, the problem states that both a and b are negative. Thus, a must equal –1.

Let's now find the values of b.

3 – 4b = 11 or 3 – 4b = –11

Subtract 3 from both sides.

–4b = 8 or –4b = –14

Divide by –4.

b = –2 or b = 7/2

Since b must also be negative, b must equal –2.

We have determined that a is –1 and b is –2. The original question asks us to find |b – a|.

|b – a| = |–2 – (–1)| = | –2 + 1 | = |–1| = 1.

The answer is 1.

In order the find the answer, you must first solve what is inside the absolute value signs.

Following order of operations, you must first multiply  which equals .

Then you must subtract  from  as shown below:

Now, you must take the absolute value of  which is positive , the correct answer.

Substitute 0.6 for  :

Substitute .