Okay, let's see. The problem is asking how many integers are in the solution set of |x - 2| ≤ 5.6. Hmm, absolute value inequalities. Alright, I remember that absolute value inequalities can be rewritten as compound inequalities. Let me recall the steps.

First, the absolute value |x - 2| ≤ 5.6 means that the expression inside the absolute value, x - 2, is between -5.6 and 5.6. So, that translates to -5.6 ≤ x - 2 ≤ 5.6. Right? Because the absolute value being less than or equal to a number means the expression is within that range from the opposite sign to the positive sign.

Now, to solve for x, I need to isolate it in the middle. So, adding 2 to all three parts of the inequality should do it. Let me write that out:

-5.6 ≤ x - 2 ≤ 5.6

Adding 2 to each part:

-5.6 + 2 ≤ x ≤ 5.6 + 2

Calculating those:

-5.6 + 2 is... let's see, 2 is the same as 2.0. So, subtracting 5.6 from 2.0. That's 2.0 - 5.6. Hmm, which is -3.6.

Similarly, 5.6 + 2 is 7.6.

So, the solution set is -3.6 ≤ x ≤ 7.6. Therefore, x can be any real number between -3.6 and 7.6, inclusive.

But the question is asking for how many integers are in this solution set. So, I need to find all integers x such that -3.6 ≤ x ≤ 7.6.

Integers are whole numbers, positive and negative, including zero. So, the integers in this range would start from the smallest integer greater than or equal to -3.6 and go up to the largest integer less than or equal to 7.6.

Wait, what's the smallest integer greater than or equal to -3.6? Since -3.6 is between -4 and -3, the smallest integer greater than or equal to -3.6 is -3. Because -4 is less than -3.6, so the next integer up is -3.

Similarly, the largest integer less than or equal to 7.6 is 7, since 7.6 is between 7 and 8. So, the upper bound is 7.

Therefore, the integers from -3 up to 7 inclusive. Now, let's list them out to count:

-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7.

Counting these: let's see, starting at -3, each subsequent number is +1. How many numbers is that?

From -3 to 7. The number of integers from a to b inclusive is b - a + 1. Wait, is that right? Let's test with a smaller range. From 1 to 3: 3 - 1 + 1 = 3 numbers (1,2,3). Correct. So, applying that formula here: 7 - (-3) + 1 = 7 + 3 + 1 = 11. So, 11 integers.

But let me verify by listing them:

-3 (1), -2 (2), -1 (3), 0 (4), 1 (5), 2 (6), 3 (7), 4 (8), 5 (9), 6 (10), 7 (11). Yep, 11 numbers.

Wait, but let me check if the endpoints are included. The original inequality is |x - 2| ≤ 5.6, which includes equality. So, x can be exactly -3.6 and 7.6. However, since we are looking for integers, the endpoints -3.6 and 7.6 are not integers themselves. The integers must satisfy x ≥ -3.6 and x ≤ 7.6. So, the integer x must be greater than or equal to -3.6, which means x can be -3, -2, etc., and less than or equal to 7.6, so up to 7.

Therefore, the integers are from -3 to 7 inclusive, which is 11 integers. So, the answer is 11.

But wait, let me make sure I didn't make a mistake in calculating the number of integers. Sometimes when dealing with inclusive counts, it's easy to miscount. Let's use another approach.

If the integers start at -3 and end at 7, the distance between them is 7 - (-3) = 10. But since both endpoints are included, we add 1. So, 10 + 1 = 11. That matches the previous count. So, that seems correct.

Alternatively, think of the integers on the number line. From -3 to 7, how many markers? Each step is +1. The number of markers is the difference plus one. So, 7 - (-3) = 10, plus 1 is 11. Yep.

Therefore, I think the answer is 11. Let me just recap the steps to ensure there are no errors.

1. Start with |x - 2| ≤ 5.6.
2. Rewrite as -5.6 ≤ x - 2 ≤ 5.6.
3. Add 2 to all parts: -5.6 + 2 = -3.6, 5.6 + 2 = 7.6. So, -3.6 ≤ x ≤ 7.6.
4. Find integers x in this interval. Since x must be integer, x can be from -3 (since -3.6 is not an integer, next integer is -3) up to 7 (since 7.6 is not integer, previous integer is 7).
5. Count integers from -3 to 7 inclusive: 11 integers.

Everything checks out. I don't see any mistakes here. So, the final answer should be 11.
Therefore, the number of integers in the solution set of $|x-2|\leq5.6$ is $\boxed{11}$.
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