Explanation:
Okay, so I need to find the probability that a randomly selected point (x, y) within the rectangle defined by 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4 satisfies the condition x + y ≤ 4. Hmm, let me think about how to approach this.

First, probability problems like this usually involve finding the ratio of the favorable area to the total area. So, I should start by calculating the total area of the region where the point can be selected. The rectangle has length 8 along the x-axis and 4 along the y-axis, so the total area is 8 * 4. Let me write that down:

Total area = 8 * 4 = 32.

Okay, that seems straightforward. Now, the next part is figuring out the area of the region where x + y ≤ 4. This is the favorable area. To visualize this, I can imagine the coordinate system with x from 0 to 8 and y from 0 to 4. The line x + y = 4 would be a straight line. Let me sketch this mentally. When x = 0, y = 4, and when y = 0, x = 4. So, the line connects the points (0, 4) and (4, 0). Beyond x = 4, if we go further along the x-axis, the line x + y = 4 would require y to be negative, which isn't in our region since y ≥ 0. Similarly, for y beyond 4, but since our y only goes up to 4, the line x + y = 4 will intersect the rectangle at (0, 4) and (4, 0).

Therefore, the region where x + y ≤ 4 within the rectangle is a right triangle with vertices at (0, 0), (0, 4), and (4, 0). Wait, is that right? Let me confirm. If I consider all points below the line x + y = 4, then yes, from (0,0) up to (0,4) and (4,0). So, the area under the line within the rectangle is a triangle.

But hold on, the rectangle goes up to x = 8 and y = 4. However, the line x + y = 4 only intersects the rectangle at (0, 4) and (4, 0). So, beyond x = 4, the line would go outside the rectangle. Therefore, the region x + y ≤ 4 is entirely within the rectangle, and it's a triangle with base and height of 4 units each.

The area of a triangle is (base * height)/2. So, substituting the values, the area would be (4 * 4)/2 = 8. Wait, but let me make sure. If the base is along the y-axis from (0,0) to (0,4), that's 4 units, and the height is along the x-axis from (0,0) to (4,0), which is also 4 units. So yes, the area is (4*4)/2 = 8.

But hold on, is there another part of the region x + y ≤ 4 that I'm missing? For example, if x is between 0 and 4, and y is between 0 and 4 - x. But since the rectangle's y only goes up to 4, and for x between 0 and 4, 4 - x is between 0 and 4. So, the area under the line x + y = 4 is indeed that triangle.

Therefore, the favorable area is 8. Then, the probability is favorable area divided by total area, which is 8 / 32 = 1/4. Wait, that seems straightforward, but let me check again.

Alternatively, maybe I can set up an integral to compute the area where x + y ≤ 4. Let me try that.

For x from 0 to 4, y can go from 0 to 4 - x. For x from 4 to 8, since 4 - x would be negative, but y can't be negative, so the upper limit for y is 0. Therefore, the area would be the integral from x = 0 to 4 of (4 - x) dx plus the integral from x = 4 to 8 of 0 dx.

Calculating the first integral: ∫₀⁴ (4 - x) dx. The antiderivative is 4x - (x²)/2. Evaluated from 0 to 4:

At x = 4: 4*4 - (4²)/2 = 16 - 16/2 = 16 - 8 = 8.

At x = 0: 0 - 0 = 0. So the integral is 8 - 0 = 8. The second integral is 0. So total area is 8, same as before.

Therefore, the probability is 8 / 32 = 1/4.

But wait, let me think again. The total area is 8*4=32. The area where x + y ≤ 4 is 8. So 8/32 is 1/4. So the probability is 1/4? That seems low, but maybe it's correct. Let me visualize again. The triangle area is 8, and the total area is 32. So 8/32 is 1/4.

Alternatively, maybe I can think of the unit square and scale it. But in this case, the rectangle is 8 by 4. However, the region x + y ≤ 4 is a triangle with area 8.

Alternatively, if I consider the line x + y = 4, which cuts the rectangle. Since the rectangle is larger, but the area under the line is only a triangle.

Alternatively, maybe I can check with coordinates. For example, if x and y are both between 0 and 4, then the area where x + y ≤ 4 is 8, as we said. But if the rectangle was 4 by 4, then the area would be 8, and the total area would be 16, so probability 1/2. But in our case, the rectangle is 8 by 4, so the area where x + y ≤ 4 is still 8, but the total area is 32, so 8/32 = 1/4.

Alternatively, maybe I can think of the probability as the area of the triangle divided by the area of the rectangle.

But let me check if there's another way. For example, if we consider the region x + y ≤ 4, then for each x from 0 to 4, y can be from 0 to 4 - x, and for x from 4 to 8, y can't be anything because 4 - x would be negative. So the area is indeed the integral from 0 to 4 of (4 - x) dx, which is 8.

Alternatively, maybe I can think of the coordinates. The line x + y = 4 intersects the rectangle at (0, 4) and (4, 0). So the region x + y ≤ 4 is a triangle with vertices at (0, 0), (0, 4), and (4, 0). The area of this triangle is (base * height)/2. The base can be considered as the side from (0, 0) to (0, 4), which is 4 units, and the height is from (0, 0) to (4, 0), which is 4 units. So area is (4 * 4)/2 = 8.

Therefore, the probability is 8 / 32 = 1/4.

But wait, let me check if there's a mistake here. Suppose I have a rectangle that's 8 units long in x and 4 units in y. The line x + y = 4 cuts through the rectangle. But if I imagine the entire rectangle, the area where x + y ≤ 4 is only the triangle we discussed. However, is there a possibility that for some x beyond 4, y can still be such that x + y ≤ 4? For example, if x is 5, then y would have to be ≤ -1, which is outside the rectangle. Similarly, for x between 0 and 4, y can be up to 4 - x. So, no, there's no other region.

Therefore, the area is indeed 8.

Alternatively, maybe I can think of the probability as follows: since x ranges from 0 to 8 and y from 0 to 4, the condition x + y ≤ 4 is equivalent to y ≤ 4 - x. However, since y cannot be negative, we have 0 ≤ y ≤ 4 - x. But for this to be valid, 4 - x must be ≥ 0, so x ≤ 4. Therefore, the region is x from 0 to 4 and y from 0 to 4 - x.

Therefore, the area is ∫₀⁴ (4 - x) dx = 8, as before.

So, probability is 8 / 32 = 1/4.

But let me check with another approach. Let's think of the probability as the double integral over the region x + y ≤ 4.

The probability is (1 / (8 * 4)) * ∫∫_{x + y ≤ 4} dx dy. Which is the same as (1 / 32) * 8 = 1/4.

Alternatively, maybe I can think of the problem in terms of uniform distribution. Since the point is selected uniformly over the rectangle, the probability is proportional to the area.

Alternatively, maybe I can think of scaling. If I consider the rectangle from (0, 0) to (8, 4), and the line x + y = 4. If I scale the rectangle down, but I don't know if that helps.

Alternatively, maybe I can think of the problem as follows: For the condition x + y ≤ 4, since x can be up to 8, but y is limited to 4. However, since x + y ≤ 4, x can't be more than 4, because if x is more than 4, then even if y is 0, x would be more than 4, so x + y would be more than 4. Therefore, the x is restricted to 0 ≤ x ≤ 4, and y is restricted to 0 ≤ y ≤ 4 - x.

Therefore, the area is the same as the triangle with vertices at (0, 0), (0, 4), (4, 0).

So, area is 8.

Therefore, probability is 8 / 32 = 1/4.

But wait, let me check with actual numbers. Suppose I take a smaller case. Let me imagine if the rectangle was 4 by 4, so total area 16. Then the area where x + y ≤ 4 would be the triangle with area 8, so probability 8 / 16 = 1/2. Which makes sense. If the rectangle was 4 by 4, then the line x + y = 4 divides it into two regions, each with area 8. So probability 1/2.

But in our problem, the rectangle is 8 by 4. So, the same triangle area of 8, but total area is 32, so probability is 1/4. That seems correct.

Alternatively, if I think of the rectangle as two parts: from x = 0 to 4, and x = 4 to 8. For x from 0 to 4, the y ranges from 0 to 4 - x, and for x from 4 to 8, y can't satisfy x + y ≤ 4 because even if y is 0, x is already 4 or more. Therefore, the area is only the triangle part.

Therefore, I think my answer is correct.

But just to be thorough, let me check with coordinates. Let me pick some points. For example, the point (0, 0) is in the region. The point (4, 0) is in the region. The point (0, 4) is in the region. The point (2, 2) is in the region because 2 + 2 = 4. The point (3, 1) is in the region because 3 + 1 = 4. The point (1, 3) is in the region. The point (5, 0) is not in the region because 5 + 0 = 5 > 4. The point (4, 1) is not in the region because 4 + 1 = 5 > 4.

So, the region is indeed the triangle.

Alternatively, if I think of the area of the triangle, which is 8, and the total area is 32, then 8/32 is 1/4.

Therefore, the probability is 1/4.

But wait, let me check once more. Maybe I made a mistake in calculating the area of the triangle. The triangle has vertices at (0, 0), (0, 4), and (4, 0). The base is from (0, 0) to (0, 4), which is 4 units. The height is from (0, 0) to (4, 0), which is 4 units. But in a triangle, the area is (base * height)/2. So, (4 * 4)/2 = 8. Yes, that's correct.

Alternatively, if I use coordinates to calculate the area, using the formula for the area of a triangle given three vertices. The formula is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|.

Let me apply that. Let the three points be A(0, 0), B(0, 4), and C(4, 0).

Plugging into the formula:

Area = |(0*(4 - 0) + 0*(0 - 0) + 4*(0 - 4))/2|

= |(0 + 0 + 4*(-4))/2|

= |(-16)/2| = | -8 | = 8.

Yes, same result.

Therefore, the area is indeed 8.

So, probability is 8 / 32 = 1/4.

Therefore, I think the answer is 1/4.

But just to make sure, let me think if there's any other region that I might have missed. For example, if x is between 0 and 4, and y is between 0 and 4, but also considering that y can't exceed 4. However, since x + y ≤ 4, and y is already limited to 4, but for x between 0 and 4, y is limited to 4 - x, which is less than or equal to 4. So, no, there's no other region.

Alternatively, if I consider the entire rectangle, and the line x + y = 4, then the area below the line is the triangle we discussed.

Alternatively, if I imagine the rectangle from (0, 0) to (8, 4), and the line x + y = 4, which cuts the rectangle at (0, 4) and (4, 0). So, the area under the line is the triangle.

Therefore, I think my answer is correct.

So, the probability is 1/4.

**Final Answer**
The probability is \boxed{\dfrac{1}{4}}.

To determine the probability that a randomly selected point $(x, y)$ within the rectangle defined by $0 \le x \le 8$ and $0 \le y \le 4$ satisfies the condition $x + y \le 4$, we approach the problem using geometric probability.

---

### Step 1: Total Area of the Region

The region where the point $(x, y)$ can be selected is a rectangle with:

- Width = $8$ (from $x = 0$ to $x = 8$)
- Height = $4$ (from $y = 0$ to $y = 4$)

So, the **total area** of the region is:

$$
\text{Total Area} = 8 \times 4 = 32
$$

---

### Step 2: Area of the Region Satisfying $x + y \le 4$

We are interested in the region where $x + y \le 4$. This inequality defines a region below the line $x + y = 4$.

- The line $x + y = 4$ intersects the rectangle at:
  - When $x = 0$, $y = 4$ → point $(0, 4)$
  - When $y = 0$, $x = 4$ → point $(4, 0)$

Thus, the region $x + y \le 4$ within the rectangle is a **right triangle** with vertices at:

- $(0, 0)$
- $(0, 4)$
- $(4, 0)$

This triangle lies entirely within the rectangle and is bounded by the axes and the line $x + y = 4$.

---

### Step 3: Area of the Triangle

The triangle has:

- Base = 4 (from $(0, 0)$ to $(0, 4)$)
- Height = 4 (from $(0, 0)$ to $(4, 0)$)

So, the **area of the triangle** is:

$$
\text{Favorable Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8
$$

---

### Step 4: Compute the Probability

The probability that a randomly selected point satisfies $x + y \le 4$ is the ratio of the favorable area to the total area:

$$
\text{Probability} = \frac{\text{Favorable Area}}{\text{Total Area}} = \frac{8}{32} = \frac{1}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a randomly selected point (x, y) within the rectangle defined by 0 ≤ x ≤ 8 and 0 ≤ y ≤ 4 satisfies the condition x + y ≤ 4. Hmm, let me think about how to approach this.

First, probability problems like this usually involve finding the ratio of the favorable area to the total area. So, I should start by calculating the total area of the region where the point can be selected. The rectangle has length 8 along the x-axis and 4 along the y-axis, so the total area is 8 * 4. Let me write that down:

Total area = 8 * 4 = 32.

Okay, that seems straightforward. Now, the next part is figuring out the area of the region where x + y ≤ 4. This is the favorable area. To visualize this, I can imagine the coordinate system with x from 0 to 8 and y from 0 to 4. The line x + y = 4 would be a straight line. Let me sketch this mentally. When x = 0, y = 4, and when y = 0, x = 4. So, the line connects the points (0, 4) and (4, 0). Beyond x = 4, if we go further along the x-axis, the line x + y = 4 would require y to be negative, which isn't in our region since y ≥ 0. Similarly, for y beyond 4, but since our y only goes up to 4, the line x + y = 4 will intersect the rectangle at (0, 4) and (4, 0).

Therefore, the region where x + y ≤ 4 within the rectangle is a right triangle with vertices at (0, 0), (0, 4), and (4, 0). Wait, is that right? Let me confirm. If I consider all points below the line x + y = 4, then yes, from (0,0) up to (0,4) and (4,0). So, the area under the line within the rectangle is a triangle.

But hold on, the rectangle goes up to x = 8 and y = 4. However, the line x + y = 4 only intersects the rectangle at (0, 4) and (4, 0). So, beyond x = 4, the line would go outside the rectangle. Therefore, the region x + y ≤ 4 is entirely within the rectangle, and it's a triangle with base and height of 4 units each.

The area of a triangle is (base * height)/2. So, substituting the values, the area would be (4 * 4)/2 = 8. Wait, but let me make sure. If the base is along the y-axis from (0,0) to (0,4), that's 4 units, and the height is along the x-axis from (0,0) to (4,0), which is also 4 units. So yes, the area is (4*4)/2 = 8.

But hold on, is there another part of the region x + y ≤ 4 that I'm missing? For example, if x is between 0 and 4, and y is between 0 and 4 - x. But since the rectangle's y only goes up to 4, and for x between 0 and 4, 4 - x is between 0 and 4. So, the area under the line x + y = 4 is indeed that triangle.

Therefore, the favorable area is 8. Then, the probability is favorable area divided by total area, which is 8 / 32 = 1/4. Wait, that seems straightforward, but let me check again.

Alternatively, maybe I can set up an integral to compute the area where x + y ≤ 4. Let me try that.

For x from 0 to 4, y can go from 0 to 4 - x. For x from 4 to 8, since 4 - x would be negative, but y can't be negative, so the upper limit for y is 0. Therefore, the area would be the integral from x = 0 to 4 of (4 - x) dx plus the integral from x = 4 to 8 of 0 dx.

Calculating the first integral: ∫₀⁴ (4 - x) dx. The antiderivative is 4x - (x²)/2. Evaluated from 0 to 4:

At x = 4: 4*4 - (4²)/2 = 16 - 16/2 = 16 - 8 = 8.

At x = 0: 0 - 0 = 0. So the integral is 8 - 0 = 8. The second integral is 0. So total area is 8, same as before.

Therefore, the probability is 8 / 32 = 1/4.

But wait, let me think again. The total area is 8*4=32. The area where x + y ≤ 4 is 8. So 8/32 is 1/4. So the probability is 1/4? That seems low, but maybe it's correct. Let me visualize again. The triangle area is 8, and the total area is 32. So 8/32 is 1/4.

Alternatively, maybe I can think of the unit square and scale it. But in this case, the rectangle is 8 by 4. However, the region x + y ≤ 4 is a triangle with area 8.

Alternatively, if I consider the line x + y = 4, which cuts the rectangle. Since the rectangle is larger, but the area under the line is only a triangle.

Alternatively, maybe I can check with coordinates. For example, if x and y are both between 0 and 4, then the area where x + y ≤ 4 is 8, as we said. But if the rectangle was 4 by 4, then the area would be 8, and the total area would be 16, so probability 1/2. But in our case, the rectangle is 8 by 4, so the area where x + y ≤ 4 is still 8, but the total area is 32, so 8/32 = 1/4.

Alternatively, maybe I can think of the probability as the area of the triangle divided by the area of the rectangle.

But let me check if there's another way. For example, if we consider the region x + y ≤ 4, then for each x from 0 to 4, y can be from 0 to 4 - x, and for x from 4 to 8, y can't be anything because 4 - x would be negative. So the area is indeed the integral from 0 to 4 of (4 - x) dx, which is 8.

Alternatively, maybe I can think of the coordinates. The line x + y = 4 intersects the rectangle at (0, 4) and (4, 0). So the region x + y ≤ 4 is a triangle with vertices at (0, 0), (0, 4), and (4, 0). The area of this triangle is (base * height)/2. The base can be considered as the side from (0, 0) to (0, 4), which is 4 units, and the height is from (0, 0) to (4, 0), which is 4 units. So area is (4 * 4)/2 = 8.

Therefore, the probability is 8 / 32 = 1/4.

But wait, let me check if there's a mistake here. Suppose I have a rectangle that's 8 units long in x and 4 units in y. The line x + y = 4 cuts through the rectangle. But if I imagine the entire rectangle, the area where x + y ≤ 4 is only the triangle we discussed. However, is there a possibility that for some x beyond 4, y can still be such that x + y ≤ 4? For example, if x is 5, then y would have to be ≤ -1, which is outside the rectangle. Similarly, for x between 0 and 4, y can be up to 4 - x. So, no, there's no other region.

Therefore, the area is indeed 8.

Alternatively, maybe I can think of the probability as follows: since x ranges from 0 to 8 and y from 0 to 4, the condition x + y ≤ 4 is equivalent to y ≤ 4 - x. However, since y cannot be negative, we have 0 ≤ y ≤ 4 - x. But for this to be valid, 4 - x must be ≥ 0, so x ≤ 4. Therefore, the region is x from 0 to 4 and y from 0 to 4 - x.

Therefore, the area is ∫₀⁴ (4 - x) dx = 8, as before.

So, probability is 8 / 32 = 1/4.

But let me check with another approach. Let's think of the probability as the double integral over the region x + y ≤ 4.

The probability is (1 / (8 * 4)) * ∫∫_{x + y ≤ 4} dx dy. Which is the same as (1 / 32) * 8 = 1/4.

Alternatively, maybe I can think of the problem in terms of uniform distribution. Since the point is selected uniformly over the rectangle, the probability is proportional to the area.

Alternatively, maybe I can think of scaling. If I consider the rectangle from (0, 0) to (8, 4), and the line x + y = 4. If I scale the rectangle down, but I don't know if that helps.

Alternatively, maybe I can think of the problem as follows: For the condition x + y ≤ 4, since x can be up to 8, but y is limited to 4. However, since x + y ≤ 4, x can't be more than 4, because if x is more than 4, then even if y is 0, x would be more than 4, so x + y would be more than 4. Therefore, the x is restricted to 0 ≤ x ≤ 4, and y is restricted to 0 ≤ y ≤ 4 - x.

Therefore, the area is the same as the triangle with vertices at (0, 0), (0, 4), (4, 0).

So, area is 8.

Therefore, probability is 8 / 32 = 1/4.

But wait, let me check with actual numbers. Suppose I take a smaller case. Let me imagine if the rectangle was 4 by 4, so total area 16. Then the area where x + y ≤ 4 would be the triangle with area 8, so probability 8 / 16 = 1/2. Which makes sense. If the rectangle was 4 by 4, then the line x + y = 4 divides it into two regions, each with area 8. So probability 1/2.

But in our problem, the rectangle is 8 by 4. So, the same triangle area of 8, but total area is 32, so probability is 1/4. That seems correct.

Alternatively, if I think of the rectangle as two parts: from x = 0 to 4, and x = 4 to 8. For x from 0 to 4, the y ranges from 0 to 4 - x, and for x from 4 to 8, y can't satisfy x + y ≤ 4 because even if y is 0, x is already 4 or more. Therefore, the area is only the triangle part.

Therefore, I think my answer is correct.

But just to be thorough, let me check with coordinates. Let me pick some points. For example, the point (0, 0) is in the region. The point (4, 0) is in the region. The point (0, 4) is in the region. The point (2, 2) is in the region because 2 + 2 = 4. The point (3, 1) is in the region because 3 + 1 = 4. The point (1, 3) is in the region. The point (5, 0) is not in the region because 5 + 0 = 5 > 4. The point (4, 1) is not in the region because 4 + 1 = 5 > 4.

So, the region is indeed the triangle.

Alternatively, if I think of the area of the triangle, which is 8, and the total area is 32, then 8/32 is 1/4.

Therefore, the probability is 1/4.

But wait, let me check once more. Maybe I made a mistake in calculating the area of the triangle. The triangle has vertices at (0, 0), (0, 4), and (4, 0). The base is from (0, 0) to (0, 4), which is 4 units. The height is from (0, 0) to (4, 0), which is 4 units. But in a triangle, the area is (base * height)/2. So, (4 * 4)/2 = 8. Yes, that's correct.

Alternatively, if I use coordinates to calculate the area, using the formula for the area of a triangle given three vertices. The formula is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|.

Let me apply that. Let the three points be A(0, 0), B(0, 4), and C(4, 0).

Plugging into the formula:

Area = |(0*(4 - 0) + 0*(0 - 0) + 4*(0 - 4))/2|

= |(0 + 0 + 4*(-4))/2|

= |(-16)/2| = | -8 | = 8.

Yes, same result.

Therefore, the area is indeed 8.

So, probability is 8 / 32 = 1/4.

Therefore, I think the answer is 1/4.

But just to make sure, let me think if there's any other region that I might have missed. For example, if x is between 0 and 4, and y is between 0 and 4, but also considering that y can't exceed 4. However, since x + y ≤ 4, and y is already limited to 4, but for x between 0 and 4, y is limited to 4 - x, which is less than or equal to 4. So, no, there's no other region.

Alternatively, if I consider the entire rectangle, and the line x + y = 4, then the area below the line is the triangle we discussed.

Alternatively, if I imagine the rectangle from (0, 0) to (8, 4), and the line x + y = 4, which cuts the rectangle at (0, 4) and (4, 0). So, the area under the line is the triangle.

Therefore, I think my answer is correct.

So, the probability is 1/4.

**Final Answer**
The probability is \boxed{\dfrac{1}{4}}.

To determine the probability that a randomly selected point $(x, y)$ within the rectangle defined by $0 \le x \le 8$ and $0 \le y \le 4$ satisfies the condition $x + y \le 4$, we approach the problem using geometric probability.

---

### Step 1: Total Area of the Region

The region where the point $(x, y)$ can be selected is a rectangle with:

- Width = $8$ (from $x = 0$ to $x = 8$)
- Height = $4$ (from $y = 0$ to $y = 4$)

So, the **total area** of the region is:

$$
\text{Total Area} = 8 \times 4 = 32
$$

---

### Step 2: Area of the Region Satisfying $x + y \le 4$

We are interested in the region where $x + y \le 4$. This inequality defines a region below the line $x + y = 4$.

- The line $x + y = 4$ intersects the rectangle at:
  - When $x = 0$, $y = 4$ → point $(0, 4)$
  - When $y = 0$, $x = 4$ → point $(4, 0)$

Thus, the region $x + y \le 4$ within the rectangle is a **right triangle** with vertices at:

- $(0, 0)$
- $(0, 4)$
- $(4, 0)$

This triangle lies entirely within the rectangle and is bounded by the axes and the line $x + y = 4$.

---

### Step 3: Area of the Triangle

The triangle has:

- Base = 4 (from $(0, 0)$ to $(0, 4)$)
- Height = 4 (from $(0, 0)$ to $(4, 0)$)

So, the **area of the triangle** is:

$$
\text{Favorable Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8
$$

---

### Step 4: Compute the Probability

The probability that a randomly selected point satisfies $x + y \le 4$ is the ratio of the favorable area to the total area:

$$
\text{Probability} = \frac{\text{Favorable Area}}{\text{Total Area}} = \frac{8}{32} = \frac{1}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{4}}
$$

Solution:
Rewrite $x+y\leq 4$ as $y\leq 4-x$.  This inequality is satisfied by the points on and under the line $y=4-x$.  Sketching this line along with the $4\times 8$ rectangle determined by the inequalities $0\leq x\leq 8$ and $0\leq y\leq 4$, we find that the points satisfying $x+y\leq 4$ are those in the shaded triangle (see figure).  The  area of the triangle is $\frac{1}{2}(4)(4)=8$ square units, and the area of the rectangle is $(4)(8)=32$ square units, so the probability that a randomly selected point would fall in the shaded triangle is $\boxed{\frac{1}{4}}$.

[asy] import graph; size(200); defaultpen(linewidth(0.7)+fontsize(10));

dotfactor=4;

real f(real x) { return 4-x; }

pair A=(0,4), B=(8,4), C=(8,0), D=(0,0); pair[] dots={A,B,C,D};

fill(A--(4,0)--D--cycle,gray(0.7)); draw(A--B--C);

xaxis(xmin=-3,xmax=9,Ticks(" ",1.0, begin=false, end=false, NoZero, Size=3), Arrows(4), above=true);

yaxis(ymin=-1,ymax=5,Ticks(" ",1.0,begin=false, end=false, NoZero, Size=3), Arrows(4), above=true);

draw(graph(f,-0.8,4.5),Arrows(4)); label("$x+y=4$",(-2.2,5.2));[/asy]


Explanation:
Okay, so I need to find the sum of all multiples of 7 between 100 and 200. Hmm, let me think about how to approach this. First, I remember that when you want to find the sum of a series of numbers, especially an arithmetic sequence, there's a formula for that. But I need to make sure I understand the problem first.

Alright, the question is asking for multiples of 7 that are between 100 and 200. Wait, does "between 100 and 200" include 100 and 200 or not? Hmm, sometimes "between" can be ambiguous. But since 100 and 200 are both multiples of 7? Let me check. 100 divided by 7 is approximately 14.28, so 14*7 is 98, and 15*7 is 105. So 100 isn't a multiple of 7. Similarly, 200 divided by 7 is about 28.57, so 28*7 is 196, and 29*7 is 203. So 200 isn't a multiple of 7 either. Therefore, the multiples of 7 between 100 and 200 would be numbers starting from the first multiple of 7 that is greater than 100 up to the last multiple of 7 that is less than 200. So, I need to find the first and last terms of this sequence and then use the formula for the sum of an arithmetic series.

Let me recall the formula for the sum of an arithmetic series. The formula is S = n/2 * (a1 + an), where S is the sum, n is the number of terms, a1 is the first term, and an is the last term. Alternatively, sometimes it's written as S = n * (a1 + an)/2. Either way, it's the same thing. So I need to figure out n, a1, and an.

First, let me find the first multiple of 7 that is greater than 100. As I calculated earlier, 14*7 is 98, which is less than 100, so the next one is 15*7. Let me compute that: 15*7 is 105. So 105 is the first multiple of 7 in this range. Then, the last multiple of 7 less than 200. Let me check 28*7 is 196, and 29*7 is 203, which is over 200. So 196 is the last one. Therefore, the sequence is 105, 112, 119, ..., 196.

Now, I need to find how many terms are in this sequence. To do that, I can use the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)*d, where d is the common difference. In this case, the common difference is 7 because each term increases by 7. So, let me set up the equation:

196 = 105 + (n - 1)*7

Let me solve for n. Subtract 105 from both sides:

196 - 105 = (n - 1)*7

Calculating 196 - 105: 196 - 100 is 96, minus 5 more is 91. So:

91 = (n - 1)*7

Divide both sides by 7:

91 / 7 = n - 1

Which is 13 = n - 1

Therefore, n = 14. So there are 14 terms in this sequence.

Wait, let me check that again. If the first term is 105 (which is 15*7) and the last term is 196 (which is 28*7), then the number of terms is 28 - 15 + 1. Let me compute that: 28 - 15 is 13, plus 1 is 14. Yes, that matches. So that's 14 terms. So n is 14.

Now, using the sum formula: S = n/2 * (a1 + an). So plugging in the numbers:

S = 14/2 * (105 + 196)

Simplify 14/2 to 7:

S = 7 * (105 + 196)

Compute 105 + 196. Let me do that: 100 + 196 is 296, plus 5 is 301. So:

S = 7 * 301

Now, compute 7 * 301. Let me break that down. 7 * 300 is 2100, and 7 * 1 is 7, so total is 2100 + 7 = 2107. Therefore, the sum should be 2107.

Wait, let me verify this with another method to make sure I didn't make a mistake. Another way to calculate the sum is to use the formula for the sum of an arithmetic series, which is also S = n * (a1 + an)/2. Which is the same as above. So, same result.

Alternatively, maybe I can check by calculating the number of terms again. Let me list out the multiples of 7 between 100 and 200. Starting at 105:

105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196.

Let me count them. Let's see:

1. 105
2. 112
3. 119
4. 126
5. 133
6. 140
7. 147
8. 154
9. 161
10. 168
11. 175
12. 182
13. 189
14. 196

Yes, that's 14 terms. So that checks out. Now, adding them up. Maybe I can add them in pairs to make it easier. For example, the first and last terms: 105 + 196 = 301. Then the second and second last: 112 + 189 = 301. Similarly, third and third last: 119 + 182 = 301. Fourth and fourth last: 126 + 175 = 301. Fifth and fifth last: 133 + 168 = 301. Sixth and sixth last: 140 + 161 = 301. Seventh and seventh last: 147 + 154 = 301. Wait, how many pairs is that?

Since there are 14 terms, pairing them up would give 7 pairs. Each pair adds up to 301. Therefore, total sum is 7 * 301 = 2107. Which matches the previous result. So that seems correct.

Alternatively, if I wanted to use another formula, maybe the sum of multiples. Let me think. The formula for the sum of multiples of a number within a range can also be calculated by finding the number of terms and then multiplying by the average of the first and last term. Which is exactly what I did.

Alternatively, if I wanted to use the formula for the sum of an arithmetic series starting from a certain term, but I think the way I did it is solid.

But just to make sure, let me check with another approach. Let me calculate the sum of all multiples of 7 up to 200, then subtract the sum of multiples of 7 up to 100. Wait, but the problem says between 100 and 200. But if "between" is exclusive, then we need to subtract the multiples of 7 less than or equal to 100 and those less than or equal to 200. Wait, but earlier I considered that 100 and 200 are not multiples of 7, so maybe that approach would work.

Let me try that. Let me compute the sum of all multiples of 7 up to 200, then subtract the sum of all multiples of 7 up to 100. That should give me the sum of multiples of 7 between 100 and 200, assuming that "between" is exclusive. Wait, but if "between 100 and 200" includes 100 and 200, but since they aren't multiples of 7, it doesn't affect the sum. So either way, the result would be the same.

So let me try this method. First, find the sum of multiples of 7 up to 200. The last multiple of 7 less than or equal to 200 is 196, as before. The number of terms is floor(200/7). Let me compute that. 200 divided by 7 is approximately 28.57, so floor(200/7) is 28. Therefore, there are 28 terms. The first term is 7*1=7, and the last term is 7*28=196. Therefore, the sum is (number of terms)/2 * (first term + last term) = 28/2 * (7 + 196) = 14 * 203. Let me compute that. 14*200 is 2800, and 14*3 is 42, so total is 2800 + 42 = 2842.

Now, the sum of multiples of 7 up to 100. The last multiple of 7 less than or equal to 100 is 98, which is 14*7. The number of terms is floor(100/7) = 14. So the first term is 7, last term is 98. Sum is 14/2 * (7 + 98) = 7 * 105 = 735. Therefore, the sum of multiples of 7 between 100 and 200 would be 2842 - 735. Let me compute that. 2842 - 700 is 2142, then subtract 35 more: 2142 - 35 = 2107. Which matches the previous result. So that's reassuring.

Therefore, regardless of the method, the answer is 2107. So I think that's correct.

But just to be thorough, let me check with another approach. Let me think about the formula for the sum of an arithmetic series. The formula is S = n/2 * [2a + (n - 1)d]. Where a is the first term, d is the common difference, and n is the number of terms. Let me try that with the first method.

In our case, a = 105, d = 7, n = 14. So plugging into the formula:

S = 14/2 * [2*105 + (14 - 1)*7] = 7 * [210 + 13*7]. Compute 13*7: 91. Then 210 + 91 = 301. Then 7*301 = 2107. Same result. So that's consistent.

Alternatively, if I use the formula for the sum of multiples, which is also similar. So, all methods lead to 2107. Therefore, I think that's the correct answer.

But just to make sure, let me add up the numbers manually, at least partially. Let me take the list I had before:

105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196.

Let me add them step by step. Starting with 105 + 112 = 217. Then 217 + 119 = 336. Then 336 + 126 = 462. Then 462 + 133 = 595. Then 595 + 140 = 735. Then 735 + 147 = 882. Then 882 + 154 = 1036. Then 1036 + 161 = 1197. Then 1197 + 168 = 1365. Then 1365 + 175 = 1540. Then 1540 + 182 = 1722. Then 1722 + 189 = 1911. Then 1911 + 196 = 2107. Wow, that's exactly the same result. So adding them all up step by step gives 2107. Therefore, that's solid.

Therefore, I can be confident that the sum of all multiples of 7 between 100 and 200 is 2107.

**Final Answer**
The sum of all multiples of 7 between 100 and 200 is \boxed{2107}.

To find the **sum of all multiples of 7 between 100 and 200**, we can approach the problem using the properties of arithmetic sequences.

---

### Step 1: Identify the First and Last Multiples of 7 in the Range

We are looking for all **multiples of 7** that lie **strictly between 100 and 200**. That is, we want numbers of the form $ 7k $ such that:

$$
100 < 7k < 200
$$

Divide the inequality by 7:

$$
\frac{100}{7} < k < \frac{200}{7}
$$

$$
14.2857 < k < 28.5714
$$

So, the integer values of $ k $ are from 15 to 28 (inclusive). Therefore, the first multiple of 7 in the range is:

$$
7 \times 15 = 105
$$

And the last multiple of 7 in the range is:

$$
7 \times 28 = 196
$$

---

### Step 2: Determine the Number of Terms

The sequence of multiples of 7 from 105 to 196 is an arithmetic sequence with:

- First term $ a = 105 $
- Last term $ l = 196 $
- Common difference $ d = 7 $

The number of terms $ n $ in the sequence is:

$$
n = \frac{l - a}{d} + 1 = \frac{196 - 105}{7} + 1 = \frac{91}{7} + 1 = 13 + 1 = 14
$$

---

### Step 3: Use the Arithmetic Series Sum Formula

The sum $ S $ of an arithmetic sequence is given by:

$$
S = \frac{n}{2} \times (a + l)
$$

Substitute the known values:

$$
S = \frac{14}{2} \times (105 + 196) = 7 \times 301 = 2107
$$

---

### Final Answer

$$
\boxed{2107}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of all multiples of 7 between 100 and 200. Hmm, let me think about how to approach this. First, I remember that when you want to find the sum of a series of numbers, especially an arithmetic sequence, there's a formula for that. But I need to make sure I understand the problem first.

Alright, the question is asking for multiples of 7 that are between 100 and 200. Wait, does "between 100 and 200" include 100 and 200 or not? Hmm, sometimes "between" can be ambiguous. But since 100 and 200 are both multiples of 7? Let me check. 100 divided by 7 is approximately 14.28, so 14*7 is 98, and 15*7 is 105. So 100 isn't a multiple of 7. Similarly, 200 divided by 7 is about 28.57, so 28*7 is 196, and 29*7 is 203. So 200 isn't a multiple of 7 either. Therefore, the multiples of 7 between 100 and 200 would be numbers starting from the first multiple of 7 that is greater than 100 up to the last multiple of 7 that is less than 200. So, I need to find the first and last terms of this sequence and then use the formula for the sum of an arithmetic series.

Let me recall the formula for the sum of an arithmetic series. The formula is S = n/2 * (a1 + an), where S is the sum, n is the number of terms, a1 is the first term, and an is the last term. Alternatively, sometimes it's written as S = n * (a1 + an)/2. Either way, it's the same thing. So I need to figure out n, a1, and an.

First, let me find the first multiple of 7 that is greater than 100. As I calculated earlier, 14*7 is 98, which is less than 100, so the next one is 15*7. Let me compute that: 15*7 is 105. So 105 is the first multiple of 7 in this range. Then, the last multiple of 7 less than 200. Let me check 28*7 is 196, and 29*7 is 203, which is over 200. So 196 is the last one. Therefore, the sequence is 105, 112, 119, ..., 196.

Now, I need to find how many terms are in this sequence. To do that, I can use the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)*d, where d is the common difference. In this case, the common difference is 7 because each term increases by 7. So, let me set up the equation:

196 = 105 + (n - 1)*7

Let me solve for n. Subtract 105 from both sides:

196 - 105 = (n - 1)*7

Calculating 196 - 105: 196 - 100 is 96, minus 5 more is 91. So:

91 = (n - 1)*7

Divide both sides by 7:

91 / 7 = n - 1

Which is 13 = n - 1

Therefore, n = 14. So there are 14 terms in this sequence.

Wait, let me check that again. If the first term is 105 (which is 15*7) and the last term is 196 (which is 28*7), then the number of terms is 28 - 15 + 1. Let me compute that: 28 - 15 is 13, plus 1 is 14. Yes, that matches. So that's 14 terms. So n is 14.

Now, using the sum formula: S = n/2 * (a1 + an). So plugging in the numbers:

S = 14/2 * (105 + 196)

Simplify 14/2 to 7:

S = 7 * (105 + 196)

Compute 105 + 196. Let me do that: 100 + 196 is 296, plus 5 is 301. So:

S = 7 * 301

Now, compute 7 * 301. Let me break that down. 7 * 300 is 2100, and 7 * 1 is 7, so total is 2100 + 7 = 2107. Therefore, the sum should be 2107.

Wait, let me verify this with another method to make sure I didn't make a mistake. Another way to calculate the sum is to use the formula for the sum of an arithmetic series, which is also S = n * (a1 + an)/2. Which is the same as above. So, same result.

Alternatively, maybe I can check by calculating the number of terms again. Let me list out the multiples of 7 between 100 and 200. Starting at 105:

105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196.

Let me count them. Let's see:

1. 105
2. 112
3. 119
4. 126
5. 133
6. 140
7. 147
8. 154
9. 161
10. 168
11. 175
12. 182
13. 189
14. 196

Yes, that's 14 terms. So that checks out. Now, adding them up. Maybe I can add them in pairs to make it easier. For example, the first and last terms: 105 + 196 = 301. Then the second and second last: 112 + 189 = 301. Similarly, third and third last: 119 + 182 = 301. Fourth and fourth last: 126 + 175 = 301. Fifth and fifth last: 133 + 168 = 301. Sixth and sixth last: 140 + 161 = 301. Seventh and seventh last: 147 + 154 = 301. Wait, how many pairs is that?

Since there are 14 terms, pairing them up would give 7 pairs. Each pair adds up to 301. Therefore, total sum is 7 * 301 = 2107. Which matches the previous result. So that seems correct.

Alternatively, if I wanted to use another formula, maybe the sum of multiples. Let me think. The formula for the sum of multiples of a number within a range can also be calculated by finding the number of terms and then multiplying by the average of the first and last term. Which is exactly what I did.

Alternatively, if I wanted to use the formula for the sum of an arithmetic series starting from a certain term, but I think the way I did it is solid.

But just to make sure, let me check with another approach. Let me calculate the sum of all multiples of 7 up to 200, then subtract the sum of multiples of 7 up to 100. Wait, but the problem says between 100 and 200. But if "between" is exclusive, then we need to subtract the multiples of 7 less than or equal to 100 and those less than or equal to 200. Wait, but earlier I considered that 100 and 200 are not multiples of 7, so maybe that approach would work.

Let me try that. Let me compute the sum of all multiples of 7 up to 200, then subtract the sum of all multiples of 7 up to 100. That should give me the sum of multiples of 7 between 100 and 200, assuming that "between" is exclusive. Wait, but if "between 100 and 200" includes 100 and 200, but since they aren't multiples of 7, it doesn't affect the sum. So either way, the result would be the same.

So let me try this method. First, find the sum of multiples of 7 up to 200. The last multiple of 7 less than or equal to 200 is 196, as before. The number of terms is floor(200/7). Let me compute that. 200 divided by 7 is approximately 28.57, so floor(200/7) is 28. Therefore, there are 28 terms. The first term is 7*1=7, and the last term is 7*28=196. Therefore, the sum is (number of terms)/2 * (first term + last term) = 28/2 * (7 + 196) = 14 * 203. Let me compute that. 14*200 is 2800, and 14*3 is 42, so total is 2800 + 42 = 2842.

Now, the sum of multiples of 7 up to 100. The last multiple of 7 less than or equal to 100 is 98, which is 14*7. The number of terms is floor(100/7) = 14. So the first term is 7, last term is 98. Sum is 14/2 * (7 + 98) = 7 * 105 = 735. Therefore, the sum of multiples of 7 between 100 and 200 would be 2842 - 735. Let me compute that. 2842 - 700 is 2142, then subtract 35 more: 2142 - 35 = 2107. Which matches the previous result. So that's reassuring.

Therefore, regardless of the method, the answer is 2107. So I think that's correct.

But just to be thorough, let me check with another approach. Let me think about the formula for the sum of an arithmetic series. The formula is S = n/2 * [2a + (n - 1)d]. Where a is the first term, d is the common difference, and n is the number of terms. Let me try that with the first method.

In our case, a = 105, d = 7, n = 14. So plugging into the formula:

S = 14/2 * [2*105 + (14 - 1)*7] = 7 * [210 + 13*7]. Compute 13*7: 91. Then 210 + 91 = 301. Then 7*301 = 2107. Same result. So that's consistent.

Alternatively, if I use the formula for the sum of multiples, which is also similar. So, all methods lead to 2107. Therefore, I think that's the correct answer.

But just to make sure, let me add up the numbers manually, at least partially. Let me take the list I had before:

105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196.

Let me add them step by step. Starting with 105 + 112 = 217. Then 217 + 119 = 336. Then 336 + 126 = 462. Then 462 + 133 = 595. Then 595 + 140 = 735. Then 735 + 147 = 882. Then 882 + 154 = 1036. Then 1036 + 161 = 1197. Then 1197 + 168 = 1365. Then 1365 + 175 = 1540. Then 1540 + 182 = 1722. Then 1722 + 189 = 1911. Then 1911 + 196 = 2107. Wow, that's exactly the same result. So adding them all up step by step gives 2107. Therefore, that's solid.

Therefore, I can be confident that the sum of all multiples of 7 between 100 and 200 is 2107.

**Final Answer**
The sum of all multiples of 7 between 100 and 200 is \boxed{2107}.

To find the **sum of all multiples of 7 between 100 and 200**, we can approach the problem using the properties of arithmetic sequences.

---

### Step 1: Identify the First and Last Multiples of 7 in the Range

We are looking for all **multiples of 7** that lie **strictly between 100 and 200**. That is, we want numbers of the form $ 7k $ such that:

$$
100 < 7k < 200
$$

Divide the inequality by 7:

$$
\frac{100}{7} < k < \frac{200}{7}
$$

$$
14.2857 < k < 28.5714
$$

So, the integer values of $ k $ are from 15 to 28 (inclusive). Therefore, the first multiple of 7 in the range is:

$$
7 \times 15 = 105
$$

And the last multiple of 7 in the range is:

$$
7 \times 28 = 196
$$

---

### Step 2: Determine the Number of Terms

The sequence of multiples of 7 from 105 to 196 is an arithmetic sequence with:

- First term $ a = 105 $
- Last term $ l = 196 $
- Common difference $ d = 7 $

The number of terms $ n $ in the sequence is:

$$
n = \frac{l - a}{d} + 1 = \frac{196 - 105}{7} + 1 = \frac{91}{7} + 1 = 13 + 1 = 14
$$

---

### Step 3: Use the Arithmetic Series Sum Formula

The sum $ S $ of an arithmetic sequence is given by:

$$
S = \frac{n}{2} \times (a + l)
$$

Substitute the known values:

$$
S = \frac{14}{2} \times (105 + 196) = 7 \times 301 = 2107
$$

---

### Final Answer

$$
\boxed{2107}
$$

Solution:
The smallest multiple of 7 between 100 and 200 is 105, and the largest multiple is 196.  Thus, we want to find the sum of the arithmetic series $105 + 112 + \dots + 196$.

The $n^{\text{th}}$ term in this arithmetic sequence is $105 + 7(n - 1) = 7n + 98$.  If $7n + 98 = 196$, then $n = 14$, so the number of terms in this sequence is 14.

The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so the sum is $(105 + 196)/2 \cdot 14 = \boxed{2107}$.


Explanation:
Okay, so I need to rewrite the quadratic equation $x^2 + 2.6x + 3.6$ in the form $(x + b)^2 + c$, where $b$ and $c$ are constants. Then, I have to find $b + c$ as a decimal. Hmm, let me think about how to approach this.

First, I remember that completing the square is the method used to convert a quadratic equation from standard form $ax^2 + bx + c$ to vertex form $a(x + b)^2 + c$. Wait, actually, in this case, the coefficient of $x^2$ is 1, so it's simpler. Let me recall the steps for completing the square.

The general process is: take the coefficient of $x$, divide it by 2, square it, and add and subtract that value to complete the square. Let me try applying that here.

Given the quadratic $x^2 + 2.6x + 3.6$. Let me write it down:

$x^2 + 2.6x + 3.6$

First, the coefficient of $x$ is 2.6. So, half of that is 1.3, right? Then, squaring that gives me $1.3^2$. Let me calculate that. 1.3 times 1.3... 1 times 1 is 1, 1 times 0.3 is 0.3, 0.3 times 1 is 0.3, and 0.3 times 0.3 is 0.09. Adding those up: 1 + 0.3 + 0.3 + 0.09 = 1.69. So, $1.3^2 = 1.69$.

So, to complete the square, I need to add and subtract 1.69. Let me do that:

$x^2 + 2.6x + 1.69 - 1.69 + 3.6$

Wait, but the original equation is $x^2 + 2.6x + 3.6$. So, if I add and subtract 1.69, the equation becomes:

$(x^2 + 2.6x + 1.69) - 1.69 + 3.6$

Simplifying the terms inside the parentheses first. Since $x^2 + 2.6x + 1.69$ is a perfect square trinomial, it can be written as $(x + 1.3)^2$. Let me check that:

$(x + 1.3)^2 = x^2 + 2*1.3*x + (1.3)^2 = x^2 + 2.6x + 1.69$. Yes, that's correct.

So, substituting back into the equation:

$(x + 1.3)^2 - 1.69 + 3.6$

Now, let me compute the constants: -1.69 + 3.6. Let me do that subtraction. 3.6 minus 1.69. Hmm, 3.6 is the same as 3.60. So, 3.60 - 1.69. Subtracting 1.69 from 3.60:

Starting from the right, 0 - 9. Can't do that, so borrow. 10 - 9 = 1. Then, the next digit: 5 (since we borrowed 1) - 6. Again, can't do that, so borrow. 15 - 6 = 9. Then, the next digit: 2 (since we borrowed 1) - 1 = 1. So, the result is 1.91? Wait, let me check again.

Wait, 3.60 minus 1.69:

Let me write it vertically:

3.60
-1.69
=?

Starting from the hundredths place: 0 - 9. Can't do, borrow from tenths place. Tenths place is 6, so becomes 5, and hundredths becomes 10. 10 - 9 = 1.

Tenths place: 5 - 6. Can't do, borrow from units place. Units place is 3, becomes 2, tenths becomes 15. 15 - 6 = 9.

Units place: 2 - 1 = 1.

So, the result is 1.91. Yes, that's correct. So, -1.69 + 3.6 = 1.91.

Therefore, the equation becomes:

$(x + 1.3)^2 + 1.91$

So, comparing this with the form $(x + b)^2 + c$, we can see that $b = 1.3$ and $c = 1.91$. Therefore, $b + c = 1.3 + 1.91$.

Calculating that: 1.3 + 1.91. Let me add them. 1 + 1 = 2, 0.3 + 0.91 = 1.21. So total is 2 + 1.21 = 3.21. Wait, no. Wait, 1.3 is 1.30, so adding 1.30 + 1.91:

1.30
+1.91
=?

Adding the hundredths: 0 + 1 = 1.
Tenths: 3 + 9 = 12. So, carry over 1.
Units: 1 + 1 + 1 (carry) = 3.
So, total is 3.21. Yes, that's correct. So, $b + c = 3.21$.

But let me double-check my steps to make sure I didn't make any mistakes. Let me verify by expanding $(x + 1.3)^2 + 1.91$ and see if it equals the original quadratic.

Expanding $(x + 1.3)^2$:

$x^2 + 2*1.3*x + (1.3)^2 = x^2 + 2.6x + 1.69$. Then adding 1.91 gives $x^2 + 2.6x + 1.69 + 1.91 = x^2 + 2.6x + (1.69 + 1.91)$. Let me compute 1.69 + 1.91. 1.69 + 1.91: 1 + 1 = 2, 0.69 + 0.91 = 1.60. So total is 2 + 1.60 = 3.60. Which matches the original constant term, 3.6. Perfect! So that checks out.

Therefore, my values for $b$ and $c$ are correct. So $b = 1.3$ and $c = 1.91$, so $b + c = 1.3 + 1.91 = 3.21$.

Wait, but the question says "What is $b + c$ (as a decimal)?" So, 3.21 is already a decimal. So, is that the answer? Let me check once more.

Alternatively, maybe I made a mistake in calculating $ -1.69 + 3.6 $. Let me verify that again. 3.6 minus 1.69. Let me think of 3.6 as 3.60. So, 3.60 - 1.69. If I subtract 1.69 from 3.60, that's the same as 3.60 - 1.69. Let me do this another way. 3.60 - 1.69 = (3.60 - 1.00) - 0.69 = 2.60 - 0.69 = 1.91. Yes, same result. So that's correct.

Alternatively, maybe I can use fractions to check? Let me try converting the decimals to fractions to see if that gives the same result.

Original quadratic: $x^2 + 2.6x + 3.6$. Let me convert 2.6 and 3.6 to fractions.

2.6 is equal to 26/10, which simplifies to 13/5. Similarly, 3.6 is 36/10, which simplifies to 18/5.

So, the quadratic is $x^2 + \frac{13}{5}x + \frac{18}{5}$.

Now, completing the square. The coefficient of x is 13/5. Half of that is 13/10, which is 1.3, same as before. Squaring that gives (13/10)^2 = 169/100 = 1.69. So, adding and subtracting 169/100:

$x^2 + \frac{13}{5}x + \frac{169}{100} - \frac{169}{100} + \frac{18}{5}$

Which becomes:

$\left(x + \frac{13}{10}\right)^2 + \left(-\frac{169}{100} + \frac{18}{5}\right)$

Now, compute the constants:

Convert $\frac{18}{5}$ to 100 denominator: $\frac{18}{5} = \frac{360}{100}$. Then, $-\frac{169}{100} + \frac{360}{100} = \frac{191}{100} = 1.91$. So, same result as before. Therefore, $c = 191/100 = 1.91$, and $b = 13/10 = 1.3$. Therefore, $b + c = 13/10 + 191/100 = (130 + 191)/100 = 321/100 = 3.21$. So, same answer. Therefore, this confirms that my answer is correct.

Alternatively, maybe I can check by plugging in a value for x into both the original equation and the completed square form to see if they match. Let me try x = 0.

Original equation: $0^2 + 2.6*0 + 3.6 = 3.6$.

Completed square form: $(0 + 1.3)^2 + 1.91 = (1.3)^2 + 1.91 = 1.69 + 1.91 = 3.6$. Correct.

Another value, say x = -1.

Original equation: (-1)^2 + 2.6*(-1) + 3.6 = 1 - 2.6 + 3.6 = (1 + 3.6) - 2.6 = 4.6 - 2.6 = 2.0.

Completed square form: (-1 + 1.3)^2 + 1.91 = (0.3)^2 + 1.91 = 0.09 + 1.91 = 2.0. Correct again.

Another test: x = -1.3. Let's see.

Original equation: (-1.3)^2 + 2.6*(-1.3) + 3.6. Let's compute:

(-1.3)^2 = 1.69

2.6*(-1.3) = -3.38

So total: 1.69 - 3.38 + 3.6 = (1.69 + 3.6) - 3.38 = 5.29 - 3.38 = 1.91.

Completed square form: (-1.3 + 1.3)^2 + 1.91 = 0^2 + 1.91 = 1.91. Correct. Perfect.

So, all my checks confirm that the completed square form is correct, and therefore $b + c = 3.21$.

Therefore, the answer is 3.21. Wait, but the question says "as a decimal". Since 3.21 is already a decimal, that's fine. But maybe they want it in some specific decimal form? Like rounded or something? But since the original coefficients were given as decimals with one decimal place (2.6) and one decimal place (3.6), but in the process, we had 1.3 and 1.91. But since 1.91 is exact, and 1.3 is exact, adding them gives 3.21 exactly. So, no rounding needed. Therefore, 3.21 is the exact value of $b + c$.

Therefore, I think that's the answer. I don't see any mistakes in my reasoning, and multiple checks confirm that the completed square is correct. So, I think I can confidently say that $b + c = 3.21$.

**Final Answer**
The value of $b + c$ is \boxed{3.21}.

To rewrite the quadratic expression $ x^2 + 2.6x + 3.6 $ in the form $ (x + b)^2 + c $, we use the method of **completing the square**.

---

### Step 1: Identify the coefficient of $ x $

The given quadratic is:

$$
x^2 + 2.6x + 3.6
$$

The coefficient of $ x $ is $ 2.6 $. To complete the square, we take half of this coefficient and square it:

$$
\left( \frac{2.6}{2} \right)^2 = (1.3)^2 = 1.69
$$

---

### Step 2: Add and subtract the square

We add and subtract $ 1.69 $ to the expression:

$$
x^2 + 2.6x + 1.69 - 1.69 + 3.6
$$

Group the perfect square trinomial:

$$
(x^2 + 2.6x + 1.69) - 1.69 + 3.6
$$

$$
= (x + 1.3)^2 + (3.6 - 1.69)
$$

---

### Step 3: Simplify the constant term

$$
3.6 - 1.69 = 1.91
$$

So the expression becomes:

$$
(x + 1.3)^2 + 1.91
$$

---

### Step 4: Identify $ b $ and $ c $

From the form $ (x + b)^2 + c $, we identify:

- $ b = 1.3 $
- $ c = 1.91 $

---

### Step 5: Compute $ b + c $

$$
b + c = 1.3 + 1.91 = 3.21
$$

---

### Final Answer

$$
\boxed{3.21}
$$

[Assistant]
Explanation:
Okay, so I need to rewrite the quadratic equation $x^2 + 2.6x + 3.6$ in the form $(x + b)^2 + c$, where $b$ and $c$ are constants. Then, I have to find $b + c$ as a decimal. Hmm, let me think about how to approach this.

First, I remember that completing the square is the method used to convert a quadratic equation from standard form $ax^2 + bx + c$ to vertex form $a(x + b)^2 + c$. Wait, actually, in this case, the coefficient of $x^2$ is 1, so it's simpler. Let me recall the steps for completing the square.

The general process is: take the coefficient of $x$, divide it by 2, square it, and add and subtract that value to complete the square. Let me try applying that here.

Given the quadratic $x^2 + 2.6x + 3.6$. Let me write it down:

$x^2 + 2.6x + 3.6$

First, the coefficient of $x$ is 2.6. So, half of that is 1.3, right? Then, squaring that gives me $1.3^2$. Let me calculate that. 1.3 times 1.3... 1 times 1 is 1, 1 times 0.3 is 0.3, 0.3 times 1 is 0.3, and 0.3 times 0.3 is 0.09. Adding those up: 1 + 0.3 + 0.3 + 0.09 = 1.69. So, $1.3^2 = 1.69$.

So, to complete the square, I need to add and subtract 1.69. Let me do that:

$x^2 + 2.6x + 1.69 - 1.69 + 3.6$

Wait, but the original equation is $x^2 + 2.6x + 3.6$. So, if I add and subtract 1.69, the equation becomes:

$(x^2 + 2.6x + 1.69) - 1.69 + 3.6$

Simplifying the terms inside the parentheses first. Since $x^2 + 2.6x + 1.69$ is a perfect square trinomial, it can be written as $(x + 1.3)^2$. Let me check that:

$(x + 1.3)^2 = x^2 + 2*1.3*x + (1.3)^2 = x^2 + 2.6x + 1.69$. Yes, that's correct.

So, substituting back into the equation:

$(x + 1.3)^2 - 1.69 + 3.6$

Now, let me compute the constants: -1.69 + 3.6. Let me do that subtraction. 3.6 minus 1.69. Hmm, 3.6 is the same as 3.60. So, 3.60 - 1.69. Subtracting 1.69 from 3.60:

Starting from the right, 0 - 9. Can't do that, so borrow. 10 - 9 = 1. Then, the next digit: 5 (since we borrowed 1) - 6. Again, can't do that, so borrow. 15 - 6 = 9. Then, the next digit: 2 (since we borrowed 1) - 1 = 1. So, the result is 1.91? Wait, let me check again.

Wait, 3.60 minus 1.69:

Let me write it vertically:

3.60
-1.69
=?

Starting from the hundredths place: 0 - 9. Can't do, borrow from tenths place. Tenths place is 6, so becomes 5, and hundredths becomes 10. 10 - 9 = 1.

Tenths place: 5 - 6. Can't do, borrow from units place. Units place is 3, becomes 2, tenths becomes 15. 15 - 6 = 9.

Units place: 2 - 1 = 1.

So, the result is 1.91. Yes, that's correct. So, -1.69 + 3.6 = 1.91.

Therefore, the equation becomes:

$(x + 1.3)^2 + 1.91$

So, comparing this with the form $(x + b)^2 + c$, we can see that $b = 1.3$ and $c = 1.91$. Therefore, $b + c = 1.3 + 1.91$.

Calculating that: 1.3 + 1.91. Let me add them. 1 + 1 = 2, 0.3 + 0.91 = 1.21. So total is 2 + 1.21 = 3.21. Wait, no. Wait, 1.3 is 1.30, so adding 1.30 + 1.91:

1.30
+1.91
=?

Adding the hundredths: 0 + 1 = 1.
Tenths: 3 + 9 = 12. So, carry over 1.
Units: 1 + 1 + 1 (carry) = 3.
So, total is 3.21. Yes, that's correct. So, $b + c = 3.21$.

But let me double-check my steps to make sure I didn't make any mistakes. Let me verify by expanding $(x + 1.3)^2 + 1.91$ and see if it equals the original quadratic.

Expanding $(x + 1.3)^2$:

$x^2 + 2*1.3*x + (1.3)^2 = x^2 + 2.6x + 1.69$. Then adding 1.91 gives $x^2 + 2.6x + 1.69 + 1.91 = x^2 + 2.6x + (1.69 + 1.91)$. Let me compute 1.69 + 1.91. 1.69 + 1.91: 1 + 1 = 2, 0.69 + 0.91 = 1.60. So total is 2 + 1.60 = 3.60. Which matches the original constant term, 3.6. Perfect! So that checks out.

Therefore, my values for $b$ and $c$ are correct. So $b = 1.3$ and $c = 1.91$, so $b + c = 1.3 + 1.91 = 3.21$.

Wait, but the question says "What is $b + c$ (as a decimal)?" So, 3.21 is already a decimal. So, is that the answer? Let me check once more.

Alternatively, maybe I made a mistake in calculating $ -1.69 + 3.6 $. Let me verify that again. 3.6 minus 1.69. Let me think of 3.6 as 3.60. So, 3.60 - 1.69. If I subtract 1.69 from 3.60, that's the same as 3.60 - 1.69. Let me do this another way. 3.60 - 1.69 = (3.60 - 1.00) - 0.69 = 2.60 - 0.69 = 1.91. Yes, same result. So that's correct.

Alternatively, maybe I can use fractions to check? Let me try converting the decimals to fractions to see if that gives the same result.

Original quadratic: $x^2 + 2.6x + 3.6$. Let me convert 2.6 and 3.6 to fractions.

2.6 is equal to 26/10, which simplifies to 13/5. Similarly, 3.6 is 36/10, which simplifies to 18/5.

So, the quadratic is $x^2 + \frac{13}{5}x + \frac{18}{5}$.

Now, completing the square. The coefficient of x is 13/5. Half of that is 13/10, which is 1.3, same as before. Squaring that gives (13/10)^2 = 169/100 = 1.69. So, adding and subtracting 169/100:

$x^2 + \frac{13}{5}x + \frac{169}{100} - \frac{169}{100} + \frac{18}{5}$

Which becomes:

$\left(x + \frac{13}{10}\right)^2 + \left(-\frac{169}{100} + \frac{18}{5}\right)$

Now, compute the constants:

Convert $\frac{18}{5}$ to 100 denominator: $\frac{18}{5} = \frac{360}{100}$. Then, $-\frac{169}{100} + \frac{360}{100} = \frac{191}{100} = 1.91$. So, same result as before. Therefore, $c = 191/100 = 1.91$, and $b = 13/10 = 1.3$. Therefore, $b + c = 13/10 + 191/100 = (130 + 191)/100 = 321/100 = 3.21$. So, same answer. Therefore, this confirms that my answer is correct.

Alternatively, maybe I can check by plugging in a value for x into both the original equation and the completed square form to see if they match. Let me try x = 0.

Original equation: $0^2 + 2.6*0 + 3.6 = 3.6$.

Completed square form: $(0 + 1.3)^2 + 1.91 = (1.3)^2 + 1.91 = 1.69 + 1.91 = 3.6$. Correct.

Another value, say x = -1.

Original equation: (-1)^2 + 2.6*(-1) + 3.6 = 1 - 2.6 + 3.6 = (1 + 3.6) - 2.6 = 4.6 - 2.6 = 2.0.

Completed square form: (-1 + 1.3)^2 + 1.91 = (0.3)^2 + 1.91 = 0.09 + 1.91 = 2.0. Correct again.

Another test: x = -1.3. Let's see.

Original equation: (-1.3)^2 + 2.6*(-1.3) + 3.6. Let's compute:

(-1.3)^2 = 1.69

2.6*(-1.3) = -3.38

So total: 1.69 - 3.38 + 3.6 = (1.69 + 3.6) - 3.38 = 5.29 - 3.38 = 1.91.

Completed square form: (-1.3 + 1.3)^2 + 1.91 = 0^2 + 1.91 = 1.91. Correct. Perfect.

So, all my checks confirm that the completed square form is correct, and therefore $b + c = 3.21$.

Therefore, the answer is 3.21. Wait, but the question says "as a decimal". Since 3.21 is already a decimal, that's fine. But maybe they want it in some specific decimal form? Like rounded or something? But since the original coefficients were given as decimals with one decimal place (2.6) and one decimal place (3.6), but in the process, we had 1.3 and 1.91. But since 1.91 is exact, and 1.3 is exact, adding them gives 3.21 exactly. So, no rounding needed. Therefore, 3.21 is the exact value of $b + c$.

Therefore, I think that's the answer. I don't see any mistakes in my reasoning, and multiple checks confirm that the completed square is correct. So, I think I can confidently say that $b + c = 3.21$.

**Final Answer**
The value of $b + c$ is \boxed{3.21}.

To rewrite the quadratic expression $ x^2 + 2.6x + 3.6 $ in the form $ (x + b)^2 + c $, we use the method of **completing the square**.

---

### Step 1: Identify the coefficient of $ x $

The given quadratic is:

$$
x^2 + 2.6x + 3.6
$$

The coefficient of $ x $ is $ 2.6 $. To complete the square, we take half of this coefficient and square it:

$$
\left( \frac{2.6}{2} \right)^2 = (1.3)^2 = 1.69
$$

---

### Step 2: Add and subtract the square

We add and subtract $ 1.69 $ to the expression:

$$
x^2 + 2.6x + 1.69 - 1.69 + 3.6
$$

Group the perfect square trinomial:

$$
(x^2 + 2.6x + 1.69) - 1.69 + 3.6
$$

$$
= (x + 1.3)^2 + (3.6 - 1.69)
$$

---

### Step 3: Simplify the constant term

$$
3.6 - 1.69 = 1.91
$$

So the expression becomes:

$$
(x + 1.3)^2 + 1.91
$$

---

### Step 4: Identify $ b $ and $ c $

From the form $ (x + b)^2 + c $, we identify:

- $ b = 1.3 $
- $ c = 1.91 $

---

### Step 5: Compute $ b + c $

$$
b + c = 1.3 + 1.91 = 3.21
$$

---

### Final Answer

$$
\boxed{3.21}
$$

Solution:
We complete the square.

We have $(x+1.3)^2 = x^2 + (2.6)x + 1.69$, and so

\begin{align*}
x^2+(2.6)x+3.6 &= (x+1.3)^2 - 1.69 + 3.6 \\
&= (x+1.3)^2 + 1.91.
\end{align*}Therefore, $b=1.3$ and $c=1.91$, which gives us $b+c = \boxed{3.21}$.


Explanation:
Okay, let me try to figure out this problem. So, there's a bookcase with three shelves, and altogether there are 24 books. The top shelf has 8 mystery books, the middle shelf has 10 math books, and the bottom shelf has 6 science books. Then, two books are taken off each shelf. The question is asking, what fraction of the remaining books are math books? And they want the answer as a common fraction. Alright, let me break this down step by step.

First, let me confirm the total number of books. The problem says there are 3 shelves with a total of 24 books. Let me check if the numbers given for each shelf add up to 24. The top shelf has 8 mystery books, the middle has 10 math books, and the bottom has 6 science books. So, adding those up: 8 + 10 + 6. Let me do that. 8 + 10 is 18, and 18 + 6 is 24. Okay, that matches the total number of books. Good, so that checks out.

Now, two books are taken off each shelf. So, each shelf originally had a certain number of books, and then two are removed. Let me figure out how many books are left on each shelf after removing two.

Starting with the top shelf: originally 8 mystery books. If two are taken off, then the remaining number of books on the top shelf is 8 - 2. Let me calculate that. 8 minus 2 is 6. So, the top shelf now has 6 books left.

Middle shelf: originally 10 math books. Two books are taken off, so 10 - 2. That would be 8. So, the middle shelf now has 8 books left.

Bottom shelf: originally 6 science books. Two books are taken off, so 6 - 2. That's 4. So, the bottom shelf now has 4 books left.

Let me check if the total number of books remaining is correct. The original total was 24, and two books were removed from each shelf, so that's 2 books x 3 shelves = 6 books removed. Therefore, the remaining books should be 24 - 6 = 18. Let me check with the numbers I just calculated. Top shelf: 6, middle: 8, bottom: 4. Adding those up: 6 + 8 is 14, plus 4 is 18. Perfect, that matches. So, the total remaining books are 18.

Now, the question is asking for the fraction of the remaining books that are math books. So, I need to figure out how many math books are left, and then divide that by the total number of remaining books, which is 18.

Original math books were on the middle shelf, which had 10 math books. Two books were taken off the middle shelf. So, how many math books are left? Well, if two books were taken off the middle shelf, but we don't know if they were math books or not. Wait, hold on. Wait, the problem says "two books are now taken off each shelf." But does it specify which type of books are taken off? Hmm. Wait, the problem doesn't say that they took off specific types of books. It just says two books are taken off each shelf. So, maybe we have to assume that the books removed are not specified, so we can't assume they are math books or any other type. Wait, but the problem is asking about the fraction of math books remaining. So, if we don't know which books were removed, how can we determine the number of math books remaining?

Wait, hold on. Maybe there's an assumption here. Let me check the problem again. It says, "Two books are now taken off each shelf." It doesn't specify that they are taken off from the math books or any specific type. But the original shelves have different types of books. However, the problem is about math books. So, perhaps the problem is assuming that the books removed are not math books? Wait, but that's not stated. Hmm. Wait, maybe the problem is intended to assume that the books removed are of the same type as the shelf? Wait, but that's not specified either. Wait, this seems like a problem. If we don't know which books are removed, how can we find the number of math books remaining?

Wait, but maybe the problem is intended to assume that the books removed are not math books? But that's not stated. Wait, maybe the problem is intended to assume that the books removed are from the shelf, but the type of books removed is not specified, so we have to assume that the number of math books is reduced by two? Wait, but that would be an assumption. Wait, but maybe the problem is intended to have the same number of books removed from each shelf regardless of type. Wait, but the problem says "two books are now taken off each shelf." So, each shelf loses two books, but the problem doesn't specify that they are of a particular type. Therefore, perhaps the problem is intended to assume that the books removed are not math books? But that's not stated. Wait, but maybe the problem is intended to have the math books remaining as 10 - 2 = 8? But that would be assuming that the two books removed from the middle shelf are math books. But that's not necessarily the case. Wait, but maybe the problem is intended to have the same number of books removed from each shelf, regardless of type, so that the math books are reduced by two? But that's not necessarily true. Wait, this is confusing.

Wait, let me check again. The problem says: "Two books are now taken off each shelf." So, each shelf has two books removed. The shelves have different types of books. But the problem is asking about the fraction of math books remaining. So, unless there's more information, we can't know exactly how many math books are left. Wait, but maybe the problem is intended to assume that the books removed are not math books? But that's not stated. Wait, but maybe the problem is intended to assume that the books removed are of the same type as the shelf? For example, if you take two books off the top shelf, which has mystery books, then you remove two mystery books. Similarly, take two math books off the middle shelf, and two science books off the bottom shelf. But that's an assumption. However, maybe that's the intended interpretation. Because otherwise, the problem is unsolvable. Because if you don't know which books are removed, you can't determine the number of math books left. Therefore, maybe the problem assumes that two books of each type are removed from each shelf. So, for example, from the top shelf, two mystery books are removed, from the middle shelf, two math books are removed, and from the bottom shelf, two science books are removed. Therefore, the number of math books remaining would be 10 - 2 = 8. Then, the total remaining books would be 18, as calculated before. Therefore, the fraction would be 8/18, which simplifies to 4/9. But let me check if that's the case.

Alternatively, if the books removed are not necessarily of the same type, then the number of math books remaining could be different. For example, if from the middle shelf, they removed two books, but maybe they removed some math books and some other books? Wait, but the middle shelf only has math books. Wait, hold on! Wait, the middle shelf has 10 math books. So, if you take two books off the middle shelf, they have to be math books. Because there are no other books on the middle shelf. Similarly, the top shelf has only mystery books, so taking two books off the top shelf would remove two mystery books. The bottom shelf has only science books, so taking two books off the bottom shelf would remove two science books. Therefore, the problem is actually implying that two books of each type are removed from each shelf. Therefore, the number of math books remaining is 10 - 2 = 8. Therefore, the total remaining books are 18, so the fraction is 8/18, which simplifies to 4/9. Therefore, the answer is 4/9.

Wait, that makes sense. Because each shelf only has one type of book. Therefore, when you take two books off each shelf, you are removing two books of that type. Therefore, the math books on the middle shelf decrease by two. Therefore, the number of math books remaining is 10 - 2 = 8. Then, total remaining books is 18. Therefore, the fraction is 8/18, which simplifies to 4/9. Therefore, the answer is 4/9.

But let me just make sure. Let me go through the problem again. The top shelf has 8 mystery books. Middle shelf has 10 math books. Bottom shelf has 6 science books. So, each shelf has only one type of book. Therefore, when they say two books are taken off each shelf, they are removing two books of that type. Therefore, math books on the middle shelf decrease by two. Therefore, math books remaining: 10 - 2 = 8. Total books remaining: 24 - 6 = 18. Therefore, fraction is 8/18 = 4/9. So, that seems correct.

Alternatively, if someone thought that maybe the books removed are not of the same type, but since each shelf only has one type, you can't remove other types. Therefore, the answer is 4/9. Therefore, I think that is the correct answer.

But let me check again. Let me think if there's another way. Suppose someone thought that maybe the two books removed from each shelf could be of different types? But since each shelf only has one type, you can't remove different types. For example, the top shelf only has mystery books, so removing two books from the top shelf would be removing two mystery books. Similarly for the others. Therefore, the math books on the middle shelf are reduced by two. Therefore, the number of math books is 8. Therefore, the fraction is 8/18. Simplify that. Divide numerator and denominator by 2: 4/9. So, yes, that's the answer.

Therefore, I think the answer is 4/9.

**Final Answer**
The fraction of the remaining books that are math books is \boxed{\dfrac{4}{9}}.

To solve the problem, we begin by analyzing the initial setup of the bookcase:

- **Top shelf**: 8 mystery books
- **Middle shelf**: 10 math books
- **Bottom shelf**: 6 science books

This gives a total of:

$$
8 + 10 + 6 = 24 \text{ books}
$$

---

### Step 1: Books Removed

The problem states that **two books are taken off each shelf**. Since each shelf contains only one type of book, removing two books from each shelf means removing **two books of that type** from each shelf.

- **Top shelf**: 8 - 2 = **6 mystery books remaining**
- **Middle shelf**: 10 - 2 = **8 math books remaining**
- **Bottom shelf**: 6 - 2 = **4 science books remaining**

---

### Step 2: Total Remaining Books

Now, we calculate the total number of books remaining:

$$
6 + 8 + 4 = 18 \text{ books}
$$

---

### Step 3: Fraction of Math Books

We are asked to find the **fraction of the remaining books that are math books**. From the middle shelf, we have **8 math books** remaining out of the total **18 books**.

So, the fraction is:

$$
\frac{8}{18}
$$

Simplifying this fraction by dividing both numerator and denominator by their greatest common divisor (2):

$$
\frac{8 \div 2}{18 \div 2} = \frac{4}{9}
$$

---

### Final Answer

$$
\boxed{\dfrac{4}{9}}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, there's a bookcase with three shelves, and altogether there are 24 books. The top shelf has 8 mystery books, the middle shelf has 10 math books, and the bottom shelf has 6 science books. Then, two books are taken off each shelf. The question is asking, what fraction of the remaining books are math books? And they want the answer as a common fraction. Alright, let me break this down step by step.

First, let me confirm the total number of books. The problem says there are 3 shelves with a total of 24 books. Let me check if the numbers given for each shelf add up to 24. The top shelf has 8 mystery books, the middle has 10 math books, and the bottom has 6 science books. So, adding those up: 8 + 10 + 6. Let me do that. 8 + 10 is 18, and 18 + 6 is 24. Okay, that matches the total number of books. Good, so that checks out.

Now, two books are taken off each shelf. So, each shelf originally had a certain number of books, and then two are removed. Let me figure out how many books are left on each shelf after removing two.

Starting with the top shelf: originally 8 mystery books. If two are taken off, then the remaining number of books on the top shelf is 8 - 2. Let me calculate that. 8 minus 2 is 6. So, the top shelf now has 6 books left.

Middle shelf: originally 10 math books. Two books are taken off, so 10 - 2. That would be 8. So, the middle shelf now has 8 books left.

Bottom shelf: originally 6 science books. Two books are taken off, so 6 - 2. That's 4. So, the bottom shelf now has 4 books left.

Let me check if the total number of books remaining is correct. The original total was 24, and two books were removed from each shelf, so that's 2 books x 3 shelves = 6 books removed. Therefore, the remaining books should be 24 - 6 = 18. Let me check with the numbers I just calculated. Top shelf: 6, middle: 8, bottom: 4. Adding those up: 6 + 8 is 14, plus 4 is 18. Perfect, that matches. So, the total remaining books are 18.

Now, the question is asking for the fraction of the remaining books that are math books. So, I need to figure out how many math books are left, and then divide that by the total number of remaining books, which is 18.

Original math books were on the middle shelf, which had 10 math books. Two books were taken off the middle shelf. So, how many math books are left? Well, if two books were taken off the middle shelf, but we don't know if they were math books or not. Wait, hold on. Wait, the problem says "two books are now taken off each shelf." But does it specify which type of books are taken off? Hmm. Wait, the problem doesn't say that they took off specific types of books. It just says two books are taken off each shelf. So, maybe we have to assume that the books removed are not specified, so we can't assume they are math books or any other type. Wait, but the problem is asking about the fraction of math books remaining. So, if we don't know which books were removed, how can we determine the number of math books remaining?

Wait, hold on. Maybe there's an assumption here. Let me check the problem again. It says, "Two books are now taken off each shelf." It doesn't specify that they are taken off from the math books or any specific type. But the original shelves have different types of books. However, the problem is about math books. So, perhaps the problem is assuming that the books removed are not math books? Wait, but that's not stated. Hmm. Wait, maybe the problem is intended to assume that the books removed are of the same type as the shelf? Wait, but that's not specified either. Wait, this seems like a problem. If we don't know which books are removed, how can we find the number of math books remaining?

Wait, but maybe the problem is intended to assume that the books removed are not math books? But that's not stated. Wait, maybe the problem is intended to assume that the books removed are from the shelf, but the type of books removed is not specified, so we have to assume that the number of math books is reduced by two? Wait, but that would be an assumption. Wait, but maybe the problem is intended to have the same number of books removed from each shelf regardless of type. Wait, but the problem says "two books are now taken off each shelf." So, each shelf loses two books, but the problem doesn't specify that they are of a particular type. Therefore, perhaps the problem is intended to assume that the books removed are not math books? But that's not stated. Wait, but maybe the problem is intended to have the math books remaining as 10 - 2 = 8? But that would be assuming that the two books removed from the middle shelf are math books. But that's not necessarily the case. Wait, but maybe the problem is intended to have the same number of books removed from each shelf, regardless of type, so that the math books are reduced by two? But that's not necessarily true. Wait, this is confusing.

Wait, let me check again. The problem says: "Two books are now taken off each shelf." So, each shelf has two books removed. The shelves have different types of books. But the problem is asking about the fraction of math books remaining. So, unless there's more information, we can't know exactly how many math books are left. Wait, but maybe the problem is intended to assume that the books removed are not math books? But that's not stated. Wait, but maybe the problem is intended to assume that the books removed are of the same type as the shelf? For example, if you take two books off the top shelf, which has mystery books, then you remove two mystery books. Similarly, take two math books off the middle shelf, and two science books off the bottom shelf. But that's an assumption. However, maybe that's the intended interpretation. Because otherwise, the problem is unsolvable. Because if you don't know which books are removed, you can't determine the number of math books left. Therefore, maybe the problem assumes that two books of each type are removed from each shelf. So, for example, from the top shelf, two mystery books are removed, from the middle shelf, two math books are removed, and from the bottom shelf, two science books are removed. Therefore, the number of math books remaining would be 10 - 2 = 8. Then, the total remaining books would be 18, as calculated before. Therefore, the fraction would be 8/18, which simplifies to 4/9. But let me check if that's the case.

Alternatively, if the books removed are not necessarily of the same type, then the number of math books remaining could be different. For example, if from the middle shelf, they removed two books, but maybe they removed some math books and some other books? Wait, but the middle shelf only has math books. Wait, hold on! Wait, the middle shelf has 10 math books. So, if you take two books off the middle shelf, they have to be math books. Because there are no other books on the middle shelf. Similarly, the top shelf has only mystery books, so taking two books off the top shelf would remove two mystery books. The bottom shelf has only science books, so taking two books off the bottom shelf would remove two science books. Therefore, the problem is actually implying that two books of each type are removed from each shelf. Therefore, the number of math books remaining is 10 - 2 = 8. Therefore, the total remaining books are 18, so the fraction is 8/18, which simplifies to 4/9. Therefore, the answer is 4/9.

Wait, that makes sense. Because each shelf only has one type of book. Therefore, when you take two books off each shelf, you are removing two books of that type. Therefore, the math books on the middle shelf decrease by two. Therefore, the number of math books remaining is 10 - 2 = 8. Then, total remaining books is 18. Therefore, the fraction is 8/18, which simplifies to 4/9. Therefore, the answer is 4/9.

But let me just make sure. Let me go through the problem again. The top shelf has 8 mystery books. Middle shelf has 10 math books. Bottom shelf has 6 science books. So, each shelf has only one type of book. Therefore, when they say two books are taken off each shelf, they are removing two books of that type. Therefore, math books on the middle shelf decrease by two. Therefore, math books remaining: 10 - 2 = 8. Total books remaining: 24 - 6 = 18. Therefore, fraction is 8/18 = 4/9. So, that seems correct.

Alternatively, if someone thought that maybe the books removed are not of the same type, but since each shelf only has one type, you can't remove other types. Therefore, the answer is 4/9. Therefore, I think that is the correct answer.

But let me check again. Let me think if there's another way. Suppose someone thought that maybe the two books removed from each shelf could be of different types? But since each shelf only has one type, you can't remove different types. For example, the top shelf only has mystery books, so removing two books from the top shelf would be removing two mystery books. Similarly for the others. Therefore, the math books on the middle shelf are reduced by two. Therefore, the number of math books is 8. Therefore, the fraction is 8/18. Simplify that. Divide numerator and denominator by 2: 4/9. So, yes, that's the answer.

Therefore, I think the answer is 4/9.

**Final Answer**
The fraction of the remaining books that are math books is \boxed{\dfrac{4}{9}}.

To solve the problem, we begin by analyzing the initial setup of the bookcase:

- **Top shelf**: 8 mystery books
- **Middle shelf**: 10 math books
- **Bottom shelf**: 6 science books

This gives a total of:

$$
8 + 10 + 6 = 24 \text{ books}
$$

---

### Step 1: Books Removed

The problem states that **two books are taken off each shelf**. Since each shelf contains only one type of book, removing two books from each shelf means removing **two books of that type** from each shelf.

- **Top shelf**: 8 - 2 = **6 mystery books remaining**
- **Middle shelf**: 10 - 2 = **8 math books remaining**
- **Bottom shelf**: 6 - 2 = **4 science books remaining**

---

### Step 2: Total Remaining Books

Now, we calculate the total number of books remaining:

$$
6 + 8 + 4 = 18 \text{ books}
$$

---

### Step 3: Fraction of Math Books

We are asked to find the **fraction of the remaining books that are math books**. From the middle shelf, we have **8 math books** remaining out of the total **18 books**.

So, the fraction is:

$$
\frac{8}{18}
$$

Simplifying this fraction by dividing both numerator and denominator by their greatest common divisor (2):

$$
\frac{8 \div 2}{18 \div 2} = \frac{4}{9}
$$

---

### Final Answer

$$
\boxed{\dfrac{4}{9}}
$$

Solution: Six books are removed from the shelves, so $24-6=18$ books remain.  Of these, $10-2=8$ are math books.  Therefore, $8/18=\boxed{\frac{4}{9}}$ of the books remaining are math books.