Explanation:
Okay, so I need to find the range of the function y equals log base 2 of the square root of sine x, where x is between 0 degrees and 180 degrees. Let me start by understanding what the function is doing here.

First, let me write the function down again to make sure I have it right: y = log₂(√(sin x)). Alright, so that's the logarithm base 2 of the square root of sine x. The domain is given as 0° < x < 180°, so I need to consider x values between 0 and 180 degrees, not including the endpoints.

But wait, before I go further, I should check the domain of the function itself. The logarithm function is only defined when its argument is positive. So, the expression inside the log, which is √(sin x), has to be positive. Also, the square root requires that the inside, sin x, is non-negative. So, sin x must be greater than or equal to zero. But since the square root is in the log, and the log is only defined for positive numbers, √(sin x) must be greater than 0. Therefore, sin x must be greater than 0.

Therefore, the domain of y is the set of x in (0°, 180°) where sin x > 0. But wait, sin x is positive in the interval (0°, 180°), right? Because between 0 and 180 degrees, sine is positive in the first and second quadrants. Specifically, sin x is positive for 0° < x < 180°, so that's good. However, we have to make sure that the square root is defined, so sin x has to be non-negative, which is already satisfied here. But since the log is defined only when the argument is positive, we need √(sin x) > 0, which implies sin x > 0. Therefore, the domain of the function is indeed 0° < x < 180°, since in this interval sin x is positive. Wait, but at x = 0° and 180°, sin x is zero, but those are excluded. So that's fine.

So, the domain is already given as 0° < x < 180°, but maybe I need to verify that there aren't any restrictions within that interval. But since sin x is positive there, so that's okay.

So, the next step is to figure out the range of y = log₂(√(sin x)) over 0° < x < 180°.

To find the range, I can think about the possible values that sin x can take in this interval, then apply the square root and the logarithm to see what values y can take.

First, let's recall that in the interval 0° < x < 180°, the sine function starts at 0 when x approaches 0°, increases to 1 at x = 90°, and then decreases back to 0 at x approaching 180°. Therefore, the range of sin x over this interval is (0, 1]. So, sin x can take any value between 0 and 1, not including 0 but including 1.

But since we have √(sin x), let's consider the square root of sin x. The square root function is increasing, so if sin x is in (0, 1], then √(sin x) is in (0, 1]. Specifically, when sin x approaches 0, √(sin x) approaches 0, and when sin x is 1, √(sin x) is 1. Therefore, √(sin x) ∈ (0, 1].

Now, the function y = log₂(√(sin x)) is the log base 2 of something in (0, 1]. Let me recall that log base 2 of a number between 0 and 1 is negative, because 2 raised to a negative power is between 0 and 1. For example, log₂(1/2) = -1, log₂(1/4) = -2, etc. Also, log₂(1) = 0.

So, since √(sin x) ranges from approaching 0 to 1, the log base 2 of that will range from approaching negative infinity up to 0. But we need to check if it's inclusive or exclusive.

Let me formalize this. Let me denote z = √(sin x). Since sin x ∈ (0, 1], then z ∈ (0, 1]. Therefore, y = log₂(z).

Since z ∈ (0, 1], log₂(z) is in (-∞, 0]. Because when z approaches 0 from the right, log₂(z) approaches -∞, and when z = 1, log₂(1) = 0. So the range of y is (-∞, 0]. However, we need to check if the upper bound is attainable. Since z can be 1 (when sin x = 1, which occurs at x = 90°), so y = log₂(1) = 0 is attainable. But the lower bound is approached as z approaches 0, which happens when sin x approaches 0, i.e., when x approaches 0° or 180°. However, since x is in open interval (0°, 180°), sin x approaches 0 but never actually reaches 0. Therefore, z approaches 0 but never equals 0, so log₂(z) approaches -∞ but never actually reaches -∞. Therefore, the range of y is (-∞, 0].

But wait, hold on. Let me check if there's any mistake here. Let me verify with specific values.

For example, if x approaches 0°, then sin x approaches 0, so √(sin x) approaches 0, so log₂(√(sin x)) approaches log₂(0+) which is -∞. Similarly, as x approaches 180°, sin x approaches 0 again, so same thing.

At x = 90°, sin x = 1, so √(sin x) = 1, so log₂(1) = 0. Therefore, the maximum value of y is 0, and the minimum approaches -∞. Therefore, the range should be all real numbers less than or equal to 0.

But let me think again. Wait, the original function is log base 2 of sqrt(sin x). Maybe I can simplify that expression first.

Let me recall that log₂(√(sin x)) can be rewritten using logarithm properties. Since √(sin x) is the same as (sin x)^(1/2), so log₂((sin x)^(1/2)) = (1/2) log₂(sin x). So, y = (1/2) log₂(sin x).

So maybe that's another way to look at it.

Therefore, since sin x ∈ (0, 1], then log₂(sin x) is in (-∞, 0], since log₂(sin x) is log base 2 of something between 0 and 1. Therefore, multiplying by 1/2, the range would be (-∞, 0].

So same result. Therefore, the range is (-∞, 0].

But let me check if there's any restriction I might have missed. For example, could there be any points where sin x is negative? But in the given interval, 0° < x < 180°, sin x is positive. So, no issues there.

Also, when I simplified the expression to (1/2) log₂(sin x), is that correct? Let me check: log_b(a^c) = c log_b(a). So yes, that's correct.

Therefore, since log₂(sin x) can take any value from -∞ up to 0 (since sin x can go from 0 to 1), then multiplying by 1/2 scales the range but doesn't change the fact that it goes from -∞ to 0.

Therefore, the range of y is all real numbers less than or equal to 0.

But let me think again. Suppose I take sin x = 1/4. Then sqrt(sin x) = 1/2. Then log₂(1/2) = -1. So y = -1. If sin x = 1/16, sqrt(sin x) = 1/4, log₂(1/4) = -2. So y = -2. So as sin x gets smaller, y becomes more negative.

Therefore, the range is from negative infinity up to 0. But since sin x can be made as small as approaching zero, y can approach negative infinity.

But wait, the problem says 0° < x < 180°, so sin x approaches 0 but doesn't reach it, so y approaches negative infinity but doesn't reach it. However, 0 is achieved at x = 90°, so the upper limit is included.

Therefore, the range is (-∞, 0].

But let me check if the question is asking for the range in terms of interval notation. So, yes, (-∞, 0].

Wait, but maybe there's a mistake here? Let me check if there's a possible maximum or minimum. For example, maybe when sin x is maximum, which is 1, so log₂(sqrt(1)) = log₂(1) = 0. When sin x approaches 0, log₂(sqrt(sin x)) approaches -infty. So yes, the range is from -infty to 0, including 0.

Therefore, the answer should be (-∞, 0].

But let me check again if there's any step I missed.

Alternatively, maybe I can consider substituting t = sin x, so t ∈ (0, 1]. Then y = log₂(√t) = (1/2) log₂ t.

So let's analyze the function y = (1/2) log₂ t where t ∈ (0, 1].

The function log₂ t is defined for t > 0. For t ∈ (0, 1), log₂ t is negative, and for t = 1, log₂ t = 0. So, as t approaches 0+, log₂ t approaches -infty, so y approaches -infty. At t = 1, y is (1/2)(0) = 0.

Therefore, the range of y is (-infty, 0].

Therefore, yes, that seems consistent.

Alternatively, maybe I can check with specific values. Let's take x = 90°, which is π/2 radians. Then sin x = 1, so y = log₂(√1) = log₂(1) = 0. That's the maximum value.

Take x approaching 0°, say x = 1°, then sin(1°) is approximately 0.01745. Then √(sin x) ≈ √0.01745 ≈ 0.132. Then log₂(0.132) ≈ log₂(1/7.56) ≈ -log₂(7.56) ≈ -2.93. So that's a negative number.

If I take x approaching 180°, say x = 179°, sin(179°) ≈ sin(180° - 1°) = sin(1°) ≈ 0.01745, same as before. Therefore, same result.

So as x approaches 0° or 180°, y approaches negative infinity.

So there's no upper bound except zero, which is achieved.

Therefore, the range is all real numbers y such that y ≤ 0, which in interval notation is (-∞, 0].

Wait, but the problem says "the range of the function", so I need to make sure that there's no mistake in the calculation. Let me check again.

Alternatively, maybe I can write the original function as follows:

y = log₂(√(sin x)) = (1/2) log₂(sin x).

Let me consider the function log₂(sin x). Since sin x ∈ (0, 1], log₂(sin x) ∈ (-∞, 0]. So multiplying by 1/2, the range becomes (-∞, 0].

Yes, that's the same conclusion.

Alternatively, if I wanted to write this in terms of y, since log₂(sin x) can be any real number less than or equal to 0, then multiplying by 1/2 just scales it, so the range remains (-∞, 0].

Therefore, I think that's correct.

But just to make sure there's no restriction due to the square root or anything else, let me think again.

The square root of sin x is defined because sin x is positive in the interval. The log is defined because the argument is positive. So, no issues.

Therefore, my conclusion is that the range is (-∞, 0].

But let me check if there's a different way that the problem might be interpreted. For example, is the square root only over the sine, or is it sqrt(sin x)? The original function is written as y = log₂(√(sin x)), so the square root is only over sin x.

Yes, that's how it's written. So, it's not log of sqrt(sinx), but log base 2 of the square root of sin x.

Alternatively, if it were log base 2 of sqrt(sinx), which is same as (1/2) log₂(sinx).

Therefore, all steps are consistent.

So, I think I can be confident that the range is all real numbers less than or equal to 0, which is written as (-∞, 0].

But let me check if maybe the problem expects the answer in degrees or radians? Wait, no, the range of a function is a set of real numbers, regardless of the angle measure. So, the answer is just in terms of y-values, so it's (-∞, 0].

Alternatively, maybe the problem expects the answer in a specific format? The question says "put your final answer within \boxed{}". Usually, for ranges, if it's an interval, we write it in interval notation. So, \boxed{(-\infty, 0]}.

But let me check if maybe there's a mistake. For example, if someone confuses the logarithm with something else.

Wait, let me verify with another value. Suppose x = 30°, sin(30°) = 0.5, so sqrt(0.5) ≈ 0.7071. Then log₂(0.7071) ≈ log₂(1/√2) = -0.5. Therefore, y = -0.5. Which is between -infty and 0. So that's correct.

Another value, x = 60°, sin(60°) = √3/2 ≈ 0.866, sqrt(sin x) ≈ sqrt(0.866) ≈ 0.93, log₂(0.93) ≈ negative value. Let me calculate it. Since log₂(0.93) is equal to ln(0.93)/ln(2) ≈ (-0.07257)/0.693 ≈ -0.1047. So, y ≈ -0.1047. Which is still between -infty and 0.

So, all these examples confirm that the range is indeed all real numbers less than or equal to zero.

Therefore, I think my answer is correct.

**Final Answer**
The range of the function is \boxed{(-\infty, 0]}.

To determine the **range** of the function
$$
y = \log_2 \left( \sqrt{\sin x} \right), \quad 0^\circ < x < 180^\circ,
$$
we will proceed step by step.

---

### **1. Domain Consideration**

The expression inside the logarithm must be **positive** because the logarithm is only defined for positive real numbers. The square root of $\sin x$ is defined only if $\sin x \geq 0$, and since we are taking the square root, it must be **positive**.
So,

- $\sin x > 0$ for $0^\circ < x < 180^\circ$, which is always true in that interval.
- Therefore, the domain of the function is valid for all $x$ values in $(0^\circ, 180^\circ)$.

---

### **2. Behavior of the Inner Function: $\sin x$**

In the interval $0^\circ < x < 180^\circ$, the sine function behaves as follows:
- $\sin x$ increases from 0 to 1 as $x$ goes from $0^\circ$ to $90^\circ$,
- then decreases back to 0 as $x$ approaches $180^\circ$.

So, the **range of $\sin x$** is:
$$
0 < \sin x \leq 1.
$$

---

### **3. Applying the Square Root**

Since $\sin x \in (0, 1]$, applying the square root yields:
$$
\sqrt{\sin x} \in (0, 1].
$$

---

### **4. Applying the Logarithm**

Let’s now analyze the logarithm function with base 2:
$$
y = \log_2 \left( \sqrt{\sin x} \right).
$$

We can rewrite this using logarithmic identities:
$$
y = \frac{1}{2} \log_2 (\sin x).
$$

We know that:
- $\log_2(\sin x)$ is defined for $0 < \sin x \leq 1$,
- Since $\sin x \in (0, 1]$, it follows that $\log_2(\sin x) \in (-\infty, 0]$, because:
  - $\log_2(x) < 0$ when $0 < x < 1$,
  - $\log_2(1) = 0$.

Thus, multiplying by $1/2$, we get:
$$
y \in (-\infty, 0].
$$

This is the **range** of the function.

---

### **5. Confirming Endpoint Attainability**

- Since $\sin x = 1$ at $x = 90^\circ$, the value of $y = \log_2(\sqrt{1}) = \log_2(1) = 0$ is achieved.
- As $x \to 0^\circ$ or $180^\circ$, $\sin x \to 0$, so $\sqrt{\sin x} \to 0$, and $\log_2(\sqrt{\sin x}) \to -\infty$.

Therefore, the value $y = 0$ is **attained** (i.e., included), but $y = -\infty$ is never reached—it is a limit approached but not attained.

---

### **Final Result**

The range of the function
$$
y = \log_2 \left( \sqrt{\sin x} \right)
$$
for $0^\circ < x < 180^\circ$ is:

$$
\boxed{(-\infty, 0]}
$$

[Assistant]
Explanation:
Okay, so I need to find the range of the function y equals log base 2 of the square root of sine x, where x is between 0 degrees and 180 degrees. Let me start by understanding what the function is doing here.

First, let me write the function down again to make sure I have it right: y = log₂(√(sin x)). Alright, so that's the logarithm base 2 of the square root of sine x. The domain is given as 0° < x < 180°, so I need to consider x values between 0 and 180 degrees, not including the endpoints.

But wait, before I go further, I should check the domain of the function itself. The logarithm function is only defined when its argument is positive. So, the expression inside the log, which is √(sin x), has to be positive. Also, the square root requires that the inside, sin x, is non-negative. So, sin x must be greater than or equal to zero. But since the square root is in the log, and the log is only defined for positive numbers, √(sin x) must be greater than 0. Therefore, sin x must be greater than 0.

Therefore, the domain of y is the set of x in (0°, 180°) where sin x > 0. But wait, sin x is positive in the interval (0°, 180°), right? Because between 0 and 180 degrees, sine is positive in the first and second quadrants. Specifically, sin x is positive for 0° < x < 180°, so that's good. However, we have to make sure that the square root is defined, so sin x has to be non-negative, which is already satisfied here. But since the log is defined only when the argument is positive, we need √(sin x) > 0, which implies sin x > 0. Therefore, the domain of the function is indeed 0° < x < 180°, since in this interval sin x is positive. Wait, but at x = 0° and 180°, sin x is zero, but those are excluded. So that's fine.

So, the domain is already given as 0° < x < 180°, but maybe I need to verify that there aren't any restrictions within that interval. But since sin x is positive there, so that's okay.

So, the next step is to figure out the range of y = log₂(√(sin x)) over 0° < x < 180°.

To find the range, I can think about the possible values that sin x can take in this interval, then apply the square root and the logarithm to see what values y can take.

First, let's recall that in the interval 0° < x < 180°, the sine function starts at 0 when x approaches 0°, increases to 1 at x = 90°, and then decreases back to 0 at x approaching 180°. Therefore, the range of sin x over this interval is (0, 1]. So, sin x can take any value between 0 and 1, not including 0 but including 1.

But since we have √(sin x), let's consider the square root of sin x. The square root function is increasing, so if sin x is in (0, 1], then √(sin x) is in (0, 1]. Specifically, when sin x approaches 0, √(sin x) approaches 0, and when sin x is 1, √(sin x) is 1. Therefore, √(sin x) ∈ (0, 1].

Now, the function y = log₂(√(sin x)) is the log base 2 of something in (0, 1]. Let me recall that log base 2 of a number between 0 and 1 is negative, because 2 raised to a negative power is between 0 and 1. For example, log₂(1/2) = -1, log₂(1/4) = -2, etc. Also, log₂(1) = 0.

So, since √(sin x) ranges from approaching 0 to 1, the log base 2 of that will range from approaching negative infinity up to 0. But we need to check if it's inclusive or exclusive.

Let me formalize this. Let me denote z = √(sin x). Since sin x ∈ (0, 1], then z ∈ (0, 1]. Therefore, y = log₂(z).

Since z ∈ (0, 1], log₂(z) is in (-∞, 0]. Because when z approaches 0 from the right, log₂(z) approaches -∞, and when z = 1, log₂(1) = 0. So the range of y is (-∞, 0]. However, we need to check if the upper bound is attainable. Since z can be 1 (when sin x = 1, which occurs at x = 90°), so y = log₂(1) = 0 is attainable. But the lower bound is approached as z approaches 0, which happens when sin x approaches 0, i.e., when x approaches 0° or 180°. However, since x is in open interval (0°, 180°), sin x approaches 0 but never actually reaches 0. Therefore, z approaches 0 but never equals 0, so log₂(z) approaches -∞ but never actually reaches -∞. Therefore, the range of y is (-∞, 0].

But wait, hold on. Let me check if there's any mistake here. Let me verify with specific values.

For example, if x approaches 0°, then sin x approaches 0, so √(sin x) approaches 0, so log₂(√(sin x)) approaches log₂(0+) which is -∞. Similarly, as x approaches 180°, sin x approaches 0 again, so same thing.

At x = 90°, sin x = 1, so √(sin x) = 1, so log₂(1) = 0. Therefore, the maximum value of y is 0, and the minimum approaches -∞. Therefore, the range should be all real numbers less than or equal to 0.

But let me think again. Wait, the original function is log base 2 of sqrt(sin x). Maybe I can simplify that expression first.

Let me recall that log₂(√(sin x)) can be rewritten using logarithm properties. Since √(sin x) is the same as (sin x)^(1/2), so log₂((sin x)^(1/2)) = (1/2) log₂(sin x). So, y = (1/2) log₂(sin x).

So maybe that's another way to look at it.

Therefore, since sin x ∈ (0, 1], then log₂(sin x) is in (-∞, 0], since log₂(sin x) is log base 2 of something between 0 and 1. Therefore, multiplying by 1/2, the range would be (-∞, 0].

So same result. Therefore, the range is (-∞, 0].

But let me check if there's any restriction I might have missed. For example, could there be any points where sin x is negative? But in the given interval, 0° < x < 180°, sin x is positive. So, no issues there.

Also, when I simplified the expression to (1/2) log₂(sin x), is that correct? Let me check: log_b(a^c) = c log_b(a). So yes, that's correct.

Therefore, since log₂(sin x) can take any value from -∞ up to 0 (since sin x can go from 0 to 1), then multiplying by 1/2 scales the range but doesn't change the fact that it goes from -∞ to 0.

Therefore, the range of y is all real numbers less than or equal to 0.

But let me think again. Suppose I take sin x = 1/4. Then sqrt(sin x) = 1/2. Then log₂(1/2) = -1. So y = -1. If sin x = 1/16, sqrt(sin x) = 1/4, log₂(1/4) = -2. So y = -2. So as sin x gets smaller, y becomes more negative.

Therefore, the range is from negative infinity up to 0. But since sin x can be made as small as approaching zero, y can approach negative infinity.

But wait, the problem says 0° < x < 180°, so sin x approaches 0 but doesn't reach it, so y approaches negative infinity but doesn't reach it. However, 0 is achieved at x = 90°, so the upper limit is included.

Therefore, the range is (-∞, 0].

But let me check if the question is asking for the range in terms of interval notation. So, yes, (-∞, 0].

Wait, but maybe there's a mistake here? Let me check if there's a possible maximum or minimum. For example, maybe when sin x is maximum, which is 1, so log₂(sqrt(1)) = log₂(1) = 0. When sin x approaches 0, log₂(sqrt(sin x)) approaches -infty. So yes, the range is from -infty to 0, including 0.

Therefore, the answer should be (-∞, 0].

But let me check again if there's any step I missed.

Alternatively, maybe I can consider substituting t = sin x, so t ∈ (0, 1]. Then y = log₂(√t) = (1/2) log₂ t.

So let's analyze the function y = (1/2) log₂ t where t ∈ (0, 1].

The function log₂ t is defined for t > 0. For t ∈ (0, 1), log₂ t is negative, and for t = 1, log₂ t = 0. So, as t approaches 0+, log₂ t approaches -infty, so y approaches -infty. At t = 1, y is (1/2)(0) = 0.

Therefore, the range of y is (-infty, 0].

Therefore, yes, that seems consistent.

Alternatively, maybe I can check with specific values. Let's take x = 90°, which is π/2 radians. Then sin x = 1, so y = log₂(√1) = log₂(1) = 0. That's the maximum value.

Take x approaching 0°, say x = 1°, then sin(1°) is approximately 0.01745. Then √(sin x) ≈ √0.01745 ≈ 0.132. Then log₂(0.132) ≈ log₂(1/7.56) ≈ -log₂(7.56) ≈ -2.93. So that's a negative number.

If I take x approaching 180°, say x = 179°, sin(179°) ≈ sin(180° - 1°) = sin(1°) ≈ 0.01745, same as before. Therefore, same result.

So as x approaches 0° or 180°, y approaches negative infinity.

So there's no upper bound except zero, which is achieved.

Therefore, the range is all real numbers y such that y ≤ 0, which in interval notation is (-∞, 0].

Wait, but the problem says "the range of the function", so I need to make sure that there's no mistake in the calculation. Let me check again.

Alternatively, maybe I can write the original function as follows:

y = log₂(√(sin x)) = (1/2) log₂(sin x).

Let me consider the function log₂(sin x). Since sin x ∈ (0, 1], log₂(sin x) ∈ (-∞, 0]. So multiplying by 1/2, the range becomes (-∞, 0].

Yes, that's the same conclusion.

Alternatively, if I wanted to write this in terms of y, since log₂(sin x) can be any real number less than or equal to 0, then multiplying by 1/2 just scales it, so the range remains (-∞, 0].

Therefore, I think that's correct.

But just to make sure there's no restriction due to the square root or anything else, let me think again.

The square root of sin x is defined because sin x is positive in the interval. The log is defined because the argument is positive. So, no issues.

Therefore, my conclusion is that the range is (-∞, 0].

But let me check if there's a different way that the problem might be interpreted. For example, is the square root only over the sine, or is it sqrt(sin x)? The original function is written as y = log₂(√(sin x)), so the square root is only over sin x.

Yes, that's how it's written. So, it's not log of sqrt(sinx), but log base 2 of the square root of sin x.

Alternatively, if it were log base 2 of sqrt(sinx), which is same as (1/2) log₂(sinx).

Therefore, all steps are consistent.

So, I think I can be confident that the range is all real numbers less than or equal to 0, which is written as (-∞, 0].

But let me check if maybe the problem expects the answer in degrees or radians? Wait, no, the range of a function is a set of real numbers, regardless of the angle measure. So, the answer is just in terms of y-values, so it's (-∞, 0].

Alternatively, maybe the problem expects the answer in a specific format? The question says "put your final answer within \boxed{}". Usually, for ranges, if it's an interval, we write it in interval notation. So, \boxed{(-\infty, 0]}.

But let me check if maybe there's a mistake. For example, if someone confuses the logarithm with something else.

Wait, let me verify with another value. Suppose x = 30°, sin(30°) = 0.5, so sqrt(0.5) ≈ 0.7071. Then log₂(0.7071) ≈ log₂(1/√2) = -0.5. Therefore, y = -0.5. Which is between -infty and 0. So that's correct.

Another value, x = 60°, sin(60°) = √3/2 ≈ 0.866, sqrt(sin x) ≈ sqrt(0.866) ≈ 0.93, log₂(0.93) ≈ negative value. Let me calculate it. Since log₂(0.93) is equal to ln(0.93)/ln(2) ≈ (-0.07257)/0.693 ≈ -0.1047. So, y ≈ -0.1047. Which is still between -infty and 0.

So, all these examples confirm that the range is indeed all real numbers less than or equal to zero.

Therefore, I think my answer is correct.

**Final Answer**
The range of the function is \boxed{(-\infty, 0]}.

To determine the **range** of the function
$$
y = \log_2 \left( \sqrt{\sin x} \right), \quad 0^\circ < x < 180^\circ,
$$
we will proceed step by step.

---

### **1. Domain Consideration**

The expression inside the logarithm must be **positive** because the logarithm is only defined for positive real numbers. The square root of $\sin x$ is defined only if $\sin x \geq 0$, and since we are taking the square root, it must be **positive**.
So,

- $\sin x > 0$ for $0^\circ < x < 180^\circ$, which is always true in that interval.
- Therefore, the domain of the function is valid for all $x$ values in $(0^\circ, 180^\circ)$.

---

### **2. Behavior of the Inner Function: $\sin x$**

In the interval $0^\circ < x < 180^\circ$, the sine function behaves as follows:
- $\sin x$ increases from 0 to 1 as $x$ goes from $0^\circ$ to $90^\circ$,
- then decreases back to 0 as $x$ approaches $180^\circ$.

So, the **range of $\sin x$** is:
$$
0 < \sin x \leq 1.
$$

---

### **3. Applying the Square Root**

Since $\sin x \in (0, 1]$, applying the square root yields:
$$
\sqrt{\sin x} \in (0, 1].
$$

---

### **4. Applying the Logarithm**

Let’s now analyze the logarithm function with base 2:
$$
y = \log_2 \left( \sqrt{\sin x} \right).
$$

We can rewrite this using logarithmic identities:
$$
y = \frac{1}{2} \log_2 (\sin x).
$$

We know that:
- $\log_2(\sin x)$ is defined for $0 < \sin x \leq 1$,
- Since $\sin x \in (0, 1]$, it follows that $\log_2(\sin x) \in (-\infty, 0]$, because:
  - $\log_2(x) < 0$ when $0 < x < 1$,
  - $\log_2(1) = 0$.

Thus, multiplying by $1/2$, we get:
$$
y \in (-\infty, 0].
$$

This is the **range** of the function.

---

### **5. Confirming Endpoint Attainability**

- Since $\sin x = 1$ at $x = 90^\circ$, the value of $y = \log_2(\sqrt{1}) = \log_2(1) = 0$ is achieved.
- As $x \to 0^\circ$ or $180^\circ$, $\sin x \to 0$, so $\sqrt{\sin x} \to 0$, and $\log_2(\sqrt{\sin x}) \to -\infty$.

Therefore, the value $y = 0$ is **attained** (i.e., included), but $y = -\infty$ is never reached—it is a limit approached but not attained.

---

### **Final Result**

The range of the function
$$
y = \log_2 \left( \sqrt{\sin x} \right)
$$
for $0^\circ < x < 180^\circ$ is:

$$
\boxed{(-\infty, 0]}
$$

Solution: Between 0 and 180 degrees, the value of $\sin x$ is between 0 (exclusive) and 1 (inclusive). Thus, the value of $\sqrt{\sin x}$ is between 0 (exclusive) and 1 (inclusive). Since the range of $\log_2 x$ for $0<x\le1$ is all non-positive numbers, the range of the entire function is all non-positive numbers, or $x \in \boxed{(-\infty, 0]}$.


Explanation:
Okay, so I need to figure out the value of this operation @ between two fractions, 7/30 and 10/21. The operation is defined as (m/n) @ (p/q) equals m times p times (q/n). Let me write that down again to make sure I have it right.

So, the definition is: if you have two fractions, the first one is m over n, and the second one is p over q, then the operation @ is calculated by multiplying m by p, then multiplying that by q divided by n. So, mathematically, that's (m * p * q)/n? Wait, let me check the original definition again.

It says (m)(p)(q/n). So, m multiplied by p multiplied by (q/n). So, that's m*p*(q/n). Which can also be written as (m*p*q)/n. Yeah, that seems correct. So, the result of the operation is (m*p*q)/n.

But wait, the problem says "for simplified fractions p/q". So, does that mean that p/q is already simplified? Or maybe the operation is defined such that the fractions are simplified? Let me think. The problem says "for simplified fractions p/q", so maybe it's just saying that the fractions are in their simplest form? So, we can assume that 7/30 and 10/21 are simplified fractions? Let me check.

7 and 30: factors of 7 are 1 and 7. 30 divided by 7 is not an integer, so yes, 7/30 is simplified. 10 and 21: 10 factors are 1, 2, 5, 10. 21 factors are 1, 3, 7, 21. No common factors except 1, so 10/21 is also simplified. So, we don't have to simplify them first.

So, given that, let's apply the operation. Let me identify m, n, p, q. The first fraction is 7/30, so m is 7 and n is 30. The second fraction is 10/21, so p is 10 and q is 21.

So, according to the definition, (m)(p)(q/n) is equal to 7 * 10 * (21 / 30). Let me compute that step by step.

First, compute 7 * 10. That's 70. Then, compute 21 divided by 30. Let me write that as a fraction: 21/30. Maybe we can simplify that before multiplying? 21 and 30 have a common factor of 3. Dividing numerator and denominator by 3 gives 7/10. So, 21/30 simplifies to 7/10.

So, then the entire expression becomes 70 multiplied by 7/10. Let me compute that.

Multiplying 70 by 7 gives 490, and then dividing by 10 gives 49.0. Wait, so is the result 49? Let me check again.

Alternatively, maybe I should do the multiplication differently. Let me see.

Original expression: 7 * 10 * (21 / 30). Let me rearrange the terms: 7 * (21 / 30) * 10. Let me compute 7 * 21 first. 7*21 is 147, then divide by 30, then multiply by 10.

So, 147 divided by 30 is... Let me compute that. 30 goes into 147 four times (30*4=120), with a remainder of 27. So, 147/30 = 4.9, which is 4 and 27/30, which simplifies to 4 and 9/10. Then multiplying by 10 gives 49.

Alternatively, maybe there's a way to cancel out factors before multiplying? Let's see. Let me write all the numbers: 7, 10, 21, and 30.

Breaking them into prime factors:

- 7 is prime: 7
- 10 is 2 * 5
- 21 is 3 * 7
- 30 is 2 * 3 * 5

So, putting it all together:

Numerator: 7 * 10 * 21 = 7 * (2*5) * (3*7)

Denominator: 30 = 2 * 3 * 5

So, combining numerator and denominator:

Numerator: 7 * 2 * 5 * 3 * 7

Denominator: 2 * 3 * 5

So, we can simplify by canceling out the common factors. Let's see:

- The denominator has 2, 3, 5.

- The numerator has 2, 3, 5, and then two 7s.

So, canceling the 2, 3, and 5 in the numerator and denominator. So, that leaves us with 7 * 7 in the numerator, and 1 in the denominator. So, total is 49.

Oh, that's a better way. So, the result is 49.

Therefore, the simplified value of (7/30) @ (10/21) is 49.

Wait, let me check once again to make sure I didn't make a mistake.

Original operation: (m/n) @ (p/q) = m * p * (q/n).

So, m = 7, n = 30, p = 10, q = 21.

So, m * p * (q/n) = 7 * 10 * (21 / 30) = 70 * (7/10) = 49.

Yes, that seems correct.

Alternatively, if I compute 21 divided by 30 first, which is 0.7, then multiply by 70: 0.7 * 70 is 49.

So, all methods lead to 49.

So, I think that's the answer.

But just to be thorough, let me check if there's any step where I might have misapplied the definition.

The operation is defined as (m/n) @ (p/q) = m*p*(q/n).

So, m is the numerator of the first fraction, n is the denominator, p is the numerator of the second fraction, q is the denominator of the second fraction.

So, substituting the numbers:

First fraction is 7/30, so m=7, n=30.

Second fraction is 10/21, so p=10, q=21.

So, m*p is 7*10=70, and then multiplied by (q/n)=21/30.

So, 70 * 21 / 30.

Which is (70/30)*21.

But 70/30 simplifies to 7/3. So, 7/3 * 21.

7/3 * 21: 21 divided by 3 is 7, so 7 * 7 = 49.

Same result.

So, another way of looking at it.

Alternatively, if I think of 70 * 21 = 1470, and then divided by 30: 1470 / 30 = 49.

Yes, 30*49 = 1470.

So, all these methods are confirming that the answer is 49.

I think that's solid. So, the simplified value is 49.

**Final Answer**
\boxed{49}

To solve the operation defined by:

$$
\frac{m}{n} @ \frac{p}{q} = m \cdot p \cdot \left( \frac{q}{n} \right)
$$

We are given the two fractions:

$$
\frac{7}{30} \quad \text{and} \quad \frac{10}{21}
$$

---

### Step 1: Identify Variables

From the definitions:
- $ m = 7 $
- $ n = 30 $
- $ p = 10 $
- $ q = 21 $

---

### Step 2: Substitute into the Operation

We substitute these values into the given formula:

$$
\frac{7}{30} @ \frac{10}{21} = 7 \cdot 10 \cdot \left( \frac{21}{30} \right)
$$

---

### Step 3: Compute the Expression

Start by calculating the product of the first two terms:

$$
7 \cdot 10 = 70
$$

Now multiply that by the fraction:

$$
70 \cdot \left( \frac{21}{30} \right)
$$

We can simplify the fraction $ \frac{21}{30} $ first by dividing numerator and denominator by their greatest common divisor, which is 3:

$$
\frac{21}{30} = \frac{7}{10}
$$

So now the expression becomes:

$$
70 \cdot \frac{7}{10}
$$

Now compute the product:

$$
70 \cdot \frac{7}{10} = \frac{70 \cdot 7}{10} = \frac{490}{10} = 49
$$

---

### Final Answer

$$
\boxed{49}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of this operation @ between two fractions, 7/30 and 10/21. The operation is defined as (m/n) @ (p/q) equals m times p times (q/n). Let me write that down again to make sure I have it right.

So, the definition is: if you have two fractions, the first one is m over n, and the second one is p over q, then the operation @ is calculated by multiplying m by p, then multiplying that by q divided by n. So, mathematically, that's (m * p * q)/n? Wait, let me check the original definition again.

It says (m)(p)(q/n). So, m multiplied by p multiplied by (q/n). So, that's m*p*(q/n). Which can also be written as (m*p*q)/n. Yeah, that seems correct. So, the result of the operation is (m*p*q)/n.

But wait, the problem says "for simplified fractions p/q". So, does that mean that p/q is already simplified? Or maybe the operation is defined such that the fractions are simplified? Let me think. The problem says "for simplified fractions p/q", so maybe it's just saying that the fractions are in their simplest form? So, we can assume that 7/30 and 10/21 are simplified fractions? Let me check.

7 and 30: factors of 7 are 1 and 7. 30 divided by 7 is not an integer, so yes, 7/30 is simplified. 10 and 21: 10 factors are 1, 2, 5, 10. 21 factors are 1, 3, 7, 21. No common factors except 1, so 10/21 is also simplified. So, we don't have to simplify them first.

So, given that, let's apply the operation. Let me identify m, n, p, q. The first fraction is 7/30, so m is 7 and n is 30. The second fraction is 10/21, so p is 10 and q is 21.

So, according to the definition, (m)(p)(q/n) is equal to 7 * 10 * (21 / 30). Let me compute that step by step.

First, compute 7 * 10. That's 70. Then, compute 21 divided by 30. Let me write that as a fraction: 21/30. Maybe we can simplify that before multiplying? 21 and 30 have a common factor of 3. Dividing numerator and denominator by 3 gives 7/10. So, 21/30 simplifies to 7/10.

So, then the entire expression becomes 70 multiplied by 7/10. Let me compute that.

Multiplying 70 by 7 gives 490, and then dividing by 10 gives 49.0. Wait, so is the result 49? Let me check again.

Alternatively, maybe I should do the multiplication differently. Let me see.

Original expression: 7 * 10 * (21 / 30). Let me rearrange the terms: 7 * (21 / 30) * 10. Let me compute 7 * 21 first. 7*21 is 147, then divide by 30, then multiply by 10.

So, 147 divided by 30 is... Let me compute that. 30 goes into 147 four times (30*4=120), with a remainder of 27. So, 147/30 = 4.9, which is 4 and 27/30, which simplifies to 4 and 9/10. Then multiplying by 10 gives 49.

Alternatively, maybe there's a way to cancel out factors before multiplying? Let's see. Let me write all the numbers: 7, 10, 21, and 30.

Breaking them into prime factors:

- 7 is prime: 7
- 10 is 2 * 5
- 21 is 3 * 7
- 30 is 2 * 3 * 5

So, putting it all together:

Numerator: 7 * 10 * 21 = 7 * (2*5) * (3*7)

Denominator: 30 = 2 * 3 * 5

So, combining numerator and denominator:

Numerator: 7 * 2 * 5 * 3 * 7

Denominator: 2 * 3 * 5

So, we can simplify by canceling out the common factors. Let's see:

- The denominator has 2, 3, 5.

- The numerator has 2, 3, 5, and then two 7s.

So, canceling the 2, 3, and 5 in the numerator and denominator. So, that leaves us with 7 * 7 in the numerator, and 1 in the denominator. So, total is 49.

Oh, that's a better way. So, the result is 49.

Therefore, the simplified value of (7/30) @ (10/21) is 49.

Wait, let me check once again to make sure I didn't make a mistake.

Original operation: (m/n) @ (p/q) = m * p * (q/n).

So, m = 7, n = 30, p = 10, q = 21.

So, m * p * (q/n) = 7 * 10 * (21 / 30) = 70 * (7/10) = 49.

Yes, that seems correct.

Alternatively, if I compute 21 divided by 30 first, which is 0.7, then multiply by 70: 0.7 * 70 is 49.

So, all methods lead to 49.

So, I think that's the answer.

But just to be thorough, let me check if there's any step where I might have misapplied the definition.

The operation is defined as (m/n) @ (p/q) = m*p*(q/n).

So, m is the numerator of the first fraction, n is the denominator, p is the numerator of the second fraction, q is the denominator of the second fraction.

So, substituting the numbers:

First fraction is 7/30, so m=7, n=30.

Second fraction is 10/21, so p=10, q=21.

So, m*p is 7*10=70, and then multiplied by (q/n)=21/30.

So, 70 * 21 / 30.

Which is (70/30)*21.

But 70/30 simplifies to 7/3. So, 7/3 * 21.

7/3 * 21: 21 divided by 3 is 7, so 7 * 7 = 49.

Same result.

So, another way of looking at it.

Alternatively, if I think of 70 * 21 = 1470, and then divided by 30: 1470 / 30 = 49.

Yes, 30*49 = 1470.

So, all these methods are confirming that the answer is 49.

I think that's solid. So, the simplified value is 49.

**Final Answer**
\boxed{49}

To solve the operation defined by:

$$
\frac{m}{n} @ \frac{p}{q} = m \cdot p \cdot \left( \frac{q}{n} \right)
$$

We are given the two fractions:

$$
\frac{7}{30} \quad \text{and} \quad \frac{10}{21}
$$

---

### Step 1: Identify Variables

From the definitions:
- $ m = 7 $
- $ n = 30 $
- $ p = 10 $
- $ q = 21 $

---

### Step 2: Substitute into the Operation

We substitute these values into the given formula:

$$
\frac{7}{30} @ \frac{10}{21} = 7 \cdot 10 \cdot \left( \frac{21}{30} \right)
$$

---

### Step 3: Compute the Expression

Start by calculating the product of the first two terms:

$$
7 \cdot 10 = 70
$$

Now multiply that by the fraction:

$$
70 \cdot \left( \frac{21}{30} \right)
$$

We can simplify the fraction $ \frac{21}{30} $ first by dividing numerator and denominator by their greatest common divisor, which is 3:

$$
\frac{21}{30} = \frac{7}{10}
$$

So now the expression becomes:

$$
70 \cdot \frac{7}{10}
$$

Now compute the product:

$$
70 \cdot \frac{7}{10} = \frac{70 \cdot 7}{10} = \frac{490}{10} = 49
$$

---

### Final Answer

$$
\boxed{49}
$$

Solution: We have $\frac{7}{30}@\frac{10}{21}=(7)(10)\left(\frac{21}{30}\right)=\boxed{49}$.


Explanation:
Okay, so I need to figure out where the point (0,0) ends up after being reflected over the vertical line x=1 and then over the line y=2. Let me start by recalling how reflections over vertical and horizontal lines work.

First, reflecting a point over a vertical line. If I remember correctly, reflecting over a vertical line x = a affects the x-coordinate of the point. The formula for reflecting a point (x, y) over the vertical line x = a should be something like (2a - x, y). Let me verify that. Suppose I have a point to the left of x = a, say (a - d, y). Its reflection should be a distance d to the right of x = a, so the x-coordinate would be a + d. Since d = a - x_original, substituting gives a + (a - x_original) = 2a - x_original. Yeah, that seems right. So the reflection over x = a is (2a - x, y). Similarly, reflecting over a horizontal line y = b would change the y-coordinate to 2b - y. So that's the formula for horizontal reflections.

So, the first step is to reflect (0,0) over x = 1. Let me apply that formula. Here, a = 1. So the new x-coordinate is 2*1 - 0 = 2. The y-coordinate remains 0. Therefore, the image after the first reflection is (2, 0). Let me visualize that. The original point is at (0,0), which is 1 unit to the left of x=1. Reflecting it over x=1 should put it 1 unit to the right of x=1, so indeed (2, 0). That makes sense.

Now, the next step is to reflect this image, which is (2, 0), over the line y = 2. Using the horizontal reflection formula, where b = 2. The y-coordinate of the new point will be 2*2 - 0 = 4. The x-coordinate remains the same, so 2. Therefore, the resulting point after both reflections should be (2, 4). Let me check this again. The original point (2,0) is 2 units below y=2. Reflecting it over y=2 would place it 2 units above y=2, so y-coordinate is 2 + 2 = 4. Yep, that works. So the final point is (2, 4).

Wait, but let me make sure I didn't mix up the order of reflections. The problem says first reflect over x=1, then over y=2. So first vertical then horizontal. So my steps were correct. Let me think if there's any chance I confused the order. Suppose I did horizontal first? But no, the problem says first x=1, then y=2. So my answer seems correct.

Alternatively, maybe there's a different approach? Maybe using coordinates transformation? Let's see. Alternatively, think about what each reflection does.

First reflection over x=1: as before, (0,0) becomes (2, 0). Then reflecting (2, 0) over y=2: since y=2 is a horizontal line, the reflection changes the y-coordinate. The formula is 2*2 - y = 4 - 0 = 4, so (2, 4). That's the same as before.

Alternatively, maybe I can think about the distance from the point to the line. For reflection over x=1: distance from (0,0) to x=1 is 1 unit. So image is same distance on the other side, so 1 unit to the right of x=1, which is at x=1 +1 = 2. Then for the second reflection over y=2: distance from (2, 0) to y=2 is 2 units. So image is 2 units above y=2, so y=2 + 2 = 4. Therefore, (2, 4). Same result.

Is there any chance that the reflections could be different? Wait, if you reflect over two different lines, sometimes the order matters. For example, reflecting over x and then y is different than reflecting over y and then x. But in this case, since the first reflection is over vertical line and the second over horizontal line, the order does matter. Let me check with an example.

Suppose I did the reverse: first reflect over y=2, then over x=1. Let me see what would happen. Original point (0,0). Reflect over y=2: y-coordinate becomes 2*2 - 0 = 4, so (0, 4). Then reflect over x=1: x-coordinate becomes 2*1 - 0 = 2, so (2, 4). Wait, same result? So in this case, reflecting over x=1 then y=2 gives same result as reflecting over y=2 then x=1? But that's only because the original point was (0,0). Maybe in general, reflecting over vertical then horizontal is same as horizontal then vertical? Wait, but in general, reflections over different axes are not commutative. For example, reflecting over x-axis then y-axis would be different than reflecting over y-axis then x-axis. But in this case, with the specific point (0,0), maybe it's the same?

Wait, let me think. If I take another point, say (1,1). Reflect over x=1 first: reflection over x=1 would be (2 -1, 1) = (1,1). Wait, same point? Because (1,1) is on the line x=1. So reflecting it over x=1 leaves it unchanged. Then reflecting over y=2 would take (1,1) to (1, 2*2 -1) = (1,3). Alternatively, if I first reflect over y=2: (1, 2*2 -1) = (1,3). Then reflect over x=1: (2*1 -1, 3) = (1,3). So same result. But if I take another point not on the axes. Let me take (0,1). Reflect over x=1: (2,1). Then over y=2: (2, 4 -1) = (2,3). Alternatively, first reflect over y=2: (0, 2*2 -1) = (0,3). Then reflect over x=1: (2*1 -0, 3) = (2,3). Same result. So for this particular point (0,1), same result. Wait, so is it that reflecting over x=a then y=b is same as reflecting over y=b then x=a? For any point?

Wait, suppose I have a general point (x, y). Reflect over x=a: (2a - x, y). Then reflect over y=b: (2a - x, 2b - y). Alternatively, reflect over y=b first: (x, 2b - y). Then reflect over x=a: (2a - x, 2b - y). So same result. Therefore, the order of reflecting over vertical line and then horizontal line gives same result as reflecting over horizontal line then vertical line. So in general, for any point, reflecting over vertical line then horizontal line is same as reflecting over horizontal line then vertical line. Therefore, for our problem, since the order is first vertical then horizontal, but in reality, since it's commutative in this case, the result is same as reversing the order. Therefore, maybe my answer is correct.

But in general, reflection over different lines may not be commutative. For example, reflecting over x-axis then y-axis is different than reflecting over y-axis then x-axis. However, in this case, because the reflections are over vertical and horizontal lines, and the order doesn't matter? Wait, but in the previous example, reflecting over x-axis then y-axis would take (x, y) to (-x, -y), same as reflecting over y-axis then x-axis. Wait, but maybe for vertical and horizontal reflections, the order doesn't matter? Let me check with a general point. Suppose we reflect over x=a then y=b: (2a - x, 2b - y). If we first reflect over y=b then x=a: (2a - x, 2b - y). So same result. Therefore, the order does not matter when reflecting over vertical line and then horizontal line. Therefore, for the problem, even though the problem says first x=1 then y=2, the result is same as first y=2 then x=1. Therefore, maybe the answer is same. However, the problem specifies the order, so even though the result is same, we need to follow the order. But in this case, the answer is same either way. Therefore, regardless of the order, the result is (2, 4).

But maybe the problem wants us to follow the order, even though the result is same. But just to make sure, my answer is (2, 4). Let me check again.

Original point (0,0). Reflect over x=1: (2, 0). Reflect over y=2: (2, 4). Yes. Alternatively, if we do the reverse, reflect over y=2 first: (0, 4). Reflect over x=1: (2, 4). Same result. Therefore, answer is (2, 4). So maybe the answer is (2, 4). Therefore, I think that's correct.

But let me think again if there's another way to approach this problem. Maybe using transformation matrices or something. But reflections are linear transformations? Wait, no, reflections over lines not through the origin are not linear. Wait, reflecting over x=1 is not a linear transformation because it involves a translation. Similarly, reflecting over y=2 is not linear. So maybe we need to think in terms of affine transformations. However, since we have two reflections, maybe we can combine them?

Alternatively, think about the composition of two reflections. But since the lines are not parallel, the composition is a reflection or a glide reflection? Wait, reflecting over two lines that are not parallel... Wait, vertical and horizontal lines are perpendicular. So reflecting over two perpendicular lines... The composition would be a rotation. Wait, let me recall. If you reflect over two perpendicular lines, the composition is a rotation by twice the angle between them. Since vertical and horizontal are perpendicular, the angle between them is 90 degrees, so the composition is rotation by 180 degrees. Wait, is that right?

Wait, reflecting over x=1 and then over y=2. If we think of these as reflections over vertical and horizontal lines. But in general, reflecting over two lines that are not parallel results in a rotation. If they are perpendicular, the rotation is by 180 degrees? Let me check with the point (0,0). If we apply two reflections, according to the composition, it should be equivalent to a rotation. Let me see. The rotation by 180 degrees around some point. If we have (0,0) mapped to (2, 4). Let me see what rotation would take (0,0) to (2,4). A 180-degree rotation around the point (a, b) would take any point (x,y) to (2a - x, 2b - y). Comparing to our case, after two reflections, we have (2, 4). So if we think that it's equivalent to a 180-degree rotation, then for (0,0), (2, 4) would be the image. So 2a - 0 = 2 => a=1, and 2b - 0 = 4 => b=2. Therefore, the composition of reflections over x=1 and y=2 is equivalent to a 180-degree rotation about the point (1, 2). That seems correct. So indeed, reflecting over x=1 then over y=2 is equivalent to rotating 180 degrees around (1, 2). So applying that to (0,0) gives (2, 4), which is the same as before. Therefore, this confirms the result.

Alternatively, if I had done the composition of reflections over two lines, the formula would be that the composition of reflections over two lines is a rotation about the intersection point of the two lines by twice the angle between them. Since x=1 and y=2 are perpendicular lines, their intersection is at (1, 2). The angle between them is 90 degrees, so the composition is a rotation by 2*90 = 180 degrees. Therefore, the result is a rotation of 180 degrees about (1,2). Therefore, (0,0) rotated 180 degrees around (1,2) would be (2*1 - 0, 2*2 - 0) = (2, 4). Same answer. Therefore, this method also gives (2, 4).

So regardless of the method, whether step-by-step reflections, considering composition of reflections as rotation, the answer is (2,4). Therefore, I feel confident that the result is indeed (2,4).

Just to make sure there's no calculation mistake. Let me go through the steps once again.

First reflection: over x=1. Original point (0,0). Distance from x=1 is 1 unit left. So reflection is 1 unit right, x=1 + 1 = 2. So new point (2, 0). Then reflection over y=2. Original point (2, 0). Distance from y=2 is 2 units down. So reflection is 2 units up, y=2 + 2 = 4. New point (2, 4). Yep, that's straightforward.

Alternatively, using formulas:

First reflection over x = a: (x', y') = (2a - x, y). So here, a=1: x' = 2*1 - 0 = 2, y' = 0. Second reflection over y = b: (x'', y'') = (x', 2b - y'). So here, b=2: y'' = 2*2 - 0 = 4. So (2, 4). Yep.

Alternatively, if I use coordinate geometry. Suppose I have to find the reflection of a point over a line. For reflection over x=1, the formula is as above. Similarly for reflection over y=2. So all steps are consistent.

Alternatively, maybe I can use vectors. The reflection over x=1 can be considered as a reflection across the vertical line. The formula for reflection across a vertical line x = a is as mentioned. Similarly for horizontal.

Alternatively, think about the reflection as follows. For the first reflection, over x=1: the line x=1 is vertical. To find the reflection of a point, we can find the perpendicular distance from the point to the line and then go the same distance on the other side. So for (0,0) to x=1, the horizontal distance is 1 unit. So reflected point is 1 unit on the other side, so (1 + 1, 0) = (2, 0). Then reflecting (2, 0) over y=2: the vertical distance is 2 units (from y=0 to y=2), so reflected point is 2 units above y=2, which is y=2 + 2 = 4. So (2, 4). Same result.

I think all methods lead to the same answer. Therefore, I can be confident that the resulting point after both reflections is (2, 4).

**Final Answer**
The resulting point is \boxed{(2, 4)}.

To determine the final position of the point $(0, 0)$ after two successive reflections, we will analyze each step carefully using the general formulas for reflecting a point over a vertical and horizontal line.

---

### Step 1: Reflection Over the Vertical Line $x = 1$

The reflection of a point $(x, y)$ over a vertical line $x = a$ yields:

$$
(x', y') = (2a - x, y)
$$

Substituting $a = 1$ and the original point $(0, 0)$:

$$
x' = 2(1) - 0 = 2, \quad y' = 0 \Rightarrow (2, 0)
$$

So, the image after the first reflection is **$(2, 0)$**.

---

### Step 2: Reflection Over the Horizontal Line $y = 2$

The reflection of a point $(x, y)$ over a horizontal line $y = b$ yields:

$$
(x', y') = (x, 2b - y)
$$

Substituting $b = 2$ and the point from the previous step $(2, 0)$:

$$
x' = 2, \quad y' = 2(2) - 0 = 4 \Rightarrow (2, 4)
$$

---

### Final Result

After reflecting over $x = 1$, then over $y = 2$, the point $(0, 0)$ ends up at the coordinate:

$$
\boxed{(2, 4)}
$$

[Assistant]
Explanation:
Okay, so I need to figure out where the point (0,0) ends up after being reflected over the vertical line x=1 and then over the line y=2. Let me start by recalling how reflections over vertical and horizontal lines work.

First, reflecting a point over a vertical line. If I remember correctly, reflecting over a vertical line x = a affects the x-coordinate of the point. The formula for reflecting a point (x, y) over the vertical line x = a should be something like (2a - x, y). Let me verify that. Suppose I have a point to the left of x = a, say (a - d, y). Its reflection should be a distance d to the right of x = a, so the x-coordinate would be a + d. Since d = a - x_original, substituting gives a + (a - x_original) = 2a - x_original. Yeah, that seems right. So the reflection over x = a is (2a - x, y). Similarly, reflecting over a horizontal line y = b would change the y-coordinate to 2b - y. So that's the formula for horizontal reflections.

So, the first step is to reflect (0,0) over x = 1. Let me apply that formula. Here, a = 1. So the new x-coordinate is 2*1 - 0 = 2. The y-coordinate remains 0. Therefore, the image after the first reflection is (2, 0). Let me visualize that. The original point is at (0,0), which is 1 unit to the left of x=1. Reflecting it over x=1 should put it 1 unit to the right of x=1, so indeed (2, 0). That makes sense.

Now, the next step is to reflect this image, which is (2, 0), over the line y = 2. Using the horizontal reflection formula, where b = 2. The y-coordinate of the new point will be 2*2 - 0 = 4. The x-coordinate remains the same, so 2. Therefore, the resulting point after both reflections should be (2, 4). Let me check this again. The original point (2,0) is 2 units below y=2. Reflecting it over y=2 would place it 2 units above y=2, so y-coordinate is 2 + 2 = 4. Yep, that works. So the final point is (2, 4).

Wait, but let me make sure I didn't mix up the order of reflections. The problem says first reflect over x=1, then over y=2. So first vertical then horizontal. So my steps were correct. Let me think if there's any chance I confused the order. Suppose I did horizontal first? But no, the problem says first x=1, then y=2. So my answer seems correct.

Alternatively, maybe there's a different approach? Maybe using coordinates transformation? Let's see. Alternatively, think about what each reflection does.

First reflection over x=1: as before, (0,0) becomes (2, 0). Then reflecting (2, 0) over y=2: since y=2 is a horizontal line, the reflection changes the y-coordinate. The formula is 2*2 - y = 4 - 0 = 4, so (2, 4). That's the same as before.

Alternatively, maybe I can think about the distance from the point to the line. For reflection over x=1: distance from (0,0) to x=1 is 1 unit. So image is same distance on the other side, so 1 unit to the right of x=1, which is at x=1 +1 = 2. Then for the second reflection over y=2: distance from (2, 0) to y=2 is 2 units. So image is 2 units above y=2, so y=2 + 2 = 4. Therefore, (2, 4). Same result.

Is there any chance that the reflections could be different? Wait, if you reflect over two different lines, sometimes the order matters. For example, reflecting over x and then y is different than reflecting over y and then x. But in this case, since the first reflection is over vertical line and the second over horizontal line, the order does matter. Let me check with an example.

Suppose I did the reverse: first reflect over y=2, then over x=1. Let me see what would happen. Original point (0,0). Reflect over y=2: y-coordinate becomes 2*2 - 0 = 4, so (0, 4). Then reflect over x=1: x-coordinate becomes 2*1 - 0 = 2, so (2, 4). Wait, same result? So in this case, reflecting over x=1 then y=2 gives same result as reflecting over y=2 then x=1? But that's only because the original point was (0,0). Maybe in general, reflecting over vertical then horizontal is same as horizontal then vertical? Wait, but in general, reflections over different axes are not commutative. For example, reflecting over x-axis then y-axis would be different than reflecting over y-axis then x-axis. But in this case, with the specific point (0,0), maybe it's the same?

Wait, let me think. If I take another point, say (1,1). Reflect over x=1 first: reflection over x=1 would be (2 -1, 1) = (1,1). Wait, same point? Because (1,1) is on the line x=1. So reflecting it over x=1 leaves it unchanged. Then reflecting over y=2 would take (1,1) to (1, 2*2 -1) = (1,3). Alternatively, if I first reflect over y=2: (1, 2*2 -1) = (1,3). Then reflect over x=1: (2*1 -1, 3) = (1,3). So same result. But if I take another point not on the axes. Let me take (0,1). Reflect over x=1: (2,1). Then over y=2: (2, 4 -1) = (2,3). Alternatively, first reflect over y=2: (0, 2*2 -1) = (0,3). Then reflect over x=1: (2*1 -0, 3) = (2,3). Same result. So for this particular point (0,1), same result. Wait, so is it that reflecting over x=a then y=b is same as reflecting over y=b then x=a? For any point?

Wait, suppose I have a general point (x, y). Reflect over x=a: (2a - x, y). Then reflect over y=b: (2a - x, 2b - y). Alternatively, reflect over y=b first: (x, 2b - y). Then reflect over x=a: (2a - x, 2b - y). So same result. Therefore, the order of reflecting over vertical line and then horizontal line gives same result as reflecting over horizontal line then vertical line. So in general, for any point, reflecting over vertical line then horizontal line is same as reflecting over horizontal line then vertical line. Therefore, for our problem, since the order is first vertical then horizontal, but in reality, since it's commutative in this case, the result is same as reversing the order. Therefore, maybe my answer is correct.

But in general, reflection over different lines may not be commutative. For example, reflecting over x-axis then y-axis is different than reflecting over y-axis then x-axis. However, in this case, because the reflections are over vertical and horizontal lines, and the order doesn't matter? Wait, but in the previous example, reflecting over x-axis then y-axis would take (x, y) to (-x, -y), same as reflecting over y-axis then x-axis. Wait, but maybe for vertical and horizontal reflections, the order doesn't matter? Let me check with a general point. Suppose we reflect over x=a then y=b: (2a - x, 2b - y). If we first reflect over y=b then x=a: (2a - x, 2b - y). So same result. Therefore, the order does not matter when reflecting over vertical line and then horizontal line. Therefore, for the problem, even though the problem says first x=1 then y=2, the result is same as first y=2 then x=1. Therefore, maybe the answer is same. However, the problem specifies the order, so even though the result is same, we need to follow the order. But in this case, the answer is same either way. Therefore, regardless of the order, the result is (2, 4).

But maybe the problem wants us to follow the order, even though the result is same. But just to make sure, my answer is (2, 4). Let me check again.

Original point (0,0). Reflect over x=1: (2, 0). Reflect over y=2: (2, 4). Yes. Alternatively, if we do the reverse, reflect over y=2 first: (0, 4). Reflect over x=1: (2, 4). Same result. Therefore, answer is (2, 4). So maybe the answer is (2, 4). Therefore, I think that's correct.

But let me think again if there's another way to approach this problem. Maybe using transformation matrices or something. But reflections are linear transformations? Wait, no, reflections over lines not through the origin are not linear. Wait, reflecting over x=1 is not a linear transformation because it involves a translation. Similarly, reflecting over y=2 is not linear. So maybe we need to think in terms of affine transformations. However, since we have two reflections, maybe we can combine them?

Alternatively, think about the composition of two reflections. But since the lines are not parallel, the composition is a reflection or a glide reflection? Wait, reflecting over two lines that are not parallel... Wait, vertical and horizontal lines are perpendicular. So reflecting over two perpendicular lines... The composition would be a rotation. Wait, let me recall. If you reflect over two perpendicular lines, the composition is a rotation by twice the angle between them. Since vertical and horizontal are perpendicular, the angle between them is 90 degrees, so the composition is rotation by 180 degrees. Wait, is that right?

Wait, reflecting over x=1 and then over y=2. If we think of these as reflections over vertical and horizontal lines. But in general, reflecting over two lines that are not parallel results in a rotation. If they are perpendicular, the rotation is by 180 degrees? Let me check with the point (0,0). If we apply two reflections, according to the composition, it should be equivalent to a rotation. Let me see. The rotation by 180 degrees around some point. If we have (0,0) mapped to (2, 4). Let me see what rotation would take (0,0) to (2,4). A 180-degree rotation around the point (a, b) would take any point (x,y) to (2a - x, 2b - y). Comparing to our case, after two reflections, we have (2, 4). So if we think that it's equivalent to a 180-degree rotation, then for (0,0), (2, 4) would be the image. So 2a - 0 = 2 => a=1, and 2b - 0 = 4 => b=2. Therefore, the composition of reflections over x=1 and y=2 is equivalent to a 180-degree rotation about the point (1, 2). That seems correct. So indeed, reflecting over x=1 then over y=2 is equivalent to rotating 180 degrees around (1, 2). So applying that to (0,0) gives (2, 4), which is the same as before. Therefore, this confirms the result.

Alternatively, if I had done the composition of reflections over two lines, the formula would be that the composition of reflections over two lines is a rotation about the intersection point of the two lines by twice the angle between them. Since x=1 and y=2 are perpendicular lines, their intersection is at (1, 2). The angle between them is 90 degrees, so the composition is a rotation by 2*90 = 180 degrees. Therefore, the result is a rotation of 180 degrees about (1,2). Therefore, (0,0) rotated 180 degrees around (1,2) would be (2*1 - 0, 2*2 - 0) = (2, 4). Same answer. Therefore, this method also gives (2, 4).

So regardless of the method, whether step-by-step reflections, considering composition of reflections as rotation, the answer is (2,4). Therefore, I feel confident that the result is indeed (2,4).

Just to make sure there's no calculation mistake. Let me go through the steps once again.

First reflection: over x=1. Original point (0,0). Distance from x=1 is 1 unit left. So reflection is 1 unit right, x=1 + 1 = 2. So new point (2, 0). Then reflection over y=2. Original point (2, 0). Distance from y=2 is 2 units down. So reflection is 2 units up, y=2 + 2 = 4. New point (2, 4). Yep, that's straightforward.

Alternatively, using formulas:

First reflection over x = a: (x', y') = (2a - x, y). So here, a=1: x' = 2*1 - 0 = 2, y' = 0. Second reflection over y = b: (x'', y'') = (x', 2b - y'). So here, b=2: y'' = 2*2 - 0 = 4. So (2, 4). Yep.

Alternatively, if I use coordinate geometry. Suppose I have to find the reflection of a point over a line. For reflection over x=1, the formula is as above. Similarly for reflection over y=2. So all steps are consistent.

Alternatively, maybe I can use vectors. The reflection over x=1 can be considered as a reflection across the vertical line. The formula for reflection across a vertical line x = a is as mentioned. Similarly for horizontal.

Alternatively, think about the reflection as follows. For the first reflection, over x=1: the line x=1 is vertical. To find the reflection of a point, we can find the perpendicular distance from the point to the line and then go the same distance on the other side. So for (0,0) to x=1, the horizontal distance is 1 unit. So reflected point is 1 unit on the other side, so (1 + 1, 0) = (2, 0). Then reflecting (2, 0) over y=2: the vertical distance is 2 units (from y=0 to y=2), so reflected point is 2 units above y=2, which is y=2 + 2 = 4. So (2, 4). Same result.

I think all methods lead to the same answer. Therefore, I can be confident that the resulting point after both reflections is (2, 4).

**Final Answer**
The resulting point is \boxed{(2, 4)}.

To determine the final position of the point $(0, 0)$ after two successive reflections, we will analyze each step carefully using the general formulas for reflecting a point over a vertical and horizontal line.

---

### Step 1: Reflection Over the Vertical Line $x = 1$

The reflection of a point $(x, y)$ over a vertical line $x = a$ yields:

$$
(x', y') = (2a - x, y)
$$

Substituting $a = 1$ and the original point $(0, 0)$:

$$
x' = 2(1) - 0 = 2, \quad y' = 0 \Rightarrow (2, 0)
$$

So, the image after the first reflection is **$(2, 0)$**.

---

### Step 2: Reflection Over the Horizontal Line $y = 2$

The reflection of a point $(x, y)$ over a horizontal line $y = b$ yields:

$$
(x', y') = (x, 2b - y)
$$

Substituting $b = 2$ and the point from the previous step $(2, 0)$:

$$
x' = 2, \quad y' = 2(2) - 0 = 4 \Rightarrow (2, 4)
$$

---

### Final Result

After reflecting over $x = 1$, then over $y = 2$, the point $(0, 0)$ ends up at the coordinate:

$$
\boxed{(2, 4)}
$$

Solution:
When $(0,0)$ is reflected over the line $x=1$, the image is $(2,0)$. [asy]
draw((-2, 0)--(6, 0), Arrow); draw((0, -2)--(0, 6), Arrow);
label("$x$", (6, 0), E); label("$y$", (0, 6), N);
label("$(0, 0)$", (0, 0), SW); label("$(2, 0)$", (2, 0), SE);
label("$(2, 4)$", (2, 4), NE);
label("$y = 2$", (6, 2), E); label("$x = 1$", (1, -2), S);
draw((-2, 2)--(6, 2), dashed); draw((1, -2)--(1, 6), dashed);
dot((0, 0)); dot((2, 0)); dot((2, 4));
[/asy] When $(2,0)$ is reflected over the line $y=2$, the image is $\boxed{(2,4)}$.


Explanation:
Okay, so I need to factor the quadratic expression 10x² - x - 24 into the form (Ax - 8)(Bx + 3), where A and B are integers. Then find AB + B. Hmm, let me think about how to approach this.

First, I remember that factoring a quadratic in the form ax² + bx + c involves finding two binomials (mx + n)(px + q) such that m*p = a, n*q = c, and m*q + n*p = b. In this case, the original quadratic is 10x² - x - 24. So, comparing to the standard form, a is 10, b is -1, and c is -24.

But the problem already gives me the form (Ax - 8)(Bx + 3). Let me expand this to see what coefficients I get. If I multiply (Ax - 8)(Bx + 3), I should get:

Ax * Bx = ABx²

Ax * 3 = 3Ax

-8 * Bx = -8Bx

-8 * 3 = -24

So combining the terms, the expanded form is ABx² + (3A - 8B)x - 24.

Comparing this to the original quadratic, which is 10x² - x - 24, so the coefficients must satisfy:

AB = 10 (from the x² term)

3A - 8B = -1 (from the x term)

and the constant term is already -24, which matches. So I need to find integers A and B such that AB = 10 and 3A - 8B = -1.

Alright, so first step: find integer pairs (A, B) such that AB = 10. Since 10 can be factored into integers in several ways. Let me list all possible pairs (A, B):

1. A = 1, B = 10

2. A = 2, B = 5

3. A = 5, B = 2

4. A = 10, B = 1

Also considering negative factors:

5. A = -1, B = -10

6. A = -2, B = -5

7. A = -5, B = -2

8. A = -10, B = -1

But since AB = 10, which is positive, A and B must be both positive or both negative. However, let's check the equation 3A - 8B = -1. If both A and B are negative, let me think. Suppose A and B are both negative. Let me denote A = -a, B = -b where a and b are positive integers. Then AB = (-a)(-b) = ab = 10. Then the equation becomes 3(-a) - 8(-b) = -3a + 8b = -1. So, 8b - 3a = -1. Let me see if there are positive integers a and b satisfying that. But maybe I can check both cases: positive and negative.

But maybe first let me check positive factors. Let me check each possible positive pair.

First pair: A = 1, B = 10.

Then 3A - 8B = 3*1 - 8*10 = 3 - 80 = -77 ≠ -1. Not good.

Second pair: A = 2, B = 5. Then 3*2 - 8*5 = 6 - 40 = -34 ≠ -1.

Third pair: A = 5, B = 2. 3*5 - 8*2 = 15 - 16 = -1. Oh! That equals -1. Perfect! So this seems to work. So A = 5, B = 2.

Wait, let me check if there are other possibilities. Let me check the fourth pair: A = 10, B = 1. Then 3*10 - 8*1 = 30 - 8 = 22 ≠ -1. Not good. So the third pair works. Let me check the negative factors.

Let me take the negative pairs. Let me do similar steps. For example, take A = -1, B = -10. Then 3A -8B = 3*(-1) -8*(-10) = -3 + 80 = 77 ≠ -1. Not good.

Next pair: A = -2, B = -5. 3*(-2) - 8*(-5) = -6 + 40 = 34 ≠ -1.

Next pair: A = -5, B = -2. 3*(-5) -8*(-2) = -15 + 16 = 1 ≠ -1. Close, but not quite.

Last pair: A = -10, B = -1. 3*(-10) -8*(-1) = -30 + 8 = -22 ≠ -1.

So negative factors don't seem to work. Hence, the only possible pair is A = 5, B = 2.

Therefore, AB + B = (5*2) + 2 = 10 + 2 = 12. Wait, but let me confirm again.

Wait, AB is 5*2 = 10. Then AB + B is 10 + 2 = 12. But maybe I should check if the original expression factors correctly with these values. Let me check:

(Ax -8)(Bx +3) = (5x -8)(2x +3). Multiply this out:

First, 5x * 2x = 10x².

Outer term: 5x * 3 = 15x.

Inner term: -8 * 2x = -16x.

Last term: -8 * 3 = -24.

Combine middle terms: 15x -16x = -x. So total expression is 10x² -x -24. Perfect! It matches the original expression. So A=5, B=2.

Therefore, AB + B is 5*2 + 2 = 10 + 2 = 12. Wait, but hold on. The question says "AB + B". So AB is 10, B is 2, so 10 + 2 = 12? Wait, AB + B is (A*B) + B? Wait, AB is A multiplied by B. So AB + B would be (A*B) + B = B*(A + 1). Wait, 5*2 + 2 = 10 + 2 = 12. So that's correct. So the answer is 12?

But let me check once again if there are other possible pairs. Wait, could there be another pair? For example, if I have AB = 10. Maybe A and B are fractions? But the problem says A and B are integers. So no, only integer factors. So I think the only possible values are A=5, B=2.

But let me just check again. Let me think if there are other integer pairs for AB=10? For example, 10 can be written as 10*1, 5*2, etc., as I did before. So there are no other integer pairs. So the answer is 12. Therefore, AB + B is 12. Wait, but maybe I made a mistake in the calculation? Let me check again:

If A=5, B=2, then AB + B is 5*2 + 2 = 10 + 2 = 12. Yes. But maybe I misread the problem? Let me check again the original question.

"The expression 10x² -x -24 can be written as (Ax -8)(Bx +3), where A and B are integers. What is AB + B?"

Yes, so AB is A times B, then add B. So 5*2 + 2 = 12. So the answer is 12.

Alternatively, maybe the question is AB + B? Maybe they meant A*B + B? Which is same as B*(A + 1). But regardless, 5*2 + 2 is 12. Alternatively, if there was a miscalculation in solving for A and B?

Let me check the equation again. AB = 10. 3A -8B = -1. When A=5, B=2, 3*5 is 15, 8*2 is 16. 15 - 16 is -1. Which matches. So that's correct. Therefore, the answer is 12.

Wait, but I have a doubt because sometimes when factoring quadratics, there might be different ways. Let me try another approach. Let me try to factor 10x² -x -24 using the standard factoring method.

To factor 10x² - x -24, we look for two numbers that multiply to (10)*(-24) = -240 and add up to -1. Wait, that's the method where you split the middle term.

So looking for two numbers m and n such that m * n = 10*(-24) = -240 and m + n = -1.

So let me think. Let me list factors of -240 that add to -1.

Hmm, since their product is negative, one is positive and the other is negative. Their sum is -1. So let me think of two numbers where one is larger in magnitude than the other, such that their difference is 1.

Let me think. Let me think of factors of 240:

Let me think 16 and 15. 16*15=240. If I take 16 and -15: their product is -240 and their sum is 1. Hmm, 16 - 15 = 1. But we need sum to be -1. So maybe -16 and 15? Their product is -240, sum is -16 + 15 = -1. Yes! Perfect.

So, then we can split the middle term using -16x and 15x.

So original expression is 10x² -x -24.

Split the -x into -16x + 15x:

10x² -16x +15x -24.

Now group:

(10x² -16x) + (15x -24).

Factor out common terms:

From first group: 2x(5x - 8).

From the second group: 3(5x - 8). Wait:

Wait, 15x -24. Let me check:

15x -24. Let me factor 3: 3*(5x - 8). Yes! Because 3*5x =15x and 3*(-8) = -24. Perfect.

So then, the expression becomes:

2x(5x - 8) + 3(5x - 8) = (2x + 3)(5x - 8).

So that's the factorization: (5x -8)(2x +3). Which is same as (Ax -8)(Bx +3) with A=5, B=2. Therefore, same as before. Hence AB + B is 5*2 + 2 = 12. So answer is 12. So I think that's correct.

Alternatively, maybe I can think of it as (Ax -8)(Bx +3) = 10x² -x -24. If I think of A and B such that AB = 10. So since 10 factors into 5 and 2, and the other factors. Also, the constants are -8 and 3, so when multiplied give -24. Which is indeed -8*3 = -24. So that's correct.

Alternatively, maybe there is another way to get different A and B? Let me think. Suppose AB = 10, so possible A and B are (1, 10), (2,5), (5,2), (10,1), (-1,-10), etc. But as we saw earlier, only A=5, B=2 works for the middle term. So that is unique. Hence answer is 12.

Therefore, I think the answer is 12. So AB + B = 12. So I can write that as the final answer.

**Final Answer**
\boxed{12}

To solve the problem, we are given the quadratic expression:

$$
10x^2 - x - 24
$$

and asked to factor it as:

$$
(Ax - 8)(Bx + 3)
$$

where $ A $ and $ B $ are integers. We are then to compute $ AB + B $.

---

### Step 1: Expand the Factored Form

Let’s expand the expression $ (Ax - 8)(Bx + 3) $:

$$
(Ax - 8)(Bx + 3) = ABx^2 + 3Ax - 8Bx - 24 = ABx^2 + (3A - 8B)x - 24
$$

Now compare this with the original quadratic expression:

$$
10x^2 - x - 24
$$

Matching the coefficients, we obtain the system of equations:

1. $ AB = 10 $ (coefficient of $ x^2 $)
2. $ 3A - 8B = -1 $ (coefficient of $ x $)
3. The constant term matches directly, so we can ignore that.

---

### Step 2: Solve the System of Equations

From $ AB = 10 $, we consider all integer factor pairs of 10:

$$
(1,10), (2,5), (5,2), (10,1)
$$

Plug each pair into the second equation $ 3A - 8B = -1 $:

- $ A = 1, B = 10 $: $ 3(1) - 8(10) = 3 - 80 = -77 $ (not equal to -1)
- $ A = 2, B = 5 $: $ 3(2) - 8(5) = 6 - 40 = -34 $ (not equal to -1)
- $ A = 5, B = 2 $: $ 3(5) - 8(2) = 15 - 16 = -1 $ ✅
- $ A = 10, B = 1 $: $ 3(10) - 8(1) = 30 - 8 = 22 $ (not equal to -1)

So, the valid solution is:

$$
A = 5, \quad B = 2
$$

---

### Step 3: Compute $ AB + B $

Now compute:

$$
AB + B = (5)(2) + 2 = 10 + 2 = \boxed{12}
$$

---

### Final Answer:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to factor the quadratic expression 10x² - x - 24 into the form (Ax - 8)(Bx + 3), where A and B are integers. Then find AB + B. Hmm, let me think about how to approach this.

First, I remember that factoring a quadratic in the form ax² + bx + c involves finding two binomials (mx + n)(px + q) such that m*p = a, n*q = c, and m*q + n*p = b. In this case, the original quadratic is 10x² - x - 24. So, comparing to the standard form, a is 10, b is -1, and c is -24.

But the problem already gives me the form (Ax - 8)(Bx + 3). Let me expand this to see what coefficients I get. If I multiply (Ax - 8)(Bx + 3), I should get:

Ax * Bx = ABx²

Ax * 3 = 3Ax

-8 * Bx = -8Bx

-8 * 3 = -24

So combining the terms, the expanded form is ABx² + (3A - 8B)x - 24.

Comparing this to the original quadratic, which is 10x² - x - 24, so the coefficients must satisfy:

AB = 10 (from the x² term)

3A - 8B = -1 (from the x term)

and the constant term is already -24, which matches. So I need to find integers A and B such that AB = 10 and 3A - 8B = -1.

Alright, so first step: find integer pairs (A, B) such that AB = 10. Since 10 can be factored into integers in several ways. Let me list all possible pairs (A, B):

1. A = 1, B = 10

2. A = 2, B = 5

3. A = 5, B = 2

4. A = 10, B = 1

Also considering negative factors:

5. A = -1, B = -10

6. A = -2, B = -5

7. A = -5, B = -2

8. A = -10, B = -1

But since AB = 10, which is positive, A and B must be both positive or both negative. However, let's check the equation 3A - 8B = -1. If both A and B are negative, let me think. Suppose A and B are both negative. Let me denote A = -a, B = -b where a and b are positive integers. Then AB = (-a)(-b) = ab = 10. Then the equation becomes 3(-a) - 8(-b) = -3a + 8b = -1. So, 8b - 3a = -1. Let me see if there are positive integers a and b satisfying that. But maybe I can check both cases: positive and negative.

But maybe first let me check positive factors. Let me check each possible positive pair.

First pair: A = 1, B = 10.

Then 3A - 8B = 3*1 - 8*10 = 3 - 80 = -77 ≠ -1. Not good.

Second pair: A = 2, B = 5. Then 3*2 - 8*5 = 6 - 40 = -34 ≠ -1.

Third pair: A = 5, B = 2. 3*5 - 8*2 = 15 - 16 = -1. Oh! That equals -1. Perfect! So this seems to work. So A = 5, B = 2.

Wait, let me check if there are other possibilities. Let me check the fourth pair: A = 10, B = 1. Then 3*10 - 8*1 = 30 - 8 = 22 ≠ -1. Not good. So the third pair works. Let me check the negative factors.

Let me take the negative pairs. Let me do similar steps. For example, take A = -1, B = -10. Then 3A -8B = 3*(-1) -8*(-10) = -3 + 80 = 77 ≠ -1. Not good.

Next pair: A = -2, B = -5. 3*(-2) - 8*(-5) = -6 + 40 = 34 ≠ -1.

Next pair: A = -5, B = -2. 3*(-5) -8*(-2) = -15 + 16 = 1 ≠ -1. Close, but not quite.

Last pair: A = -10, B = -1. 3*(-10) -8*(-1) = -30 + 8 = -22 ≠ -1.

So negative factors don't seem to work. Hence, the only possible pair is A = 5, B = 2.

Therefore, AB + B = (5*2) + 2 = 10 + 2 = 12. Wait, but let me confirm again.

Wait, AB is 5*2 = 10. Then AB + B is 10 + 2 = 12. But maybe I should check if the original expression factors correctly with these values. Let me check:

(Ax -8)(Bx +3) = (5x -8)(2x +3). Multiply this out:

First, 5x * 2x = 10x².

Outer term: 5x * 3 = 15x.

Inner term: -8 * 2x = -16x.

Last term: -8 * 3 = -24.

Combine middle terms: 15x -16x = -x. So total expression is 10x² -x -24. Perfect! It matches the original expression. So A=5, B=2.

Therefore, AB + B is 5*2 + 2 = 10 + 2 = 12. Wait, but hold on. The question says "AB + B". So AB is 10, B is 2, so 10 + 2 = 12? Wait, AB + B is (A*B) + B? Wait, AB is A multiplied by B. So AB + B would be (A*B) + B = B*(A + 1). Wait, 5*2 + 2 = 10 + 2 = 12. So that's correct. So the answer is 12?

But let me check once again if there are other possible pairs. Wait, could there be another pair? For example, if I have AB = 10. Maybe A and B are fractions? But the problem says A and B are integers. So no, only integer factors. So I think the only possible values are A=5, B=2.

But let me just check again. Let me think if there are other integer pairs for AB=10? For example, 10 can be written as 10*1, 5*2, etc., as I did before. So there are no other integer pairs. So the answer is 12. Therefore, AB + B is 12. Wait, but maybe I made a mistake in the calculation? Let me check again:

If A=5, B=2, then AB + B is 5*2 + 2 = 10 + 2 = 12. Yes. But maybe I misread the problem? Let me check again the original question.

"The expression 10x² -x -24 can be written as (Ax -8)(Bx +3), where A and B are integers. What is AB + B?"

Yes, so AB is A times B, then add B. So 5*2 + 2 = 12. So the answer is 12.

Alternatively, maybe the question is AB + B? Maybe they meant A*B + B? Which is same as B*(A + 1). But regardless, 5*2 + 2 is 12. Alternatively, if there was a miscalculation in solving for A and B?

Let me check the equation again. AB = 10. 3A -8B = -1. When A=5, B=2, 3*5 is 15, 8*2 is 16. 15 - 16 is -1. Which matches. So that's correct. Therefore, the answer is 12.

Wait, but I have a doubt because sometimes when factoring quadratics, there might be different ways. Let me try another approach. Let me try to factor 10x² -x -24 using the standard factoring method.

To factor 10x² - x -24, we look for two numbers that multiply to (10)*(-24) = -240 and add up to -1. Wait, that's the method where you split the middle term.

So looking for two numbers m and n such that m * n = 10*(-24) = -240 and m + n = -1.

So let me think. Let me list factors of -240 that add to -1.

Hmm, since their product is negative, one is positive and the other is negative. Their sum is -1. So let me think of two numbers where one is larger in magnitude than the other, such that their difference is 1.

Let me think. Let me think of factors of 240:

Let me think 16 and 15. 16*15=240. If I take 16 and -15: their product is -240 and their sum is 1. Hmm, 16 - 15 = 1. But we need sum to be -1. So maybe -16 and 15? Their product is -240, sum is -16 + 15 = -1. Yes! Perfect.

So, then we can split the middle term using -16x and 15x.

So original expression is 10x² -x -24.

Split the -x into -16x + 15x:

10x² -16x +15x -24.

Now group:

(10x² -16x) + (15x -24).

Factor out common terms:

From first group: 2x(5x - 8).

From the second group: 3(5x - 8). Wait:

Wait, 15x -24. Let me check:

15x -24. Let me factor 3: 3*(5x - 8). Yes! Because 3*5x =15x and 3*(-8) = -24. Perfect.

So then, the expression becomes:

2x(5x - 8) + 3(5x - 8) = (2x + 3)(5x - 8).

So that's the factorization: (5x -8)(2x +3). Which is same as (Ax -8)(Bx +3) with A=5, B=2. Therefore, same as before. Hence AB + B is 5*2 + 2 = 12. So answer is 12. So I think that's correct.

Alternatively, maybe I can think of it as (Ax -8)(Bx +3) = 10x² -x -24. If I think of A and B such that AB = 10. So since 10 factors into 5 and 2, and the other factors. Also, the constants are -8 and 3, so when multiplied give -24. Which is indeed -8*3 = -24. So that's correct.

Alternatively, maybe there is another way to get different A and B? Let me think. Suppose AB = 10, so possible A and B are (1, 10), (2,5), (5,2), (10,1), (-1,-10), etc. But as we saw earlier, only A=5, B=2 works for the middle term. So that is unique. Hence answer is 12.

Therefore, I think the answer is 12. So AB + B = 12. So I can write that as the final answer.

**Final Answer**
\boxed{12}

To solve the problem, we are given the quadratic expression:

$$
10x^2 - x - 24
$$

and asked to factor it as:

$$
(Ax - 8)(Bx + 3)
$$

where $ A $ and $ B $ are integers. We are then to compute $ AB + B $.

---

### Step 1: Expand the Factored Form

Let’s expand the expression $ (Ax - 8)(Bx + 3) $:

$$
(Ax - 8)(Bx + 3) = ABx^2 + 3Ax - 8Bx - 24 = ABx^2 + (3A - 8B)x - 24
$$

Now compare this with the original quadratic expression:

$$
10x^2 - x - 24
$$

Matching the coefficients, we obtain the system of equations:

1. $ AB = 10 $ (coefficient of $ x^2 $)
2. $ 3A - 8B = -1 $ (coefficient of $ x $)
3. The constant term matches directly, so we can ignore that.

---

### Step 2: Solve the System of Equations

From $ AB = 10 $, we consider all integer factor pairs of 10:

$$
(1,10), (2,5), (5,2), (10,1)
$$

Plug each pair into the second equation $ 3A - 8B = -1 $:

- $ A = 1, B = 10 $: $ 3(1) - 8(10) = 3 - 80 = -77 $ (not equal to -1)
- $ A = 2, B = 5 $: $ 3(2) - 8(5) = 6 - 40 = -34 $ (not equal to -1)
- $ A = 5, B = 2 $: $ 3(5) - 8(2) = 15 - 16 = -1 $ ✅
- $ A = 10, B = 1 $: $ 3(10) - 8(1) = 30 - 8 = 22 $ (not equal to -1)

So, the valid solution is:

$$
A = 5, \quad B = 2
$$

---

### Step 3: Compute $ AB + B $

Now compute:

$$
AB + B = (5)(2) + 2 = 10 + 2 = \boxed{12}
$$

---

### Final Answer:

$$
\boxed{12}
$$

Solution: We see that $10x^2-x-24=(5x-8)(2x+3)$, thus $A = 5$ and $B = 2$. Hence, $AB + B = \boxed{12}.$


Explanation:
Okay, so I need to combine these two fractions into a single fraction. The expression is (4 + 6a)/5 minus (1 + 3a)/4. Hmm, let me think. To combine fractions, they need a common denominator. The denominators here are 5 and 4. What's the least common denominator for 5 and 4? Well, 5 and 4 don't have any common factors except 1, so the least common denominator should be 5 times 4, which is 20. Alright, so I need to convert both fractions to have 20 as the denominator.

Starting with the first fraction, (4 + 6a)/5. To get from 5 to 20, I multiply the denominator by 4. So, I have to do the same to the numerator. That means multiplying the numerator (4 + 6a) by 4. Let me write that out: 4*(4 + 6a). Let me calculate that. 4 times 4 is 16, and 4 times 6a is 24a. So, the first fraction becomes (16 + 24a)/20.

Now the second fraction is (1 + 3a)/4. To convert the denominator from 4 to 20, I multiply by 5. So again, I have to multiply the numerator by 5. Let me do that: 5*(1 + 3a). Calculating that, 5 times 1 is 5, and 5 times 3a is 15a. So, the second fraction becomes (5 + 15a)/20.

Wait, but hold on. The original expression is the first fraction minus the second fraction. So, after converting both fractions to have denominator 20, it should be (16 + 24a)/20 minus (5 + 15a)/20. Since the denominators are the same, I can subtract the numerators directly. Let me write that out:

[ (16 + 24a) - (5 + 15a) ] / 20.

Now I need to simplify the numerator. Let me distribute the subtraction across the second numerator. So, 16 + 24a - 5 - 15a. Let me combine like terms. The constants: 16 - 5 is 11. Then the terms with a: 24a - 15a is 9a. So, the numerator simplifies to 11 + 9a.

Therefore, the combined fraction is (11 + 9a)/20. Let me check if this can be simplified further. The numerator is 11 + 9a. 11 and 9 have no common factors besides 1, and there's no common factor with the denominator 20. So, I think that's as simplified as it gets. Wait, but let me make sure I didn't make a mistake in my calculations.

Let me go back through each step. Starting with the original expression: (4 + 6a)/5 - (1 + 3a)/4. Converted first fraction: multiply numerator and denominator by 4. 4*(4 + 6a) = 16 + 24a, denominator 20. Second fraction: multiply numerator and denominator by 5. 5*(1 + 3a) = 5 + 15a, denominator 20. Subtracting them gives [16 + 24a - 5 - 15a]/20. Combining constants: 16 - 5 = 11. Combining a terms: 24a - 15a = 9a. So numerator is 11 + 9a, denominator 20. Yes, that seems right.

Alternatively, maybe I can check with specific numbers. Let me pick a value for a, compute the original expression and the simplified expression to see if they match. Let me choose a = 0. Then original expression:

(4 + 0)/5 - (1 + 0)/4 = 4/5 - 1/4. Converting to common denominator 20: 16/20 - 5/20 = 11/20. Which matches the simplified expression (11 + 0)/20 = 11/20. Good.

Another test: let a = 1. Original expression: (4 + 6*1)/5 - (1 + 3*1)/4 = (10)/5 - (4)/4 = 2 - 1 = 1. Simplified expression: (11 + 9*1)/20 = 20/20 = 1. Perfect, same result.

Another test: a = 2. Original: (4 + 12)/5 - (1 + 6)/4 = 16/5 - 7/4. Convert to 20 denominator: 16/5 = 64/20, 7/4 = 35/20. So 64 - 35 = 29. 29/20. Simplified expression: (11 + 18)/20 = 29/20. Correct again.

So, seems like my answer is correct. Therefore, the expression as a single fraction is (11 + 9a)/20. I can also write it as (9a + 11)/20, which is the same thing. Depending on the preferred order of terms, but both are correct.

Alternatively, if I wanted to write it as a fraction with the variable term first, it's (9a + 11)/20. But since the original problem didn't specify, both forms are acceptable. But maybe the question expects the terms in the order they were given? In the original expression, the first fraction had 4 + 6a, so maybe 11 + 9a is the way to go. But I think either way is fine. Let me check if the problem has any particular preference. The problem says "as a single fraction," and doesn't specify the order. So both are correct. However, sometimes people prefer writing the terms in alphabetical order or by degree, but since a is the only variable, it's just 9a + 11. But again, same thing. So I think (11 + 9a)/20 is acceptable.

So, I think I've verified that multiple times. The answer is (11 + 9a)/20.

**Final Answer**
The expression as a single fraction is \boxed{\dfrac{9a + 11}{20}}.

To combine the expression
$$
\frac{4 + 6a}{5} - \frac{1 + 3a}{4}
$$
into a single fraction, follow these steps:

---

### **Step 1: Find a Common Denominator**

The denominators are 5 and 4. The least common denominator (LCD) of these numbers is
$$
\text{LCD} = 5 \times 4 = 20
$$

---

### **Step 2: Convert Each Fraction to Have Denominator 20**

Convert each term:

- For $ \frac{4 + 6a}{5} $:
  Multiply numerator and denominator by 4:
  $$
  \frac{4 + 6a}{5} = \frac{4(4 + 6a)}{20} = \frac{16 + 24a}{20}
  $$

- For $ \frac{1 + 3a}{4} $:
  Multiply numerator and denominator by 5:
  $$
  \frac{1 + 3a}{4} = \frac{5(1 + 3a)}{20} = \frac{5 + 15a}{20}
  $$

---

### **Step 3: Subtract the Fractions**

Now subtract the two fractions:
$$
\frac{16 + 24a}{20} - \frac{5 + 15a}{20} = \frac{(16 + 24a) - (5 + 15a)}{20}
$$

Simplify the numerator:

$$
(16 - 5) + (24a - 15a) = 11 + 9a
$$

So, the combined fraction is:
$$
\frac{11 + 9a}{20}
$$

---

### **Final Answer**

$$
\boxed{\frac{9a + 11}{20}}
$$

[Assistant]
Explanation:
Okay, so I need to combine these two fractions into a single fraction. The expression is (4 + 6a)/5 minus (1 + 3a)/4. Hmm, let me think. To combine fractions, they need a common denominator. The denominators here are 5 and 4. What's the least common denominator for 5 and 4? Well, 5 and 4 don't have any common factors except 1, so the least common denominator should be 5 times 4, which is 20. Alright, so I need to convert both fractions to have 20 as the denominator.

Starting with the first fraction, (4 + 6a)/5. To get from 5 to 20, I multiply the denominator by 4. So, I have to do the same to the numerator. That means multiplying the numerator (4 + 6a) by 4. Let me write that out: 4*(4 + 6a). Let me calculate that. 4 times 4 is 16, and 4 times 6a is 24a. So, the first fraction becomes (16 + 24a)/20.

Now the second fraction is (1 + 3a)/4. To convert the denominator from 4 to 20, I multiply by 5. So again, I have to multiply the numerator by 5. Let me do that: 5*(1 + 3a). Calculating that, 5 times 1 is 5, and 5 times 3a is 15a. So, the second fraction becomes (5 + 15a)/20.

Wait, but hold on. The original expression is the first fraction minus the second fraction. So, after converting both fractions to have denominator 20, it should be (16 + 24a)/20 minus (5 + 15a)/20. Since the denominators are the same, I can subtract the numerators directly. Let me write that out:

[ (16 + 24a) - (5 + 15a) ] / 20.

Now I need to simplify the numerator. Let me distribute the subtraction across the second numerator. So, 16 + 24a - 5 - 15a. Let me combine like terms. The constants: 16 - 5 is 11. Then the terms with a: 24a - 15a is 9a. So, the numerator simplifies to 11 + 9a.

Therefore, the combined fraction is (11 + 9a)/20. Let me check if this can be simplified further. The numerator is 11 + 9a. 11 and 9 have no common factors besides 1, and there's no common factor with the denominator 20. So, I think that's as simplified as it gets. Wait, but let me make sure I didn't make a mistake in my calculations.

Let me go back through each step. Starting with the original expression: (4 + 6a)/5 - (1 + 3a)/4. Converted first fraction: multiply numerator and denominator by 4. 4*(4 + 6a) = 16 + 24a, denominator 20. Second fraction: multiply numerator and denominator by 5. 5*(1 + 3a) = 5 + 15a, denominator 20. Subtracting them gives [16 + 24a - 5 - 15a]/20. Combining constants: 16 - 5 = 11. Combining a terms: 24a - 15a = 9a. So numerator is 11 + 9a, denominator 20. Yes, that seems right.

Alternatively, maybe I can check with specific numbers. Let me pick a value for a, compute the original expression and the simplified expression to see if they match. Let me choose a = 0. Then original expression:

(4 + 0)/5 - (1 + 0)/4 = 4/5 - 1/4. Converting to common denominator 20: 16/20 - 5/20 = 11/20. Which matches the simplified expression (11 + 0)/20 = 11/20. Good.

Another test: let a = 1. Original expression: (4 + 6*1)/5 - (1 + 3*1)/4 = (10)/5 - (4)/4 = 2 - 1 = 1. Simplified expression: (11 + 9*1)/20 = 20/20 = 1. Perfect, same result.

Another test: a = 2. Original: (4 + 12)/5 - (1 + 6)/4 = 16/5 - 7/4. Convert to 20 denominator: 16/5 = 64/20, 7/4 = 35/20. So 64 - 35 = 29. 29/20. Simplified expression: (11 + 18)/20 = 29/20. Correct again.

So, seems like my answer is correct. Therefore, the expression as a single fraction is (11 + 9a)/20. I can also write it as (9a + 11)/20, which is the same thing. Depending on the preferred order of terms, but both are correct.

Alternatively, if I wanted to write it as a fraction with the variable term first, it's (9a + 11)/20. But since the original problem didn't specify, both forms are acceptable. But maybe the question expects the terms in the order they were given? In the original expression, the first fraction had 4 + 6a, so maybe 11 + 9a is the way to go. But I think either way is fine. Let me check if the problem has any particular preference. The problem says "as a single fraction," and doesn't specify the order. So both are correct. However, sometimes people prefer writing the terms in alphabetical order or by degree, but since a is the only variable, it's just 9a + 11. But again, same thing. So I think (11 + 9a)/20 is acceptable.

So, I think I've verified that multiple times. The answer is (11 + 9a)/20.

**Final Answer**
The expression as a single fraction is \boxed{\dfrac{9a + 11}{20}}.

To combine the expression
$$
\frac{4 + 6a}{5} - \frac{1 + 3a}{4}
$$
into a single fraction, follow these steps:

---

### **Step 1: Find a Common Denominator**

The denominators are 5 and 4. The least common denominator (LCD) of these numbers is
$$
\text{LCD} = 5 \times 4 = 20
$$

---

### **Step 2: Convert Each Fraction to Have Denominator 20**

Convert each term:

- For $ \frac{4 + 6a}{5} $:
  Multiply numerator and denominator by 4:
  $$
  \frac{4 + 6a}{5} = \frac{4(4 + 6a)}{20} = \frac{16 + 24a}{20}
  $$

- For $ \frac{1 + 3a}{4} $:
  Multiply numerator and denominator by 5:
  $$
  \frac{1 + 3a}{4} = \frac{5(1 + 3a)}{20} = \frac{5 + 15a}{20}
  $$

---

### **Step 3: Subtract the Fractions**

Now subtract the two fractions:
$$
\frac{16 + 24a}{20} - \frac{5 + 15a}{20} = \frac{(16 + 24a) - (5 + 15a)}{20}
$$

Simplify the numerator:

$$
(16 - 5) + (24a - 15a) = 11 + 9a
$$

So, the combined fraction is:
$$
\frac{11 + 9a}{20}
$$

---

### **Final Answer**

$$
\boxed{\frac{9a + 11}{20}}
$$

Solution: The common denominator of $5$ and $4$ is $20$, so we multiply top and bottom of the first fraction by $4$ and multiply top and bottom of the second fraction by $5$. We get \[\frac{4(4+6a)}{4 \cdot 5} - \frac{5(1+3a)}{4 \cdot 5} = \frac{16+24a}{20}-\frac{5+15a}{20}.\] We combine the fractions, being careful to put the numerator of the second fraction in parentheses (as we are subtracting the entire numerator), yielding \[\frac{16+24a-(5+15a)}{20} = \frac{16+24a-5-15a}{20}=\boxed{\frac{11+9a}{20}}.\]