Explanation:
Okay, so Charlie is riding his unicycle, and I need to find out how fast he's going in inches per second. The problem says the wheel has a radius of 9 inches and makes 2 revolutions every 3 seconds. Hmm, let me think about how to approach this.

First, I remember that the speed of a wheel is related to its circumference and the number of revolutions it makes. So maybe I need to find the circumference of the wheel first? Yeah, circumference is the distance around the edge of the wheel, right? The formula for circumference is π times diameter, or 2π times radius. Since they gave me the radius, which is 9 inches, I can use that.

Let me write that down. Circumference C = 2πr. Plugging in the radius: C = 2 * π * 9 inches. Let me calculate that. 2 times 9 is 18, so the circumference is 18π inches. Okay, so each full revolution, the wheel covers 18π inches.

Now, the wheel makes 2 revolutions every 3 seconds. So, if I can figure out how much distance is covered in one second, that would be the speed. Let me think. If 2 revolutions take 3 seconds, then the distance covered in 3 seconds is 2 times the circumference. So that would be 2 * 18π inches. Let me compute that. 2 * 18 is 36, so 36π inches in 3 seconds.

To find the speed per second, I need to divide that distance by the time, which is 3 seconds. So speed = 36π inches / 3 seconds. Dividing 36 by 3 gives 12. So the speed is 12π inches per second. Wait, is that right? Let me check my steps again.

First, circumference: radius 9 inches, so diameter is 18 inches. Circumference is π*diameter, so π*18 inches. That's correct. Then, 2 revolutions per 3 seconds. So in 3 seconds, the wheel goes around twice, so total distance is 2 * circumference, which is 2*18π = 36π inches. Then divide by 3 seconds to get inches per second: 36π / 3 = 12π. Yeah, that seems right.

But maybe I should convert π into a numerical value to get a decimal? Wait, the question says "what is the unicycle's speed in inches per second?" It doesn't specify whether to leave it in terms of π or give a numerical approximation. Since they didn't specify, maybe I should leave it as 12π inches per second. But let me check if there's a standard way. In math problems, unless asked for a decimal, leaving it in terms of π is usually acceptable.

Alternatively, maybe I should compute the numerical value. Let me see. π is approximately 3.1416, so 12 * 3.1416 is... 12 * 3 is 36, 12 * 0.1416 is about 1.6992, so total is approximately 37.6992 inches per second. But again, unless they ask for a decimal, maybe 12π is better.

Wait, let me check the problem again. It says "what is the unicycle's speed in inches per second?" and the initial data is given in inches and seconds. The answer could be either, but since π is exact, maybe they want the exact value. Let me check if there's any possible mistake in my calculation.

Alternatively, maybe I can approach it using angular velocity? Let me try that method to verify. Angular velocity ω is revolutions per second. The wheel makes 2 revolutions every 3 seconds, so ω = 2/3 revolutions per second. Then, linear speed v is ω multiplied by the radius? Wait, no, linear speed is ω multiplied by the radius? Wait, no, actually, linear speed is ω multiplied by the radius? Wait, hold on. Wait, angular velocity is in radians per second. So if I convert revolutions to radians, then yes.

So 1 revolution is 2π radians. So ω = 2 revolutions / 3 seconds * 2π radians / 1 revolution = (4π)/3 radians per second. Then linear speed v = ω * r. So that would be (4π/3) * 9 inches. Let me compute that. 4π/3 * 9 = 4π * 3 = 12π inches per second. Oh, same answer! So that confirms it. So whether I do it through circumference and distance over time, or through angular velocity, I get 12π inches per second.

Therefore, I think that's the correct answer. So the unicycle's speed is 12π inches per second.

But just to make sure, let me think again. If the wheel makes 2 revolutions in 3 seconds, then in one second, it makes 2/3 revolutions. Each revolution is 18π inches, so per second, it's 2/3 * 18π inches. Let me calculate that. 2/3 * 18 is (2*18)/3 = 36/3 = 12. So 12π inches per second. Yep, same result.

So all methods lead to 12π. Therefore, I think that's the answer. Unless there's some trick I'm missing here. For example, is the unicycle's speed the same as the wheel's speed? Well, assuming that the unicycle is moving without slipping, which is a standard assumption unless stated otherwise. So the distance the unicycle moves forward is equal to the arc length covered by the wheel. So yes, the speed of the unicycle is equal to the linear speed of the wheel. So that's correct.

Therefore, I think I can confidently say the answer is 12π inches per second.

**Final Answer**
The unicycle's speed is \boxed{12\pi} inches per second.

To determine the unicycle's speed in inches per second, we begin by analyzing the motion of the wheel and how it relates to the unicycle's forward movement.

---

### Step 1: Calculate the Circumference of the Wheel

The unicycle's wheel has a **radius of 9 inches**. The **circumference** of a circle is given by the formula:

$$
C = 2\pi r
$$

Substituting the radius:

$$
C = 2\pi \times 9 = 18\pi \text{ inches}
$$

This means that with each full revolution, the wheel travels **18π inches** forward.

---

### Step 2: Determine the Distance Traveled in a Given Time

The wheel makes **2 revolutions every 3 seconds**. Therefore, in 3 seconds, the total distance traveled is:

$$
\text{Distance} = 2 \times 18\pi = 36\pi \text{ inches}
$$

---

### Step 3: Calculate the Speed

Speed is defined as **distance divided by time**. So, the speed in inches per second is:

$$
\text{Speed} = \frac{36\pi}{3} = 12\pi \text{ inches per second}
$$

---

### Final Answer

$$
\boxed{12\pi}
$$

This is the unicycle's speed in inches per second.

[Assistant]
Explanation:
Okay, so Charlie is riding his unicycle, and I need to find out how fast he's going in inches per second. The problem says the wheel has a radius of 9 inches and makes 2 revolutions every 3 seconds. Hmm, let me think about how to approach this.

First, I remember that the speed of a wheel is related to its circumference and the number of revolutions it makes. So maybe I need to find the circumference of the wheel first? Yeah, circumference is the distance around the edge of the wheel, right? The formula for circumference is π times diameter, or 2π times radius. Since they gave me the radius, which is 9 inches, I can use that.

Let me write that down. Circumference C = 2πr. Plugging in the radius: C = 2 * π * 9 inches. Let me calculate that. 2 times 9 is 18, so the circumference is 18π inches. Okay, so each full revolution, the wheel covers 18π inches.

Now, the wheel makes 2 revolutions every 3 seconds. So, if I can figure out how much distance is covered in one second, that would be the speed. Let me think. If 2 revolutions take 3 seconds, then the distance covered in 3 seconds is 2 times the circumference. So that would be 2 * 18π inches. Let me compute that. 2 * 18 is 36, so 36π inches in 3 seconds.

To find the speed per second, I need to divide that distance by the time, which is 3 seconds. So speed = 36π inches / 3 seconds. Dividing 36 by 3 gives 12. So the speed is 12π inches per second. Wait, is that right? Let me check my steps again.

First, circumference: radius 9 inches, so diameter is 18 inches. Circumference is π*diameter, so π*18 inches. That's correct. Then, 2 revolutions per 3 seconds. So in 3 seconds, the wheel goes around twice, so total distance is 2 * circumference, which is 2*18π = 36π inches. Then divide by 3 seconds to get inches per second: 36π / 3 = 12π. Yeah, that seems right.

But maybe I should convert π into a numerical value to get a decimal? Wait, the question says "what is the unicycle's speed in inches per second?" It doesn't specify whether to leave it in terms of π or give a numerical approximation. Since they didn't specify, maybe I should leave it as 12π inches per second. But let me check if there's a standard way. In math problems, unless asked for a decimal, leaving it in terms of π is usually acceptable.

Alternatively, maybe I should compute the numerical value. Let me see. π is approximately 3.1416, so 12 * 3.1416 is... 12 * 3 is 36, 12 * 0.1416 is about 1.6992, so total is approximately 37.6992 inches per second. But again, unless they ask for a decimal, maybe 12π is better.

Wait, let me check the problem again. It says "what is the unicycle's speed in inches per second?" and the initial data is given in inches and seconds. The answer could be either, but since π is exact, maybe they want the exact value. Let me check if there's any possible mistake in my calculation.

Alternatively, maybe I can approach it using angular velocity? Let me try that method to verify. Angular velocity ω is revolutions per second. The wheel makes 2 revolutions every 3 seconds, so ω = 2/3 revolutions per second. Then, linear speed v is ω multiplied by the radius? Wait, no, linear speed is ω multiplied by the radius? Wait, no, actually, linear speed is ω multiplied by the radius? Wait, hold on. Wait, angular velocity is in radians per second. So if I convert revolutions to radians, then yes.

So 1 revolution is 2π radians. So ω = 2 revolutions / 3 seconds * 2π radians / 1 revolution = (4π)/3 radians per second. Then linear speed v = ω * r. So that would be (4π/3) * 9 inches. Let me compute that. 4π/3 * 9 = 4π * 3 = 12π inches per second. Oh, same answer! So that confirms it. So whether I do it through circumference and distance over time, or through angular velocity, I get 12π inches per second.

Therefore, I think that's the correct answer. So the unicycle's speed is 12π inches per second.

But just to make sure, let me think again. If the wheel makes 2 revolutions in 3 seconds, then in one second, it makes 2/3 revolutions. Each revolution is 18π inches, so per second, it's 2/3 * 18π inches. Let me calculate that. 2/3 * 18 is (2*18)/3 = 36/3 = 12. So 12π inches per second. Yep, same result.

So all methods lead to 12π. Therefore, I think that's the answer. Unless there's some trick I'm missing here. For example, is the unicycle's speed the same as the wheel's speed? Well, assuming that the unicycle is moving without slipping, which is a standard assumption unless stated otherwise. So the distance the unicycle moves forward is equal to the arc length covered by the wheel. So yes, the speed of the unicycle is equal to the linear speed of the wheel. So that's correct.

Therefore, I think I can confidently say the answer is 12π inches per second.

**Final Answer**
The unicycle's speed is \boxed{12\pi} inches per second.

To determine the unicycle's speed in inches per second, we begin by analyzing the motion of the wheel and how it relates to the unicycle's forward movement.

---

### Step 1: Calculate the Circumference of the Wheel

The unicycle's wheel has a **radius of 9 inches**. The **circumference** of a circle is given by the formula:

$$
C = 2\pi r
$$

Substituting the radius:

$$
C = 2\pi \times 9 = 18\pi \text{ inches}
$$

This means that with each full revolution, the wheel travels **18π inches** forward.

---

### Step 2: Determine the Distance Traveled in a Given Time

The wheel makes **2 revolutions every 3 seconds**. Therefore, in 3 seconds, the total distance traveled is:

$$
\text{Distance} = 2 \times 18\pi = 36\pi \text{ inches}
$$

---

### Step 3: Calculate the Speed

Speed is defined as **distance divided by time**. So, the speed in inches per second is:

$$
\text{Speed} = \frac{36\pi}{3} = 12\pi \text{ inches per second}
$$

---

### Final Answer

$$
\boxed{12\pi}
$$

This is the unicycle's speed in inches per second.

Solution: We need to know how many inches are in one revolution of the unicycle's wheel. In other words, we need to know the circumference of the wheel. Since the radius is $9$ inches, the circumference is $18\pi$ inches. So, if there are $18\pi$ inches in one revolution, and the wheel makes $2$ revolutions every $3$ seconds, the unicycle's speed is $18\pi\cdot\frac{2}{3} = \boxed{12\pi}$ inches per second.


Explanation:
Okay, so I need to find the degree of the polynomial p(x) given by that expression. Let me first write down the problem again to make sure I have it right:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)].

And a, b, c are distinct real numbers. Alright, so the question is asking for the degree of this polynomial. Let me think about how to approach this.

First, I know that when you have a polynomial expressed as a sum of terms like this, each term is a rational function, but since the final result is supposed to be a polynomial, maybe there's some cancellation happening? Or maybe each term is a polynomial, and when you add them up, the result is a polynomial of some degree. Let me check.

Each numerator is a quadratic polynomial: (x - b)(x - c) is quadratic, same with the others. The denominators are constants, since a, b, c are constants. So each term is a quadratic divided by a constant, so each term is a quadratic polynomial. Therefore, adding three quadratic polynomials together would give a polynomial of degree at most 2. But maybe the leading coefficients cancel out? So maybe the degree is less than 2? Or maybe not?

Wait, but the problem says "Find the degree of the polynomial p(x)", so maybe it's a quadratic polynomial? But maybe it's of lower degree? Let me check with specific values. Maybe plugging in some numbers for a, b, c to see what p(x) looks like.

Alternatively, maybe there's a smarter way. Let me recall that these kinds of expressions look like Lagrange interpolation polynomials. Let me think. The Lagrange interpolation formula for a quadratic polynomial passing through three points. Let me recall that if you have three points (a, f(a)), (b, f(b)), (c, f(c)), then the interpolating polynomial is given by the sum over each term [f(x_i) * product_{j≠i} (x - x_j)/(x_i - x_j)]. Wait, so if we take f(x) = 1, then the interpolating polynomial would be the sum of [1 * product terms], which is exactly the expression given here. Wait, so p(x) is the Lagrange interpolating polynomial for the function f(x) = 1, passing through the points (a, 1), (b, 1), (c, 1). But if we interpolate a constant function, then the interpolating polynomial should be the constant function 1, right? Because a constant function is already a polynomial of degree 0, and the interpolating polynomial should match it. Therefore, p(x) must be equal to 1? But then that would mean p(x) is a constant polynomial, so degree 0. But that seems conflicting with my initial thought that each term is quadratic. Wait, but maybe there's cancellation?

Wait, let me check with specific numbers. Let me choose specific values for a, b, c. Let me take a = 0, b = 1, c = 2. Then compute p(x).

So, let's compute each term:

First term: [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x - 1)(x - 2)] / [(-1)(-2)] = [(x^2 - 3x + 2)] / 2.

Second term: [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [(1)(-1)] = [x^2 - 2x]/(-1) = -x^2 + 2x.

Third term: [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)] / [(2)(1)] = [x^2 - x]/2.

So adding them up:

First term: (x² - 3x + 2)/2

Second term: -x² + 2x

Third term: (x² - x)/2

Let me combine them:

Let me convert all terms to have denominator 2:

First term: (x² - 3x + 2)/2

Second term: (-2x² + 4x)/2

Third term: (x² - x)/2

Now add them together:

[(x² - 3x + 2) + (-2x² + 4x) + (x² - x)] / 2

Compute numerator:

x² - 3x + 2 - 2x² + 4x + x² - x

Combine like terms:

x² - 2x² + x² = 0

-3x + 4x - x = 0

Then +2 remains.

So numerator is 2, so overall p(x) = 2 / 2 = 1. So p(x) is 1. Which is a constant polynomial. Therefore, degree 0. But that seems to contradict my initial thought that each term is quadratic. But in reality, when you add them up, they cancel out to a constant. Therefore, the degree is 0. But wait, the problem says "Find the degree of the polynomial p(x)", so maybe the answer is 0? But let me check with another example to make sure.

Let me take different a, b, c. Let me take a = 1, b = 2, c = 3.

Compute p(x):

First term: [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = [(x² -5x +6)] / [(-1)(-2)] = (x² -5x +6)/2

Second term: [(x -1)(x -3)] / [(2 -1)(2 -3)] = [(x² -4x +3)] / [(1)(-1)] = -(x² -4x +3) = -x² +4x -3

Third term: [(x -1)(x -2)] / [(3 -1)(3 -2)] = [(x² -3x +2)] / [(2)(1)] = (x² -3x +2)/2

Adding them up:

First term: (x² -5x +6)/2

Second term: -x² +4x -3

Third term: (x² -3x +2)/2

Convert all to denominator 2:

First term: (x² -5x +6)/2

Second term: (-2x² +8x -6)/2

Third term: (x² -3x +2)/2

Adding numerators:

(x² -5x +6) + (-2x² +8x -6) + (x² -3x +2) =

x² -2x² +x² = 0

-5x +8x -3x = 0

6 -6 +2 = 2

So again, numerator is 2, so p(x) = 2/2 = 1. So again, p(x) is 1. So that seems to confirm that p(x) is the constant polynomial 1. Therefore, degree 0. But wait, the problem says "Find the degree of the polynomial p(x)", so answer is 0? But why does the problem mention that a, b, c are distinct? Because if they are not distinct, then denominators would be zero. But since they are distinct, denominators are non-zero. But in any case, the result is always 1? So regardless of a, b, c (as long as distinct), p(x) is 1. Therefore, degree 0. But let me think again.

Wait, but maybe I made a mistake here? Let me check with another example where maybe the result isn't 1? Wait, but according to the Lagrange interpolation, if you interpolate the constant function 1 at three points, then the interpolating polynomial is 1. So regardless of the points, as long as they are distinct, the interpolating polynomial is 1. Therefore, p(x) is 1. Therefore, degree 0. Therefore, answer is 0. But maybe I need to check another case where maybe the calculation is different?

Wait, let me try with a = 0, b = 1, c = -1. Let me compute p(x):

First term: [(x - 1)(x - (-1))]/[(0 - 1)(0 - (-1))] = [(x -1)(x +1)] / [(-1)(1)] = (x² -1)/(-1) = -x² +1

Second term: [(x - 0)(x - (-1))]/[(1 - 0)(1 - (-1))] = [x(x +1)] / [1 * 2] = (x² +x)/2

Third term: [(x - 0)(x -1)] / [(-1 - 0)(-1 -1)] = [x(x -1)] / [(-1)(-2)] = [x² -x]/2

Adding them up:

First term: -x² +1

Second term: (x² +x)/2

Third term: (x² -x)/2

Combine second and third terms: [(x² +x) + (x² -x)] / 2 = (2x²)/2 = x²

So total p(x) = -x² +1 + x² = 1. Again, 1. So same result. Therefore, seems like p(x) is always 1. Therefore, degree 0. Therefore, answer is 0. But why does the problem give such a complicated expression? Maybe I need to think again.

Alternatively, maybe there's a mistake in my reasoning? Let me think again. The Lagrange interpolation formula for three points gives a quadratic polynomial that passes through those three points. But if the function we are interpolating is a constant function, then the interpolating polynomial must be that constant function. Therefore, p(x) is 1. Therefore, degree 0. Therefore, answer is 0. But maybe the problem is designed to trick us into thinking it's quadratic, but actually it's constant. Therefore, the answer is 0. But let me check another way.

Alternatively, maybe I can consider the general case. Let me suppose that p(x) is a quadratic polynomial. Let me denote p(x) = Ax² + Bx + C. Then, since p(x) is equal to 1 for all x, then A = 0, B = 0, C = 1. Therefore, degree 0. But how can I show that p(x) is equal to 1?

Alternatively, maybe evaluate p(x) at specific points. For example, let me plug in x = a. Then:

First term: [(a - b)(a - c)] / [(a - b)(a - c)] = 1

Second term: [(a - a)(a - c)] / [(b - a)(b - c)] = 0 / ... = 0

Third term: [(a - a)(a - b)] / [(c - a)(c - b)] = 0 / ... = 0

Therefore, p(a) = 1 + 0 + 0 = 1.

Similarly, p(b) = 1, and p(c) = 1. Therefore, p(x) is a polynomial that takes the value 1 at three distinct points a, b, c. If p(x) is a quadratic polynomial, then the only quadratic polynomial that takes the value 1 at three distinct points is the constant polynomial 1. Because if a quadratic polynomial takes the same value at three distinct points, then it must be constant. Wait, but a quadratic polynomial can't take the same value at three distinct points unless it's constant. Because if p(x) - 1 has three roots, but if p(x) is quadratic, then p(x) - 1 is quadratic, which can have at most two roots unless it's the zero polynomial. Therefore, p(x) - 1 = 0 for all x, so p(x) = 1. Therefore, p(x) is constant. Therefore, degree 0. Therefore, answer is 0. Therefore, the degree is 0. Therefore, the answer is \boxed{0}.

But let me check again. Suppose that p(x) is a quadratic polynomial. Then, since p(a) = p(b) = p(c) = 1, then p(x) - 1 has three roots, but since p(x) - 1 is quadratic, it can have at most two roots unless it's identically zero. Therefore, p(x) - 1 = 0, so p(x) = 1. Therefore, degree 0. Therefore, answer is 0. Therefore, the answer is 0. Therefore, I think that's the correct answer.

But maybe I need to check if p(x) is indeed a polynomial. Because each term is a rational function, but when you add them up, they might not have denominators. Wait, but in the problem statement, it says "the polynomial p(x)", so it's given that p(x) is a polynomial. Therefore, the denominators must divide the numerators. But in our previous examples, when we added them up, they canceled out. Therefore, the expression simplifies to 1, which is a polynomial. Therefore, the degree is 0. Therefore, answer is 0.

Alternatively, maybe there's another way to think about this. Let me think about the general form. Let me consider that the given expression is the Lagrange interpolation polynomial for the function f(x) = 1, which is indeed 1. Therefore, p(x) = 1. Therefore, degree 0. Therefore, answer is 0.

Alternatively, maybe I can think of the expression as a combination of basis polynomials. For example, the Lagrange basis polynomials are defined as L_a(x) = [(x - b)(x - c)] / [(a - b)(a - c)], and similarly for L_b(x) and L_c(x). Then, the sum L_a(x) + L_b(x) + L_c(x) would be the interpolating polynomial for the function f(x) = 1, which is 1. Therefore, p(x) = 1. Therefore, degree 0. Therefore, answer is 0.

Therefore, I think the answer is 0. But maybe I need to check if there's a case where p(x) isn't 1? Let me try with a different function. Wait, but the problem is specifically for f(x) = 1. So if I take f(x) = 1, then the interpolating polynomial is 1. Therefore, regardless of the points, as long as they are distinct, the interpolating polynomial is 1. Therefore, p(x) is 1. Therefore, degree 0. Therefore, answer is 0.

But maybe the problem is designed to think that each term is quadratic, so adding three quadratics would give degree 2, but in reality, they cancel out. Therefore, the answer is 0. Therefore, I think the answer is 0. Therefore, I will go with \boxed{0}.

**Final Answer**
\boxed{0}

To determine the degree of the polynomial

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)},
$$

we begin by analyzing the structure of the expression.

---

### Step 1: Recognize the Form

Each term in the sum is of the form:

$$
\frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ x_i, x_j, x_k $ are distinct real numbers (in this case, $ a, b, c $). This is the **Lagrange basis polynomial** for the point $ x_i $, used in **Lagrange interpolation**.

In general, the Lagrange interpolation polynomial for a function $ f(x) $ that passes through the points $ (a, f(a)), (b, f(b)), (c, f(c)) $ is given by:

$$
P(x) = f(a) \cdot \frac{(x - b)(x - c)}{(a - b)(a - c)} + f(b) \cdot \frac{(x - a)(x - c)}{(b - a)(b - c)} + f(c) \cdot \frac{(x - a)(x - b)}{(c - a)(c - b)}.
$$

In our case, the function being interpolated is the constant function $ f(x) = 1 $. Therefore, the interpolating polynomial becomes:

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)}.
$$

This is exactly the expression given in the problem.

---

### Step 2: Understand the Result of the Interpolation

Since the function $ f(x) = 1 $ is a constant function, the interpolating polynomial must also be constant. That is, the polynomial $ p(x) $ must be equal to 1 for all $ x $, because it passes through the points $ (a, 1), (b, 1), (c, 1) $.

This is a key insight: **a constant function is the only polynomial that takes the same value at three distinct points** unless it is constant. Therefore, the expression simplifies to:

$$
p(x) = 1.
$$

---

### Step 3: Determine the Degree

The polynomial $ p(x) = 1 $ is a **constant polynomial**, which has degree **0**.

---

### Final Answer

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to find the degree of the polynomial p(x) given by that expression. Let me first write down the problem again to make sure I have it right:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)].

And a, b, c are distinct real numbers. Alright, so the question is asking for the degree of this polynomial. Let me think about how to approach this.

First, I know that when you have a polynomial expressed as a sum of terms like this, each term is a rational function, but since the final result is supposed to be a polynomial, maybe there's some cancellation happening? Or maybe each term is a polynomial, and when you add them up, the result is a polynomial of some degree. Let me check.

Each numerator is a quadratic polynomial: (x - b)(x - c) is quadratic, same with the others. The denominators are constants, since a, b, c are constants. So each term is a quadratic divided by a constant, so each term is a quadratic polynomial. Therefore, adding three quadratic polynomials together would give a polynomial of degree at most 2. But maybe the leading coefficients cancel out? So maybe the degree is less than 2? Or maybe not?

Wait, but the problem says "Find the degree of the polynomial p(x)", so maybe it's a quadratic polynomial? But maybe it's of lower degree? Let me check with specific values. Maybe plugging in some numbers for a, b, c to see what p(x) looks like.

Alternatively, maybe there's a smarter way. Let me recall that these kinds of expressions look like Lagrange interpolation polynomials. Let me think. The Lagrange interpolation formula for a quadratic polynomial passing through three points. Let me recall that if you have three points (a, f(a)), (b, f(b)), (c, f(c)), then the interpolating polynomial is given by the sum over each term [f(x_i) * product_{j≠i} (x - x_j)/(x_i - x_j)]. Wait, so if we take f(x) = 1, then the interpolating polynomial would be the sum of [1 * product terms], which is exactly the expression given here. Wait, so p(x) is the Lagrange interpolating polynomial for the function f(x) = 1, passing through the points (a, 1), (b, 1), (c, 1). But if we interpolate a constant function, then the interpolating polynomial should be the constant function 1, right? Because a constant function is already a polynomial of degree 0, and the interpolating polynomial should match it. Therefore, p(x) must be equal to 1? But then that would mean p(x) is a constant polynomial, so degree 0. But that seems conflicting with my initial thought that each term is quadratic. Wait, but maybe there's cancellation?

Wait, let me check with specific numbers. Let me choose specific values for a, b, c. Let me take a = 0, b = 1, c = 2. Then compute p(x).

So, let's compute each term:

First term: [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x - 1)(x - 2)] / [(-1)(-2)] = [(x^2 - 3x + 2)] / 2.

Second term: [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [(1)(-1)] = [x^2 - 2x]/(-1) = -x^2 + 2x.

Third term: [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)] / [(2)(1)] = [x^2 - x]/2.

So adding them up:

First term: (x² - 3x + 2)/2

Second term: -x² + 2x

Third term: (x² - x)/2

Let me combine them:

Let me convert all terms to have denominator 2:

First term: (x² - 3x + 2)/2

Second term: (-2x² + 4x)/2

Third term: (x² - x)/2

Now add them together:

[(x² - 3x + 2) + (-2x² + 4x) + (x² - x)] / 2

Compute numerator:

x² - 3x + 2 - 2x² + 4x + x² - x

Combine like terms:

x² - 2x² + x² = 0

-3x + 4x - x = 0

Then +2 remains.

So numerator is 2, so overall p(x) = 2 / 2 = 1. So p(x) is 1. Which is a constant polynomial. Therefore, degree 0. But that seems to contradict my initial thought that each term is quadratic. But in reality, when you add them up, they cancel out to a constant. Therefore, the degree is 0. But wait, the problem says "Find the degree of the polynomial p(x)", so maybe the answer is 0? But let me check with another example to make sure.

Let me take different a, b, c. Let me take a = 1, b = 2, c = 3.

Compute p(x):

First term: [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = [(x² -5x +6)] / [(-1)(-2)] = (x² -5x +6)/2

Second term: [(x -1)(x -3)] / [(2 -1)(2 -3)] = [(x² -4x +3)] / [(1)(-1)] = -(x² -4x +3) = -x² +4x -3

Third term: [(x -1)(x -2)] / [(3 -1)(3 -2)] = [(x² -3x +2)] / [(2)(1)] = (x² -3x +2)/2

Adding them up:

First term: (x² -5x +6)/2

Second term: -x² +4x -3

Third term: (x² -3x +2)/2

Convert all to denominator 2:

First term: (x² -5x +6)/2

Second term: (-2x² +8x -6)/2

Third term: (x² -3x +2)/2

Adding numerators:

(x² -5x +6) + (-2x² +8x -6) + (x² -3x +2) =

x² -2x² +x² = 0

-5x +8x -3x = 0

6 -6 +2 = 2

So again, numerator is 2, so p(x) = 2/2 = 1. So again, p(x) is 1. So that seems to confirm that p(x) is the constant polynomial 1. Therefore, degree 0. But wait, the problem says "Find the degree of the polynomial p(x)", so answer is 0? But why does the problem mention that a, b, c are distinct? Because if they are not distinct, then denominators would be zero. But since they are distinct, denominators are non-zero. But in any case, the result is always 1? So regardless of a, b, c (as long as distinct), p(x) is 1. Therefore, degree 0. But let me think again.

Wait, but maybe I made a mistake here? Let me check with another example where maybe the result isn't 1? Wait, but according to the Lagrange interpolation, if you interpolate the constant function 1 at three points, then the interpolating polynomial is 1. So regardless of the points, as long as they are distinct, the interpolating polynomial is 1. Therefore, p(x) is 1. Therefore, degree 0. Therefore, answer is 0. But maybe I need to check another case where maybe the calculation is different?

Wait, let me try with a = 0, b = 1, c = -1. Let me compute p(x):

First term: [(x - 1)(x - (-1))]/[(0 - 1)(0 - (-1))] = [(x -1)(x +1)] / [(-1)(1)] = (x² -1)/(-1) = -x² +1

Second term: [(x - 0)(x - (-1))]/[(1 - 0)(1 - (-1))] = [x(x +1)] / [1 * 2] = (x² +x)/2

Third term: [(x - 0)(x -1)] / [(-1 - 0)(-1 -1)] = [x(x -1)] / [(-1)(-2)] = [x² -x]/2

Adding them up:

First term: -x² +1

Second term: (x² +x)/2

Third term: (x² -x)/2

Combine second and third terms: [(x² +x) + (x² -x)] / 2 = (2x²)/2 = x²

So total p(x) = -x² +1 + x² = 1. Again, 1. So same result. Therefore, seems like p(x) is always 1. Therefore, degree 0. Therefore, answer is 0. But why does the problem give such a complicated expression? Maybe I need to think again.

Alternatively, maybe there's a mistake in my reasoning? Let me think again. The Lagrange interpolation formula for three points gives a quadratic polynomial that passes through those three points. But if the function we are interpolating is a constant function, then the interpolating polynomial must be that constant function. Therefore, p(x) is 1. Therefore, degree 0. Therefore, answer is 0. But maybe the problem is designed to trick us into thinking it's quadratic, but actually it's constant. Therefore, the answer is 0. But let me check another way.

Alternatively, maybe I can consider the general case. Let me suppose that p(x) is a quadratic polynomial. Let me denote p(x) = Ax² + Bx + C. Then, since p(x) is equal to 1 for all x, then A = 0, B = 0, C = 1. Therefore, degree 0. But how can I show that p(x) is equal to 1?

Alternatively, maybe evaluate p(x) at specific points. For example, let me plug in x = a. Then:

First term: [(a - b)(a - c)] / [(a - b)(a - c)] = 1

Second term: [(a - a)(a - c)] / [(b - a)(b - c)] = 0 / ... = 0

Third term: [(a - a)(a - b)] / [(c - a)(c - b)] = 0 / ... = 0

Therefore, p(a) = 1 + 0 + 0 = 1.

Similarly, p(b) = 1, and p(c) = 1. Therefore, p(x) is a polynomial that takes the value 1 at three distinct points a, b, c. If p(x) is a quadratic polynomial, then the only quadratic polynomial that takes the value 1 at three distinct points is the constant polynomial 1. Because if a quadratic polynomial takes the same value at three distinct points, then it must be constant. Wait, but a quadratic polynomial can't take the same value at three distinct points unless it's constant. Because if p(x) - 1 has three roots, but if p(x) is quadratic, then p(x) - 1 is quadratic, which can have at most two roots unless it's the zero polynomial. Therefore, p(x) - 1 = 0 for all x, so p(x) = 1. Therefore, p(x) is constant. Therefore, degree 0. Therefore, answer is 0. Therefore, the degree is 0. Therefore, the answer is \boxed{0}.

But let me check again. Suppose that p(x) is a quadratic polynomial. Then, since p(a) = p(b) = p(c) = 1, then p(x) - 1 has three roots, but since p(x) - 1 is quadratic, it can have at most two roots unless it's identically zero. Therefore, p(x) - 1 = 0, so p(x) = 1. Therefore, degree 0. Therefore, answer is 0. Therefore, the answer is 0. Therefore, I think that's the correct answer.

But maybe I need to check if p(x) is indeed a polynomial. Because each term is a rational function, but when you add them up, they might not have denominators. Wait, but in the problem statement, it says "the polynomial p(x)", so it's given that p(x) is a polynomial. Therefore, the denominators must divide the numerators. But in our previous examples, when we added them up, they canceled out. Therefore, the expression simplifies to 1, which is a polynomial. Therefore, the degree is 0. Therefore, answer is 0.

Alternatively, maybe there's another way to think about this. Let me think about the general form. Let me consider that the given expression is the Lagrange interpolation polynomial for the function f(x) = 1, which is indeed 1. Therefore, p(x) = 1. Therefore, degree 0. Therefore, answer is 0.

Alternatively, maybe I can think of the expression as a combination of basis polynomials. For example, the Lagrange basis polynomials are defined as L_a(x) = [(x - b)(x - c)] / [(a - b)(a - c)], and similarly for L_b(x) and L_c(x). Then, the sum L_a(x) + L_b(x) + L_c(x) would be the interpolating polynomial for the function f(x) = 1, which is 1. Therefore, p(x) = 1. Therefore, degree 0. Therefore, answer is 0.

Therefore, I think the answer is 0. But maybe I need to check if there's a case where p(x) isn't 1? Let me try with a different function. Wait, but the problem is specifically for f(x) = 1. So if I take f(x) = 1, then the interpolating polynomial is 1. Therefore, regardless of the points, as long as they are distinct, the interpolating polynomial is 1. Therefore, p(x) is 1. Therefore, degree 0. Therefore, answer is 0.

But maybe the problem is designed to think that each term is quadratic, so adding three quadratics would give degree 2, but in reality, they cancel out. Therefore, the answer is 0. Therefore, I think the answer is 0. Therefore, I will go with \boxed{0}.

**Final Answer**
\boxed{0}

To determine the degree of the polynomial

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)},
$$

we begin by analyzing the structure of the expression.

---

### Step 1: Recognize the Form

Each term in the sum is of the form:

$$
\frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ x_i, x_j, x_k $ are distinct real numbers (in this case, $ a, b, c $). This is the **Lagrange basis polynomial** for the point $ x_i $, used in **Lagrange interpolation**.

In general, the Lagrange interpolation polynomial for a function $ f(x) $ that passes through the points $ (a, f(a)), (b, f(b)), (c, f(c)) $ is given by:

$$
P(x) = f(a) \cdot \frac{(x - b)(x - c)}{(a - b)(a - c)} + f(b) \cdot \frac{(x - a)(x - c)}{(b - a)(b - c)} + f(c) \cdot \frac{(x - a)(x - b)}{(c - a)(c - b)}.
$$

In our case, the function being interpolated is the constant function $ f(x) = 1 $. Therefore, the interpolating polynomial becomes:

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)}.
$$

This is exactly the expression given in the problem.

---

### Step 2: Understand the Result of the Interpolation

Since the function $ f(x) = 1 $ is a constant function, the interpolating polynomial must also be constant. That is, the polynomial $ p(x) $ must be equal to 1 for all $ x $, because it passes through the points $ (a, 1), (b, 1), (c, 1) $.

This is a key insight: **a constant function is the only polynomial that takes the same value at three distinct points** unless it is constant. Therefore, the expression simplifies to:

$$
p(x) = 1.
$$

---

### Step 3: Determine the Degree

The polynomial $ p(x) = 1 $ is a **constant polynomial**, which has degree **0**.

---

### Final Answer

$$
\boxed{0}
$$

Solution:
Note that $p(x)$ has degree at most 2.  Also, $p(a) = p(b) = p(c) = 1.$  Thus, the polynomials $p(x)$ and 1 agree at three different values, so by the Identity Theorem, they are the same polynomial.  Hence, the degree of $p(x)$ (which is the constant polynomial 1) is $\boxed{0}.$

You can check manually that
\[p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)}\]simplifies to 1.


Explanation:
Okay, so I need to find cos(angle RPS) given that sin(angle RPQ) is 7/25. Let me try to visualize the diagram first. There's a point P at the origin, Q at (2,0), and S at (-2,0). R is some point above the x-axis, rotated by an angle whose sine is 7/25. The diagram shows lines Q to S, P to R, and probably other connections.

First, let me recall that angle RPQ is the angle at point P between points R and Q. So, angle RPQ is formed by the lines PR and PQ. Similarly, angle RPS is the angle at point P between points R and S. Since S is on the opposite side of P from Q, maybe these angles are supplementary or something? Wait, but I need to be careful here.

Let me try to sketch mentally. Points Q and S are on the x-axis, symmetric with respect to P, which is at the origin. Q is at (2,0), S is at (-2,0). Point R is somewhere above the x-axis, connected to P. The Asymptote code says R is obtained by rotating the point (1.5, 0) by an angle whose sine is 7/25. So, rotating the point (1.5, 0) around P (which is at (0,0)) by some angle θ where sinθ = 7/25. That would place R somewhere in the upper half-plane.

So, if I can figure out coordinates of R, maybe I can compute the required cosine. Let me try that approach.

First, let me find the coordinates of R. The original point before rotation is (1.5, 0). Rotating this point by angle θ around P (origin) gives R. The rotation matrix is:

[cosθ, -sinθ]
[sinθ, cosθ]

So, applying this to (1.5, 0):

x-coordinate: 1.5*cosθ - 0*sinθ = 1.5*cosθ

y-coordinate: 1.5*sinθ + 0*cosθ = 1.5*sinθ

But we know that sinθ = 7/25. Therefore, we can find cosθ. Since sin²θ + cos²θ = 1, cosθ = sqrt(1 - (49/625)) = sqrt(576/625) = 24/25. Since θ is an angle of rotation, and R is above the x-axis, θ is between 0 and π, so cosθ is positive. Therefore, cosθ = 24/25.

Therefore, coordinates of R are:

x = 1.5*(24/25) = (3/2)*(24/25) = (72)/50 = 36/25 = 1.44

y = 1.5*(7/25) = (3/2)*(7/25) = 21/50 = 0.42

So R is at (36/25, 21/50). Let me confirm that. Yes, 1.5 is 3/2, so multiplying by 24/25 gives 72/50 = 36/25, and 7/25 times 3/2 is 21/50. Okay.

Now, we need to find cos(angle RPS). Let me figure out what angle RPS is. It's the angle at point P between points R and S. So, vectors PR and PS. Since P is at (0,0), vector PR is from P to R, which is (36/25, 21/50), and vector PS is from P to S, which is (-2, 0). Wait, S is at (-2, 0), so vector PS is (-2, 0).

Therefore, angle between vectors PR and PS. To find the cosine of that angle, we can use the dot product formula:

cos(angle) = (PR · PS) / (|PR| |PS|)

First, compute PR · PS. That's (36/25)*(-2) + (21/50)*0 = -72/25 + 0 = -72/25.

Then, compute |PR|. The magnitude of PR is sqrt( (36/25)^2 + (21/50)^2 ). Let me compute that.

First, square of 36/25 is (1296)/625.

Square of 21/50 is (441)/2500.

So total is 1296/625 + 441/2500. Let me convert to common denominator:

Multiply numerator and denominator of 1296/625 by 4: 5184/2500.

So total is 5184/2500 + 441/2500 = 5625/2500.

Simplify 5625/2500: divide numerator and denominator by 25: 225/100, then divide by 25 again: 9/4? Wait, 5625 ÷ 25 = 225, 2500 ÷25=100. Then 225/100 = 9/4? Wait, 225 divided by 25 is 9, 100 divided by 25 is 4. Yes. So sqrt(9/4) = 3/2. Wait, that's interesting. So |PR| is 3/2. Wait, that's the same as the original point before rotation, which was (1.5, 0). Makes sense because rotating a point doesn't change its distance from the origin. So |PR| is 1.5, which is 3/2. So that checks out.

Then |PS| is the distance from P to S, which is sqrt( (-2)^2 + 0^2 ) = sqrt(4) = 2.

Therefore, cos(angle RPS) = (-72/25) / ( (3/2)*2 ) = (-72/25) / 3 = (-72/25)*(1/3) = -24/25.

Wait, that seems straightforward. But let me check again.

Wait, the denominator is |PR| * |PS| = (3/2)*2 = 3. So yes, the denominator is 3. The numerator is -72/25. Therefore, cos(angle RPS) = (-72/25)/3 = -24/25. So the answer is -24/25? But let me think again if I did everything correctly.

Wait, angle RPS is between PR and PS. Since PS is pointing to the left along the x-axis, and PR is pointing to the upper right. So the angle between them is obtuse? Because the dot product is negative, so cosine is negative, which would mean angle is greater than 90 degrees. That seems plausible.

But let me check if there's another way to approach this. Maybe using the given sin(angle RPQ) = 7/25. Let me see if that's consistent with what I found.

Angle RPQ is the angle at P between PR and PQ. PQ is from P(0,0) to Q(2,0), so along the positive x-axis. So angle RPQ is the angle between PR and the positive x-axis. Wait, but earlier we said that R was obtained by rotating the point (1.5, 0) by angle θ where sinθ = 7/25. Therefore, angle between PR and the x-axis is θ, so sinθ = 7/25, which matches the given. Therefore, angle RPQ is θ, so sin(angle RPQ) = 7/25. Therefore, angle RPS is the angle between PR and PS, which is the angle between PR and the negative x-axis. So if angle between PR and positive x-axis is θ, then angle between PR and negative x-axis is π - θ. Therefore, cos(angle RPS) = cos(π - θ) = -cosθ. Since cosθ is 24/25, then cos(angle RPS) is -24/25. Which matches the previous result. So that's a good check.

Therefore, the answer is -24/25. Wait, but the problem says "What is cos ∠RPS?" So is that negative? Let me check once more.

Yes, because angle RPS is between PR and PS (which is the negative x-axis). Since PR is in the first quadrant, making an angle θ with the positive x-axis, then the angle between PR and the negative x-axis is π - θ, whose cosine is -cosθ. Therefore, cos(angle RPS) = -cosθ = -24/25. So that's correct.

Alternatively, using coordinates, we found the same result. So I think that's solid.

But let me check once again the coordinates. Coordinates of R: (36/25, 21/50). Let me compute the angle between PR and PS. Vector PR is (36/25, 21/50), vector PS is (-2, 0). The dot product is (36/25)(-2) + (21/50)(0) = -72/25. The magnitude of PR is 3/2, magnitude of PS is 2. So the cosine is (-72/25)/( (3/2)*2 ) = (-72/25)/3 = -24/25. Yep. So that's correct.

Therefore, the answer is -24/25. But let me check if the question is maybe expecting a positive value? But no, because angle RPS is obtuse, so cosine is negative. So I think that's correct.

Wait, but maybe I made a mistake in interpreting the angle? Let me check the Asymptote code again. The Asymptote code defines SS as (-2,0), P is (0,0), Q is (2,0), R is rotate(aSin(7/25))*(1.5,0). So rotating the point (1.5, 0) by angle whose sine is 7/25. So that angle is θ where sinθ = 7/25, so θ is in the first quadrant. Therefore, R is in the first quadrant. Then angle RPQ is θ, which is between PR and PQ (positive x-axis). Then angle RPS is between PR and PS (negative x-axis). So that angle is π - θ, so cosine is -cosθ. So yes, -24/25.

Alternatively, maybe there's a different approach using triangle relations? Let me think. For example, in triangle RPQ, we know sin(angle RPQ) = 7/25. If we can find some sides or other angles, maybe relate to triangle RPS?

But since we already have coordinate-based answer, and verified with two methods, I think that's solid. So I think the answer is -24/25.

But let me check once more for calculation errors. For example, when I calculated |PR|, I said that it's 3/2. Since the original point was (1.5, 0), which is 3/2, and rotation doesn't change the distance, so yes, |PR| is 3/2. Then |PS| is 2. Then the dot product is (3/2)*(-2) + (21/50)*0 = -3. Wait, hold on! Wait, earlier I said PR · PS is -72/25. Wait, but if PR is (36/25, 21/50), and PS is (-2, 0), then the dot product is (36/25)(-2) + (21/50)(0) = -72/25 + 0 = -72/25. But if |PR| is 3/2, then why is the dot product -72/25? Let me check:

Wait, 3/2 times 2 is 3. So the denominator is 3. So -72/25 divided by 3 is -24/25. But if I think of PR · PS as |PR||PS|cos(angle), which is (3/2)(2)cos(angle) = 3 cos(angle). Therefore, 3 cos(angle) = -72/25 => cos(angle) = -24/25. So that's correct. So that's consistent. So no problem here.

Alternatively, if I compute the dot product as (36/25)(-2) + (21/50)(0) = -72/25. But if I compute |PR| * |PS| * cos(angle) = (3/2)(2)cos(angle) = 3 cos(angle). Therefore, equate to -72/25. So 3 cos(angle) = -72/25 => cos(angle) = -24/25. Correct. So that's correct.

Therefore, I think the answer is -24/25. So I think that's the answer. Let me check if the problem says anything about the diagram. The Asymptote code draws Q--SS (which is from Q(2,0) to S(-2,0)), and P--R. So the diagram is symmetric with respect to P. So angle RPS is indeed between PR and PS. So I think that's correct.

Therefore, I think the answer is \boxed{-\dfrac{24}{25}}.

**Final Answer**
\boxed{-\dfrac{24}{25}}

To solve the problem, we are given that:

$$
\sin(\angle RPQ) = \frac{7}{25}
$$

and we are asked to find:

$$
\cos(\angle RPS)
$$

---

### Step 1: Understand the Geometry

From the Asymptote code and the description, we can deduce the following:

- Point $ P $ is at the origin: $ (0, 0) $
- Point $ Q $ is at $ (2, 0) $
- Point $ S $ is at $ (-2, 0) $
- Point $ R $ is obtained by rotating the point $ (1.5, 0) $ around $ P $ by an angle $ \theta $ such that $ \sin(\theta) = \frac{7}{25} $

This means that $ \angle RPQ $ is the angle between the positive x-axis (from $ P $ to $ Q $) and the vector $ \vec{PR} $, which is the rotated vector from $ P $ to $ R $.

---

### Step 2: Use Trigonometric Identities

We are given:

$$
\sin(\theta) = \frac{7}{25}
$$

Using the Pythagorean identity:

$$
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

So, $ \cos(\theta) = \frac{24}{25} $

---

### Step 3: Determine $ \angle RPS $

We are asked to find $ \cos(\angle RPS) $, which is the angle between vectors $ \vec{PR} $ and $ \vec{PS} $.

- $ \vec{PS} $ is the vector from $ P $ to $ S $, which is $ (-2, 0) $
- $ \vec{PR} $ is the vector from $ P $ to $ R $, which is obtained by rotating $ (1.5, 0) $ by angle $ \theta $

From the rotation, the coordinates of $ R $ are:

$$
R = (1.5 \cos\theta, 1.5 \sin\theta) = \left( \frac{36}{25}, \frac{21}{50} \right)
$$

So, $ \vec{PR} = \left( \frac{36}{25}, \frac{21}{50} \right) $

---

### Step 4: Use the Dot Product to Find $ \cos(\angle RPS) $

We use the dot product formula:

$$
\cos(\angle RPS) = \frac{\vec{PR} \cdot \vec{PS}}{|\vec{PR}| \cdot |\vec{PS}|}
$$

- $ \vec{PR} \cdot \vec{PS} = \left( \frac{36}{25} \right)(-2) + \left( \frac{21}{50} \right)(0) = -\frac{72}{25} $
- $ |\vec{PR}| = \sqrt{\left( \frac{36}{25} \right)^2 + \left( \frac{21}{50} \right)^2} = \frac{3}{2} $ (since the original point was $ (1.5, 0) $, and rotation preserves distance)
- $ |\vec{PS}| = \sqrt{(-2)^2 + 0^2} = 2 $

So:

$$
\cos(\angle RPS) = \frac{-\frac{72}{25}}{\frac{3}{2} \cdot 2} = \frac{-\frac{72}{25}}{3} = -\frac{24}{25}
$$

---

### Step 5: Confirm with Angle Relationship

Alternatively, since $ \angle RPS $ is the angle between $ \vec{PR} $ and the negative x-axis, and $ \angle RPQ $ is the angle between $ \vec{PR} $ and the positive x-axis, we can say:

$$
\angle RPS = \pi - \angle RPQ
$$

Then:

$$
\cos(\angle RPS) = \cos(\pi - \theta) = -\cos(\theta) = -\frac{24}{25}
$$

This confirms our earlier result.

---

### Final Answer

$$
\boxed{-\dfrac{24}{25}}
$$

[Assistant]
Explanation:
Okay, so I need to find cos(angle RPS) given that sin(angle RPQ) is 7/25. Let me try to visualize the diagram first. There's a point P at the origin, Q at (2,0), and S at (-2,0). R is some point above the x-axis, rotated by an angle whose sine is 7/25. The diagram shows lines Q to S, P to R, and probably other connections.

First, let me recall that angle RPQ is the angle at point P between points R and Q. So, angle RPQ is formed by the lines PR and PQ. Similarly, angle RPS is the angle at point P between points R and S. Since S is on the opposite side of P from Q, maybe these angles are supplementary or something? Wait, but I need to be careful here.

Let me try to sketch mentally. Points Q and S are on the x-axis, symmetric with respect to P, which is at the origin. Q is at (2,0), S is at (-2,0). Point R is somewhere above the x-axis, connected to P. The Asymptote code says R is obtained by rotating the point (1.5, 0) by an angle whose sine is 7/25. So, rotating the point (1.5, 0) around P (which is at (0,0)) by some angle θ where sinθ = 7/25. That would place R somewhere in the upper half-plane.

So, if I can figure out coordinates of R, maybe I can compute the required cosine. Let me try that approach.

First, let me find the coordinates of R. The original point before rotation is (1.5, 0). Rotating this point by angle θ around P (origin) gives R. The rotation matrix is:

[cosθ, -sinθ]
[sinθ, cosθ]

So, applying this to (1.5, 0):

x-coordinate: 1.5*cosθ - 0*sinθ = 1.5*cosθ

y-coordinate: 1.5*sinθ + 0*cosθ = 1.5*sinθ

But we know that sinθ = 7/25. Therefore, we can find cosθ. Since sin²θ + cos²θ = 1, cosθ = sqrt(1 - (49/625)) = sqrt(576/625) = 24/25. Since θ is an angle of rotation, and R is above the x-axis, θ is between 0 and π, so cosθ is positive. Therefore, cosθ = 24/25.

Therefore, coordinates of R are:

x = 1.5*(24/25) = (3/2)*(24/25) = (72)/50 = 36/25 = 1.44

y = 1.5*(7/25) = (3/2)*(7/25) = 21/50 = 0.42

So R is at (36/25, 21/50). Let me confirm that. Yes, 1.5 is 3/2, so multiplying by 24/25 gives 72/50 = 36/25, and 7/25 times 3/2 is 21/50. Okay.

Now, we need to find cos(angle RPS). Let me figure out what angle RPS is. It's the angle at point P between points R and S. So, vectors PR and PS. Since P is at (0,0), vector PR is from P to R, which is (36/25, 21/50), and vector PS is from P to S, which is (-2, 0). Wait, S is at (-2, 0), so vector PS is (-2, 0).

Therefore, angle between vectors PR and PS. To find the cosine of that angle, we can use the dot product formula:

cos(angle) = (PR · PS) / (|PR| |PS|)

First, compute PR · PS. That's (36/25)*(-2) + (21/50)*0 = -72/25 + 0 = -72/25.

Then, compute |PR|. The magnitude of PR is sqrt( (36/25)^2 + (21/50)^2 ). Let me compute that.

First, square of 36/25 is (1296)/625.

Square of 21/50 is (441)/2500.

So total is 1296/625 + 441/2500. Let me convert to common denominator:

Multiply numerator and denominator of 1296/625 by 4: 5184/2500.

So total is 5184/2500 + 441/2500 = 5625/2500.

Simplify 5625/2500: divide numerator and denominator by 25: 225/100, then divide by 25 again: 9/4? Wait, 5625 ÷ 25 = 225, 2500 ÷25=100. Then 225/100 = 9/4? Wait, 225 divided by 25 is 9, 100 divided by 25 is 4. Yes. So sqrt(9/4) = 3/2. Wait, that's interesting. So |PR| is 3/2. Wait, that's the same as the original point before rotation, which was (1.5, 0). Makes sense because rotating a point doesn't change its distance from the origin. So |PR| is 1.5, which is 3/2. So that checks out.

Then |PS| is the distance from P to S, which is sqrt( (-2)^2 + 0^2 ) = sqrt(4) = 2.

Therefore, cos(angle RPS) = (-72/25) / ( (3/2)*2 ) = (-72/25) / 3 = (-72/25)*(1/3) = -24/25.

Wait, that seems straightforward. But let me check again.

Wait, the denominator is |PR| * |PS| = (3/2)*2 = 3. So yes, the denominator is 3. The numerator is -72/25. Therefore, cos(angle RPS) = (-72/25)/3 = -24/25. So the answer is -24/25? But let me think again if I did everything correctly.

Wait, angle RPS is between PR and PS. Since PS is pointing to the left along the x-axis, and PR is pointing to the upper right. So the angle between them is obtuse? Because the dot product is negative, so cosine is negative, which would mean angle is greater than 90 degrees. That seems plausible.

But let me check if there's another way to approach this. Maybe using the given sin(angle RPQ) = 7/25. Let me see if that's consistent with what I found.

Angle RPQ is the angle at P between PR and PQ. PQ is from P(0,0) to Q(2,0), so along the positive x-axis. So angle RPQ is the angle between PR and the positive x-axis. Wait, but earlier we said that R was obtained by rotating the point (1.5, 0) by angle θ where sinθ = 7/25. Therefore, angle between PR and the x-axis is θ, so sinθ = 7/25, which matches the given. Therefore, angle RPQ is θ, so sin(angle RPQ) = 7/25. Therefore, angle RPS is the angle between PR and PS, which is the angle between PR and the negative x-axis. So if angle between PR and positive x-axis is θ, then angle between PR and negative x-axis is π - θ. Therefore, cos(angle RPS) = cos(π - θ) = -cosθ. Since cosθ is 24/25, then cos(angle RPS) is -24/25. Which matches the previous result. So that's a good check.

Therefore, the answer is -24/25. Wait, but the problem says "What is cos ∠RPS?" So is that negative? Let me check once more.

Yes, because angle RPS is between PR and PS (which is the negative x-axis). Since PR is in the first quadrant, making an angle θ with the positive x-axis, then the angle between PR and the negative x-axis is π - θ, whose cosine is -cosθ. Therefore, cos(angle RPS) = -cosθ = -24/25. So that's correct.

Alternatively, using coordinates, we found the same result. So I think that's solid.

But let me check once again the coordinates. Coordinates of R: (36/25, 21/50). Let me compute the angle between PR and PS. Vector PR is (36/25, 21/50), vector PS is (-2, 0). The dot product is (36/25)(-2) + (21/50)(0) = -72/25. The magnitude of PR is 3/2, magnitude of PS is 2. So the cosine is (-72/25)/( (3/2)*2 ) = (-72/25)/3 = -24/25. Yep. So that's correct.

Therefore, the answer is -24/25. But let me check if the question is maybe expecting a positive value? But no, because angle RPS is obtuse, so cosine is negative. So I think that's correct.

Wait, but maybe I made a mistake in interpreting the angle? Let me check the Asymptote code again. The Asymptote code defines SS as (-2,0), P is (0,0), Q is (2,0), R is rotate(aSin(7/25))*(1.5,0). So rotating the point (1.5, 0) by angle whose sine is 7/25. So that angle is θ where sinθ = 7/25, so θ is in the first quadrant. Therefore, R is in the first quadrant. Then angle RPQ is θ, which is between PR and PQ (positive x-axis). Then angle RPS is between PR and PS (negative x-axis). So that angle is π - θ, so cosine is -cosθ. So yes, -24/25.

Alternatively, maybe there's a different approach using triangle relations? Let me think. For example, in triangle RPQ, we know sin(angle RPQ) = 7/25. If we can find some sides or other angles, maybe relate to triangle RPS?

But since we already have coordinate-based answer, and verified with two methods, I think that's solid. So I think the answer is -24/25.

But let me check once more for calculation errors. For example, when I calculated |PR|, I said that it's 3/2. Since the original point was (1.5, 0), which is 3/2, and rotation doesn't change the distance, so yes, |PR| is 3/2. Then |PS| is 2. Then the dot product is (3/2)*(-2) + (21/50)*0 = -3. Wait, hold on! Wait, earlier I said PR · PS is -72/25. Wait, but if PR is (36/25, 21/50), and PS is (-2, 0), then the dot product is (36/25)(-2) + (21/50)(0) = -72/25 + 0 = -72/25. But if |PR| is 3/2, then why is the dot product -72/25? Let me check:

Wait, 3/2 times 2 is 3. So the denominator is 3. So -72/25 divided by 3 is -24/25. But if I think of PR · PS as |PR||PS|cos(angle), which is (3/2)(2)cos(angle) = 3 cos(angle). Therefore, 3 cos(angle) = -72/25 => cos(angle) = -24/25. So that's correct. So that's consistent. So no problem here.

Alternatively, if I compute the dot product as (36/25)(-2) + (21/50)(0) = -72/25. But if I compute |PR| * |PS| * cos(angle) = (3/2)(2)cos(angle) = 3 cos(angle). Therefore, equate to -72/25. So 3 cos(angle) = -72/25 => cos(angle) = -24/25. Correct. So that's correct.

Therefore, I think the answer is -24/25. So I think that's the answer. Let me check if the problem says anything about the diagram. The Asymptote code draws Q--SS (which is from Q(2,0) to S(-2,0)), and P--R. So the diagram is symmetric with respect to P. So angle RPS is indeed between PR and PS. So I think that's correct.

Therefore, I think the answer is \boxed{-\dfrac{24}{25}}.

**Final Answer**
\boxed{-\dfrac{24}{25}}

To solve the problem, we are given that:

$$
\sin(\angle RPQ) = \frac{7}{25}
$$

and we are asked to find:

$$
\cos(\angle RPS)
$$

---

### Step 1: Understand the Geometry

From the Asymptote code and the description, we can deduce the following:

- Point $ P $ is at the origin: $ (0, 0) $
- Point $ Q $ is at $ (2, 0) $
- Point $ S $ is at $ (-2, 0) $
- Point $ R $ is obtained by rotating the point $ (1.5, 0) $ around $ P $ by an angle $ \theta $ such that $ \sin(\theta) = \frac{7}{25} $

This means that $ \angle RPQ $ is the angle between the positive x-axis (from $ P $ to $ Q $) and the vector $ \vec{PR} $, which is the rotated vector from $ P $ to $ R $.

---

### Step 2: Use Trigonometric Identities

We are given:

$$
\sin(\theta) = \frac{7}{25}
$$

Using the Pythagorean identity:

$$
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

So, $ \cos(\theta) = \frac{24}{25} $

---

### Step 3: Determine $ \angle RPS $

We are asked to find $ \cos(\angle RPS) $, which is the angle between vectors $ \vec{PR} $ and $ \vec{PS} $.

- $ \vec{PS} $ is the vector from $ P $ to $ S $, which is $ (-2, 0) $
- $ \vec{PR} $ is the vector from $ P $ to $ R $, which is obtained by rotating $ (1.5, 0) $ by angle $ \theta $

From the rotation, the coordinates of $ R $ are:

$$
R = (1.5 \cos\theta, 1.5 \sin\theta) = \left( \frac{36}{25}, \frac{21}{50} \right)
$$

So, $ \vec{PR} = \left( \frac{36}{25}, \frac{21}{50} \right) $

---

### Step 4: Use the Dot Product to Find $ \cos(\angle RPS) $

We use the dot product formula:

$$
\cos(\angle RPS) = \frac{\vec{PR} \cdot \vec{PS}}{|\vec{PR}| \cdot |\vec{PS}|}
$$

- $ \vec{PR} \cdot \vec{PS} = \left( \frac{36}{25} \right)(-2) + \left( \frac{21}{50} \right)(0) = -\frac{72}{25} $
- $ |\vec{PR}| = \sqrt{\left( \frac{36}{25} \right)^2 + \left( \frac{21}{50} \right)^2} = \frac{3}{2} $ (since the original point was $ (1.5, 0) $, and rotation preserves distance)
- $ |\vec{PS}| = \sqrt{(-2)^2 + 0^2} = 2 $

So:

$$
\cos(\angle RPS) = \frac{-\frac{72}{25}}{\frac{3}{2} \cdot 2} = \frac{-\frac{72}{25}}{3} = -\frac{24}{25}
$$

---

### Step 5: Confirm with Angle Relationship

Alternatively, since $ \angle RPS $ is the angle between $ \vec{PR} $ and the negative x-axis, and $ \angle RPQ $ is the angle between $ \vec{PR} $ and the positive x-axis, we can say:

$$
\angle RPS = \pi - \angle RPQ
$$

Then:

$$
\cos(\angle RPS) = \cos(\pi - \theta) = -\cos(\theta) = -\frac{24}{25}
$$

This confirms our earlier result.

---

### Final Answer

$$
\boxed{-\dfrac{24}{25}}
$$

Solution:
For any angle $x$, we have $\cos(180^\circ - x)=-\cos x$, so $\cos \angle RPS = \cos(180^\circ - \angle RPQ) =- \cos\angle RPQ$.

Since $\sin^2 \angle RPQ + \cos^2 \angle RPQ = 1$, we have $\cos^2\angle RPQ = 1 - \left(\frac{7}{25}\right)^2 = \frac{576}{625}$.  Since $\angle RPQ$ is acute, we have $\cos\angle RPQ = \frac{24}{25}$, which gives us $\cos\angle RPS = -\cos\angle RPQ = \boxed{-\frac{24}{25}}$.


Explanation:
Okay, so there's this robot that moves at a constant speed. Let me try to figure out how long the hallway is. The problem says that the robot takes 2.5 hours to travel 1 kilometer. Then, moving at the same speed, it takes 90 seconds to travel the length of a hallway. We need to find out how many meters long that hallway is.

First, I need to find the robot's speed. Since speed is distance divided by time, right? So maybe I can calculate the speed using the first piece of information: 1 kilometer in 2.5 hours. Let me write that down.

Speed = Distance / Time. So, the distance here is 1 kilometer, and the time is 2.5 hours. But wait, the time is in hours and the distance is in kilometers. Maybe I should convert the time into seconds or the distance into meters? Hmm, but maybe I can just keep it in kilometers per hour for now and then convert later? Let me think.

Alternatively, since the second part of the problem gives time in seconds, maybe I should convert the robot's speed into meters per second? That might make it easier to calculate the hallway length in meters. Let me try that approach.

First, let's convert the robot's speed from kilometers per hour to meters per second.

So, 1 kilometer is 1000 meters. 2.5 hours is how many seconds? Well, 1 hour is 60 minutes, and 1 minute is 60 seconds. So 2.5 hours is 2.5 * 60 * 60 seconds. Let me calculate that.

First, 2.5 hours * 60 minutes/hour = 150 minutes. Then, 150 minutes * 60 seconds/minute = 9000 seconds. So, 2.5 hours is 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, its speed is 1000 meters / 9000 seconds. Let me compute that.

Dividing numerator and denominator by 1000, that's 1/9 meters per second. Wait, 1000 divided by 9000 is 1/9? Let me check. 9000 divided by 1000 is 9, so yes, 1000/9000 = 1/9. So the speed is 1/9 m/s.

Alternatively, maybe I can check that again. If it takes 2.5 hours to go 1 km, then in one hour it would go 1 / 2.5 km. Which is 0.4 km per hour. Then converting that to meters per second: 0.4 km/h is 400 meters per hour. Then divide by 3600 seconds to get meters per second. 400 / 3600 = 1/9 m/s. Yep, same result. So that's correct.

So the robot's speed is 1/9 meters per second.

Now, the problem says that it takes 90 seconds to travel the length of the hallway. So, if I can find the distance, which is the length of the hallway, using the formula: distance = speed * time.

So, distance = (1/9 m/s) * 90 seconds. Let me compute that.

Multiplying 1/9 by 90: 90 divided by 9 is 10. So, 10 meters? Wait, that seems short for a hallway, but maybe it's possible. Let me check my steps again to make sure I didn't make a mistake.

First, converting 2.5 hours to seconds: 2.5 * 60 * 60. 2.5 * 60 is 150, 150 * 60 is 9000. Correct. Then, 1 km is 1000 meters. So speed is 1000 meters / 9000 seconds = 1/9 m/s. Correct. Then, time taken for hallway is 90 seconds. So distance is speed * time = (1/9) * 90 = 10 meters.

Hmm. So according to this, the hallway is 10 meters long. But maybe I made a mistake in unit conversion? Let me check again.

Alternatively, maybe I should approach it using ratios? Since the robot's speed is constant, the time taken is proportional to the distance.

So, if 1 km takes 2.5 hours, then how long does it take to travel x kilometers? Well, but the time given is 90 seconds. Let me convert 2.5 hours to seconds to make the units match.

As before, 2.5 hours is 9000 seconds. So, 1 km takes 9000 seconds. Therefore, the time per meter is 9000 seconds / 1000 meters = 9 seconds per meter. Therefore, if it takes 90 seconds to travel the hallway, then the length is 90 seconds / 9 seconds per meter = 10 meters.

Same answer. So that seems to confirm it.

Alternatively, maybe I can think in terms of proportions.

If 1 km (which is 1000 meters) takes 2.5 hours, then how many meters does it travel in 90 seconds?

First, let's convert 2.5 hours to seconds: 2.5 * 3600 = 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, the number of meters it travels in 1 second is 1000 / 9000 = 1/9 meters per second. Then, in 90 seconds, it travels 1/9 * 90 = 10 meters.

Same result again.

So, all methods lead to 10 meters. Maybe the hallway is indeed 10 meters long.

But just to make sure, let me check if there's another way someone might approach this problem and perhaps make a mistake.

For example, someone might think that since 2.5 hours is 2 hours and 30 minutes, and then try to convert 90 seconds into hours? Let me see.

90 seconds is 90/3600 hours, which is 1/40 hours. Then, if the robot's speed is 1 km per 2.5 hours, then in 1/40 hours, it would travel (1 km / 2.5 hours) * (1/40 hours) = (1 / 2.5) * (1 / 40) km.

Calculating that: 1 divided by 2.5 is 0.4, then 0.4 divided by 40 is 0.01 km. Then, converting 0.01 km to meters is 0.01 * 1000 = 10 meters.

Same answer again.

So regardless of the method, the answer is 10 meters.

Therefore, I think the hallway is 10 meters long.

But just to make sure, let me check if I messed up any unit conversions.

Original data: 2.5 hours for 1 km.

Convert 2.5 hours to seconds: 2.5 * 60 * 60.

Yes, 2.5 * 60 is 150 minutes, 150 * 60 is 9000 seconds.

So 1 km is 1000 meters.

So speed is 1000 meters / 9000 seconds = 1/9 m/s.

Then, time for hallway is 90 seconds.

Distance = speed * time = (1/9) * 90 = 10 meters.

Yes.

Alternatively, if I wanted to use km and hours, but then convert the time to hours.

So, 90 seconds is 90 / 3600 hours = 0.025 hours.

Then, since the robot's speed is 1 km / 2.5 hours, then distance = speed * time = (1 / 2.5) * 0.025 km.

Calculating that: 1 divided by 2.5 is 0.4, multiplied by 0.025 is 0.01 km.

Convert to meters: 0.01 * 1000 = 10 meters.

Same answer.

So, all methods confirm that the hallway is 10 meters long.

Therefore, I think the answer is 10 meters.

**Final Answer**
The hallway is \boxed{10} meters long.

To determine the length of the hallway, we begin by analyzing the robot's constant speed based on the given information:

---

### **Step 1: Determine the Robot's Speed**

The robot travels **1 kilometer (1000 meters)** in **2.5 hours**. To find its speed in **meters per second**, we first convert the time from hours to seconds:

$$
2.5 \text{ hours} = 2.5 \times 60 \times 60 = 9000 \text{ seconds}
$$

Now, calculate the speed:

$$
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1000 \text{ meters}}{9000 \text{ seconds}} = \frac{1}{9} \text{ meters per second}
$$

---

### **Step 2: Use the Speed to Find the Hallway Length**

We are told that the robot takes **90 seconds** to travel the length of the hallway. Using the formula:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

Substitute the known values:

$$
\text{Distance} = \frac{1}{9} \text{ m/s} \times 90 \text{ s} = 10 \text{ meters}
$$

---

### **Conclusion**

After verifying through multiple approaches (unit conversions, proportionality, and direct calculation), we consistently find that the hallway is:

$$
\boxed{10}
$$

meters long.

[Assistant]
Explanation:
Okay, so there's this robot that moves at a constant speed. Let me try to figure out how long the hallway is. The problem says that the robot takes 2.5 hours to travel 1 kilometer. Then, moving at the same speed, it takes 90 seconds to travel the length of a hallway. We need to find out how many meters long that hallway is.

First, I need to find the robot's speed. Since speed is distance divided by time, right? So maybe I can calculate the speed using the first piece of information: 1 kilometer in 2.5 hours. Let me write that down.

Speed = Distance / Time. So, the distance here is 1 kilometer, and the time is 2.5 hours. But wait, the time is in hours and the distance is in kilometers. Maybe I should convert the time into seconds or the distance into meters? Hmm, but maybe I can just keep it in kilometers per hour for now and then convert later? Let me think.

Alternatively, since the second part of the problem gives time in seconds, maybe I should convert the robot's speed into meters per second? That might make it easier to calculate the hallway length in meters. Let me try that approach.

First, let's convert the robot's speed from kilometers per hour to meters per second.

So, 1 kilometer is 1000 meters. 2.5 hours is how many seconds? Well, 1 hour is 60 minutes, and 1 minute is 60 seconds. So 2.5 hours is 2.5 * 60 * 60 seconds. Let me calculate that.

First, 2.5 hours * 60 minutes/hour = 150 minutes. Then, 150 minutes * 60 seconds/minute = 9000 seconds. So, 2.5 hours is 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, its speed is 1000 meters / 9000 seconds. Let me compute that.

Dividing numerator and denominator by 1000, that's 1/9 meters per second. Wait, 1000 divided by 9000 is 1/9? Let me check. 9000 divided by 1000 is 9, so yes, 1000/9000 = 1/9. So the speed is 1/9 m/s.

Alternatively, maybe I can check that again. If it takes 2.5 hours to go 1 km, then in one hour it would go 1 / 2.5 km. Which is 0.4 km per hour. Then converting that to meters per second: 0.4 km/h is 400 meters per hour. Then divide by 3600 seconds to get meters per second. 400 / 3600 = 1/9 m/s. Yep, same result. So that's correct.

So the robot's speed is 1/9 meters per second.

Now, the problem says that it takes 90 seconds to travel the length of the hallway. So, if I can find the distance, which is the length of the hallway, using the formula: distance = speed * time.

So, distance = (1/9 m/s) * 90 seconds. Let me compute that.

Multiplying 1/9 by 90: 90 divided by 9 is 10. So, 10 meters? Wait, that seems short for a hallway, but maybe it's possible. Let me check my steps again to make sure I didn't make a mistake.

First, converting 2.5 hours to seconds: 2.5 * 60 * 60. 2.5 * 60 is 150, 150 * 60 is 9000. Correct. Then, 1 km is 1000 meters. So speed is 1000 meters / 9000 seconds = 1/9 m/s. Correct. Then, time taken for hallway is 90 seconds. So distance is speed * time = (1/9) * 90 = 10 meters.

Hmm. So according to this, the hallway is 10 meters long. But maybe I made a mistake in unit conversion? Let me check again.

Alternatively, maybe I should approach it using ratios? Since the robot's speed is constant, the time taken is proportional to the distance.

So, if 1 km takes 2.5 hours, then how long does it take to travel x kilometers? Well, but the time given is 90 seconds. Let me convert 2.5 hours to seconds to make the units match.

As before, 2.5 hours is 9000 seconds. So, 1 km takes 9000 seconds. Therefore, the time per meter is 9000 seconds / 1000 meters = 9 seconds per meter. Therefore, if it takes 90 seconds to travel the hallway, then the length is 90 seconds / 9 seconds per meter = 10 meters.

Same answer. So that seems to confirm it.

Alternatively, maybe I can think in terms of proportions.

If 1 km (which is 1000 meters) takes 2.5 hours, then how many meters does it travel in 90 seconds?

First, let's convert 2.5 hours to seconds: 2.5 * 3600 = 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, the number of meters it travels in 1 second is 1000 / 9000 = 1/9 meters per second. Then, in 90 seconds, it travels 1/9 * 90 = 10 meters.

Same result again.

So, all methods lead to 10 meters. Maybe the hallway is indeed 10 meters long.

But just to make sure, let me check if there's another way someone might approach this problem and perhaps make a mistake.

For example, someone might think that since 2.5 hours is 2 hours and 30 minutes, and then try to convert 90 seconds into hours? Let me see.

90 seconds is 90/3600 hours, which is 1/40 hours. Then, if the robot's speed is 1 km per 2.5 hours, then in 1/40 hours, it would travel (1 km / 2.5 hours) * (1/40 hours) = (1 / 2.5) * (1 / 40) km.

Calculating that: 1 divided by 2.5 is 0.4, then 0.4 divided by 40 is 0.01 km. Then, converting 0.01 km to meters is 0.01 * 1000 = 10 meters.

Same answer again.

So regardless of the method, the answer is 10 meters.

Therefore, I think the hallway is 10 meters long.

But just to make sure, let me check if I messed up any unit conversions.

Original data: 2.5 hours for 1 km.

Convert 2.5 hours to seconds: 2.5 * 60 * 60.

Yes, 2.5 * 60 is 150 minutes, 150 * 60 is 9000 seconds.

So 1 km is 1000 meters.

So speed is 1000 meters / 9000 seconds = 1/9 m/s.

Then, time for hallway is 90 seconds.

Distance = speed * time = (1/9) * 90 = 10 meters.

Yes.

Alternatively, if I wanted to use km and hours, but then convert the time to hours.

So, 90 seconds is 90 / 3600 hours = 0.025 hours.

Then, since the robot's speed is 1 km / 2.5 hours, then distance = speed * time = (1 / 2.5) * 0.025 km.

Calculating that: 1 divided by 2.5 is 0.4, multiplied by 0.025 is 0.01 km.

Convert to meters: 0.01 * 1000 = 10 meters.

Same answer.

So, all methods confirm that the hallway is 10 meters long.

Therefore, I think the answer is 10 meters.

**Final Answer**
The hallway is \boxed{10} meters long.

To determine the length of the hallway, we begin by analyzing the robot's constant speed based on the given information:

---

### **Step 1: Determine the Robot's Speed**

The robot travels **1 kilometer (1000 meters)** in **2.5 hours**. To find its speed in **meters per second**, we first convert the time from hours to seconds:

$$
2.5 \text{ hours} = 2.5 \times 60 \times 60 = 9000 \text{ seconds}
$$

Now, calculate the speed:

$$
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1000 \text{ meters}}{9000 \text{ seconds}} = \frac{1}{9} \text{ meters per second}
$$

---

### **Step 2: Use the Speed to Find the Hallway Length**

We are told that the robot takes **90 seconds** to travel the length of the hallway. Using the formula:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

Substitute the known values:

$$
\text{Distance} = \frac{1}{9} \text{ m/s} \times 90 \text{ s} = 10 \text{ meters}
$$

---

### **Conclusion**

After verifying through multiple approaches (unit conversions, proportionality, and direct calculation), we consistently find that the hallway is:

$$
\boxed{10}
$$

meters long.

Solution: We see that 2.5 hours is the same as $2.5\cdot 60 = 150$ minutes, or $150\cdot 60 = 9000$ seconds.  This is 100 times longer than the robot was traveling the hallway, meaning the hallway is $\frac{1}{100}$ kilometers, or $\frac{1000}{100} = \boxed{10}$ meters long.