[User] Evaluate the integral $$\int_{0}^{\frac{1}{2}} \frac{dx}{x\ln^2(x)}.$$

[Assistant]
To evaluate the integral \(\int_{0}^{\frac{1}{2}} \frac{dx}{x \ln^2(x)}\), we use the substitution \( u = \ln x \). This substitution simplifies the integrand significantly.

1. **Substitution**:
   \[
   u = \ln x \implies du = \frac{1}{x} dx
   \]
   When \( x = \frac{1}{2} \), \( u = \ln \left(\frac{1}{2}\right) = -\ln 2 \).
   When \( x \to 0^+ \), \( u \to -\infty \).

   The integral transforms as follows:
   \[
   \int_{0}^{\frac{1}{2}} \frac{dx}{x \ln^2(x)} = \int_{-\infty}^{-\ln 2} \frac{1}{u^2} \, du
   \]

2. **Evaluate the Integral**:
   The antiderivative of \( \frac{1}{u^2} \) is \( -\frac{1}{u} \). Therefore,
   \[
   \int \frac{1}{u^2} \, du = -\frac{1}{u}
   \]

3. **Apply the Limits**:
   Evaluate the antiderivative from \( -\infty \) to \( -\ln 2 \):
   \[
   \left[ -\frac{1}{u} \right]_{-\infty}^{-\ln 2} = -\frac{1}{-\ln 2} - \left( -\frac{1}{-\infty} \right)
   \]
   Simplifying this, we get:
   \[
   \frac{1}{\ln 2} - 0 = \frac{1}{\ln 2}
   \]

Thus, the value of the integral is:
\[
\boxed{\frac{1}{\ln 2}}
\]

System: You are a helpful assistant.


[User] What is the minimum number of questions you need to ask, where each question can be answered with Yes, No, or Maybe, to determine a randomly selected card from a standard 52-card deck?

[Assistant]
To determine the minimum number of questions needed to identify a randomly selected card from a standard 52-card deck, where each question can be answered with Yes, No, or Maybe, we can use information theory.

1. **Understanding the Problem:**
   - A standard deck has 52 cards.
   - Each question can have 3 possible answers: Yes, No, or Maybe.
   - We need to find the minimum number of questions \( n \) such that \( 3^n \geq 52 \).

2. **Using Logarithms:**
   - We need to solve the inequality \( 3^n \geq 52 \).
   - Taking the logarithm base 3 of both sides, we get:
     \[
     n \geq \log_3(52)
     \]
   - Using the change of base formula, \(\log_3(52) = \frac{\ln(52)}{\ln(3)}\).

3. **Calculating the Logarithm:**
   - Calculate \(\ln(52)\) and \(\ln(3)\):
     \[
     \ln(52) \approx 3.95124371858
     \]
     \[
     \ln(3) \approx 1.09861228867
     \]
   - Therefore:
     \[
     \log_3(52) \approx \frac{3.95124371858}{1.09861228867} \approx 3.595
     \]

4. **Determining the Minimum Number of Questions:**
   - Since \( n \) must be an integer, we take the ceiling of \( \log_3(52) \):
     \[
     n = \lceil 3.595 \rceil = 4
     \]

5. **Verification:**
   - With 3 questions, the number of possible outcomes is \( 3^3 = 27 \), which is insufficient.
   - With 4 questions, the number of possible outcomes is \( 3^4 = 81 \), which is sufficient.

Therefore, the minimum number of questions needed to identify a randomly selected card from a standard 52-card deck is \(\boxed{4}\).

System: You are a helpful assistant.


[User] Find the line through the point (3,1,-2) that intersects and is perpendicular to the line given by the parametric equations \( x = -1 + t \), \( y = -2 + t \), \( z = -1 + t \).

[Assistant]
To find the line through the point \((3,1,-2)\) that intersects and is perpendicular to the line given by the parametric equations \( x = -1 + t \), \( y = -2 + t \), \( z = -1 + t \), we follow these steps:

1. **Identify the direction vector of the given line:**
   The parametric equations \( x = -1 + t \), \( y = -2 + t \), \( z = -1 + t \) indicate that the direction vector of the given line is \(\langle 1, 1, 1 \rangle\).

2. **Determine the point of intersection \(P\):**
   Let \(P\) be the point of intersection of the two lines. Since \(P\) lies on the given line, its coordinates can be written as:
   \[
   P = (-1 + s, -2 + s, -1 + s)
   \]
   for some parameter \(s\).

3. **Find the direction vector of the desired line:**
   The desired line passes through the point \((3,1,-2)\) and \(P\). The direction vector of this line is:
   \[
   \langle -1 + s - 3, -2 + s - 1, -1 + s + 2 \rangle = \langle s - 4, s - 3, s + 1 \rangle
   \]

4. **Ensure the lines are perpendicular:**
   The direction vectors of the two lines must have a dot product of zero. The dot product of \(\langle 1, 1, 1 \rangle\) and \(\langle s - 4, s - 3, s + 1 \rangle\) is:
   \[
   1 \cdot (s - 4) + 1 \cdot (s - 3) + 1 \cdot (s + 1) = s - 4 + s - 3 + s + 1 = 3s - 6
   \]
   Setting this equal to zero:
   \[
   3s - 6 = 0 \implies s = 2
   \]

5. **Determine the coordinates of \(P\):**
   Substituting \(s = 2\) into the parametric equations of the given line:
   \[
   P = (-1 + 2, -2 + 2, -1 + 2) = (1, 0, 1)
   \]

6. **Find the direction vector of the desired line:**
   The direction vector from \((3,1,-2)\) to \(P = (1,0,1)\) is:
   \[
   \langle 1 - 3, 0 - 1, 1 + 2 \rangle = \langle -2, -1, 3 \rangle
   \]

7. **Write the parametric equations of the desired line:**
   Using the point \((3,1,-2)\) and the direction vector \(\langle -2, -1, 3 \rangle\), the parametric equations of the desired line are:
   \[
   \begin{cases}
   x = 3 - 2k \\
   y = 1 - k \\
   z = -2 + 3k
   \end{cases}
   \]
   where \(k\) is a parameter.

Thus, the final answer is:
\[
\boxed{
\begin{cases}
x = 3 - 2k \\
y = 1 - k \\
z = -2 + 3k
\end{cases}
}
\]

System: You are a helpful assistant.


[User]
A city has a system of bus routes with the following properties:
- Each route has exactly 11 bus stops.
- It is possible to travel between any two bus stops without changing routes.
- Any two bus routes have exactly one bus stop in common.
What is the number of bus routes in the city?

[Assistant]
To solve the problem of determining the number of bus routes in the city, we need to analyze the given conditions and apply the appropriate combinatorial design theory. The conditions are:

1. Each route has exactly 11 bus stops.
2. It is possible to travel between any two bus stops without changing routes.
3. Any two bus routes have exactly one bus stop in common.

These conditions suggest a symmetric Balanced Incomplete Block Design (BIBD) with parameters \( (v, k, \lambda) = (v, 11, 1) \), where:
- \( v \) is the number of bus stops.
- \( k \) is the number of bus stops per route.
- \( \lambda \) is the number of routes that any two bus stops share.

In a symmetric BIBD, the number of blocks (routes) \( b \) equals the number of elements (stops) \( v \), and each element appears in \( r \) blocks, with each block containing \( k \) elements. The relations in a symmetric BIBD are given by:
1. \( \lambda(v - 1) = r(k - 1) \)
2. \( bk = vr \)

Given \( k = 11 \) and \( \lambda = 1 \):
- From the relation \( \lambda(v - 1) = r(k - 1) \), substituting the values gives:
  \[
  1(v - 1) = r(11 - 1) \implies v - 1 = 10r
  \]
- Since the design is symmetric, \( b = v \) and \( r = k \), leading to:
  \[
  r = 11
  \]
- Substituting \( r = 11 \) into the equation \( v - 1 = 10r \) gives:
  \[
  v - 1 = 10 \times 11 \implies v - 1 = 110 \implies v = 111
  \]

Thus, the number of bus routes (which equals the number of stops \( v \)) is \(\boxed{111}\).

System: You are a helpful assistant.


[User] A department store offers two promotions. Promotion A says, "Buy one pair of shoes, get the second pair for half the price." Promotion B says, "Buy one pair of shoes, get $\$10$ off the second pair." Jane wants to buy two pairs of shoes that cost $\$30$ each. She can only use one of the promotions, A or B. Jane decides to use the promotion that will save her the most money. How many dollars does Jane save by picking one promotion over the other? (For example, if Jane spends $\$150$ on her purchase by using one promotion and $\$100$ on her purchase by using the other promotion, she saves $150-100=50$ dollars by using the second promotion over the first.)

[Assistant]
To determine how much Jane saves by choosing the better promotion, we need to calculate the total cost under each promotion and then find the difference between the two.

First, let's calculate the total cost under Promotion A:
- Promotion A: "Buy one pair of shoes, get the second pair for half the price."
- The first pair costs \$30.
- The second pair costs half of \$30, which is \$15.
- Therefore, the total cost under Promotion A is:
  \[
  30 + 15 = 45 \text{ dollars}
  \]

Next, let's calculate the total cost under Promotion B:
- Promotion B: "Buy one pair of shoes, get \$10 off the second pair."
- The first pair costs \$30.
- The second pair costs \$30 minus \$10, which is \$20.
- Therefore, the total cost under Promotion B is:
  \[
  30 + 20 = 50 \text{ dollars}
  \]

Now, we compare the total costs under both promotions:
- Cost under Promotion A: \$45
- Cost under Promotion B: \$50

The difference in cost between the two promotions is:
\[
50 - 45 = 5 \text{ dollars}
\]

Therefore, Jane saves \(\boxed{5}\) dollars by choosing Promotion A over Promotion B.

System: You are a helpful assistant.


[User] At a particular school with 43 students, each student takes chemistry, biology, or both. The chemistry class is three times as large as the biology class, and 5 students are taking both classes. How many people are in the chemistry class?

[Assistant]
To solve the problem, we need to determine the number of students in the chemistry class given the following conditions:

1. There are 43 students in total.
2. Each student takes chemistry, biology, or both.
3. The chemistry class is three times as large as the biology class.
4. 5 students are taking both classes.

Let \( B \) represent the number of students in the biology class. Since the chemistry class is three times as large as the biology class, the number of students in the chemistry class is \( 3B \).

Using the principle of inclusion-exclusion, the total number of students can be expressed as:
\[
\text{Total} = (\text{Chemistry}) + (\text{Biology}) - (\text{Both})
\]

Substituting the given values:
\[
43 = 3B + B - 5
\]

Simplify the equation:
\[
43 = 4B - 5
\]

Add 5 to both sides to isolate the term with \( B \):
\[
48 = 4B
\]

Divide both sides by 4 to solve for \( B \):
\[
B = 12
\]

Now, calculate the number of students in the chemistry class:
\[
\text{Chemistry} = 3B = 3 \times 12 = 36
\]

Thus, the number of students in the chemistry class is:
\[
\boxed{36}
\]

System: You are a helpful assistant.


[User] A circle passes through vertices A and D and touches side BC of a square with side length 2. What is the radius of the circle?

[Assistant]
Given a square with side length 2, vertices \( A(0,0) \), \( B(2,0) \), \( C(2,2) \), and \( D(0,2) \), a circle passes through vertices \( A \) and \( D \) and touches side \( BC \). We need to find the radius of the circle.

1. **Assign coordinates and set up the circle equation**:
   - The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\).
   - Since the circle passes through \( A(0,0) \) and \( D(0,2) \), we substitute these points into the circle's equation:
     \[
     (0 - h)^2 + (0 - k)^2 = r^2 \implies h^2 + k^2 = r^2
     \]
     \[
     (0 - h)^2 + (2 - k)^2 = r^2 \implies h^2 + (2 - k)^2 = r^2
     \]

2. **Solve for the y-coordinate \( k \)**:
   - Equate the two expressions for \( r^2 \):
     \[
     h^2 + k^2 = h^2 + (2 - k)^2
     \]
   - Simplify the equation:
     \[
     k^2 = (2 - k)^2 \implies k^2 = 4 - 4k + k^2 \implies 0 = 4 - 4k \implies k = 1
     \]

3. **Determine the radius using the tangent condition**:
   - The circle is tangent to the line \( x = 2 \). The distance from the center \((h, 1)\) to this line is \( |2 - h| \), which equals the radius \( r \).
   - Using the radius from point \( A \):
     \[
     h^2 + 1 = r^2
     \]
   - Substitute \( r = |2 - h| \) into the equation:
     \[
     h^2 + 1 = (2 - h)^2 \implies h^2 + 1 = 4 - 4h + h^2 \implies 1 = 4 - 4h \implies -3 = -4h \implies h = \frac{3}{4}
     \]

4. **Calculate the radius**:
   - The radius \( r = |2 - h| = |2 - \frac{3}{4}| = \frac{5}{4} \).

Thus, the radius of the circle is \(\boxed{\frac{5}{4}}\).

System: You are a helpful assistant.


[User] In a class of thirty students, one prize was awarded for English, another for Physics, and a third for Math. In how many ways can the recipients be chosen?

[Assistant]
To solve the problem of determining the number of ways to award three distinct prizes (English, Physics, and Math) to 30 students, we need to consider that the order in which the prizes are awarded matters. This is a permutation problem because each prize is unique.

1. **First Prize (English):** There are 30 possible students who can receive the English prize.
2. **Second Prize (Physics):** After awarding the English prize, 29 students remain. Thus, there are 29 possible students who can receive the Physics prize.
3. **Third Prize (Math):** After awarding the English and Physics prizes, 28 students remain. Thus, there are 28 possible students who can receive the Math prize.

To find the total number of ways to assign the three prizes, we multiply the number of choices for each prize:

\[
30 \times 29 \times 28
\]

Now, we perform the multiplication step by step:

\[
30 \times 29 = 870
\]

\[
870 \times 28 = 870 \times (20 + 8) = 870 \times 20 + 870 \times 8
\]

\[
870 \times 20 = 17,\!400
\]

\[
870 \times 8 = 6,\!960
\]

\[
17,\!400 + 6,\!960 = 24,\!360
\]

Therefore, the total number of ways to choose the recipients for the three prizes is:

\[
\boxed{24360}
\]

System: You are a helpful assistant.


[User] Calculate the sum of the geometric series $1+\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \dots$. Express your answer as a common fraction.

[Assistant]
To find the sum of the infinite geometric series \(1 + \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \dots\), we use the formula for the sum of an infinite geometric series:

\[
S = \frac{a_1}{1 - r}
\]

where \(a_1\) is the first term and \(r\) is the common ratio. For this series:
- The first term \(a_1 = 1\)
- The common ratio \(r = \frac{1}{3}\)

Since \(|r| = \left|\frac{1}{3}\right| < 1\), the series converges, and we can apply the formula:

\[
S = \frac{1}{1 - \frac{1}{3}}
\]

First, simplify the denominator:

\[
1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}
\]

Thus, the sum \(S\) is:

\[
S = \frac{1}{\frac{2}{3}} = 1 \cdot \frac{3}{2} = \frac{3}{2}
\]

Therefore, the sum of the infinite geometric series is:

\[
\boxed{\frac{3}{2}}
\]

System: You are a helpful assistant.


[User] Three-fourths of the parrots on Bird Island are green, and the remainder are blue. If there are 92 parrots total on Bird Island, how many of those parrots are blue?

[Assistant]
To determine the number of blue parrots on Bird Island, we start by noting that three-fourths of the parrots are green, and the remainder are blue. Given that the total number of parrots is 92, we can calculate the number of blue parrots as follows:

1. Since three-fourths of the parrots are green, the remaining one-fourth must be blue.
2. To find one-fourth of the total number of parrots, we divide 92 by 4:
   \[
   \frac{92}{4} = 23
   \]
3. Therefore, the number of blue parrots is 23.

To verify, we can check that the number of green parrots plus the number of blue parrots equals the total number of parrots:
\[
\frac{3}{4} \times 92 = 69 \quad \text{(number of green parrots)}
\]
\[
69 + 23 = 92
\]
This confirms that our solution is correct.

Thus, the number of blue parrots is \(\boxed{23}\).

System: You are a helpful assistant.


[User] What is the slope of the spiral $r = \theta$ at $\theta = \pi/4$?

[Assistant]
To find the slope of the spiral \( r = \theta \) at \( \theta = \frac{\pi}{4} \), we start by using the formula for the slope in polar coordinates:

\[
\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}
\]

Given \( r = \theta \), we have \( \frac{dr}{d\theta} = 1 \). Substituting \( r = \theta \) and \( \frac{dr}{d\theta} = 1 \) into the formula, we get:

\[
\frac{dy}{dx} = \frac{\sin \theta + \theta \cos \theta}{\cos \theta - \theta \sin \theta}
\]

We need to evaluate this expression at \( \theta = \frac{\pi}{4} \). At \( \theta = \frac{\pi}{4} \):

\[
\sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}
\]

Substituting these values into the numerator and denominator, we get:

\[
\text{Numerator} = \sin \left( \frac{\pi}{4} \right) + \frac{\pi}{4} \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} + \frac{\pi}{4} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \left( 1 + \frac{\pi}{4} \right)
\]

\[
\text{Denominator} = \cos \left( \frac{\pi}{4} \right) - \frac{\pi}{4} \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} - \frac{\pi}{4} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \left( 1 - \frac{\pi}{4} \right)
\]

Thus, the slope simplifies to:

\[
\frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} \left( 1 + \frac{\pi}{4} \right)}{\frac{\sqrt{2}}{2} \left( 1 - \frac{\pi}{4} \right)} = \frac{1 + \frac{\pi}{4}}{1 - \frac{\pi}{4}} = \frac{4 + \pi}{4 - \pi}
\]

Therefore, the slope of the spiral at \( \theta = \frac{\pi}{4} \) is:

\[
\boxed{\frac{4 + \pi}{4 - \pi}}
\]

System: You are a helpful assistant.


[User] If $2^{10} \cdot 2^{15}$ is expressed as some integer to the fifth power, what is that integer?

[Assistant]
To solve the problem of expressing \(2^{10} \cdot 2^{15}\) as an integer to the fifth power, we start by simplifying the expression using the properties of exponents. Specifically, when multiplying exponents with the same base, we add the exponents:

\[
2^{10} \cdot 2^{15} = 2^{10 + 15} = 2^{25}
\]

Next, we need to express \(2^{25}\) as an integer raised to the fifth power. We can rewrite \(2^{25}\) as:

\[
2^{25} = (2^5)^5
\]

Now, we calculate \(2^5\):

\[
2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32
\]

Therefore, we have:

\[
2^{25} = 32^5
\]

Thus, the integer that, when raised to the fifth power, equals \(2^{25}\) is:

\[
\boxed{32}
\]

System: You are a helpful assistant.


[User] If every computer is used by two people, there are three students who do not get a computer. On the other hand, if each computer is used by three people, there are three unused computers. How many students will use the computer?

[Assistant]
To determine the number of students using the computers, we start by defining the variables:
- Let \( C \) be the number of computers.
- Let \( S \) be the number of students.

We are given two scenarios:

1. **First Scenario**: If every computer is used by two people, there are three students who do not get a computer. This can be expressed as:
   \[
   S = 2C + 3
   \]

2. **Second Scenario**: If each computer is used by three people, there are three unused computers. This can be expressed as:
   \[
   S = 3(C - 3)
   \]
   Simplifying the second equation:
   \[
   S = 3C - 9
   \]

Since both equations represent the total number of students \( S \), we can set them equal to each other:
\[
2C + 3 = 3C - 9
\]

Next, we solve for \( C \):
\[
2C + 3 = 3C - 9
\]
Subtract \( 2C \) from both sides:
\[
3 = C - 9
\]
Add 9 to both sides:
\[
C = 12
\]

Now, substitute \( C = 12 \) back into one of the original equations to find \( S \). Using the first equation:
\[
S = 2C + 3
\]
\[
S = 2(12) + 3
\]
\[
S = 24 + 3
\]
\[
S = 27
\]

To verify, we check both scenarios:
- **First Scenario**: With 12 computers and 2 students per computer, \( 12 \times 2 = 24 \) students are accommodated, leaving \( 27 - 24 = 3 \) students without a computer.
- **Second Scenario**: With 27 students and 3 students per computer, \( 27 / 3 = 9 \) computers are used, leaving \( 12 - 9 = 3 \) computers unused.

Both scenarios are consistent with the given conditions.

Thus, the number of students is \(\boxed{27}\).

System: You are a helpful assistant.


[User] In how many ways can the number 259,200 be written as a product of 3 numbers, where the order of the numbers matters?

[Assistant]
To determine the number of ways to write 259,200 as a product of three ordered numbers, we start by factorizing 259,200 into its prime factors:

\[
259,200 = 2^7 \times 3^4 \times 5^2
\]

For each prime factor, we need to find the number of ways to distribute its exponent among the three numbers. This can be done using the "stars and bars" theorem, which states that the number of ways to distribute \( n \) indistinguishable items (the exponent) into \( k \) distinguishable bins (the three numbers) is given by the combination formula \( \binom{n + k - 1}{k - 1} \).

- For the prime 2 with exponent 7:
  \[
  \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36
  \]

- For the prime 3 with exponent 4:
  \[
  \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15
  \]

- For the prime 5 with exponent 2:
  \[
  \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
  \]

Multiplying these results gives the total number of ordered triples:

\[
36 \times 15 \times 6 = 36 \times 90 = 3240
\]

Thus, the number of ways to write 259,200 as a product of three ordered numbers is \(\boxed{3240}\).

System: You are a helpful assistant.


[User] Compute $\sin 270^\circ$.

[Assistant]
To compute \(\sin 270^\circ\), we use the unit circle, which is a circle with a radius of 1 centered at the origin of the coordinate system. The angle \(270^\circ\) is measured counterclockwise from the positive x-axis.

1. **Unit Circle Coordinates:**
   - The angle \(270^\circ\) corresponds to three-quarters of a full circle (which is \(360^\circ\)).
   - Starting from \(0^\circ\) at the point \((1,0)\):
     - \(90^\circ\) is at \((0,1)\),
     - \(180^\circ\) is at \((-1,0)\),
     - \(270^\circ\) is at \((0,-1)\).

2. **Sine Function on the Unit Circle:**
   - The sine of an angle in the unit circle is the y-coordinate of the corresponding point on the circle.
   - For \(270^\circ\), the coordinates are \((0, -1)\).
   - Therefore, \(\sin 270^\circ = -1\).

3. **Alternative Verification:**
   - Using the periodicity of the sine function, \(\sin(\theta) = \sin(\theta + 360^\circ k)\):
     - \(\sin(270^\circ) = \sin(270^\circ - 360^\circ) = \sin(-90^\circ)\).
   - Using the property of sine for negative angles, \(\sin(-\theta) = -\sin(\theta)\):
     - \(\sin(-90^\circ) = -\sin(90^\circ) = -1\).

Thus, the value of \(\sin 270^\circ\) is \(\boxed{-1}\).

System: You are a helpful assistant.


[User] In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?

[Assistant]
To determine the total number of people in the line, we need to use the given information that the 1013th person from the left is also the 1010th person from the right.

Let's denote the total number of people in the line by \( N \).

1. The 1013th person from the left implies there are 1012 people to their left.
2. The 1010th person from the right implies there are 1009 people to their right.

Since the person in question is counted in both the left and right counts, the total number of people in the line can be calculated as:
\[ N = \text{(people to the left)} + \text{(the person)} + \text{(people to the right)} \]
\[ N = 1012 + 1 + 1009 \]

Adding these values:
\[ N = 1012 + 1 + 1009 = 2022 \]

Alternatively, we can use the formula:
\[ N = \text{(position from the left)} + \text{(position from the right)} - 1 \]
\[ N = 1013 + 1010 - 1 = 2022 \]

Thus, the total number of people in the line is:
\[
\boxed{2022}
\]

System: You are a helpful assistant.


[User] What is the indefinite integral of $r e^{r}$ with respect to $r$?

[Assistant]
To find the indefinite integral of \( r e^{r} \) with respect to \( r \), we use integration by parts. The formula for integration by parts is given by:

\[
\int u \, dv = uv - \int v \, du
\]

We choose \( u = r \) and \( dv = e^{r} \, dr \). Then, we compute the differentials:

\[
du = dr \quad \text{and} \quad v = \int e^{r} \, dr = e^{r}
\]

Applying the integration by parts formula:

\[
\int r e^{r} \, dr = r e^{r} - \int e^{r} \, dr
\]

The remaining integral is straightforward:

\[
\int e^{r} \, dr = e^{r}
\]

Thus, we have:

\[
\int r e^{r} \, dr = r e^{r} - e^{r} + C
\]

We can factor out \( e^{r} \) from the expression:

\[
\int r e^{r} \, dr = e^{r} (r - 1) + C
\]

Therefore, the indefinite integral of \( r e^{r} \) with respect to \( r \) is:

\[
\boxed{e^{r} (r - 1) + C}
\]

System: You are a helpful assistant.


[User] Find a binomial coefficient equal to the expression $\binom{n}{k} + 3\binom{n}{k-1} + 3\binom{n}{k-2} + \binom{n}{k-3}$.

[Assistant]
To solve the problem of finding a binomial coefficient equal to the expression \(\binom{n}{k} + 3\binom{n}{k-1} + 3\binom{n}{k-2} + \binom{n}{k-3}\), we can use generating functions.

1. **Generating Functions Approach:**
   - The generating function for \(\binom{n}{k}\) is \((1 + x)^n\).
   - The polynomial \(1 + 3x + 3x^2 + x^3\) is the expansion of \((1 + x)^3\).

2. **Combining Generating Functions:**
   - The product of the generating functions \((1 + x)^n\) and \((1 + x)^3\) gives \((1 + x)^{n+3}\).
   - The coefficient of \(x^k\) in \((1 + x)^{n+3}\) is \(\binom{n+3}{k}\).

3. **Verification with Examples:**
   - For \(n = 5\) and \(k = 3\):
     \[
     \binom{5}{3} + 3\binom{5}{2} + 3\binom{5}{1} + \binom{5}{0} = 10 + 3 \cdot 10 + 3 \cdot 5 + 1 = 56
     \]
     \[
     \binom{8}{3} = 56
     \]
   - For \(n = 2\) and \(k = 2\):
     \[
     \binom{2}{2} + 3\binom{2}{1} + 3\binom{2}{0} + \binom{2}{-1} = 1 + 3 \cdot 2 + 3 \cdot 1 + 0 = 10
     \]
     \[
     \binom{5}{2} = 10
     \]
   - For \(n = 3\) and \(k = 1\):
     \[
     \binom{3}{1} + 3\binom{3}{0} + 3\binom{3}{-1} + \binom{3}{-2} = 3 + 3 \cdot 1 + 0 + 0 = 6
     \]
     \[
     \binom{6}{1} = 6
     \]

These examples confirm that the expression \(\binom{n}{k} + 3\binom{n}{k-1} + 3\binom{n}{k-2} + \binom{n}{k-3}\) simplifies to \(\binom{n+3}{k}\).

Therefore, the binomial coefficient is \(\boxed{\binom{n+3}{k}}\).

System: You are a helpful assistant.


[User] What is the least common multiple of 12, 18 and 30?

[Assistant]
To find the least common multiple (LCM) of 12, 18, and 30, we start by determining the prime factorizations of each number:

- The prime factorization of 12 is \(2^2 \times 3^1\).
- The prime factorization of 18 is \(2^1 \times 3^2\).
- The prime factorization of 30 is \(2^1 \times 3^1 \times 5^1\).

Next, we identify the highest power of each prime number that appears in these factorizations:

- The highest power of 2 is \(2^2\) (from 12).
- The highest power of 3 is \(3^2\) (from 18).
- The highest power of 5 is \(5^1\) (from 30).

The LCM is obtained by multiplying these highest powers together:

\[
LCM = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 36 \times 5 = 180
\]

To verify, we check that 180 is divisible by 12, 18, and 30:

- \(180 \div 12 = 15\)
- \(180 \div 18 = 10\)
- \(180 \div 30 = 6\)

Since 180 is divisible by all three numbers, we confirm that the LCM of 12, 18, and 30 is indeed \(\boxed{180}\).

System: You are a helpful assistant.


[User] Consider a rectangle with length \( x \) and perimeter 36 feet. Express the area \( a \) as a function of \( x \) and determine the domain of the function.

[Assistant]
To express the area \( a \) of a rectangle as a function of its length \( x \) given that the perimeter is 36 feet, we start with the formula for the perimeter of a rectangle:

\[ P = 2 \times (\text{length} + \text{width}) \]

Given the perimeter \( P = 36 \) feet and the length \( x \), let the width be \( w \). Thus, we have:

\[ 36 = 2(x + w) \]

Dividing both sides by 2, we get:

\[ 18 = x + w \]

Solving for \( w \):

\[ w = 18 - x \]

The area \( a \) of the rectangle is given by the product of its length and width:

\[ a = x \times w \]

Substituting \( w = 18 - x \) into the area formula:

\[ a = x(18 - x) \]

Simplifying this expression:

\[ a = 18x - x^2 \]

Thus, the area \( a \) as a function of \( x \) is:

\[ a(x) = 18x - x^2 \]

Next, we determine the domain of this function. Since \( x \) represents the length and \( w \) represents the width of the rectangle, both must be positive:

\[ x > 0 \]
\[ w = 18 - x > 0 \]

Solving the second inequality:

\[ 18 - x > 0 \]
\[ x < 18 \]

Combining these two inequalities, we get:

\[ 0 < x < 18 \]

Therefore, the domain of the function \( a(x) = 18x - x^2 \) is:

\[ 0 < x < 18 \]

In interval notation, the domain is:

\[ (0, 18) \]

Thus, the final answer is:

\[ \boxed{a(x) = 18x - x^2 \text{ with domain } (0, 18)} \]

System: You are a helpful assistant.


[User] Carbon-10 is a radioactive isotope with a half-life of 19.255 seconds. Suppose you start with 1 gram of carbon-10. Determine the function \( f(t) \) that represents the amount of carbon-10 left after \( t \) seconds. Use a formula of the form \( f(t) = A \cdot 2^{-t/T} \) with suitable values for \( A \) and \( T \).

[Assistant]
To determine the function \( f(t) \) that represents the amount of Carbon-10 left after \( t \) seconds, we use the exponential decay formula given by:

\[
f(t) = A \cdot 2^{-t/T}
\]

where:
- \( A \) is the initial amount of Carbon-10,
- \( T \) is the half-life of Carbon-10.

Given:
- The initial amount \( A = 1 \) gram,
- The half-life \( T = 19.255 \) seconds.

Substituting these values into the formula, we get:

\[
f(t) = 1 \cdot 2^{-t/19.255}
\]

Since the initial amount \( A = 1 \) can be omitted for simplicity, the function becomes:

\[
f(t) = 2^{-t/19.255}
\]

Thus, the function representing the amount of Carbon-10 remaining after \( t \) seconds is:

\[
\boxed{f(t) = 2^{-t/19.255}}
\]

System: You are a helpful assistant.


[User] The three roots of the equation \[ax^3+bx^2+cx+d=0\]are $1,$ $2,$ and $3.$ Compute $\frac{c}{d}.$

[Assistant]
To find \(\frac{c}{d}\) for the cubic equation \(ax^3 + bx^2 + cx + d = 0\) with roots 1, 2, and 3, we use Vieta's formulas.

According to Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\) with roots \(r, s, t\):
1. The sum of the roots: \(r + s + t = -\frac{b}{a}\)
2. The sum of the products of the roots taken two at a time: \(rs + rt + st = \frac{c}{a}\)
3. The product of the roots: \(rst = -\frac{d}{a}\)

Given the roots \(1, 2, 3\):
- The sum of the products of the roots taken two at a time is:
  \[
  1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3 = 2 + 3 + 6 = 11
  \]
  Therefore, \(\frac{c}{a} = 11\), so \(c = 11a\).

- The product of the roots is:
  \[
  1 \cdot 2 \cdot 3 = 6
  \]
  Therefore, \(-\frac{d}{a} = 6\), so \(d = -6a\).

Now, we need to find \(\frac{c}{d}\):
\[
\frac{c}{d} = \frac{11a}{-6a} = -\frac{11}{6}
\]

Thus, the value of \(\frac{c}{d}\) is:
\[
\boxed{-\frac{11}{6}}
\]

System: You are a helpful assistant.


[User] George is about to get a certain amount of change less than one dollar from the cash register. If he gets the most quarters possible and the rest in pennies, he would need to receive 3 pennies to meet the amount. If he gets the most dimes possible and the rest in pennies, he would need to receive 8 pennies to meet the amount. What is the sum, in cents, of the possible amounts of change that he is trying to get?

[Assistant]
To solve the problem, we need to determine the possible amounts of change that George could receive, given the constraints involving quarters and dimes.

1. **Determine the possible amounts using quarters:**
   - Let \( q \) be the number of quarters.
   - The amount can be expressed as \( 25q + 3 \).
   - Since the amount must be less than 100 cents, we have:
     \[
     25q + 3 < 100
     \]
   - Solving for \( q \):
     \[
     25q < 97 \implies q < \frac{97}{25} \implies q < 3.88
     \]
   - Therefore, \( q \) can be 0, 1, 2, or 3.
   - The possible amounts are:
     \[
     25 \cdot 0 + 3 = 3 \quad (\text{not valid since it does not satisfy dime condition})
     \]
     \[
     25 \cdot 1 + 3 = 28
     \]
     \[
     25 \cdot 2 + 3 = 53 \quad (\text{not valid since it does not satisfy dime condition})
     \]
     \[
     25 \cdot 3 + 3 = 78
     \]
   - Valid amounts from quarters are: \( 28 \) and \( 78 \).

2. **Determine the possible amounts using dimes:**
   - Let \( d \) be the number of dimes.
   - The amount can be expressed as \( 10d + 8 \).
   - Since the amount must be less than 100 cents, we have:
     \[
     10d + 8 < 100
     \]
   - Solving for \( d \):
     \[
     10d < 92 \implies d < \frac{92}{10} \implies d < 9.2
     \]
   - Therefore, \( d \) can be 0, 1, 2, ..., 9.
   - The possible amounts are:
     \[
     10 \cdot 0 + 8 = 8 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 1 + 8 = 18 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 2 + 8 = 28
     \]
     \[
     10 \cdot 3 + 8 = 38 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 4 + 8 = 48 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 5 + 8 = 58 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 6 + 8 = 68 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 7 + 8 = 78
     \]
     \[
     10 \cdot 8 + 8 = 88 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
     \[
     10 \cdot 9 + 8 = 98 \quad (\text{not valid since it does not satisfy quarter condition})
     \]
   - Valid amounts from dimes are: \( 28 \) and \( 78 \).

3. **Find the common valid amounts:**
   - The common valid amounts from both conditions are \( 28 \) and \( 78 \).

4. **Sum the valid amounts:**
   \[
   28 + 78 = 106
   \]

Thus, the sum of the possible amounts of change George is trying to get is:
\[
\boxed{106}
\]

System: You are a helpful assistant.


[User] The size of the increase from each term to the next in the list $a, 225\frac{1}{2}, c, d, 284$ is always the same. What is the value of $a$?

[Assistant]
To solve for the value of \( a \) in the arithmetic sequence \( a, 225.5, c, d, 284 \), we need to determine the common difference \( k \) and then use it to find \( a \).

1. Identify the common difference \( k \):
   - The sequence is arithmetic, so the difference between consecutive terms is constant.
   - The second term is \( 225.5 \) and the fifth term is \( 284 \).
   - The difference between the fifth term and the second term is \( 3k \) because there are three intervals between the second and fifth terms.
   - Set up the equation:
     \[
     225.5 + 3k = 284
     \]
   - Solve for \( k \):
     \[
     3k = 284 - 225.5 = 58.5 \\
     k = \frac{58.5}{3} = 19.5
     \]

2. Determine the first term \( a \):
   - The second term is \( a + k \), and it is given as \( 225.5 \).
   - Substitute \( k = 19.5 \) into the equation:
     \[
     a + 19.5 = 225.5
     \]
   - Solve for \( a \):
     \[
     a = 225.5 - 19.5 = 206
     \]

3. Verify the sequence:
   - The first term is \( a = 206 \).
   - The second term is \( 206 + 19.5 = 225.5 \).
   - The third term is \( 225.5 + 19.5 = 245 \).
   - The fourth term is \( 245 + 19.5 = 264.5 \).
   - The fifth term is \( 264.5 + 19.5 = 284 \).

The value of \( a \) is \(\boxed{206}\).

System: You are a helpful assistant.


[User] A board game spinner is divided into three parts labeled $A$, $B$  and $C$. The probability of the spinner landing on $A$ is $\frac{1}{3}$ and the probability of the spinner landing on $B$ is $\frac{5}{12}$.  What is the probability of the spinner landing on $C$? Express your answer as a common fraction.

[Assistant]
To find the probability of the spinner landing on \( C \), we start by noting that the total probability of the spinner landing on any section must sum to 1. Given the probabilities for landing on \( A \) and \( B \):

\[
P(A) = \frac{1}{3} \quad \text{and} \quad P(B) = \frac{5}{12}
\]

First, we convert \( P(A) \) to a fraction with a denominator of 12:

\[
P(A) = \frac{1}{3} = \frac{4}{12}
\]

Next, we add the probabilities of landing on \( A \) and \( B \):

\[
P(A) + P(B) = \frac{4}{12} + \frac{5}{12} = \frac{9}{12}
\]

The probability of landing on \( C \) is the remaining probability after accounting for \( A \) and \( B \):

\[
P(C) = 1 - \left( P(A) + P(B) \right) = 1 - \frac{9}{12}
\]

Since \( 1 = \frac{12}{12} \), we have:

\[
P(C) = \frac{12}{12} - \frac{9}{12} = \frac{3}{12}
\]

Simplifying the fraction:

\[
P(C) = \frac{3}{12} = \frac{1}{4}
\]

Thus, the probability of the spinner landing on \( C \) is \(\boxed{\frac{1}{4}}\).

System: You are a helpful assistant.


[User] When $f(x) = ax^3 - 6x^2 + bx - 5$ is divided by $x - 1,$ the remainder is $-5.$  When $f(x)$ is divided by $x + 2,$ the remainder is $-53.$  Find the ordered pair $(a,b).$

[Assistant]
To solve the problem, we use the Remainder Theorem which states that the remainder of the division of a polynomial \( f(x) \) by \( x - c \) is \( f(c) \).

Given the polynomial \( f(x) = ax^3 - 6x^2 + bx - 5 \), we know:
1. When \( f(x) \) is divided by \( x - 1 \), the remainder is \(-5\).
2. When \( f(x) \) is divided by \( x + 2 \), the remainder is \(-53\).

Using the Remainder Theorem, we can set up the following equations:
1. \( f(1) = -5 \)
2. \( f(-2) = -53 \)

First, we substitute \( x = 1 \) into the polynomial:
\[
f(1) = a(1)^3 - 6(1)^2 + b(1) - 5 = a - 6 + b - 5 = a + b - 11
\]
Since \( f(1) = -5 \), we have:
\[
a + b - 11 = -5 \implies a + b = 6 \quad \text{(Equation 1)}
\]

Next, we substitute \( x = -2 \) into the polynomial:
\[
f(-2) = a(-2)^3 - 6(-2)^2 + b(-2) - 5 = -8a - 24 - 2b - 5 = -8a - 2b - 29
\]
Since \( f(-2) = -53 \), we have:
\[
-8a - 2b - 29 = -53 \implies -8a - 2b = -24 \implies 4a + b = 12 \quad \text{(Equation 2)}
\]

Now we solve the system of linear equations:
1. \( a + b = 6 \)
2. \( 4a + b = 12 \)

Subtract Equation 1 from Equation 2:
\[
(4a + b) - (a + b) = 12 - 6 \implies 3a = 6 \implies a = 2
\]

Substitute \( a = 2 \) back into Equation 1:
\[
2 + b = 6 \implies b = 4
\]

Thus, the ordered pair \((a, b)\) is \(\boxed{(2, 4)}\).

To verify, we check the remainders:
1. For \( x = 1 \):
\[
f(1) = 2(1)^3 - 6(1)^2 + 4(1) - 5 = 2 - 6 + 4 - 5 = -5
\]
2. For \( x = -2 \):
\[
f(-2) = 2(-2)^3 - 6(-2)^2 + 4(-2) - 5 = -16 - 24 - 8 - 5 = -53
\]

Both conditions are satisfied, confirming the solution \(\boxed{(2, 4)}\).

System: You are a helpful assistant.


[User] A square has an area of $25x^2 + 20x + 4$. Find the restrictions on $x$ such that the side length of the square is positive.

[Assistant]
To find the restrictions on \( x \) such that the side length of the square is positive, given that the area of the square is \( 25x^2 + 20x + 4 \), we can proceed as follows:

1. **Identify the Side Length:**
   The area of a square is given by the square of its side length. Therefore, if the area is \( 25x^2 + 20x + 4 \), we need to determine if this expression can be written as a perfect square.

2. **Check for Perfect Square:**
   We observe that \( 25x^2 + 20x + 4 \) can be factored as a perfect square trinomial:
   \[
   25x^2 + 20x + 4 = (5x + 2)^2
   \]
   This means the side length of the square is \( 5x + 2 \).

3. **Ensure Positive Side Length:**
   For the side length to be positive, we need:
   \[
   5x + 2 > 0
   \]

4. **Solve the Inequality:**
   Solving the inequality \( 5x + 2 > 0 \):
   \[
   5x + 2 > 0 \implies 5x > -2 \implies x > -\frac{2}{5}
   \]

5. **Conclusion:**
   The restriction on \( x \) such that the side length of the square is positive is:
   \[
   \boxed{x > -\frac{2}{5}}
   \]

System: You are a helpful assistant.


[User] What is the remainder of $8^6 + 7^7 + 6^8$ divided by 5?

[Assistant]
To find the remainder when \(8^6 + 7^7 + 6^8\) is divided by 5, we will use modular arithmetic properties to simplify each term individually.

1. Calculate \(8^6 \mod 5\):
   \[
   8 \equiv 3 \mod 5
   \]
   Therefore,
   \[
   8^6 \equiv 3^6 \mod 5
   \]
   We can simplify \(3^6\) by noting that \(3^4 \equiv 1 \mod 5\):
   \[
   3^6 = 3^{4+2} = (3^4) \cdot (3^2) \equiv 1 \cdot 9 \equiv 9 \equiv 4 \mod 5
   \]
   So,
   \[
   8^6 \equiv 4 \mod 5
   \]

2. Calculate \(7^7 \mod 5\):
   \[
   7 \equiv 2 \mod 5
   \]
   Therefore,
   \[
   7^7 \equiv 2^7 \mod 5
   \]
   We can simplify \(2^7\) by noting that \(2^4 \equiv 1 \mod 5\):
   \[
   2^7 = 2^{4+3} = (2^4) \cdot (2^3) \equiv 1 \cdot 8 \equiv 8 \equiv 3 \mod 5
   \]
   So,
   \[
   7^7 \equiv 3 \mod 5
   \]

3. Calculate \(6^8 \mod 5\):
   \[
   6 \equiv 1 \mod 5
   \]
   Therefore,
   \[
   6^8 \equiv 1^8 \equiv 1 \mod 5
   \]

Now, we sum the remainders:
\[
8^6 + 7^7 + 6^8 \equiv 4 + 3 + 1 \equiv 8 \mod 5
\]
Finally, we find the remainder when 8 is divided by 5:
\[
8 \equiv 3 \mod 5
\]

Thus, the remainder when \(8^6 + 7^7 + 6^8\) is divided by 5 is \(\boxed{3}\).

System: You are a helpful assistant.


[User] Evaluate the integral $\int_{2}^{3} \frac{x^{2}-2}{(x-1)^{3}}\  dx$.

[Assistant]
To evaluate the integral \(\int_{2}^{3} \frac{x^2 - 2}{(x-1)^3} \, dx\), we will use the substitution \( t = x - 1 \). This substitution gives \( dx = dt \), and the limits of integration change as follows:
- When \( x = 2 \), \( t = 1 \)
- When \( x = 3 \), \( t = 2 \)

Rewriting the integral in terms of \( t \):

\[
\int_{2}^{3} \frac{x^2 - 2}{(x-1)^3} \, dx = \int_{1}^{2} \frac{(t+1)^2 - 2}{t^3} \, dt
\]

Expanding the numerator:

\[
(t+1)^2 - 2 = t^2 + 2t + 1 - 2 = t^2 + 2t - 1
\]

Thus, the integral becomes:

\[
\int_{1}^{2} \frac{t^2 + 2t - 1}{t^3} \, dt = \int_{1}^{2} \left( \frac{t^2}{t^3} + \frac{2t}{t^3} - \frac{1}{t^3} \right) \, dt = \int_{1}^{2} \left( \frac{1}{t} + \frac{2}{t^2} - \frac{1}{t^3} \right) \, dt
\]

We can now integrate each term separately:

\[
\int_{1}^{2} \frac{1}{t} \, dt = \left[ \ln |t| \right]_{1}^{2} = \ln 2 - \ln 1 = \ln 2
\]

\[
\int_{1}^{2} \frac{2}{t^2} \, dt = 2 \int_{1}^{2} t^{-2} \, dt = 2 \left[ -t^{-1} \right]_{1}^{2} = 2 \left( -\frac{1}{2} + 1 \right) = 2 \left( \frac{1}{2} \right) = 1
\]

\[
\int_{1}^{2} \frac{1}{t^3} \, dt = \int_{1}^{2} t^{-3} \, dt = \left[ -\frac{1}{2t^2} \right]_{1}^{2} = -\frac{1}{2} \left( \frac{1}{4} - 1 \right) = -\frac{1}{2} \left( -\frac{3}{4} \right) = \frac{3}{8}
\]

Combining these results, we get:

\[
\int_{1}^{2} \left( \frac{1}{t} + \frac{2}{t^2} - \frac{1}{t^3} \right) \, dt = \left[ \ln |t| - \frac{2}{t} + \frac{1}{2t^2} \right]_{1}^{2}
\]

Evaluating at the limits:

\[
\left[ \ln 2 - \frac{2}{2} + \frac{1}{2 \cdot 2^2} \right] - \left[ \ln 1 - \frac{2}{1} + \frac{1}{2 \cdot 1^2} \right] = \left( \ln 2 - 1 + \frac{1}{8} \right) - \left( 0 - 2 + \frac{1}{2} \right)
\]

Simplifying the constants:

\[
\left( \ln 2 - 1 + \frac{1}{8} \right) - \left( -2 + \frac{1}{2} \right) = \ln 2 - 1 + \frac{1}{8} + 2 - \frac{1}{2}
\]

Combining the constants:

\[
-1 + 2 - \frac{1}{2} + \frac{1}{8} = 1 - \frac{4}{8} + \frac{1}{8} = 1 - \frac{3}{8} = \frac{5}{8}
\]

Thus, the value of the integral is:

\[
\boxed{\ln 2 + \frac{5}{8}}
\]

System: You are a helpful assistant.


[User] Evaluate the integral \(\int \frac{x-1}{\sqrt{xe^x}} \, dx\).

[Assistant]
To evaluate the integral \(\int \frac{x-1}{\sqrt{xe^x}} \, dx\), we start by simplifying the integrand. The expression inside the integral can be rewritten as follows:

\[
\frac{x-1}{\sqrt{xe^x}} = \frac{x-1}{\sqrt{x} \cdot \sqrt{e^x}} = \frac{x-1}{x^{1/2} e^{x/2}} = \frac{x}{x^{1/2} e^{x/2}} - \frac{1}{x^{1/2} e^{x/2}} = x^{1/2} e^{-x/2} - x^{-1/2} e^{-x/2}.
\]

Thus, the integral becomes:

\[
\int \left( x^{1/2} e^{-x/2} - x^{-1/2} e^{-x/2} \right) \, dx = \int x^{1/2} e^{-x/2} \, dx - \int x^{-1/2} e^{-x/2} \, dx.
\]

We will evaluate each integral separately. First, consider the integral \(\int x^{1/2} e^{-x/2} \, dx\). We use integration by parts with \(u = x^{1/2}\) and \(dv = e^{-x/2} \, dx\). Then, \(du = \frac{1}{2} x^{-1/2} \, dx\) and \(v = -2 e^{-x/2}\). Applying integration by parts, we get:

\[
\int x^{1/2} e^{-x/2} \, dx = -2 x^{1/2} e^{-x/2} - \int -2 e^{-x/2} \cdot \frac{1}{2} x^{-1/2} \, dx = -2 x^{1/2} e^{-x/2} + \int x^{-1/2} e^{-x/2} \, dx.
\]

Now, we notice that the remaining integral \(\int x^{-1/2} e^{-x/2} \, dx\) is the same as the second integral in our original split. Therefore, we can write:

\[
\int x^{1/2} e^{-x/2} \, dx - \int x^{-1/2} e^{-x/2} \, dx = \left( -2 x^{1/2} e^{-x/2} + \int x^{-1/2} e^{-x/2} \, dx \right) - \int x^{-1/2} e^{-x/2} \, dx = -2 x^{1/2} e^{-x/2}.
\]

Thus, the integral evaluates to:

\[
\int \frac{x-1}{\sqrt{xe^x}} \, dx = -2 x^{1/2} e^{-x/2} + C = -2 \sqrt{\frac{x}{e^x}} + C.
\]

Therefore, the final answer is:

\[
\boxed{-2 \sqrt{\frac{x}{e^x}} + C}.
\]

System: You are a helpful assistant.


[User] Suppose that $x^4$ is a prime number. Find all $a, b, c, d \in \mathbb{Q}$ such that $ax^3 + bx^2 + cx + d = 0$.

[Assistant]
Given that \( x^4 \) is a prime number \( p \), we have \( x^4 = p \). This implies that \( x \) is a root of the polynomial \( t^4 - p \). To determine the minimal polynomial of \( x \) over \( \mathbb{Q} \), we apply Eisenstein's criterion to the polynomial \( t^4 - p \).

Eisenstein's criterion states that a polynomial \( f(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_1t + a_0 \) is irreducible over \( \mathbb{Q} \) if there exists a prime \( p \) such that:
1. \( p \) divides each \( a_i \) for \( i < n \),
2. \( p \) does not divide the leading coefficient (which is 1 here),
3. \( p^2 \) does not divide the constant term \( a_0 \).

For the polynomial \( t^4 - p \), we have:
- The leading coefficient is 1, which is not divisible by \( p \).
- The constant term is \(-p\), which is divisible by \( p \) but not by \( p^2 \).
- The coefficients of \( t^3, t^2, t \) are all 0, which are trivially divisible by \( p \).

Thus, Eisenstein's criterion applies, and \( t^4 - p \) is irreducible over \( \mathbb{Q} \). Therefore, the minimal polynomial of \( x \) over \( \mathbb{Q} \) is \( t^4 - p \), and the degree of the extension \( \mathbb{Q}(x) \) over \( \mathbb{Q} \) is 4.

Since \( x \) is algebraic of degree 4 over \( \mathbb{Q} \), the elements \( 1, x, x^2, x^3 \) are linearly independent over \( \mathbb{Q} \). This means that any polynomial equation \( ax^3 + bx^2 + cx + d = 0 \) with rational coefficients \( a, b, c, d \) must have all coefficients equal to zero to avoid contradicting the linear independence of \( 1, x, x^2, x^3 \).

Therefore, the only solution is:
\[
\boxed{a = 0, \, b = 0, \, c = 0, \, d = 0}
\]

System: You are a helpful assistant.


[User] Given \( x = y^2 \) and \( y = x^2 \), where \( x \) and \( y \) are complex numbers, find the least possible real value of \( x^2 + x + 1 \).

[Assistant]
Given the equations \( x = y^2 \) and \( y = x^2 \), we start by substituting \( y = x^2 \) into \( x = y^2 \):

\[
x = (x^2)^2 = x^4
\]

This simplifies to the equation:

\[
x^4 - x = 0
\]

Factoring out \( x \), we get:

\[
x(x^3 - 1) = 0
\]

The solutions to this equation are:

\[
x = 0 \quad \text{or} \quad x^3 = 1
\]

The solutions to \( x^3 = 1 \) are the cube roots of unity: \( x = 1 \), \( x = \omega \), and \( x = \omega^2 \), where \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

Next, we evaluate \( x^2 + x + 1 \) for each solution:

1. If \( x = 0 \):

\[
x^2 + x + 1 = 0^2 + 0 + 1 = 1
\]

2. If \( x = 1 \):

\[
x^2 + x + 1 = 1^2 + 1 + 1 = 3
\]

3. If \( x = \omega \) or \( x = \omega^2 \):

Since \( \omega \) and \( \omega^2 \) are roots of the equation \( x^2 + x + 1 = 0 \):

\[
x^2 + x + 1 = 0
\]

The least possible real value of \( x^2 + x + 1 \) among these solutions is \( 0 \).

Thus, the least possible real value of \( x^2 + x + 1 \) is:

\[
\boxed{0}
\]

System: You are a helpful assistant.


[User] How many vertical asymptotes does the function \( f(x) = \frac{x^2 - 6x - 16}{x^3 - 5x^2 - 2x + 24} \) have?

[Assistant]
To determine the number of vertical asymptotes of the function \( f(x) = \frac{x^2 - 6x - 16}{x^3 - 5x^2 - 2x + 24} \), we need to factor both the numerator and the denominator and identify the points where the denominator is zero but the numerator is not zero.

1. **Factor the numerator:**
   \[
   x^2 - 6x - 16 = (x - 8)(x + 2)
   \]
   The roots of the numerator are \( x = 8 \) and \( x = -2 \).

2. **Factor the denominator:**
   We use the Rational Root Theorem to find the roots of the cubic polynomial \( x^3 - 5x^2 - 2x + 24 \). Testing possible rational roots, we find:
   \[
   x = 3 \quad \text{(since } 3^3 - 5 \cdot 3^2 - 2 \cdot 3 + 24 = 0 \text{)}
   \]
   Using synthetic division or polynomial division, we can factor \( x^3 - 5x^2 - 2x + 24 \) as:
   \[
   x^3 - 5x^2 - 2x + 24 = (x - 3)(x^2 - 2x - 8)
   \]
   Further factoring the quadratic:
   \[
   x^2 - 2x - 8 = (x - 4)(x + 2)
   \]
   Thus, the denominator factors to:
   \[
   x^3 - 5x^2 - 2x + 24 = (x - 3)(x - 4)(x + 2)
   \]
   The roots of the denominator are \( x = 3 \), \( x = 4 \), and \( x = -2 \).

3. **Simplify the function:**
   \[
   f(x) = \frac{(x - 8)(x + 2)}{(x - 3)(x - 4)(x + 2)}
   \]
   We can cancel the common factor \( (x + 2) \) from the numerator and the denominator, provided \( x \neq -2 \):
   \[
   f(x) = \frac{x - 8}{(x - 3)(x - 4)}, \quad \text{for } x \neq -2
   \]

4. **Identify vertical asymptotes:**
   Vertical asymptotes occur where the denominator of the simplified function is zero, but the numerator is not zero. The simplified denominator \( (x - 3)(x - 4) \) is zero at \( x = 3 \) and \( x = 4 \). At these points, the numerator \( (x - 8) \) is not zero.

   The factor \( x + 2 \) was canceled, indicating a hole at \( x = -2 \), not a vertical asymptote.

Therefore, the vertical asymptotes are at \( x = 3 \) and \( x = 4 \).

\[
\boxed{2}
\]

System: You are a helpful assistant.


[User] What is the value of $x$ if $|x-1| = |x-2|$? Express your answer as a common fraction.

[Assistant]
To solve the equation \( |x - 1| = |x - 2| \), we need to consider the definition of absolute value and the fact that the equation implies that the distance from \( x \) to 1 is equal to the distance from \( x \) to 2. This occurs when \( x \) is the midpoint between 1 and 2.

We can solve this algebraically by considering the two cases for the absolute value equation:

1. **Case 1: \( x - 1 = x - 2 \)**
   \[
   x - 1 = x - 2
   \]
   Subtract \( x \) from both sides:
   \[
   -1 = -2
   \]
   This is a contradiction, so there is no solution in this case.

2. **Case 2: \( x - 1 = -(x - 2) \)**
   \[
   x - 1 = -x + 2
   \]
   Add \( x \) to both sides:
   \[
   2x - 1 = 2
   \]
   Add 1 to both sides:
   \[
   2x = 3
   \]
   Divide both sides by 2:
   \[
   x = \frac{3}{2}
   \]

To verify, substitute \( x = \frac{3}{2} \) back into the original equation:
\[
\left| \frac{3}{2} - 1 \right| = \left| \frac{3}{2} - 2 \right|
\]
Simplify both sides:
\[
\left| \frac{3}{2} - \frac{2}{2} \right| = \left| \frac{3}{2} - \frac{4}{2} \right|
\]
\[
\left| \frac{1}{2} \right| = \left| -\frac{1}{2} \right|
\]
\[
\frac{1}{2} = \frac{1}{2}
\]
This confirms that \( x = \frac{3}{2} \) is indeed a solution.

Thus, the value of \( x \) is \(\boxed{\frac{3}{2}}\).

System: You are a helpful assistant.


[User] Given a positive integer in its base-10 representation, if you repeatedly add it to its reverse, will you always eventually reach a palindrome? If so, provide a bound on the number of iterations required. If not, find or describe a number that never palindromizes.

[Assistant]
To determine whether repeatedly adding a positive integer to its reverse will always result in a palindrome, we need to investigate the properties of such a process. This process is known as the "196 algorithm" or part of the "Lychrel process."

### Analysis:
1. **Example with Small Numbers:**
   - Consider the number 56. Its reverse is 65. Adding them: \(56 + 65 = 121\), which is a palindrome.
   - Consider the number 57. Its reverse is 75. Adding them: \(57 + 75 = 132\). The reverse of 132 is 231. Adding them: \(132 + 231 = 363\), which is a palindrome.

2. **Example with Larger Numbers:**
   - Consider the number 89. Its reverse is 98. Adding them: \(89 + 98 = 187\). The reverse of 187 is 781. Adding them: \(187 + 781 = 968\). The reverse of 968 is 869. Adding them: \(968 + 869 = 1837\). Continuing this process, it takes 24 iterations for 89 to become a palindrome.

3. **Candidate Lychrel Number:**
   - The number 196 is a well-known candidate for a Lychrel number. Despite extensive computational efforts, no palindrome has been found for 196 through this process. It is conjectured that 196 never becomes a palindrome, but this has not been proven.

### Conclusion:
- Not all positive integers will eventually become palindromes when repeatedly added to their reverse.
- A well-known example of a number that is conjectured to never become a palindrome is 196.
- Since there are numbers that do not form a palindrome through this process, a universal bound on the number of iterations required to reach a palindrome cannot be provided.

Therefore, the answer to the problem is:

\[
\boxed{196}
\]

System: You are a helpful assistant.


[User] If the value of $2^x + 2^{-x}$ is $\sqrt[3]{3}$, find the exact value of $8^x + 8^{-x}$. Give the final result in its simplest form.

[Assistant]
Given that \(2^x + 2^{-x} = \sqrt[3]{3}\), we need to find the value of \(8^x + 8^{-x}\).

First, note that:
\[ 8^x = (2^3)^x = 2^{3x} \quad \text{and} \quad 8^{-x} = (2^3)^{-x} = 2^{-3x}. \]
Therefore:
\[ 8^x + 8^{-x} = 2^{3x} + 2^{-3x}. \]

Let \( y = 2^x + 2^{-x} = \sqrt[3]{3} \).

We use the identity for the cube of a sum:
\[ (2^x + 2^{-x})^3 = (2^{3x} + 2^{-3x}) + 3 \cdot (2^x \cdot 2^{-x}) \cdot (2^x + 2^{-x}). \]

Since \( 2^x \cdot 2^{-x} = 1 \), the equation simplifies to:
\[ y^3 = (2^{3x} + 2^{-3x}) + 3y. \]

Rearranging to solve for \(2^{3x} + 2^{-3x}\), we get:
\[ 2^{3x} + 2^{-3x} = y^3 - 3y. \]

Substituting \( y = \sqrt[3]{3} \) into the equation:
\[ 2^{3x} + 2^{-3x} = (\sqrt[3]{3})^3 - 3 \cdot \sqrt[3]{3}. \]

Since \((\sqrt[3]{3})^3 = 3\), this simplifies to:
\[ 2^{3x} + 2^{-3x} = 3 - 3\sqrt[3]{3}. \]

Thus, the exact value of \(8^x + 8^{-x}\) is:
\[ \boxed{3 - 3\sqrt[3]{3}}. \]

System: You are a helpful assistant.


[User] How do you factor $x^n - 1$, where $n \geq 3$, into irreducibles in $\mathbb{Z}[x]$?

[Assistant]
To factor the polynomial \( x^n - 1 \) into irreducibles in \( \mathbb{Z}[x] \) for \( n \geq 3 \), we use the concept of cyclotomic polynomials. The \( n \)-th cyclotomic polynomial, denoted \( \Phi_n(x) \), is the minimal polynomial over the integers for the primitive \( n \)-th roots of unity. The polynomial \( x^n - 1 \) can be factored as the product of cyclotomic polynomials corresponding to the divisors of \( n \).

The factorization is given by:

\[
x^n - 1 = \prod_{d \mid n} \Phi_d(x),
\]

where the product is taken over all positive divisors \( d \) of \( n \). Each \( \Phi_d(x) \) is irreducible over \( \mathbb{Z}[x] \).

To verify this, consider the following steps:

1. **Definition of Cyclotomic Polynomials**: The \( d \)-th cyclotomic polynomial \( \Phi_d(x) \) is defined as the polynomial whose roots are the primitive \( d \)-th roots of unity. These roots are the complex numbers \( e^{2\pi i k/d} \) where \( k \) is coprime to \( d \).

2. **Factorization Property**: The polynomial \( x^n - 1 \) can be written as the product of cyclotomic polynomials for all divisors \( d \) of \( n \). This is a well-known result in algebra and number theory.

3. **Irreducibility of Cyclotomic Polynomials**: Each \( \Phi_d(x) \) is irreducible over \( \mathbb{Z}[x] \). This can be proven using properties of roots of unity and the minimality of the polynomial.

For example, consider \( n = 6 \). The divisors of 6 are 1, 2, 3, and 6. The corresponding cyclotomic polynomials are:

\[
\Phi_1(x) = x - 1,
\]
\[
\Phi_2(x) = x + 1,
\]
\[
\Phi_3(x) = x^2 + x + 1,
\]
\[
\Phi_6(x) = x^2 - x + 1.
\]

Thus, the factorization of \( x^6 - 1 \) is:

\[
x^6 - 1 = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1).
\]

Each of these polynomials is irreducible over \( \mathbb{Z}[x] \).

Therefore, the factorization of \( x^n - 1 \) into irreducibles in \( \mathbb{Z}[x] \) is:

\[
\boxed{x^n - 1 = \prod_{d \mid n} \Phi_d(x)}.
\]

System: You are a helpful assistant.


[User] Find the derivative $\frac{d}{dx} \sqrt{2x-1}$.

[Assistant]
To find the derivative of \(\sqrt{2x - 1}\) with respect to \(x\), we will use the chain rule. The chain rule states that if \(y = f(g(x))\), then the derivative \(y'\) is given by:

\[
y' = f'(g(x)) \cdot g'(x)
\]

Here, we can rewrite \(\sqrt{2x - 1}\) as \((2x - 1)^{1/2}\). Let \(u = 2x - 1\). Then the function becomes \(u^{1/2}\).

1. **Find the derivative of the outer function with respect to \(u\):**

   The outer function is \(u^{1/2}\). Using the power rule, the derivative of \(u^{1/2}\) with respect to \(u\) is:

   \[
   \frac{d}{du} u^{1/2} = \frac{1}{2} u^{-1/2}
   \]

2. **Find the derivative of the inner function with respect to \(x\):**

   The inner function is \(u = 2x - 1\). The derivative of \(u = 2x - 1\) with respect to \(x\) is:

   \[
   \frac{d}{dx} (2x - 1) = 2
   \]

3. **Apply the chain rule:**

   According to the chain rule, we multiply the derivative of the outer function by the derivative of the inner function:

   \[
   \frac{d}{dx} (2x - 1)^{1/2} = \frac{1}{2} (2x - 1)^{-1/2} \cdot 2
   \]

4. **Simplify the expression:**

   Simplifying the expression, we get:

   \[
   \frac{d}{dx} (2x - 1)^{1/2} = \frac{1}{2} \cdot 2 \cdot (2x - 1)^{-1/2} = 1 \cdot (2x - 1)^{-1/2} = \frac{1}{\sqrt{2x - 1}}
   \]

Therefore, the derivative of \(\sqrt{2x - 1}\) with respect to \(x\) is:

\[
\boxed{\frac{1}{\sqrt{2x - 1}}}
\]

System: You are a helpful assistant.


[User] Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?

[Assistant]
To solve the problem, we need to determine how many more 25-cent coins Joe has than 5-cent coins, given the total number of coins and their values.

Let's define the variables:
- Let \( x \) be the number of 5-cent coins.
- Let \( y \) be the number of 25-cent coins.
- The number of 10-cent coins is \( x + 3 \) because Joe has 3 more 10-cent coins than 5-cent coins.

We are given two key pieces of information:
1. The total number of coins is 23.
2. The total value of the coins is 320 cents.

From the first piece of information, we can write the equation:
\[ x + (x + 3) + y = 23 \]
Simplifying this, we get:
\[ 2x + y + 3 = 23 \]
\[ 2x + y = 20 \quad \text{(Equation 1)} \]

From the second piece of information, we can write the equation for the total value:
\[ 5x + 10(x + 3) + 25y = 320 \]
Expanding and combining like terms, we get:
\[ 5x + 10x + 30 + 25y = 320 \]
\[ 15x + 25y + 30 = 320 \]
\[ 15x + 25y = 290 \quad \text{(Equation 2)} \]

Now, we have a system of linear equations:
1. \( 2x + y = 20 \)
2. \( 15x + 25y = 290 \)

We can solve this system using substitution or elimination. Let's use substitution. From Equation 1, solve for \( y \):
\[ y = 20 - 2x \]

Substitute \( y = 20 - 2x \) into Equation 2:
\[ 15x + 25(20 - 2x) = 290 \]
\[ 15x + 500 - 50x = 290 \]
\[ -35x + 500 = 290 \]
\[ -35x = 290 - 500 \]
\[ -35x = -210 \]
\[ x = \frac{-210}{-35} \]
\[ x = 6 \]

Now, substitute \( x = 6 \) back into the equation \( y = 20 - 2x \):
\[ y = 20 - 2(6) \]
\[ y = 20 - 12 \]
\[ y = 8 \]

So, Joe has:
- 6 five-cent coins,
- \( 6 + 3 = 9 \) ten-cent coins,
- 8 twenty-five-cent coins.

Finally, we need to find how many more 25-cent coins Joe has than 5-cent coins:
\[ y - x = 8 - 6 = 2 \]

Thus, Joe has \(\boxed{2}\) more 25-cent coins than 5-cent coins.

System: You are a helpful assistant.


[User] How many variations of sundaes are there if a daily sundae consists of three scoops of ice cream chosen from four flavors: Vanilla, Pistachio, Black Walnut, and Strawberry?

[Assistant]
To determine the number of variations of sundaes with three scoops of ice cream chosen from four flavors (Vanilla, Pistachio, Black Walnut, and Strawberry), we need to consider that the order of the scoops does not matter and repetition of flavors is allowed. This scenario is a combination problem with repetition.

The formula for combinations with repetition is given by:
\[
\binom{n + k - 1}{k}
\]
where \( n \) is the number of flavors and \( k \) is the number of scoops.

In this problem:
- \( n = 4 \) (Vanilla, Pistachio, Black Walnut, Strawberry)
- \( k = 3 \) (three scoops)

Substituting these values into the formula, we get:
\[
\binom{4 + 3 - 1}{3} = \binom{6}{3}
\]

Next, we calculate \( \binom{6}{3} \):
\[
\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3! \cdot 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]

Thus, the number of variations of sundaes is \(\boxed{20}\).

System: You are a helpful assistant.


[User] How many ways are there to color the vertices of a square with six colors if rotations are considered the same?

[Assistant]
To determine the number of distinct ways to color the vertices of a square using six colors, considering that rotations are considered the same, we can use Burnside's Lemma. Burnside's Lemma states that the number of distinct colorings under group actions (such as rotations) is equal to the average number of colorings fixed by each group element.

The square's rotational symmetry group, \(C_4\), has four elements: rotations by 0°, 90°, 180°, and 270°. We need to count the number of colorings fixed by each rotation and then average these counts.

1. **0° Rotation (Identity Rotation):**
   - Every coloring is fixed by this rotation.
   - Number of fixed colorings: \(6^4 = 1296\).

2. **90° Rotation:**
   - For a coloring to be fixed, all four vertices must be the same color.
   - Number of fixed colorings: \(6\).

3. **180° Rotation:**
   - This swaps vertices 1 and 3, and vertices 2 and 4.
   - Each pair of vertices must be the same color.
   - Number of fixed colorings: \(6 \times 6 = 36\).

4. **270° Rotation:**
   - Similar to the 90° rotation, all four vertices must be the same color.
   - Number of fixed colorings: \(6\).

Now, we apply Burnside's Lemma to find the average number of fixed colorings:

\[
\text{Number of distinct colorings} = \frac{1}{4} \left(1296 + 6 + 36 + 6\right) = \frac{1344}{4} = 336
\]

Thus, the number of distinct colorings of the vertices of a square with six colors, considering rotations as the same, is:

\[
\boxed{336}
\]

System: You are a helpful assistant.


[User] Find all positive values of $c$ so that the inequality $x^2-6x+c<0$ has real solutions for $x$. Express your answer in interval notation.

[Assistant]
To determine the positive values of \( c \) such that the inequality \( x^2 - 6x + c < 0 \) has real solutions, we need to ensure that the quadratic equation \( x^2 - 6x + c = 0 \) has two distinct real roots. This occurs when the discriminant is positive.

The discriminant \( D \) of the quadratic equation \( x^2 - 6x + c = 0 \) is given by:
\[
D = (-6)^2 - 4 \cdot 1 \cdot c = 36 - 4c
\]

For the quadratic to have two distinct real roots, the discriminant must be positive:
\[
36 - 4c > 0
\]

Solving this inequality:
\[
36 > 4c \implies c < 9
\]

Since we are looking for positive values of \( c \), we combine this with \( c > 0 \). Therefore, the values of \( c \) must lie in the interval \( (0, 9) \).

To verify, consider the quadratic \( x^2 - 6x + c \). For \( c \) in the interval \( (0, 9) \), the discriminant is positive, ensuring two distinct real roots. The quadratic will be negative between these roots, satisfying the inequality \( x^2 - 6x + c < 0 \).

For \( c = 9 \), the discriminant is zero, and the quadratic becomes a perfect square \((x-3)^2\), which is non-negative and does not satisfy the inequality.

For \( c \geq 9 \), the discriminant is non-positive, and the quadratic does not have two distinct real roots, meaning the inequality \( x^2 - 6x + c < 0 \) has no solutions.

Thus, the positive values of \( c \) that satisfy the inequality \( x^2 - 6x + c < 0 \) are:
\[
\boxed{(0, 9)}
\]

System: You are a helpful assistant.


[User] Goldilocks has a $1/2$ chance of choosing the correct path at each fork. If she chooses the wrong path more than two times, she gets lost. What is the probability she gets lost if she encounters five forks?

[Assistant]
To determine the probability that Goldilocks gets lost, we need to calculate the probability that she chooses the wrong path more than two times out of five forks. This is a binomial probability problem with \( n = 5 \) trials and a probability of success (choosing the wrong path) \( p = \frac{1}{2} \).

The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

We need to find the probability that Goldilocks makes three, four, or five wrong choices, i.e., \( P(X \geq 3) \).

First, we calculate the individual probabilities for \( k = 3 \), \( k = 4 \), and \( k = 5 \):

1. For \( k = 3 \):
\[
P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \cdot \left(\frac{1}{2}\right)^5 = 10 \cdot \frac{1}{32} = \frac{10}{32}
\]

2. For \( k = 4 \):
\[
P(X = 4) = \binom{5}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^1 = 5 \cdot \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32}
\]

3. For \( k = 5 \):
\[
P(X = 5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^0 = 1 \cdot \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32}
\]

Next, we sum these probabilities to find \( P(X \geq 3) \):
\[
P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}
\]

Thus, the probability that Goldilocks gets lost is \(\boxed{\frac{1}{2}}\).

System: You are a helpful assistant.


[User] The perimeter of a rectangular garden is 60 feet. If the length of the field is twice the width, what is the area of the field, in square feet?

[Assistant]
To find the area of the rectangular garden, we start by using the given information about the perimeter and the relationship between the length and the width.

1. Let the width of the garden be \( w \).
2. Since the length is twice the width, the length is \( 2w \).
3. The perimeter of a rectangle is given by the formula \( P = 2 \times (\text{length} + \text{width}) \). Substituting the given perimeter and the expressions for length and width, we get:
   \[
   2 \times (2w + w) = 60
   \]
4. Simplify the equation:
   \[
   2 \times 3w = 60 \implies 6w = 60
   \]
5. Solve for \( w \):
   \[
   w = \frac{60}{6} = 10
   \]
6. Now that we have the width, we can find the length:
   \[
   \text{length} = 2w = 2 \times 10 = 20
   \]
7. The area of the rectangle is given by the product of the length and the width:
   \[
   \text{Area} = \text{length} \times \text{width} = 20 \times 10 = 200
   \]

Thus, the area of the garden is \(\boxed{200}\) square feet.

System: You are a helpful assistant.


[User] For what value of \( x \) does \( \frac{3x}{2x} = 0 \)?

[Assistant]
To solve the equation \(\frac{3x}{2x} = 0\), we start by analyzing the expression.

1. **Identify the condition for the expression to be defined:**
   The expression \(\frac{3x}{2x}\) is undefined when the denominator is zero. Therefore, \(2x \neq 0\), which implies \(x \neq 0\).

2. **Simplify the expression:**
   Assuming \(x \neq 0\), we can simplify \(\frac{3x}{2x}\) by canceling the common factor \(x\) in the numerator and the denominator:
   \[
   \frac{3x}{2x} = \frac{3}{2}
   \]
   Thus, the equation becomes:
   \[
   \frac{3}{2} = 0
   \]

3. **Analyze the simplified equation:**
   The equation \(\frac{3}{2} = 0\) is a contradiction because \(\frac{3}{2}\) is a non-zero constant (specifically, 1.5), and it cannot equal zero.

4. **Consider the case \(x = 0\):**
   If \(x = 0\), the original expression \(\frac{3x}{2x}\) becomes \(\frac{0}{0}\), which is undefined.

Since the simplified form \(\frac{3}{2} = 0\) is a contradiction and the expression is undefined when \(x = 0\), there is no value of \(x\) that satisfies the original equation.

Therefore, the solution to the equation \(\frac{3x}{2x} = 0\) is:
\[
\boxed{\text{No solution}}
\]

System: You are a helpful assistant.


[User] Compute \(\sigma(12)\), the sum of the divisors of 12.

[Assistant]
To compute \(\sigma(12)\), the sum of the divisors of 12, we can use the prime factorization method. First, we factorize 12 into its prime factors:

\[ 12 = 2^2 \times 3^1 \]

The formula for the sum of the divisors \(\sigma(n)\) when \(n\) is factored into primes as \(p_1^{k_1} \times p_2^{k_2} \times \cdots \times p_r^{k_r}\) is:

\[ \sigma(n) = (p_1^0 + p_1^1 + \cdots + p_1^{k_1}) \times (p_2^0 + p_2^1 + \cdots + p_2^{k_2}) \times \cdots \times (p_r^0 + p_r^1 + \cdots + p_r^{k_r}) \]

For \(12 = 2^2 \times 3^1\), we compute each part of the formula separately:

\[ \sigma(2^2) = 2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7 \]

\[ \sigma(3^1) = 3^0 + 3^1 = 1 + 3 = 4 \]

Now, we multiply these results together:

\[ \sigma(12) = \sigma(2^2) \times \sigma(3^1) = 7 \times 4 = 28 \]

Thus, the sum of the divisors of 12 is:

\[
\boxed{28}
\]

System: You are a helpful assistant.


[User] Give an example of a function that is both odd and bounded.

System: You are a helpful assistant.


[User] What is the probability that each drawn pair is of different colors (one Black and one White) when drawing two balls at a time from a bag containing 4 Black balls and 4 White balls until the bag is empty?

[Assistant]
To find the probability that each drawn pair is of different colors (one Black and one White) when drawing two balls at a time from a bag containing 4 Black balls and 4 White balls until the bag is empty, we can proceed as follows:

### Step-by-Step Solution:

1. **First Draw:**
   - Total ways to draw 2 balls from 8: \( \binom{8}{2} = 28 \).
   - Favorable outcomes (1 black, 1 white): \( \binom{4}{1} \times \binom{4}{1} = 16 \).
   - Probability: \( \frac{16}{28} = \frac{4}{7} \).

2. **Second Draw (after removing 1 black and 1 white):**
   - Remaining balls: 3 black, 3 white (total 6).
   - Total ways: \( \binom{6}{2} = 15 \).
   - Favorable outcomes: \( \binom{3}{1} \times \binom{3}{1} = 9 \).
   - Probability: \( \frac{9}{15} = \frac{3}{5} \).

3. **Third Draw (after removing another pair):**
   - Remaining balls: 2 black, 2 white (total 4).
   - Total ways: \( \binom{4}{2} = 6 \).
   - Favorable outcomes: \( \binom{2}{1} \times \binom{2}{1} = 4 \).
   - Probability: \( \frac{4}{6} = \frac{2}{3} \).

4. **Fourth Draw (final pair):**
   - Remaining balls: 1 black, 1 white.
   - Only one valid outcome (1 black and 1 white).
   - Probability: \( 1 \).

### Total Probability:
The total probability is the product of the probabilities of each successive draw:
\[
\frac{4}{7} \times \frac{3}{5} \times \frac{2}{3} \times 1 = \frac{4 \times 3 \times 2}{7 \times 5 \times 3} = \frac{24}{105} = \frac{8}{35}.
\]

### Verification:
Alternatively, the total number of ways to pair 8 balls into 4 pairs is \( \frac{8!}{2^4 \times 4!} = 105 \). The number of valid pairings (each black with a white) is \( 4! = 24 \). Thus, the probability is \( \frac{24}{105} = \frac{8}{35} \).

Therefore, the probability that each drawn pair is of different colors is:
\[
\boxed{\frac{8}{35}}
\]

System: You are a helpful assistant.


[User] In a triangle with vertices $A, B,$ and $C$, and respective opposite sides of lengths $a, b,$ and $c$, if $a = 1$, $b = 3$, and $\angle C = 60^\circ$, then what is $\sin^2{B}$?

[Assistant]
To find \(\sin^2 B\) in a triangle with vertices \(A, B,\) and \(C\) and respective opposite sides of lengths \(a, b,\) and \(c\), where \(a = 1\), \(b = 3\), and \(\angle C = 60^\circ\), we can proceed as follows:

1. **Use the Law of Cosines to find side \(c\):**
   The Law of Cosines states:
   \[
   c^2 = a^2 + b^2 - 2ab \cos C
   \]
   Plugging in the given values:
   \[
   c^2 = 1^2 + 3^2 - 2 \cdot 1 \cdot 3 \cdot \cos 60^\circ
   \]
   Since \(\cos 60^\circ = \frac{1}{2}\):
   \[
   c^2 = 1 + 9 - 2 \cdot 1 \cdot 3 \cdot \frac{1}{2} = 1 + 9 - 3 = 7
   \]
   Therefore, \(c = \sqrt{7}\).

2. **Use the Law of Sines to find \(\sin B\):**
   The Law of Sines states:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
   \]
   We know \(b = 3\), \(c = \sqrt{7}\), and \(\sin C = \sin 60^\circ = \frac{\sqrt{3}}{2}\). Thus:
   \[
   \frac{3}{\sin B} = \frac{\sqrt{7}}{\frac{\sqrt{3}}{2}}
   \]
   Solving for \(\sin B\):
   \[
   \frac{3}{\sin B} = \frac{2\sqrt{7}}{\sqrt{3}}
   \]
   \[
   \sin B = \frac{3 \cdot \sqrt{3}}{2 \cdot \sqrt{7}} = \frac{3\sqrt{3}}{2\sqrt{7}}
   \]
   Simplifying:
   \[
   \sin B = \frac{3\sqrt{3}}{2\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{21}}{14}
   \]

3. **Compute \(\sin^2 B\):**
   \[
   \sin^2 B = \left( \frac{3\sqrt{21}}{14} \right)^2 = \frac{9 \cdot 21}{196} = \frac{189}{196} = \frac{27}{28}
   \]

Thus, the value of \(\sin^2 B\) is \(\boxed{\frac{27}{28}}\).

System: You are a helpful assistant.


[User] What is the area, in square inches, of a right triangle with a 24-inch leg and a 25-inch hypotenuse?

[Assistant]
To find the area of a right triangle with one leg of 24 inches and a hypotenuse of 25 inches, we first need to determine the length of the other leg. We can use the Pythagorean theorem, which states that for a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\), the relationship is given by:

\[ a^2 + b^2 = c^2 \]

Here, one leg \(a = 24\) inches and the hypotenuse \(c = 25\) inches. Let the unknown leg be \(b\). Plugging in the known values, we get:

\[ 24^2 + b^2 = 25^2 \]

Calculating the squares, we have:

\[ 576 + b^2 = 625 \]

To find \(b^2\), we subtract 576 from both sides:

\[ b^2 = 625 - 576 \]
\[ b^2 = 49 \]

Taking the square root of both sides, we get:

\[ b = \sqrt{49} \]
\[ b = 7 \]

So, the other leg of the triangle is 7 inches. Now, we can find the area of the right triangle using the formula for the area of a triangle, which is:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base and height are the two legs of the triangle, 24 inches and 7 inches. Therefore, the area is:

\[ \text{Area} = \frac{1}{2} \times 24 \times 7 \]
\[ \text{Area} = \frac{1}{2} \times 168 \]
\[ \text{Area} = 84 \]

Thus, the area of the right triangle is \(\boxed{84}\) square inches.

System: You are a helpful assistant.


[User] What is the value of $(x - y)(x + y)$ if $x = 10$ and $y = 15$?

[Assistant]
To find the value of \((x - y)(x + y)\) when \(x = 10\) and \(y = 15\), we can use the difference of squares formula, which states that \((x - y)(x + y) = x^2 - y^2\).

First, we calculate \(x^2\):
\[
x^2 = 10^2 = 100
\]

Next, we calculate \(y^2\):
\[
y^2 = 15^2 = 225
\]

Now, we substitute these values into the difference of squares formula:
\[
(x - y)(x + y) = x^2 - y^2 = 100 - 225 = -125
\]

Therefore, the value of \((x - y)(x + y)\) is \(\boxed{-125}\).

System: You are a helpful assistant.


[User] Compute $\sum_{r=1}^{n} r(r-1) \binom{n}{r}$.

[Assistant]
To compute the sum \(\sum_{r=1}^{n} r(r-1) \binom{n}{r}\), we start by expressing \(r(r-1) \binom{n}{r}\) in a more convenient form.

First, recall the binomial coefficient \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). We can rewrite \(r(r-1) \binom{n}{r}\) as follows:
\[
r(r-1) \binom{n}{r} = r(r-1) \cdot \frac{n!}{r!(n-r)!}.
\]
Notice that \(r(r-1) = r(r-1) \cdot 1 = r(r-1) \cdot \frac{(r-2)!}{(r-2)!} = \frac{r(r-1)(r-2)!}{(r-2)!} = \frac{r!}{(r-2)!}\). Thus,
\[
r(r-1) \binom{n}{r} = \frac{r!}{(r-2)!} \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-2)!(n-r)!} = n(n-1) \cdot \frac{(n-2)!}{(r-2)!(n-r)!} = n(n-1) \binom{n-2}{r-2}.
\]

Therefore, the original sum can be rewritten as:
\[
\sum_{r=1}^{n} r(r-1) \binom{n}{r} = \sum_{r=1}^{n} n(n-1) \binom{n-2}{r-2}.
\]
We can factor out the constant \(n(n-1)\) from the sum:
\[
n(n-1) \sum_{r=1}^{n} \binom{n-2}{r-2}.
\]
Next, we shift the index of summation by letting \(k = r-2\). When \(r = 1\), \(k = -1\) (but \(\binom{n-2}{-1} = 0\)), and when \(r = n\), \(k = n-2\). Therefore, the sum becomes:
\[
n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k}.
\]
The sum \(\sum_{k=0}^{n-2} \binom{n-2}{k}\) is the binomial expansion of \((1+1)^{n-2} = 2^{n-2}\). Thus, we have:
\[
n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k} = n(n-1) \cdot 2^{n-2}.
\]

Therefore, the value of the sum is:
\[
\boxed{n(n-1)2^{n-2}}.
\]

System: You are a helpful assistant.


[User] What is the point-slope form of the equation of a line with slope \( \frac{3}{5} \) that passes through the point \((10, -2)\)?

[Assistant]
To find the point-slope form of the equation of a line with a slope of \(\frac{3}{5}\) that passes through the point \((10, -2)\), we use the point-slope form formula:

\[ y - y_1 = m(x - x_1) \]

Here, \( m \) is the slope, and \((x_1, y_1)\) is the point through which the line passes. Given:
\[ m = \frac{3}{5} \]
\[ (x_1, y_1) = (10, -2) \]

Substitute these values into the point-slope form formula:

\[ y - (-2) = \frac{3}{5}(x - 10) \]

Simplify the left side of the equation:

\[ y + 2 = \frac{3}{5}(x - 10) \]

Thus, the point-slope form of the equation of the line is:

\[
\boxed{y + 2 = \frac{3}{5}(x - 10)}
\]

System: You are a helpful assistant.


[User] In triangle PQR, PQ has length 6, QR has length 15, and angle PQR is 60 degrees. S is a point on PR such that QS bisects angle PQR. Find the length of PS.

[Assistant]
To find the length of \( PS \) in triangle \( PQR \) where \( PQ = 6 \), \( QR = 15 \), and \( \angle PQR = 60^\circ \), and \( S \) is a point on \( PR \) such that \( QS \) bisects \( \angle PQR \), we can use the Angle Bisector Theorem and the Law of Cosines.

First, apply the Law of Cosines to find the length of \( PR \):

\[
PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\angle PQR)
\]

Substitute the given values:

\[
PR^2 = 6^2 + 15^2 - 2 \cdot 6 \cdot 15 \cdot \cos(60^\circ)
\]

Since \( \cos(60^\circ) = 0.5 \):

\[
PR^2 = 36 + 225 - 2 \cdot 6 \cdot 15 \cdot 0.5
\]

Simplify the expression:

\[
PR^2 = 36 + 225 - 90 = 171
\]

Thus,

\[
PR = \sqrt{171} = 3\sqrt{19}
\]

Next, use the Angle Bisector Theorem, which states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. Therefore:

\[
\frac{PS}{SR} = \frac{PQ}{QR} = \frac{6}{15} = \frac{2}{5}
\]

Let \( PS = 2x \) and \( SR = 5x \). Since \( PS + SR = PR \):

\[
2x + 5x = 3\sqrt{19}
\]

\[
7x = 3\sqrt{19}
\]

Solve for \( x \):

\[
x = \frac{3\sqrt{19}}{7}
\]

Thus, the length of \( PS \) is:

\[
PS = 2x = 2 \cdot \frac{3\sqrt{19}}{7} = \frac{6\sqrt{19}}{7}
\]

The length of \( PS \) is:

\[
\boxed{\frac{6\sqrt{19}}{7}}
\]

System: You are a helpful assistant.


[User] Evaluate the limit $\lim_{n \to \infty }(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{3n})$.

[Assistant]
To evaluate the limit \(\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \cdots + \frac{1}{3n} \right)\), we can use the concept of Riemann sums.

First, rewrite the sum in a more convenient form:
\[
\sum_{k=n+1}^{3n} \frac{1}{k}.
\]
This sum has \(2n\) terms. We can factor out \(\frac{1}{n}\) from each term:
\[
\sum_{k=n+1}^{3n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{1}{n+m} = \frac{1}{n} \sum_{m=1}^{2n} \frac{1}{1 + \frac{m}{n}}.
\]
This expression is a Riemann sum for the function \(f(x) = \frac{1}{1+x}\) over the interval \([0, 2]\). Specifically, it is the right Riemann sum with \(2n\) subintervals of width \(\frac{1}{n}\):
\[
\frac{1}{n} \sum_{m=1}^{2n} \frac{1}{1 + \frac{m}{n}} \approx \int_0^2 \frac{1}{1+x} \, dx.
\]
As \(n \to \infty\), the Riemann sum converges to the integral:
\[
\lim_{n \to \infty} \frac{1}{n} \sum_{m=1}^{2n} \frac{1}{1 + \frac{m}{n}} = \int_0^2 \frac{1}{1+x} \, dx.
\]
We can evaluate this integral using the antiderivative of \(\frac{1}{1+x}\):
\[
\int_0^2 \frac{1}{1+x} \, dx = \left[ \ln(1+x) \right]_0^2 = \ln(3) - \ln(1) = \ln(3).
\]
Therefore, the limit is:
\[
\boxed{\ln 3}.
\]

System: You are a helpful assistant.


[User] Find the area of the ellipse given by $x^2 + 6x + 4y^2 - 8y + 9 = 0.$

[Assistant]
To find the area of the ellipse given by the equation \( x^2 + 6x + 4y^2 - 8y + 9 = 0 \), we first rewrite the equation in standard form by completing the square.

1. **Complete the square for the \( x \)-terms:**
   \[
   x^2 + 6x = (x + 3)^2 - 9
   \]

2. **Complete the square for the \( y \)-terms:**
   \[
   4y^2 - 8y = 4(y^2 - 2y) = 4[(y - 1)^2 - 1] = 4(y - 1)^2 - 4
   \]

3. **Substitute back into the original equation:**
   \[
   (x + 3)^2 - 9 + 4(y - 1)^2 - 4 + 9 = 0
   \]
   Simplify the constants:
   \[
   (x + 3)^2 + 4(y - 1)^2 - 4 = 0 \implies (x + 3)^2 + 4(y - 1)^2 = 4
   \]

4. **Divide by 4 to convert to standard form:**
   \[
   \frac{(x + 3)^2}{4} + \frac{(y - 1)^2}{1} = 1
   \]

From the standard form of the ellipse \(\frac{(x + 3)^2}{4} + \frac{(y - 1)^2}{1} = 1\), we identify the semi-major axis \( a \) and the semi-minor axis \( b \):
- The semi-major axis \( a \) is \( \sqrt{4} = 2 \).
- The semi-minor axis \( b \) is \( \sqrt{1} = 1 \).

The area \( A \) of the ellipse is given by the formula:
\[
A = \pi \times a \times b
\]

Substituting the values of \( a \) and \( b \):
\[
A = \pi \times 2 \times 1 = 2\pi
\]

Thus, the area of the ellipse is \(\boxed{2\pi}\).

System: You are a helpful assistant.


[User] Find the number of triplets $(x, y, z)$ of positive integers that satisfy the equation $x + y + z = 2013$.

[Assistant]
To find the number of triplets \((x, y, z)\) of positive integers that satisfy the equation \(x + y + z = 2013\), we can use the stars and bars method with a slight adjustment to account for the fact that each variable must be at least 1.

1. **Transform the Problem:**
   Since \(x, y, z\) are positive integers, we can transform the variables by setting \(x' = x - 1\), \(y' = y - 1\), and \(z' = z - 1\). This transformation ensures that \(x', y', z'\) are non-negative integers. The equation then becomes:
   \[
   (x' + 1) + (y' + 1) + (z' + 1) = 2013
   \]
   Simplifying, we get:
   \[
   x' + y' + z' = 2010
   \]

2. **Apply the Stars and Bars Theorem:**
   The number of non-negative integer solutions to the equation \(x' + y' + z' = 2010\) is given by the binomial coefficient \(\binom{n + k - 1}{k - 1}\), where \(n = 2010\) and \(k = 3\). Therefore, we need to calculate:
   \[
   \binom{2010 + 3 - 1}{3 - 1} = \binom{2012}{2}
   \]

3. **Calculate the Binomial Coefficient:**
   The binomial coefficient \(\binom{2012}{2}\) is calculated as follows:
   \[
   \binom{2012}{2} = \frac{2012 \times 2011}{2}
   \]
   Performing the multiplication and division:
   \[
   \frac{2012 \times 2011}{2} = \frac{4046132}{2} = 2023066
   \]

Thus, the number of triplets \((x, y, z)\) of positive integers that satisfy the equation \(x + y + z = 2013\) is \(\boxed{2023066}\).

System: You are a helpful assistant.


[User] Find a simple expression for the sum \(\sum_{k=1}^{\infty} \frac{1}{k^2 + 2}\) in terms of elementary functions.

[Assistant]
To find a simple expression for the sum \(\sum_{k=1}^{\infty} \frac{1}{k^2 + 2}\), we can use the known result for the series \(\sum_{k=1}^{\infty} \frac{1}{k^2 + a^2}\). The formula for this series is given by:

\[
\sum_{k=1}^{\infty} \frac{1}{k^2 + a^2} = \frac{\pi}{2a} \coth(\pi a) - \frac{1}{2a^2}
\]

In our problem, we have \(a = \sqrt{2}\). Substituting \(a = \sqrt{2}\) into the formula, we get:

\[
\sum_{k=1}^{\infty} \frac{1}{k^2 + 2} = \frac{\pi}{2\sqrt{2}} \coth(\pi \sqrt{2}) - \frac{1}{2(\sqrt{2})^2}
\]

Simplifying the expression, we have:

\[
\sum_{k=1}^{\infty} \frac{1}{k^2 + 2} = \frac{\pi}{2\sqrt{2}} \coth(\pi \sqrt{2}) - \frac{1}{4}
\]

Thus, the sum \(\sum_{k=1}^{\infty} \frac{1}{k^2 + 2}\) can be expressed as:

\[
\boxed{\frac{\pi}{2\sqrt{2}} \coth(\pi \sqrt{2}) - \frac{1}{4}}
\]

System: You are a helpful assistant.


[User] Points \( X \) and \( Y \) lie on a circle of radius 2 and center at \( O \), and the measure of \( \angle XOY \) is \( 30^\circ \). What is the area of the smaller region bounded by the circle and the chord joining \( X \) to \( Y \)? Express your answer in terms of \( \pi \).

[Assistant]
To find the area of the smaller region bounded by the circle and the chord joining points \( X \) and \( Y \) on a circle with radius 2 and center \( O \), where \( \angle XOY = 30^\circ \), we can use the formula for the area of a segment. The area of the segment is given by the area of the sector minus the area of the triangle formed by the two radii and the chord.

1. **Sector Area**: The area of the sector formed by \( \angle XOY \) is calculated as:
   \[
   \text{Sector Area} = \left( \frac{30}{360} \right) \pi r^2 = \left( \frac{1}{12} \right) \pi (2)^2 = \frac{4\pi}{12} = \frac{\pi}{3}
   \]

2. **Triangle Area**: The area of the triangle \( \triangle XOY \) is calculated using the formula for the area of a triangle with two sides and the included angle:
   \[
   \text{Triangle Area} = \frac{1}{2} r^2 \sin \theta = \frac{1}{2} (2)^2 \sin 30^\circ = \frac{1}{2} \cdot 4 \cdot \frac{1}{2} = 1
   \]

3. **Segment Area**: The area of the segment is the difference between the area of the sector and the area of the triangle:
   \[
   \text{Segment Area} = \text{Sector Area} - \text{Triangle Area} = \frac{\pi}{3} - 1
   \]

Therefore, the area of the smaller region bounded by the circle and the chord joining \( X \) to \( Y \) is \(\boxed{\dfrac{\pi}{3} - 1}\).

System: You are a helpful assistant.


[User] A right cylinder with a height of 5 inches has a radius of 2 inches. What is the area of the curved surface of the cylinder, in square inches? Express your answer in terms of $\pi$.

[Assistant]
To find the area of the curved surface of a right cylinder, we use the formula for the lateral surface area, which is given by:

\[
2\pi r h
\]

where \( r \) is the radius of the base of the cylinder and \( h \) is the height of the cylinder.

Given:
- The radius \( r = 2 \) inches
- The height \( h = 5 \) inches

Substituting these values into the formula, we get:

\[
2\pi \times 2 \times 5
\]

First, we calculate the product of the constants:

\[
2 \times 2 = 4
\]

Next, we multiply this result by the height:

\[
4 \times 5 = 20
\]

Thus, the area of the curved surface is:

\[
20\pi \text{ square inches}
\]

So, the area of the curved surface of the cylinder is:

\[
\boxed{20\pi}
\]

System: You are a helpful assistant.


[User] Let \( f_n \) denote the number of ways to toss a fair coin \( n \) times such that no two consecutive heads occur. Find the recursion for \( f_n \).

[Assistant]
To find the recursion for \( f_n \), which denotes the number of ways to toss a fair coin \( n \) times such that no two consecutive heads occur, we can proceed as follows:

1. **Define the problem**: We need to count the number of valid sequences of coin tosses of length \( n \) where no two consecutive heads appear.

2. **Base cases**:
   - For \( n = 1 \): The sequences are H and T. There are no consecutive heads, so \( f_1 = 2 \).
   - For \( n = 2 \): The sequences are HT, TH, and TT. The sequence HH is invalid. So, \( f_2 = 3 \).

3. **Recursive relation**:
   - Consider a valid sequence of length \( n \). The sequence can start with either T or H.
   - If the sequence starts with T, the remaining \( n-1 \) tosses must form a valid sequence of length \( n-1 \). There are \( f_{n-1} \) such sequences.
   - If the sequence starts with H, the next toss must be T to avoid two consecutive heads. The remaining \( n-2 \) tosses must form a valid sequence of length \( n-2 \). There are \( f_{n-2} \) such sequences.
   - Therefore, the total number of valid sequences of length \( n \) is the sum of the valid sequences starting with T and those starting with HT:
     \[
     f_n = f_{n-1} + f_{n-2}
     \]

4. **Conclusion**:
   - The recursion for \( f_n \) is given by:
     \[
     f_n = f_{n-1} + f_{n-2}
     \]
   - This relation holds for \( n \geq 3 \), with the base cases \( f_1 = 2 \) and \( f_2 = 3 \).

Thus, the recursion for \( f_n \) is:
\[
\boxed{f_n = f_{n-1} + f_{n-2}}
\]

System: You are a helpful assistant.


[User] A circle has a radius of \( \log_{10}(a^2) \) and a circumference of \( \log_{10}(b^4) \). What is \( \log_ab \)?

[Assistant]
To solve the problem, we start by using the given information about the circle's radius and circumference.

The radius \( r \) of the circle is given by:
\[
r = \log_{10}(a^2)
\]
Using the logarithm property \( \log_b(x^n) = n \log_b(x) \), we can simplify the radius:
\[
r = 2 \log_{10}(a)
\]

The circumference \( C \) of the circle is given by:
\[
C = \log_{10}(b^4)
\]
Again, using the logarithm property, we simplify the circumference:
\[
C = 4 \log_{10}(b)
\]

The formula for the circumference of a circle is:
\[
C = 2\pi r
\]
Substituting the expressions for \( C \) and \( r \) into this formula, we get:
\[
4 \log_{10}(b) = 2\pi \cdot 2 \log_{10}(a)
\]
Simplifying the right-hand side:
\[
4 \log_{10}(b) = 4\pi \log_{10}(a)
\]
Dividing both sides by 4:
\[
\log_{10}(b) = \pi \log_{10}(a)
\]

To find \( \log_a b \), we use the change of base formula for logarithms:
\[
\log_a b = \frac{\log_{10}(b)}{\log_{10}(a)}
\]
Substituting \( \log_{10}(b) = \pi \log_{10}(a) \) into the change of base formula:
\[
\log_a b = \frac{\pi \log_{10}(a)}{\log_{10}(a)} = \pi
\]

Thus, the value of \( \log_a b \) is:
\[
\boxed{\pi}
\]

System: You are a helpful assistant.


[User] In an \( m \times n \) rectangular grid, how many unit squares does the diagonal connecting two opposite vertices pass through? Express your answer in terms of \( m \) and \( n \).

[Assistant]
To determine the number of unit squares a diagonal passes through in an \( m \times n \) rectangular grid, we can use the formula \( m + n - \gcd(m, n) \). Here's a detailed derivation of this formula:

1. **Understanding the Grid and the Diagonal:**
   - Consider an \( m \times n \) grid, where \( m \) is the number of rows and \( n \) is the number of columns.
   - The diagonal starts at the top-left corner \((0,0)\) and ends at the bottom-right corner \((n,m)\).

2. **Counting the Crossings:**
   - As the diagonal moves from \((0,0)\) to \((n,m)\), it crosses \( n \) vertical lines and \( m \) horizontal lines.
   - Each time the diagonal crosses a vertical or horizontal line, it enters a new unit square.

3. **Handling Overlaps:**
   - The diagonal will cross a grid point (where both a vertical and a horizontal line meet) if and only if the coordinates are multiples of the greatest common divisor (gcd) of \( m \) and \( n \).
   - The number of such grid points (excluding the starting point) is \( \gcd(m, n) - 1 \).

4. **Calculating the Total Number of Squares:**
   - The total number of vertical and horizontal line crossings is \( n + m \).
   - However, each time the diagonal crosses a grid point, it enters a new square only once, not twice. Therefore, we need to subtract the number of grid points (excluding the starting point) from the total crossings.
   - The number of grid points is \( \gcd(m, n) \), but we exclude the starting point, so we subtract \( \gcd(m, n) - 1 \).

5. **Final Formula:**
   - The total number of unit squares the diagonal passes through is given by:
     \[
     m + n - \gcd(m, n)
     \]

Thus, the number of unit squares a diagonal passes through in an \( m \times n \) grid is:
\[
\boxed{m + n - \gcd(m, n)}
\]

System: You are a helpful assistant.


[User] A sphere has a surface area of \(676\pi\). What is its volume? Express your answer in terms of \(\pi\).

[Assistant]
To find the volume of a sphere given its surface area, we start with the formula for the surface area of a sphere:

\[
4\pi r^2
\]

We are given that the surface area is \(676\pi\). Therefore, we set up the equation:

\[
4\pi r^2 = 676\pi
\]

Next, we solve for \(r^2\) by dividing both sides by \(4\pi\):

\[
r^2 = \frac{676\pi}{4\pi} = \frac{676}{4} = 169
\]

Taking the square root of both sides, we find:

\[
r = \sqrt{169} = 13
\]

Now that we have the radius \(r = 13\), we use the formula for the volume of a sphere:

\[
V = \frac{4}{3}\pi r^3
\]

Substituting \(r = 13\) into the volume formula, we get:

\[
V = \frac{4}{3}\pi (13)^3
\]

Next, we calculate \(13^3\):

\[
13^3 = 13 \times 13 \times 13 = 169 \times 13 = 2197
\]

So, the volume is:

\[
V = \frac{4}{3}\pi \times 2197 = \frac{4 \times 2197}{3}\pi = \frac{8788}{3}\pi
\]

Thus, the volume of the sphere is:

\[
\boxed{\frac{8788}{3}\pi}
\]

System: You are a helpful assistant.


[User] In a business school, 75% of students adopted marketing, 62% adopted finance, and 88% adopted HR as electives. What is the minimum number of students specializing in all three subjects?

[Assistant]
To determine the minimum number of students specializing in all three subjects (marketing, finance, and HR), we use the principle of inclusion-exclusion for three sets. The formula for the minimum overlap is:

\[
\text{Minimum number of students in all three} = M + F + H - 2 \times 100\%
\]

where \(M\), \(F\), and \(H\) are the percentages of students in marketing, finance, and HR respectively. Substituting the given percentages:

\[
M = 75\%, \quad F = 62\%, \quad H = 88\%
\]

We calculate the minimum overlap as follows:

\[
75\% + 62\% + 88\% - 2 \times 100\% = 225\% - 200\% = 25\%
\]

Assuming the total number of students is 100 (a common base for percentage problems), the minimum number of students specializing in all three subjects is 25% of 100, which is 25.

Thus, the minimum number of students specializing in all three subjects is:

\[
\boxed{25}
\]

System: You are a helpful assistant.


[User] Given matrices $\boldsymbol{A}=\begin{bmatrix} 1 & 3 & -4 \\ -1 & -3 & 4 \end{bmatrix}$ and $\boldsymbol{B}=\begin{bmatrix} 1 & 3 \\ -1 & 8 \end{bmatrix}$, find the product $AB$.

[Assistant]
To find the product \( \boldsymbol{AB} \), we need to check if the dimensions of the matrices allow for multiplication. Specifically, the number of columns in matrix \( \boldsymbol{A} \) must equal the number of rows in matrix \( \boldsymbol{B} \).

Given:
\[ \boldsymbol{A} = \begin{bmatrix} 1 & 3 & -4 \\ -1 & -3 & 4 \end{bmatrix} \]
\[ \boldsymbol{B} = \begin{bmatrix} 1 & 3 \\ -1 & 8 \end{bmatrix} \]

Matrix \( \boldsymbol{A} \) is a \( 2 \times 3 \) matrix (2 rows and 3 columns).
Matrix \( \boldsymbol{B} \) is a \( 2 \times 2 \) matrix (2 rows and 2 columns).

For the product \( \boldsymbol{AB} \) to be defined, the number of columns in \( \boldsymbol{A} \) (which is 3) must equal the number of rows in \( \boldsymbol{B} \) (which is 2). Since 3 does not equal 2, the product \( \boldsymbol{AB} \) is undefined.

Therefore, the final answer is:
\[
\boxed{\text{undefined}}
\]

System: You are a helpful assistant.


[User] Write the expression $\frac{4+3c}{7}+2$ as a single fraction.

[Assistant]
To combine the expression \(\frac{4+3c}{7} + 2\) into a single fraction, follow these steps:

1. **Identify the common denominator**: The first term \(\frac{4+3c}{7}\) already has a denominator of 7. The second term, 2, can be written as \(\frac{2}{1}\). To add these fractions, we need a common denominator. The least common denominator of 7 and 1 is 7.

2. **Convert the second term to a fraction with the common denominator**: Convert 2 to a fraction with a denominator of 7:
   \[
   2 = \frac{2 \times 7}{1 \times 7} = \frac{14}{7}
   \]

3. **Add the fractions**: Now that both terms have the same denominator, we can add the numerators:
   \[
   \frac{4 + 3c}{7} + \frac{14}{7} = \frac{(4 + 3c) + 14}{7}
   \]

4. **Combine like terms in the numerator**: Simplify the numerator by combining the constant terms:
   \[
   (4 + 3c) + 14 = 4 + 14 + 3c = 18 + 3c
   \]

5. **Write the final combined fraction**: The expression as a single fraction is:
   \[
   \frac{18 + 3c}{7}
   \]

6. **Alternative form (if factoring is allowed)**: If factoring is allowed, you can factor out a 3 from the numerator:
   \[
   \frac{3c + 18}{7} = \frac{3(c + 6)}{7}
   \]

However, the problem only asks to write the expression as a single fraction. Therefore, the final answer is:
\[
\boxed{\frac{3c + 18}{7}}
\]

System: You are a helpful assistant.


[User] Compute the value of $ \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}$.

[Assistant]
To compute the value of \( \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}} \), we will simplify the expression step by step from the innermost fraction outward.

1. Start with the innermost fraction:
   \[
   \frac{1}{2} = 0.5
   \]
   So, the next layer is:
   \[
   2 + \frac{1}{2} = 2 + 0.5 = 2.5 = \frac{5}{2}
   \]

2. Now, consider the next layer:
   \[
   \frac{1}{\frac{5}{2}} = \frac{2}{5}
   \]
   So, the next layer is:
   \[
   2 + \frac{2}{5} = \frac{10}{5} + \frac{2}{5} = \frac{12}{5}
   \]

3. Now, consider the next layer:
   \[
   \frac{1}{\frac{12}{5}} = \frac{5}{12}
   \]
   So, the next layer is:
   \[
   2 + \frac{5}{12} = \frac{24}{12} + \frac{5}{12} = \frac{29}{12}
   \]

4. Finally, the entire expression is:
   \[
   \frac{1}{\frac{29}{12}} = \frac{12}{29}
   \]

Thus, the value of the given expression is:
\[
\boxed{\frac{12}{29}}
\]

System: You are a helpful assistant.


[User] Evaluate $\lim_{x\rightarrow 7}\sqrt{x-10}$.

[Assistant]
To evaluate the limit \(\lim_{x \to 7} \sqrt{x - 10}\), we need to consider the domain of the function \(\sqrt{x - 10}\). The square root function \(\sqrt{y}\) is defined only for \(y \geq 0\). Therefore, for \(\sqrt{x - 10}\) to be defined, we must have \(x - 10 \geq 0\), which simplifies to \(x \geq 10\).

Since \(x = 7\) is less than 10, the expression inside the square root, \(x - 10\), is negative when \(x = 7\). Specifically, \(7 - 10 = -3\), and \(\sqrt{-3}\) is not a real number.

Moreover, for the limit \(\lim_{x \to 7} \sqrt{x - 10}\) to exist, the function \(\sqrt{x - 10}\) must be defined in some neighborhood around \(x = 7\). However, there is no interval around \(x = 7\) where \(x - 10 \geq 0\). Therefore, the function \(\sqrt{x - 10}\) is not defined for any \(x\) in any neighborhood of 7.

Since the function is not defined in any neighborhood of \(x = 7\), the limit does not exist in the real numbers.

\[
\boxed{\text{Does Not Exist}}
\]

System: You are a helpful assistant.


[User] Convert $10101_3$ to a base 10 integer.

[Assistant]
To convert the base 3 number \(10101_3\) to a base 10 integer, we need to evaluate the value of each digit based on its position. The positions of the digits are as follows, starting from the rightmost digit (which is the least significant):

\[
1 \times 3^4 + 0 \times 3^3 + 1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0
\]

Let's calculate each term step-by-step:

1. The rightmost digit is \(1\) and its position is \(3^0\):
   \[
   1 \times 3^0 = 1 \times 1 = 1
   \]

2. The next digit to the left is \(0\) and its position is \(3^1\):
   \[
   0 \times 3^1 = 0 \times 3 = 0
   \]

3. The next digit to the left is \(1\) and its position is \(3^2\):
   \[
   1 \times 3^2 = 1 \times 9 = 9
   \]

4. The next digit to the left is \(0\) and its position is \(3^3\):
   \[
   0 \times 3^3 = 0 \times 27 = 0
   \]

5. The leftmost digit is \(1\) and its position is \(3^4\):
   \[
   1 \times 3^4 = 1 \times 81 = 81
   \]

Now, we sum all these values together:

\[
81 + 0 + 9 + 0 + 1 = 91
\]

Thus, the base 10 equivalent of the base 3 number \(10101_3\) is \(\boxed{91}\).

System: You are a helpful assistant.