[User] Find all pairs of positive integers \( (x, y) \) such that \( x^6 = y^2 + 127 \).

[Assistant]
To find all pairs of positive integers \((x, y)\) that satisfy the equation \(x^6 = y^2 + 127\), we start by rewriting the equation as a difference of squares:

\[
x^6 - y^2 = 127
\]

We can factor the left-hand side as:

\[
(x^3 - y)(x^3 + y) = 127
\]

Since 127 is a prime number, its only positive divisors are 1 and 127. Therefore, we have two cases to consider:

1. \(x^3 - y = 1\) and \(x^3 + y = 127\)
2. \(x^3 - y = 127\) and \(x^3 + y = 1\)

We discard the second case because it would imply \(x^3\) is negative, which is not possible for positive integers \(x\) and \(y\). Thus, we focus on the first case:

\[
\begin{cases}
x^3 - y = 1 \\
x^3 + y = 127
\end{cases}
\]

Adding these two equations, we get:

\[
2x^3 = 128 \implies x^3 = 64 \implies x = 4
\]

Substituting \(x = 4\) into the first equation:

\[
4^3 - y = 1 \implies 64 - y = 1 \implies y = 63
\]

Thus, we have the solution \((x, y) = (4, 63)\). To verify, we substitute back into the original equation:

\[
4^6 = 4096 \quad \text{and} \quad 63^2 + 127 = 3969 + 127 = 4096
\]

Both sides are equal, confirming that \((4, 63)\) is indeed a solution.

To ensure there are no other solutions, we consider larger values of \(x\). For \(x > 4\), \(x^6\) grows rapidly, and \(y^2 = x^6 - 127\) would not be a perfect square. For example, for \(x = 5\):

\[
5^6 = 15625 \quad \text{and} \quad y^2 = 15625 - 127 = 15498
\]

Since 15498 is not a perfect square, there are no solutions for \(x > 4\).

Therefore, the only solution is \(\boxed{(4, 63)}\).

System: You are a helpful assistant.


[User] Let \( f: \mathbb{N} \rightarrow \mathbb{N} \) be a function such that \( f(f(n)) + 2f(n) = 3n + 4 \) for all \( n \in \mathbb{N} \). Find \( f(2006) \).

[Assistant]
To solve for \( f(2006) \) given the functional equation \( f(f(n)) + 2f(n) = 3n + 4 \) for all \( n \in \mathbb{N} \), we start by assuming a linear form for \( f(n) \). Let \( f(n) = an + b \), where \( a \) and \( b \) are constants.

Substituting \( f(n) = an + b \) into the equation, we get:
\[
f(f(n)) = f(an + b) = a(an + b) + b = a^2n + ab + b
\]
Thus, the equation becomes:
\[
a^2n + ab + b + 2(an + b) = 3n + 4
\]
Simplifying, we equate coefficients of \( n \) and the constant terms:
\[
(a^2 + 2a)n + (ab + 3b) = 3n + 4
\]
This gives us two equations:
1. \( a^2 + 2a = 3 \)
2. \( ab + 3b = 4 \)

Solving the quadratic equation \( a^2 + 2a - 3 = 0 \), we factorize:
\[
a^2 + 2a - 3 = (a + 3)(a - 1) = 0
\]
The roots are \( a = -3 \) and \( a = 1 \). Since \( f: \mathbb{N} \rightarrow \mathbb{N} \), \( a \) must be a positive integer. Therefore, \( a = 1 \).

Substituting \( a = 1 \) into the second equation:
\[
1 \cdot b + 3b = 4 \implies 4b = 4 \implies b = 1
\]

Thus, the function is \( f(n) = n + 1 \).

To verify, we substitute \( f(n) = n + 1 \) back into the original equation:
\[
f(f(n)) + 2f(n) = f(n + 1) + 2(n + 1) = (n + 1 + 1) + 2(n + 1) = n + 2 + 2n + 2 = 3n + 4
\]
The equation holds true, confirming that \( f(n) = n + 1 \) is indeed a solution.

Finally, substituting \( n = 2006 \):
\[
f(2006) = 2006 + 1 = 2007
\]

Therefore, the answer is \(\boxed{2007}\).

System: You are a helpful assistant.


[User] Evaluate the product \(\sin\left(\frac{\pi}{199}\right) \cdot \sin\left(\frac{2\pi}{199}\right) \cdot \sin\left(\frac{3\pi}{199}\right) \cdots \sin\left(\frac{198\pi}{199}\right)\).

[Assistant]
To evaluate the product \(\sin\left(\frac{\pi}{199}\right) \cdot \sin\left(\frac{2\pi}{199}\right) \cdot \sin\left(\frac{3\pi}{199}\right) \cdots \sin\left(\frac{198\pi}{199}\right)\), we use the known formula for the product of sines of the form \(\sin\left(\frac{k\pi}{n}\right)\) for \(k\) from 1 to \(n-1\).

The formula states that the product of \(\sin\left(\frac{k\pi}{n}\right)\) from \(k = 1\) to \(n-1\) is given by:
\[
\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}
\]

For \(n = 199\), we need to find the product:
\[
\prod_{k=1}^{198} \sin\left(\frac{k\pi}{199}\right)
\]

Using the formula, we substitute \(n = 199\):
\[
\prod_{k=1}^{198} \sin\left(\frac{k\pi}{199}\right) = \frac{199}{2^{198}}
\]

Thus, the value of the product is:
\[
\boxed{\dfrac{199}{2^{198}}}
\]

System: You are a helpful assistant.


[User] Given the polynomial \( P(x) \) with the conditions \( P(1) = 0 \), \( P(2) = 1 \), and \( P(x^2) = x^4 + mx^2 + n \), find \( P(0) \).

[Assistant]
Given the polynomial \( P(x) \) with the conditions \( P(1) = 0 \), \( P(2) = 1 \), and \( P(x^2) = x^4 + mx^2 + n \), we need to find \( P(0) \).

First, we recognize that \( P(x) \) must be a quadratic polynomial because \( P(x^2) = x^4 + mx^2 + n \) implies \( P(y) = y^2 + my + n \) where \( y = x^2 \). Therefore, we can express \( P(x) \) as:
\[ P(x) = x^2 + mx + n \]

Using the given conditions:
1. \( P(1) = 0 \):
   \[
   1^2 + m \cdot 1 + n = 0 \implies 1 + m + n = 0 \implies m + n = -1
   \]

2. \( P(2) = 1 \):
   \[
   2^2 + m \cdot 2 + n = 1 \implies 4 + 2m + n = 1 \implies 2m + n = -3
   \]

We solve the system of equations:
1. \( m + n = -1 \)
2. \( 2m + n = -3 \)

Subtracting the first equation from the second:
\[
(2m + n) - (m + n) = -3 - (-1) \implies m = -2
\]

Substituting \( m = -2 \) into the first equation:
\[
-2 + n = -1 \implies n = 1
\]

Thus, the polynomial is \( P(x) = x^2 - 2x + 1 \).

To find \( P(0) \):
\[
P(0) = 0^2 - 2 \cdot 0 + 1 = 1
\]

Therefore, the value of \( P(0) \) is \(\boxed{1}\).

System: You are a helpful assistant.


[User] Find a general solution for \( n \) such that \( x^n + x - 1 \equiv 0 \pmod{x^2 - x + 1} \).

[Assistant]
To solve the problem of finding all integers \( n \) such that \( x^n + x - 1 \equiv 0 \pmod{x^2 - x + 1} \), we start by noting that we need \( x^n \equiv -x + 1 \pmod{x^2 - x + 1} \).

First, we reduce higher powers of \( x \) modulo \( x^2 - x + 1 \) using the relation \( x^2 \equiv x - 1 \pmod{x^2 - x + 1} \). We calculate the powers of \( x \) modulo \( x^2 - x + 1 \):

\[
\begin{aligned}
x^0 &\equiv 1 \pmod{x^2 - x + 1}, \\
x^1 &\equiv x \pmod{x^2 - x + 1}, \\
x^2 &\equiv x - 1 \pmod{x^2 - x + 1}, \\
x^3 &\equiv x \cdot x^2 \equiv x \cdot (x - 1) \equiv x^2 - x \equiv (x - 1) - x \equiv -1 \pmod{x^2 - x + 1}, \\
x^4 &\equiv x \cdot x^3 \equiv x \cdot (-1) \equiv -x \pmod{x^2 - x + 1}, \\
x^5 &\equiv x \cdot x^4 \equiv x \cdot (-x) \equiv -x^2 \equiv -(x - 1) \equiv -x + 1 \pmod{x^2 - x + 1}, \\
x^6 &\equiv x \cdot x^5 \equiv x \cdot (-x + 1) \equiv -x^2 + x \equiv -(x - 1) + x \equiv 1 \pmod{x^2 - x + 1}.
\end{aligned}
\]

We observe that the powers of \( x \) repeat every 6 steps. Therefore, \( x^n \equiv x^{n \mod 6} \pmod{x^2 - x + 1} \).

From the calculations, we see that \( x^5 \equiv -x + 1 \pmod{x^2 - x + 1} \). Substituting this into the original equation, we get:

\[
x^5 + x - 1 \equiv (-x + 1) + x - 1 \equiv 0 \pmod{x^2 - x + 1}.
\]

Thus, \( n \) must satisfy \( n \equiv 5 \pmod{6} \). The general solution is \( n = 6k + 5 \) for integers \( k \).

Therefore, the general solution for \( n \) is \(\boxed{n \equiv 5 \pmod{6}}\).

System: You are a helpful assistant.


[User] Compute the limit of the sequence $\{z_n\}$ where $z_n = i^{n!}$.

System: You are a helpful assistant.


[User] Given \( f(t) = t + \ln(t) \), find \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \). Provide exact values.

[Assistant]
To find \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \) for the function \( f(t) = t + \ln(t) \), we follow these steps:

1. **Verify that \( f \) is invertible:**
   - The domain of \( f \) is \( t > 0 \) because of the \( \ln(t) \) term.
   - Compute the derivative \( f'(t) = 1 + \frac{1}{t} \).
   - Since \( t > 0 \), \( \frac{1}{t} > 0 \), so \( f'(t) = 1 + \frac{1}{t} > 0 \) for all \( t > 0 \).
   - Therefore, \( f \) is strictly increasing and hence invertible.

2. **Find \( f^{-1}(1 + e) \):**
   - Let \( a = f^{-1}(1 + e) \). Then \( f(a) = 1 + e \).
   - This gives the equation \( a + \ln(a) = 1 + e \).
   - Testing \( a = e \):
     \[
     e + \ln(e) = e + 1 = 1 + e
     \]
   - Therefore, \( f^{-1}(1 + e) = e \).

3. **Compute \( f'(e) \):**
   - The derivative of \( f \) is \( f'(t) = 1 + \frac{1}{t} \).
   - Evaluate at \( t = e \):
     \[
     f'(e) = 1 + \frac{1}{e}
     \]

4. **Find \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \):**
   - Use the formula for the derivative of the inverse function:
     \[
     \frac{d(f^{-1})}{dt} = \frac{1}{f'(f^{-1}(t))}
     \]
   - Substitute \( t = 1 + e \) and \( f^{-1}(1 + e) = e \):
     \[
     \frac{d(f^{-1})}{dt} \bigg|_{t=1+e} = \frac{1}{f'(e)} = \frac{1}{1 + \frac{1}{e}}
     \]
   - Simplify the expression:
     \[
     \frac{1}{1 + \frac{1}{e}} = \frac{1}{\frac{e + 1}{e}} = \frac{e}{e + 1}
     \]

Thus, the derivative of the inverse function \( \frac{d(f^{-1})}{dt} \) at \( t = 1 + e \) is:
\[
\boxed{\frac{e}{e + 1}}
\]

System: You are a helpful assistant.


[User] Compute $\cos 210^\circ$.

[Assistant]
To compute \(\cos 210^\circ\), we follow these steps:

1. **Determine the Quadrant**: The angle \(210^\circ\) is in the third quadrant because it is between \(180^\circ\) and \(270^\circ\). In the third quadrant, the cosine values are negative.

2. **Find the Reference Angle**: The reference angle for \(210^\circ\) is calculated by subtracting \(180^\circ\) from \(210^\circ\):
   \[
   210^\circ - 180^\circ = 30^\circ
   \]

3. **Use the Reference Angle**: The reference angle \(30^\circ\) is a standard angle. We know from the unit circle that:
   \[
   \cos 30^\circ = \frac{\sqrt{3}}{2}
   \]

4. **Adjust for the Quadrant**: Since \(210^\circ\) is in the third quadrant and cosine values are negative in this quadrant, we take the negative of the cosine of the reference angle:
   \[
   \cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}
   \]

Thus, the value of \(\cos 210^\circ\) is:
\[
\boxed{-\frac{\sqrt{3}}{2}}
\]

System: You are a helpful assistant.


[User] What is the value of $T_1 + T_2 + \ldots + T_{15}$, where $T_n$ denotes the $n$th triangular number?

[Assistant]
To find the value of \( T_1 + T_2 + \ldots + T_{15} \), where \( T_n \) denotes the \( n \)-th triangular number, we use the formula for the sum of the first \( k \) triangular numbers, which is given by:

\[
\sum_{i=1}^{k} T_i = \frac{k(k + 1)(k + 2)}{6}
\]

For \( k = 15 \):

\[
\sum_{i=1}^{15} T_i = \frac{15 \cdot 16 \cdot 17}{6}
\]

We break down the calculation step-by-step:

1. Compute \( 15 \cdot 16 \):
   \[
   15 \cdot 16 = 240
   \]

2. Multiply the result by 17:
   \[
   240 \cdot 17 = 4080
   \]

3. Divide by 6:
   \[
   \frac{4080}{6} = 680
   \]

Thus, the value of \( T_1 + T_2 + \ldots + T_{15} \) is \(\boxed{680}\).

System: You are a helpful assistant.


[User] How many ways are there to distribute 28 identical apples to 4 people such that each person receives at least one apple?

[Assistant]
To determine the number of ways to distribute 28 identical apples to 4 people such that each person receives at least one apple, we can use the stars and bars method with an adjustment for the constraint that each person gets at least one apple.

1. **Adjust for the constraint**: Since each person must receive at least one apple, we first give each person one apple. This uses up \(4\) apples, leaving us with \(28 - 4 = 24\) apples to distribute freely among the 4 people.

2. **Formulate the problem**: The problem now reduces to finding the number of non-negative integer solutions to the equation \(x_1 + x_2 + x_3 + x_4 = 24\), where \(x_i\) represents the number of additional apples the \(i\)-th person receives.

3. **Apply the stars and bars formula**: The number of non-negative integer solutions to the equation \(x_1 + x_2 + x_3 + x_4 = 24\) is given by the combination formula \(\binom{n + k - 1}{k - 1}\), where \(n = 24\) (remaining apples) and \(k = 4\) (people).

\[
\binom{24 + 4 - 1}{4 - 1} = \binom{27}{3}
\]

4. **Calculate the combination**:

\[
\binom{27}{3} = \frac{27 \times 26 \times 25}{3 \times 2 \times 1} = \frac{17550}{6} = 2925
\]

Thus, the number of ways to distribute the 28 apples to 4 people such that each person receives at least one apple is \(\boxed{2925}\).

System: You are a helpful assistant.


[User] Calculate the sum $1 + 3 + 5 + \cdots + 15 + 17$.

[Assistant]
To calculate the sum of the sequence \(1 + 3 + 5 + \cdots + 15 + 17\), we can use the formula for the sum of an arithmetic sequence. The sequence is an arithmetic sequence where the first term \(a_1 = 1\), the last term \(a_n = 17\), and the common difference \(d = 2\).

First, we need to determine the number of terms in the sequence. The \(n\)-th term of an arithmetic sequence is given by:
\[
a_n = a_1 + (n-1)d
\]
Setting \(a_n = 17\), we solve for \(n\):
\[
17 = 1 + (n-1) \cdot 2
\]
\[
17 = 1 + 2n - 2
\]
\[
17 = 2n - 1
\]
\[
18 = 2n
\]
\[
n = 9
\]
So, there are 9 terms in the sequence.

Next, we use the sum formula for an arithmetic sequence:
\[
S = \frac{n}{2} \times (a_1 + a_n)
\]
Substituting the known values:
\[
S = \frac{9}{2} \times (1 + 17) = \frac{9}{2} \times 18 = 9 \times 9 = 81
\]

Alternatively, we can use the fact that the sum of the first \(n\) odd numbers is \(n^2\). Since there are 9 terms, which are the first 9 odd numbers:
\[
1 + 3 + 5 + \cdots + 17 = 9^2 = 81
\]

Thus, the sum of the sequence \(1 + 3 + 5 + \cdots + 15 + 17\) is:
\[
\boxed{81}
\]

System: You are a helpful assistant.


[User] An investment of $\$$10,000 is made in a government bond that will pay 6$\%$ interest compounded annually. At the end of five years, what is the total number of dollars in this investment? Express your answer to the nearest whole number.

[Assistant]
To determine the total amount of an investment of \$10,000 after 5 years with 6% annual compound interest, we use the compound interest formula:

\[ A = P (1 + r)^t \]

where:
- \( P \) is the principal amount (\$10,000),
- \( r \) is the annual interest rate (6%, or 0.06),
- \( t \) is the number of years (5).

First, we calculate the multiplier:

\[ (1 + r)^t = (1 + 0.06)^5 = 1.06^5 \]

Next, we compute \( 1.06^5 \):

\[ 1.06^1 = 1.06 \]
\[ 1.06^2 = 1.06 \times 1.06 = 1.1236 \]
\[ 1.06^3 = 1.1236 \times 1.06 = 1.191016 \]
\[ 1.06^4 = 1.191016 \times 1.06 = 1.26247696 \]
\[ 1.06^5 = 1.26247696 \times 1.06 = 1.3382255776 \]

Now, we multiply the principal by this multiplier:

\[ A = 10,000 \times 1.3382255776 = 13,382.255776 \]

Finally, we round the result to the nearest whole number:

\[ 13,382.255776 \approx 13,382 \]

Therefore, the total amount after 5 years is:

\[
\boxed{13382}
\]

System: You are a helpful assistant.


[User] Points A and B are on different sides of a line \( l \). The distance from point A to line \( l \) is 10 inches, and the distance from point B to line \( l \) is 4 inches. What is the distance between the midpoint O of segment AB and the line \( l \)?

[Assistant]
To find the distance between the midpoint \( O \) of segment \( AB \) and the line \( l \), we start by considering the vertical distances of points \( A \) and \( B \) from the line \( l \).

Given:
- The distance from point \( A \) to line \( l \) is 10 inches.
- The distance from point \( B \) to line \( l \) is 4 inches.
- Points \( A \) and \( B \) are on different sides of the line \( l \).

We can represent the vertical coordinates of points \( A \) and \( B \) relative to the line \( l \) as follows:
- Point \( A \) is 10 inches above \( l \), so its vertical coordinate is \( 10 \).
- Point \( B \) is 4 inches below \( l \), so its vertical coordinate is \( -4 \).

The midpoint \( O \) of segment \( AB \) has a vertical coordinate that is the average of the vertical coordinates of \( A \) and \( B \):
\[
\text{Vertical coordinate of } O = \frac{10 + (-4)}{2} = \frac{6}{2} = 3
\]

The distance from the midpoint \( O \) to the line \( l \) is the absolute value of this vertical coordinate:
\[
\text{Distance from } O \text{ to } l = |3| = 3 \text{ inches}
\]

Thus, the distance from the midpoint \( O \) of segment \( AB \) to the line \( l \) is \(\boxed{3}\) inches.

System: You are a helpful assistant.


[User] Two years ago there were 20 trailer homes on Elm Street with an average age of 18 years. At that time, a group of brand new trailer homes was then added to Elm Street. Today, the average age of all the trailer homes on Elm Street is 14 years. How many new trailer homes were added two years ago?

[Assistant]
To solve the problem, we need to determine how many new trailer homes were added to Elm Street two years ago.

First, let's establish the given information:
- Two years ago, there were 20 trailer homes with an average age of 18 years.
- At that time, a group of brand new trailer homes was added.
- Today, the average age of all trailer homes is 14 years.

Let \( x \) be the number of new trailer homes added two years ago.

### Step-by-Step Solution:

1. **Determine the total age of the original trailer homes two years ago:**
   - Two years ago, the average age of the 20 original trailer homes was 18 years.
   - Therefore, the total age of these 20 trailer homes two years ago was:
     \[
     20 \times 18 = 360 \text{ years}
     \]

2. **Calculate the current total age of the original trailer homes:**
   - Since two years have passed, each of the original 20 trailer homes has aged by 2 years.
   - Thus, the current total age of the original trailer homes is:
     \[
     360 + 20 \times 2 = 360 + 40 = 400 \text{ years}
     \]

3. **Calculate the current total age of the new trailer homes:**
   - The new trailer homes were added two years ago and are brand new at that time, so each is now 2 years old.
   - Therefore, the total age of the \( x \) new trailer homes is:
     \[
     x \times 2 = 2x \text{ years}
     \]

4. **Determine the combined total age of all trailer homes today:**
   - The total number of trailer homes today is \( 20 + x \).
   - The combined total age of all trailer homes today is:
     \[
     400 + 2x \text{ years}
     \]

5. **Set up the equation for the average age:**
   - The average age of all trailer homes today is 14 years.
   - Therefore, we can set up the equation:
     \[
     \frac{400 + 2x}{20 + x} = 14
     \]

6. **Solve the equation for \( x \):**
   - Multiply both sides by \( 20 + x \) to clear the denominator:
     \[
     400 + 2x = 14(20 + x)
     \]
   - Expand the right-hand side:
     \[
     400 + 2x = 280 + 14x
     \]
   - Subtract \( 2x \) from both sides:
     \[
     400 = 280 + 12x
     \]
   - Subtract 280 from both sides:
     \[
     120 = 12x
     \]
   - Divide both sides by 12:
     \[
     x = 10
     \]

Thus, the number of new trailer homes added two years ago is \(\boxed{10}\).

System: You are a helpful assistant.


[User] The sum of two numbers is $45$. Their difference is $3$. What is the lesser of the two numbers?

[Assistant]
To solve the problem where the sum of two numbers is 45 and their difference is 3, we can use a system of equations. Let the two numbers be \( x \) (the larger number) and \( y \) (the smaller number). We can set up the following system of equations based on the given information:

\[
\begin{cases}
x + y = 45 \\
x - y = 3
\end{cases}
\]

First, we add the two equations together to eliminate \( y \):

\[
(x + y) + (x - y) = 45 + 3
\]

This simplifies to:

\[
2x = 48
\]

Next, we solve for \( x \):

\[
x = \frac{48}{2} = 24
\]

Now that we have \( x = 24 \), we substitute this value back into the first equation to find \( y \):

\[
24 + y = 45
\]

Solving for \( y \):

\[
y = 45 - 24 = 21
\]

Thus, the two numbers are 24 and 21. The problem asks for the lesser of the two numbers, which is \( y \).

Therefore, the lesser of the two numbers is:

\[
\boxed{21}
\]

System: You are a helpful assistant.


[User] Morgan walks \( x \) yards north, then walks \( (2x-1) \) yards west. She is now \( (2x+1) \) yards from her starting point. What is the distance, in yards, she travels?

[Assistant]
To solve the problem, we start by noting that Morgan's path forms a right triangle. The northward distance is \( x \) yards, the westward distance is \( (2x - 1) \) yards, and the straight-line distance from her starting point is \( (2x + 1) \) yards. Using the Pythagorean theorem, we can set up the following equation:

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

First, we expand the terms:

\[
(2x - 1)^2 = 4x^2 - 4x + 1
\]
\[
(2x + 1)^2 = 4x^2 + 4x + 1
\]

Substituting these into the equation, we get:

\[
x^2 + 4x^2 - 4x + 1 = 4x^2 + 4x + 1
\]

Combining like terms on the left side:

\[
5x^2 - 4x + 1 = 4x^2 + 4x + 1
\]

Next, we move all terms to one side of the equation:

\[
5x^2 - 4x + 1 - 4x^2 - 4x - 1 = 0
\]

Simplifying:

\[
x^2 - 8x = 0
\]

Factoring out \( x \):

\[
x(x - 8) = 0
\]

This gives us two potential solutions: \( x = 0 \) or \( x = 8 \). Since \( x = 0 \) does not make sense in the context of the problem (as it would imply no movement), we have:

\[
x = 8
\]

Now, we calculate the total distance Morgan traveled. The northward distance is \( x = 8 \) yards, and the westward distance is \( 2x - 1 = 2(8) - 1 = 15 \) yards. Therefore, the total distance traveled is:

\[
x + (2x - 1) = 8 + 15 = 23 \text{ yards}
\]

Thus, the distance Morgan travels is:

\[
\boxed{23}
\]

System: You are a helpful assistant.


[User]
Find the critical points of the system of differential equations given by:
\[ x' = y(x^2 + 1) \]
\[ y' = 2xy^2 \]

[Assistant]
To find the critical points of the system of differential equations given by:
\[ x' = y(x^2 + 1) \]
\[ y' = 2xy^2 \]

we need to determine the points \((x, y)\) where both \(x'\) and \(y'\) are zero. This involves solving the system of equations:
\[ y(x^2 + 1) = 0 \]
\[ 2xy^2 = 0 \]

### Step-by-Step Solution:

1. **Solve the first equation \( y(x^2 + 1) = 0 \):**
   - The product \( y(x^2 + 1) \) is zero if either \( y = 0 \) or \( x^2 + 1 = 0 \).
   - Since \( x^2 + 1 \) is always positive for all real \( x \), the only solution is \( y = 0 \).

2. **Substitute \( y = 0 \) into the second equation \( 2xy^2 = 0 \):**
   - When \( y = 0 \), the equation becomes \( 2x(0)^2 = 0 \), which simplifies to \( 0 = 0 \).
   - This equation holds true for any value of \( x \).

### Conclusion:
- The first equation \( y(x^2 + 1) = 0 \) implies \( y = 0 \).
- Substituting \( y = 0 \) into the second equation \( 2xy^2 = 0 \) shows that the equation is satisfied for any \( x \).

Therefore, the critical points of the system are all points of the form \((x, 0)\) where \( x \) can be any real number.

\[
\boxed{(x, 0) \text{ for all } x \in \mathbb{R}}
\]

System: You are a helpful assistant.


[User] When the square root of $x$ is cubed, the answer is 64. What is the value of $x$?

[Assistant]
To solve the problem, we start with the equation given in the problem statement:

\[
(\sqrt{x})^3 = 64
\]

We can rewrite the square root as an exponent:

\[
(x^{1/2})^3 = 64
\]

Using the property of exponents \((a^m)^n = a^{mn}\), we simplify the left side:

\[
x^{3/2} = 64
\]

To isolate \(x\), we raise both sides of the equation to the power of \(\frac{2}{3}\):

\[
(x^{3/2})^{2/3} = 64^{2/3}
\]

Simplifying the left side, we get:

\[
x = 64^{2/3}
\]

Next, we need to compute \(64^{2/3}\). We can break this down into two steps:

1. Compute the cube root of 64:

\[
64^{1/3} = 4
\]

2. Square the result:

\[
4^2 = 16
\]

Thus, we have:

\[
x = 16
\]

To verify, we substitute \(x = 16\) back into the original equation:

\[
(\sqrt{16})^3 = 4^3 = 64
\]

This confirms that our solution is correct. Therefore, the value of \(x\) is:

\[
\boxed{16}
\]

System: You are a helpful assistant.


[User] Express $2^{6}(2^{7}-1)$ as the sum of powers of 2.

[Assistant]
To express \(2^6(2^7 - 1)\) as the sum of powers of 2, we start by simplifying the expression.

1. **Expand the expression:**
   \[
   2^6(2^7 - 1) = 2^6 \cdot 2^7 - 2^6 \cdot 1
   \]
   Using the property of exponents \(a^m \cdot a^n = a^{m+n}\), we get:
   \[
   2^6 \cdot 2^7 = 2^{6+7} = 2^{13}
   \]
   Therefore:
   \[
   2^6(2^7 - 1) = 2^{13} - 2^6
   \]

2. **Express \(2^{13} - 2^6\) as a sum of powers of 2:**
   Notice that \(2^{13} - 2^6\) can be written as the sum of consecutive powers of 2 from \(2^6\) to \(2^{12}\). This is because:
   \[
   2^{13} - 2^6 = 2^6 + 2^7 + 2^8 + 2^9 + 2^{10} + 2^{11} + 2^{12}
   \]
   This follows from the geometric series sum formula, where the sum of \(2^k\) from \(k = 6\) to \(k = 12\) is:
   \[
   \sum_{k=6}^{12} 2^k = 2^6 + 2^7 + 2^8 + 2^9 + 2^{10} + 2^{11} + 2^{12}
   \]

3. **Final answer:**
   The expression \(2^6(2^7 - 1)\) can be written as:
   \[
   \boxed{2^6 + 2^7 + 2^8 + 2^9 + 2^{10} + 2^{11} + 2^{12}}
   \]

System: You are a helpful assistant.


[User] Three bags of jelly beans contain 26, 28, and 30 beans. The ratios of yellow beans to all beans in each of these bags are 50%, 25%, and 20%, respectively. All three bags of candy are dumped into one bowl. What is the closest ratio of yellow jelly beans to all beans in the bowl?

[Assistant]
To determine the ratio of yellow jelly beans to all beans in the bowl, we first calculate the number of yellow beans in each bag:

1. **First bag**: 50% of 26 beans:
   \[
   0.5 \times 26 = 13 \text{ yellow beans}
   \]

2. **Second bag**: 25% of 28 beans:
   \[
   0.25 \times 28 = 7 \text{ yellow beans}
   \]

3. **Third bag**: 20% of 30 beans:
   \[
   0.2 \times 30 = 6 \text{ yellow beans}
   \]

Adding these together gives the total number of yellow beans:
\[
13 + 7 + 6 = 26 \text{ yellow beans}
\]

Next, we calculate the total number of beans in the bowl:
\[
26 + 28 + 30 = 84 \text{ total beans}
\]

The ratio of yellow beans to the total number of beans is:
\[
\frac{26}{84}
\]

Simplifying this fraction by dividing both the numerator and the denominator by their greatest common divisor (2):
\[
\frac{26 \div 2}{84 \div 2} = \frac{13}{42}
\]

Thus, the closest ratio of yellow jelly beans to all beans in the bowl is:
\[
\boxed{\frac{13}{42}}
\]

System: You are a helpful assistant.


[User] The area of a rectangle is 432 square centimeters. What would the new area be if the length of the rectangle was decreased by $10\%$ and the width of the rectangle was increased by $10\%$? Express your answer to the nearest whole number.

[Assistant]
To solve the problem, we start by noting that the area of the original rectangle is given by:

\[ A = L \times W = 432 \text{ square centimeters} \]

where \( L \) is the length and \( W \) is the width of the rectangle.

According to the problem, the length is decreased by 10%, and the width is increased by 10%. Therefore, the new length and new width are:

\[ \text{New length} = 0.9L \]
\[ \text{New width} = 1.1W \]

The new area of the rectangle can be calculated by multiplying the new length and the new width:

\[ \text{New area} = (0.9L) \times (1.1W) \]

We can simplify this expression:

\[ \text{New area} = 0.9 \times 1.1 \times L \times W \]

Since \( L \times W = 432 \):

\[ \text{New area} = 0.9 \times 1.1 \times 432 \]

Next, we calculate the product of 0.9 and 1.1:

\[ 0.9 \times 1.1 = 0.99 \]

So the new area is:

\[ \text{New area} = 0.99 \times 432 \]

Now, we perform the multiplication:

\[ 0.99 \times 432 = 427.68 \]

Finally, we round 427.68 to the nearest whole number:

\[ 427.68 \approx 428 \]

Therefore, the new area of the rectangle, rounded to the nearest whole number, is:

\[
\boxed{428}
\]

System: You are a helpful assistant.


[User] Suppose \(\gcd(p, q) = 1\), \(\gcd(p, 10) = 1\), and \(\gcd(q, 10) = 1\). If the decimal expansion of \(1/p\) has period \(r\) and the decimal expansion of \(1/q\) has period \(s\), find the period of the decimal expansion of \(1/(pq)\).

[Assistant]
To find the period of the decimal expansion of \( \frac{1}{pq} \) given that \( \gcd(p, q) = 1 \), \( \gcd(p, 10) = 1 \), and \( \gcd(q, 10) = 1 \), and knowing that the periods of \( \frac{1}{p} \) and \( \frac{1}{q} \) are \( r \) and \( s \) respectively, we proceed as follows:

1. **Understanding the Period of Decimal Expansion**:
   - The period of the decimal expansion of \( \frac{1}{n} \) when \( n \) is coprime with 10 is the smallest positive integer \( k \) such that \( 10^k \equiv 1 \pmod{n} \). This \( k \) is the multiplicative order of 10 modulo \( n \).

2. **Periods of \( \frac{1}{p} \) and \( \frac{1}{q} \)**:
   - The period of \( \frac{1}{p} \) is \( r \), meaning \( 10^r \equiv 1 \pmod{p} \).
   - The period of \( \frac{1}{q} \) is \( s \), meaning \( 10^s \equiv 1 \pmod{q} \).

3. **Multiplicative Order Modulo \( pq \)**:
   - Since \( p \) and \( q \) are coprime, by the Chinese Remainder Theorem, the multiplicative order of 10 modulo \( pq \) is the least common multiple (LCM) of the orders modulo \( p \) and modulo \( q \).
   - Therefore, the smallest \( k \) such that \( 10^k \equiv 1 \pmod{pq} \) is \( \text{lcm}(r, s) \).

4. **Conclusion**:
   - The period of the decimal expansion of \( \frac{1}{pq} \) is the least common multiple of the periods of \( \frac{1}{p} \) and \( \frac{1}{q} \).

Thus, the period of the decimal expansion of \( \frac{1}{pq} \) is \( \boxed{\text{lcm}(r, s)} \).

System: You are a helpful assistant.


[User] In order to compute the area of a particular circle, Juan first measures the length of its diameter. The actual diameter is 20 cm, but Juan's measurement has an error of up to $20\%$. What is the largest possible percent error, in percent, in Juan's computed area of the circle?

[Assistant]
To determine the largest possible percent error in Juan's computed area of the circle, we start by recalling the formula for the area of a circle, which is \( A = \pi r^2 \). Given that the diameter \( d \) is twice the radius, we can express the area in terms of the diameter as \( A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \).

The actual diameter of the circle is 20 cm. Juan's measurement of the diameter has an error of up to 20%. Therefore, the measured diameter \( d' \) can be either 20% more or 20% less than the actual diameter. This gives us two possible measured diameters:
\[ d' = 20 \times 1.2 = 24 \text{ cm} \]
\[ d' = 20 \times 0.8 = 16 \text{ cm} \]

Next, we calculate the area using these measured diameters. The actual area of the circle is:
\[ A_{\text{actual}} = \frac{\pi (20)^2}{4} = 100\pi \text{ cm}^2 \]

For the measured diameter of 24 cm, the computed area is:
\[ A_{24} = \frac{\pi (24)^2}{4} = 144\pi \text{ cm}^2 \]

For the measured diameter of 16 cm, the computed area is:
\[ A_{16} = \frac{\pi (16)^2}{4} = 64\pi \text{ cm}^2 \]

Now, we calculate the percent error for each case. The percent error is given by:
\[ \text{Percent Error} = \left| \frac{A_{\text{computed}} - A_{\text{actual}}}{A_{\text{actual}}} \right| \times 100\% \]

For the measured diameter of 24 cm:
\[ \text{Percent Error}_{24} = \left| \frac{144\pi - 100\pi}{100\pi} \right| \times 100\% = \left| \frac{44\pi}{100\pi} \right| \times 100\% = 44\% \]

For the measured diameter of 16 cm:
\[ \text{Percent Error}_{16} = \left| \frac{64\pi - 100\pi}{100\pi} \right| \times 100\% = \left| \frac{-36\pi}{100\pi} \right| \times 100\% = 36\% \]

The largest possible percent error is the maximum of these two values, which is 44%.

Therefore, the largest possible percent error in Juan's computed area is \(\boxed{44\%}\).

System: You are a helpful assistant.


[User] Calculate \( \frac{d}{d \epsilon } \left( \| \dot{ \mathbf{x} } ^{ \epsilon } \| ^{2} \right) \) where \( \dot{ \mathbf{x} } ^{ \epsilon } = \frac{ \mathbf{x}(t) + \epsilon \mathbf{z} (t) }{ \| \mathbf{x} (t) + \epsilon \mathbf{z} (t) \| } \) at the point where \( \epsilon = 0 \).

[Assistant]
To calculate the derivative \( \frac{d}{d \epsilon} \left( \| \dot{\mathbf{x}}^\epsilon \|^2 \right) \) at \( \epsilon = 0 \), we start with the given vector \( \dot{\mathbf{x}}^\epsilon \):

\[
\dot{\mathbf{x}}^\epsilon = \frac{\mathbf{x}(t) + \epsilon \mathbf{z}(t)}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|}
\]

The norm squared of \( \dot{\mathbf{x}}^\epsilon \) is:

\[
\| \dot{\mathbf{x}}^\epsilon \|^2 = \left( \frac{\mathbf{x}(t) + \epsilon \mathbf{z}(t)}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|} \right) \cdot \left( \frac{\mathbf{x}(t) + \epsilon \mathbf{z}(t)}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|} \right)
\]

Simplifying the expression, we get:

\[
\| \dot{\mathbf{x}}^\epsilon \|^2 = \frac{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|^2}{\| \mathbf{x}(t) + \epsilon \mathbf{z}(t) \|^2} = 1
\]

Since the norm squared of \( \dot{\mathbf{x}}^\epsilon \) is always 1 for all \( \epsilon \) (as long as \( \mathbf{x}(t) + \epsilon \mathbf{z}(t) \neq \mathbf{0} \)), the function \( \| \dot{\mathbf{x}}^\epsilon \|^2 \) is a constant function. The derivative of a constant function with respect to \( \epsilon \) is zero. Therefore, evaluating the derivative at \( \epsilon = 0 \) gives:

\[
\frac{d}{d \epsilon} \left( \| \dot{\mathbf{x}}^\epsilon \|^2 \right) \bigg|_{\epsilon = 0} = 0
\]

Thus, the final answer is:

\[
\boxed{0}
\]

System: You are a helpful assistant.


[User] Evaluate the limit $\lim_{x \to u} \frac{\sin 6x - \sin 6u}{2x - 2u}$.

[Assistant]
To evaluate the limit \(\lim_{x \to u} \frac{\sin 6x - \sin 6u}{2x - 2u}\), we can use the definition of the derivative or trigonometric identities. Here, we will present both approaches in a clean and detailed manner.

### Approach 1: Using the Definition of the Derivative

1. Rewrite the expression:
   \[
   \lim_{x \to u} \frac{\sin 6x - \sin 6u}{2x - 2u} = \frac{1}{2} \lim_{x \to u} \frac{\sin 6x - \sin 6u}{x - u}.
   \]

2. Recognize that \(\frac{\sin 6x - \sin 6u}{x - u}\) is the definition of the derivative of \(\sin 6x\) evaluated at \(x = u\):
   \[
   \frac{d}{dx} (\sin 6x) = 6 \cos 6x.
   \]

3. Evaluate the derivative at \(x = u\):
   \[
   \frac{d}{dx} (\sin 6x) \bigg|_{x = u} = 6 \cos 6u.
   \]

4. Substitute back into the limit:
   \[
   \frac{1}{2} \cdot 6 \cos 6u = 3 \cos 6u.
   \]

Thus, the limit is:
\[
\boxed{3 \cos 6u}.
\]

### Approach 2: Using Trigonometric Identities

1. Use the trigonometric identity for the difference of sines:
   \[
   \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right).
   \]
   Here, \(A = 6x\) and \(B = 6u\):
   \[
   \sin 6x - \sin 6u = 2 \cos \left( \frac{6x + 6u}{2} \right) \sin \left( \frac{6x - 6u}{2} \right).
   \]

2. Simplify the arguments:
   \[
   \sin 6x - \sin 6u = 2 \cos (3(x + u)) \sin (3(x - u)).
   \]

3. Substitute back into the limit:
   \[
   \lim_{x \to u} \frac{2 \cos (3(x + u)) \sin (3(x - u))}{2(x - u)} = \lim_{x \to u} \cos (3(x + u)) \cdot \frac{\sin (3(x - u))}{x - u}.
   \]

4. Recognize that as \(x \to u\), \(3(x - u) \to 0\), and use the limit property \(\lim_{h \to 0} \frac{\sin h}{h} = 1\):
   \[
   \frac{\sin (3(x - u))}{x - u} = 3 \cdot \frac{\sin (3(x - u))}{3(x - u)} \to 3 \cdot 1 = 3.
   \]

5. Substitute back into the limit:
   \[
   \cos (3(u + u)) \cdot 3 = \cos (6u) \cdot 3 = 3 \cos 6u.
   \]

Thus, the limit is:
\[
\boxed{3 \cos 6u}.
\]

System: You are a helpful assistant.


[User] Evaluate the integral \(\int \frac{3x-6}{\sqrt{x^2-4x+11}} \, dx\).

[Assistant]
To evaluate the integral \(\int \frac{3x-6}{\sqrt{x^2-4x+11}} \, dx\), we start by simplifying the expression under the square root in the denominator. We complete the square for \(x^2 - 4x + 11\):

\[
x^2 - 4x + 11 = (x-2)^2 + 7.
\]

Thus, the integral becomes:

\[
\int \frac{3x-6}{\sqrt{(x-2)^2 + 7}} \, dx.
\]

Next, we factor the numerator \(3x - 6\) as \(3(x-2)\):

\[
\int \frac{3(x-2)}{\sqrt{(x-2)^2 + 7}} \, dx.
\]

We use the substitution \(u = (x-2)^2 + 7\). Then, we compute the differential \(du\):

\[
du = 2(x-2) \, dx \implies (x-2) \, dx = \frac{1}{2} du.
\]

Substituting \(u\) and \(du\) into the integral, we get:

\[
\int \frac{3(x-2)}{\sqrt{(x-2)^2 + 7}} \, dx = 3 \int \frac{(x-2)}{\sqrt{u}} \, dx = 3 \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{3}{2} \int u^{-1/2} \, du.
\]

The integral of \(u^{-1/2}\) is \(2u^{1/2}\):

\[
\frac{3}{2} \int u^{-1/2} \, du = \frac{3}{2} \cdot 2u^{1/2} + C = 3u^{1/2} + C.
\]

Substituting back \(u = (x-2)^2 + 7\), we obtain:

\[
3u^{1/2} + C = 3\sqrt{(x-2)^2 + 7} + C.
\]

Since \((x-2)^2 + 7 = x^2 - 4x + 11\), we can write the final answer as:

\[
3\sqrt{x^2 - 4x + 11} + C.
\]

Thus, the final answer is:

\[
\boxed{3\sqrt{x^2 - 4x + 11} + C}.
\]

System: You are a helpful assistant.


[User] Given parallelogram ABCD, with AB = 10, AD = 12, and BD = $2\sqrt{13}$, find AC.

[Assistant]
To find the length of \( AC \) in parallelogram \( ABCD \) where \( AB = 10 \), \( AD = 12 \), and \( BD = 2\sqrt{13} \), we can use the property of parallelograms that states the sum of the squares of the diagonals is equal to the sum of the squares of all the sides. The formula is:

\[
AC^2 + BD^2 = 2(AB^2 + AD^2)
\]

First, we compute \( BD^2 \):

\[
BD = 2\sqrt{13} \implies BD^2 = (2\sqrt{13})^2 = 4 \times 13 = 52
\]

Next, we compute \( AB^2 \) and \( AD^2 \):

\[
AB = 10 \implies AB^2 = 10^2 = 100
\]
\[
AD = 12 \implies AD^2 = 12^2 = 144
\]

Now, we substitute these values into the formula:

\[
AC^2 + 52 = 2(100 + 144)
\]

Calculate the sum inside the parentheses:

\[
100 + 144 = 244
\]

Then multiply by 2:

\[
2 \times 244 = 488
\]

So the equation becomes:

\[
AC^2 + 52 = 488
\]

Subtract 52 from both sides to isolate \( AC^2 \):

\[
AC^2 = 488 - 52 = 436
\]

Finally, take the square root of both sides to find \( AC \):

\[
AC = \sqrt{436}
\]

Factor 436 to simplify the square root:

\[
436 = 4 \times 109 \implies \sqrt{436} = \sqrt{4 \times 109} = 2\sqrt{109}
\]

Therefore, the length of \( AC \) is:

\[
\boxed{2\sqrt{109}}
\]

System: You are a helpful assistant.


[User] If the sides of a cyclic quadrilateral are the roots of the polynomial \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), express the sum of the sides \(S\) in terms of \(p\), \(q\), \(s\), and \(t\).

[Assistant]
To find the sum of the sides \(S\) of a cyclic quadrilateral whose sides are the roots of the polynomial \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), we can use Vieta's formulas.

Given the polynomial \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), let the roots be \(a\), \(b\), \(c\), and \(d\). According to Vieta's formulas, the sum of the roots of the polynomial is given by the coefficient of the \(x^3\) term with a sign change. Specifically, for the polynomial \(x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0\), the sum of the roots is \(-a_3\).

In our case, the polynomial is \(x^4 - 2sx^3 + tx^2 - qx + 2p = 0\), so the coefficient of \(x^3\) is \(-2s\). Therefore, the sum of the roots \(a + b + c + d\) is:

\[
a + b + c + d = -(-2s) = 2s
\]

Thus, the sum of the sides \(S\) of the cyclic quadrilateral is:

\[
\boxed{2s}
\]

System: You are a helpful assistant.


[User] How many multiples of 10 are between 9 and 101?

[Assistant]
To determine how many multiples of 10 are between 9 and 101, we first identify the smallest and largest multiples of 10 within this range.

The smallest multiple of 10 greater than 9 is 10.
The largest multiple of 10 less than or equal to 101 is 100.

The multiples of 10 in this range form an arithmetic sequence: 10, 20, 30, ..., 100.

The formula for the number of terms in an arithmetic sequence is given by:
\[
\text{Number of terms} = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1
\]

Here, the first term (\(a\)) is 10, the last term (\(l\)) is 100, and the common difference (\(d\)) is 10.

Plugging in these values:
\[
\text{Number of terms} = \frac{100 - 10}{10} + 1 = \frac{90}{10} + 1 = 9 + 1 = 10
\]

Thus, there are 10 multiples of 10 between 9 and 101.

\[
\boxed{10}
\]

System: You are a helpful assistant.


[User] If $\sqrt{5 + x} + \sqrt{20 - x} = 7$, what is the value of $(5 + x)(20 - x)$?

[Assistant]
To solve the problem, we start with the given equation:

\[
\sqrt{5 + x} + \sqrt{20 - x} = 7
\]

Let \( A = \sqrt{5 + x} \) and \( B = \sqrt{20 - x} \). Therefore, the equation becomes:

\[
A + B = 7
\]

We are asked to find the value of \( (5 + x)(20 - x) \). Notice that:

\[
(5 + x) = A^2 \quad \text{and} \quad (20 - x) = B^2
\]

Thus, we need to find \( A^2 B^2 \), which is \( (AB)^2 \).

First, square both sides of the equation \( A + B = 7 \):

\[
(A + B)^2 = 7^2
\]

This expands to:

\[
A^2 + 2AB + B^2 = 49
\]

We know that:

\[
A^2 + B^2 = (5 + x) + (20 - x) = 25
\]

Substitute \( A^2 + B^2 = 25 \) into the expanded equation:

\[
25 + 2AB = 49
\]

Solve for \( 2AB \):

\[
2AB = 49 - 25 = 24
\]

Thus,

\[
AB = \frac{24}{2} = 12
\]

Now, we need to find \( (AB)^2 \):

\[
(AB)^2 = 12^2 = 144
\]

Therefore, the value of \( (5 + x)(20 - x) \) is:

\[
\boxed{144}
\]

System: You are a helpful assistant.


[User] A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.5 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?

[Assistant]
To solve the problem, we need to determine the rate at which the angle between the ladder and the ground is changing when the bottom of the ladder is 8 feet from the wall. Here are the detailed steps:

1. **Define Variables and Given Information:**
   - Let \( x \) be the distance from the bottom of the ladder to the wall.
   - Let \( \theta \) be the angle between the ladder and the ground.
   - The ladder has a constant length of 10 feet.
   - The bottom of the ladder slides away from the wall at a rate of \( \frac{dx}{dt} = 1.5 \) ft/s.
   - We need to find \( \frac{d\theta}{dt} \) when \( x = 8 \) ft.

2. **Relate \( x \) and \( \theta \):**
   - In the right triangle formed by the ladder, wall, and ground, we have:
     \[
     \cos \theta = \frac{x}{10}
     \]

3. **Differentiate Both Sides with Respect to Time \( t \):**
   - Using the chain rule:
     \[
     \frac{d}{dt} (\cos \theta) = \frac{d}{dt} \left( \frac{x}{10} \right)
     \]
   - The left side becomes:
     \[
     -\sin \theta \cdot \frac{d\theta}{dt}
     \]
   - The right side becomes:
     \[
     \frac{1}{10} \cdot \frac{dx}{dt}
     \]
   - Therefore:
     \[
     -\sin \theta \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot \frac{dx}{dt}
     \]

4. **Solve for \( \frac{d\theta}{dt} \):**
   - Rearrange the equation to isolate \( \frac{d\theta}{dt} \):
     \[
     \frac{d\theta}{dt} = -\frac{1}{10 \sin \theta} \cdot \frac{dx}{dt}
     \]

5. **Find \( \sin \theta \) When \( x = 8 \) ft:**
   - When \( x = 8 \) ft, the other leg of the triangle (the height) is:
     \[
     \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \text{ ft}
     \]
   - Therefore:
     \[
     \sin \theta = \frac{6}{10} = \frac{3}{5}
     \]

6. **Substitute the Known Values:**
   - Substitute \( \sin \theta = \frac{3}{5} \) and \( \frac{dx}{dt} = 1.5 \) ft/s into the equation:
     \[
     \frac{d\theta}{dt} = -\frac{1}{10 \cdot \frac{3}{5}} \cdot 1.5 = -\frac{1}{6} \cdot 1.5 = -\frac{1.5}{6} = -\frac{3}{20} \text{ rad/s}
     \]

Therefore, the rate at which the angle between the ladder and the ground is changing when the bottom of the ladder is 8 feet from the wall is \(\boxed{-\frac{3}{20} \text{ rad/s}}\).

System: You are a helpful assistant.


[User] How many pages are there in a book if it takes 852 digits to number the pages consecutively?

[Assistant]
To determine the number of pages in a book given that it takes 852 digits to number the pages consecutively, we can break down the problem into different ranges of page numbers and calculate the number of digits used in each range.

1. **Pages 1 to 9:**
   - There are 9 pages.
   - Each page uses 1 digit.
   - Total digits used: \( 9 \times 1 = 9 \) digits.

2. **Pages 10 to 99:**
   - There are \( 99 - 10 + 1 = 90 \) pages.
   - Each page uses 2 digits.
   - Total digits used: \( 90 \times 2 = 180 \) digits.

3. **Total digits used for pages 1 to 99:**
   - Sum of digits from 1 to 9 and 10 to 99: \( 9 + 180 = 189 \) digits.

4. **Remaining digits:**
   - Total digits given: 852.
   - Digits used for pages 1 to 99: 189.
   - Remaining digits: \( 852 - 189 = 663 \) digits.

5. **Pages 100 onwards:**
   - Each page uses 3 digits.
   - Number of pages: \( \frac{663}{3} = 221 \) pages.

6. **Total number of pages:**
   - Pages from 1 to 99: 99 pages.
   - Pages from 100 onwards: 221 pages.
   - Total pages: \( 99 + 221 = 320 \) pages.

Thus, the number of pages in the book is \(\boxed{320}\).

System: You are a helpful assistant.


[User] Margaret and her younger sister Wylie are both between 10 and 20 years old. The sum of their ages has a units digit of 6, and the difference between their ages is 2. If Wylie's age is an even number, how old is Margaret?

[Assistant]
To solve the problem, we need to determine Margaret's age given the following conditions:
1. Both Margaret (M) and Wylie (W) are between 10 and 20 years old.
2. The sum of their ages has a units digit of 6.
3. The difference between their ages is 2.
4. Wylie's age is an even number.

Let's denote Margaret's age as \( M \) and Wylie's age as \( W \). From the conditions, we have:
\[ M = W + 2 \]
\[ 10 \leq W < M \leq 20 \]
\[ M + W \equiv 6 \pmod{10} \]
\[ W \text{ is even} \]

First, substitute \( M = W + 2 \) into the sum condition:
\[ M + W = (W + 2) + W = 2W + 2 \]
We need \( 2W + 2 \equiv 6 \pmod{10} \):
\[ 2W + 2 \equiv 6 \pmod{10} \]
Subtract 2 from both sides:
\[ 2W \equiv 4 \pmod{10} \]
Divide both sides by 2:
\[ W \equiv 2 \pmod{5} \]

Since \( W \) is an even number between 10 and 20, the possible values for \( W \) are 10, 12, 14, 16, and 18. We need to find which of these values satisfy \( W \equiv 2 \pmod{5} \):

- \( 10 \div 5 = 2 \) remainder 0, so \( 10 \not\equiv 2 \pmod{5} \)
- \( 12 \div 5 = 2 \) remainder 2, so \( 12 \equiv 2 \pmod{5} \)
- \( 14 \div 5 = 2 \) remainder 4, so \( 14 \not\equiv 2 \pmod{5} \)
- \( 16 \div 5 = 3 \) remainder 1, so \( 16 \not\equiv 2 \pmod{5} \)
- \( 18 \div 5 = 3 \) remainder 3, so \( 18 \not\equiv 2 \pmod{5} \)

Thus, the only value for \( W \) that satisfies all conditions is \( W = 12 \).

Now, calculate Margaret's age:
\[ M = W + 2 = 12 + 2 = 14 \]

Verify the conditions:
- The sum of their ages: \( M + W = 14 + 12 = 26 \), which has a units digit of 6.
- The difference between their ages: \( M - W = 14 - 12 = 2 \).
- Wylie's age is even (12).

Therefore, the correct age for Margaret is:
\[
\boxed{14}
\]

System: You are a helpful assistant.


[User] A bowling ball is a solid ball with a spherical surface and diameter 30 cm. To custom fit a bowling ball for each bowler, three holes are drilled in the ball. Bowler Kris has holes drilled that are 8 cm deep and have diameters of 2 cm, 2 cm, and 3 cm. Assuming the three holes are right circular cylinders, find the number of cubic centimeters in the volume of the fitted bowling ball. Express your answer in terms of $\pi$.

[Assistant]
To find the volume of the fitted bowling ball after the holes are drilled, we need to calculate the volume of the original sphere and then subtract the volumes of the three cylindrical holes.

1. **Volume of the original sphere:**
   The diameter of the sphere is 30 cm, so the radius \( r \) is 15 cm. The volume \( V \) of a sphere is given by the formula:
   \[
   V = \frac{4}{3} \pi r^3
   \]
   Substituting \( r = 15 \) cm:
   \[
   V = \frac{4}{3} \pi (15)^3 = \frac{4}{3} \pi (3375) = 4500 \pi \text{ cm}^3
   \]

2. **Volume of the cylindrical holes:**
   - The first two holes have a diameter of 2 cm, so the radius is 1 cm, and the depth is 8 cm. The volume \( V \) of a cylinder is given by the formula:
     \[
     V = \pi r^2 h
     \]
     For each of these holes:
     \[
     V = \pi (1)^2 (8) = 8 \pi \text{ cm}^3
     \]
     Since there are two such holes, the total volume for these two holes is:
     \[
     2 \times 8 \pi = 16 \pi \text{ cm}^3
     \]

   - The third hole has a diameter of 3 cm, so the radius is 1.5 cm, and the depth is 8 cm. The volume of this hole is:
     \[
     V = \pi (1.5)^2 (8) = \pi (2.25)(8) = 18 \pi \text{ cm}^3
     \]

3. **Total volume of the holes:**
   Summing the volumes of all three holes:
   \[
   16 \pi + 18 \pi = 34 \pi \text{ cm}^3
   \]

4. **Volume of the fitted bowling ball:**
   Subtract the total volume of the holes from the volume of the original sphere:
   \[
   4500 \pi - 34 \pi = 4466 \pi \text{ cm}^3
   \]

Thus, the volume of the fitted bowling ball is:
\[
\boxed{4466\pi}
\]

System: You are a helpful assistant.