[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] Evaluate $\log_{\sqrt{6}} (216\sqrt{6})$.

[Assistant]
To evaluate \(\log_{\sqrt{6}} (216\sqrt{6})\), we start by expressing the base and the argument in terms of base 6.

First, note that \(\sqrt{6} = 6^{1/2}\). Therefore, the logarithmic expression becomes:
\[
\log_{6^{1/2}} (216 \sqrt{6})
\]

Next, we express 216 and \(\sqrt{6}\) in terms of base 6. We know that:
\[
216 = 6^3 \quad \text{and} \quad \sqrt{6} = 6^{1/2}
\]
Thus, the argument \(216 \sqrt{6}\) can be written as:
\[
216 \sqrt{6} = 6^3 \cdot 6^{1/2} = 6^{3 + 1/2} = 6^{7/2}
\]

Now, the logarithmic expression is:
\[
\log_{6^{1/2}} (6^{7/2})
\]

Using the property of logarithms \(\log_b (b^y) = y\), we can simplify this expression. Here, the base is \(6^{1/2}\) and the argument is \(6^{7/2}\). Let \(x\) be the value of the logarithm:
\[
(6^{1/2})^x = 6^{7/2}
\]

This implies:
\[
6^{x/2} = 6^{7/2}
\]

Since the bases are the same, we can equate the exponents:
\[
\frac{x}{2} = \frac{7}{2}
\]

Solving for \(x\), we multiply both sides by 2:
\[
x = 7
\]

Therefore, the value of the logarithm is:
\[
\boxed{7}
\]

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] If \( \log_6 x = 2.5 \), what is the value of \( x \)?

[Assistant]
To solve the equation \( \log_6 x = 2.5 \), we start by converting it to its exponential form. The equation \( \log_6 x = 2.5 \) implies that \( x = 6^{2.5} \).

Next, we recognize that \( 2.5 \) can be written as \( \frac{5}{2} \). Therefore, we have:
\[
x = 6^{5/2}
\]

We can rewrite \( 6^{5/2} \) as:
\[
6^{5/2} = \left(6^{1/2}\right)^5 = \left(\sqrt{6}\right)^5
\]

Alternatively, we can express \( 6^{5/2} \) as:
\[
6^{5/2} = \sqrt{6^5}
\]

Now, we compute \( 6^5 \):
\[
6^5 = 6 \times 6 \times 6 \times 6 \times 6 = 7776
\]

Thus,
\[
6^{5/2} = \sqrt{7776}
\]

We simplify \( \sqrt{7776} \) by factoring it:
\[
7776 = 36 \times 216
\]

Since \( 36 = 6^2 \) and \( 216 = 6^3 \), we have:
\[
\sqrt{7776} = \sqrt{36 \times 216} = \sqrt{36} \times \sqrt{216} = 6 \times \sqrt{216}
\]

We further simplify \( \sqrt{216} \):
\[
216 = 6^3 \implies \sqrt{216} = \sqrt{6^3} = 6^{3/2} = 6 \times \sqrt{6}
\]

Therefore,
\[
\sqrt{7776} = 6 \times 6 \times \sqrt{6} = 36 \sqrt{6}
\]

So, the value of \( x \) is:
\[
x = 36 \sqrt{6}
\]

To verify, we check:
\[
\log_6 (36 \sqrt{6}) = \log_6 (36) + \log_6 (\sqrt{6}) = \log_6 (6^2) + \log_6 (6^{1/2}) = 2 + \frac{1}{2} = 2.5
\]

Thus, the solution is correct. The value of \( x \) is:
\[
\boxed{36 \sqrt{6}}
\]

System: You are a helpful assistant.


[User] Solve the equation \( e^{1-x} = 17.5 \).

[Assistant]
To solve the equation \( e^{1-x} = 17.5 \), we start by taking the natural logarithm of both sides of the equation. This gives:

\[
\ln(e^{1-x}) = \ln(17.5)
\]

Since the natural logarithm and the exponential function are inverses, we can simplify the left side:

\[
1 - x = \ln(17.5)
\]

Next, we solve for \( x \) by isolating it on one side of the equation. Subtract \( \ln(17.5) \) from both sides:

\[
1 - \ln(17.5) = x
\]

Therefore, the solution to the equation is:

\[
x = 1 - \ln(17.5)
\]

To verify, we substitute \( x = 1 - \ln(17.5) \) back into the original equation:

\[
e^{1 - (1 - \ln(17.5))} = e^{\ln(17.5)} = 17.5
\]

This confirms that the solution is correct. Thus, the final answer is:

\[
\boxed{1 - \ln(17.5)}
\]

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 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] 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] 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] If the perimeter of rectangle ABCD is 20 inches, what is the least value of diagonal AC in inches?

[Assistant]
To find the least value of the diagonal \( AC \) in a rectangle \( ABCD \) with a perimeter of 20 inches, we start by noting the basic formulas and constraints.

1. **Perimeter Constraint:**
   The perimeter of the rectangle is given by:
   \[
   P = 2(l + w) = 20
   \]
   Dividing both sides by 2, we get:
   \[
   l + w = 10
   \]
   where \( l \) is the length and \( w \) is the width.

2. **Diagonal Calculation:**
   The diagonal \( AC \) of the rectangle can be found using the Pythagorean theorem:
   \[
   AC = \sqrt{l^2 + w^2}
   \]
   We need to minimize \( \sqrt{l^2 + w^2} \). Since the square root function is monotonically increasing, minimizing \( \sqrt{l^2 + w^2} \) is equivalent to minimizing \( l^2 + w^2 \).

3. **Substitution:**
   Using the constraint \( l + w = 10 \), we can express \( w \) in terms of \( l \):
   \[
   w = 10 - l
   \]
   Substitute \( w \) into the expression for the diagonal squared:
   \[
   l^2 + w^2 = l^2 + (10 - l)^2
   \]
   Expanding the expression:
   \[
   l^2 + (10 - l)^2 = l^2 + (100 - 20l + l^2) = 2l^2 - 20l + 100
   \]

4. **Minimizing the Quadratic Expression:**
   To minimize \( 2l^2 - 20l + 100 \), we find the vertex of the parabola. The vertex form for a quadratic equation \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 2 \) and \( b = -20 \):
   \[
   l = -\frac{-20}{2 \cdot 2} = \frac{20}{4} = 5
   \]
   Therefore, \( l = 5 \) inches. Substituting \( l = 5 \) back into the constraint \( l + w = 10 \):
   \[
   w = 10 - 5 = 5
   \]

5. **Diagonal Calculation:**
   With \( l = 5 \) and \( w = 5 \), the rectangle is a square. The diagonal \( AC \) is:
   \[
   AC = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
   \]

Therefore, the least value of the diagonal \( AC \) is:
\[
\boxed{5\sqrt{2}}
\]

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] Find the supremum and infimum of the set \( S = \left\{ x \mid x \text{ is rational and } x^2 \leq 7 \right\} \).

[Assistant]
To find the supremum and infimum of the set \( S = \left\{ x \mid x \text{ is rational and } x^2 \leq 7 \right\} \), we proceed as follows:

1. **Define the Set \( S \)**:
   The set \( S \) consists of all rational numbers \( x \) such that \( x^2 \leq 7 \). This can be rewritten as:
   \[
   S = \{ x \in \mathbb{Q} \mid -\sqrt{7} \leq x \leq \sqrt{7} \}
   \]

2. **Identify the Interval**:
   The set \( S \) is the intersection of the interval \([- \sqrt{7}, \sqrt{7}]\) with the set of rational numbers \( \mathbb{Q} \). Since the rationals are dense in the reals, \( S \) is a dense subset of the interval \([- \sqrt{7}, \sqrt{7}]\).

3. **Determine the Supremum**:
   The supremum of \( S \) is the least upper bound of \( S \). Since \( S \) contains all rational numbers \( x \) such that \( x \leq \sqrt{7} \), and since \( \sqrt{7} \) is an upper bound for \( S \), we need to check if there is any smaller upper bound. Suppose there is some \( y < \sqrt{7} \) that is an upper bound for \( S \). Then, there would be rational numbers arbitrarily close to \( \sqrt{7} \) that are still in \( S \), contradicting \( y \) being an upper bound. Therefore, the supremum of \( S \) is \( \sqrt{7} \).

4. **Determine the Infimum**:
   The infimum of \( S \) is the greatest lower bound of \( S \). Since \( S \) contains all rational numbers \( x \) such that \( x \geq -\sqrt{7} \), and since \( -\sqrt{7} \) is a lower bound for \( S \), we need to check if there is any larger lower bound. Suppose there is some \( y > -\sqrt{7} \) that is a lower bound for \( S \). Then, there would be rational numbers arbitrarily close to \( -\sqrt{7} \) that are still in \( S \), contradicting \( y \) being a lower bound. Therefore, the infimum of \( S \) is \( -\sqrt{7} \).

5. **Conclusion**:
   The supremum of \( S \) is \( \sqrt{7} \) and the infimum of \( S \) is \( -\sqrt{7} \).

Thus, the final answer is:
\[
\boxed{\sup S = \sqrt{7} \text{ and } \inf S = -\sqrt{7}}
\]

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] An equilateral triangle with side length 6 is circumscribed about a circle, which is circumscribed about a square. Determine the area of the square.

[Assistant]
To determine the area of the square inscribed in the circle which is itself inscribed in an equilateral triangle with side length 6, we can follow these steps:

1. **Calculate the inradius of the equilateral triangle:**
   The formula for the inradius \( r \) of an equilateral triangle with side length \( a \) is:
   \[
   r = \frac{a \sqrt{3}}{6}
   \]
   Given \( a = 6 \):
   \[
   r = \frac{6 \sqrt{3}}{6} = \sqrt{3}
   \]
   So, the radius of the circle inscribed in the equilateral triangle is \( \sqrt{3} \).

2. **Determine the relationship between the circle and the square:**
   The circle circumscribes the square, meaning the diagonal of the square is equal to the diameter of the circle. The diameter of the circle is:
   \[
   2r = 2 \sqrt{3}
   \]
   Let \( s \) be the side length of the square. The diagonal of the square can be expressed as:
   \[
   s \sqrt{2}
   \]
   Setting the diagonal of the square equal to the diameter of the circle:
   \[
   s \sqrt{2} = 2 \sqrt{3}
   \]

3. **Solve for the side length of the square:**
   \[
   s = \frac{2 \sqrt{3}}{\sqrt{2}} = \frac{2 \sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2 \sqrt{6}}{2} = \sqrt{6}
   \]

4. **Calculate the area of the square:**
   The area \( A \) of the square is given by the square of its side length:
   \[
   A = s^2 = (\sqrt{6})^2 = 6
   \]

Thus, the area of the square is \(\boxed{6}\).

System: You are a helpful assistant.


[User] Evaluate the limit $\underset{x\to \infty }{\mathop{\lim }}\,\left( \sin \left( x+1 \right)-\sin x \right)$.

[Assistant]
To evaluate the limit \(\lim_{x \to \infty} \left( \sin(x+1) - \sin(x) \right)\), we will use a trigonometric identity to simplify the expression and analyze its behavior as \(x\) approaches infinity.

1. **Apply the Sine Difference Identity:**
   The sine difference identity states:
   \[
   \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
   \]
   Here, let \(A = x + 1\) and \(B = x\). Applying the identity, we get:
   \[
   \sin(x+1) - \sin(x) = 2 \cos\left(\frac{(x+1) + x}{2}\right) \sin\left(\frac{(x+1) - x}{2}\right)
   \]
   Simplify the arguments:
   \[
   \sin(x+1) - \sin(x) = 2 \cos\left(\frac{2x + 1}{2}\right) \sin\left(\frac{1}{2}\right)
   \]

2. **Simplify the Expression:**
   Notice that \(\sin\left(\frac{1}{2}\right)\) is a constant. Let \(C = 2 \sin\left(\frac{1}{2}\right)\). Thus, the expression becomes:
   \[
   \sin(x+1) - \sin(x) = C \cos\left(x + \frac{1}{2}\right)
   \]
   where \(C = 2 \sin\left(\frac{1}{2}\right)\).

3. **Analyze the Behavior as \(x \to \infty\):**
   The function \(\cos\left(x + \frac{1}{2}\right)\) oscillates between \(-1\) and \(1\) as \(x\) approaches infinity. Since \(C\) is a constant, the expression \(C \cos\left(x + \frac{1}{2}\right)\) will also oscillate between \(-C\) and \(C\).

4. **Conclusion:**
   Because \(\cos\left(x + \frac{1}{2}\right)\) does not approach a single value as \(x\) approaches infinity, the expression \(C \cos\left(x + \frac{1}{2}\right)\) will continue to oscillate between \(-C\) and \(C\). Therefore, the limit does not exist.

Thus, the final answer is:
\[
\boxed{\text{DNE}}
\]

System: You are a helpful assistant.


[User] Let $a$ and $b$ be real numbers such that $a + b = 8$. Let $c$ be the minimum possible value of $x^2 + ax + b$ over all real numbers $x$. Find the maximum possible value of $c$ over all such $a$ and $b$.

[Assistant]
To find the maximum possible value of \(c\), where \(c\) is the minimum value of the quadratic function \(x^2 + ax + b\) over all real numbers \(x\) and given that \(a + b = 8\), we proceed as follows:

1. **Determine the minimum value of the quadratic function \(x^2 + ax + b\):**
   The vertex of the quadratic function \(x^2 + ax + b\) occurs at \(x = -\frac{a}{2}\). Substituting \(x = -\frac{a}{2}\) into the function, we get:
   \[
   c = \left(-\frac{a}{2}\right)^2 + a\left(-\frac{a}{2}\right) + b = \frac{a^2}{4} - \frac{a^2}{2} + b = -\frac{a^2}{4} + b
   \]

2. **Express \(b\) in terms of \(a\):**
   Using the constraint \(a + b = 8\), we have:
   \[
   b = 8 - a
   \]
   Substituting \(b = 8 - a\) into the expression for \(c\):
   \[
   c = -\frac{a^2}{4} + (8 - a) = -\frac{a^2}{4} - a + 8
   \]

3. **Maximize the quadratic function \(c = -\frac{a^2}{4} - a + 8\):**
   The quadratic function \(c = -\frac{a^2}{4} - a + 8\) is a downward-opening parabola (since the coefficient of \(a^2\) is negative). The maximum value occurs at the vertex of this parabola, which is given by:
   \[
   a = -\frac{B}{2A}
   \]
   where \(A = -\frac{1}{4}\) and \(B = -1\). Therefore:
   \[
   a = -\frac{-1}{2 \left(-\frac{1}{4}\right)} = -\frac{1}{-\frac{1}{2}} = -2
   \]

4. **Substitute \(a = -2\) back into the expression for \(c\):**
   \[
   c = -\frac{(-2)^2}{4} - (-2) + 8 = -\frac{4}{4} + 2 + 8 = -1 + 2 + 8 = 9
   \]

Thus, the maximum possible value of \(c\) is \(\boxed{9}\).

System: You are a helpful assistant.


[User] Nathan will roll two six-sided dice. What is the probability that he will roll a number less than three on the first die and a number greater than three on the second die? Express your answer as a common fraction.

[Assistant]
To find the probability that Nathan rolls a number less than three on the first die and a number greater than three on the second die, we can break the problem into two parts and then combine the probabilities.

1. **First Die:**
   - The possible outcomes are 1, 2, 3, 4, 5, and 6.
   - The favorable outcomes for the first die are 1 and 2 (since we want a number less than three).
   - There are 2 favorable outcomes out of 6 possible outcomes.
   - Therefore, the probability for the first die is:
     \[
     \frac{2}{6} = \frac{1}{3}
     \]

2. **Second Die:**
   - The possible outcomes are 1, 2, 3, 4, 5, and 6.
   - The favorable outcomes for the second die are 4, 5, and 6 (since we want a number greater than three).
   - There are 3 favorable outcomes out of 6 possible outcomes.
   - Therefore, the probability for the second die is:
     \[
     \frac{3}{6} = \frac{1}{2}
     \]

Since the two dice rolls are independent events, the combined probability is the product of the individual probabilities:
\[
\frac{2}{6} \times \frac{3}{6} = \frac{6}{36} = \frac{1}{6}
\]

Thus, the probability that Nathan rolls a number less than three on the first die and a number greater than three on the second die is:
\[
\boxed{\frac{1}{6}}
\]

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] What is the remainder when $6x^3-15x^2+21x-23$ is divided by $3x-6$?

[Assistant]
To find the remainder when \(6x^3 - 15x^2 + 21x - 23\) is divided by \(3x - 6\), we can use the Remainder Theorem. The Remainder Theorem states that the remainder of a polynomial \(f(x)\) when divided by \(mx - n\) is \(f\left(\frac{n}{m}\right)\).

Here, the divisor is \(3x - 6\), so we set \(3x - 6 = 0\) to find \(x = \frac{6}{3} = 2\). Therefore, we need to evaluate the polynomial \(f(x) = 6x^3 - 15x^2 + 21x - 23\) at \(x = 2\).

First, we substitute \(x = 2\) into the polynomial:
\[
f(2) = 6(2)^3 - 15(2)^2 + 21(2) - 23
\]
Next, we compute each term:
\[
6(2)^3 = 6 \cdot 8 = 48
\]
\[
-15(2)^2 = -15 \cdot 4 = -60
\]
\[
21(2) = 21 \cdot 2 = 42
\]
\[
-23 = -23
\]
Now, we sum these values:
\[
48 - 60 + 42 - 23
\]
First, compute \(48 - 60\):
\[
48 - 60 = -12
\]
Next, add 42:
\[
-12 + 42 = 30
\]
Finally, subtract 23:
\[
30 - 23 = 7
\]

Therefore, the remainder when \(6x^3 - 15x^2 + 21x - 23\) is divided by \(3x - 6\) is \(\boxed{7}\).

System: You are a helpful assistant.


[User] Given that \( m \) and \( n \) are positive integers and \( m^2 + 5n^2 = 36 \), find the value of \( m + 5n \).

[Assistant]
To solve the equation \( m^2 + 5n^2 = 36 \) for positive integers \( m \) and \( n \), we will systematically test possible values for \( n \) and determine if the corresponding \( m \) is an integer.

1. **Testing \( n = 1 \):**
   \[
   5n^2 = 5(1)^2 = 5
   \]
   \[
   m^2 = 36 - 5 = 31
   \]
   Since 31 is not a perfect square, \( m \) is not an integer. Thus, \( n = 1 \) is not a solution.

2. **Testing \( n = 2 \):**
   \[
   5n^2 = 5(2)^2 = 20
   \]
   \[
   m^2 = 36 - 20 = 16
   \]
   Since \( 16 \) is a perfect square, \( m = 4 \). Thus, \( n = 2 \) and \( m = 4 \) is a solution.

3. **Testing \( n \geq 3 \):**
   \[
   5n^2 \geq 5(3)^2 = 45
   \]
   Since \( 5n^2 \) exceeds 36 for \( n \geq 3 \), \( m^2 \) would be negative, which is not possible for positive integers.

Thus, the only valid solution is \( m = 4 \) and \( n = 2 \).

Finally, we calculate \( m + 5n \):
\[
m + 5n = 4 + 5(2) = 4 + 10 = 14
\]

Therefore, the value of \( m + 5n \) is \(\boxed{14}\).

System: You are a helpful assistant.


[User] How can you solve for \( x \) in the congruence \( 2x \equiv 4 \pmod{5} \)?

[Assistant]
To solve the congruence \( 2x \equiv 4 \pmod{5} \), we need to isolate \( x \). This involves finding the modular inverse of 2 modulo 5, which is a number \( y \) such that \( 2y \equiv 1 \pmod{5} \).

First, we determine the modular inverse of 2 modulo 5. We need to find \( y \) such that:
\[
2y \equiv 1 \pmod{5}
\]

By testing values, we find:
\[
2 \times 1 = 2 \equiv 2 \pmod{5} \quad (\text{not } 1)
\]
\[
2 \times 2 = 4 \equiv 4 \pmod{5} \quad (\text{not } 1)
\]
\[
2 \times 3 = 6 \equiv 1 \pmod{5} \quad (\text{yes, this works})
\]

Thus, the modular inverse of 2 modulo 5 is 3. We now multiply both sides of the original congruence \( 2x \equiv 4 \pmod{5} \) by 3:
\[
3 \times 2x \equiv 3 \times 4 \pmod{5}
\]
\[
6x \equiv 12 \pmod{5}
\]

Since \( 6 \equiv 1 \pmod{5} \), we have:
\[
x \equiv 12 \pmod{5}
\]

Simplifying \( 12 \mod 5 \):
\[
12 \equiv 2 \pmod{5}
\]

Therefore, the solution to the congruence is:
\[
x \equiv 2 \pmod{5}
\]

We can verify this solution by substituting \( x = 2 \) back into the original congruence:
\[
2 \times 2 = 4 \equiv 4 \pmod{5}
\]

Thus, the solution is confirmed. The final answer is:
\[
\boxed{2}
\]

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] Find the largest value of $c$ such that $-2$ is in the range of  $f(x)=x^2+3x+c$.

[Assistant]
To find the largest value of \( c \) such that \(-2\) is in the range of the function \( f(x) = x^2 + 3x + c \), we need to determine the minimum value of the quadratic function and ensure that this minimum value is at most \(-2\).

1. **Identify the vertex of the parabola**:
   The vertex form of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). For \( f(x) = x^2 + 3x + c \), we have \( a = 1 \) and \( b = 3 \). Thus, the \( x \)-coordinate of the vertex is:
   \[
   x = -\frac{3}{2 \cdot 1} = -\frac{3}{2} = -1.5
   \]

2. **Calculate the minimum value of the function**:
   Substitute \( x = -1.5 \) into the function to find the minimum value:
   \[
   f(-1.5) = (-1.5)^2 + 3(-1.5) + c
   \]
   Simplify the expression:
   \[
   f(-1.5) = 2.25 - 4.5 + c = -2.25 + c
   \]
   Therefore, the minimum value of the function is \( c - 2.25 \).

3. **Ensure \(-2\) is in the range**:
   For \(-2\) to be in the range, the minimum value must be less than or equal to \(-2\):
   \[
   c - 2.25 \leq -2
   \]
   Solving for \( c \):
   \[
   c \leq -2 + 2.25
   \]
   \[
   c \leq 0.25
   \]

4. **Determine the largest possible value of \( c \)**:
   The largest value of \( c \) that satisfies the inequality is \( c = 0.25 \).

5. **Verify the solution**:
   Substitute \( c = 0.25 \) into the function and check if \(-2\) is in the range:
   \[
   f(x) = x^2 + 3x + 0.25
   \]
   The minimum value is:
   \[
   f(-1.5) = (-1.5)^2 + 3(-1.5) + 0.25 = 2.25 - 4.5 + 0.25 = -2
   \]
   This confirms that \(-2\) is indeed in the range when \( c = 0.25 \).

Therefore, the largest value of \( c \) such that \(-2\) is in the range of \( f(x) = x^2 + 3x + c \) is \(\boxed{\frac{1}{4}}\).

System: You are a helpful assistant.


[User] A pirate is counting the loot he has stolen from S.S. AOPS, on which everything is counted in base 5.  A treasure chest contains $3124_{5}$ dollars worth of jewelry, $3122_{5}$ dollars worth of gold coins, and $124_{5}$ dollars worth of rubbing alcohol.  What is the total dollar amount that this pirate has accumulated? Express your answer in base 10.

[Assistant]
To find the total dollar amount the pirate has accumulated, we need to convert each base 5 value to base 10 and then sum the results.

1. **Convert Jewelry (3124₅) to base 10:**
   \[
   3 \times 5^3 + 1 \times 5^2 + 2 \times 5^1 + 4 \times 5^0 = 3 \times 125 + 1 \times 25 + 2 \times 5 + 4 \times 1 = 375 + 25 + 10 + 4 = 414_{10}
   \]

2. **Convert Gold Coins (3122₅) to base 10:**
   \[
   3 \times 5^3 + 1 \times 5^2 + 2 \times 5^1 + 2 \times 5^0 = 3 \times 125 + 1 \times 25 + 2 \times 5 + 2 \times 1 = 375 + 25 + 10 + 2 = 412_{10}
   \]

3. **Convert Rubbing Alcohol (124₅) to base 10:**
   \[
   1 \times 5^2 + 2 \times 5^1 + 4 \times 5^0 = 1 \times 25 + 2 \times 5 + 4 \times 1 = 25 + 10 + 4 = 39_{10}
   \]

Next, we sum the converted values in base 10:
\[
414 + 412 + 39 = 865
\]

Thus, the total dollar amount the pirate has accumulated is \(\boxed{865}\).

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] Simplify the expression \(\frac{2a}{\sqrt{a-b} + \sqrt{a+b}}\).

[Assistant]
To simplify the expression \(\frac{2a}{\sqrt{a-b} + \sqrt{a+b}}\), we start by rationalizing the denominator. This involves multiplying both the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{a-b} - \sqrt{a+b}\).

First, we write the expression and multiply by the conjugate:

\[
\frac{2a}{\sqrt{a-b} + \sqrt{a+b}} \cdot \frac{\sqrt{a-b} - \sqrt{a+b}}{\sqrt{a-b} - \sqrt{a+b}}
\]

The denominator simplifies using the difference of squares formula:

\[
(\sqrt{a-b} + \sqrt{a+b})(\sqrt{a-b} - \sqrt{a+b}) = (\sqrt{a-b})^2 - (\sqrt{a+b})^2 = (a - b) - (a + b) = -2b
\]

Next, we simplify the numerator:

\[
2a(\sqrt{a-b} - \sqrt{a+b})
\]

Putting it all together, we have:

\[
\frac{2a(\sqrt{a-b} - \sqrt{a+b})}{-2b}
\]

We can simplify this fraction by factoring out the common factor of 2 in the numerator and the denominator:

\[
\frac{2a(\sqrt{a-b} - \sqrt{a+b})}{-2b} = \frac{a(\sqrt{a-b} - \sqrt{a+b})}{-b} = \frac{a(\sqrt{a+b} - \sqrt{a-b})}{b}
\]

Thus, the simplified expression is:

\[
\boxed{\frac{a(\sqrt{a+b} - \sqrt{a-b})}{b}}
\]

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] In how many ways can one choose 6 candies from 8 available brands?

[Assistant]
To solve the problem of determining the number of ways to choose 6 candies from 8 available brands, we use the concept of combinations with repetition. The formula for combinations with repetition is given by:

\[
C(n + k - 1, k)
\]

where \( n \) is the number of types or brands, and \( k \) is the number of items to choose. In this problem, \( n = 8 \) and \( k = 6 \). Therefore, we need to compute \( C(8 + 6 - 1, 6) \), which simplifies to \( C(13, 6) \).

The formula for combinations \( C(n, k) \) is:

\[
C(n, k) = \frac{n!}{k!(n-k)!}
\]

Applying this to our specific case:

\[
C(13, 6) = \frac{13!}{6! \cdot 7!}
\]

We can simplify the factorial expressions by canceling out \( 7! \) from the numerator and the denominator:

\[
C(13, 6) = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3 \times 2 \times 1}
\]

Now, we compute the numerator and the denominator separately:

Numerator:
\[
13 \times 12 = 156
\]
\[
156 \times 11 = 1716
\]
\[
1716 \times 10 = 17160
\]
\[
17160 \times 9 = 154440
\]
\[
154440 \times 8 = 1235520
\]

Denominator:
\[
6 \times 5 = 30
\]
\[
30 \times 4 = 120
\]
\[
120 \times 3 = 360
\]
\[
360 \times 2 = 720
\]
\[
720 \times 1 = 720
\]

Now, we divide the numerator by the denominator:

\[
\frac{1235520}{720} = 1716
\]

Thus, the number of ways to choose 6 candies from 8 available brands is:

\[
\boxed{1716}
\]

System: You are a helpful assistant.


[User] Evaluate the limit $\lim _{ n\rightarrow \infty  }{ \frac { { 1 }^{ 99 }+{ 2 }^{ 99 }+{ 3 }^{ 99 }+{ 4 }^{ 99 }+.................+{ n }^{ 99 } }{ { n }^{ 100 } }  }$.

[Assistant]
To evaluate the limit \(\lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \cdots + n^{99}}{n^{100}}\), we can use the concept of Riemann sums or the asymptotic behavior of polynomial sums.

### Approach Using Riemann Sums

1. **Rewrite the Sum**:
   Consider the sum \(1^{99} + 2^{99} + \cdots + n^{99}\). We can rewrite it as:
   \[
   \sum_{k=1}^n k^{99}
   \]

2. **Factor Out \(n^{99}\)**:
   Factor out \(n^{99}\) from each term in the sum:
   \[
   \sum_{k=1}^n k^{99} = n^{99} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}
   \]

3. **Form the Limit**:
   Now, divide by \(n^{100}\):
   \[
   \frac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}} = \frac{n^{99} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}}{n^{100}} = \frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}
   \]

4. **Recognize the Riemann Sum**:
   The expression \(\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99}\) is a Riemann sum for the function \(f(x) = x^{99}\) over the interval \([0, 1]\).

5. **Evaluate the Integral**:
   As \(n \to \infty\), the Riemann sum converges to the integral:
   \[
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{99} = \int_0^1 x^{99} \, dx
   \]

6. **Compute the Integral**:
   Evaluate the integral:
   \[
   \int_0^1 x^{99} \, dx = \left[ \frac{x^{100}}{100} \right]_0^1 = \frac{1^{100}}{100} - \frac{0^{100}}{100} = \frac{1}{100}
   \]

### Conclusion

Thus, the limit is:
\[
\lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \cdots + n^{99}}{n^{100}} = \boxed{\frac{1}{100}}
\]

System: You are a helpful assistant.


[User] A new US flag has seven rows of stars, with the middle row having the most stars and the number of stars decreasing by one each time you go up or down from the middle. If the entire flag has 51 stars, how many stars does the middle row have?

[Assistant]
To determine the number of stars in the middle row of a new US flag with seven rows, where the middle row has the most stars and the number of stars decreases by one each time you go up or down from the middle, we can proceed as follows:

1. **Define the number of stars in the middle row:**
   Let \( x \) be the number of stars in the middle row (Row 4).

2. **Determine the number of stars in each row:**
   - Row 1: \( x - 3 \)
   - Row 2: \( x - 2 \)
   - Row 3: \( x - 1 \)
   - Row 4: \( x \)
   - Row 5: \( x - 1 \)
   - Row 6: \( x - 2 \)
   - Row 7: \( x - 3 \)

3. **Sum the number of stars in all rows:**
   The total number of stars is given by the sum of the stars in each row:
   \[
   (x - 3) + (x - 2) + (x - 1) + x + (x - 1) + (x - 2) + (x - 3)
   \]

4. **Simplify the sum:**
   Combine like terms:
   \[
   (x - 3) + (x - 2) + (x - 1) + x + (x - 1) + (x - 2) + (x - 3) = 7x - 12
   \]

5. **Set up the equation:**
   Given that the total number of stars is 51, we have:
   \[
   7x - 12 = 51
   \]

6. **Solve for \( x \):**
   \[
   7x - 12 = 51
   \]
   \[
   7x = 51 + 12
   \]
   \[
   7x = 63
   \]
   \[
   x = \frac{63}{7}
   \]
   \[
   x = 9
   \]

7. **Verification:**
   Verify the solution by summing the stars in each row with \( x = 9 \):
   - Row 1: \( 9 - 3 = 6 \)
   - Row 2: \( 9 - 2 = 7 \)
   - Row 3: \( 9 - 1 = 8 \)
   - Row 4: \( 9 \)
   - Row 5: \( 9 - 1 = 8 \)
   - Row 6: \( 9 - 2 = 7 \)
   - Row 7: \( 9 - 3 = 6 \)

   Summing these values:
   \[
   6 + 7 + 8 + 9 + 8 + 7 + 6 = 51
   \]

Thus, the number of stars in the middle row is \(\boxed{9}\).

System: You are a helpful assistant.


[User] What is the expected value of the roll of a standard 6-sided die?

[Assistant]
To find the expected value of rolling a standard 6-sided die, we need to consider each possible outcome and its associated probability. A standard 6-sided die has outcomes \(1, 2, 3, 4, 5,\) and \(6\), each with an equal probability of \(\frac{1}{6}\).

The expected value \(E\) is calculated by summing the products of each outcome and its probability:

\[
E = \sum_{i=1}^{6} i \cdot \frac{1}{6}
\]

Breaking this down:

\[
E = \left(1 \cdot \frac{1}{6}\right) + \left(2 \cdot \frac{1}{6}\right) + \left(3 \cdot \frac{1}{6}\right) + \left(4 \cdot \frac{1}{6}\right) + \left(5 \cdot \frac{1}{6}\right) + \left(6 \cdot \frac{1}{6}\right)
\]

Simplify each term:

\[
E = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6}
\]

Combine the fractions:

\[
E = \frac{1 + 2 + 3 + 4 + 5 + 6}{6}
\]

Calculate the sum of the numerators:

\[
1 + 2 + 3 + 4 + 5 + 6 = 21
\]

Thus:

\[
E = \frac{21}{6} = 3.5
\]

Therefore, the expected value of the roll of a standard 6-sided die is \(\boxed{3.5}\).

System: You are a helpful assistant.