So I need to compute the sample variance of the set {4, -19}. I remember that sample variance is a measure of how spread out the numbers are in a dataset. The formula for sample variance is:

\[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]

where \( n \) is the number of observations, \( x_i \) are the individual observations, and \( \bar{x} \) is the sample mean.

First, I need to find the sample mean, \( \bar{x} \). The sample mean is calculated by summing all the observations and dividing by the number of observations.

Given data: 4 and -19

Number of observations, \( n = 2 \)

Sample mean, \( \bar{x} = \frac{4 + (-19)}{2} = \frac{-15}{2} = -7.5 \)

Next, I need to calculate the squared differences from the mean for each observation.

For \( x_1 = 4 \):

\( (4 - (-7.5))^2 = (4 + 7.5)^2 = 11.5^2 = 132.25 \)

For \( x_2 = -19 \):

\( (-19 - (-7.5))^2 = (-19 + 7.5)^2 = (-11.5)^2 = 132.25 \)

Now, sum these squared differences:

\( \sum_{i=1}^{n} (x_i - \bar{x})^2 = 132.25 + 132.25 = 264.5 \)

Finally, divide this sum by \( n-1 \), which is \( 2-1 = 1 \):

\( s^2 = \frac{264.5}{1} = 264.5 \)

Wait a minute, but when I recall the formula, it's \( \frac{1}{n-1} \), so yes, dividing by \( n-1 \) gives me the unbiased estimate of the population variance.

But let me double-check my calculations to make sure I didn't make any mistakes.

First, the mean was correctly calculated as \( -7.5 \).

Then, the squared differences:

For 4: \( 4 - (-7.5) = 4 + 7.5 = 11.5 \), squared is 132.25

For -19: \( -19 - (-7.5) = -19 + 7.5 = -11.5 \), squared is also 132.25

Sum is indeed 264.5

Dividing by \( n-1 = 1 \) gives 264.5

Hmm, that seems correct.

Alternatively, I know that variance can be thought of as the average of the squared deviations from the mean. So, with only two points, it makes sense that the variance would be relatively high since they're quite far apart.

Let me see if there's another way to approach this to confirm.

I could use the formula for sample variance directly:

\( s^2 = \frac{n}{n-1} \left( \frac{\sum x_i^2}{n} - \bar{x}^2 \right) \)

Plugging in the values:

\( n = 2 \)

\( \sum x_i^2 = 4^2 + (-19)^2 = 16 + 361 = 377 \)

\( \bar{x}^2 = (-7.5)^2 = 56.25 \)

So,

\( s^2 = \frac{2}{1} \left( \frac{377}{2} - 56.25 \right) = 2 \left( 188.5 - 56.25 \right) = 2 \times 132.25 = 264.5 \)

Same result as before. Good, that matches.

Alternatively, I've heard of using the computational formula for variance, which is:

\( s^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 \)

Using this formula:

\( \sum x_i^2 = 377 \)

\( \bar{x}^2 = 56.25 \)

\( n = 2 \)

So,

\( s^2 = \frac{377}{2} - 56.25 = 188.5 - 56.25 = 132.25 \)

Wait, that's half of what I got earlier. Hmm, that doesn't match. Did I make a mistake here?

Wait, no. Actually, in the first approach, I had \( s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2 = \frac{264.5}{1} = 264.5 \)

In this second approach, I have \( s^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{377}{2} - 56.25 = 188.5 - 56.25 = 132.25 \)

Wait a minute, these should be equal because \( \sum (x_i - \bar{x})^2 = \sum x_i^2 - 2n\bar{x}\bar{x} + n\bar{x}^2 = \sum x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 = \sum x_i^2 - n\bar{x}^2 \)

Therefore, \( s^2 = \frac{\sum x_i^2 - n\bar{x}^2}{n} = \frac{\sum x_i^2}{n} - \bar{x}^2 \)

So, both methods should give the same result. But in this case, I'm getting 264.5 vs 132.25, which suggests I made an error somewhere.

Wait, perhaps I misapplied the formula. Let's re-examine.

The correct formula for sample variance is:

\( s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2 \)

Which expands to:

\( s^2 = \frac{1}{n-1} \left( \sum x_i^2 - n\bar{x}^2 \right) \)

Yes, that's consistent with what I used earlier.

So, plugging in the numbers:

\( s^2 = \frac{1}{1} \left( 377 - 2 \times 56.25 \right) = 377 - 112.5 = 264.5 \)

Ah, I see where I went wrong in my initial calculation. When I used the formula \( s^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 \), I forgot to multiply by \( n \) inside the parentheses.

Correcting that:

\( s^2 = \frac{377}{2} - 56.25 = 188.5 - 56.25 = 132.25 \)

Wait, no, that's not right. Because according to the corrected formula, it should be \( s^2 = \frac{\sum x_i^2 - n\bar{x}^2}{n} = \frac{377 - 2 \times 56.25}{2} = \frac{377 - 112.5}{2} = \frac{264.5}{2} = 132.25 \)

Okay, now it matches the previous result. So, the correct sample variance is 132.25.

I must have made a mistake in my initial application of the formula. It's good to double-check these calculations to ensure accuracy.

Therefore, the sample variance of the set {4, -19} is \( 132.25 \).

Putting it in LaTeX format as requested:

\[ \boxed{132.25} \]

**Final Answer**

\[ \boxed{132.25} \]


So I need to find the characteristic polynomial of this 4x4 matrix. The characteristic polynomial is found by computing the determinant of the matrix minus lambda times the identity matrix, right? So first, let's write down the matrix:

A = \(\begin{pmatrix}
0 & -2 & -11 & -13 \\
4 & 3 & 5 & 2 \\
-8 & -1 & -6 & -6 \\
-2 & -11 & -14 & 8
\end{pmatrix}\)

The characteristic polynomial is given by det(A - λI), where I is the 4x4 identity matrix.

So, A - λI = \(\begin{pmatrix}
0 - λ & -2 & -11 & -13 \\
4 & 3 - λ & 5 & 2 \\
-8 & -1 & -6 - λ & -6 \\
-2 & -11 & -14 & 8 - λ
\end{pmatrix}\)

Now, I need to compute the determinant of this matrix. Computing the determinant of a 4x4 matrix can be a bit tedious, but I'll proceed step by step.

One way to compute the determinant is to use cofactor expansion along any row or column. Let's choose the first row for expansion, since it has some zeros that might simplify calculations.

So, det(A - λI) = (-λ) * C11 + (-2) * C12 + (-11) * C13 + (-13) * C14

Where Cij is the cofactor of the element at row i, column j.

First, I need to find the minors M11, M12, M13, M14, and then the cofactors C11, C12, C13, C14.

Let's start with M11, which is the minor of the element at position (1,1).

M11 is the determinant of the 3x3 submatrix obtained by removing the first row and first column.

So, M11 = det(\(\begin{pmatrix}
3 - λ & 5 & 2 \\
-1 & -6 - λ & -6 \\
-11 & -14 & 8 - λ
\end{pmatrix}\))

Similarly,

M12 = det(\(\begin{pmatrix}
4 & 5 & 2 \\
-8 & -6 - λ & -6 \\
-2 & -14 & 8 - λ
\end{pmatrix}\))

M13 = det(\(\begin{pmatrix}
4 & 3 - λ & 2 \\
-8 & -1 & -6 \\
-2 & -11 & 8 - λ
\end{pmatrix}\))

M14 = det(\(\begin{pmatrix}
4 & 3 - λ & 5 \\
-8 & -1 & -6 - λ \\
-2 & -11 & -14
\end{pmatrix}\))

Computing these 3x3 determinants will still require some work.

Let me recall the formula for the determinant of a 3x3 matrix:

det(B) = b11*(b22*b33 - b23*b32) - b12*(b21*b33 - b23*b31) + b13*(b21*b32 - b22*b31)

But maybe there's a better way to organize my work here. Perhaps expanding along a different row or column would be easier, but since the first row has a zero in the first entry, it's already somewhat simplified.

Alternatively, I could try to perform row operations to introduce more zeros before computing the determinant, but remember that row operations affect the determinant, so I have to keep track of those changes.

Wait, actually, since we're dealing with A - λI, which is singular only when λ is such that det(A - λI) = 0, perhaps I should consider using properties of block matrices if possible, but this doesn't seem like a block diagonal matrix.

Another thought: maybe I can expand along the second row instead, since it has a positive 4 and a negative -1, which might not introduce too many complications.

Let me try expanding along the second row.

So, det(A - λI) = 4 * C21 + 3 - λ * C22 + 5 * C23 + 2 * C24

This seems similar to the first approach, just choosing a different row.

Hmm.

Maybe I should stick with the first choice and plow through the calculations.

Let's focus on computing M11, M12, M13, and M14 one by one.

Starting with M11:

M11 = det(\(\begin{pmatrix}
3 - λ & 5 & 2 \\
-1 & -6 - λ & -6 \\
-11 & -14 & 8 - λ
\end{pmatrix}\))

Using the formula for the determinant of a 3x3 matrix:

det(M11) = (3 - λ)*[(-6 - λ)(8 - λ) - (-6)(-14)] - 5*[(-1)(8 - λ) - (-6)(-11)] + 2*[(-1)(-14) - (-6)(-11)]

Let's compute each term step by step.

First, (-6 - λ)(8 - λ) - (-6)(-14):

= (-6)(8) + (-6)(-λ) + (-λ)(8) + (-λ)(-λ) - (6*14)

= -48 + 6λ - 8λ + λ² - 84

= λ² - 2λ - 132

Next, (-1)(8 - λ) - (-6)(-11):

= -8 + λ - 66

= λ - 74

Then, (-1)(-14) - (-6)(-11):

= 14 - 66

= -52

Putting it all together:

det(M11) = (3 - λ)(λ² - 2λ - 132) - 5(λ - 74) + 2(-52)

= (3 - λ)(λ² - 2λ - 132) - 5λ + 370 - 104

= (3 - λ)(λ² - 2λ - 132) - 5λ + 266

Now, expand (3 - λ)(λ² - 2λ - 132):

= 3*λ² - 3*2λ - 3*132 - λ*λ² + λ*2λ + λ*132

= 3λ² - 6λ - 396 - λ³ + 2λ² + 132λ

= -λ³ + 5λ² + 126λ - 396

So, det(M11) = -λ³ + 5λ² + 126λ - 396 - 5λ + 266

= -λ³ + 5λ² + 121λ - 130

Alright, that's M11.

Now, M12:

M12 = det(\(\begin{pmatrix}
4 & 5 & 2 \\
-8 & -6 - λ & -6 \\
-2 & -14 & 8 - λ
\end{pmatrix}\))

Again, using the determinant formula for 3x3:

det(M12) = 4*[(-6 - λ)(8 - λ) - (-6)(-14)] - 5*[(-8)(8 - λ) - (-6)(-2)] + 2*[(-8)(-14) - (-6 - λ)(-2)]

Compute each part:

First, (-6 - λ)(8 - λ) - (-6)(-14):

We already computed this as λ² - 2λ - 132 earlier.

Second, (-8)(8 - λ) - (-6)(-2):

= -64 + 8λ - 12

= 8λ - 76

Third, (-8)(-14) - (-6 - λ)(-2):

= 112 - [12 + 2λ]

= 112 - 12 - 2λ

= 100 - 2λ

So, det(M12) = 4(λ² - 2λ - 132) - 5(8λ - 76) + 2(100 - 2λ)

= 4λ² - 8λ - 528 - 40λ + 380 + 200 - 4λ

= 4λ² - 52λ + 52

Wait, let's check that again.

4λ² - 8λ - 528 - 40λ + 380 + 200 - 4λ

Combine like terms:

4λ² + (-8λ - 40λ - 4λ) + (-528 + 380 + 200)

= 4λ² - 52λ - 148

Yes, that's correct.

Moving on to M13:

M13 = det(\(\begin{pmatrix}
4 & 3 - λ & 2 \\
-8 & -1 & -6 \\
-2 & -11 & 8 - λ
\end{pmatrix}\))

Determinant:

= 4*[(-1)(8 - λ) - (-6)(-11)] - (3 - λ)*[(-8)(8 - λ) - (-6)(-2)] + 2*[(-8)(-11) - (-1)(-2)]

Compute each part:

First, (-1)(8 - λ) - (-6)(-11):

= -8 + λ - 66

= λ - 74

Second, (-8)(8 - λ) - (-6)(-2):

= -64 + 8λ - 12

= 8λ - 76

Third, (-8)(-11) - (-1)(-2):

= 88 - 2

= 86

So, det(M13) = 4(λ - 74) - (3 - λ)(8λ - 76) + 2*86

= 4λ - 296 - (24λ - 228 - 8λ² + 76λ) + 172

= 4λ - 296 - ( -8λ² + 100λ - 228 ) + 172

= 4λ - 296 + 8λ² - 100λ + 228 + 172

= 8λ² - 96λ + 104

Wait, let's double-check that.

Combining terms:

8λ² + (4λ - 100λ) + (-296 + 228 + 172)

= 8λ² - 96λ + 104

Yes, that's correct.

Lastly, M14:

M14 = det(\(\begin{pmatrix}
4 & 3 - λ & 5 \\
-8 & -1 & -6 - λ \\
-2 & -11 & -14
\end{pmatrix}\))

Determinant:

= 4*[(-1)(-14) - (-6 - λ)(-11)] - (3 - λ)*[(-8)(-14) - (-6 - λ)(-2)] + 5*[(-8)(-11) - (-1)(-2)]

Compute each part:

First, (-1)(-14) - (-6 - λ)(-11):

= 14 - [66 + 11λ]

= 14 - 66 - 11λ

= -52 - 11λ

Second, (-8)(-14) - (-6 - λ)(-2):

= 112 - [12 + 2λ]

= 112 - 12 - 2λ

= 100 - 2λ

Third, (-8)(-11) - (-1)(-2):

= 88 - 2

= 86

So, det(M14) = 4(-52 - 11λ) - (3 - λ)(100 - 2λ) + 5*86

= -208 - 44λ - (300 - 6λ - 100λ + 2λ²) + 430

= -208 - 44λ - (2λ² - 106λ + 300) + 430

= -208 - 44λ - 2λ² + 106λ - 300 + 430

= -2λ² + 62λ - 78

Alright, now that I have all the minors, I can plug them back into the expression for det(A - λI):

det(A - λI) = (-λ)(-λ³ + 5λ² + 121λ - 130) + (-2)(4λ² - 52λ + 52) + (-11)(8λ² - 96λ + 104) + (-13)(-2λ² + 62λ - 78)

Let's expand each term:

First term: (-λ)(-λ³ + 5λ² + 121λ - 130) = λ⁴ - 5λ³ - 121λ² + 130λ

Second term: (-2)(4λ² - 52λ + 52) = -8λ² + 104λ - 104

Third term: (-11)(8λ² - 96λ + 104) = -88λ² + 1056λ - 1144

Fourth term: (-13)(-2λ² + 62λ - 78) = 26λ² - 806λ + 1014

Now, sum all these up:

λ⁴ - 5λ³ - 121λ² + 130λ - 8λ² + 104λ - 104 - 88λ² + 1056λ - 1144 + 26λ² - 806λ + 1014

Combine like terms:

λ⁴ - 5λ³ + (-121λ² - 8λ² - 88λ² + 26λ²) + (130λ + 104λ + 1056λ - 806λ) + (-104 - 1144 + 1014)

Calculate coefficients:

For λ⁴: 1

For λ³: -5

For λ²: -121 - 8 - 88 + 26 = -201

For λ: 130 + 104 + 1056 - 806 = 484

For constants: -104 - 1144 + 1014 = -234

So, the characteristic polynomial is:

λ⁴ - 5λ³ - 201λ² + 484λ - 234

Wait a minute, I should verify these calculations because they look pretty involved, and it's easy to make mistakes in addition and subtraction.

Let me double-check the coefficient for λ²:

-121 - 8 = -129

-129 - 88 = -217

-217 + 26 = -191

Hmm, earlier I had -201, but according to this, it's -191. There must be a mistake somewhere.

Wait, perhaps I miscalculated one of the terms.

Looking back, M11 was -λ³ + 5λ² + 121λ - 130

M12 was 4λ² - 52λ + 52

M13 was 8λ² - 96λ + 104

M14 was -2λ² + 62λ - 78

When combining them:

λ⁴ - 5λ³ + (-121λ² - 8λ² - 88λ² + 26λ²) + (130λ + 104λ + 1056λ - 806λ) + (-104 - 1144 + 1014)

Calculating again:

-121 - 8 = -129

-129 - 88 = -217

-217 + 26 = -191

So, the coefficient for λ² is indeed -191, not -201. It seems I made a mistake earlier.

Therefore, the characteristic polynomial is:

λ⁴ - 5λ³ - 191λ² + 484λ - 234

I think that's the final answer.

**Final Answer**

\[ \boxed{\lambda^4 - 5\lambda^3 - 191\lambda^2 + 484\lambda - 234} \]


So I need to find the greatest common divisor (GCD) of two numbers: 22176 and 17556. The GCD is the largest number that divides both of them without leaving a remainder. There are a few methods to find it, but I think using the Euclidean algorithm would be efficient here.

First, I recall that the Euclidean algorithm involves repeated division and taking remainders. Specifically, for two numbers a and b, where a > b, the GCD(a, b) is equal to GCD(b, a mod b). We keep applying this until we get a remainder of zero; the last non-zero remainder is the GCD.

So, let's set a = 22176 and b = 17556.

Step 1:
Compute a mod b, which is 22176 mod 17556.

To calculate 22176 mod 17556, I'll divide 22176 by 17556 and find the remainder.

22176 ÷ 17556 ≈ 1.258

The integer part is 1, so:

22176 - 1 * 17556 = 22176 - 17556 = 4620

So, 22176 mod 17556 = 4620.

Now, apply the Euclidean algorithm:

GCD(22176, 17556) = GCD(17556, 4620)

Step 2:
Repeat the process with a = 17556 and b = 4620.

Compute 17556 mod 4620.

17556 ÷ 4620 ≈ 3.819

Integer part is 3:

17556 - 3 * 4620 = 17556 - 13860 = 3696

So, 17556 mod 4620 = 3696.

Apply the Euclidean algorithm again:

GCD(17556, 4620) = GCD(4620, 3696)

Step 3:
Continue with a = 4620 and b = 3696.

Compute 4620 mod 3696.

4620 ÷ 3696 ≈ 1.253

Integer part is 1:

4620 - 1 * 3696 = 4620 - 3696 = 924

So, 4620 mod 3696 = 924.

Apply the Euclidean algorithm once more:

GCD(4620, 3696) = GCD(3696, 924)

Step 4:
With a = 3696 and b = 924.

Compute 3696 mod 924.

3696 ÷ 924 ≈ 4

Since 924 * 4 = 3696, the remainder is 0.

So, 3696 mod 924 = 0.

Now, according to the Euclidean algorithm, when the remainder reaches 0, the GCD is the divisor at that step, which is 924.

Therefore, the GCD of 22176 and 17556 is 924.

Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes.

Starting with:

a = 22176

b = 17556

First step:

22176 ÷ 17556 = 1 (since 17556 * 1 = 17556)

Remainder: 22176 - 17556 = 4620

Second step:

17556 ÷ 4620 = 3 (since 4620 * 3 = 13860)

Remainder: 17556 - 13860 = 3696

Third step:

4620 ÷ 3696 = 1 (since 3696 * 1 = 3696)

Remainder: 4620 - 3696 = 924

Fourth step:

3696 ÷ 924 = 4 (since 924 * 4 = 3696)

Remainder: 3696 - 3696 = 0

Indeed, the GCD is 924.

Alternatively, I can verify this by checking if 924 divides both 22176 and 17556 exactly.

Let's check:

22176 ÷ 924 = ?

Well, I know that 924 * 24 = 924 * 20 + 924 * 4 = 18480 + 3696 = 22176

Yes, 924 * 24 = 22176

Similarly, 17556 ÷ 924 = ?

924 * 19 = 924 * 10 + 924 * 9 = 9240 + 8316 = 17556

Yes, 924 * 19 = 17556

So, indeed, 924 is a common divisor of both numbers.

Also, since the Euclidean algorithm gives us the greatest such divisor, there shouldn't be a larger one.

Therefore, the greatest common divisor of {22176, 17556} is \boxed{924}.


I have this integral to solve:

$$\int -\frac{2 e^{\frac{15}{2}-\frac{17 x}{2}} (102 x + 313)}{(6 x + 17)^3} \, dx$$

Wow, that looks pretty complicated. Let me see if I can simplify it or find a substitution that makes it easier.

First, I notice there's an exponential function with a linear exponent, and a rational function involving polynomials. The denominator is $(6x + 17)^3$, which suggests that maybe substituting for the denominator would be useful.

Let me try letting $u = 6x + 17$. Then, $du/dx = 6$, so $dx = du/6$.

Substituting $u$ into the integral:

$$\int -\frac{2 e^{\frac{15}{2} - \frac{17(x)}{2}} (102x + 313)}{u^3} \cdot \frac{1}{6} \, du$$

Hmm, but now I need to express everything in terms of $u$. That seems tricky because of the $x$ terms inside the exponential and the numerator.

Wait a minute, let's look at the exponent of $e$: $\frac{15}{2} - \frac{17x}{2}$. If I substitute $u = 6x + 17$, then $x = \frac{u - 17}{6}$.

So, let's compute $- \frac{17x}{2}$ in terms of $u$:

$$- \frac{17}{2} \left( \frac{u - 17}{6} \right) = - \frac{17(u - 17)}{12} = - \frac{17u - 289}{12}$$

Therefore, the exponential becomes:

$$e^{\frac{15}{2} - \frac{17u}{12}}$$

Now, let's look at the numerator: $102x + 313$. Substitute $x = \frac{u - 17}{6}$:

$$102 \left( \frac{u - 17}{6} \right) + 313 = 17(u - 17) + 313 = 17u - 289 + 313 = 17u + 24$$

So, putting it all together, the integral becomes:

$$\int -\frac{2 e^{\frac{15}{2} - \frac{17u}{12}} (17u + 24)}{u^3} \cdot \frac{1}{6} \, du$$

Simplify the constants:

$$= -\frac{2}{6} \int \frac{e^{\frac{15}{2} - \frac{17u}{12}} (17u + 24)}{u^3} \, du$$

$$= -\frac{1}{3} \int \frac{e^{\frac{15}{2} - \frac{17u}{12}} (17u + 24)}{u^3} \, du$$

This still looks complicated. Maybe there's another approach.

Let me consider integration by parts. Integration by parts is useful when you have a product of functions. In this case, we have an exponential times a rational function.

Recall that integration by parts formula:

$$\int u \, dv = uv - \int v \, du$$

I need to choose $u$ and $dv$ wisely. Let's set:

$$u = e^{\frac{15}{2} - \frac{17x}{2}} \quad \text{and} \quad dv = -\frac{2(102x + 313)}{(6x + 17)^3} \, dx$$

Then,

$$du = -\frac{17}{2} e^{\frac{15}{2} - \frac{17x}{2}} \, dx$$

And I need to find $v = \int dv$.

But finding $v$ seems as difficult as the original integral. Maybe this isn't the best choice.

Alternatively, perhaps I should look for a substitution that simplifies both the exponential and the rational part simultaneously.

Wait, going back to my earlier substitution $u = 6x + 17$, I tried to express everything in terms of $u$, but it got messy. Maybe I should persist with that approach.

Let me write down the integral again after substitution:

$$-\frac{1}{3} \int \frac{e^{\frac{15}{2} - \frac{17u}{12}} (17u + 24)}{u^3} \, du$$

This looks like it might be amenable to integration by parts, where I can set:

$$u = \frac{17u + 24}{u^3} = \frac{17u + 24}{u^3} = \frac{17}{u^2} + \frac{24}{u^3}$$

And

$$dv = e^{\frac{15}{2} - \frac{17u}{12}} \, du$$

Then,

$$du = \left( -\frac{34}{u^3} - \frac{72}{u^4} \right) du$$

And

$$v = \int e^{\frac{15}{2} - \frac{17u}{12}} \, du = e^{\frac{15}{2} - \frac{17u}{12}} \cdot \frac{12}{17}$$

Applying integration by parts:

$$\int u \, dv = uv - \int v \, du$$

$$= \left( \frac{17}{u^2} + \frac{24}{u^3} \right) e^{\frac{15}{2} - \frac{17u}{12}} \cdot \frac{12}{17} - \int e^{\frac{15}{2} - \frac{17u}{12}} \left( -\frac{34}{u^3} - \frac{72}{u^4} \right) \, du$$

This seems even more complicated than before. Maybe integration by parts isn't the way to go here.

Let me consider another approach. Perhaps there's a way to manipulate the integrand algebraically to make it easier to integrate.

Looking back at the original integrand:

$$-\frac{2 e^{\frac{15}{2} - \frac{17 x}{2}} (102 x + 313)}{(6 x + 17)^3}$$

I wonder if the numerator is related to the derivative of the denominator or something similar.

Compute the derivative of the denominator:

$$\frac{d}{dx}(6x + 17)^3 = 3(6x + 17)^2 \cdot 6 = 18(6x + 17)^2$$

Not sure if that helps directly.

Alternatively, perhaps I can factor out some terms from the numerator to relate it to the derivative of the denominator.

Wait, maybe I can use substitution combined with recognizing a pattern that resembles the derivative of a quotient.

Let me recall that the derivative of $\frac{f(x)}{g(x)}$ is $\frac{f' g - f g'}{g^2}$.

In this case, if I let $f(x) = e^{\frac{15}{2} - \frac{17x}{2}}$ and $g(x) = (6x + 17)^n$, then perhaps choosing $n$ appropriately could lead somewhere.

But this seems too vague. Maybe I need to think differently.

Another thought: since the denominator is $(6x + 17)^3$, and the numerator has a term involving $x$, perhaps expressing everything in terms of $6x + 17$ could help.

Let me try to write the entire integrand in terms of $u = 6x + 17$.

We already did that earlier, but it got messy. Maybe I need to accept that and proceed carefully.

Given that, the integral becomes:

$$-\frac{1}{3} \int \frac{e^{\frac{15}{2} - \frac{17u}{12}} (17u + 24)}{u^3} \, du$$

Perhaps I can split the fraction into two separate fractions:

$$-\frac{1}{3} \int \left( \frac{17u}{u^3} + \frac{24}{u^3} \right) e^{\frac{15}{2} - \frac{17u}{12}} \, du$$

$$= -\frac{1}{3} \int \left( \frac{17}{u^2} + \frac{24}{u^3} \right) e^{\frac{15}{2} - \frac{17u}{12}} \, du$$

$$= -\frac{1}{3} \left[ \int \frac{17}{u^2} e^{\frac{15}{2} - \frac{17u}{12}} \, du + \int \frac{24}{u^3} e^{\frac{15}{2} - \frac{17u}{12}} \, du \right]$$

These two integrals still look challenging, but maybe they can be solved using substitution or integration by parts.

Let me tackle the first one:

$$\int \frac{17}{u^2} e^{\frac{15}{2} - \frac{17u}{12}} \, du$$

Let me set $w = \frac{15}{2} - \frac{17u}{12}$, then $dw/du = -\frac{17}{12}$, so $du = -\frac{12}{17} dw$.

Substituting:

$$\int \frac{17}{u^2} e^{w} \left( -\frac{12}{17} \right) dw$$

$$= -12 \int \frac{1}{u^2} e^{w} dw$$

But I still have $u$ in the denominator, which is problematic. Since $w = \frac{15}{2} - \frac{17u}{12}$, I can express $u$ in terms of $w$:

$$u = \frac{15 - 12w}{17}$$

So,

$$\int \frac{1}{u^2} e^{w} dw = \int \frac{1}{\left( \frac{15 - 12w}{17} \right)^2} e^{w} \left( -\frac{12}{17} \right) dw$$

$$= -\frac{17}{12} \int \frac{1}{(15 - 12w)^2} e^{w} dw$$

This doesn't seem any better. Maybe this substitution isn't helping.

Perhaps I should consider integrating by parts instead.

Let me set:

$$u = \frac{17}{u^2}, \quad dv = e^{\frac{15}{2} - \frac{17u}{12}} du$$

Then,

$$du = -\frac{34}{u^3} du, \quad v = \int e^{\frac{15}{2} - \frac{17u}{12}} du = e^{\frac{15}{2} - \frac{17u}{12}} \cdot \frac{12}{17}$$

Applying integration by parts:

$$\int u \, dv = uv - \int v \, du$$

$$= \frac{17}{u^2} e^{\frac{15}{2} - \frac{17u}{12}} \cdot \frac{12}{17} - \int e^{\frac{15}{2} - \frac{17u}{12}} \left( -\frac{34}{u^3} \right) du$$

$$= \frac{12}{u^2} e^{\frac{15}{2} - \frac{17u}{12}} + 34 \int \frac{e^{\frac{15}{2} - \frac{17u}{12}}}{u^3} du$$

Now, the remaining integral is:

$$34 \int \frac{e^{\frac{15}{2} - \frac{17u}{12}}}{u^3} du$$

This still looks complicated. Maybe I need to find another substitution or look for a different approach altogether.

Let me consider that the original integrand has a form that might suggest a particular substitution or recognition of a known integral.

Alternatively, perhaps there's a way to recognize that the integrand is the result of the product rule applied to some function.

Wait, maybe I should look for a function whose derivative matches the given integrand.

Suppose there exists a function $F(x)$ such that:

$$F'(x) = -\frac{2 e^{\frac{15}{2} - \frac{17 x}{2}} (102 x + 313)}{(6 x + 17)^3}$$

If I can guess what $F(x)$ might be, that could help me find the integral without having to do all the work.

Given the complexity of the integrand, perhaps $F(x)$ involves terms like $e^{\frac{15}{2} - \frac{17x}{2}}$ multiplied by some power of $(6x + 17)$.

Let me assume that $F(x) = k e^{\frac{15}{2} - \frac{17x}{2}} (6x + 17)^m$, where $k$ and $m$ are constants to be determined.

Then,

$$F'(x) = k \left[ -\frac{17}{2} e^{\frac{15}{2} - \frac{17x}{2}} (6x + 17)^m + e^{\frac{15}{2} - \frac{17x}{2}} \cdot m \cdot 6 \cdot (6x + 17)^{m-1} \right]$$

$$= k e^{\frac{15}{2} - \frac{17x}{2}} \left[ -\frac{17}{2} (6x + 17)^m + 6m (6x + 17)^{m-1} \right]$$

Comparing this to the integrand:

$$-\frac{2 e^{\frac{15}{2} - \frac{17 x}{2}} (102 x + 313)}{(6 x + 17)^3}$$

I need to match coefficients and exponents.

First, note that $(102x + 313)$ can be expressed in terms of $(6x + 17)$.

Let me solve for $x$ in terms of $u = 6x + 17$:

$$u = 6x + 17 \implies x = \frac{u - 17}{6}$$

Thus,

$$102x + 313 = 102 \left( \frac{u - 17}{6} \right) + 313 = 17(u - 17) + 313 = 17u - 289 + 313 = 17u + 24$$

Similarly,

$$(6x + 17)^3 = u^3$$

So, the integrand becomes:

$$-\frac{2 e^{\frac{15}{2} - \frac{17 x}{2}} (17u + 24)}{u^3}$$

Now, comparing this to $F'(x) = k e^{\frac{15}{2} - \frac{17x}{2}} \left[ -\frac{17}{2} (6x + 17)^m + 6m (6x + 17)^{m-1} \right]$

We have:

$$-\frac{2 e^{\frac{15}{2} - \frac{17 x}{2}} (17u + 24)}{u^3} = k e^{\frac{15}{2} - \frac{17 x}{2}} \left[ -\frac{17}{2} u^m + 6m u^{m-1} \right]$$

Dividing both sides by $e^{\frac{15}{2} - \frac{17x}{2}}$:

$$-\frac{2 (17u + 24)}{u^3} = k \left[ -\frac{17}{2} u^m + 6m u^{m-1} \right]$$

Simplify the left side:

$$-\frac{2(17u + 24)}{u^3} = -\frac{34u + 48}{u^3} = -\frac{34}{u^2} - \frac{48}{u^3}$$

So,

$$-\frac{34}{u^2} - \frac{48}{u^3} = k \left( -\frac{17}{2} u^m + 6m u^{m-1} \right)$$

For these two expressions to be equal for all $u$, their corresponding powers of $u$ must be equal, and their coefficients must match.

Therefore, we have:

1. For $u^{-2}$: coefficient on the right is zero, so $-\frac{17}{2} u^m + 6m u^{m-1} = 0$

2. For $u^{-3}$: coefficient on the right is $-k$, so $-\frac{17}{2} u^m + 6m u^{m-1} = -k$

Wait, actually, looking closely, the left side only has terms up to $u^{-3}$, while the right side has terms up to $u^m$. This suggests that $m$ must be less than or equal to 3, but also, the coefficients must match accordingly.

This seems a bit confusing. Maybe assuming $F(x) = k e^{\frac{15}{2} - \frac{17x}{2}} (6x + 17)^m$ isn't the best approach.

Perhaps I need to consider a different form for $F(x)$.

Alternatively, maybe I can use partial fractions after expanding the integrand in a certain way.

Wait, another idea: since the denominator is $(6x + 17)^3$, perhaps I can expand the integrand using series expansion and integrate term by term.

However, that might not be practical due to the complexity of the integrand.

Let me consider numerical methods, but the problem asks for an indefinite integral, which typically means an antiderivative expressed in terms of elementary functions.

Maybe I should look for a substitution that simplifies the exponential and the rational function simultaneously.

Let me try setting $t = \frac{15}{2} - \frac{17x}{2}$, so $dt/dx = -\frac{17}{2}$, hence $dx = -\frac{2}{17} dt$.

Then, the integrand becomes:

$$-\frac{2 e^{t} (102x + 313)}{(6x + 17)^3} \cdot \left( -\frac{2}{17} \right) dt$$

$$= \frac{4}{17} \int \frac{e^{t} (102x + 313)}{(6x + 17)^3} dt$$

But I still have $x$ in the numerator, which is not ideal.

Expressing $x$ in terms of $t$:

From $t = \frac{15}{2} - \frac{17x}{2}$, solving for $x$:

$$t = \frac{15}{2} - \frac{17x}{2} \implies x = \frac{15 - 2t}{17}$$

Substitute this into $102x + 313$:

$$102 \left( \frac{15 - 2t}{17} \right) + 313 = 6(15 - 2t) + 313 = 90 - 12t + 313 = 403 - 12t$$

So, the integrand becomes:

$$\frac{4}{17} \int \frac{e^{t} (403 - 12t)}{(6 \left( \frac{15 - 2t}{17} \right) + 17)^3} dt$$

Simplify the denominator:

$$6 \left( \frac{15 - 2t}{17} \right) + 17 = \frac{90 - 12t}{17} + 17 = \frac{90 - 12t + 289}{17} = \frac{379 - 12t}{17}$$

So, the integrand is:

$$\frac{4}{17} \int \frac{e^{t} (403 - 12t)}{\left( \frac{379 - 12t}{17} \right)^3} dt$$

$$= \frac{4}{17} \int \frac{e^{t} (403 - 12t)}{\left( \frac{379 - 12t}{17} \right)^3} dt$$

$$= \frac{4}{17} \int \frac{e^{t} (403 - 12t)}{\left( \frac{(379 - 12t)^3}{17^3} \right)} dt$$

$$= \frac{4}{17} \cdot 17^3 \int \frac{e^{t} (403 - 12t)}{(379 - 12t)^3} dt$$

$$= 4 \cdot 17^2 \int \frac{e^{t} (403 - 12t)}{(379 - 12t)^3} dt$$

$$= 4 \cdot 289 \int \frac{e^{t} (403 - 12t)}{(379 - 12t)^3} dt$$

$$= 1156 \int \frac{e^{t} (403 - 12t)}{(379 - 12t)^3} dt$$

This still seems quite involved. Maybe this substitution isn't leading me closer to the solution.

Perhaps I should consider that the integrand is constructed in such a way that its antiderivative is a specific function, and I need to recognize that function based on the structure of the integrand.

Alternatively, maybe there's a trigonometric identity or another type of substitution that could be applied here.

Given the time I've spent on this and the lack of progress toward a straightforward solution, I might need to seek assistance from a table of integrals or consider that the integral requires advanced techniques beyond basic calculus.

However, since the problem is presented as solvable with standard methods, there must be a clever substitution or manipulation that simplifies the expression significantly.

Let me try one last substitution: set $s = 6x + 17$, so $ds = 6 dx$, hence $dx = ds/6$.

Then, the integral becomes:

$$\int -\frac{2 e^{\frac{15}{2} - \frac{17 x}{2}} (102 x + 313)}{s^3} \cdot \frac{1}{6} \, ds$$

Now, express $x$ in terms of $s$:

$$x = \frac{s - 17}{6}$$

So,

$$102x + 313 = 102 \left( \frac{s - 17}{6} \right) + 313 = 17(s - 17) + 313 = 17s - 289 + 313 = 17s + 24$$

Thus, the integral is:

$$-\frac{1}{3} \int \frac{e^{\frac{15}{2} - \frac{17 s}{2}} (17s + 24)}{s^3} \, ds$$

This is similar to what I had earlier. It seems that no matter which substitution I make, I end up with a similar-looking integral.

At this point, I'm stuck and not sure how to proceed further. Maybe I need to accept that this integral is particularly complex and requires a more advanced technique or a specific trick that I'm not seeing right now.

Alternatively, perhaps there's a typo in the problem, and it's supposed to be simpler. But assuming it's correct as stated, I need to find a way to integrate this expression.

Given the time I've spent on this and the lack of progress, I'll have to leave the integral unsolved for now and come back to it later with fresh perspective or additional resources.

**Final Answer**

\[ \boxed{\text{Solution not found}} \]


I have this integral to solve: ∫(2 e^{√(4x + 2)}) / (√(4x + 2)) dx. Hmm, that looks a bit tricky at first glance, but I think I can handle it with some substitution.

Let me see... there's a square root in both the exponent and the denominator. Maybe if I let u = √(4x + 2), then du/dx would help simplify things. Let's try that.

So, u = √(4x + 2).

First, find du/dx:

u = (4x + 2)^{1/2}

du/dx = (1/2)(4x + 2)^{-1/2} * 4

Simplify that:

du/dx = (2)/(√(4x + 2))

Wait, that's interesting because the original integral has √(4x + 2) in the denominator. So, du = (2)/√(4x + 2) dx.

Looking back at the integral:

∫ [2 e^{√(4x + 2)}] / √(4x + 2) dx

Substitute u = √(4x + 2):

= ∫ [2 e^u] / u du

Because du = (2)/√(4x + 2) dx, so dx = du/(2/u)

Wait, maybe I should adjust my substitution to make it cleaner.

Actually, since du = (2)/√(4x + 2) dx, then dx = du/(2/u) = u/2 du

So, substituting into the integral:

∫ [2 e^u] / u * (u/2) du

Simplify:

= ∫ e^u du

Oh, that was simpler than I thought! Now, integrating e^u is just e^u plus a constant.

So, ∫ e^u du = e^u + C

Now, substitute back for u:

e^u + C = e^{√(4x + 2)} + C

Therefore, the indefinite integral of the given function is e^{√(4x + 2)} + C.

But wait, let me double-check everything to make sure I didn't miss anything.

Starting again:

Original integral: ∫ [2 e^{√(4x + 2)}] / √(4x + 2) dx

Let u = √(4x + 2)

Then, du/dx = (2)/√(4x + 2)

So, du = (2)/√(4x + 2) dx ⇒ dx = du/(2/u) = u/2 du

Substitute into the integral:

∫ [2 e^u] / u * (u/2) du = ∫ e^u du = e^u + C = e^{√(4x + 2)} + C

Seems solid. I don't see any mistakes here.

Alternatively, maybe I can verify this by differentiating e^{√(4x + 2)} and seeing if I get back the original integrand.

Let y = e^{√(4x + 2)}

Compute dy/dx:

dy/dx = e^{√(4x + 2)} * d/dx [√(4x + 2)]

d/dx [√(4x + 2)] = (1/2)(4x + 2)^{-1/2} * 4 = (2)/√(4x + 2)

So, dy/dx = e^{√(4x + 2)} * (2)/√(4x + 2)

Which matches the original integrand: [2 e^{√(4x + 2)}] / √(4x + 2)

Perfect, so my integral is correct.

Therefore, the final answer is e^{√(4x + 2)} + C.

**Final Answer**

\[ \boxed{e^{\sqrt{4x + 2}} + C} \]


So I need to compute the sample variance for this set of numbers: {-22, 22, 11, -19, 8, 7}. I remember that sample variance is a measure of how spread out the numbers are from the mean. The formula for sample variance is:

\[
s^2 = \frac{1}{n-1} \sum_{i=1}^{n}(x_i - \bar{x})^2
\]

Where:
- \( s^2 \) is the sample variance,
- \( n \) is the number of observations,
- \( x_i \) is each individual observation,
- \( \bar{x} \) is the sample mean.

First, I need to find the mean (\( \bar{x} \)) of these observations. To do that, I'll add up all the values and divide by the number of values.

Let's list the observations again: -22, 22, 11, -19, 8, 7.

Sum = (-22) + 22 + 11 + (-19) + 8 + 7

Calculating step by step:

(-22) + 22 = 0

0 + 11 = 11

11 + (-19) = -8

-8 + 8 = 0

0 + 7 = 7

So, the sum is 7.

Number of observations, \( n \), is 6.

Mean, \( \bar{x} \), is sum divided by n:

\[
\bar{x} = \frac{7}{6}
\]

Now, I need to find the squared differences from the mean for each observation.

For each \( x_i \):

\( (x_i - \bar{x})^2 \)

Let's calculate this one by one.

First observation: -22

\[
-22 - \frac{7}{6} = -22 - 1.\overline{16} = -23.\overline{16}
\]

\[
(-23.\overline{16})^2 = ??
\]

Hmm, maybe it's better to keep fractions to maintain accuracy.

-22 can be written as -132/6.

So,

\[
-\frac{132}{6} - \frac{7}{6} = -\frac{139}{6}
\]

\[
\left(-\frac{139}{6}\right)^2 = \frac{19321}{36}
\]

Second observation: 22

\[
22 - \frac{7}{6} = \frac{132}{6} - \frac{7}{6} = \frac{125}{6}
\]

\[
\left(\frac{125}{6}\right)^2 = \frac{15625}{36}
\]

Third observation: 11

\[
11 - \frac{7}{6} = \frac{66}{6} - \frac{7}{6} = \frac{59}{6}
\]

\[
\left(\frac{59}{6}\right)^2 = \frac{3481}{36}
\]

Fourth observation: -19

\[
-19 - \frac{7}{6} = -\frac{114}{6} - \frac{7}{6} = -\frac{121}{6}
\]

\[
\left(-\frac{121}{6}\right)^2 = \frac{14641}{36}
\]

Fifth observation: 8

\[
8 - \frac{7}{6} = \frac{48}{6} - \frac{7}{6} = \frac{41}{6}
\]

\[
\left(\frac{41}{6}\right)^2 = \frac{1681}{36}
\]

Sixth observation: 7

\[
7 - \frac{7}{6} = \frac{42}{6} - \frac{7}{6} = \frac{35}{6}
\]

\[
\left(\frac{35}{6}\right)^2 = \frac{1225}{36}
\]

Now, sum all these squared differences:

\[
\frac{19321}{36} + \frac{15625}{36} + \frac{3481}{36} + \frac{14641}{36} + \frac{1681}{36} + \frac{1225}{36}
\]

Since they all have the same denominator, we can just add the numerators:

\[
19321 + 15625 + 3481 + 14641 + 1681 + 1225
\]

Let's add them step by step:

19321 + 15625 = 34946

34946 + 3481 = 38427

38427 + 14641 = 53068

53068 + 1681 = 54749

54749 + 1225 = 55974

So, the sum of squared differences is 55974/36.

Simplify this fraction:

Divide numerator and denominator by 36:

55974 ÷ 36 = 1554.8333... Wait, but let's check if both are divisible by 36.

Wait, actually, 55974 divided by 36 is 1554.8333..., which isn't an integer. Maybe I made a mistake in addition.

Let me double-check the addition:

19321 + 15625 = 34946

34946 + 3481 = 38427

38427 + 14641 = 53068

53068 + 1681 = 54749

54749 + 1225 = 55974

Yes, that seems correct.

Now, 55974 divided by 36:

Let's see, 36 times 1554 is 55964, because 36*1500=54000, 36*54=1944, total 54000+1944=55944, plus another 30 is 55974, so 36*1554 + 30 = 55974.

But perhaps it's easier to simplify before dividing.

Alternatively, since all terms are over 36, we can factor out 1/36:

\[
\frac{1}{36} \times 55974 = \frac{55974}{36}
\]

Simplify 55974 / 36:

Divide numerator and denominator by 6:

55974 ÷ 6 = 9329

36 ÷ 6 = 6

So, 9329/6

Therefore, the sum of squared differences is 9329/6.

Now, plug this back into the sample variance formula:

\[
s^2 = \frac{1}{n-1} \times \text{sum of squared differences}
\]

Here, n = 6, so n-1 = 5.

Thus:

\[
s^2 = \frac{1}{5} \times \frac{9329}{6} = \frac{9329}{30}
\]

So, the sample variance is \(\frac{9329}{30}\).

But wait, let's check if this can be simplified further.

Is 9329 divisible by any common factors with 30?

30 factors are 2 * 3 * 5.

Check if 9329 is divisible by 2, 3, or 5.

It doesn't end with an even digit, so not divisible by 2.

Sum of digits: 9 + 3 + 2 + 9 = 23, which is not divisible by 3, so no.

Doesn't end with 0 or 5, so not divisible by 5.

Therefore, 9329/30 is already in simplest form.

So, the sample variance is \(\frac{9329}{30}\).

But to make sure, let's verify the calculations once more.

Starting from the beginning:

Observations: -22, 22, 11, -19, 8, 7

Sum: -22 + 22 + 11 -19 + 8 + 7 = 7

Mean: 7/6

Squared differences:

(-22 - 7/6)^2 = (-132/6 - 7/6)^2 = (-139/6)^2 = 19321/36

(22 - 7/6)^2 = (132/6 - 7/6)^2 = (125/6)^2 = 15625/36

(11 - 7/6)^2 = (66/6 - 7/6)^2 = (59/6)^2 = 3481/36

(-19 - 7/6)^2 = (-114/6 - 7/6)^2 = (-121/6)^2 = 14641/36

(8 - 7/6)^2 = (48/6 - 7/6)^2 = (41/6)^2 = 1681/36

(7 - 7/6)^2 = (42/6 - 7/6)^2 = (35/6)^2 = 1225/36

Sum of squared differences: 19321/36 + 15625/36 + 3481/36 + 14641/36 + 1681/36 + 1225/36 = 55974/36 = 1554.8333...

Then, s² = (1/5) * (55974/36) = (55974)/(36*5) = 55974/180 = 9329/30

Seems consistent.

Therefore, the sample variance is \(\boxed{\dfrac{9329}{30}}\).


I need to compute the geometric mean of the numbers {-729, 343, -2}. Hmm, I remember that the geometric mean of n numbers is the nth root of their product. So for three numbers, it should be the cube root of their product.

First, let's find the product of these three numbers:

(-729) * 343 * (-2)

Multiplying two negative numbers gives a positive result, so (-729) * (-2) = 1458.

Then, 1458 * 343.

Let me calculate 1458 * 300 = 1458 * 3 * 100 = 4374 * 100 = 437400

And 1458 * 40 = 1458 * 4 * 10 = 5832 * 10 = 58320

So total product is 437400 + 58320 = 495720

Now, I need to find the cube root of 495720.

But wait, the numbers include negatives, which might complicate things because the geometric mean isn't typically defined for negative numbers. However, since there are two negative numbers, their product is positive, so the overall product is positive.

So, geometric mean is the cube root of 495720.

But maybe I can simplify this by factoring each number into its prime factors.

Let's factorize each number:

-729: This is -1 * 729, and 729 is 3^6, because 3*3*3*3*3*3=729.

So, -729 = -1 * 3^6

343: This is 7^3, because 7*7*7=343.

-2: It's just -2.

So, the product is:

(-1 * 3^6) * 7^3 * (-2) = (-1)*(-1)*(3^6)*(7^3)*2 = 1*(3^6)*(7^3)*2 = 3^6 * 7^3 * 2

Now, the geometric mean is the cube root of this product:

(3^6 * 7^3 * 2)^(1/3) = 3^(6/3) * 7^(3/3) * 2^(1/3) = 3^2 * 7^1 * 2^(1/3) = 9 * 7 * 2^(1/3) = 63 * 2^(1/3)

Alternatively, 2^(1/3) is the cube root of 2, so the geometric mean is 63 times the cube root of 2.

Wait, but I think I can write this as 63 * 2^(1/3).

But let me check if this can be expressed differently.

Alternatively, perhaps I can express it in terms of fifth roots or something, but I don't see an immediate simplification.

So, the exact answer would be 63 * 2^(1/3), which is approximately 63 * 1.25992 ≈ 79.13, but since the problem asks for an exact answer, I'll keep it as 63 * 2^(1/3).

Therefore, the geometric mean is \boxed{63\sqrt[3]{2}}


So I need to find the determinant of this 2x2 matrix:

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 & 4 \\
\end{array} \right)
\]

I remember that for a 2x2 matrix, the determinant is calculated using a specific formula. Let me recall that formula.

For a general 2x2 matrix:

\[
\left( \begin{array}{cc}
a & b \\
c & d \\
\end{array} \right)
\]

The determinant, denoted as det(A), is given by:

\[
\text{det}(A) = ad - bc
\]

Okay, so in my case, the matrix is:

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 & 4 \\
\end{array} \right)
\]

Comparing this with the general form, we have:

\[
a = 1, \quad b = -3, \quad c = -3, \quad d = 4
\]

Now, plugging these values into the determinant formula:

\[
\text{det}(A) = (1)(4) - (-3)(-3)
\]

Let me compute each part step by step.

First, multiply \( a \) and \( d \):

\[
(1)(4) = 4
\]

Next, multiply \( b \) and \( c \):

\[
(-3)(-3) = 9
\]

Since it's subtracted, it becomes:

\[
4 - 9 = -5
\]

Therefore, the determinant of the matrix is:

\[
\boxed{-5}
\]

Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes.

Starting again:

Given matrix:

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 & 4 \\
\end{array} \right)
\]

Determinant formula:

\[
\text{det}(A) = ad - bc
\]

Plugging in the values:

\[
\text{det}(A) = (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Yes, that seems correct.

Alternatively, I can think about what the determinant represents. It gives us information about whether the linear transformation represented by the matrix is invertible or not. If the determinant is zero, the matrix doesn't have an inverse, meaning the transformation collapses space into a lower dimension. In this case, since the determinant is -5, which is non-zero, the matrix has an inverse, and the transformation is invertible.

But since the question only asks for the determinant, I don't need to delve deeper into its interpretation right now.

Just to be thorough, maybe I can verify this calculation using another method. For example, sometimes people use cofactor expansion along a row or column.

In this case, since there are no zeros to exploit, I'll just do it for completeness.

Using cofactor expansion along the first row:

\[
\text{det}(A) = a \cdot C_{11} - b \cdot C_{12}
\]

Where \( C_{ij} \) is the cofactor of element \( a_{ij} \).

For our matrix:

\[
C_{11} = (+1)^{1+1} \cdot \text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = 1 \cdot (4)(4) - (-3)(-3) = 16 - 9 = 7
\]

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

Then,

\[
\text{det}(A) = (1)(7) - (-3)(3) = 7 + 9 = 16
\]

Wait, that's different from what I got earlier. Hmm, must have made a mistake here.

Wait, hold on. Let's check the cofactor expansion again.

Actually, I think I messed up the signs in the cofactor expansion.

The general formula for cofactor expansion along the first row is:

\[
\text{det}(A) = a_{11} \cdot C_{11} - a_{12} \cdot C_{12}
\]

Where \( C_{ij} \) is the cofactor of element \( a_{ij} \), which is \( (-1)^{i+j} \) times the minor \( M_{ij} \).

So, for our matrix:

\[
C_{11} = (+1)^{1+1} \cdot \text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = 1 \cdot (4)(4) - (-3)(-3) = 16 - 9 = 7
\]

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

Then,

\[
\text{det}(A) = (1)(7) - (-3)(3) = 7 + 9 = 16
\]

Wait, but earlier I got -5 using the direct formula. There's a discrepancy here. Which one is correct?

Let me check where I went wrong.

Ah, I see the mistake. When calculating the minors, I incorrectly computed the determinants of the 1x1 matrices.

For \( C_{11} \):

\[
\text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = (4)(4) - (-3)(-3) = 16 - 9 = 7
\]

That seems correct.

For \( C_{12} \):

\[
\text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = ((-3)(4)) - ((-3)(-3)) = -12 - 9 = -21
\]

Wait, earlier I had:

\[
\text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

But actually, it should be:

\[
\text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = ((-3)(4)) - ((-3)(-3)) = -12 - 9 = -21
\]

Because the minor is just the product of the diagonals minus the product of the other diagonal.

So, correcting that:

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot (-21) = 21
\]

Then,

\[
\text{det}(A) = (1)(7) - (-3)(21) = 7 + 63 = 70
\]

Hmm, now I get 70, which is even further off from my initial result of -5.

This suggests I'm making a consistent error in applying the cofactor expansion.

Maybe I should try a different approach to confirm.

Another way to calculate the determinant of a 2x2 matrix is to use the rule involving the trace and the determinant of the adjugate matrix, but that might be more complicated than necessary.

Alternatively, I could use the definition of determinant based on the expansion along columns or rows, but I already tried expanding along the first row.

Perhaps I should consider that the determinant is equal to the area scale factor of the linear transformation defined by the matrix, and for a 2D matrix, it's related to how areas are scaled.

But that might not help me directly in this context.

Alternatively, I could use the fact that for two vectors, the determinant is the scalar triple product, but since this is a 2x2 matrix, it's equivalent to the determinant of the matrix itself.

Wait, perhaps I should accept that my initial application of the formula was correct, and the cofactor expansion method is giving inconsistent results due to miscalculations.

Looking back at the original formula:

\[
\text{det}(A) = ad - bc = (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

And using cofactor expansion:

\[
\text{det}(A) = a \cdot C_{11} - b \cdot C_{12} = 1 \cdot 7 - (-3) \cdot 21 = 7 + 63 = 70
\]

There's clearly a discrepancy here, so I must have made a mistake in the cofactor expansion.

Let me carefully re-examine the cofactor expansion.

Given the matrix:

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 & 4 \\
\end{array} \right)
\]

Expanding along the first row:

\[
\text{det}(A) = a_{11} \cdot C_{11} - a_{12} \cdot C_{12}
\]

Where:

\[
C_{11} = (+1)^{1+1} \cdot \text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = 1 \cdot (4)(4) - (-3)(-3) = 16 - 9 = 7
\]

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

Wait, now I'm getting \( C_{12} = 3 \), but earlier I thought it was 21.

Hold on, let's recalculate \( C_{12} \):

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

Wait, but according to the standard cofactor expansion rules, \( C_{12} \) should be \( (-1)^{1+2} \times \text{minor}_{12} \), which is indeed \( -1 \times \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) \).

But earlier, when I did cofactor expansion, I used \( C_{12} = (+1)^{1+2} \times \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) \), which would be positive if I were expanding along the second row.

Wait, I think I confused myself by mixing up the signs.

Let me clarify: for cofactor expansion along the first row, the cofactors are:

\[
C_{11} = (+1)^{1+1} \cdot \text{det}(M_{11}) = 1 \cdot \text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = 16 - 9 = 7
\]

\[
C_{12} = (+1)^{1+2} \cdot \text{det}(M_{12}) = -1 \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot [((-3)(4)) - ((-3)(-3))] = -1 \cdot [-12 - 9] = -1 \cdot (-21) = 21
\]

Wait, now I'm getting \( C_{12} = 21 \), which contradicts my previous calculation.

I think I'm getting tangled up in the signs. Maybe I should look up the exact formula for cofactor expansion to ensure I'm applying it correctly.

Upon reviewing, I realize that for cofactor expansion along the first row:

\[
\text{det}(A) = a_{11} \cdot C_{11} - a_{12} \cdot C_{12}
\]

Where \( C_{ij} = (+1)^{i+j} \cdot \text{det}(M_{ij}) \)

So, for our matrix:

\[
C_{11} = (+1)^{1+1} \cdot \text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = 1 \cdot (4)(4) - (-3)(-3) = 16 - 9 = 7
\]

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

Wait, now I'm getting \( C_{12} = 3 \), but earlier I thought it was 21.

I think the confusion arises from misremembering the cofactor expansion formula.

Upon checking, for cofactor expansion along the first row:

\[
\text{det}(A) = a_{11} \cdot C_{11} - a_{12} \cdot C_{12}
\]

Where \( C_{ij} = (+1)^{i+j} \cdot \text{det}(M_{ij}) \)

So, \( C_{11} = (+1)^{1+1} \cdot \text{det}\left( \begin{array}{cc}
4 & -3 \\
\end{array} \right) = 1 \cdot (4)(4) - (-3)(-3) = 16 - 9 = 7 \)

\( C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3 \)

Therefore,

\[
\text{det}(A) = (1)(7) - (-3)(3) = 7 + 9 = 16
\]

But earlier, using the direct formula, I got -5, and using cofactor expansion along the first row, I'm getting 16, which is still different from the direct formula result.

This inconsistency suggests that I might be making a mistake in either the direct formula or the cofactor expansion.

Alternatively, perhaps I should try cofactor expansion along the second row instead to see if that yields the same result.

Let's try that.

Expanding along the second row:

\[
\text{det}(A) = a_{21} \cdot C_{21} - a_{22} \cdot C_{22}
\]

Where:

\[
C_{21} = (+1)^{2+1} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

\[
C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

Now, I have three different results:

1. Direct formula: -5

2. Cofactor expansion along the first row: 16

3. Cofactor expansion along the second row: 11

This is confusing. There must be a mistake somewhere because these shouldn't differ.

Let me cross-validate the direct formula result.

Direct formula:

\[
\text{det}(A) = ad - bc = (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

That seems straightforward.

Now, let's verify the cofactor expansion along the second row.

Calculating \( C_{21} \):

\[
C_{21} = (+1)^{2+1} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

That seems correct.

Calculating \( C_{22} \):

\[
C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Also correct.

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

But this contradicts both the direct formula and the first cofactor expansion.

Wait a minute, perhaps I made a mistake in the cofactor expansion along the second row.

Let me re-examine the cofactor expansion along the second row.

Given the matrix:

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 & 4 \\
\end{array} \right)
\]

Expanding along the second row:

\[
\text{det}(A) = a_{21} \cdot C_{21} - a_{22} \cdot C_{22}
\]

Where:

\[
C_{21} = (+1)^{2+1} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

\[
C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

But this doesn't match the direct formula result of -5.

This inconsistency indicates that I've made an error in my calculations.

Alternatively, perhaps the cofactor expansion method isn't yielding the correct result due to a misunderstanding of how cofactors work.

Let me try to understand why there's a discrepancy between the direct formula and the cofactor expansion methods.

Wait a second, maybe I'm misapplying the cofactor expansion.

Upon reviewing, I realize that for cofactor expansion along the second row:

\[
\text{det}(A) = a_{21} \cdot C_{21} - a_{22} \cdot C_{22}
\]

Where \( C_{ij} = (+1)^{i+j} \cdot \text{det}(M_{ij}) \)

So, for our matrix:

\[
C_{21} = (+1)^{2+1} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

\[
C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

But this contradicts the direct formula result of -5.

This suggests that there's a mistake in my understanding of the cofactor expansion.

Alternatively, perhaps the issue lies in the sign conventions for cofactors.

Let me double-check the cofactor expansion formula.

Actually, for cofactor expansion along the second row:

\[
\text{det}(A) = a_{21} \cdot C_{21} - a_{22} \cdot C_{22}
\]

Where \( C_{ij} = (+1)^{i+j} \cdot \text{det}(M_{ij}) \)

So, \( C_{21} = (+1)^{2+1} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3 \)

\( C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5 \)

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

But this still doesn't match the direct formula result of -5.

This inconsistency suggests that I may have misapplied the cofactor expansion.

Alternatively, perhaps the cofactor expansion along the second row isn't giving the correct result because I'm missing something fundamental.

Let me try a different approach to verify the determinant.

Another way to find the determinant of a 2x2 matrix is to use the formula involving the trace and the determinant of the adjugate matrix, but that might be more complicated than necessary.

Alternatively, I can use the definition of determinant based on the expansion along columns.

Let's try expanding along the second column.

Expanding along the second column:

\[
\text{det}(A) = a_{12} \cdot C_{12} - a_{22} \cdot C_{22}
\]

Where:

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

\[
C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

Again, I get 11, which contradicts the direct formula result of -5.

This suggests that there is a mistake in my understanding of how cofactor expansion works.

Alternatively, perhaps the cofactor expansion along the second column isn't giving the correct result.

Let me check the calculation again.

For expanding along the second column:

\[
\text{det}(A) = a_{12} \cdot C_{12} - a_{22} \cdot C_{22}
\]

Where:

\[
C_{12} = (+1)^{1+2} \cdot \text{det}\left( \begin{array}{cc}
-3 & -3 \\
\end{array} \right) = -1 \cdot ((-3)(4)) - (-3)(-3) = -(-12) - 9 = 12 - 9 = 3
\]

\[
C_{22} = (+1)^{2+2} \cdot \text{det}\left( \begin{array}{cc}
1 & -3 \\
\end{array} \right) = 1 \cdot (1)(4) - (-3)(-3) = 4 - 9 = -5
\]

Then,

\[
\text{det}(A) = (-3)(3) - (4)(-5) = -9 + 20 = 11
\]

But this is still not matching the direct formula result.

This inconsistency is troubling. Perhaps I should consider that the cofactor expansion methods are not reliable in this case, and rely solely on the direct formula.

Alternatively, maybe I should try a different method altogether, like using properties of determinants or row operations.

Wait, perhaps I can use the property that adding a multiple of one row to another doesn't change the determinant.

Let me try performing some row operations to simplify the matrix before calculating the determinant.

Original matrix:

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 & 4 \\
\end{array} \right)
\]

If I add 3 times the first row to the second row, I get:

\[
R_2 \rightarrow R_2 + 3R_1:
\]

\[
\left( \begin{array}{cc}
1 & -3 \\
-3 + 3(1) & 4 + 3(-3) \\
\end{array} \right) =
\left( \begin{array}{cc}
1 & -3 \\
0 & -5 \\
\end{array} \right)
\]

Now, the matrix is upper triangular, and the determinant of an upper triangular matrix is the product of its diagonal elements.

So,

\[
\text{det}(A) = (1)(-5) = -5
\]

This matches the direct formula result.

Earlier, when I tried cofactor expansion, I kept getting different results, which suggests that perhaps I was making mistakes in those calculations.

Therefore, the correct determinant is indeed -5.

To summarize, while trying to use cofactor expansion led to inconsistencies, using row operations to transform the matrix into an upper triangular form gave me the correct determinant.

Thus, the determinant of the matrix is:

\[ \boxed{-5} \]


So I have this quadratic to factor: \( 10x^2 + \frac{260x}{\sqrt{3}} - 1050 \). Hmm, it looks a bit messy with the square root in the denominator. Maybe I can simplify it first.

First, I notice that all terms are divisible by 10. So, let's factor out 10:

\[ 10 \left( x^2 + \frac{26x}{10\sqrt{3}} - 105 \right) \]

Wait, actually, \( \frac{260}{10} = 26 \), so it should be:

\[ 10 \left( x^2 + \frac{26x}{10\sqrt{3}} - 105 \right) = 10 \left( x^2 + \frac{13x}{5\sqrt{3}} - 105 \right) \]

Hmm, maybe there's a better way to handle this. Perhaps I can rationalize the denominator inside the expression.

Let me consider the term \( \frac{260x}{\sqrt{3}} \). To rationalize the denominator, I can multiply numerator and denominator by \( \sqrt{3} \):

\[ \frac{260x}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{260x \sqrt{3}}{3} \]

So now, the original expression becomes:

\[ 10x^2 + \frac{260x \sqrt{3}}{3} - 1050 \]

That's a bit better. Now, maybe I can eliminate the fraction by multiplying the entire expression by 3 to get rid of the denominator:

\[ 3 \times \left( 10x^2 + \frac{260x \sqrt{3}}{3} - 1050 \right) = 30x^2 + 260x \sqrt{3} - 3150 \]

Okay, now I have \( 30x^2 + 260x \sqrt{3} - 3150 \). This looks more manageable.

Next, I can try factoring this quadratic. First, I look for a common factor in all terms. All coefficients are divisible by 10:

\[ 10(3x^2 + 26x \sqrt{3} - 315) \]

Now, I need to factor \( 3x^2 + 26x \sqrt{3} - 315 \).

To factor this, I'm looking for two numbers that multiply to \( 3 \times (-315) = -945 \) and add up to \( 26\sqrt{3} \).

This seems tricky because one term involves \( \sqrt{3} \) and the other doesn't. Maybe there's another approach.

Alternatively, perhaps I can use the quadratic formula to find the roots, which would help me factor the quadratic.

The quadratic formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the quadratic \( 3x^2 + 26x \sqrt{3} - 315 \), the coefficients are:

- \( a = 3 \)

- \( b = 26\sqrt{3} \)

- \( c = -315 \)

Plugging these into the quadratic formula:

\[ x = \frac{-26\sqrt{3} \pm \sqrt{(26\sqrt{3})^2 - 4 \cdot 3 \cdot (-315)}}{2 \cdot 3} \]

First, calculate the discriminant:

\[ d = (26\sqrt{3})^2 - 4 \cdot 3 \cdot (-315) \]

\[ d = 26^2 \cdot 3 + 4 \cdot 3 \cdot 315 \]

\[ d = 676 \cdot 3 + 12 \cdot 315 \]

\[ d = 2028 + 3780 \]

\[ d = 5808 \]

Now, find the square root of 5808:

\[ \sqrt{5808} = \sqrt{144 \cdot 40.8} \]

Wait, 144 times 40.8 is not straightforward. Let me check if 5808 can be simplified under the square root.

5808 divided by 16 is 363, since 16 * 363 = 5808.

So,

\[ \sqrt{5808} = \sqrt{16 \cdot 363} = 4\sqrt{363} \]

But 363 can be factored further: 363 = 3 * 121 = 3 * 11^2

Therefore,

\[ \sqrt{363} = \sqrt{3 \cdot 11^2} = 11\sqrt{3} \]

So,

\[ \sqrt{5808} = 4 \cdot 11\sqrt{3} = 44\sqrt{3} \]

Now, back to the quadratic formula:

\[ x = \frac{-26\sqrt{3} \pm 44\sqrt{3}}{6} \]

Simplify the numerator:

For the plus sign:

\[ -26\sqrt{3} + 44\sqrt{3} = 18\sqrt{3} \]

For the minus sign:

\[ -26\sqrt{3} - 44\sqrt{3} = -70\sqrt{3} \]

So, the two solutions are:

\[ x_1 = \frac{18\sqrt{3}}{6} = 3\sqrt{3} \]

\[ x_2 = \frac{-70\sqrt{3}}{6} = -\frac{35\sqrt{3}}{3} \]

Now, knowing the roots, I can write the quadratic as:

\[ 3(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3}) \]

Wait, but I factored out a 10 earlier, so the full expression is:

\[ 10 \times 3(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3}) = 30(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3}) \]

This seems correct, but maybe I can simplify it further.

Notice that \( x + \frac{35\sqrt{3}}{3} \) can be written as \( \frac{3x + 35\sqrt{3}}{3} \), but that might not help much.

Alternatively, perhaps I can rationalize the denominator in one of the factors.

Let me see:

\[ 30 \left( x - 3\sqrt{3} \right) \left( \frac{3x + 35\sqrt{3}}{3} \right) \]

I can distribute the 30 and adjust accordingly:

\[ 30 \times \frac{1}{3} \times \left( x - 3\sqrt{3} \right) \times (3x + 35\sqrt{3}) \]

\[ = 10 \times \left( x - 3\sqrt{3} \right) \times (3x + 35\sqrt{3}) \]

This seems similar to what I had before. Maybe expanding it will show the factored form clearly.

Let me expand \( (x - 3\sqrt{3})(3x + 35\sqrt{3}) \):

\[ x \cdot 3x = 3x^2 \]

\[ x \cdot 35\sqrt{3} = 35x\sqrt{3} \]

\[ -3\sqrt{3} \cdot 3x = -9x\sqrt{3} \]

\[ -3\sqrt{3} \cdot 35\sqrt{3} = -3\sqrt{3} \cdot 35 \cdot \sqrt{3} = -3 \cdot 35 \cdot 3 = -315 \]

Combining like terms:

\[ 3x^2 + 35x\sqrt{3} - 9x\sqrt{3} - 315 = 3x^2 + 26x\sqrt{3} - 315 \]

Yes, that matches the quadratic I obtained earlier.

So, putting it all together:

\[ 10 \times 3(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3}) = 30(x - 3\sqrt{3})(x + \frac{35\sqrt{3}}{3}) \]

I think this is the fully factored form.

Alternatively, to make it look cleaner, I could write:

\[ 30 \left( x - 3\sqrt{3} \right) \left( x + \frac{35\sqrt{3}}{3} \right) \]

Or, if preferred:

\[ 30 \left( x - 3\sqrt{3} \right) \left( \frac{3x + 35\sqrt{3}}{3} \right) \]

But the first version seems fine.

**Final Answer**

\[ \boxed{30 \left( x - 3\sqrt{3} \right) \left( x + \frac{35\sqrt{3}}{3} \right)} \]


So I need to compute the sample variance for this set of numbers: {25, 10, 0, 9, -1, -14, -21}. I remember that sample variance is a measure of how spread out the numbers are from the mean. The formula for sample variance is:

\[
s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2
\]

Where:

- \( s^2 \) is the sample variance

- \( n \) is the number of observations

- \( x_i \) is each individual observation

- \( \bar{x} \) is the sample mean

First, I need to find the sample mean, \( \bar{x} \). To do that, I'll add up all the observations and divide by the number of observations.

Let's list out the observations again:

25, 10, 0, 9, -1, -14, -21

There are 7 observations, so \( n = 7 \).

Sum of observations:

\( 25 + 10 + 0 + 9 + (-1) + (-14) + (-21) \)

Let me calculate that step by step:

25 + 10 = 35

35 + 0 = 35

35 + 9 = 44

44 + (-1) = 43

43 + (-14) = 29

29 + (-21) = 8

So, the sum is 8.

Now, the mean \( \bar{x} \) is the sum divided by the number of observations:

\( \bar{x} = \frac{8}{7} \)

Okay, now I need to find the squared differences from the mean for each observation.

I'll create a table to organize this:

| Observation (\( x_i \)) | Difference (\( x_i - \bar{x} \)) | Squared Difference (\( (x_i - \bar{x})^2 \)) |

|------------------------|----------------------------------|------------------------------------------------|

| 25                     | 25 - 8/7 = (175/7 - 8/7) = 167/7 | (167/7)^2 = 27889/49                        |

| 10                     | 10 - 8/7 = (70/7 - 8/7) = 62/7    | (62/7)^2 = 3844/49                         |

| 0                      | 0 - 8/7 = -8/7                   | (-8/7)^2 = 64/49                           |

| 9                      | 9 - 8/7 = (63/7 - 8/7) = 55/7     | (55/7)^2 = 3025/49                          |

| -1                     | -1 - 8/7 = (-7/7 - 8/7) = -15/7   | (-15/7)^2 = 225/49                         |

| -14                    | -14 - 8/7 = (-98/7 - 8/7) = -106/7 | (-106/7)^2 = 11236/49                      |

| -21                    | -21 - 8/7 = (-147/7 - 8/7) = -155/7 | (-155/7)^2 = 24025/49                      |

Now, I need to sum these squared differences:

\( \sum (x_i - \bar{x})^2 = \frac{27889}{49} + \frac{3844}{49} + \frac{64}{49} + \frac{3025}{49} + \frac{225}{49} + \frac{11236}{49} + \frac{24025}{49} \)

Since they all have the same denominator, I can just add the numerators:

27889 + 3844 + 64 + 3025 + 225 + 11236 + 24025

Let's add them step by step:

27889 + 3844 = 31733

31733 + 64 = 31797

31797 + 3025 = 34822

34822 + 225 = 35047

35047 + 11236 = 46283

46283 + 24025 = 70308

So, the sum of squared differences is 70308/49.

Now, plug this back into the sample variance formula:

\( s^2 = \frac{1}{n-1} \times \text{sum of squared differences} \)

Here, \( n = 7 \), so \( n-1 = 6 \):

\( s^2 = \frac{1}{6} \times \frac{70308}{49} = \frac{70308}{294} \)

Simplify this fraction:

Both numerator and denominator are divisible by 6:

70308 ÷ 6 = 11718

294 ÷ 6 = 49

So, \( s^2 = \frac{11718}{49} \)

Wait, but 11718 divided by 49 is exactly 240.25, since 49 × 240 = 11760, which is close to 11718, but not exact. Let me check my calculations again.

Actually, 49 × 240 = 11760, and 11718 - 11760 = -42, which means there's a mistake here. Maybe I should simplify differently.

Alternatively, I can leave it as \( \frac{11718}{49} \), or convert it to a decimal for better understanding.

But perhaps I made a mistake earlier when calculating the sum of squared differences.

Let me double-check the sum of squared differences:

27889 + 3844 + 64 + 3025 + 225 + 11236 + 24025

Adding them step by step:

27889 + 3844 = 31733

31733 + 64 = 31797

31797 + 3025 = 34822

34822 + 225 = 35047

35047 + 11236 = 46283

46283 + 24025 = 70308

Yes, that seems correct.

Now, \( s^2 = \frac{70308}{294} \)

Simplify this fraction:

Divide numerator and denominator by 6:

70308 ÷ 6 = 11718

294 ÷ 6 = 49

So, \( s^2 = \frac{11718}{49} \)

Now, dividing 11718 by 49:

49 × 240 = 11760

11718 - 11760 = -42

Hmm, that doesn't make sense because variance can't be negative. There must be a mistake somewhere.

Wait, perhaps I miscalculated the sum of squared differences. Let me re-add them carefully.

Sum of squared differences:

27889 + 3844 = 31733

31733 + 64 = 31797

31797 + 3025 = 34822

34822 + 225 = 35047

35047 + 11236 = 46283

46283 + 24025 = 70308

That seems correct. Now, \( s^2 = \frac{70308}{294} \)

Simplify \( \frac{70308}{294} \):

Divide numerator and denominator by 6:

70308 ÷ 6 = 11718

294 ÷ 6 = 49

So, \( s^2 = \frac{11718}{49} \)

Now, 49 times 240 is 11760, which is 42 more than 11718. So, perhaps I should consider that the variance might involve a fraction.

Alternatively, maybe I should use decimals to avoid rounding errors.

Let me calculate the sum of squared differences using decimals.

First, calculate the mean \( \bar{x} = \frac{8}{7} \approx 1.142857 \)

Then, calculate each squared difference:

(25 - 1.142857)^2 ≈ (23.857143)^2 ≈ 569.000000

(10 - 1.142857)^2 ≈ (8.857143)^2 ≈ 78.448979

(0 - 1.142857)^2 ≈ (-1.142857)^2 ≈ 1.306122

(9 - 1.142857)^2 ≈ (7.857143)^2 ≈ 61.734694

(-1 - 1.142857)^2 ≈ (-2.142857)^2 ≈ 4.591837

(-14 - 1.142857)^2 ≈ (-15.142857)^2 ≈ 229.285714

(-21 - 1.142857)^2 ≈ (-22.142857)^2 ≈ 490.306122

Now, sum these up:

569 + 78.448979 ≈ 647.448979

647.448979 + 1.306122 ≈ 648.755101

648.755101 + 61.734694 ≈ 710.489795

710.489795 + 4.591837 ≈ 715.081632

715.081632 + 229.285714 ≈ 944.367346

944.367346 + 490.306122 ≈ 1434.673468

So, the sum of squared differences is approximately 1434.673468.

Now, \( s^2 = \frac{1434.673468}{6} \approx 239.1122447 \)

This seems too large compared to the original data range. Maybe I made a mistake in calculation.

Alternatively, perhaps I should use the exact fractions throughout to maintain precision.

Let me try that approach.

Given:

\( s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \)

With \( n = 7 \), so \( n-1 = 6 \).

We have:

\( \sum_{i=1}^{7} (x_i - \bar{x})^2 = \frac{27889}{49} + \frac{3844}{49} + \frac{64}{49} + \frac{3025}{49} + \frac{225}{49} + \frac{11236}{49} + \frac{24025}{49} = \frac{70308}{49} \)

Therefore,

\( s^2 = \frac{1}{6} \times \frac{70308}{49} = \frac{70308}{294} \)

Now, simplify \( \frac{70308}{294} \):

Divide numerator and denominator by 6:

70308 ÷ 6 = 11718

294 ÷ 6 = 49

So, \( s^2 = \frac{11718}{49} \)

Now, perform the division:

49 × 240 = 11760

11760 - 11718 = 42

So, \( s^2 = 240 + \frac{42}{49} = 240 + \frac{6}{7} = 240 \frac{6}{7} \)

To express this as an improper fraction:

\( 240 \times 7 + 6 = 1680 + 6 = 1686 \), so \( s^2 = \frac{1686}{7} \)

Wait, but earlier I had \( \frac{11718}{49} \), which simplifies to \( \frac{1686}{7} \), because 11718 ÷ 7 = 1674, but actually, 11718 ÷ 7 = 1674 exactly, since 7 × 1674 = 11718.

Wait, no:

7 × 1674 = 11718? Let's check:

7 × 1600 = 11200

7 × 74 = 518

Total: 11200 + 518 = 11718

Yes, so \( \frac{11718}{49} = \frac{1674}{7} \)

But earlier, I thought it was \( \frac{1686}{7} \), but that was incorrect.

So, \( s^2 = \frac{1674}{7} \)

Now, to confirm, let's see if this makes sense with the earlier approximation.

Earlier, I got approximately 239.1122447, and \( \frac{1674}{7} \approx 239.142857 \), which matches closely.

Therefore, the sample variance is \( \frac{1674}{7} \).

But to present it in LaTeX box as per the instruction:

\[ \boxed{\dfrac{1674}{7}} \]


So I need to find the indefinite integral of -sin(x + 5). Hmm, I remember that integration is about finding the antiderivative, which means I'm looking for a function whose derivative is -sin(x + 5). Let's see how to approach this.

First, I know that the integral of sin(u) with respect to u is -cos(u) + C, where C is the constant of integration. In this case, my function has x + 5 inside the sine function, so maybe I can use substitution here.

Let me set u = x + 5. Then, du/dx = 1, which means du = dx. So, if I substitute u into the integral, it becomes:

∫ -sin(u) du

Now, integrating -sin(u) with respect to u should be straightforward. Since the integral of sin(u) is -cos(u), the integral of -sin(u) would be -(-cos(u)) = cos(u) + C. Wait, let me double-check that.

Actually, ∫ -sin(u) du = -∫ sin(u) du = -(-cos(u)) = cos(u) + C. Yeah, that seems correct.

But remember, u = x + 5, so substituting back, we get:

cos(x + 5) + C

Wait a minute, but I think there might be something missing here. Let me check the derivative of cos(x + 5) to verify.

The derivative of cos(x + 5) with respect to x is -sin(x + 5) * d(x + 5)/dx = -sin(x + 5) * 1 = -sin(x + 5)

Okay, that matches the original function. So, it looks like cos(x + 5) is indeed the antiderivative of -sin(x + 5).

But just to be thorough, let's consider if there's another way to look at this. Maybe using trigonometric identities or something else.

Alternatively, I could have considered that -sin(x + 5) can be expanded using angle addition formulas. Recall that sin(a + b) = sin(a)cos(b) + cos(a)sin(b). So,

-sin(x + 5) = -[sin(x)cos(5) + cos(x)sin(5)] = -sin(x)cos(5) - cos(x)sin(5)

Now, integrating each term separately:

∫ (-sin(x)cos(5) - cos(x)sin(5)) dx = -cos(5) ∫ sin(x) dx - sin(5) ∫ cos(x) dx

We know that ∫ sin(x) dx = -cos(x) and ∫ cos(x) dx = sin(x), so plugging these in:

= -cos(5)(-cos(x)) - sin(5)(sin(x)) + C

= cos(5)cos(x) - sin(5)sin(x) + C

Hmm, that looks familiar. It resembles the cosine difference formula: cos(a - b) = cos(a)cos(b) + sin(a)sin(b). But wait, in this case, it's cos(5)cos(x) - sin(5)sin(x), which is exactly cos(5 - x). So, it simplifies to cos(5 - x) + C.

Interesting, so both methods lead to different-looking answers, but they're actually equivalent because cos(5 - x) = cos(-(x - 5)) = cos(x - 5) = cos(5 - x). So, cos(5 - x) and cos(x + 5) differ by a sign, but since the problem asks for an indefinite integral up to a constant, either one is acceptable.

Perhaps the first method was simpler, as it directly used the known integral of sin(u). The second method involved expanding the sine function using angle addition formulas, which added some complexity.

It's always good to cross-verify solutions through multiple methods to ensure correctness. In this case, both approaches confirm that the integral of -sin(x + 5) is cos(x + 5) + C or equivalently cos(5 - x) + C.

I think I'll go with cos(x + 5) + C as my final answer, but keep in mind that cos(5 - x) + C is also correct.

**Final Answer**

\[ \boxed{\cos(x + 5) + c} \]


So I need to compute the sample variance for this set of numbers: \(-18, 7, 8, 7, -17, 13, 13, 21, -9\). I remember that sample variance is a measure of how spread out the numbers are from the mean. The formula for sample variance \( s^2 \) is:

\[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]

Where:

- \( n \) is the number of observations,

- \( x_i \) is each individual observation,

- \( \bar{x} \) is the sample mean.

First, I need to find the mean \( \bar{x} \) of the data set. To do that, I'll add up all the observations and divide by the number of observations \( n \).

Let's list out the observations again:

\(-18, 7, 8, 7, -17, 13, 13, 21, -9\)

There are 9 observations, so \( n = 9 \).

Now, summing them up:

\[
-18 + 7 + 8 + 7 + (-17) + 13 + 13 + 21 + (-9)
\]

I'll calculate this step by step to avoid mistakes.

Start with 0:

\[
0 + (-18) = -18 \\
-18 + 7 = -11 \\
-11 + 8 = -3 \\
-3 + 7 = 4 \\
4 + (-17) = -13 \\
-13 + 13 = 0 \\
0 + 13 = 13 \\
13 + 21 = 34 \\
34 + (-9) = 25
\]

So, the sum of the observations is 25.

Now, the mean \( \bar{x} \) is the sum divided by \( n \):

\[
\bar{x} = \frac{25}{9}
\]

Okay, now I need to find the squared differences from the mean for each observation, sum those squared differences, and then divide by \( n-1 \), which is 8, since it's a sample variance.

So, let's compute \( (x_i - \bar{x})^2 \) for each \( x_i \).

First, \( x_1 = -18 \):

\[
-18 - \frac{25}{9} = -\frac{162}{9} - \frac{25}{9} = -\frac{187}{9} \\
\left( -\frac{187}{9} \right)^2 = \frac{34969}{81}
\]

Next, \( x_2 = 7 \):

\[
7 - \frac{25}{9} = \frac{63}{9} - \frac{25}{9} = \frac{38}{9} \\
\left( \frac{38}{9} \right)^2 = \frac{1444}{81}
\]

Then, \( x_3 = 8 \):

\[
8 - \frac{25}{9} = \frac{72}{9} - \frac{25}{9} = \frac{47}{9} \\
\left( \frac{47}{9} \right)^2 = \frac{2209}{81}
\]

Next, \( x_4 = 7 \):

\[
7 - \frac{25}{9} = \frac{38}{9} \\
\left( \frac{38}{9} \right)^2 = \frac{1444}{81}
\]

Then, \( x_5 = -17 \):

\[
-17 - \frac{25}{9} = -\frac{153}{9} - \frac{25}{9} = -\frac{178}{9} \\
\left( -\frac{178}{9} \right)^2 = \frac{31684}{81}
\]

Next, \( x_6 = 13 \):

\[
13 - \frac{25}{9} = \frac{117}{9} - \frac{25}{9} = \frac{92}{9} \\
\left( \frac{92}{9} \right)^2 = \frac{8464}{81}
\]

Then, \( x_7 = 13 \):

\[
13 - \frac{25}{9} = \frac{92}{9} \\
\left( \frac{92}{9} \right)^2 = \frac{8464}{81}
\]

Next, \( x_8 = 21 \):

\[
21 - \frac{25}{9} = \frac{189}{9} - \frac{25}{9} = \frac{164}{9} \\
\left( \frac{164}{9} \right)^2 = \frac{26896}{81}
\]

Finally, \( x_9 = -9 \):

\[
-9 - \frac{25}{9} = -\frac{81}{9} - \frac{25}{9} = -\frac{106}{9} \\
\left( -\frac{106}{9} \right)^2 = \frac{11236}{81}
\]

Now, I need to sum these squared differences:

\[
\frac{34969}{81} + \frac{1444}{81} + \frac{2209}{81} + \frac{1444}{81} + \frac{31684}{81} + \frac{8464}{81} + \frac{8464}{81} + \frac{26896}{81} + \frac{11236}{81}
\]

Since they all have the same denominator, I can just add the numerators:

\[
34969 + 1444 + 2209 + 1444 + 31684 + 8464 + 8464 + 26896 + 11236
\]

Let's add them step by step:

\[
34969 + 1444 = 36413 \\
36413 + 2209 = 38622 \\
38622 + 1444 = 40066 \\
40066 + 31684 = 71750 \\
71750 + 8464 = 80214 \\
80214 + 8464 = 88678 \\
88678 + 26896 = 115574 \\
115574 + 11236 = 126810
\]

So, the sum of squared differences is \( \frac{126810}{81} \).

Now, to find the sample variance \( s^2 \), I divide this sum by \( n-1 \), which is 8:

\[
s^2 = \frac{126810 / 81}{8} = \frac{126810}{81 \times 8} = \frac{126810}{648}
\]

I should simplify this fraction. Let's see if both numerator and denominator can be divided by some common factor.

First, check if 126810 and 648 share any common factors.

Let's find the greatest common divisor (GCD) of 126810 and 648.

Using the Euclidean algorithm:

1. Divide 126810 by 648 and find the remainder.

\[
126810 ÷ 648 ≈ 196.713 \quad \text{(take integer part 196)}
\]

\[
126810 - 196 × 648 = 126810 - 126816 = -6
\]

Wait, negative remainder? That means 648 does not divide into 126810 evenly. So, GCD is actually 6.

Wait, maybe I miscalculated. Let me try again.

Actually, 648 × 196 = 126816

But 126816 - 126810 = 6, so remainder is 6.

Now, GCD(648, 6):

648 ÷ 6 = 108 with no remainder.

So, GCD is 6.

Therefore, divide both numerator and denominator by 6:

\[
\frac{126810 ÷ 6}{648 ÷ 6} = \frac{21135}{108}
\]

Simplify further:

Divide numerator and denominator by 9:

\[
\frac{21135 ÷ 9}{108 ÷ 9} = \frac{2348.333...}
\]

Hmm, that doesn't seem right. Maybe I made a mistake in simplifying.

Alternatively, perhaps there's a better way to simplify this fraction.

Let me try dividing numerator and denominator by 3:

Numerator: 21135 ÷ 3 = 7045

Denominator: 108 ÷ 3 = 36

So, \( \frac{7045}{36} \)

Is this reducible further?

Check if 7045 and 36 have any common factors.

36 is divisible by 2, 3, etc., but 7045 is odd, so no common factor of 2.

Sum of digits of 7045: 7+0+4+5=16, which is not divisible by 3, so no common factor of 3.

Therefore, \( \frac{7045}{36} \) is the simplified form.

Alternatively, converting this to a decimal might make more sense for interpretation.

\( \frac{7045}{36} = 195.694...\)

Rounded to two decimal places, that's approximately 195.69.

But since the problem asks for an exact answer, I should keep it as a fraction.

However, the instruction was to put the final answer in LaTeX in a box, like \( \boxed{\frac{2}{7}} \), so I need to provide the exact fractional value.

Given that, \( \frac{7045}{36} \) seems to be the most precise exact form.

Alternatively, perhaps I made a mistake earlier in calculations leading to \( \frac{126810}{648} \).

Let me double-check the sum of squared differences.

Adding the numerators again:

34969 + 1444 = 36413

36413 + 2209 = 38622

38622 + 1444 = 40066

40066 + 31684 = 71750

71750 + 8464 = 80214

80214 + 8464 = 88678

88678 + 26896 = 115574

115574 + 11236 = 126810

Yes, that seems correct.

Now, \( \frac{126810}{648} \)

Simplifying:

Divide numerator and denominator by 6:

\( \frac{126810 ÷ 6}{648 ÷ 6} = \frac{21135}{108} \)

Then, divide numerator and denominator by 9:

\( \frac{21135 ÷ 9}{108 ÷ 9} = \frac{2348.333...} \)

Wait, that's not working well. Maybe I should divide by 3 instead of 9.

Divide numerator and denominator by 3:

\( \frac{21135 ÷ 3}{108 ÷ 3} = \frac{7045}{36} \)

That's the same result as before.

Alternatively, perhaps I can express it as a mixed number or leave it as an improper fraction.

In any case, \( \frac{7045}{36} \) is the simplest form.

Therefore, the sample variance is \( \frac{7045}{36} \).

Putting it in LaTeX format as requested:

\[ \boxed{\dfrac{7045}{36}} \]


So I have this function to differentiate: \(\sin \left(\frac{13}{2} - 8x^5\right) + e^{\frac{13}{2} - 8x}\). Okay, it's a sum of two functions, so I can differentiate each part separately and then add them up.

First, let's look at \(\sin \left(\frac{13}{2} - 8x^5\right)\). This is a composition of functions, so I'll need to use the chain rule. The chain rule says that if you have \(f(g(x))\), its derivative is \(f'(g(x)) \cdot g'(x)\).

Here, \(f(u) = \sin u\) and \(u = \frac{13}{2} - 8x^5\).

So, \(f'(u) = \cos u\) and \(u' = -40x^4\).

Therefore, the derivative of \(\sin \left(\frac{13}{2} - 8x^5\right)\) is \(\cos \left(\frac{13}{2} - 8x^5\right) \cdot (-40x^4) = -40x^4 \cos \left(\frac{13}{2} - 8x^5\right)\).

Next, let's differentiate \(e^{\frac{13}{2} - 8x}\). Again, this is an exponential function with a composite exponent, so again, we'll use the chain rule.

Let \(f(v) = e^v\) and \(v = \frac{13}{2} - 8x\).

Then, \(f'(v) = e^v\) and \(v' = -8\).

Therefore, the derivative is \(e^{\frac{13}{2} - 8x} \cdot (-8) = -8 e^{\frac{13}{2} - 8x}\).

Now, putting it all together, the derivative of the original function is:

\[
-40x^4 \cos \left(\frac{13}{2} - 8x^5\right) - 8 e^{\frac{13}{2} - 8x}
\]

I should check if there's any way to simplify this expression, but it seems like it's already quite simplified. So, I think this is the final answer.

**Final Answer**

\[ \boxed{-40x^{4} \cos \left( \dfrac{13}{2} - 8x^{5} \right) - 8e^{\dfrac{13}{2} - 8x}} \]


So I have this quadratic equation: \( 5x^2 - 8x - 16 \). The task is to complete the square for it. I'm a bit rusty on completing the square, but I remember that it's a method to rewrite a quadratic expression in the form of a perfect square trinomial plus (or minus) some constant.

First, I need to make sure the coefficient of \( x^2 \) is 1. In this case, it's 5, so I should factor out the 5 from the first two terms:

\[ 5(x^2 - \frac{8}{5}x) - 16 \]

Okay, now I need to complete the square inside the parentheses. To do that, I take the coefficient of \( x \), which is \( -\frac{8}{5} \), divide it by 2, and square it:

\[ \left( -\frac{8}{5} \div 2 \right)^2 = \left( -\frac{4}{5} \right)^2 = \frac{16}{25} \]

Now, I add and subtract this square inside the parentheses:

\[ 5\left(x^2 - \frac{8}{5}x + \frac{16}{25} - \frac{16}{25}\right) - 16 \]

I can group the perfect square trinomial and the remaining term:

\[ 5\left(\left(x^2 - \frac{8}{5}x + \frac{16}{25}\right) - \frac{16}{25}\right) - 16 \]

The expression inside the first set of parentheses is now a perfect square:

\[ 5\left(\left(x - \frac{4}{5}\right)^2 - \frac{16}{25}\right) - 16 \]

Now, distribute the 5 through the parentheses:

\[ 5\left(x - \frac{4}{5}\right)^2 - 5 \times \frac{16}{25} - 16 \]

Simplify the multiplication:

\[ 5\left(x - \frac{4}{5}\right)^2 - \frac{80}{25} - 16 \]

Reduce \( \frac{80}{25} \):

\[ 5\left(x - \frac{4}{5}\right)^2 - \frac{16}{5} - 16 \]

Now, convert 16 to fifths to combine the constants:

\[ 5\left(x - \frac{4}{5}\right)^2 - \frac{16}{5} - \frac{80}{5} \]

Combine the fractions:

\[ 5\left(x - \frac{4}{5}\right)^2 - \frac{96}{5} \]

So, the completed square form of the quadratic \( 5x^2 - 8x - 16 \) is:

\[ \boxed{5\left(x - \frac{4}{5}\right)^2 - \frac{96}{5}} \]