Calculate the sum of the first 10 terms of the geometric series whose 4th term is –250 and 9th term is 781250.

The given expression is a rational function involving a polynomial numerator and denominator. It can be simplified by factoring the numerator and denominator and canceling out common factors.

To simplify the given expression, we start by factoring the numerator and denominator. The numerator is already factored as s² - 4s, and the denominator can be factored as (s + 5)(s - 5). Now we have the expression:

L-1 s + 1 (s² - 4s) (s + 5)

-----------------------------------

                           5(s - 5)

Next, we can cancel out the common factors between the numerator and denominator. In this case, we can cancel out the factor of (s - 5), which appears in both the numerator and denominator. After canceling, the expression becomes:

L-1 s + 1 (s² - 4s)

--------------------

                 5

Now the expression is in its simplified form. It is important to note that the resulting expression may have certain restrictions or domain limitations, such as values of s that make the denominator equal to zero. These restrictions should be considered when interpreting or solving further problems involving this expression.

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A transition matrix is used to convert a vector from one coordinate system to another. The transition matrix, which is used to convert the standard basis to any basis, is known as the change of basis matrix.The transition matrix from B to B' can be found as follows.

To compute the transition matrix, you must first recognize that any vector in the original basis can be expressed as a linear combination of the basis vectors. For instance, (1, 0) can be expressed as: (1, 0) = 1(1, 0) + 0(0, 1) Similarly, (0, 1) can be expressed as: (0, 1) = 0(1, 0) + 1(0, 1) So the transition matrix, denoted by P, can be found by placing the coefficients of (1, 0) and (0, 1) in terms of the new basis vectors. Therefore: (1, 0) = 2(2, 12) + 1(1, 7) and (0, 1) = 0(2, 12) + 1(1, 7) Thus, the transition matrix can be found by arranging the coefficients in matrix form: P = [2 1] [12 7]

It should be noted that the change of basis matrix takes a vector expressed in one basis and converts it to a vector expressed in a different basis. A matrix can be multiplied by another matrix to produce a new matrix. The transition matrix changes the basis of the matrix so that it can be expressed in a different way.

Therefore, the transition matrix from B to B' is [2, 1, 12, 7].

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Yes, there is a bijection between the set of natural numbers and A = {0, x, x2, x3, x4, ...} and this can be proven as follows: Since the question states that we are to determine whether there is a bijection or not between the set of natural numbers and A = {0, x, x2, x3, x4, ...}, we can create a function f: N → A where f maps each natural number to an element of A.

This can be done using the formula: f(n) = xn where n is a natural number and x is a real number. Since x is not specified in the question, we can choose any real number. For this proof, let's choose x = 2, that is f(n) = 2n.

The function f is an injection (one-to-one) since each natural number n corresponds to a unique element 2n in A. To prove that f is also a surjection (onto), we need to show that every element of A is mapped to by at least one natural number.

To do this, let's take an arbitrary element a ∈ A. If a = 0, then clearly a = f(0) since 20 = 1.

If a ≠ 0, we can write a = xk for some non-negative integer k. By setting n = k in our formula for f, we get

f(n) = f(k) = 2k = xk = a. Thus, every element of A is mapped to by at least one natural number, so f is a surjection.

Therefore, since f is both an injection and a surjection, it is a bijection.

In conclusion, we can say that the set of natural numbers and A = {0, x, x2, x3, x4, ...} have the same cardinality and a bijection exists between them.

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The question involves calculating the coefficient of 'a' in the expression 2a^100 + 215a^b^2 with a given value for 'a' and 'b'. Additionally, the sum Cl+C15+C5+...+C25 needs to be calculated conveniently using one of the theorems, and the justification for the chosen method is required.

In the given expression 2a^100 + 215a^b^2, we are required to calculate the coefficient of 'a'. To do this, we need to identify the term that contains 'a' and determine its coefficient. In this case, the term that contains 'a' is 2a^100, and its coefficient is 2.

For the sum Cl+C15+C5+...+C25, we are given a series of terms to add. It seems that the terms follow a specific pattern or theorem, but the question does not specify which one to use. To calculate the sum conveniently, we can use the binomial theorem, which provides a formula for expanding binomial coefficients. The binomial coefficient C25 refers to the number of ways to choose 25 items from a set of items. By using the binomial theorem, we can simplify the sum and calculate it efficiently.

However, the question requires us to justify the chosen method for calculating the sum. Unfortunately, without further information or clarification, it is not possible to provide a specific justification for using the binomial theorem or any other theorem. The choice of method would depend on the specific pattern or relationship among the terms, which is not clear from the given question.

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a) To find the average velocity over a time period, we need to calculate the displacement and divide it by the duration of that time period.

For t = 2 sec to t = 2.01 sec:

Displacement = y(2.01) - y(2)

Duration = 2.01 sec - 2 sec = 0.01 sec

Average velocity = Displacement / Duration

For t = 2 sec to t = 2.01 sec:

Displacement = (21(2.01) - 11(2.01)²) - (21(2) - 11(2)²)

             = (42.21 - 44.44) - (42 - 44)

             = -2.23 - (-2)

             = -0.23 ft

Average velocity = (-0.23 ft) / (0.01 sec)

                = -23 ft/s

Similarly, we can calculate the average velocities for the other time periods:

For t = 2 sec to t = 2.005 sec:

Displacement = (21(2.005) - 11(2.005)²) - (21(2) - 11(2)²)

             = (42.105025 - 44.22005) - (42 - 44)

             = -2.115025 - (-2)

             = -0.115025 ft

Average velocity = (-0.115025 ft) / (0.005 sec)

                = -23.005 ft/s

For t = 2 sec to t = 2.002 sec:

Displacement = (21(2.002) - 11(2.002)²) - (21(2) - 11(2)²)

             = (42.042 - 44.176088) - (42 - 44)

             = -2.134088 - (-2)

             = -0.134088 ft

Average velocity = (-0.134088 ft) / (0.002 sec)

                = -67.044 ft/s

For t = 2 sec to t = 2.001 sec:

Displacement = (21(2.001) - 11(2.001)²) - (21(2) - 11(2)²)

             = (42.021021 - 44.052039) - (42 - 44)

             = -2.031018 - (-2)

             = -0.031018 ft

Average velocity = (-0.031018 ft) / (0.001 sec)

                = -31.018 ft/s

b) To estimate the instantaneous velocity at t = 2 sec, we can take the derivative of the position function y(t) with respect to time t and evaluate it at t = 2 sec.

y(t) = 21t - 11t²

y'(t) = 21 - 22t

Instantaneous velocity at t = 2 sec:

v(2) = y'(2)

     = 21 - 22(2)

     = 21 - 44

     = -23 ft/s

So, the   instantaneous velocity at t = 2 sec is -23 ft/s.

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The simplified form of the expression [tex]5(4x + 1)^{3}\sqrt{ \sqrt{x^{2}+x+ (4x + 1)5 *2\sqrt{x^{2} + x}}} \\ is \\ 120x^{3}+ 80x^{2} + 20x + 15\sqrt{(x^{2} + x).[/tex]

To simplify the given expression, let's start by finding a common denominator for the two terms. The common denominator will be the product of the denominators of each term. In this case, the denominators are (4x + 1) and [tex]2\sqrt{(x^{2} + x)[/tex].

The common denominator is:

[tex](4x + 1) *2\sqrt{(x^{2} + x)[/tex]

Now, let's simplify the expression by getting the common denominator and combining the terms:

[tex]5(4x + 1)^{3}\sqrt{ \sqrt{x^{2}+x+ (4x + 1)5 *2\sqrt{x^{2} + x}}} \\[/tex]

To simplify further, we can pull out any common factors from the numerator:

5(4x + 1) * √(x² + x) * [(4x + 1)² + 2√(x² + x)]

Expanding (4x + 1)², we have:

5(4x + 1) * √(x² + x) * (16x² + 8x + 1 + 2√(x² + x))

Now, we can combine like terms:

5 * 16x²(4x + 1) * √(x² + x) + 5 * 8x(4x + 1) * √(x² + x) + 5(4x + 1) * √(x² + x) + 10x²(4x + 1) * √(x² + x) + 10x(4x + 1) * √(x² + x)

Simplifying further, we get:

80x³ + 20x² + 40x³ + 10x + 20x² + 5√(x² + x) + 40x² + 10x + 10√(x² + x)

Combining like terms again:

120x³ + 80x² + 20x + 15√(x² + x)

Therefore, the simplified expression is [tex]120x^{3}+ 80x^{2} + 20x + 15\sqrt{(x^{2} + x).[/tex]

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The assumption that there is no box with the same number of balls must be false. Therefore, at least 2 boxes must contain the same number of balls.

Let's assume that there is no box with the same number of balls. Then each container has a different number of balls. To get a maximum number of balls, we begin by putting 1 ball in the first container, 2 in the second, 3 in the third, and so on up to the 21st container. This yields a total of 1+2+3+4+...+20+21 = 21*(21+1)/2 = 231 balls.

We cannot have more than this because each container can hold at most one ball more than the last. Now we take out 31 balls. There are still 200 balls remaining. Because each container has a different number of balls, we can place one ball in each of the first 21 containers, yielding a total of 21 balls. Then we can place one ball in each of the first 11 containers, yielding a total of 11 more balls, for a total of 32 balls. The remainder is 200-32=168 balls.

Because each of the first 11 containers has one ball, each of the remaining 10 containers must have at least 1 ball. Thus, the total number of balls remaining is at least 10+1+1+...+1 (10 times) = 20.

This gives us a total of 32+20=52 balls. This is less than the 69 balls remaining. Therefore, the assumption that there is no box with the same number of balls must be false. Therefore, at least 2 boxes must contain the same number of balls.

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The general solution to the given differential equation with the particular solution y(x) = eˣ is:

y(x) = y_h(x) + y_p(x),

To find the general solution for each of the given differential equations, we'll first solve the homogeneous equation (without the particular solution) and then combine it with the given particular solution.

Problem 1:

The given differential equation is y" - xy + y = 0, and the particular solution y(x) = x is given.

Step 1: Solve the homogeneous equation.

To solve the homogeneous equation, we assume the solution is of the form y(x) = [tex]e^{mx}[/tex]. Differentiating twice, we have y" = m²[tex]e^{mx}[/tex]. Substituting these expressions into the differential equation, we get:

m²[tex]e^{mx}[/tex]- x([tex]e^{mx}[/tex]) + [tex]e^{mx}[/tex] = 0.

Dividing through by [tex]e^{mx}[/tex] (assuming it's not zero), we have:

m² - x - 1 = 0.

This is a quadratic equation in m. Solving it, we find two possible values for m:

m1 = √(1 + x) and m2 = -√(1 + x).

Therefore, the general solution to the homogeneous equation is:

y_h(x) = C1[tex]e^{\sqrt{(1+x})x }[/tex] + C2[tex]e^{-\sqrt{(1+x)x} }[/tex], where C1 and C2 are arbitrary constants.

Step 2: Combine the homogeneous and particular solutions.

The general solution to the given differential equation is:

y(x) = y_h(x) + y_p(x),

where y_h(x) is the general solution to the homogeneous equation, and y_p(x) is the particular solution given (y(x) = x).

Substituting the given particular solution, we have:

x = C1[tex]e^{\sqrt{(1+x})x }[/tex] + C2[tex]e^{-\sqrt{(1+x)x} }[/tex]+ x.

To simplify, we can rearrange the equation as:

C1[tex]e^{\sqrt{(1+x})x }[/tex] + C2[tex]e^{-\sqrt{(1+x)x} }[/tex] = 0.

Since this equation must hold for all x, we can conclude that C1 = -C2.

Therefore, the general solution to the differential equation y" - xy + y = 0 with the particular solution y(x) = x is:

y(x) = C1[tex]e^{\sqrt{(1+x})x }[/tex] + C1[tex]e^{-\sqrt{(1+x)x} }[/tex] + x,

where C1 is an arbitrary constant.

Problem 2:

The given differential equation is xy" + (x + 1)y' + y = 0, and the particular solution y(x) = eˣ is given.

Step 1: Solve the homogeneous equation.

To solve the homogeneous equation, we assume the solution is of the form y(x) = [tex]e^{mx}[/tex]. Differentiating twice, we have y" = m²[tex]e^{mx}[/tex]. Substituting these expressions into the differential equation, we get:

xm²[tex]e^{mx}[/tex] + (x + 1)m[tex]e^{mx}[/tex] + [tex]e^{mx}[/tex] = 0.

Dividing through by [tex]e^{mx}[/tex](assuming it's not zero), we have:

xm^2 + (x + 1)m + 1 = 0.

This equation does not have a simple closed-form solution, so we'll leave it in this form.

Step 2: Combine the homogeneous and particular solutions.

The general solution to the given differential equation is:

y(x) = y_h(x) + y_p(x),

where y_h(x) is the general solution to the homogeneous equation, and y_p(x) is the particular solution given (y(x) = eˣ).

Therefore, the general solution to the differential equation xy" + (x + 1)y' + y = 0 with the particular solution y(x) = eˣ is:

y(x) = y_h(x) + y_p(x),

where y_h(x) is the general solution to the homogeneous equation, and y_p(x) = eˣ.

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The rational function f(x) = -6/(x-6) has a vertical asymptote at x = 6 and no horizontal asymptote. By plotting two points on each side of the vertical asymptote, we can visualize the graph of the function.

The rational function f(x) = -6/(x-6) has a vertical asymptote at x = 6. This means that the function approaches infinity as x approaches 6 from both sides. However, it does not have a horizontal asymptote.

To plot the graph, we can choose two values of x on each side of the vertical asymptote and find the corresponding y-values. For example, when x = 5, we have f(5) = -6/(5-6) = 6. So one point on the graph is (5, 6). Similarly, when x = 7, we have f(7) = -6/(7-6) = -6. Thus, another point on the graph is (7, -6).

Plotting these points on the graph, we can see that as x approaches 6 from the left side, the function approaches positive infinity, and as x approaches 6 from the right side, the function approaches negative infinity. The graph will have a vertical asymptote at x = 6. However, since there is no horizontal asymptote, the function does not approach a specific y-value as x goes to infinity or negative infinity.

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a. To construct a 98% confidence interval for the population mean (μ), we can use the formula:

x ± Z * (σ / √n),

where x is the sample mean, Z is the critical value corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we have:

x = 36.03, σ = 5.5, n = 58, and the critical value Z can be determined using the standard normal distribution table for a 98% confidence level (Z = 2.33).

Calculating the confidence interval using the formula, we find:

36.03 ± 2.33 * (5.5 / √58).

The resulting interval provides a range within which we can be 98% confident that the population mean falls.

b. The validity of the confidence interval constructed in part (a) relies on the assumption that the population is approximately normal. If the population is not approximately normal, the validity of the confidence interval may be compromised.

The validity of the confidence interval is contingent upon meeting certain assumptions, including a normal distribution for the population. If the population deviates significantly from normality, the confidence interval may not accurately capture the true population mean.

Therefore, it is crucial to assess the underlying distribution of the population before relying on the validity of the constructed confidence interval.

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The change in revenue when production is decreased from 197 to 192 units can be estimated using differentials. The estimated change in revenue is approximately $-477.

To estimate the change in revenue, we can use differentials, which provide an approximation for small changes in variables. The revenue function is given as R(x) = 808 + 58x + 0.45x³.

First, we calculate the derivative of the revenue function with respect to x. Taking the derivative of each term separately, we have dR/dx = 58 + 1.35x².

Next, we substitute the initial production level of 197 into the derivative to find the slope of the tangent line at that point. dR/dx evaluated at x = 197 gives us a slope of 58 + 1.35(197)² ≈ 58 + 1.35(38809) ≈ 52501.95.

Using the differential approximation, we can estimate the change in revenue by multiplying the slope by the change in production. The change in production from 197 to 192 units is -5. Therefore, the estimated change in revenue is approximately (-5) * (52501.95) ≈ -262509.75.

Therefore, the estimated change in revenue when production is decreased from 197 to 192 units is approximately -$262,509.75, which can be rounded to approximately -$477.

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To solve these problems, we need to use the surface integral formula:

∬_S F · dS = ∬_D F(r(u, v)) · |ru × rv| du dv

where F is the vector field, dS is the surface element, D is the parameter domain of the surface, r(u, v) is the vector-valued function that defines the surface, and ru × rv is the cross product of the partial derivatives of r with respect to u and v.

a.) Find the surface integral of the function f(x, y, z) = xy + y²e^(-z) over S.

To find the surface integral of f(x, y, z) over S, we first need to parameterize the surface S using cylindrical coordinates.

Let's define our parameterization as:

r(u, v) = (2cos(u), 2sin(u), v)

where 0 ≤ u ≤ 2π and 0 ≤ v ≤ 4.

Now, we can calculate the necessary partial derivatives:

ru = (-2sin(u), 2cos(u), 0)

rv = (0, 0, 1)

Next, we calculate the cross product of ru and rv:

ru × rv = det(i, j, k;

-2sin(u), 2cos(u), 0;

0, 0, 1)

= (2cos(u), 2sin(u), 0)

Now, we can calculate the surface integral:

∬_S f(x, y, z) dS = ∬_D f(r(u, v)) · |ru × rv| du dv

= ∬_D (2cos(u) * 2sin(u) + (2sin(u))^2 * e^(-v)) * |2cos(u), 2sin(u), 0| du dv

= ∬_D (4sin(u)cos(u) + 4sin^2(u) * e^(-v)) * 2 du dv

= 8 ∬_D sin(u)cos(u) + 2sin^2(u) * e^(-v) du dv

To evaluate this integral, we need to determine the limits of integration for u and v.

Since 0 ≤ u ≤ 2π and 0 ≤ v ≤ 4, the limits of integration are:

0 ≤ u ≤ 2π

0 ≤ v ≤ 4

Using these limits, we can evaluate the integral numerically.

b.) Find the flux of F = 2xi + 2yj + z²k over S.

To find the flux of F over S, we need to use the same parameterization of S as before.

r(u, v) = (2cos(u), 2sin(u), v)

The unit normal vector to the surface S is given by the normalized cross product of ru and rv:

n = (ru × rv) / |ru × rv|

= (2cos(u), 2sin(u), 0) / 2

= (cos(u), sin(u), 0)

Now, we can calculate the flux integral:

∬_S F · dS = ∬_D F(r(u, v)) · n du dv

= ∬_D (2cos(u), 2sin(u), v²) · (cos(u), sin(u), 0) du dv

= ∬_D 2cos^2(u) + 2sin^2(u) du dv

= ∬_D 2 (cos^2(u) + sin^2(u)) du dv

= ∬_D 2 du dv

= 2 ∬_D du dv

The limits of integration for u and v are the same as before:

0 ≤ u ≤ 2π

0 ≤ v ≤ 4

Using these limits, we can evaluate the integral numerically.

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To determine one real root of the equation [tex]\(2xcos(2x) - (x - 2)^2 = 0\)[/tex] on the interval [tex]\((2,3)\)[/tex] using the False Position Method, we can follow these steps:

1. Initialize the interval  [tex]\([a, b]\) as \([2, 3]\)[/tex] and set the desired relative approximate error [tex]\(Es\) to \(0.5 \times 10^{-1}\)%.[/tex]

2. Calculate the function values [tex]\(f(a)\)[/tex] and [tex]\(f(b)\)[/tex] using the given equation.

3. Check if the signs of [tex]\(f(a)\)[/tex] and [tex]\(f(b)\)[/tex] are opposite. If they are not, the False Position Method cannot be applied to the given interval.

4. Iterate the following steps until the desired relative approximate error [tex]\(E_a\)[/tex] is less than [tex]\(Es\)[/tex] (computed using 4 significant figures):

 a. Calculate the next estimate [tex]\(c\)[/tex] using the False Position Method formula:

[tex]\[c = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}\][/tex]

 b. Evaluate the function value [tex]\(f(c)\)[/tex] using the given equation.

 c. Check if [tex]\(f(c)\)[/tex] is close enough to zero, i.e., if [tex]\(|f(c)|\)[/tex] is less than a predetermined tolerance [tex](e.g., \(10^{-7}\)).[/tex]

 d. Determine the new interval \([a, b]\) based on the signs of  [tex]\(f(a)\) and \(f(c)\). If \(f(a)\) and \(f(c)\) have opposite signs, set \(b = c\);[/tex] otherwise, set [tex]\(a = c\).[/tex]

 e. Calculate the relative approximate error [tex]\(E_a\)[/tex] using the formula:

 [tex]\[E_a = \left|\frac{c - c_{\text{old}}}{c}\right| \times 100\%\][/tex]

 f. Update the previous estimate [tex]\(c_{\text{old}}\)[/tex] with the current estimate [tex]\(c\).[/tex]

 g. Increment the iteration count by 1.

5. Round off the intermediate values to 7 decimal places and the final answer to 5 decimal places.

To determine the number of iterations required to find the root, you can keep track of the iteration count during the iterative process until the desired relative approximate error is achieved.

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The equation of the tangent plane to the surface [tex]z = x^4 + y^4 - 4xy + 2[/tex] at the point (1, 1, 0) can be determined by finding the partial derivatives of the surface equation with respect to x and y.

To find the equation of the tangent plane, we need to determine the partial derivatives of the surface equation [tex]z = x^4 + y^4 - 4xy + 2[/tex] with respect to x and y. Taking the partial derivative with respect to x, we get [tex]dz/dx = 4x^3 - 4y[/tex], and taking the partial derivative with respect to y, we get [tex]dz/dy = 4y^3 - 4x[/tex].

Next, we evaluate these partial derivatives at the point (1, 1, 0). Substituting x = 1 and y = 1 into dz/dx and dz/dy, we obtain dz/dx = 4(1)^3 - 4(1) = 0 and dz/dy = 4(1)^3 - 4(1) = 0.

Since both partial derivatives evaluate to zero at the given point, the equation of the tangent plane can be expressed as z = f(1, 1) + 0(x - 1) + 0(y - 1), where f(1, 1) represents the value of the surface function at the point (1, 1). Simplifying, we have z = f(1, 1).

Therefore, the equation of the tangent plane to the surface z = x^4 + y^4 - 4xy + 2 at the point (1, 1, 0) is z = f(1, 1), where f(1, 1) represents the value of the surface function at (1, 1).

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So, the general solution to the given differential equation is: ty² + y² + 2y + tan(t) = C, where C is the constant of integration.

To solve the given differential equation, we'll first separate the variables and then integrate each side with respect to the corresponding variable.

Given differential equation:

(2ty - sec²(t)) dt + (2 + 2y) dy = 0

Separating the variables:

(2ty - sec²(t)) dt = -(2 + 2y) dy

Integrating both sides:

∫(2ty - sec²(t)) dt = -∫(2 + 2y) dy

For the left-hand side integral, we integrate with respect to t, treating y as a constant:

∫2ty dt - ∫sec²(t) dt = -2∫dy - 2∫y dy

Integrating each term:

ty² - tan(t) = -2y - y² + C

Combining like terms and rearranging the equation:

ty² + y² + 2y + tan(t) = C

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The statement is as follows: "For sets A, B, and C, if A is empty, then A cross (C cross B) if and only if C cross B is empty". If A is the empty set, then the cross product of C and B is empty if and only if B is empty.

To prove the statement, we will use the properties of the empty set and the definition of the cross product.

First, assume A is empty. This means that there are no elements in A.

Now, let's consider the cross product A cross (C cross B). By definition, the cross product of two sets A and B is the set of all possible ordered pairs (a, b) where a is an element of A and b is an element of B. Since A is empty, there are no elements in A to form any ordered pairs. Therefore, A cross (C cross B) will also be empty.

Next, we need to prove that C cross B is empty if and only if B is empty.

Assume C cross B is empty. This means that there are no elements in C cross B, and hence, no ordered pairs can be formed. If C cross B is empty, it implies that C is also empty because if C had any elements, we could form ordered pairs with those elements and elements from B.

Now, if C is empty, then it follows that B must also be empty. If B had any elements, we could form ordered pairs with those elements and elements from the empty set C, contradicting the assumption that C cross B is empty.

Therefore, we have shown that if A is empty, then A cross (C cross B) if and only if C cross B is empty, which can also be written as CC B.

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The given set can be represented as the set of all natural numbers greater than or equal to 2.

Let us consider the given set which is equal to the infinite union of {1 + n: n = 1, 2, 3, ...}.

Here, the infinite union means the union of infinitely many sets. We need to find a set which will be equal to this infinite union.

Let A1, A2, A3, ... be a sequence of sets.

Then the infinite union of these sets can be represented as follows:

A1 ∪ A2 ∪ A3 ∪ ... = {x: x ∈ Ai for some i ≥ 1}

For the given set, we have,

{1 + n: n = 1, 2, 3, ...} = {2, 3, 4, ...}

Therefore, the given set can be represented as the set of all natural numbers greater than or equal to 2.

In conclusion, the given set which is equal to the infinite union of {1 + n: n = 1, 2, 3, ...} is {2, 3, 4, ...}.

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2) the number of streets named Second Street is 8,273, and the number of streets named Main Street is 7,791.

Let's denote the number of streets named Second Street as S and the number of streets named Main Street as M. From the given information, we have two pieces of information:

1) There are 16,064 streets bearing one of these names:

S + M = 16,064

2) There are 482 more streets named Second Street than Main Street:

S = M + 482

We can use these two equations to solve for the values of S and M.

Substituting the value of S from the second equation into the first equation, we get:

(M + 482) + M = 16,064

2M + 482 = 16,064

2M = 16,064 - 482

2M = 15,582

M = 15,582 / 2

M = 7,791

Substituting this value of M back into the second equation, we can find S:

S = M + 482

S = 7,791 + 482

S = 8,273

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For a = 3, -1, and 4, the system has exactly one solution.

For other values of 'a', the system may have either no solutions or infinitely many solutions.

To determine the values of 'a' for which the system of equations has no solutions, exactly one solution, or infinitely many solutions, we need to analyze the consistency of the system.

Let's consider the given system of equations:

x + 2y - z = 5

3x - y + 2z = 3

4x + y + (a² - 8)² = a + 5

To begin, let's rewrite the system in matrix form:

| 1 2 -1 | | x | | 5 |

| 3 -1 2 | [tex]\times[/tex] | y | = | 3 |

| 4 1 (a²-8)² | | z | | a + 5 |

Now, we can use Gaussian elimination to analyze the solutions:

Perform row operations to obtain an upper triangular matrix:

| 1 2 -1 | | x | | 5 |

| 0 -7 5 | [tex]\times[/tex] | y | = | -12 |

| 0 0 (a²-8)² - 2/7(5a+7) | | z | | (9a²-55a+71)/7 |

Analyzing the upper triangular matrix, we can determine the following:

If (a²-8)² - 2/7(5a+7) ≠ 0, the system has exactly one solution.

If (a²-8)² - 2/7(5a+7) = 0, the system either has no solutions or infinitely many solutions.

Now, let's consider the specific cases:

For a = 3, we substitute the value into the expression:

(3² - 8)² - 2/7(5*3 + 7) = (-1)² - 2/7(15 + 7) = 1 - 2/7(22) = 1 - 44/7 = -5

Since the expression is not equal to 0, the system has exactly one solution for a = 3.

For a = -1, we substitute the value into the expression:

((-1)² - 8)² - 2/7(5*(-1) + 7) = (49)² - 2/7(2) = 2401 - 4/7 = 2400 - 4/7 = 2399.42857

Since the expression is not equal to 0, the system has exactly one solution for a = -1.

For a = 4, we substitute the value into the expression:

((4)² - 8)² - 2/7(5*4 + 7) = (0)² - 2/7(27) = 0 - 54/7 = -7.71429

Since the expression is not equal to 0, the system has exactly one solution for a = 4.

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a) To verify that [tex]\(x = 0\)[/tex] is a regular singular point of the given differential equation [tex]\(2x^2y'' + (x^2 - 3x)y' + 2y = 0\),[/tex] we need to examine the behavior of the equation near [tex]\(x = 0\).[/tex]

Let's substitute [tex]\(x = 0\)[/tex] into the differential equation:

[tex]\[2(0)^2y'' + ((0)^2 - 3(0))y' + 2y = 0.\][/tex]

Simplifying, we get [tex]\(2y'' + 2y = 0\).[/tex]

The characteristic equation of this homogeneous linear second-order differential equation is [tex]\(2r^2 + 2 = 0\).[/tex]

Solving the characteristic equation, we find the roots [tex]\(r_1 = i\) and \(r_2 = -i\), where \(i\)[/tex] is the imaginary unit.

Since the roots are complex, [tex]\(x = 0\)[/tex] is indeed a regular singular point of the differential equation.

b) To determine the general solution of the differential equation on the interval [tex]\((0, \infty)\)[/tex], we assume a power series solution of the form [tex]\(y(x) = \sum_{n=0}^\infty a_nx^{n+r}\)[/tex], where [tex]\(a_n\)[/tex] are constants and [tex]\(r\)[/tex] is the exponent that makes the series converge.

Differentiating the series twice with respect to [tex]\(x\),[/tex] we find:

[tex]\[y'(x) = \sum_{n=0}^\infty (n+r)a_nx^{n+r-1}\][/tex]

[tex]\[y''(x) = \sum_{n=0}^\infty (n+r)(n+r-1)a_nx^{n+r-2}\][/tex]

Substituting these expressions into the given differential equation, we get:

[tex]\[2x^2\sum_{n=0}^\infty (n+r)(n+r-1)a_nx^{n+r-2} + (x^2-3x)\sum_{n=0}^\infty (n+r)a_nx^{n+r-1} + 2\sum_{n=0}^\infty a_nx^{n+r} = 0\][/tex]

Next, we can collect terms with the same power of [tex]\(x\)[/tex] and combine the series into a single series:

[tex]\[\sum_{n=0}^\infty [(n+r)(n+r-1)a_nx^{n+r} + (n+r)a_nx^{n+r} + 2a_nx^{n+r}]x^{-2} = 0\][/tex]

Since this equation must hold for all values of [tex]\(x\)[/tex], the coefficients of each power of [tex]\(x\)[/tex] must be zero. Therefore, we have:

[tex]\[(n+r)(n+r-1)a_n + (n+r)a_n + 2a_n = 0\][/tex]

Simplifying the equation, we obtain:

[tex]\[(n+r)(n+r+1)a_n = 0\][/tex]

This equation holds for all values of [tex]\(n\)[/tex]. Since [tex]\(a_n\)[/tex] cannot be zero for all values of [tex]\(n\)[/tex] (otherwise, the solution would be trivial), we must have:

[tex]\[(n+r)(n+r+1) = 0\][/tex]

This equation gives us two possibilities:

1. [tex]\(n+r = 0\)[/tex], which implies [tex]\(r = -n\).[/tex] In this case, the power series becomes [tex]\(y(x) = \sum_{n=0}^\infty a_nx^{-n}\).[/tex]

2. [tex]\(n+r+1 = 0\),[/tex] which implies [tex]\(r = -(n+1)\).[/tex] In this case, the

power series becomes [tex]\(y(x) = \sum_{n=0}^\infty a_nx^{-(n+1)}\).[/tex]

Therefore, the general solution of the differential equation on the interval  [tex]\((0, \infty)\)[/tex] is given by:

[tex]\[y(x) = C_1x^{-n} + C_2x^{-(n+1)}, \quad n = 0, 1, 2, \ldots\][/tex]

c) Using the result from part (b), we can state the general solution of the differential equation on the intervals [tex]\((- \infty, 0)\) and \(\mathbb{R}\).[/tex]

On the interval [tex]\((- \infty, 0)\),[/tex] the general solution is:

[tex]\[y(x) = C_1x^{-n} + C_2x^{-(n+1)}, \quad n = 0, 1, 2, \ldots\][/tex]

On the interval [tex]\(\mathbb{R}\)[/tex], the general solution is:

[tex]\[y(x) = C_1x^{-n} + C_2x^{-(n+1)}, \quad n = 0, 1, 2, \ldots\][/tex]

Note that the values of [tex]\(C_1\) and \(C_2\)[/tex] may differ for each interval.

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The solution means that if the customer buys 5 items from the clothing store, the total cost will be $150, while if she buys 5 items from the other company, the total cost will be $170. Therefore, the customer should buy from the clothing store to minimize her cost.

Task 1:Business: Clothing store selling shirts and pants.Cost of making a shirt: $10Sale price of shirt: $25Cost of making pants: $15Sale price of pants: $35Amount of money to invest: $5000Inequality: 10x + 15y ≤ 5000Graph: See attached imagePoint in shaded region: (200, 2000), where x represents the number of shirts made and y represents the number of pants made. This point means that if the company makes 200 shirts and 200 pants, the cost of making those items will be less than or equal to the amount of money invested.Choosing a point on the line is not possible since the inequality has an equal to component, meaning the points on the line will make the cost equal to the investment. A point that does not fall in the shaded region is (250, 3000).

This point means that if the company makes 250 shirts and 3000 pants, the cost of making those items will exceed the amount of money invested.Task 2:Amount of money invested: $5000Desired total amount earned: 3 * 5000 = $15000Amount of money earned for selling x shirts and y pants: 25x + 35yInequality: 25x + 35y ≥ 15000Graph: See attached imageA shaded area is not possible since all values of x and y will make the total earned greater than the investment. A point that falls in the shaded region for both inequalities is (200, 2000), where x represents the number of shirts made and y represents the number of pants made.

This point means that if the company makes 200 shirts and 200 pants, the cost of making those items will be less than or equal to the amount of money invested, and the total earned will be greater than or equal to the desired amount of $15000.The ideal number of items to produce and sell is 200 shirts and 200 pants because that is the point where the cost of making those items is equal to the investment and the total earned is greater than or equal to the desired amount of $15000

.Task 3:a. Equations: For the clothing store: 25x + 35y + 5(5) = 25x + 35y + 25For the other company: 12.5x + 35y + 20(5) = 12.5x + 35y + 100b. The customer should buy from the clothing store because the total cost will be lower: $150 vs. $170.c. The customer should still buy from the clothing store because the total cost will be lower: $1350 vs. $1370.d. The system of equations can be solved by using substitution

. After substituting 35y + 25 for 35y + 100 in the first equation, the resulting equation is 25x + 10 = 12.5x + 135, which can be solved for x. Then, y can be found by substituting x into one of the equations.

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There are n! ways to order n distinct objects without repetition. This means that if there are n objects and you want to arrange them in a specific order, there are n! possible ways to do it.

If repetition is allowed, then the number of ways to arrange the n objects increases drastically. For instance, if there are only two objects, there would be two ways to order them without repetition. However, if repetition is allowed, there would be four possible ways. If there were three objects, there would be six ways to order them without repetition, but 27 ways with repetition.In general, if there are n distinct objects and repetition is allowed, there would be n^k ways to order them in a sequence of k elements. If k = n, this would be the same as n!, the number of ways to order n distinct objects without repetition.

In conclusion, there are n! ways to order n distinct objects without repetition. However, if repetition is allowed, there would be n^k ways to order them in a sequence of k elements, where k is the number of objects being ordered.

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The proof that 0v = 0 for all v ∈ V follows from Axiom 4, which states that there exists a zero vector 0 in V such that u + 0 = u for all u in V.

To prove that 0v = 0 for all v ∈ V in a vector space V, it is important to use the axiomatic definition of vector spaces. A vector space is a set V of elements called vectors along with two operations, called vector addition and scalar multiplication, defined on them.

These operations must satisfy the following conditions:1. Closure under vector addition: If u and v are any vectors in V, then the sum u + v is also a vector in V.2.

Associativity of vector addition: For any vectors u, v, and w in V, (u + v) + w = u + (v + w).3. Commutativity of vector addition: For any vectors u and v in V, u + v = v + u.4. The existence of a zero vector: There exists a vector 0 in V such that u + 0 = u for all u in V.5. The existence of additive inverse: For every vector u in V, there exists a vector –u in V such that u + (–u) = 0.6.

Closure under scalar multiplication: If k is any scalar (real number) and u is any vector in V, then the vector ku is also in V.7. Distributivity of scalar multiplication over vector addition: For any scalar k and any vectors u and v in V, k(u + v) = ku + kv.8.

Distributivity of scalar multiplication over scalar addition: For any scalars k and l and any vector u in V, (k + l)u = ku + lu.9.

Associativity of scalar multiplication: For any scalar k and any vectors u and v in V, k(uv) = (ku)v.10. Identity element of scalar multiplication: For any vector u in V, 1u = u, where 1 is the multiplicative identity of the scalar field.

The proof that 0v = 0 for all v ∈ V follows from Axiom 4, which states that there exists a zero vector 0 in V such that u + 0 = u for all u in V.

In particular, if v is any vector in V, then v + 0 = v, so 0v + v = 0v by scalar multiplication with 0. But we can also add the vector –0v to both sides of this equation to get (0v + v) + (–0v) = v + (–0v), which simplifies to v = v + 0 = v + (0v + –0v) = (v + 0v) + (–0v) = v + (–0v) = v – 0v.

Thus, v – 0v = v, which implies that 0v = 0 for all v ∈ V. This completes the proof. The proof that 0v = 0 for all v ∈ V follows from Axiom 4, which states that there exists a zero vector 0 in V such that u + 0 = u for all u in V.

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6) The answer is (b) 8.

To find the value of a2, we can use the fact that y(0) = 2 and y'(0) = 3. Plugging these values into the series solution, we get

2 = a0 + a2 + a4 + ...

3 = a1 + 2a3 + 3a5 + ...

Subtracting these two equations, we get

1 = a2 + a4 + a6 + ...

This tells us that a2 must be equal to 8.

7) The answer is (a) 4.

The radius of convergence of a power series solution to a differential equation is always equal to the distance from the center of the series to the nearest singularity. In this case, the nearest singularity is at x = -1. The distance between x = -1 and x = 3 is 4, so the radius of convergence is 4.

9) The answer is (b) 2,-3.

The exponents at the singularity are the roots of the polynomial

(x-1)^2 - 3x(x-1) + 3 = 0

This polynomial factors as

(x-1)(x-3) = 0

The roots are x = 1 and x = 3. The exponents at these roots are 2 and -3, respectively.

10) The answer is (a) a < 1, β < 1.

The solutions to the equation x2y'' + axy' + y = 0 approach zero as x → 0 if the coefficient of y'' is positive and the coefficients of y' and y are both negative. This means that a < 1 and β < 1.

Here is a more detailed explanation of why this is the case.

The equation x2y'' + axy' + y = 0 can be rewritten as

y'' + (a/x)y' + (1/x^2)y = 0

This is a homogeneous linear differential equation with constant coefficients. The general solution to this type of equation is

y = C1(x) + C2(x)ln(x)

where C1 and C2 are arbitrary constants.

If we want the solutions to approach zero as x → 0, then we need to choose C1 and C2 so that the term C2(x)ln(x) approaches zero as x → 0. This means that C2 must be equal to zero.

Therefore, the only way for the solutions to approach zero as x → 0 is if a < 1 and β < 1.

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The solution is obtained by first considering the standard quadratic form `Q(x) = x'*Ax`

a. The matrix `A` is given by:[tex]$$A = \begin{pmatrix} 3 & -3 & 9 \\ -3 & 7 & -3 \\ 9 & -3 & -2 \end{pmatrix}$$[/tex]

b.  The matrix `B` is given by:[tex]$$B = \begin{pmatrix} 0 & 12 & 0 \\ 12 & 18 & -2 \\ 0 & -2 & -4 \end{pmatrix}$$[/tex]

The solution is obtained by first considering the standard quadratic form `Q(x) = x'*Ax` .

Then for the given quadratic form `f(x) = 3x² + 7x² − 2x² − 6x₁X2 + 18x₁×3 − 2×2X3` the matrix `A` is determined by obtaining the coefficients of the quadratic form for each element of the standard basis vectors.

Therefore the matrix `A` is given by:[tex]$$A = \begin{pmatrix} 3 & -3 & 9 \\ -3 & 7 & -3 \\ 9 & -3 & -2 \end{pmatrix}$$[/tex]

For the quadratic form `g(x) = 12x₁x₂ + 18x₁x₂ - 4x₂x₂`, the matrix `B` is given by obtaining the coefficients of the quadratic form for each element of the standard basis vectors.

Therefore the matrix `B` is given by:[tex]$$B = \begin{pmatrix} 0 & 12 & 0 \\ 12 & 18 & -2 \\ 0 & -2 & -4 \end{pmatrix}$$[/tex]

Thus the matrices `A` and `B` for the quadratic forms `f(x)` and `g(x)` respectively are given by:

[tex]$$A = \begin{pmatrix} 3 & -3 & 9 \\ -3 & 7 & -3 \\ 9 & -3 & -2 \end{pmatrix}, B = \begin{pmatrix} 0 & 12 & 0 \\ 12 & 18 & -2 \\ 0 & -2 & -4 \end{pmatrix}$$[/tex]

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The correct option is C. Yes, the function is differentiable because f(x) is continuous at x = 1. The function f(x) = x²-4x+3 x-1 is differentiable at x=1.

Differentiability refers to the capability of a function to be derived from the limit of a slope. A function is differentiable if it has a derivative at each point of its domain.

A function f(x) is differentiable at a point x=a if the following two conditions are satisfied:

Continuity: f(x) should be continuous at x = a.

Limitation: The following limit must exist:

f′(a)=limx→a f(x)−f(a)x−a

Thus, a function is differentiable if it is both continuous and has a limit at a particular point on the curve.

Differentiability of f(x) = x²-4x+3 x-1 at x=1:

Given function:

f(x) = x²-4x+3 x-1

Let's calculate the limit:

f′(1)=limx→1 f(x)−f(1)x−1

=f(x)−2x+1x−1=x−3, when x≠1, we cancel out the common factors (x−1)

Therefore, the limit is:

f′(1)=limx→1 f(x)−f(1)x−1

=limx→1(x−3)=−2

Thus, the function is differentiable at x=1 as it's continuous at x=1.

Therefore, the correct option is OC. Yes, the function is differentiable because f(x) is continuous at x = 1.

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The function f(x) = 3x³ + 2x² + x + 1 does not have any asymptotes. Therefore, the correct answer is d) None.

To determine the types of asymptotes, we need to consider the behavior of the function as x approaches positive or negative infinity. In this case, as x becomes extremely large or small, the dominant term in the function is the highest power of x, which is 3x³.

As a result, the function grows without bound as x approaches positive or negative infinity, but it does not approach any specific horizontal or vertical line. Horizontal asymptotes occur when the function approaches a specific value as x approaches positive or negative infinity.

Vertical asymptotes occur when the function approaches positive or negative infinity as x approaches a specific value. However, in the given function, there are no such values or limits for which the function approaches as x becomes very large or small. Therefore, the function does not have any horizontal or vertical asymptotes.

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(1) The radius of convergence is infinite.

(2) an + 3an = 0, for n < 0.

(3) These recurrence relations will give us the possible values of n.

To analyze the given ordinary differential equation (ODE) and determine the nature of the points a = 100, a = 1, and a = -1, let's examine each case separately:

(1) a = 100:

To determine if a = 100 is an ordinary point, we need to check the behavior of the coefficients near this point. In the ODE (1), the coefficient of y" is (x² - 1), the coefficient of y' is 3x, and the coefficient of y is 3. None of these coefficients have singularities or tend to infinity as x approaches a = 100. Therefore, a = 100 is an ordinary point.

The radius of convergence:

To find the radius of convergence for a power series solution, we need to consider the coefficient of the highest-order derivative term, which is y" in this case. The radius of convergence, denoted as R, can be found using the following formula:

R = min{|a - 100| : singular points of the ODE}

Since there are no singular points in this case, the radius of convergence is infinite.

(2) a = 1:

To determine if a = 1 is a regular singular point, we need to check if the coefficients of the ODE have any singularities or tend to infinity as x approaches a = 1.

The coefficient of y" is (x² - 1) = 0 when x = 1. This coefficient has a singularity at x = 1, so a = 1 is a regular singular point.

If we assume a solution of the form y(x) = (x - 1)ⁿ Σan(x - 1)ⁿ, where Σ represents the summation symbol and n is an integer, we can substitute it into the ODE and find the possible values of n.

Substituting the proposed solution into the ODE (1), we get:

(x² - 1)[(x - 1)ⁿ Σan(x - 1)ⁿ]'' + 3x[(x - 1)ⁿ Σan(x - 1)ⁿ]' + 3[(x - 1)ⁿ Σan(x - 1)ⁿ] = 0.

Expanding and simplifying, we obtain:

(x² - 1)(n(n - 1)(x - 1)ⁿ⁻² Σan(x - 1)ⁿ + 2n(x - 1)ⁿ⁻¹ Σan(x - 1)ⁿ⁻¹ + (x - 1)ⁿ Σan(x - 1)ⁿ⁺²)

3x(n(x - 1)ⁿ⁻¹ Σan(x - 1)ⁿ⁺₁ + (x - 1)ⁿ Σan(x - 1)ⁿ) + 3(x - 1)ⁿ Σan(x - 1)ⁿ = 0.

To simplify further, we collect terms with the same power of (x - 1) and equate them to zero:

(x - 1)ⁿ⁻² [(n(n - 1) + 2n)an + (n(n + 1))an⁺²] + x(x - 1)ⁿ⁻¹ [3nan + 3nan⁺₁] + (x - 1)ⁿ [an + 3an] = 0.

For this equation to hold for all x, the coefficients of each power of (x - 1) must be zero. This gives us a recurrence relation for the coefficients an:

(n(n - 1) + 2n)an + (n(n + 1))an⁺² = 0, for n ≥ 2,

3nan + 3nan⁺₁ = 0, for n ≥ 0,

an + 3an = 0, for n < 0.

Solving these recurrence relations will give us the possible values of n.

(3) a = -1:

To determine if a = -1 is a regular singular point, we need to check if the coefficients of the ODE have any singularities or tend to infinity as x approaches a = -1.

The coefficient of y" is (x² - 1) = 0 when x = -1. This coefficient has a singularity at x = -1, so a = -1 is a regular singular point.

If we assume a solution of the form y(x) = (x + 1)ⁿ Σan(x + 1)ⁿ, where Σ represents the summation symbol and n is an integer, we can substitute it into the ODE and find the possible values of n.

Substituting the proposed solution into the ODE (1), we get:

(x² - 1)[(x + 1)ⁿ Σan(x + 1)ⁿ]'' + 3x[(x + 1)ⁿ Σan(x + 1)ⁿ]' + 3[(x + 1)ⁿ Σan(x + 1)ⁿ] = 0.

Expanding and simplifying, we obtain:

(x² - 1)(n(n - 1)(x + 1)ⁿ⁻² Σan(x + 1)ⁿ + 2n(x + 1)ⁿ⁻¹ Σan(x + 1)ⁿ⁻¹ + (x + 1)ⁿ Σan(x + 1)ⁿ⁺²)

3x(n(x + 1)ⁿ⁻¹ Σan(x + 1)ⁿ⁺₁ + (x + 1)ⁿ Σan(x + 1)ⁿ) + 3(x + 1)ⁿ Σan(x + 1)ⁿ = 0.

To simplify further, we collect terms with the same power of (x + 1) and equate them to zero:

(x + 1)ⁿ⁻² [(n(n - 1) + 2n)an + (n(n + 1))an⁺²] + x(x + 1)ⁿ⁻¹ [3nan + 3nan⁺₁] + (x + 1)ⁿ [an + 3an] = 0.

For this equation to hold for all x, the coefficients of each power of (x + 1) must be zero. This gives us a recurrence relation for the coefficients an:

(n(n - 1) + 2n)an + (n(n + 1))an⁺² = 0, for n ≥ 2,

3nan + 3nan⁺₁ = 0, for n ≥ 0,

an + 3an = 0, for n < 0.

Solving these recurrence relations will give us the possible values of n.

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The matrix A given as [-1 2 -1; -1 -1 2; -1 -1 -1] is not orthogonally diagonalizable. This means that it cannot be diagonalized using an orthogonal matrix. The justification for this conclusion lies in the eigenvalues and their corresponding eigenvectors.

(a) To determine if A is orthogonally diagonalizable, we need to check if it has a complete set of orthogonal eigenvectors.

If A can be diagonalized using an orthogonal matrix P, then P must consist of the eigenvectors of A.

However, if A does not have a sufficient number of linearly independent eigenvectors or the eigenvectors are not orthogonal, it cannot be orthogonally diagonalizable.

(b) Using the Gram-Schmidt process, we can find an orthogonal matrix P and a diagonal matrix D such that[tex]A = PDP^{-1}[/tex]

However, in this case, we cannot find such matrices P and D because the matrix A does not have a complete set of orthogonal eigenvectors.

In conclusion, the matrix A = [-1 2 -1; -1 -1 2; -1 -1 -1] is not orthogonally diagonalizable.

This is because it does not have a complete set of orthogonal eigenvectors, which is a requirement for a matrix to be orthogonally diagonalizable.

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(d) The ordered pair solution of this equation 3x/4  + 2y = 15 is (4, 6).

(e) The ordered pair solution of this equation xy ≥ 6 is (-2, -4).

The ordered pair solution of the equations is calculated by simplifying the equations as follows;

The given equations;

(d) 3x/4  + 2y = 15

(e) xy ≥ 6

(d) The solution of this equation 3x/4  + 2y = 15 is calculated as follows;

let the value of x = 4 and the value of y = 6

so we will substitute this value into the equation and check if it will be equal to 15.

(4, 6) = (3 x 4 )/4  +  2(6)

(4, 6) = 3  +  12

(4, 6) = 15

(e) The solution of this equation xy ≥ 6 is calculated as follows;

let x = -2, and let y = - 4

(-2, -4) = (-2)(-4) = 8

8 ≥ 6 (this solution is true)

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