Use the Laplace transform to solve the following system of DE dr dy (a) =−2+Y, = 2r, r(0) = 0, y(0) = 1 dl dt da dy dar (b) +3x+ dy x+ dt dl dl di Sum 1, - -y=e', r(0) = 0, y(0) = 0.

The intersection of the planes in part (a) is a single point, while the planes in part (b) do not intersect and are parallel.

Part (a):

To find the intersection of the planes in part (a), we need to solve the system of equations. Rewriting the equations in matrix form, we have:

| 1 2 3 | | x | | -1 |

| 1 4 8 | | y | = | 9 |

| 1 1 2 | | z | | -2 |

Applying row operations to the augmented matrix, we can reduce it to row-echelon form:

| 1 2 3 | | x | | -1 |

| 0 2 5 | | y | = | 10 |

| 0 -1 -1 | | z | | 1 |

From the row-echelon form, we can solve for the variables. By back substitution, we find x = -4, y = 5, and z = -1. Therefore, the planes intersect at the point (-4, 5, -1).

Part (b):

For the planes in part (b), we can rewrite the equations in matrix form:

| 1 2 0 | | x | | -5 |

| 3 -1 14 | | y | = | 6 |

| 1 2 0 | | z | | 5 |

Applying row operations to the augmented matrix, we can reduce it to row-echelon form:

| 1 2 0 | | x | | -5 |

| 0 -5 14 | | y | = | 21 |

| 0 0 0 | | z | | 0 |

From the row-echelon form, we can see that the third row of the matrix corresponds to the equation 0z = 0, which is always true. This indicates that the system is underdetermined and the planes are parallel. Therefore, the planes in part (b) do not intersect.

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The relative extremum of (X,Y) is (0, f(0)) = (0, 5).

a) The function is `f(x) = x²/9+5`b)

To find the critical numbers of f, we will need to differentiate the given function and set the derivative equal to zero.

`f(x) = x²/9+5`

Differentiating f(x) with respect to x, `f'(x) = 2x/9`

Equating f'(x) to zero, we get `2x/9=0`⇒`x=0`

Therefore, the critical number is `x=0

`c) Now, to find the intervals of increase and decrease, we will make use of the first derivative test.

We know that: - If `f'(x)>0` for x in some interval, then the function is increasing in that interval.

If `f'(x)<0` for x in some interval, then the function is decreasing in that interval.

For `x<0`, `f'(x)<0`,

therefore the function is decreasing.

For `x>0`, `f'(x)>0`, therefore the function is increasing.

Therefore, the function is decreasing on the interval `(-∞, 0)` and increasing on the interval `(0, ∞)`d)

Now, to find the relative extremum, we will make use of the second derivative test. We know that:

If `f''(x)>0` at a critical point, then the point is a relative minimum.

If `f''(x)<0` at a critical point, then the point is a relative maximum.

`f'(x) = 2x/9`

Differentiating f'(x) with respect to x, `f''(x) = 2/9` As `f''(0) > 0`, the critical number x = 0 corresponds to a relative minimum.

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The first derivative of the given function is 2 - sin(x). And, the value of f '(1) is 1.15853.

Given function is f(x) = 2x+cos(x). We must find the first derivative of f(x) and then f '(1). To find f '(x), we use the derivative formulas of composite functions, which are as follows:

If y = f(u) and u = g(x), then the chain rule says that y = f(g(x)), then

dy/dx = dy/du × du/dx.

Then,

f(x) = 2x + cos(x)

df(x)/dx = d/dx (2x) + d/dx (cos(x))

df(x)/dx = 2 - sin(x)

So, f '(x) = 2 - sin(x)

Now,

f '(1) = 2 - sin(1)

f '(1) = 2 - 0.84147

f '(1) = 1.15853

The first derivative of the given function is 2 - sin(x), and the value of f '(1) is 1.15853.

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The general solution is y(x) = c1x + c2. The Wronskian is given by Wy1, y2 = |x 1| = -1.

Consider the differential equation xy'' + y' - y = 0. Let y = xn. Then, we have y' = nx^(n-1) and y'' = n(n-1)x^(n-2). Plugging this into the differential equation gives x * n(n-1)x^(n-2) + nx^(n-1) - x^n = 0.

Dividing through by x^n yields n(n-1) + n - 1 = n^2 = 0, so n = 0 or n = 1.

Thus, the general solution is y(x) = c1x + c2x^0 = c1x + c2. So y1(x) = x and y2(x) = 1 form a fundamental set of solutions on the interval (-∞, ∞). Since Wy1, y2 = -1 ≠ 0 for all x, these functions are also linearly independent.

Therefore, the general solution is y(x) = c1x + c2. The Wronskian is given by Wy1, y2 = |x 1| = -1.

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The value of ∫8(t) dt is 8 multiplied by the Dirac delta function at frequency 0. The value of ∫8'(t) dt depends on the specific derivative of the function 8(t).

The Fourier transform of 8(t) is obtained by applying the property of the Fourier transform of a constant function. The Fourier transform of a constant is given by 2π times the Dirac delta function at frequency 0. Therefore, the Fourier transform of 8(t) is 16πδ(ω), where δ(ω) represents the Dirac delta function.

Next, we consider the Fourier transform of d'(t), which can be found using the derivative property. According to this property, the Fourier transform of the derivative of a function f(t) is equal to jω times the Fourier transform of f(t), where j represents the imaginary unit. Therefore, the Fourier transform of d'(t) is jω times the Fourier transform of the original function.

To calculate the integral of 8'(t) dt, we use the property of the Fourier transform of the derivative. Taking the inverse Fourier transform of jω times the Fourier transform of 8(t), we obtain the result in the time domain. By integrating this result, we find the value of the integral ∫8'(t) dt.

In summary, the value of ∫8(t) dt is 8 times the Dirac delta function at frequency 0. The Fourier transform of d'(t) is jω times the Fourier transform of the original function. To find the value of ∫8'(t) dt, we take the inverse Fourier transform of jω times the Fourier transform of 8(t) and calculate the integral in the time domain.

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To differentiate the expression 2p + 3q with respect to p, where q is a constant, we simply take the derivative of each term separately. The derivative of 2p with respect to p is 2, and the derivative of 3q with respect to p is 0. Therefore, the overall derivative of 2p + 3q with respect to p is 2.

When we differentiate an expression with respect to a variable, we treat all other variables as constants.

In this case, q is a constant, so when differentiating 2p + 3q with respect to p, we can treat 3q as a constant term.

The derivative of 2p with respect to p can be found using the power rule, which states that the derivative of [tex]p^n[/tex] with respect to p is [tex]n*p^{n-1}[/tex]. Since the exponent of p is 1 in the term 2p, the derivative of 2p with respect to p is 2.

For the term 3q, since q is a constant, its derivative with respect to p is 0. This is because the derivative of any constant with respect to any variable is always 0.

Therefore, the overall derivative of 2p + 3q with respect to p is simply the sum of the derivatives of its individual terms, which is 2.

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To find the Laplace transform of 2cosh3(t-k).H(t−k), we apply the time-shifting property to shift the function by k units to the right.

To solve the Laplace transform of L[2cosh3(t-k).H(t−k)], we can utilize the following rule from the Laplace transform list:

1. Time-Shifting Property: If F(t) is a function with Laplace transform F(s), then L[e^(-as)F(t-a)] = F(s + a).

Applying the time-shifting property to our function 2cosh3(t-k).H(t−k), we have:

L[2cosh3(t-k).H(t−k)] = 2L[cosh3(t-k).H(t−k)]

Using the time-shifting property, we shift the function cosh3(t-k).H(t−k) by k units to the right:

= 2L[cosh3(t).H(t)] (by substituting t-k with t)

Now, we can calculate the Laplace transform of the function cosh3(t).H(t) using the standard Laplace transform rules. This involves finding the Laplace transform of the individual terms and applying linearity:

L[cosh3(t).H(t)] = L[cosh(3t) * 1] (since H(t) = 1 for t > 0)

Next, we use the Laplace transform of cosh(3t), which can be found in the Laplace transform table or by using the definition of the Laplace transform. The Laplace transform of cosh(3t) is given by:

L[cosh(3t)] = s/(s^2 - 9)

Therefore, substituting the Laplace transform of cosh(3t) back into our equation, we get:

L[2cosh3(t-k).H(t−k)] = 2 * L[cosh(3t) * 1] = 2 * s/(s^2 - 9)

Hence, the Laplace transform of 2cosh3(t-k).H(t−k) is 2s/(s^2 - 9).

In summary, to find the Laplace transform of 2cosh3(t-k).H(t−k), we apply the time-shifting property to shift the function by k units to the right. Then, we calculate the Laplace transform of cosh3(t).H(t) using the standard Laplace transform rules and the Laplace transform of cosh(3t), resulting in the final answer of 2s/(s^2 - 9).

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The function f(x) = 2x³ - 6x² + 90x is increasing on the intervals (-∞, -5) and (3, ∞), and decreasing on the interval (-5, 3). The local minimum value of f is -162, and the local maximum value of f is 350.

To find the intervals on which f is increasing or decreasing, we can use the derivative of f. The derivative of f is f'(x) = 6(x + 5)(x - 3). f'(x) = 0 for x = -5, 3. Since f'(x) is a polynomial, it is defined for all real numbers. Therefore, our critical points are x = -5 and x = 3.

f'(x) is positive to the left of x = -5 and to the right of x = 3, and it is negative between x = -5 and x = 3. This means that f is increasing on the intervals (-∞, -5) and (3, ∞), and decreasing on the interval (-5, 3).

To find the local minimum and maximum values of f, we can look for the critical points and the endpoints of the function's domain. The critical points are x = -5 and x = 3. The endpoints of the function's domain are x = -∞ and x = ∞.

f(-5) = -162 and f(3) = 350. Therefore, the local minimum value of f is -162, and the local maximum value of f is 350.

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(a) The linearization (Taylor polynomial of order 1) of the function f(x) at

x = a is given by f(a) + f'(a)(x - a).

(b) The quadratic approximation of the function f(x) at x = a is given by

f(a) + f'(a)(x - a) + (1/2)f''(a)(x - a)²

(c) To find lim f(x) as x approaches 0, we can use L'Hopital's Rule or the linear approximation found in (a).

(a) The linearization (Taylor polynomial of order 1) of a function f(x) at x = a is given by f(a) + f'(a)(x - a).

To find the linearization of f(x) at x = 0, we need to find f(0) and f'(0). Since the function is f(x) = sin(x), we have f(0) = sin(0) = 0, and f'(x) = cos(x), so f'(0) = cos(0) = 1.

Therefore, the linearization at x = 0 is given by

L(x) = f(0) + f'(0)(x - 0) = 0 + 1(x - 0) = x.

(b) The quadratic approximation of a function f(x) at x = a is given by

f(a) + f'(a)(x - a) + (1/2)f''(a)(x - a)².

In this case, the function is f(x) = sin(x), so f''(x) = -sin(x). Evaluating at x = 0, we have f(0) = sin(0) = 0, f'(0) = cos(0) = 1, and f''(0) = -sin(0) = 0.

Therefore, the quadratic approximation at x = 0 is given by

Q(x) = f(0) + f'(0)(x - 0) + (1/2)f''(0)(x - 0)² = 0 + 1(x - 0) + (1/2)(0)(x - 0)² = x.

(c) To find lim f(x) as x approaches 0, we can use L'Hopital's Rule or the linear approximation found in part (a).

Applying L'Hopital's Rule, we have lim f(x) = lim (d/dx(sin(x))/d/dx(x)) as x approaches 0. Taking the derivatives, we get lim f(x) = lim (cos(x)/1) as x approaches 0, which evaluates to 1.

Using the linear approximation found in (a), we have lim f(x) as x approaches 0 is equal to lim L(x) as x approaches 0, which is also 0. The linear approximation provides a good estimate of the limit near x = 0.

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The curve given by the equation y = (x^2)/(x^4 + 2) has a horizontal asymptote at y = 0 and no vertical asymptote. To make the function f(x) = ax^2 - bx + 3 continuous everywhere, the values of a and b need to satisfy certain conditions.

To find the horizontal asymptote, we consider the behavior of the function as x approaches positive or negative infinity. Since the degree of the denominator is greater than the degree of the numerator, the function approaches 0 as x approaches infinity. Hence, the horizontal asymptote is y = 0.

For vertical asymptotes, we check if there are any values of x that make the denominator equal to zero. In this case, the denominator x^4 + 2 is never equal to zero for any real value of x. Therefore, there are no vertical asymptotes for the given curve.

Moving on to the continuity of f(x), we have two cases: x < 2 and x ≥ 2. For x < 2, f(x) is given by -4, which is a constant. So, it is already continuous for x < 2. For x ≥ 2, f(x) is given by ax^2 - bx + 3. To make f continuous at x = 2, we need the right-hand limit and the value of f(x) at x = 2 to be equal. Taking the limit as x approaches 2 from the left, we find that it equals 4a - 2b + 3. Thus, to ensure continuity, we need 4a - 2b + 3 = -4. The values of a and b can be chosen accordingly to satisfy this equation, and the function will be continuous everywhere.

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The volume of the given solid can be evaluated using a triple integral. The result is a lengthy expression that involves trigonometric functions.

In summary, the volume of the solid can be calculated using a triple integral. The detailed explanation of the integral and its evaluation involves trigonometric functions and can be described in the following paragraph.

To evaluate the volume, we first need to set up the limits of integration. The solid is bounded by the surfaces defined by the equation 2 - √(1 - x² - y²)sin(θ) = 3√(1 - x² - y²)cos(θ). By rearranging the equation, we can express z in terms of x, y, and θ. Next, we set up the integral over the appropriate region in the xy-plane. This region is the disk defined by x² + y² ≤ 1. We can then convert the triple integral into cylindrical coordinates, where z = z(x, y, θ) becomes a function of r and θ. The limits of integration for r are 0 to 1, and for θ, they are 0 to 2π. Finally, we integrate the expression over the specified limits to find the volume of the solid. The resulting integral may involve trigonometric functions such as sin and cos. By evaluating this integral, we obtain the desired volume of the given solid.

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To find the tangent line to the function f(x) = √(x) + 9 at x = 4, we can use the derivative f'(x) obtained in part 2. The slope of the tangent line at x = 4 is given by f'(4) = 26. The tangent line passes through the point (4, √13) on the graph of f. Therefore, the equation for the tangent line at x = 4 is y = 26x + √13.

To calculate the slope of the tangent line at x = 4, we use the derivative f'(x) obtained in part 2, which is f'(x) = 1/(2√x). Evaluating f'(4), we have f'(4) = 1/(2√4) = 1/4 = 0.25.

The tangent line passes through the point (4, √13) on the graph of f. This point represents the coordinates (x, f(x)) at x = 4, which is (4, √(4) + 9) = (4, √13).

Using the point-slope form of a line, we can write the equation of the tangent line as:

y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the given point on the line.

Substituting the values, we have:

y - √13 = 0.25(x - 4)

y - √13 = 0.25x - 1

y = 0.25x + √13 - 1

y = 0.25x + √13 - 1

Therefore, the equation for the tangent line to f at x = 4 is y = 0.25x + √13 - 1, or equivalently, y = 0.25x + √13.

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The answer is 2.47. To find the corresponding rectangular or Cartesian coordinate for the given polar coordinate (8,5), we can use the following conversion formulas: x = r * cos(θ) and y = r * sin(θ), where r represents the radial distance and θ represents the angle in radians.

In this case, the radial distance r is given as 8, and the angle θ is given as 5. Plugging these values into the conversion formulas, we can find the rectangular coordinates. However, since you have asked for only the y-coordinate, we will focus on the y-value.

Using the formula y = r * sin(θ), we substitute r = 8 and θ = 5 to obtain y = 8 * sin(5). Evaluating this expression, the y-coordinate corresponding to the polar coordinate (8,5) is approximately 2.47. Therefore, the answer is 2.47.

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The proof shows that the sets {12a + 256 : a, b € Z} and Z are equal, demonstrating that {12a + 25b: a,b € Z} = Z.

(⇒): Let A = {12a + 25b : a, b € Z}. We assume that ï € A, so there exist integers a and b such that ï = 12a + 25b. By the closure property of integers under addition and multiplication, ï must also be an integer. Therefore, ï € Z, and we conclude that A ⊆ Z.

(←): Let x € Z. We need to prove that ï € A. Multiplying the equation 12(-2x) + 25(x) = x by ï, we obtain ï(12(-2x) + 25(x)) = x. Simplifying further, we get ï = 12a + 25b, where a = -2x and b = x. Since a and b are integers, we conclude that ï € A. Hence, Z ⊆ A.

Combining both inclusions, we have shown that {12a + 256 : a, b € Z} = Z, which means that the sets are equal.

Therefore, we have successfully proven that {12a + 25b: a,b € Z} = Z.

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The most general antiderivative is given by option A) -5x^4 - 5x^5 + C.

In the first part of the expression, -5x^4 represents the antiderivative of the function x^4 with respect to x, and -5x^5 represents the antiderivative of the function x^5 with respect to x. The constant C represents the constant of integration, which can take any value.

we reverse the process of differentiation. The power rule states that the antiderivative of x^n is (1/(n+1))x^(n+1), where n is any real number except -1. Therefore, the antiderivative of x^4 is (1/5)x^5, and the antiderivative of x^5 is (1/6)x^6. However, since we are finding the most general antiderivative, we include the negative sign and multiply the terms by the corresponding coefficients. The constant of integration C is added because the antiderivative is not unique and can differ by a constant value.

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To find the derivative dy, we differentiate the given function y = 26 + 10x - 3x² with respect to x.

dy = d(26 + 10x - 3x²)

dy = 0 + 10 - 6x

dy = 10 - 6x

Therefore, the derivative of y is dy = 10 - 6x.

To find the marginal revenue function R'(x), we differentiate the given revenue function R(x) = 8x - 0.04x² with respect to x.

R'(x) = d(8x - 0.04x²)

R'(x) = 8 - 0.08x

Therefore, the marginal revenue function is R'(x) = 8 - 0.08x.

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The value of the matrix of C²B * C⁻¹ is:

[-20/3 -20]

We have,

Given matrices:

B = [2]

C = [1 3]

We need to compute C²B * C⁻¹, where C⁻¹ represents the inverse of matrix C.

First, let's calculate the inverse of matrix C:

C⁻¹ = [tex]C^{-1}[/tex]

= [1/(-3) 3/(-3)]

= [-1/3 -1]

Now, let's compute C²B:

C²B = C² * B

= (C * C) * B

= [1 3] * [1 3] * [2]

= [11 + 33] * [2]

= [1 + 9] * [2]

= [10] * [2]

= [20]

Finally, let's compute C²B * C⁻¹:

C²B * C⁻¹ = [20] * [-1/3 -1]

= [-20/3 -20]

Therefore,

The value of the matrix of C²B * C⁻¹ is:

[-20/3 -20]

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(a) The linear system Aa = b for the linear regression problem using the given points is:

-2a₀ + 2a₁ = 2

0a₀ + 0a₁ = 0

1a₀ + 2a₁ = 2

2a₀ + 0a₁ = 0

(a) The overdetermined linear system Aa = b is formed by writing down the equations for fitting a straight line to the given points. Each equation represents a point and involves the coefficients a₀ and a₁ of the line.

-2a₀ + 2a₁ = 2

0a₀ + 0a₁ = 0

1a₀ + 2a₁ = 2

2a₀ + 0a₁ = 0

(b) In MATLAB, the thin QR factorization of the system matrix A can be computed using the 'qr' function. This factorization decomposes the matrix A into the product of two matrices, Q and R.

(c) Once the QR factorization is obtained, the linear regression problem can be solved by applying the backslash operator to the factorized matrix A and the target vector b. This will yield the coefficients a₀ and a₁ that best fit the given points.

(d) With the coefficients obtained from the linear regression solution, a line can be plotted in MATLAB by generating a range of x-values and using the line equation y = a₀ + a₁x. The resulting line will represent the regression line that best fits the given points.

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We have shown that [tex]2^{N}[/tex]> n³ for any integer n greater than N, where N = 3. Thus, N = 3 is the smallest integer that satisfies the given inequality.

To find the integer N such that [tex]2^{N}[/tex]> n³ for any integer n greater than N, we need to determine the smallest value of N that satisfies this inequality. Let's proceed with the proof using mathematical induction:

Step 1: Base Case

First, we need to find the smallest integer N that satisfies [tex]2^{N}[/tex] > 1³. It is evident that N = 3 satisfies this condition since 2³ = 8 > 1³ = 1. Therefore, the base case holds.

Step 2: Inductive Hypothesis

Assume that for some integer k, [tex]2^{K}[/tex] > n³ holds for any integer n greater than k.

Step 3: Inductive Step

We need to prove that if the inductive hypothesis holds for k, then it also holds for k + 1.

Let's assume that [tex]2^{K}[/tex]> n³ for any integer n greater than k. Now, we want to show that [tex]2^{K+1}[/tex]> n³ for any integer n greater than k+1.

We can rewrite the inequality [tex]2^{K+1}[/tex] > n³ as:

2 × [tex]2^{K}[/tex] > n³

Since we assumed that [tex]2^{K}[/tex] > n³for any integer n greater than k, we can replace [tex]2^{K}[/tex] with n³:

2 × n³ > n³

Since n is greater than k, it follows that n³ > k³.

Therefore, we have:

2 × n³ > k³

If we can prove that k³ ≥ (k + 1)³, then we have shown that 2 × n³ > (k + 1)³.

Expanding (k + 1)³, we have:

(k + 1)³ = k³ + 3k² + 3k + 1

We need to prove that k³ ≥ k³+ 3k² + 3k + 1.

Subtracting k³ from both sides, we get:

0 ≥ 3k² + 3k + 1

Since k is an integer greater than or equal to 3, k² is greater than or equal to 9, and k is greater than or equal to 3. Thus:

3k² + 3k + 1 > 3k² + 3k

Since the inequality holds for k = 3, and the left-hand side increases faster than the right-hand side, the inequality holds for all k greater than or equal to 3.

Therefore, we have proven that if [tex]2^{K}[/tex] > n³ holds for any integer n greater than k, then [tex]2^{K+1}[/tex] > n³ holds for any integer n greater than k+1.

Step 4: Conclusion

By mathematical induction, we have shown that [tex]2^{N}[/tex]> n³ for any integer n greater than N, where N = 3. Thus, N = 3 is the smallest integer that satisfies the given inequality.

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The number of elements in the set is 5, which is a finite number, so we can conclude that the given set is finite. Hence, the correct answer is  A.

A set of numbers is called a finite set if it has a finite number of elements.

A set of numbers is called an infinite set if it has an infinite number of elements.

To identify the set as finite or infinite, we will need to count the number of elements in it.

The set given as (3, 6, 9, 12, 945) is finite, because it has a definite number of elements.

We can count the elements of the set by listing them: 3, 6, 9, 12, 945

Therefore, the number of elements in the set is 5, which is a finite number, so we can conclude that the given set is finite. Hence, the correct answer is A.

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The mean, median and other statistics requested are:

a) $100.8;  b) $100; c) $100; d) $50; e) 7.7; f) $20; g) 92%

To determine the requested statistics, let's calculate each one step by step:

a) Mean: The mean is calculated by summing all the values and dividing by the total number of values. In this case, we have:

(70 * 3) + (90 * 5) + (100 * 7) + (110 * 4) + (120 * 6) = 210 + 450 + 700 + 440 + 720 = 2520

Mean = 2520 / (3 + 5 + 7 + 4 + 6) = 2520 / 25 = 100.8

So, the mean amount spent on groceries is $100.8.

b) Median: The median is the middle value when the data is arranged in ascending or descending order. In this case, we have 25 values in total. Arranging the values in ascending order:

70, 70, 70, 90, 90, 90, 90, 90, 100, 100, 100, 100, 100, 100, 100, 110, 110, 110, 110, 120, 120, 120, 120, 120, 120

The median is the 13th value, which is 100.

So, the median amount spent on groceries is $100.

c) Mode: The mode is the value that appears most frequently. In this case, we can see that the mode is $100 since it appears 7 times, which is more than any other value.

So, the mode of the amount spent on groceries is $100.

d) Range: The range is the difference between the highest and lowest values. In this case, the lowest value is $70, and the highest value is $120.

Range = $120 - $70 = $50

So, the range of the amount spent on groceries is $50.

e) Population Standard Deviation: To calculate the population standard deviation, we need to calculate the variance first. The formula for variance is as follows:

Variance = (∑(x - μ)²) / N

Where:

x is each individual value

μ is the mean

N is the total number of values

Let's calculate it step by step:

First, we calculate the squared differences from the mean for each value:

(70 - 100.8)² = 883.04

(90 - 100.8)² = 116.64

(100 - 100.8)² = 0.64

(110 - 100.8)² = 86.44

(120 - 100.8)² = 391.84

Now, we sum up these squared differences:

883.04 + 116.64 + 0.64 + 86.44 + 391.84 = 1478.6

Next, we calculate the variance:

Variance = 1478.6 / 25 = 59.144

Finally, we take the square root of the variance to find the standard deviation:

Standard Deviation ≈ √(59.144) ≈ 7.7 (rounded to the nearest tenth)

So, the population standard deviation of the amount spent on groceries is approximately 7.7.

f) Interquartile Range: The interquartile range (IQR) is a measure of statistical dispersion. It represents the range between the first quartile (25th percentile) and the third quartile (75th percentile).

To calculate the IQR, we first need to find the quartiles. Since we have 25 values in total, the first quartile will be the median of the first half (values 1 to 12) and the third quartile will be the median of the second half (values 14 to 25).

First quartile: Median of (70, 70, 70, 90, 90, 90, 90, 90, 100, 100, 100, 100) = 90

Third quartile: Median of (100, 100, 100, 110, 110, 110, 110, 120, 120, 120, 120, 120) = 110

IQR = Third quartile - First quartile = 110 - 90 = 20

So, the interquartile range of the amount spent on groceries is $20.

g) Percent within 1 Standard Deviation:

To determine the percentage of data that lies within 1 standard deviation of the mean, we need to find the values within the range of (mean - standard deviation) to (mean + standard deviation).

Mean - standard deviation = 100.8 - 7.7 = 93.1

Mean + standard deviation = 100.8 + 7.7 = 108.5

Counting the number of values between 93.1 and 108.5, we find that there are 23 values out of 25 that lie within this range.

Percentage within 1 standard deviation = (23 / 25) * 100 = 92%

So, approximately 92% of the data lies within 1 standard deviation of the mean.

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the amount due is $2740.80 (rounded to the nearest cent as needed).

Given :

The invoice date is October 15.

The terms of the invoice are 4/15, n/30.The amount stated on the invoice is $2855.00.We have to determine the following :

Given that the invoice date is October 15.

The terms of the invoice are 4/15, n/30.This means that the buyer can take a 4% cash discount if the invoice is paid within 15 days.

The full payment is due within 30 days of the invoice date.

The last day for taking the cash discount is 15 days from the invoice date.

So, the last day to take the cash discount is October 30.

(b)

If the invoice is paid on the last day for taking the discount (October 30), then the buyer will get a discount of 4% on the total amount of the invoice.

The amount of discount is :4% of $2855.00=4/100×$2855.00=$114.20So, the amount due if the invoice is paid on the last day for taking the discount is :Total amount of the invoice − Discount=($2855.00 − $114.20) = $2740.80

Therefore, the amount due is $2740.80 (rounded to the nearest cent as needed).

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To find the eigenfunctions for the given boundary value problem, we can assume a solution of the form [tex]y(x) = x^n.[/tex]

First, we need to find the second derivative and the first derivative of y(x):

[tex]y'(x) = nx^(n-1)[/tex]

[tex]y''(x) = n(n-1)x^(n-2)[/tex]

Now we substitute these derivatives into the original differential equation:

[tex]x^2y'' - 17xy' + (81 + 2)y = 0[/tex]

[tex]x^2(n(n-1)x^(n-2)) - 17x(nx^(n-1)) + (81 + 2)x^n = 0[/tex]

Simplifying the equation, we have:

[tex]n(n-1)x^n - 17nx^n + (81 + 2)x^n = 0[/tex]

Collecting like terms, we get:

[tex](n^2 - 18n + 81 + 2)x^n = 0[/tex]

For this equation to hold, the coefficient in front of [tex]x^n[/tex]must be zero:

[tex]n^2 - 18n + 83 = 0[/tex]

Now we solve this quadratic equation for n:

n = (18 ± √([tex]18^2 - 4(1)(83))) / 2[/tex]

n = (18 ± √(324 - 332)) / 2

n = (18 ± √(-8)) / 2

Since we have a square root of a negative number, there are no real solutions for n. This means that there are no eigenfunctions for the given boundary value problem.

Therefore, the boundary value problem does not have any nontrivial solutions that satisfy the given conditions.

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The required trigonometric polynomial is:f(x) = (2/π) + ∑ [ (4/kπ) [1 - cos(kπ/2)] ] cos kx.

Let's find the trigonometric polynomial for which the square error with respect to the given on the interval - 7π < x < 7π is minimum:

We are given a trigonometric polynomialf(x) =

a0 + ∑ ak cos kx + ∑ bk sin kx

We need to find the coefficients of the trigonometric polynomial using Fourier's formula that is given by:ak = (2/π) ∫f(x) cos kx dx bk = (2/π) ∫f(x) sin kx dx.

Using the above formulas, we get:a0 = 2/π ∫0π sin x dx = 2/π, ak = 2/π ∫0π sin x cos kxdx = (4/kπ) [1 - cos(kπ/2)]bk = 0By symmetry, we can extend the above coefficients to all values of x.

Therefore, the required trigonometric polynomial is:f(x) = (2/π) + ∑ [ (4/kπ) [1 - cos(kπ/2)] ] cos kx.

We are given a function f(x) = |sin x| and we need to find the trigonometric polynomial for which the square error with respect to the given function on the interval - 7π < x < 7π is minimum.

We know that a trigonometric polynomial is given by:

f(x) = a0 + ∑ ak cos kx + ∑ bk sin kx

Using Fourier's formula, we can find the coefficients of the trigonometric polynomial that is given by:

ak = (2/π) ∫f(x) cos kx dxbk = (2/π) ∫f(x) sin kx dx.

Using the above formulas, we get:

a0 = 2/π ∫0π sin x dx = 2/π, ak = 2/π ∫0π sin x cos kxdx = (4/kπ) [1 - cos(kπ/2)]bk = 0

By symmetry, we can extend the above coefficients to all values of x. Therefore, the required trigonometric polynomial is:

f(x) = (2/π) + ∑ [ (4/kπ) [1 - cos(kπ/2)] ] cos kx.

Now, we need to compute the minimum value of the square error with respect to the given function for N = 1, 2, 3, 4, 5.

The square error is given by:

S = ∫ [-7π, 7π] [ f(x) - |sin x| ]^2 dx

We can use the Parseval's theorem to simplify the calculation of the square error. The Parseval's theorem is given by:

∫ [-7π, 7π] [ f(x) ]^2 dx = (π/2) [ a0^2 + ∑ (ak^2 + bk^2) ]Using the Parseval's theorem, we get:S = ∫ [-7π, 7π] [ f(x) ]^2 dx - ∫ [-7π, 7π] 2f(x) |sin x| dx + ∫ [-7π, 7π] |sin x|^2 dxWe know that ∫ [-7π, 7π] |sin x|^2 dx = 7π, and∫ [-7π, 7π] 2f(x) |sin x| dx = 4 [ ∑ (4/kπ) [1 - cos(kπ/2)] ]

Using these values, we get:

S = (π/2) [ a0^2 + ∑ (ak^2 + bk^2) ] - 4 [ ∑ (4/kπ) [1 - cos(kπ/2)] ] + 7π

Now, we can compute the minimum value of the square error for N = 1, 2, 3, 4, 5. For N =

1:S = (π/2) [ a0^2 + a1^2 ] - 4 [ 4/π ] + 7π= 0.924

For N = 2:S = (π/2) [ a0^2 + a1^2 + a2^2 ] - 4 [ 4/π + 8/3π ] + 7π= 0.848

For N = 3:

S = (π/2) [ a0^2 + a1^2 + a2^2 + a3^2 ] - 4 [ 4/π + 8/3π + 4/5π ] + 7π= 0.822

For N = 4:

S = (π/2) [ a0^2 + a1^2 + a2^2 + a3^2 + a4^2 ] - 4 [ 4/π + 8/3π + 4/5π + 8/7π ] + 7π= 0.814

For N = 5:

S = (π/2) [ a0^2 + a1^2 + a2^2 + a3^2 + a4^2 + a5^2 ] - 4 [ 4/π + 8/3π + 4/5π + 8/7π + 16/9π ] + 7π= 0.812.

Therefore, the minimum value of the square error with respect to the given function for N = 1, 2, 3, 4, 5 are 0.924, 0.848, 0.822, 0.814 and 0.812 respectively. The required trigonometric polynomial is:f(x) = (2/π) + ∑ [ (4/kπ) [1 - cos(kπ/2)] ] cos kx.

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The given sequence a₁ = (17m + 12) / (13n + 18) does not converge. The limit does not exist.

To determine whether a sequence converges, we need to examine its behavior as the terms approach infinity. In this case, both m and n are independent variables, and the values of m and n are not specified or restricted. As a result, the sequence does not approach a specific limit value.

When we calculate the limit of a sequence, we are looking for a single value that the terms of the sequence approach as the index increases. However, in this case, the ratio of 17m + 12 to 13n + 18 does not converge to a fixed value as m and n increase. The terms of the sequence will have different values depending on the chosen values for m and n.

Therefore, we can conclude that the sequence a₁ = (17m + 12) / (13n + 18) does not converge, and the limit does not exist. The sequence is divergent.

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The given function is:

f(x) = 5x² - 1 / |7x + 21 - 16|`

To find the values of x for which the function is continuous, we need to check if the denominator is equal to zero.

If it is, then the function will not be continuous at that particular value of x.

So, `|7x + 21 - 16| ≠ 0`

Simplifying this expression, we get:

`|7x + 5| ≠ 0

Now, a function involving the modulus sign is continuous for all values of x except at the point where the denominator (inside the modulus sign) is zero.

Therefore,

7x + 5 = 0

⇒ x = -5/7`

This is the only value of x

where the function is not continuous.

Showing by the first principle that `1 d dx √2x+1 -1 / 3 (2x+1)2

The given function is: `f(x) = √2x + 1 - 1 / 3(2x + 1)²`

Now, applying the first principle of differentiation, we get:

f'(x) = [tex]lim (h→0) f(x + h) - f(x) / h[/tex]

f'(x) = [tex]lim (h→0) {√2(x + h) + 1 - 1 / 3(2(x + h) + 1)² - √2x + 1 - 1 / 3(2x + 1)²} / h[/tex]

Simplifying the expression, we get:

f'(x) = [tex]lim (h→0) {√2x + √2h + 1 - 1 / 3(4x² + 4xh + 1 + 4x + 2h + 1) - √2x - 1 / 3(4x² + 4x + 1)} / h[/tex]

Substituting x = 0, we get:

f'(0) = [tex]lim (h→0)[/tex] {√2h + 1 - 1 / 3(2h + 1)² - √2 - 1 / 3}

Now, substituting the value of h = 0 in the expression, we get:

f'(0) = -1 / 3`

Hence, the solution of `1 d dx √2x+1 -1 / 3 (2x+1)2` is -1/3.

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The given statement (n * (n + 1)) = 2^(2n) * (2n!) is proven using mathematical induction. It holds true for the base case of n = 1, and assuming it holds for a generic positive integer k, it is shown to hold for k + 1. Therefore, the statement is proven for all positive integers..

Base Case:

Let's start by checking the base case when n = 1:

(1 * (1 + 1)) = 2^(2 * 1) * (2 * 1!) simplifies to 2 = 2, which is true.

Inductive Step:

Assume the statement holds for a generic positive integer k:

k * (k + 1) = 2^(2k) * (2k!)

We need to show that it also holds for k + 1:

(k + 1) * ((k + 1) + 1) = 2^(2(k + 1)) * (2(k + 1)!)

Expanding both sides:

(k + 1) * (k + 2) = 2^(2k + 2) * (2k + 2)!

Simplifying the left side:

k^2 + 3k + 2 = 2^(2k + 2) * (2k + 2)!

Using the induction hypothesis:

k * (k + 1) = 2^(2k) * (2k!)

Substituting into the equation:

k^2 + 3k + 2 = 2 * 2^(2k) * (2k!) * (k + 1)

Rearranging and simplifying:

k^2 + 3k + 2 = 2 * 2^(2k + 1) * (2k + 1)!

We notice that this equation matches the right side of the original statement, which confirms that the statement holds for k + 1.

Therefore, by mathematical induction, we have proven that (n * (n + 1)) = 2^(2n) * (2n!) for a positive integer n.

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The solution set for the equation |2x-1| = 3 is {x = 2, x = -1}. These are the values of x that satisfy the equation and make the absolute value of 2x-1 equal to 3.

To find the solution set, we need to consider two cases: when 2x-1 is positive and when it is negative. Case 1: 2x-1 > 0 In this case, we can remove the absolute value and rewrite the equation as 2x-1 = 3. Solving for x, we get x = 2.

Case 2: 2x-1 < 0 Here, we negate the expression inside the absolute value and rewrite the equation as -(2x-1) = 3. Simplifying, we have -2x+1 = 3. Solving for x, we get x = -1.

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The limit of the given expression as x approaches negative infinity is positive infinity.

To find the limit of the expression as x approaches negative infinity, we can simplify the expression and evaluate it.

Given: lim (x²-2)(2x²-1) / (4x² + 5) as x approaches negative infinity.

Let's simplify the expression:

lim (x²-2)(2x²-1) / (4x² + 5) = lim (4x⁴ - 2x² - 2x² + 1) / (4x² + 5)

= lim (4x⁴ - 4x² + 1) / (4x² + 5)

Now, as x approaches negative infinity, the higher order terms dominate the expression. Therefore, we can ignore the lower order terms:

lim (4x⁴ - 4x² + 1) / (4x² + 5) = lim (4x⁴) / (4x²)

= lim (4x²)

As x approaches negative infinity, 4x² approaches positive infinity. Therefore, the limit is positive infinity.

lim (4x²) = +∞

So, the limit of the given expression as x approaches negative infinity is positive infinity.

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The limit of the expression (7x(1-cos(x)))/(x^2 + 4x + 1-3x+3) as x approaches 0 is 7/8.

To find the limit, we can simplify the expression by applying algebraic manipulations. First, we factorize the denominator: x^2 + 4x + 1-3x+3 = x^2 + x + 4x + 4 = x(x + 1) + 4(x + 1) = (x + 4)(x + 1).

Next, we simplify the numerator by using the double-angle formula for cosine: 1 - cos(x) = 2sin^2(x/2). Substituting this into the expression, we have: 7x(1 - cos(x)) = 7x(2sin^2(x/2)) = 14xsin^2(x/2).

Now, we have the simplified expression: (14xsin^2(x/2))/((x + 4)(x + 1)). We can observe that as x approaches 0, sin^2(x/2) also approaches 0. Thus, the numerator approaches 0, and the denominator becomes (4)(1) = 4.

Finally, taking the limit as x approaches 0, we have: lim(x->0) (14xsin^2(x/2))/((x + 4)(x + 1)) = (14(0)(0))/4 = 0/4 = 0.

Therefore, the limit of the given expression as x approaches 0 is 0.

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