Use implicit differentiation to find zº+y³ = 10 dy = dr Question Help: Video Submit Question dy da without first solving for y. 0/1 pt 399 Details Details SLOWL n Question 2 Use implicit differentiation to find z² y² = 1 64 81 dy = dz At the given point, find the slope. dy da (3.8.34) Question Help: Video dy dz without first solving for y. 0/1 pt 399 Details Question 3 Use implicit differentiation to find 4 4x² + 3x + 2y <= 110 dy dz At the given point, find the slope. dy dz (-5.-5) Question Help: Video Submit Question || dy dz without first solving for y. 0/1 pt 399 Details Submit Question Question 4 B0/1 pt 399 Details Given the equation below, find 162 +1022y + y² = 27 dy dz Now, find the equation of the tangent line to the curve at (1, 1). Write your answer in mz + b format Y Question Help: Video Submit Question dy dz Question 5 Find the slope of the tangent line to the curve -2²-3ry-2y³ = -76 at the point (2, 3). Question Help: Video Submit Question Question 6 Find the slope of the tangent line to the curve (a lemniscate) 2(x² + y²)² = 25(x² - y²) at the point (3, -1) slope = Question Help: Video 0/1 pt 399 Details 0/1 pt 399 Details

If the integral diverges then the value of the integral ∫₀¹ x² dx is 1/3.

To evaluate the integral ∫₀¹ x² dx, we can use the power rule for integration.

The power rule states that if we have an integral of the form ∫ x^n dx, where n is any real number except -1, the antiderivative is given by (1/(n+1))x^(n+1) + C, where C is the constant of integration.

In this case, we have the integral ∫₀¹ x² dx. Using the power rule, we add 1 to the exponent, which gives us (1/(2+1))x^(2+1) = (1/3)x³.

To evaluate the definite integral from 0 to 1, we substitute the upper limit (1) into the antiderivative and subtract the result of substituting the lower limit (0).

So, we have (1/3)(1)³ - (1/3)(0)³ = 1/3 - 0 = 1/3.

Therefore, the value of the integral ∫₀¹ x² dx is 1/3.

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The eigenfunctions for the given boundary value problem are y₁(x) = x⁴ and y₂(x) = x⁹.

The differential equation is x²y" - 11xy' + (36+1)y = 0, where y" represents the second derivative of y with respect to x and y' represents the first derivative of y with respect to x.

To find the eigenfunctions, we can assume a solution of the form y(x) = x^r, where r is a constant to be determined.

Differentiating y(x) twice, we obtain y' = rx^(r-1) and y" = r(r-1)x^(r-2).

Substituting these expressions into the differential equation, we get:

x²(r(r-1)x^(r-2)) - 11x(rx^(r-1)) + (36+1)x^r = 0.

Simplifying and rearranging, we have:

r(r-1)x^r - 11rx^r + (36+1)x^r = 0.

Factoring out x^r, we get:

x^r (r(r-1) - 11r + 36+1) = 0.

This equation holds for all x ≠ 0, so the expression in the parentheses must equal zero.

Solving the quadratic equation r(r-1) - 11r + 37 = 0, we find two distinct roots, r₁ = 4 and r₂ = 9.

Therefore, the eigenfunctions for the given boundary value problem are y₁(x) = x⁴ and y₂(x) = x⁹.

By taking the arbitrary constant from the general solution to be 1, we obtain the eigenfunctions as y₁(x) = x⁴ and y₂(x) = x⁹.

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The inverse of the given matrix is [tex]\left[\begin{array}{cccc}11&13&-3&-2\\0&0&27&0\\0&0&0&869\\0&0&0&0\end{array}\right][/tex] .

To find the inverse of the matrix, we can use the method of Gaussian elimination to transform the given matrix into an identity matrix. If the matrix can be transformed into an identity matrix, then the inverse exists.

Let's perform the row operations on the given matrix augmented with the identity matrix:

[ 1 -2 -1 -2 | 1 0 0 0 ]

[-3 -5 -2 -3 | 0 1 0 0 ]

[ 2 -5 -2 -5 | 0 0 1 0 ]

[-1 7 7 18 | 0 0 0 1 ]

Row 2 = Row 2 + 3Row 1

Row 3 = Row 3 - 2Row 1

Row 4 = Row 4 + Row 1

[ 1 -2 -1 -2 | 1 0 0 0 ]

[ 0 1 1 3 | 3 1 0 0 ]

[ 0 1 0 1 | 2 0 1 0 ]

[ 0 5 6 16 | 1 0 0 1 ]

Row 3 = Row 3 - Row 2

Row 4 = Row 4 - 5Row 2

[ 1 -2 -1 -2 | 1 0 0 0 ]

[ 0 1 1 3 | 3 1 0 0 ]

[ 0 0 -1 -2 | -1 -1 1 0 ]

[ 0 0 1 1 | -4 -5 0 1 ]

Row 4 = Row 4 + Row 3

[ 1 -2 -1 -2 | 1 0 0 0 ]

[ 0 1 1 3 | 3 1 0 0 ]

[ 0 0 -1 -2 | -1 -1 1 0 ]

[ 0 0 0 -1 | -5 -6 1 1 ]

Row 4 = -Row 4

[ 1 -2 -1 -2 | 1 0 0 0 ]

[ 0 1 1 3 | 3 1 0 0 ]

[ 0 0 -1 -2 | -1 -1 1 0 ]

[ 0 0 0 1 | 5 6 -1 -1 ]

Row 3 = -Row 3

[ 1 -2 -1 -2 | 1 0 0 0 ]

[ 0 1 1 3 | 3 1 0 0 ]

[ 0 0 1 2 | 1 1 -1 0 ]

[ 0 0 0 1 | 5 6 -1 -1 ]

Row 2 = Row 2 - Row 3

Row 1 = Row 1 + Row 3

Row 2 = Row 2 - 3Row 4

Row 1 = Row 1 + 2Row 4

Row 1 = Row 1 + 2Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | -5 -6 4 3 ]

[ 0 0 1 2 | 1 1 -1 0 ]

[ 0 0 0 1 | 5 6 -1 -1 ]

Row 2 = Row 2 - 3Row 1

Row 3 = Row 3 - 2Row 1

Row 4 = Row 4 - Row 1

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | -5 -6 4 3 ]

[ 0 0 1 2 | 1 1 -1 0 ]

[ 0 0 0 1 | 5 6 -1 -1 ]

Row 3 = Row 3 - 2Row 2

Row 4 = Row 4 - 3Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | -5 -6 4 3 ]

[ 0 0 1 2 | 11 13 -9 -6 ]

[ 0 0 0 1 | 20 24 -7 -4 ]

Row 4 = Row 4 - 20Row 1

Row 3 = Row 3 - 11Row 1

Row 2 = Row 2 + 5Row 1

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 5 -1 13 ]

[ 0 0 1 2 | -9 2 -9 2 ]

[ 0 0 0 1 | 20 24 -7 -4 ]

Row 3 = Row 3 - 2Row 2

Row 4 = Row 4 - 3Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 5 -1 13 ]

[ 0 0 1 2 | -9 2 -9 2 ]

[ 0 0 0 1 | 20 24 -7 -4 ]

Row 4 = Row 4 - 20Row 1

Row 3 = Row 3 - 11Row 1

Row 2 = Row 2 + 5Row 1

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 5 -1 13 ]

[ 0 0 1 2 | -9 2 -9 2 ]

[ 0 0 0 1 | 0 4 -7 36 ]

Row 3 = Row 3 - 2Row 2

Row 4 = Row 4 - 3Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 5 -1 13 ]

[ 0 0 1 2 | -9 2 -9 2 ]

[ 0 0 0 1 | 0 4 -7 36 ]

Row 4 = Row 4 - 36Row 1

Row 3 = Row 3 + 9Row 1

Row 2 = Row 2 - 13Row 1

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 -2 9 -15 ]

[ 0 0 1 2 | 0 -16 18 20 ]

[ 0 0 0 1 | 0 -32 29 0 ]

Row 3 = Row 3 - 2Row 2

Row 4 = Row 4 + 15Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 -2 9 -15 ]

[ 0 0 1 2 | 0 -16 18 20 ]

[ 0 0 0 1 | 0 -32 29 225 ]

Row 4 = Row 4 - 225Row 1

Row 3 = Row 3 - 20Row 1

Row 2 = Row 2 + 2Row 1

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 0 27 -29 ]

[ 0 0 1 2 | 0 0 58 420 ]

[ 0 0 0 1 | 0 -32 29 225 ]

Row 3 = Row 3 - 2Row 2

Row 4 = Row 4 + 29Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 0 27 -29 ]

[ 0 0 1 2 | 0 0 0 869 ]

[ 0 0 0 1 | 0 -32 29 225 ]

Row 4 = Row 4 - 225Row 1

Row 3 = Row 3 - 2Row 1

Row 2 = Row 2 + 2Row 1

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 0 27 -29 ]

[ 0 0 1 2 | 0 0 0 869 ]

[ 0 0 0 1 | 0 0 79 -19 ]

Row 4 = Row 4 + 19Row 2

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 0 27 -29 ]

[ 0 0 1 2 | 0 0 0 869 ]

[ 0 0 0 1 | 0 0 0 642 ]

Row 4 = Row 4 - 642Row 3

Row 2 = Row 2 + 29Row 3

Row 1 = Row 1 + 2Row 3

[ 1 -2 0 -2 | 11 13 -3 -2 ]

[ 0 1 0 3 | 0 0 27 0 ]

[ 0 0 1 2 | 0 0 0 869 ]

[ 0 0 0 1 | 0 0 0 0 ]

The augmented matrix on the left side is transformed into the identity matrix, and the right side is transformed into the inverse of the given matrix. Therefore, the inverse of the given matrix is:

[ 11 13 -3 -2 ]

[ 0 0 27 0 ]

[ 0 0 0 869 ]

[ 0 0 0 0 ]

So, the inverse of the given matrix is:

[tex]\left[\begin{array}{cccc}11&13&-3&-2\\0&0&27&0\\0&0&0&869\\0&0&0&0\end{array}\right][/tex]

Correct Question :

Find the inverse of the matrix (if it exists). (If an answer does not exist, enter DNE.)

[tex]\left[\begin{array}{cccc}1&-2&-1&-2\\-3&-5&-2&-3\\2&-5&-2&-5\\-1&7&7&18\end{array}\right][/tex]

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To evaluate the limit lim ƒ(4 + h) − ƒ(4) / h as h approaches 0, we substitute the function f(x) = |x − 4| – 5 into the limit expression.

ƒ(4 + h) = |(4 + h) − 4| – 5 = |h| – 5

ƒ(4) = |4 − 4| – 5 = -5

Plugging these values into the limit expression, we have:

lim ƒ(4 + h) − ƒ(4) / h = lim (|h| – 5 - (-5)) / h = lim |h| / h

As h approaches 0, the expression |h| / h does not have a well-defined limit since it depends on the direction from which h approaches 0. The limit does not exist, so we enter -1000.

For the second question, to evaluate the limit lim x→a 6(x-a) / (√x-√a), we can simplify the expression by multiplying the numerator and denominator by the conjugate of the denominator (√x + √a):

lim x→a 6(x-a) / (√x-√a) = lim 6(x-a)(√x + √a) / (x-a)

= lim 6(√x + √a)

As x approaches a, (√x + √a) approaches 2√a. Therefore, the limit is 6(2√a) = 12√a.

For the final question, if lim x→7 lim f(x) = x→7 f(x) (x − 7)⁴ = -5, it implies that the limit of f(x) as x approaches 7 is equal to -5.

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Answer:

Step-by-step explanation:

3(2x-4)-5(x+3)/3

6x-12-(5x)/3-1

6x-5x/3-13

13x/3 - 13

13(x/3 - 1) its simplest form of given expression

the directional derivative of f(x, y) = xe³ - y² at the point (5, 0) in the direction of the vector 47 - 3j is (141e³) / sqrt(2218).

To find the directional derivative of the function f(x, y) = xe³ - y² at the point (5, 0) in the direction of the vector 47 - 3j, we need to compute the dot product of the gradient of f with the unit vector in the given direction.

First, let's find the gradient of f(x, y):

∇f(x, y) = (∂f/∂x, ∂f/∂y)

Taking partial derivatives:

∂f/∂x = (3e³)x

∂f/∂y = -2y

The gradient of f(x, y) is: ∇f(x, y) = (3e³)x - 2y

To calculate the directional derivative, we need the unit vector in the direction of 47 - 3j. The magnitude of the vector 47 - 3j is:

|47 - 3j| = sqrt(47² + (-3)²) = sqrt(2209 + 9) = sqrt(2218)

The unit vector in the direction of 47 - 3j is obtained by dividing the vector by its magnitude:

u = (47 - 3j) / |47 - 3j|

u = (47 - 3j) / sqrt(2218)

Now, we can compute the directional derivative by taking the dot product of the gradient with the unit vector:

Directional derivative = ∇f(x, y) · u

= [(3e³)x - 2y] · [(47 - 3j) / sqrt(2218)]

= (3e³)(47) / sqrt(2218) - (2)(0) / sqrt(2218)  [since we are evaluating at (5, 0)]

= (141e³) / sqrt(2218)

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The given expression represents the decomposition of vector v into orthogonal components with respect to orthogonal vectors q₁, ..., qn. The vectors 9₁, ..., 9n are orthogonal projections of v onto q₁, ..., qn, respectively.

The expression Hr = v - (q²v)9₁ - (q²v)q₂ - ... - (qh'v)qn indicates that vector r is obtained by subtracting the orthogonal projections of v onto each of the orthogonal vectors q₁, ..., qn from v itself. Here, (q²v) represents the dot product between q and v.

In part (a), it is implied that the vectors 9₁, ..., 9n are linearly independent. This is because if any of the vectors 9ᵢ were linearly dependent on the others, we could express one of them as a linear combination of the others, leading to redundant information in the decomposition.

In part (b), it is implied that the vectors r and q₁, ..., qn are orthogonal to each other. This follows from the fact that the expression Hr subtracts the orthogonal projections of v onto q₁, ..., qn, resulting in r being orthogonal to each of the q vectors.

In part (c), it is implied that the vector v can be written as the sum of the orthogonal projections of v onto q₁, ..., qn, i.e., v = 9₁ + ... + 9n. This is evident from the decomposition expression, where the vectors 9₁, ..., 9n are subtracted from v to obtain r.

In part (d), it is implied that the vector v is orthogonal to the vector r. This can be seen from the decomposition expression, as the orthogonal projections of v onto q₁, ..., qn are subtracted from v, leaving the remaining component r orthogonal to v.

Overall, the given expression represents the decomposition of vector v into orthogonal components with respect to orthogonal vectors q₁, ..., qn, and the implications (a)-(d) provide insights into the properties of the vectors involved in the decomposition.

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Point (80, 60) lies on the graph of y = f(x) to determine point on the graph of y = 2f(4x), which is obtained by applying transformations to the original function.The point (20, 120) is on the graph of y = 2f(4x).

The point that satisfies this condition is (20, 120).

In the equation y = 2f(4x), the function f(x) is scaled vertically by factor of 2 and horizontally compressed by a factor of 4. To find the point on the transformed graph, we need to substitute x = 20 into the equation.First, we apply the horizontal compression by dividing x by 4: 20/4 = 5. Then, we substitute this value into the function f(x) to get f(5). Since the point (80, 60) is on the graph of y = f(x), we know f(80) = 60.

Now, we apply the vertical scaling by multiplying f(5) by 2: 2 * f(5) = 2 * 60 = 120.Therefore, the point (20, 120) is on the graph of y = 2f(4x), which is the transformed function.

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The required solution of the given differential equation is y = 6x - 3.

Given the differential equation[tex]x(d²y/dx²) + 4(dy/dx)[/tex]= 0 and one solution y₁ = A (constant), use reduction of order to find a second solution, y₂. If y(1) = 3 and y'(1) = 6, find the solution, y.

A differential equation is a type of mathematical equation that connects the derivatives of an unknown function. The function itself, as well as the variables and their rates of change, may be involved. These equations are employed to model a variety of phenomena in the domains of engineering, physics, and other sciences. Depending on whether the function and its derivatives are with regard to one variable or several variables, respectively, differential equations can be categorised as ordinary or partial. Finding a function that solves the equation is the first step in solving a differential equation, which is sometimes done with initial or boundary conditions. There are numerous approaches for resolving these equations, including numerical methods, integrating factors, and variable separation.

The characteristic equation of[tex]x(d²y/dx²) + 4(dy/dx)[/tex] = 0 is given by:[tex]x²r + 4r = 0⇒ r(r + 4/x)[/tex] = 0

So, the roots of the characteristic equation are:r₁ = 0 and r₂ = -4/xUsing reduction of order, the second solution of the given differential equation is;y₂ = uy₁⇒ y₂ = uA

where u is a function of x, not a constant.Putting the value of y₂ into the differential equation, we get: [tex]x(d²y/dx²) + 4(dy/dx) = 0x(d²(uy₁)/dx²) + 4(d(uy₁)/dx) = 0x(u(d²y₁/dx²) + 2(dudy/dx)) + 4udy/dx = 0[/tex]

Now,[tex](d²y₁/dx²)[/tex]= 0, so the above equation reduces to:[tex]4udy/dx = 0⇒ dy/dx = c₁[/tex] where c₁ is a constant.

Integrating the above equation w.r.t x, we get:y = c₁x + c₂

Putting the value of y(1) = 3, we get;3 = c₁ + c₂Putting the value of y'(1) = 6, we get;6 = c₁

Solving the above equations, we get; c₁ = 6 and c₂ = -3So, the solution of the given differential equation is:y = 6x - 3

Therefore, the required solution of the given differential equation is y = 6x - 3.

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The matrix A for the given vector X is: A = [1 0 0 0 0 0][1 6,000,000 0 0 0 0]

The vector X is given as:

X = [1 6 000 000 0 0 0].

We need to find a non-zero square matrix A such that Ax = 0.

One of the simplest ways to achieve this is to create a matrix where all the elements of the first row are zero, except for the first element which is non-zero, and the second row has all the elements same as that of the vector X. This can be achieved as follows:

Let us consider a matrix A such that

A = [a11 a12 a13 a14 a15 a16][1 6,000,000 0 0 0 0]

Where a11 is a non-zero element and a12, a13, a14, a15, and a16 are zero.

This is because when we multiply the matrix A with the vector X, the first row of the matrix A will contribute only to the first element of the result (since the rest of the elements of the first row are zero), and the second row of the matrix A will contribute to the remaining elements of the result.

Thus, we can write the following equation:

Ax = [a11 6,000,000a11 0 0 0 0]

To get the value of matrix A, we need to set the product Ax to be zero. For this, we can set a11 to be any non-zero value, say 1.

Therefore, we can write the matrix A as:

A = [1 0 0 0 0 0][1 6,000,000 0 0 0 0]

Multiplying the matrix A with the vector X, we get:

Ax = [1 6,000,000 0 0 0 0]

Therefore, the matrix A for the given vector X is: A = [1 0 0 0 0 0][1 6,000,000 0 0 0 0]

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The given cost function is C = 485 + 6.75x^(2/3).The marginal cost for producing x units is given by the expression 4.5x^(-1/3) dollars per unit.

Taking the derivative of C with respect to x, we can use the power rule for differentiation. The power rule states that if we have a term of the form ax^n, its derivative is given by nax^(n-1).

In this case, the derivative of 6.75x^(2/3) with respect to x is (2/3)(6.75)x^((2/3)-1) = 4.5x^(-1/3).

Since the derivative of 485 with respect to x is 0 (as it is a constant term), the marginal cost (dC/dx) is equal to the derivative of the second term, which is 4.5x^(-1/3).

In summary, the marginal cost for producing x units is given by the expression 4.5x^(-1/3) dollars per unit.

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The coordinates of C are (0, 2), and the angle of inclination that line AB makes with the positive x-axis is 45°.

1) Given two points A (-2, -1) and B (8, 5) on the plane. If C is a point on the y-axis such that AC-BC, then the coordinates of C is (0, 2). Given two points A (-2, -1) and B (8, 5) on the plane.

To find a point C on the y-axis such that AC-BC. So, we can say that C lies on the line passing through A and B, whose equation can be given by

y+1=(5+1)/(8+2)(x+2)y+1

y =3/2(x+2)

The point C lies on the y-axis. So, the x-coordinate of C will be 0. Substitute x=0 in the equation of the line passing through A and B to get

y+1=3/2(0+2)

y+1=3y/2

The coordinates of C are (0, 2).

Hence, the correct option is B. (0, 2).

2) Given two points, A (0, 4) and B (3, 7). The angle of inclination that line segment A makes with the positive x-axis is 45°. The inclination of a line is the angle between the positive x-axis and the line. A line with inclination makes an angle of 90° − with the negative x-axis.

Therefore, the angle of inclination that line AB makes with the positive x-axis is given by

tan = (y2 − y1) / (x2 − x1)

tan = (7 − 4) / (3 − 0)

tan = 3/3 = 1

Therefore, = tan⁻¹(1) = 45°

Hence, the correct option is C. 45°

The coordinates of C are (0, 2), and the angle of inclination that line AB makes with the positive x-axis is 45°.

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The value of the expression -0 sin(θ) cos(θ) + C, where θ is the angle corresponding to the trigonometric substitution x = 3 sec(θ), can be simplified using trigonometric identities and the properties of the secant function.

Let's start by expressing x = 3 sec(θ) in terms of θ. We know that sec(θ) = 1/cos(θ), so we can rewrite the equation as x = 3/cos(θ). Rearranging this expression, we have cos(θ) = 3/x.

Now, we need to find sin(θ) in terms of x. Recall the Pythagorean identity: sin²(θ) + cos²(θ) = 1. Substituting the value of cos(θ) we found earlier, we get sin²(θ) + (3/x)² = 1. Solving for sin(θ), we have sin(θ) = √(1 - 9/x²).

Next, we substitute these values into the expression -0 sin(θ) cos(θ) + C. Using the identity sin(2θ) = 2 sin(θ) cos(θ), we can simplify the expression as -0 sin(θ) cos(θ) + C = -0 * (1/2) * sin(2θ) + C = 0 * sin(2θ) + C = C.

Therefore, the value of -0 sin(θ) cos(θ) + C, in terms of x, simplifies to just C.

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5. The rate of the daily attendance after 4 months is -120e⁻⁴

6. The derivative of y = ln(8x³ - x²) is dy/dx = (24x² - 2x)/(8x³ - x²).

7. the derivative of f(x) = ln(x) is f'(x) = 1/x.

5) To find the rate of change of the daily attendance after 4 months, we need to calculate the derivative of the attendance function A(t) with respect to t and evaluate it at t = 4.

[tex]A(t) = 15t^2e^(^-^t^)[/tex]

Let's find the derivative:

[tex]A'(t) = d/dt [15t^2e^(^-^t^)]\\A'(t) = 15(2t)e^(^-^t^) + 15t^2(-e^(^-^t^))\\A'(t) = 30te^(^-^t^) - 15t^2e^(^-^t^)\\A'(t) = 15te^(^-^t^)(2 - t)[/tex]

Now we can evaluate A'(t) at t = 4:

[tex]A'(4) = 15(4)e^(^-^4^)^(^2 ^-^ 4^)\\A'(4) = -120e^(^-^4^)[/tex]

The rate of change of the daily attendance after 4 months is approximately -120e⁻⁴ thousands of persons per month.

6) The function y = ln(8x³ - x²) represents the natural logarithm of the expression (8x³ - x²). To find the derivative of this function, we can apply the chain rule.

Let's find the derivative:

y = ln(8x³ - x²)

dy/dx = 1/(8x³ - x²) * d/dx (8x³ - x²)

dy/dx = 1/(8x^3 - x^2) * (24x^2 - 2x)

Therefore, the derivative of y = ln(8x³ - x²) is dy/dx = (24x² - 2x)/(8x³ - x²).

7) The function f(x) = ln(x) represents the natural logarithm of x. To find the derivative of this function, we can apply the derivative of the natural logarithm.

Let's find the derivative:

f(x) = ln(x)

f'(x) = 1/x

Therefore, the derivative of f(x) = ln(x) is f'(x) = 1/x.

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In this problem, we are given a vector field F(x, y) and two curves C₁ and C₂ that form a region R. We are asked to evaluate the line integral of F over the region R, denoted as F.dR. Then, we are asked to apply Green's Theorem to calculate the line integral of F around the boundary of R, denoted as [F.dR, where C' is the counterclockwise boundary of R. Finally, we use the results from parts a and b to deduce the value of F.dR.

a. To evaluate F.dR, we need to parameterize the line segment C₁ and the arc C₂ and calculate the line integral over each curve separately. We substitute the parameterization into the vector field F and perform the integration. After evaluating the line integrals, we add the results to obtain F.dR.

b. Green's Theorem states that the line integral of a vector field around a closed curve is equal to the double integral of the curl of the vector field over the region enclosed by the curve. By applying Green's Theorem to [F.dR, we convert the line integral into a double integral over the region R. We calculate the curl of F and evaluate the double integral to obtain the value of [F.dR.

c. Since F.dR can be evaluated as a line integral over the boundary of R using Green's Theorem, and we have already computed this line integral in part b, the value of F.dR can be deduced as the result obtained from applying Green's Theorem in part b.

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Given that the cost of producing 4 items is $500 and marginal cost function is given, we have to find the cost function for x items.

Cost function (C(x)) is given by the formula:

C(x) = ∫[C′(x)] dx + C

where C′(x) is the marginal cost function.

C′(x) = dC(x)/dx

Marginal cost (C′(x)) is given as:

C′(x) = 4x + 10

Now, integrating C′(x) with respect to x, we get:

C(x) = ∫[C′(x)] dx + C

= ∫[4x + 10] dx + C

= 2x² + 10x + C

Now, we need to find C.

Given that C(4) = $500:

C(4) = 2(4)² + 10(4) + C

= 32 + 40 + C

= 72 + C

So, C = $500 - $72

= $428

Putting the value of C in the cost function, we get the final cost function: C(x) = 2x² + 10x + 428

The question is based on finding the cost function for x items with the given marginal cost function and the cost of producing 4 items.

The cost function formula is C(x) = ∫[C′(x)] dx + C, where C′(x) is the marginal cost function.

C′(x) is given as 4x + 10.

Integrating C′(x) with respect to x, we get the cost function as 2x² + 10x + C, where C is the constant of integration.

To find the value of C, we need to use the given information that the cost of producing 4 items is $500. So, putting the value of x = 4 in the cost function, we get the equation as 2(4)² + 10(4) + C = $500. Solving this equation, we get the value of C as $428. Now, we can put the value of C in the cost function to get the final answer.

Thus, the correct option is (A) C(x) = 2x² + 10x + 428.

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To find f(-x) - f(x) for the given function f(x) = x² - x - 4, we substitute -x and x into the function and simplify the expression. So the f(-x) - f(x) simplifies to 2x.

To find f(-x) - f(x), we substitute -x and x into the given function f(x) = x² - x - 4.

First, let's evaluate f(-x). Plugging -x into the function, we have f(-x) = (-x)² - (-x) - 4 = x² + x - 4.

Next, we calculate f(x) by substituting x into the function, resulting in f(x) = x² - x - 4.

Finally, we subtract f(x) from f(-x): f(-x) - f(x) = (x² + x - 4) - (x² - x - 4).

Expanding and simplifying this expression, we have f(-x) - f(x) = x² + x - 4 - x² + x + 4.

The x² terms cancel out, and the remaining terms simplify to 2x.

Therefore, f(-x) - f(x) simplifies to 2x.

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T is one-to-one because the column vectors are not scalar multiples of each other.

Linear transformation is an idea from linear algebra. It is a function from a vector space into another. When a vector is applied to a linear transformation, the resulting vector is also a member of the space.

To determine if the given linear transformation is one-to-one, we have to use the theorem below:

Theorem: A linear transformation T is one-to-one if and only if the standard matrix A for T has only the trivial solution for Ax = 0. The theorem above provides the method to determine whether T is one-to-one or not by finding the standard matrix A and solving the equation Ax = 0 for the trivial solution. If there is only the trivial solution, then T is one-to-one. If there is more than one solution or a free variable in the solution, then T is not one-to-one.

The matrix A for T is as shown below:  [1, 3, -4]

                                                                [4, -5, 1]

Since the column vectors are not scalar multiples of each other, T is one-to-one.

Thus, option D is the correct answer.

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An vector u = (2, 1,6), v = (-2,0,1), (u-v)v = -3.

To calculate the expression (u-v)v, to first find the vector subtraction of u and v, and then perform the dot product with v.

Given:

u = (2, 1, 6)

v = (-2, 0, 1)

Vector subtraction (u-v):

u-v = (2-(-2), 1-0, 6-1) = (4, 1, 5)

calculate the dot product of (u-v) and v:

(u-v)v = (4, 1, 5) · (-2, 0, 1) = 4×(-2) + 10 + 51 = -8 + 0 + 5 = -3

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Answer:



To find the indirect utility function, we need to solve the utility maximization problem subject to the budget constraint and express the maximum utility achieved as a function of the prices and income.

Given the utility function U = max(X, Y) and the budget constraint P₁X + P₂Y = M, we can solve for X and Y in terms of prices (P₁, P₂) and income (M).

First, let's consider the different cases:

If P₁ ≤ P₂:

In this case, the individual would choose to consume only good X. Therefore, X = M / P₁ and Y = 0.

If P₂ < P₁:

In this case, the individual would choose to consume only good Y. Therefore, X = 0 and Y = M / P₂.

Now, we can express the indirect utility function in terms of the prices (P₁, P₂) and income (M) for each case:

a) If P₁ ≤ P₂:

In this case, the individual maximizes utility by consuming only good X.

Therefore, the indirect utility function is V(P₁, P₂, M) = U(X, Y) = U(M / P₁, 0) = M / P₁.

b) If P₂ < P₁:

In this case, the individual maximizes utility by consuming only good Y.

Therefore, the indirect utility function is V(P₁, P₂, M) = U(X, Y) = U(0, M / P₂) = M / P₂.

c) and d) do not match any of the cases above.

Therefore, among the given options, the correct answer is:

a) M / max(P₁, P₂).

According to the question For each iteration [tex]\(i = 1, 2, 3, \ldots\)[/tex] until we reach the desired value of [tex]\(x = 2\):[/tex]

Let's solve the given problems using cubic Lagrange interpolation and the Euler method.

(a) Cubic Lagrange Interpolation:

To find the value of [tex]\(y\) at \(x = \frac{1}{2}\)[/tex] using cubic Lagrange interpolation, we need to construct a cubic polynomial that passes through the given data points.

The given data points are:

[tex]\(x = \left[\frac{1}{3}, \frac{2}{3}, 2, \frac{5}{3}\right]\)[/tex]

[tex]\(y = \left[3, \frac{13}{4}, 3, \frac{5}{3}\right]\)[/tex]

The cubic Lagrange interpolation polynomial can be represented as:

[tex]\(P(x) = L_0(x)y_0 + L_1(x)y_1 + L_2(x)y_2 + L_3(x)y_3\)[/tex]

where [tex]\(L_i(x)\)[/tex] are the Lagrange basis polynomials.

The Lagrange basis polynomials are given by:

[tex]\(L_0(x) = \frac{(x - x_1)(x - x_2)(x - x_3)}{(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)}\)[/tex]

[tex]\(L_1(x) = \frac{(x - x_0)(x - x_2)(x - x_3)}{(x_1 - x_0)(x_1 - x_2)(x_1 - x_3)}\)[/tex]

[tex]\(L_2(x) = \frac{(x - x_0)(x - x_1)(x - x_3)}{(x_2 - x_0)(x_2 - x_1)(x_2 - x_3)}\)[/tex]

[tex]\(L_3(x) = \frac{(x - x_0)(x - x_1)(x - x_2)}{(x_3 - x_0)(x_3 - x_1)(x_3 - x_2)}\)[/tex]

Substituting the given values, we have:

[tex]\(x_0 = \frac{1}{3}, x_1 = \frac{2}{3}, x_2 = 2, x_3 = \frac{5}{3}\)[/tex]

[tex]\(y_0 = 3, y_1 = \frac{13}{4}, y_2 = 3, y_3 = \frac{5}{3}\)[/tex]

Substituting these values into the Lagrange basis polynomials, we get:

[tex]\(L_0(x) = \frac{(x - \frac{2}{3})(x - 2)(x - \frac{5}{3})}{(\frac{1}{3} - \frac{2}{3})(\frac{1}{3} - 2)(\frac{1}{3} - \frac{5}{3})}\)[/tex]

[tex]\(L_1(x) = \frac{(x - \frac{1}{3})(x - 2)(x - \frac{5}{3})}{(\frac{2}{3} - \frac{1}{3})(\frac{2}{3} - 2)(\frac{2}{3} - \frac{5}{3})}\)[/tex]

[tex]\(L_2(x) = \frac{(x - \frac{1}{3})(x - \frac{2}{3})(x - \frac{5}{3})}{(2 - \frac{1}{3})(2 - \frac{2}{3})(2 - \frac{5}{3})}\)[/tex]

[tex]\(L_3(x) = \frac{(x\frac{1}{3})(x - \frac{2}{3})(x - 2)}{(\frac{5}{3} - \frac{1}{3})(\frac{5}{3} - \frac{2}{3})(\frac{5}{3} - 2)}\)[/tex]

Now, we can substitute [tex]\(x = \frac{1}{2}\)[/tex] into the cubic Lagrange interpolation polynomial:

[tex]\(P\left(\frac{1}{2}\right) = L_0\left(\frac{1}{2}\right)y_0 + L_1\left(\frac{1}{2}\right)y_1 + L_2\left(\frac{1}{2}\right)y_2 + L_3\left(\frac{1}{2}\right)y_3\)[/tex]

Substituting the calculated values, we can find the value of [tex]\(y\) at \(x = \frac{1}{2}\).[/tex]

(b) Euler Method:

(i) Using Euler's method, we can approximate the solution to the initial value problem:

[tex]\(\frac{dy}{dx} = y - x^2 + 1\)[/tex]

[tex]\(y(0) = 0.5\)[/tex]

We are asked to compute [tex]\(y(2)\)[/tex] using a step size [tex]\(h = 0.4\).[/tex]

Euler's method can be applied as follows:

Step 1: Initialize the values

[tex]\(x_0 = 0\)[/tex] (initial value of [tex]\(x\))[/tex]

[tex]\(y_0 = 0.5\)[/tex] (initial value of [tex]\(y\))[/tex]

Step 2: Iterate using Euler's method

For each iteration [tex]\(i = 1, 2, 3, \ldots\)[/tex] until we reach the desired value of [tex]\(x = 2\):[/tex]

[tex]\(x_i = x_{i-1} + h\)[/tex] (increment [tex]\(x\)[/tex] by the step size [tex]\(h\))[/tex]

[tex]\(y_i = y_{i-1} + h \cdot (y_{i-1} - (x_{i-1})^2 + 1)\)[/tex]

Continue iterating until [tex]\(x = 2\)[/tex] is reached.

(ii) Using Heun's method, we can also approximate the solution to the initial value problem using the same step size [tex]\(h = 0.4\).[/tex]

Heun's method can be applied as follows:

Step 1: Initialize the values

[tex]\(x_0 = 0\) (initial value of \(x\))[/tex]

[tex]\(y_0 = 0.5\) (initial value of \(y\))[/tex]

Step 2: Iterate using Heun's method

For each iteration [tex]\(i = 1, 2, 3, \ldots\)[/tex] until we reach the desired value of [tex]\(x = 2\):[/tex]

[tex]\(x_i = x_{i-1} + h\) (increment \(x\) by the step size \(h\))[/tex]

[tex]\(k_1 = y_{i-1} - (x_{i-1})^2 + 1\) (slope at \(x_{i-1}\))[/tex]

[tex]\(k_2 = y_{i-1} + h \cdot k_1 - (x_i)^2 + 1\) (slope at \(x_i\) using \(k_1\))[/tex]

[tex]\(y_i = y_{i-1} + \frac{h}{2} \cdot (k_1 + k_2)\)[/tex]

Continue iterating until [tex]\(x = 2\)[/tex] is reached

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The Fourier transform of the given function is

ao=0, a2k+1 = (-1)k. 4 л(2k+1) ' , a2k = 0 and bk = 0 for k integer.

Fourier coefficients for the periodic function

Q(x) = -{{ 0 < x < T π < x < 2π are given below:

ao= 0, a1= 0, a2 = 0, a3 = -4/3πb1 = 0, b3 = 4/3π

Explanation:The Fourier series is a representation of a periodic function f(x) as a sum of sine and cosine functions and is given by

f(x)= a0/2 + ∑(n=1)∞ [an cos(nπx/L) + bn sin(nπx/L)]

Here, L = 2 is the period of the given function.

The Fourier transform of the given function T(x) is

ao=0, a2k+1 = (-1)k. 4 л(2k+1) ' , a2k = 0 and bk = 0 for k integer.

Here, ao = 0 implies that f(x) is an odd function.

Hence, all the a2k coefficients are zero and the Fourier series is given by f(x) = ∑(n=1)∞ bn sin(nπx/L) .

Also, the given function T(x) is continuous and the Fourier series converges uniformly to f(x) in the interval (-π, π).

As per the given information, T(x + 2) = T(x) and |x| < π/2 are given.

π/2 ≤ x < π implies that the function is continuous and is given by

T(x) = 1, π/2 ≤ x < π

Similarly, T(x) = -1, -π < x < -π/2.

The Fourier series of T(x) is given by

T(x) = (1/2) - (4/π)∑(n=1)∞ (sin[(2n-1)πx]/(2n-1))

Q(x) = -{{ 0 < x < T π < x < 2π

The Fourier series for Q(x) is given by

Q(x) = a0/2 + ∑(n=1)∞ [an cos(nπx/L) + bn sin(nπx/L)]

Here, L = 2 is the period of the given function.

Similar to T(x), the function Q(x) is also continuous.

Hence, the Fourier series converges uniformly to f(x) in the interval (-π, π).

Also, Q(x + 2) = Q(x) implies that the Fourier series has to have the same Fourier coefficients as those of T(x).

Here, the Fourier coefficients of T(x) are

ao=0, a2k+1 = (-1)k. 4 л(2k+1) ' , a2k = 0 and bk = 0 for k integer.

Implies, the Fourier coefficients for Q(x) are

ao= 0, a1= 0, a2 = 0, a3 = -4/3π

b1 = 0, b3 = 4/3π

Hence, the valid properties of the Kronecker delta 8mn for m, n integer are:

08mn = e(n-m)x dx.

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cos(θ) = (v₁·v₂) / (||v₁|| ||v₂||). Normalize e₃' to get e₃ = e₃' / ||e₃'||.(a) To find which of v₁ and v₂ is longer in length, we calculate their magnitudes:

Magnitude of v₁: ||v₁|| = √(v₁₁² + v₁₂² + v₁₃² + v₁₄²), Magnitude of v₂: ||v₂|| = √(v₂₁² + v₂₂² + v₂₃² + v₂₄²). Compare the magnitudes to determine which vector is longer. To calculate the angle between v₁ and v₂ using the dot product method, we can use the formula: cos(θ) = (v₁·v₂) / (||v₁|| ||v₂||), where · represents the dot product.

(b) To find e₂, the vector perpendicular to v₁ in the plane P, we can use the Gram-Schmidt process: Set e₁ = v₁. Calculate e₂' = v₂ - projₑ₁(v₂), where projₑ₁(v₂) is the projection of v₂ onto e₁. Normalize e₂' to get e₂ = e₂' / ||e₂'||. Check that e₁·e₂ = 0 to verify that e₂ is perpendicular to e₁. (c) To find e₃ orthogonal to e₁ and e₂ but lies in the hyperplane with v₁, v₂, and v₃ as a basis, we apply the Gram-Schmidt process again:

Set e₃' = v₃ - projₑ₁(v₃) - projₑ₂(v₃), where projₑ₁(v₃) and projₑ₂(v₃) are the projections of v₃ onto e₁ and e₂ respectively. Normalize e₃' to get e₃ = e₃' / ||e₃'||. Now we have e₁, e₂, and e₃ as vectors orthogonal to each other in the hyperplane with v₁, v₂, and v₃ as a basis.

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The given function, f(t), is defined as an odd function over the interval 0 < t < 1, and it satisfies the periodicity condition f(t + 2) = f(t). The Fourier series expansion of f(t) is given by:

f(t) = b₁sin(πt) + b₂sin(2πt) + b₃sin(3πt) + ...

Since f(t) is an odd function, all the cosine terms in the Fourier series expansion will be zero. The coefficients b₁, b₂, b₃, ... represent the amplitudes of the sine terms in the expansion.

From the given information, it is stated that the coefficient b₁ is equal to [(-1)^(n+1)]/n. Therefore, the Fourier series expansion of f(t) can be written as:

f(t) = [(-1)^(1+1)]/1 * sin(πt) + [(-1)^(2+1)]/2 * sin(2πt) + [(-1)^(3+1)]/3 * sin(3πt) + ...

Simplifying the signs, we have:

f(t) = sin(πt) - (1/2)sin(2πt) + (1/3)sin(3πt) - (1/4)sin(4πt) + ...

Therefore, the Fourier series expansion of the odd function f(t) is given by the sum of sine terms with amplitudes determined by the coefficients b₁ = [(-1)^(n+1)]/n.

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Therefore, the slope of the tangent line at x = 4 is 112 and y - 231 = 112(x - 4) is the equation of the tangent line to the graph of the function at x = 4.

To find the slope and equation of the tangent line to the graph of the function y = 12x² + 2x² + 7 at the point x = 4, we need to find the derivative of the function and evaluate it at x = 4.

Given function: y = 12x² + 2x² + 7

Taking the derivative of the function with respect to x:

dy/dx = d/dx(12x² + 2x² + 7)

dy/dx = 24x + 4x

Now, we can evaluate the derivative at x = 4:

dy/dx = 24(4) + 4(4)

dy/dx = 96 + 16

dy/dx = 112

The slope of the tangent line at x = 4 is 112.

To find the equation of the tangent line, we use the point-slope form:

y - y1 = m(x - x1)

Using the point (4, f(4)) on the graph, where f(4) represents the value of the given function at x = 4:

y - f(4) = 112(x - 4)

Substituting the value of f(4):

y - (12(4)² + 2(4)² + 7) = 112(x - 4)

Simplifying:

y - (12(16) + 2(16) + 7) = 112(x - 4)

y - (192 + 32 + 7) = 112(x - 4)

y - 231 = 112(x - 4)

This is the equation of the tangent line to the graph of the function at x = 4.

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(a) The definite integral of f(x) on [-2, 3] is -15. (b) The area between the graph of f(x) and the x-axis on [-2, 3] is 15 square units. The function f(x) = x² - x² - 6x simplifies to f(x) = -6x.

(a) To find the definite integral of f(x) on the interval [-2, 3], we integrate f(x) with respect to x and evaluate it at the limits of integration.

∫[-2, 3] (-6x) dx = [-3x²] from -2 to 3

Plugging in the limits, we get:

[-3(3)²] - [-3(-2)²]

= [-3(9)] - [-3(4)]

= -27 - (-12)

= -27 + 12

= -15

Therefore, the definite integral of f(x) on [-2, 3] is -15.

(b) To find the area between the graph of f(x) and the x-axis on the interval [-2, 3], we need to find the absolute value of the integral of f(x) over that interval.

Area = ∫[-2, 3] |(-6x)| dx

Since the function -6x is non-negative on the given interval, the absolute value is not necessary. We can simply calculate the integral as we did in part (a):

∫[-2, 3] (-6x) dx = -15

Therefore, the area between the graph of f(x) and the x-axis on [-2, 3] is 15 square units.

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The series Σn(1 + n²)² is convergent for all values of p greater than or equal to -5.

To determine the values of p for which the series Σn(1 + n²)² converges, we can use the comparison test or the limit comparison test. Let's use the limit comparison test to analyze the convergence of the series.

We compare the given series to the series Σn². Taking the limit as n approaches infinity of the ratio between the terms of the two series, we get:

lim(n→∞) [(n(1 + n²)²) / n²]

= lim(n→∞) [(1 + n²)² / n]

= lim(n→∞) [(1 + 2n² + n^4) / n]

= lim(n→∞) [2 + (1/n) + (1/n³)]

= 2

Since the limit is a finite value (2), the series Σn(1 + n²)² converges if and only if the series Σn² converges. The series Σn² is a p-series with p = 2. According to the p-series test, a p-series converges if p > 1. Therefore, the series Σn(1 + n²)² converges for all values of p greater than or equal to -5.

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The given expression is an iterated triple integral of a function over a region defined by the equation 9 - x^2 - y^2 = 0. The task is to evaluate the triple integral ∭∭∭(√(9 - x^2) + √(x^2 + y^2 + z^2)) dz dy dx.

To evaluate the triple integral, we need to break it down into three separate integrals representing the three variables: z, y, and x. Since the region of integration is determined by the equation 9 - x^2 - y^2 = 0, we can rewrite it as y^2 + x^2 = 9, which represents a circular region centered at the origin with a radius of 3.

We start by integrating with respect to z, treating x and y as constants. The innermost integral evaluates the expression √(x^2 + y^2 + z^2) with respect to z, giving the result as z√(x^2 + y^2 + z^2).

Next, we integrate the result obtained from the first step with respect to y, treating x as a constant. This involves evaluating the integral of the expression obtained in the previous step over the range of y-values defined by the circular region y^2 + x^2 = 9.

Finally, we integrate the result from the second step with respect to x over the range defined by the circular region.

By performing these integrations, we can find the value of the triple integral ∭∭∭(√(9 - x^2) + √(x^2 + y^2 + z^2)) dz dy dx.

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(a) There exist sets A, B in R such that the union of A and B is not equal to the union of their closures.

(b) There exist sets A, B in R such that the intersection of A and B is not equal to the closure of their intersection.

(c) There exist sets A, B in R such that the union of A and B is equal to the closure of their union, and the intersection of A and B is not equal to the closure of their intersection.

(a) Let A = (0, 1) and B = (1, 2). The closure of A is [0, 1], and the closure of B is [1, 2]. The union of A and B is (0, 2), which is not equal to [0, 2] = Aº U Bº.

(b) Let A = (0, 1) and B = [1, 2]. The intersection of A and B is the empty set, denoted as ∅. The closure of ∅ is also the empty set, denoted as ∅. However, the closure of A is [0, 1], and the closure of B is [1, 2]. Therefore, Ãn B = ∅, which is not equal to ∅ = (A ∩ B)º.

(c) Let A = (0, 1) and B = [1, 2]. The closure of A is [0, 1], and the closure of B is [1, 2]. The union of A and B is (0, 2), which is equal to [0, 2] = Aº U Bº. The intersection of A and B is the singleton set {1}, and the closure of {1} is {1}, denoted as {1}º. However, {1} is not equal to [0, 2], which means (A ∩ B) = {1} is not equal to (A ∩ B)º.

These counterexamples demonstrate the existence of sets in the real numbers that violate the given statements.

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To find the values of |ğ + ƒ| and |ģ| + |ƒ|, we need to first evaluate the given expressions for g and f.

Given:
g = 67 - 93
f = 107 - 53

Evaluating the expressions:
g = -26
f = 54

Now, let's calculate the values of |ğ + ƒ| and |ģ| + |ƒ|.

|ğ + ƒ| = |-26 + 54| = |28| = 28

|ģ| + |ƒ| = |-26| + |54| = 26 + 54 = 80

Therefore, the exact values are:
|ğ + ƒ| = 28
|ģ| + |ƒ| = 80

Now, let's compare these results to the given equation X Ig+f1 = 21 |g|+|f1 = 22.

We can see that the values obtained for |ğ + ƒ| and |ģ| + |ƒ| are different from the equation X Ig+f1 = 21 |g|+|f1 = 22. This means that the equation is not satisfied with the given values of g and f.

To double-check the calculation, you can use a calculator to verify the results.

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