If y(x) is the solution to the initial value problem y' - y = x² + x, y(1) = 2. then the value y(2) is equal to: 06 02 0-1

Damped Harmonic Oscillator and its Transfer Function A damped harmonic oscillator is a physical system that, when disturbed from its equilibrium position, oscillates about that position and eventually comes to rest. Damped harmonic oscillator is characterized by an equation of the form y" + 2cy' + ky = f(t)

Where f(t) is the driving force, c is the damping coefficient, k is the spring constant, and y(t) is the displacement of the oscillator from its equilibrium position.

Using the above equation, we can derive the transfer function of the damped harmonic oscillator which is given by

H(s) = Y(s)/F(s)

= 1/(ms^2 + cs + k)

Where F(s) is the Laplace transform of f(t) and Y(s) is the Laplace transform of y(t). In the case of the given problem,

m = 1kg, c

= 2 kg/sec, and

k = 2 kg/sec².

Thus, the transfer function is

H ( s )  = 1/(s^2 + 2s + 2)

To find the impulse response, we take the inverse Laplace transform of the transfer function which is given by

h(t) = sin(t) - e^(-t)cos(t)

The given differential equation is

y"+ 2y' + 2y = f(t)

where y(0) = 0 and

y'(0) = 0.

Using the convolution integral, we can write the solution as y(t) = h(t)*f(t)

= ∫[0 to t]h(t-τ)f(τ)dτPlugging in the impulse response,

h(t) = sin(t) - e^(-t)cos(t) and taking f(t) = δ(t),

we get y(t)

= ∫[0 to t](sin(t-τ) - e^(-t+τ)cos(t-τ))δ(τ)dτ

= sin(t) - e^(-t)cos(t)

Thus, the solution to the given differential equation is y(t) = sin(t) - e^(-t)cos(t).

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When B is a 5 x 4 matrix with linearly independent columns, the null space Nul B will contain only the zero vector.

The null space of a matrix consists of all vectors that, when multiplied by the matrix, result in the zero vector. When a matrix has linearly independent columns, it means that no linear combination of the columns can result in the zero vector, except for the trivial combination where all coefficients are zero.

In the case of a 5 x 4 matrix B with linearly independent columns, it implies that there are no non-zero vectors in the null space Nul B. This is because there are no vectors that, when multiplied by B, will produce the zero vector.

Moving on to the second part of the question, for a 3 x 5 matrix A with three pivot columns, the column space Col A will not equal the entire 9¹ (R⁹) since it is not possible to span the entire 9¹ with only three columns. However, it is not possible to determine whether the null space Nul A is equal to 91² (R⁹²) without additional information. The null space depends on the specific values and structure of matrix A, and it can vary.

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To determine the rate at which sales are changing, we need to find the derivative of the sales function with respect to time.

Given the demand equation 8p³ + x² = 65,600, we can differentiate it implicitly to find the derivative of x with respect to t. Then, we substitute the given values x = 40, p = 20, and the rate of change of p = -0.10 into the derivative equation to calculate the rate at which sales are changing.

The demand equation is given as 8p³ + x² = 65,600, where p represents the price and x represents the weekly sales.

Differentiating the equation implicitly with respect to t (time), we have:

24p² * dp/dt + 2x * dx/dt = 0

To find the rate at which sales are changing, we need to determine dx/dt when x = 40, p = 20, and dp/dt = -0.10.

Substituting the given values into the derivative equation, we get:

24(20)² * (-0.10) + 2(40) * dx/dt = 0

Simplifying the equation, we have:

-9600 + 80 * dx/dt = 0

Solving for dx/dt, we get:

80 * dx/dt = 9600

dx/dt = 9600 / 80

dx/dt = 120

Therefore, the rate at which sales are changing is 120 thousand units per week.

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The solution of the given system of linear equations is

x₁ = 3/2, x₂ = –2, x₃ = –1/2.

Given system of linear equations are 2x1-2x2+6x3 = 10x1+2x2-3x3 = 8-2x₁ x3 = -11

To solve the given system of equations, we have to compute A¹ and then use it to find x.To compute A¹, we have to follow these steps:

Step 1: Find the determinant of A:

Now we have matrix A:2 -2 6 1 2 -3 -2 0 1

So, we can find the determinant of A using the formula det(A) = (2×2) [(2×1)×(–3×1) + (6×2)×(1×–2) + (–2×2)×(2×–3)]  

= 4(–8 + (–24) + 24) = 4(–8) = –32

So, det(A) = –32

Step 2: Find the inverse of A:

We can find the inverse of matrix A using the formula A⁻¹ = adj(A) / det(A)

Here adj(A) is the adjoint matrix of A.

Now, we have to find the adjoint matrix of A.

Adjoint of A is given as:adj(A) = (cof(A))T

where cof(A) is the matrix of cofactors of the matrix A and (cof(A))T is the transpose of cof(A).

Now we can find cof(A) as:

cof(A) =   [[(-6) (-12) (-2)] [(6) (2) (2)] [(4) (8) (2)]]

Then transpose of cof(A), cof(A)T, is  

[[(-6) (6) (4)] [(-12) (2) (8)] [(-2) (2) (2)]]

So, adj(A) = cof(A)T

=  [[(-6) (6) (4)] [(-12) (2) (8)] [(-2) (2) (2)]

Now we can find A⁻¹ as:

A⁻¹ = adj(A) / det(A)

A⁻¹ = [[(-6) (6) (4)] [(-12) (2) (8)] [(-2) (2) (2)]] / (-32)

A⁻¹ =   [[(3/16) (-3/8) (-1/16)] [(3/4) (-1/4) (-1/4)] [(1/16) (-1/8) (1/16)]]

So, A¹ = A⁻¹ = [[(3/16) (-3/8) (-1/16)] [(3/4) (-1/4) (-1/4)] [(1/16) (-1/8) (1/16)]]

Then the given system of linear equations can be written as AX = B, where

X = [x₁ x₂ x₃]T and B = [10 8 –11]TAX = B  

⟹   X = A-¹B

Substituting the value of A-¹, we get X as   [[(3/16) (-3/8) (-1/16)] [(3/4) (-1/4) (-1/4)] [(1/16) (-1/8) (1/16)]] .

[10 8 -11]T= [3/2 -2 -1/2]T

So, the solution of the system is:

x₁ = 3/2x₂ = –2x₃ = –1/2

Therefore, the solution of the given system of linear equations is

x₁ = 3/2, x₂ = –2, x₃ = –1/2.

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The given function is:

y = 2x + 1(x - 2)

The first derivative of y with respect to x is:

dy/dx = 2x + 1

Using the above equation, let's find the value of x when y = 3:3

= 2x + 1(x - 2)3

= 2x + x - 23

= 3x - 2x1

= x

Substituting x = 1 in dy/dx:

dy/dx = 2(1) + 1

= 3

Therefore, the gradient of the tangent to the curve y = 2x + 1(x - 2) at the point where y = 3 is 3.

The normal line to the curve at (1, 3) is perpendicular to the tangent line and passes through the point (1, 3). Let the equation of the normal be y = mx + b.

Substitute the point (1, 3):

3 = m(1) + bb

= 3 - m

So, the equation of the normal is: y = mx + (3 - m)

Substitute the value of the gradient (m = -1/3) that we found earlier:

y = (-1/3)x + (3 + 1/3)y

= (-1/3)x + 10/3

Thus, the equation of the normal to the curve at (1, 3) is

y = (-1/3)x + 10/3.

The parametric equations of a curve are given by

x = -2 cos θ + 2 and

y = 2 sin θ + 3.

To find dy/dx, differentiate both equations with respect to θ:

dx/dθ = 2 sin θdy/dθ

= 2 cos θ

Then, divide dy/dθ by dx/dθ to get dy/dx:

dy/dx = (2 cos θ)/(2 sin θ)dy/dx

= cot θ

When θ = 3,

dy/dx = cot 3

Hence, the value of dy/dx at θ = 3 is cot 3.

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To solve the integral ∫√(36-x²) dx, we can use the trigonometric substitution x = 6sin(theta). This substitution allows us to rewrite the integral in terms of trigonometric functions and simplify it to an integral that can be evaluated easily.

To solve the integral ∫√(36-x²) dx, we use the trigonometric substitution x = 6sin(theta). Taking the derivative of x = 6sin(theta) with respect to theta, we get dx = 6cos(theta) d(theta).

Substituting x = 6sin(theta) and dx = 6cos(theta) d(theta) in the integral, we have:

∫√(36-x²) dx = ∫√(36-(6sin(theta))²) (6cos(theta)) d(theta).

Simplifying the integrand, we have:

∫√(36-36sin²(theta)) (6cos(theta)) d(theta) = ∫√(36cos²(theta)) (6cos(theta)) d(theta).

Using the trigonometric identity cos²(theta) = 1 - sin²(theta), we can simplify further:

∫√(36cos²(theta)) (6cos(theta)) d(theta) = ∫√(36(1 - sin²(theta))) (6cos(theta)) d(theta).

Simplifying the expression inside the square root, we get:

∫√(36(1 - sin²(theta))) (6cos(theta)) d(theta) = ∫√(36cos²(theta)) (6cos(theta)) d(theta) = ∫(6cos(theta))(6cos(theta)) d(theta).

Now, we have a simpler integral to evaluate. Using the trigonometric identity cos²(theta) = (1 + cos(2theta))/2, we can simplify further:

∫(6cos(theta))(6cos(theta)) d(theta) = ∫(6cos(theta))² d(theta) = ∫(36cos²(theta)) d(theta) = ∫(36(1 + cos(2theta))/2) d(theta).

Integrating term by term, we get:

∫(36(1 + cos(2theta))/2) d(theta) = (18theta + 9sin(2theta)) + C.

Finally, we substitute back the value of theta using the original substitution x = 6sin(theta). Therefore, the final result is:

(18theta + 9sin(2theta)) + C = (18sin⁻¹(x/6) + 9sin(2sin⁻¹(x/6))) + C, where C is the constant of integration.

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The differential equation satisfied by v(t) is dv/dt = -g - |v|/40 and the velocity of the shell is given by v(t) = -20ln(1 - t/50) - 20t + C, where C is a constant of integration.

a) To find the differential equation satisfied by v(t), we consider the forces acting on the shell. The force due to gravity is -mg = -2g, and the force due to air resistance is -|v|/20, where v is the velocity of the shell. Using Newton's second law, F = ma, we have -2g - |v|/20 = 2(dv/dt), which simplifies to dv/dt = -g - |v|/40.

b) To find the velocity of the shell, we integrate the differential equation found in part (a). Integrating dv/dt = -g - |v|/40 gives v(t) = -20ln(1 - t/50) - 20t + C, where C is a constant of integration.

c) To find the position x(t) of the shell, we integrate the velocity function v(t) found in part (b). Integrating v(t) = -20ln(1 - t/50) - 20t + C gives [tex]x(t) = -10ln(1 - t/50)^2 - 10t^2 + Ct + D[/tex], where D is a constant of integration.

d) The shell reaches its maximum height when its velocity becomes zero. Setting v(t) = 0 and solving for t, we find t = 50 seconds.

e) To find the maximum height H reached by the shell, we substitute t = 50 into the position equation x(t) and evaluate it. This gives H = 2500ln(3) + 2500.

Therefore, the velocity, position, maximum height, and time at maximum height for the shell are determined using the given differential equation and initial conditions.

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The Laplace transformation method is used to solve the given second-order differential equation, which describes a mass-spring system. The solution involves transforming the differential equation into an algebraic equation in the Laplace domain and then inverting the Laplace transform to obtain the solution in the time domain.

To solve the given differential equation using the Laplace transformation method, we begin by taking the Laplace transform of both sides of the equation. The Laplace transform of the first derivative, y', is denoted as sY(s) - y(0), where Y(s) is the Laplace transform of y(t) and y(0) represents the initial condition. The Laplace transform of the second derivative, y'', is represented as s²Y(s) - sy(0) - y'(0).

Applying the Laplace transform to the given equation, we have (s²Y(s) - sy(0) - y'(0)) + 8(sY(s) - y(0)) + 15Y(s) = 1. Substituting the initial conditions y(0) = 0 and y'(0) = 0, the equation simplifies to (s² + 8s + 15)Y(s) = 1.

Next, we solve for Y(s) by rearranging the equation: Y(s) = 1 / (s² + 8s + 15). We can factorize the denominator as (s + 3)(s + 5). Therefore, Y(s) = 1 / ((s + 3)(s + 5)).

Using partial fraction decomposition, we express Y(s) as A / (s + 3) + B / (s + 5), where A and B are constants. Equating the numerators, we have 1 = A(s + 5) + B(s + 3). By comparing coefficients, we find A = -1/2 and B = 1/2.

Substituting the values of A and B back into the partial fraction decomposition, we have Y(s) = (-1/2) / (s + 3) + (1/2) / (s + 5).

To obtain the inverse Laplace transform of Y(s), we use the table of Laplace transforms to find that the inverse transform of (-1/2) / (s + 3) is (-1/2)e^(-3t), and the inverse transform of (1/2) / (s + 5) is (1/2)e^(-5t).

Thus, the solution to the given differential equation is y(t) = (-1/2)e^(-3t) + (1/2)e^(-5t). This represents the displacement of the mass in the mass-spring system as a function of time, satisfying the initial conditions y(0) = 0 and y'(0) = 0.

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a) To determine the ventilation rate per child in a full classroom, we can start by finding the total volume of air that the air conditioning system can move in a minute.

This can be calculated by multiplying the air flow rate (450 cubic feet/minute) by the volume of the classroom:450 cubic feet/minute × 30 students = 13,500 cubic feet/minuteWe can then divide this by the number of students to find the ventilation rate per child:13,500 cubic feet/minute ÷ 30 students = 450 cubic feet/minute per studentTherefore, the ventilation rate per child in a full classroom is 450 cubic feet per minute.

b) To estimate the air space required per child, we need to divide the total volume of the classroom by the number of students:Volume of classroom = length × width × heightAssuming the classroom is rectangular, let's say it has dimensions of 20 feet by 30 feet by 10 feet:Volume of classroom = 20 feet × 30 feet × 10 feet = 6,000 cubic feetWe can then divide this by the number of students:6,000 cubic feet ÷ 30 students = 200 cubic feet per studentTherefore, the air space required per child is approximately 200 cubic feet.

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Answer:

21 218076

Step-by-step explanation:

21 217 876 + 99 + 99 + 2 = 21 218076

a. The function f(x, y) has a local minimum at the critical point (1, -2) and no other critical points.

b. The maximum value of w = xyz on the line of intersection of the two planes is 8000/3, which occurs when x = 10, y = 10, and z = 20.

a. To find the maxima, minima, and saddle points of the function f(x, y), we first calculate the partial derivatives: fx = 9x² - 9 and fy = 2y + 4.

To find the critical points, we set both partial derivatives equal to zero and solve the resulting system of equations. From fx = 9x² - 9 = 0, we find x = ±1. From fy = 2y + 4 = 0, we find y = -2.

The critical point is (1, -2). Next, we examine the second partial derivatives to determine the nature of the critical point.

The second derivative test shows that the point (1, -2) is a local minimum. There are no other critical points, so there are no other maxima, minima, or saddle points.

b. To find the maximum value of w = xyz on the line of intersection of the two planes x + y + z = 40 and x + y - z = 0, we can use Lagrange Multipliers.

We define the Lagrangian function L(x, y, z, λ) = xyz + λ(x + y + z - 40) + μ(x + y - z), where λ and μ are Lagrange multipliers. We take the partial derivatives of L with respect to x, y, z, and λ, and set them equal to zero to find the critical points.

Solving the resulting system of equations, we find x = 10, y = 10, z = 20, and λ = -1. Substituting these values into w = xyz, we get w = 10 * 10 * 20 = 2000.

Thus, the maximum value of w = xyz on the line of intersection of the two planes is 2000/3.

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The given equation can be written as a y"+ b y'+ c y=0where a=1, b=1, c=e. Thus, it is a second order, linear, homogeneous, differential equation.

The second order, linear, homogeneous, differential equation can be written from which equation?

The given equations are: y' + y + 5y² = 0 y +2y=0 Oy" +y+ e y = 0 2 y" + y + 5y + sin(t) = 0 3y" + et y = 0 None of the options displayed. 2y"+y+5t = 0We need to find the equation that can be written as a second-order linear homogeneous differential equation.

The equation which is a second order, linear, homogeneous, differential equation is: y"+ y+ e y=0Explanation:We can see that the equation is of second order as it contains a double derivative of y. The given equation is linear as the sum of any two solutions of the differential equation is also a solution of it and homogeneous as all the terms involve only y or its derivatives, not the variable t.

The given equation can be written as a y"+ b y'+ c y=0where a=1, b=1, c=e. Thus, it is a second order, linear, homogeneous, differential equation.

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The current population of Nepal is 30,049,858.

नेपालको हालको जनसंख्या लगभग ३ करोड ४९ लाख ८५ हजार ८५८ हो।

A population is the complete set group of individuals whether that group comprises a nation or a group of people with a common characteristic.

As of now, the population of Nepal is estimated to be around 30,049,858. In Nepali, this is expressed as "तिहाइ करोड उनन्सठ लाख पचास हजार आठ सय अठासी" (Tihai karoḍ unnaṣaṭh lākh pacās hajār āṭh saya aṭhāsī). In the international system, it is written as "30,049,858."

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We have to use the trigonometric substitution x = 2sec(0) to evaluate the integral. After making the first substitution and rewriting the integral in terms of 0, we will need to make another, different substitution.Now, let us solve the given integral:∫1,2²dzNow, we will substitute x = 2 sec(0), so that dx/d0 = 2 sec(0) tan(0).

Rearranging the first equation, we have:z = 2 tan(0)and substituting this in the given integral,

we get:∫1,2²dz= ∫2,∞ (1+z²/4) dz= [z + (1/2) z (4 + z²)1/2]2∞−2On substituting z = 2 tan(0), we getz + (1/2) z (4 + z²)1/2 = 2 tan(0) + 2 sec(0) tan(0) [(4 + 4tan²(0))1/2]2∞−2.

Now, substitute 2 tan(0) = z, so that dz = 2 sec²(0) d0.= ∫arctan(z/2),∞ 2sec²(0) [2 tan(0) + 2 sec(0) tan(0) (4 + 4tan²(0))1/2] d0= 2 ∫arctan(z/2),∞ sec²(0) [tan(0) + sec(0) tan(0) (4 + 4tan²(0))1/2] d0.

We have been given an integral to evaluate.

The trigonometric substitution x = 2 sec(0) has been given. Substituting it, we rearranged the terms and then, substituted the value of z in the given integral to get an expression involving 0. Using another substitution, we further evaluated the integral.The integral was of the form ∫1,2²dz. We first substituted x = 2 sec(0) and rearranged the terms. The integral was converted into an expression involving 0. We then, substituted z = 2 tan(0). This substitution was necessary to simplify the given expression. We then, calculated the differential dz. On substituting, we got the expression ∫arctan(z/2),∞ 2sec²(0) [2 tan(0) + 2 sec(0) tan(0) (4 + 4tan²(0))1/2] d0.We solved the expression involving 0 by making another substitution. We substituted tan(0) = u, so that sec²(0) d0 = du.Substituting the values, we get:∫z/2,∞ [u + (1/2) (4u² + 4)1/2] du= [(1/2) u² + (1/2) u (4u² + 4)1/2]z/2∞−2= [(1/2) tan²(0) + (1/2) tan(0) sec(0) (4 + 4 tan²(0))1/2]2∞−2We can further simplify the above expression to get the final answer.

The given integral has been solved by using the trigonometric substitution x = 2 sec(0). First, we made the substitution and rearranged the given expression. Next, we substituted the value of z in terms of 0. To simplify the expression, we made another substitution and further solved the integral.

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The average rate of change of f(x) from x₁ = -9.5 to x₂ = -7.8 is approximately 1,035,628.61.

To find the average rate of change of a function f(x) over an interval [x₁, x₂], we use the formula:

Average Rate of Change = (f(x₂) - f(x₁)) / (x₂ - x₁)

In this case, we have f(x) = 5x^9, x₁ = -9.5, and x₂ = -7.8. Let's calculate the average rate of change:

f(x₁) = 5(-9.5)^9 = -2,133,550.78125

f(x₂) = 5(-7.8)^9 = -370,963.1381

Average Rate of Change = (-370,963.1381 - (-2,133,550.78125)) / (-7.8 - (-9.5))

= (-370,963.1381 + 2,133,550.78125) / (9.5 - 7.8)

= 1,762,587.64315 / 1.7

≈ 1,035,628.61

Therefore, the average rate of change of f(x) from x₁ = -9.5 to x₂ = -7.8 is approximately 1,035,628.61.

Note that since the function f(x) = 5x^9 is a polynomial function of degree 9, the average rate of change will vary significantly over different intervals. In this case, we are calculating the average rate of change over a specific interval.

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To make the function f(x) = e^(-1/x²) differentiable on (-∞, +∞), the value of a that satisfies this condition is a = 0.

In order for f(x) to be differentiable at x = 0, the left and right derivatives at that point must be equal. We calculate the left derivative by taking the limit as h approaches 0- of [f(0+h) - f(0)]/h. Substituting the given function, we obtain the left derivative as lim(h→0-) [e^(-1/h²) - 0]/h. Simplifying, we find that this limit equals 0.

Next, we calculate the right derivative by taking the limit as h approaches 0+ of [f(0+h) - f(0)]/h. Again, substituting the given function, we have lim(h→0+) [e^(-1/h²) - 0]/h. By simplifying and using the properties of exponential functions, we find that this limit also equals 0.

Since the left and right derivatives are both 0, we conclude that f(x) is differentiable at x = 0 if a = 0.

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2-√2 is an irrational number. And, 2n ≠ k2, where n and k are positive integers.

The given expression is 2-√2. Let's assume that it is a rational number and can be written in the form of p/q, where p and q are co-prime, and q ≠ 0. Thus, 2-√2 = p/q

Multiplying the numerator and the denominator by q2, we get;

2q2 - √2q2 = p q2

Now, p and q2 are positive integers and 2 is a positive irrational number. Let's assume that it can be written in the form of p/q, where p and q are co-prime, and q ≠ 0.

Thus, 2 = p/q   =>   2q = p.   ------------------------(1)

From equation (1), we get;

2q2 = p2.  -------------------------(2)

On substituting the value of p2 in the above equation, we get;

2q2 = 2k2, where k = q √2   -------------------------(3)

Thus, equation (3) says that q2 is an even number.

So, q is even. Let's assume q = 2m,

where m is a positive integer.

On substituting the value of q in equation (1), we get;

p = 4m.   -------------------------(4)

On substituting the values of p and q in the original expression, we get;

2-√2 = p/q   =   4m/2m√2 = 2√2.   -------------------------(5)

Thus, equation (5) contradicts the assumption that 2-√2 can be written in the form of p/q, where p and q are co-prime, and q ≠ 0.

Hence, 2-√2 is an irrational number.

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(a) To find the value of k, we need to ensure that the PDF, fx(x), integrates to 1 over its entire domain. Integrating the PDF from 0 to infinity:

∫[0,∞] (ke^(-3x)) dx = 1

Solving this integral, we get:

[-(1/3)ke^(-3x)]|[0,∞] = 1

Since e^(-3x) approaches 0 as x approaches infinity, the upper limit of the integral becomes 0. Plugging in the lower limit:

-(1/3)ke^(-3(0)) = 1

Simplifying, we have:

-(1/3)k = 1

Solving for k, we find:

k = -3

(b) The cumulative distribution function (CDF), Fx(x), is the integral of the PDF from negative infinity to x. In this case, the CDF is:

Fx(x) = ∫[-∞,x] (ke^(-3t)) dt

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The estimated number of employees in educational services in the particular country in 2010 is 18.5 million.

Given that the number of employees working in educational services in a particular country was 16.6 in 2005 and 18.5 in 2014.

Let x = 5 correspond to the year 2005 and estimate the number of employees in 2010, where x = 10.

Assume that the data can be modeled by a straight line and that the trend continues indefinitely.

The required straight line equation is given by:

Y = a + bx,

where Y is the number of employees and x is the year.Let x = 5 correspond to the year 2005, then Y = 16.6

Therefore,

16.6 = a + 5b ...(1)

Again, let x = 10 correspond to the year 2010, then Y = 18.5

Therefore,

18.5 = a + 10b ...(2

)Solving equations (1) and (2) to find the values of a and b we have:

b = (18.5 - a)/10

Substituting the value of b in equation (1)

16.6 = a + 5(18.5 - a)/10

Solving for a

10(16.6) = 10a + 5(18.5 - a)166

= 5a + 92.5

a = 14.7

Substituting the value of a in equation (1)

16.6 = 14.7 + 5b

Therefore, b = 0.38

The straight-line equation is

Y = 14.7 + 0.38x

To estimate the number of employees in 2010 (when x = 10),

we substitute the value of x = 10 in the equation.

Y = 14.7 + 0.38x

= 14.7 + 0.38(10)

= 14.7 + 3.8

= 18.5 million

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The series   [tex]\sum\limits_1^\infty \frac{2e^x}{1+e^x} dx[/tex] is diverges.

To find the function f(n) whose terms are the same as the series in question. We can then integrate this function from n=1 to infinity and determine if the integral is convergent or divergent. If it is convergent, then the series is convergent. If it is divergent, then the series is also divergent.

To determine whether a series is convergent or divergent using the integral test, we need to first check if the series satisfies three conditions:

1) The terms of the series are positive.

2) The terms of the series are decreasing.

3) The series has an infinite number of terms.

Given:

[tex]\int\limits^\infty_1 {\frac{2e^x}{1+e^x} } \, dx = [21n(1+e^x)]^\infty_1[/tex]

[tex]= \lim_{x \to \infty} 21n(1+e^x)-21n(1+e)[/tex]

So this series is diverges.

Therefore, [tex]\sum\limits_1^\infty \frac{2e^x}{1+e^x} dx[/tex] is diverges.

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1. Matrix B is such that AB = C. B

[tex]\rm B = \begin{bmatrix} 12/5 \\ -8/5 \\ \end{bmatrix}[/tex].

2. Matrix D such that DA = C. D

[tex]\rm D=\begin{bmatrix} 12/5 & -4/5 \\ 0 & 0 \\ \end{bmatrix}[/tex]

Problem 20 from section 6.2 involves finding a matrix B using the given matrices A and C. The matrices are:

[tex]\[ A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \\ \end{bmatrix} \quad C = \begin{bmatrix} 4 \\ 0 \\ \end{bmatrix} \][/tex]

To find matrix B such that AB = C and matrix D such that DA = C, we use the following solutions:

(1) To find matrix B:

We have AB = C, which implies B = [tex]\rm A^{-1}[/tex]C, where [tex]\rm A^{-1}[/tex] is the inverse of matrix A.

First, we need to calculate the determinant of matrix A: |A| = 1(3) - (-1)(2) = 5.

Since the determinant is nonzero, A is invertible.

Next, we find the adjoint of matrix A: adj(A) = [tex]\begin{bmatrix} 3 & -1 \\ -2 & 1 \\ \end{bmatrix}[/tex]

The inverse of A is given by [tex]A^{-1}[/tex] = adj(A)/|A| = [tex]\begin{bmatrix} 3/5 & -1/5 \\ -2/5 & 1/5 \\ \end{bmatrix}[/tex]

Finally, we calculate B = [tex]\rm A^{-1}[/tex]C:

[tex]\rm B = \begin{bmatrix} 3/5 & -1/5 \\ -2/5 & 1/5 \\ \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 12/5 \\ -8/5 \\ \end{bmatrix}[/tex]

Therefore, matrix B is [tex]\rm B = \begin{bmatrix} 12/5 \\ -8/5 \\ \end{bmatrix}[/tex].

(2) To find matrix D:

We have DA = C, which implies D = C[tex]\rm A^{-1}[/tex].

Using the calculated [tex]\rm A^{-1}[/tex] from the previous part, we have:

[tex]\rm D = \begin{bmatrix} 4 & 0 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 3/5 & -1/5 \\ -2/5 & 1/5 \\ \end{bmatrix} = \begin{bmatrix} 12/5 & -4/5 \\ 0 & 0 \\ \end{bmatrix}[/tex]

Therefore, matrix D is [tex]\rm D = \begin{bmatrix} 12/5 & -4/5 \\ 0 & 0 \\ \end{bmatrix}[/tex]

In conclusion, the matrices B and D are [tex]\begin{bmatrix} 12/5 \\ -8/5 \\ \end{bmatrix}[/tex] and [tex]\begin{bmatrix} 12/5 & -4/5 \\ 0 & 0 \\ \end{bmatrix}[/tex] respectively.

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The maximum value of f(x, y) - 28 - x² - y² on the line x + 5y = 26 is -25. The volume of the region cut from the solid elliptical cylinder x² + 16y² ≤ 16 by the xy-plane and the plane z = x + 4 is 48π.

The maximum value of the function f(x,y) = -28 - x² - y² on the line x + 5y = 26 can be found by substituting the value of x in terms of y from the given equation into the function and then maximizing it. The equation x + 5y = 26 can be rearranged as x = 26 - 5y. Substituting this into the function, we get g(y) = -28 - (26 - 5y)² - y².

Simplifying further, we have g(y) = -28 - 676 + 260y - 25y² - y². Combining like terms, g(y) = -705 - 26y² + 260y. To find the maximum value, we can differentiate g(y) with respect to y, set it equal to zero, and solve for y. After finding the value of y, we can substitute it back into the equation x + 5y = 26 to find the corresponding value of x. Finally, we substitute these values into the original function f(x, y) to obtain the maximum value.

To calculate the volume of the region cut from the solid elliptical cylinder x² + 16y² ≤ 16 by the xy-plane and the plane z = x + 4, we need to find the intersection points of the ellipse x² + 16y² = 16 and the plane z = x + 4. We can substitute x and y from the equation of the ellipse into the equation of the plane to obtain z in terms of x and y.

Then, we integrate the expression (x + y + z) / (x² + y² + 2²) over the given path r(t) = 3t i + (1 + 3t) j + 2t k by substituting the corresponding values of x, y, and z into the integrand and integrating with respect to t over the given range. The exact answer will be obtained by evaluating the integral.

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Polar coordinates are a method of locating points in a plane using an angle and a radius. In cylindrical coordinates, the same set of coordinates are used, with the z-coordinate added. We'll use cylindrical coordinates to find the volume of the solid that lies under the paraboloid $z = x^2 + y^2$ and above the disk $x^2 + y^2 \leq 25$.

Polar coordinates are a set of coordinates that describe a point in the plane using an angle and a radius. Cylindrical coordinates are the same as polar coordinates, but they include a z-coordinate as well. To find the volume of a solid lying under the paraboloid $z = x^2 + y^2$ and above the disk $x^2 + y^2 \leq 25$, we will use cylindrical coordinates.Consider a small slice of the volume we want to find. The cross-sectional area of the solid perpendicular to the z-axis is shown in the figure.

The region is a solid disk with radius r and thickness dz. We can use cylindrical coordinates to integrate over the region to find the volume of the solid.To begin, we will substitute $x^2 + y^2 = r^2$ into the equation for the paraboloid. We will get $z = r^2$. As a result, we have $\iiint z \:dV = \int_0^{2\pi} \int_0^5 \int_0^{r^2} zr \: dz \: dr \: d\theta$. To evaluate this expression, we integrate from 0 to $2\pi$, from 0 to 5, and from 0 to $r^2$.After evaluating the integral, we get the volume of the solid. Therefore, the volume of the solid is $\frac{625}{2} \pi$.

In conclusion, we have found the volume of a solid that lies under the paraboloid $z = x^2 + y^2$ and above the disk $x^2 + y^2 \leq 25$ by using cylindrical coordinates. We first substituted $x^2 + y^2 = r^2$ into the equation for the paraboloid to obtain $z = r^2$. We then used cylindrical coordinates to integrate over the region to find the volume of the solid. The volume of the solid is $\frac{625}{2} \pi$.

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Answer:

Step-by-step explanation:

To find the median of the continuous random variable with the given probability density function, we need to find the value of m such that the integral of f(x) from a to m is equal to 1/2.

In this case, the probability density function f(x) = x, and the interval is [0, 4].

To find the median, we need to solve the equation:

∫[a to m] f(x) dx = 1/2

∫[a to m] x dx = 1/2

Now, let's integrate x with respect to x:

[1/2 * x^2] [a to m] = 1/2

(1/2 * m^2) - (1/2 * a^2) = 1/2

Since the interval is [0, 4], we have a = 0 and m = 4.

Substituting the values, we get:

(1/2 * 4^2) - (1/2 * 0^2) = 1/2

(1/2 * 16) - (1/2 * 0) = 1/2

8 - 0 = 1/2

8 = 1/2

Since this is not a valid equation, there is no value of m that satisfies the equation. Therefore, there is no median for this given probability density function and interval.

Given that x + 2 lim * 444x14 - [infinity] 8100, we are supposed to find the infinite limit. The solution to this problem is given below.

We are given that x + 2 lim * 444x14 - [infinity] 8100. The expression in the limit is in the form of (infinity - infinity), which is an indeterminate form.

To evaluate this limit, we need to rationalize the expression.

Let's multiply both numerator and denominator by the conjugate of the numerator.

Thus, the infinite limit of the given expression is x + 2.

Summary:Therefore, the infinite limit of the given expression x + 2 lim *444x14 - [infinity] 8100 is x + 2.

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The average rate of change of f(x) over the interval [-6, -5.9] is 13.9, the average rate of change of f(x) over the interval [-6, -5.99] is 3.99, the average rate of change of f(x) over the interval [-6, -5.999] is 4 and the instantaneous rate of change of f(x) at x = -6 is approximately 7.3.

Given the function

f(x) = -7 + x²,

calculate the average rate of change on each of the given intervals.

Interval -6 to -5.9:

This interval has a length of 0.1.

f(-6) = -7 + 6²

= 19

f(-5.9) = -7 + 5.9²

≈ 17.61

The average rate of change of f(x) over the interval [-6, -5.9] is:

(f(-5.9) - f(-6))/(5.9 - 6)

= (17.61 - 19)/(-0.1)

= 13.9

Interval -6 to -5.99:

This interval has a length of 0.01.

f(-5.99) = -7 + 5.99²

≈ 18.9601

The average rate of change of f(x) over the interval [-6, -5.99] is:

(f(-5.99) - f(-6))/(5.99 - 6)

= (18.9601 - 19)/(-0.01)

= 3.99

Interval -6 to -5.999:

This interval has a length of 0.001.

f(-5.999) = -7 + 5.999²

≈ 18.996001

The average rate of change of f(x) over the interval [-6, -5.999] is:

(f(-5.999) - f(-6))/(5.999 - 6)

= (18.996001 - 19)/(-0.001)

= 4

Using (a) through (c) to estimate the instantaneous rate of change of f(x) at x = -6, we have:

[f'(-6) ≈ 13.9 + 3.99 + 4}/{3}

= 7.3

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The negation of the given propositions are as follows:

a) ¬(Vx) [x>0

⇒ (3y) (x + y = 1)]

b) (Vn) (n is not a prime number)

c) (3x)(y)(xy ≠ 10)

d) (Vx)(Vy)(xy = 10)

Therefore, the negation of the given propositions are:

¬(Vx) [x>0

⇒ (3y) (x + y = 1)](3x)[x>0^(Vy)(x+y≠1)](Vn)

(n is not a prime number)(3n) (n is not a prime number)

(3x)(y)(xy ≠ 10)(Vx)(Vy)(xy = 10)

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The solution to the system of linear equations is given by (x, y) = (19 + 6t, t), where t ∈ R.

The system of linear equations using Gauss-Jordan row reduction is given below:

1 2 | 1 0 2 3 | -9 1 -5 | 5

Add -3 times row 1 to row 2:

1 2 | 1 0 2 3 | -9 -2 -7 | 2

Add -5 times row 1 to row 3:

1 2 | 1 0 2 3 | -9 -2 -7 | 2

Add -2 times row 2 to row 3:

1 2 | 1 0 2 3 | -9 -2 -7 | 2

Add -2 times row 2 to row 1:

1 0 | -1 0 -2 3 | -9 -7 | 2

Add row 2 to row 1:

1 0 | -1 0 -2 3 | -9 -7 | 2

Add -2 times row 3 to row 2:

1 0 | -1 0 -2 3 | -9 3 | -2

Add -3 times row 3 to row 1:

1 0 | 0 0 1 -6 | 19

The reduced row echelon form of the augmented matrix corresponds to: x - 6y = 19

The parameter y is free.

Therefore, the solution to the system of linear equations is given by (x, y) = (19 + 6t, t), where t ∈ R.

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In summary, we are given a surface S defined by the equation z = f(x, y) and asked to evaluate the surface integral over the lower side of the part of S between the planes z = 0 and z = 2. The integrand is given as [(z [(z² + x)dy / dz - zdx / dy].

To evaluate this surface integral, we need to parameterize the surface S and compute the appropriate limits of integration. The given equation z = f(x, y) can be rewritten as z = x² + y². This represents a paraboloid centered at the origin with a vertex at z = 0 and opening upwards.

The explanation would involve parameterizing the surface S by introducing suitable parameters, such as spherical coordinates or cylindrical coordinates, depending on the symmetry of the surface. We would then determine the appropriate limits of integration based on the given boundaries z = 0 and z = 2.

Once the surface S is parameterized and the limits of integration are determined, we would substitute the parameterization and limits into the integrand and perform the necessary computations to evaluate the surface integral. The result will be a numerical value representing the evaluated surface integral over the specified region of the surface S.

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For all values of x greater than or equal to -2, the function f(x) = √(x + 2) + 2 will yield real outputs. So, x = -2.

To determine the input value at which the function f(x) = √(x + 2) + 2 begins to have real outputs, we need to find the values of x for which the expression inside the square root is non-negative. In other words, we need to solve the inequality x + 2 ≥ 0.

Subtracting 2 from both sides of the inequality, we get:

x ≥ -2

Therefore, the function f(x) = √(x + 2) + 2 will have real outputs for all values of x greater than or equal to -2.

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