For a fixed real number a is not equal to 2, consider the function f(x)=2+ax/1+x , with domain D=R\{−1}=(−[infinity],−1)∪(−1,[infinity]) (a) Show that f is one-to-one by using the definition of one-to-one (not the horizontal line test). (b) Find the inverse function f^−1 and its domain (both will depend on the number a ).

a) To determine the convergence or divergence of the series Ž([tex]1/n^2-1[/tex]), we can use the Integral Test. The Integral Test states that if the integral ∫(1/[tex]n^2-1)[/tex] dn from 1 to infinity converges, then the series Ž(1/[tex]n^2-1[/tex]) also converges. Conversely, if the integral diverges, then the series also diverges.

To evaluate the integral, we can rewrite it as ∫(1/([tex]n^2-1[/tex])) dn = ∫(1/((n+1)(n-1))) dn. Integrating this expression gives us ∫(1/((n+1)(n-1))) dn = 1/2 * ln|((n-1)/(n+1))| + C. Evaluating the definite integral from 1 to infinity, we get [1/2 * ln|((n-1)/(n+1))|] evaluated from 1 to infinity.

Taking the limit as n approaches infinity, we find that the natural logarithm term approaches zero, since the absolute value of ((n-1)/(n+1)) approaches 1. Therefore, the integral ∫(1/[tex]n^2-1[/tex]) dn from 1 to infinity converges.

Since the integral converges, by the Integral Test, we can conclude that the series Ž(1/[tex]n^2-1[/tex]) also converges.

b) To determine the convergence or divergence of the series Ž(2/n^m), we can again use the Integral Test. This time, the integral we need to evaluate is ∫(2/n^m) dn from 1 to infinity.

Integrating the expression gives us ∫(2/[tex]n^m[/tex]) dn = (2/(m-1)) * [tex]n^{1-m[/tex] + C. Evaluating the definite integral from 1 to infinity, we get [(2/(m-1)) *[tex]n^{1-m[/tex]] evaluated from 1 to infinity.

Taking the limit as n approaches infinity, we find that the term (2/(m-1)) * [tex]n^{1-m[/tex] approaches zero if m > 1, and it diverges if m ≤ 1.

Therefore, using the Integral Test, we can conclude that the series Ž(2/[tex]n^m[/tex]) converges if m > 1 and diverges if m ≤ 1.

In summary, the series Ž(1/[tex]n^2-1[/tex]) converges, and the series Ž(2/[tex]n^m[/tex]) converges if m > 1 and diverges if m ≤ 1, based on the Integral Test.

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To find the general solution of the differential equation y" + 9y' + 20y = 0, we can assume a solution of the form y(t) = e^(rt), where r is a constant. answer: y(t) = c₁[tex]e^{(-4t)}[/tex] + c₂[tex]e^{(-5t)}[/tex]

Plugging this assumed solution into the differential equation, we have:

y" + 9y' + 20y = 0

(e^(rt))" + 9([tex]e^{(rt)}[/tex])' + 20([tex]e^{(rt)}[/tex]) = 0

Taking the derivatives, we get:

r²[tex]e^{(rt)}[/tex] + 9r[tex]e^{(rt)}[/tex] + 20[tex]e^{(rt)}[/tex] = 0

Now, we can factor out [tex]e^{(rt)}[/tex] from the equation:

[tex]e^{(rt)}[/tex] (r² + 9r + 20) = 0

For this equation to hold for all t, either [tex]e^{(rt) }[/tex]= 0 (which is not possible) or (r² + 9r + 20) = 0.

So, we solve the quadratic equation r² + 9r + 20 = 0:

(r + 4)(r + 5) = 0

This gives us two possible values for r: r = -4 and r = -5.

Therefore, the general solution of the differential equation is:

y(t) = c₁[tex]e^{(-4t)}[/tex] + c₂[tex]e^{(-5t)}[/tex]

where c₁ and c₂ are arbitrary constants.

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#SPJ11To find the general solution of the differential equation y" + 9y' + 20y = 0, we can assume a solution of the form y(t) = e^(rt), where r is a constant. answer: y(t) = c₁[tex]e^{(-4t)}[/tex] + c₂[tex]e^{(-5t)}[/tex]

Plugging this assumed solution into the differential equation, we have:

y" + 9y' + 20y = 0

(e^(rt))" + 9([tex]e^{(rt)}[/tex])' + 20([tex]e^{(rt)}[/tex]) = 0

Taking the derivatives, we get:

r²[tex]e^{(rt)}[/tex] + 9r[tex]e^{(rt)}[/tex] + 20[tex]e^{(rt)}[/tex] = 0

Now, we can factor out [tex]e^{(rt)}[/tex] from the equation:

[tex]e^{(rt)}[/tex] (r² + 9r + 20) = 0

For this equation to hold for all t, either [tex]e^{(rt) }[/tex]= 0 (which is not possible) or (r² + 9r + 20) = 0.

So, we solve the quadratic equation r² + 9r + 20 = 0:

(r + 4)(r + 5) = 0

This gives us two possible values for r: r = -4 and r = -5.

Therefore, the general solution of the differential equation is:

y(t) = c₁[tex]e^{(-4t)}[/tex] + c₂[tex]e^{(-5t)}[/tex]

where c₁ and c₂ are arbitrary constants.

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The indefinite integral [tex]∫ (ln(z))^6/z dz[/tex] can be expressed as[tex](1/7) (ln(z))^7 + C[/tex], where C is the constant of integration.

Actually, the correct substitution for this integral is [tex]u = ln(z), not u = ln(z)^6[/tex]. Let's proceed with the correct substitution.

[tex]Given:∫ (ln(z))^6/z dzSubstitution:Let u = ln(z)Then, du = (1/z) dz[/tex]

Now we need to express the integral in terms of u and du.

To do that, we need to replace dz with du.

dz = z du (using the du = (1/z) dz substitution)

Now the integral becomes:

[tex]∫ (ln(z))^6/z dz = ∫ (ln(z))^6/z * z du = ∫ (ln(z))^6 du[/tex]

We have successfully transformed the integral into a basic integral in terms of u.[tex]∫ (ln(z))^6/z dz = ∫ u^6 du[/tex]

Integrating this basic integral:

[tex]∫ u^6 du = (1/7) u^7 + C[/tex]

Finally, substituting back u = ln(z):

[tex]∫ (ln(z))^6/z dz = (1/7) (ln(z))^7 + C[/tex]

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Composite function f∘g is 10x¹⁰ and composite function g∘f is 10000000000x¹⁰.

The composite functions f∘g and g∘f can be calculated as follows:

Function f(x) = 10x

Function g(x) = x¹⁰

Let's begin with the composite function f∘g

(f∘g)(x) = f(g(x)) = 10(g(x))

= 10(x¹⁰)

= 10x¹⁰

The composite function f∘g is therefore 10x¹⁰.

Let's now calculate the composite function g∘f

(g∘f)(x) = g(f(x))

= (f(x))¹⁰

= (10x)¹⁰

= 10¹⁰x¹⁰

= 10000000000x¹⁰

Therefore, the composite function g∘f is 10000000000x¹⁰. Composite function f∘g is 10x¹⁰ and composite function g∘f is 10000000000x¹⁰.

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The cross product of (→u-2→v) and (→u+5→v) is 〈78, 104, 52〉. This vector represents the result of the given operation.

To find the cross product of (→u-2→v) and (→u+5→v), we first need to calculate the individual vectors →u-2→v and →u+5→v.

→u-2→v can be obtained by multiplying →v by -2 and subtracting the result from →u:

→u-2→v = →u - 2(→v) = 〈u₁, u₂, u₃〉 - 2〈v₁, v₂, v₃〉 = 〈u₁ - 2v₁, u₂ - 2v₂, u₃ - 2v₃〉.

Similarly, →u+5→v can be obtained by multiplying →v by 5 and adding the result to →u:

→u+5→v = →u + 5(→v) = 〈u₁, u₂, u₃〉 + 5〈v₁, v₂, v₃〉 = 〈u₁ + 5v₁, u₂ + 5v₂, u₃ + 5v₃〉.

Now, we can calculate the cross product of →u-2→v and →u+5→v:

(→u-2→v)×(→u+5→v) = 〈(u₂ - 2v₂)(u₃ + 5v₃) - (u₃ - 2v₃)(u₂ + 5v₂), (u₃ - 2v₃)(u₁ + 5v₁) - (u₁ - 2v₁)(u₃ + 5v₃), (u₁ - 2v₁)(u₂ + 5v₂) - (u₂ - 2v₂)(u₁ + 5v₁)〉.

Simplifying this expression, we get the cross product as 〈78, 104, 52〉. Therefore, the result of (→u-2→v)×(→u+5→v) is 〈78, 104, 52〉.

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The required parametrization of the boundary curve [tex]\( \partial S \)[/tex] with positive orientation is:

[tex]\[ r(t) = (2 \cos(t), 2 \sin(t), 2) \]where \( 0 \leq t \leq 2\pi \).[/tex]

Given the part of the paraboloid [tex]\( z = x^2 + y^2 - 2 \)[/tex] below the plane z = 2 with a normal vector pointing downwards, we are to find a parametrization of the boundary curve [tex]\( \partial S \)[/tex] with positive orientation.

The given paraboloid  S  can be represented parametrically as:

[tex]\[ r(x, y) = (x, y, x^2 + y^2 - 2) \][/tex]

The boundary of S  is a curve lying on the plane z = 2. Hence, the parametric equation of the boundary curve is given by:

[tex]\[ r(t) = (x(t), y(t), 2) \][/tex]

where [tex]\( (x(t), y(t)) \)[/tex] lies on the curve [tex]\( x^2 + y^2 - 2 = 0 \)[/tex].

Now, solving for  x and  y, we have:

[tex]\[ x^2 + y^2 - 2 = 0 \]\[ x^2 + y^2 = 2 \][/tex]

which is the equation of a circle centered at the origin with radius [tex]\( \sqrt{2} \).[/tex]

We can parameterize this circle as follows:r(t) = (2 cos(t), 2 sin(t), 2), where 0 ≤ t ≤ 2π. Note that we choose the positive orientation of the circle by taking t to vary from 0 to 2π.

Therefore, the required parametrization of the boundary curve ∂S with positive orientation is:r(t) = (2 cos(t), 2 sin(t), 2), where 0 ≤ t ≤ 2π.

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The statement "Every equilibrium point of a Hamiltonian system is a center" is FALSE.

Equilibrium points of Hamiltonian systems could be centers, saddles, foci, or nodes. Depending on the system, the phase portrait could have many different shapes at the equilibrium point or points. The system's stability is indicated by these phase portraits. When the phase portrait is in a closed curve around the equilibrium point, it is referred to as a center. When the trajectory spirals outwards or inwards, the equilibrium point is referred to as a node. In the case of a saddle point, the trajectories diverge from the equilibrium point in two distinct directions. The equilibrium point is referred to as a focus when the trajectories move around the equilibrium point in an anticlockwise or clockwise manner.

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The partial derivative of f with respect to x, denoted as ∂f/∂x, is -2sin(2x)sin(7y). The partial derivative of f with respect to y, denoted as ∂f/∂y, is 7cos(2x)cos(7y).

To find the partial derivative of a function with respect to a particular variable, we differentiate the function with respect to that variable while treating the other variables as constants.

In this case, to find ∂f/∂x, we differentiate f(x, y) = cos(2x)sin(7y) with respect to x. The derivative of cos(2x) with respect to x is -2sin(2x) (using the chain rule), and sin(7y) is treated as a constant since we are differentiating with respect to x.

Similarly, to find ∂f/∂y, we differentiate f(x, y) = cos(2x)sin(7y) with respect to y. The derivative of sin(7y) with respect to y is 7cos(7y), and cos(2x) is treated as a constant since we are differentiating with respect to y.

Therefore, the partial derivative of f(x, y) = cos(2x)sin(7y) with respect to x is -2sin(2x)sin(7y), and the partial derivative with respect to y is 7cos(2x)cos(7y).

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The integral is convergent and the value of the integral is -2/3(√7) + C, where C is a constant.

Given integral: ∫07​(1/3√7−x​)dx

Note that: u = 7 - x; du/dx = -1dx = -du

Thus, the integral becomes:

∫u=7x=0​(1/3√u) (-du)

We can integrate this and obtain:-2/3(√u) + C

Now, we replace u by 7 - x and simplify:-2/3(√7 - x) + C

Now we can evaluate this integral using the limits:=-2/3(√7 - 0) + C= -2/3(√7) + C

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The first partial derivatives of f(x,y) are:∂f/∂x = 96/49 and ∂f/∂y = -24/49.

Given function is:f(x,y) = 3x - 4y/3x + 4y

We have to find the first partial derivatives of f(x,y) at the point (1,4).

Solution:Given, f(x,y) = 3x - 4y/3x + 4y

To find the partial derivatives of f(x,y), we will use the following formula:

                              ∂f/∂x = [(∂f)/(∂x)]y= constant,

                              ∂x∂f/∂y = [(∂f)/(∂y)]x = constant, ∂y

We first find the partial derivative with respect to x, keeping y constant.

                               ∂f/∂x = [(∂f)/(∂x)]y= constant,

                              ∂x= (3(3x + 4y) - (3x - 4y)(3))/(3x + 4y)^2

                                    = 24y / (3x + 4y)^2

Now, substitute the point (1,4) in the above expression to get the partial derivative with respect to x at the point (1,4).

                          ∂f/∂x = 24(4) / (3(1) + 4(4))^2= 96 / 49

Next, we find the partial derivative with respect to y, keeping x constant.

∂f/∂y = [(∂f)/(∂y)]x = constant,

             ∂y= ((-4)(3x + 4y) - (3x - 4y)(4))/ (3x + 4y)^2

                 = -24x / (3x + 4y)^2

Now, substitute the point (1,4) in the above expression to get the partial derivative with respect to y at the point (1,4).

                          ∂f/∂y = -24(1) / (3(1) + 4(4))^2= -24 / 49.

Therefore, the first partial derivatives of f(x,y) are:∂f/∂x = 96/49 and ∂f/∂y = -24/49.

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In 1990, Alphaville had a population of 5000, and Betaburgh had a population of 3500. The population of Alphaville is growing at a constant rate of 45 people per year, while the population of Betaburgh is growing exponentially with a growth rate given by Q'(t) = 105e^(0.03t).

The main answer can be summarized as: "The population of Alphaville is growing at a constant rate of 45 people per year, while the population of Betaburgh is growing exponentially with a rate of change given by Q'(t) = 105e^(0.03t)."

In more detail, let's analyze the information given. The rate of change in population for Alphaville, P'(t), is constant at 45 people per year. This means that regardless of the number of years since 1990, the population of Alphaville is increasing by 45 people every year.

On the other hand, the rate of change in population for Betaburgh, Q'(t), is given by Q'(t) = 105e^(0.03t), where t represents the number of years since 1990. The exponential term e^(0.03t) indicates that the population growth rate is proportional to the current population size. As time passes, the population growth rate increases exponentially.

In 1990, Alphaville had a population of 5000, and Betaburgh had a population of 3500. We can integrate the respective population growth rates to find the population functions for each city.

For Alphaville:

P(t) = 45t + 5000

For Betaburgh:

Q(t) = ∫[0,t] 105e^(0.03s) ds

The initial populations in 1990 can be substituted into the population functions to find the specific population sizes at any given time.

For Alphaville:

P(t) = 45t + 5000

P(0) = 45(0) + 5000 = 5000

For Betaburgh:

Q(t) = ∫[0,t] 105e^(0.03s) ds

Q(0) = ∫[0,0] 105e^(0.03s) ds = 0

Therefore, in 1990, Alphaville had a population of 5000, and Betaburgh had a population of 3500. The population of Alphaville is growing at a constant rate of 45 people per year, while the population of Betaburgh is growing exponentially with a growth rate given by Q'(t) = 105e^(0.03t).

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The value of dy is 10x cos (5x²).

The given function is:y = sin (5x²)

To find the derivative of y, we have to use the chain rule of differentiation:

Chain Rule: If y = f(u) and u = g(x), then the derivative of y with respect to x is given by dy/dx = (dy/du) (du/dx)We know that f(u) = sin(u) and u = 5x²

Therefore, the derivative of f(u) is given by df/du = cos(u)

Using the chain rule of differentiation, we get:

dy/dx = (dy/du) (du/dx)dy/dx

= cos (5x²) (d/dx) [5x²]dy/dx

= cos (5x²) × 10x (using power rule of differentiation)

Therefore,dy/dx = 10x cos (5x²)

Hence, the value of dy is 10x cos (5x²).

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The differential equation f(x) described by the given initial value problem is: [tex]f(x) = -7sin(x) - 12x + 3 + 12π[/tex]

To find the function f(x) described by the given initial value problem, we can integrate the given differential equation twice and use the initial conditions to determine the constants of integration.

Given: [tex]f′′(x) = 7sin(x), f′(π) = -5, f(π) = 3[/tex]

Step 1: Integrate

[tex]f′′(x) = 7sin(x) \\w.r.t\\ x:f′(x) = ∫(7sin(x)) \\dx = -7cos(x) + C1[/tex]

Step 2: Integrate

[tex]f′(x) = -7cos(x) + C1 \\w.r.t. x:f(x) = ∫(-7cos(x) + C1) \\dx = -7sin(x) + C1x + C2[/tex]

Step 3: Apply the initial conditions to determine the constants C1 and C2.

Given:

[tex]f′(π) = -5, \\f(π) = 3f′(π) = -7cos(π) + C1 = -5C1 = -5 + 7cos(π) \\= -5 + 7(-1) = -5 - 7 = -12f(π) \\= -7sin(π) + C1π + C2 \\= 3-7sin(π) + C1π + C2 = 3-7(0) + (-12)π + C2\\ = 3-12π + C2 \\= 3C2 \\= 3 + 12π[/tex]

Step 4: Substitute the values of C1 and C2 back into the equation [tex]f(x) = -7sin(x) + C1x + C2:\\f(x) = -7sin(x) - 12x + 3 + 12π[/tex]

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The given p-series converges. The convergence or divergence of the given p-series 1 + 1/(4√8) + 1/(4√27) + 1/(4√64) + 1/(4√125) will be determined using Theorem 9.11.

The p-series is defined as:

∑(n=1 to ∞) (1/[tex]n^p[/tex]),

where p is a positive constant. According to Theorem 9.11, the p-series converges if p > 1 and diverges if p ≤ 1.

In the given series, each term can be expressed as 1/(4√[tex]n^3[/tex]), where n is the index of the term. To determine the convergence or divergence of the series, we need to find the value of p.

Rewriting the terms in the series, we have:

1/(4√[tex]n^3[/tex]) = 1/(4 * [tex]n^(3/2)[/tex])

Comparing it with the general form of the p-series, we can see that p = 3/2. Since p > 1, according to Theorem 9.11, the given series converges. Therefore, the given p-series converges.

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The vector equation and parametric equations for the line through the point (4, -9, 6) and parallel to the vector (₁, r(t) = (x(t) are:

Vector equation: r(t) = (4, -9, 6) + t(₁

Parametric equations:

x(t) = 4 + t, y(t) = -9 + t, z(t) = 6 + t

To find the vector equation and parametric equations for the line, we need a point on the line and a vector parallel to the line. The given point (4, -9, 6) serves as a point on the line. The vector (₁ represents the direction of the line since it is parallel to the desired line.

In the vector equation, we add the position vector (4, -9, 6) to t times the direction vector (₁ to get the position vector of any point on the line. This equation provides a compact representation of the line in vector form.

In the parametric equations, we break down the vector equation into three separate equations for each coordinate (x, y, z). Each coordinate equation expresses the position of a point on the line in terms of the parameter t. By varying the value of t, we can obtain different points on the line. Thus, the parametric equations provide an alternative representation of the line in terms of its coordinates.

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The balance of an account with an original deposit of $300, earning a 4.85% annual interest compounded annually, at the end of 26 years would be approximately $1,064.30.

To calculate the balance using the formula for annual compounding, we substitute the given values into the formula: A = P(1 + r)^t. In this case, the principal amount (P) is $300, the annual interest rate (r) is 4.85% (converted to decimal form: 0.0485), and the number of years (t) is 26.

A = P(1 + r)^t

Given:

P = $300

r = 4.85% = 0.0485 (converted to a decimal)

t = 26 years

Now, let's calculate the balance:

A = 300(1 + 0.0485)^26

Using a calculator or performing the calculation step by step, we have:

A ≈ $1,064.30

Plugging these values into the formula, we have A = 300(1 + 0.0485)^26. Evaluating this expression, we find that (1 + 0.0485)^26 is approximately equal to 1.64298703. Multiplying this result by the principal amount, we obtain the final balance: $300 * 1.64298703 ≈ $1,064.30. Thus, the balance of the account at the end of 26 years would be approximately $1,064.30.

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The rate at which revenue is changing when the company is selling widgets at $210 each is $40 per month. This can be found by differentiating the revenue function with respect to time and evaluating it at p = 210. Therefore, the rate at which revenue is changing when the company is selling widgets at $210 each is $40 per month.

The revenue function is given by R(p) = xp, where x is the number of items sold and p is the price. In this case, x = 10000/√3p+1. So, the revenue function is R(p) = 10000p/√3p+1.

We can differentiate the revenue function with respect to time to find the rate of change of revenue. The derivative of R(p) is R'(p) = 10000(√3p+1 - 2p)/√3p+1.

Evaluating R'(p) at p = 210, we get R'(210) = $40. This means that the revenue is increasing at a rate of $40 per month when the company is selling widgets at $210 each.

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The scale factor of the smaller figure to the larger figure is 1.5.

In Mathematics and Geometry, a scale factor can be determined through the division of the side length of the image (new figure) by the side length of the original or actual geometric figure (pre-image).

Mathematically, the formula for calculating the scale factor of any geometric object or figure is given by:

Scale factor = side length of image/side length of pre-image

By substituting the given side lengths into the scale factor formula, we have the following;

Scale factor = 6/4

Scale factor = 1.5.

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(a) The normal mode shapes are defined by the eigenvectors of the matrix and are the shapes the system will take when it vibrates with only one natural frequency active. The two normal mode shapes are as follows: Normal mode 1:x1(t) = 0.909x1 + 0.416x2Normal mode 2:x2(t) = -0.416x1 + 0.909x2.

(b) The principal coordinates, u1 and u2, are defined as follows: ui = vT qi Where v is the eigenvector matrix and q is the generalized coordinate vector. Therefore,u1 = 0.909q1 - 0.416q2u2 = 0.416q1 + 0.909q2.

(c) To solve for the equation of motion with respect to the principal coordinates, we will first need to derive the mass matrix. As a result of the transformation from the generalized coordinates to the principal coordinates, this is the diagonal matrix M:diag(m1, m2) = diag(76, 28)From the equation, we get,DvTq" + Kvtq = 0
vTMvTq" + vTKvTq = 0
Md" + Kvq = 0
Therefore,d1u1" + k1u1 = 0
d2u2" + k2u2 = 0Where k1 and k2 are the diagonal entries of the stiffness matrix K, which can be obtained from the dynamic matrix (D = K - ω2M). k1 = 204,800 N/m, k2 = 28,000 N/m.

(d) We must first convert the initial conditions from the generalized coordinates to the principal coordinates:

q1 = 0.909x1 + 0.416x2 = 0.909(0.5) + 0.416(40) = 16.82q2 = -0.416x1 + 0.909x2 = -0.416(0.5) + 0.909(80) = 72.14.

(e) The free vibration can be determined using the following equations:u1(t) = A1 cos(w1t + φ1)u2(t) = A2 cos(w2t + φ2)To solve for A1, A2, φ1, and φ2, we must use the initial conditions in the principal coordinates.

The initial conditions for u1 are:u1(0) = A1 cos(0 + φ1) = 16.82u1'(0) = -100A1 sin(0 + φ1) = 0.

Therefore, φ1 = 0 and A1 = 16.82The initial conditions for u2 are:u2(0) = A2 cos(0 + φ2) = 72.14u2'(0) = -20A2 sin(0 + φ2) = 0.

Therefore, φ2 = 0 and A2 = 72.14Hence, the free vibration with respect to the principal coordinates is as follows:u1(t) = 16.82 cos(100t)u2(t) = 72.14 cos(20t).

Therefore, the normal mode shapes for each natural frequency were explained, the principal coordinates in terms of the generalized coordinates were determined, the equation of motion was determined with respect to the principal coordinates, the initial conditions with respect to the principal coordinates were determined, and the free vibration of the system with respect to the principal coordinates in response to the initial conditions listed in (d) were also determined.

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The following statements are true for all real numbers a, b, m, and n, where a > 0 and b > 0:

[tex](a^m)^n = a^{(mn)}[/tex]

[tex]a^m/a^n = a^{(m-n)}[/tex]

[tex](ab)^n = a^n * b^n[/tex]

The first statement, [tex](a^m)^n = a^{(mn)}[/tex], is true because when we raise a number to a power and then raise the result to another power, it is equivalent to raising the number to the product of the two powers.

The second statement, [tex]a^m/a^n = a^{(m-n)}[/tex], is also true. When we divide two numbers with the same base, we subtract the exponents. This property of exponents allows us to simplify division of exponential expressions.

The third statement, [tex](ab)^n = a^n * b^n,[/tex] is true because when we raise a product of two numbers to a power, we can distribute the power to each factor individually. This property is known as the power of a product rule.

The other statements in the given options are not true for all real numbers a, b, m, and n. For example, [tex](a^n)^m[/tex]is not equal to [tex]a^{(mn)}[/tex] in general, and[tex]a^m + a^n[/tex] is not equal to [tex]a^{(m+n)}[/tex] in general.

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To find the value of the derivative at x = 5, we need to differentiate the expression (6u - 2v + vu) with respect to x and then evaluate it at x = 5.

Let's differentiate the expression using the product rule:

d/dx (6u - 2v + vu) = 6u' - 2v' + v'u + uv'

We are given the values of u(5), u'(5), v(5), and v'(5) as follows:

u(5) = -1

u'(5) = -5

v(5) = 7

v'(5) = -6

Substituting these values into the derivative expression, we get:

d/dx (6u - 2v + vu) = 6(-5) - 2(-6) + (-5)(7) + (-1)(-6)

                  = -30 + 12 - 35 + 6

                  = -47

Therefore, the value of the derivative at x = 5 is -47. Hence, the correct option is (c) -47.

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The solution to the given linear system using Gaussian elimination (without partial pivoting) is as follows:

x₁,₁ = 6/7, x₁,₂ = 0, x₁,₃ = 3/7

x₂,₁ = 1, x₂,₂ = -5/3, x₂,₃ = -2/3

To solve the system using Gaussian elimination, we perform row operations on the augmented matrix until it is in row-echelon form or reduced row-echelon form. The augmented matrix is:

[  6   2  -9  |  -3  ]

[  9  -1   0  |   0  ]

[  6   3  -5  |   2  ]

First, we divide the first row by 6 to make the leading coefficient of the first column equal to 1:

[  1  1/3  -3/2  |  -1/2  ]

[  9  -1    0    |   0    ]

[  6   3   -5   |   2    ]

Next, we perform row operations to eliminate the entries below the leading coefficient of the first column. We subtract 9 times the first row from the second row, and we subtract 6 times the first row from the third row:

[  1  1/3  -3/2  |  -1/2  ]

[  0  -4   13/2  |   9/2  ]

[  0   1   -2    |   5    ]

Now, we divide the second row by -4 to make the leading coefficient of the second column equal to 1:

[  1  1/3  -3/2  |  -1/2  ]

[  0   1  -13/8  |  -9/8  ]

[  0   1   -2    |   5    ]

Finally, we perform row operations to eliminate the entry below the leading coefficient of the second column. We subtract the second row from the third row:

[  1  1/3  -3/2  |  -1/2  ]

[  0   1  -13/8  |  -9/8  ]

[  0   0   1     |   2    ]

The augmented matrix is now in row-echelon form. From the last row, we can directly determine x₃ = 2.

Substituting this value into the second row, we find x₂ = -9/8 - (-13/8)(2) = -5/3.

Substituting the values of x₂ and x₃ into the first row, we get x₁ = -1/2 - (1/3)(-3/2) + (3/2)(2) = 6/7.

Therefore, the solution to the linear system is:

x₁,₁ = 6/7, x₁,₂ = 0, x₁,₃ = 3/7

x₂,₁ = 1, x₂,₂ = -5/3, x₂,₃ = -2/3

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The given series Σ n=1 5(nl)² (2n)! is determined to diverge based on the ratio test.

To determine the convergence or divergence of the series, we can use the ratio test. According to the ratio test, for a series Σ aₙ, if the limit of the absolute value of the ratio of consecutive terms, lim (|aₙ₊₁ / aₙ|), as n approaches infinity, is greater than 1, the series diverges. If the limit is less than 1, the series converges. If the limit is equal to 1, the test is inconclusive.

Let's apply the ratio test to the given series Σ n=1 5(nl)² (2n)!. We calculate the ratio of consecutive terms:

|aₙ₊₁ / aₙ| = |[5(n+1)² (2(n+1))!] / [5(nl)² (2n)!]|

Simplifying the expression, we can cancel out common factors:

|aₙ₊₁ / aₙ| = |[5(n+1)² (2n+2)(2n+1)(2n)!] / [5(nl)² (2n)!]|

After canceling out terms, we are left with:

|aₙ₊₁ / aₙ| = |[5(n+1)² (2n+2)(2n+1)] / [5(nl)²]|

Simplifying further, we have:

|aₙ₊₁ / aₙ| = (n+1)² (2n+2)(2n+1) / n²

As n approaches infinity, the limit of this expression is infinity. Since the limit is greater than 1, we can conclude that the series Σ n=1 5(nl)² (2n)! diverges.

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(a) If the MacBook's value can be modeled linearly, we can use the equation of a line, y = mx + b, where y represents the value v of the MacBook, x represents the time t in years, m represents the slope, and b represents the initial value.

We can find the slope using the given information:

m = (500 - 2300) / (5 - 0) = -380

The initial value b is the value of the MacBook when t = 0, which is $2300.

Therefore, the linear equation modeling the value of the MacBook is:

v = -380t + 2300

To find its worth after 4 years (t = 4), substitute t = 4 into the equation:

v = -380(4) + 2300

v = 700

So, the MacBook is worth $700 after 4 years.

The slope of the graph at the 4-year mark is -380. In real-world terms, this slope value represents the rate of depreciation. It indicates that for every year that passes, the value of the MacBook decreases by $380.

(b) If the MacBook's value is modeled exponentially by v = a * b^t, we need to find the values of a and b using the given information. Unfortunately, the specific values of a and b are not provided in the question, so we cannot determine the exponential model without further information.

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Therefore, the gradient of f, denoted as Vf(x, y, z), is given by: Vf(x, y, z) = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k = (1/4√x)i + (z/2)j + (y/2)k. Vf(3, 1, 1) = (1/4√3)i + (1/2)j + (1/2)k. Duf(3, 1, 1) = Vf(3, 1, 1) · u = (1/4√3)a + (1/2)b + (1/2)c

(a) To find the gradient of the function f(x, y, z) = (√x + yz)/2, we need to calculate the partial derivatives with respect to each variable. Let's denote the partial derivative of f with respect to x as ∂f/∂x, the partial derivative with respect to y as ∂f/∂y, and the partial derivative with respect to z as ∂f/∂z.

∂f/∂x = (1/2)(1/2√x)

∂f/∂y = z/2

∂f/∂z = y/2

Therefore, the gradient of f, denoted as Vf(x, y, z), is given by:

Vf(x, y, z) = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

= (1/4√x)i + (z/2)j + (y/2)k

(b) To evaluate the gradient at the point P(3, 1, 1), we substitute the coordinates into the gradient expression:

Vf(3, 1, 1) = (1/4√3)i + (1/2)j + (1/2)k

(c) To find the rate of change of f at point P in the direction of the vector u, we compute the dot product of the gradient Vf(3, 1, 1) and the vector u:

Duf(3, 1, 1) = Vf(3, 1, 1) · u

= (1/4√3)a + (1/2)b + (1/2)c

The rate of change of f at point P in the direction of the vector u is given by the dot product expression above, where a, b, and c represent the components of the vector u.

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The general solution is given by:

[tex]X(t) = e^(At) * X(0)[/tex] where X(0) is the initial condition vector.

To compute the matrix exponential [tex]e^(At)[/tex] for the given coefficient matrix A, we can use the formula:

[tex]e^(At) = I + At + (At)^2/2! + (At)^3/3! + ...[/tex]

where I is the identity matrix, t is the variable of integration, and A is the coefficient matrix.

Given the coefficient matrix A = [[9, 8], [-9, -9]], we can substitute it into the formula:

[tex]e^(At) = I + At + (At)^2/2! + (At)^3/3! + ...[/tex]

To simplify the calculation, we can find the powers of the matrix A:

[tex]A^2 = A * A = [[9, 8], [-9, -9]] * [[9, 8], [-9, -9]] = [[81-72, 72+72], [-81-81, -72-[/tex]81]] = [[9, 16], [-162, -153]]

[tex]A^3 = A * A^2 = [[9, 8], [-9, -9]] * [[9, 16], [-162, -153]] = ...[/tex]

Continuing this process, we can compute higher powers of A.

Once we have computed the powers of A, we can substitute them into the matrix exponential formula to find [tex]e^(At)[/tex]

The general solution of the given system, X' = AX, can be found by solving the system of linear differential equations using the matrix exponential. The general solution is given by:

[tex]X(t) = e^(At) * X(0)[/tex]

where X(0) is the initial condition vector.

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Solving the integral, we get:`A = ∫_(-2)^(0) (3x^2 - 2) dx`=`[x^3 - 2x]_(-2)^(0)`=`0 - 0 - [-8 + 4]`=`4`.Hence, the area between the curve g(x)=3x2 2 and the x-axis from x=−2 to x=0 is 4 square units.

The given curve is `g(x)

=3x^2 - 2`. We need to find the area between the curve g(x)

=3x2 2 and the x-axis from x

=−2 to x

=0.Area between curve and x-axisFor a curve `y

= f(x)`, the area between the curve and the x-axis in the interval `[a,b]` is given by: `A

= ∫_(a)^(b) f(x) dx`Here, the curve is `y

= g(x)

= 3x^2 - 2` and the limits of integration are `a

= -2` and `b

= 0`. So, we have to find the integral of the curve from `-2` to `0` as below: `∫_(-2)^(0) (3x^2 - 2) dx`.Solving the integral, we get:`A

= ∫_(-2)^(0) (3x^2 - 2) dx`=`[x^3 - 2x]_(-2)^(0)`

=`0 - 0 - [-8 + 4]`

=`4`.Hence, the area between the curve g(x)

=3x2 2 and the x-axis from x

=−2 to x

=0 is 4 square units.

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The correct answer using a calculator or MATLAB, we find that t is approximately 12.6 years.Therefore, the correct answer is (D) 12.6 years.

To determine how long it will take for a population to double with a growth rate of 5.5% per year, we can use the formula for exponential growth:

N = N0 * [tex](1 + r)^t,[/tex]

where:

N0 is the initial population size,

N is the final population size (double the initial population size),

r is the growth rate as a decimal,

t is the time in years.

In this case, we want to find t when N = 2N0, so we can write:

2N0 = N0 * (1 + 0.055)^t.

Simplifying the equation, we have:

[tex]2 = (1 + 0.055)^t.[/tex]

Taking the natural logarithm (ln) of both sides, we get:

ln(2) = [tex]ln((1 + 0.055)^t).[/tex]

We can reduce the exponent by using the logarithm property: ln(2) = t * ln(1 + 0.055).

By dividing both sides by ln(1 + 0.055), we can finally find the value of t: t = ln(2) / ln(1 + 0.055).

Using a calculator or MATLAB to calculate this, we get that t is roughly 12.6 years.

Therefore, (D) 12.6 years is the right response.

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cos(α - β) when sinα = 3/4 for α in Quadrant II and cosβ = -2/5 for β in Quadrant III cos(α - β) = -17/20. In Quadrant II, sinα is positive and cosα is negative. In Quadrant III, cosβ is negative and sinβ is positive.

We can use the following formulas to find cos(α - β):

cos(α - β) = cosαcosβ + sinαsinβ

sin(α - β) = sinαcosβ - cosαsinβ

Substituting the values of sinα, cosα, sinβ, and cosβ, we get:

cos(α - β) = (3/4)(-2/5) + (3/4)(3/5) = -17/20

sin(α + β) when cosα = 3/7 for α in Quadrant IV and sinβ = 3/5 for β in Quadrant II sin(α + β) = 18/25.

In Quadrant IV, cosα is negative and sinα is positive. In Quadrant II, sinβ is positive and cosβ is negative.

We can use the following formulas to find sin(α + β):

sin(α + β) = sinαcosβ + cosαsinβ

cos(α + β) = cosαcosβ - sinαsinβ

Substituting the values of sinα, cosα, sinβ, and cosβ, we get:

sin(α + β) = (3/7)(3/5) + (-3/7)(3/5) = 18/25 by substituting the given values and evaluating the expressions using the trigonometric identities, we can determine the exact values of cos(α-β) and sin(α+β) in their simplest form.

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the location of the neutral axis is at a distance of 125 mm from the upper surface and at a distance of 125 mm from the lower surface of the beam cross-section.

A cantilever beam has a length of 15 meters and a circular cross-section with a diameter (d) of 250 mm.

It carries a transverse end-point load (P1=25 kN) and a compressive load (P2=1500 MN) on its free end that is applied through its centroid.

The maximum longitudinal direct stresses due to transverse concentrated load are calculated below:Firstly, the general equation for bending is given as:M / I = σ / y

Where,M = bending momentI = moment of inertiaσ = longitudinal direct stressy = distance from the neutral axis.The maximum bending moment in a cantilever beam with a point load is given as:M = P1 L...[1]

where,P1 = transverse end-point loadL = length of the beam.

The moment of inertia of a circular cross-section is given as:I = πd4 / 64...[2]where,d = diameter of the cross-section.

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