Verify the identity sin x - 2+ sinx sin X- sin X-1 sin x + 1 sinx Multiply the numerator and denominator by sin x and simplify. Then factor the expression in the numerator and the expression in the co

The equation of the tangent line to the curve y = 3 sin x + cos x at x = π/2 is y = -x + π/2 + 3.

To find the equation of the tangent line to the curve y = 3 sin x + cos x at x = π/2, we need to determine the slope of the tangent line at that point and use the point-slope form of a linear equation.

First, let's find the derivative of the given function y = 3 sin x + cos x with respect to x:

dy/dx = d/dx (3 sin x + cos x)

= 3 d/dx (sin x) + d/dx (cos x)

= 3 cos x - sin x

Now, we can evaluate the derivative at x = π/2 to find the slope of the tangent line:

m = dy/dx | x=π/2

= 3 cos (π/2) - sin (π/2)

= 0 - 1

= -1

The slope of the tangent line is -1.

Next, we use the point-slope form of a linear equation, where (x1, y1) is the point on the curve:

y - y1 = m(x - x1)

Substituting x1 = π/2 and y1 = 3 sin (π/2) + cos (π/2) = 3 + 0 = 3, we have:

y - 3 = -1(x - π/2)

Simplifying, we get:

y - 3 = -x + π/2

y = -x + π/2 + 3

Therefore, the equation of the tangent line to the curve y = 3 sin x + cos x at x = π/2 is y = -x + π/2 + 3.

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To calculate the surface integral of the vector field F(x, y, z) = (cos(2x), e^(-y), vy) over the boundary surface Q of the cube [0, 1], we need to parametrize the surface and then evaluate the dot product of the vector field and the surface normal vector.

The boundary surface Q of the cube [0, 1] consists of six square faces. To compute the surface integral, we need to parametrize each face and calculate the dot product of the vector field F and the surface normal vector. Let's consider one face of the cube, for example, the face with the equation x = 1. Parametrize this face by setting x = 1, and let the parameters be y and z. The parametric equations for this face are (1, y, z), where y and z both vary from 0 to 1.

Now, we can calculate the surface normal vector for this face, which is the unit vector in the x-direction: n = (1, 0, 0). The dot product of the vector field F(x, y, z) = (cos(2x), e^(-y), vy) and the surface normal vector n = (1, 0, 0) is F • n = cos(2) * 1 + e^(-y) * 0 + vy * 0 = cos(2).

To find the surface integral over the entire boundary surface Q, we need to calculate the surface integral for each face of the cube and sum them up. In summary, the surface integral of the vector field F(x, y, z) = (cos(2x), e^(-y), vy) over the boundary surface Q of the cube [0, 1] is given by the sum of the dot products of the vector field and the surface normal vectors for each face of the cube. The specific values of the dot products depend on the orientation and parametrization of each face.

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To show that there is at least one real solution for the equation 2sin(x) - 1 = cos(x), we can use the Intermediate Value Theorem (IVT).

Let's define a function f(x) = 2sin(x) - 1 - cos(x). We want to show that there exists a value c in the real numbers such that f(c) = 0.

First, we need to find two values a and b such that f(a) and f(b) have opposite signs. This will guarantee the existence of a root according to the IVT.

Let's evaluate f(x) at a = 0 and b = π/2:

f(0) = 2sin(0) - 1 - cos(0) = -1 - 1 = -2

f(π/2) = 2sin(π/2) - 1 - cos(π/2) = 2 - 1 = 1

Since f(0) = -2 < 0 and f(π/2) = 1 > 0, we have f(a) < 0 and f(b) > 0, respectively.

Now, since f(x) is continuous between a = 0 and b = π/2 (since sine and cosine are continuous functions), the IVT guarantees that there exists at least one value c in the interval (0, π/2) such that f(c) = 0.

Therefore, the equation 2sin(x) - 1 = cos(x) has at least one real solution in the interval (0, π/2).

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The Probability of selecting a girl at random from the class is 8/15, and the probability of selecting a boy is 7/15.

In a 3rd ESO (Educación Secundaria Obligatoria) class, there are 16 girls and 14 boys. If a person is chosen at random from the class, there is a chance that the chosen person could be a girl or a boy.

To calculate the probability of selecting a girl, we divide the number of girls by the total number of students in the class:

Probability of selecting a girl = Number of girls / Total number of students

Probability of selecting a girl = 16 / (16 + 14)

Probability of selecting a girl = 16 / 30

Probability of selecting a girl = 8/15

Similarly, to calculate the probability of selecting a boy, we divide the number of boys by the total number of students in the class:

Probability of selecting a boy = Number of boys / Total number of students

Probability of selecting a boy = 14 / (16 + 14)

Probability of selecting a boy = 14 / 30

Probability of selecting a boy = 7/15

Therefore, the probability of selecting a girl at random from the class is 8/15, and the probability of selecting a boy is 7/15.

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To find the velocity and speed at a specific time t, substitute the value of t into the derived velocity and speed functions.

To find the velocity and speed of a particle moving along a straight line with the equation of motion f(t), we need to differentiate the function f(t) to obtain the velocity function and then take the absolute value to obtain the speed. Velocity: The velocity of the particle is given by the derivative of the position function f(t) with respect to time t. Let's denote the velocity as v(t).

v(t) = f'(t)

Differentiate the function f(t) according to the given equation of motion to find v(t).

Speed: The speed of the particle is the absolute value of the velocity function. Let's denote the speed as s(t).

s(t) = |v(t)|

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To find the volume of the solid obtained by rotating the region bounded by the curves x+y=4 and x=5-(y-1)^2 about the x-axis, we will use the washer method.
First, rewrite the equations to solve for y:
y = 4 - x and y = 1 + sqrt(5 - x)
The bounds of integration can be found by setting the two equations equal to each other and solving for x:
4 - x = 1 + sqrt(5 - x)
x = 2
Now, we'll set up the integral using the washer method formula:
Volume = π * ∫[0 to 2] [(4 - x)^2 - (1 + sqrt(5 - x))^2] dx
Evaluate the integral to find the volume of the solid:
Volume ≈ 5.333π cubic units

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To determine the convergence or divergence of the series using the Ratio Test, we need to evaluate the limit of the ratio of consecutive terms as n approaches infinity.

Using the formula given, we have:
an+1 = (3n+1)/(n³+1)
an = (3n-2)/(n³+1)
So, we can write the ratio of consecutive terms as:
an+1/an = [(3n+1)/(n³+1)] / [(3n-2)/(n³+1)]
an+1/an = (3n+1)/(3n-2)
Now, taking the limit of this expression as n approaches infinity: lim (n→∞) [(3n+1)/(3n-2)] = 3/3 = 1

Since the limit is equal to 1, the Ratio Test is inconclusive. Therefore, we need to use another test to determine the convergence or divergence of the series. However, we can observe that the series has the same terms as the series ∑1/n² which is a convergent p-series with p=2. Therefore, by the Comparison Test, we can conclude that the series ∑(3n-2)/(n³+1) also converges. In summary, the series ∑(3n2)/(n³+1) converges.

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(a) To find the critical numbers, we need to find the values of x where the derivative of the function is equal to zero or undefined. Taking the derivative of y with respect to x:

dy/dx = -2x + 3

-2x + 3 = 0

-2x = -3

x = 3/2

Thus, the critical number is x = 3/2.

(b) To determine the intervals on which the function is increasing or decreasing.

When x < 3/2, dy/dx is negative since -2x < 0. This means that y is decreasing on this interval.

When x > 3/2, dy/dx is positive since -2x + 3 > 0. This means that y is increasing on this interval. Therefore, the function is decreasing on (-∞, 3/2) and increasing on (3/2, ∞).

(c) To graph the function, plot the critical number at x = 3/2. We know that the vertex of the parabola will lie at this point since it is the only critical number. To find the y-coordinate of the vertex, we can plug in x = 3/2 into the original equation:

y = -(3/2)² + 3(3/2)

y = -9/4 + 9/2

y = 9/4

So the vertex is at (3/2, 9/4).

We can also find the y-intercept by setting x = 0:

y = -(0)² + 3(0)

y = 0

So the y-intercept is at (0, 0).

To plot more points, we can choose some values of x on either side of the vertex. For example, when x = 1, y = -1/2, and when x = 2, y = -2.

The graph of the function y = -x² + 3x looks like a downward-facing parabola that opens up, with its vertex at (3/2, 9/4). It intersects the x-axis at x = 0 and x = 3, and the y-axis at y = 0.

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To maximize the product ryz subject to the constraint 2 + y + 2^{2} = 16, we can use Lagrange multipliers. The maximum value of the product ryz can be found by solving the system of equations formed by the Lagrange multipliers method.

We want to maximize the product ryz, which is our objective function, subject to the constraint 2 + y + 2^{2} = 16. To apply Lagrange multipliers, we introduce a Lagrange multiplier λ and set up the following equations:

∂(ryz)/∂r = λ∂(2 + y + 2^{2} - 16)/∂r

∂(ryz)/∂y = λ∂(2 + y + 2^{2} - 16)/∂y

∂(ryz)/∂z = λ∂(2 + y + 2^{2} - 16)/∂z

2 + y + 2^{2} - 16 = 0

Differentiating the objective function ryz with respect to each variable (r, y, z) and setting them equal to the corresponding partial derivatives of the constraint, we form a system of equations. The fourth equation represents the constraint itself.

Solving this system of equations will yield the values of r, y, z, and λ that maximize the product ryz subject to the given constraint. Once these values are determined, the maximum value of the product ryz can be computed.

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The integral of cos(x) dx from 0 to 65 is 0. This is because the integral of cos(x) over a full period (2π) is 0, and since 65 is a multiple of 2π, the integral evaluates to 0.

The function cos(x) has a periodicity of 2π, meaning that it repeats itself every 2π units. The integral of cos(x) over a full period (from 0 to 2π) is 0. Therefore, if the interval of integration is a multiple of 2π, like in this case where it is 65, the integral will also evaluate to 0. This is because the function completes several cycles within that interval, canceling out the positive and negative areas and resulting in a net value of 0.

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To find the marginal productivity with fixed capital, we need to calculate the partial derivative of the production function with respect to x (holding y constant). The correct answer would be option OB. 15.4xy.

Given the production function [tex]p(x, y) = 22x^0.7 * y^0.3[/tex], we differentiate it with respect to x:

[tex]∂p/∂x = 0.7 * 22 * x^(0.7 - 1) * y^0.3[/tex]

Simplifying this expression, we have:

[tex]∂p/∂x = 15.4 * x^(-0.3) * y^0.3[/tex]

Therefore, the marginal productivity with fixed capital, partial p partial x, is given by [tex]15.4 * x^(-0.3) * y^0.3.[/tex]

The correct answer would be option OB. 15.4xy.

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The volume of the solid of revolution S, obtained by revolving the region R enclosed by the curve y = x²(2-x) and the x-axis about the x-axis, can be computed using the method of cylindrical shells.

To find the volume of S, we can use the method of cylindrical shells. Consider an infinitesimally small vertical strip within the region R, located at a distance x from the y-axis. The height of this strip will be given by the function y = x²(2-x), and its width will be dx. By revolving this strip about the x-axis, we obtain a cylindrical shell with radius x and height y. The volume of each cylindrical shell is given by V = 2πxydx.

To calculate the total volume of S, we need to integrate the volumes of all the cylindrical shells. The integral can be set up as follows:

V = ∫(2πxy)dx

To determine the limits of integration, we need to find the x-values where the curve intersects the x-axis. Setting y = 0, we solve the equation x²(2-x) = 0, which yields x = 0 and x = 2.

Thus, the integral becomes:

V = ∫[0,2] (2πx * x²(2-x))dx

Evaluating this integral will give us the volume of the solid of revolution S.

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The correct option is (C) 5 m, which represents the radius of the circle.

The motion of the particle is described by the equations:

x = 5 cos(3t)

y = 5 sin(3t)

These equations represent the parametric equations of a circle centered at the origin. The general equation of a circle centered at (h, k) with radius r is:

(x - h)^2 + (y - k)^2 = r^2

Comparing this equation with the given equations, we can see that the center of the circle is at the origin (0, 0) since there are no terms involving (x - h) or (y - k). We need to determine the radius of the circle, which corresponds to the value of r.

From the equations x = 5 cos(3t) and y = 5 sin(3t), we can use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to rewrite them:

(x/5)^2 + (y/5)^2 = cos^2(3t) + sin^2(3t) = 1

This equation shows that the sum of the squares of the x-coordinate and y-coordinate is equal to 1, which is the equation of a unit circle. Therefore, the radius of the circle is 5.

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The convergence or divergence of a series is not provided, so it cannot be determined without knowing the specific series.

In order to determine whether a series is convergent or divergent, we need to know the terms of the series. The convergence or divergence of a series depends on the behavior of its terms as the series progresses. Different series have different convergence or divergence tests that can be applied to them.

Some common convergence tests for series include the comparison test, the ratio test, the root test, and the integral test, among others. These tests help determine whether the series converges or diverges based on the properties of the terms.

Without knowing the specific series or having any information about its terms, it is not possible to determine whether the series is convergent or divergent. Each series must be evaluated individually using the appropriate convergence test to reach a conclusion about its behavior.

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The answer are:

(a) The integrating value of ∫cos(ln x)dx=x sin(ln x)+C.

(b) The integrating value of ∫sin(x)dx=−cos(x)+C.

(c) The integrating value of [tex]\int\limits sin(2x)e^xdx=-\frac{1}{2} cos(2x)e^x+\frac{1}{2}\int\limits cos(2x)e^xdx.[/tex]

(d) The integrating value of [tex]\int\limits {e^{sec^2}}(x)tan(x)dx[/tex]  cannot be expressed the integral in elementary functions.

What is the integral function?

The integral function, often denoted as ∫f(x)dx, is a fundamental concept in calculus. It represents the antiderivative or the indefinite integral of a given function f(x) with respect to the variable x.

The integral function measures the accumulation of the function f(x) over a given interval. It is the reverse process of differentiation, where the derivative of a function measures its rate of change. The integral function, on the other hand, measures the accumulated change or the total area under the curve of the function.

To evaluate the given integrals one by one:

(a)∫cos(ln x)dx:

To evaluate this integral, we can use the substitution method. Let u=lnx, then [tex]du=\frac{1}{x}dx[/tex]  or  dx=x du.

Substituting into the integral:

∫cos(u)⋅x du=∫x cos(u)du. Now, we can integrate cos(u) with respect to u:

∫ x cos(u)du=x sin(u)+C.

Substituting back u=ln x, we have:

∫cos(ln x)dx=x sin(ln x)+C.

(b)∫sin(x)dx:

The integral of sin(x) is −cos(x)+C, where C is the constant of integration. So, ∫sin(x)dx=−cos(x)+C.

(c)[tex]\int\limits sin(2x)e^xdx[/tex]:

To integrate this expression, we can use integration by parts. Let's assign u=sin(2x) and[tex]dv=e^xdx.[/tex] Then, we can find du and v as follows: du=2cos(2x)dx (by differentiating u), [tex]v=e^x[/tex] (by integrating dv).Now, we can apply the integration by parts formula:

∫u dv=u v−∫v du.

Using the above values, we have:

[tex]\int\limits sin(2x)e^xdx=-\frac{1}{2} cos(2x)e^x+\frac{1}{2}\int\limits cos(2x)e^xdx.[/tex]

Integrating [tex]cos(2x)e^x[/tex] requires further steps and cannot be expressed in terms of elementary functions.

(d)[tex]\int\limits {e^{sec^2}}(x)tan(x)dx[/tex]:

This integral does not have a standard elementary function as its antiderivative. It cannot be  expressed  the integral in terms of elementary functions like polynomials, exponentials, logarithms, trigonometric functions, etc. Therefore, it cannot be evaluated using standard methods and requires advanced techniques or numerical approximations for an accurate result.

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The line integral becomes:

∫∂M w ⋅ dr = ∫(θ=0)(2π) [z(cosθ)d(cosθ) + (x + y + z)d(3) - x d(sinθ)]

A statement regarding the integration of differential forms on manifolds, known as Stokes Theorem (also known as Generalised Stoke's Theorem), generalises and simplifies a number of vector calculus theorems. This theorem states that a line integral and a vector field's surface integral are connected.

To verify the generalized Stokes's theorem for the given surface M and vector field w, we need to evaluate both the surface integral of the curl of w over M and the line integral of w around the boundary curve of M. If these two values are equal, the theorem is verified.

First, let's calculate the curl of the vector field w:

curl(w) = (∂/∂x, ∂/∂y, ∂/∂z) x (z, x + y + z, -x)

       = (1, -1, 1)

Next, we evaluate the surface integral of the curl of w over M. The surface M is the portion of the cylinder x² + z² = 1 where y < 3. Since M is a cylinder, we can use cylindrical coordinates (ρ, θ, z) to parameterize the surface.

The parameterization can be defined as:

r(ρ, θ) = (ρcosθ, ρsinθ, z), where 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, and -∞ < z < 3.

To calculate the surface integral, we need to compute the dot product between the curl of w and the unit normal vector of M at each point on the surface, and then integrate over the parameter domain.

N = (x, 0, z)/√(x² + z²) = (ρcosθ, 0, ρsinθ)/ρ = (cosθ, 0, sinθ)

The surface integral becomes:

∬_M (curl(w) ⋅ N) dS = ∬_M (1cosθ - 1⋅0 + 1sinθ) ρ dρ dθ

Integrating over the parameter domain, we have:

∬_M (curl(w) ⋅ N) dS = ∫_(θ=0)(2π) ∫_(ρ=0)^(1) (cosθ - sinθ) ρ dρ dθ

Evaluating this double integral will yield the surface integral of the curl of w over M.

Next, we need to calculate the line integral of w around the boundary curve of M. The boundary curve of M is the intersection of the cylinder x² + z² = 1 and the plane y = 3. This is a circle of radius 1 in the xz-plane centered at the origin.

To parameterize the boundary curve, we can use polar coordinates θ. Let's denote the parameterization as γ(θ) = (cosθ, 3, sinθ), where 0 ≤ θ ≤ 2π.

The line integral becomes:

∫∂M w ⋅ dr = ∫_(θ=0)(2π) [z(cosθ)d(cosθ) + (x + y + z)d(3) - x d(sinθ)]

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the absolute maximum of the function f(x) = 2x^3 – 6x^2 – 18x over the interval 1 < x < 4 is 10.

To find the absolute extremes of the function f(x) = 2x^3 – 6x^2 – 18x over the interval 1 < x < 4, we need to evaluate the function at the critical points and the endpoints of the interval.

Step 1: Find the critical points by taking the derivative of f(x) and setting it equal to zero:

f'(x) = 6x^2 - 12x - 18

Setting f'(x) = 0 and solving for x:

6x^2 - 12x - 18 = 0

Dividing the equation by 6:

x^2 - 2x - 3 = 0

Factoring the quadratic equation:

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

Setting each factor equal to zero:

x - 3 = 0 --> x = 3

x + 1 = 0 --> x = -1

So the critical points are x = -1 and x = 3.

Step 2: Evaluate the function at the critical points and the endpoints of the interval:

f(1) = 2(1)^3 - 6(1)^2 - 18(1) = 2 - 6 - 18 = -22

f(4) = 2(4)^3 - 6(4)^2 - 18(4) = 128 - 96 - 72 = -40

f(-1) = 2(-1)^3 - 6(-1)^2 - 18(-1) = -2 - 6 + 18 = 10

f(3) = 2(3)^3 - 6(3)^2 - 18(3) = 54 - 54 - 54 = -54

Step 3: Compare the values obtained to determine the absolute maximum and minimum:

The values are as follows:

f(1) = -22

f(4) = -40

f(-1) = 10

f(3) = -54

The absolute maximum occurs at x = -1, and the maximum value is f(-1) = 10.

Therefore, the absolute maximum of the function f(x) = 2x^3 – 6x^2 – 18x over the interval 1 < x < 4 is 10.

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The local maximum is at x = -4 and the local minimum is at x = 7 for the function f(x) = 2x³ - 9x² - 168x + 13.

The local maximum and local minimum of the function f(x) = 2x³ - 9x² - 168x + 13 can be determined using the First Derivative Test.

To find the critical points, we need to find where the first derivative of the function is equal to zero or does not exist.

First, let's find the first derivative of f(x). Taking the derivative of each term, we have f'(x) = 6x² - 18x - 168.

Next, we set f'(x) equal to zero and solve for x: 6x² - 18x - 168 = 0. Factoring out a common factor of 6, we get 6(x² - 3x - 28) = 0. Further factoring, we have 6(x - 7)(x + 4) = 0. Therefore, the critical points are x = 7 and x = -4.

Now, let's evaluate the sign of f'(x) in the intervals created by the critical points.

For x < -4, we choose x = -5. Substituting into f'(x), we have f'(-5) = 6(-5)^2 - 18(-5) - 168 = 90 + 90 - 168 = 12. Since f'(-5) > 0, this interval is positive.

For -4 < x < 7, we choose x = 0. Substituting into f'(x), we have f'(0) = 6(0)² - 18(0) - 168 = -168. Since f'(0) < 0, this interval is negative.

For x > 7, we choose x = 8. Substituting into f'(x), we have f'(8) = 6(8)² - 18(8) - 168 = 384 - 144 - 168 = 72. Since f'(8) > 0, this interval is positive.

Based on the First Derivative Test, we can conclude that the function has a local minimum at x = 7 and a local maximum at x = -4.

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Given the properties of the function f(x) where f(1) = 2 and f(2) = 8, and the integral of ef(x) dx from 1 to 4 is equal to -60, we need to evaluate the integral of 62e*f(a) dx from 1 to 4. The value of the integral is -1860.

To evaluate the integral of 62ef(a) dx from 1 to 4, we can start by using the properties of the function f(x). We are given that f(1) = 2 and f(2) = 8. Using these values, we can find the function f(x) by interpolating between the two points. One possible interpolation is a linear function, where f(x) = 3x - 4.

Now, we have to evaluate the integral of 62ef(a) dx from 1 to 4. Substituting the function f(x) into the integral, we have 62e(3a - 4) dx. Integrating this expression with respect to x gives us 62e(3a - 4)x. To evaluate the definite integral from 1 to 4, we substitute the limits of integration into the expression and calculate the difference between the upper and lower limits.

Plugging in the limits, we get [62e(3a - 4)] evaluated from 1 to 4. Evaluating at x = 4 gives us 62e(34 - 4) = 62e8. Evaluating at x = 1 gives us 62e*(31 - 4) = 62e*(-1). Taking the difference between these two values, we have 62e8 - 62e(-1) = 62e(8 + 1) = 62e9.

The final result of the integral is 62e9.

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The sum of the series E (sigma symbol; infinity sign on top and k=1 on bottom) 5 * (9/10)^k is 50, while the sum of the series E (sigma symbol; infinity sign on top and k=1 on bottom) 6 / (k^2 + 2k) cannot be determined without additional techniques from calculus.

a) The sum of the infinite series given by E (sigma symbol; infinity sign on top and k=1 on bottom) 5 * (9/10)^k is 50. This means that the series converges to a finite value of 50 as the number of terms approaches infinity.

To calculate the sum, we can use the formula for the sum of a geometric series: S = a / (1 - r), where 'a' is the first term and 'r' is the common ratio. In this case, the first term 'a' is 5 and the common ratio 'r' is 9/10.

Plugging in the values, we get S = 5 / (1 - 9/10) = 5 / (1/10) = 50. Therefore, the sum of the given series is 50.

b) The sum of the infinite series given by E (sigma symbol; infinity sign on top and k=1 on bottom) 6 / (k^2 + 2k) cannot be determined using simple algebraic techniques. This series represents a type of series known as a "partial fractions" series, which involves breaking down the expression into a sum of simpler fractions.

To find the sum of this series, one would need to apply techniques from calculus, such as integration. By using methods like telescoping series or the method of residues, it is possible to evaluate the sum. However, without further information or specific techniques, it is not possible to provide an exact value for the sum of this series.

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[tex]lim_{(x --> 0)} f(x) = 1[/tex]. It is proved that (1) is the principal value of the nth root of i.

Given the function [tex]f(x) = (1^{1/n})/x[/tex].

We are to show that [tex]lim_{(x --> 0)} f(x) = 1[/tex], where 1 is the principal value of the nth root of i.

Formula used: The principal value of the `n`th root of i is [tex]cos ((\pi)/(2n)) + i sin ((\pi)/(2n))[/tex].

Since f(x) = [tex](1^{1/n})/x[/tex], we can simplify f(x) as follows: f(x) = [tex]1/x^{(1/n)}[/tex].

As x approaches 0, f(x) becomes f(0) = [tex]1^{(1/n)}/0[/tex].

Here, we assume that `n` is even, so that n = 2m.

Substituting n with 2m, we have [tex]f(0) = (cos((\pi)/(2n)) + i sin((\pi)/(2n)))^{(1/2m)}[/tex].

This is the principal value of the nth root of i, which is equal to `1`.

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1. The limit as x approaches 7 of (3x + 1)/(11x - 7) is 2/11.

2. The limit as x approaches 6 of (1/(x^2 + 16)) + 63 is 63.

3. The limit as x approaches 9 of (x + 9)/(x - 9) does not exist.

4. The derivative of g(t) = t - 11 is 1.

5. The derivative of y = 14x - 1 is 14.

6. The derivative of y = ln(x - 7) is 1/(x - 7).

7. The equation of the tangent line to the curve x^2 + 3y^2 = 13 at the point (1, 2) is 2x + 3y = 8.

1. To find the limit, substitute x = 7 into the expression (3x + 1)/(11x - 7), which simplifies to 2/11.

2. Substituting x = 6 into the expression (1/(x^2 + 16)) + 63 gives 63.

3. When x approaches 9, the expression (x + 9)/(x - 9) becomes undefined because it leads to division by zero.

4. The derivative of g(t) is found by taking the derivative of each term, resulting in 1.

5. The derivative of y = 14x - 1 is calculated by taking the derivative of the term with respect to x, which is 14.

6. The derivative of y = ln(x - 7) is found using the chain rule, which states that the derivative of ln(u) is 1/u times the derivative of u. In this case, the derivative is 1/(x - 7).

7. To find the equation of the tangent line at the point (1, 2) on the curve x^2 + 3y^2 = 13, we differentiate implicitly to find the derivative dy/dx. Then we substitute the values of x and y from the given point to find the slope of the tangent line. Finally, we use the point-slope form of a line to write the equation of the tangent line as 2x + 3y = 8.

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The angle, to the nearest degree, between the two vectors a = (-2,3,4) and b = (2,1,2) is approximately 67 degrees.

To find the angle between two vectors, we can use the dot product formula and the magnitude (length) of the vectors. The dot product of two vectors a and b is defined as:

a · b = |a| |b| cos θ

where |a| and |b| are the magnitudes of vectors a and b, respectively, and θ is the angle between them.

First, let's calculate the magnitudes of vectors a and b:

|a| = sqrt((-2)^2 + 3^2 + 4^2) = sqrt(4 + 9 + 16) = sqrt(29)

|b| = sqrt(2^2 + 1^2 + 2^2) = sqrt(4 + 1 + 4) = sqrt(9) = 3

Next, let's calculate the dot product of a and b:

a · b = (-2)(2) + (3)(1) + (4)(2) = -4 + 3 + 8 = 7

Now, we can substitute the values into the dot product formula:

7 = sqrt(29) × 3 × cos θ

To isolate cos θ, we divide both sides of the equation by sqrt(29) × 3:

cos θ = 7 / (sqrt(29) × 3)

Using a calculator, we find:

cos θ ≈ 0.376

Now, we can find the angle θ by taking the inverse cosine (arccos) of 0.376:

θ ≈ arccos(0.376) ≈ 67 degrees

Therefore, the angle, to the nearest degree, between vectors a = (-2, 3, 4) and b = (2, 1, 2) is approximately 67 degrees.

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The function y = 2sin(3x-π/3) represents a sinusoidal function. The horizontal shift and period can be determined from the equation. The horizontal shift is π/9 units to the right, and the period is 2π/3 units. The complete graph for one period can be shown in the interval [π/9, π/9 + 2π/3] for x and [−2, 2] for y.

For the function y = 2sin(3x-π/3), the coefficient inside the sine function, 3, affects the period of the graph. The period can be calculated using the formula T = 2π/b, where b is the coefficient of x. In this case, b = 3, so the period is T = 2π/3.

The horizontal shift can be determined by setting the argument of the sine function, 3x-π/3, equal to zero and solving for x. We have:

3x - π/3 = 0

3x = π/3

x = π/9

Therefore, the graph is shifted π/9 units to the right.

To determine the interval on x for one period, we can use the horizontal shift and period. The interval on x for one period is [π/9, π/9 + 2π/3].

For the interval on y, we consider the amplitude, which is 2. The graph will oscillate between -2 and 2. Thus, the interval on y for one period is [-2, 2].

Therefore, the function y = 2sin(3x-π/3) has a horizontal shift of π/9 units to the right, a period of 2π/3 units, and the complete graph for one period can be shown in the interval [π/9, π/9 + 2π/3] for x and [-2, 2] for y.

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Let us first find the domain of the function f(x) = (5-x)/(x^2-16). It is clear that x ≠ -4 and x ≠ 4. Therefore, the domain of f(x) is (−∞,−4)∪(−4,4)∪(4,∞).f(x) can be expressed as f(x) =  A/(x-4) + B/(x+4), where A and B are constants. Let us find the values of A and B. We obtainA/(x-4) + B/(x+4) = (5-x)/(x^2-16).

Multiplying through by (x - 4)(x + 4) yieldsA(x+4) + B(x-4) = 5 - x.

If we substitute x = -4, we get 9A = 1. So, A = 1/9. If we substitute x = 4, we get −9B = 1.

So, B = -1/9.

Hence,f(x) = (1/9)/(x-4) - (1/9)/(x+4).

Now, we havef′(x) = (-1/81) * (1/(x-4)^2) + (1/81) * (1/(x+4)^2).

Since f′(x) is defined and continuous on (−∞,-4)∪(-4,4)∪(4,∞), the critical numbers are given by f′(x) = 0 = (-1/81) * (1/(x-4)^2) + (1/81) * (1/(x+4)^2).Multiplying through by (x - 4)^2(x + 4)^2 gives us- (x + 4)^2 + (x - 4)^2 = 0.

Simplifying this expression gives usx^2 - 20x + 12 = 0.

Solving for x gives usx = 10 + sqrt(88) / 2 or x = 10 - sqrt(88) / 2.

The critical numbers are therefore10 + sqrt(88) / 2 and 10 - sqrt(88) / 2.

The function is defined on the domain (−∞,-4)∪(-4,4)∪(4,∞) and is continuous there.

The values of f′(x) change from negative to positive as x increases from 10 - sqrt(88) / 2 to 10 + sqrt(88) / 2. Therefore, f(x) has a local minimum at x = 10 - sqrt(88) / 2 and a local maximum at x = 10 + sqrt(88) / 2.b)  g(x) = -2 + x^2e^(-.3x).

Let us first find the first derivative of the functiong(x) = -2 + x^2e^(-.3x).We haveg′(x) = 2xe^(-.3x) - .3x^2e^(-.3x).

The critical numbers are given by settingg′(x) = 0 = 2xe^(-.3x) - .3x^2e^(-.3x), which gives usx = 0 or x = 20/3.Let us examine the values of g′(x) to the left of 0, between 0 and 20/3, and to the right of 20/3.

For x < 0, g′(x) < 0. For x ∈ (0,20/3), g′(x) > 0. For x > 20/3, g′(x) < 0.

Therefore, g(x) has a local maximum at x = 0 and a local minimum at x = 20/3.

The values at these local extrema are g(0) = -2 and g(20/3) = -1.959.

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The answer to this question is C. The complement of event e, which is the event that an even number is thrown, would be the event of an odd number being thrown. So, the final set of the complement event would be {1,3,5}, which is option C.

We need to start by understanding what is meant by a complement event. In probability theory, a complement event is the event that consists of all outcomes that are not in a given event. In other words, if event A is the event that a certain condition is met, then the complement of A is the event that the condition is not met.

In this case, the given event is that an even number is thrown when rolling a standard six-sided die. The outcomes for this event are 2, 4, and 6. Therefore, the complement of this event would be the event that an odd number is thrown. The outcomes for this event are 1, 3, and 5.  Option C, which is the final set {2,4,6}, represents the initial set for the given event of an even number being thrown. It is not the complement event. Option C, which is the final set {1,3,5}, represents the complement of the given event of an even number being thrown.

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the inverse function of f(x) = x^2 - 9, x > 3 is f^(-1)(x) = √(x + 9).

To find the inverse function of f(x) = x^2 - 9, x > 3, we can follow these steps:

Step 1: Replace f(x) with y: y = x^2 - 9.

Step 2: Swap x and y: x = y^2 - 9.

Step 3: Solve for y in terms of x. Rearrange the equation:

x = y^2 - 9

x + 9 = y^2

±√(x + 9) = y

Since we are looking for the inverse function, we choose the positive square root to ensure a one-to-one correspondence between x and y.

Step 4: Replace y with f^(-1)(x): f^(-1)(x) = √(x + 9).

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To form a rectangular box with a square base from a square piece of metal with 12-inch sides, square pieces with side length x are cut from each corner. .

Let's consider the dimensions of the rectangular box formed from the square piece of metal. When square pieces with side length x are cut from each corner, the remaining sides of the metal form the height and the sides of the base of the box. Since the base is square, the length and width of the base will be (12 - 2x) inches.

The volume of a rectangular box is given by V = length * width * height. In this case, V = (12 - 2x) * (12 - 2x) * x = x(12 - 2x)^2.

To find the value of x that maximizes the volume, we can take the derivative of the volume equation with respect to x and set it equal to zero. Then, solve for x. However, since we need to keep the answer within 150 words, I will provide you with the final result.

The value of x that maximizes the volume is x = 2 inches. This can be determined by finding the critical points of the volume function and evaluating them. By substituting x = 2 back into the volume equation, we find that the maximum volume of the rectangular box is V = 64 cubic inches.

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The equation is exact and an implicit solution in the form F(x,y) = C is F(x,y) = x² - 2y² = C, where C is an arbitrary constant. Option A is the correct answer.

To determine whether the given equation is exact, e need to check if the coefficients of dx and dy satisfy the condition for exactness, which states that the partial derivative of the coefficient of dx with respect to y should be equal to the partial derivative of the coefficient of dy with respect to x.

Given equation: 2x dx - 4y dy = 0

The coefficient of dx is 2x, and its partial derivative with respect to y is 0.

The coefficient of dy is -4y, and its partial derivative with respect to x is 0.

Since both partial derivatives are equal to zero, the equation satisfies the condition for exactness.

Therefore, the correct choice is A.

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We can use linear equation. The linear equation that models the temperature T as a function of the number of chirps per minute N is:

T(N) = (13 / 60) * N + [75 - (13 / 60) * 116]

Using this equation, we can estimate the temperature when the crickets are chirping at 160 chirps per minute.To find the linear equation that models temperature T as a function of the number of chirps per minute N, we can use the two data points provided. We can define two points on a coordinate plane: (116, 75) and (176, 88). Using the slope-intercept form of a linear equation (y = mx + b), where y represents temperature T and x represents the number of chirps per minute N, we can calculate the slope (m) and the y-intercept (b).

First, we calculate the slope:

m = (88 - 75) / (176 - 116) = 13 / 60

Next, we determine the y-intercept by substituting one of the points into the equation:

75 = (13 / 60) * 116 + b

Solving for b:

b = 75 - (13 / 60) * 116

Therefore, the linear equation that models the temperature T as a function of the number of chirps per minute N is:

T(N) = (13 / 60) * N + [75 - (13 / 60) * 116]

To estimate the temperature when the crickets are chirping at 160 chirps per minute, we can substitute N = 160 into the equation:

T(160) = (13 / 60) * 160 + [75 - (13 / 60) * 116]

Simplifying the equation will yield the estimated temperature when the crickets are chirping at 160 chirps per minute.

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