the two figures are similar. find the values of x and y, then use that to find the ratio (larger to smaller) of the perimeter and the area

1. Equation of f: S: z = x, y = 4(z^2 - 1)

The surface S is a hyperboloid of one sheet.

2. Equations of the traces on each coordinate plane:

Trace on the xy-plane: y = -4 (horizontal line)

Trace on the xz-plane: z = x (diagonal line passing through the origin)

Trace on the yz-plane: y = 4(z² - 1) (vertical parabola)

c. The graph of the given equation is given in the attachment.

To determine the equation of the surface S generated by revolving the curve y = 4(x² - 1) about the y-axis, we can start by rewriting the equation of the curve in terms of x and z (since the surface S is in three dimensions).

1. Equation of f:

Let's substitute x = z and y = y to obtain the equation in terms of x and z:

x = z

y = 4(x² - 1) = 4(z² - 1)

The equation of the surface S is then given by:

S: z = x

y = 4(z² - 1)

The surface S is a quadric surface known as a hyperboloid of one sheet.

Equations of the traces on each coordinate plane:

2. To find the traces of S on the coordinate planes, we set one of the variables (x, y, or z) to zero and solve for the remaining variables.

Trace on the xy-plane (z = 0):

Substituting z = 0 into the equation of S, we get:

x = 0

y = 4(0²- 1) = -4

The equation of the trace on the xy-plane is y = -4, which represents a horizontal line.

Trace on the xz-plane (y = 0):

Substituting y = 0 into the equation of S, we have:

x = z

0 = 4(z² - 1)

Solving this equation, we find two values for z:

z = 1 and z = -1

Therefore, the equation of the trace on the xz-plane is z = x, which represents a diagonal line passing through the origin.

Trace on the yz-plane (x = 0):

Substituting x = 0 into the equation of S, we get:

0 = z

y = 4(z² - 1)

Solving this equation, we find two values for z:

z = 1 and z = -1

Therefore, the equation of the trace on the yz-plane is y = 4(z² - 1), which represents a vertical parabola.

3. The trace on the xy-plane is a horizontal line at y = -4.

The trace on the xz-plane is a diagonal line passing through the origin.

The trace on the yz-plane is a vertical parabola.

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The ball is 80 ft off the ground 5 seconds after it has been fired. Hence, the correct answer is option a. 80 ft.

When the baseball is fired straight up into the air at a velocity of 96 ft/s, it experiences only the force of gravity acting on it. The acceleration due to gravity is approximately 32 ft/[tex]s^2[/tex]. Since the ball is moving upward initially, it slows down due to the gravitational force until it reaches its highest point where its velocity becomes zero. After that, it starts descending back towards the ground.

To determine the height of the ball 5 seconds after it has been fired, we can use the kinematic equation:

h = h₀ + v₀t - 0.5[tex]gt^2[/tex]

Here, h is the height, h₀ is the initial height (which is zero in this case since the ball is fired from the ground), v₀ is the initial velocity (96 ft/s), t is the time (5 seconds), and g is the acceleration due to gravity (32 ft/[tex]s^2[/tex]).

Plugging in the values, we get:

[tex]h = 0 + (96 ft/s)(5 s) - 0.5(32 ft/s^2)(5 s)^2\\h = 0 + 480 ft - 0.5(32 ft/s^2)(25 s^2)\\h = 0 + 480 ft - 400 ft\\h = 80 ft\\[/tex]

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Upon differentiation, the value of [tex]\( f_{y}(9,0) \)[/tex] is 117.

To find [tex]\( f_{y}(9,0) \)[/tex] (the partial derivative of f with respect to y at the point (9,0), we need to differentiate the function [tex]\( f(x, y) = 13xe^y \)[/tex] with respect to y while treating x as a constant.

To differentiate f(x, y) with respect to y, we use the rules of differentiation. The derivative of [tex]\( e^y \)[/tex] with respect to y is simply [tex]\( e^y \)[/tex], and the derivative of x with respect to y (treating it as a constant) is 0.

Therefore, we have:

[tex]\[ f_y(x, y) = 13x \cdot e^y \][/tex]

Substituting x = 9 and y = 0, we get:

[tex]\[ f_y(9, 0) = 13 \cdot 9 \cdot e^0 = 13 \cdot 9 \cdot 1 = 117 \][/tex]

So, [tex]\( f_y(9, 0) = 117 \)[/tex].

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Complete Question:
Find [tex]\( f_{y}(9,0) \)[/tex] if [tex]\( f(x, y)=13 x e^{y} \)[/tex] [tex]\[ f_{y}(9,0)= \][/tex]?

The area between the curves (y = ln x) and (y = ln 2x) from (x = 1) to (x = 4) is [tex]\(A = 3\ln 2\).[/tex]

To find the area between the curves[tex]\(y = \ln x\)[/tex]and [tex]\(y = \ln 2x\)[/tex] from (x = 1) to (x = 4), we need to calculate the definite integral of the difference between the two curves over that interval:

[tex]\[A = \int_{1}^{4} (\ln 2x - \ln x) \, dx\][/tex]

Using the properties of logarithms, we can simplify the integrand:

[tex]\[A = \int_{1}^{4} \ln \left(\frac{2x}{x}\right) \, dx = \int_{1}^{4} \ln 2 \, dx\][/tex]

Since [tex]\(\ln 2\)[/tex] is a constant, we can move it outside the integral:

[tex]\[A = \ln 2 \int_{1}^{4} \, dx\][/tex]

Now we can evaluate the definite integral:

[tex]\[A = \ln 2 \left[x\right]_{1}^{4} = \ln 2 \cdot (4-1) = \ln 2 \cdot 3 = \boxed{3\ln 2}\][/tex]

Therefore, the area between the curves (y = ln x) and (y = ln 2x) from (x = 1) to (x = 4) is [tex]\(A = 3\ln 2\).[/tex]

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The lift-to-drag ratio of the aircraft is 46.57.

An aircraft with a total mass of 8,300 kg is required to maintain a constant speed of 985 km/h at an altitude of 14,000 meters. If the total drag is equal to 1750 N and the lift coefficient is equal to 0.803, find the surface area of the wing and the lift to drag ratio of this aircraft.

The aircraft is said to be in steady level flight when the weight is equal to the lift generated by the wing and the drag is equal to the thrust required to maintain constant speed.

During steady flight, the weight of the aircraft is balanced by the lift force generated by the wings and the drag force of the aircraft is balanced by the thrust force required to maintain the constant speed.

The lift coefficient, which depends on the angle of attack and the shape of the wing, is a dimensionless quantity that provides an indication of the lift force generated per unit area of wing.

The lift coefficient can be calculated using the following equation:

L = 0.5ρV2SCL

where: L = Lift force

ρ = Air density

V = Velocity

S = Wing surface area

CL = Lift coefficient

Given that the lift coefficient, CL, is 0.803,

and the total mass of the aircraft is 8,300 kg, the lift force, L, can be calculated as follows:

L = W = 8300g

where: g = Acceleration due to gravity

= 9.81 m/s2

= 81423.6 N

The velocity of the aircraft is given as 985 km/h, which is equivalent to 273.6 m/s.

At an altitude of 14,000 meters, the air density is approximately 0.304 kg/m3.

Using the lift equation, L = 0.5ρV2SCL

The surface area of the wing, S, can be calculated as:

[tex]S = L / (0.5\rho V2CL)S \\= 81423.6 / (0.5 \times 0.304 \times 273.6 \times 273.6 \times 0.803)S \\= 50.03 m2[/tex]

Therefore, the surface area of the wing is 50.03 m2.

The lift-to-drag ratio, L/D, is an indicator of the efficiency of the aircraft. It is defined as the ratio of the lift force to the drag force, L/D = L/D.

The drag force of the aircraft is given as 1750 N.

Using the lift-to-drag ratio equation,

[tex]L/D = L / DL/D \\= 81423.6 / 1750\\L/D = 46.57[/tex]

Therefore, the lift-to-drag ratio of the aircraft is 46.57.

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The rate of change of profit with respect to time when x = 40 units and dx/dt = 5 is 1000 units per time unit (e.g. dollars per hour, if time is measured in hours).

To find the rate of change of profit with respect to time, we need to use the formula:

Profit = Revenue - Cost

Also, use the chain rule of differentiation, which states that if y is a function of u, and u is a function of x, then:

[tex]dy/dx = dy/du * du/dx[/tex]

In this case, Profit is a function of x, and x is a function of time (t), so we can write:

[tex]dP/dt = dP/dx * dx/dt[/tex]

where P is the profit function.

We are given the revenue and cost functions, so we have

[tex]Revenue = R(x) = 250x \\

Cost = C(x) = 50x + 3000[/tex]

where x is the number of units produced and sold.

Using the formula for profit, we have,

[tex]Profit = P(x) = R(x) - C(x) \\

Profit = 250x - 50x - 3000 \\

Profit = 200x - 3000[/tex]

To find the rate of change of profit with respect to time, differentiate P(x) with respect to x and then multiply by dx/dt: by

[tex]dP/dt = dP/dx * dx/dt \\

dP/dt = (d/dx)(200x - 3000) * 5 [/tex]

(assuming dx/dt = 5 when x = 40)

dP/dt = 200 * 5

dP/dt = 1000

Hence, the rate of change of profit with respect to time when x = 40 units and dx/dt = 5 is 1000 units per time unit (e.g. dollars per hour, if time is measured in hours).

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The last term is 300

The sum of all the terms is 15150.0

The average of all the terms is 151.5

Here is an example solution in Python that follows the given pseudocode:

# Prompt for and read the first term

first_term = int(input("Enter first term: "))

# Prompt for and read the common difference

common_difference = int(input("Enter common difference: "))

# Prompt for and read the number of terms

number_of_terms = int(input("Enter number of terms: "))

# Calculate the last term

last_term = first_term + (number_of_terms - 1) * common_difference

# Calculate the sum of all the terms

sum_of_terms = number_of_terms * (first_term + last_term) / 2

# Calculate the average of all the terms

average_of_terms = sum_of_terms / number_of_terms

# Display the results

print("The last term is", last_term)

print("The sum of all the terms is", sum_of_terms)

print("The average of all the terms is", average_of_terms)

If you run this code and enter the values from the sample run (first term: 3, common difference: 3, number of terms: 100), it will produce the following output:

The last term is 300

The sum of all the terms is 15150.0

The average of all the terms is 151.5

The program prompts the user for the first term, common difference, and number of terms. Then it calculates the last term using the given formula. Next, it calculates the sum of all the terms and the average of all the terms using the provided formulas. Finally, it displays the calculated results.

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The particle undergoes oscillatory motion along the x and y axes, completing one full oscillation in its trajectory, as described by the equations x = 4 sin t and y = 5 cos t in the interval [0, 2π].

The given equations describe the position of a particle in terms of its coordinates (x, y) as x = 4 sin t and y = 5 cos t, where t varies in the interval [0, 2π].

To describe the motion of the particle, we analyze the equations and interpret the behavior of x and y as t changes.

The equation represents oscillatory motion along the x-axis. The amplitude of the oscillation is 4, and the particle moves between the maximum position at x = 4 and the minimum position at x = -4. As t varies from 0 to 2π, the particle completes one full oscillation along the x-axis.

Similarly, the equation represents oscillatory motion along the y-axis. The amplitude of the oscillation is 5, and the particle moves between the maximum position at y = 5 and the minimum position at y = -5. As t varies from 0 to 2π, the particle completes one full oscillation along the y-axis.

Combining the motions along both axes, we can describe the complete motion of the particle as follows:

The concept used in solving this problem is the understanding of trigonometric functions and their graphical representations. The sine and cosine functions describe periodic motion, and by applying them to the equations x = 4 sin t and y = 5 cos t, we can interpret the motion of the particle in terms of oscillations along the x and y axes.

Therefore, the motion of the particle can be described as a combination of oscillatory motion along the x and y axes, with the particle completing one full oscillation in its motion.

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To use cylindrical coordinates, we can express the equations of the sphere and cone in terms of (r), (\theta), and (z). The equation of the sphere is:

[x^2 + y^2 + z^2 = 4]

In cylindrical coordinates, this becomes:

[r^2 + z^2 = 4]

The equation of the cone is:

[z^2 = x^2 + y^2]

Substituting for (x) and (y) in terms of (r) and (\theta):

[z^2 = r^2\cos^2\theta + r^2\sin^2\theta = r^2]

So the equation of the cone in cylindrical coordinates is simply:

[z^2 = r^2]

To find the upper nappe of the cone, we need to restrict (z) to be positive. So the solid is given by:

[r^2 + z^2 \leq 4,\quad z \geq \sqrt{r^2}]

Simplifying the second inequality gives:

[z \geq r]

Now we can set up the integral for the volume:

[V = \iiint_D dV]

where (D) is the region defined by the above inequalities. In cylindrical coordinates, the volume element is (dV = r,dr,d\theta,dz), so the integral becomes:

[V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{\sqrt{4-r^2}} r,dz,dr,d\theta]

Evaluating this integral gives:

[V = \frac{8}{3}\pi]

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The given series is converging by Limit Comparison Test because it can be compared to the converging p-series ∑n=1[infinity]​1/n, which is a p-series with p=1. The limit of bn​ is zero and the bn+1​bn​ ratio test is satisfied.

Given series: ∑n=1[infinity]​(−1)n−1bn​=5​1​−6​1​+7​1​−8​1​+9​1​−⋯

We need to identify bn​ and test whether the given series is converging or diverging. The given series is an alternating series since the sign of the terms alternate between positive and negative. Alternating Series Test is used to determine whether the alternating series converges or not.

In order to use the Alternating Series Test, it is necessary to check that the series is decreasing. It means that, as n increases, each term is smaller than its predecessor or simply, bn+1​ ≤ bn​ for all n. Let us identify the bn​ term of the series: We can observe that, bn​=1 for n=1,

bn​=1/6 for n=2,

bn​=1/7 for n=3,

bn​=1/8 for n=4,

bn​=1/9 for n=5, ...

Hence, bn​=1/(n+4) for n≥1.

Using Limit Comparison Test, limn→∞​bn​​/1/n=limn→∞​n/(n+4)=1.

The given series is converging by Limit Comparison Test

because it can be compared to the converging p-series ∑n=1[infinity]​1/n, which is a p-series with p=1. The limit of bn​ is zero and the bn+1​bn​ ratio test is satisfied.

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No. the sum of any two lengths of the triangle is greater than the third length.

Option (A) is correct.

Since, A triangle is a polygon with three edges and three vertices. In Euclidean geometry, a triangle is completely determined by the three points that make up its vertices.

Kathy wants to place a printer 9ft from the filing cabinet. Can the other two distances are shown to be 7ft and 2ft.

Hence, No the sum of any two lengths of the triangle should be greater than the third side.

but 7 + 2 = 9,

so the condition for a triangle is false.

Hence, the correct explanation would be "No. the sum of any two lengths of the triangle is greater than the third length."

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Complete question is,

Kathy wants to place a printer 9ft from the filing cabinet. Can the other two distances shown be 7ft and 2ft? Identify the answer with the correct explanation. Printer Desk Filing Cabinet

O Yes; the sum of any two lengths is greater than the third length.

O No; the sum of any two lengths is not greater than the third length.

O Yes; the sum of any two lengths is equal to the third length.

O No; the sum of any two lengths is greater than the third length

the area of the region bounded by the curves[tex]y = x^2 - 2x[/tex] and y = 0 on the interval [-2, 1] is 16/3 square units.

To determine the area of the region bounded by the curves y = x^2 - 2x and y = 0 on the interval [-2, 1], find the area under the curve between these two points.

First, let's find the points of intersection between the two curves:

[tex]x^2 - 2x = 0[/tex]

Factoring out x,

x(x - 2) = 0

So, x = 0 or x = 2.

The points of intersection are (0, 0) and (2, 0).

To find the area, we need to integrate the curve [tex]y = x^2 - 2x[/tex] between x = -2 and x = 2:

Area = ∫[-2,2] (x^2 - 2x) dx

Integrating,

[tex]Area = [ (1/3)x^3 - x^2 ][-2,2][/tex]

Substituting the upper and lower limits,

[tex]Area = (1/3)(2^3 - (-2)^3) - (2^2 - (-2)^2)[/tex]

[tex]Area = (1/3)(8 + 8) - (4 - 4)[/tex]

[tex]Area = (1/3)(16) - 0[/tex]

Area = 16/3

Therefore, the area of the region bounded by the curves[tex]y = x^2 - 2x[/tex] and y = 0 on the interval [-2, 1] is 16/3 square units.

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The ball's height changed by approximately -83.2 feet between 1 second and 6 seconds.

To determine the change in height of the ball between 1 second and 6 seconds, we need to integrate the velocity function over the given interval.

The velocity function is given as f(t) = -30t + 288. To find the height function h(t), we integrate f(t) with respect to time:

h(t) =[tex]∫(-30t + 288) dt = -15t^2 + 288t + C,[/tex]

where C is the constant of integration. To find C, we use the initial condition that the ball is launched from a height of 5 feet at t = 0:

h(0) = [tex]-15(0)^2 + 288(0) + C = 5.[/tex]

C = 5.

Therefore, the height function is h(t) = [tex]-15t^2 + 288t + 5.[/tex]

To calculate the change in height between 1 second and 6 seconds, we subtract the height at t = 1 from the height at t = 6:

Δh = h(6) - h(1) =[tex](-15(6)^2 + 288(6) + 5) - (-15(1)^2 + 288(1) + 5).[/tex]

Calculating this expression gives us the change in height of the ball between 1 second and 6 seconds.

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Answer:

1/3

Step-by-step explanation:

This answer is unintuitive, but stick with me.

If a family has 2 children, and each can be a boy (B) or a girl (G), there are 4 possible combinations:

BB BG GB GG

We know that this family doesn't have 2 girls, leaving us with 3 options:

BB BG GB

Because the order is unspecified, all 3 of these options are possible. Only one has two biys out of 3 possibilities, so it is 1/3.

The  value of the integral is (\iint_D (4x+y) dA = \frac{16}{3}).

We can evaluate the integral using polar coordinates. In polar coordinates, the region D is defined by (0\leq r \leq 2) and (0\leq \theta \leq \pi/2).

The integrand is given by (4x+y = 4r\cos\theta + r\sin\theta). The differential element of area in polar coordinates is (dA = r dr d\theta). Therefore, we have:

(\iint_D (4x+y) dA = \int_{0}^{\pi/2}\int_{0}^{2} (4r\cos\theta + r\sin\theta) r dr d\theta)

Integrating with respect to r first, we get:

(\int_{0}^{\pi/2}\int_{0}^{2} (4r\cos\theta + r\sin\theta) r dr d\theta = \int_{0}^{\pi/2}\int_{0}^{2} (4r^2\cos\theta + r^2\sin\theta) dr d\theta)

Evaluating the inner integral gives:

(\int_{0}^{2} (4r^2\cos\theta + r^2\sin\theta) dr = [\frac{4}{3}r^3\cos\theta + \frac{1}{3}r^3\sin\theta]_0^2 = \frac{16}{3}\cos\theta + \frac{8}{3}\sin\theta)

Substituting this expression back into the original integral and integrating with respect to theta, we get:

(\int_{0}^{\pi/2}\frac{16}{3}\cos\theta + \frac{8}{3}\sin\theta d\theta = [\frac{16}{3}\sin\theta - \frac{8}{3}\cos\theta]_0^{\pi/2} = \frac{16}{3})

Therefore, the value of the integral is (\iint_D (4x+y) dA = \frac{16}{3}).

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The amount of each coupon payment for the given bond value is given by first option $36.35.

Bond par value is equal to $1,000

Current yield percent is equal to 7.17%

= 7.05 /100

= 0.0705

Bond quoted value is  103.12

Payment method

= Semi-annual coupon payment

Calculation of annual coupon amount is equal to,

Current yield = Annual coupon / (Bond value × Bond quoted)

⇒0.0705 = Annual coupon / [($1,000 × 103.12)/100]

⇒0.0705 = Annual coupon / ($1,031.2)

⇒ Annual coupon = 0.0705 × $1,031.2

⇒ Annual coupon = $72.6996

Computation of each coupon payment is equal to

Each coupon payment = Annual coupon amount / 2

⇒ Each coupon payment  = $72.6996 / 2

⇒ Each coupon payment = $36.3498

⇒Each coupon payment = $36.35

Therefore, the amount of each coupon payment is equal to first option $36.35.

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Answer:

  9. (a) -4.58872425

  10. (a) -18.22610955

Step-by-step explanation:

You want the approximate value of f'(5.245) if f(x) = ln(3x) +sin(5x -4) -3 and the approximate value of f''(2.156) if f(x) = 2tan(x) +cos(2x) using h = 0.025 and 0.003, respectively.

The approximate value of f'(x) is the difference quotient ...

  f'(x) ≈ (f(x+h) -f(x))/h

For x=5.245 and h=0.025, this is ...

  f'(x) ≈ ((f(5.245 +0.025) -f(5.245))/0.025

The calculator screen in the first attachment shows the value of this is about ...

  f'(5.245) ≈ -4.58872425

For the second derivative, we use ...

  f'(x) ≈ ((f(x +h) -f(x))/h

  f''(x) = (f'(x -h) -f'(x))/h

The calculator screen in the second attachment shows the calculations and the final value of f''(2.156) as ...

  f''(2.156) ≈ -18.22610955

__

Additional comment

Note that the calculator must be set to radians mode.

<95141404393>

The concept of the fundamental period is important in the analysis of periodic signals and is used to obtain the Fourier series representation of a periodic signal. The fundamental period of the signal is π.

Given that, x[n] = 4A + sin(BπA + Bn), where A = 2 and B = 1 (based on the section number and the last digit of the student ID). Hence, the expression for the signal x[n] can be written as x[n] = 4(2) + sin(π + n).x[n] = 8 + sin(π + n).The given signal is a periodic signal with a period of T.

Hence, x[n] = x[n + T].x[n + T] = x[n]⇒ 8 + sin(π + n + T) = 8 + sin(π + n).⇒ sin(π + n + T) = sin(π + n). This is true for all values of n if (π + n + T) - (π + n) = 2πk, where k is an integer.⇒ T = 2πk. From the above expression, it is evident that the value of T is a function of k.

In order to obtain the fundamental period, we need to find the smallest possible value of T that satisfies the above equation.We know that sin(π + n) = -sin(n).Hence, sin(π + n + T) = -sin(n + T).⇒ -sin(n + T) = sin(n). We know that the maximum value of sin(n) is 1 and the minimum value is -1.

Therefore, sin(n + T) = sin(n) should hold true for all values of n from -∞ to ∞. Hence, the smallest possible value of T can be obtained by finding the smallest value of T for which sin(T) = sin(0) = 0. Since sin(T) = 0, T = mπ, where m is an integer.

Substituting this value of T in the given equation, we get, sin(π + n + T) = sin(π + n + mπ) = -sin(n + mπ) = -sin(n). Therefore, the fundamental period of the given signal is T = mπ = π. Hence, the fundamental period of the signal is π.

The fundamental period of a periodic signal is the smallest possible period for which the signal repeats itself. In other words, the fundamental period is the smallest possible value of T for which x[n] = x[n + T] holds true for all values of n.

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Based on the provided histogram, we can determine the number of students who took greater than 6.5 but less than 9.5 minutes to run a mile by examining the corresponding bars on the histogram.

The histogram has a horizontal axis labeled from 4.5 to 9.5 in increments of 1. The approximate heights of the bars are given as follows:

For the bar at 6.5 on the horizontal axis, the approximate height is 6.

For the bar at 7.5 on the horizontal axis, the approximate height is 8.

For the bar at 8.5 on the horizontal axis, the approximate height is 7.

To find the number of students who took greater than 6.5 but less than 9.5 minutes, we need to add up the heights of the bars for the corresponding range.

From the provided histogram, we can see that there is one bar between 6.5 and 7.5, with a height of 6. Then, there are two bars between 7.5 and 8.5, with heights of 8 and 7 respectively.

Adding up the heights: 6 + 8 + 7 = 21.

Therefore, the number of students who took greater than 6.5 but less than 9.5 minutes to run a mile is 21.

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DeMorgan's laws hold true in Boolean algebra based on the truth tables for the "and" and "or" operations.

DeMorgan’s laws are a set of principles that explain the equivalence between certain Boolean operations.

DeMorgan's Laws state that the negation of a conjunction is the disjunction of the negations of the two conjunctions, and that the negation of a disjunction is the conjunction of the negations of the two disjuncts.

These principles are based on truth tables for the “and” and “or” operations in Boolean algebra.

Truth Tables for And and Or

The Boolean operator "and" produces a true result only if both inputs are true. A truth table for the “and” operation looks like this:

p    q      p and q

T     T           T

T     F           F

F     T           F

F     F           F

The Boolean operator "or" produces a true result if either of its inputs is true. A truth table for the “or” operation looks like this:

p    q     p or q

T     T          T

T     F          T

F     T          T

F     F          F

DeMorgan's Laws in And and Or

DeMorgan's first law states that the negation of a conjunction is the disjunction of the negations of the two conjunctions. In other words:

NOT (p AND q) is equivalent to (NOT p) OR (NOT q)

A truth table for this operation looks like this:

p    q    (p and q)  ~(p and q)  ~p   ~q  ~p OR ~q

T     T        T                F         F      F     F

T     F        F                T        F     T     T

F     T        F                T        T     F     T

F     F        F                T        T     T     T

DeMorgan's second law states that the negation of a disjunction is the conjunction of the negations of the two disjuncts. In other words:

NOT (p OR q) is equivalent to (NOT p) AND (NOT q)

A truth table for this operation looks like this:

p    q    (p or q)  ~(p or q)  ~p   ~q  ~p AND ~q

T     T        T                F         F      F     F

T     F        T                F         F      T     F

F     T        T                F         T      F     F

F     F        F                T         T      T     T

Thus, DeMorgan's laws hold true in Boolean algebra based on the truth tables for the "and" and "or" operations.

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[tex]\boxed{P=\begin{bmatrix}\frac{16}{17}&-\frac{4}{17}\\-\frac{4}{17}&\frac{1}{17}\end{bmatrix}}$$[/tex]

That is the solution to the problem with the given conditions.

Given the following:

y=[9,2]

[tex]u_1=\left[\frac{4}{\sqrt{17}},\frac{-1}{\sqrt{17}}\right]$$[/tex]

[tex]w=\text{span}\{u_1\}$$[/tex]

Part a) U is a 2x1 matrix with only one column. Thus,

U=[u_1]

Therefore,

[tex]U^\top U=[u_1]^\top[u_1]\end{bmatrix}$$[/tex]

[tex]=\begin{bmatrix}\frac{4}{\sqrt{17}}&\frac{-1}{\sqrt{17}}\end{bmatrix}\begin{bmatrix}\frac{4}{\sqrt{17}}\\\frac{-1}{\sqrt{17}}\end{bmatrix}[/tex]

=1

Also,

[tex]U^\top=\begin{bmatrix}\frac{4}{\sqrt{17}}\\\frac{-1}{\sqrt{17}}\end{bmatrix}$$[/tex]

[tex]$$U^\top U=[1]$$[/tex]

[tex]$$U^\top=\begin{bmatrix}\frac{4}{\sqrt{17}}\\\frac{-1}{\sqrt{17}}\end{bmatrix}$$[/tex]

[tex]$$\boxed{U^\top U=[1]}$$[/tex]

b)Since [tex]$w=\text{span}\{u_1\}$[/tex] is a one-dimensional subspace of [tex]$\mathbb{R}^2$[/tex], the matrix P that projects onto w is

[tex]P=\frac{uu^\top}{u^\top u}$$[/tex]

where u is any nonzero vector in w and [tex]$u^\top$[/tex]is the transpose of u. Since u_1 is a basis for w, we can use it to find the projection matrix. Thus,

[tex]P=\frac{uu^\top}{u^\top u}\end{bmatrix}$$[/tex]

[tex]=\frac{u_1u_1^\top}{\lVert u_1\rVert^2}[/tex]

[tex]=\frac{1}{17}\begin{bmatrix}16&-4\\-4&1\end{bmatrix}[/tex]

Thus, [tex]\boxed{P=\begin{bmatrix}\frac{16}{17}&-\frac{4}{17}\\-\frac{4}{17}&\frac{1}{17}\end{bmatrix}}$$[/tex]

That is the solution to the problem with the given conditions.

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The set of solutions can also be written as an ordered pair:

[tex]$(\frac{1}{4}, -\frac{3}{4}, \frac{5}{3}, -5)$[/tex]

The given linear system of equations is as follows:

[tex]$$3x_1 + x_4 = -15$$\\$$-4x_2 + x_3 = 20$$[/tex]

In matrix form, the above equations are:

[tex]$$\begin{pmatrix}3 & 0 & 0 & 1\\0 & -4 & 1 & 0\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = \begin{pmatrix}-15\\20\end{pmatrix}$$[/tex]

The matrix form is of the type Ax = b, where A is the coefficient matrix, x is the vector of variables, and b is the vector of constants.

Thus, to find the solution to this system of equations, we need to find the inverse of matrix A, and multiply it with the vector b, i.e., x = A⁻¹b.

Here is how we can solve this system of equations using matrix multiplication:

[tex]$$\begin{pmatrix}3 & 0 & 0 & 1\\0 & -4 & 1 & 0\end{pmatrix}^{-1} = \frac{1}{12}\begin{pmatrix}4 & 0\\3 & -1\\0 & 4\\-12 & 0\end{pmatrix}$$[/tex]

Multiplying this inverse matrix with the vector b, we get:

[tex]$$\begin{pmatrix}\frac{1}{4}\\-\frac{3}{4}\\\frac{5}{3}\\-5\end{pmatrix}$$[/tex]

Thus, the solution to the given system of equations is:

[tex]$$x_1 = \frac{1}{4}, x_2 = -\frac{3}{4}, x_3 = \frac{5}{3}, x_4 = -5$$[/tex]

This set of solutions can also be written as an ordered pair:

[tex]$(\frac{1}{4}, -\frac{3}{4}, \frac{5}{3}, -5)$[/tex]

The solution has four variables and, therefore, we need to provide four values.

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We can find this area by subtracting the area of the trapezoid below the curve from the area of the rectangle with height f(4) - f(2) = (2(4) + 3) - (2(2) + 3) = 5 and width 2:Area of region = (f(4) - f(2)) * 2 - Area of trapezoid= (2(4) + 3) - (2(2) + 3)) * 2 - 18= (11 - 7) * 2 - 18= -2 units^2.So the integral of (2x +3) with respect to x from 2 to 4 is -2.

A trapezoid is a quadrilateral with one pair of parallel sides that are referred to as the bases. The parallel sides are named b1 and b2, and the distance between them is referred to as the height (h).The formula for calculating the area of a trapezoid is as follows:Area of trapezoid = 1/2 * (b1 + b2) * h, where b1 and b2 are the lengths of the parallel sides and h is the distance between them.Now, to evaluate the integral of (2x +3) with respect to x from 2 to 4, we can use the formula for the area of a trapezoid. Let's first find the area of the trapezoid below the curve y = (2x +3) from x = 2 to x = 4.To do this, we need to find the lengths of the two bases and the height. Let's take a look at the graph of y = (2x +3) from x = 2 to x = 4:So the length of the first base (b1) is f(2) = 2(2) + 3 = 7, and the length of the second base (b2) is f(4) = 2(4) + 3 = 11. The height of the trapezoid (h) is the horizontal distance between the two bases, which is 4 - 2 = 2 units.Using the formula for the area of a trapezoid, we get:Area of trapezoid = 1/2 * (b1 + b2) * h= 1/2 * (7 + 11) * 2= 18 units^2.Now, to evaluate the integral of (2x +3) with respect to x from 2 to 4, we need to find the area of the region bounded by the curve y = (2x +3), the x-axis, and the vertical lines x = 2 and x = 4.

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The given function is 2x + 3 which is to be integrated with respect to x from 2 to 4.

The formula to find the area of a trapezoid is:

A = 1/2(b1+b2)h

where, b1 and b2 are the parallel bases and h is the height of the trapezoid.

Since the trapezoid is not given and the function is a polynomial function, we can integrate it using the definite integral formula:

∫_a^b▒f(x)dxwhere a and b are the lower and upper limits of the function.

The function to be integrated is 2x+3.

Therefore, ∫_2^4▒(2x+3)dx= [x² + 3x]_2^4= [(4² + 3(4)) - (2² + 3(2))] = 22

Therefore, the value of the integral of (2x +3) with respect to x from 2 to 4 is 22.

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The option a. 2/13 is the correct answer.

Given:In a clinic, when 130 patients examined90 had heart trouble50 had diabetes30 had both diseases

We have to find the probability that the patient did not have heart trouble or diabetes using the formula of conditional probability:P(A or B) = P(A) + P(B) - P(A and B)

The probability of a patient not having heart trouble or diabetes is given by:

Probability (not have heart trouble or diabetes) = P(neither heart trouble nor diabetes) = P(only neither)

The number of patients having only neither heart trouble nor diabetes can be found by subtracting the patients having heart trouble,

the patients having diabetes, and the patients having both diseases from the total number of patients.

Hence, the number of patients having only neither heart trouble nor diabetes is:(130 - 90 - 50 + 30) = 20

Probability (not have heart trouble or diabetes) = P(only neither) = 20/130 = 2/13

Therefore, the option a. 2/13 is the correct answer.

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The probability that the patient did not have heart trouble or diabetes is 2/13.

The correct option is a) 2/13.

Step 1: Find the probability of having both heart trouble and diabetes:

P(Heart trouble and Diabetes) = Number of patients with both conditions / Total number of patients

                           = 30 / 130

                           = 3/13

Step 2: Find the probability of having only heart trouble:

P(Only Heart trouble) = Number of patients with heart trouble - Number of patients with both conditions / Total number of patients

                     = (90 - 30) / 130

                     = 60/130

                     = 6/13

Step 3: Find the probability of having only diabetes:

P(Only Diabetes) = Number of patients with diabetes - Number of patients with both conditions / Total number of patients

                = (50 - 30) / 130

                = 20/130

                = 2/13

Step 4: Find the probability of not having heart trouble or diabetes:

P(Not Heart trouble or Diabetes) = 1 - P(Only Heart trouble) - P(Only Diabetes) - P(Heart trouble and Diabetes)

                              = 1 - (6/13) - (2/13) - (3/13)

                              = 1 - 11/13

                              = 2/13

Therefore, the probability that the patient did not have heart trouble or diabetes is 2/13.

The correct option is a) 2/13.

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The actual dimension of the distance is approximately 0.0046667 cm.

To compute the actual dimension of a distance given a drawing measurement and a scale, you can use the following formula:

Actual Dimension = Drawing Measurement × Scale Factor

In this case, the given drawing measurement is 28 cm, and the scale is 1:60 m.

To calculate the scale factor, we need to convert the scale to the same unit as the drawing measurement. Since the drawing measurement is in centimeters (cm), we need to convert the scale from meters (m) to centimeters (cm).

1 meter = 100 centimeters

So, the scale factor is:

Scale Factor = 1:60 m = 1 cm : 60 cm = 1 : 6000 cm

Now we can calculate the actual dimension:

Actual Dimension = Drawing Measurement × Scale Factor

Actual Dimension = 28 cm × 1/6000

Actual Dimension = 28/6000 cm

Simplifying the fraction, we get:

Actual Dimension ≈ 0.0046667 cm

Therefore, the actual dimension of the distance is approximately 0.0046667 cm.

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The required answers are:

a) Constant c = 2, probability density functions PDF [tex]f(x) = x^3 / 4[/tex] for 0 < x < 2, CDF F(x) = [tex]x^4 / 16[/tex], Mean and Variance need specific integration limits. The PDF is bounded.

b) Constant c = 2, probability density functions PDF [tex]f(x) = (3/16)x^2[/tex]for -2 < x < 2, CDF [tex]F(x) = (x^3 + 8) / 16[/tex], Mean and Variance need specific integration limits. The probability density functions PDF is bounded.

c) The probability density functions PDF [tex]f(x) = c/\sqrt x[/tex] for 0 < x < 1 is unbounded at the lower boundary x = 0.

a) For the function [tex]f(x) = x^3 / 4[/tex], we need to find the constant c such that f(x) is a probability density function (PDF) by ensuring that the integral of f(x) over its entire range is equal to 1.

i) Constant c: To find the value of c, we need to evaluate the integral:

[tex]\int_0 ^ c (x^3 / 4) \,dx = 1[/tex]

Integrating the function gives us:

[tex][(x^4 / 16) ]_0^ c = 1\\(c^4 / 16) - 0 = 1\\c^4 / 16 = 1[/tex]

Solving for c, we have:

[tex]c^4 = 16\\c = 2[/tex]

Therefore, the constant c is 2.

ii) CDF (Cumulative Distribution Function), F(x): The cumulative distribution function can be obtained by integrating the pdf:[tex]F(x) = \int_0 ^ x(x^3 / 4) \,dx[/tex]

Evaluating the integral, we get:

[tex]F(x) = (x^4 / 16)_0 ^ x\\F(x) = (x^4 / 16) - 0\\F(x) = x^4 / 16[/tex]

iii) Graphs: The PDF and CDF graphs can be sketched using the obtained equations. The PDF, [tex]f(x) = x^3 / 4[/tex], will be a curve that starts from the origin, increases, and reaches its maximum at x = 2 (the value of c). After x = 2, the curve will decline. The CDF, [tex]F(x) = x^4 / 16[/tex], will be a curve that starts from 0, increases monotonically, and approaches 1 as x approaches positive infinity.

iv) Mean and Variance: To find the mean and variance, we need to use the formulas:

Mean [tex]\mu=\int_0^ c x * f(x)\, dx[/tex]

Variance [tex]\sigma^2=\int_0^ c {(x-\mu)}^2*f(x)\, dx[/tex]

Using the obtained PDF, [tex]f(x) = x^3 / 4[/tex], we can calculate the mean and variance using these formulas. However, without specific integration limits, we cannot provide the exact values of the mean and variance.

b) For the function [tex]f(x) = (3/16)x^2[/tex]  for -c < x < c, we need to find the constant c such that f(x) is a PDF.

i) Constant c: Since the function is symmetric around x = 0, we only need to find the constant for the positive range:

[tex]\int_0^ c (3/16)x^2\, dx = 1[/tex]

Integrating the function gives us:

[tex][(x^3 / 16) ] _0 ^c= 1\\(c^3 / 16) - 0 = 1\\(c^3 / 16) = 1[/tex]

Solving for c, we have:

[tex]c^3 = 16\\c = 2[/tex]

Therefore, the constant c is 2.

ii) CDF (Cumulative Distribution Function), F(x): The cumulative distribution function can be obtained by integrating the pdf:

[tex]F(x) = \int_{-c}^ x ((3/16)x^2) \,dx[/tex]

Evaluating the integral, we get:

[tex]F(x) = [(x^3 / 16) ] _{-c} ^x\\F(x) = (x^3 / 16) - ((-c)^3 / 16)\\F(x) = (x^3 + c^3) / 16[/tex]

iii) Graphs: The PDF and CDF graphs can be sketched using the obtained equations. The PDF, [tex]f(x) = (3/16)x^2[/tex], will be a symmetric curve centered around x = 0, reaching its maximum at x = 2 (the value of c), and then declining. The CDF,[tex]F(x) = (x^3 + c^3) / 16[/tex], will be a curve that starts from 0, increases monotonically, and approaches 1 as x approaches positive and negative infinity.

iv) Mean and Variance: To find the mean and variance, we need to use the formulas:

Mean [tex]\mu = \int_{-c} ^c (x * f(x)) \,dx[/tex]

Variance [tex]\sigma^2 = \int_{-c} ^c (x-\mu)^2 * f(x) \,dx[/tex]

Using the obtained PDF,[tex]f(x) = (3/16)x^2[/tex], we can calculate the mean and variance using these formulas. However, without specific integration limits, we cannot provide the exact values of the mean and variance.

c) For the function [tex]f(x) = c/\sqrt x[/tex] for 0 < x < 1, we need to determine if this PDF is bounded.

To check if the PDF is bounded, we need to examine its behavior as x approaches the boundaries of the interval (0 and 1).

As x approaches 0, the denominator [tex]\sqrt x[/tex] approaches 0, which causes the PDF to approach infinity. Therefore, the PDF is unbounded at the lower boundary x = 0.

As x approaches 1, the denominator  [tex]\sqrt x[/tex] approaches 1, and the PDF remains finite. Hence, the PDF is bounded at the upper boundary x = 1.

In conclusion, the PDF f(x) = c/ [tex]\sqrt x[/tex]  for 0 < x < 1 is bounded at the upper boundary x = 1, but unbounded at the lower boundary x = 0.

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The mean for the probability density function \(f(x) = 3x^2\) when \(x\) ranges from 0 to 1 is \(\frac{3}{4}\).

To find the mean for the probability density function [tex]\(f(x) = 3x^2\)[/tex] when [tex]\(x\)[/tex] ranges from 0 to 1, we need to calculate the integral of [tex]\(xf(x)\)[/tex] over the given range and divide it by the total probability.

The integral of \(xf(x)\) can be calculated as follows:

[tex]\(\int_{0}^{1} x(3x^2) dx\)[/tex]

Simplifying the integrand:

\(\int_{0}^{1} 3x^3 dx\)

Integrating with respect to \(x\):

\(\left[\frac{3}{4}x^4\right]_0^1\)

Substituting the limits:

\(\left(\frac{3}{4}(1)^4\right) - \left(\frac{3}{4}(0)^4\right)\)

Simplifying:

\(\frac{3}{4}\)

Since the probability density function is defined as \(f(x)\), we need to normalize it by dividing by the total probability.

To find the total probability, we integrate \(f(x)\) over the entire range:

\(\int_{0}^{1} 3x^2 dx\)

Integrating with respect to \(x\):

\(\left[x^3\right]_0^1\)

Substituting the limits:

\(1^3 - 0^3 = 1\)

Now, we can calculate the mean by dividing the integral of \(xf(x)\) by the total probability:

\(\frac{\frac{3}{4}}{1} = \frac{3}{4}\)

Therefore, the mean for the probability density function \(f(x) = 3x^2\) when \(x\) ranges from 0 to 1 is \(\frac{3}{4}\).

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The first left endpoint of any partition of an interval [a, b] is the value of 'a'.

An interval [a, b] represents a range of real numbers between the values of 'a' and 'b', including both endpoints.

The left endpoint, 'a', is the smaller value of the two, and the right endpoint, 'b', is the larger value.

Now, let's consider the concept of partitioning an interval.

Partitioning an interval means dividing it into smaller subintervals.

In this context, typically refer to a partition as a set of points that divides the interval into subintervals.

For example, let's consider the interval [a, b] and a partition P = {x₁, x₂, x₃, ..., xₙ}.

This partition divides the interval into subintervals [a, x₁], [x₁, x₂], [x₂, x₃], ..., [xₙ₋₁, xₙ], [xₙ, b].

Each subinterval represents a smaller range within the larger interval.

Now, coming to the first left endpoint of any partition, it refers to the leftmost point of the first subinterval in the partition.

Since the interval [a, b] includes the left endpoint 'a', the first left endpoint of any partition will always be the value of 'a'.

Therefore, the first left endpoint of an interval [a, b] for any partition is the value 'a', as it represents leftmost point of first subinterval in partition.

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The equation of the plane through the intersection of the plane x+y+z=1 and 2x+3y+4z=5 is (x+y+z−1)+λ(2x+3y+4z−5)=0.

Here, (2λ+1)x+(3λ+1)y+(4λ+1)z−(5λ+1)=0......(1)

The direction ratios a₁, b₁, c₁ of this plane are (2λ+1),(3λ+1) and (4λ+1)

The plane in equation (1) is perpendicular to x-y+z=0

Its direction ratios a₂, b₂, c₂ are 1, -1 and 1

Since the planes are perpendicular,

a₁a₂+b₁b₂+c₁c₂ =0

⇒ (2λ+1)−(3λ+1)+(4λ+1)=0

⇒ 3λ+1=0

⇒ λ= -1/3

Substituting λ= -1/3​  in equation (1), we obtain

1/3 x− 1/3 z+ 2/3=0

⇒ x−z+2=0

This is the required equation of the plane.

Therefore, the equation of the plane through the intersection of the plane x+y+z=1 and 2x+3y+4z=5 is (x+y+z−1)+λ(2x+3y+4z−5)=0.

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"Your question is incomplete, probably the complete question/missing part is:"

Find the equation of the plane the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0.

we have:`∫C​(x^2 + y) ds = 64π/3`This is the required value of integral.

The formula for evaluating the given integral is:`∫C​(x^2+y)dS`where `C` is given as the part of the circle `x^2 + y^2 = 16` between the points `(-4, 0)` and `(0, -4)`.

Integration of the given function with respect to `ds` gives:`∫C​(x^2+y)dS = ∫C​x^2ds + ∫C​yds`The first integral can be computed by changing to polar coordinates.

We obtain:`∫C​x^2ds = ∫θ1θ2∫0^4 r^4cos^2θ r dr dθ`After evaluating the inner integral, we get:`∫C​x^2ds = 32(θ2 - θ1)/3`The second integral can be computed in the same way.

We obtain:`∫C​yds = ∫θ1θ2∫0^4 r^4sin^2θ r dr dθ`After evaluating the inner integral, we get:`∫C​yds = 32(θ2 - θ1)/3`.

Therefore, the main answer is:`∫C​(x^2+y)dS = 64(θ2 - θ1)/3`The points `(-4, 0)` and `(0, -4)` are opposite ends of a diameter of the circle.

Thus, they divide the circle into two equal arcs of length `πr`. Therefore, the length of the arc joining these two points is:`L = 2πr/2 = πr = 4π`The angle `θ` subtended by this arc at the center of the circle is `π`.

Thus, we have:`θ2 - θ1 = π`Substituting this value in the above expression, we obtain:`∫C​(x^2+y)dS = 64(π)/3`.

The given question can be solved by applying Green's theorem which states that for a vector field `F = P i + Q j`, and a region `D` bounded by a simple, closed, piecewise-smooth curve `C` oriented counterclockwise,

we have:`∫C​F.ds = ∫∫D (∂Q/∂x - ∂P/∂y) dA`where `ds` is a line element on the curve `C`, `dA` is an area element in the region `D`, and `P` and `Q` are the partial derivatives of `F` with respect to `x` and `y`, respectively.Using this theorem, we can evaluate the given integral by defining the vector field `F = (x^2 + y)i` and applying Green's theorem.

We have:`∫C​F.ds = ∫∫D (∂(x^2 + y)/∂x) dA`Since the curve `C` is given as the part of the circle `x^2 + y^2 = 16` between the points `(-4, 0)` and `(0, -4)`, we can define the region `D` as the part of the disk `x^2 + y^2 ≤ 16` that lies below the line joining these two points. In polar coordinates, we have `r^2 = x^2 + y^2` and `y = -x`.

Substituting this in the equation of the circle, we get:`x^2 + (-x)^2 = 16`which gives:`x = ±2sqrt(2)` and `y = ∓2sqrt(2)`Thus, the points `(-4, 0)` and `(0, -4)` correspond to `(-2sqrt(2), -2sqrt(2))` and `(2sqrt(2), 2sqrt(2))`, respectively. Since `y = -x`, we can restrict `D` to the region defined by `-2sqrt(2) ≤ x ≤ 2sqrt(2)` and `0 ≤ r ≤ 4`.

Thus, we have:`∫C​F.ds = ∫∫D (2x) r dr dθ`where `θ` varies from `3π/4` to `π/4`. The limits of `r` are `0` and `4`.After evaluating the double integral, we get:`∫C​F.ds = 64π/3`Comparing this with the previous result, we see that they are the same.

Thus, we have:`∫C​(x^2 + y) ds = 64π/3`This is the required answer.

Therefore, the main answer to this question is ∫C​(x^2+y)dS = 64(θ2 - θ1)/3. The given question can be solved by applying Green's theorem which states that for a vector field `F = P i + Q j`, and a region `D` bounded by a simple, closed, piecewise-smooth curve `C` oriented counterclockwise.

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