General Motors stock fell from $41 per share in 2013 to $24.98 per share during 2016.
a. If you bought and then sold 300 shares at these prices, what was your loss?

b. Express your loss as a percent of the purchase price. Round to the nearest tenth of a percent.

The given differential equation y'' + 8y' + 16y = 0 represents a damped spring system with a damping coefficient of 8 and a spring constant of 16.

To describe the spring system represented by the differential equation y'' + 8y' + 16y = 0, we will be using the given terms.

1. Differential equation (DE): The given DE is a second-order linear homogeneous differential equation with constant coefficients. It represents the motion of a damped spring system, where y'' denotes the acceleration, y' denotes the velocity, and y denotes the displacement of the mass.

2. Damping: The term 8y' represents the damping in the spring system. It is proportional to the velocity (y') of the mass, and acts to oppose the motion, thus slowing down the oscillation.

3. Spring constant: The term 16y represents the restoring force exerted by the spring, which is proportional to the displacement (y) of the mass. The spring constant is 16.

4. Natural frequency: The natural frequency of the spring system can be found by considering the undamped case (i.e., without the 8y' term). In this case, the DE becomes y'' + 16y = 0. The natural frequency (ω_n) can be calculated as the square root of the spring constant divided by the mass (ω_n = √(k/m)). We don't have the mass value, so we can only state that ω_n = √(16/m).

5. Damping coefficient: The damping coefficient is the constant proportionality factor for the damping term. In this case, it is 8.

6. Damped frequency: Damped frequency (ω_d) is the frequency of oscillation when damping is present. It can be found using the natural frequency and the damping ratio (ζ). However, we do not have enough information to calculate the damping ratio or the damped frequency in this case.

In summary, the given differential equation y'' + 8y' + 16y = 0 represents a damped spring system with a damping coefficient of 8 and a spring constant of 16. The natural frequency depends on the mass, but the damped frequency cannot be calculated without additional information.

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The length of the curve y = x^4 for 0≤ x ≤1 is approximately 1.082.

To find the length of the curve y = x^4 for 0≤ x ≤1, you'll need to use the arc length formula:

Arc length = ∫√(1 + (dy/dx)^2) dx from a to b, where a = 0 and b = 1.

First, find the derivative of y with respect to x:
y = x^4
dy/dx = 4x^3

Now, square the derivative and add 1:
(4x^3)^2 + 1 = 16x^6 + 1

Next, find the square root of the result:
√(16x^6 + 1)

Now, integrate the expression with respect to x from 0 to 1:
∫(√(16x^6 + 1)) dx from 0 to 1

Unfortunately, this integral doesn't have a closed-form solution, so we'll need to use numerical methods, such as Simpson's rule or a numerical integration calculator, to approximate the length.

Using a numerical integration calculator, the length of the curve y = x^4 for 0≤ x ≤1 is approximately 1.082.

Your answer: 1.082

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The length of the curve y = x^4 for 0≤ x ≤1 is approximately 1.082.

To find the length of the curve y = x^4 for 0≤ x ≤1, you'll need to use the arc length formula:

Arc length = ∫√(1 + (dy/dx)^2) dx from a to b, where a = 0 and b = 1.

First, find the derivative of y with respect to x:
y = x^4
dy/dx = 4x^3

Now, square the derivative and add 1:
(4x^3)^2 + 1 = 16x^6 + 1

Next, find the square root of the result:
√(16x^6 + 1)

Now, integrate the expression with respect to x from 0 to 1:
∫(√(16x^6 + 1)) dx from 0 to 1

Unfortunately, this integral doesn't have a closed-form solution, so we'll need to use numerical methods, such as Simpson's rule or a numerical integration calculator, to approximate the length.

Using a numerical integration calculator, the length of the curve y = x^4 for 0≤ x ≤1 is approximately 1.082.

Your answer: 1.082

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To compute the average velocity over each time interval, we use the formula: average velocity = (h(t2) - h(t1))/(t2 - t1), where h(t) is the height function.

Using the given height function, h(t) = 34t - 4.9t^2, we calculate the average velocities:
1. [3, 3.01]:
Average Velocity = (h(3.01) - h(3))/(3.01 - 3) ≈ -17.147 m/s
2. [3, 3.001]:
Average Velocity = (h(3.001) - h(3))/(3.001 - 3) ≈ -17.194 m/s
3. [3, 3.0001]:
Average Velocity = (h(3.0001) - h(3))/(3.0001 - 3) ≈ -17.199 m/s
4. [2.99, 3]:
Average Velocity = (h(3) - h(2.99))/(3 - 2.99) ≈ -17.243 m/s
5. [2.999, 3]:
Average Velocity = (h(3) - h(2.999))/(3 - 2.999) ≈ -17.205 m/s
6. [2.9999, 3]:
Average Velocity = (h(3) - h(2.9999))/(3 - 2.9999) ≈ -17.200 m/s
To estimate the instantaneous velocity at t=3, observe the average velocities as the time intervals approach t=3:
As the intervals get closer to t=3, the average velocities appear to approach -17.2 m/s. Thus, the estimated instantaneous velocity at t=3 is:
V ≈ -17.2 m/s

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School districts preferred hiring unmarried women as teachers in the late nineteenth and early part of the twentieth century due to societal beliefs that married women were expected to prioritize their roles as wives and mothers, leaving little time or energy for teaching responsibilities.

During the late nineteenth and early twentieth centuries, societal beliefs placed a strong emphasis on women's domestic roles as wives and mothers. This resulted in a bias against hiring married women as teachers, as it was assumed that they would prioritize their family responsibilities over their teaching duties.

In contrast, unmarried women were seen as more dedicated and committed to their profession, as they were not expected to balance their professional and domestic responsibilities.

Furthermore, teaching was considered an appropriate profession for unmarried women, as it was viewed as an extension of their nurturing and caretaking roles within the family. This stereotype was reinforced by the fact that many female teachers were required to remain single in order to keep their teaching positions.

Overall, the preference for hiring unmarried women as teachers was a reflection of societal beliefs about gender roles and expectations during this time period.

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Answer:

B

Step-by-step explanation:

28/35, cos is always the adjacent side to the longest side (hypotenuse)

Answer:

1. 2(x+3.60) = 19.40

Divide both sides by 2:

x+3.60 = 9.70

Subtract 3.60 from both sides:

x = 6.10

Answer: 6.10

2. 45.93 + 112 + (−61.24)

45.93 + 112 = 157.93

157.93 - 61.24 = 96.69

Answer: 96.69

3. 20x + 2 > −98

Subtract 2 from both sides:

20x > −100

Divide both sides by 20:

x > −5

Answer: x > −5

4. 2/5 (4x - 8)

= 8x/5 - 16/5

Answer: 8x/5 - 16/5

5. On a school field trip, the number of students (y) is always proportional to the number of adults (x). In one group there are 96 students and 8 adults. What is the constant of proportionality between this relationship?

The constant of proportionality is the number that, when multiplied by the number of adults, gives the number of students. In this case, the constant of proportionality is 96/8 = 12.

Answer: 12

According to this range , The right response is hence A.

Range in mathematics is a statistical indicator of dispersion, or how widely spaced a given data collection is from smallest to largest. The range of a piece of data is the distinction between the largest and lowest value.

The range of the data is 9, and it is almost symmetric. This could occur because the bookstore offers a discount on all books costing more than $6.

Data that is symmetrical is uniformly distributed around the mean. In other words, the distribution's left side is the right side's mirror image. We can infer that the mean is roughly in the middle of the price range in this situation because the data is almost symmetric.

The difference between the largest and smallest numbers in a piece of data is known as the range of the data.

The range in this instance is 9, as there are nine dollars between the highest price ($9) and the lowest price ($0).

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The order of integration does not affect the final answer, but may affect the complexity of the integrals.

To calculate the volume of the region W using triple integrals, we need to determine the bounds for each variable.

First, we can see that the planes z=0 and y=1 bound the region in the z and y directions, respectively.

Next, to find the bounds for x, we need to find the intersection of the two cylinders. Solving for y in the equation [tex]z=1-y^2[/tex], we get y = ±sqrt(1-z). Substituting this into the equation [tex]y=x^2[/tex], we get [tex]x^2[/tex] = ±sqrt(1-z), or x = ±sqrt(sqrt(1-z)). So the bounds for x are -sqrt(sqrt(1-z)) to sqrt(sqrt(1-z)).

Now we can set up the triple integrals in the three orders:

Note that the order of integration does not affect the final answer, but may affect the complexity of the integrals.

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The derivative of f(x) at x = c does not exist.

To find the derivative of f(x) at x = c using the alternative form of the derivative, we first need to calculate the derivative of f(x) with respect to x.

Given that f(x) = x^3 - 2x^2 + 9, we can find the derivative of f(x) using the power rule and the constant multiple rule. The power rule states that the derivative of x^n, where n is a constant, is n*x^(n-1). The constant multiple rule states that the derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function.

Applying the power rule and constant multiple rule to f(x), we get:

f'(x) = 3x^2 - 4x

Now, we can evaluate f'(x) at x = c, which in this case is x = -2:

f'(-2) = 3(-2)^2 - 4(-2)

= 3(4) + 8

= 12 + 8

= 20

So, the derivative of f(x) at x = -2 is 20. However, we are asked to find the derivative at x = c = -2 using the alternative form of the derivative.

The alternative form of the derivative states that the derivative of a function at a specific point is equal to the limit of the difference quotient as x approaches the given point. In other words, the derivative at x = c is equal to the limit of (f(x) - f(c))/(x - c) as x approaches c.

Substituting c = -2 into the alternative form of the derivative, we get:

f'(-2) = lim(x->-2) (f(x) - f(-2))/(x - (-2))

However, if we try to evaluate this limit, we get an indeterminate form of 0/0. This means that the derivative of f(x) at x = -2 does not exist, as the limit of the difference quotient is undefined. Therefore, the main answer is that the derivative of f(x) at x = c does not exist.

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Answer: 26 child tickets were sold that day.

Step-by-step explanation:

Let's say the number of child tickets sold is "x".

According to the problem, the number of adult tickets sold is four times the number of child tickets sold. So, the number of adult tickets sold would be 4x.

6.10x + 9.90(4x) = 1188.20

6.10x + 39.60x = 1188.20

45.70x = 1188.20

x = 26

At the given rate, the athlete could run 10.584 miles in 1.5 hours.

To determine how many miles the athlete could run in 1.5 hours at the given rate, follow these steps:

Step 1: Calculate the athlete's speed in miles per minute.

The athlete can run 6 miles in 51 minutes, so their speed is:

Speed = Distance ÷ Time = 6 miles ÷ 51 minutes ≈ 0.1176 miles per minute.

Step 2: Convert 1.5 hours to minutes.

1.5 hours = 1.5 × 60 = 90 minutes.

Step 3: Calculate the distance the athlete can run in 1.5 hours.

Distance = Speed × Time = 0.1176 miles per minute × 90 minutes ≈ 10.584 miles.

Therefore, at the given rate, the athlete could run approximately 10.584 miles in 1.5 hours.

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The particle is at rest when the velocity is zero.

To find the values of t, you need to calculate the first derivatives of the parametric equations and set them equal to zero.

Main answer: The particle is at rest for t = 0 and t = 2.


1. Calculate the first derivatives of x(t) and y(t):
dx/dt = 3t² - 6t
dy/dt = 6t² - 6t - 12

2. Set the derivatives equal to zero and solve for t:
3t² - 6t = 0
6t² - 6t - 12 = 0

3. Factor the equations:
t(3t - 6) = 0
6(t² - t - 2) = 0

4. Solve for t:
t = 0, (3t - 6) = 0
t² - t - 2 = 0

5. From the first equation, t = 0 or t = 2.
From the second equation, use the quadratic formula:
t = (1 ± √(1 + 8))/2
t ≈ 1.41, -1.41

6. The particle is at rest for t = 0 and t = 2. The other values do not correspond to a stationary point.

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the derivative of the function, then evaluate it at x=1 and finally multiply it by δx.

δy = -0.04 and f'(x)δx = -0.04.

An example of a differentiable function is f, and its derivative is f ′. If f has a derivative, it is denoted by the symbol f ′ and is known as f's second derivative. Similar to the second derivative, the third derivative of f is the derivative of the second derivative, if it exists. By carrying on with this method, the nth derivative can be defined, if it exists, as the derivative of the (n1)th derivative.

To find δy and f'(x)δx for the function y=f(x)=x−x^3 with x=1 and δx=0.02, we'll first find the derivative of the function, then evaluate it at x=1, and finally multiply it by δx.

1. The derivative of f(x)=x−x³ is f'(x)=1-3x²
2. Evaluating f'(x) at x=1, we get f'(1)=1-3(1)²=1-3=-2.
3. Now, we'll multiply f'(x) by δx: f'(1)δx = (-2)(0.02)=-0.04.

So, δy = -0.04 and f'(x)δx = -0.04.

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Currently Jane is 24 years old. Let's start by using variables to represent the ages of Jane and Amy. Let j be Jane's current age and a be Amy's current age.

From the first sentence of the problem, we know that j = a + 8. Now, let's focus on the second sentence of the problem. It says that Amy is now twice as old as Jane was at one-third Jane's current age.

Let's break down this sentence into smaller pieces. "Jane was at one-third Jane's current age" means that Jane's age at that time was j/3. "Amy is now twice as old as Jane was at one-third Jane's current age" means that:

a = 2(j/3)

3/2 × a = j

Now we have two equations that relate the ages of Jane and Amy:

j = a + 8

3/2 × a = j

We can substitute the first equation into the second equation to get an equation that only has one variable:

1/2 × a = 8

a = 16

So Amy = 16 years old. We can use the first equation to find Jane's age:

j = a + 8

j = 16 + 8

j = 24

Currently Jane is 24 years old.

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The factored form of polynomial P(x) is [tex]P(x) = 1(x + 2)(x^2 - x + 9).[/tex]

To factor [tex]P(x) = x^3 + 2x^2 + 9x + 18,[/tex]we need to first look for any common factors that we can factor out. In this case, we can factor out a 1, so:

[tex]P(x) = 1(x^3 + 2x^2 + 9x + 18)[/tex]

Next, we can try to find the roots of the polynomial by using the Rational Root Theorem, which states that if a polynomial has integer coefficients, then any rational root of the polynomial must have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In this case, the constant term is 18 and the leading coefficient is 1, so the possible rational roots are:

±1, ±2, ±3, ±6, ±9, ±18

We can try these roots by using synthetic division or long division to see if they are roots of the polynomial. After trying a few of these roots, we find that -2 is a root of the polynomial, so we can factor out (x + 2):

[tex]P(x) = 1(x^3 + 2x^2 + 9x + 18)\\     = 1(x + 2)(x^2 + ax + b)[/tex]

where a and b are coefficients that we need to find. To find a and b, we can use the fact that the coefficient of x^2 in the factored form should be equal to the coefficient of x^2 in the original polynomial. That is,

2 + 2a = 2

Solving for a, we get a = -1. Next, we can expand the factor (x^2 - x + b) and equate the coefficients of x and the constant term to the corresponding coefficients in the original polynomial. That is,

2a + b = 9
2b = 18

Solving for b, we get b = 9. Therefore, we have:

[tex]P(x) = 1(x + 2)(x^2 - x + 9)[/tex]

So the factored form of P(x) is [tex]P(x) = 1(x + 2)(x^2 - x + 9).[/tex]

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The factored form of polynomial P(x) is [tex]P(x) = 1(x + 2)(x^2 - x + 9).[/tex]

To factor [tex]P(x) = x^3 + 2x^2 + 9x + 18,[/tex]we need to first look for any common factors that we can factor out. In this case, we can factor out a 1, so:

[tex]P(x) = 1(x^3 + 2x^2 + 9x + 18)[/tex]

Next, we can try to find the roots of the polynomial by using the Rational Root Theorem, which states that if a polynomial has integer coefficients, then any rational root of the polynomial must have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In this case, the constant term is 18 and the leading coefficient is 1, so the possible rational roots are:

±1, ±2, ±3, ±6, ±9, ±18

We can try these roots by using synthetic division or long division to see if they are roots of the polynomial. After trying a few of these roots, we find that -2 is a root of the polynomial, so we can factor out (x + 2):

[tex]P(x) = 1(x^3 + 2x^2 + 9x + 18)\\     = 1(x + 2)(x^2 + ax + b)[/tex]

where a and b are coefficients that we need to find. To find a and b, we can use the fact that the coefficient of x^2 in the factored form should be equal to the coefficient of x^2 in the original polynomial. That is,

2 + 2a = 2

Solving for a, we get a = -1. Next, we can expand the factor (x^2 - x + b) and equate the coefficients of x and the constant term to the corresponding coefficients in the original polynomial. That is,

2a + b = 9
2b = 18

Solving for b, we get b = 9. Therefore, we have:

[tex]P(x) = 1(x + 2)(x^2 - x + 9)[/tex]

So the factored form of P(x) is [tex]P(x) = 1(x + 2)(x^2 - x + 9).[/tex]

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The numeric variable equivalent to equivalent to (x > 5) is, !(x < 5). The answer is D.

The original statement is "x > 5". The negation of this statement is "not (x > 5)", which is equivalent to "x <= 5". However, option A is the opposite of the correct answer since it says "x < 5", not "x <= 5". Option B says "not (x >= 5)", which is equivalent to "x < 5", but again, it is not the correct answer since it uses the "not greater than or equal to" symbol.

Option C says "not (x <= 5)", which is equivalent to "x > 5", but this is the opposite of the original statement. Therefore, the correct answer is D. !(x < 5), which is equivalent to "not (x is less than 5)", or "x is greater than or equal to 5". Hence, option D is correct.

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The mean (expected value) of the random variable x associated with the probability density function f(x) = (1/72)x^2 over the interval [0, 6] is 4.5.

To find the mean (expected value) of the random variable x associated with the probability density function f(x) = 1/72 x^2 over the interval [0, 6], we use the formula:

E(x) = ∫[0,6] x f(x) dx

= ∫[0,6] x (1/72 x^2) dx

= (1/72) ∫[0,6] x^3 dx

= (1/72) [(1/4) x^4] [0,6]

= (1/72) [(1/4) (6^4 - 0^4)]

= (1/72) (6^4/4)

= (1/72) (324)

= 4.5

To find the mean (expected value) of the random variable x associated with the probability density function f(x) = (1/72)x^2 over the interval [0, 6], we need to integrate the product of x and the probability density function over the given interval.

Mean (expected value) = E(x) = ∫(x * f(x)) dx, over the interval [0, 6]

E(x) = ∫(x * (1/72)x^2) dx from 0 to 6
E(x) = (1/72) * ∫(x^3) dx from 0 to 6

Now, integrate x^3 with respect to x:

E(x) = (1/72) * (x^4 / 4) | from 0 to 6

Now, evaluate the integral at the limits:

E(x) = (1/72) * ((6^4 / 4) - (0^4 / 4))
E(x) = (1/72) * (1296 / 4)
E(x) = (1/72) * 324

Finally, multiply the result:

E(x) = 4.5

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Required function g(h(x)) is 6 x - 9.

A function is a relationship between a set of outputs referred to as the range and a set of  inputs referred to as the domain, with the condition that each input is contain  to exactly one output. An input x corresponding to a function f output, which is represented by f(x).

We can combine two functions so that the outputs of one function become the inputs of the other if we have two functions is known as composite function . A composite function is defined by this action,that the function g f(x) = g(f(x)) is known as a composite function. This is occasionally referred to as a function of a function. g f can also be written as g o f instead.

We have, h(x)=3 x-5 and g(x)=2 x+1.

So, g(h(x)) = g(3 x - 5) = 2(3 x - 5) + 1 = 6 x - 9.

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Answer:

108 ft ^2

Step-by-step explanation:

12^2 - 6^2 = 108

The correct answer to the above question is Option C. divergent i.e., The series [infinity] ∑ sin(n) + 3^n is divergent.

To determine the convergence of the series, we need to check both the convergence of sin(n) and 3^n series.

Since both series diverge, their sum will also diverge, and the given series is therefore divergent.

In summary, the given series [infinity] ∑ sin(n) + 3^n is divergent as both the sin(n) and 3^n series diverge.

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The area of shaded polygons in RSM with 5 and 7 measurements having a blue form have a surface area of 0 square units.

In RSM, the area of the shaded polygons can be calculated.

They have provided 5 and 7 measurements in this instance, which we can use to determine how long the sides of the blue object should be.

The rectangle measures 7 units long by 5 units wide.

The bases of the two right triangles are 5 units and their heights are 2 units.

We apply the algorithm to determine the rectangle's area.

A = l x w,

Where,

A is denoted as the area,

l is  denoted as  the length and

w is denoted as the width.

A = 7 x 5 = 35 square units.
The shaded polygons' areas should be added.

The combined area of the shaded polygons in the RSM is calculated by adding the areas of each polygon.

Total Area = A1 + A2 and so on.

The area of one of the right angle triangles, we use the formula,

A = [tex]\frac{1}{2}[/tex] x b x h, [tex]\frac{1}{2}[/tex]

Where,

A is denoted as the area,

b is denoted as the base and

h is  denoted as the height.

Plugging in the values we get

A =  x 5 x 2 = [tex]5^{2}[/tex] units.

Since there are two right triangles the total area is

2 x 5 = 10 square units.
Therefore,

The area of the blue shape is

35 + 10 = 45 square units.

The rectangle's area and the areas of the two right triangles are,

35 + 10 = [tex]54^{2}[/tex] units.

Consequently, the shaded polygons' area is

45 - 45 = 0 square units.
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The shapes involved in the first figure is a triangle and a trapezium, with an area of 139.5. The shapes involved in the second figure is a triangle and a rectangle, with an area of 22 square units.

The first figure can be observed to be made up of a triangle and a trapezium. While the second is a triangle and a rectangle, so we shall calculate for the area and sum the results to get the total area of the composite figures as follows:

First figure:

area of the triangle = 1/2 × 9 × 6 = 27 square units

area of the trapezium = 1/2 × (6 + 9) × 15 = 112.5 square units

area of the first figure = 27 + 112.5 = 139.5 square units

Second figure:

area of the triangle = 1/2 × 4 × 2 = 4 square units

area of the rectangle = 9 × 2 = 18 square

area of the second figure = 4 + 18 = 22 square units.

Therefore, the shapes involved in the first figure is a triangle and a trapezium, with an area of 139.5. The shapes involved in the second figure is a triangle and a rectangle, with an area of 22 square units.

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By the rank-nullity theorem, we know that for any matrix A, the rank of A plus the dimension of the null space of A is equal to the number of columns of A. If the null space of a 7 times 9 matrix is 3-dimensional, Rank A = 6, Dim Row A = 6, Dim Col A = 6

we know that for any matrix A, the rank of A plus the dimension of the null space of A is equal to the number of columns of A. That is:

Rank A + Dim Null A = # of columns of A

In this case, we are given that the null space of the 7x9 matrix A is 3-dimensional. Therefore, we have:

Rank A + 3 = 9

Solving for Rank A, we get:

Rank A = 6

Now, we also know that the rank of a matrix is equal to the dimension of its row space and the dimension of its column space. That is:

Rank A = Dim Row A = Dim Col A

Therefore, we have:

Rank A = Dim Row A = Dim Col A = 6

So the correct option is: Rank A = 6, Dim Row A = 6, Dim Col A = 6

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Answer:

0

Step-by-step explanation:

ny^2 + 2y - 4 = 0

ny^2 + 2y = 4

y(ny + 2) = 4

y = 4

ny + 2 = 4

ny = 2, 0 = 2

The only possible solution to make this expression incorrect is if 0 = 2, so n is equal to 0.

Check the picture below.

[tex]\cfrac{1^2}{2^2}=\cfrac{110}{A}\implies \cfrac{1}{4}=\cfrac{110}{A}\implies A=440~in^2[/tex]

The following parts can be answered by the concept of Differential equation.

(a) For the differential equation 4x²y'' + 4xy' - (64x² - 9)y = 0, we can rewrite it as:

y'' + (1/x)y' - (16 - 9/x²)y = 0

This is a Bessel's equation of order ν = 3. The general solution is given by:

y(x) = c_1 J_3(2√2x) + c_2 Y_3(2√2x)

where c_1 and c_2 are constants, J_3 is the Bessel function of the first kind of order 3, and Y_3 is the Bessel function of the second kind of order 3.

(b) For the differential equation x²y'' + xy' - (36x² - 9)y = 0, we can rewrite it as:

y'' + (1/x)y' - (36 - 9/x²)y = 0

This is also a Bessel's equation, but with order ν = 3/2. The general solution is given by:

y(x) = c_1 J_(3/2)(6x) + c_2 Y_(3/2)(6x)

where c_1 and c_2 are constants, J_(3/2) is the Bessel function of the first kind of order 3/2, and Y_(3/2) is the Bessel function of the second kind of order 3/2.

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Answer:

3

Step-by-step explanation:

gd=14cm

dc=17cm

then,

gd-dc

14cm-17cm

0=14cm-17cm

0=-3

0+3

3

The solution to the differential equation that satisfies the initial conditions y(3) = y0 and y(8) = y1 is:

y(t) = (2/3)t - (1/3)cos(t) + (1/3)sin(t) + y1 + (1/3)sin(3) - (2

The given differential equation is:

3y''+2y'y-y-tan(t)=0

To check the existence and uniqueness of a solution, we need to verify if the function p(t) and q(t) satisfy the conditions of the Existence and Uniqueness Theorem.

The Existence and Uniqueness Theorem states that if the functions p(t) and q(t) are continuous on an interval containing a point t0 and if p(t) is not equal to zero at t0, then there exists a unique solution to the differential equation y'' + p(t) y' + q(t) y = g(t) that satisfies the initial conditions y(t0) = y0 and y'(t0) = y1.

Comparing the given differential equation with the standard form of the Existence and Uniqueness Theorem, we get:

p(t) = 2y(t)

q(t) = -t - tan(t)

g(t) = 0

To find the interval of existence, we need to check the continuity of p(t) and q(t) and also the value of p(t) at t0.

Here, p(t) is continuous everywhere and q(t) is continuous on the interval (3, 8). To check the value of p(t) at t0, we need to find y(t) that satisfies the initial conditions y(3) = y0 and y(8) = y1.

Let's assume that y(t) = A(t) + B(t), where A(t) satisfies y(3) = y0 and A'(3) = 0 and B(t) satisfies y(8) = y1 and B'(8) = 0.

Solving the differential equation for A(t), we get:

A(t) = c1 cos(sqrt(3)(t-3)) + c2 sin(sqrt(3)(t-3)) + (2/3)t - (1/3)cos(t) + (1/3)sin(t) + (1/3)sin(3)

Using the initial conditions y(3) = y0 and A'(3) = 0, we get:

A(t) = (2/3)t - (1/3)cos(t) + (1/3)sin(t) + (1/3)sin(3) - (2/3)cos(3) - y0

Solving the differential equation for B(t), we get:

B(t) = c3 cos(sqrt(3)(t-8)) + c4 sin(sqrt(3)(t-8)) + (2/3)t - (1/3)cos(t) + (1/3)sin(t) - (1/3)sin(3)

Using the initial conditions y(8) = y1 and B'(8) = 0, we get:

B(t) = (2/3)t - (1/3)cos(t) + (1/3)sin(t) - (1/3)sin(3) + (2/3)cos(3) + y1

Therefore, the solution to the differential equation that satisfies the initial conditions y(3) = y0 and y(8) = y1 is:

y(t) = (2/3)t - (1/3)cos(t) + (1/3)sin(t) + y1 + (1/3)sin(3) - (2)

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The composite figure has an area of 24 square units.

In this question we find the representation of a composite figure formed by the combination of four figures, a triangle and three rectangles, whose area formulas are listed below:

Rectangle

A = b · h

Triangle

A = 0.5 · b · h

Where:

Now we proceed to determine the area of the composite figure:

A = 2 · 3 + 0.5 · 2 · 1 + 7 · 2 + 1 · 3

A = 6 + 1 + 14 + 3

A = 24

The area of the composite figure is equal to 24 square units.  

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