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For a the system is observable,In control theory, a system is observable if it is possible to determine the internal state of the system by observing its inputs and outputs. This means that the system's internal state can be reconstructed by observing its external behavior. An observer is a system that estimates the internal state of a system using its inputs and outputs. An observer whose dynamics can be characterized by a second-order system is called a second-order observer.

a.To check the observability of the system, we need to find the observability matrix. The observability matrix for a second-order system is given by:
O = [C ; CA]
where C is the output matrix and A is the state matrix.
In this case, the state matrix is:  A = [0 1 ; -4 0]
and the output matrix is:  C = [1 0]
So the observability matrix is:  O = [1 0 ; 0 1 ; 0 -4 ; -4 0]
We can check the rank of this matrix, which should be equal to the number of states (i.e., 2) for the system to be observable.
Rank(O) = 4, which means the system is observable.
b. We need to design an observer with a natural frequency of 10 and a damping ratio of 0.8.
The observer dynamics can be modeled by a second-order system:  ẋo = Ao xo + Bo y,  yo = Co xo
where xo is the observer state vector, y is the measured output, and Ao, Bo, and Co are matrices to be determined.
The characteristic equation of the observer system is:
s^2 + 2ζωn s + ωn^2 = 0
where ζ is the damping ratio, ωn is the natural frequency, and s is the Laplace variable.
Plugging in the given values, we get:  s^2 + 1.6s + 100 = 0
Solving for the roots of this equation, we get:  s1 = -0.8 + 9.9499i,  s2 = -0.8 - 9.9499i
So the observer poles are located at these two complex conjugate values.
We can use the pole-placement method to design the observer matrices Ao, Bo, and Co.
First, let's choose a suitable observer gain L:  L = [l1; l2]
where l1 and l2 are scalars to be determined.
The observer matrices can then be calculated as:  Ao = A - LC,  Bo = [B ; 0], and Co = [0 0 1 0]
where C is the output matrix of the original system.
Plugging in the given values, we get:  Ao = [0 1 ; -4 0] - l1[1 0] - l2[0 1], Bo = [0 ; 1], and Co = [0 0 1 0]
We can use the pole-placement method to find suitable values of l1 and l2 that give us the desired observer poles.
The characteristic equation of the observer system can be written as:  det(sI - Ao) = (s + 0.8 - 9.9499i)(s + 0.8 + 9.9499i)
Expanding the determinant and equating the coefficients of s^2 and s, we get: l1 = -9.6, and l2 = 40.2
So the observer gain matrix is: L = [-9.6 ; 40.2]
Plugging in the observer gain into the observer matrices, we get: Ao = [0 1 ; -4 -9.6], Bo = [0 ; 1], and Co = [0 0 1 0]
And that's it! We have designed an observer with the desired natural frequency and damping ratio for the given second-order system.

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the general solution for the given differential equation is: y = c1e^x + c2e^-4x + c3 + (1/12)e^-2x + y_p(x),

For the first equation y"' – 2y" – 5y' + 6y = e^x + x^2, we assume a particular solution of the form y_p = Ae^x + Bx^2 + Cx + D. Taking the first, second and third derivatives, we get:

y_p' = Ae^x + 2Bx + C

y_p'' = Ae^x + 2B

y_p''' = Ae^x

Substituting these back into the original equation, we get:

Ae^x - 4Be^x - 5Ae^x - 10Bx - 5C + 6Ae^x + 6Bx^2 + 6Cx + 6D = e^x + x^2

Simplifying and comparing coefficients, we get:

A - 5A + 6A = 1, which gives A = 1

-4B + 2B + 6B = 0, which gives B = 0

-5A - 5C + 6C = 0, which gives C = 5/2

6D = 0, which gives D = 0

Therefore, the particular solution is y_p = e^x + (5/2)x.

For the second equation y"' + y" – 5y' + 3y = e^-x +sinx, we assume a particular solution of the form y_p = Ae^-x + Bsinx + Ccosx + Dx + E. Taking the first, second and third derivatives, we get:

y_p' = -Ae^-x + Bcosx - Csinx + D

y_p'' = Ae^-x - Bsinx - Ccosx

y_p''' = -Ae^-x - Bcosx + Csinx

Substituting these back into the original equation, we get:

-Ae^-x - Bcosx + Csinx - Ae^-x + Bsinx + Ccosx - 5(-Ae^-x + Bcosx - Csinx) + 3(Ae^-x - Bsinx - Ccosx) = e^-x + sinx

Simplifying and comparing coefficients, we get:

-2A - 4B + 4C = 1, which gives A = -1/2, B = 0 and C = 1/4

-2B - 2C + 5A + 3B - 3C = 0, which gives A = -1/2

-2C - 5A + 3C = 0, which gives C = 5/14

D = 0

E = 0

Therefore, the particular solution is y_p = (-1/2)e^-x + (5/14)cosx.

For the third equation y''' + 3y" – 4y = e^-2x, we assume a particular solution of the form y_p = Ae^-2x. Taking the first, second and third derivatives, we get:

y_p' = -2Ae^-2x

y_p'' = 4Ae^-2x

y_p''' = -8Ae^-2x

Substituting these back into the original equation, we get:

-8Ae^-2x + 12Ae^-2x - 4Ae^-2x = e^-2x

Simplifying and comparing coefficients, we get:

A = 1/12

Therefore, the particular solution is y_p = (1/12)e^-2x.

Hence, the general solution for the differential equation y''' + 3y" – 4y = e^-2x is given by:

y = y_h + y_p,

where y_h is the homogeneous solution and y_p is the particular solution.

The homogeneous solution is obtained by assuming y_h = e^mx and solving the characteristic equation:

m^3 + 3m^2 - 4m = 0.

Factoring out m, we get:

m(m^2 + 3m - 4) = 0.

Solving for the roots of the quadratic factor, we get:

m = 1, -4, or 0.

Therefore, the homogeneous solution is:

y_h = c1e^x + c2e^-4x + c3,

where c1, c2 and c3 are constants determined by the initial or boundary conditions.

Hence, the general solution for the given differential equation is:

y = c1e^x + c2e^-4x + c3 + (1/12)e^-2x + y_p(x),

where y_p(x) is the particular solution determined using the method of undetermined coefficients.

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The fraction of stars in the milky way having an earth-size planet in the habitable zone are around 10-20%.

Milky way is the galaxy comprising our solar system and therefore marking Earth's location in the universe. The milky way galaxy is just one among billions of different galaxies in the universe.

It contains numerous stars and the planets revolving it. Thus, the milky way galaxy can said to contain thousands of planetary systems. The planets in turn can be habitable or non-habitable owing to atmospheric conditions.

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Answer:678.24 :)

Step-by-step explanation:

v=sh

=pi(r^2-v^2)*h

=pi(5^-2-4^2)*24

=pi-9*24

=678.24 in

A. We have shown that a = b = c = 0, and f, g, h are linearly independent.

B. the orthonormal basis for U is {u1, u2, u3} = {√x, (2x - 1)/√3, √(10/7)(x² - 1/3)}.

C. we have: 1( p,u1 ) = ∫0 x³√x dx = 2/5.

What is function?

In mathematics, a function is a relationship between two sets of elements, called the domain and the range, such that each element in the domain is associated with a unique element in the range.

Let's solve each part of the problem:

(a) To show that f, g, h are linearly independent, we need to show that the only solution to the equation:

c1f(x) + c2g(x) + c3h(x) = 0

for all x in [0,1] is c1=c2=c3=0.

Using the definitions of f(x), g(x), and h(x), we have:

c1√x + c2x + c3x^2 = 0

To show that this equation only holds for c1=c2=c3=0, we need to choose three distinct values of x, say x=0, x=1/4, and x=1, and solve the resulting system of linear equations:

c1√0 + c2(0) + c3(0)^2 = 0

c1√(1/4) + c2(1/4) + c3(1/4)^2 = 0

c1√1 + c2(1) + c3(1)^2 = 0

The first equation reduces to c1=0. Substituting this into the remaining two equations yields:

c2/2 + c3/16 = 0

c2 + c3 = 0

This system of linear equations has only the trivial solution c1=c2=c3=0, so f, g, h are linearly independent.

(b) To find an orthonormal basis for U, we can use the Gram-Schmidt process. Let u1 = f(x) = √x, and define v1 = u1/||u1||, where ||u1|| = √(1/3):

v1(x) = √3√x

Next, let u2 = g(x) = x, and subtract the projection of u2 onto v1:

u2 - proj_v1(u2) = x - ∫0^1 (x√3√t)/t dt

= x - 2/(3√3)

Let v2 = u2/||u2||, where ||u2|| = √(1/3 - 4/(9√3)):

v2(x) = √(9x^2 - 4)/(3√3)

Finally, let u3 = h(x) = x^2, and subtract the projections onto v1 and v2:

u3 - proj_v1(u3) - proj_v2(u3) = x^2 - (2/3)x - 1/5

Let v3 = u3/||u3||, where ||u3|| = √(1/3 - 4/(9√3) - 4/75):

v3(x) = √(225x^4 - 120x^3 + 40x^2 - 15x + 4)/(5√15)

Therefore, an orthonormal basis for U is {v1, v2, v3}.

(c) To find the closest approximation of p(x) = x^3 in U, we need to find the projection of p(x) onto U:

proj_U(p(x)) = (p(x), v1)v1 + (p(x), v2)v2 + (p(x), v3)v3

Using the inner product, we have:

(p(x), v1) = ∫0^1 x^3√3√x dx = 2/5√3

(p(x), v2) =

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The simplified expression is -4mv.

We have,

To simplify the expression:

-20(m²v)(-v³) / 5(-v²)(-m^4)

We can first cancel out common factors in the numerator and denominator.

-20(m²v)(-v³) / 5(-v²)(-m^4)

= -4m²v² / vm²

Next, we can simplify further by canceling out a factor of v in the numerator and denominator:

-4m²v² / vm² = -4mv

Therefore,

The simplified expression is -4mv.

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Answer:

median = 78.5

Step-by-step explanation:

the median is the middle value of the data set arranged in ascending order. If there is no exact middle value then it the average of the values either side of the middle.

arrange scores in ascending order

45 , 63 , 77 , 80 , 86 , 93

                  ↑ middle

median = (77 + 80) ÷ 2 = 157 ÷ 2 = 78.5