## Eigenvalues & Eigenvectors Assignment Help

Eigenvalues & Eigenvectors

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Eigenvalues & Eigenvectors

Problem 1

The matrix has eigenvalues1=3 and 2=−2. Let’s find the eigenvectors corresponding to 1=3. Let v=v2v1. Then(A−3I)v=0 gives us

2−3−1−4−1−3v1v2=00 from which we obtain the duplicate equations

v1−4v2−v1−4v2=0

If we let v2=t, thenv1=−4t. All eigenvectors corresponding to1=3are multiples of1−4 and thus the eigenspace corresponding to1=3is given by the span of1−4. That is,1−4is a basis of the eigenspace corresponding to 1=3.

Repeating this process with 2=−2, we find that

4v1−4V2−v1+v2=0

If we let v2=t then v1=t as well. Thus, an eigenvector corresponding to2=−2 is 11 and the eigenspace corresponding to 2=−2 is given by the span of11. 11is a basis for the eigenspace corresponding to 2=−2.

Problem 3

Problem 5

Problem 14

Problem 22

Problem 25

Matrices A2 and A3 cannot be diagonized because for a square matrix A, wherever A is similar to diagonal matrix then the matrix is diagonizable.

Problem 15-24: Eigenvalues & Eigenvectors Matrices

Problem 15

Since all entries are ≥ 0 and each column sums to 1, this A is a Markov matrix. Thus we know that λ1 = 1 is an eigenvalue. Since tr(A) = λ1 + λ2 = 3/2, we conclude λ2 = 1/2 is another eigenvalue. We diagonalize it using the matrix S of eigenvectors:

= → =

This last matrix product equals

Problem 19

1. False
2. True
3. True
4. False

References

Bhatti, M. A. (2012). Practical Optimization Methods with Mathematica Applications. New York, NY: Springer.

Edwards, C. H., & David E. Penney, D. E. (2009). Differential Equations: Computing and Modeling. Upper Saddle River, CA: Pearson Education, Inc.

Shores, T. S. (2007). Applied linear algebra and matrix analysis. New York, NY: Springer Science+Business Media, LLC.

Strang, G. (2003). Introduction to linear algebra. Wellesley, MA:  Wellesley-Cambridge Press

Strang, G. (2006). Linear algebra and its applications. Belmont, CA:  Thomson, Brooks/Cole Publishers.

Strang, G. (2009). Eigenvalues and Eigenvectors. Boston, MA: Lord Foundation of Massachusetts.

Zhang, F. (2009). Linear Algebra: Challenging Problems for Students, Baltimore, MA: The Johns Hopkins University Press.http://nsuworks.nova.edu/cnso_math_facbooks/3/

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## Linear algebra Term Paper Available Here

Linear algebra

Chapter 3

Question 1

If the above data points actually lay on a straight line y = C+Dt, we have:

Call the matrix A and the vector on the right-hand side. Of course this system is inconsistent, but we want to find  such that  is as close as possible to . As we’ve seen, the correct choice of  is given by:

To compute this, first note that

Therefore,

And so

Therefore, the best-fit line for the data is:

Here are the data points and the best-fit line on the same graph:

Question 13

Let  Then, we set  and

Question 14

We now compute If we denote then by the Gram-Schmidt process,

Question 16

Chapter 4

Question 5

Question 6

Let be any basis of a -dimensional subspace of Then by Gram-Schmidt orthogonalisation process, we get an orthonormal set with and for

Question 7

Question 16

Question 17

Question 24

Question 25

Suppose we have matrix of dimension with Then by the application of the Gram-Schmidt orthogonalisation process yields a set of orthonormal vectors of In this case, for each we have

Question 14

Question 3

References

Bretscher, O. (2004). Linear Algebra with Applications, (3rd ed.).  New York, NY: Prentice Hall.

Farin, G., & Hansford, D. (2004). Practical Linear Algebra: A Geometry Toolbox. London: AK Peters.

Friedberg, S. H., Insel, A. J., & Spence, L. E. (2002). Linear Algebra, (4th ed.). New York, NY: Prentice Hall.

Leon, S. J. (2006). Linear Algebra with Applications, (7th ed.). New York, NY: Pearson Prentice Hall.

McMahon, D. (2005). Linear Algebra Demystified. New York, NY: McGraw–Hill Professional.

Zhang, F. (2009). Linear Algebra: Challenging Problems for Students. Baltimore, MA: The Johns Hopkins University Press.

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## Matrices Assignment Help Available Here

Matrices

Order Instructions:

Hi,

Please mention the explanation into the answer, on how did you solve the questions. And also please put the question along with the answer.

=> add -1 times the 1st row to the 3rd row  =>

=> add -2 times the 2nd row to the 1st row=>

=> add -4 times the 1st row to the 2nd row  =>

=> add -7 times the 1st row to the 3rd row =>

=> multiply the 2nd row by -1/3 =>

=> add 6 times the 2nd row to the 3rd row =>

=> add -2 times the 2nd row to the 1st row =>

3rd row is free for both A and B.

Consider the augmented matrix

We can convert this to reduced echelon form by subtracting twice row 1 from row 3 and

subtracting 3 times row 2 from row 3:

In order for this system to be consistent, it must be the case that

There are no constraints on b1 and b2, the possible right-hand sides of the equation are vectors of the form

for any real numbers b1 and b2. In other words, the column space of A is the plane containing the vectors and .  Looking at the reduced echelon form of the matrix, we see that it is of rank 2 and that a particular solution of the given equation is .

=> Switch rows 1 and 3 =>

Divide row 1 by 2, row 2 by 3 =>

Multiply row 2 by 3 and subtract from row 1 =>

=> Switch rows 1 and 3 =>

Divide row 1 by 2, row 2 by 3 =>

Multiply row 2 by 3 and subtract from row 1 =>

=> Switch rows 1 and 3 =>

Divide row 1 by 2, row 2 by 3 =>

Multiply row 2 by 3 and subtract from row 1 =>

Let .  Then . Thus .

So .

By the Rank-Nullity Theorem,

and

So it suffices to show that, therefore,

Form the augmented matrix [A b]:

Then subtracting row 1 from rows 2 and 3 and multiplying row 1 by 1 2 yields

Next, subtracting twice row 2 from row 1 and adding row 2 to row 3 gives

This is now in reduced echelon form, so we can answer the question. Notice that the pivot columns are the first and second columns; hence, the column space of A is the span of the first two columns of A, namely  and .

Geometrically, this is just the plane containing these two vectors. Returning the the reduced echelon form of the augmented matrix, notice that we must have

x1 = 4 − x3 + 2x4

x2 = −1 − x3 − 2x4

so the special solutions are of the form

for some real numbers x3 and x4. Hence, the nullspace of A consists precisely of such linear combinations.

Finally, all solutions to the equation Ax = b are of the form

where the first term is a particular solution and the latter two terms comprise the special (or homogeneous) solutions.

1. Solution:

Solution: Any four vectors in R3 are dependent (since the dimension of R3 is 3), and the vectors v1, v2, v3 are independent because they are columns of an invertible 3 by 3 matrix. Solving the system Ax = 0 is finding the nullspace of the matrix

Special solution here is x4 = 1, x3 = −4, x2 = 1, x1 = 1, so the nullspace consists of all (c1, c2, c3, c4) = x4(1, 1, −4, 1).

1. Solution:

Since v4 = v2 − v1, v5 = v3 − v1, and v6 = v3 − v2, there are at most three independent vectors among these: furthermore, applying row reduction to the matrix [v1 v2 v3] gives three pivots, showing that v1, v2, and v3 are independent.

The echelon matrix  has only a single pivot, in the second column . The second column  is therefore a basis for the column space . (The third column of  is equal to four times the second column.) The dimension of  is 1 (same as the rank of ).

, all vectors with the last coordinate equal to 0.

, all vectors with the first coordinate equal to 0.

, all vectors with the last coordinate equal to 0.

, all vectors with the first coordinate equal to 0.

transformed into echelon form

=>  add -1 times the 1st row to the 2nd row =>

=>  add -1 times the 2nd row to the 3rd row =>

=>  add -3 times the 2nd row to the 1st row =>

The second and the fourth variables are the pivots. The first, third, and the fifth variables are free variables. The row operation matrix E (R = EA) is

Nullspace: The null space is as follow:

for .

Column space: Since the pivot columns (the second and the fourth) of A span the column space, we have

for

Row space: it is the same as the row space of R as R is obtained by invertible row operations. So

for

Left nullspace: It has a basis given by the rows of E for which the corresponding rows of R are all zero. That is to say, we need to take the last row of E. Thus,

for

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## Triangular Factors and Row Exchanges

Triangular Factors and Row Exchanges

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coursework.

1.5. Triangular Factors and Row Exchanges

Question 1

Upper triangular matrix is non-singular when no entry on the main diagonal is zero.

Question 2

Solution:

Hence elimination will subtract 4 times row 2 from row 3.

Looking at the U matrix, we see the pivots along the diagonal of the matrix:

To find out if a row exchange will be needed or not, first we determine A

After carrying the first elimination, we get:

Hence, there would not be a need for a row exchange.

Question 3

Question 7

Question 13

Question 23

1.6. Inverses and Transposes

Question 1

Question 2

• When is applied to a matrix A its effect on A is to replace the first row of by the 3rd row of, the 3rd row of, and to replace the 3rd row by the first row at the same time. As an example. Hence to reverse this effect, we need to perform the same operation again, i.e. replace the 1st row with the 3rd row and replaced the 3rd row by the 1st row, this is P1

Hence,

Therefore,

Is an indication to replace 1st row by the 3rd row and to replace the 2nd row by the 1st row, and to replace the 2nd row by the 3rd row. For example; hence, in order to reverse it, we need to replace the 1st row by the 2nd row, and to replace the 2nd row by the 3rd row and to replace the 3rd row by the 1st row at the same time.

Hence,

• In a permutation matrix P, each row will have at most one non-zero entry with value of 1. Consider the entry Pi, j = This entry will cause row i to be replaced by row j. Hence to reverse the effect, we need to replace row j by row i, or in other words, we need to have the entry (j, i) in the inverse matrix be 1. But this is the same as transposing P, since in a transposed matrix the entry (i, j) goes to (j, Hence. Now, where  , are permutation matrices (in other words, each row of  is all zeros, except for one entry with value of 1.)

Hence the entry C (i, i) will be 1 whenever A (i, j) = B (j, i) = 1, this is from the definition of matrix multiplication, element by element view, since:

For all entries except when the entry, and at the same time, but since   is the transpose of P, then whenever then  only when.

Hence this leads to  with all other entries in C being zero, i.e.

Question 4

(a)

If A is invertible and AB = AC, then B = C is true.

Because;

This means that; IB = IC, and subsequently B = C

(b)

Suppose:

AC = AB; then:

0 = AB – AC = A (B-C)

If  , Note that if we define

Then you can find by multiplying it out that:

Therefore, if we set    we find that:
So for this choice of A, just pick any 2×2 matrix for B, and define C = D + B; and automatically, you will find that: AC = A (D + B) = AD + AB = 0 + AB = AB; but C is not = B.

Question 6

Then the only row on the left that doesn’t already look like the identity matrix is the second row; we just need subtract rows 1 and 3 from row 2, which gives:

Hence,

Replace the 1st row by half of itself and add half of the 1st row to the 2nd row:

Next, add a third of the second row to the first, add 2/3 the second row to the third, and multiply the second row by 2/3:

Finally, multiply the third row by 3/4, and then add 1/3 of the result to row 1 and add 2/3 of the result to row 2:

Thus,

First, switch rows 1 and 3:

Now, subtract row 2 from row 1 and subtract row 3 from row 2:

Thus,

Question 10

For the first choice of A, we write the augmented matrix [A I]:

Then subtracting two times row 1 from row 2 and subtracting three times row 3 from row 2 yields

Hence,

For the second choice of A, write the augmented matrix [A I]:

Subtracting row 1 from rows 2 and 3 yields:

In turn, subtracting row 2 from rows 1 and 3 yields:

Finally, subtracting row 3 from row 2 yields:

Hence,

Question 13

Question 17

• The inverse of a lower (upper) triangular matrix is still lower (upper) triangular. Multiplying lower (upper) triangular matrices gives a lower (upper) triangular matrix.
• The main diagonal of and  are the same as those of  and   , so we have. By comparing the off-diagonals of, both matrices must be diagonal. ,  is invertible so.

Then

Question 18

To make the pivot actually occur at f, switch rows 1 and 3:

Now, subtract d/j times row 1 from row 2 and subtract a/f times row 1 from row 3; note:

If e = 0 then the 2nd row is all zeros, which means that there can’t be a pivot in that row. Thus, when A is invertible, it must be the case that e 0. Therefore, a pivot is in the 2nd column, and the entry below can be eliminated through subtraction of  times on row 2 from row 3:

If A is to be invertible, it must be the case that c 0. Therefore, the conditions which ensure that A is invertible are:

Turning to B, the matrix becomes:

Then, in order to have a pivot in the 2nd row, it must be the case that

Or, equivalently,

On the other hand, if, so we can switch rows 1 and 2 to get

Then we can eliminate a by subtracting  times row 1 from row 2 (note:

Again, if we are to have a pivot in the 2nd row, it must be the case that:

Or, equivalently,

Therefore, either

Question 22

Question 43

The 5 by 5  also has 1s on the diagonal and super-diagonal.

Review Exercises

Question 1.2

Question 1.4

Question 1.10

Has L = I and;

A = LU has U = A (pivots on the diagonal);

A = LDU has   with 1s on the diagonal.

Question 1.17

Question 1.28

2 by 2; d = 0 not allowed;

d = 1, e = 1; then l = 1, f = 0 is not allowed.

Vector Spaces

Question 2

The answer is: (a), (b) and (e).

Question 3

is the x-axis;  is the line through (1, 1);  is ;  is the line through (-2, 1, 0);  is the point (0, 0) in ; the null space  is .

Question 14

The subspaces of  are  itself, lines through (0, 0), and the point (0, 0).

The subspaces of  are  itself, three-dimensional planes , two-dimensional subspaces , one-dimensional lines through (0, 0, 0), and (0, 0, 0) alone indicate that the smallest subspace containing P and L is either P or .

Question 24

The extra column b enlarges the column space, unless b is already in that space,

Reference

Strang, G. (2013). Linear Algebra and its Applications, (4th ed.). New York, NY: McGraw-Hill Publishers.

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## Acoustic principles using mathematical analysis

Acoustic principles using mathematical analysis

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Main Assignment Questions (90% of unit weighting)
AR 50345

The objective of the course work is to demonstrate an understanding of the acoustic principles discussed during the unit week. This can be done in the form of mathematical analysis, an essay style answer or as bullet points. Marks against each section are
indicated as are word lengths, as appropriate.
Q1 a) Please outline your understanding of the following acoustic terms, (using 50 to 100 words each). Credit will be given for each key requirement or unique feature included. Additionally, please indicate where each of these terms are used in the field of acoustics: (10 marks)
Free field
Diffused field
dB
dBA
NR
W-weighting
Rw
R
0
w
(apparent sound reduction)
Dw
Dn
T
w
calculating the following.
(15 marks)
dBA for the following spectra:
32Hz 64Hz 125Hz 250Hz 500Hz 1kHz 2kHz 4kHz 8kHz
30 30 30 – – – – – — – – 30 35 36 50 36 35
65 83 90 75 43 35 – – — 10 10 10 10 10 10 10 -NR:
32Hz 64Hz 125Hz 250Hz 500Hz 1kHz 2kHz 4kHz 8kHz
30 30 30 – – – – – — – – 30 35 36 50 36 35
65 83 90 75 43 35 – – — 10 10 10 10 10 10 10 -Rw
(spread sheet only) for the following spectra:
125Hz 250Hz 500Hz 1kHz 2kHz
16 24 33 42 43
39 44 56 67 74
38 37 48 64 78
34 39 51 60 68
2 of 4 AR50345
Q2 a) Please explain, or undertake mathematically, how one determines
the composite sound reduction of the following two elements. The area of element 1 should be taken as 9m 2 with an Rw of 40dB.
The second element has an area of 2m 2 and an R w of 30dB.
(5 marks)
b) Use the above explanation to show how the 10dB rule works, i.e.
“There is no benefit in increasing the sound reduction of the larger element more than 10 dB over the smaller element which has a lesser sound reduction.” (5 marks)

Q3 a) Calculate the sound reduction of a facade comprising of: (6 marks)
10m 2 of glass/windows 20m 2 of brickwork and a 5m
2 spandrel panel
The area of the room should be taken as 200m 2
. The reverberation time in this room is equal to 0.7 seconds. Please state the sound reduction used for each element making up the facade and the source of this information.
b) Repeat this exercise but now include a plot of the combined sound reduction when an open window is placed in the facade.
This window has an effective open area of 1m by 0.01m; this window is then further opened in increments to a final effective open area of 1mby 0.25m. The area of the window should be taken out of the glazed element within the facade. (4 marks)

Q4 Sustainability and Acoustics. (20 marks)
Acoustics and sustainability are not often referred to in tandem. On the other hand, it has been shown that there are many links between these two subjects. Please provide your thoughts with respect to the links between acoustics and sustainable building
design.
Please also provide a review of products which are available to enhance the acoustic performance of a vented facade. This information can also be complemented with acoustics details relating to cross ventilation. (500 words) 3 of 4 AR50345

Q5 Reverberation time calculation. (15 marks)
Compose an Excel spreadsheet that calculates the absorption coefficient of the sound absorbing panels in the East Building classroom from 125 Hz to 4000 Hz (octave bands). To do this you should use the test data from the practical session for the empty room (with tables and chairs removed).                                    This data is as follows:
Octave Band Centre Frequency, Hz
125 250 500 1000 2000 4000
Measured Average RT Empty
(Seconds)
0.9 0.8 0.9 0.7 0.7 0.6
The room dimensions may be taken as 13.2W x 5.2D x 3.2H.
There are 65 absorptive panels each with an area of 0.45mx 0.6m. You will need to estimate the absorption coefficients for the other surfaces (e.g. carpet, concrete soffit, glass facade, studwork
walls and doors and use these to calculate the absorption coefficient of the panels. What effect would distributing the panels throughout the room have compared with locating the panels in one large area?

Q6 A building has two items of plant on the roof, (a kitchen extract fan and a kitchen supply fan). The sound power levels of the units are as follows: (20 marks)
Frequency 63Hz 125Hz 250Hz 500Hz 1kHz 2kHz 4kHz
Kitchen Extract 84 85 83 80 78 76 72
Kitchen Supply 76 78 78 75 73 70 68
Calculate the A-weighted sound pressure level (dBA) at the nearest house, which is approximately 25m from the plant. You may assume that the kitchen extract fan is screened by a parapet that just obscures the line of sight. The kitchen supply fan is un-screened. You should clearly identify each step in your calculation or submit a spreadsheet.

Question 1: Part B

dBA for the following spectra:

 32 Hz 64 Hz 125 Hz 250 Hz 500 Hz 1 kHz 2 kHz 4 kHz 8 kHz 30 30 30 45 20 30 20 25 25 35 53 60 30 35 30 50 36 35 65 83 90 75 43 35 30 45 55 10 10 10 10 10 10 10 10 10

NR for the following spectra:

 32 Hz 64 Hz 125 Hz 250 Hz 500 Hz 1 kHz 2 kHz 4 kHz 8 kHz 30 30 25 35 30 45 20 35 30 30 50 65 30 35 30 50 36 35 65 83 90 75 43 35 25 40 50 10 10 10 10 10 10 10 10 10

Question 2: Part A

The composite sound reduction of two elements can be used using the weighted sound reduction index formula given below:

Where;

D = Level difference

S = Area of the test element through which the noise is transmitted

A = Total sound absorption by the element

Therefore, the composite sound reduction of the two elements can be determined by using the above formulae which can be used to the absorption factors of the two elements given by [ ]. With the absorption factors of the two elements and the average sound pressure of the source and receiving area, the composite sound reduction can be determined for the two elements.

Question 2: Part B

Using the explanation given above it is possible to show that the 10 dB rule works, i.e. “There is no benefit in increasing the sound reduction of the larger element more that 10 dB over the smaller element which has a lesser sound reduction.”

Let’s now consider the parameters of the two elements;

Element 1: Area = 9 m2 and Rw = 40 dB

Element 2: Area = 2 m2 and Rw = 30 dB

The composite sound reduction equation provided above can be used to proof that the 10 dB rule works, by computing the absorption factors for each element, which in the case of these two elements can be calculated from the given details of the two elements. The calculations for absorption factors for the two elements are shown below:

Element 1

Element 2

The above results for the absorption factors of both elements proves the 10 dB rule since their values are close despite the difference of 10 dB between the two elements.

Question 3: Part A

Sound Reduction is given by the formula shown below:

• Glass/windows

• Brick wall

• Spandrel panels

Question 3: Part B

The plot is shown below:

Question 4

Acoustics and sustainability are the two key considerations when designing sustainable or green buildings mainly because both are positively related in terms of the selected specifications, where those of one aspect of a building i.e. acoustics are dependent of the specifications of the building’s sustainability. This is highly attributable to the fact that, acoustics which are majorly concerned with sound insulation have become a very important factor when designing sustainable buildings as our environments continue to become noisier. The reason why acoustic has become a highly influential factor is because sound insulation has a significant impact on the extents of embodied energy in a particular building, which makes it essential to clearly understand the effect of sound insulation on embodied energy levels right from the beginning of the designing of sustainable buildings.

The close connection between acoustic (sound insulation) and sustainability is how the former influences the latter on basis of embodied energy, especially because embodied energy in a particular building can be effectively reduced by re-using an existing building through refurbishments that enables acoustic specifications are fulfilled in a manner that is very cost effective. This is in most cases achieved through timely up front acoustic testing of an existing building to understand its performance prior to refurbishment. Hence, upon establishing the acoustic performance of an existing building as well as understanding its restrictions and limitations, the sustainability of the building can be significantly improved by alleviating these restrictions or limitations.

When designing sustainable buildings, acoustics have become a very significant consideration because it determines the type of materials used in the construction of the sustainable building which is an indication of how the two components i.e. sustainability and acoustics are interrelated. Therefore, early acoustic testing is in most cases carried out as an effective an method to make sure that designs of sustainable buildings are sufficient, and that there is high quality construction in which sustainability is given priority. As a result, the architects usually make decisions of the construction materials to be used in a sustainable building either lightweight or heavyweight materials based of the intended quality of acoustics (sound insulation) as well as its subsequent sustainability. For instance, high/heavy weight materials are in most cases favored for enhanced acoustics; however lightweight materials such as timber often offer better or equal acoustic performance. This means that sustainability of buildings can be improved by using lightweight construction due to their less embodied energy compared to heavyweight construction materials apart from achieving higher levels of acoustic performance do not lead to desired sustainability.

Therefore, as it is theoretically know, a ±6 dB change in acoustic performance or sound insulation is equivalent to a doubling or halving of mass of construction materials for a particular building. This implies that, from the perspective of sustainability when acoustic parameters are over specified there will be an eventual significant impact on the created waste. This indicates that acoustic performance considerations or specifications or parameters in a particular building have a direct impact on its sustainability, hence necessitating the need to always make sure that a balance has been achieved between acoustic and sustainability at the design level of a building.

Furthermore, there are various products that are used for the enhancement of the acoustic of a vented façade. For instance, glass fiber reinforced concrete (GFRC) acoustic panels are one of the products often used in constructing vented facades for enhanced acoustic performance because they not only lead to sound insulation, but also improves sustainability of the building. Mineral wool is the other product used within partitions of vented facades for the purpose of acoustic dampening within stud walls, and it results to significant enhancement in acoustic performance in addition to being sustainable and cost effective method.

Question 5

A copy of the Excel spreadsheet is attached below:

When the absorptive panels are spread throughout the room more sound will be absorbed because they will be in proximity to the source of sound leading to significant sound reduction; however when they are placed in one big area there will be less sound reduction since they will be overlying each other.

Question 6

Where;

Lw = Sound power level 10-12 w(dB)

Lp = Sound pressure level 20 mPa (dB)

D= distance from fan in meters

dB

D = 25 meters

References

Fahy, F. and Gardonio, P. (2007), Sound and Structural Vibration: Radiation, Transmission and Response, (2nd edition). London: Academic Press.

Lohse, D., Schmitz, B. and Versluis, M. (2001), “Snapping shrimp make flashing bubbles”. Nature , Vol. 413 Issue: 6855, pp. 477–478.

Pohlmann, K. (2010), Principles of Digital Audio, (6th edition). New York, NY: McGraw Hill Professional. p. 336-339.

Raichel, D. R. (2006), The Science and Applications of Acoustics, (2nd edition). London: Springer.

Wilson, C. E. (2006), Noise Control, (Revised edition). Malabar, FL: Krieger Publishing Company.

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## Matrices Assignment Help Available Here

Matrices

Order Instructions:

I am attaching the questions to the email. I am looking for the correct answers for all the questions. Please send me the answers by 2 PM CST.

### Matrices

Question 1: Compute the products

• Graphical drawings of the column vectors

Question 3: Find two inner products and a matrix product

Two inner products (-14, -14)

Matrix product =

Two inner products (-14, 15)

Matrix product =

Two inner products (-14, 15)

Matrix product =

Question 5: Solution for x

Question 6: Solving a 2 by 2 matrices

Solution: since  indicates the entry in A which is in the ith row and in the jth column, we see that:

Likewise

Therefore,

Also,

Question 10: True or false statements

• True
• True
• True
• False

Question 24:

Question 30:

Gaussian Elimination

Question1:

Question 3:

2x – 3 y = 3 2x – 3 y = 3 x = 3 Subtract 2 x row 1 from row 2

y + z = 1 gives y + z = 1 and y = 1 Subtract 1 x row 1 from row 3

2y 3z = 2    -5z = 0   z = 0 Subtract 2 x row 2 from row 3

Question 10:

Question 11:

Question 17:

Question 28:

• True
• False
• True

The Geometry of Linear Equations

Question 1:

The lines intersect at (x, y) = (3, 1). Then 3(column 1) + 1(column 2) = (4, 4).

Question 9:

Column 3 = 2(column 2) – column 1. Ifb = (0, 0, 0), then (u, v, w) = (c, -2c, c)

Question 13:

The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) +

2(-2, 1) = 4(column 1) + 2(column 2) = right-hand side (0, 6).

Matrices

Question 1: Compute the products

• Graphical drawings of the column vectors

Question 3: Find two inner products and a matrix product

Two inner products (-14, -14)

Matrix product

Two inner products (-14, 15)

Matrix product =

Two inner products (-14, 15)

Matrix product =

Question 5: Solution for x

Question 6: Solving a 2 by 2 matrices

Solution: since  indicates the entry in A which is in the ith row and in the jth column, we see that:

Likewise

Therefore,

Also,

Question 10: True or false statements

• True
• True
• True
• False

Question 24:

Question 30:

Gaussian Elimination

Question1:

Question 3:

2x – 3 y = 3 2x – 3 y = 3 x = 3 Subtract 2 x row 1 from row 2

y + z = 1 gives y + z = 1 and y = 1 Subtract 1 x row 1 from row 3

2y 3z = 2    -5z = 0   z = 0 Subtract 2 x row 2 from row 3

Question 10:

Question 11:

Question 17:

Question 28:

• True
• False
• True

The Geometry of Linear Equations

Question 1:

The lines intersect at (x, y) = (3, 1). Then 3(column 1) + 1(column 2) = (4, 4).

Question 9:

Column 3 = 2(column 2) – column 1. Ifb = (0, 0, 0), then (u, v, w) = (c, -2c, c)

Question 13:

The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) +

2(-2, 1) = 4(column 1) + 2(column 2) = right-hand side (0, 6).

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## Mathematical problems Assignment Help

Mathematical problems

Order Instructions:

My student number is some of the problem require the last 2 digits of the student number

Question 1

• For point B:

Within point in a stressed body, there exist three mutually perpendicular planes and the resultant stress is always normal stress. The mutually perpendicular planes are known as principal planes, and the resultant normal stresses acting on them are called principal stresses.

For point A:

Question 2

For pure shear to occur, the stresses σ1 and σ1 must be equal in magnitude and opposite in sign

σ1= Pr/t and σ2= Pr/2t-F/A= Pr/2t-F/2πrt

Pr/t=- (Pr/2t-F/2πrt) collecting like terms together

Pr/t+ Pr/2t= F/2πrt solving for F, we get F=3π Pr2=3* π*0.12*12000000Pa

F=360 kN

Required thickness is given by solving for t in equations above. Using the allowable normal stress we have 110 MPa= Pr/2t-F/2πrt, therefore,

t= (P*r2*π-F)/220πr = (12000000*0.12* π-190000)/ (220000000* π*0.1) = 0.00271 m

Question 3

Pipe diameter, outer =110mm=0.11m, inner diameter=90mm=0.09m

Wind pressure =1.5 kPa

Solution

Area of the sign board=2*1=2m2, resultant force W= wind pressure * area

W=2*1.5=3KN

Torque T=resultant force W* distance b, T=3*1.05=3.15 kN-M

Bending moment M=W* height=3*3.5=10.5 kN-M

Shear force V=W =3 KN

Bending moment M produces a tensile stress σa at point A and tensile stress

σa = (M*d2/2)/ [π(d24-d14)/64]= (10.5 *0.11/2)/ [π(0.114-0.094)/64]=0.5775/(3.96626*10^-6)=145603.163 MPa

Torque produces shear stresses ҭ1 at point A and B

Ҭtorsion == (T*d2/2)/ [π (d24-d14)/32]=(3.15 *0.11/2)/ [π(0.114-0.094)/32]= 0.17325/(7.932522*10^-6)=21840.4689

Maximum plane shear stress Ҭmax=√ [( σx – σy)2/2 + (Ҭxy)2]

At point A the shear stress is zero

At point B σx=0, σy= σa and Ҭxy= Ҭ1=21840.4689 Pa, therefore, Maximum plane shear stress

Ҭmax =√ [(0 – 145603.163)2/2 + (21840.4689)2]= √(1.107715*10^10) =105248.04 Pa

At point C, σy= σx=0 and Ҭxy= Ҭ1+ Ҭ2, Ҭ2 =2V/A= (2*3000)/2=3000 N/m2, hence Maximum plane shear stress   Ҭmax =√ [(0 – 0)2/2 + (3000+21840.4689)2] =24840.4689 Pa

Question four

Solution

Tensile stress = tensile force P/ area A=34000/[ (π*0.052)/4]

=34000/0.001963496=17.316 MPa

Shear torsion= Tr/Ip = (34000 *0.05/2)/[(0.05)4*π/32]=850/(6.135922*10^-7)=1385.285 MPa

Principal stresses are σ1,2  = [( σx + σy)2/2 ±√[( σxy)2/2+ (ҭxy)2]

σ1,2  = (0 +17.316)/2  ±√[(( 0 -17.316)/2)2+ (1385.285)2]

σ1= 8.658+√1919089.49=1393.97 MPa

σ2= -1376.6541 MPa

Maximum in plane shear stress

= √ [((0 -17.316)/2)2+ (1385.285)2] = 1385.31 MPa

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## Curriculum development;Math Term Paper

Curriculum development

Order Instructions:

a literature and Internet search for information about curriculum development and change in your content area of interest(math). Create a timeline that includes at least eight events that shaped curriculum development from a historical perspective in your selected content area. Write 100- to 200-words describing the importance of each event. Include an APA-formatted reference page

The curriculum development was shaped by the use of convectional approach that emphasized on the notion that entailed usage of mathematical ideas, and techniques that could aid in solving problems in an easier way. Despite all this, there was the need of tutor to give guidance where required. This method was an advancement from the pedagogical consideration that was earlier used.

Classical education was known to shape the future of curriculum in mathematics field due to the fact that it encouraged use of deductive reasoning mainly the paradigm teaching in days before. Rote learning was a new advancement in the curriculum where one could repeat what has been taught or even memorize it at his own free time (Scholarstic, 2014). Rote learning was initially known to be the multiple tables and formulas used in the math’s field. The new rote learning was now supported by mathematical reasoning to find solutions to problems.

Exercisers was a new method enforced for learners to understand learning skills better. Problem solving was an event that shaped the curriculum where math’s ingenuity was solved through setting students to be open-ended in tackling them. The student’s prior understanding is more build through the use of math’s problem solving technique.

The new math changed the curriculum following that it encouraged the use of set theories in solving problems, a method known to have been introduced by Tom Lehrer’s. In older days, learners were up to getting the right answer until when new math technique was introduced as it aimed at ensuring that a person’s knows what he is doing rather than getting the right answer. There was the use of conventional approach used in math’s solving problems. Relational approach was used in the curriculum and it advocated for usage of class topics in problem solving. Unlike before, this method enlightens students to apply math to real live happenings outside classroom work as events take place on a daily basis. The standards-based approach was more concerned on helping students to understand ideas and procedures whereby, it was most applied to college students. Unlike before where this method was not formalized, it was now made formal by the national council of all math’s teachers in the whole of United Kingdom (Scholarstic, 2014). This is to show that curriculum in mathematics has well been shaped to date.

References

Scholarstic. (2014). Subject Leader Guide for Maths – Key Stage 1 – 3 (National Curriculum Handbook). Scholastic Press

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