solutionA= 1 1 0 4 3 1 2 0 .pdf

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solution: A= 1 1 0 4 3 1 2 0 5 Finding the determinant of the matrix ? det A = -3 The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse matrix was calculated the minors and cofactors of the matrix ? M1,1 = 3 1 0 5 = 15 C1,1 = (-1)1+1M1,1 = 15 M1,2 = 4 1 2 5 = 18 C1,2 = (-1)1+2M1,2 = -18 M1,3 = 4 3 2 0 = -6 C1,3 = (-1)1+3M1,3 = -6 M2,1 = 1 0 0 5 = 5 C2,1 = (-1)2+1M2,1 = -5 M2,2 = 1 0 2 5 = 5 C2,2 = (-1)2+2M2,2 = 5 M2,3 = 1 1 2 0 = -2 C2,3 = (-1)2+3M2,3 = 2 M3,1 = 1 0 3 1 = 1 C3,1 = (-1)3+1M3,1 = 1 M3,2 = 1 0 4 1 = 1 C3,2 = (-1)3+2M3,2 = -1 M3,3 = 1 1 4 3 = -1 C3,3 = (-1)3+3M3,3 = -1 Matrix of cofactors: C = 15 -18 -6 -5 5 2 1 -1 -1 Transpose of cofactor matrix: CT = 15 -5 1 -18 5 -1 -6 2 -1 Finding the inverse matrix. A-1 = CT = det A -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 B) First, let\'s add the Identity Matrix to the right of our matrix 1 1 0 1 0 0 4 3 1 0 1 0 2 0 5 0 0 1 Now, let\'s do Gauss-Jordan Elimination on our new matrix... Add (-4 * row1) to row2 1 1 0 1 0 0 0 -1 1 -4 1 0 2 0 5 0 0 1 Add (-2 * row1) to row3 1 1 0 1 0 0 0 -1 1 -4 1 0 0 -2 5 -2 0 1 Divide row2 by -1 1 1 0 1 0 0 0 1 -1 4 -1 0 0 -2 5 -2 0 1 Add (2 * row2) to row3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 3 6 -2 1 Divide row3 by 3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 1 2 -2/3 1/3 Add (1 * row3) to row2 1 1 0 1 0 0 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 Add (-1 * row2) to row1 1 0 0 -5 5/3 -1/3 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 The inverse matrix can now be found in the right 3 columns of our reduced row echelon matrix Here is the inverse matrix: -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 A= 1 1 0 4 3 1 2 0 5 Solution solution: A= 1 1 0 4 3 1 2 0 5 Finding the determinant of the matrix ? det A = -3 The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse matrix was calculated the minors and cofactors of the matrix ? M1,1 = 3 1 0 5 = 15 C1,1 = (-1)1+1M1,1 = 15 M1,2 = 4 1 2 5 = 18 C1,2 = (-1)1+2M1,2 = -18 M1,3 = 4 3 2 0 = -6 C1,3 = (-1)1+3M1,3 = -6 M2,1 = 1 0 0 5 = 5 C2,1 = (-1)2+1M2,1 = -5 M2,2 = 1 0 2 5 = 5 C2,2 = (-1)2+2M2,2 = 5 M2,3 = 1 1 2 0 = -2 C2,3 = (-1)2+3M2,3 = 2 M3,1 = 1 0 3 1 = 1 C3,1 = (-1)3+1M3,1 = 1 M3,2 = 1 0 4 1 = 1 C3,2 = (-1)3+2M3,2 = -1 M3,3 = 1 1 4 3 = -1 C3,3 = (-1)3+3M3,3 = -1 Matrix of cofactors: C = 15 -18 -6 -5 5 2 1 -1 -1 Transpose of cofactor matrix: CT = 15 -5 1 -18 5 -1 -6 2 -1 Finding the inverse matrix. A-1 = CT = det A -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 B) First, let\'s add the Identity Matrix to the right of our matrix 1 1 0 1 0 0 4 3 1 0 1 0 2 0 5 0 0 1 Now, let\'s do Gauss-Jordan Elimination on our new matrix... Add (-4 * row1) to row2 1 1 0 1 0 0 0 -1 1 -4 1 0 2 0 5 0 0 1 Add (-2 * row1) to row3 1 1 0 1 0 0 0 -1 1 -4 1 0 0 -2 5 -2 0 1 Divide row2 by -1 1 1 0 1 0 0 0 1 -1 4 -1 0 0 -2 5 -2 0 1 Add (2 * row2) to row3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 3 6 -2 1 Divide row3 by 3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 1 2 -2/3 1/3 Add (1 * row3) to row2 1 1 0 1 0 0 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 Add (-1 * row2) to row1 1.

Education

C1,2 = (-1)1+2M1,2 =
-18
M1,3 =
4
3
2
0
=
-6
C1,3 = (-1)1+3M1,3 =
-6
M2,1 =
1
0
0
5
=
5
C2,1 = (-1)2+1M2,1 =
-5
M2,2 =
1
0
2
5
=
5
C2,2 = (-1)2+2M2,2 =
5
M2,3 =
1
1
2
0
=
-2

C2,3 = (-1)2+3M2,3 =
2
M3,1 =
1
0
3
1
=
1
C3,1 = (-1)3+1M3,1 =
1
M3,2 =
1
0
4
1
=
1
C3,2 = (-1)3+2M3,2 =
-1
M3,3 =
1
1
4
3
=
-1
C3,3 = (-1)3+3M3,3 =
-1
Matrix of cofactors:
C =
15
-18
-6
-5
5

2
1
-1
-1
Transpose of cofactor matrix:
CT =
15
-5
1
-18
5
-1
-6
2
-1
Finding the inverse matrix.
A-1 =
CT
=
det A
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
B)
First, let's add the Identity Matrix to the right of our matrix
1
1

0
1
0
0
4
3
1
0
1
0
2
0
5
0
0
1
Now, let's do Gauss-Jordan Elimination on our new matrix...
Add (-4 * row1) to row2
1
1
0
1
0
0
0
-1
1
-4
1
0
2
0
5
0

0
1
Add (-2 * row1) to row3
1
1
0
1
0
0
0
-1
1
-4
1
0
0
-2
5
-2
0
1
Divide row2 by -1
1
1
0
1
0
0
0
1
-1
4
-1
0

0
-2
5
-2
0
1
Add (2 * row2) to row3
1
1
0
1
0
0
0
1
-1
4
-1
0
0
0
3
6
-2
1
Divide row3 by 3
1
1
0
1
0
0
0
1

-1
4
-1
0
0
0
1
2
-2/3
1/3
Add (1 * row3) to row2
1
1
0
1
0
0
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
Add (-1 * row2) to row1
1
0
0
-5

5/3
-1/3
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
The inverse matrix can now be found in the right 3 columns of our reduced row echelon matrix
Here is the inverse matrix:
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
A=
1
1
0
4
3
1
2
0

5
Solution
solution:
A=
1
1
0
4
3
1
2
0
5
Finding the determinant of the matrix ?
det A = -3
The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse
matrix was calculated the minors and cofactors of the matrix ?
M1,1 =
3
1
0
5
=
15
C1,1 = (-1)1+1M1,1 =
15
M1,2 =
4
1
2

5
=
18
C1,2 = (-1)1+2M1,2 =
-18
M1,3 =
4
3
2
0
=
-6
C1,3 = (-1)1+3M1,3 =
-6
M2,1 =
1
0
0
5
=
5
C2,1 = (-1)2+1M2,1 =
-5
M2,2 =
1
0
2
5
=
5
C2,2 = (-1)2+2M2,2 =
5
M2,3 =
1
1
2

0
=
-2
C2,3 = (-1)2+3M2,3 =
2
M3,1 =
1
0
3
1
=
1
C3,1 = (-1)3+1M3,1 =
1
M3,2 =
1
0
4
1
=
1
C3,2 = (-1)3+2M3,2 =
-1
M3,3 =
1
1
4
3
=
-1
C3,3 = (-1)3+3M3,3 =
-1
Matrix of cofactors:
C =
15
-18

-6
-5
5
2
1
-1
-1
Transpose of cofactor matrix:
CT =
15
-5
1
-18
5
-1
-6
2
-1
Finding the inverse matrix.
A-1 =
CT
=
det A
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
B)

First, let's add the Identity Matrix to the right of our matrix
1
1
0
1
0
0
4
3
1
0
1
0
2
0
5
0
0
1
Now, let's do Gauss-Jordan Elimination on our new matrix...
Add (-4 * row1) to row2
1
1
0
1
0
0
0
-1
1
-4
1
0
2

0
5
0
0
1
Add (-2 * row1) to row3
1
1
0
1
0
0
0
-1
1
-4
1
0
0
-2
5
-2
0
1
Divide row2 by -1
1
1
0
1
0
0
0
1
-1

4
-1
0
0
-2
5
-2
0
1
Add (2 * row2) to row3
1
1
0
1
0
0
0
1
-1
4
-1
0
0
0
3
6
-2
1
Divide row3 by 3
1
1
0
1
0

0
0
1
-1
4
-1
0
0
0
1
2
-2/3
1/3
Add (1 * row3) to row2
1
1
0
1
0
0
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
Add (-1 * row2) to row1
1

0
0
-5
5/3
-1/3
0
1
0
6
-5/3
1/3
0
0
1
2
-2/3
1/3
The inverse matrix can now be found in the right 3 columns of our reduced row echelon matrix
Here is the inverse matrix:
-5
5/3
-1/3
6
-5/3
1/3
2
-2/3
1/3
A=
1
1
0
4
3

Title of this process is The Project Life Cycle (Phases)The pr.pdf

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The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription. because it is composed with 12 subunits and due to incapable of specific promoter recognition it can\'t initiate transcription. Solution The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription. because it is composed with 12 subunits and due to incapable of specific promoter recognition it can\'t initiate transcription..

The answer is A) Pol II, with twelve subunits on its own, is capable.pdf

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Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active investigation. All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy include, 1). Diseases with known genetic defects or mutations 2). Diseases with point mutations on one or more genes 3). The possibility of delivering genes to the affected tissues 4). If the substitution of defective gene correct the disease status Solution Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active investigation. All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy include, 1). Diseases with known genetic defects or mutations 2). Diseases with point mutations on one or more genes 3). The possibility of delivering genes to the affected tissues 4). If the substitution of defective gene correct the disease status.

Q1). Gene therapy is an experimental approach to treat the disease b.pdf

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Solution : Three modes of DNA replication 1) Semi conservative: => Each daughter strand is composed of one old DNA strand and one new DNA strand. 2) Conservative: => The original double helix remains as complete unit. => The new DNA double helix is produced as a single unit. => The old DNA is completly conserved. 3) Dispersive mode: =>Separation of the individual covalent phosphodiester bonds is required for this mode of replication..

Solution Three modes of DNA replication 1) Semi conservative.pdf

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package s3; // Copy paste this Java Template and save it as \"EmergencyRoom.java\" import java.util.*; import java.io.*; // write your matric number here: // write your name here: // write list of collaborators here: // year 2016 hash code: XAbyuzR78fXeaMHBdLan (do NOT delete this line) class patientlist { private String PatientName; private int emergencyLvl; public patientlist(String PatientName, int emergencyLvl) { this.PatientName = PatientName; this.emergencyLvl = emergencyLvl; } public String getPatientName() { return PatientName; } public void setPatientName(String PatientName) { this.PatientName = PatientName; } public int getEmergencyLvl() { return emergencyLvl; } public void setEmergencyLvl(int emergencyLvl) { this.emergencyLvl = emergencyLvl; } } class EmergencyRoom { // if needed, declare a private data structure here that // is accessible to all methods in this class private List patientlist; public EmergencyRoom() { // Write necessary code during construction patientlist = new ArrayList(); } void ArriveAtHospital(String patientName, int emergencyLvl) { // You have to insert the information (patientName, emergencyLvl) // into your chosen data structure if (patientName.length() > 15 || patientName.length() < 1) { System.out.println(\"patient name is either too long or too short.\ Please enter a name between 1 to 15 characters.\"); } else if (emergencyLvl > 100 || emergencyLvl < 30) { System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid level between 30 to 100.\"); } else { int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { System.out.println(\"Patient already admitted\"); break; } } if (i == patientlist.size()) { patientlist p = new patientlist(patientName.toUpperCase(), emergencyLvl); patientlist.add(p); } } } void UpdateEmergencyLvl(String patientName, int incEmergencyLvl) { // You have to update the emergencyLvl of patientName to // emergencyLvl += incEmergencyLvl // and modify your chosen data structure (if needed) int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { //System.out.println(\"Patient already admitted\"); //break; if (incEmergencyLvl > 70 || incEmergencyLvl < 0) { System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid increment between 0 to 70.\"); break; } else { patientlist.get(i).setEmergencyLvl(patientlist.get(i).getEmergencyLvl() + incEmergencyLvl); } } } } void Treat(String patientName) { // This patientName is treated by the doctor // remove him/her from your chosen data structure int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { patientlist.remove(i); } } } String Query() { String ans = \"The emergency room is empty\"; int i; if(patientlist.isEmpty()) return ans; else{ int max=patientlist.get(0).getEme.

package s3; Copy paste this Java Template and save it as Emer.pdf

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Our body is having two line defence system against pathogens.Pathogens are micro-organisms such like virus,bactria,parasite,fungi. First line of defence system:- It is a outside defence system which protect body from the outside ,it includes skin ,tear,mucus,cilia,stomach acid,urine flow. Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to your body unless your skin get brokken.The pathogens can get enter through your nose ,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop pathogen and break down their cell wall because they contain enzyme.those that are not killed immediately are trapped in mucus swallowed. cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine flow help to flush out pathogens from the bladder.stomach acid is present in stomach and helps to kill pathogens.this is the fine defence. Second line of defence system :- once pathogen cross your first line of defence second line of defence active for example ,through the cut pathogen enter into body and develope infection then second line defence active .Through a sequence of steps called immune response ,the immune system attacks these pathogens. In the second line of deffence system is a group of cells ,tissues,organs work together to protect the body .This is called immune system. The cells involved in the immune system called white blood cells is helps to destoys the desease causing organisms.there are different types of white blood cells but function is same that protect you from the organisms. They are neutrophiles,T-helper cells,Cytotoxic cells,marcophases,B-cells. These cells work together and kill the micro organism. Tissues and Organs :- This are involved in the immune system are the lymphatic system ,lymph nodes and lymp fluid.This are having specific function. Lymphatic system is a system of thin tube present through out the body .These tubes are called lymph vessels.they contain lymph. Lymph is a fluid that contains white blood cells. lymph helps to protect body form micro organisms. This is a second line of defence system. Solution Our body is having two line defence system against pathogens.Pathogens are micro-organisms such like virus,bactria,parasite,fungi. First line of defence system:- It is a outside defence system which protect body from the outside ,it includes skin ,tear,mucus,cilia,stomach acid,urine flow. Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to your body unless your skin get brokken.The pathogens can get enter through your nose ,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop pathogen and break down their cell wall because they contain enzyme.those that are not killed immediately are trapped in mucus swallowed. cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine flow help to flush out.

Our body is having two line defence system against pathogens.Pathoge.pdf

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Once neurons are produced, they migrate and modify to form six layered structure of cortex. The earliest formed neurons form the preplate which is then split into marginal plate and subplate by invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and ganglionic eminences. The corticular neurons exhibit radial migration and form projection neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both types arrange in inside out sequence and the early neurons form the deep layers while the late born neuron form the outer layers of cortex. Solution Once neurons are produced, they migrate and modify to form six layered structure of cortex. The earliest formed neurons form the preplate which is then split into marginal plate and subplate by invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and ganglionic eminences. The corticular neurons exhibit radial migration and form projection neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both types arrange in inside out sequence and the early neurons form the deep layers while the late born neuron form the outer layers of cortex..

Once neurons are produced, they migrate and modify to form six layer.pdf

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PDU is called Protocol Data Unit.It consists of user data and protoc.pdf

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1.A hole is the absence of an electron in a particular place in an atom.Electrons and holes do not move as rapidly through the solid materials as electrons do in a vaccum.The holes move more slowly than the electrons.The holes and electrons move in opposite direction. 2.Here if the electric current moving from left to right ,than the holes will be moving in opposite direction that is from right to left. Regardless of the semiconductor type ,with the potencial applied there will be net effect.Both electrons and holes contribute to total current flow,altough they are moving in opposite directions,they are of opposite charge. Solution 1.A hole is the absence of an electron in a particular place in an atom.Electrons and holes do not move as rapidly through the solid materials as electrons do in a vaccum.The holes move more slowly than the electrons.The holes and electrons move in opposite direction. 2.Here if the electric current moving from left to right ,than the holes will be moving in opposite direction that is from right to left. Regardless of the semiconductor type ,with the potencial applied there will be net effect.Both electrons and holes contribute to total current flow,altough they are moving in opposite directions,they are of opposite charge..

1.A hole is the absence of an electron in a particular place in an a.pdf

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