SolutionPreparing the T-Accounts for the Above TransactionPrep.pdf
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This document provides the T-accounts and trial balance for a business transaction. It lists cash, common stock, accounts receivable, supplies, salaries expense, accounts payable, dividends, and the ending balance as debit and credit accounts with their corresponding values totaling $20,350 for the accounting equation.
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A. In order to survive inside the body of the host, a pathogen must .pdf
A. In order to survive inside the body of the host, a pathogen must have ability to survive in the
environment inside host body that is entirely different from outside world. To do so, a pathogen
must develop different types adaptation in itself to successfully reside and reproduce in this
novel environment.
The first adaptation it must have is that it should develop a mechanism in itself that can help it to
evade immune surveillance of host. It must have a defence system of enzymes and proteins that
will enable it to protect itself from immune response and phagocytic cells of host . Another
adaptation it needs to have is the ability to tolerate different types of changes in temperature, pH
, osmotic conditions etc because environment inside host body can have different kinds of these
environment conditions. It also should have ability to perform anaerobic respiration if the need
arises.
B) Factors that makes it difficult, containing an pathogen and developing an effective vaccine
are;
1. Ability of a pathogen to reproduce at an alarming rate .
2. Ability of pathogen to continously change its surface receptors that makes it difficult to design
and effective vaccine against it.
3. Ability of a pathogen to rapidly evolve into different strains .
C) HIV is a Zoonotic pathogen. It is believed to have evolved from a tyoe of Simian virus that
infected chimpanzees. This Virus evolved into HIV and infected humans.
Solution
A. In order to survive inside the body of the host, a pathogen must have ability to survive in the
environment inside host body that is entirely different from outside world. To do so, a pathogen
must develop different types adaptation in itself to successfully reside and reproduce in this
novel environment.
The first adaptation it must have is that it should develop a mechanism in itself that can help it to
evade immune surveillance of host. It must have a defence system of enzymes and proteins that
will enable it to protect itself from immune response and phagocytic cells of host . Another
adaptation it needs to have is the ability to tolerate different types of changes in temperature, pH
, osmotic conditions etc because environment inside host body can have different kinds of these
environment conditions. It also should have ability to perform anaerobic respiration if the need
arises.
B) Factors that makes it difficult, containing an pathogen and developing an effective vaccine
are;
1. Ability of a pathogen to reproduce at an alarming rate .
2. Ability of pathogen to continously change its surface receptors that makes it difficult to design
and effective vaccine against it.
3. Ability of a pathogen to rapidly evolve into different strains .
C) HIV is a Zoonotic pathogen. It is believed to have evolved from a tyoe of Simian virus that
infected chimpanzees. This Virus evolved into HIV and infected humans..
10) The Correct answer is improvements in learning through hypnosis .pdf
10) The Correct answer is improvements in learning through hypnosis is due to improved
attention and motivation.
hence a) is the correct answer
Solution
10) The Correct answer is improvements in learning through hypnosis is due to improved
attention and motivation.
hence a) is the correct answer.
1.1.The words ‘data’ and ‘information’ are often used as though they.pdf
1.1.The words ‘data’ and ‘information’ are often used as though they mean exactly the same
thing, but there is a difference. Data are the undigested facts and figures that are collected on
innumerable subjects. You may gather data yourself or use data that have been gathered by other
people.
1.2..Strengths, weaknesses and suggestions. for the main other information sources reviewed .
1.3.The data must be collected carefully because organisation decision-making processes are
based on the generated information from this data. While selecting data and information for
decision-making we must apply some criteria to this selection such as accuracy, validity, clarity
etc.
1.4.
However simple or complex your data set, think about what you might need to do to ensure that
your management of the data respects the terms of your consent, and in particular, the
confidentiality and anonymity that participants were promised.
Take advice from relevant staff in your institution. Your Data Protection manager can advise you
on protocols for handling personal data. Your computing or information services department
should be able to advise you on setting up secure databases for the different forms of data that
will be generated by your research.
As with everything in this guidebook, the earlier you can start to think about these issues, the
better. When you are preparing your research proposal, you need to plan for data management -
this is a requirement for ESRC applications, and increasingly for other funders. If your work will
generate complex or sensitive datasets, you may need to plan and cost some time for a database
manager or information specialist to develop and manage the systems that you need to keep your
data secure.
Do you have suitable arrangements in place for archiving data? Befor you access or collect your
data, you should check institution what requirements they have in place for data storage, and
what facilities are available (e.g. for data archiving).
2.2. Identify a problem or opportunity, Gather information, Analyze the situation,Develop
options, Evaluate alternatives, Select a preferred alternative,Act on the decision.
2.3.There are numerous ways of presenting data to a client but you have to ask yourself several
questions. Who is the client? To what will they respond best? What response do you require?
Will the information and the decisions you made be transferrable between presentational styles?
Will it be a formalpresentation or can you have some fun?
Once you have answered these questions you can then set about shaping your presentation.
Personally if you are able to do so (as in whether the environment within which you will be
presenting is accommodating and data allows you do) I would recommend the use of pictures,
charts and visual stimuli as much as possible; a picture paints a thousand words after all. Of
course if the decisions you have made cannot be transferred into numerical data any pie charts,
bar graphs or statistical analysi.
1. Osmotic pressure = imRTwhere m = molality of solute, R = molar.pdf
1. Osmotic pressure = imRT
where m = molality of solute, R = molar gas constant, T = temperature and i = van\'t Hoff factor.
i is a measure of how many particles each particle of solute dissociates into.
ethylene glycol which is a non-electrolyte, i = 1
acetic acid (weak electrolyte with partial dissociation int H+ and acetate), 1 < i < 2
and NaCl (strong electrolyte with complete dissociation intp Na+ and Cl-), i = 2
So the osmotic pressure of ethylene glycol < acetic acid < NaCl for the same molality of solute
2. Vapor pressure depends on the strength of the intermolecular bonding between molecules. The
weaker the intermolecular bonds, the more easily the molecules escape into the gas phase and the
higher the vapor pressure.
For BrF3. PF5 and CF4 are all molecular compounds held together by intermolecular van del
Waal\'s (dispersion) forces, which increase with molecular weight.
So in terms of intermolecular bond strength, BrF3 > PF5 > CF4, and the vapor pressures are
BrF3 < PF5
Solution
1. Osmotic pressure = imRT
where m = molality of solute, R = molar gas constant, T = temperature and i = van\'t Hoff factor.
i is a measure of how many particles each particle of solute dissociates into.
ethylene glycol which is a non-electrolyte, i = 1
acetic acid (weak electrolyte with partial dissociation int H+ and acetate), 1 < i < 2
and NaCl (strong electrolyte with complete dissociation intp Na+ and Cl-), i = 2
So the osmotic pressure of ethylene glycol < acetic acid < NaCl for the same molality of solute
2. Vapor pressure depends on the strength of the intermolecular bonding between molecules. The
weaker the intermolecular bonds, the more easily the molecules escape into the gas phase and the
higher the vapor pressure.
For BrF3. PF5 and CF4 are all molecular compounds held together by intermolecular van del
Waal\'s (dispersion) forces, which increase with molecular weight.
So in terms of intermolecular bond strength, BrF3 > PF5 > CF4, and the vapor pressures are
BrF3 < PF5.
1-Children are more susceptible to Trichurus species be it male or f.pdf
1-Children are more susceptible to Trichurus species be it male or female.Hence the correct
answer is option c and d.
2-Evolutionary predecessor of insect is thought to be annelida. Earthworm too is annelida.
Hence, the correct answer is option a and b.
Solution
1-Children are more susceptible to Trichurus species be it male or female.Hence the correct
answer is option c and d.
2-Evolutionary predecessor of insect is thought to be annelida. Earthworm too is annelida.
Hence, the correct answer is option a and b..
Javac code for the manager of the Jeter County softball teamimp.pdf
// Javac code for the manager of the Jeter County softball team
import java.util.Scanner;
import java.util.ArrayList;
public class Hiitters {
public static void main(String args[]){
System.out.println(\"Jeter County softball team\ \ \");
Scanner sc = new Scanner(System.in);
// araay list storing the jersy number, hiits , by-bats and average of each player
ArrayList jersy_number = new ArrayList();
ArrayList hiits = new ArrayList();
ArrayList bats = new ArrayList();
ArrayList average = new ArrayList();
while(true)
{
// prompt the user for a player jersey number,
// the number of hiits, and the
// number of at-bats
System.out.print(\"\ Enter jersy number: \");
int number = sc.nextInt();
// program accepts players continuously until 0 is entered for the jersey number
if(number == 0)
{
System.out.println( \"\ End of job\");
break;
}
System.out.print(\"Enter hiits: \");
int h = sc.nextInt();
System.out.print(\"Enter at-bats: \");
int b = sc.nextInt();
// compute average and make it to 3 decimal place
double avg = (double)h/b;
avg = avg*1000;
avg = Math.round(avg);
avg = avg /1000;
// add the values to the corresponding arraylist
jersy_number.add(number);
hiits.add(h);
bats.add(b);
average.add(avg);
}
int size = jersy_number.size();
// print the desired result depending on the average
for(int i = 0; i < size; i++)
{
if(average.get(i) > .400)
{
System.out.println( \"\ Excellent Hiitter\");
System.out.println( \"Jersy Number: \" + jersy_number.get(i));
System.out.println( \"hiits: \" + hiits.get(i));
System.out.println( \"At-Bats: \" + bats.get(i));
System.out.println( \"Average: \" + average.get(i));
}
else
{
System.out.println( \"\ Jersy Number: \" + jersy_number.get(i));
System.out.println( \"Average: \" + average.get(i));
}
}
}
}
/*
output:
Jeter County softball team
Enter jersy number: 10
Enter hiits: 10
Enter at-bats: 30
Enter jersy number: 7
Enter hiits: 4
Enter at-bats: 7
Enter jersy number: 20
Enter hiits: 7
Enter at-bats: 20
Enter jersy number: 45
Enter hiits: 7
Enter at-bats: 15
Enter jersy number: 4
Enter hiits: 5
Enter at-bats: 20
Enter jersy number: 0
Jersy Number: 10
Average: 0.333
Excellent Hiitter
Jersy Number: 7
hiits: 4
At-Bats: 7
Average: 0.571
Jersy Number: 20
Average: 0.35
Excellent Hiitter
Jersy Number: 45
hiits: 7
At-Bats: 15
Average: 0.467
Jersy Number: 4
Average: 0.25
End of job
*/
Solution
// Javac code for the manager of the Jeter County softball team
import java.util.Scanner;
import java.util.ArrayList;
public class Hiitters {
public static void main(String args[]){
System.out.println(\"Jeter County softball team\ \ \");
Scanner sc = new Scanner(System.in);
// araay list storing the jersy number, hiits , by-bats and average of each player
ArrayList jersy_number = new ArrayList();
ArrayList hiits = new ArrayList();
ArrayList bats = new ArrayList();
ArrayList average = new ArrayList();
while(true)
{
// prompt the user for a player jersey number,
// the number of hiits, and the
// number of at-bats
System.out.print(\"\ Enter .
1) Mitochondria are known as power houses of cell. The mitochondria .pdf
1) Mitochondria are known as power houses of cell. The mitochondria are the double membrane
structure that can grow, fuse with one another and capable of dividing. During the replication of
mitochondrial DNA the mitochondria become double in size. They are oval in structure with
plain outer membrane and the inner membrane exhibits folding known as cristae. The inner
membrane encloses mitochondrial matrix. The matrix contains the DNA and enzymes involved
in citric acid cycle or Krebs cycle.
The cristae lodge the enzymes required in the electron transport chain to carry out the process of
oxidative phosphorylation. The energy is produced through a series of steps in which the
electrons are transported and the protons are pumped out of the matrix into the inner space
between the two membranes creating the electron potential gradient across the membranes. The
protons flow back to the matrix through an enzyme complex known as ATP synthase that is
found in cristae and utilizes the energy produced by the flow of protons to phosphorylate the
ADP forming the ATP.
2) uncoupler or uncoupling protein (UCP) is a mitochondrial protein located in the inner
membrane. These are involved in inhibition of the coupling between the electron transport and
phosphorylation reactions. As a result, synthesis of ATP is inhibited without affecting the
respiratory chain and ATP synthase (H+ -ATPase).
3) Under availability of oxygen the cell is more efficient in the energy. Under anaerobic
conditions the process of glucose breakdown cannot proceed further after glycolysis. The
glycolysis takes place in the cytosol producing 2 ATP molecules per each glucose molecule.
In presence of oxygen or under aerobic conditions the process of glycolysis further proceeds for
mitochondrial processing including the citric acid cycle and the electron transport system. This
process supplies enough energy to produce 30 more ATP molecules comprising to the total yield
of 32 ATP molecules per glucose molecule at the end of aerobic respiration. Thus, the ATP
molecules produced are 2 by glycolysis 2 by citric acid cycle and 28 by oxidative
phosphorylation. Hence, aerobic respiration is an energy efficient process than anaerobic
respiration.
Solution
1) Mitochondria are known as power houses of cell. The mitochondria are the double membrane
structure that can grow, fuse with one another and capable of dividing. During the replication of
mitochondrial DNA the mitochondria become double in size. They are oval in structure with
plain outer membrane and the inner membrane exhibits folding known as cristae. The inner
membrane encloses mitochondrial matrix. The matrix contains the DNA and enzymes involved
in citric acid cycle or Krebs cycle.
The cristae lodge the enzymes required in the electron transport chain to carry out the process of
oxidative phosphorylation. The energy is produced through a series of steps in which the
electrons are transported and the protons are pumped out of the matrix into .
the molecule is 2,2-dichloro-ethanol all carbons.pdf
the molecule is 2,2-dichloro-ethanol all carbons are sp3 hybridized all carbons have
a tetrahedral geometry only available electron pairs on chlorine and oxygen and they will be
oriented in such a fashion that repulsive forces are balanced or atleast minimum
Solution
the molecule is 2,2-dichloro-ethanol all carbons are sp3 hybridized all carbons have
a tetrahedral geometry only available electron pairs on chlorine and oxygen and they will be
oriented in such a fashion that repulsive forces are balanced or atleast minimum.
Recommended
A. In order to survive inside the body of the host, a pathogen must .pdf
A. In order to survive inside the body of the host, a pathogen must have ability to survive in the
environment inside host body that is entirely different from outside world. To do so, a pathogen
must develop different types adaptation in itself to successfully reside and reproduce in this
novel environment.
The first adaptation it must have is that it should develop a mechanism in itself that can help it to
evade immune surveillance of host. It must have a defence system of enzymes and proteins that
will enable it to protect itself from immune response and phagocytic cells of host . Another
adaptation it needs to have is the ability to tolerate different types of changes in temperature, pH
, osmotic conditions etc because environment inside host body can have different kinds of these
environment conditions. It also should have ability to perform anaerobic respiration if the need
arises.
B) Factors that makes it difficult, containing an pathogen and developing an effective vaccine
are;
1. Ability of a pathogen to reproduce at an alarming rate .
2. Ability of pathogen to continously change its surface receptors that makes it difficult to design
and effective vaccine against it.
3. Ability of a pathogen to rapidly evolve into different strains .
C) HIV is a Zoonotic pathogen. It is believed to have evolved from a tyoe of Simian virus that
infected chimpanzees. This Virus evolved into HIV and infected humans.
Solution
A. In order to survive inside the body of the host, a pathogen must have ability to survive in the
environment inside host body that is entirely different from outside world. To do so, a pathogen
must develop different types adaptation in itself to successfully reside and reproduce in this
novel environment.
The first adaptation it must have is that it should develop a mechanism in itself that can help it to
evade immune surveillance of host. It must have a defence system of enzymes and proteins that
will enable it to protect itself from immune response and phagocytic cells of host . Another
adaptation it needs to have is the ability to tolerate different types of changes in temperature, pH
, osmotic conditions etc because environment inside host body can have different kinds of these
environment conditions. It also should have ability to perform anaerobic respiration if the need
arises.
B) Factors that makes it difficult, containing an pathogen and developing an effective vaccine
are;
1. Ability of a pathogen to reproduce at an alarming rate .
2. Ability of pathogen to continously change its surface receptors that makes it difficult to design
and effective vaccine against it.
3. Ability of a pathogen to rapidly evolve into different strains .
C) HIV is a Zoonotic pathogen. It is believed to have evolved from a tyoe of Simian virus that
infected chimpanzees. This Virus evolved into HIV and infected humans..
10) The Correct answer is improvements in learning through hypnosis .pdf
10) The Correct answer is improvements in learning through hypnosis is due to improved
attention and motivation.
hence a) is the correct answer
Solution
10) The Correct answer is improvements in learning through hypnosis is due to improved
attention and motivation.
hence a) is the correct answer.
1.1.The words ‘data’ and ‘information’ are often used as though they.pdf
1.1.The words ‘data’ and ‘information’ are often used as though they mean exactly the same
thing, but there is a difference. Data are the undigested facts and figures that are collected on
innumerable subjects. You may gather data yourself or use data that have been gathered by other
people.
1.2..Strengths, weaknesses and suggestions. for the main other information sources reviewed .
1.3.The data must be collected carefully because organisation decision-making processes are
based on the generated information from this data. While selecting data and information for
decision-making we must apply some criteria to this selection such as accuracy, validity, clarity
etc.
1.4.
However simple or complex your data set, think about what you might need to do to ensure that
your management of the data respects the terms of your consent, and in particular, the
confidentiality and anonymity that participants were promised.
Take advice from relevant staff in your institution. Your Data Protection manager can advise you
on protocols for handling personal data. Your computing or information services department
should be able to advise you on setting up secure databases for the different forms of data that
will be generated by your research.
As with everything in this guidebook, the earlier you can start to think about these issues, the
better. When you are preparing your research proposal, you need to plan for data management -
this is a requirement for ESRC applications, and increasingly for other funders. If your work will
generate complex or sensitive datasets, you may need to plan and cost some time for a database
manager or information specialist to develop and manage the systems that you need to keep your
data secure.
Do you have suitable arrangements in place for archiving data? Befor you access or collect your
data, you should check institution what requirements they have in place for data storage, and
what facilities are available (e.g. for data archiving).
2.2. Identify a problem or opportunity, Gather information, Analyze the situation,Develop
options, Evaluate alternatives, Select a preferred alternative,Act on the decision.
2.3.There are numerous ways of presenting data to a client but you have to ask yourself several
questions. Who is the client? To what will they respond best? What response do you require?
Will the information and the decisions you made be transferrable between presentational styles?
Will it be a formalpresentation or can you have some fun?
Once you have answered these questions you can then set about shaping your presentation.
Personally if you are able to do so (as in whether the environment within which you will be
presenting is accommodating and data allows you do) I would recommend the use of pictures,
charts and visual stimuli as much as possible; a picture paints a thousand words after all. Of
course if the decisions you have made cannot be transferred into numerical data any pie charts,
bar graphs or statistical analysi.
1. Osmotic pressure = imRTwhere m = molality of solute, R = molar.pdf
1. Osmotic pressure = imRT
where m = molality of solute, R = molar gas constant, T = temperature and i = van\'t Hoff factor.
i is a measure of how many particles each particle of solute dissociates into.
ethylene glycol which is a non-electrolyte, i = 1
acetic acid (weak electrolyte with partial dissociation int H+ and acetate), 1 < i < 2
and NaCl (strong electrolyte with complete dissociation intp Na+ and Cl-), i = 2
So the osmotic pressure of ethylene glycol < acetic acid < NaCl for the same molality of solute
2. Vapor pressure depends on the strength of the intermolecular bonding between molecules. The
weaker the intermolecular bonds, the more easily the molecules escape into the gas phase and the
higher the vapor pressure.
For BrF3. PF5 and CF4 are all molecular compounds held together by intermolecular van del
Waal\'s (dispersion) forces, which increase with molecular weight.
So in terms of intermolecular bond strength, BrF3 > PF5 > CF4, and the vapor pressures are
BrF3 < PF5
Solution
1. Osmotic pressure = imRT
where m = molality of solute, R = molar gas constant, T = temperature and i = van\'t Hoff factor.
i is a measure of how many particles each particle of solute dissociates into.
ethylene glycol which is a non-electrolyte, i = 1
acetic acid (weak electrolyte with partial dissociation int H+ and acetate), 1 < i < 2
and NaCl (strong electrolyte with complete dissociation intp Na+ and Cl-), i = 2
So the osmotic pressure of ethylene glycol < acetic acid < NaCl for the same molality of solute
2. Vapor pressure depends on the strength of the intermolecular bonding between molecules. The
weaker the intermolecular bonds, the more easily the molecules escape into the gas phase and the
higher the vapor pressure.
For BrF3. PF5 and CF4 are all molecular compounds held together by intermolecular van del
Waal\'s (dispersion) forces, which increase with molecular weight.
So in terms of intermolecular bond strength, BrF3 > PF5 > CF4, and the vapor pressures are
BrF3 < PF5.
1-Children are more susceptible to Trichurus species be it male or f.pdf
1-Children are more susceptible to Trichurus species be it male or female.Hence the correct
answer is option c and d.
2-Evolutionary predecessor of insect is thought to be annelida. Earthworm too is annelida.
Hence, the correct answer is option a and b.
Solution
1-Children are more susceptible to Trichurus species be it male or female.Hence the correct
answer is option c and d.
2-Evolutionary predecessor of insect is thought to be annelida. Earthworm too is annelida.
Hence, the correct answer is option a and b..
Javac code for the manager of the Jeter County softball teamimp.pdf
// Javac code for the manager of the Jeter County softball team
import java.util.Scanner;
import java.util.ArrayList;
public class Hiitters {
public static void main(String args[]){
System.out.println(\"Jeter County softball team\ \ \");
Scanner sc = new Scanner(System.in);
// araay list storing the jersy number, hiits , by-bats and average of each player
ArrayList jersy_number = new ArrayList();
ArrayList hiits = new ArrayList();
ArrayList bats = new ArrayList();
ArrayList average = new ArrayList();
while(true)
{
// prompt the user for a player jersey number,
// the number of hiits, and the
// number of at-bats
System.out.print(\"\ Enter jersy number: \");
int number = sc.nextInt();
// program accepts players continuously until 0 is entered for the jersey number
if(number == 0)
{
System.out.println( \"\ End of job\");
break;
}
System.out.print(\"Enter hiits: \");
int h = sc.nextInt();
System.out.print(\"Enter at-bats: \");
int b = sc.nextInt();
// compute average and make it to 3 decimal place
double avg = (double)h/b;
avg = avg*1000;
avg = Math.round(avg);
avg = avg /1000;
// add the values to the corresponding arraylist
jersy_number.add(number);
hiits.add(h);
bats.add(b);
average.add(avg);
}
int size = jersy_number.size();
// print the desired result depending on the average
for(int i = 0; i < size; i++)
{
if(average.get(i) > .400)
{
System.out.println( \"\ Excellent Hiitter\");
System.out.println( \"Jersy Number: \" + jersy_number.get(i));
System.out.println( \"hiits: \" + hiits.get(i));
System.out.println( \"At-Bats: \" + bats.get(i));
System.out.println( \"Average: \" + average.get(i));
}
else
{
System.out.println( \"\ Jersy Number: \" + jersy_number.get(i));
System.out.println( \"Average: \" + average.get(i));
}
}
}
}
/*
output:
Jeter County softball team
Enter jersy number: 10
Enter hiits: 10
Enter at-bats: 30
Enter jersy number: 7
Enter hiits: 4
Enter at-bats: 7
Enter jersy number: 20
Enter hiits: 7
Enter at-bats: 20
Enter jersy number: 45
Enter hiits: 7
Enter at-bats: 15
Enter jersy number: 4
Enter hiits: 5
Enter at-bats: 20
Enter jersy number: 0
Jersy Number: 10
Average: 0.333
Excellent Hiitter
Jersy Number: 7
hiits: 4
At-Bats: 7
Average: 0.571
Jersy Number: 20
Average: 0.35
Excellent Hiitter
Jersy Number: 45
hiits: 7
At-Bats: 15
Average: 0.467
Jersy Number: 4
Average: 0.25
End of job
*/
Solution
// Javac code for the manager of the Jeter County softball team
import java.util.Scanner;
import java.util.ArrayList;
public class Hiitters {
public static void main(String args[]){
System.out.println(\"Jeter County softball team\ \ \");
Scanner sc = new Scanner(System.in);
// araay list storing the jersy number, hiits , by-bats and average of each player
ArrayList jersy_number = new ArrayList();
ArrayList hiits = new ArrayList();
ArrayList bats = new ArrayList();
ArrayList average = new ArrayList();
while(true)
{
// prompt the user for a player jersey number,
// the number of hiits, and the
// number of at-bats
System.out.print(\"\ Enter .
1) Mitochondria are known as power houses of cell. The mitochondria .pdf
1) Mitochondria are known as power houses of cell. The mitochondria are the double membrane
structure that can grow, fuse with one another and capable of dividing. During the replication of
mitochondrial DNA the mitochondria become double in size. They are oval in structure with
plain outer membrane and the inner membrane exhibits folding known as cristae. The inner
membrane encloses mitochondrial matrix. The matrix contains the DNA and enzymes involved
in citric acid cycle or Krebs cycle.
The cristae lodge the enzymes required in the electron transport chain to carry out the process of
oxidative phosphorylation. The energy is produced through a series of steps in which the
electrons are transported and the protons are pumped out of the matrix into the inner space
between the two membranes creating the electron potential gradient across the membranes. The
protons flow back to the matrix through an enzyme complex known as ATP synthase that is
found in cristae and utilizes the energy produced by the flow of protons to phosphorylate the
ADP forming the ATP.
2) uncoupler or uncoupling protein (UCP) is a mitochondrial protein located in the inner
membrane. These are involved in inhibition of the coupling between the electron transport and
phosphorylation reactions. As a result, synthesis of ATP is inhibited without affecting the
respiratory chain and ATP synthase (H+ -ATPase).
3) Under availability of oxygen the cell is more efficient in the energy. Under anaerobic
conditions the process of glucose breakdown cannot proceed further after glycolysis. The
glycolysis takes place in the cytosol producing 2 ATP molecules per each glucose molecule.
In presence of oxygen or under aerobic conditions the process of glycolysis further proceeds for
mitochondrial processing including the citric acid cycle and the electron transport system. This
process supplies enough energy to produce 30 more ATP molecules comprising to the total yield
of 32 ATP molecules per glucose molecule at the end of aerobic respiration. Thus, the ATP
molecules produced are 2 by glycolysis 2 by citric acid cycle and 28 by oxidative
phosphorylation. Hence, aerobic respiration is an energy efficient process than anaerobic
respiration.
Solution
1) Mitochondria are known as power houses of cell. The mitochondria are the double membrane
structure that can grow, fuse with one another and capable of dividing. During the replication of
mitochondrial DNA the mitochondria become double in size. They are oval in structure with
plain outer membrane and the inner membrane exhibits folding known as cristae. The inner
membrane encloses mitochondrial matrix. The matrix contains the DNA and enzymes involved
in citric acid cycle or Krebs cycle.
The cristae lodge the enzymes required in the electron transport chain to carry out the process of
oxidative phosphorylation. The energy is produced through a series of steps in which the
electrons are transported and the protons are pumped out of the matrix into .
the molecule is 2,2-dichloro-ethanol all carbons.pdf
the molecule is 2,2-dichloro-ethanol all carbons are sp3 hybridized all carbons have
a tetrahedral geometry only available electron pairs on chlorine and oxygen and they will be
oriented in such a fashion that repulsive forces are balanced or atleast minimum
Solution
the molecule is 2,2-dichloro-ethanol all carbons are sp3 hybridized all carbons have
a tetrahedral geometry only available electron pairs on chlorine and oxygen and they will be
oriented in such a fashion that repulsive forces are balanced or atleast minimum.
Increases. Gases have higher entropy than liquids.pdf
Increases. Gases have higher entropy than liquids, which have higher entropy than
solids. Whenever a phase change occurs from a lower entropy phase to a higher entropy phase,
the entropy increases.
Solution
Increases. Gases have higher entropy than liquids, which have higher entropy than
solids. Whenever a phase change occurs from a lower entropy phase to a higher entropy phase,
the entropy increases..
Simple C++ program to multiply two polynomials#include iostrea.pdf
// Simple C++ program to multiply two polynomials
#include
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i
Solution
// Simple C++ program to multiply two polynomials
#include
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i.
design and deliver an confirmation based academic and instruction sy.pdf
design and deliver an confirmation based academic and instruction system design and deliver
an confirmation based positive behavioral support system increase our parent outreach and
involvement Finally, design and bring this integrated, combined educational system through a
deliberate planning and organizational growth process comprehensive reforms are very hard and
involve, at times, the path of most confrontation comprehensive reform needs advanced and
continuing efforts to make constituencies that are give to long-term efforts, to deliberate
planning, and to the growth of systems that can maintain the change process over time and
during changes in leadership Comprehensive reform is not for each community, nor is each
community at a willingness level to start this process Comprehensive reform needs a merge of
outside technical support and local obligation, leadership, planning, funding, and assessment
those results in local rights and self-restitution. Comprehensive reform requires refurbish,
revision, reconsideration, and recommitment. Significant change is not interpreted as a sign of
malfunction. Comprehensive reform frequently requires the development of completely new
systems and ways of being. The change of existing systems or the execution of new systems built
alongside old systems frequently will not direct to real change and enduring outcomes.
Solution
design and deliver an confirmation based academic and instruction system design and deliver
an confirmation based positive behavioral support system increase our parent outreach and
involvement Finally, design and bring this integrated, combined educational system through a
deliberate planning and organizational growth process comprehensive reforms are very hard and
involve, at times, the path of most confrontation comprehensive reform needs advanced and
continuing efforts to make constituencies that are give to long-term efforts, to deliberate
planning, and to the growth of systems that can maintain the change process over time and
during changes in leadership Comprehensive reform is not for each community, nor is each
community at a willingness level to start this process Comprehensive reform needs a merge of
outside technical support and local obligation, leadership, planning, funding, and assessment
those results in local rights and self-restitution. Comprehensive reform requires refurbish,
revision, reconsideration, and recommitment. Significant change is not interpreted as a sign of
malfunction. Comprehensive reform frequently requires the development of completely new
systems and ways of being. The change of existing systems or the execution of new systems built
alongside old systems frequently will not direct to real change and enduring outcomes..
OK, let me try to talk you through this. The arro.pdf
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
separates the H from the O. The curved arrow at the bottom takes away the bond between the
two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon.
Solution
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
separates the H from the O. The curved arrow at the bottom takes away the bond between the
two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon..
x=input(Enter the first sequence ); l1=input(Enter the lowe.pdf
x=input(\'Enter the first sequence: \');
l1=input(\'Enter the lower limit: \');
u1=input(\'Enter the upper limit: \');
x1=l1:1:u1;
h=input(\'Enter the second sequence: \');
l2=input(\'Enter the lower limit: \');
u2=input(\'Enter the upper limit: \');
h1=l2:1:u2;
l=l1+l2;
u=u1+u2;
n=l:1:u;
s=numel(n);
i=1;
for i=1:s
y(i)=0;
for k=1:numel(x)
if (i+1-k)<=0
y(i)=y(i)+(x(k)*0);
else if (i+1-k)>numel(h)
y(i)=y(i)+(x(k)*0);
else
y(i)=y(i)+(x(k)*h(i+1-k));
k=k+1;
end
end
end
i=i+1;
end
disp(y);
subplot(2,2,1);stem(x1,x);
title(\'First sequence\');xlabel(\'n\');ylabel(\'x(n)\');
subplot(2,2,2);stem(h1,h);
title(\'Second Sequence\');xlabel(\'n\');ylabel(\'h(n)\');
subplot(2,2,[3 4]);stem(n,y);
title(\'Convoluted sequence\');xlabel(\'n\');ylabel(\'y(n)\');
Comment only
X = input(\'Enter x: \'); %input vector X and H
H = input(\'Enter h: \') ;
LenX = length(X); %defining their lenghts
LenH = length(H);
y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size
% lenth of X + length of H
t = zeros(1,LenH); % definign a vector t of same length as H
for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1
if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin
t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1)
for j = 1:LenH % in the if condition a for loop from 1 to length of H
y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first
iteration
% i.e. for i=1 only firt multiplication would
% be non zero rest all zeroes.
end
for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values
in t
% this cycle would continue until i is lesser then
% length X i.e. overlap increasing every iteration less
% and less non zero vales in t every iteration
t(k) = t(k-1);
end
else % now when all of the T is non zero which means 100% overlap in else overlap would start
to decrease between T and H
% T is basically X
t(1)= 0;
for j = 1:LenH % Now we start filling up Zeroes in T i.e. overlap began to decrease now and
each iteration it would decrease
% i.e T moving to left until there is no more
% over lap
y(i) = y(i) + (H(j)*t(j)); % in this for loop we multiply all respective vales of h and t and add the
% putting it at y(1) or y(2) or Y(i) in first iteration
end
for k = LenH:-1:2 %% here just like similar loop above t where we were filling up t with vales of
x
%now we are filling up zeroos in t i.e. over lap decreasing
t(k) = t(k-1);
end
end
end
ly=length(y)
indices=[ly]
for i=1:ly
indices(i)=i;
end
disp (y); %displays vector y.
disp (indices); % displays vector indices.
stem(y);
ylabel(\'Y[n]\');
xlabel(\'[n]\');
title(\'Convolution without conv function\');
x=input(\'Enter the first sequence: \');
l1=input(\'Enter the lower limit: \');
u1=input(\'Enter the upper limit: \');
x1=l1:1:u1;
h=input(\'Enter the second sequence: \');
l2=input(\'Enter the lower limit: \');
u2=input(\'Enter the upper limit: \');
h1=l2:1:u2;
l=l.
Yes , we can view .Xmzx files are encoded files that take to be un.pdf
Yes , we can view .
Xmzx files are encoded files that take to be unlocked using Examsoff/Softest software
application. As they\'re coded you can\'t just visible it without the encryption key (username and
password)
Solution
Yes , we can view .
Xmzx files are encoded files that take to be unlocked using Examsoff/Softest software
application. As they\'re coded you can\'t just visible it without the encryption key (username and
password).
The middle Carbon is the SP hybridized atom.SP hybrid orbitals con.pdf
The middle Carbon is the SP hybridized atom.
SP hybrid orbitals consist of one S and one P orbital combining to form two sp hybrid orbitals.
We can see the centre carbon having two sigma bonds with the adjoining atoms.
The pi bonds are formed by p orbital overlap. So the center carbon is the SP hybridized atom.
Solution
The middle Carbon is the SP hybridized atom.
SP hybrid orbitals consist of one S and one P orbital combining to form two sp hybrid orbitals.
We can see the centre carbon having two sigma bonds with the adjoining atoms.
The pi bonds are formed by p orbital overlap. So the center carbon is the SP hybridized atom..
The link provided by your teacher is a popular article based on the .pdf
The link provided by your teacher is a popular article based on the findings of an original
research[1].
Briefly, the study has used mice and has used the following tools/ techniques: Generation of the
Sox4-IRES-GFP/Luciferase Mouse Model; histopathology and immunohistochemistry;
estimation of lean mass, fat content, and bone mineral density with DEXA; keratocyte
clonogenic assay; and, microarray-based transcriptional profiling. For more details, you may
refer to the original research article.
Reference:
1. Miguel Foronda, Paula Martínez, Stefan Schoeftner, Gonzalo Gómez-López, Ralph Schneider,
Juana M. Flores, David G. Pisano, Maria A. Blasco. Sox4 Links Tumor Suppression to
Accelerated Aging in Mice by Modulating Stem Cell Activation. Cell Reports, 2014; DOI:
10.1016/j.celrep.2014.06.031
Solution
The link provided by your teacher is a popular article based on the findings of an original
research[1].
Briefly, the study has used mice and has used the following tools/ techniques: Generation of the
Sox4-IRES-GFP/Luciferase Mouse Model; histopathology and immunohistochemistry;
estimation of lean mass, fat content, and bone mineral density with DEXA; keratocyte
clonogenic assay; and, microarray-based transcriptional profiling. For more details, you may
refer to the original research article.
Reference:
1. Miguel Foronda, Paula Martínez, Stefan Schoeftner, Gonzalo Gómez-López, Ralph Schneider,
Juana M. Flores, David G. Pisano, Maria A. Blasco. Sox4 Links Tumor Suppression to
Accelerated Aging in Mice by Modulating Stem Cell Activation. Cell Reports, 2014; DOI:
10.1016/j.celrep.2014.06.031.
EPO, hematopoietin, or hemopoietin, is a glycoprotein hormone that c.pdf
EPO, hematopoietin, or hemopoietin, is a glycoprotein hormone that controls erythropoiesis, or
red platelet creation. It is a cytokine (protein flagging particle) for erythrocyte (red platelet)
antecedents in the bone marrow. Human EPO has an atomic weight of 34 kDa. Erythropoietin is
delivered by interstitial fibroblasts in the kidney in close relationship with peritubular narrow and
proximal convoluted tubule. It is likewise created in perisinusoidal cells in the liver. While liver
generation prevails in the fetal and perinatal period, renal creation is overwhelming amid
adulthood. Exogenous erythropoietin can be given to individuals whose kidneys can\'t make
enough. Recombinant human erythropoietin (rhEPO) is created by recombinant DNA innovation
in cell culture. A few distinctive pharmaceutical specialists are accessible with an assortment of
glycosylation examples and are by and large called erythropoiesis-invigorating operators (ESA).
Real illustrations are epoetin alfa and epoetin beta. The particular subtle elements for named
utilize shift between the bundle embeds, yet ESAs have been utilized as a part of the treatment of
iron deficiency in ceaseless kidney ailment, frailty in myelodysplasia, and in pallor from
malignancy chemotherapy. Boxed notices incorporate a danger of death, myocardial dead tissue,
stroke, venous thromboembolism, and tumor recurrence.[6] rhEPO has been utilized unlawfully
as an execution improving drug;[7] it can regularly be distinguished in blood, because of slight
contrasts from the endogenous protein, for instance, in components of posttranslational change.
Answer : then net reabsorption of the X
The reabsorption or discharge rate is the contrast between the sum separated over the glomerular
vessels and the sum discharged in pee. It is figured with the accompanying conditions:
Filtered load = GFR x [plasma] Excretion rate = V x [urine] Reabsorption rate = Filtered stack -
Excretion rate Secretion rate = Excretion rate - Filtered stack
- If the separated load is more prominent than the discharge rate, then net reabsorption of the
substance has happened. On the off chance that the sifted load is not exactly the discharge rate,
then net emission of the substance has happened.
Insulin is used in studies of kidney function it is
Reabsorbed and but not secreted
Insulin, a 51-amino corrosive peptide hormone with an atomic weight of roughly 6000 Da, is
incorporated by pancreatic islet beta cells and systematized by a quality situated in the short arm
of chromosome 11 . Half-life (t1/2) of insulin is short ., and it is not bound to plasma proteins.
Under fasting conditions, insulin discharge is nonstop with an emission rate of roughly 0.5–1
unit/h. Insulin discharge increments 3–10 times with sustenance ingestion, putting the aggregate
day by day insulin emission at around 18–32 units ..Insulin follows up on the kidney by
authoritative with receptors situated in proximal tubular cells at both luminal and contralumina.
The answer is D) IIA buffer system must contain a weak acid and i.pdf
The answer is: D) II
A buffer system must contain a weak acid and its conjgate base, or a weak base and its conjugate
acid.
I. is NOT a buffer as Ca(OH)2 and CaCl2 are not a conjugate acid/base pair
II IS a buffer as NH3 and NH4+ are a conjugate acid/base pair
III. is not a buffer as HClO and HBr are not a conjugate acid/base pair
IV. is not a buffer as BaI2 and NaI are not a conjugate acid/base pair
V. is not a buffer as HCl and NaCl are not a conjugate acid/base pair
Solution
The answer is: D) II
A buffer system must contain a weak acid and its conjgate base, or a weak base and its conjugate
acid.
I. is NOT a buffer as Ca(OH)2 and CaCl2 are not a conjugate acid/base pair
II IS a buffer as NH3 and NH4+ are a conjugate acid/base pair
III. is not a buffer as HClO and HBr are not a conjugate acid/base pair
IV. is not a buffer as BaI2 and NaI are not a conjugate acid/base pair
V. is not a buffer as HCl and NaCl are not a conjugate acid/base pair.
c. underwent morphogenesis between the early and late gastrula stage.pdf
c. underwent morphogenesis between the early and late gastrula stages.
[Morphogenesis of cell is a biological process that determines the fate of cell in accordance with
the environment of surrounding cells and soluble molecules known as morphogens.
In early stage of gastrula, the germ cells forming epidermis are yet uncommited and their cell
surface receptors or proteins are are not yet destined for a specialized function. The
transplantation of cells destined to become neural tissue in this region are modulated in
accordance to the surrounding cells in early gastrula. On the other hand the, the germ cells in late
gastrula are commited to a specific lineage and express a specific pattern of receptors or proteins.
This kind of environment can not modulate the cellular fate of transplanted cells.]
Solution
c. underwent morphogenesis between the early and late gastrula stages.
[Morphogenesis of cell is a biological process that determines the fate of cell in accordance with
the environment of surrounding cells and soluble molecules known as morphogens.
In early stage of gastrula, the germ cells forming epidermis are yet uncommited and their cell
surface receptors or proteins are are not yet destined for a specialized function. The
transplantation of cells destined to become neural tissue in this region are modulated in
accordance to the surrounding cells in early gastrula. On the other hand the, the germ cells in late
gastrula are commited to a specific lineage and express a specific pattern of receptors or proteins.
This kind of environment can not modulate the cellular fate of transplanted cells.].
Text Component FeaturesThe JTextComponent class is the foundation .pdf
Text Component Features
The JTextComponent class is the foundation for Swing text components. This class provides the
following customizable features for all of its descendants:
A model, known as a document, that manages the component\'s content.
A view, which displays the component on screen.
A controller, known as an editor kit, that reads and writes text and implements editing
capabilities with actions.
upport for infinite undo and redo.
A pluggable caret and support for caret change listeners and navigation filters.
A layout manager encapsulates an algorithm for positioning and sizing of GUI components.
Rather than building the layout algorithms into your code and watching for window resizing and
other layout-modifying events, the algorithm is kept separate. This allows a layout algorithm to
be reused for several applications, while simplifying your application code.
LayoutManager is an interface in the Java class libraries that describes how a Container and a
layout manager communicate. It describes several methods which:
Component Size Methods
The Component class defines several size accessor methods to assist the layout process. Each of
these methods returns a Dimension object describing the requested size. These methods are as
follows:
An additional interface, LayoutManager2, was added in JDK 1.1, which adds a few more
positioning and validation methods.
Layout managers will use these sizing methods when they are figuring out where to place
components and what size they should be. Layout managers can respect or ignore as much or as
little of this information as they see fit. Each layout manager has its own algorithm and may or
may not use this information when deciding component placement. Which of these methods is
respected or ignored is very important information and should be documented carefully when
creating your own layout manager.
Controlling the sizes returned for your own components can be accomplished in two ways,
depending on whether you are using the JFC Project Swing components.
If your component is a JFC Project Swing component, you inherit three methods,
setPreferredSize(Dimension), setMinimumSize(Dimension) and setMaximumSize(Dimension).
You can call these methods directly to explicitly set the size information for a component. For
example:
Layout Managers and Containers
A layout manager must be associated with a Container object to perform its work. If a container
does not have an associated layout manager, the container simply places components wherever
specified by using the setBounds(), setLocation() and/or setSize() methods.
If a container has an associated layout manager, the container asks that layout manager to
position and size its components before they are painted. The layout manager itself does
notperform the painting; it simply decides what size and position each component should occupy
and calls setBounds(), setLocation() and/or setSize() on each of those components.
A LayoutManager is associated with a C.
operating system linux,ubuntu,Mac Geometri.pdf
//operating system linux,ubuntu,Mac
/*********************GeometricObject.java**********************/
public abstract class GeometricObject {
private String color = \"white\";
private boolean filled;
// default constructure
public GeometricObject() {
super();
// TODO Auto-generated constructor stub
}
// construct a Geometric Object
// parameterized constructure
public GeometricObject(String color, boolean filled) {
this.color = color;
this.filled = filled;
}
/** Getter method for color */
public String getColor() {
return color;
}
/** Setter method for color */
public void setColor(String color) {
this.color = color;
}
/**
* Getter method for filled. Since filled is boolean, so the gret method
* name is isFilled
*/
public boolean isFilled() {
return filled;
}
/** Setter method for filled */
public void setFilled(boolean filled) {
this.filled = filled;
}
/** Abstract method for FindArea */
public abstract double findArea();
/** Abstract method for findPerimeter */
public abstract double findPerimeter();
}
/*****************************Octagon.java**************/
public class Octagon extends GeometricObject implements Cloneable, Comparable {
private double side;
/** construct a Octagon with specified side */
public Octagon(double side) {
super();
this.side = side;
}
/** Implement the abstract method findArea in GeometricObject */
@Override
public double findArea() {
double area = (2 + 4 / Math.sqrt(2)) * side * side;
return area;
}
/** Implement the abstract method findArea in findPerimeter */
@Override
public double findPerimeter() {
double perimeter = 8 * side;
return perimeter;
}
/** Implement the compareTo method in Comparable interface */
@Override
public int compareTo(Object o) {
// TODO Auto-generated method stub
return 0;
}
public Object clone() throws CloneNotSupportedException {
return super.clone();
}
}
/*******************App.java*******************/
public class App {
public static void main(String[] args) {
Octagon a1 = new Octagon(5);//creating object of octagon
System.out.println(\"Area of a1 is \" + a1.findArea());
System.out.println(\"Perimeter of a1 is \" + a1.findPerimeter());
Object a2;
try {
a2 = a1.clone();
String result = (a1.compareTo(a2) == 0) ? \"a1 and its clone a2 have the same area\"
: \"a1 and its clone a2 have different areas\";
System.out.println(\"Compare a1 and its clone a2:\ \\t \" + result);
System.out.println(\"Hashcode of a1: \" + a1.hashCode());
System.out.println(\"Hashcode of a2: \" + a2.hashCode());
System.out.println(\"Displaying a1: \" + a1);
System.out.println(\"Displaying a2: \" + a2);
} catch (CloneNotSupportedException e) {
e.printStackTrace();
}
}
}
/*****************output***************/
gopal@gopal:~/Desktop/chegg$ javac GeometricObject.java
gopal@gopal:~/Desktop/chegg$ javac Octagon.java
gopal@gopal:~/Desktop/chegg$ javac App.java
gopal@gopal:~/Desktop/chegg$ java App
Area of a1 is 120.71067811865476
Perimeter of a1 is 40.0
Compare a1 and its clone a2:
a1 and its clone a2 have the same area
.
NO TILL FARMINGNo-till farming (also called zero tillage or direct.pdf
NO TILL FARMING
No-till farming (also called zero tillage or direct drilling) is a way of growing crops or pasture
from year to year without disturbing the soil through tillage. No-till is an agricultural technique
which increases the amount of water that infiltrates into the soil and increases organic matter
retention and cycling of nutrients in the soil. In many agricultural regions it can reduce or
eliminate soil erosion. It increases the amount and variety of life in and on the soil, including
disease-causing organisms and disease suppression organisms. The most powerful benefit of no-
tillage is improvement in soil biological fertility, making soils more resilient. Farm operations
are made much more efficient, particularly improved time of sowing and better trafficability of
farm operations.
TILLAGE SYSTEM
There are a variety of tillage systems available for crop production. While tillage operations are
performed for various reasons, producers must evaluate the need for each and every field
operation conducted in order to improve profitability. In addition, the effects of the tillage
operations on the soil system and the environment must be considered. More information is
available on the following tillage systems:
NITROGEN LEACHING
Nitrogen is a common element in nature and an essential plant nutrient. Approximately
78% of Earth\'s atmosphere is nitrogen (N2). The strong bond between the atoms of N2 makes
this gas quite inert and not directly usable by plants and animals. As nitrogen naturally cycles
through the air, water and soil it undergoes various chemical and biological transformations.
Nitrogen promotes plant growth. Livestock then eat the crops producing manure, which is
returned to the soil, adding organic and mineral forms of nitrogen. The cycle is complete when
the next crop uses the amended soil. To increase food production, fertilizers, such as nitrate
(NO3) and ammonium (NH4), which are easily absorbed by plants, are introduced to the plant
root zone. However, soils do not absorb the excess NO3 ions, which then move downward freely
with drainage water, and are leached into groundwater, streams and oceans. The degree of
leaching is affected by:
The level of nitrous oxide (N2O) in the Earth\'s atmosphere is increasing at a rate of 0.2 to 0.3%
annually. Anthropogenic sources of nitrogen are 50% greater than from natural sources, such as,
soils and oceans. Leached agricultural inputs, i.e. fertilizers and manures, accounts for 75% of
the anthropogenic source of nitrogen. The Food and Agriculture Organization of the United
Nations (FAO) estimates world demand for nitrogen fertilizers will increase by 1.7% annually
between 2011 and 2015. An increase of 7.5 million tonnes. Regional increases of nitrogen
fertilizer use are expected to be 67% by Asia, 18% by the Americas, 10% by Europe, 3% by
Africa,and 1% by Oceania.
Solution
NO TILL FARMING
No-till farming (also called zero tillage or direct drilling) is a way of growing crops or.
Lions use sounds with different intensities, senses such as tough an.pdf
Lions use sounds with different intensities, senses such as tough and behaviours for
communication. Pheromones are the chemicals released by an animal to communicate with its
mate and this also occurs in lions. Jacobson’s organ is an olfactory organ located at the roof of
the lion’s mouth. This organ perceive the smell and send impulses to the olfactory center in the
brain.
Solution
Lions use sounds with different intensities, senses such as tough and behaviours for
communication. Pheromones are the chemicals released by an animal to communicate with its
mate and this also occurs in lions. Jacobson’s organ is an olfactory organ located at the roof of
the lion’s mouth. This organ perceive the smell and send impulses to the olfactory center in the
brain..
let R be ringI be a subgroup under additionif I be idealfor a,.pdf
let R be ring
I be a subgroup under addition
if I be ideal
for a,b in R
(a+I)+(b+I)=a+I+b+I
=a+(I+b)+I
=a+b+I+I
=a+b+I
and
(a+I)(b+I)=ab+aI+Ib+I
=ab+I+I+I since aI left ideal and Ib is right ideal
=ab+I
if (a+I)+(b+I)= a+b+I
and (a+I)(b+I)=ab+I
R is ring and I be subgroup
a,b in R
imples a,b in I
and a,b in I
then a+b in I
thus I is closed under addition
and a, 0 be in I
(a+I)+(0+I)=a+0+I
=a+I
thus 0 is zero element of R
-a be additive inverse
a,-a be in R
(a+I)+(-a+I)=a-a+I=0+I
thus I is closed under additive inverse
(a+I)(b+I)= (ab+I)
1 be multiplicative inverse
1+I be ideal
(a+I)(1+I)=a*1+I
=a+I
thus 1+I be multiplicate inverse
thus I be ideal
Solution
let R be ring
I be a subgroup under addition
if I be ideal
for a,b in R
(a+I)+(b+I)=a+I+b+I
=a+(I+b)+I
=a+b+I+I
=a+b+I
and
(a+I)(b+I)=ab+aI+Ib+I
=ab+I+I+I since aI left ideal and Ib is right ideal
=ab+I
if (a+I)+(b+I)= a+b+I
and (a+I)(b+I)=ab+I
R is ring and I be subgroup
a,b in R
imples a,b in I
and a,b in I
then a+b in I
thus I is closed under addition
and a, 0 be in I
(a+I)+(0+I)=a+0+I
=a+I
thus 0 is zero element of R
-a be additive inverse
a,-a be in R
(a+I)+(-a+I)=a-a+I=0+I
thus I is closed under additive inverse
(a+I)(b+I)= (ab+I)
1 be multiplicative inverse
1+I be ideal
(a+I)(1+I)=a*1+I
=a+I
thus 1+I be multiplicate inverse
thus I be ideal.
If we use evaporation or recrystallization, we would only remove the.pdf
If we use evaporation or recrystallization, we would only remove the water, leaving behind the
barium sulfate and the sodium chloride. We want to separate them , not get a mixture of barium
sulfate and sodium chloride. Also, any unreacted barium nitrate and sodium sulfate will become
solids and contaminate the barium sulphate solid that we will obtain.
We use filtration, because this method allows the soluble salts to pass through the filter, leaving
behind the insoluble part, the barium sulfate. Even unreacted barium nitrate and sodium sulfate is
allowed to pass through the filter.
BaSO4 is a solid precipitate and NaCl in dissolved in the water. If you evaporated the water, you
would be left with a mixture of NaCl and BaSO4.
To recrystallize BaSO4, you would first have to isolate it (by filtration), then dissolve it and
allow it to recrystallize. Why not just filter was wash out any adhering NaCl (or unreacted
reactant).
Solution
If we use evaporation or recrystallization, we would only remove the water, leaving behind the
barium sulfate and the sodium chloride. We want to separate them , not get a mixture of barium
sulfate and sodium chloride. Also, any unreacted barium nitrate and sodium sulfate will become
solids and contaminate the barium sulphate solid that we will obtain.
We use filtration, because this method allows the soluble salts to pass through the filter, leaving
behind the insoluble part, the barium sulfate. Even unreacted barium nitrate and sodium sulfate is
allowed to pass through the filter.
BaSO4 is a solid precipitate and NaCl in dissolved in the water. If you evaporated the water, you
would be left with a mixture of NaCl and BaSO4.
To recrystallize BaSO4, you would first have to isolate it (by filtration), then dissolve it and
allow it to recrystallize. Why not just filter was wash out any adhering NaCl (or unreacted
reactant)..
import java.util.Random; public class randomCalls { .pdf
import java.util.Random;
public class randomCalls {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Random rn = new Random();
int n;
for(int i=1;i<=720;i++)
{
n = rn.nextInt(6) +5;
System.out.print(n);
System.out.print(\' \');
System.out.println(i);
}
}
}
Solution
import java.util.Random;
public class randomCalls {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Random rn = new Random();
int n;
for(int i=1;i<=720;i++)
{
n = rn.nextInt(6) +5;
System.out.print(n);
System.out.print(\' \');
System.out.println(i);
}
}
}.
I am considering the case where you have multiple name in same file .pdf
I am considering the case where you have multiple name in same file and you want to replace all
of them.
Step1: open terminal in linux operating system
Step2: move to desired file location using below command
Cd /home/beth/project/
Step3: Open desired file in current location
vim 1stqtr and press [enter]
you can see your file content on terminal
step 4: Now go to command mode in VM using below command
press [esc key]
step 5: Now you have to search your old name and replace with new name. For this use below
command structure.
:s/Search/Replace/CommandFlag
Explanation: s -> it is used to command Vi editor that I am going to use search and replace
command
Search-> It is a word which you want to replace
Replace-> It is a word from which you want to replace
Command Flag-> It is command flag which provides extra features like case
sensitive etc.
Say your old name is Ain
:%s/Smith/Oliver/g
To find and replace all occurrences of case insensitive ‘Smith with ‘Oliver, enter:
:%s/Smith/Oliver/gi
To find and replace all occurrences of ‘eth1’ with ‘br1’, but ask for confirmation first, enter:
:%s/Smith/Oliver/gc
Step 6: In the last enter below command to save and exit.
:wq[enter]
Solution
I am considering the case where you have multiple name in same file and you want to replace all
of them.
Step1: open terminal in linux operating system
Step2: move to desired file location using below command
Cd /home/beth/project/
Step3: Open desired file in current location
vim 1stqtr and press [enter]
you can see your file content on terminal
step 4: Now go to command mode in VM using below command
press [esc key]
step 5: Now you have to search your old name and replace with new name. For this use below
command structure.
:s/Search/Replace/CommandFlag
Explanation: s -> it is used to command Vi editor that I am going to use search and replace
command
Search-> It is a word which you want to replace
Replace-> It is a word from which you want to replace
Command Flag-> It is command flag which provides extra features like case
sensitive etc.
Say your old name is Ain
:%s/Smith/Oliver/g
To find and replace all occurrences of case insensitive ‘Smith with ‘Oliver, enter:
:%s/Smith/Oliver/gi
To find and replace all occurrences of ‘eth1’ with ‘br1’, but ask for confirmation first, enter:
:%s/Smith/Oliver/gc
Step 6: In the last enter below command to save and exit.
:wq[enter].
Hormone A chemical substance produced in the body by ductless gland.pdf
Hormone: A chemical substance produced in the body by ductless glands called endocrine
glands they control and regulate the activity of certain cells or organs of our body. The pathway
is:
· Neurosecretory neurons are stimulated (1)
· Releasing hormones are secreted from the hypothalamus (2)
· Endocrine cell secretes hormone (3)
· Hormone travels through the bloodstream (4)
· Cells utilize receptors to provide specificity of the hormone (5)
Solution
Hormone: A chemical substance produced in the body by ductless glands called endocrine
glands they control and regulate the activity of certain cells or organs of our body. The pathway
is:
· Neurosecretory neurons are stimulated (1)
· Releasing hormones are secreted from the hypothalamus (2)
· Endocrine cell secretes hormone (3)
· Hormone travels through the bloodstream (4)
· Cells utilize receptors to provide specificity of the hormone (5).
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
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Increases. Gases have higher entropy than liquids.pdf
Increases. Gases have higher entropy than liquids, which have higher entropy than
solids. Whenever a phase change occurs from a lower entropy phase to a higher entropy phase,
the entropy increases.
Solution
Increases. Gases have higher entropy than liquids, which have higher entropy than
solids. Whenever a phase change occurs from a lower entropy phase to a higher entropy phase,
the entropy increases..
Simple C++ program to multiply two polynomials#include iostrea.pdf
// Simple C++ program to multiply two polynomials
#include
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i
Solution
// Simple C++ program to multiply two polynomials
#include
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i
using namespace std;
// A[] represents coefficients of first polynomial
// B[] represents coefficients of second polynomial
// m and n are sizes of A[] and B[] respectively
int *multiply(int A[], int B[], int m, int n)
{
int *prod = new int[m+n-1];
// Initialize the porduct polynomial
for (int i = 0; i.
design and deliver an confirmation based academic and instruction sy.pdf
design and deliver an confirmation based academic and instruction system design and deliver
an confirmation based positive behavioral support system increase our parent outreach and
involvement Finally, design and bring this integrated, combined educational system through a
deliberate planning and organizational growth process comprehensive reforms are very hard and
involve, at times, the path of most confrontation comprehensive reform needs advanced and
continuing efforts to make constituencies that are give to long-term efforts, to deliberate
planning, and to the growth of systems that can maintain the change process over time and
during changes in leadership Comprehensive reform is not for each community, nor is each
community at a willingness level to start this process Comprehensive reform needs a merge of
outside technical support and local obligation, leadership, planning, funding, and assessment
those results in local rights and self-restitution. Comprehensive reform requires refurbish,
revision, reconsideration, and recommitment. Significant change is not interpreted as a sign of
malfunction. Comprehensive reform frequently requires the development of completely new
systems and ways of being. The change of existing systems or the execution of new systems built
alongside old systems frequently will not direct to real change and enduring outcomes.
Solution
design and deliver an confirmation based academic and instruction system design and deliver
an confirmation based positive behavioral support system increase our parent outreach and
involvement Finally, design and bring this integrated, combined educational system through a
deliberate planning and organizational growth process comprehensive reforms are very hard and
involve, at times, the path of most confrontation comprehensive reform needs advanced and
continuing efforts to make constituencies that are give to long-term efforts, to deliberate
planning, and to the growth of systems that can maintain the change process over time and
during changes in leadership Comprehensive reform is not for each community, nor is each
community at a willingness level to start this process Comprehensive reform needs a merge of
outside technical support and local obligation, leadership, planning, funding, and assessment
those results in local rights and self-restitution. Comprehensive reform requires refurbish,
revision, reconsideration, and recommitment. Significant change is not interpreted as a sign of
malfunction. Comprehensive reform frequently requires the development of completely new
systems and ways of being. The change of existing systems or the execution of new systems built
alongside old systems frequently will not direct to real change and enduring outcomes..
OK, let me try to talk you through this. The arro.pdf
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
separates the H from the O. The curved arrow at the bottom takes away the bond between the
two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon.
Solution
OK, let me try to talk you through this. The arrow pointing straight up takes a
double bond between the O and the C (all intersections of lines are understood to be carbon
atoms) and forms an O-H bond and leaves a single O-C bond. The curved arrow on the right,
straddling the O takes away the H-O bond and forms a second bond between the O and C. This
separates the H from the O. The curved arrow at the bottom takes away the bond between the
two carbons, separating the molecule. We are left with O=C=O which is carbon dioxide. Carbon
dioxide has two double bonds and is a linear compound. This is the species with the least (we
should say fewest since we can count the atoms) atoms. Then for the first part we have
everything else A central carbon with a single bond to each of two -OH groups and a double
bond to one carbon and that carbon has single bonds to two other carbons. So the central carbon
has for bonds 1 to O, 2 to O, 3 and 4 to C. Then that C has four bonds, 1 and 2 to the central
carbon and then 3 to a carbon and 4 to a carbon..
x=input(Enter the first sequence ); l1=input(Enter the lowe.pdf
x=input(\'Enter the first sequence: \');
l1=input(\'Enter the lower limit: \');
u1=input(\'Enter the upper limit: \');
x1=l1:1:u1;
h=input(\'Enter the second sequence: \');
l2=input(\'Enter the lower limit: \');
u2=input(\'Enter the upper limit: \');
h1=l2:1:u2;
l=l1+l2;
u=u1+u2;
n=l:1:u;
s=numel(n);
i=1;
for i=1:s
y(i)=0;
for k=1:numel(x)
if (i+1-k)<=0
y(i)=y(i)+(x(k)*0);
else if (i+1-k)>numel(h)
y(i)=y(i)+(x(k)*0);
else
y(i)=y(i)+(x(k)*h(i+1-k));
k=k+1;
end
end
end
i=i+1;
end
disp(y);
subplot(2,2,1);stem(x1,x);
title(\'First sequence\');xlabel(\'n\');ylabel(\'x(n)\');
subplot(2,2,2);stem(h1,h);
title(\'Second Sequence\');xlabel(\'n\');ylabel(\'h(n)\');
subplot(2,2,[3 4]);stem(n,y);
title(\'Convoluted sequence\');xlabel(\'n\');ylabel(\'y(n)\');
Comment only
X = input(\'Enter x: \'); %input vector X and H
H = input(\'Enter h: \') ;
LenX = length(X); %defining their lenghts
LenH = length(H);
y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size
% lenth of X + length of H
t = zeros(1,LenH); % definign a vector t of same length as H
for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1
if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin
t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1)
for j = 1:LenH % in the if condition a for loop from 1 to length of H
y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first
iteration
% i.e. for i=1 only firt multiplication would
% be non zero rest all zeroes.
end
for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values
in t
% this cycle would continue until i is lesser then
% length X i.e. overlap increasing every iteration less
% and less non zero vales in t every iteration
t(k) = t(k-1);
end
else % now when all of the T is non zero which means 100% overlap in else overlap would start
to decrease between T and H
% T is basically X
t(1)= 0;
for j = 1:LenH % Now we start filling up Zeroes in T i.e. overlap began to decrease now and
each iteration it would decrease
% i.e T moving to left until there is no more
% over lap
y(i) = y(i) + (H(j)*t(j)); % in this for loop we multiply all respective vales of h and t and add the
% putting it at y(1) or y(2) or Y(i) in first iteration
end
for k = LenH:-1:2 %% here just like similar loop above t where we were filling up t with vales of
x
%now we are filling up zeroos in t i.e. over lap decreasing
t(k) = t(k-1);
end
end
end
ly=length(y)
indices=[ly]
for i=1:ly
indices(i)=i;
end
disp (y); %displays vector y.
disp (indices); % displays vector indices.
stem(y);
ylabel(\'Y[n]\');
xlabel(\'[n]\');
title(\'Convolution without conv function\');
x=input(\'Enter the first sequence: \');
l1=input(\'Enter the lower limit: \');
u1=input(\'Enter the upper limit: \');
x1=l1:1:u1;
h=input(\'Enter the second sequence: \');
l2=input(\'Enter the lower limit: \');
u2=input(\'Enter the upper limit: \');
h1=l2:1:u2;
l=l.
Yes , we can view .Xmzx files are encoded files that take to be un.pdf
Yes , we can view .
Xmzx files are encoded files that take to be unlocked using Examsoff/Softest software
application. As they\'re coded you can\'t just visible it without the encryption key (username and
password)
Solution
Yes , we can view .
Xmzx files are encoded files that take to be unlocked using Examsoff/Softest software
application. As they\'re coded you can\'t just visible it without the encryption key (username and
password).
The middle Carbon is the SP hybridized atom.SP hybrid orbitals con.pdf
The middle Carbon is the SP hybridized atom.
SP hybrid orbitals consist of one S and one P orbital combining to form two sp hybrid orbitals.
We can see the centre carbon having two sigma bonds with the adjoining atoms.
The pi bonds are formed by p orbital overlap. So the center carbon is the SP hybridized atom.
Solution
The middle Carbon is the SP hybridized atom.
SP hybrid orbitals consist of one S and one P orbital combining to form two sp hybrid orbitals.
We can see the centre carbon having two sigma bonds with the adjoining atoms.
The pi bonds are formed by p orbital overlap. So the center carbon is the SP hybridized atom..
The link provided by your teacher is a popular article based on the .pdf
The link provided by your teacher is a popular article based on the findings of an original
research[1].
Briefly, the study has used mice and has used the following tools/ techniques: Generation of the
Sox4-IRES-GFP/Luciferase Mouse Model; histopathology and immunohistochemistry;
estimation of lean mass, fat content, and bone mineral density with DEXA; keratocyte
clonogenic assay; and, microarray-based transcriptional profiling. For more details, you may
refer to the original research article.
Reference:
1. Miguel Foronda, Paula Martínez, Stefan Schoeftner, Gonzalo Gómez-López, Ralph Schneider,
Juana M. Flores, David G. Pisano, Maria A. Blasco. Sox4 Links Tumor Suppression to
Accelerated Aging in Mice by Modulating Stem Cell Activation. Cell Reports, 2014; DOI:
10.1016/j.celrep.2014.06.031
Solution
The link provided by your teacher is a popular article based on the findings of an original
research[1].
Briefly, the study has used mice and has used the following tools/ techniques: Generation of the
Sox4-IRES-GFP/Luciferase Mouse Model; histopathology and immunohistochemistry;
estimation of lean mass, fat content, and bone mineral density with DEXA; keratocyte
clonogenic assay; and, microarray-based transcriptional profiling. For more details, you may
refer to the original research article.
Reference:
1. Miguel Foronda, Paula Martínez, Stefan Schoeftner, Gonzalo Gómez-López, Ralph Schneider,
Juana M. Flores, David G. Pisano, Maria A. Blasco. Sox4 Links Tumor Suppression to
Accelerated Aging in Mice by Modulating Stem Cell Activation. Cell Reports, 2014; DOI:
10.1016/j.celrep.2014.06.031.
EPO, hematopoietin, or hemopoietin, is a glycoprotein hormone that c.pdf
EPO, hematopoietin, or hemopoietin, is a glycoprotein hormone that controls erythropoiesis, or
red platelet creation. It is a cytokine (protein flagging particle) for erythrocyte (red platelet)
antecedents in the bone marrow. Human EPO has an atomic weight of 34 kDa. Erythropoietin is
delivered by interstitial fibroblasts in the kidney in close relationship with peritubular narrow and
proximal convoluted tubule. It is likewise created in perisinusoidal cells in the liver. While liver
generation prevails in the fetal and perinatal period, renal creation is overwhelming amid
adulthood. Exogenous erythropoietin can be given to individuals whose kidneys can\'t make
enough. Recombinant human erythropoietin (rhEPO) is created by recombinant DNA innovation
in cell culture. A few distinctive pharmaceutical specialists are accessible with an assortment of
glycosylation examples and are by and large called erythropoiesis-invigorating operators (ESA).
Real illustrations are epoetin alfa and epoetin beta. The particular subtle elements for named
utilize shift between the bundle embeds, yet ESAs have been utilized as a part of the treatment of
iron deficiency in ceaseless kidney ailment, frailty in myelodysplasia, and in pallor from
malignancy chemotherapy. Boxed notices incorporate a danger of death, myocardial dead tissue,
stroke, venous thromboembolism, and tumor recurrence.[6] rhEPO has been utilized unlawfully
as an execution improving drug;[7] it can regularly be distinguished in blood, because of slight
contrasts from the endogenous protein, for instance, in components of posttranslational change.
Answer : then net reabsorption of the X
The reabsorption or discharge rate is the contrast between the sum separated over the glomerular
vessels and the sum discharged in pee. It is figured with the accompanying conditions:
Filtered load = GFR x [plasma] Excretion rate = V x [urine] Reabsorption rate = Filtered stack -
Excretion rate Secretion rate = Excretion rate - Filtered stack
- If the separated load is more prominent than the discharge rate, then net reabsorption of the
substance has happened. On the off chance that the sifted load is not exactly the discharge rate,
then net emission of the substance has happened.
Insulin is used in studies of kidney function it is
Reabsorbed and but not secreted
Insulin, a 51-amino corrosive peptide hormone with an atomic weight of roughly 6000 Da, is
incorporated by pancreatic islet beta cells and systematized by a quality situated in the short arm
of chromosome 11 . Half-life (t1/2) of insulin is short ., and it is not bound to plasma proteins.
Under fasting conditions, insulin discharge is nonstop with an emission rate of roughly 0.5–1
unit/h. Insulin discharge increments 3–10 times with sustenance ingestion, putting the aggregate
day by day insulin emission at around 18–32 units ..Insulin follows up on the kidney by
authoritative with receptors situated in proximal tubular cells at both luminal and contralumina.
The answer is D) IIA buffer system must contain a weak acid and i.pdf
The answer is: D) II
A buffer system must contain a weak acid and its conjgate base, or a weak base and its conjugate
acid.
I. is NOT a buffer as Ca(OH)2 and CaCl2 are not a conjugate acid/base pair
II IS a buffer as NH3 and NH4+ are a conjugate acid/base pair
III. is not a buffer as HClO and HBr are not a conjugate acid/base pair
IV. is not a buffer as BaI2 and NaI are not a conjugate acid/base pair
V. is not a buffer as HCl and NaCl are not a conjugate acid/base pair
Solution
The answer is: D) II
A buffer system must contain a weak acid and its conjgate base, or a weak base and its conjugate
acid.
I. is NOT a buffer as Ca(OH)2 and CaCl2 are not a conjugate acid/base pair
II IS a buffer as NH3 and NH4+ are a conjugate acid/base pair
III. is not a buffer as HClO and HBr are not a conjugate acid/base pair
IV. is not a buffer as BaI2 and NaI are not a conjugate acid/base pair
V. is not a buffer as HCl and NaCl are not a conjugate acid/base pair.
c. underwent morphogenesis between the early and late gastrula stage.pdf
c. underwent morphogenesis between the early and late gastrula stages.
[Morphogenesis of cell is a biological process that determines the fate of cell in accordance with
the environment of surrounding cells and soluble molecules known as morphogens.
In early stage of gastrula, the germ cells forming epidermis are yet uncommited and their cell
surface receptors or proteins are are not yet destined for a specialized function. The
transplantation of cells destined to become neural tissue in this region are modulated in
accordance to the surrounding cells in early gastrula. On the other hand the, the germ cells in late
gastrula are commited to a specific lineage and express a specific pattern of receptors or proteins.
This kind of environment can not modulate the cellular fate of transplanted cells.]
Solution
c. underwent morphogenesis between the early and late gastrula stages.
[Morphogenesis of cell is a biological process that determines the fate of cell in accordance with
the environment of surrounding cells and soluble molecules known as morphogens.
In early stage of gastrula, the germ cells forming epidermis are yet uncommited and their cell
surface receptors or proteins are are not yet destined for a specialized function. The
transplantation of cells destined to become neural tissue in this region are modulated in
accordance to the surrounding cells in early gastrula. On the other hand the, the germ cells in late
gastrula are commited to a specific lineage and express a specific pattern of receptors or proteins.
This kind of environment can not modulate the cellular fate of transplanted cells.].
Text Component FeaturesThe JTextComponent class is the foundation .pdf
Text Component Features
The JTextComponent class is the foundation for Swing text components. This class provides the
following customizable features for all of its descendants:
A model, known as a document, that manages the component\'s content.
A view, which displays the component on screen.
A controller, known as an editor kit, that reads and writes text and implements editing
capabilities with actions.
upport for infinite undo and redo.
A pluggable caret and support for caret change listeners and navigation filters.
A layout manager encapsulates an algorithm for positioning and sizing of GUI components.
Rather than building the layout algorithms into your code and watching for window resizing and
other layout-modifying events, the algorithm is kept separate. This allows a layout algorithm to
be reused for several applications, while simplifying your application code.
LayoutManager is an interface in the Java class libraries that describes how a Container and a
layout manager communicate. It describes several methods which:
Component Size Methods
The Component class defines several size accessor methods to assist the layout process. Each of
these methods returns a Dimension object describing the requested size. These methods are as
follows:
An additional interface, LayoutManager2, was added in JDK 1.1, which adds a few more
positioning and validation methods.
Layout managers will use these sizing methods when they are figuring out where to place
components and what size they should be. Layout managers can respect or ignore as much or as
little of this information as they see fit. Each layout manager has its own algorithm and may or
may not use this information when deciding component placement. Which of these methods is
respected or ignored is very important information and should be documented carefully when
creating your own layout manager.
Controlling the sizes returned for your own components can be accomplished in two ways,
depending on whether you are using the JFC Project Swing components.
If your component is a JFC Project Swing component, you inherit three methods,
setPreferredSize(Dimension), setMinimumSize(Dimension) and setMaximumSize(Dimension).
You can call these methods directly to explicitly set the size information for a component. For
example:
Layout Managers and Containers
A layout manager must be associated with a Container object to perform its work. If a container
does not have an associated layout manager, the container simply places components wherever
specified by using the setBounds(), setLocation() and/or setSize() methods.
If a container has an associated layout manager, the container asks that layout manager to
position and size its components before they are painted. The layout manager itself does
notperform the painting; it simply decides what size and position each component should occupy
and calls setBounds(), setLocation() and/or setSize() on each of those components.
A LayoutManager is associated with a C.
operating system linux,ubuntu,Mac Geometri.pdf
//operating system linux,ubuntu,Mac
/*********************GeometricObject.java**********************/
public abstract class GeometricObject {
private String color = \"white\";
private boolean filled;
// default constructure
public GeometricObject() {
super();
// TODO Auto-generated constructor stub
}
// construct a Geometric Object
// parameterized constructure
public GeometricObject(String color, boolean filled) {
this.color = color;
this.filled = filled;
}
/** Getter method for color */
public String getColor() {
return color;
}
/** Setter method for color */
public void setColor(String color) {
this.color = color;
}
/**
* Getter method for filled. Since filled is boolean, so the gret method
* name is isFilled
*/
public boolean isFilled() {
return filled;
}
/** Setter method for filled */
public void setFilled(boolean filled) {
this.filled = filled;
}
/** Abstract method for FindArea */
public abstract double findArea();
/** Abstract method for findPerimeter */
public abstract double findPerimeter();
}
/*****************************Octagon.java**************/
public class Octagon extends GeometricObject implements Cloneable, Comparable {
private double side;
/** construct a Octagon with specified side */
public Octagon(double side) {
super();
this.side = side;
}
/** Implement the abstract method findArea in GeometricObject */
@Override
public double findArea() {
double area = (2 + 4 / Math.sqrt(2)) * side * side;
return area;
}
/** Implement the abstract method findArea in findPerimeter */
@Override
public double findPerimeter() {
double perimeter = 8 * side;
return perimeter;
}
/** Implement the compareTo method in Comparable interface */
@Override
public int compareTo(Object o) {
// TODO Auto-generated method stub
return 0;
}
public Object clone() throws CloneNotSupportedException {
return super.clone();
}
}
/*******************App.java*******************/
public class App {
public static void main(String[] args) {
Octagon a1 = new Octagon(5);//creating object of octagon
System.out.println(\"Area of a1 is \" + a1.findArea());
System.out.println(\"Perimeter of a1 is \" + a1.findPerimeter());
Object a2;
try {
a2 = a1.clone();
String result = (a1.compareTo(a2) == 0) ? \"a1 and its clone a2 have the same area\"
: \"a1 and its clone a2 have different areas\";
System.out.println(\"Compare a1 and its clone a2:\ \\t \" + result);
System.out.println(\"Hashcode of a1: \" + a1.hashCode());
System.out.println(\"Hashcode of a2: \" + a2.hashCode());
System.out.println(\"Displaying a1: \" + a1);
System.out.println(\"Displaying a2: \" + a2);
} catch (CloneNotSupportedException e) {
e.printStackTrace();
}
}
}
/*****************output***************/
gopal@gopal:~/Desktop/chegg$ javac GeometricObject.java
gopal@gopal:~/Desktop/chegg$ javac Octagon.java
gopal@gopal:~/Desktop/chegg$ javac App.java
gopal@gopal:~/Desktop/chegg$ java App
Area of a1 is 120.71067811865476
Perimeter of a1 is 40.0
Compare a1 and its clone a2:
a1 and its clone a2 have the same area
.
NO TILL FARMINGNo-till farming (also called zero tillage or direct.pdf
NO TILL FARMING
No-till farming (also called zero tillage or direct drilling) is a way of growing crops or pasture
from year to year without disturbing the soil through tillage. No-till is an agricultural technique
which increases the amount of water that infiltrates into the soil and increases organic matter
retention and cycling of nutrients in the soil. In many agricultural regions it can reduce or
eliminate soil erosion. It increases the amount and variety of life in and on the soil, including
disease-causing organisms and disease suppression organisms. The most powerful benefit of no-
tillage is improvement in soil biological fertility, making soils more resilient. Farm operations
are made much more efficient, particularly improved time of sowing and better trafficability of
farm operations.
TILLAGE SYSTEM
There are a variety of tillage systems available for crop production. While tillage operations are
performed for various reasons, producers must evaluate the need for each and every field
operation conducted in order to improve profitability. In addition, the effects of the tillage
operations on the soil system and the environment must be considered. More information is
available on the following tillage systems:
NITROGEN LEACHING
Nitrogen is a common element in nature and an essential plant nutrient. Approximately
78% of Earth\'s atmosphere is nitrogen (N2). The strong bond between the atoms of N2 makes
this gas quite inert and not directly usable by plants and animals. As nitrogen naturally cycles
through the air, water and soil it undergoes various chemical and biological transformations.
Nitrogen promotes plant growth. Livestock then eat the crops producing manure, which is
returned to the soil, adding organic and mineral forms of nitrogen. The cycle is complete when
the next crop uses the amended soil. To increase food production, fertilizers, such as nitrate
(NO3) and ammonium (NH4), which are easily absorbed by plants, are introduced to the plant
root zone. However, soils do not absorb the excess NO3 ions, which then move downward freely
with drainage water, and are leached into groundwater, streams and oceans. The degree of
leaching is affected by:
The level of nitrous oxide (N2O) in the Earth\'s atmosphere is increasing at a rate of 0.2 to 0.3%
annually. Anthropogenic sources of nitrogen are 50% greater than from natural sources, such as,
soils and oceans. Leached agricultural inputs, i.e. fertilizers and manures, accounts for 75% of
the anthropogenic source of nitrogen. The Food and Agriculture Organization of the United
Nations (FAO) estimates world demand for nitrogen fertilizers will increase by 1.7% annually
between 2011 and 2015. An increase of 7.5 million tonnes. Regional increases of nitrogen
fertilizer use are expected to be 67% by Asia, 18% by the Americas, 10% by Europe, 3% by
Africa,and 1% by Oceania.
Solution
NO TILL FARMING
No-till farming (also called zero tillage or direct drilling) is a way of growing crops or.
Lions use sounds with different intensities, senses such as tough an.pdf
Lions use sounds with different intensities, senses such as tough and behaviours for
communication. Pheromones are the chemicals released by an animal to communicate with its
mate and this also occurs in lions. Jacobson’s organ is an olfactory organ located at the roof of
the lion’s mouth. This organ perceive the smell and send impulses to the olfactory center in the
brain.
Solution
Lions use sounds with different intensities, senses such as tough and behaviours for
communication. Pheromones are the chemicals released by an animal to communicate with its
mate and this also occurs in lions. Jacobson’s organ is an olfactory organ located at the roof of
the lion’s mouth. This organ perceive the smell and send impulses to the olfactory center in the
brain..
let R be ringI be a subgroup under additionif I be idealfor a,.pdf
let R be ring
I be a subgroup under addition
if I be ideal
for a,b in R
(a+I)+(b+I)=a+I+b+I
=a+(I+b)+I
=a+b+I+I
=a+b+I
and
(a+I)(b+I)=ab+aI+Ib+I
=ab+I+I+I since aI left ideal and Ib is right ideal
=ab+I
if (a+I)+(b+I)= a+b+I
and (a+I)(b+I)=ab+I
R is ring and I be subgroup
a,b in R
imples a,b in I
and a,b in I
then a+b in I
thus I is closed under addition
and a, 0 be in I
(a+I)+(0+I)=a+0+I
=a+I
thus 0 is zero element of R
-a be additive inverse
a,-a be in R
(a+I)+(-a+I)=a-a+I=0+I
thus I is closed under additive inverse
(a+I)(b+I)= (ab+I)
1 be multiplicative inverse
1+I be ideal
(a+I)(1+I)=a*1+I
=a+I
thus 1+I be multiplicate inverse
thus I be ideal
Solution
let R be ring
I be a subgroup under addition
if I be ideal
for a,b in R
(a+I)+(b+I)=a+I+b+I
=a+(I+b)+I
=a+b+I+I
=a+b+I
and
(a+I)(b+I)=ab+aI+Ib+I
=ab+I+I+I since aI left ideal and Ib is right ideal
=ab+I
if (a+I)+(b+I)= a+b+I
and (a+I)(b+I)=ab+I
R is ring and I be subgroup
a,b in R
imples a,b in I
and a,b in I
then a+b in I
thus I is closed under addition
and a, 0 be in I
(a+I)+(0+I)=a+0+I
=a+I
thus 0 is zero element of R
-a be additive inverse
a,-a be in R
(a+I)+(-a+I)=a-a+I=0+I
thus I is closed under additive inverse
(a+I)(b+I)= (ab+I)
1 be multiplicative inverse
1+I be ideal
(a+I)(1+I)=a*1+I
=a+I
thus 1+I be multiplicate inverse
thus I be ideal.
If we use evaporation or recrystallization, we would only remove the.pdf
If we use evaporation or recrystallization, we would only remove the water, leaving behind the
barium sulfate and the sodium chloride. We want to separate them , not get a mixture of barium
sulfate and sodium chloride. Also, any unreacted barium nitrate and sodium sulfate will become
solids and contaminate the barium sulphate solid that we will obtain.
We use filtration, because this method allows the soluble salts to pass through the filter, leaving
behind the insoluble part, the barium sulfate. Even unreacted barium nitrate and sodium sulfate is
allowed to pass through the filter.
BaSO4 is a solid precipitate and NaCl in dissolved in the water. If you evaporated the water, you
would be left with a mixture of NaCl and BaSO4.
To recrystallize BaSO4, you would first have to isolate it (by filtration), then dissolve it and
allow it to recrystallize. Why not just filter was wash out any adhering NaCl (or unreacted
reactant).
Solution
If we use evaporation or recrystallization, we would only remove the water, leaving behind the
barium sulfate and the sodium chloride. We want to separate them , not get a mixture of barium
sulfate and sodium chloride. Also, any unreacted barium nitrate and sodium sulfate will become
solids and contaminate the barium sulphate solid that we will obtain.
We use filtration, because this method allows the soluble salts to pass through the filter, leaving
behind the insoluble part, the barium sulfate. Even unreacted barium nitrate and sodium sulfate is
allowed to pass through the filter.
BaSO4 is a solid precipitate and NaCl in dissolved in the water. If you evaporated the water, you
would be left with a mixture of NaCl and BaSO4.
To recrystallize BaSO4, you would first have to isolate it (by filtration), then dissolve it and
allow it to recrystallize. Why not just filter was wash out any adhering NaCl (or unreacted
reactant)..
import java.util.Random; public class randomCalls { .pdf
import java.util.Random;
public class randomCalls {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Random rn = new Random();
int n;
for(int i=1;i<=720;i++)
{
n = rn.nextInt(6) +5;
System.out.print(n);
System.out.print(\' \');
System.out.println(i);
}
}
}
Solution
import java.util.Random;
public class randomCalls {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Random rn = new Random();
int n;
for(int i=1;i<=720;i++)
{
n = rn.nextInt(6) +5;
System.out.print(n);
System.out.print(\' \');
System.out.println(i);
}
}
}.
I am considering the case where you have multiple name in same file .pdf
I am considering the case where you have multiple name in same file and you want to replace all
of them.
Step1: open terminal in linux operating system
Step2: move to desired file location using below command
Cd /home/beth/project/
Step3: Open desired file in current location
vim 1stqtr and press [enter]
you can see your file content on terminal
step 4: Now go to command mode in VM using below command
press [esc key]
step 5: Now you have to search your old name and replace with new name. For this use below
command structure.
:s/Search/Replace/CommandFlag
Explanation: s -> it is used to command Vi editor that I am going to use search and replace
command
Search-> It is a word which you want to replace
Replace-> It is a word from which you want to replace
Command Flag-> It is command flag which provides extra features like case
sensitive etc.
Say your old name is Ain
:%s/Smith/Oliver/g
To find and replace all occurrences of case insensitive ‘Smith with ‘Oliver, enter:
:%s/Smith/Oliver/gi
To find and replace all occurrences of ‘eth1’ with ‘br1’, but ask for confirmation first, enter:
:%s/Smith/Oliver/gc
Step 6: In the last enter below command to save and exit.
:wq[enter]
Solution
I am considering the case where you have multiple name in same file and you want to replace all
of them.
Step1: open terminal in linux operating system
Step2: move to desired file location using below command
Cd /home/beth/project/
Step3: Open desired file in current location
vim 1stqtr and press [enter]
you can see your file content on terminal
step 4: Now go to command mode in VM using below command
press [esc key]
step 5: Now you have to search your old name and replace with new name. For this use below
command structure.
:s/Search/Replace/CommandFlag
Explanation: s -> it is used to command Vi editor that I am going to use search and replace
command
Search-> It is a word which you want to replace
Replace-> It is a word from which you want to replace
Command Flag-> It is command flag which provides extra features like case
sensitive etc.
Say your old name is Ain
:%s/Smith/Oliver/g
To find and replace all occurrences of case insensitive ‘Smith with ‘Oliver, enter:
:%s/Smith/Oliver/gi
To find and replace all occurrences of ‘eth1’ with ‘br1’, but ask for confirmation first, enter:
:%s/Smith/Oliver/gc
Step 6: In the last enter below command to save and exit.
:wq[enter].
Hormone A chemical substance produced in the body by ductless gland.pdf
Hormone: A chemical substance produced in the body by ductless glands called endocrine
glands they control and regulate the activity of certain cells or organs of our body. The pathway
is:
· Neurosecretory neurons are stimulated (1)
· Releasing hormones are secreted from the hypothalamus (2)
· Endocrine cell secretes hormone (3)
· Hormone travels through the bloodstream (4)
· Cells utilize receptors to provide specificity of the hormone (5)
Solution
Hormone: A chemical substance produced in the body by ductless glands called endocrine
glands they control and regulate the activity of certain cells or organs of our body. The pathway
is:
· Neurosecretory neurons are stimulated (1)
· Releasing hormones are secreted from the hypothalamus (2)
· Endocrine cell secretes hormone (3)
· Hormone travels through the bloodstream (4)
· Cells utilize receptors to provide specificity of the hormone (5).
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