The pathway would be overactive.The GTPase activity of the alpha s.pdf
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The pathway would be overactive. The GTPase activity of the alpha subunit will be disrupted which leads to continuous input of the signal and constant generation of second messenger. Solution The pathway would be overactive. The GTPase activity of the alpha subunit will be disrupted which leads to continuous input of the signal and constant generation of second messenger..
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a) It is given E is the event of getting 2. We can observe that 2 is.pdf
a) It is given E is the event of getting 2. We can observe that 2 is an even number.
It is given F is the event of getting an even number.
Event E is giving an even number and event F is also giving an even number.That means both
are occuring at the same time. so they are not mutually exclusive.
b)The events E and F to be complementary their probabilities should add up to 1.
P(E) = 1/6
P(F) = 3/6
P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3
2/3 is not equal to one.
so events E and F are not complementary.
Solution
a) It is given E is the event of getting 2. We can observe that 2 is an even number.
It is given F is the event of getting an even number.
Event E is giving an even number and event F is also giving an even number.That means both
are occuring at the same time. so they are not mutually exclusive.
b)The events E and F to be complementary their probabilities should add up to 1.
P(E) = 1/6
P(F) = 3/6
P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3
2/3 is not equal to one.
so events E and F are not complementary..
4.1 (a) SoftwoodsThese generally come from the gymnosperm trees wh.pdf
4.1 (a) Softwoods
These generally come from the gymnosperm trees which have needles and cones or simply
called the coniferous trees. These trees have needles in place of leaves and remain green
throughout the year. All the softwoods are not softer than the hard woods, there are certain
hardwoods that are softer. They grow primarly in pacific northwest and south of united states.
Due to their faster growth, softwoods are cheaper than the hardwoods. However they are less fire
resistant compared to the hardwoods.
Examples of softwoods: Pines, redwood, fir, juniper, cedar and hemlock.
Uses:
Although they are classified as softwoods, but they are sufficiently dense and hard to be used for
various construction purposes. They are generally used for making the framework of houses.
About 80% of the timber comes from the softwoods. Various building components such as
windows, doors, and furniture, medium-density fiberboard (MDF), paper, Christmas trees, and
many more things use softwoods.
4.1 (b) Hardwoods
Hardwood is extracted from te deciduous trees or angiosperm trees which are not monocots.
These trees have broad leaves which they shed in the autumn or winter. Most hardwoods have
higher density than the softwoods. They grow slower than the softwoods, so they are expensive.
examples of hardwoods: oak, maple, mahogany, cherry, walnut, rosewood, teak
Uses: mostly used to manufacture the high quality furniture,decks, flooring and construction that
needs to last for longer periods.
4.1 (c)
Most of the structural timber (80%) comes from the softwoods, but the hardwoods are used for
the high quality furniture, decks, flooring and constructions which needs to last for longer
periods.
Solution
4.1 (a) Softwoods
These generally come from the gymnosperm trees which have needles and cones or simply
called the coniferous trees. These trees have needles in place of leaves and remain green
throughout the year. All the softwoods are not softer than the hard woods, there are certain
hardwoods that are softer. They grow primarly in pacific northwest and south of united states.
Due to their faster growth, softwoods are cheaper than the hardwoods. However they are less fire
resistant compared to the hardwoods.
Examples of softwoods: Pines, redwood, fir, juniper, cedar and hemlock.
Uses:
Although they are classified as softwoods, but they are sufficiently dense and hard to be used for
various construction purposes. They are generally used for making the framework of houses.
About 80% of the timber comes from the softwoods. Various building components such as
windows, doors, and furniture, medium-density fiberboard (MDF), paper, Christmas trees, and
many more things use softwoods.
4.1 (b) Hardwoods
Hardwood is extracted from te deciduous trees or angiosperm trees which are not monocots.
These trees have broad leaves which they shed in the autumn or winter. Most hardwoods have
higher density than the softwoods. They grow slower than the softwoods, so they are expensive.
.
A special type of IP address is the limited broadcast address 255.25.pdf
A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast
involves delivering a message from one sender to many recipients. Senders direct an IP
broadcast to 255.255.255.255 to indicate all other nodes on the local network (LAN) should pick
up that message. This broadcast is \'limited in that it does not reach every node on the Internet,
only nodes on the LAN
Solution
A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast
involves delivering a message from one sender to many recipients. Senders direct an IP
broadcast to 255.255.255.255 to indicate all other nodes on the local network (LAN) should pick
up that message. This broadcast is \'limited in that it does not reach every node on the Internet,
only nodes on the LAN.
10 leading causes of death in the United States In the recent sta.pdf
10 leading causes of death in the United States :
In the recent statistics the causes of death in the US is heart disease and cancer.
1. HEART DISEASE (614,348) : Heart disease is the leading causes of death for both men and
women in US.Major symptoms are chest pain,breathlessness,nausea,cold sweats etc.Causes of
heart disease are high cholesterol, high blood pressure,obesity,smoking,lack of physical
activity,sleep apnea,stress.
2. CANCER (591,699 ) : Cancer effects people of all gender,ages.Estimated cancer related
deaths (male) in 2015 are lung and bronchus, prostate,colon and rectum,pancreas,liver,leukemia
etc and for female lung and bronchus,breast,colon and rectum,pancreas,ovary,leukemia,uterine
corpous.
3. CHRONIC LOWER RESPIRATORY DISEASE (147,101) :Causes of chronic lower
repiratory disease is mainly smoking,air pollution and dust,genetics.It also includes the deaths
from chronic bronchitis,emphysema and asthma.
4. ACCIDENTS (136,053) : Motor vehicle accidents,accidental falls,accidental
drowning,accidental exposure to fire etc.The raod vehicle accidents can be avoided by not
drinking and driving.
5. STROKE (133,103) : Stroke is not considerd by heart disease.Cerebrocvascular disease is
developes with the complication of blood vessels.The type of cerebrovascular disease are
Stroke,Transient ischemic attack,Vascular dementia.Symptoms are numbness on face and
arm,confusion,trouble in speaking,vision problem,headache.Stroke can be prevented by eating
healthy diet,maintaining weight,physical activity and exercise,avoid smoking and
alcohol,maintain cholesterol and blood pressure level.
6. ALZHEIMER\'S DISEASE ( 93,514) : This disease develops by the progressive mental
deteriorataion in the old age.5.3 millions Americans may have alzheimer\'s disease in
2015.Symptoms are memory loss,difficulty in family tasks and solving problems,confusion with
time and place,changes mood and personality.
7. DIABETES/DIABETES MELLITUS (76,488) : Diabetes caused by the deficiency of insulin
and failure to metabolize sugar and starch in the body,diabetes can cause serious helath
complications like heart disease,kidney failure,vision problem etc.There are two types of
diabetes a) type 1: Insulin-depedent diabetes mellitus and b) type 2 : Non-insulin-dependent
diabetes mellitus. Symptoms are frequent urination,excessive thirst,extreme hunger,vision
change,feeling tired,dry skin.
8. INFLUENZA AND PNEUMONIA (55,227) : Influenza is highly contagious viral infection
and most severe in winter season.The cause of Pneumonia are virus,chemicals,mycoplasmas and
infectious fungi.Symptoms are fever,headache,cough,chills,sore throat,nasal
congestion,wheezing.
9. KIDNEY DISEASE (48,146) : Nephritis,nephrotic syndrome,nephrosis are the conditions of
kidney diseases.Symptoms are loss of appetite,fatigue,headache,nausea etc.
10. SUICIDE AND SELF-HARM (42,773) : Causes of suicude are mental depression and
stress,substance abuse,family violence including sexual abuse,prior suici.
C Program to shuffle a given Aay#include stdio.h #include .pdf
// C Program to shuffle a given Aay
#include
#include
#include
// A utility function to print an Aay
void printArray (int A[], int n)
{
int i;
for (i = 0; i < n; i++)
printf(\"%d \", A[i]);
printf(\"\ \");
}
int* shuffle(int A[], int B[])
{
int n1 = 5;
int n2 = 2;
int i = 0, j = 0, k = 0;
int *C;
C = (int*)malloc(7 * sizeof(int));
// Merging starts
while (i < n1 && j < n2)
{
if (k%2 == 0)
{
C[k] = A[i];
i++;
k++;
}
else
{
C[k] = B[j];
k++;
j++;
}
}
/* elements in array \'A\' are still remaining
where as the array \'B\' is exhausted */
while (i < n1)
{
C[k] = A[i];
i++;
k++;
}
/* elements in array \'B\' are still remaining
where as the array \'A\' is exhausted */
while (j < n2)
{
C[k] = B[j];
k++;
j++;
}
return C;
}
int main()
{
int A[] = {1, 2, 3, 4, 5};
int B[] = {10, 20};
int n1 = 5;
int n2 = 2;
int nC = 7;
int *C;
C = (int*)malloc(7 * sizeof(int));
printf(\"Array A: \");
printArray(A, n1);
printf(\"Array B: \");
printArray(B, n2);
C = shuffle (A, B);
printf(\"\ Shuffled: \");
printArray(C,nC);
return 0;
}
/*
output:
Array A: 1 2 3 4 5
Array B: 10 20
Shuffled: 1 10 2 20 3 4 5
*/
Solution
// C Program to shuffle a given Aay
#include
#include
#include
// A utility function to print an Aay
void printArray (int A[], int n)
{
int i;
for (i = 0; i < n; i++)
printf(\"%d \", A[i]);
printf(\"\ \");
}
int* shuffle(int A[], int B[])
{
int n1 = 5;
int n2 = 2;
int i = 0, j = 0, k = 0;
int *C;
C = (int*)malloc(7 * sizeof(int));
// Merging starts
while (i < n1 && j < n2)
{
if (k%2 == 0)
{
C[k] = A[i];
i++;
k++;
}
else
{
C[k] = B[j];
k++;
j++;
}
}
/* elements in array \'A\' are still remaining
where as the array \'B\' is exhausted */
while (i < n1)
{
C[k] = A[i];
i++;
k++;
}
/* elements in array \'B\' are still remaining
where as the array \'A\' is exhausted */
while (j < n2)
{
C[k] = B[j];
k++;
j++;
}
return C;
}
int main()
{
int A[] = {1, 2, 3, 4, 5};
int B[] = {10, 20};
int n1 = 5;
int n2 = 2;
int nC = 7;
int *C;
C = (int*)malloc(7 * sizeof(int));
printf(\"Array A: \");
printArray(A, n1);
printf(\"Array B: \");
printArray(B, n2);
C = shuffle (A, B);
printf(\"\ Shuffled: \");
printArray(C,nC);
return 0;
}
/*
output:
Array A: 1 2 3 4 5
Array B: 10 20
Shuffled: 1 10 2 20 3 4 5
*/.
1.Router 1.It is a machine which aheads datagrams over the networks.pdf
1.Router: 1.It is a machine which aheads datagrams over the networks.
2.This router is present at a node which connects two or more networks.
3.It avails of headers and tables to establish the path.
4.We can filter the data through this router in very small.
5.in order to communicate one another this router uses a protocol.
2.Hubs: 1.a hub is a frequent connection point .
2.It contains no of ports.
3.The data type in hub is frames.
4.hubs are present as an association for the networks.
5.It is some what slow.
6.we can use it as a stand alone device.
3.Servers: 1.A server is a device that manages all the network( printers ,hard disks...) devices.
2.generally this servers sre many types as follows
1.Database server:A system which operate through queries.
2.Print server :A system which directs one or more printer.
3.File server and network server which directs the network traffic.
4.It offers many functionality frequently called \"servers.\"
5.Selecting the computer as \"server class hardware\" leads to more reliable and powerful than a
simple computer.
4.Network cabling: 1.It is used to join the one network device to another network device.
2.It the agency to travel from one netork to another network
3.Depending on their sizeand topology the cables are different as folllows.
1 .co-axial cable
2.optical fibre cable
3.twisted pair cacle
4.For acheving the good network we need to know the charecteristics of a cable at first.
5.Sound masking upgrades the isolation.
Solution
1.Router: 1.It is a machine which aheads datagrams over the networks.
2.This router is present at a node which connects two or more networks.
3.It avails of headers and tables to establish the path.
4.We can filter the data through this router in very small.
5.in order to communicate one another this router uses a protocol.
2.Hubs: 1.a hub is a frequent connection point .
2.It contains no of ports.
3.The data type in hub is frames.
4.hubs are present as an association for the networks.
5.It is some what slow.
6.we can use it as a stand alone device.
3.Servers: 1.A server is a device that manages all the network( printers ,hard disks...) devices.
2.generally this servers sre many types as follows
1.Database server:A system which operate through queries.
2.Print server :A system which directs one or more printer.
3.File server and network server which directs the network traffic.
4.It offers many functionality frequently called \"servers.\"
5.Selecting the computer as \"server class hardware\" leads to more reliable and powerful than a
simple computer.
4.Network cabling: 1.It is used to join the one network device to another network device.
2.It the agency to travel from one netork to another network
3.Depending on their sizeand topology the cables are different as folllows.
1 .co-axial cable
2.optical fibre cable
3.twisted pair cacle
4.For acheving the good network we need to know the charecteristics of a cable at first.
5.Sound masking upgrades the isolation..
1) List currently running jobsANS) see currently runningcommand.pdf
1) List currently running jobs
ANS) see currently running
command: ps
examples:
$ps
2.Kill one of the running jobs
ANS) a.kill any process by PID
command:
kill {PID}
example:
$kill 1012
3.Run the ls command with –help option
ANS) ls command:
ls command is used to list the contents of a directory
Syntax:
ls [-a] [-A] [-b] [-c] [-C] [-d] [-f] [-F] [-g] [-i] [-l] [-L] [-m] [-o] [-p][-q] [-r] [-R] [-s] [-t] [-u] [-x]
[pathnames]
Parameters OR Switches
• -a Shows All The Files, Including Hidden File Which Begin With ADot.
• -A Shows All The Files Including Hidden Files. Does Not Displays The
Working Directory (.) OR The Parent Directory (..)
• -b Force Printing of Non-Printable Characters To Be in Octal \\ddd
Notation.
• -c Use Time of Last Modification of The i-Node For Sorting (-t) OR For
Printing (-l OR -n)
• -C Multi-Column Output With Entries Sorted Down The Columns.
Which is Generally The Default Option.
• -d Lists Only The Directory Names But Not its Contents.
• -f Force Each Argument To Be Interpreted As A Directory And List
The Name Found in Each Slot. -f Option Turns off -l, -t, -s, And -r, And
Turns on –a, The List Order is The Order in Which Entries Appear in The
Directory.
• -F Mark Directories With A Trailing Slash (/), DOORS With A Trailing
Greater-Than Sign (>), Executable Files With A Trailing Asterisk (*),
FIFOs With A Trailing Vertical Bar (|), Symbolic Links With A Trailing
Equals Sign (=).
• -g Same As -l Parameter, Except The Owner is Not Printed.
• -i For Each File, Prints The i-Node Number in The First Column.
• -l Shows The User With Huge Amounts of Information With
Permissions, Owners, Size, And When Last Modified.)
• -L If An Argument is A Symbolic Link, List The File OR Directory To
Which The Link References Rather Than The Link Itself.
• -m Stream Output Format, In This Files Are Listed Across The Page,
Separated By Commas.
• -n Same As –l Parameter, Except That The Owner\'s UID And Group\'s
GID Numbers Are Printed, Rather Than The Associated Character
Strings.
-o The Same As -l, Except That The Group is Not Printed.
-p Displays A Forward Slash (/) in Front of All Directories.
-q Force Printing of Non-Printable Characters in File Names As The
Character Question Mark (?).
-r Reverses The Order of How The Files Are Displayed.
-R Includes The Contents of Subdirectories.
-s Give Size in Blocks, Including Indirect Blocks, For Each Entry.
-t Shows The End User The Files Modification Time.
-u Use Time of Last Access Instead of Last Modification For Sorting
Use With The -t Option OR For Printing Use With The -l Option.
-x Displays Files in Columns.
-1 Prints One Entry Per Line of Output.
pathnames File OR Directory To List.
4.Run the help command on echo
ANS) echo command display to the Screen.
“echo” is Useful For
• Producing Diagnostics in Command Files
• For Sending Known Data into A Pipe
• For Displaying The Contents of Environment Variables
syntax:
echo [-n] text
5.Cat an existing file using a command that misspells one c.
Step1 Total pressure = partial pressure of Hydrog.pdf
Step1 Total pressure = partial pressure of Hydrogen + Vapour pressure of water at
the given temperature. Step2 Partial pressure of Hydrogen = 767 - Vapour pressure of water at
295.3 K Step3 = 767.0- 20.2 = 746.8 torr
Solution
Step1 Total pressure = partial pressure of Hydrogen + Vapour pressure of water at
the given temperature. Step2 Partial pressure of Hydrogen = 767 - Vapour pressure of water at
295.3 K Step3 = 767.0- 20.2 = 746.8 torr.
Recommended
a) It is given E is the event of getting 2. We can observe that 2 is.pdf
a) It is given E is the event of getting 2. We can observe that 2 is an even number.
It is given F is the event of getting an even number.
Event E is giving an even number and event F is also giving an even number.That means both
are occuring at the same time. so they are not mutually exclusive.
b)The events E and F to be complementary their probabilities should add up to 1.
P(E) = 1/6
P(F) = 3/6
P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3
2/3 is not equal to one.
so events E and F are not complementary.
Solution
a) It is given E is the event of getting 2. We can observe that 2 is an even number.
It is given F is the event of getting an even number.
Event E is giving an even number and event F is also giving an even number.That means both
are occuring at the same time. so they are not mutually exclusive.
b)The events E and F to be complementary their probabilities should add up to 1.
P(E) = 1/6
P(F) = 3/6
P(E)+P(F) = 1/6 + 3/6 = 4/6 = 2/3
2/3 is not equal to one.
so events E and F are not complementary..
4.1 (a) SoftwoodsThese generally come from the gymnosperm trees wh.pdf
4.1 (a) Softwoods
These generally come from the gymnosperm trees which have needles and cones or simply
called the coniferous trees. These trees have needles in place of leaves and remain green
throughout the year. All the softwoods are not softer than the hard woods, there are certain
hardwoods that are softer. They grow primarly in pacific northwest and south of united states.
Due to their faster growth, softwoods are cheaper than the hardwoods. However they are less fire
resistant compared to the hardwoods.
Examples of softwoods: Pines, redwood, fir, juniper, cedar and hemlock.
Uses:
Although they are classified as softwoods, but they are sufficiently dense and hard to be used for
various construction purposes. They are generally used for making the framework of houses.
About 80% of the timber comes from the softwoods. Various building components such as
windows, doors, and furniture, medium-density fiberboard (MDF), paper, Christmas trees, and
many more things use softwoods.
4.1 (b) Hardwoods
Hardwood is extracted from te deciduous trees or angiosperm trees which are not monocots.
These trees have broad leaves which they shed in the autumn or winter. Most hardwoods have
higher density than the softwoods. They grow slower than the softwoods, so they are expensive.
examples of hardwoods: oak, maple, mahogany, cherry, walnut, rosewood, teak
Uses: mostly used to manufacture the high quality furniture,decks, flooring and construction that
needs to last for longer periods.
4.1 (c)
Most of the structural timber (80%) comes from the softwoods, but the hardwoods are used for
the high quality furniture, decks, flooring and constructions which needs to last for longer
periods.
Solution
4.1 (a) Softwoods
These generally come from the gymnosperm trees which have needles and cones or simply
called the coniferous trees. These trees have needles in place of leaves and remain green
throughout the year. All the softwoods are not softer than the hard woods, there are certain
hardwoods that are softer. They grow primarly in pacific northwest and south of united states.
Due to their faster growth, softwoods are cheaper than the hardwoods. However they are less fire
resistant compared to the hardwoods.
Examples of softwoods: Pines, redwood, fir, juniper, cedar and hemlock.
Uses:
Although they are classified as softwoods, but they are sufficiently dense and hard to be used for
various construction purposes. They are generally used for making the framework of houses.
About 80% of the timber comes from the softwoods. Various building components such as
windows, doors, and furniture, medium-density fiberboard (MDF), paper, Christmas trees, and
many more things use softwoods.
4.1 (b) Hardwoods
Hardwood is extracted from te deciduous trees or angiosperm trees which are not monocots.
These trees have broad leaves which they shed in the autumn or winter. Most hardwoods have
higher density than the softwoods. They grow slower than the softwoods, so they are expensive.
.
A special type of IP address is the limited broadcast address 255.25.pdf
A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast
involves delivering a message from one sender to many recipients. Senders direct an IP
broadcast to 255.255.255.255 to indicate all other nodes on the local network (LAN) should pick
up that message. This broadcast is \'limited in that it does not reach every node on the Internet,
only nodes on the LAN
Solution
A special type of IP address is the limited broadcast address 255.255.255.255. A broadcast
involves delivering a message from one sender to many recipients. Senders direct an IP
broadcast to 255.255.255.255 to indicate all other nodes on the local network (LAN) should pick
up that message. This broadcast is \'limited in that it does not reach every node on the Internet,
only nodes on the LAN.
10 leading causes of death in the United States In the recent sta.pdf
10 leading causes of death in the United States :
In the recent statistics the causes of death in the US is heart disease and cancer.
1. HEART DISEASE (614,348) : Heart disease is the leading causes of death for both men and
women in US.Major symptoms are chest pain,breathlessness,nausea,cold sweats etc.Causes of
heart disease are high cholesterol, high blood pressure,obesity,smoking,lack of physical
activity,sleep apnea,stress.
2. CANCER (591,699 ) : Cancer effects people of all gender,ages.Estimated cancer related
deaths (male) in 2015 are lung and bronchus, prostate,colon and rectum,pancreas,liver,leukemia
etc and for female lung and bronchus,breast,colon and rectum,pancreas,ovary,leukemia,uterine
corpous.
3. CHRONIC LOWER RESPIRATORY DISEASE (147,101) :Causes of chronic lower
repiratory disease is mainly smoking,air pollution and dust,genetics.It also includes the deaths
from chronic bronchitis,emphysema and asthma.
4. ACCIDENTS (136,053) : Motor vehicle accidents,accidental falls,accidental
drowning,accidental exposure to fire etc.The raod vehicle accidents can be avoided by not
drinking and driving.
5. STROKE (133,103) : Stroke is not considerd by heart disease.Cerebrocvascular disease is
developes with the complication of blood vessels.The type of cerebrovascular disease are
Stroke,Transient ischemic attack,Vascular dementia.Symptoms are numbness on face and
arm,confusion,trouble in speaking,vision problem,headache.Stroke can be prevented by eating
healthy diet,maintaining weight,physical activity and exercise,avoid smoking and
alcohol,maintain cholesterol and blood pressure level.
6. ALZHEIMER\'S DISEASE ( 93,514) : This disease develops by the progressive mental
deteriorataion in the old age.5.3 millions Americans may have alzheimer\'s disease in
2015.Symptoms are memory loss,difficulty in family tasks and solving problems,confusion with
time and place,changes mood and personality.
7. DIABETES/DIABETES MELLITUS (76,488) : Diabetes caused by the deficiency of insulin
and failure to metabolize sugar and starch in the body,diabetes can cause serious helath
complications like heart disease,kidney failure,vision problem etc.There are two types of
diabetes a) type 1: Insulin-depedent diabetes mellitus and b) type 2 : Non-insulin-dependent
diabetes mellitus. Symptoms are frequent urination,excessive thirst,extreme hunger,vision
change,feeling tired,dry skin.
8. INFLUENZA AND PNEUMONIA (55,227) : Influenza is highly contagious viral infection
and most severe in winter season.The cause of Pneumonia are virus,chemicals,mycoplasmas and
infectious fungi.Symptoms are fever,headache,cough,chills,sore throat,nasal
congestion,wheezing.
9. KIDNEY DISEASE (48,146) : Nephritis,nephrotic syndrome,nephrosis are the conditions of
kidney diseases.Symptoms are loss of appetite,fatigue,headache,nausea etc.
10. SUICIDE AND SELF-HARM (42,773) : Causes of suicude are mental depression and
stress,substance abuse,family violence including sexual abuse,prior suici.
C Program to shuffle a given Aay#include stdio.h #include .pdf
// C Program to shuffle a given Aay
#include
#include
#include
// A utility function to print an Aay
void printArray (int A[], int n)
{
int i;
for (i = 0; i < n; i++)
printf(\"%d \", A[i]);
printf(\"\ \");
}
int* shuffle(int A[], int B[])
{
int n1 = 5;
int n2 = 2;
int i = 0, j = 0, k = 0;
int *C;
C = (int*)malloc(7 * sizeof(int));
// Merging starts
while (i < n1 && j < n2)
{
if (k%2 == 0)
{
C[k] = A[i];
i++;
k++;
}
else
{
C[k] = B[j];
k++;
j++;
}
}
/* elements in array \'A\' are still remaining
where as the array \'B\' is exhausted */
while (i < n1)
{
C[k] = A[i];
i++;
k++;
}
/* elements in array \'B\' are still remaining
where as the array \'A\' is exhausted */
while (j < n2)
{
C[k] = B[j];
k++;
j++;
}
return C;
}
int main()
{
int A[] = {1, 2, 3, 4, 5};
int B[] = {10, 20};
int n1 = 5;
int n2 = 2;
int nC = 7;
int *C;
C = (int*)malloc(7 * sizeof(int));
printf(\"Array A: \");
printArray(A, n1);
printf(\"Array B: \");
printArray(B, n2);
C = shuffle (A, B);
printf(\"\ Shuffled: \");
printArray(C,nC);
return 0;
}
/*
output:
Array A: 1 2 3 4 5
Array B: 10 20
Shuffled: 1 10 2 20 3 4 5
*/
Solution
// C Program to shuffle a given Aay
#include
#include
#include
// A utility function to print an Aay
void printArray (int A[], int n)
{
int i;
for (i = 0; i < n; i++)
printf(\"%d \", A[i]);
printf(\"\ \");
}
int* shuffle(int A[], int B[])
{
int n1 = 5;
int n2 = 2;
int i = 0, j = 0, k = 0;
int *C;
C = (int*)malloc(7 * sizeof(int));
// Merging starts
while (i < n1 && j < n2)
{
if (k%2 == 0)
{
C[k] = A[i];
i++;
k++;
}
else
{
C[k] = B[j];
k++;
j++;
}
}
/* elements in array \'A\' are still remaining
where as the array \'B\' is exhausted */
while (i < n1)
{
C[k] = A[i];
i++;
k++;
}
/* elements in array \'B\' are still remaining
where as the array \'A\' is exhausted */
while (j < n2)
{
C[k] = B[j];
k++;
j++;
}
return C;
}
int main()
{
int A[] = {1, 2, 3, 4, 5};
int B[] = {10, 20};
int n1 = 5;
int n2 = 2;
int nC = 7;
int *C;
C = (int*)malloc(7 * sizeof(int));
printf(\"Array A: \");
printArray(A, n1);
printf(\"Array B: \");
printArray(B, n2);
C = shuffle (A, B);
printf(\"\ Shuffled: \");
printArray(C,nC);
return 0;
}
/*
output:
Array A: 1 2 3 4 5
Array B: 10 20
Shuffled: 1 10 2 20 3 4 5
*/.
1.Router 1.It is a machine which aheads datagrams over the networks.pdf
1.Router: 1.It is a machine which aheads datagrams over the networks.
2.This router is present at a node which connects two or more networks.
3.It avails of headers and tables to establish the path.
4.We can filter the data through this router in very small.
5.in order to communicate one another this router uses a protocol.
2.Hubs: 1.a hub is a frequent connection point .
2.It contains no of ports.
3.The data type in hub is frames.
4.hubs are present as an association for the networks.
5.It is some what slow.
6.we can use it as a stand alone device.
3.Servers: 1.A server is a device that manages all the network( printers ,hard disks...) devices.
2.generally this servers sre many types as follows
1.Database server:A system which operate through queries.
2.Print server :A system which directs one or more printer.
3.File server and network server which directs the network traffic.
4.It offers many functionality frequently called \"servers.\"
5.Selecting the computer as \"server class hardware\" leads to more reliable and powerful than a
simple computer.
4.Network cabling: 1.It is used to join the one network device to another network device.
2.It the agency to travel from one netork to another network
3.Depending on their sizeand topology the cables are different as folllows.
1 .co-axial cable
2.optical fibre cable
3.twisted pair cacle
4.For acheving the good network we need to know the charecteristics of a cable at first.
5.Sound masking upgrades the isolation.
Solution
1.Router: 1.It is a machine which aheads datagrams over the networks.
2.This router is present at a node which connects two or more networks.
3.It avails of headers and tables to establish the path.
4.We can filter the data through this router in very small.
5.in order to communicate one another this router uses a protocol.
2.Hubs: 1.a hub is a frequent connection point .
2.It contains no of ports.
3.The data type in hub is frames.
4.hubs are present as an association for the networks.
5.It is some what slow.
6.we can use it as a stand alone device.
3.Servers: 1.A server is a device that manages all the network( printers ,hard disks...) devices.
2.generally this servers sre many types as follows
1.Database server:A system which operate through queries.
2.Print server :A system which directs one or more printer.
3.File server and network server which directs the network traffic.
4.It offers many functionality frequently called \"servers.\"
5.Selecting the computer as \"server class hardware\" leads to more reliable and powerful than a
simple computer.
4.Network cabling: 1.It is used to join the one network device to another network device.
2.It the agency to travel from one netork to another network
3.Depending on their sizeand topology the cables are different as folllows.
1 .co-axial cable
2.optical fibre cable
3.twisted pair cacle
4.For acheving the good network we need to know the charecteristics of a cable at first.
5.Sound masking upgrades the isolation..
1) List currently running jobsANS) see currently runningcommand.pdf
1) List currently running jobs
ANS) see currently running
command: ps
examples:
$ps
2.Kill one of the running jobs
ANS) a.kill any process by PID
command:
kill {PID}
example:
$kill 1012
3.Run the ls command with –help option
ANS) ls command:
ls command is used to list the contents of a directory
Syntax:
ls [-a] [-A] [-b] [-c] [-C] [-d] [-f] [-F] [-g] [-i] [-l] [-L] [-m] [-o] [-p][-q] [-r] [-R] [-s] [-t] [-u] [-x]
[pathnames]
Parameters OR Switches
• -a Shows All The Files, Including Hidden File Which Begin With ADot.
• -A Shows All The Files Including Hidden Files. Does Not Displays The
Working Directory (.) OR The Parent Directory (..)
• -b Force Printing of Non-Printable Characters To Be in Octal \\ddd
Notation.
• -c Use Time of Last Modification of The i-Node For Sorting (-t) OR For
Printing (-l OR -n)
• -C Multi-Column Output With Entries Sorted Down The Columns.
Which is Generally The Default Option.
• -d Lists Only The Directory Names But Not its Contents.
• -f Force Each Argument To Be Interpreted As A Directory And List
The Name Found in Each Slot. -f Option Turns off -l, -t, -s, And -r, And
Turns on –a, The List Order is The Order in Which Entries Appear in The
Directory.
• -F Mark Directories With A Trailing Slash (/), DOORS With A Trailing
Greater-Than Sign (>), Executable Files With A Trailing Asterisk (*),
FIFOs With A Trailing Vertical Bar (|), Symbolic Links With A Trailing
Equals Sign (=).
• -g Same As -l Parameter, Except The Owner is Not Printed.
• -i For Each File, Prints The i-Node Number in The First Column.
• -l Shows The User With Huge Amounts of Information With
Permissions, Owners, Size, And When Last Modified.)
• -L If An Argument is A Symbolic Link, List The File OR Directory To
Which The Link References Rather Than The Link Itself.
• -m Stream Output Format, In This Files Are Listed Across The Page,
Separated By Commas.
• -n Same As –l Parameter, Except That The Owner\'s UID And Group\'s
GID Numbers Are Printed, Rather Than The Associated Character
Strings.
-o The Same As -l, Except That The Group is Not Printed.
-p Displays A Forward Slash (/) in Front of All Directories.
-q Force Printing of Non-Printable Characters in File Names As The
Character Question Mark (?).
-r Reverses The Order of How The Files Are Displayed.
-R Includes The Contents of Subdirectories.
-s Give Size in Blocks, Including Indirect Blocks, For Each Entry.
-t Shows The End User The Files Modification Time.
-u Use Time of Last Access Instead of Last Modification For Sorting
Use With The -t Option OR For Printing Use With The -l Option.
-x Displays Files in Columns.
-1 Prints One Entry Per Line of Output.
pathnames File OR Directory To List.
4.Run the help command on echo
ANS) echo command display to the Screen.
“echo” is Useful For
• Producing Diagnostics in Command Files
• For Sending Known Data into A Pipe
• For Displaying The Contents of Environment Variables
syntax:
echo [-n] text
5.Cat an existing file using a command that misspells one c.
Step1 Total pressure = partial pressure of Hydrog.pdf
Step1 Total pressure = partial pressure of Hydrogen + Vapour pressure of water at
the given temperature. Step2 Partial pressure of Hydrogen = 767 - Vapour pressure of water at
295.3 K Step3 = 767.0- 20.2 = 746.8 torr
Solution
Step1 Total pressure = partial pressure of Hydrogen + Vapour pressure of water at
the given temperature. Step2 Partial pressure of Hydrogen = 767 - Vapour pressure of water at
295.3 K Step3 = 767.0- 20.2 = 746.8 torr.
I would say very many elements exista as monoatom.pdf
I would say very many elements exista as monoatomic molecules.All the metals,for
example.Don\'t get confused by the fact that if metals have a crystaline structure,then the
molecules would be composed by more than one atoms... An Element Is A Single Atom. No, an
element is any material that is composed of only a single kind of atom (ie: all its atoms are
identical) While A Molecule Is A Group Of Atoms That Are Bond Together To Form Another
Compounds... A molecule is a group of atoms that are bonded together, but they do not form a
compound. A compound is is something whose molecules contain more than one kind of atom.
Some Elements Exist As A Diatomic Atom. Some elements are a collection of diatomic
molecules. You can not have a diatomic atom - that\'s self-contradictory. Example Of These Are
Hydrogen And The Halogens Family( Chlorine. Fluorine Etc.) In Case Of Hydrogen. Hydrogen
Is A Diatomic Atom. If The Hydrogen Bonded To Other Elements This Is An Example Of
Molecule... Hydrogen is made up of diatomic molecules. If hydrogen bonds with atoms of
another element you get a compound. In Short Element Is A Single Atom While Molecule
Composed Of 2 Or More Atoms That Are Bonded Together. You can not compare an element
and a molecule. You can compare an element and a compound. An element is made of only a
single kind of atom - whether as a crystal (Fe), or isolated atoms (He) or as molecules (H2), does
not matter. A compound is made up of molecules (or crystals) with more than one kind of atom
(NH3, H2O, NaCl)
Solution
I would say very many elements exista as monoatomic molecules.All the metals,for
example.Don\'t get confused by the fact that if metals have a crystaline structure,then the
molecules would be composed by more than one atoms... An Element Is A Single Atom. No, an
element is any material that is composed of only a single kind of atom (ie: all its atoms are
identical) While A Molecule Is A Group Of Atoms That Are Bond Together To Form Another
Compounds... A molecule is a group of atoms that are bonded together, but they do not form a
compound. A compound is is something whose molecules contain more than one kind of atom.
Some Elements Exist As A Diatomic Atom. Some elements are a collection of diatomic
molecules. You can not have a diatomic atom - that\'s self-contradictory. Example Of These Are
Hydrogen And The Halogens Family( Chlorine. Fluorine Etc.) In Case Of Hydrogen. Hydrogen
Is A Diatomic Atom. If The Hydrogen Bonded To Other Elements This Is An Example Of
Molecule... Hydrogen is made up of diatomic molecules. If hydrogen bonds with atoms of
another element you get a compound. In Short Element Is A Single Atom While Molecule
Composed Of 2 Or More Atoms That Are Bonded Together. You can not compare an element
and a molecule. You can compare an element and a compound. An element is made of only a
single kind of atom - whether as a crystal (Fe), or isolated atoms (He) or as molecules (H2), does
not matter. A compound is made up of .
E is simply hexane...no polarity at all...nothing.pdf
E is simply hexane...no polarity at all...nothing to speak of has extra electron
density. A is a benzene ring, and if you play with the resonance you will see that sometimes the
electrons can be in many different places which will make it polar at those times. :) nate
Solution
E is simply hexane...no polarity at all...nothing to speak of has extra electron
density. A is a benzene ring, and if you play with the resonance you will see that sometimes the
electrons can be in many different places which will make it polar at those times. :) nate.
Ferroxyl contains the chemical compound Potassium.pdf
Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with
the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and
creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn
blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic
solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation
that leads to rust. Same thing basically for the OH.
Solution
Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with
the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and
creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn
blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic
solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation
that leads to rust. Same thing basically for the OH..
There are two different types of mechanisms. Ther.pdf
There are two different types of mechanisms. There are elimination and substitution
reactions. The elimination reaction is when there is a proton transfer and loss of leaving group
taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic
attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI
and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would
have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than
SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is
unclear.
Solution
There are two different types of mechanisms. There are elimination and substitution
reactions. The elimination reaction is when there is a proton transfer and loss of leaving group
taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic
attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI
and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would
have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than
SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is
unclear..
What must be true about a set of data when its standard deviation is.pdf
What must be true about a set of data when its standard deviation is zero (s= 0)? 1) All values of
the data appear with the same frequency. 2) All of the data have the same value. 3) The mean
value of the data is also zero. 4) All of the above are correct. 5) None of the above are correct.
Solution
5) None of the above are correct..
When one drives from sea level to trailhead in high Sierras, altitud.pdf
When one drives from sea level to trailhead in high Sierras, altitude increases. At higher
altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of
enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This
is detrimental as it can even stop our brain due to lack of oxygen.
These symptoms are collectively called Altitude sickness which reduces, as one spends some
time in such region as our body gets acclimatized to such environment.
Solution
When one drives from sea level to trailhead in high Sierras, altitude increases. At higher
altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of
enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This
is detrimental as it can even stop our brain due to lack of oxygen.
These symptoms are collectively called Altitude sickness which reduces, as one spends some
time in such region as our body gets acclimatized to such environment..
the reaction as follows- 14N7 + 0n1 -- 4He2 + 1.pdf
the reaction as follows:- 14N7 + 0n1 --> 4He2 + 10C6 so, the nuclide produced
When Nitrogen-14 is bombarded by (andabsorbs) a proton, a new nuclide is produced, plus an
alphaparticle is Carbon.
Solution
the reaction as follows:- 14N7 + 0n1 --> 4He2 + 10C6 so, the nuclide produced
When Nitrogen-14 is bombarded by (andabsorbs) a proton, a new nuclide is produced, plus an
alphaparticle is Carbon..
The important thing here is to understand the Standard Electrode Pot.pdf
The important thing here is to understand the Standard Electrode Potential for each of our metals
involved.
Whichever of the metals has a larger negative number is the one that will be our anode, the other
will be our cathode.
Therefore, for this reaction, we know that Aluminum is going to be our anode.
With this reasoning, we know that where the anode is, so will oxidation occur. If we think about
it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and
feeding it to the other side.
Since we know the salt bridge is there to balance out the lost electrons from the oxidation side,
we know that the NO3- are going to flow to the left hand side to make up for the electrons that
are going from the left hand side to the right hand side.
Your salt bridge sends anions to the left hand side!
Hope this helps!
Solution
The important thing here is to understand the Standard Electrode Potential for each of our metals
involved.
Whichever of the metals has a larger negative number is the one that will be our anode, the other
will be our cathode.
Therefore, for this reaction, we know that Aluminum is going to be our anode.
With this reasoning, we know that where the anode is, so will oxidation occur. If we think about
it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and
feeding it to the other side.
Since we know the salt bridge is there to balance out the lost electrons from the oxidation side,
we know that the NO3- are going to flow to the left hand side to make up for the electrons that
are going from the left hand side to the right hand side.
Your salt bridge sends anions to the left hand side!
Hope this helps!.
The metal ion is Pb2+.The white precipitate PbCl2 is insoluble in .pdf
The metal ion is Pb2+.
The white precipitate PbCl2 is insoluble in cold water but soluble in hot water.
The precipitation reaction is:
Pb2+(aq) + 2 HCl(aq) => PbCl2(s) + 2 H+(aq)
Solution
The metal ion is Pb2+.
The white precipitate PbCl2 is insoluble in cold water but soluble in hot water.
The precipitation reaction is:
Pb2+(aq) + 2 HCl(aq) => PbCl2(s) + 2 H+(aq).
The electron carriers that feed into the ETC are NADH and FADH2.NA.pdf
The electron carriers that feed into the ETC are NADH and FADH2.
NADH enters the ETC through complex I of the ETC.
FADH2 enters the ETC through complex II of the ETC.
Solution
The electron carriers that feed into the ETC are NADH and FADH2.
NADH enters the ETC through complex I of the ETC.
FADH2 enters the ETC through complex II of the ETC..
The developmental potential, or potency of a cell, describes the ran.pdf
The developmental potential, or potency of a cell, describes the range of different cell types it
can become. Therefore it involves.
The determination means a cell restricts itself choosing a particular fate.
Differentiation follows determination and the cell becomes a specific cell type ultimately.
Solution
The developmental potential, or potency of a cell, describes the range of different cell types it
can become. Therefore it involves.
The determination means a cell restricts itself choosing a particular fate.
Differentiation follows determination and the cell becomes a specific cell type ultimately..
Tested on linux,ubuntu#include stdlib.h #include systypes..pdf
//Tested on linux,ubuntu
#include
#include
#include
#include
#include
/*Main function start*/
int main(int argc, char *argv[])
{
/*variable for buffer*/
struct stat statbuf;
/*pointer variable declaration*/
DIR *dp;
struct dirent *dirp;
/*if we are not providing file path then it will take current directory as name*/
if(argc==1)
{
dp = opendir(\"./\");
}
else
{
dp = opendir(argv[1]);
}
while ( (dirp=readdir(dp)) != NULL)
{
/*lstat - get symbolic link status
* Please readabout lstat more*/
lstat(dirp->d_name, &statbuf);
/*checking it is directory or not*/
if(S_ISDIR(statbuf.st_mode))
{
/*it is printing directory name with /*/
printf(\"%s/ \ \", dirp->d_name);
}
else
{
/*If it is file then we are printing only name of file*/
printf(\"%s\ \", dirp->d_name);
}
}
/*cosing directory pointer*/
closedir(dp);
exit(0);
}
/*main function end*/
/**********output**********/
gopal@gopal:~/Desktop/chegg$ gcc directory.c
gopal@gopal:~/Desktop/chegg$ ./a.out
dLinkedList.c
random.c
jasmin/
LinkedStack.cpp
VideoGame.java
LinkedStack.cpp
Client.class
web.xml
../
TemplateMain.cpp
Energy.py
PrecondViolatedExcept.h
CSA.cpp
BoxofProduce.cpp
./
Unanswere/
Tree.c
target/
readFile.c
inventory.txt
goodQuestion
interest.c
VideoGameManager.java
Thanks a lot
Solution
//Tested on linux,ubuntu
#include
#include
#include
#include
#include
/*Main function start*/
int main(int argc, char *argv[])
{
/*variable for buffer*/
struct stat statbuf;
/*pointer variable declaration*/
DIR *dp;
struct dirent *dirp;
/*if we are not providing file path then it will take current directory as name*/
if(argc==1)
{
dp = opendir(\"./\");
}
else
{
dp = opendir(argv[1]);
}
while ( (dirp=readdir(dp)) != NULL)
{
/*lstat - get symbolic link status
* Please readabout lstat more*/
lstat(dirp->d_name, &statbuf);
/*checking it is directory or not*/
if(S_ISDIR(statbuf.st_mode))
{
/*it is printing directory name with /*/
printf(\"%s/ \ \", dirp->d_name);
}
else
{
/*If it is file then we are printing only name of file*/
printf(\"%s\ \", dirp->d_name);
}
}
/*cosing directory pointer*/
closedir(dp);
exit(0);
}
/*main function end*/
/**********output**********/
gopal@gopal:~/Desktop/chegg$ gcc directory.c
gopal@gopal:~/Desktop/chegg$ ./a.out
dLinkedList.c
random.c
jasmin/
LinkedStack.cpp
VideoGame.java
LinkedStack.cpp
Client.class
web.xml
../
TemplateMain.cpp
Energy.py
PrecondViolatedExcept.h
CSA.cpp
BoxofProduce.cpp
./
Unanswere/
Tree.c
target/
readFile.c
inventory.txt
goodQuestion
interest.c
VideoGameManager.java
Thanks a lot.
Since HEPES is a zwitterionic organic chemical buffering agent.At .pdf
Since HEPES is a zwitterionic organic chemical buffering agent.
At pH<7.0 it exists as cation (a positive ion , as negative charge is compensated by H+ present in
the solution) to bring its pH to 7.45 (basic region) we must add NaOH.
pH will increase when [H+] decreases. To decrease [H+] we must add NaOH ( a base).
Solution
Since HEPES is a zwitterionic organic chemical buffering agent.
At pH<7.0 it exists as cation (a positive ion , as negative charge is compensated by H+ present in
the solution) to bring its pH to 7.45 (basic region) we must add NaOH.
pH will increase when [H+] decreases. To decrease [H+] we must add NaOH ( a base)..
public static class LinkedDequeT implements DequeADTT{ priv.pdf
public static class LinkedDeque implements DequeADT{
private int size;
private Node first;
private Node last;
public LinkedDeque() {
}
// checking empty method or not
public boolean isEmpty() {
return first == null;
}
// checking size
public int size() {
return size;
}
// enqueing front
public void enqueueFront(T num) {
checkNull(num);
Node tempNode = new Node();
tempNode.num = num;
if (first != null) { // check root is not null
tempNode.next = first;
first.previous = tempNode;
}
first = tempNode;
if (last == null) last = first;
size++;
}
// removing front
public Item dequeueFront() {
CheckEmpty();
Node oldFirst = first;
first = first.next;
if (first == null)
last = null;
else
first.previous = null;
size--;
return oldFirst.item;
}
// inseeting element at end of list
public void enqueueBack(Item item) {
checkNull(item);
Node newLast = new Node();
newLast.item = item;
if (last != null) {
newLast.previous = last;
last.next = newLast;
}
last = newLast;
if (first == null) first = last;
size++;
}
// removing element from end of the list
public Item dequeueBack() {
CheckEmpty();
Node oldLast = last;
last = oldLast.previous;
if (last == null)
first = null;
else
last.next = null;
size--;
return oldLast.item;
}
private void CheckEmpty() {
if (first == null)
throw new NoSuchElementException();
}
private void checkNull(Item item) {
if (item == null)
throw new NullPointerException();
}
}
Solution
public static class LinkedDeque implements DequeADT{
private int size;
private Node first;
private Node last;
public LinkedDeque() {
}
// checking empty method or not
public boolean isEmpty() {
return first == null;
}
// checking size
public int size() {
return size;
}
// enqueing front
public void enqueueFront(T num) {
checkNull(num);
Node tempNode = new Node();
tempNode.num = num;
if (first != null) { // check root is not null
tempNode.next = first;
first.previous = tempNode;
}
first = tempNode;
if (last == null) last = first;
size++;
}
// removing front
public Item dequeueFront() {
CheckEmpty();
Node oldFirst = first;
first = first.next;
if (first == null)
last = null;
else
first.previous = null;
size--;
return oldFirst.item;
}
// inseeting element at end of list
public void enqueueBack(Item item) {
checkNull(item);
Node newLast = new Node();
newLast.item = item;
if (last != null) {
newLast.previous = last;
last.next = newLast;
}
last = newLast;
if (first == null) first = last;
size++;
}
// removing element from end of the list
public Item dequeueBack() {
CheckEmpty();
Node oldLast = last;
last = oldLast.previous;
if (last == null)
first = null;
else
last.next = null;
size--;
return oldLast.item;
}
private void CheckEmpty() {
if (first == null)
throw new NoSuchElementException();
}
private void checkNull(Item item) {
if (item == null)
throw new NullPointerException();
}
}.
No matter. you can write either the H (on the C with the Br) and Cl .pdf
No matter. you can write either the H (on the C with the Br) and Cl (on the C with the H) are on
the wedges or the Br and H instead of the H and Cl, be drawn on wedges. Both will be same that
is trans isomers. If he asks about R, S configuration in that case they may be different.
Solution
No matter. you can write either the H (on the C with the Br) and Cl (on the C with the H) are on
the wedges or the Br and H instead of the H and Cl, be drawn on wedges. Both will be same that
is trans isomers. If he asks about R, S configuration in that case they may be different..
Moles of N2O5 = number of moleculesAvogadros number= 4.04 x 1020.pdf
Moles of N2O5 = number of molecules/Avogadros number
= 4.04 x 1020/6.022 x 1023
= 0.00067087 mol
Mass of N2O5 = moles x molar mass of N2O5
= 0.00067087 x 108.01
= 0.0725 g = 7.25 x 10-2 g
Solution
Moles of N2O5 = number of molecules/Avogadros number
= 4.04 x 1020/6.022 x 1023
= 0.00067087 mol
Mass of N2O5 = moles x molar mass of N2O5
= 0.00067087 x 108.01
= 0.0725 g = 7.25 x 10-2 g.
LOOPSFor LoopFor loop executes group of Java statements as long.pdf
LOOPS
For Loop:
For loop executes group of Java statements as long as the boolean condition evaluates to true.
For loop combines three elements which we generally use: initialization statement, boolean
expression and increment or decrement statement.
For loop syntax:
for( ; ; )
{
;
}
The initialization statement is executed before the loop starts. It is generally used to initialize the
loop variable.
Condition statement is evaluated before each time the block of statements are executed. Block of
statements are executed only if the boolean condition evaluates to true.
Statement is executed after the loop body is done. Generally it is being used to increment or
decrement the loop variable.
Following example shows use of simple for loop.
for(int i=0 ; i < 5 ; i++)
{
System.out.println(“i is : “ + i);
}
It is possible to initialize multiple variable in the initialization block of the for loop by separating
it by comma as given in the below example.
for(int i=0, j =5 ; i < 5 ; i++)
It is also possible to have more than one increment or decrement section as well as given below.
for(int i=0; i < 5 ; i++, j--)
However it is not possible to include declaration and initialization in the initialization block of
the for loop.
While Loop
While loop executes group of Java statements as long as the boolean condition evaluates to true.
While loop syntax
while(){
;
}
Block of statements is any valid Java code. Boolean condition is any valid Java expression that
evaluates to boolean value. Braces are options if there is only one statement to be executed.
While loop evaluates the boolean expression, and if it is true, it executes the block of statements.
It continuously executes the block of statements until the boolean condition evaluates to false
value. After that loop is terminated and statement immediately after the loop body will be
executed.
Do While Loop
Do while loop executes group of Java statements as long as the boolean condition evaluates to
true.
It is possible that the statement block associated with While loop never get executed because
While loop tests the boolean condition before executing the block of statements associated with
it.
In Java programming, sometime it\'s necessary to execute the block of statements at least once
before evaluating the boolean condition. Do while loop is similar to the While loop, but it
evaluates the boolean condition after executing the block of the statement.
Do While loop syntax
Do{
;
}while();
Arithmetic Assignment Operators Example
/*
Arithmetic Assignment Operators Example
This example shows how to use Java airthmetic assignment operators like
+=, -=, *= and /=.
*/
public class ArithmeticAssignmentOperatorExample {
public static void main(String[] args) {
/*
* Java provides short cut operators to combine an Arithmetic
* operator and assignment operator.
*
* For example,
*
* i = i + 5;
*
* can also be rewritten as
*
* i+= 4;
*/
int i = 5;
int j = 10;
i += 5; //same as i = i + 5
j -= 2; //same as j = j - 2
Syst.
Intermediate levels of disturbance often lead to higher diversity be.pdf
Intermediate levels of disturbance often lead to higher diversity because:
Both species that do well at low disturbance and those that do well at high disturbance are
usually present.
Ecological succession refers to:
Fairly predictable changes in species composition over time in a particular location.
Solution
Intermediate levels of disturbance often lead to higher diversity because:
Both species that do well at low disturbance and those that do well at high disturbance are
usually present.
Ecological succession refers to:
Fairly predictable changes in species composition over time in a particular location..
In the life cycle of a sexually reproducing organism, regulation of .pdf
In the life cycle of a sexually reproducing organism, regulation of the cell cycle is involved the
development of specialised structures and cell types in the organism.
Solution
In the life cycle of a sexually reproducing organism, regulation of the cell cycle is involved the
development of specialised structures and cell types in the organism..
Initial volume = 10 LDecrease in Volume = 25 of 10 = 104 = 2.5.pdf
Initial volume = 10 L
Decrease in Volume = 25% of 10 = 10/4 = 2.5 L
Final Volume = 10 - 2.5 = 7.5 L
Initial Pressure = 1.5 bar
PV = nRt
PV = Constant {Because Temperature remains Constant}
Let the final pressure is P
1.5/7.5 = P/10
P = 2 bar
So %age increase = (2-1.5)/1.5 * 100 = 33.33 %
Solution
Initial volume = 10 L
Decrease in Volume = 25% of 10 = 10/4 = 2.5 L
Final Volume = 10 - 2.5 = 7.5 L
Initial Pressure = 1.5 bar
PV = nRt
PV = Constant {Because Temperature remains Constant}
Let the final pressure is P
1.5/7.5 = P/10
P = 2 bar
So %age increase = (2-1.5)/1.5 * 100 = 33.33 %.
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
More Related Content
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I would say very many elements exista as monoatom.pdf
I would say very many elements exista as monoatomic molecules.All the metals,for
example.Don\'t get confused by the fact that if metals have a crystaline structure,then the
molecules would be composed by more than one atoms... An Element Is A Single Atom. No, an
element is any material that is composed of only a single kind of atom (ie: all its atoms are
identical) While A Molecule Is A Group Of Atoms That Are Bond Together To Form Another
Compounds... A molecule is a group of atoms that are bonded together, but they do not form a
compound. A compound is is something whose molecules contain more than one kind of atom.
Some Elements Exist As A Diatomic Atom. Some elements are a collection of diatomic
molecules. You can not have a diatomic atom - that\'s self-contradictory. Example Of These Are
Hydrogen And The Halogens Family( Chlorine. Fluorine Etc.) In Case Of Hydrogen. Hydrogen
Is A Diatomic Atom. If The Hydrogen Bonded To Other Elements This Is An Example Of
Molecule... Hydrogen is made up of diatomic molecules. If hydrogen bonds with atoms of
another element you get a compound. In Short Element Is A Single Atom While Molecule
Composed Of 2 Or More Atoms That Are Bonded Together. You can not compare an element
and a molecule. You can compare an element and a compound. An element is made of only a
single kind of atom - whether as a crystal (Fe), or isolated atoms (He) or as molecules (H2), does
not matter. A compound is made up of molecules (or crystals) with more than one kind of atom
(NH3, H2O, NaCl)
Solution
I would say very many elements exista as monoatomic molecules.All the metals,for
example.Don\'t get confused by the fact that if metals have a crystaline structure,then the
molecules would be composed by more than one atoms... An Element Is A Single Atom. No, an
element is any material that is composed of only a single kind of atom (ie: all its atoms are
identical) While A Molecule Is A Group Of Atoms That Are Bond Together To Form Another
Compounds... A molecule is a group of atoms that are bonded together, but they do not form a
compound. A compound is is something whose molecules contain more than one kind of atom.
Some Elements Exist As A Diatomic Atom. Some elements are a collection of diatomic
molecules. You can not have a diatomic atom - that\'s self-contradictory. Example Of These Are
Hydrogen And The Halogens Family( Chlorine. Fluorine Etc.) In Case Of Hydrogen. Hydrogen
Is A Diatomic Atom. If The Hydrogen Bonded To Other Elements This Is An Example Of
Molecule... Hydrogen is made up of diatomic molecules. If hydrogen bonds with atoms of
another element you get a compound. In Short Element Is A Single Atom While Molecule
Composed Of 2 Or More Atoms That Are Bonded Together. You can not compare an element
and a molecule. You can compare an element and a compound. An element is made of only a
single kind of atom - whether as a crystal (Fe), or isolated atoms (He) or as molecules (H2), does
not matter. A compound is made up of .
E is simply hexane...no polarity at all...nothing.pdf
E is simply hexane...no polarity at all...nothing to speak of has extra electron
density. A is a benzene ring, and if you play with the resonance you will see that sometimes the
electrons can be in many different places which will make it polar at those times. :) nate
Solution
E is simply hexane...no polarity at all...nothing to speak of has extra electron
density. A is a benzene ring, and if you play with the resonance you will see that sometimes the
electrons can be in many different places which will make it polar at those times. :) nate.
Ferroxyl contains the chemical compound Potassium.pdf
Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with
the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and
creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn
blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic
solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation
that leads to rust. Same thing basically for the OH.
Solution
Ferroxyl contains the chemical compound Potassium ferricyanide which reacts with
the ferrous ions to produce the color \"Prussian blue\". Iron is being oxidized at the anode and
creating Ferrous (Fe+2) ions. Therefore, with the introduction of Ferroxyl, the anode will turn
blue as the Ferroxyl reacts with the Ferrous ions. It is also worth noting that this is a acidic
solution and the blue pigment is insoluble in water. This is useful in detecting metal oxidation
that leads to rust. Same thing basically for the OH..
There are two different types of mechanisms. Ther.pdf
There are two different types of mechanisms. There are elimination and substitution
reactions. The elimination reaction is when there is a proton transfer and loss of leaving group
taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic
attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI
and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would
have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than
SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is
unclear.
Solution
There are two different types of mechanisms. There are elimination and substitution
reactions. The elimination reaction is when there is a proton transfer and loss of leaving group
taking place. The substitution reaction is when there is a loss of leaving group and nucleophlic
attack. From what we see in this reaction, we see that there is a nucleophilic attack with the MeI
and OH. So this is a substitution reaction. Now to determine whether it is SN1 or SN2, we would
have to look at the reaction. There is a tertary compound, which favors SN1 reactions better than
SN2. The final answer would be SN1. I hope this makes sense. Let me know if anything is
unclear..
What must be true about a set of data when its standard deviation is.pdf
What must be true about a set of data when its standard deviation is zero (s= 0)? 1) All values of
the data appear with the same frequency. 2) All of the data have the same value. 3) The mean
value of the data is also zero. 4) All of the above are correct. 5) None of the above are correct.
Solution
5) None of the above are correct..
When one drives from sea level to trailhead in high Sierras, altitud.pdf
When one drives from sea level to trailhead in high Sierras, altitude increases. At higher
altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of
enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This
is detrimental as it can even stop our brain due to lack of oxygen.
These symptoms are collectively called Altitude sickness which reduces, as one spends some
time in such region as our body gets acclimatized to such environment.
Solution
When one drives from sea level to trailhead in high Sierras, altitude increases. At higher
altitudes, atmospheric pressure drops and so does the oxygen content of the air. We run out of
enough oxygen in blood for our brain and body that leads to dizziness and heavy breathing. This
is detrimental as it can even stop our brain due to lack of oxygen.
These symptoms are collectively called Altitude sickness which reduces, as one spends some
time in such region as our body gets acclimatized to such environment..
the reaction as follows- 14N7 + 0n1 -- 4He2 + 1.pdf
the reaction as follows:- 14N7 + 0n1 --> 4He2 + 10C6 so, the nuclide produced
When Nitrogen-14 is bombarded by (andabsorbs) a proton, a new nuclide is produced, plus an
alphaparticle is Carbon.
Solution
the reaction as follows:- 14N7 + 0n1 --> 4He2 + 10C6 so, the nuclide produced
When Nitrogen-14 is bombarded by (andabsorbs) a proton, a new nuclide is produced, plus an
alphaparticle is Carbon..
The important thing here is to understand the Standard Electrode Pot.pdf
The important thing here is to understand the Standard Electrode Potential for each of our metals
involved.
Whichever of the metals has a larger negative number is the one that will be our anode, the other
will be our cathode.
Therefore, for this reaction, we know that Aluminum is going to be our anode.
With this reasoning, we know that where the anode is, so will oxidation occur. If we think about
it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and
feeding it to the other side.
Since we know the salt bridge is there to balance out the lost electrons from the oxidation side,
we know that the NO3- are going to flow to the left hand side to make up for the electrons that
are going from the left hand side to the right hand side.
Your salt bridge sends anions to the left hand side!
Hope this helps!
Solution
The important thing here is to understand the Standard Electrode Potential for each of our metals
involved.
Whichever of the metals has a larger negative number is the one that will be our anode, the other
will be our cathode.
Therefore, for this reaction, we know that Aluminum is going to be our anode.
With this reasoning, we know that where the anode is, so will oxidation occur. If we think about
it, where oxidation is occurring, the metal (in this case Aluminum) is giving up electrons and
feeding it to the other side.
Since we know the salt bridge is there to balance out the lost electrons from the oxidation side,
we know that the NO3- are going to flow to the left hand side to make up for the electrons that
are going from the left hand side to the right hand side.
Your salt bridge sends anions to the left hand side!
Hope this helps!.
The metal ion is Pb2+.The white precipitate PbCl2 is insoluble in .pdf
The metal ion is Pb2+.
The white precipitate PbCl2 is insoluble in cold water but soluble in hot water.
The precipitation reaction is:
Pb2+(aq) + 2 HCl(aq) => PbCl2(s) + 2 H+(aq)
Solution
The metal ion is Pb2+.
The white precipitate PbCl2 is insoluble in cold water but soluble in hot water.
The precipitation reaction is:
Pb2+(aq) + 2 HCl(aq) => PbCl2(s) + 2 H+(aq).
The electron carriers that feed into the ETC are NADH and FADH2.NA.pdf
The electron carriers that feed into the ETC are NADH and FADH2.
NADH enters the ETC through complex I of the ETC.
FADH2 enters the ETC through complex II of the ETC.
Solution
The electron carriers that feed into the ETC are NADH and FADH2.
NADH enters the ETC through complex I of the ETC.
FADH2 enters the ETC through complex II of the ETC..
The developmental potential, or potency of a cell, describes the ran.pdf
The developmental potential, or potency of a cell, describes the range of different cell types it
can become. Therefore it involves.
The determination means a cell restricts itself choosing a particular fate.
Differentiation follows determination and the cell becomes a specific cell type ultimately.
Solution
The developmental potential, or potency of a cell, describes the range of different cell types it
can become. Therefore it involves.
The determination means a cell restricts itself choosing a particular fate.
Differentiation follows determination and the cell becomes a specific cell type ultimately..
Tested on linux,ubuntu#include stdlib.h #include systypes..pdf
//Tested on linux,ubuntu
#include
#include
#include
#include
#include
/*Main function start*/
int main(int argc, char *argv[])
{
/*variable for buffer*/
struct stat statbuf;
/*pointer variable declaration*/
DIR *dp;
struct dirent *dirp;
/*if we are not providing file path then it will take current directory as name*/
if(argc==1)
{
dp = opendir(\"./\");
}
else
{
dp = opendir(argv[1]);
}
while ( (dirp=readdir(dp)) != NULL)
{
/*lstat - get symbolic link status
* Please readabout lstat more*/
lstat(dirp->d_name, &statbuf);
/*checking it is directory or not*/
if(S_ISDIR(statbuf.st_mode))
{
/*it is printing directory name with /*/
printf(\"%s/ \ \", dirp->d_name);
}
else
{
/*If it is file then we are printing only name of file*/
printf(\"%s\ \", dirp->d_name);
}
}
/*cosing directory pointer*/
closedir(dp);
exit(0);
}
/*main function end*/
/**********output**********/
gopal@gopal:~/Desktop/chegg$ gcc directory.c
gopal@gopal:~/Desktop/chegg$ ./a.out
dLinkedList.c
random.c
jasmin/
LinkedStack.cpp
VideoGame.java
LinkedStack.cpp
Client.class
web.xml
../
TemplateMain.cpp
Energy.py
PrecondViolatedExcept.h
CSA.cpp
BoxofProduce.cpp
./
Unanswere/
Tree.c
target/
readFile.c
inventory.txt
goodQuestion
interest.c
VideoGameManager.java
Thanks a lot
Solution
//Tested on linux,ubuntu
#include
#include
#include
#include
#include
/*Main function start*/
int main(int argc, char *argv[])
{
/*variable for buffer*/
struct stat statbuf;
/*pointer variable declaration*/
DIR *dp;
struct dirent *dirp;
/*if we are not providing file path then it will take current directory as name*/
if(argc==1)
{
dp = opendir(\"./\");
}
else
{
dp = opendir(argv[1]);
}
while ( (dirp=readdir(dp)) != NULL)
{
/*lstat - get symbolic link status
* Please readabout lstat more*/
lstat(dirp->d_name, &statbuf);
/*checking it is directory or not*/
if(S_ISDIR(statbuf.st_mode))
{
/*it is printing directory name with /*/
printf(\"%s/ \ \", dirp->d_name);
}
else
{
/*If it is file then we are printing only name of file*/
printf(\"%s\ \", dirp->d_name);
}
}
/*cosing directory pointer*/
closedir(dp);
exit(0);
}
/*main function end*/
/**********output**********/
gopal@gopal:~/Desktop/chegg$ gcc directory.c
gopal@gopal:~/Desktop/chegg$ ./a.out
dLinkedList.c
random.c
jasmin/
LinkedStack.cpp
VideoGame.java
LinkedStack.cpp
Client.class
web.xml
../
TemplateMain.cpp
Energy.py
PrecondViolatedExcept.h
CSA.cpp
BoxofProduce.cpp
./
Unanswere/
Tree.c
target/
readFile.c
inventory.txt
goodQuestion
interest.c
VideoGameManager.java
Thanks a lot.
Since HEPES is a zwitterionic organic chemical buffering agent.At .pdf
Since HEPES is a zwitterionic organic chemical buffering agent.
At pH<7.0 it exists as cation (a positive ion , as negative charge is compensated by H+ present in
the solution) to bring its pH to 7.45 (basic region) we must add NaOH.
pH will increase when [H+] decreases. To decrease [H+] we must add NaOH ( a base).
Solution
Since HEPES is a zwitterionic organic chemical buffering agent.
At pH<7.0 it exists as cation (a positive ion , as negative charge is compensated by H+ present in
the solution) to bring its pH to 7.45 (basic region) we must add NaOH.
pH will increase when [H+] decreases. To decrease [H+] we must add NaOH ( a base)..
public static class LinkedDequeT implements DequeADTT{ priv.pdf
public static class LinkedDeque implements DequeADT{
private int size;
private Node first;
private Node last;
public LinkedDeque() {
}
// checking empty method or not
public boolean isEmpty() {
return first == null;
}
// checking size
public int size() {
return size;
}
// enqueing front
public void enqueueFront(T num) {
checkNull(num);
Node tempNode = new Node();
tempNode.num = num;
if (first != null) { // check root is not null
tempNode.next = first;
first.previous = tempNode;
}
first = tempNode;
if (last == null) last = first;
size++;
}
// removing front
public Item dequeueFront() {
CheckEmpty();
Node oldFirst = first;
first = first.next;
if (first == null)
last = null;
else
first.previous = null;
size--;
return oldFirst.item;
}
// inseeting element at end of list
public void enqueueBack(Item item) {
checkNull(item);
Node newLast = new Node();
newLast.item = item;
if (last != null) {
newLast.previous = last;
last.next = newLast;
}
last = newLast;
if (first == null) first = last;
size++;
}
// removing element from end of the list
public Item dequeueBack() {
CheckEmpty();
Node oldLast = last;
last = oldLast.previous;
if (last == null)
first = null;
else
last.next = null;
size--;
return oldLast.item;
}
private void CheckEmpty() {
if (first == null)
throw new NoSuchElementException();
}
private void checkNull(Item item) {
if (item == null)
throw new NullPointerException();
}
}
Solution
public static class LinkedDeque implements DequeADT{
private int size;
private Node first;
private Node last;
public LinkedDeque() {
}
// checking empty method or not
public boolean isEmpty() {
return first == null;
}
// checking size
public int size() {
return size;
}
// enqueing front
public void enqueueFront(T num) {
checkNull(num);
Node tempNode = new Node();
tempNode.num = num;
if (first != null) { // check root is not null
tempNode.next = first;
first.previous = tempNode;
}
first = tempNode;
if (last == null) last = first;
size++;
}
// removing front
public Item dequeueFront() {
CheckEmpty();
Node oldFirst = first;
first = first.next;
if (first == null)
last = null;
else
first.previous = null;
size--;
return oldFirst.item;
}
// inseeting element at end of list
public void enqueueBack(Item item) {
checkNull(item);
Node newLast = new Node();
newLast.item = item;
if (last != null) {
newLast.previous = last;
last.next = newLast;
}
last = newLast;
if (first == null) first = last;
size++;
}
// removing element from end of the list
public Item dequeueBack() {
CheckEmpty();
Node oldLast = last;
last = oldLast.previous;
if (last == null)
first = null;
else
last.next = null;
size--;
return oldLast.item;
}
private void CheckEmpty() {
if (first == null)
throw new NoSuchElementException();
}
private void checkNull(Item item) {
if (item == null)
throw new NullPointerException();
}
}.
No matter. you can write either the H (on the C with the Br) and Cl .pdf
No matter. you can write either the H (on the C with the Br) and Cl (on the C with the H) are on
the wedges or the Br and H instead of the H and Cl, be drawn on wedges. Both will be same that
is trans isomers. If he asks about R, S configuration in that case they may be different.
Solution
No matter. you can write either the H (on the C with the Br) and Cl (on the C with the H) are on
the wedges or the Br and H instead of the H and Cl, be drawn on wedges. Both will be same that
is trans isomers. If he asks about R, S configuration in that case they may be different..
Moles of N2O5 = number of moleculesAvogadros number= 4.04 x 1020.pdf
Moles of N2O5 = number of molecules/Avogadros number
= 4.04 x 1020/6.022 x 1023
= 0.00067087 mol
Mass of N2O5 = moles x molar mass of N2O5
= 0.00067087 x 108.01
= 0.0725 g = 7.25 x 10-2 g
Solution
Moles of N2O5 = number of molecules/Avogadros number
= 4.04 x 1020/6.022 x 1023
= 0.00067087 mol
Mass of N2O5 = moles x molar mass of N2O5
= 0.00067087 x 108.01
= 0.0725 g = 7.25 x 10-2 g.
LOOPSFor LoopFor loop executes group of Java statements as long.pdf
LOOPS
For Loop:
For loop executes group of Java statements as long as the boolean condition evaluates to true.
For loop combines three elements which we generally use: initialization statement, boolean
expression and increment or decrement statement.
For loop syntax:
for( ; ; )
{
;
}
The initialization statement is executed before the loop starts. It is generally used to initialize the
loop variable.
Condition statement is evaluated before each time the block of statements are executed. Block of
statements are executed only if the boolean condition evaluates to true.
Statement is executed after the loop body is done. Generally it is being used to increment or
decrement the loop variable.
Following example shows use of simple for loop.
for(int i=0 ; i < 5 ; i++)
{
System.out.println(“i is : “ + i);
}
It is possible to initialize multiple variable in the initialization block of the for loop by separating
it by comma as given in the below example.
for(int i=0, j =5 ; i < 5 ; i++)
It is also possible to have more than one increment or decrement section as well as given below.
for(int i=0; i < 5 ; i++, j--)
However it is not possible to include declaration and initialization in the initialization block of
the for loop.
While Loop
While loop executes group of Java statements as long as the boolean condition evaluates to true.
While loop syntax
while(){
;
}
Block of statements is any valid Java code. Boolean condition is any valid Java expression that
evaluates to boolean value. Braces are options if there is only one statement to be executed.
While loop evaluates the boolean expression, and if it is true, it executes the block of statements.
It continuously executes the block of statements until the boolean condition evaluates to false
value. After that loop is terminated and statement immediately after the loop body will be
executed.
Do While Loop
Do while loop executes group of Java statements as long as the boolean condition evaluates to
true.
It is possible that the statement block associated with While loop never get executed because
While loop tests the boolean condition before executing the block of statements associated with
it.
In Java programming, sometime it\'s necessary to execute the block of statements at least once
before evaluating the boolean condition. Do while loop is similar to the While loop, but it
evaluates the boolean condition after executing the block of the statement.
Do While loop syntax
Do{
;
}while();
Arithmetic Assignment Operators Example
/*
Arithmetic Assignment Operators Example
This example shows how to use Java airthmetic assignment operators like
+=, -=, *= and /=.
*/
public class ArithmeticAssignmentOperatorExample {
public static void main(String[] args) {
/*
* Java provides short cut operators to combine an Arithmetic
* operator and assignment operator.
*
* For example,
*
* i = i + 5;
*
* can also be rewritten as
*
* i+= 4;
*/
int i = 5;
int j = 10;
i += 5; //same as i = i + 5
j -= 2; //same as j = j - 2
Syst.
Intermediate levels of disturbance often lead to higher diversity be.pdf
Intermediate levels of disturbance often lead to higher diversity because:
Both species that do well at low disturbance and those that do well at high disturbance are
usually present.
Ecological succession refers to:
Fairly predictable changes in species composition over time in a particular location.
Solution
Intermediate levels of disturbance often lead to higher diversity because:
Both species that do well at low disturbance and those that do well at high disturbance are
usually present.
Ecological succession refers to:
Fairly predictable changes in species composition over time in a particular location..
In the life cycle of a sexually reproducing organism, regulation of .pdf
In the life cycle of a sexually reproducing organism, regulation of the cell cycle is involved the
development of specialised structures and cell types in the organism.
Solution
In the life cycle of a sexually reproducing organism, regulation of the cell cycle is involved the
development of specialised structures and cell types in the organism..
Initial volume = 10 LDecrease in Volume = 25 of 10 = 104 = 2.5.pdf
Initial volume = 10 L
Decrease in Volume = 25% of 10 = 10/4 = 2.5 L
Final Volume = 10 - 2.5 = 7.5 L
Initial Pressure = 1.5 bar
PV = nRt
PV = Constant {Because Temperature remains Constant}
Let the final pressure is P
1.5/7.5 = P/10
P = 2 bar
So %age increase = (2-1.5)/1.5 * 100 = 33.33 %
Solution
Initial volume = 10 L
Decrease in Volume = 25% of 10 = 10/4 = 2.5 L
Final Volume = 10 - 2.5 = 7.5 L
Initial Pressure = 1.5 bar
PV = nRt
PV = Constant {Because Temperature remains Constant}
Let the final pressure is P
1.5/7.5 = P/10
P = 2 bar
So %age increase = (2-1.5)/1.5 * 100 = 33.33 %.
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