They can form sulfides, but the metal sulfides th.pdf
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They can form sulfides, but the metal sulfides that form will notprecipitate. Solution They can form sulfides, but the metal sulfides that form will notprecipitate..
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5 steps (make sure to press See full answer)- make observat.pdf
5 steps: (make sure to press \"See full answer\")
- make observation and gathering all available data
Solution
5 steps: (make sure to press \"See full answer\")
- make observation and gathering all available data.
2cosx=(cosxtanx+sinx)(tanx)taking RHS cosxtanx+sinx)(tanx)= .pdf
2cosx=(cosxtanx+sinx)/(tanx)
taking RHS : cosxtanx+sinx)/(tanx)
= cosxtanx/tanx + sinx/tanx
= cosx + sinx *cosx/sinx
= cosx + cosx
= 2cosx
= LHS
hence LHS = RHS
Solution
2cosx=(cosxtanx+sinx)/(tanx)
taking RHS : cosxtanx+sinx)/(tanx)
= cosxtanx/tanx + sinx/tanx
= cosx + sinx *cosx/sinx
= cosx + cosx
= 2cosx
= LHS
hence LHS = RHS.
1.5 X 10-8Solution1.5 X 10-8.pdf
This document contains a single numerical value of 1.5 X 10-8. No other context or information is provided about what this number represents or relates to.
#includestdio.h#includestdlib.hint encrypt(void);int decry.pdf
#include
#include
int encrypt(void);
int decrypt(void);
int encrypt_view(void);
int decrypt_view(void);
FILE *fp1, *fp2;
char ch;
int main()
{
int choice;
while(1)
{
printf(\"Select One of the Following:\ \");
printf(\"\ 1. Encrypt\ \");
printf(\"2. Decrypt\ \");
printf(\"3. View The Encypted File\ \");
printf(\"4. View The Decrypted File\ \");
printf(\"5. Exit\ \");
printf(\"\ Enter Your Choice:\\t\");
scanf(\"%d\", &choice);
switch(choice)
{
case 1: encrypt();
break;
case 2: decrypt();
break;
case 3: encrypt_view();
break;
case 4: decrypt_view();
break;
case 5: exit(1);
}
}
printf(\"\ \");
return 0;
}
int encrypt()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\");
if(fp1 == NULL)
{
printf(\"Source File Could Not Be Found\ \");
}
fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\");
if(fp2 == NULL)
{
printf(\"Target File Could Not Be Found\ \");
}
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
printf(\"\ End Of File\ \");
break;
}
else
{
ch = ch - (8 * 5 - 3);
fputc(ch, fp2);
}
}
fclose(fp1);
fclose(fp2);
printf(\"\ \");
return 0;
}
int decrypt()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\");
if(fp1 == NULL)
{
printf(\"Source File Could Not Be Found\ \");
}
fp2 = fopen(\"/home/tusharsoni/Desktop/Source\",\"w\");
if(fp2 == NULL)
{
printf(\"Target File Could Not Be Found\ \");
}
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
printf(\"\ End Of File\ \");
break;
}
else
{
ch = ch + (8 * 5 - 3);
fputc(ch, fp2);
}
}
fclose(fp1);
fclose(fp2);
printf(\"\ \");
return 0;
}
int encrypt_view()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\");
if(fp1 == NULL)
{
printf(\"No File Found\ \");
exit(1);
}
else
{
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
break;
}
else
{
printf(\"%c\", ch);
}
}
printf(\"\ \");
fclose(fp1);
}
printf(\"\ \");
return 0;
}
int decrypt_view()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\");
if(fp1 == NULL)
{
printf(\"No File Found\ \");
exit(1);
}
else
{
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
break;
}
else
{
printf(\"%c\", ch);
}
}
printf(\"\ \");
fclose(fp1);
}
return 0;
printf(\"\ \");
}
Solution
#include
#include
int encrypt(void);
int decrypt(void);
int encrypt_view(void);
int decrypt_view(void);
FILE *fp1, *fp2;
char ch;
int main()
{
int choice;
while(1)
{
printf(\"Select One of the Following:\ \");
printf(\"\ 1. Encrypt\ \");
printf(\"2. Decrypt\ \");
printf(\"3. View The Encypted File\ \");
printf(\"4. View The Decrypted File\ \");
printf(\"5. Exit\ \");
printf(\"\ Enter Your Choice:\\t\");
scanf(\"%d\", &choice);
switch(choice)
{
case 1: encrypt();
break;
case 2: decrypt();
break;
case 3: encrypt_view();
break;
case 4: decrypt_view();
break;
case 5: exit(1);
}
}
printf(\"\ \");
return 0;
}
int encrypt()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\");
if(fp1 == NULL)
{
printf(\"Source File Could Not Be Found\ \");
}
fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\");
if(fp2 == NULL)
{
printf(\"Ta.
Neuman System Model helps in studying the stressors, find out the de.pdf
Neuman System Model helps in studying the stressors, find out the defense, gives the feedback
and helps in preventing the occurrence through assessment, diagnosis, planning, implementation
and evaluation taking the person as a whole. Neuman System Model says that individual person
as whole is in interaction with environment. Neuman System Model helps in diagnosis and
prevents the stressors. Neuman System Model implements and improves the immunity. Neuman
System Model takes the help of social elements such as relatives and neighbors to protect the
person.
Solution
Neuman System Model helps in studying the stressors, find out the defense, gives the feedback
and helps in preventing the occurrence through assessment, diagnosis, planning, implementation
and evaluation taking the person as a whole. Neuman System Model says that individual person
as whole is in interaction with environment. Neuman System Model helps in diagnosis and
prevents the stressors. Neuman System Model implements and improves the immunity. Neuman
System Model takes the help of social elements such as relatives and neighbors to protect the
person..
NaOH will completely react will acetic acid and t.pdf
NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH
left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037
Solution
NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH
left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037.
the bonding between the carban and the nitrogen i.pdf
the bonding between the carban and the nitrogen in hydrogen cyanide or
hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the
molecule
Solution
the bonding between the carban and the nitrogen in hydrogen cyanide or
hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the
molecule.
The salts are more ionic, more soluble they are i.pdf
The salts are more ionic, more soluble they are in water since H2O is a polar
solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic
bonded.
Solution
The salts are more ionic, more soluble they are in water since H2O is a polar
solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic
bonded..
Recommended
5 steps (make sure to press See full answer)- make observat.pdf
5 steps: (make sure to press \"See full answer\")
- make observation and gathering all available data
Solution
5 steps: (make sure to press \"See full answer\")
- make observation and gathering all available data.
2cosx=(cosxtanx+sinx)(tanx)taking RHS cosxtanx+sinx)(tanx)= .pdf
2cosx=(cosxtanx+sinx)/(tanx)
taking RHS : cosxtanx+sinx)/(tanx)
= cosxtanx/tanx + sinx/tanx
= cosx + sinx *cosx/sinx
= cosx + cosx
= 2cosx
= LHS
hence LHS = RHS
Solution
2cosx=(cosxtanx+sinx)/(tanx)
taking RHS : cosxtanx+sinx)/(tanx)
= cosxtanx/tanx + sinx/tanx
= cosx + sinx *cosx/sinx
= cosx + cosx
= 2cosx
= LHS
hence LHS = RHS.
1.5 X 10-8Solution1.5 X 10-8.pdf
This document contains a single numerical value of 1.5 X 10-8. No other context or information is provided about what this number represents or relates to.
#includestdio.h#includestdlib.hint encrypt(void);int decry.pdf
#include
#include
int encrypt(void);
int decrypt(void);
int encrypt_view(void);
int decrypt_view(void);
FILE *fp1, *fp2;
char ch;
int main()
{
int choice;
while(1)
{
printf(\"Select One of the Following:\ \");
printf(\"\ 1. Encrypt\ \");
printf(\"2. Decrypt\ \");
printf(\"3. View The Encypted File\ \");
printf(\"4. View The Decrypted File\ \");
printf(\"5. Exit\ \");
printf(\"\ Enter Your Choice:\\t\");
scanf(\"%d\", &choice);
switch(choice)
{
case 1: encrypt();
break;
case 2: decrypt();
break;
case 3: encrypt_view();
break;
case 4: decrypt_view();
break;
case 5: exit(1);
}
}
printf(\"\ \");
return 0;
}
int encrypt()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\");
if(fp1 == NULL)
{
printf(\"Source File Could Not Be Found\ \");
}
fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\");
if(fp2 == NULL)
{
printf(\"Target File Could Not Be Found\ \");
}
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
printf(\"\ End Of File\ \");
break;
}
else
{
ch = ch - (8 * 5 - 3);
fputc(ch, fp2);
}
}
fclose(fp1);
fclose(fp2);
printf(\"\ \");
return 0;
}
int decrypt()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\");
if(fp1 == NULL)
{
printf(\"Source File Could Not Be Found\ \");
}
fp2 = fopen(\"/home/tusharsoni/Desktop/Source\",\"w\");
if(fp2 == NULL)
{
printf(\"Target File Could Not Be Found\ \");
}
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
printf(\"\ End Of File\ \");
break;
}
else
{
ch = ch + (8 * 5 - 3);
fputc(ch, fp2);
}
}
fclose(fp1);
fclose(fp2);
printf(\"\ \");
return 0;
}
int encrypt_view()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\");
if(fp1 == NULL)
{
printf(\"No File Found\ \");
exit(1);
}
else
{
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
break;
}
else
{
printf(\"%c\", ch);
}
}
printf(\"\ \");
fclose(fp1);
}
printf(\"\ \");
return 0;
}
int decrypt_view()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\");
if(fp1 == NULL)
{
printf(\"No File Found\ \");
exit(1);
}
else
{
while(1)
{
ch = fgetc(fp1);
if(ch == EOF)
{
break;
}
else
{
printf(\"%c\", ch);
}
}
printf(\"\ \");
fclose(fp1);
}
return 0;
printf(\"\ \");
}
Solution
#include
#include
int encrypt(void);
int decrypt(void);
int encrypt_view(void);
int decrypt_view(void);
FILE *fp1, *fp2;
char ch;
int main()
{
int choice;
while(1)
{
printf(\"Select One of the Following:\ \");
printf(\"\ 1. Encrypt\ \");
printf(\"2. Decrypt\ \");
printf(\"3. View The Encypted File\ \");
printf(\"4. View The Decrypted File\ \");
printf(\"5. Exit\ \");
printf(\"\ Enter Your Choice:\\t\");
scanf(\"%d\", &choice);
switch(choice)
{
case 1: encrypt();
break;
case 2: decrypt();
break;
case 3: encrypt_view();
break;
case 4: decrypt_view();
break;
case 5: exit(1);
}
}
printf(\"\ \");
return 0;
}
int encrypt()
{
printf(\"\ \");
fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\");
if(fp1 == NULL)
{
printf(\"Source File Could Not Be Found\ \");
}
fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\");
if(fp2 == NULL)
{
printf(\"Ta.
Neuman System Model helps in studying the stressors, find out the de.pdf
Neuman System Model helps in studying the stressors, find out the defense, gives the feedback
and helps in preventing the occurrence through assessment, diagnosis, planning, implementation
and evaluation taking the person as a whole. Neuman System Model says that individual person
as whole is in interaction with environment. Neuman System Model helps in diagnosis and
prevents the stressors. Neuman System Model implements and improves the immunity. Neuman
System Model takes the help of social elements such as relatives and neighbors to protect the
person.
Solution
Neuman System Model helps in studying the stressors, find out the defense, gives the feedback
and helps in preventing the occurrence through assessment, diagnosis, planning, implementation
and evaluation taking the person as a whole. Neuman System Model says that individual person
as whole is in interaction with environment. Neuman System Model helps in diagnosis and
prevents the stressors. Neuman System Model implements and improves the immunity. Neuman
System Model takes the help of social elements such as relatives and neighbors to protect the
person..
NaOH will completely react will acetic acid and t.pdf
NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH
left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037
Solution
NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH
left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037.
the bonding between the carban and the nitrogen i.pdf
the bonding between the carban and the nitrogen in hydrogen cyanide or
hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the
molecule
Solution
the bonding between the carban and the nitrogen in hydrogen cyanide or
hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the
molecule.
The salts are more ionic, more soluble they are i.pdf
The salts are more ionic, more soluble they are in water since H2O is a polar
solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic
bonded.
Solution
The salts are more ionic, more soluble they are in water since H2O is a polar
solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic
bonded..
Look at the position of Pb in the periodic table .pdf
Look at the position of Pb in the periodic table and the postion of the Alkali Earth
metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged
cations. Lead however is on the right hand side, and would like to have more electrons to get into
that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in
nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now
since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more
easily than the same compounds with lead, since usually the lattice energy of lead compounds is
much higher (In case of insoluble compounds higher than the energy that would be released upon
solvation, which thus doesn\'t occur).
Solution
Look at the position of Pb in the periodic table and the postion of the Alkali Earth
metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged
cations. Lead however is on the right hand side, and would like to have more electrons to get into
that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in
nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now
since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more
easily than the same compounds with lead, since usually the lattice energy of lead compounds is
much higher (In case of insoluble compounds higher than the energy that would be released upon
solvation, which thus doesn\'t occur)..
Yes. A tree which has a root and nothing else.SolutionYe.pdf
A tree with only a root node and no other nodes is considered an empty tree. It has the minimum requirements to be defined as a tree with a single root element but no child or leaf nodes extending from that root. Such a degenerate tree structure would contain just one node with no branches.
First you will want to convert the molarity of ea.pdf
First you will want to convert the molarity of each to moles. To do this multiply the
molarity by the volume in liters. Next add the moles together and divide that by the total volume
of the solution. Molarity = moles/liters
Solution
First you will want to convert the molarity of each to moles. To do this multiply the
molarity by the volume in liters. Next add the moles together and divide that by the total volume
of the solution. Molarity = moles/liters.
Will be uploaded shortlySolutionWill be uploaded shortly.pdf
No summary can be provided as the document was empty and did not contain any content to summarize.
The process of sperm competition is defined as ‘the competition with.pdf
The process of sperm competition is defined as ‘the competition within a single female between
the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when
females mate at least two males, whose sperm must exhibit spatial and temporal overlap.
Successful sperm must be good at gaining dominance over those already in store and must be
able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection
via sperm competition are much better understood in invertebrates, especially in insects. They
often use copulatory plugs that are inserted immediately after a male copulates with a female
reducing the possibility of fertilization by subsequent copulations from another male.eg Indian
mealmoth (Plodia interpunctella), Bumblebee.
In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory
gland proteins), from their accessory glands to impede the female from participating in future
copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia),
crickets (Teleogryllus oceanicus).
There are certain offensive behaviours for suceeding followed by antlers.
an example of marine invertebrate showing gamete compatibility and sperm competition is
sympatric Asterias sea stars.
Solution
The process of sperm competition is defined as ‘the competition within a single female between
the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when
females mate at least two males, whose sperm must exhibit spatial and temporal overlap.
Successful sperm must be good at gaining dominance over those already in store and must be
able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection
via sperm competition are much better understood in invertebrates, especially in insects. They
often use copulatory plugs that are inserted immediately after a male copulates with a female
reducing the possibility of fertilization by subsequent copulations from another male.eg Indian
mealmoth (Plodia interpunctella), Bumblebee.
In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory
gland proteins), from their accessory glands to impede the female from participating in future
copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia),
crickets (Teleogryllus oceanicus).
There are certain offensive behaviours for suceeding followed by antlers.
an example of marine invertebrate showing gamete compatibility and sperm competition is
sympatric Asterias sea stars..
The answer isA. Public, PrivateThe public and private community .pdf
The answer is
A. Public, Private
The public and private community strings are used within SNMP to read and write respectively.
The option C and D are not correct because they do not exist as community strings.
Solution
The answer is
A. Public, Private
The public and private community strings are used within SNMP to read and write respectively.
The option C and D are not correct because they do not exist as community strings..
TCO is the assessment of all life time costs from owning ertain kind.pdf
TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three
major types of costs initial cost to obtain hardware/software second one is operational expenses
for the assets suh as installation and managment of assets third one is indirect cost that affects the
bussiness for instance the cost of system downtime from an outage.
virtualization can reduce tco in many ways.for example hardware virtualization can reduce the
cost associated with owing multiple physical servers due to ability to have them
virtualized.Virtualization enables multiple operating systems to run on the same physical
platform.
It is not unusual to achieve 10:1 virtual to physical machine consolidation. This means that ten
server applications can be run on a single machine that had required as many physical computers
to provide the unique operating system and technical specification environments in order to
operate. Server utilization is optimized and legacy software can maintain old OS configurations
while new applications are running in VMs with updated platforms.
Use of a VM enables rapid deployment by isolating the application in a known and controlled
environment. Unknown factors such as mixed libraries caused by numerous installs can be
eliminated. Severe crashes that required hours of reinstallation now take moments by simply
copying a virtual image.
As server workloads vary, virtualization provides the ability for virtual machines that are over
utilizing the resources of a server to be moved to underutilized servers. This dynamic load
balancing creates efficient utilization of server resources.
Disaster recovery is a critical component for IT, as system crashes can create huge economic
losses. Virtualization technology enables a virtual image on a machine to be instantly re-imaged
on another server if a machine failure occurs.
Multinational flexibility provides seamless transitions between different operating systems on a
single machine reducing desktop footprint and hardware expenditure.
Virtualization of systems helps prevent system crashes due to memory corruption caused by
software like device drivers. VT-d for Directed I/O Architecture provides methods to better
control system devices by defining the architecture for DMA and interrupt remapping to ensure
improved isolation of I/O resources for greater reliability, security, and availability.
Solution
TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three
major types of costs initial cost to obtain hardware/software second one is operational expenses
for the assets suh as installation and managment of assets third one is indirect cost that affects the
bussiness for instance the cost of system downtime from an outage.
virtualization can reduce tco in many ways.for example hardware virtualization can reduce the
cost associated with owing multiple physical servers due to ability to have them
virtualized.Virtualization enables multiple operat.
ROA = Net Income (2015) [Assets (2014) + Assets (2015)] 2= 220.pdf
ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2
= 220 / (500 + 400) / 2 = 48.89%
ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2
= 220 / (324 + 120) /2 = 99.10%
Solution
ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2
= 220 / (500 + 400) / 2 = 48.89%
ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2
= 220 / (324 + 120) /2 = 99.10%.
Reply if you have any doubts.public class Life { public stati.pdf
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells to nex gen states
this.cells = TempCells;
}
/* getNumberOfNeighbors is a stub; note that it\'s private. */
private int getNumberOfNeighbors(int row, int col)
{
int neighbourX[] = new int[8];
int neighbourY[] = new int[8];
neighbourX[0] = row - 1; neighbourY[0] = col - 1;
neighbourX[1] = row - 1; neighbourY[1] = col;
neighbourX[2] = row - 1; neighbourY[2] = col + 1;
neighbourX[3] = row; neighbourY[3] = col - 1;
neighbourX[4] = row; neighbourY[4] = col + 1;
neighbourX[5] = row + 1; neighbourY[5] = col - 1;
neighbourX[6] = row + 1; neighbourY[6] = col;
neighbourX[7] = row + 1; neighbourY[7] = col + 1;
for(int i=0;i<8;i++)
{
if(neighbourX[i] < 0)
neighbourX[i] = size - 1;
if(neighbourY[i] < 0)
neighbourY[i] = size - 1;
if(neighbourX[i] >= size)
neighbourX[i] = 0;
if(neighbourY[i] >= size)
neighbourY[i] = 0;
}
int count = 0;
for(int i=0;i<8;i++)
{
count += this.cells[neighbourX[i]][neighbourY[i]];
}
return count;
}
/* The main only instantiates the objects and sets them working. */
public static void main(String[] args) {
Life life = new Life(25);
CellWorld world = new CellWorld(life);
}
}
Solution
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells.
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.pdf
pH + pOH = 14
1. 14-12.2= 1.8pOH
pOH = -log(kb)
Kb = 10^-1.8
2. -log(6.3*10^-5) = pOH
pOH = 4.2
Solution
pH + pOH = 14
1. 14-12.2= 1.8pOH
pOH = -log(kb)
Kb = 10^-1.8
2. -log(6.3*10^-5) = pOH
pOH = 4.2.
Outear includes the pinna (also called auricle), the ear canal, and .pdf
Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer
of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for
hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum).
Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones
or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the
Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes
the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles
lead to an increase in the force applied to the stapes footplate then to the malleus.
Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ
labyrinth or vestibular apparatus.
Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which
presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of
Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of
the brain that is eighth cranial nerve.
In brief \"The outer ear, collestc osund waves. It then goes to the middle ear, whcih transimits the
signal to the ear durm. After that it goes throguh the inner ear. On the other side of your ear
drum, is the hammer, the anvil, and the stirrup. These rae the three smallest bones in your body.
These bones transimit the signal to the cochlea. The cohclea has nerve hairs that tune differtn
sounds. These nerve hairs, send the different souns to the brain, where the sound is interpreted\"
Solution
Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer
of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for
hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum).
Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones
or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the
Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes
the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles
lead to an increase in the force applied to the stapes footplate then to the malleus.
Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ
labyrinth or vestibular apparatus.
Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which
presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of
Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of
the brain that is eighth cranial nerve.
In brief \"The outer ear, collestc osund waves. It then goes to the m.
Lewis base is a donor of an electron pair (i.e., lone pair of ele.pdf
Lewis base:
is a donor of an electron pair (i.e., lone pair of electrons)
For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons.
Thus, all these three may behavior as Lewis bases.
Solution
Lewis base:
is a donor of an electron pair (i.e., lone pair of electrons)
For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons.
Thus, all these three may behavior as Lewis bases..
It will be B)- Planning ,directiing and controllingManagement is n.pdf
It will be B)- Planning ,directiing and controlling
Management is not directly involved in manufacturing and selling
Solution
It will be B)- Planning ,directiing and controlling
Management is not directly involved in manufacturing and selling.
Internet worm is a useless code which effect the code in the softwar.pdf
Internet worm is a useless code which effect the code in the software. It mainly infects the code
that are going to effect the code with some vulnerabilities, there have been several worms that
have given the practice of creating internet worms popularity. Efforts have been made to model
the behaviour of worms as they propagate. when the desighner knows about this worms it will be
very easy to take it off and run the program very good. The worm spread itself the own
vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid
the effectiveness of the internet worm. The user must be aware of the internet worms that are
cited because there are many tools that are available to find the vulnerabilities thar are occuring
through internet.
Solution
Internet worm is a useless code which effect the code in the software. It mainly infects the code
that are going to effect the code with some vulnerabilities, there have been several worms that
have given the practice of creating internet worms popularity. Efforts have been made to model
the behaviour of worms as they propagate. when the desighner knows about this worms it will be
very easy to take it off and run the program very good. The worm spread itself the own
vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid
the effectiveness of the internet worm. The user must be aware of the internet worms that are
cited because there are many tools that are available to find the vulnerabilities thar are occuring
through internet..
Here we have the compound V3(PO4)2 Separating it into its ions.pdf
Here we have the compound V3(PO4)2
Separating it into it\'s ions we have:
V2+ and PO43-
since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the
name of the compound is Vanadium (II) Phosphate.
Please don\'t forget to rate, good luck studying!
Solution
Here we have the compound V3(PO4)2
Separating it into it\'s ions we have:
V2+ and PO43-
since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the
name of the compound is Vanadium (II) Phosphate.
Please don\'t forget to rate, good luck studying!.
F+ cells have fertility factors while F- cells lack fertility factor.pdf
F+ cells have fertility factors while F- cells lack fertility factors
F+ is a donor cells while F- is recipient cells
F+ cells has circular plasmid separate from chromosome
F- cell does not have before receiving plasmid
Cell connect through T-pilus
Tra K and tra I were required for efficient inter plasmid recombination at ori T
whole DNA is transferred to recipient F-
Solution
F+ cells have fertility factors while F- cells lack fertility factors
F+ is a donor cells while F- is recipient cells
F+ cells has circular plasmid separate from chromosome
F- cell does not have before receiving plasmid
Cell connect through T-pilus
Tra K and tra I were required for efficient inter plasmid recombination at ori T
whole DNA is transferred to recipient F-.
D is correct.D. transport input ssh in vty Line Configuration mode.pdf
D is correct.
D. transport input ssh in vty Line Configuration mode
This is the correct command, and must be entered in Line Configuration mode.
Solution
D is correct.
D. transport input ssh in vty Line Configuration mode
This is the correct command, and must be entered in Line Configuration mode..
Dissociation of 2 protons = dissociation of 2 H+ ionsThe answer is.pdf
Dissociation of 2 protons = dissociation of 2 H+ ions
The answer is: C. H3PO4 --> 2H+ + HPO42-
Solution
Dissociation of 2 protons = dissociation of 2 H+ ions
The answer is: C. H3PO4 --> 2H+ + HPO42-.
CASE ANSWERS1.C) Bacteria typically transmitted by airborne dropl.pdf
CASE ANSWERS:
1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can
spread via the droplets expelled from a Sterptococcus pyogenes infected person.
2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the
white blood cells that are in the first line of defence as an innate immune response that comes to
site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the
streptococcus pyogenes.
3.The following statements about the antibody productions is TRUE :
B) Antibodies are produced by B cells with the help of T helper 2 cell.
Solution
CASE ANSWERS:
1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can
spread via the droplets expelled from a Sterptococcus pyogenes infected person.
2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the
white blood cells that are in the first line of defence as an innate immune response that comes to
site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the
streptococcus pyogenes.
3.The following statements about the antibody productions is TRUE :
B) Antibodies are produced by B cells with the help of T helper 2 cell..
Bacteria May have peptidoglycan cell wallHave ester linked lipid.pdf
Bacteria
May have peptidoglycan cell wall
Have ester linked lipids
Methanogens
Methane-producing archaea (methanogens) are strictly anaerobic
convert substrate to methane
Archae
Have RNA polymerase similar to eukaryotes
Have ether linked lipids in their cell membranes
Solution
Bacteria
May have peptidoglycan cell wall
Have ester linked lipids
Methanogens
Methane-producing archaea (methanogens) are strictly anaerobic
convert substrate to methane
Archae
Have RNA polymerase similar to eukaryotes
Have ether linked lipids in their cell membranes.
clinical examination of hip joint (1).pdf
described clinical examination all orthopeadic conditions .
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Look at the position of Pb in the periodic table .pdf
Look at the position of Pb in the periodic table and the postion of the Alkali Earth
metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged
cations. Lead however is on the right hand side, and would like to have more electrons to get into
that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in
nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now
since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more
easily than the same compounds with lead, since usually the lattice energy of lead compounds is
much higher (In case of insoluble compounds higher than the energy that would be released upon
solvation, which thus doesn\'t occur).
Solution
Look at the position of Pb in the periodic table and the postion of the Alkali Earth
metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged
cations. Lead however is on the right hand side, and would like to have more electrons to get into
that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in
nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now
since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more
easily than the same compounds with lead, since usually the lattice energy of lead compounds is
much higher (In case of insoluble compounds higher than the energy that would be released upon
solvation, which thus doesn\'t occur)..
Yes. A tree which has a root and nothing else.SolutionYe.pdf
A tree with only a root node and no other nodes is considered an empty tree. It has the minimum requirements to be defined as a tree with a single root element but no child or leaf nodes extending from that root. Such a degenerate tree structure would contain just one node with no branches.
First you will want to convert the molarity of ea.pdf
First you will want to convert the molarity of each to moles. To do this multiply the
molarity by the volume in liters. Next add the moles together and divide that by the total volume
of the solution. Molarity = moles/liters
Solution
First you will want to convert the molarity of each to moles. To do this multiply the
molarity by the volume in liters. Next add the moles together and divide that by the total volume
of the solution. Molarity = moles/liters.
Will be uploaded shortlySolutionWill be uploaded shortly.pdf
No summary can be provided as the document was empty and did not contain any content to summarize.
The process of sperm competition is defined as ‘the competition with.pdf
The process of sperm competition is defined as ‘the competition within a single female between
the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when
females mate at least two males, whose sperm must exhibit spatial and temporal overlap.
Successful sperm must be good at gaining dominance over those already in store and must be
able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection
via sperm competition are much better understood in invertebrates, especially in insects. They
often use copulatory plugs that are inserted immediately after a male copulates with a female
reducing the possibility of fertilization by subsequent copulations from another male.eg Indian
mealmoth (Plodia interpunctella), Bumblebee.
In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory
gland proteins), from their accessory glands to impede the female from participating in future
copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia),
crickets (Teleogryllus oceanicus).
There are certain offensive behaviours for suceeding followed by antlers.
an example of marine invertebrate showing gamete compatibility and sperm competition is
sympatric Asterias sea stars.
Solution
The process of sperm competition is defined as ‘the competition within a single female between
the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when
females mate at least two males, whose sperm must exhibit spatial and temporal overlap.
Successful sperm must be good at gaining dominance over those already in store and must be
able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection
via sperm competition are much better understood in invertebrates, especially in insects. They
often use copulatory plugs that are inserted immediately after a male copulates with a female
reducing the possibility of fertilization by subsequent copulations from another male.eg Indian
mealmoth (Plodia interpunctella), Bumblebee.
In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory
gland proteins), from their accessory glands to impede the female from participating in future
copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia),
crickets (Teleogryllus oceanicus).
There are certain offensive behaviours for suceeding followed by antlers.
an example of marine invertebrate showing gamete compatibility and sperm competition is
sympatric Asterias sea stars..
The answer isA. Public, PrivateThe public and private community .pdf
The answer is
A. Public, Private
The public and private community strings are used within SNMP to read and write respectively.
The option C and D are not correct because they do not exist as community strings.
Solution
The answer is
A. Public, Private
The public and private community strings are used within SNMP to read and write respectively.
The option C and D are not correct because they do not exist as community strings..
TCO is the assessment of all life time costs from owning ertain kind.pdf
TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three
major types of costs initial cost to obtain hardware/software second one is operational expenses
for the assets suh as installation and managment of assets third one is indirect cost that affects the
bussiness for instance the cost of system downtime from an outage.
virtualization can reduce tco in many ways.for example hardware virtualization can reduce the
cost associated with owing multiple physical servers due to ability to have them
virtualized.Virtualization enables multiple operating systems to run on the same physical
platform.
It is not unusual to achieve 10:1 virtual to physical machine consolidation. This means that ten
server applications can be run on a single machine that had required as many physical computers
to provide the unique operating system and technical specification environments in order to
operate. Server utilization is optimized and legacy software can maintain old OS configurations
while new applications are running in VMs with updated platforms.
Use of a VM enables rapid deployment by isolating the application in a known and controlled
environment. Unknown factors such as mixed libraries caused by numerous installs can be
eliminated. Severe crashes that required hours of reinstallation now take moments by simply
copying a virtual image.
As server workloads vary, virtualization provides the ability for virtual machines that are over
utilizing the resources of a server to be moved to underutilized servers. This dynamic load
balancing creates efficient utilization of server resources.
Disaster recovery is a critical component for IT, as system crashes can create huge economic
losses. Virtualization technology enables a virtual image on a machine to be instantly re-imaged
on another server if a machine failure occurs.
Multinational flexibility provides seamless transitions between different operating systems on a
single machine reducing desktop footprint and hardware expenditure.
Virtualization of systems helps prevent system crashes due to memory corruption caused by
software like device drivers. VT-d for Directed I/O Architecture provides methods to better
control system devices by defining the architecture for DMA and interrupt remapping to ensure
improved isolation of I/O resources for greater reliability, security, and availability.
Solution
TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three
major types of costs initial cost to obtain hardware/software second one is operational expenses
for the assets suh as installation and managment of assets third one is indirect cost that affects the
bussiness for instance the cost of system downtime from an outage.
virtualization can reduce tco in many ways.for example hardware virtualization can reduce the
cost associated with owing multiple physical servers due to ability to have them
virtualized.Virtualization enables multiple operat.
ROA = Net Income (2015) [Assets (2014) + Assets (2015)] 2= 220.pdf
ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2
= 220 / (500 + 400) / 2 = 48.89%
ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2
= 220 / (324 + 120) /2 = 99.10%
Solution
ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2
= 220 / (500 + 400) / 2 = 48.89%
ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2
= 220 / (324 + 120) /2 = 99.10%.
Reply if you have any doubts.public class Life { public stati.pdf
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells to nex gen states
this.cells = TempCells;
}
/* getNumberOfNeighbors is a stub; note that it\'s private. */
private int getNumberOfNeighbors(int row, int col)
{
int neighbourX[] = new int[8];
int neighbourY[] = new int[8];
neighbourX[0] = row - 1; neighbourY[0] = col - 1;
neighbourX[1] = row - 1; neighbourY[1] = col;
neighbourX[2] = row - 1; neighbourY[2] = col + 1;
neighbourX[3] = row; neighbourY[3] = col - 1;
neighbourX[4] = row; neighbourY[4] = col + 1;
neighbourX[5] = row + 1; neighbourY[5] = col - 1;
neighbourX[6] = row + 1; neighbourY[6] = col;
neighbourX[7] = row + 1; neighbourY[7] = col + 1;
for(int i=0;i<8;i++)
{
if(neighbourX[i] < 0)
neighbourX[i] = size - 1;
if(neighbourY[i] < 0)
neighbourY[i] = size - 1;
if(neighbourX[i] >= size)
neighbourX[i] = 0;
if(neighbourY[i] >= size)
neighbourY[i] = 0;
}
int count = 0;
for(int i=0;i<8;i++)
{
count += this.cells[neighbourX[i]][neighbourY[i]];
}
return count;
}
/* The main only instantiates the objects and sets them working. */
public static void main(String[] args) {
Life life = new Life(25);
CellWorld world = new CellWorld(life);
}
}
Solution
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells.
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.pdf
pH + pOH = 14
1. 14-12.2= 1.8pOH
pOH = -log(kb)
Kb = 10^-1.8
2. -log(6.3*10^-5) = pOH
pOH = 4.2
Solution
pH + pOH = 14
1. 14-12.2= 1.8pOH
pOH = -log(kb)
Kb = 10^-1.8
2. -log(6.3*10^-5) = pOH
pOH = 4.2.
Outear includes the pinna (also called auricle), the ear canal, and .pdf
Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer
of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for
hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum).
Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones
or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the
Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes
the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles
lead to an increase in the force applied to the stapes footplate then to the malleus.
Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ
labyrinth or vestibular apparatus.
Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which
presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of
Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of
the brain that is eighth cranial nerve.
In brief \"The outer ear, collestc osund waves. It then goes to the middle ear, whcih transimits the
signal to the ear durm. After that it goes throguh the inner ear. On the other side of your ear
drum, is the hammer, the anvil, and the stirrup. These rae the three smallest bones in your body.
These bones transimit the signal to the cochlea. The cohclea has nerve hairs that tune differtn
sounds. These nerve hairs, send the different souns to the brain, where the sound is interpreted\"
Solution
Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer
of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for
hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum).
Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones
or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the
Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes
the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles
lead to an increase in the force applied to the stapes footplate then to the malleus.
Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ
labyrinth or vestibular apparatus.
Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which
presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of
Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of
the brain that is eighth cranial nerve.
In brief \"The outer ear, collestc osund waves. It then goes to the m.
Lewis base is a donor of an electron pair (i.e., lone pair of ele.pdf
Lewis base:
is a donor of an electron pair (i.e., lone pair of electrons)
For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons.
Thus, all these three may behavior as Lewis bases.
Solution
Lewis base:
is a donor of an electron pair (i.e., lone pair of electrons)
For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons.
Thus, all these three may behavior as Lewis bases..
It will be B)- Planning ,directiing and controllingManagement is n.pdf
It will be B)- Planning ,directiing and controlling
Management is not directly involved in manufacturing and selling
Solution
It will be B)- Planning ,directiing and controlling
Management is not directly involved in manufacturing and selling.
Internet worm is a useless code which effect the code in the softwar.pdf
Internet worm is a useless code which effect the code in the software. It mainly infects the code
that are going to effect the code with some vulnerabilities, there have been several worms that
have given the practice of creating internet worms popularity. Efforts have been made to model
the behaviour of worms as they propagate. when the desighner knows about this worms it will be
very easy to take it off and run the program very good. The worm spread itself the own
vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid
the effectiveness of the internet worm. The user must be aware of the internet worms that are
cited because there are many tools that are available to find the vulnerabilities thar are occuring
through internet.
Solution
Internet worm is a useless code which effect the code in the software. It mainly infects the code
that are going to effect the code with some vulnerabilities, there have been several worms that
have given the practice of creating internet worms popularity. Efforts have been made to model
the behaviour of worms as they propagate. when the desighner knows about this worms it will be
very easy to take it off and run the program very good. The worm spread itself the own
vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid
the effectiveness of the internet worm. The user must be aware of the internet worms that are
cited because there are many tools that are available to find the vulnerabilities thar are occuring
through internet..
Here we have the compound V3(PO4)2 Separating it into its ions.pdf
Here we have the compound V3(PO4)2
Separating it into it\'s ions we have:
V2+ and PO43-
since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the
name of the compound is Vanadium (II) Phosphate.
Please don\'t forget to rate, good luck studying!
Solution
Here we have the compound V3(PO4)2
Separating it into it\'s ions we have:
V2+ and PO43-
since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the
name of the compound is Vanadium (II) Phosphate.
Please don\'t forget to rate, good luck studying!.
F+ cells have fertility factors while F- cells lack fertility factor.pdf
F+ cells have fertility factors while F- cells lack fertility factors
F+ is a donor cells while F- is recipient cells
F+ cells has circular plasmid separate from chromosome
F- cell does not have before receiving plasmid
Cell connect through T-pilus
Tra K and tra I were required for efficient inter plasmid recombination at ori T
whole DNA is transferred to recipient F-
Solution
F+ cells have fertility factors while F- cells lack fertility factors
F+ is a donor cells while F- is recipient cells
F+ cells has circular plasmid separate from chromosome
F- cell does not have before receiving plasmid
Cell connect through T-pilus
Tra K and tra I were required for efficient inter plasmid recombination at ori T
whole DNA is transferred to recipient F-.
D is correct.D. transport input ssh in vty Line Configuration mode.pdf
D is correct.
D. transport input ssh in vty Line Configuration mode
This is the correct command, and must be entered in Line Configuration mode.
Solution
D is correct.
D. transport input ssh in vty Line Configuration mode
This is the correct command, and must be entered in Line Configuration mode..
Dissociation of 2 protons = dissociation of 2 H+ ionsThe answer is.pdf
Dissociation of 2 protons = dissociation of 2 H+ ions
The answer is: C. H3PO4 --> 2H+ + HPO42-
Solution
Dissociation of 2 protons = dissociation of 2 H+ ions
The answer is: C. H3PO4 --> 2H+ + HPO42-.
CASE ANSWERS1.C) Bacteria typically transmitted by airborne dropl.pdf
CASE ANSWERS:
1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can
spread via the droplets expelled from a Sterptococcus pyogenes infected person.
2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the
white blood cells that are in the first line of defence as an innate immune response that comes to
site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the
streptococcus pyogenes.
3.The following statements about the antibody productions is TRUE :
B) Antibodies are produced by B cells with the help of T helper 2 cell.
Solution
CASE ANSWERS:
1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can
spread via the droplets expelled from a Sterptococcus pyogenes infected person.
2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the
white blood cells that are in the first line of defence as an innate immune response that comes to
site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the
streptococcus pyogenes.
3.The following statements about the antibody productions is TRUE :
B) Antibodies are produced by B cells with the help of T helper 2 cell..
Bacteria May have peptidoglycan cell wallHave ester linked lipid.pdf
Bacteria
May have peptidoglycan cell wall
Have ester linked lipids
Methanogens
Methane-producing archaea (methanogens) are strictly anaerobic
convert substrate to methane
Archae
Have RNA polymerase similar to eukaryotes
Have ether linked lipids in their cell membranes
Solution
Bacteria
May have peptidoglycan cell wall
Have ester linked lipids
Methanogens
Methane-producing archaea (methanogens) are strictly anaerobic
convert substrate to methane
Archae
Have RNA polymerase similar to eukaryotes
Have ether linked lipids in their cell membranes.
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