Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
•
0 likes•4 views
First substance = Ionic solid Second substance = Covalent solid Solution First substance = Ionic solid Second substance = Covalent solid.
Report
Share
Report
Share
Download to read offline
Recommended
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4
Solution
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
Computer graphics are pictures and movies produced use computers fre.pdf
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years
Solution
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate
Solution
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate.
C is correct. Only the host that the unicast message is addressed to.pdf
mol L-1 s-1 is the answer
Solution
mol L-1 s-1 is the answer.
According to the seriousness level, the following organs areThymu.pdf
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14
Solution
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
AdvantagesThe main objective of business combination is to elimina.pdf
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O
Solution
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
coordination no is the no of ligands in the coordination sphere and in this case it is 6
Solution
coordination no is the no of ligands in the coordination sphere and in this case it is 6.
A is correct. The packets to be filtered would be heading into the r.pdf
At the beginning of the 11th year, there would be 1,024,000 rabbits.
Solution
At the beginning of the 11th year, there would be 1,024,000 rabbits..
Recommended
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4
Solution
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
Computer graphics are pictures and movies produced use computers fre.pdf
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years
Solution
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate
Solution
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate.
C is correct. Only the host that the unicast message is addressed to.pdf
mol L-1 s-1 is the answer
Solution
mol L-1 s-1 is the answer.
According to the seriousness level, the following organs areThymu.pdf
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14
Solution
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
AdvantagesThe main objective of business combination is to elimina.pdf
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O
Solution
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
coordination no is the no of ligands in the coordination sphere and in this case it is 6
Solution
coordination no is the no of ligands in the coordination sphere and in this case it is 6.
A is correct. The packets to be filtered would be heading into the r.pdf
At the beginning of the 11th year, there would be 1,024,000 rabbits.
Solution
At the beginning of the 11th year, there would be 1,024,000 rabbits..
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body.
Solution
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body..
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the
biological evidence. This is useful where small amount of male DNA is found amidst a large
amount of female DNA , as in the sexual assault cases.It can also help in the missing person
investigation as it extends the range of potential reference samples. Sine the Y chromosome is
passed on from the fathers to their sone (unchanged, except in the case of some mutations),
therefore all the males of the same paternal lineage will possess common Y chromosome
haplotype.
A set of Short tandem repeats (STR) are found on the male specific Y chromosome.The coding
genes found on the short arm of the Ychromosome are vital to male sex determination,
spermatogenesis andother male related functions. The Y-STRs are polymorphic among the
nonrelated males.
Sometimes when sepoarating the mal ecomponent and the female component becomes difficult,
then the YSTRs help in the determination. Since The Y-STRs are not present in the females,
therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs
are present the male component is easily detected as only this part will be amplified in the PCR.
The YSTRs are also helpful when a number of males are involved. As regular STRs cause
masking effect the YSTR will help to identify all the males who have contributed to the
evidence.
ADVANTAGES-
1. extremely discriminating.
2. Extremely sensitive.e. low amounts of DNA can be used for determination.
3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev
been left can be used for DNA profiling.
4. Multiple testing systems ar available, which increases the likelyhood of successfully
generating DNA profiles.
5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling.
6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of
a particular individual can be profiled.
LIMITATIONS
1. Random match probability cannot be excluded.
2. Sometimes even when crime is not committed, we do get ,low levels of DNA on surfaces.
3.The important question while dealing with low DNA quantity is that whether the DNA that
was deposited was deposited during the crime or at some other time.
4.Searching evidence is challenging.
5.Sometimes finding the evidence is altogether impossibl.
6. In case of mtDNA /T STRs , as they are passed on from generation to generation either from
fathers to sons or mothers to daughters and son, therefore sometimes unrelated persons are also
share the same profile.
7. DNA mixtures are very complex.
8.Interpretation is required in many cases.
16- from Cyanobacterial to humans, many terristrial organisms have acquired the circadian
rhythms to adapt to sunlight to increase their survival rate.Circadian Rhythm is a biological clock
that displays an endogenous, entrainable oscillation of 24 hours.Thesse oscillations are driven by
circadian clocks.
PULSE CHASE EXPERIME.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=node;
return list;
}
else
{
link_node *prev=list;
while(temp!=NULL&&compare_node(node,temp)>=0)
{
prev=temp;
temp=temp->next;
}
prev->next=node;
node->next=temp;
return list;
}
}
//method definition for displaying the list
void display_list( struct link_node *head )
{
link_node *temp;
while(head)
{
printf(\"%s\ \",head->node_str);
temp=head;
head=head->next;
free(temp);
}
}
Sample output:
Enter the string:
Geneva
America
Solution
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=no.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74
Solution
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global
count def display_board( state ): #print \"-------------\" print \" %i %i %i \" %
(state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1],
state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5],
state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index =
new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1]
new_state[index - 1] = new_state[index] new_state[index] = temp return
new_state else: return None def move_down( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1]
new_state[index + 1] = new_state[index] new_state[index] = temp
return new_state else: return None def move_left( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index
- 3] new_state[index - 3] = new_state[index] new_state[index] = temp
return new_state else: return None def move_right( state ): new_state =
state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp =
new_state[index + 3] new_state[index + 3] = new_state[index]
new_state[index] = temp return new_state else: return None def
create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator,
depth, cost ) def expand_node( node, nodes ): expanded_nodes = []
expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) )
expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1,
0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth
+ 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\",
node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move
function returned None) expanded_nodes = [node for node in expanded_nodes if node.state
!= None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if
node not in nodes] #list comprehension! #expanded_nodes = [node for node in
expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension!
expanded_nodes1 = [] for node in expanded_nodes: temp=node
temp=temp.parent temp=temp.parent if temp==None:
expanded_nodes1.extend(expanded_nodes) break if node.state
!=temp.state: expanded_nodes1.append(node) return
expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue
with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while
True: # We\'ve run out of states, no solution. if len( nodes ) == 0:
return None # take the node from the front of the queue node = nodes.pop(0)
count=count+1 #display_board(node.state) # Append the move we made to
moves # if this node is the goal, return the moves it took to get here. if node.state
== goal: #display_board(node.state) moves = []
temp = node print cou.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is linear
Solution
7 steps are required as 9 is the 7th element and the search is linear.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present
Solution
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present.
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions.
Solution
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions..
Use Daltons Law of partial pressures. P(Total).pdf
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm
Solution
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm.
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear.
Solution
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear..
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4
Solution
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4.
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved.
Solution
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it.
Solution
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it..
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08
Solution
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08.
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g)
Solution
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
Solution!DOCTYPE html html head style type=textcss.pdf
Solution
IDENameCostOpen SourceDestinationNETBEANSFreeyescross platform IDEINTELLIJ
IDEA$199noit has great moderen UIECLIPSEfreeyesit features huge library of
pluginsJDEVELOPERfreeyesit has fully loades tool set for creating javaDR. JAVAFreeyesit
supports cross platformBLUEJFreeyesit suitable for small software projectsJCREATORfreeyesit
is fast and efficient ideJGRASPfreeyesit is super lightweight ideGREENFOOTfreeyesit supports
designing for 2D games CODEENVY$1(month)nocloud based ide.
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os.
Solution
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os..
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
More Related Content
More from ANJALIENTERPRISES1
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body.
Solution
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body..
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the
biological evidence. This is useful where small amount of male DNA is found amidst a large
amount of female DNA , as in the sexual assault cases.It can also help in the missing person
investigation as it extends the range of potential reference samples. Sine the Y chromosome is
passed on from the fathers to their sone (unchanged, except in the case of some mutations),
therefore all the males of the same paternal lineage will possess common Y chromosome
haplotype.
A set of Short tandem repeats (STR) are found on the male specific Y chromosome.The coding
genes found on the short arm of the Ychromosome are vital to male sex determination,
spermatogenesis andother male related functions. The Y-STRs are polymorphic among the
nonrelated males.
Sometimes when sepoarating the mal ecomponent and the female component becomes difficult,
then the YSTRs help in the determination. Since The Y-STRs are not present in the females,
therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs
are present the male component is easily detected as only this part will be amplified in the PCR.
The YSTRs are also helpful when a number of males are involved. As regular STRs cause
masking effect the YSTR will help to identify all the males who have contributed to the
evidence.
ADVANTAGES-
1. extremely discriminating.
2. Extremely sensitive.e. low amounts of DNA can be used for determination.
3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev
been left can be used for DNA profiling.
4. Multiple testing systems ar available, which increases the likelyhood of successfully
generating DNA profiles.
5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling.
6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of
a particular individual can be profiled.
LIMITATIONS
1. Random match probability cannot be excluded.
2. Sometimes even when crime is not committed, we do get ,low levels of DNA on surfaces.
3.The important question while dealing with low DNA quantity is that whether the DNA that
was deposited was deposited during the crime or at some other time.
4.Searching evidence is challenging.
5.Sometimes finding the evidence is altogether impossibl.
6. In case of mtDNA /T STRs , as they are passed on from generation to generation either from
fathers to sons or mothers to daughters and son, therefore sometimes unrelated persons are also
share the same profile.
7. DNA mixtures are very complex.
8.Interpretation is required in many cases.
16- from Cyanobacterial to humans, many terristrial organisms have acquired the circadian
rhythms to adapt to sunlight to increase their survival rate.Circadian Rhythm is a biological clock
that displays an endogenous, entrainable oscillation of 24 hours.Thesse oscillations are driven by
circadian clocks.
PULSE CHASE EXPERIME.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=node;
return list;
}
else
{
link_node *prev=list;
while(temp!=NULL&&compare_node(node,temp)>=0)
{
prev=temp;
temp=temp->next;
}
prev->next=node;
node->next=temp;
return list;
}
}
//method definition for displaying the list
void display_list( struct link_node *head )
{
link_node *temp;
while(head)
{
printf(\"%s\ \",head->node_str);
temp=head;
head=head->next;
free(temp);
}
}
Sample output:
Enter the string:
Geneva
America
Solution
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=no.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74
Solution
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global
count def display_board( state ): #print \"-------------\" print \" %i %i %i \" %
(state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1],
state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5],
state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index =
new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1]
new_state[index - 1] = new_state[index] new_state[index] = temp return
new_state else: return None def move_down( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1]
new_state[index + 1] = new_state[index] new_state[index] = temp
return new_state else: return None def move_left( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index
- 3] new_state[index - 3] = new_state[index] new_state[index] = temp
return new_state else: return None def move_right( state ): new_state =
state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp =
new_state[index + 3] new_state[index + 3] = new_state[index]
new_state[index] = temp return new_state else: return None def
create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator,
depth, cost ) def expand_node( node, nodes ): expanded_nodes = []
expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) )
expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1,
0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth
+ 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\",
node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move
function returned None) expanded_nodes = [node for node in expanded_nodes if node.state
!= None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if
node not in nodes] #list comprehension! #expanded_nodes = [node for node in
expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension!
expanded_nodes1 = [] for node in expanded_nodes: temp=node
temp=temp.parent temp=temp.parent if temp==None:
expanded_nodes1.extend(expanded_nodes) break if node.state
!=temp.state: expanded_nodes1.append(node) return
expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue
with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while
True: # We\'ve run out of states, no solution. if len( nodes ) == 0:
return None # take the node from the front of the queue node = nodes.pop(0)
count=count+1 #display_board(node.state) # Append the move we made to
moves # if this node is the goal, return the moves it took to get here. if node.state
== goal: #display_board(node.state) moves = []
temp = node print cou.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is linear
Solution
7 steps are required as 9 is the 7th element and the search is linear.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present
Solution
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present.
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions.
Solution
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions..
Use Daltons Law of partial pressures. P(Total).pdf
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm
Solution
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm.
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear.
Solution
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear..
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4
Solution
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4.
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved.
Solution
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it.
Solution
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it..
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08
Solution
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08.
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g)
Solution
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
Solution!DOCTYPE html html head style type=textcss.pdf
Solution
IDENameCostOpen SourceDestinationNETBEANSFreeyescross platform IDEINTELLIJ
IDEA$199noit has great moderen UIECLIPSEfreeyesit features huge library of
pluginsJDEVELOPERfreeyesit has fully loades tool set for creating javaDR. JAVAFreeyesit
supports cross platformBLUEJFreeyesit suitable for small software projectsJCREATORfreeyesit
is fast and efficient ideJGRASPfreeyesit is super lightweight ideGREENFOOTfreeyesit supports
designing for 2D games CODEENVY$1(month)nocloud based ide.
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os.
Solution
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os..
More from ANJALIENTERPRISES1 (20)
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as pH .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is line.pdf
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of mol.pdf
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occ.pdf
Use Daltons Law of partial pressures. P(Total).pdf
Use Daltons Law of partial pressures. P(Total).pdf
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an ex.pdf
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by mul.pdf
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at min.pdf
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to.pdf
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22the frequen.pdf
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
Solution!DOCTYPE html html head style type=textcss.pdf
Solution!DOCTYPE html html head style type=textcss.pdf
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
Recently uploaded
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...Nguyen Thanh Tu Collection
https://app.box.com/s/hqnndn05v4q5a4k4jd597rkdbda0fniiBest Digital Marketing Institute In NOIDA
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
Normal Labour/ Stages of Labour/ Mechanism of Labour
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Chapter -12, Antibiotics (One Page Notes).pdf
This is a notes for the D.Pharm students and related to the antibiotic drugs.
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
How to Make a Field invisible in Odoo 17
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH ANSWERS.
The Challenger.pdf DNHS Official Publication
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Recently uploaded (20)
The French Revolution Class 9 Study Material pdf free download
The French Revolution Class 9 Study Material pdf free download
Group Presentation 2 Economics.Ariana Buscigliopptx
Group Presentation 2 Economics.Ariana Buscigliopptx
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
Thesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.ppt
Normal Labour/ Stages of Labour/ Mechanism of Labour
Normal Labour/ Stages of Labour/ Mechanism of Labour
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
Unit 8 - Information and Communication Technology (Paper I).pdf
Unit 8 - Information and Communication Technology (Paper I).pdf
Multithreading_in_C++ - std::thread, race condition
Multithreading_in_C++ - std::thread, race condition
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
- 1. Among 40 subjects randomly choose 20 subjects and assign them Solution Among 40 subjects randomly choose 20 subjects and assign them