answer is N2 note; Nitrogen atoms are bigger an.pdf
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answer is: N2 note; Nitrogen atoms are bigger and so nitrogen molecules should be more soluble in water due to greater dipole-induced dipole attractions. Solution answer is: N2 note; Nitrogen atoms are bigger and so nitrogen molecules should be more soluble in water due to greater dipole-induced dipole attractions..
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#includestdio.h#includestdlib.h structure of a stack node .pdf
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
enQueue(q, 1);
enQueue(q, 2);
enQueue(q, 3);
/* Dequeue items */
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
getchar();
}
Solution
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct s.
Solution Since we know that the subshell with .pdf
Solution
: Since we know that the subshell with : n=2 and l=1 is the 2p subshell; Similarly : if n=3 and
l=0, it is the 3s subshell and so on that\'s why it is n=3, l=2 and m_l=0 Please rate.
let V be nonempty open subset of X then V inters.pdf
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X
Solution
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X.
Look at the periodic table. When Be loses its f.pdf
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be
Solution
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be.
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))....pdf
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14])
Solution
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14]).
the salt is neutral in charge. OS of Ca is +2, so.pdf
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2
Solution
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2.
nonmetals, metals, metalloids, and gases. They ar.pdf
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life
Solution
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life.
Recommended
#includestdio.h#includestdlib.h structure of a stack node .pdf
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
enQueue(q, 1);
enQueue(q, 2);
enQueue(q, 3);
/* Dequeue items */
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
getchar();
}
Solution
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct s.
Solution Since we know that the subshell with .pdf
Solution
: Since we know that the subshell with : n=2 and l=1 is the 2p subshell; Similarly : if n=3 and
l=0, it is the 3s subshell and so on that\'s why it is n=3, l=2 and m_l=0 Please rate.
let V be nonempty open subset of X then V inters.pdf
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X
Solution
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X.
Look at the periodic table. When Be loses its f.pdf
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be
Solution
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be.
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))....pdf
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14])
Solution
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14]).
the salt is neutral in charge. OS of Ca is +2, so.pdf
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2
Solution
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2.
nonmetals, metals, metalloids, and gases. They ar.pdf
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life
Solution
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life.
there are two types that are presented as ch3 and ch...Solution.pdf
there are two types that are presented as ch3 and ch...
Solution
there are two types that are presented as ch3 and ch....
In thermodynamics, a state function, function of .pdf
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni)
Solution
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni).
The process of protein synthesis and secretion starts from the endop.pdf
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis).
Solution
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis)..
The IUPAC names of the given compounds is 2-chloro-1-butanol.pdf
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol
Solution
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol.
Solution - (c) - 20001 SharesCalculation of Number of shares to be.pdf
Solution
- (c) - 20001 Shares
Calculation of Number of shares to be controlled to guarantee the election to be board.
= [1/(Number of Seats of Board of Directors + 1)*Shares Outstanding] + 1
= [1/(5+1)*120000 + 1]
= [(1/6)*120000] + 1
= 20001 Shares..
NOBecause here we are not working on past experience of data.S.pdf
NO
Because here we are not working on past experience of data.
Solution
NO
Because here we are not working on past experience of data..
Move decimal to behind the first number4.71Round4.714.71 X 1.pdf
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal)
Solution
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal).
Lewis acid is the electron pair acceptorLewis base is the electron.pdf
Lewis acid is the electron pair acceptor
Lewis base is the electron pair donar
a) OH- + Al(OH)3 -----> Al(OH)4-
Al(OH)3 is the lewis acid
OH- is the lewis base
b) SO3 + H2O --------> H2SO4
H2O is the lewis acid
SO3 is the lewis base
H2O is the lewis acid
SO3 is the lewis base
c) Co3+ + 6NH3 -----> Co(NH3)63+
Co 3+ is the lewis acid
NH3 is the lewis base
Co 3+ is the lewis acid
NH3 is the lewis base
Solution
Lewis acid is the electron pair acceptor
Lewis base is the electron pair donar
a) OH- + Al(OH)3 -----> Al(OH)4-
Al(OH)3 is the lewis acid
OH- is the lewis base
b) SO3 + H2O --------> H2SO4
H2O is the lewis acid
SO3 is the lewis base
H2O is the lewis acid
SO3 is the lewis base
c) Co3+ + 6NH3 -----> Co(NH3)63+
Co 3+ is the lewis acid
NH3 is the lewis base
Co 3+ is the lewis acid
NH3 is the lewis base.
first treat with alcoholic KOH to get cyclo hexene and then treat it.pdf
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol
Solution
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol.
An azeotrope is a mixture of two or more liquids .pdf
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4
Solution
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4.
Here we have sodium bisulfate NaHSO4, as you have shown the ions pr.pdf
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :)
Solution
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :).
Frontal lobe is the region of forebrain which comprises of various c.pdf
Frontal lobe is the region of forebrain which comprises of various centres for motor and concious
functions, including intelligence and decision making. Some specific areas are defined in the
frontal lobe according to their specifc functions. These are:
1. The motor strip: It is a part of the primary motor cortex of the frontal and prefrontal cortex of
the brain and maintains essential functions related to motion and movement. It is also known as
the precentral gyrus.
2. The precentral gyrus: This area structurally forms the posterior lobes of the frontal area cortex
and is important for the activity of motor neurons of limbs.
3. Broca\'s speech area: This region resides in a major part of the frontal lobe and is essential for
learning ability of languages and speech.
Importantly, Wernicke\'s speech area is an integral part of the temporal lobe of the brain near the
ears. It is not a part of the frontal lobe. Thus, the correct answers is choice A, WERNICKE\'S
SPEECH AREA.
Solution
Frontal lobe is the region of forebrain which comprises of various centres for motor and concious
functions, including intelligence and decision making. Some specific areas are defined in the
frontal lobe according to their specifc functions. These are:
1. The motor strip: It is a part of the primary motor cortex of the frontal and prefrontal cortex of
the brain and maintains essential functions related to motion and movement. It is also known as
the precentral gyrus.
2. The precentral gyrus: This area structurally forms the posterior lobes of the frontal area cortex
and is important for the activity of motor neurons of limbs.
3. Broca\'s speech area: This region resides in a major part of the frontal lobe and is essential for
learning ability of languages and speech.
Importantly, Wernicke\'s speech area is an integral part of the temporal lobe of the brain near the
ears. It is not a part of the frontal lobe. Thus, the correct answers is choice A, WERNICKE\'S
SPEECH AREA..
def deepReverse(L) List=[]; List.append(L); if(le.pdf
def deepReverse(L):
List=[];
List.append(L);
if(len(List)==1):
return List[0];
else:
temp=[];
for i in range(0,len(List)):
temp.append(List[i]);
return deepReverse(temp).append(List[0]);
Solution
def deepReverse(L):
List=[];
List.append(L);
if(len(List)==1):
return List[0];
else:
temp=[];
for i in range(0,len(List)):
temp.append(List[i]);
return deepReverse(temp).append(List[0]);.
Entropy can be transferred to or from a system in two forms heat tr.pdf
Entropy can be transferred to or from a system in two forms: heat transfer and mass flow (in
contrast, energy is transferred by work also). Entropy transfer is recognized at the system
boundary as entropy crosses the boundary, and it represents the entropy gained or lost by a
system during a process. The only form of entropy interaction associated with a fixed mass or
closed system is heat transfer, and thus the entropy transfer for an adiabatic closed system is
zero. Heat Transfer Heat is, in essence, a form of disorganized energy, and some disorganization
(entropy) will flow with heat. Heat transfer to a system increases the entropy of that system and
thus the level of molecular disorder or randomness, and heat 4 - 13 transfer from a system
decreases it. In fact, heat rejection is the only way the entropy of a fixed mass can be decreased.
The ratio of the heat transfer Q at a location to the absolute temperature T at that location is
called the entropy flow or entropy transfer, and is expressed as Entropy transfer with heat Sheat=
Q /T 4-14 The quantity Q/T represents the entropy transfer accompanied by heat transfer, and the
direction of entropy transfer is the same as the direction of heat transfer since absolute
temperature T is always a positive quantity. Therefore, the sign of entropy transfer is the same as
the sign of heat transfer positive if into the system, and negative if out of the system. • When two
systems are in contact, the entropy transfer from the warmer system is equal to the entropy
transfer into the cooler one at the point of contact. That is, no entropy can be created or destroyed
at the boundary since the boundary has no thickness and occupies no volume. • Note that work is
entropy-free, and no entropy is transferred with work. Energy is transferred with both heat and
work whereas entropy is transferred only with heat. • The first law of thermodynamics makes no
distinction between heat transfer and work; it considers them as equals. • The distinction between
heat transfer and work is brought out by the second law: an energy interaction which is
accompanied by entropy transfer is heat transfer, and an energy interaction which is not
accompanied by entropy transfer is work. That is, no entropy is exchanged during a work
interaction between a system and its surroundings. Thus only energy is exchanged during work
interaction whereas both energy and entropy are exchanged during heat transfer
Solution
Entropy can be transferred to or from a system in two forms: heat transfer and mass flow (in
contrast, energy is transferred by work also). Entropy transfer is recognized at the system
boundary as entropy crosses the boundary, and it represents the entropy gained or lost by a
system during a process. The only form of entropy interaction associated with a fixed mass or
closed system is heat transfer, and thus the entropy transfer for an adiabatic closed system is
zero. Heat Transfer Heat is, in essence, a form of disorganized e.
Cones in gymnosprems are homologous to gametophyte of seedless plant.pdf
Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens.
Solution
Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens..
c.12d.-3c.11d.11Solution.cpp Example program #include .pdf
c.12
d.-3
c.11
d.11
Solution
.cpp
// Example program
#include //header fr input output function
using namespace std;//it tells the compiler to link std namespace
int main()//main function
{
double x=2.5;
double y=-1.5;
int m=18;
int n=4;
int c;
c=5*x-1/5;
cout<.
Answer1)Responsive design is the idea where all the developed pag.pdf
Answer:
1)Responsive design is the idea where all the developed pages are embedded with model,view
and controller in the same page of the design so that it is easy for the user to transfer the data
from model to view and view to model.This also reduces the burden on the server when the user
makes any request all the form elements will not reach the server in responsive design but makes
the specific action element to reach the server and gives the output to the user.
2)Application templates in android involves both the layouts and UI components which are used
to build the design pages of the front end of the page.The common layouts which are used are
Gird Layout,Flow Layout,Relative Layout and UI components such as
listview,gridview,spinner,togglebuttons,radiobuttons,progressbar etc are used in any of the front
end design of the android for application to develop.We use XML for the design of the front end
in design in the application.
Android Application :
Note : Deploy the below files in Eclipse Id or Android Studio and run
MainActivity.java
package com.example.listviewcustom;
import java.io.File;
import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.Toast;
public class MainActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
/*
String path=\"/storage/sdcard0/sample_images/\";
File f=new File(path);
if(f.exists()){
String[] files=f.list();
ArrayAdapter adapter=new ArrayAdapter(getApplicationContext(),
android.R.layout.simple_spinner_dropdown_item,files );
*/
ListView lView=(ListView)findViewById(R.id.listView1);
lView.setAdapter(new MyAdapter(this));
/*
}else{
Toast.makeText(getApplicationContext(), \"Path is not available ....\",2000).show();
}
*/
}
public void reload(){
ListView lView=(ListView)findViewById(R.id.listView1);
lView.setAdapter(new MyAdapter(this));
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.activity_main, menu);
return true;
}
}
MyAdapter.java
package com.example.listviewcustom;
import java.io.File;
import android.net.Uri;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.webkit.WebView.FindListener;
import android.widget.BaseAdapter;
import android.widget.ImageView;
import android.widget.TextView;
public class MyAdapter extends BaseAdapter{
//String path=\"/storage/sdcard0/sample_images/\";
String path=\"/mnt/sdcard/sample_images/\";
File f=new File(path);
String[] files=f.list();
MainActivity activity;
public MyAdapter(MainActivity mActivity){
this.activity=mActivity;
}
@Override
public int getCount() {
// TODO Auto-generated method stub
return files.length;
}
@Override
public Object getItem(int position) {
// TODO Auto-.
Answer B is the correct answer. It is the only IP address shown in t.pdf
Answer B is the correct answer. It is the only IP address shown in the four options that is a
routable IP address. The address groups listed in answers A, C, and D are reserved for private
networks and are not routed by the Internet, so those answers are incorrect. You will need to
memorize the address groups reserved for private IP networks.
Solution
Answer B is the correct answer. It is the only IP address shown in the four options that is a
routable IP address. The address groups listed in answers A, C, and D are reserved for private
networks and are not routed by the Internet, so those answers are incorrect. You will need to
memorize the address groups reserved for private IP networks..
Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
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there are two types that are presented as ch3 and ch...Solution.pdf
there are two types that are presented as ch3 and ch...
Solution
there are two types that are presented as ch3 and ch....
In thermodynamics, a state function, function of .pdf
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni)
Solution
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni).
The process of protein synthesis and secretion starts from the endop.pdf
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis).
Solution
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis)..
The IUPAC names of the given compounds is 2-chloro-1-butanol.pdf
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol
Solution
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol.
Solution - (c) - 20001 SharesCalculation of Number of shares to be.pdf
Solution
- (c) - 20001 Shares
Calculation of Number of shares to be controlled to guarantee the election to be board.
= [1/(Number of Seats of Board of Directors + 1)*Shares Outstanding] + 1
= [1/(5+1)*120000 + 1]
= [(1/6)*120000] + 1
= 20001 Shares..
NOBecause here we are not working on past experience of data.S.pdf
NO
Because here we are not working on past experience of data.
Solution
NO
Because here we are not working on past experience of data..
Move decimal to behind the first number4.71Round4.714.71 X 1.pdf
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal)
Solution
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal).
Lewis acid is the electron pair acceptorLewis base is the electron.pdf
Lewis acid is the electron pair acceptor
Lewis base is the electron pair donar
a) OH- + Al(OH)3 -----> Al(OH)4-
Al(OH)3 is the lewis acid
OH- is the lewis base
b) SO3 + H2O --------> H2SO4
H2O is the lewis acid
SO3 is the lewis base
H2O is the lewis acid
SO3 is the lewis base
c) Co3+ + 6NH3 -----> Co(NH3)63+
Co 3+ is the lewis acid
NH3 is the lewis base
Co 3+ is the lewis acid
NH3 is the lewis base
Solution
Lewis acid is the electron pair acceptor
Lewis base is the electron pair donar
a) OH- + Al(OH)3 -----> Al(OH)4-
Al(OH)3 is the lewis acid
OH- is the lewis base
b) SO3 + H2O --------> H2SO4
H2O is the lewis acid
SO3 is the lewis base
H2O is the lewis acid
SO3 is the lewis base
c) Co3+ + 6NH3 -----> Co(NH3)63+
Co 3+ is the lewis acid
NH3 is the lewis base
Co 3+ is the lewis acid
NH3 is the lewis base.
first treat with alcoholic KOH to get cyclo hexene and then treat it.pdf
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol
Solution
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol.
An azeotrope is a mixture of two or more liquids .pdf
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4
Solution
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4.
Here we have sodium bisulfate NaHSO4, as you have shown the ions pr.pdf
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :)
Solution
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :).
Frontal lobe is the region of forebrain which comprises of various c.pdf
Frontal lobe is the region of forebrain which comprises of various centres for motor and concious
functions, including intelligence and decision making. Some specific areas are defined in the
frontal lobe according to their specifc functions. These are:
1. The motor strip: It is a part of the primary motor cortex of the frontal and prefrontal cortex of
the brain and maintains essential functions related to motion and movement. It is also known as
the precentral gyrus.
2. The precentral gyrus: This area structurally forms the posterior lobes of the frontal area cortex
and is important for the activity of motor neurons of limbs.
3. Broca\'s speech area: This region resides in a major part of the frontal lobe and is essential for
learning ability of languages and speech.
Importantly, Wernicke\'s speech area is an integral part of the temporal lobe of the brain near the
ears. It is not a part of the frontal lobe. Thus, the correct answers is choice A, WERNICKE\'S
SPEECH AREA.
Solution
Frontal lobe is the region of forebrain which comprises of various centres for motor and concious
functions, including intelligence and decision making. Some specific areas are defined in the
frontal lobe according to their specifc functions. These are:
1. The motor strip: It is a part of the primary motor cortex of the frontal and prefrontal cortex of
the brain and maintains essential functions related to motion and movement. It is also known as
the precentral gyrus.
2. The precentral gyrus: This area structurally forms the posterior lobes of the frontal area cortex
and is important for the activity of motor neurons of limbs.
3. Broca\'s speech area: This region resides in a major part of the frontal lobe and is essential for
learning ability of languages and speech.
Importantly, Wernicke\'s speech area is an integral part of the temporal lobe of the brain near the
ears. It is not a part of the frontal lobe. Thus, the correct answers is choice A, WERNICKE\'S
SPEECH AREA..
def deepReverse(L) List=[]; List.append(L); if(le.pdf
def deepReverse(L):
List=[];
List.append(L);
if(len(List)==1):
return List[0];
else:
temp=[];
for i in range(0,len(List)):
temp.append(List[i]);
return deepReverse(temp).append(List[0]);
Solution
def deepReverse(L):
List=[];
List.append(L);
if(len(List)==1):
return List[0];
else:
temp=[];
for i in range(0,len(List)):
temp.append(List[i]);
return deepReverse(temp).append(List[0]);.
Entropy can be transferred to or from a system in two forms heat tr.pdf
Entropy can be transferred to or from a system in two forms: heat transfer and mass flow (in
contrast, energy is transferred by work also). Entropy transfer is recognized at the system
boundary as entropy crosses the boundary, and it represents the entropy gained or lost by a
system during a process. The only form of entropy interaction associated with a fixed mass or
closed system is heat transfer, and thus the entropy transfer for an adiabatic closed system is
zero. Heat Transfer Heat is, in essence, a form of disorganized energy, and some disorganization
(entropy) will flow with heat. Heat transfer to a system increases the entropy of that system and
thus the level of molecular disorder or randomness, and heat 4 - 13 transfer from a system
decreases it. In fact, heat rejection is the only way the entropy of a fixed mass can be decreased.
The ratio of the heat transfer Q at a location to the absolute temperature T at that location is
called the entropy flow or entropy transfer, and is expressed as Entropy transfer with heat Sheat=
Q /T 4-14 The quantity Q/T represents the entropy transfer accompanied by heat transfer, and the
direction of entropy transfer is the same as the direction of heat transfer since absolute
temperature T is always a positive quantity. Therefore, the sign of entropy transfer is the same as
the sign of heat transfer positive if into the system, and negative if out of the system. • When two
systems are in contact, the entropy transfer from the warmer system is equal to the entropy
transfer into the cooler one at the point of contact. That is, no entropy can be created or destroyed
at the boundary since the boundary has no thickness and occupies no volume. • Note that work is
entropy-free, and no entropy is transferred with work. Energy is transferred with both heat and
work whereas entropy is transferred only with heat. • The first law of thermodynamics makes no
distinction between heat transfer and work; it considers them as equals. • The distinction between
heat transfer and work is brought out by the second law: an energy interaction which is
accompanied by entropy transfer is heat transfer, and an energy interaction which is not
accompanied by entropy transfer is work. That is, no entropy is exchanged during a work
interaction between a system and its surroundings. Thus only energy is exchanged during work
interaction whereas both energy and entropy are exchanged during heat transfer
Solution
Entropy can be transferred to or from a system in two forms: heat transfer and mass flow (in
contrast, energy is transferred by work also). Entropy transfer is recognized at the system
boundary as entropy crosses the boundary, and it represents the entropy gained or lost by a
system during a process. The only form of entropy interaction associated with a fixed mass or
closed system is heat transfer, and thus the entropy transfer for an adiabatic closed system is
zero. Heat Transfer Heat is, in essence, a form of disorganized e.
Cones in gymnosprems are homologous to gametophyte of seedless plant.pdf
Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens.
Solution
Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens..
c.12d.-3c.11d.11Solution.cpp Example program #include .pdf
c.12
d.-3
c.11
d.11
Solution
.cpp
// Example program
#include //header fr input output function
using namespace std;//it tells the compiler to link std namespace
int main()//main function
{
double x=2.5;
double y=-1.5;
int m=18;
int n=4;
int c;
c=5*x-1/5;
cout<.
Answer1)Responsive design is the idea where all the developed pag.pdf
Answer:
1)Responsive design is the idea where all the developed pages are embedded with model,view
and controller in the same page of the design so that it is easy for the user to transfer the data
from model to view and view to model.This also reduces the burden on the server when the user
makes any request all the form elements will not reach the server in responsive design but makes
the specific action element to reach the server and gives the output to the user.
2)Application templates in android involves both the layouts and UI components which are used
to build the design pages of the front end of the page.The common layouts which are used are
Gird Layout,Flow Layout,Relative Layout and UI components such as
listview,gridview,spinner,togglebuttons,radiobuttons,progressbar etc are used in any of the front
end design of the android for application to develop.We use XML for the design of the front end
in design in the application.
Android Application :
Note : Deploy the below files in Eclipse Id or Android Studio and run
MainActivity.java
package com.example.listviewcustom;
import java.io.File;
import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.Toast;
public class MainActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
/*
String path=\"/storage/sdcard0/sample_images/\";
File f=new File(path);
if(f.exists()){
String[] files=f.list();
ArrayAdapter adapter=new ArrayAdapter(getApplicationContext(),
android.R.layout.simple_spinner_dropdown_item,files );
*/
ListView lView=(ListView)findViewById(R.id.listView1);
lView.setAdapter(new MyAdapter(this));
/*
}else{
Toast.makeText(getApplicationContext(), \"Path is not available ....\",2000).show();
}
*/
}
public void reload(){
ListView lView=(ListView)findViewById(R.id.listView1);
lView.setAdapter(new MyAdapter(this));
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.activity_main, menu);
return true;
}
}
MyAdapter.java
package com.example.listviewcustom;
import java.io.File;
import android.net.Uri;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.webkit.WebView.FindListener;
import android.widget.BaseAdapter;
import android.widget.ImageView;
import android.widget.TextView;
public class MyAdapter extends BaseAdapter{
//String path=\"/storage/sdcard0/sample_images/\";
String path=\"/mnt/sdcard/sample_images/\";
File f=new File(path);
String[] files=f.list();
MainActivity activity;
public MyAdapter(MainActivity mActivity){
this.activity=mActivity;
}
@Override
public int getCount() {
// TODO Auto-generated method stub
return files.length;
}
@Override
public Object getItem(int position) {
// TODO Auto-.
Answer B is the correct answer. It is the only IP address shown in t.pdf
Answer B is the correct answer. It is the only IP address shown in the four options that is a
routable IP address. The address groups listed in answers A, C, and D are reserved for private
networks and are not routed by the Internet, so those answers are incorrect. You will need to
memorize the address groups reserved for private IP networks.
Solution
Answer B is the correct answer. It is the only IP address shown in the four options that is a
routable IP address. The address groups listed in answers A, C, and D are reserved for private
networks and are not routed by the Internet, so those answers are incorrect. You will need to
memorize the address groups reserved for private IP networks..
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