Pollinationis theprocessby whichpollenis transferred in the reproduc.pdf
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Pollinationis theprocessby whichpollenis transferred in the reproduction ofplants, thereby enablingfertilisationandsexual reproduction. flower can to self pollination by spreading the seeds through various media such as air water etc Solution Pollinationis theprocessby whichpollenis transferred in the reproduction ofplants, thereby enablingfertilisationandsexual reproduction. flower can to self pollination by spreading the seeds through various media such as air water etc.
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S1 = S0 is fluorescence. S1 = T1 is intersyste.pdf
S1 => S0 is fluorescence. S1 => T1 is intersystem crossing.
Solution
S1 => S0 is fluorescence. S1 => T1 is intersystem crossing..
let x be concentration of SO2 and O2 2x+x=6.4 x=2.pdf
let x be concentration of SO2 and O2 2x+x=6.4 x=2.133 concentration of
SO2=2.133M
Solution
let x be concentration of SO2 and O2 2x+x=6.4 x=2.133 concentration of
SO2=2.133M.
isotopes have same atomic number but different ma.pdf
isotopes have same atomic number but different mass number.
Solution
isotopes have same atomic number but different mass number..
i ) Lead orthophosphate ii) dinitrogen pentasulfi.pdf
i ) Lead orthophosphate ii) dinitrogen pentasulfide formulae i) Au(CN)3 ii) PF3
Solution
i ) Lead orthophosphate ii) dinitrogen pentasulfide formulae i) Au(CN)3 ii) PF3.
TThe function is increasing in the interval [1,5)The function is n.pdf
TThe function is increasing in the interval [1,5)
The function is neither decreasing nor constant. It is increasing in the interval [1,5)
Solution
TThe function is increasing in the interval [1,5)
The function is neither decreasing nor constant. It is increasing in the interval [1,5).
This is the code of shellsortvoid Shellsort( ElementType A[ ], int.pdf
This is the code of shellsort
void Shellsort( ElementType A[ ], int N )
{
int i, j, Increment; ElementType Tmp;
for( Increment = N/2; Increment > 0; Increment /= 2 )
for( i = Increment; i < N; i++ )
{
Tmp=A[i];
for(j=i;j>=Increment; j -= Increment )
if(Tmp
Solution
This is the code of shellsort
void Shellsort( ElementType A[ ], int N )
{
int i, j, Increment; ElementType Tmp;
for( Increment = N/2; Increment > 0; Increment /= 2 )
for( i = Increment; i < N; i++ )
{
Tmp=A[i];
for(j=i;j>=Increment; j -= Increment )
if(Tmp.
The balanced ionic Equation in basic medium is H2O + 2 CrO 42 --.pdf
The balanced ionic Equation in basic medium is :
H2O + 2 CrO 42 -- + 3 HSnO2- ----> 2 CrO2- + 2 OH- + 3 HSnO3-
The sum of the coefficients is = 1+2 + 3 + 2 + 2 + 3 = 13
Solution
The balanced ionic Equation in basic medium is :
H2O + 2 CrO 42 -- + 3 HSnO2- ----> 2 CrO2- + 2 OH- + 3 HSnO3-
The sum of the coefficients is = 1+2 + 3 + 2 + 2 + 3 = 13.
The acidic hydrogen attaches to the carbon bonded to the least alkyl.pdf
The acidic hydrogen attaches to the carbon bonded to the least alkyl substituents
(Markovnikov\'s Rule), resulting in a tertiary carbocation. The chloride ion then attacks the
electrophilic carbon, forming 1-chloro-1-methylcyclohexane.
Solution
The acidic hydrogen attaches to the carbon bonded to the least alkyl substituents
(Markovnikov\'s Rule), resulting in a tertiary carbocation. The chloride ion then attacks the
electrophilic carbon, forming 1-chloro-1-methylcyclohexane..
Recommended
S1 = S0 is fluorescence. S1 = T1 is intersyste.pdf
S1 => S0 is fluorescence. S1 => T1 is intersystem crossing.
Solution
S1 => S0 is fluorescence. S1 => T1 is intersystem crossing..
let x be concentration of SO2 and O2 2x+x=6.4 x=2.pdf
let x be concentration of SO2 and O2 2x+x=6.4 x=2.133 concentration of
SO2=2.133M
Solution
let x be concentration of SO2 and O2 2x+x=6.4 x=2.133 concentration of
SO2=2.133M.
isotopes have same atomic number but different ma.pdf
isotopes have same atomic number but different mass number.
Solution
isotopes have same atomic number but different mass number..
i ) Lead orthophosphate ii) dinitrogen pentasulfi.pdf
i ) Lead orthophosphate ii) dinitrogen pentasulfide formulae i) Au(CN)3 ii) PF3
Solution
i ) Lead orthophosphate ii) dinitrogen pentasulfide formulae i) Au(CN)3 ii) PF3.
TThe function is increasing in the interval [1,5)The function is n.pdf
TThe function is increasing in the interval [1,5)
The function is neither decreasing nor constant. It is increasing in the interval [1,5)
Solution
TThe function is increasing in the interval [1,5)
The function is neither decreasing nor constant. It is increasing in the interval [1,5).
This is the code of shellsortvoid Shellsort( ElementType A[ ], int.pdf
This is the code of shellsort
void Shellsort( ElementType A[ ], int N )
{
int i, j, Increment; ElementType Tmp;
for( Increment = N/2; Increment > 0; Increment /= 2 )
for( i = Increment; i < N; i++ )
{
Tmp=A[i];
for(j=i;j>=Increment; j -= Increment )
if(Tmp
Solution
This is the code of shellsort
void Shellsort( ElementType A[ ], int N )
{
int i, j, Increment; ElementType Tmp;
for( Increment = N/2; Increment > 0; Increment /= 2 )
for( i = Increment; i < N; i++ )
{
Tmp=A[i];
for(j=i;j>=Increment; j -= Increment )
if(Tmp.
The balanced ionic Equation in basic medium is H2O + 2 CrO 42 --.pdf
The balanced ionic Equation in basic medium is :
H2O + 2 CrO 42 -- + 3 HSnO2- ----> 2 CrO2- + 2 OH- + 3 HSnO3-
The sum of the coefficients is = 1+2 + 3 + 2 + 2 + 3 = 13
Solution
The balanced ionic Equation in basic medium is :
H2O + 2 CrO 42 -- + 3 HSnO2- ----> 2 CrO2- + 2 OH- + 3 HSnO3-
The sum of the coefficients is = 1+2 + 3 + 2 + 2 + 3 = 13.
The acidic hydrogen attaches to the carbon bonded to the least alkyl.pdf
The acidic hydrogen attaches to the carbon bonded to the least alkyl substituents
(Markovnikov\'s Rule), resulting in a tertiary carbocation. The chloride ion then attacks the
electrophilic carbon, forming 1-chloro-1-methylcyclohexane.
Solution
The acidic hydrogen attaches to the carbon bonded to the least alkyl substituents
(Markovnikov\'s Rule), resulting in a tertiary carbocation. The chloride ion then attacks the
electrophilic carbon, forming 1-chloro-1-methylcyclohexane..
Propanoic acid will react (in an acid-base reaction) with aqueous ba.pdf
Propanoic acid will react (in an acid-base reaction) with aqueous base to form the soluble
propanoate anion and water:
CH3CH2COOH(l) + OH-(aq) => CH3CH2COO-(aq) + H2O(l)
Solution
Propanoic acid will react (in an acid-base reaction) with aqueous base to form the soluble
propanoate anion and water:
CH3CH2COOH(l) + OH-(aq) => CH3CH2COO-(aq) + H2O(l).
A.B.(C)Journal Entries for the year 2014DateAccount title and ex.pdf
A.
B.
(C)Journal Entries for the year 2014DateAccount title and explanationDebitCredit1-
JanInvestment in 7% Bonds1300000 To Cash1300000(Being 10 Year 7% bond bought
directly from Javier)1-JulCash45500 To Interest income45500(being semi annual interest
received)31-DecAccrued Interest45500 To Interest income45500(Being semi annual
interest due)
Solution
A.
B.
(C)Journal Entries for the year 2014DateAccount title and explanationDebitCredit1-
JanInvestment in 7% Bonds1300000 To Cash1300000(Being 10 Year 7% bond bought
directly from Javier)1-JulCash45500 To Interest income45500(being semi annual interest
received)31-DecAccrued Interest45500 To Interest income45500(Being semi annual
interest due).
c) increases the melting point .pdf
This 1 sentence document states that adding a solute to a solvent increases the melting point of the solution.
G is Hamiltonian graph.Hence a path exists which visits each verte.pdf
G is Hamiltonian graph.
Hence a path exists which visits each vertex exactly once.
If possible let G has a cycle of length more than n/2
Recall the definition A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a
cycle that visits each vertex exactly once (except for the vertex that is both the start and end,
which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph
Consider the cycle path which has length more than n/2.
We know each edge connects two vertices. And as the graph is hamilton and we have taken a
cycle path, the path can visit each vertex exactly once except start point
Hence if n is odd, (n-1)/2 edges will connect all the vertices
Or if n is even n/2-1 edges will connect all the vertices.
Hence cycle can not have length more than n/2
Thus proved
Solution
G is Hamiltonian graph.
Hence a path exists which visits each vertex exactly once.
If possible let G has a cycle of length more than n/2
Recall the definition A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a
cycle that visits each vertex exactly once (except for the vertex that is both the start and end,
which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph
Consider the cycle path which has length more than n/2.
We know each edge connects two vertices. And as the graph is hamilton and we have taken a
cycle path, the path can visit each vertex exactly once except start point
Hence if n is odd, (n-1)/2 edges will connect all the vertices
Or if n is even n/2-1 edges will connect all the vertices.
Hence cycle can not have length more than n/2
Thus proved.
Enumerable A set is one whose members can be enumerated is called e.pdf
Enumerable: A set is one whose members can be enumerated is called enumerable, or countable.
There are arranged in a single list with a first entry, a second entry, and so on, so that every
member of
the set appears sooner or later on the list.
Examples: the set P of positive integers is enumerated by the list.
If a member of a list is appeared in the list, it is absolutely correct.
The only requirement is that member should atleast showed up once. We don\'t care if the list is
redundant. we only need is that list must be complete.
if the list is redundant, we can thin out it to get a irredundant list. If the list has gaps it is also
correct because one could go out and fill them. The only requiremnt is that every element in the
set that is enumerated be associated with some positive number, and not that every positive
member of the set associated with it.
Solution
Enumerable: A set is one whose members can be enumerated is called enumerable, or countable.
There are arranged in a single list with a first entry, a second entry, and so on, so that every
member of
the set appears sooner or later on the list.
Examples: the set P of positive integers is enumerated by the list.
If a member of a list is appeared in the list, it is absolutely correct.
The only requirement is that member should atleast showed up once. We don\'t care if the list is
redundant. we only need is that list must be complete.
if the list is redundant, we can thin out it to get a irredundant list. If the list has gaps it is also
correct because one could go out and fill them. The only requiremnt is that every element in the
set that is enumerated be associated with some positive number, and not that every positive
member of the set associated with it..
C.There is some evidence against H0, and a study using a larger samp.pdf
C.There is some evidence against H0, and a study using a larger sample size may be worthwhile.
Solution
C.There is some evidence against H0, and a study using a larger sample size may be worthwhile..
Bohr orbits have fixed sizes and fixed energiesSolutionBohr or.pdf
Bohr orbits have fixed sizes and fixed energies
Solution
Bohr orbits have fixed sizes and fixed energies.
c) At high altitudes the partial pressure of carbon dioxide as well .pdf
c) At high altitudes the partial pressure of carbon dioxide as well as oxygen is low. It increases
the body\'s ventilation in enabling the body to take up more oxygen and blowing off large
amounts of carbon dioxide. The plasma bicarbonate concentration (or we can say dissolved
carbon dioxide concentration) keeps on dropping and then reaches a stable lower position
d) This specific condition is called hypoxic pulmonary vasoconstriction in which in presence of
low oxygen levels, the respective alveoli detecting these levels get constricted so that the blood
in them gets squeezed out and gets re-distributed to other alveolar regions where the oxygen
concentration might be higher. This is an adaptive mechanism and it reduces the amount of blood
that goes through the lungs without being oxygenated.
c) Two major factors contribute to pulmonary edema which are hypoxic pulmonary
vasoconstriction and elevated pulmonary artery pressure. The pressure in the artery forces the
plasma/water into the constricted alveoli cells and causes edema. The chances for this increases
at higher altitudes due to persistent hypoxic conditions in the alveoli leading to their constriction.
3. a) Asthmatics have an increased smooth muscle tone during the attacks. The suffering patients
are able to breathe in easily but breathing out completely is not possible due to muscular spasms
in the lungs. This traps a lot of gas in the lungs and is also aided by mucuous production.
Resultingly the residual volume of the lungs gets increased as it is not expelled out. Another
change is the reduction in inspiratory volume in asthama patients as there is already some gas
present in the lungs.
Solution
c) At high altitudes the partial pressure of carbon dioxide as well as oxygen is low. It increases
the body\'s ventilation in enabling the body to take up more oxygen and blowing off large
amounts of carbon dioxide. The plasma bicarbonate concentration (or we can say dissolved
carbon dioxide concentration) keeps on dropping and then reaches a stable lower position
d) This specific condition is called hypoxic pulmonary vasoconstriction in which in presence of
low oxygen levels, the respective alveoli detecting these levels get constricted so that the blood
in them gets squeezed out and gets re-distributed to other alveolar regions where the oxygen
concentration might be higher. This is an adaptive mechanism and it reduces the amount of blood
that goes through the lungs without being oxygenated.
c) Two major factors contribute to pulmonary edema which are hypoxic pulmonary
vasoconstriction and elevated pulmonary artery pressure. The pressure in the artery forces the
plasma/water into the constricted alveoli cells and causes edema. The chances for this increases
at higher altitudes due to persistent hypoxic conditions in the alveoli leading to their constriction.
3. a) Asthmatics have an increased smooth muscle tone during the attacks. The suffering patients
are able to breathe in easily but.
AnswerFrogs have a pair of large eyes in the head which help in e.pdf
Answer:
Frogs have a pair of large eyes in the head which help in excellent night vision and depth
perception.
They not have outer ears instead it has a pair of eardrums called tympanum which is the hearing
organ.
frog uses his two nostrils to smell odours, while in water it uses a second type of olfactory organ
called Jacobson’s organ which aids in detecting chemicals in the water.
frogs produce sounds by a small sac in their throats that vibrates the air as they slowly let it out.
Solution
Answer:
Frogs have a pair of large eyes in the head which help in excellent night vision and depth
perception.
They not have outer ears instead it has a pair of eardrums called tympanum which is the hearing
organ.
frog uses his two nostrils to smell odours, while in water it uses a second type of olfactory organ
called Jacobson’s organ which aids in detecting chemicals in the water.
frogs produce sounds by a small sac in their throats that vibrates the air as they slowly let it out..
Answer1. Benefits of separating header files from the cpp files ar.pdf
Answer
1. Benefits of separating header files from the cpp files are:
a. reusability, as header files are most commonly used files included in the cpp files, they are
inclusively used for almost all the functions. To avoid mentioning them again and again in each
and every files in a project they are separated from all the cpp files and are clubed together in a
single header file which is used during the execution of the project and and is linked with the
other cpp files during the runtime.
b. compilation becomes faster, as linking will be needed only once unlike each time for each cpp
file
c. organized code, header file acts similar to book index and helps you to locate specific function
d. helps in separating interface from implementation
2.
It\'s not that a programmer should minimize the use of #include in header file, they need to be
careful while using #include in the header file and always use them within the #ifnotdefined and
#ifnotdefinedend block in order to avoid clash and error of multiple inclusion.
3.
include guard... guard against multiple incidental includes
it is needed to prevent multiple inclusion error
4.
const is the keyword for making any value/element constant.
Example:
const int a = 5;
const int *p;
etc.
Solution
Answer
1. Benefits of separating header files from the cpp files are:
a. reusability, as header files are most commonly used files included in the cpp files, they are
inclusively used for almost all the functions. To avoid mentioning them again and again in each
and every files in a project they are separated from all the cpp files and are clubed together in a
single header file which is used during the execution of the project and and is linked with the
other cpp files during the runtime.
b. compilation becomes faster, as linking will be needed only once unlike each time for each cpp
file
c. organized code, header file acts similar to book index and helps you to locate specific function
d. helps in separating interface from implementation
2.
It\'s not that a programmer should minimize the use of #include in header file, they need to be
careful while using #include in the header file and always use them within the #ifnotdefined and
#ifnotdefinedend block in order to avoid clash and error of multiple inclusion.
3.
include guard... guard against multiple incidental includes
it is needed to prevent multiple inclusion error
4.
const is the keyword for making any value/element constant.
Example:
const int a = 5;
const int *p;
etc..
a) can be anionic b) can be polydentate c) can be Lewis base d.pdf
a) can be anionic
b) can be polydentate
c) can be Lewis base
d) can be neutral
e)must contain a lone pair of e-
f)can be monodentate
g)can contain more than one donar atom
Solution
a) can be anionic
b) can be polydentate
c) can be Lewis base
d) can be neutral
e)must contain a lone pair of e-
f)can be monodentate
g)can contain more than one donar atom.
7Solution7.pdf
The document provides a single percentage, 7%, without any additional context or explanation. It also includes the word "Solution" but does not elaborate on what problem or issue the 7% refers to solving. The limited information does not allow for a multi-sentence summary.
A base must contain OH in its chemical formul.pdf
A base must contain \"OH\" in its chemical formula. A base would increase the
concentration of H3O+ when it is dissolved in water
Solution
A base must contain \"OH\" in its chemical formula. A base would increase the
concentration of H3O+ when it is dissolved in water.
(a)PreOrder, PostOrder and Inorder are three ways you can iterate .pdf
(a)
PreOrder, PostOrder and Inorder are three ways you can iterate a binary tree. The way these
three are different is in order of root,left child and right child.
Pre Order: root is printed first then lefy child and in last right child.
Given: TreeNode is class which have data part, left child and right child.
TreeNode root is root element of binary tree
Recursive algorithm of Preorder traversal
(A)void printPreorder(TreeNode root)
{
if (root == null) //Exit case if Tree is empty
return;
/* first print data of root because in preorder root is printed first */
System.out.println(root.value ); //recursive calling
/* then recur on left subtree of root*/
printPreorder(root.left); //recursive calling
/* now recur on right subtree of root*/
printPreorder(root.right); //recursive calling
}
Explanation: Here first we check If tree is exist or not. If exist we print root of tree and then
move to left part of current node and apply same process again untill left part is completed. After
that move to right sub tree and do same process again.
(b)
T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of root and n-k-1 on the other side.
Let’s do analysis of boundary conditions
Case 1:left subtree is empty then all the n-1 nodes present in right part only. SO k=0
T(n) = T(0) + T(n-1) + c -(1)
Putting value of T(n-1) in equation 1
T(n) = 2T(0) + T(n-2) + 2c -(2)
Similarly puting value of T(n-3) in equation 2
T(n) = 3T(0) + T(n-3) + 3c
Similarly
T(n) = 4T(0) + T(n-4) + 4c
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0)-T(0) + (n)c-c
Value of T(0) will be some constant say d.
T(n)= n(d+c)-(c+d)
T(n)=(n-1)(c+d)
T(n) = Theta of n
Case 2: Both left and right subtrees have equal number of nodes.
T(n) = 2T(n/2) + c
which clearly shows T(n)= theta(n)
(c)
void printPreorder(TreeNode root)
{
if (root == null) //Exit case if Tree is empty
return;
/*print left subtree of root*/
printPreorder(root.left); //recursive calling
/* print data of root because in Inorder root is printed in middle */
System.out.println(root.value ); //recursive calling
/* now recur on right subtree of root*/
printPreorder(root.right); //recursive calling
}
(d)
void printPreorder(TreeNode root)
{
if (root == null) //Exit case if Tree is empty
return;
/*print left subtree of root*/
printPreorder(root.left); //recursive calling
/* now recur on right subtree of root*/
printPreorder(root.right); //recursive calling
/* print data of root because in postorder root is printed in last */
System.out.println(root.value ); //recursive calling
}
Solution
(a)
PreOrder, PostOrder and Inorder are three ways you can iterate a binary tree. The way these
three are different is in order of root,left child and right child.
Pre Order: root is printed first then lefy child and in last right child.
Given: TreeNode is class which have data part, left child and right child.
TreeNode root is root element of binary tree
Recursive algorithm of Preorder traversal
(A)void printPreorder(TreeNode root)
{
if .
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...
BÀI TẬP BỔ TRỢ TIẾNG ANH 8 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2023-2024 (CÓ FI...Nguyen Thanh Tu Collection
https://app.box.com/s/y977uz6bpd3af4qsebv7r9b7s21935vdLAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
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Propanoic acid will react (in an acid-base reaction) with aqueous ba.pdf
Propanoic acid will react (in an acid-base reaction) with aqueous base to form the soluble
propanoate anion and water:
CH3CH2COOH(l) + OH-(aq) => CH3CH2COO-(aq) + H2O(l)
Solution
Propanoic acid will react (in an acid-base reaction) with aqueous base to form the soluble
propanoate anion and water:
CH3CH2COOH(l) + OH-(aq) => CH3CH2COO-(aq) + H2O(l).
A.B.(C)Journal Entries for the year 2014DateAccount title and ex.pdf
A.
B.
(C)Journal Entries for the year 2014DateAccount title and explanationDebitCredit1-
JanInvestment in 7% Bonds1300000 To Cash1300000(Being 10 Year 7% bond bought
directly from Javier)1-JulCash45500 To Interest income45500(being semi annual interest
received)31-DecAccrued Interest45500 To Interest income45500(Being semi annual
interest due)
Solution
A.
B.
(C)Journal Entries for the year 2014DateAccount title and explanationDebitCredit1-
JanInvestment in 7% Bonds1300000 To Cash1300000(Being 10 Year 7% bond bought
directly from Javier)1-JulCash45500 To Interest income45500(being semi annual interest
received)31-DecAccrued Interest45500 To Interest income45500(Being semi annual
interest due).
c) increases the melting point .pdf
This 1 sentence document states that adding a solute to a solvent increases the melting point of the solution.
G is Hamiltonian graph.Hence a path exists which visits each verte.pdf
G is Hamiltonian graph.
Hence a path exists which visits each vertex exactly once.
If possible let G has a cycle of length more than n/2
Recall the definition A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a
cycle that visits each vertex exactly once (except for the vertex that is both the start and end,
which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph
Consider the cycle path which has length more than n/2.
We know each edge connects two vertices. And as the graph is hamilton and we have taken a
cycle path, the path can visit each vertex exactly once except start point
Hence if n is odd, (n-1)/2 edges will connect all the vertices
Or if n is even n/2-1 edges will connect all the vertices.
Hence cycle can not have length more than n/2
Thus proved
Solution
G is Hamiltonian graph.
Hence a path exists which visits each vertex exactly once.
If possible let G has a cycle of length more than n/2
Recall the definition A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a
cycle that visits each vertex exactly once (except for the vertex that is both the start and end,
which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph
Consider the cycle path which has length more than n/2.
We know each edge connects two vertices. And as the graph is hamilton and we have taken a
cycle path, the path can visit each vertex exactly once except start point
Hence if n is odd, (n-1)/2 edges will connect all the vertices
Or if n is even n/2-1 edges will connect all the vertices.
Hence cycle can not have length more than n/2
Thus proved.
Enumerable A set is one whose members can be enumerated is called e.pdf
Enumerable: A set is one whose members can be enumerated is called enumerable, or countable.
There are arranged in a single list with a first entry, a second entry, and so on, so that every
member of
the set appears sooner or later on the list.
Examples: the set P of positive integers is enumerated by the list.
If a member of a list is appeared in the list, it is absolutely correct.
The only requirement is that member should atleast showed up once. We don\'t care if the list is
redundant. we only need is that list must be complete.
if the list is redundant, we can thin out it to get a irredundant list. If the list has gaps it is also
correct because one could go out and fill them. The only requiremnt is that every element in the
set that is enumerated be associated with some positive number, and not that every positive
member of the set associated with it.
Solution
Enumerable: A set is one whose members can be enumerated is called enumerable, or countable.
There are arranged in a single list with a first entry, a second entry, and so on, so that every
member of
the set appears sooner or later on the list.
Examples: the set P of positive integers is enumerated by the list.
If a member of a list is appeared in the list, it is absolutely correct.
The only requirement is that member should atleast showed up once. We don\'t care if the list is
redundant. we only need is that list must be complete.
if the list is redundant, we can thin out it to get a irredundant list. If the list has gaps it is also
correct because one could go out and fill them. The only requiremnt is that every element in the
set that is enumerated be associated with some positive number, and not that every positive
member of the set associated with it..
C.There is some evidence against H0, and a study using a larger samp.pdf
C.There is some evidence against H0, and a study using a larger sample size may be worthwhile.
Solution
C.There is some evidence against H0, and a study using a larger sample size may be worthwhile..
Bohr orbits have fixed sizes and fixed energiesSolutionBohr or.pdf
Bohr orbits have fixed sizes and fixed energies
Solution
Bohr orbits have fixed sizes and fixed energies.
c) At high altitudes the partial pressure of carbon dioxide as well .pdf
c) At high altitudes the partial pressure of carbon dioxide as well as oxygen is low. It increases
the body\'s ventilation in enabling the body to take up more oxygen and blowing off large
amounts of carbon dioxide. The plasma bicarbonate concentration (or we can say dissolved
carbon dioxide concentration) keeps on dropping and then reaches a stable lower position
d) This specific condition is called hypoxic pulmonary vasoconstriction in which in presence of
low oxygen levels, the respective alveoli detecting these levels get constricted so that the blood
in them gets squeezed out and gets re-distributed to other alveolar regions where the oxygen
concentration might be higher. This is an adaptive mechanism and it reduces the amount of blood
that goes through the lungs without being oxygenated.
c) Two major factors contribute to pulmonary edema which are hypoxic pulmonary
vasoconstriction and elevated pulmonary artery pressure. The pressure in the artery forces the
plasma/water into the constricted alveoli cells and causes edema. The chances for this increases
at higher altitudes due to persistent hypoxic conditions in the alveoli leading to their constriction.
3. a) Asthmatics have an increased smooth muscle tone during the attacks. The suffering patients
are able to breathe in easily but breathing out completely is not possible due to muscular spasms
in the lungs. This traps a lot of gas in the lungs and is also aided by mucuous production.
Resultingly the residual volume of the lungs gets increased as it is not expelled out. Another
change is the reduction in inspiratory volume in asthama patients as there is already some gas
present in the lungs.
Solution
c) At high altitudes the partial pressure of carbon dioxide as well as oxygen is low. It increases
the body\'s ventilation in enabling the body to take up more oxygen and blowing off large
amounts of carbon dioxide. The plasma bicarbonate concentration (or we can say dissolved
carbon dioxide concentration) keeps on dropping and then reaches a stable lower position
d) This specific condition is called hypoxic pulmonary vasoconstriction in which in presence of
low oxygen levels, the respective alveoli detecting these levels get constricted so that the blood
in them gets squeezed out and gets re-distributed to other alveolar regions where the oxygen
concentration might be higher. This is an adaptive mechanism and it reduces the amount of blood
that goes through the lungs without being oxygenated.
c) Two major factors contribute to pulmonary edema which are hypoxic pulmonary
vasoconstriction and elevated pulmonary artery pressure. The pressure in the artery forces the
plasma/water into the constricted alveoli cells and causes edema. The chances for this increases
at higher altitudes due to persistent hypoxic conditions in the alveoli leading to their constriction.
3. a) Asthmatics have an increased smooth muscle tone during the attacks. The suffering patients
are able to breathe in easily but.
AnswerFrogs have a pair of large eyes in the head which help in e.pdf
Answer:
Frogs have a pair of large eyes in the head which help in excellent night vision and depth
perception.
They not have outer ears instead it has a pair of eardrums called tympanum which is the hearing
organ.
frog uses his two nostrils to smell odours, while in water it uses a second type of olfactory organ
called Jacobson’s organ which aids in detecting chemicals in the water.
frogs produce sounds by a small sac in their throats that vibrates the air as they slowly let it out.
Solution
Answer:
Frogs have a pair of large eyes in the head which help in excellent night vision and depth
perception.
They not have outer ears instead it has a pair of eardrums called tympanum which is the hearing
organ.
frog uses his two nostrils to smell odours, while in water it uses a second type of olfactory organ
called Jacobson’s organ which aids in detecting chemicals in the water.
frogs produce sounds by a small sac in their throats that vibrates the air as they slowly let it out..
Answer1. Benefits of separating header files from the cpp files ar.pdf
Answer
1. Benefits of separating header files from the cpp files are:
a. reusability, as header files are most commonly used files included in the cpp files, they are
inclusively used for almost all the functions. To avoid mentioning them again and again in each
and every files in a project they are separated from all the cpp files and are clubed together in a
single header file which is used during the execution of the project and and is linked with the
other cpp files during the runtime.
b. compilation becomes faster, as linking will be needed only once unlike each time for each cpp
file
c. organized code, header file acts similar to book index and helps you to locate specific function
d. helps in separating interface from implementation
2.
It\'s not that a programmer should minimize the use of #include in header file, they need to be
careful while using #include in the header file and always use them within the #ifnotdefined and
#ifnotdefinedend block in order to avoid clash and error of multiple inclusion.
3.
include guard... guard against multiple incidental includes
it is needed to prevent multiple inclusion error
4.
const is the keyword for making any value/element constant.
Example:
const int a = 5;
const int *p;
etc.
Solution
Answer
1. Benefits of separating header files from the cpp files are:
a. reusability, as header files are most commonly used files included in the cpp files, they are
inclusively used for almost all the functions. To avoid mentioning them again and again in each
and every files in a project they are separated from all the cpp files and are clubed together in a
single header file which is used during the execution of the project and and is linked with the
other cpp files during the runtime.
b. compilation becomes faster, as linking will be needed only once unlike each time for each cpp
file
c. organized code, header file acts similar to book index and helps you to locate specific function
d. helps in separating interface from implementation
2.
It\'s not that a programmer should minimize the use of #include in header file, they need to be
careful while using #include in the header file and always use them within the #ifnotdefined and
#ifnotdefinedend block in order to avoid clash and error of multiple inclusion.
3.
include guard... guard against multiple incidental includes
it is needed to prevent multiple inclusion error
4.
const is the keyword for making any value/element constant.
Example:
const int a = 5;
const int *p;
etc..
a) can be anionic b) can be polydentate c) can be Lewis base d.pdf
a) can be anionic
b) can be polydentate
c) can be Lewis base
d) can be neutral
e)must contain a lone pair of e-
f)can be monodentate
g)can contain more than one donar atom
Solution
a) can be anionic
b) can be polydentate
c) can be Lewis base
d) can be neutral
e)must contain a lone pair of e-
f)can be monodentate
g)can contain more than one donar atom.
7Solution7.pdf
The document provides a single percentage, 7%, without any additional context or explanation. It also includes the word "Solution" but does not elaborate on what problem or issue the 7% refers to solving. The limited information does not allow for a multi-sentence summary.
A base must contain OH in its chemical formul.pdf
A base must contain \"OH\" in its chemical formula. A base would increase the
concentration of H3O+ when it is dissolved in water
Solution
A base must contain \"OH\" in its chemical formula. A base would increase the
concentration of H3O+ when it is dissolved in water.
(a)PreOrder, PostOrder and Inorder are three ways you can iterate .pdf
(a)
PreOrder, PostOrder and Inorder are three ways you can iterate a binary tree. The way these
three are different is in order of root,left child and right child.
Pre Order: root is printed first then lefy child and in last right child.
Given: TreeNode is class which have data part, left child and right child.
TreeNode root is root element of binary tree
Recursive algorithm of Preorder traversal
(A)void printPreorder(TreeNode root)
{
if (root == null) //Exit case if Tree is empty
return;
/* first print data of root because in preorder root is printed first */
System.out.println(root.value ); //recursive calling
/* then recur on left subtree of root*/
printPreorder(root.left); //recursive calling
/* now recur on right subtree of root*/
printPreorder(root.right); //recursive calling
}
Explanation: Here first we check If tree is exist or not. If exist we print root of tree and then
move to left part of current node and apply same process again untill left part is completed. After
that move to right sub tree and do same process again.
(b)
T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of root and n-k-1 on the other side.
Let’s do analysis of boundary conditions
Case 1:left subtree is empty then all the n-1 nodes present in right part only. SO k=0
T(n) = T(0) + T(n-1) + c -(1)
Putting value of T(n-1) in equation 1
T(n) = 2T(0) + T(n-2) + 2c -(2)
Similarly puting value of T(n-3) in equation 2
T(n) = 3T(0) + T(n-3) + 3c
Similarly
T(n) = 4T(0) + T(n-4) + 4c
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0)-T(0) + (n)c-c
Value of T(0) will be some constant say d.
T(n)= n(d+c)-(c+d)
T(n)=(n-1)(c+d)
T(n) = Theta of n
Case 2: Both left and right subtrees have equal number of nodes.
T(n) = 2T(n/2) + c
which clearly shows T(n)= theta(n)
(c)
void printPreorder(TreeNode root)
{
if (root == null) //Exit case if Tree is empty
return;
/*print left subtree of root*/
printPreorder(root.left); //recursive calling
/* print data of root because in Inorder root is printed in middle */
System.out.println(root.value ); //recursive calling
/* now recur on right subtree of root*/
printPreorder(root.right); //recursive calling
}
(d)
void printPreorder(TreeNode root)
{
if (root == null) //Exit case if Tree is empty
return;
/*print left subtree of root*/
printPreorder(root.left); //recursive calling
/* now recur on right subtree of root*/
printPreorder(root.right); //recursive calling
/* print data of root because in postorder root is printed in last */
System.out.println(root.value ); //recursive calling
}
Solution
(a)
PreOrder, PostOrder and Inorder are three ways you can iterate a binary tree. The way these
three are different is in order of root,left child and right child.
Pre Order: root is printed first then lefy child and in last right child.
Given: TreeNode is class which have data part, left child and right child.
TreeNode root is root element of binary tree
Recursive algorithm of Preorder traversal
(A)void printPreorder(TreeNode root)
{
if .
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