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The document discusses a solution but provides no details about the problem or type of solution. It contains a single word, "Solution", but no other information. There are no sentences to summarize from this very short document.
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0.0.176Solution0.0.176.pdf
The document provides a solution but does not contain any details about the problem or solution. It only includes the identifier "0.0.176" repeated three times with the word "Solution" in the middle.
#include iostream #include fstream #include stack #inclu.pdf
#include
#include
#include
#include
using namespace std;
int main()
{
ifstream infile;
char f[100];
cout << \"Enter the name of the file you wish to open.\" << endl;
cin >> f;
infile.open(f);
//cout << f << endl;
if(infile.fail())
{
cout << \"Cannot find file.\" << endl;
}
else
{
char c;
std::stack mystack;
int flag=0;
while(infile>>c)
{
if(c==\'(\'||c==\'[\'||c==\'{\')
{
mystack.push(c);
}
if(!mystack.empty()&& (c==\')\' || c==\']\'||c==\'}\'))
{
if(
(mystack.top()==\'(\'&&c==\')\'||mystack.top()==\'{\'&&c==\'}\'||mystack.top()==\'[\'&&c==\']\'))
{
mystack.pop();
}
else
{
flag==1;
break;
}
}
else
{
flag=1;
break;
}
}
if(mystack.empty()&&flag==0)
{
cout<<\"File data is balanced\";
}
else
{
cout<<\"File data is not balanced\";
}
}
}
Solution
#include
#include
#include
#include
using namespace std;
int main()
{
ifstream infile;
char f[100];
cout << \"Enter the name of the file you wish to open.\" << endl;
cin >> f;
infile.open(f);
//cout << f << endl;
if(infile.fail())
{
cout << \"Cannot find file.\" << endl;
}
else
{
char c;
std::stack mystack;
int flag=0;
while(infile>>c)
{
if(c==\'(\'||c==\'[\'||c==\'{\')
{
mystack.push(c);
}
if(!mystack.empty()&& (c==\')\' || c==\']\'||c==\'}\'))
{
if(
(mystack.top()==\'(\'&&c==\')\'||mystack.top()==\'{\'&&c==\'}\'||mystack.top()==\'[\'&&c==\']\'))
{
mystack.pop();
}
else
{
flag==1;
break;
}
}
else
{
flag=1;
break;
}
}
if(mystack.empty()&&flag==0)
{
cout<<\"File data is balanced\";
}
else
{
cout<<\"File data is not balanced\";
}
}
}.
Initial investment = $2,600,000+80,000+120,000+100,000= $2,900,00.pdf
Initial investment = $2,600,000+80,000+120,000+100,000= $2,900,000
Gross profit = sales - cost = 125,000* (40-30) = $1,250,000
Gross profit - net effect on introducing new wine = 1,250,000-20,000 = 1,230,000
Cash flow in year 5 = Profit after tax + depreciation + changes in working capital
= 877,500+ 210,000 + 100,000 = $1,187,500
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000)
(2,900,000)
1
1,230,000
(990,000)
240,000
60,000
180,000
1,170,000
0.909091
1,063,636.36
(1,836,363.64)
2
1,230,000
(1,350,000)
(120,000)
-
(120,000)
1,230,000
0.826446
1,016,528.93
(819,834.71)
3
1,230,000
(450,000)
780,000
195,000
585,000
1,035,000
0.751315
777,610.82
(42,223.89)
4
1,380,000
(210,000)
1,170,000
292,500
877,500
1,187,500
0.683013
811,078.48
768,854.59
NPV
768,855
IRR
22%
MIRR
17%
Payback period
3.05
Years
Yes, the project should be under taken because the NPV is positive and IRR and MIRR are more
than the cost of capital.
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000)
(2,900,000)
1
1,230,000
(990,000)
240,000
60,000
180,000
1,170,000
0.909091
1,063,636.36
(1,836,363.64)
2
1,230,000
(1,350,000)
(120,000)
-
(120,000)
1,230,000
0.826446
1,016,528.93
(819,834.71)
3
1,230,000
(450,000)
780,000
195,000
585,000
1,035,000
0.751315
777,610.82
(42,223.89)
4
1,380,000
(210,000)
1,170,000
292,500
877,500
1,187,500
0.683013
811,078.48
768,854.59
NPV
768,855
IRR
22%
MIRR
17%
Payback period
3.05
Years
Solution
Initial investment = $2,600,000+80,000+120,000+100,000= $2,900,000
Gross profit = sales - cost = 125,000* (40-30) = $1,250,000
Gross profit - net effect on introducing new wine = 1,250,000-20,000 = 1,230,000
Cash flow in year 5 = Profit after tax + depreciation + changes in working capital
= 877,500+ 210,000 + 100,000 = $1,187,500
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000)
(2,900,000)
1
1,230,000
(990,000)
240,000
60,000
180,000
1,170,000
0.909091
1,063,636.36
(1,836,363.64)
2
1,230,000
(1,350,000)
(120,000)
-
(120,000)
1,230,000
0.826446
1,016,528.93
(819,834.71)
3
1,230,000
(450,000)
780,000
195,000
585,000
1,035,000
0.751315
777,610.82
(42,223.89)
4
1,380,000
(210,000)
1,170,000
292,500
877,500
1,187,500
0.683013
811,078.48
768,854.59
NPV
768,855
IRR
22%
MIRR
17%
Payback period
3.05
Years
Yes, the project should be under taken because the NPV is positive and IRR and MIRR are more
than the cost of capital.
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000.
the concentration increases .pdf
The document discusses a solution where the concentration increases. No other details are provided about what is being concentrated or the nature of the solution. The document is very brief and does not contain much contextual information that could be summarized.
It is an anhydride its struture like maleic anh.pdf
It is an anhydride it\'s struture like maleic anhydride the only difference is the
carbon-carbon bond in the ring is a single bond
Solution
It is an anhydride it\'s struture like maleic anhydride the only difference is the
carbon-carbon bond in the ring is a single bond.
In most of the invertebrates, these structures ar.pdf
In most of the invertebrates, these structures are simple tubular forms whereas
vertebrates have complex tubular organs called kidneys. Some of these structures are mentioned
here. Protonephridia or flame cells are the excretory structures in Platyhelminthes (Flatworms,
e.g., Planaria), rotifers, some annelids and the cephalochordate
Solution
In most of the invertebrates, these structures are simple tubular forms whereas
vertebrates have complex tubular organs called kidneys. Some of these structures are mentioned
here. Protonephridia or flame cells are the excretory structures in Platyhelminthes (Flatworms,
e.g., Planaria), rotifers, some annelids and the cephalochordate.
difficult to understand. I tried but your questio.pdf
difficult to understand. I tried but your question cannnot be understood. Please give
lifesaver/helpful
Solution
difficult to understand. I tried but your question cannnot be understood. Please give
lifesaver/helpful.
Cl- has the larger ionization energy since it onl.pdf
Cl- has the larger ionization energy since it only has one extra electron instead of
two. S will gladly give up one of its two extra electrons, so the energy will be lower.
Solution
Cl- has the larger ionization energy since it only has one extra electron instead of
two. S will gladly give up one of its two extra electrons, so the energy will be lower..
Recommended
0.0.176Solution0.0.176.pdf
The document provides a solution but does not contain any details about the problem or solution. It only includes the identifier "0.0.176" repeated three times with the word "Solution" in the middle.
#include iostream #include fstream #include stack #inclu.pdf
#include
#include
#include
#include
using namespace std;
int main()
{
ifstream infile;
char f[100];
cout << \"Enter the name of the file you wish to open.\" << endl;
cin >> f;
infile.open(f);
//cout << f << endl;
if(infile.fail())
{
cout << \"Cannot find file.\" << endl;
}
else
{
char c;
std::stack mystack;
int flag=0;
while(infile>>c)
{
if(c==\'(\'||c==\'[\'||c==\'{\')
{
mystack.push(c);
}
if(!mystack.empty()&& (c==\')\' || c==\']\'||c==\'}\'))
{
if(
(mystack.top()==\'(\'&&c==\')\'||mystack.top()==\'{\'&&c==\'}\'||mystack.top()==\'[\'&&c==\']\'))
{
mystack.pop();
}
else
{
flag==1;
break;
}
}
else
{
flag=1;
break;
}
}
if(mystack.empty()&&flag==0)
{
cout<<\"File data is balanced\";
}
else
{
cout<<\"File data is not balanced\";
}
}
}
Solution
#include
#include
#include
#include
using namespace std;
int main()
{
ifstream infile;
char f[100];
cout << \"Enter the name of the file you wish to open.\" << endl;
cin >> f;
infile.open(f);
//cout << f << endl;
if(infile.fail())
{
cout << \"Cannot find file.\" << endl;
}
else
{
char c;
std::stack mystack;
int flag=0;
while(infile>>c)
{
if(c==\'(\'||c==\'[\'||c==\'{\')
{
mystack.push(c);
}
if(!mystack.empty()&& (c==\')\' || c==\']\'||c==\'}\'))
{
if(
(mystack.top()==\'(\'&&c==\')\'||mystack.top()==\'{\'&&c==\'}\'||mystack.top()==\'[\'&&c==\']\'))
{
mystack.pop();
}
else
{
flag==1;
break;
}
}
else
{
flag=1;
break;
}
}
if(mystack.empty()&&flag==0)
{
cout<<\"File data is balanced\";
}
else
{
cout<<\"File data is not balanced\";
}
}
}.
Initial investment = $2,600,000+80,000+120,000+100,000= $2,900,00.pdf
Initial investment = $2,600,000+80,000+120,000+100,000= $2,900,000
Gross profit = sales - cost = 125,000* (40-30) = $1,250,000
Gross profit - net effect on introducing new wine = 1,250,000-20,000 = 1,230,000
Cash flow in year 5 = Profit after tax + depreciation + changes in working capital
= 877,500+ 210,000 + 100,000 = $1,187,500
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000)
(2,900,000)
1
1,230,000
(990,000)
240,000
60,000
180,000
1,170,000
0.909091
1,063,636.36
(1,836,363.64)
2
1,230,000
(1,350,000)
(120,000)
-
(120,000)
1,230,000
0.826446
1,016,528.93
(819,834.71)
3
1,230,000
(450,000)
780,000
195,000
585,000
1,035,000
0.751315
777,610.82
(42,223.89)
4
1,380,000
(210,000)
1,170,000
292,500
877,500
1,187,500
0.683013
811,078.48
768,854.59
NPV
768,855
IRR
22%
MIRR
17%
Payback period
3.05
Years
Yes, the project should be under taken because the NPV is positive and IRR and MIRR are more
than the cost of capital.
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000)
(2,900,000)
1
1,230,000
(990,000)
240,000
60,000
180,000
1,170,000
0.909091
1,063,636.36
(1,836,363.64)
2
1,230,000
(1,350,000)
(120,000)
-
(120,000)
1,230,000
0.826446
1,016,528.93
(819,834.71)
3
1,230,000
(450,000)
780,000
195,000
585,000
1,035,000
0.751315
777,610.82
(42,223.89)
4
1,380,000
(210,000)
1,170,000
292,500
877,500
1,187,500
0.683013
811,078.48
768,854.59
NPV
768,855
IRR
22%
MIRR
17%
Payback period
3.05
Years
Solution
Initial investment = $2,600,000+80,000+120,000+100,000= $2,900,000
Gross profit = sales - cost = 125,000* (40-30) = $1,250,000
Gross profit - net effect on introducing new wine = 1,250,000-20,000 = 1,230,000
Cash flow in year 5 = Profit after tax + depreciation + changes in working capital
= 877,500+ 210,000 + 100,000 = $1,187,500
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000)
(2,900,000)
1
1,230,000
(990,000)
240,000
60,000
180,000
1,170,000
0.909091
1,063,636.36
(1,836,363.64)
2
1,230,000
(1,350,000)
(120,000)
-
(120,000)
1,230,000
0.826446
1,016,528.93
(819,834.71)
3
1,230,000
(450,000)
780,000
195,000
585,000
1,035,000
0.751315
777,610.82
(42,223.89)
4
1,380,000
(210,000)
1,170,000
292,500
877,500
1,187,500
0.683013
811,078.48
768,854.59
NPV
768,855
IRR
22%
MIRR
17%
Payback period
3.05
Years
Yes, the project should be under taken because the NPV is positive and IRR and MIRR are more
than the cost of capital.
Year
Cash flows
Depreciation
Profit after depreciation
Tax at 25%
Profit after tax
Cash flows
Discounting factor at 10%
PV of cash flows
cumulative cash flows
0
(2,900,000)
-
(2,900,000)
1
(2,900,000.
the concentration increases .pdf
The document discusses a solution where the concentration increases. No other details are provided about what is being concentrated or the nature of the solution. The document is very brief and does not contain much contextual information that could be summarized.
It is an anhydride its struture like maleic anh.pdf
It is an anhydride it\'s struture like maleic anhydride the only difference is the
carbon-carbon bond in the ring is a single bond
Solution
It is an anhydride it\'s struture like maleic anhydride the only difference is the
carbon-carbon bond in the ring is a single bond.
In most of the invertebrates, these structures ar.pdf
In most of the invertebrates, these structures are simple tubular forms whereas
vertebrates have complex tubular organs called kidneys. Some of these structures are mentioned
here. Protonephridia or flame cells are the excretory structures in Platyhelminthes (Flatworms,
e.g., Planaria), rotifers, some annelids and the cephalochordate
Solution
In most of the invertebrates, these structures are simple tubular forms whereas
vertebrates have complex tubular organs called kidneys. Some of these structures are mentioned
here. Protonephridia or flame cells are the excretory structures in Platyhelminthes (Flatworms,
e.g., Planaria), rotifers, some annelids and the cephalochordate.
difficult to understand. I tried but your questio.pdf
difficult to understand. I tried but your question cannnot be understood. Please give
lifesaver/helpful
Solution
difficult to understand. I tried but your question cannnot be understood. Please give
lifesaver/helpful.
Cl- has the larger ionization energy since it onl.pdf
Cl- has the larger ionization energy since it only has one extra electron instead of
two. S will gladly give up one of its two extra electrons, so the energy will be lower.
Solution
Cl- has the larger ionization energy since it only has one extra electron instead of
two. S will gladly give up one of its two extra electrons, so the energy will be lower..
Strategic planning is a method for composing a plan which relates th.pdf
Strategic planning is a method for composing a plan which relates the overall vision of a
company or organization down to the individual programs and activities necessary to accomplish
it. Strategic planning is best defined as looking at where you want to go in the future and putting
together the resources, assets, and personnel to get there. What differentiates strategic from other
forms of planning is the focus on a broader goal for the future, a vision of what place in the
market the company wishes to pursue. The overall purpose of strategic planning is “to determine
where your market is going and find a way to continue servicing it (Your 2007 Strategic Plan
2006).” To accomplish this, a strategic plan relates an organization’s vision, its overall goals,
down to the individual programs, activities, and technologies which will attempt to accomplish
the vision. Once these programs are identified, a strategic plan will encompass an overall
business plan, or roadmap, which provides a structured timeline for accomplishing these
programs as well as the organizational vision.Regardless of which level of the organization is
performing the planning there are certain elements which it must possess in order to be
successful.These elements which include the following: long range direction, a framework for
linking decisions, being a living document, and being organizationally specific.
Long Range Direction :- The most important aspect of strategic planning is the focus on the long
term vision of the organization. Vision is the organization’s view of where it will be in future in
terms of its market, size, and accomplishments. Setting goals and milestones to be reached
within a certain timeframe will keep the organization focused on what it wants to accomplish.
This long range planning is essential to strategic planning because it sets a target. As Molvig
says, “long range planning looks at where you want to be in five years, strategic planning focuses
on how to position yourself to get there.” (Molvig 1993) In addition to analyzing where the
company wishes to be in the future, strategic planning involves determining what outside forces
may impact the vision. The actions of competitors, technological breakthroughs on the horizon,
and threats from changes in the world environment are just a few of the external forces which
should be considered in a good plan.
Framework for Linking Decisions:- Decisions are made at every level of an organization. From
the CEO down to the individual program managers resources are allocated, budgets developed
and schedules planned and carried out. Minztberg states, “The assumption of strategic planning
seems to be that objectives are decided upon the top management for the entire organization,
which in turn evoke the process of formulating strategy.” (Mintzberg, The Fall and Rise of
Strategic Planning 1994) Many times, each tier has complete autonomy and little direction
provided to link the decisions into a serious, well-c.
HAsO4- as it can accept an electron to form HAsO4.pdf
HAsO4- as it can accept an electron to form HAsO42-
Solution
HAsO4- as it can accept an electron to form HAsO42-.
Solution.cpp #include iostreamheader file for input output f.pdf
Solution
.cpp
#include //header file for input output function
using namespace std;//it tells the compiler to link std namespace
int main()
{//main function
int num,rows;//variable declaration
cout<<\"How many numbers to output?\"<>num;//key board inputting
cout<<\"how many rows? \"<>rows;//key board inputting
for(int i=1;i<=rows;i++)
{//nested for loop logic
for(int j=i;j<=num;j=j+14)
{
cout<.
pH of the solution = 4.5 -log [H+] = 4.5 .pdf
pH of the solution = 4.5
-log [H+] = 4.5
log[H+] = -4.5
[H+] = 10-4.5
= 3.16*10-5
[H+ ][OH- ] = 10-14
[OH- ] = 10-14/[H+ ]
= 10-14/3.16*10-5
=3.16*10-10
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of OH- = 3.16(10-10 ) M
Solution
pH of the solution = 4.5
-log [H+] = 4.5
log[H+] = -4.5
[H+] = 10-4.5
= 3.16*10-5
[H+ ][OH- ] = 10-14
[OH- ] = 10-14/[H+ ]
= 10-14/3.16*10-5
=3.16*10-10
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of OH- = 3.16(10-10 ) M.
Essentially, treatment with LDATHF will produce .pdf
Essentially, treatment with LDA/THF will produce the kinetic enolate meaning that
the major organic product will be choice IV.
Solution
Essentially, treatment with LDA/THF will produce the kinetic enolate meaning that
the major organic product will be choice IV..
no real roots existsSolutionno real roots exists.pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. The document consists of the phrase "no real roots exists" repeated twice without any other details.
Kp for the new reaction = 1ZSolutionKp for the new reaction =.pdf
The document discusses calculating the equilibrium constant Kp for a new reaction. It states that Kp for the new reaction equals 1 divided by Z. No other details are provided about the reaction or values of Kp or Z.
In the given source code the Person class do not implement the Compa.pdf
In the given source code the Person class do not implement the Comparable interface and two
persons cannot be compared by the compareTo method.
Solution
In the given source code the Person class do not implement the Comparable interface and two
persons cannot be compared by the compareTo method..
Hello!Lets look at the balanced reaction equation starting with th.pdf
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : )
Solution
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : ).
HereVoume of NaOH left Unreacted = 7000.10 - 0.20300= 10 mlH.pdf
Here
Voume of NaOH left Unreacted = 700*0.10 - 0.20*300
= 10 ml
Here
Now the total Volume of
Solution
= 700 + 300 = 1000 ml
So
we have 10 ml of NaOH in the 1000 ml.
a) Write the recursive function in C++ to sort a set of data using M.pdf
a) Write the recursive function in C++ to sort a set of data using Merge Sort
b) Given a set of data as below:
5 10 15 29 34 42 79 87
Ans:)
merge sort:
merge sort is a one of the sorting technicis a sorting technique which divides the array into 2 sub
arrays which have size 2 and merge adjacent pair. so that after first process you have n/2 array of
size 2. this process is repeated until only one array remaining of size n.
a)
ans)
Source code: C++ program
// Program that sorts an array using merge sort
# include
# include
const int MAX = 4 ;
class array
{
private :
int *arr ;
int size ;
int count ;
public :
array( ) ;
array ( int sz ) ;
void add ( int num ) ;
void display( ) ;
static void sort ( int *a, int sz ) ;
void merge ( array &a, array &b ) ;
~array( ) ;
} ;
// initializes data member
array :: array( )
{
count = size = 0 ;
arr = NULL ;
}
// initialises data member
array :: array( int sz )
{
count = 0 ;
size = sz ;
arr = new int[sz] ;
}
// adds a new element to the array
void array :: add ( int num )
{
if ( count < size )
{
arr[count] = num ;
count++ ;
}
else
cout << \"\ Array is full\" << endl ;
}
// displays elements in an array
void array :: display( )
{
for ( int i = 0 ; i < count ; i++ )
cout << arr[i] << \"\\t\" ;
cout << endl ;
}
// merges two arrays
void array :: merge ( array &a, array &b )
{
sort ( a.arr, a.size ) ;
sort ( b.arr, b.size ) ;
size = a.count + b.count ;
arr = new int[size] ;
int i, j, k ;
for ( i = j = k = 0 ; j < a.count || k < b.count ; )
{
if ( a.arr[j] <= b.arr[k] )
arr[i++] = a.arr[j++] ;
else
arr[i++] = b.arr[k++] ;
count++ ;
if ( j == a.count || k == b.count )
break ;
}
for ( ; j < a.count ; )
{
arr[i++] = a.arr[j++] ;
count++ ;
}
for ( ; k < b.count ; )
{
arr[i++] = b.arr[k++] ;
count++ ;
}
}
// sorts an array
void array :: sort ( int *a, int sz )
{
int temp ;
for ( int i = 0 ; i <= sz - 2 ; i++ )
{
for ( int j = i + 1 ; j <= sz - 1 ; j++ )
{
if ( a[i] > a[j] )
{
temp = a[i] ;
a[i] = a[j] ;
a[j] = temp ;
}
}
}
}
// deallocates memory
array :: ~array( )
{
delete arr ;
}
void main( )
{
array a ( MAX ) ;
clrscr();
a.add ( 29 ) ;
a.add ( 5) ;
a.add ( 10) ;
a.add ( 15 ) ;
cout << \"\ Merge sort.\ \" ;
cout << \"\ First array: \" << endl ;
a.display( ) ;
array b ( MAX ) ;
b.add ( 34 ) ;
b.add ( 87 ) ;
b.add ( 79 ) ;
b.add ( 42) ;
cout << \"\ Second array: \" << endl ;
b.display( ) ;
array c ;
c.merge ( a, b ) ;
cout << \"\ Array after sorting: \" << endl ;
c.display( ) ;
getch();
}
i)insert
// initializes data member
array :: array( )
{
count = size = 0 ;
arr = NULL ;
}
// initialises data member
array :: array( int sz )
{
count = 0 ;
size = sz ;
arr = new int[sz] ;
}
// adds a new element to the array
void array :: add ( int num )
{
if ( count < size )
{
arr[count] = num ;
count++ ;
}
else
cout << \"\ Array is full\" << endl ;
}
the above code is used to inset of data in array.
ii. Selection
// displays elements in an array
void array :: display( )
{
for ( int i = 0 ; i < count ; i++ )
cout << arr[i] << \"\\t\.
B H AS H oxidied himselb to H+ AND REDUCING HNO3.pdf
B H AS H oxidied himselb to H+ AND REDUCING HNO3 TO NO
Solution
B H AS H oxidied himselb to H+ AND REDUCING HNO3 TO NO.
f(x) 6sec(4pix +pi4)phase shift 4pix +pi4 =04x = -14x = .pdf
f(x) 6sec(4pi*x +pi/4)
phase shift : 4pix +pi/4 =0
4x = -1/4
x = -1/16
So, horizontal shift of 1/16 to left
Option d)
Solution
f(x) 6sec(4pi*x +pi/4)
phase shift : 4pix +pi/4 =0
4x = -1/4
x = -1/16
So, horizontal shift of 1/16 to left
Option d).
Benzoic is more soluble in dichlormethane than wa.pdf
Benzoic is more soluble in dichlormethane than water. Benzoic is nonpolar
Solution
Benzoic is more soluble in dichlormethane than water. Benzoic is nonpolar.
Duties of a LPN (Licensed practical nurse)They are Professional n.pdf
Duties of a LPN (Licensed practical nurse):
They are Professional nurses. Educated 2 years in in a college programme.
Responsible for asses, plan, implement and evaluate nursing carrier.
LPN Work (clinical and non-clinical):
Direct care: Acute care, continuing care, community settings.
Education: In practical nurse program or in health care settings.
Research: Involved in research at clinical level.
Administration: Many non-traditional roles available today.
LPN’s are responsible and accountable for their own practice.
LPN’s are expected to follow the standards and guidelines of their profession.
Restricted activities a nurse consents to supervise.
Make decisions about invasive producers and other care that requires nursing judgments.
Provide consultation and guidance.
To inert or remove instruments, devices, fingers and hands.
LPN must maintain effective communication with the HCA.
Solution
Duties of a LPN (Licensed practical nurse):
They are Professional nurses. Educated 2 years in in a college programme.
Responsible for asses, plan, implement and evaluate nursing carrier.
LPN Work (clinical and non-clinical):
Direct care: Acute care, continuing care, community settings.
Education: In practical nurse program or in health care settings.
Research: Involved in research at clinical level.
Administration: Many non-traditional roles available today.
LPN’s are responsible and accountable for their own practice.
LPN’s are expected to follow the standards and guidelines of their profession.
Restricted activities a nurse consents to supervise.
Make decisions about invasive producers and other care that requires nursing judgments.
Provide consultation and guidance.
To inert or remove instruments, devices, fingers and hands.
LPN must maintain effective communication with the HCA..
a. hypertonic b. hypertonic c. isotonic d. hypert.pdf
a. hypertonic b. hypertonic c. isotonic d. hypertonic (K2SO4 ionizes into two
potassium and one sulfate ions, so three parts, and NaCl only into two parts) e. isotonic f.
hypertonic hypertonic = more \"stuff\"/particles/molecules floating around in the solution, so
more % would mean hypertonic compared to the cell. Also, M means molar, which means moles
per liter. Moles are an amount of chemical, so the more M per liter, means the more particles per
liter. Also, we assume cells are floating in a water-like medium, so ionic compounds will
dissociate into their ions, and make more particles. I hope this helps!! :)
Solution
a. hypertonic b. hypertonic c. isotonic d. hypertonic (K2SO4 ionizes into two
potassium and one sulfate ions, so three parts, and NaCl only into two parts) e. isotonic f.
hypertonic hypertonic = more \"stuff\"/particles/molecules floating around in the solution, so
more % would mean hypertonic compared to the cell. Also, M means molar, which means moles
per liter. Moles are an amount of chemical, so the more M per liter, means the more particles per
liter. Also, we assume cells are floating in a water-like medium, so ionic compounds will
dissociate into their ions, and make more particles. I hope this helps!! :).
Ans. Blades of brown algae are like the leaves of the seed plants in.pdf
Ans. Blades of brown algae are like the leaves of the seed plants intemperate region of the world
in following ways-
1. Both the blades of red algae and leaves of seed plants are flattened structures to maximize
absorption of solar radiation during photosynthesis.
2. Both the structures carry our most of the photosynthetic activities in their respective places.
Solution
Ans. Blades of brown algae are like the leaves of the seed plants intemperate region of the world
in following ways-
1. Both the blades of red algae and leaves of seed plants are flattened structures to maximize
absorption of solar radiation during photosynthesis.
2. Both the structures carry our most of the photosynthetic activities in their respective places..
an exception object is created and thrown. and execution control m.pdf
an exception object is created and thrown.
and execution control moves to corresponding catch block
Solution
an exception object is created and thrown.
and execution control moves to corresponding catch block.
Answer 11.A. Phosphoenolpyruvate- 13.5B. Dihydroxyacetone phosph.pdf
Answer 11.
A. Phosphoenolpyruvate- 13.5
B. Dihydroxyacetone phosphate- 16
D. Pyruvate- 12
E. Lactate- 14.5
Answer 12.
The equilibrium ratio of phosphoenolpyruvate to pyruvate under standard conditions when
[ATP]/[ADP] = 10 will be 3.06 × 10–5.
Solution
Answer 11.
A. Phosphoenolpyruvate- 13.5
B. Dihydroxyacetone phosphate- 16
D. Pyruvate- 12
E. Lactate- 14.5
Answer 12.
The equilibrium ratio of phosphoenolpyruvate to pyruvate under standard conditions when
[ATP]/[ADP] = 10 will be 3.06 × 10–5..
Suppose S is an ordered set with the least upper bound property, B.pdf
: Suppose S is an ordered set with the least upper bound property, B S, B is not empty, and B is
bounded below. Let L be the set of all lower bounds of B. Then, = supL exists in S and = infB.
Proof: Let L = { y S|y is a lower bound of B}. Since B is bounded below, there exists at least
one y S such that y is a lower bound of B. Thus, L is a non empty subset of S. And, since each y
L is a lower bound of B, we have that, y x for every x B. Thus, each x in B is an upper bound of
L. Hence, L is bounded above. Since S has the l.u.b. property, this implies sup L exists in S. Let
= supL. Now, to show that = infB. We first show that is a lower bound of B. 1 That is to show
that x for every x B. If not, there must be a B such that < . Since is the least upper bound,
cannot be an upper bound of L. Hence / B. (Why?) This is a contradiction. Hence, x for every x
B and is hence a lower bound of B. (Thus, L.) Now, to show that is the greatest lower bound
of B. If there is a S such that < , then such a / L. (Why?) Thus, is an lower bound of B but no
> is a lower bound of B. Hence, = infB.
Solution
: Suppose S is an ordered set with the least upper bound property, B S, B is not empty, and B is
bounded below. Let L be the set of all lower bounds of B. Then, = supL exists in S and = infB.
Proof: Let L = { y S|y is a lower bound of B}. Since B is bounded below, there exists at least
one y S such that y is a lower bound of B. Thus, L is a non empty subset of S. And, since each y
L is a lower bound of B, we have that, y x for every x B. Thus, each x in B is an upper bound of
L. Hence, L is bounded above. Since S has the l.u.b. property, this implies sup L exists in S. Let
= supL. Now, to show that = infB. We first show that is a lower bound of B. 1 That is to show
that x for every x B. If not, there must be a B such that < . Since is the least upper bound,
cannot be an upper bound of L. Hence / B. (Why?) This is a contradiction. Hence, x for every x
B and is hence a lower bound of B. (Thus, L.) Now, to show that is the greatest lower bound
of B. If there is a S such that < , then such a / L. (Why?) Thus, is an lower bound of B but no
> is a lower bound of B. Hence, = infB..
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HAsO4- as it can accept an electron to form HAsO4.pdf
HAsO4- as it can accept an electron to form HAsO42-
Solution
HAsO4- as it can accept an electron to form HAsO42-.
Solution.cpp #include iostreamheader file for input output f.pdf
Solution
.cpp
#include //header file for input output function
using namespace std;//it tells the compiler to link std namespace
int main()
{//main function
int num,rows;//variable declaration
cout<<\"How many numbers to output?\"<>num;//key board inputting
cout<<\"how many rows? \"<>rows;//key board inputting
for(int i=1;i<=rows;i++)
{//nested for loop logic
for(int j=i;j<=num;j=j+14)
{
cout<.
pH of the solution = 4.5 -log [H+] = 4.5 .pdf
pH of the solution = 4.5
-log [H+] = 4.5
log[H+] = -4.5
[H+] = 10-4.5
= 3.16*10-5
[H+ ][OH- ] = 10-14
[OH- ] = 10-14/[H+ ]
= 10-14/3.16*10-5
=3.16*10-10
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of OH- = 3.16(10-10 ) M
Solution
pH of the solution = 4.5
-log [H+] = 4.5
log[H+] = -4.5
[H+] = 10-4.5
= 3.16*10-5
[H+ ][OH- ] = 10-14
[OH- ] = 10-14/[H+ ]
= 10-14/3.16*10-5
=3.16*10-10
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of H+ = 3.16(10-5 ) M
molar concentration of OH- = 3.16(10-10 ) M
molar concentration of OH- = 3.16(10-10 ) M.
Essentially, treatment with LDATHF will produce .pdf
Essentially, treatment with LDA/THF will produce the kinetic enolate meaning that
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Solution
Essentially, treatment with LDA/THF will produce the kinetic enolate meaning that
the major organic product will be choice IV..
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In the given source code the Person class do not implement the Compa.pdf
In the given source code the Person class do not implement the Comparable interface and two
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Solution
In the given source code the Person class do not implement the Comparable interface and two
persons cannot be compared by the compareTo method..
Hello!Lets look at the balanced reaction equation starting with th.pdf
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : )
Solution
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : ).
HereVoume of NaOH left Unreacted = 7000.10 - 0.20300= 10 mlH.pdf
Here
Voume of NaOH left Unreacted = 700*0.10 - 0.20*300
= 10 ml
Here
Now the total Volume of
Solution
= 700 + 300 = 1000 ml
So
we have 10 ml of NaOH in the 1000 ml.
a) Write the recursive function in C++ to sort a set of data using M.pdf
a) Write the recursive function in C++ to sort a set of data using Merge Sort
b) Given a set of data as below:
5 10 15 29 34 42 79 87
Ans:)
merge sort:
merge sort is a one of the sorting technicis a sorting technique which divides the array into 2 sub
arrays which have size 2 and merge adjacent pair. so that after first process you have n/2 array of
size 2. this process is repeated until only one array remaining of size n.
a)
ans)
Source code: C++ program
// Program that sorts an array using merge sort
# include
# include
const int MAX = 4 ;
class array
{
private :
int *arr ;
int size ;
int count ;
public :
array( ) ;
array ( int sz ) ;
void add ( int num ) ;
void display( ) ;
static void sort ( int *a, int sz ) ;
void merge ( array &a, array &b ) ;
~array( ) ;
} ;
// initializes data member
array :: array( )
{
count = size = 0 ;
arr = NULL ;
}
// initialises data member
array :: array( int sz )
{
count = 0 ;
size = sz ;
arr = new int[sz] ;
}
// adds a new element to the array
void array :: add ( int num )
{
if ( count < size )
{
arr[count] = num ;
count++ ;
}
else
cout << \"\ Array is full\" << endl ;
}
// displays elements in an array
void array :: display( )
{
for ( int i = 0 ; i < count ; i++ )
cout << arr[i] << \"\\t\" ;
cout << endl ;
}
// merges two arrays
void array :: merge ( array &a, array &b )
{
sort ( a.arr, a.size ) ;
sort ( b.arr, b.size ) ;
size = a.count + b.count ;
arr = new int[size] ;
int i, j, k ;
for ( i = j = k = 0 ; j < a.count || k < b.count ; )
{
if ( a.arr[j] <= b.arr[k] )
arr[i++] = a.arr[j++] ;
else
arr[i++] = b.arr[k++] ;
count++ ;
if ( j == a.count || k == b.count )
break ;
}
for ( ; j < a.count ; )
{
arr[i++] = a.arr[j++] ;
count++ ;
}
for ( ; k < b.count ; )
{
arr[i++] = b.arr[k++] ;
count++ ;
}
}
// sorts an array
void array :: sort ( int *a, int sz )
{
int temp ;
for ( int i = 0 ; i <= sz - 2 ; i++ )
{
for ( int j = i + 1 ; j <= sz - 1 ; j++ )
{
if ( a[i] > a[j] )
{
temp = a[i] ;
a[i] = a[j] ;
a[j] = temp ;
}
}
}
}
// deallocates memory
array :: ~array( )
{
delete arr ;
}
void main( )
{
array a ( MAX ) ;
clrscr();
a.add ( 29 ) ;
a.add ( 5) ;
a.add ( 10) ;
a.add ( 15 ) ;
cout << \"\ Merge sort.\ \" ;
cout << \"\ First array: \" << endl ;
a.display( ) ;
array b ( MAX ) ;
b.add ( 34 ) ;
b.add ( 87 ) ;
b.add ( 79 ) ;
b.add ( 42) ;
cout << \"\ Second array: \" << endl ;
b.display( ) ;
array c ;
c.merge ( a, b ) ;
cout << \"\ Array after sorting: \" << endl ;
c.display( ) ;
getch();
}
i)insert
// initializes data member
array :: array( )
{
count = size = 0 ;
arr = NULL ;
}
// initialises data member
array :: array( int sz )
{
count = 0 ;
size = sz ;
arr = new int[sz] ;
}
// adds a new element to the array
void array :: add ( int num )
{
if ( count < size )
{
arr[count] = num ;
count++ ;
}
else
cout << \"\ Array is full\" << endl ;
}
the above code is used to inset of data in array.
ii. Selection
// displays elements in an array
void array :: display( )
{
for ( int i = 0 ; i < count ; i++ )
cout << arr[i] << \"\\t\.
B H AS H oxidied himselb to H+ AND REDUCING HNO3.pdf
B H AS H oxidied himselb to H+ AND REDUCING HNO3 TO NO
Solution
B H AS H oxidied himselb to H+ AND REDUCING HNO3 TO NO.
f(x) 6sec(4pix +pi4)phase shift 4pix +pi4 =04x = -14x = .pdf
f(x) 6sec(4pi*x +pi/4)
phase shift : 4pix +pi/4 =0
4x = -1/4
x = -1/16
So, horizontal shift of 1/16 to left
Option d)
Solution
f(x) 6sec(4pi*x +pi/4)
phase shift : 4pix +pi/4 =0
4x = -1/4
x = -1/16
So, horizontal shift of 1/16 to left
Option d).
Benzoic is more soluble in dichlormethane than wa.pdf
Benzoic is more soluble in dichlormethane than water. Benzoic is nonpolar
Solution
Benzoic is more soluble in dichlormethane than water. Benzoic is nonpolar.
Duties of a LPN (Licensed practical nurse)They are Professional n.pdf
Duties of a LPN (Licensed practical nurse):
They are Professional nurses. Educated 2 years in in a college programme.
Responsible for asses, plan, implement and evaluate nursing carrier.
LPN Work (clinical and non-clinical):
Direct care: Acute care, continuing care, community settings.
Education: In practical nurse program or in health care settings.
Research: Involved in research at clinical level.
Administration: Many non-traditional roles available today.
LPN’s are responsible and accountable for their own practice.
LPN’s are expected to follow the standards and guidelines of their profession.
Restricted activities a nurse consents to supervise.
Make decisions about invasive producers and other care that requires nursing judgments.
Provide consultation and guidance.
To inert or remove instruments, devices, fingers and hands.
LPN must maintain effective communication with the HCA.
Solution
Duties of a LPN (Licensed practical nurse):
They are Professional nurses. Educated 2 years in in a college programme.
Responsible for asses, plan, implement and evaluate nursing carrier.
LPN Work (clinical and non-clinical):
Direct care: Acute care, continuing care, community settings.
Education: In practical nurse program or in health care settings.
Research: Involved in research at clinical level.
Administration: Many non-traditional roles available today.
LPN’s are responsible and accountable for their own practice.
LPN’s are expected to follow the standards and guidelines of their profession.
Restricted activities a nurse consents to supervise.
Make decisions about invasive producers and other care that requires nursing judgments.
Provide consultation and guidance.
To inert or remove instruments, devices, fingers and hands.
LPN must maintain effective communication with the HCA..
a. hypertonic b. hypertonic c. isotonic d. hypert.pdf
a. hypertonic b. hypertonic c. isotonic d. hypertonic (K2SO4 ionizes into two
potassium and one sulfate ions, so three parts, and NaCl only into two parts) e. isotonic f.
hypertonic hypertonic = more \"stuff\"/particles/molecules floating around in the solution, so
more % would mean hypertonic compared to the cell. Also, M means molar, which means moles
per liter. Moles are an amount of chemical, so the more M per liter, means the more particles per
liter. Also, we assume cells are floating in a water-like medium, so ionic compounds will
dissociate into their ions, and make more particles. I hope this helps!! :)
Solution
a. hypertonic b. hypertonic c. isotonic d. hypertonic (K2SO4 ionizes into two
potassium and one sulfate ions, so three parts, and NaCl only into two parts) e. isotonic f.
hypertonic hypertonic = more \"stuff\"/particles/molecules floating around in the solution, so
more % would mean hypertonic compared to the cell. Also, M means molar, which means moles
per liter. Moles are an amount of chemical, so the more M per liter, means the more particles per
liter. Also, we assume cells are floating in a water-like medium, so ionic compounds will
dissociate into their ions, and make more particles. I hope this helps!! :).
Ans. Blades of brown algae are like the leaves of the seed plants in.pdf
Ans. Blades of brown algae are like the leaves of the seed plants intemperate region of the world
in following ways-
1. Both the blades of red algae and leaves of seed plants are flattened structures to maximize
absorption of solar radiation during photosynthesis.
2. Both the structures carry our most of the photosynthetic activities in their respective places.
Solution
Ans. Blades of brown algae are like the leaves of the seed plants intemperate region of the world
in following ways-
1. Both the blades of red algae and leaves of seed plants are flattened structures to maximize
absorption of solar radiation during photosynthesis.
2. Both the structures carry our most of the photosynthetic activities in their respective places..
an exception object is created and thrown. and execution control m.pdf
an exception object is created and thrown.
and execution control moves to corresponding catch block
Solution
an exception object is created and thrown.
and execution control moves to corresponding catch block.
Answer 11.A. Phosphoenolpyruvate- 13.5B. Dihydroxyacetone phosph.pdf
Answer 11.
A. Phosphoenolpyruvate- 13.5
B. Dihydroxyacetone phosphate- 16
D. Pyruvate- 12
E. Lactate- 14.5
Answer 12.
The equilibrium ratio of phosphoenolpyruvate to pyruvate under standard conditions when
[ATP]/[ADP] = 10 will be 3.06 × 10–5.
Solution
Answer 11.
A. Phosphoenolpyruvate- 13.5
B. Dihydroxyacetone phosphate- 16
D. Pyruvate- 12
E. Lactate- 14.5
Answer 12.
The equilibrium ratio of phosphoenolpyruvate to pyruvate under standard conditions when
[ATP]/[ADP] = 10 will be 3.06 × 10–5..
Suppose S is an ordered set with the least upper bound property, B.pdf
: Suppose S is an ordered set with the least upper bound property, B S, B is not empty, and B is
bounded below. Let L be the set of all lower bounds of B. Then, = supL exists in S and = infB.
Proof: Let L = { y S|y is a lower bound of B}. Since B is bounded below, there exists at least
one y S such that y is a lower bound of B. Thus, L is a non empty subset of S. And, since each y
L is a lower bound of B, we have that, y x for every x B. Thus, each x in B is an upper bound of
L. Hence, L is bounded above. Since S has the l.u.b. property, this implies sup L exists in S. Let
= supL. Now, to show that = infB. We first show that is a lower bound of B. 1 That is to show
that x for every x B. If not, there must be a B such that < . Since is the least upper bound,
cannot be an upper bound of L. Hence / B. (Why?) This is a contradiction. Hence, x for every x
B and is hence a lower bound of B. (Thus, L.) Now, to show that is the greatest lower bound
of B. If there is a S such that < , then such a / L. (Why?) Thus, is an lower bound of B but no
> is a lower bound of B. Hence, = infB.
Solution
: Suppose S is an ordered set with the least upper bound property, B S, B is not empty, and B is
bounded below. Let L be the set of all lower bounds of B. Then, = supL exists in S and = infB.
Proof: Let L = { y S|y is a lower bound of B}. Since B is bounded below, there exists at least
one y S such that y is a lower bound of B. Thus, L is a non empty subset of S. And, since each y
L is a lower bound of B, we have that, y x for every x B. Thus, each x in B is an upper bound of
L. Hence, L is bounded above. Since S has the l.u.b. property, this implies sup L exists in S. Let
= supL. Now, to show that = infB. We first show that is a lower bound of B. 1 That is to show
that x for every x B. If not, there must be a B such that < . Since is the least upper bound,
cannot be an upper bound of L. Hence / B. (Why?) This is a contradiction. Hence, x for every x
B and is hence a lower bound of B. (Thus, L.) Now, to show that is the greatest lower bound
of B. If there is a S such that < , then such a / L. (Why?) Thus, is an lower bound of B but no
> is a lower bound of B. Hence, = infB..
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