1. True2. I am having trouble understanding the phrasing. A precip.pdf
•
0 likes•2 views
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells which activate B cells to produce antibodies. 3. The answer is: MHC class II markers MHC class II markers are needed for T helper cell binding to other immune cells. Solution 1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells which activate B cells to produce antibodies. 3. The answer is: MHC class II markers MHC class II markers are needed for T helper cell binding to other immune cells..
Report
Share
Report
Share
Download to read offline
Recommended
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
There are two ways to solve this problem. I will .pdf
The ‘nuclei effect’ Cesium formate brine has a natural high density. The weight
comes from cesium ions rather than added particles, such as barite. In traditional OBMs, the
weighting agent is stopped at the well bore wall and forms a ?lter cake. With cesium formate the
problem of ?lter cakes are eliminated, however one needs to correct for the drilling ?uid that
?ltrates into the near formation. “The traditional logging tool algorithm is only good for elements
with atomic numbers less than roughly 24, which means that the element has one neutron for
every proton. Cesium, with an atomic number of 55, has considerably more neutrons than
protons in its nucleus. Therefore when measuring electron density, which is then used to
calculate bulk density, the algorithm should show an under-estimation of density due to this large
‘nuclei effect,’” explains Pedersen. “The only problem was that bulk density was over-
estimated,” he continues. So what was the explanation? “When going through the data I noticed
an intriguing correlation between the gamma logs and the photoelectric factor. To cut a long
story short, higher-numbered atomic elements, such as cesium, produce a dramatically higher
photoelectric effect. This increases the absorption of gamma rays in the formation so that the
electron density appears higher than it is and physical density is therefore over estimated. This
far outweighs the ‘nuclei effect’ previously mentioned,” says Pedersen. A simple solution Now
StatoilHydro understood why the logging results were wrong when interpreted using
methodology based on drilling ?uids with lower atomic numbers, the challenge was to ?nd a way
of interpreting the numbers accurately. The simplicity of the solution was stunning. After close
examination of the calibration data, a linear relationship was found between the photoelectric
factor (Pef) and the over-estimation of density. Consequently, by consistently adjusting bulk
density by the measured Pef, porosity was correctly calculated. “The beauty of this correction is
that it relies only on the Pef measurements, which are logged simultaneously with bulk density,
and can therefore apply to any cesium formate concentration,” says Pedersen. The perfect model
The logged photoelectric factor isn’t only useful for correcting the density curve and estimating
saturation of the invading ?ltrate, but is also ideal for de?ning permeable sands. Combining the
photoelectric factor and its high vertical resolution with resistivity measurements from both the
drill pass and ream pass produces a very reliable and accurate net reservoir de?nition.
Furthermore, by applying a conductive drilling ?uid based on cesium formate brine in all
production wells high-quality resistivity image logs can be run. These logs provide important
information on structural dip, depositional environment, sedimentary features, facies and
geological correlations, all used by geomodelers to produce better reservoir models. “The results
match cor.
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
This is a titration of a weak base w a strong ac.pdf
This equation makes it able for you to plug-and-play the numbers.
Let\'s classify the variables first.
n is the number of people
p is the profit acquired
If you plug in p=0 (the situation when the profit is 0), you get
0 = 10(120-n)-50n
Now all you have to do is solve for n.
Solution
This equation makes it able for you to plug-and-play the numbers.
Let\'s classify the variables first.
n is the number of people
p is the profit acquired
If you plug in p=0 (the situation when the profit is 0), you get
0 = 10(120-n)-50n
Now all you have to do is solve for n..
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th..
This is a titration of a weak base w a strong ac.pdf
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation.
Solution
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation...
Recommended
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
There are two ways to solve this problem. I will .pdf
The ‘nuclei effect’ Cesium formate brine has a natural high density. The weight
comes from cesium ions rather than added particles, such as barite. In traditional OBMs, the
weighting agent is stopped at the well bore wall and forms a ?lter cake. With cesium formate the
problem of ?lter cakes are eliminated, however one needs to correct for the drilling ?uid that
?ltrates into the near formation. “The traditional logging tool algorithm is only good for elements
with atomic numbers less than roughly 24, which means that the element has one neutron for
every proton. Cesium, with an atomic number of 55, has considerably more neutrons than
protons in its nucleus. Therefore when measuring electron density, which is then used to
calculate bulk density, the algorithm should show an under-estimation of density due to this large
‘nuclei effect,’” explains Pedersen. “The only problem was that bulk density was over-
estimated,” he continues. So what was the explanation? “When going through the data I noticed
an intriguing correlation between the gamma logs and the photoelectric factor. To cut a long
story short, higher-numbered atomic elements, such as cesium, produce a dramatically higher
photoelectric effect. This increases the absorption of gamma rays in the formation so that the
electron density appears higher than it is and physical density is therefore over estimated. This
far outweighs the ‘nuclei effect’ previously mentioned,” says Pedersen. A simple solution Now
StatoilHydro understood why the logging results were wrong when interpreted using
methodology based on drilling ?uids with lower atomic numbers, the challenge was to ?nd a way
of interpreting the numbers accurately. The simplicity of the solution was stunning. After close
examination of the calibration data, a linear relationship was found between the photoelectric
factor (Pef) and the over-estimation of density. Consequently, by consistently adjusting bulk
density by the measured Pef, porosity was correctly calculated. “The beauty of this correction is
that it relies only on the Pef measurements, which are logged simultaneously with bulk density,
and can therefore apply to any cesium formate concentration,” says Pedersen. The perfect model
The logged photoelectric factor isn’t only useful for correcting the density curve and estimating
saturation of the invading ?ltrate, but is also ideal for de?ning permeable sands. Combining the
photoelectric factor and its high vertical resolution with resistivity measurements from both the
drill pass and ream pass produces a very reliable and accurate net reservoir de?nition.
Furthermore, by applying a conductive drilling ?uid based on cesium formate brine in all
production wells high-quality resistivity image logs can be run. These logs provide important
information on structural dip, depositional environment, sedimentary features, facies and
geological correlations, all used by geomodelers to produce better reservoir models. “The results
match cor.
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
This is a titration of a weak base w a strong ac.pdf
This equation makes it able for you to plug-and-play the numbers.
Let\'s classify the variables first.
n is the number of people
p is the profit acquired
If you plug in p=0 (the situation when the profit is 0), you get
0 = 10(120-n)-50n
Now all you have to do is solve for n.
Solution
This equation makes it able for you to plug-and-play the numbers.
Let\'s classify the variables first.
n is the number of people
p is the profit acquired
If you plug in p=0 (the situation when the profit is 0), you get
0 = 10(120-n)-50n
Now all you have to do is solve for n..
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th..
This is a titration of a weak base w a strong ac.pdf
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation.
Solution
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation...
There are two ways to solve this problem. I will .pdf
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)
Solution
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)..
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of elements in a molecule, not the actual number of atoms. The empirical formula for the molecule described is A1Br3, indicating it contains aluminum, bromine, and the subscripts show their relative amounts in the molecule. The empirical formula provides the simplest whole number ratio of the elements in a compound.
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
Gender and Mental Health - Counselling and Family Therapy Applications and In...
A proprietary approach developed by bringing together the best of learning theories from Psychology, design principles from the world of visualization, and pedagogical methods from over a decade of training experience, that enables you to: Learn better, faster!
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
Physical pharmaceutics notes for B.pharm students
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...Nguyen Thanh Tu Collection
https://app.box.com/s/tacvl9ekroe9hqupdnjruiypvm9rdaneWhat is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Aberdeen
Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...
Andreas Schleicher, Director of Education and Skills at the OECD presents at the launch of PISA 2022 Volume III - Creative Minds, Creative Schools on 18 June 2024.
How Barcodes Can Be Leveraged Within Odoo 17
In this presentation, we will explore how barcodes can be leveraged within Odoo 17 to streamline our manufacturing processes. We will cover the configuration steps, how to utilize barcodes in different manufacturing scenarios, and the overall benefits of implementing this technology.
How to Predict Vendor Bill Product in Odoo 17
This slide will guide us through the process of predicting vendor bill products based on previous purchases from the vendor in Odoo 17.
More Related Content
More from ssuseraef9da
There are two ways to solve this problem. I will .pdf
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)
Solution
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)..
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of elements in a molecule, not the actual number of atoms. The empirical formula for the molecule described is A1Br3, indicating it contains aluminum, bromine, and the subscripts show their relative amounts in the molecule. The empirical formula provides the simplest whole number ratio of the elements in a compound.
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
More from ssuseraef9da (10)
There are two ways to solve this problem. I will .pdf
There are two ways to solve this problem. I will .pdf
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of the elem.pdf
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of the elem.pdf
If an oxide has the formula MO, then since O is normally found in th.pdf
If an oxide has the formula MO, then since O is normally found in th.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
#include algorithm #include vector #include iostream usi.pdf
Recently uploaded
Gender and Mental Health - Counselling and Family Therapy Applications and In...
A proprietary approach developed by bringing together the best of learning theories from Psychology, design principles from the world of visualization, and pedagogical methods from over a decade of training experience, that enables you to: Learn better, faster!
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
Physical pharmaceutics notes for B.pharm students
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...Nguyen Thanh Tu Collection
https://app.box.com/s/tacvl9ekroe9hqupdnjruiypvm9rdaneWhat is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Aberdeen
Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...
Andreas Schleicher, Director of Education and Skills at the OECD presents at the launch of PISA 2022 Volume III - Creative Minds, Creative Schools on 18 June 2024.
How Barcodes Can Be Leveraged Within Odoo 17
In this presentation, we will explore how barcodes can be leveraged within Odoo 17 to streamline our manufacturing processes. We will cover the configuration steps, how to utilize barcodes in different manufacturing scenarios, and the overall benefits of implementing this technology.
How to Predict Vendor Bill Product in Odoo 17
This slide will guide us through the process of predicting vendor bill products based on previous purchases from the vendor in Odoo 17.
skeleton System.pdf (skeleton system wow)
🔥🔥🔥🔥🔥🔥🔥🔥🔥
إضغ بين إيديكم من أقوى الملازم التي صممتها
ملزمة تشريح الجهاز الهيكلي (نظري 3)
💀💀💀💀💀💀💀💀💀💀
تتميز هذهِ الملزمة بعِدة مُميزات :
1- مُترجمة ترجمة تُناسب جميع المستويات
2- تحتوي على 78 رسم توضيحي لكل كلمة موجودة بالملزمة (لكل كلمة !!!!)
#فهم_ماكو_درخ
3- دقة الكتابة والصور عالية جداً جداً جداً
4- هُنالك بعض المعلومات تم توضيحها بشكل تفصيلي جداً (تُعتبر لدى الطالب أو الطالبة بإنها معلومات مُبهمة ومع ذلك تم توضيح هذهِ المعلومات المُبهمة بشكل تفصيلي جداً
5- الملزمة تشرح نفسها ب نفسها بس تكلك تعال اقراني
6- تحتوي الملزمة في اول سلايد على خارطة تتضمن جميع تفرُعات معلومات الجهاز الهيكلي المذكورة في هذهِ الملزمة
واخيراً هذهِ الملزمة حلالٌ عليكم وإتمنى منكم إن تدعولي بالخير والصحة والعافية فقط
كل التوفيق زملائي وزميلاتي ، زميلكم محمد الذهبي 💊💊
🔥🔥🔥🔥🔥🔥🔥🔥🔥
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"National Information Standards Organization (NISO)
This presentation was provided by Racquel Jemison, Ph.D., Christina MacLaughlin, Ph.D., and Paulomi Majumder. Ph.D., all of the American Chemical Society, for the second session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session Two: 'Expanding Pathways to Publishing Careers,' was held June 13, 2024.Educational Technology in the Health Sciences
Plenary presentation at the NTTC Inter-university Workshop, 18 June 2024, Manila Prince Hotel.
Geography as a Discipline Chapter 1 __ Class 11 Geography NCERT _ Class Notes...
Geography as discipline
Recently uploaded (20)
Gender and Mental Health - Counselling and Family Therapy Applications and In...
Gender and Mental Health - Counselling and Family Therapy Applications and In...
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...
Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Geography as a Discipline Chapter 1 __ Class 11 Geography NCERT _ Class Notes...
Geography as a Discipline Chapter 1 __ Class 11 Geography NCERT _ Class Notes...
Juneteenth Freedom Day 2024 David Douglas School District
Juneteenth Freedom Day 2024 David Douglas School District
1. True2. I am having trouble understanding the phrasing. A precip.pdf
- 1. 1. True 2. I am having trouble understanding the phrasing. A precipitate will fall out of solution, i.e., it will settle on the bottom, forming two layers: the supernatant and the precipitate. In that sense, true, precipitates separate from the solution where they are formed. 3. False, all nitrates are soluable. 4. True, all carbonates are insoluable except for those with 1As or NH4^1+ 5. False, all 1A elements are soluable. Solution 1. True 2. I am having trouble understanding the phrasing. A precipitate will fall out of solution, i.e., it will settle on the bottom, forming two layers: the supernatant and the precipitate. In that sense, true, precipitates separate from the solution where they are formed. 3. False, all nitrates are soluable. 4. True, all carbonates are insoluable except for those with 1As or NH4^1+ 5. False, all 1A elements are soluable.