Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area.
Solution
Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area..
I am Kennedy G. I am a Single Variable Calculus Assignment Solver at mathhomeworksolver.com. I hold a Master's in Mathematics, from Indiana, USA. I have been helping students with their assignments for the past 15 years. I solved assignments related to Single Variable Calculus.
Visit mathhomeworksolver.com or email support@mathhomeworksolver.com. You can also call on +1 678 648 4277 for any assistance with Single Variable Calculus Assignment.
I am Kennedy G. I am a Single Variable Calculus Assignment Solver at mathhomeworksolver.com. I hold a Master's in Mathematics, from Indiana, USA. I have been helping students with their assignments for the past 15 years. I solved assignments related to Single Variable Calculus.
Visit mathhomeworksolver.com or email support@mathhomeworksolver.com. You can also call on +1 678 648 4277 for any assistance with Single Variable Calculus Assignment.
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}
Solution
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}.
Wilhelm Johannsen first proposed the distinction between genotype a.pdfanupamele
Wilhelm Johannsen first proposed the distinction between genotype and phenotype in the study
of heredity while working in Denmark in 1909. The distinction is between the hereditary
dispositions of organisms (their genotypes) and the ways in which those dispositions manifest
themselves in the physical characteristics of those organisms (their phenotypes). This distinction
was an outgrowth of Johannsen
Solution
Wilhelm Johannsen first proposed the distinction between genotype and phenotype in the study
of heredity while working in Denmark in 1909. The distinction is between the hereditary
dispositions of organisms (their genotypes) and the ways in which those dispositions manifest
themselves in the physical characteristics of those organisms (their phenotypes). This distinction
was an outgrowth of Johannsen.
True, they are identical Thats just the bottom vi.pdfanupamele
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate.
Solution
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate..
The NH2OH is in acidic solution as a reactant and.pdfanupamele
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome.
Solution
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome..
The atom is a basic unit of matter consisting of .pdfanupamele
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each other, forming a
molecule. An atom containing an equal number of protons and electrons is electrically neutral,
otherwise it has a positive charge if there are fewer electrons (electron deficiency) or negative
charge if there are more electrons (electron excess). A positively or negatively charged atom is
known as an ion. An atom is classified according to the number of protons and neutrons in its
nucleus: the number of protons determines the chemical element, and the number of neutrons
determines the isotope of the element.[1] The name atom comes from the Greek ?t?µ?? (atomos,
\"indivisible\") from ?- (a-, \"not\") and t?µ?? (temno, \"I cut\")[2], which means uncuttable, or
indivisible, something that cannot be divided further.[3] The concept of an atom as an indivisible
component of matter was first proposed by early Indian and Greek philosophers. In the 17th and
18th centuries, chemists provided a physical basis for this idea by showing that certain
substances could not be further broken down by chemical methods. During the late 19th and
early 20th centuries, physicists discovered subatomic components and structure inside the atom,
thereby demonstrating that the \'atom\' was divisible. The principles of quantum mechanics were
used to successfully model the atom.[4][5] Atoms are minuscule objects with proportionately
tiny masses. Atoms can only be observed individually using special instruments such as the
scanning tunneling microscope. Over 99.94% of an atom\'s mass is concentrated in the
nucleus,[note 1] with protons and neutrons having roughly equal mass. Each element has at least
one isotope with unstable nuclei that can undergo radioactive decay. This can result in a
transmutation that changes the number of protons or neutrons in a nucleus.[6] Electrons that are
bound to atoms possess a set of stable energy levels, or orbitals, and can undergo transitions
between them by absorbing or emitting photons that match the energy differences between the
levels. The electrons determine the chemical properties of an element, and strongly influence an
atom\'s magnetic properties.
Solution
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each .
see the inner one is union of events from 1-inf. .pdfanupamele
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on
Solution
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on.
No you wouldnt include the intermediates in the.pdfanupamele
No you wouldn\'t include the intermediates in the rate law but you need to identify
them
Solution
No you wouldn\'t include the intermediates in the rate law but you need to identify
them.
mole of NaOH= 1240 =0.3 so molarity = 0.31.8 = .pdfanupamele
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167
Solution
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167.
millimoles of HBr=500.04=2 millimoles of KOH=25.pdfanupamele
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53
Solution
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53.
initially 2 moles of HCl per L of water suppose x.pdfanupamele
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L
Solution
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L.
i cant see the question.. please post the complet.pdfanupamele
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer.....
Solution
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer......
FT-Raman spectra of some calcium phosphates, calc.pdfanupamele
FT-Raman spectra of some calcium phosphates, calcium carbonate and calcium
hydroxide are reported. Comparing with dispersive Raman spectra, it is shown that some strong
bands in the FT-Raman spectra of these compounds which have been erroneously assigned to
Ramanbands, are in fact fluorescence emission bands. The origin of these fluorescencebands
have been discussed in terms of rare earth impurities in a particular structure. It is further
concluded that in using FT-Raman spectroscopy of minerals, care should be taken in not
confusing the fluorescencebands with Raman scattering bands. Among instrumental techniques,
fluorescence spectroscopy is recognized as one of the more sensitive. In fluorescence, the
intensity of the emission of the sample is measured. The reason for the high sensitivity of
fluorescence techniques is that the emission signal is measured above a low background level.
This is inherently more sensitive than comparing two relatively large signals as in absorption
spectroscopy. The sensitivity of fluorescence techniques is as much as 1000 times more sensitive
than absorption spectroscopy. A spectrofluorometer with high sensitivity is an asset for the
researcher. It is obvious that an instrument with high sensitivity will be able to perform
experiments with the fluorescing species at a low concentration. For analytical measurements,
sensitivity determines the detection limit of a material. For research in life science, it may be
important to work at concentration levels that are very low. This is almost always the case in
biomedical research. A more subtle requirement is the need to measure changes in fluorescence
during the course of an experiment. Even though the gross signal levels may be high, the changes
in the signal may be quite small. The ability to reliably measure these small changes is directly
dependent upon the sensitivity of the fluorometer. The measurement of sensitivity has been
discussed in the literature [1]. Unfortunately, there is little consensus on a standard procedure for
determining this important capability. This paper discusses the measurement of sensitivity and
details two methods that are applicable to any instrument. The Spectrofluorometer The
sensitivity of fluorescence is dependent on both the fluorophore and the instrument. The response
of a fluorophore will depend on the molar absorptivity and the quantum yield. These factors are,
in general, beyond the control of the analyst. The sensitivity of a spectrofluorometer depends on
a number of factors. Instrumental contributions to sensitivity are described below. 1. Source
intensity. In general, a brighter excitation source will result in brighter emission. The source for
most fluorometers is a xenon arc, which has a high intensity between 200-900 nm, the spectral
region where most fluorescence experiments are performed. While a high power arc lamp is
good for highest intensity, a more important criteria is the brightness of the lamp. The brightness.
More Related Content
Similar to Well its quite an easy one. The given curve is a cardioid. The limit.pdf
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}
Solution
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}.
Wilhelm Johannsen first proposed the distinction between genotype a.pdfanupamele
Wilhelm Johannsen first proposed the distinction between genotype and phenotype in the study
of heredity while working in Denmark in 1909. The distinction is between the hereditary
dispositions of organisms (their genotypes) and the ways in which those dispositions manifest
themselves in the physical characteristics of those organisms (their phenotypes). This distinction
was an outgrowth of Johannsen
Solution
Wilhelm Johannsen first proposed the distinction between genotype and phenotype in the study
of heredity while working in Denmark in 1909. The distinction is between the hereditary
dispositions of organisms (their genotypes) and the ways in which those dispositions manifest
themselves in the physical characteristics of those organisms (their phenotypes). This distinction
was an outgrowth of Johannsen.
True, they are identical Thats just the bottom vi.pdfanupamele
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate.
Solution
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate..
The NH2OH is in acidic solution as a reactant and.pdfanupamele
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome.
Solution
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome..
The atom is a basic unit of matter consisting of .pdfanupamele
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each other, forming a
molecule. An atom containing an equal number of protons and electrons is electrically neutral,
otherwise it has a positive charge if there are fewer electrons (electron deficiency) or negative
charge if there are more electrons (electron excess). A positively or negatively charged atom is
known as an ion. An atom is classified according to the number of protons and neutrons in its
nucleus: the number of protons determines the chemical element, and the number of neutrons
determines the isotope of the element.[1] The name atom comes from the Greek ?t?µ?? (atomos,
\"indivisible\") from ?- (a-, \"not\") and t?µ?? (temno, \"I cut\")[2], which means uncuttable, or
indivisible, something that cannot be divided further.[3] The concept of an atom as an indivisible
component of matter was first proposed by early Indian and Greek philosophers. In the 17th and
18th centuries, chemists provided a physical basis for this idea by showing that certain
substances could not be further broken down by chemical methods. During the late 19th and
early 20th centuries, physicists discovered subatomic components and structure inside the atom,
thereby demonstrating that the \'atom\' was divisible. The principles of quantum mechanics were
used to successfully model the atom.[4][5] Atoms are minuscule objects with proportionately
tiny masses. Atoms can only be observed individually using special instruments such as the
scanning tunneling microscope. Over 99.94% of an atom\'s mass is concentrated in the
nucleus,[note 1] with protons and neutrons having roughly equal mass. Each element has at least
one isotope with unstable nuclei that can undergo radioactive decay. This can result in a
transmutation that changes the number of protons or neutrons in a nucleus.[6] Electrons that are
bound to atoms possess a set of stable energy levels, or orbitals, and can undergo transitions
between them by absorbing or emitting photons that match the energy differences between the
levels. The electrons determine the chemical properties of an element, and strongly influence an
atom\'s magnetic properties.
Solution
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each .
see the inner one is union of events from 1-inf. .pdfanupamele
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on
Solution
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on.
No you wouldnt include the intermediates in the.pdfanupamele
No you wouldn\'t include the intermediates in the rate law but you need to identify
them
Solution
No you wouldn\'t include the intermediates in the rate law but you need to identify
them.
mole of NaOH= 1240 =0.3 so molarity = 0.31.8 = .pdfanupamele
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167
Solution
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167.
millimoles of HBr=500.04=2 millimoles of KOH=25.pdfanupamele
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53
Solution
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53.
initially 2 moles of HCl per L of water suppose x.pdfanupamele
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L
Solution
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L.
i cant see the question.. please post the complet.pdfanupamele
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer.....
Solution
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer......
FT-Raman spectra of some calcium phosphates, calc.pdfanupamele
FT-Raman spectra of some calcium phosphates, calcium carbonate and calcium
hydroxide are reported. Comparing with dispersive Raman spectra, it is shown that some strong
bands in the FT-Raman spectra of these compounds which have been erroneously assigned to
Ramanbands, are in fact fluorescence emission bands. The origin of these fluorescencebands
have been discussed in terms of rare earth impurities in a particular structure. It is further
concluded that in using FT-Raman spectroscopy of minerals, care should be taken in not
confusing the fluorescencebands with Raman scattering bands. Among instrumental techniques,
fluorescence spectroscopy is recognized as one of the more sensitive. In fluorescence, the
intensity of the emission of the sample is measured. The reason for the high sensitivity of
fluorescence techniques is that the emission signal is measured above a low background level.
This is inherently more sensitive than comparing two relatively large signals as in absorption
spectroscopy. The sensitivity of fluorescence techniques is as much as 1000 times more sensitive
than absorption spectroscopy. A spectrofluorometer with high sensitivity is an asset for the
researcher. It is obvious that an instrument with high sensitivity will be able to perform
experiments with the fluorescing species at a low concentration. For analytical measurements,
sensitivity determines the detection limit of a material. For research in life science, it may be
important to work at concentration levels that are very low. This is almost always the case in
biomedical research. A more subtle requirement is the need to measure changes in fluorescence
during the course of an experiment. Even though the gross signal levels may be high, the changes
in the signal may be quite small. The ability to reliably measure these small changes is directly
dependent upon the sensitivity of the fluorometer. The measurement of sensitivity has been
discussed in the literature [1]. Unfortunately, there is little consensus on a standard procedure for
determining this important capability. This paper discusses the measurement of sensitivity and
details two methods that are applicable to any instrument. The Spectrofluorometer The
sensitivity of fluorescence is dependent on both the fluorophore and the instrument. The response
of a fluorophore will depend on the molar absorptivity and the quantum yield. These factors are,
in general, beyond the control of the analyst. The sensitivity of a spectrofluorometer depends on
a number of factors. Instrumental contributions to sensitivity are described below. 1. Source
intensity. In general, a brighter excitation source will result in brighter emission. The source for
most fluorometers is a xenon arc, which has a high intensity between 200-900 nm, the spectral
region where most fluorescence experiments are performed. While a high power arc lamp is
good for highest intensity, a more important criteria is the brightness of the lamp. The brightness.
H is oxidised here no O2 is not oxidised as it g.pdfanupamele
H is oxidised here no O2 is not oxidised as it gains electrons yes Cu is reduced
here
Solution
H is oxidised here no O2 is not oxidised as it gains electrons yes Cu is reduced
here.
YES the news stories in the Illustrated American sympathetic or non-.pdfanupamele
YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of
the strikers.
The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but
run by Henry Clay Frick. Labor went on strike because of pay rates that were continually
adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and
Frick was the man on the spot and in charge. A miscommunication between these two resulted in
Frick using too much force to suppress the strikers. Labor was responsible for the strike, but
Frick (and maybe Carnegie) were resposible for subsequent events and deaths.
The Homestead strike was organized and purposeful, a harbinger of the type of strike which
would mark the modern age of labor relations in the United States. The AA strike at the
Homestead steel mill in 1892 was different from previous large-scale strikes in American history
such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier
strikes had been largely leaderless and disorganized mass uprisings of workers.
The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale
crossing of the picket line occurred, first among Eastern European immigrants and then among
all workers. The strike had collapsed so much that the state militia pulled out on October 13,
ending the 95-day occupation. The AA was nearly bankrupted by the job action. Weekly Union
relief for a member averaged $6.25 but totalled a staggering $10,000 per week when including
1,600 strikers. With only 192 out of more than 3,800 strikers in attendance, the Homestead
chapter of the AA voted, 101 to 91, to return to work on November 20, 1892.
Solution
YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of
the strikers.
The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but
run by Henry Clay Frick. Labor went on strike because of pay rates that were continually
adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and
Frick was the man on the spot and in charge. A miscommunication between these two resulted in
Frick using too much force to suppress the strikers. Labor was responsible for the strike, but
Frick (and maybe Carnegie) were resposible for subsequent events and deaths.
The Homestead strike was organized and purposeful, a harbinger of the type of strike which
would mark the modern age of labor relations in the United States. The AA strike at the
Homestead steel mill in 1892 was different from previous large-scale strikes in American history
such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier
strikes had been largely leaderless and disorganized mass uprisings of workers.
The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale
crossing of the picket line occurred, first among Eastern European i.
AnswerSOCS are the supressor of cytokine signaling proteins and a.pdfanupamele
Answer:
SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of
JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT
pathway proteins.
As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation
in SOCS family proteins could be one of the mechanism of abnormal activation of STAT .
The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of
SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine
signaling or inhibit cytokine signaling by
(i) Blocking STAT recruitment to the cytokine receptor
(ii) may target protein for degradation
(iii) bind to JAK and inhibit their kinase
(iv) target JAK for degradation via proteosome pathway
Solution
Answer:
SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of
JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT
pathway proteins.
As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation
in SOCS family proteins could be one of the mechanism of abnormal activation of STAT .
The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of
SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine
signaling or inhibit cytokine signaling by
(i) Blocking STAT recruitment to the cytokine receptor
(ii) may target protein for degradation
(iii) bind to JAK and inhibit their kinase
(iv) target JAK for degradation via proteosome pathway.
The pre-mRNA is converted into mature mRNA by modifications known as.pdfanupamele
The pre-mRNA is converted into mature mRNA by modifications known as post-transcriptional
modifications. The three main modifications involved are
Solution
The pre-mRNA is converted into mature mRNA by modifications known as post-transcriptional
modifications. The three main modifications involved are.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
Sectors of the Indian Economy - Class 10 Study Notes pdf
Well its quite an easy one. The given curve is a cardioid. The limit.pdf
1. Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area.
Solution
Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area.
