1.KClO(3) ---- Potassium chlorate 2.H(2)SO(2) ------ sulphurous acid.pdf
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1.KClO(3) ---- Potassium chlorate 2.H(2)SO(2) ------ sulphurous acid 3.SrF(2) ----- strontium fluoride 5.Hg(NO(3))(2) ----- Mercuric nitrate 6.SnCl(4) ---- Tinchloride 7.Se(3)Cl(2) ---- tri selenium dichloride 8.Hg(2)O ---- mercury (I) oxide 9.NCl(3) ---- Nitrogen trichloride Solution 1.KClO(3) ---- Potassium chlorate 2.H(2)SO(2) ------ sulphurous acid 3.SrF(2) ----- strontium fluoride 5.Hg(NO(3))(2) ----- Mercuric nitrate 6.SnCl(4) ---- Tinchloride 7.Se(3)Cl(2) ---- tri selenium dichloride 8.Hg(2)O ---- mercury (I) oxide 9.NCl(3) ---- Nitrogen trichloride.
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Seven Requirements of CTC 1 Age 2 Relationship .pdf
Seven Requirements of CTC: 1 Age 2 Relationship 3 Support 4 Dependent
Status 5 Citizenship 6 Length of Residency 7 Family Income 1. Age test
To qualify, a child must have been under age 17 (i.e., 16 years old or younger) at the end of the
tax year for which you claim the credit. 2. Relationship test The child must be your own
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child is always treated as your own child. (\"An adopted child\" includes a child lawfully placed
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also claim your brother or sister, stepbrother, stepsister. And you can claim descendants of any of
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the solvents that are used to combine together to.pdf
the solvents that are used to combine together to make the solvent pair should be
miscible. Hexane is nonpolar solvent, while water is a polar solvent. Thus, they are not miscible
and will stay as two layers after they are combined together. Thus, they cannot work together as
a solvent pair in recrystallization
Solution
the solvents that are used to combine together to make the solvent pair should be
miscible. Hexane is nonpolar solvent, while water is a polar solvent. Thus, they are not miscible
and will stay as two layers after they are combined together. Thus, they cannot work together as
a solvent pair in recrystallization.
The most simple property to calculate this is Ka.pdf
The most simple property to calculate this is Ka*Kb=e-14 -->Ka=(e-14)/Kb Ka for
NH4+: Kb of NH3=(1.8e-5) --> Ka=(e-14)/(1.8e-5)=5.56e-10 Ka for CH3NH3+: Kb for
CH3NH2=4.4e-4 --> Ka=(e-14)/(4.4e-4)=2.27e-11 Kb for F-: Ka for HF=6.8e-4 --> Kb=(e-
14)/(6.8e-4)=1.47e-11
Solution
The most simple property to calculate this is Ka*Kb=e-14 -->Ka=(e-14)/Kb Ka for
NH4+: Kb of NH3=(1.8e-5) --> Ka=(e-14)/(1.8e-5)=5.56e-10 Ka for CH3NH3+: Kb for
CH3NH2=4.4e-4 --> Ka=(e-14)/(4.4e-4)=2.27e-11 Kb for F-: Ka for HF=6.8e-4 --> Kb=(e-
14)/(6.8e-4)=1.47e-11.
The images of the compounds are not visible. Kin.pdf
The images of the compounds are not visible. Kindly post it in other question and i
will let you know.
Solution
The images of the compounds are not visible. Kindly post it in other question and i
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Step1 Moles of water produced =18 Step2 The init.pdf
Step1 Moles of water produced =18 Step2 The initial concentrations of octane and
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Sorry. SnO2 is amphoteric. This one is a tricky one..
SO2 Solu.pdf
This very short document appears to be discussing sulfur dioxide (SO2) as it contains the chemical formula twice with the word "Solution" in between. It seems to be stating that sulfur dioxide is the subject being discussed and that there may be a solution related to it, but provides no other details in the 3 word document.
scientific synonyms for height may be peak or tal.pdf
scientific synonyms for height may be peak or tallness sometimes length(in case of
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scientific synonyms for height may be peak or tallness sometimes length(in case of
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tax purposes, the term \"U.S. national\" refers to individuals who were born in American Samoa
or in the Commonwealth of the Northern Mariana Islands.) 6. Residence test The child
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threshold is $55,000 for married couples filing separately; $75,000 for single, head of household,
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the solvents that are used to combine together to.pdf
the solvents that are used to combine together to make the solvent pair should be
miscible. Hexane is nonpolar solvent, while water is a polar solvent. Thus, they are not miscible
and will stay as two layers after they are combined together. Thus, they cannot work together as
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the solvents that are used to combine together to make the solvent pair should be
miscible. Hexane is nonpolar solvent, while water is a polar solvent. Thus, they are not miscible
and will stay as two layers after they are combined together. Thus, they cannot work together as
a solvent pair in recrystallization.
The most simple property to calculate this is Ka.pdf
The most simple property to calculate this is Ka*Kb=e-14 -->Ka=(e-14)/Kb Ka for
NH4+: Kb of NH3=(1.8e-5) --> Ka=(e-14)/(1.8e-5)=5.56e-10 Ka for CH3NH3+: Kb for
CH3NH2=4.4e-4 --> Ka=(e-14)/(4.4e-4)=2.27e-11 Kb for F-: Ka for HF=6.8e-4 --> Kb=(e-
14)/(6.8e-4)=1.47e-11
Solution
The most simple property to calculate this is Ka*Kb=e-14 -->Ka=(e-14)/Kb Ka for
NH4+: Kb of NH3=(1.8e-5) --> Ka=(e-14)/(1.8e-5)=5.56e-10 Ka for CH3NH3+: Kb for
CH3NH2=4.4e-4 --> Ka=(e-14)/(4.4e-4)=2.27e-11 Kb for F-: Ka for HF=6.8e-4 --> Kb=(e-
14)/(6.8e-4)=1.47e-11.
The images of the compounds are not visible. Kin.pdf
The images of the compounds are not visible. Kindly post it in other question and i
will let you know.
Solution
The images of the compounds are not visible. Kindly post it in other question and i
will let you know..
Step1 Moles of water produced =18 Step2 The init.pdf
Step1 Moles of water produced =18 Step2 The initial concentrations of octane and
O2 are not known ; If known ,conc of octane can be found out.
Solution
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O2 are not known ; If known ,conc of octane can be found out..
Sorry. SnO2 is amphoteric. This one is a tricky o.pdf
Sorry. SnO2 is amphoteric. This one is a tricky one.
Solution
Sorry. SnO2 is amphoteric. This one is a tricky one..
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This very short document appears to be discussing sulfur dioxide (SO2) as it contains the chemical formula twice with the word "Solution" in between. It seems to be stating that sulfur dioxide is the subject being discussed and that there may be a solution related to it, but provides no other details in the 3 word document.
scientific synonyms for height may be peak or tal.pdf
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Solution
scientific synonyms for height may be peak or tallness sometimes length(in case of
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Solution
In general, they are alkenes attached to electron-withdrawing groups (esters,
ketones, nitriles, nitros). so the ans is D..
I, II S.pdf
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Group 11 has three elements namely, Cu, Ag and Au.pdf
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Group 11 has three elements namely, Cu, Ag and Au out of these elements Cu
exhibits +1 n +2 oxidation states Ag exhibits only +1 n Au exhibits +3.
Cu, (H2), Zn, Mg Activity series basedon hyd.pdf
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cis- So.pdf
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When a compound is completely oxidized with oxygen, the products wil.pdf
When a compound is completely oxidized with oxygen, the products will be H2O and CO2
Solution
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Calcium metal reacts with liquid water to produce.pdf
Calcium metal reacts with liquid water to produce calcium hydroxidesolution and
hydrogen gas Ca (s) + 2 H2O (l) > Ca(OH)2 (l) + H2 (g) Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
Solution
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hydrogen gas Ca (s) + 2 H2O (l) > Ca(OH)2 (l) + H2 (g) Ca(OH)2 + 2HCl --> CaCl2 + 2H2O.
two;Solutiontwo;.pdf
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The mineralogical and structural adjustment of solid rocks to physic.pdf
The mineralogical and structural adjustment of solid rocks to physical and chemical conditions
that have been imposed at depths below the near surface zones of weathering and diagenesis and
which differ from conditions under which the rocks in question originated.
The word \"Metamorphism\" comes from the Greek: meta = change, morph = form, so
metamorphism means to change form. In geology this refers to the changes in mineral
assemblage and texture that result from subjecting a rock to conditions such pressures,
temperatures, and chemical environments different from those under which the rock originally
formed.
There are two major kinds of metamorphism: regional and contact.
Regional metamorphism. Most metamorphic rocks are the result of regional metamorphism (also
called dynamothermal metamorphism). These rocks were typically exposed to tectonic forces
and associated high pressures and temperatures. They are usually foliated and deformed and
thought to be remnants of ancient mountain ranges.
Contact metamorphism. Contact metamorphism (also called thermal metamorphism) is the
process by which the country rock that surrounds a hot magma intrusion is metamorphosed by
the high heat flow coming from the intrusion. The zone of metamorphism that surrounds the
intrusion is called the halo (or aureole) and rarely extends more than 100 meters into the country
rock. Geostatic pressure is usually a minor factor, since contact metamorphism generally takes
place less than 10 kilometers from the surface.Metamorphism is defined as follows:
The mineralogical and structural adjustment of solid rocks to physical and chemical conditions
that have been imposed at depths below the near surface zones of weathering and diagenesis
and which differ from conditions under which the rocks in question originated.
The word \"Metamorphism\" comes from the Greek: meta = change, morph = form, so
metamorphism means to change form. In geology this refers to the changes in mineral
assemblage and texture that result from subjecting a rock to conditions such pressures,
temperatures, and chemical environments different from those under which the rock originally
formed.
There are two major kinds of metamorphism: regional and contact.
Regional metamorphism. Most metamorphic rocks are the result of regional metamorphism
(also called dynamothermal metamorphism). These rocks were typically exposed to tectonic
forces and associated high pressures and temperatures. They are usually foliated and deformed
and thought to be remnants of ancient mountain ranges.
Contact metamorphism. Contact metamorphism (also called thermal metamorphism) is the
process by which the country rock that surrounds a hot magma intrusion is metamorphosed by
the high heat flow coming from the intrusion. The zone of metamorphism that surrounds the
intrusion is called the halo (or aureole) and rarely extends more than 100 meters into the
country rock. Geostatic pressure is usually a minor factor, since contact metamorp.
The general idea is to find a root of the polynomial and then apply .pdf
The general idea is to find a root of the polynomial and then apply Horner\'s method to remove
the corresponding factor according to the Ruffini rule.
This iterative scheme is numerically unstable; the approximation errors accumulate during the
successive factorizations, so that the last roots are determined with a polynomial that deviates
widely from a factor of the original polynomial. To reduce this error, it is advisable to find the
roots in increasing order of magnitude.
Wilkinson\'s polynomial illustrates that high precision may be necessary when computing the
roots of a polynomial given its coefficients: the problem of finding the roots from the
coefficients is in general ill-conditioned.
The most simple method to find a single root fast is Newton\'s method. One can use Horner\'s
method twice to efficiently evaluate the value of the polynomial function and its first derivative;
this combination is called Birge–Vieta\'s method. This method provides quadratic convergence
for simple roots at the cost of two polynomial evaluations per step.
Closely related to Newton\'s method are Halley\'s method and Laguerre\'s method. Using one
additional Horner evaluation, the value of the second derivative is used to obtain methods of
cubic convergence order for simple roots. If one starts from a point x close to a root and uses the
same complexity of 6 function evaluations, these methods perform two steps with a residual of
O(|f(x)|9), compared to three steps of Newton\'s method with a reduction O(|f(x)|8), giving a
slight advantage to these methods.
When applying these methods to polynomials with real coefficients and real starting points,
Newton\'s and Halley\'s method stay inside the real number line. One has to choose complex
starting points to find complex roots. In contrast, the Laguerre method with a square root in its
evaluation will leave the real axis of its own accord.
Another class of methods is based on translating the problem of finding polynomial roots to the
problem of finding eigenvalues of the companion matrix of the polynomial. In principle, one can
use any eigenvalue algorithm to find the roots of the polynomial. However, for efficiency
reasons one prefers methods that employ the structure of the matrix, that is, can be implemented
in matrix-free form. Among these methods are the power method, whose application to the
transpose of the companion matrix is the classical Bernoulli\'s method to find the root of greatest
modulus. The inverse power method with shifts, which finds some smallest root first, is what
drives the complex (cpoly) variant of the Jenkins–Traub method and gives it its numerical
stability. Additionally, it is insensitive to multiple roots and has fast convergence with order
1
+
2.6
{\\displaystyle 1+\\Phi \\approx 2.6}
even in the presence of clustered roots. This fast convergence comes with a cost of three Horner
evaluations per step, resulting in a residual of O(|f(x)|2+3), which is slower than three steps of
New.
The answer is d. XThe most active nonmetal has the highest electro.pdf
The answer is d. X
The most active nonmetal has the highest electronegativity value as it should most readily accept
electrons.
Solution
The answer is d. X
The most active nonmetal has the highest electronegativity value as it should most readily accept
electrons..
The correct order is116124310139Rest of the statem.pdf
The correct order is
11
6
12
4
3
10
13
9
Rest of the statements are not correct as per the proof
Solution
The correct order is
11
6
12
4
3
10
13
9
Rest of the statements are not correct as per the proof.
B yellow to blue as the amount of hydronium ion .pdf
B yellow to blue as the amount of hydronium ion increase
Solution
B yellow to blue as the amount of hydronium ion increase.
Simply note that SO(n) is the kernel of the group homomorphism det.pdf
Simply note that SO(n) is the kernel of the group homomorphism
det:O(n)R AdetA
Recall that every kernel is a normal subgroup.
Solution
Simply note that SO(n) is the kernel of the group homomorphism
det:O(n)R AdetA
Recall that every kernel is a normal subgroup..
All liquids in containers have a concave meniscus.pdf
All liquids in containers have a concave meniscus. This is because surface
molecules being attracted to each other and also to the sides of the containers. Lakes also have a
concave meniscus, but they are so large we don\'t notice it.
Solution
All liquids in containers have a concave meniscus. This is because surface
molecules being attracted to each other and also to the sides of the containers. Lakes also have a
concave meniscus, but they are so large we don\'t notice it..
Mutants of the Drosophila dunce and rutabaga genes, which encode a c.pdf
Mutants of the Drosophila dunce and rutabaga genes, which encode a cAMP-specific
phosphodiesterase and a calcium/calmodulin-responsive adenylyl cyclase, respectively, are
deficient in short-term memory. Altered synaptic plasticity has been demonstrated at
neuromuscular junctions in these mutants, but little is known about how their central neurons are
affected. This problem was examined by using the \"giant\" neuron culture, which offers a
unique opportunity to analyze mutational effects on neuronal activity and the underlying ionic
currents in Drosophila. On the basis of instantaneous frequency and first latency of spikes
evoked by current steps, four categories of firing patterns (tonic, adaptive, delayed, and
interrupted) were identified in wild-type neurons, revealing interesting parallels to those
commonly observed in vertebrate CNS neurons.
The distinct firing patterns are correlated with expression of different ratios of 4-aminopyridine-
and tetra ethylammonium-sensitive K+ currents. Subsets of dnc and rut neurons display
abnormal spontaneous spikes and altered firing patterns. Altered frequency coding in mutant
neurons was demonstrated further by using stimulation protocols involving conditioning with
previous activity. Abnormal spike activity and reduced K+ current remain in double-mutant
neurons, suggesting that the opposite effects on cAMP metabolism by dnc and rut do not
counterbalance the mutual functional defects. The aberrant spontaneous activity and altered
frequency coding in different stimulus paradigms may present problems in the stability and
reliability of neural circuits for information processing during certain behavioral tasks, raising
the possibility of modulation in neuronal excitability as a cellular mechanism underlying learning
and memory (Zhao, 1997).
Not all memory genes first identified in other contexts, however, play a significant role in place
memory. The DopEcR gene has been implicated in several behaviors, including a 30 min
memory after courtship conditioning. This G-protein linked receptor is responsive to both
dopamine and the steroid hormone ecdysone. Remarkably, DopEcR has been shown to interact
with the cAMP cascade through double mutant and pharmacological tests. Using conditions that
induce a robust and lasting place memory, the DopEcR mutant flies do not show a defect in
memory directly after training or at 1 h post-training. This is despite the fact that the rut and
cAMP-phosphodiesterase genes (dunce) are critical for place memory. It may be that DopEcR is
not required for this type of learning and would be consistent with the independence of place
memory from dopamine signaling. Alternatively, other redundant pathways may compensate for
the reduction in DopEcR function caused by the DopEcRPB1 allele. One might further speculate
that other types of behavioral plasticity, such as reversal learning or memory enhancement after
unpredicted high temperature exposures in the heat-box might be more sen.
Mass of gold = 0.017100 x 1000= 0.17 kg = 170 gSolutionMass.pdf
Mass of gold = 0.017/100 x 1000
= 0.17 kg = 170 g
Solution
Mass of gold = 0.017/100 x 1000
= 0.17 kg = 170 g.
#include stdio.h#include stdlib.h#include string.hstruct.pdf
#include
#include
#include
struct product
{
int id;
char name;
int price;
int qty;
};
struct Bill
{
int pid;
char pname;
int pprice;
};
char mygetch();
int getid();
int billfileno();
void manageproduct();
void purchaseproduct();
void generatebill();
void addproduct();
void displayallproducts();
struct product findproduct(int id);
void updateproduct(int id, int qty);
char fproduct[]={\"product.txt\"};
char fbill[]={\"bill.txt\"};
int total=0;
int main()
{
File *fp;
int ch;
while(1)
{
printf(\"welcome product management system\ \");
printf(\"manage product\ \");
printf(\"\\purchase product\ \");
printf(\"generate bill\ \");
printf(\"exit\ \");
printf(\"enter your choice\");
scanf(\"%d\",&ch);
switch(ch)
{
case1:manageproduct();
break;
case2:purchaseproduct();
break;
case3:generate bill();
break;
case4:exit();
}
mygetch();
}
return 0;
int getid()
{
file *fp;
int value=0;
fp=fopen(\"product.txt\",\"r\");
if(fp==null)
{
fp=fopen(\"product.txt\",\"w\");
fprint(fp,\"%d\",0);
fclose(fp);
fp=fopen(\"product.txt\",\"r\");
}
fscanf(fp,\"%d\",&value);
fclose(fp);
fp=fopen(\"product.txt\",\"w\");
fprintf(\"fp,\"%d\",value+1);
fclose(fp);return value+1;
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{
int ch
init back=0;
while(1)
{
printf(\"welcome product management system\ \");
printf(\"1.add new product\ \");
printf(\"2.display all products\ \");
printf(\"back\ \");
printf(\"enter your choice\ \");
scanf(\"%d\",&ch);
switch(ch)
{
case1:addproduct();
break;
case2:displayallproduct();
break;
case3:(back==1)
break;
else
{
mygetch();
}
}
}
void addproduct()
{
file *fp;
struct product t1;
fp=fopen(fproduct,\"ab\");
t1.id=getid();
printf(\"enter product name\");
scanf(\"%s\",&t1.name);
printf(\"enter product price\");
scanf(\"%d\",&t1.price);
printf(\"enter product qty\");
scanf(\"%d\",&t1.qty);
fwrite(&t1,sizeof(t1),1,fp);
fclose(fp);
void dis[playallproducts()
{
file *fp;
struct product t;
int id,found=0;
fp=fopen(fproduct,\"rb\");
printf(\"product details\ \");
printf(\"id\\t name \\tqty\\tprice\ \ \");
while(1)
{
fread(&t,sizeof(t),1,fp);
if(feof(fp))
{
break;
}
printf(\"%d\\t\",t.id);
printf(\"%s\\t\",t.name);
printf(\"%d\\t\",t.qty);
printf(\"%d\\t\",t.price);
fclose(fp);
}
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char ch1,ch2;
int id;
while(1)
{
displayallproducts();
_fpurge(stdin);
printf(\"do u want to purchase y/n\");
scanf(\"%c\",&ch1);
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fp=fopen(fbill,\"ab\");
printf(\"enter the product id to purchase\");
scanf(\"%d\",&id);
t2=findproduct(id);
t1.pid=t2.id;
strcpy(t1.pname,t2.name);
t1.pprice=t2.price;
fwrite(&t1,sizeof(t1),1,fp);
fclose(fp);
}
_fpurge(stdin);
printf(\"do u want to continue y/n\");
scanf(\"%c\",&ch2);
if(ch2==\'n\')
{
break;
}
mygetch();
}
}
struct product findproduct(int id)
{
file *fp;
struct product t;
fp=fopen(fproduct,\"rb\");
if(feof(fp))
{
break;
}
if(t.id==id)
{
updateproduct(id,t.qty-1);
break;
}
}
fclose(fp);
return t;
}
void updateproduct(int id,int qty)
{
file *fp,*fp1;
struct product t,t1;
int found=.
Advanced Java[Extra Concepts, Not Difficult].docx
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Your Skill Boost Masterclass: Strategies for Effective Upskilling
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R = k[NOCl2][NO] Molecularity = 2 .pdf
This reaction has a rate law of R = k[NOCl2][NO] with a molecularity of 2. The rate (R) of the reaction is directly proportional to the concentrations of NOCl2 and NO, with the proportionality constant being the rate constant (k). This reaction requires collisions between two reactant molecules, NOCl2 and NO, in order for a reaction to occur.
In general, they are alkenes attached to electron.pdf
In general, they are alkenes attached to electron-withdrawing groups (esters,
ketones, nitriles, nitros). so the ans is D.
Solution
In general, they are alkenes attached to electron-withdrawing groups (esters,
ketones, nitriles, nitros). so the ans is D..
I, II S.pdf
This short document appears to be discussing a solution to a problem or issue numbered as I and II. It briefly mentions I and II at the beginning and end but does not provide any other context or details about the topic being discussed. The middle section indicates there is a solution presented but without more information it is difficult to understand what problem is being addressed or what the proposed solution entails.
Group 11 has three elements namely, Cu, Ag and Au.pdf
Group 11 has three elements namely, Cu, Ag and Au out of these elements Cu
exhibits +1 n +2 oxidation states Ag exhibits only +1 n Au exhibits +3
Solution
Group 11 has three elements namely, Cu, Ag and Au out of these elements Cu
exhibits +1 n +2 oxidation states Ag exhibits only +1 n Au exhibits +3.
Cu, (H2), Zn, Mg Activity series basedon hyd.pdf
Cu, (H2)**, Zn, Mg **Activity series basedon hydrogen standard
Solution
Cu, (H2)**, Zn, Mg **Activity series basedon hydrogen standard.
answewrSolutionanswewr.pdf
This very short document does not contain any substantive information to summarize in 3 sentences or less. It consists of only 3 words with no context provided.
cis- So.pdf
This very short document does not contain enough substantive information to generate a multi-sentence summary. It appears to be an incomplete or placeholder document.
When a compound is completely oxidized with oxygen, the products wil.pdf
When a compound is completely oxidized with oxygen, the products will be H2O and CO2
Solution
When a compound is completely oxidized with oxygen, the products will be H2O and CO2.
Calcium metal reacts with liquid water to produce.pdf
Calcium metal reacts with liquid water to produce calcium hydroxidesolution and
hydrogen gas Ca (s) + 2 H2O (l) > Ca(OH)2 (l) + H2 (g) Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
Solution
Calcium metal reacts with liquid water to produce calcium hydroxidesolution and
hydrogen gas Ca (s) + 2 H2O (l) > Ca(OH)2 (l) + H2 (g) Ca(OH)2 + 2HCl --> CaCl2 + 2H2O.
two;Solutiontwo;.pdf
This very short document contains a single word - "Solution" - with no other context provided. It is not possible to provide a meaningful 3 sentence summary as there is insufficient information and content in the original document.
The mineralogical and structural adjustment of solid rocks to physic.pdf
The mineralogical and structural adjustment of solid rocks to physical and chemical conditions
that have been imposed at depths below the near surface zones of weathering and diagenesis and
which differ from conditions under which the rocks in question originated.
The word \"Metamorphism\" comes from the Greek: meta = change, morph = form, so
metamorphism means to change form. In geology this refers to the changes in mineral
assemblage and texture that result from subjecting a rock to conditions such pressures,
temperatures, and chemical environments different from those under which the rock originally
formed.
There are two major kinds of metamorphism: regional and contact.
Regional metamorphism. Most metamorphic rocks are the result of regional metamorphism (also
called dynamothermal metamorphism). These rocks were typically exposed to tectonic forces
and associated high pressures and temperatures. They are usually foliated and deformed and
thought to be remnants of ancient mountain ranges.
Contact metamorphism. Contact metamorphism (also called thermal metamorphism) is the
process by which the country rock that surrounds a hot magma intrusion is metamorphosed by
the high heat flow coming from the intrusion. The zone of metamorphism that surrounds the
intrusion is called the halo (or aureole) and rarely extends more than 100 meters into the country
rock. Geostatic pressure is usually a minor factor, since contact metamorphism generally takes
place less than 10 kilometers from the surface.Metamorphism is defined as follows:
The mineralogical and structural adjustment of solid rocks to physical and chemical conditions
that have been imposed at depths below the near surface zones of weathering and diagenesis
and which differ from conditions under which the rocks in question originated.
The word \"Metamorphism\" comes from the Greek: meta = change, morph = form, so
metamorphism means to change form. In geology this refers to the changes in mineral
assemblage and texture that result from subjecting a rock to conditions such pressures,
temperatures, and chemical environments different from those under which the rock originally
formed.
There are two major kinds of metamorphism: regional and contact.
Regional metamorphism. Most metamorphic rocks are the result of regional metamorphism
(also called dynamothermal metamorphism). These rocks were typically exposed to tectonic
forces and associated high pressures and temperatures. They are usually foliated and deformed
and thought to be remnants of ancient mountain ranges.
Contact metamorphism. Contact metamorphism (also called thermal metamorphism) is the
process by which the country rock that surrounds a hot magma intrusion is metamorphosed by
the high heat flow coming from the intrusion. The zone of metamorphism that surrounds the
intrusion is called the halo (or aureole) and rarely extends more than 100 meters into the
country rock. Geostatic pressure is usually a minor factor, since contact metamorp.
The general idea is to find a root of the polynomial and then apply .pdf
The general idea is to find a root of the polynomial and then apply Horner\'s method to remove
the corresponding factor according to the Ruffini rule.
This iterative scheme is numerically unstable; the approximation errors accumulate during the
successive factorizations, so that the last roots are determined with a polynomial that deviates
widely from a factor of the original polynomial. To reduce this error, it is advisable to find the
roots in increasing order of magnitude.
Wilkinson\'s polynomial illustrates that high precision may be necessary when computing the
roots of a polynomial given its coefficients: the problem of finding the roots from the
coefficients is in general ill-conditioned.
The most simple method to find a single root fast is Newton\'s method. One can use Horner\'s
method twice to efficiently evaluate the value of the polynomial function and its first derivative;
this combination is called Birge–Vieta\'s method. This method provides quadratic convergence
for simple roots at the cost of two polynomial evaluations per step.
Closely related to Newton\'s method are Halley\'s method and Laguerre\'s method. Using one
additional Horner evaluation, the value of the second derivative is used to obtain methods of
cubic convergence order for simple roots. If one starts from a point x close to a root and uses the
same complexity of 6 function evaluations, these methods perform two steps with a residual of
O(|f(x)|9), compared to three steps of Newton\'s method with a reduction O(|f(x)|8), giving a
slight advantage to these methods.
When applying these methods to polynomials with real coefficients and real starting points,
Newton\'s and Halley\'s method stay inside the real number line. One has to choose complex
starting points to find complex roots. In contrast, the Laguerre method with a square root in its
evaluation will leave the real axis of its own accord.
Another class of methods is based on translating the problem of finding polynomial roots to the
problem of finding eigenvalues of the companion matrix of the polynomial. In principle, one can
use any eigenvalue algorithm to find the roots of the polynomial. However, for efficiency
reasons one prefers methods that employ the structure of the matrix, that is, can be implemented
in matrix-free form. Among these methods are the power method, whose application to the
transpose of the companion matrix is the classical Bernoulli\'s method to find the root of greatest
modulus. The inverse power method with shifts, which finds some smallest root first, is what
drives the complex (cpoly) variant of the Jenkins–Traub method and gives it its numerical
stability. Additionally, it is insensitive to multiple roots and has fast convergence with order
1
+
2.6
{\\displaystyle 1+\\Phi \\approx 2.6}
even in the presence of clustered roots. This fast convergence comes with a cost of three Horner
evaluations per step, resulting in a residual of O(|f(x)|2+3), which is slower than three steps of
New.
The answer is d. XThe most active nonmetal has the highest electro.pdf
The answer is d. X
The most active nonmetal has the highest electronegativity value as it should most readily accept
electrons.
Solution
The answer is d. X
The most active nonmetal has the highest electronegativity value as it should most readily accept
electrons..
The correct order is116124310139Rest of the statem.pdf
The correct order is
11
6
12
4
3
10
13
9
Rest of the statements are not correct as per the proof
Solution
The correct order is
11
6
12
4
3
10
13
9
Rest of the statements are not correct as per the proof.
B yellow to blue as the amount of hydronium ion .pdf
B yellow to blue as the amount of hydronium ion increase
Solution
B yellow to blue as the amount of hydronium ion increase.
Simply note that SO(n) is the kernel of the group homomorphism det.pdf
Simply note that SO(n) is the kernel of the group homomorphism
det:O(n)R AdetA
Recall that every kernel is a normal subgroup.
Solution
Simply note that SO(n) is the kernel of the group homomorphism
det:O(n)R AdetA
Recall that every kernel is a normal subgroup..
All liquids in containers have a concave meniscus.pdf
All liquids in containers have a concave meniscus. This is because surface
molecules being attracted to each other and also to the sides of the containers. Lakes also have a
concave meniscus, but they are so large we don\'t notice it.
Solution
All liquids in containers have a concave meniscus. This is because surface
molecules being attracted to each other and also to the sides of the containers. Lakes also have a
concave meniscus, but they are so large we don\'t notice it..
Mutants of the Drosophila dunce and rutabaga genes, which encode a c.pdf
Mutants of the Drosophila dunce and rutabaga genes, which encode a cAMP-specific
phosphodiesterase and a calcium/calmodulin-responsive adenylyl cyclase, respectively, are
deficient in short-term memory. Altered synaptic plasticity has been demonstrated at
neuromuscular junctions in these mutants, but little is known about how their central neurons are
affected. This problem was examined by using the \"giant\" neuron culture, which offers a
unique opportunity to analyze mutational effects on neuronal activity and the underlying ionic
currents in Drosophila. On the basis of instantaneous frequency and first latency of spikes
evoked by current steps, four categories of firing patterns (tonic, adaptive, delayed, and
interrupted) were identified in wild-type neurons, revealing interesting parallels to those
commonly observed in vertebrate CNS neurons.
The distinct firing patterns are correlated with expression of different ratios of 4-aminopyridine-
and tetra ethylammonium-sensitive K+ currents. Subsets of dnc and rut neurons display
abnormal spontaneous spikes and altered firing patterns. Altered frequency coding in mutant
neurons was demonstrated further by using stimulation protocols involving conditioning with
previous activity. Abnormal spike activity and reduced K+ current remain in double-mutant
neurons, suggesting that the opposite effects on cAMP metabolism by dnc and rut do not
counterbalance the mutual functional defects. The aberrant spontaneous activity and altered
frequency coding in different stimulus paradigms may present problems in the stability and
reliability of neural circuits for information processing during certain behavioral tasks, raising
the possibility of modulation in neuronal excitability as a cellular mechanism underlying learning
and memory (Zhao, 1997).
Not all memory genes first identified in other contexts, however, play a significant role in place
memory. The DopEcR gene has been implicated in several behaviors, including a 30 min
memory after courtship conditioning. This G-protein linked receptor is responsive to both
dopamine and the steroid hormone ecdysone. Remarkably, DopEcR has been shown to interact
with the cAMP cascade through double mutant and pharmacological tests. Using conditions that
induce a robust and lasting place memory, the DopEcR mutant flies do not show a defect in
memory directly after training or at 1 h post-training. This is despite the fact that the rut and
cAMP-phosphodiesterase genes (dunce) are critical for place memory. It may be that DopEcR is
not required for this type of learning and would be consistent with the independence of place
memory from dopamine signaling. Alternatively, other redundant pathways may compensate for
the reduction in DopEcR function caused by the DopEcRPB1 allele. One might further speculate
that other types of behavioral plasticity, such as reversal learning or memory enhancement after
unpredicted high temperature exposures in the heat-box might be more sen.
Mass of gold = 0.017100 x 1000= 0.17 kg = 170 gSolutionMass.pdf
Mass of gold = 0.017/100 x 1000
= 0.17 kg = 170 g
Solution
Mass of gold = 0.017/100 x 1000
= 0.17 kg = 170 g.
#include stdio.h#include stdlib.h#include string.hstruct.pdf
#include
#include
#include
struct product
{
int id;
char name;
int price;
int qty;
};
struct Bill
{
int pid;
char pname;
int pprice;
};
char mygetch();
int getid();
int billfileno();
void manageproduct();
void purchaseproduct();
void generatebill();
void addproduct();
void displayallproducts();
struct product findproduct(int id);
void updateproduct(int id, int qty);
char fproduct[]={\"product.txt\"};
char fbill[]={\"bill.txt\"};
int total=0;
int main()
{
File *fp;
int ch;
while(1)
{
printf(\"welcome product management system\ \");
printf(\"manage product\ \");
printf(\"\\purchase product\ \");
printf(\"generate bill\ \");
printf(\"exit\ \");
printf(\"enter your choice\");
scanf(\"%d\",&ch);
switch(ch)
{
case1:manageproduct();
break;
case2:purchaseproduct();
break;
case3:generate bill();
break;
case4:exit();
}
mygetch();
}
return 0;
int getid()
{
file *fp;
int value=0;
fp=fopen(\"product.txt\",\"r\");
if(fp==null)
{
fp=fopen(\"product.txt\",\"w\");
fprint(fp,\"%d\",0);
fclose(fp);
fp=fopen(\"product.txt\",\"r\");
}
fscanf(fp,\"%d\",&value);
fclose(fp);
fp=fopen(\"product.txt\",\"w\");
fprintf(\"fp,\"%d\",value+1);
fclose(fp);return value+1;
void manageproduct()
{
int ch
init back=0;
while(1)
{
printf(\"welcome product management system\ \");
printf(\"1.add new product\ \");
printf(\"2.display all products\ \");
printf(\"back\ \");
printf(\"enter your choice\ \");
scanf(\"%d\",&ch);
switch(ch)
{
case1:addproduct();
break;
case2:displayallproduct();
break;
case3:(back==1)
break;
else
{
mygetch();
}
}
}
void addproduct()
{
file *fp;
struct product t1;
fp=fopen(fproduct,\"ab\");
t1.id=getid();
printf(\"enter product name\");
scanf(\"%s\",&t1.name);
printf(\"enter product price\");
scanf(\"%d\",&t1.price);
printf(\"enter product qty\");
scanf(\"%d\",&t1.qty);
fwrite(&t1,sizeof(t1),1,fp);
fclose(fp);
void dis[playallproducts()
{
file *fp;
struct product t;
int id,found=0;
fp=fopen(fproduct,\"rb\");
printf(\"product details\ \");
printf(\"id\\t name \\tqty\\tprice\ \ \");
while(1)
{
fread(&t,sizeof(t),1,fp);
if(feof(fp))
{
break;
}
printf(\"%d\\t\",t.id);
printf(\"%s\\t\",t.name);
printf(\"%d\\t\",t.qty);
printf(\"%d\\t\",t.price);
fclose(fp);
}
void purchase product()
char ch1,ch2;
int id;
while(1)
{
displayallproducts();
_fpurge(stdin);
printf(\"do u want to purchase y/n\");
scanf(\"%c\",&ch1);
if(ch1==\'y\')
{
file *fp;
struct bill t1;
struct product t2;
fp=fopen(fbill,\"ab\");
printf(\"enter the product id to purchase\");
scanf(\"%d\",&id);
t2=findproduct(id);
t1.pid=t2.id;
strcpy(t1.pname,t2.name);
t1.pprice=t2.price;
fwrite(&t1,sizeof(t1),1,fp);
fclose(fp);
}
_fpurge(stdin);
printf(\"do u want to continue y/n\");
scanf(\"%c\",&ch2);
if(ch2==\'n\')
{
break;
}
mygetch();
}
}
struct product findproduct(int id)
{
file *fp;
struct product t;
fp=fopen(fproduct,\"rb\");
if(feof(fp))
{
break;
}
if(t.id==id)
{
updateproduct(id,t.qty-1);
break;
}
}
fclose(fp);
return t;
}
void updateproduct(int id,int qty)
{
file *fp,*fp1;
struct product t,t1;
int found=.
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