It will take time. So, wait for some time.I will send it.Sol.pdf
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It will take time. So, wait for some time. I will send it. Solution It will take time. So, wait for some time. I will send it..
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Cu(H2O)42+(aq) + 4CN-(aq) = Cu(CN)42-(aq) +4 H2O(l)Cu+2 is a Lew.pdf
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor.
Solution
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor..
C) CS2 for sureA, B, D and E are all ionic compoundsSolution.pdf
C) CS2 for sure
A, B, D and E are all ionic compounds
Solution
C) CS2 for sure
A, B, D and E are all ionic compounds.
Before getting public issues he should consider are 1. Increase i.pdf
Before getting public issues he should consider are :
1. Increase in scrutiny by regulatory authorities
2. Become answerable to shareholders.
3. Take consideration on any big decision from new shareholders.
4. Bearing additional cost of filing financial IPO
Solution
Before getting public issues he should consider are :
1. Increase in scrutiny by regulatory authorities
2. Become answerable to shareholders.
3. Take consideration on any big decision from new shareholders.
4. Bearing additional cost of filing financial IPO.
Assets=Liabilities+Stockholders EquityCash given $1000 on the sa.pdf
Assets=Liabilities+Stockholder\'s Equity
Cash given $1000 on the same day and $1000 is Accounts Payable
Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000
Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000
Solution
Assets=Liabilities+Stockholder\'s Equity
Cash given $1000 on the same day and $1000 is Accounts Payable
Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000
Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000.
1)Interest per year 8106= 135Simple interest = Amount RateTime.pdf
1)Interest per year 810/6= 135
Simple interest = Amount * Rate*Time
135 = 3750 *r * 1
135 = 3750*r
r = 135/3750
= .036 or 3.6%
2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year]
Term (number of quarters) = IN[Future value/present value]/IN(1+r)
= IN[8600/2000]/IN[1+.01225]
= IN 4.3/ IN 1.01225
= 1.45862/ .012176
= 119.79468
Number of years =119.79468 / 4
= 29.95 years
Term : 29.95
Solution
1)Interest per year 810/6= 135
Simple interest = Amount * Rate*Time
135 = 3750 *r * 1
135 = 3750*r
r = 135/3750
= .036 or 3.6%
2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year]
Term (number of quarters) = IN[Future value/present value]/IN(1+r)
= IN[8600/2000]/IN[1+.01225]
= IN 4.3/ IN 1.01225
= 1.45862/ .012176
= 119.79468
Number of years =119.79468 / 4
= 29.95 years
Term : 29.95
.
AnswerAnswer1. The body of a fungus is in the form of solid sh.pdf
Answer:
Answer:
1. The body of a fungus is in the form of solid sheets: True
The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf.
2. Yeasts are a kind of fungus: True
Yeast are unicellular fungus
3. False
Spore are haploid, the need to fuse with another to produce diploid, multicellular individual.
4. False
Phloem is alive at maturity because movement of material within the phloem requires energy.
5. True
6. True
7. True
Ferns have motile sperms and non motile egg.
8. False
Most of the plant species are angiosperms.
9. False
Angiosperms are divided into monocots and dicots
10. Charophyte---e. not a plant
11. fern---d. seedless vascular plant
12. pine tree—b. gymnosperm
13. moss—c. non-vascular plant
14. palm tree—a. angiosperm
Solution
Answer:
Answer:
1. The body of a fungus is in the form of solid sheets: True
The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf.
2. Yeasts are a kind of fungus: True
Yeast are unicellular fungus
3. False
Spore are haploid, the need to fuse with another to produce diploid, multicellular individual.
4. False
Phloem is alive at maturity because movement of material within the phloem requires energy.
5. True
6. True
7. True
Ferns have motile sperms and non motile egg.
8. False
Most of the plant species are angiosperms.
9. False
Angiosperms are divided into monocots and dicots
10. Charophyte---e. not a plant
11. fern---d. seedless vascular plant
12. pine tree—b. gymnosperm
13. moss—c. non-vascular plant
14. palm tree—a. angiosperm.
A. This constitutes part of a gene that encodes information for RNA .pdf
A. This constitutes part of a gene that encodes information for RNA Polymerase : Operator.
B. An activator and a repressor are the examples of : Transcription factors.
C. regulation of gene expression leads to changes in cell functions and in multicellular
organisms,can also lead to changes in cellular phenotype.
D. In bacteria, related genes can be organised in units called Operon controlled by a single
Promoter.
E. Lactose is an example of a compound that can induce lac operon.
F.Repressors are proteins which bind to DNA and alter gene expression.
G. The lac operon can be turned ON by the accumulation of the CRP-cAMP complex. Once
tghis complex binds to the DNA , it causes a change in the conformation of the DNA double
helix , making the operator, for the lac operon more attractive to RNA Polymerase.
H. Inhibition of a gene\'s expression often called Negative control , is due to the activity of a
negative regulator (Repressor) which can act to block RNA polymaerase from assessing the
promoter of a gene or inhibit RNA Polymerase from physically moving down the DNA strand.
I. The lac operon is an example of an Inducible operon because it can be induced to turn ON by
the presence of Lactose.
J. activation of a gene\'s expression often called, Positive control is due to the activity of the
Positive regulator (activator), to attract RNA polymerase to the Promoter for the gene.
K. People who cannot digest lactose, because they fail to produce the enzyme lactase, are said to
be lactose intolerant.
Solution
A. This constitutes part of a gene that encodes information for RNA Polymerase : Operator.
B. An activator and a repressor are the examples of : Transcription factors.
C. regulation of gene expression leads to changes in cell functions and in multicellular
organisms,can also lead to changes in cellular phenotype.
D. In bacteria, related genes can be organised in units called Operon controlled by a single
Promoter.
E. Lactose is an example of a compound that can induce lac operon.
F.Repressors are proteins which bind to DNA and alter gene expression.
G. The lac operon can be turned ON by the accumulation of the CRP-cAMP complex. Once
tghis complex binds to the DNA , it causes a change in the conformation of the DNA double
helix , making the operator, for the lac operon more attractive to RNA Polymerase.
H. Inhibition of a gene\'s expression often called Negative control , is due to the activity of a
negative regulator (Repressor) which can act to block RNA polymaerase from assessing the
promoter of a gene or inhibit RNA Polymerase from physically moving down the DNA strand.
I. The lac operon is an example of an Inducible operon because it can be induced to turn ON by
the presence of Lactose.
J. activation of a gene\'s expression often called, Positive control is due to the activity of the
Positive regulator (activator), to attract RNA polymerase to the Promoter for the gene.
K. People who cannot digest lactose, because they fail to produce the en.
A. ANSAutosomal dominant (a trait could passed down from previous.pdf
A. ANS:
Autosomal dominant (a trait could passed down from previous generation), X – linked recessive
(in this disease the dominant gene is not expressed in female, it only expressed in male) and
Autosomal recessive in female (Females are only carriers) effects would be included according
to pedigree.
B. ANS:
Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in
male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect
causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited
disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision.
C. ANS:
The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be
normal (According to the pedigree in offspring 4 (Female) is normal).
D. ANS:
The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected).
Solution
A. ANS:
Autosomal dominant (a trait could passed down from previous generation), X – linked recessive
(in this disease the dominant gene is not expressed in female, it only expressed in male) and
Autosomal recessive in female (Females are only carriers) effects would be included according
to pedigree.
B. ANS:
Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in
male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect
causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited
disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision.
C. ANS:
The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be
normal (According to the pedigree in offspring 4 (Female) is normal).
D. ANS:
The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected)..
Recommended
Cu(H2O)42+(aq) + 4CN-(aq) = Cu(CN)42-(aq) +4 H2O(l)Cu+2 is a Lew.pdf
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor.
Solution
Cu(H2O)42+(aq) + 4CN-(aq) <=> Cu(CN)42-(aq) +4 H2O(l)
Cu+2 is a Lewis acid as it has vacantd-orbitals that can accomodate an electron pair(Electron
pairacceptor)
CN- is a Lewis base as it has lone pair ofelectron that is donated to Cu+2 in the reaction(Electron
pair donor)
H2O is a Lewis base as the Oxygen atom haslone pair of electrons and therefre, it acts as an
electronpair donor..
C) CS2 for sureA, B, D and E are all ionic compoundsSolution.pdf
C) CS2 for sure
A, B, D and E are all ionic compounds
Solution
C) CS2 for sure
A, B, D and E are all ionic compounds.
Before getting public issues he should consider are 1. Increase i.pdf
Before getting public issues he should consider are :
1. Increase in scrutiny by regulatory authorities
2. Become answerable to shareholders.
3. Take consideration on any big decision from new shareholders.
4. Bearing additional cost of filing financial IPO
Solution
Before getting public issues he should consider are :
1. Increase in scrutiny by regulatory authorities
2. Become answerable to shareholders.
3. Take consideration on any big decision from new shareholders.
4. Bearing additional cost of filing financial IPO.
Assets=Liabilities+Stockholders EquityCash given $1000 on the sa.pdf
Assets=Liabilities+Stockholder\'s Equity
Cash given $1000 on the same day and $1000 is Accounts Payable
Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000
Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000
Solution
Assets=Liabilities+Stockholder\'s Equity
Cash given $1000 on the same day and $1000 is Accounts Payable
Cash given $1000 on the same day accounting equation -$1000=Liabilities-$1000
Accounts payable for remaining half of the expenses accounting equation Assets= +$1000-$1000.
1)Interest per year 8106= 135Simple interest = Amount RateTime.pdf
1)Interest per year 810/6= 135
Simple interest = Amount * Rate*Time
135 = 3750 *r * 1
135 = 3750*r
r = 135/3750
= .036 or 3.6%
2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year]
Term (number of quarters) = IN[Future value/present value]/IN(1+r)
= IN[8600/2000]/IN[1+.01225]
= IN 4.3/ IN 1.01225
= 1.45862/ .012176
= 119.79468
Number of years =119.79468 / 4
= 29.95 years
Term : 29.95
Solution
1)Interest per year 810/6= 135
Simple interest = Amount * Rate*Time
135 = 3750 *r * 1
135 = 3750*r
r = 135/3750
= .036 or 3.6%
2) Quarterly rate = 4.9/4 = 1.225% [4 quarters in a year]
Term (number of quarters) = IN[Future value/present value]/IN(1+r)
= IN[8600/2000]/IN[1+.01225]
= IN 4.3/ IN 1.01225
= 1.45862/ .012176
= 119.79468
Number of years =119.79468 / 4
= 29.95 years
Term : 29.95
.
AnswerAnswer1. The body of a fungus is in the form of solid sh.pdf
Answer:
Answer:
1. The body of a fungus is in the form of solid sheets: True
The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf.
2. Yeasts are a kind of fungus: True
Yeast are unicellular fungus
3. False
Spore are haploid, the need to fuse with another to produce diploid, multicellular individual.
4. False
Phloem is alive at maturity because movement of material within the phloem requires energy.
5. True
6. True
7. True
Ferns have motile sperms and non motile egg.
8. False
Most of the plant species are angiosperms.
9. False
Angiosperms are divided into monocots and dicots
10. Charophyte---e. not a plant
11. fern---d. seedless vascular plant
12. pine tree—b. gymnosperm
13. moss—c. non-vascular plant
14. palm tree—a. angiosperm
Solution
Answer:
Answer:
1. The body of a fungus is in the form of solid sheets: True
The body of fungus is called mycelium, it is not differentiated into roots, stem or leaf.
2. Yeasts are a kind of fungus: True
Yeast are unicellular fungus
3. False
Spore are haploid, the need to fuse with another to produce diploid, multicellular individual.
4. False
Phloem is alive at maturity because movement of material within the phloem requires energy.
5. True
6. True
7. True
Ferns have motile sperms and non motile egg.
8. False
Most of the plant species are angiosperms.
9. False
Angiosperms are divided into monocots and dicots
10. Charophyte---e. not a plant
11. fern---d. seedless vascular plant
12. pine tree—b. gymnosperm
13. moss—c. non-vascular plant
14. palm tree—a. angiosperm.
A. This constitutes part of a gene that encodes information for RNA .pdf
A. This constitutes part of a gene that encodes information for RNA Polymerase : Operator.
B. An activator and a repressor are the examples of : Transcription factors.
C. regulation of gene expression leads to changes in cell functions and in multicellular
organisms,can also lead to changes in cellular phenotype.
D. In bacteria, related genes can be organised in units called Operon controlled by a single
Promoter.
E. Lactose is an example of a compound that can induce lac operon.
F.Repressors are proteins which bind to DNA and alter gene expression.
G. The lac operon can be turned ON by the accumulation of the CRP-cAMP complex. Once
tghis complex binds to the DNA , it causes a change in the conformation of the DNA double
helix , making the operator, for the lac operon more attractive to RNA Polymerase.
H. Inhibition of a gene\'s expression often called Negative control , is due to the activity of a
negative regulator (Repressor) which can act to block RNA polymaerase from assessing the
promoter of a gene or inhibit RNA Polymerase from physically moving down the DNA strand.
I. The lac operon is an example of an Inducible operon because it can be induced to turn ON by
the presence of Lactose.
J. activation of a gene\'s expression often called, Positive control is due to the activity of the
Positive regulator (activator), to attract RNA polymerase to the Promoter for the gene.
K. People who cannot digest lactose, because they fail to produce the enzyme lactase, are said to
be lactose intolerant.
Solution
A. This constitutes part of a gene that encodes information for RNA Polymerase : Operator.
B. An activator and a repressor are the examples of : Transcription factors.
C. regulation of gene expression leads to changes in cell functions and in multicellular
organisms,can also lead to changes in cellular phenotype.
D. In bacteria, related genes can be organised in units called Operon controlled by a single
Promoter.
E. Lactose is an example of a compound that can induce lac operon.
F.Repressors are proteins which bind to DNA and alter gene expression.
G. The lac operon can be turned ON by the accumulation of the CRP-cAMP complex. Once
tghis complex binds to the DNA , it causes a change in the conformation of the DNA double
helix , making the operator, for the lac operon more attractive to RNA Polymerase.
H. Inhibition of a gene\'s expression often called Negative control , is due to the activity of a
negative regulator (Repressor) which can act to block RNA polymaerase from assessing the
promoter of a gene or inhibit RNA Polymerase from physically moving down the DNA strand.
I. The lac operon is an example of an Inducible operon because it can be induced to turn ON by
the presence of Lactose.
J. activation of a gene\'s expression often called, Positive control is due to the activity of the
Positive regulator (activator), to attract RNA polymerase to the Promoter for the gene.
K. People who cannot digest lactose, because they fail to produce the en.
A. ANSAutosomal dominant (a trait could passed down from previous.pdf
A. ANS:
Autosomal dominant (a trait could passed down from previous generation), X – linked recessive
(in this disease the dominant gene is not expressed in female, it only expressed in male) and
Autosomal recessive in female (Females are only carriers) effects would be included according
to pedigree.
B. ANS:
Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in
male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect
causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited
disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision.
C. ANS:
The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be
normal (According to the pedigree in offspring 4 (Female) is normal).
D. ANS:
The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected).
Solution
A. ANS:
Autosomal dominant (a trait could passed down from previous generation), X – linked recessive
(in this disease the dominant gene is not expressed in female, it only expressed in male) and
Autosomal recessive in female (Females are only carriers) effects would be included according
to pedigree.
B. ANS:
Leber hereditary optic atrophy which causes blindness in adulthood predominantly occurs in
male, in this the inheritance a point mutation was arisen in mitochondrial DNA and this affect
causing gene is directly transmitted from mother to offspring. It is mitochondrial inherited
disease in which a retinal ganglion cell was degenerated, and it leads to lose of central vision.
C. ANS:
The individuals II 3 (Male) and II 4 (Female) both are normal, so the offspring always should be
normal (According to the pedigree in offspring 4 (Female) is normal).
D. ANS:
The individuals II 2 (Male) and II 1 (Female) both are not in normal condition. According to the
pedigree male is normal and the female is not normal so the entire offspring is affected in this
disease. (According to the pedigree in offspring 1 (Female) is carrier and 2 & 3 (Males) are
affected)..
888678123317368915236 #include stdio.h #include stdlib.h.pdf
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}
Solution
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}.
18) One device is attached to a physical interface The switch c.pdf
18) One device is attached to a physical interface
The switch can be remotely managed
The default VIan has been configured
19) Flow control
20) Encapsulation
21) HTTP,TCP,IP,Ethernet
22) They define how messages are exchanged between the source and the destination
Solution
18) One device is attached to a physical interface
The switch can be remotely managed
The default VIan has been configured
19) Flow control
20) Encapsulation
21) HTTP,TCP,IP,Ethernet
22) They define how messages are exchanged between the source and the destination.
1.Joseph could not restrain himself before all those who stood by hi.pdf
1.Joseph could not restrain himself before all those who stood by him, and he cried out, “Make
everyone go out from me!” So no one stood with him while Joseph made himself known to his
brothers. And he wept aloud, and the Egyptians and the house of Pharaoh heard it. Then Joseph
said to his brothers, “I am Joseph; does my father still live?” But his brothers could not answer
him, for they were dismayed in his presence.
a. Joseph could not restrain himself before all those who stood by him: Joseph ordered all the
Egyptians out of the room and was then alone with his brothers. His great emotion showed that
Joseph did not cruelly manipulate his brothers. He was directed by God to make these
arrangements and it hurt him to do it.
b. Joseph made himself known to his brothers: This perhaps means that Joseph told them he was
Joseph and showed his brothers that he was circumcised. Jewish legend says the brothers could
never believe this high Egyptian official was Joseph unless he showed he was circumcised.
c. But his brothers could not answer him, for they were dismayed in his presence: Because of the
punishment they anticipated, the great emotion of Joseph, his manner of revelation, and the total
shock of learning Joseph was not only alive but right in front of them, the brothers were
dismayed.
Pharaoh and Joseph send the brothers home with many gifts.
Joseph gave them carts, according to the command of Pharaoh, and he gave them provisions for
the journey: The sons of Israel received transportation, provision, garments, and riches because
of who their favored brother was. Pharaoh blessed the sons of Jacob for Joseph’s sake.
i. “To return to Canaan with ‘carts from Egypt’ was the cultural equivalent of landing a jumbo jet
among a tribe of isolated savages. It would be the stuff legends are made of.” (Boice)
b. See that you do not become troubled along the way: The idea behind the words “become
troubled” is literally become angry or quarrel. Joseph knew as soon as these men left his
presence they would be tempted to act in selfish, unspiritual ways. They had to anticipate and
guard against this.
Jacob hears the good news – that Joseph lives.
Then they went up out of Egypt, and came to the land of Canaan to Jacob their father. And they
told him, saying, “Joseph is still alive, and he is governor over all the land of Egypt.” And
Jacob’s heart stood still, because he did not believe them. But when they told him all the words
which Joseph had said to them, and when he saw the carts which Joseph had sent to carry him,
the spirit of Jacob their father revived. Then Israel said, “It is enough. Joseph my son is still
alive. I will go and see him before I die.”
a. He did not believe them: Jacob was told Joseph was dead and believed it. Then he was told
Joseph was alive, and he did not believe it until his sons told him the words of Joseph and
showed him the blessings that came to them through Joseph. Then he believed Joseph was alive,
though he had not yet seen.
(a) Total amount of I2 = 501000 x 0.010 = 0.0005 mol = 0.5 mmolLe.pdf
(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol
Let a be the number of millimoles of I2 extracted
Millimoles of I2 left in aqueous phase = 0.5 - a
Distribution coefficient D = [I2](organic)/[I2](aqueous)
= [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)]
= [(a/50)]/[(0.5 - a)/50] = 85
a/(0.5 - a) = 85
a = 0.4942 mmol
Thus amount of I2 extracted = 0.4942 mmol
(b) Amount of I2 left in aqueous phase
= 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol
Concentration of I2 in aqueous phase = moles/volume (in L)
= 5.8 x 10-6/0.050 = 1.16 x 10-4 M
Solution
(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol
Let a be the number of millimoles of I2 extracted
Millimoles of I2 left in aqueous phase = 0.5 - a
Distribution coefficient D = [I2](organic)/[I2](aqueous)
= [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)]
= [(a/50)]/[(0.5 - a)/50] = 85
a/(0.5 - a) = 85
a = 0.4942 mmol
Thus amount of I2 extracted = 0.4942 mmol
(b) Amount of I2 left in aqueous phase
= 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol
Concentration of I2 in aqueous phase = moles/volume (in L)
= 5.8 x 10-6/0.050 = 1.16 x 10-4 M.
Nursing is a healthcare profession in which they serve the patients .pdf
Nursing is a healthcare profession in which they serve the patients and ought to have medical
knowledge. Nurses generally are more in contact with the patients and have knowledge of
patient\'s relatives, behaviour, experience in dealing with patients, communication skills. They
are the most important connection link between the doctors and the patient. Nurses also have
knowledge about the medical cases, projects related to community and cultural awareness.Hence,
they form integral part of the medical system. In genetic testing, various people from higher to
lower classes, different caste and creed come for checkup. The nurses are supposed to inform
and consult the patients about the necessities of genetic testing. The informed consent is
necessary to be obtained by explaining the importance of genetic testing. Also, the ethical issues
are to be followed by each and every medical institution. Since the public has trust in nurses,
they are supposed to convince the patients about the genetic testing advantages. They help the
people to understand the basics of genetic testing and the cons of it if not performed before the
delivery. Some set of population think that carrying out these tests are against the will of nature
or sometimes economical issues arise due to higher charges of these tests. Hence it is necessary
to make people aware of their misconceptions and provide them good medical facilities.
Solution
Nursing is a healthcare profession in which they serve the patients and ought to have medical
knowledge. Nurses generally are more in contact with the patients and have knowledge of
patient\'s relatives, behaviour, experience in dealing with patients, communication skills. They
are the most important connection link between the doctors and the patient. Nurses also have
knowledge about the medical cases, projects related to community and cultural awareness.Hence,
they form integral part of the medical system. In genetic testing, various people from higher to
lower classes, different caste and creed come for checkup. The nurses are supposed to inform
and consult the patients about the necessities of genetic testing. The informed consent is
necessary to be obtained by explaining the importance of genetic testing. Also, the ethical issues
are to be followed by each and every medical institution. Since the public has trust in nurses,
they are supposed to convince the patients about the genetic testing advantages. They help the
people to understand the basics of genetic testing and the cons of it if not performed before the
delivery. Some set of population think that carrying out these tests are against the will of nature
or sometimes economical issues arise due to higher charges of these tests. Hence it is necessary
to make people aware of their misconceptions and provide them good medical facilities..
yes.because the deposted money can be multiply with 10 and it beco.pdf
yes.
because the deposted money can be multiply with 10 and it becomes 100*10= $1000
Solution
yes.
because the deposted money can be multiply with 10 and it becomes 100*10= $1000.
When the substituent groups are oriented in the s.pdf
When the substituent groups are oriented in the same direction, the diastereomer is
referred to as cis, whereas, when the substituents are oriented in opposing directions, the
diastereomer is referred to as trans. An example of a small hydrocarbon displaying cis/trans
isomerism is 2-butene. Alicyclic compounds can also display cis/trans isomerism. As an example
of a geometric isomer due to a ring structure, consider 1,2-dichlorocyclohexane: is/trans
isomerism (also known as geometric isomerism, configuration isomerism, or E/Z isomerism) is a
form of stereoisomerism describing the orientation of functional groups within a molecule. In
general, such isomers contain double bonds, which cannot rotate, but they can also arise from
ring structures, wherein the rotation of bonds is greatly restricted. Cis and trans isomers occur
both in organic molecules and in inorganic coordination complexes. The terms cis and trans are
from Latin, in which cis means \"on the same side\" and trans means \"on the other side\" or
\"across\". The term \"geometric isomerism\" is considered an obsolete synonym of \"cis/trans
isomerism\" by IUPAC.[1] It is sometimes used as a synonym for general stereoisomerism (e.g.,
optical isomerism being called geometric isomerism); the correct term for non-optical
stereoisomerism is diastereomerism.
Solution
When the substituent groups are oriented in the same direction, the diastereomer is
referred to as cis, whereas, when the substituents are oriented in opposing directions, the
diastereomer is referred to as trans. An example of a small hydrocarbon displaying cis/trans
isomerism is 2-butene. Alicyclic compounds can also display cis/trans isomerism. As an example
of a geometric isomer due to a ring structure, consider 1,2-dichlorocyclohexane: is/trans
isomerism (also known as geometric isomerism, configuration isomerism, or E/Z isomerism) is a
form of stereoisomerism describing the orientation of functional groups within a molecule. In
general, such isomers contain double bonds, which cannot rotate, but they can also arise from
ring structures, wherein the rotation of bonds is greatly restricted. Cis and trans isomers occur
both in organic molecules and in inorganic coordination complexes. The terms cis and trans are
from Latin, in which cis means \"on the same side\" and trans means \"on the other side\" or
\"across\". The term \"geometric isomerism\" is considered an obsolete synonym of \"cis/trans
isomerism\" by IUPAC.[1] It is sometimes used as a synonym for general stereoisomerism (e.g.,
optical isomerism being called geometric isomerism); the correct term for non-optical
stereoisomerism is diastereomerism..
using System;public category take a look at one.00m; Console.W.pdf
using System;
public category take a look at
one.00m;
Console.WriteLine (d);
}
}
When I initial ran the on top of (or one thing similar) I expected it to output simply one (which
is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort
does not normalize itself - it remembers what percentage decimal digits it\'s (by maintaining the
exponent wherever possible) and on info, zero could also be counted as a major digit. i do not
understand the precise nature of what exponent is chosen (where there\'s a choice) once 2 totally
different decimals ar increased, divided, added etc, however you will realize it attention-grabbing
to manipulate with programs like the following:
using System;
public category take a look at
zero.00000000000010000m;
whereas (d != 0m)
}
}
Solution
using System;
public category take a look at
one.00m;
Console.WriteLine (d);
}
}
When I initial ran the on top of (or one thing similar) I expected it to output simply one (which
is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort
does not normalize itself - it remembers what percentage decimal digits it\'s (by maintaining the
exponent wherever possible) and on info, zero could also be counted as a major digit. i do not
understand the precise nature of what exponent is chosen (where there\'s a choice) once 2 totally
different decimals ar increased, divided, added etc, however you will realize it attention-grabbing
to manipulate with programs like the following:
using System;
public category take a look at
zero.00000000000010000m;
whereas (d != 0m)
}
}.
Tunnel through the IPv4--Internet traffic is expected to be carr.pdf
Tunnel through the IPv4:-
->Internet traffic is expected to be carried via tunnels as the Internet infrastructure migrates from
IPv4, the current version of the Internet protocol, to the long-anticipated upgrade known as IPv6.
->IPv4 uses 32-bit addresses and can support 4.3 billion devices connected directly to the
Internet.
->Internet:-Will probably move to IPv6 “from the edges in”
-> IPv6 will be adopted:-First by smaller organizations with greater flexibility and higher
tolerance for difficulties of pioneering
-> IPv6 packet is formed normally:- Sent to a router capable of encapsulating it in an IPv4
packet.
->Used to allow IPv6 network nodes to send packets over an IPv4 network infrastructure
->Presents a challenge for IPv6 header construction
-> Source node determines which packets must be encapsulated
->Based on the routing information the node maintains in its own routing table.
Types of Tunnels:-
1)RFC 2893 originally specified two different tunneling types
–Configured and automatic
2)RFC 4213, which made RFC 2893 obsolete
– Removed references to automatic tunneling
3)Configured tunnels
–Require that end point addresses be determined in the encapsulator device
4) From configuration data stored for each tunnel.
Solution
Tunnel through the IPv4:-
->Internet traffic is expected to be carried via tunnels as the Internet infrastructure migrates from
IPv4, the current version of the Internet protocol, to the long-anticipated upgrade known as IPv6.
->IPv4 uses 32-bit addresses and can support 4.3 billion devices connected directly to the
Internet.
->Internet:-Will probably move to IPv6 “from the edges in”
-> IPv6 will be adopted:-First by smaller organizations with greater flexibility and higher
tolerance for difficulties of pioneering
-> IPv6 packet is formed normally:- Sent to a router capable of encapsulating it in an IPv4
packet.
->Used to allow IPv6 network nodes to send packets over an IPv4 network infrastructure
->Presents a challenge for IPv6 header construction
-> Source node determines which packets must be encapsulated
->Based on the routing information the node maintains in its own routing table.
Types of Tunnels:-
1)RFC 2893 originally specified two different tunneling types
–Configured and automatic
2)RFC 4213, which made RFC 2893 obsolete
– Removed references to automatic tunneling
3)Configured tunnels
–Require that end point addresses be determined in the encapsulator device
4) From configuration data stored for each tunnel..
The key activity areas for securities firms are i. Investing Se.pdf
The key activity areas for securities firms are :
i. Investing: Security firms act as agents for investors and provide risk diversification and
liquidity services to holders of mutual funds. The major risks are underperforming stock market
whereby investors flee to other avenues of investing.
ii. Investment Banking: These functions relate to underwriting both debt and equity. The major
risk is unsold securities and money on the table due to ineffective pricing.
iii. Market Making: Market making involves the creation of a secondary market in an asset by a
securities firm or investment bank facilitating the smooth transactions of trading in the market. In
case of a crash this can lead to unsold securities and selling at a significant discount thus losing a
lot.
iv. Trading: Securities trader takes an active net position in an underlying instrument or asset.
The risk are snall bid ask spread and high fluctuating prices leading to loss on trading positions
v. Cash Management: Security firms bank deposit-like cash management accounts to individual
investors. Again the proper pricing is needed else there may be a significant loss on the position.
vi. Mergers and Acquisitions: Investment banks provide advice on mergers and acquisitions. The
risk are improper pricing leading to loss or not realizing proper synergies
vi. Other Service Functions: include custody and escrow services, clearance and settlement
services and research and advisory services. They may result in losses if there is a credit risk
present which usually is common.
Solution
The key activity areas for securities firms are :
i. Investing: Security firms act as agents for investors and provide risk diversification and
liquidity services to holders of mutual funds. The major risks are underperforming stock market
whereby investors flee to other avenues of investing.
ii. Investment Banking: These functions relate to underwriting both debt and equity. The major
risk is unsold securities and money on the table due to ineffective pricing.
iii. Market Making: Market making involves the creation of a secondary market in an asset by a
securities firm or investment bank facilitating the smooth transactions of trading in the market. In
case of a crash this can lead to unsold securities and selling at a significant discount thus losing a
lot.
iv. Trading: Securities trader takes an active net position in an underlying instrument or asset.
The risk are snall bid ask spread and high fluctuating prices leading to loss on trading positions
v. Cash Management: Security firms bank deposit-like cash management accounts to individual
investors. Again the proper pricing is needed else there may be a significant loss on the position.
vi. Mergers and Acquisitions: Investment banks provide advice on mergers and acquisitions. The
risk are improper pricing leading to loss or not realizing proper synergies
vi. Other Service Functions: include custody and escrow services, clearance and settlement
services and re.
The finger like projections of intestine known as villi increases su.pdf
The finger like projections of intestine known as villi increases surface area of small intestines.
The surface of small intestine is lined up with epithelial cells containing brush borders or
microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body
cells.
Due to the presence of specialized proteins within the microvilli known as porters, the nutrients
move into the cells.
Human gut flora contains several types of bacterial species in the small intestines that are
involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics.
So, correct option is d.transferring nutrients across intestinal membranes
Solution
The finger like projections of intestine known as villi increases surface area of small intestines.
The surface of small intestine is lined up with epithelial cells containing brush borders or
microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body
cells.
Due to the presence of specialized proteins within the microvilli known as porters, the nutrients
move into the cells.
Human gut flora contains several types of bacterial species in the small intestines that are
involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics.
So, correct option is d.transferring nutrients across intestinal membranes.
The answer isC. 12In establishing paternity by dna fingerprintin.pdf
The answer is:C. 1/2
In establishing paternity by dna fingerprinting, 50% of microsatellite repeat alleles in a child
come from the father.
Solution
The answer is:C. 1/2
In establishing paternity by dna fingerprinting, 50% of microsatellite repeat alleles in a child
come from the father..
Technically, micro RNA or miRNA are defined as small non-coding (tra.pdf
Technically, micro RNA or miRNA are defined as small non-coding (translationally inactive)
RNA molecules which have a major role in expression of other genes by the means of gene-
silencing or post-transcriptional modifications. The miRNA are found in a variety of organisms
including eukaryotes, plants, viruses and humans. Functionally, the miRNAs are actively
engaged in modulating gene expression in organisms.
The major functions of miRNAs can be discussed as below:
1. Gene/RNA silencing: The function of gene silencing using miRNAs can be brought about at
both transcriptional or post-transcriptional levels. The miRNA can either cleave the newly
synthesized mRNA strand into two or more smaller units or prevent its access to the translatinal
machinery by blocking the nucleotides by binding to it.
2. Post-trancriptional regulation of mRNA: This function of miRNA is very unique and
important in nature. The miRNA are actively engaged in removing the 3\' polyadenylation cap of
the untranslated region of the mRNA which deteriorates the structural stability of the nascent
mRNA molecule. This prevents structural stabilization of the newly synthesized mRNA
molecule from destruction and also impairs the post-transcriptional modification by the
ubiquitination protein assembly. Binding of miRNA to the mRNA at specific sites also makes it
prone to cleavage and thus prevents the appropriate post-transcriptional modifications.
Thus, the major fucntions of miRNA remains in regulation of gene expression/silencing and
modulating post-transcriptional modifications. Owing to these crucial moleular features,
miRNAs are actively being used in pre-clinical research as potential therapeutics.
Solution
Technically, micro RNA or miRNA are defined as small non-coding (translationally inactive)
RNA molecules which have a major role in expression of other genes by the means of gene-
silencing or post-transcriptional modifications. The miRNA are found in a variety of organisms
including eukaryotes, plants, viruses and humans. Functionally, the miRNAs are actively
engaged in modulating gene expression in organisms.
The major functions of miRNAs can be discussed as below:
1. Gene/RNA silencing: The function of gene silencing using miRNAs can be brought about at
both transcriptional or post-transcriptional levels. The miRNA can either cleave the newly
synthesized mRNA strand into two or more smaller units or prevent its access to the translatinal
machinery by blocking the nucleotides by binding to it.
2. Post-trancriptional regulation of mRNA: This function of miRNA is very unique and
important in nature. The miRNA are actively engaged in removing the 3\' polyadenylation cap of
the untranslated region of the mRNA which deteriorates the structural stability of the nascent
mRNA molecule. This prevents structural stabilization of the newly synthesized mRNA
molecule from destruction and also impairs the post-transcriptional modification by the
ubiquitination protein assembly. Bi.
The activation of macrophages by the TH1 effector cells is a defect .pdf
The activation of macrophages by the TH1 effector cells is a defect because of lack of immune
system.
Solution
The activation of macrophages by the TH1 effector cells is a defect because of lack of immune
system..
Solution(1a)Let A=(a,x)where a= distance from origin to y axis.pdf
Solution
(1a):
Let A=(a,x)
where a= distance from origin to y axis
B=(x+h,b)
where x+h=distance from origin to x axis
b=distance from origin to y axis.
Theyre called molecular compounds. Solutio.pdf
They\'re called molecular compounds.
Solution
They\'re called molecular compounds..
Polygon.javapublic class Polygon { private int numSides; priva.pdf
Polygon.java
public class Polygon {
private int numSides;
private double sideLength;
private double xCoord;
private double yCoord;
private double apothem;
private double perimeter;
private double area;
//Constructor
public Polygon (double xCentCoord, double yCentCoord, int countSides, double measLength,
double measApothem, double measPerimeter) {
xCoord = xCentCoord;
yCoord = yCentCoord;
numSides = countSides;
sideLength = measLength;
apothem = measApothem;
perimeter = measPerimeter;
}
//No argument constructor
public Polygon () {
xCoord = 0.0;
yCoord = 0.0;
numSides = 4;
sideLength = 10.0;
apothem = 5.0;
perimeter = 20.0;
}
//Setter Methods
//setX()
public void setXCoord(double xCentCoord) {
xCoord = xCentCoord;
}
//setY()
public void setYCoord(double yCentCoord) {
yCoord = yCentCoord;
}
//setNumSides()
public void setNumSides(int countSides) {
numSides = countSides;
}
//setSideLength()
public void setSideLength(double measLength) {
sideLength = measLength;
}
//setApothem()
public void setApothem(double measApothem) {
apothem = measApothem;
}
//Getter methods
//getXCoord
public double getXCoord() {
return xCoord;
}
//getYCoord
public double getYCoord() {
return yCoord;
}
//getNumSides
public int getNumSides() {
return numSides;
}
//getSideLength()
public double getSideLength() {
return sideLength;
}
//getApothem()
public double getApothem() {
return apothem;
}
//getPerimeter()
public double getPerimeter() {
return perimeter;
}
//getArea()
public double getArea() {
return area;
}
//Calculate for perimeter
public double getPerimeter(Polygon multiple){
double perimeter = multiple.getNumSides() * multiple.getSideLength();
return perimeter;
}
//Calculate for area
public double getArea(Polygon multiple) {
double area = 0.5 * multiple.getApothem() * multiple.getPerimeter();
return area;
}
//toString method
public String toString() {
String str = \"numsides= \" + numSides + \", sideLength= \" + sideLength;
return str;
}
}
TestPolygon.java
//import Scanner
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
//Construct a point with x = 1.0, y = 1.0
Polygon multiple = new Polygon();
//Display the values using toString
System.out.println(\"toString() results: \" + multiple.toString());
//Call the getter methods
double xCoord2 = multiple.getXCoord();
double yCoord2 = multiple.getYCoord();
int numSides2 = multiple.getNumSides();
double sideLength2 = multiple.getSideLength();
double apothem2 = multiple.getApothem();
double perimeter2 = multiple.getPerimeter();
double area2 = multiple.getArea();
//Print results
System.out.println(\"xCoord= \" + xCoord2 + \", yCoord= \" + yCoord2 + \", apothem= \"
+apothem2);
System.out.println(\"getNumsides() results: \" + numSides2);
System.out.println(\"getSideLength() results: \" + sideLength2);
System.out.println(\"getXCoord() results: \" + xCoord2);
System.out.println(\"getYCoord() results: \" + yCoord2);
System.out.println(\"getApothem() results: \" + apothem2);
System.out.pr.
ph = 7ln (0.120.1);ph = 1.27625 ----ANSWERSolutionph = 7.pdf
ph = 7*ln (0.12/0.1);
ph = 1.27625 ---->ANSWER
Solution
ph = 7*ln (0.12/0.1);
ph = 1.27625 ---->ANSWER.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Overview on Edible Vaccine: Pros & Cons with Mechanism
This ppt include the description of the edible vaccine i.e. a new concept over the traditional vaccine administered by injection.
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888678123317368915236 #include stdio.h #include stdlib.h.pdf
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}
Solution
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}.
18) One device is attached to a physical interface The switch c.pdf
18) One device is attached to a physical interface
The switch can be remotely managed
The default VIan has been configured
19) Flow control
20) Encapsulation
21) HTTP,TCP,IP,Ethernet
22) They define how messages are exchanged between the source and the destination
Solution
18) One device is attached to a physical interface
The switch can be remotely managed
The default VIan has been configured
19) Flow control
20) Encapsulation
21) HTTP,TCP,IP,Ethernet
22) They define how messages are exchanged between the source and the destination.
1.Joseph could not restrain himself before all those who stood by hi.pdf
1.Joseph could not restrain himself before all those who stood by him, and he cried out, “Make
everyone go out from me!” So no one stood with him while Joseph made himself known to his
brothers. And he wept aloud, and the Egyptians and the house of Pharaoh heard it. Then Joseph
said to his brothers, “I am Joseph; does my father still live?” But his brothers could not answer
him, for they were dismayed in his presence.
a. Joseph could not restrain himself before all those who stood by him: Joseph ordered all the
Egyptians out of the room and was then alone with his brothers. His great emotion showed that
Joseph did not cruelly manipulate his brothers. He was directed by God to make these
arrangements and it hurt him to do it.
b. Joseph made himself known to his brothers: This perhaps means that Joseph told them he was
Joseph and showed his brothers that he was circumcised. Jewish legend says the brothers could
never believe this high Egyptian official was Joseph unless he showed he was circumcised.
c. But his brothers could not answer him, for they were dismayed in his presence: Because of the
punishment they anticipated, the great emotion of Joseph, his manner of revelation, and the total
shock of learning Joseph was not only alive but right in front of them, the brothers were
dismayed.
Pharaoh and Joseph send the brothers home with many gifts.
Joseph gave them carts, according to the command of Pharaoh, and he gave them provisions for
the journey: The sons of Israel received transportation, provision, garments, and riches because
of who their favored brother was. Pharaoh blessed the sons of Jacob for Joseph’s sake.
i. “To return to Canaan with ‘carts from Egypt’ was the cultural equivalent of landing a jumbo jet
among a tribe of isolated savages. It would be the stuff legends are made of.” (Boice)
b. See that you do not become troubled along the way: The idea behind the words “become
troubled” is literally become angry or quarrel. Joseph knew as soon as these men left his
presence they would be tempted to act in selfish, unspiritual ways. They had to anticipate and
guard against this.
Jacob hears the good news – that Joseph lives.
Then they went up out of Egypt, and came to the land of Canaan to Jacob their father. And they
told him, saying, “Joseph is still alive, and he is governor over all the land of Egypt.” And
Jacob’s heart stood still, because he did not believe them. But when they told him all the words
which Joseph had said to them, and when he saw the carts which Joseph had sent to carry him,
the spirit of Jacob their father revived. Then Israel said, “It is enough. Joseph my son is still
alive. I will go and see him before I die.”
a. He did not believe them: Jacob was told Joseph was dead and believed it. Then he was told
Joseph was alive, and he did not believe it until his sons told him the words of Joseph and
showed him the blessings that came to them through Joseph. Then he believed Joseph was alive,
though he had not yet seen.
(a) Total amount of I2 = 501000 x 0.010 = 0.0005 mol = 0.5 mmolLe.pdf
(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol
Let a be the number of millimoles of I2 extracted
Millimoles of I2 left in aqueous phase = 0.5 - a
Distribution coefficient D = [I2](organic)/[I2](aqueous)
= [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)]
= [(a/50)]/[(0.5 - a)/50] = 85
a/(0.5 - a) = 85
a = 0.4942 mmol
Thus amount of I2 extracted = 0.4942 mmol
(b) Amount of I2 left in aqueous phase
= 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol
Concentration of I2 in aqueous phase = moles/volume (in L)
= 5.8 x 10-6/0.050 = 1.16 x 10-4 M
Solution
(a) Total amount of I2 = 50/1000 x 0.010 = 0.0005 mol = 0.5 mmol
Let a be the number of millimoles of I2 extracted
Millimoles of I2 left in aqueous phase = 0.5 - a
Distribution coefficient D = [I2](organic)/[I2](aqueous)
= [(mmol of I2)(organic)/V(organic)]/[mmol of I2)(aqueous)/V(aqueous)]
= [(a/50)]/[(0.5 - a)/50] = 85
a/(0.5 - a) = 85
a = 0.4942 mmol
Thus amount of I2 extracted = 0.4942 mmol
(b) Amount of I2 left in aqueous phase
= 0.5 - 0.4942 = 0.0058 mmol = 5.8 x 10-6 mol
Concentration of I2 in aqueous phase = moles/volume (in L)
= 5.8 x 10-6/0.050 = 1.16 x 10-4 M.
Nursing is a healthcare profession in which they serve the patients .pdf
Nursing is a healthcare profession in which they serve the patients and ought to have medical
knowledge. Nurses generally are more in contact with the patients and have knowledge of
patient\'s relatives, behaviour, experience in dealing with patients, communication skills. They
are the most important connection link between the doctors and the patient. Nurses also have
knowledge about the medical cases, projects related to community and cultural awareness.Hence,
they form integral part of the medical system. In genetic testing, various people from higher to
lower classes, different caste and creed come for checkup. The nurses are supposed to inform
and consult the patients about the necessities of genetic testing. The informed consent is
necessary to be obtained by explaining the importance of genetic testing. Also, the ethical issues
are to be followed by each and every medical institution. Since the public has trust in nurses,
they are supposed to convince the patients about the genetic testing advantages. They help the
people to understand the basics of genetic testing and the cons of it if not performed before the
delivery. Some set of population think that carrying out these tests are against the will of nature
or sometimes economical issues arise due to higher charges of these tests. Hence it is necessary
to make people aware of their misconceptions and provide them good medical facilities.
Solution
Nursing is a healthcare profession in which they serve the patients and ought to have medical
knowledge. Nurses generally are more in contact with the patients and have knowledge of
patient\'s relatives, behaviour, experience in dealing with patients, communication skills. They
are the most important connection link between the doctors and the patient. Nurses also have
knowledge about the medical cases, projects related to community and cultural awareness.Hence,
they form integral part of the medical system. In genetic testing, various people from higher to
lower classes, different caste and creed come for checkup. The nurses are supposed to inform
and consult the patients about the necessities of genetic testing. The informed consent is
necessary to be obtained by explaining the importance of genetic testing. Also, the ethical issues
are to be followed by each and every medical institution. Since the public has trust in nurses,
they are supposed to convince the patients about the genetic testing advantages. They help the
people to understand the basics of genetic testing and the cons of it if not performed before the
delivery. Some set of population think that carrying out these tests are against the will of nature
or sometimes economical issues arise due to higher charges of these tests. Hence it is necessary
to make people aware of their misconceptions and provide them good medical facilities..
yes.because the deposted money can be multiply with 10 and it beco.pdf
yes.
because the deposted money can be multiply with 10 and it becomes 100*10= $1000
Solution
yes.
because the deposted money can be multiply with 10 and it becomes 100*10= $1000.
When the substituent groups are oriented in the s.pdf
When the substituent groups are oriented in the same direction, the diastereomer is
referred to as cis, whereas, when the substituents are oriented in opposing directions, the
diastereomer is referred to as trans. An example of a small hydrocarbon displaying cis/trans
isomerism is 2-butene. Alicyclic compounds can also display cis/trans isomerism. As an example
of a geometric isomer due to a ring structure, consider 1,2-dichlorocyclohexane: is/trans
isomerism (also known as geometric isomerism, configuration isomerism, or E/Z isomerism) is a
form of stereoisomerism describing the orientation of functional groups within a molecule. In
general, such isomers contain double bonds, which cannot rotate, but they can also arise from
ring structures, wherein the rotation of bonds is greatly restricted. Cis and trans isomers occur
both in organic molecules and in inorganic coordination complexes. The terms cis and trans are
from Latin, in which cis means \"on the same side\" and trans means \"on the other side\" or
\"across\". The term \"geometric isomerism\" is considered an obsolete synonym of \"cis/trans
isomerism\" by IUPAC.[1] It is sometimes used as a synonym for general stereoisomerism (e.g.,
optical isomerism being called geometric isomerism); the correct term for non-optical
stereoisomerism is diastereomerism.
Solution
When the substituent groups are oriented in the same direction, the diastereomer is
referred to as cis, whereas, when the substituents are oriented in opposing directions, the
diastereomer is referred to as trans. An example of a small hydrocarbon displaying cis/trans
isomerism is 2-butene. Alicyclic compounds can also display cis/trans isomerism. As an example
of a geometric isomer due to a ring structure, consider 1,2-dichlorocyclohexane: is/trans
isomerism (also known as geometric isomerism, configuration isomerism, or E/Z isomerism) is a
form of stereoisomerism describing the orientation of functional groups within a molecule. In
general, such isomers contain double bonds, which cannot rotate, but they can also arise from
ring structures, wherein the rotation of bonds is greatly restricted. Cis and trans isomers occur
both in organic molecules and in inorganic coordination complexes. The terms cis and trans are
from Latin, in which cis means \"on the same side\" and trans means \"on the other side\" or
\"across\". The term \"geometric isomerism\" is considered an obsolete synonym of \"cis/trans
isomerism\" by IUPAC.[1] It is sometimes used as a synonym for general stereoisomerism (e.g.,
optical isomerism being called geometric isomerism); the correct term for non-optical
stereoisomerism is diastereomerism..
using System;public category take a look at one.00m; Console.W.pdf
using System;
public category take a look at
one.00m;
Console.WriteLine (d);
}
}
When I initial ran the on top of (or one thing similar) I expected it to output simply one (which
is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort
does not normalize itself - it remembers what percentage decimal digits it\'s (by maintaining the
exponent wherever possible) and on info, zero could also be counted as a major digit. i do not
understand the precise nature of what exponent is chosen (where there\'s a choice) once 2 totally
different decimals ar increased, divided, added etc, however you will realize it attention-grabbing
to manipulate with programs like the following:
using System;
public category take a look at
zero.00000000000010000m;
whereas (d != 0m)
}
}
Solution
using System;
public category take a look at
one.00m;
Console.WriteLine (d);
}
}
When I initial ran the on top of (or one thing similar) I expected it to output simply one (which
is what it might are on .NET 1.0) - however in reality, the output was one.00. The decimal sort
does not normalize itself - it remembers what percentage decimal digits it\'s (by maintaining the
exponent wherever possible) and on info, zero could also be counted as a major digit. i do not
understand the precise nature of what exponent is chosen (where there\'s a choice) once 2 totally
different decimals ar increased, divided, added etc, however you will realize it attention-grabbing
to manipulate with programs like the following:
using System;
public category take a look at
zero.00000000000010000m;
whereas (d != 0m)
}
}.
Tunnel through the IPv4--Internet traffic is expected to be carr.pdf
Tunnel through the IPv4:-
->Internet traffic is expected to be carried via tunnels as the Internet infrastructure migrates from
IPv4, the current version of the Internet protocol, to the long-anticipated upgrade known as IPv6.
->IPv4 uses 32-bit addresses and can support 4.3 billion devices connected directly to the
Internet.
->Internet:-Will probably move to IPv6 “from the edges in”
-> IPv6 will be adopted:-First by smaller organizations with greater flexibility and higher
tolerance for difficulties of pioneering
-> IPv6 packet is formed normally:- Sent to a router capable of encapsulating it in an IPv4
packet.
->Used to allow IPv6 network nodes to send packets over an IPv4 network infrastructure
->Presents a challenge for IPv6 header construction
-> Source node determines which packets must be encapsulated
->Based on the routing information the node maintains in its own routing table.
Types of Tunnels:-
1)RFC 2893 originally specified two different tunneling types
–Configured and automatic
2)RFC 4213, which made RFC 2893 obsolete
– Removed references to automatic tunneling
3)Configured tunnels
–Require that end point addresses be determined in the encapsulator device
4) From configuration data stored for each tunnel.
Solution
Tunnel through the IPv4:-
->Internet traffic is expected to be carried via tunnels as the Internet infrastructure migrates from
IPv4, the current version of the Internet protocol, to the long-anticipated upgrade known as IPv6.
->IPv4 uses 32-bit addresses and can support 4.3 billion devices connected directly to the
Internet.
->Internet:-Will probably move to IPv6 “from the edges in”
-> IPv6 will be adopted:-First by smaller organizations with greater flexibility and higher
tolerance for difficulties of pioneering
-> IPv6 packet is formed normally:- Sent to a router capable of encapsulating it in an IPv4
packet.
->Used to allow IPv6 network nodes to send packets over an IPv4 network infrastructure
->Presents a challenge for IPv6 header construction
-> Source node determines which packets must be encapsulated
->Based on the routing information the node maintains in its own routing table.
Types of Tunnels:-
1)RFC 2893 originally specified two different tunneling types
–Configured and automatic
2)RFC 4213, which made RFC 2893 obsolete
– Removed references to automatic tunneling
3)Configured tunnels
–Require that end point addresses be determined in the encapsulator device
4) From configuration data stored for each tunnel..
The key activity areas for securities firms are i. Investing Se.pdf
The key activity areas for securities firms are :
i. Investing: Security firms act as agents for investors and provide risk diversification and
liquidity services to holders of mutual funds. The major risks are underperforming stock market
whereby investors flee to other avenues of investing.
ii. Investment Banking: These functions relate to underwriting both debt and equity. The major
risk is unsold securities and money on the table due to ineffective pricing.
iii. Market Making: Market making involves the creation of a secondary market in an asset by a
securities firm or investment bank facilitating the smooth transactions of trading in the market. In
case of a crash this can lead to unsold securities and selling at a significant discount thus losing a
lot.
iv. Trading: Securities trader takes an active net position in an underlying instrument or asset.
The risk are snall bid ask spread and high fluctuating prices leading to loss on trading positions
v. Cash Management: Security firms bank deposit-like cash management accounts to individual
investors. Again the proper pricing is needed else there may be a significant loss on the position.
vi. Mergers and Acquisitions: Investment banks provide advice on mergers and acquisitions. The
risk are improper pricing leading to loss or not realizing proper synergies
vi. Other Service Functions: include custody and escrow services, clearance and settlement
services and research and advisory services. They may result in losses if there is a credit risk
present which usually is common.
Solution
The key activity areas for securities firms are :
i. Investing: Security firms act as agents for investors and provide risk diversification and
liquidity services to holders of mutual funds. The major risks are underperforming stock market
whereby investors flee to other avenues of investing.
ii. Investment Banking: These functions relate to underwriting both debt and equity. The major
risk is unsold securities and money on the table due to ineffective pricing.
iii. Market Making: Market making involves the creation of a secondary market in an asset by a
securities firm or investment bank facilitating the smooth transactions of trading in the market. In
case of a crash this can lead to unsold securities and selling at a significant discount thus losing a
lot.
iv. Trading: Securities trader takes an active net position in an underlying instrument or asset.
The risk are snall bid ask spread and high fluctuating prices leading to loss on trading positions
v. Cash Management: Security firms bank deposit-like cash management accounts to individual
investors. Again the proper pricing is needed else there may be a significant loss on the position.
vi. Mergers and Acquisitions: Investment banks provide advice on mergers and acquisitions. The
risk are improper pricing leading to loss or not realizing proper synergies
vi. Other Service Functions: include custody and escrow services, clearance and settlement
services and re.
The finger like projections of intestine known as villi increases su.pdf
The finger like projections of intestine known as villi increases surface area of small intestines.
The surface of small intestine is lined up with epithelial cells containing brush borders or
microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body
cells.
Due to the presence of specialized proteins within the microvilli known as porters, the nutrients
move into the cells.
Human gut flora contains several types of bacterial species in the small intestines that are
involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics.
So, correct option is d.transferring nutrients across intestinal membranes
Solution
The finger like projections of intestine known as villi increases surface area of small intestines.
The surface of small intestine is lined up with epithelial cells containing brush borders or
microvilli that help to absorb nutrients into the blood or lymph, which later moves to the body
cells.
Due to the presence of specialized proteins within the microvilli known as porters, the nutrients
move into the cells.
Human gut flora contains several types of bacterial species in the small intestines that are
involved in the production of vitamin K, B as well as metabolism of bile acid and xenobiotics.
So, correct option is d.transferring nutrients across intestinal membranes.
The answer isC. 12In establishing paternity by dna fingerprintin.pdf
The answer is:C. 1/2
In establishing paternity by dna fingerprinting, 50% of microsatellite repeat alleles in a child
come from the father.
Solution
The answer is:C. 1/2
In establishing paternity by dna fingerprinting, 50% of microsatellite repeat alleles in a child
come from the father..
Technically, micro RNA or miRNA are defined as small non-coding (tra.pdf
Technically, micro RNA or miRNA are defined as small non-coding (translationally inactive)
RNA molecules which have a major role in expression of other genes by the means of gene-
silencing or post-transcriptional modifications. The miRNA are found in a variety of organisms
including eukaryotes, plants, viruses and humans. Functionally, the miRNAs are actively
engaged in modulating gene expression in organisms.
The major functions of miRNAs can be discussed as below:
1. Gene/RNA silencing: The function of gene silencing using miRNAs can be brought about at
both transcriptional or post-transcriptional levels. The miRNA can either cleave the newly
synthesized mRNA strand into two or more smaller units or prevent its access to the translatinal
machinery by blocking the nucleotides by binding to it.
2. Post-trancriptional regulation of mRNA: This function of miRNA is very unique and
important in nature. The miRNA are actively engaged in removing the 3\' polyadenylation cap of
the untranslated region of the mRNA which deteriorates the structural stability of the nascent
mRNA molecule. This prevents structural stabilization of the newly synthesized mRNA
molecule from destruction and also impairs the post-transcriptional modification by the
ubiquitination protein assembly. Binding of miRNA to the mRNA at specific sites also makes it
prone to cleavage and thus prevents the appropriate post-transcriptional modifications.
Thus, the major fucntions of miRNA remains in regulation of gene expression/silencing and
modulating post-transcriptional modifications. Owing to these crucial moleular features,
miRNAs are actively being used in pre-clinical research as potential therapeutics.
Solution
Technically, micro RNA or miRNA are defined as small non-coding (translationally inactive)
RNA molecules which have a major role in expression of other genes by the means of gene-
silencing or post-transcriptional modifications. The miRNA are found in a variety of organisms
including eukaryotes, plants, viruses and humans. Functionally, the miRNAs are actively
engaged in modulating gene expression in organisms.
The major functions of miRNAs can be discussed as below:
1. Gene/RNA silencing: The function of gene silencing using miRNAs can be brought about at
both transcriptional or post-transcriptional levels. The miRNA can either cleave the newly
synthesized mRNA strand into two or more smaller units or prevent its access to the translatinal
machinery by blocking the nucleotides by binding to it.
2. Post-trancriptional regulation of mRNA: This function of miRNA is very unique and
important in nature. The miRNA are actively engaged in removing the 3\' polyadenylation cap of
the untranslated region of the mRNA which deteriorates the structural stability of the nascent
mRNA molecule. This prevents structural stabilization of the newly synthesized mRNA
molecule from destruction and also impairs the post-transcriptional modification by the
ubiquitination protein assembly. Bi.
The activation of macrophages by the TH1 effector cells is a defect .pdf
The activation of macrophages by the TH1 effector cells is a defect because of lack of immune
system.
Solution
The activation of macrophages by the TH1 effector cells is a defect because of lack of immune
system..
Solution(1a)Let A=(a,x)where a= distance from origin to y axis.pdf
Solution
(1a):
Let A=(a,x)
where a= distance from origin to y axis
B=(x+h,b)
where x+h=distance from origin to x axis
b=distance from origin to y axis.
Theyre called molecular compounds. Solutio.pdf
They\'re called molecular compounds.
Solution
They\'re called molecular compounds..
Polygon.javapublic class Polygon { private int numSides; priva.pdf
Polygon.java
public class Polygon {
private int numSides;
private double sideLength;
private double xCoord;
private double yCoord;
private double apothem;
private double perimeter;
private double area;
//Constructor
public Polygon (double xCentCoord, double yCentCoord, int countSides, double measLength,
double measApothem, double measPerimeter) {
xCoord = xCentCoord;
yCoord = yCentCoord;
numSides = countSides;
sideLength = measLength;
apothem = measApothem;
perimeter = measPerimeter;
}
//No argument constructor
public Polygon () {
xCoord = 0.0;
yCoord = 0.0;
numSides = 4;
sideLength = 10.0;
apothem = 5.0;
perimeter = 20.0;
}
//Setter Methods
//setX()
public void setXCoord(double xCentCoord) {
xCoord = xCentCoord;
}
//setY()
public void setYCoord(double yCentCoord) {
yCoord = yCentCoord;
}
//setNumSides()
public void setNumSides(int countSides) {
numSides = countSides;
}
//setSideLength()
public void setSideLength(double measLength) {
sideLength = measLength;
}
//setApothem()
public void setApothem(double measApothem) {
apothem = measApothem;
}
//Getter methods
//getXCoord
public double getXCoord() {
return xCoord;
}
//getYCoord
public double getYCoord() {
return yCoord;
}
//getNumSides
public int getNumSides() {
return numSides;
}
//getSideLength()
public double getSideLength() {
return sideLength;
}
//getApothem()
public double getApothem() {
return apothem;
}
//getPerimeter()
public double getPerimeter() {
return perimeter;
}
//getArea()
public double getArea() {
return area;
}
//Calculate for perimeter
public double getPerimeter(Polygon multiple){
double perimeter = multiple.getNumSides() * multiple.getSideLength();
return perimeter;
}
//Calculate for area
public double getArea(Polygon multiple) {
double area = 0.5 * multiple.getApothem() * multiple.getPerimeter();
return area;
}
//toString method
public String toString() {
String str = \"numsides= \" + numSides + \", sideLength= \" + sideLength;
return str;
}
}
TestPolygon.java
//import Scanner
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
//Construct a point with x = 1.0, y = 1.0
Polygon multiple = new Polygon();
//Display the values using toString
System.out.println(\"toString() results: \" + multiple.toString());
//Call the getter methods
double xCoord2 = multiple.getXCoord();
double yCoord2 = multiple.getYCoord();
int numSides2 = multiple.getNumSides();
double sideLength2 = multiple.getSideLength();
double apothem2 = multiple.getApothem();
double perimeter2 = multiple.getPerimeter();
double area2 = multiple.getArea();
//Print results
System.out.println(\"xCoord= \" + xCoord2 + \", yCoord= \" + yCoord2 + \", apothem= \"
+apothem2);
System.out.println(\"getNumsides() results: \" + numSides2);
System.out.println(\"getSideLength() results: \" + sideLength2);
System.out.println(\"getXCoord() results: \" + xCoord2);
System.out.println(\"getYCoord() results: \" + yCoord2);
System.out.println(\"getApothem() results: \" + apothem2);
System.out.pr.
ph = 7ln (0.120.1);ph = 1.27625 ----ANSWERSolutionph = 7.pdf
ph = 7*ln (0.12/0.1);
ph = 1.27625 ---->ANSWER
Solution
ph = 7*ln (0.12/0.1);
ph = 1.27625 ---->ANSWER.
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