Using gordon growth model Value of stock=Dividend paid(1+growt.pdfapexcomputer54
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5
Solution
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5.
The document discusses an issue with signing in. It provides a brief description of the problem but no solution. The document is only two sentences and does not contain enough information to generate a multi-sentence summary.
moles of NH3 = 3417 = 2 moles of O2 = 4032 = 1..pdfapexcomputer54
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0
Solution
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0.
It would be . like for hydrated compound of.pdfapexcomputer54
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O)
Solution
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O).
NO has got a strong bond because it has got an co.pdfapexcomputer54
NO has got a strong bond because it has got an coordinate covalent bond....
Solution
NO has got a strong bond because it has got an coordinate covalent bond.....
total number of moles of liquid increase, .pdfapexcomputer54
This short document discusses how the total number of moles of liquid can increase. It states that the total number of moles of liquid increase but provides no further details or context about what is causing the increase or any other related information.
what this meansSolutionwhat this means.pdfapexcomputer54
This very short document does not provide enough contextual information to generate a meaningful 3 sentence summary. It contains only a heading and subheading with no further details or body content to summarize.
Using gordon growth model Value of stock=Dividend paid(1+growt.pdfapexcomputer54
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5
Solution
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5.
The document discusses an issue with signing in. It provides a brief description of the problem but no solution. The document is only two sentences and does not contain enough information to generate a multi-sentence summary.
moles of NH3 = 3417 = 2 moles of O2 = 4032 = 1..pdfapexcomputer54
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0
Solution
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0.
It would be . like for hydrated compound of.pdfapexcomputer54
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O)
Solution
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O).
NO has got a strong bond because it has got an co.pdfapexcomputer54
NO has got a strong bond because it has got an coordinate covalent bond....
Solution
NO has got a strong bond because it has got an coordinate covalent bond.....
total number of moles of liquid increase, .pdfapexcomputer54
This short document discusses how the total number of moles of liquid can increase. It states that the total number of moles of liquid increase but provides no further details or context about what is causing the increase or any other related information.
what this meansSolutionwhat this means.pdfapexcomputer54
This very short document does not provide enough contextual information to generate a meaningful 3 sentence summary. It contains only a heading and subheading with no further details or body content to summarize.
we will use tables of normal distribution ( Z distribution)P( z .pdfapexcomputer54
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587
Solution
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587.
The Northeast blackout of 2003 was a widespread power outage that oc.pdfapexcomputer54
The Northeast blackout of 2003 was a widespread power outage that occurred throughout parts
of the Northeasternand Midwestern United States and the Canadian province of Ontario on
Thursday, August 14, 2003, just after 4:10 p.m. All told, 50 million people lost power for up to
two days in the biggest blackout in North American history. The event contributed to at least 11
deaths and cost an estimated $6 billion. A surge of electricity to western New York and Canada
touched off a series of power failures and enforced blackouts that left parts of at least eight states
in the Northeast and the Midwest without electricity.Transmission system operators scattered
across some 300 control centers nationwide monitor voltage and current data from SCADA
(supervisory control and data acquisition) systems placed at transformers, generators and other
critical points.The widespread failures provoked the evacuation of office buildings, stranded
thousands of commuters. As circuit breakers tripped at generating stations from New York to
Michigan and into Canada, millions of people were instantly caught up in the largest blackout in
American history. In New York City, power was shut off by officials struggling to head off a
wider blackout. Cleveland and Detroit went dark, as did Toronto and sections of New Jersey,
Pennsylvania, Connecticut and Massachusetts. Officials worked into the night to put the grid
back in operation and restore electric service. Mayor Michael R. Bloomberg said that that the
power was back on in parts of Brooklyn, the Bronx and Queens by 11 p.m. -- but not Manhattan.
The blackout began just after the stock exchanges had closed for the day, a slow summer day of
relatively light trading, as thousands of workers were about to head home. Office workers who
were still at their desks watched their computer monitors blink off without warning on a hot and
hazy afternoon. Soon hospitals and government buildings were switching on backup generators
to keep essential equipment operating, and the police were evacuating people trapped in
elevators. Airports throughout the affected states suffered serious disruptions, including the three
major airports in the New York metropolitan region, but did not close. Thousands of subway
passengers in New York City had to be evacuated from tunnels, and commuter trains also came
to a halt. Gov.George E. Pataki said that 600 trains were stranded.The office of the Canadian
prime minister, Jean Chrétien, initially said the power problems were caused by lightning in New
York State but later retracted that. Canadian officials later expressed uncertainty about the exact
cause but continued to insist the problem began on the United States side of the border. The
Nuclear Regulatory Commission said that the seven nuclear plants in New York and New Jersey
and two in the Midwest had shut down automatically when the failure occurred.Telephone
service was disrupted, especially calls to and from cellular phones.Officials said the.
The chemical reaction responsible for formation of water2H2+ O2--.pdfapexcomputer54
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water.
Solution
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water..
SolutionDomain name system is used to determine the IP address of.pdfapexcomputer54
Solution
:
Domain name system is used to determine the IP address of the host/domain name. DNS is very
useful to every user where it helps retrieve the website IP address. Without DNS we cannot
retrieve the website IP address. This process is known as forward DNS resolution.
DNS should be registered by some companies. If any attacker has details then the dns will be
controlled by the attacker. when we are sending the e-mail the dns data will be used then it can
be catches anywhere in the network.
So it is very important that a DNS require reliable communications..
sample mean = population mean = 2.11standard deviation = popul.pdfapexcomputer54
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137
Solution
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137.
Polymerase chain reaction (PCR) is a process used to replicate DNA i.pdfapexcomputer54
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication starts at single
origin (ori C). It do not occur in PCR.
Prokaryotic DNA replication is followed by the three basic step:-
1. initiation, 2. elongation, and 3. termination.
The process of PCR is followed by the three basic step:-
1. Denaturation, 2.,Annealing, and 3. Primer Extension
1) the process of forming single stranded DNA, :-In PCR the DNA denatures at high tempreture,
at 95oC to form its single-stranded. During the prokaryotic DNA Replication, Ori C unwinds the
DNA by DNA B or helicase, and extension of ssDNA for copying during initiation step.
2) priming of the DNA polymerase,:-In PCR primer binds with the complementry strand of DNA
during the Annealing step. In prokaryotic replcation, primase form RNA primer extended by
DNA polymerase III,during the elongation step. Synthesis ofleading strand and lagging strand
takes place in this step. It do not occur in PCR.
3) the DNA polymerase itself:-In prokaryotic DNA replication DNA polymerase III synthesize
the segment of DNA for both leading and lagging strand. In PCR, DNA polymerase extends the
primer, Taq polymerase, is used in this process and extension is performed at 72oC.
4) the end product:- As a PCR two ds DNA is formed. And in prokaryotic DNA replication 2
circular DNA is formed , which remain interlinked and topoisomerase II make them seprate by
making cut over them.
Protien involed in prokaryotic DNA replication:-
Dna A protein-It initiate DNA replication by the binding with the Dna A boxes.
Dna helicase- it seprates the ds DNA.
Topoisomerase- From the replication fork, it removes positive supercoiling.
Single stranded binding protein-to prevent the reformation of ds DNA, it binds with single
stranded structure,
primase - synthesis the short RNA primer.
DNA polymerase III-In the leading and lagging strands, it synthesize the DNA.
DNA polymerase I- Removes RNA primer
DNA ligase- it joins the fragments.
Tus;-binds with ter sequence.
DNA replication different than PCR:--PCR generates large quantities of DNA from very small
amount of DNA, PCR is a INVITRO technique, while DNA replication occurs inside the iving
system, it is a invivo technique. The whole DNA is replicated, in DNA replication, while in PCR
the amplification of segment of DNA takes place.
Solution
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication st.
Glutamic acid and Lycine Hydrogen bonding could .pdfapexcomputer54
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring.
Solution
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring..
Need a clear pictureSolutionNeed a clear picture.pdfapexcomputer54
This very short document discusses needing a clear picture but does not provide any details on the topic or solution. It repeats the phrase "Need a clear picture" but does not elaborate on what is needed, why it is needed, or how any solution might help provide a clear picture.
This one sentence document contains the word "Li2S" repeated three times. It appears to be referring to lithium sulfide, a chemical compound with the formula Li2S, but provides no other context or information.
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = .pdfapexcomputer54
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1
Solution
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1.
let an=(-1)^(2n+1)and L=-1let us ahow that L=lim anindeed if e.pdfapexcomputer54
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1
Solution
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1.
D) more, vibrational A vibration happens in any .pdfapexcomputer54
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy
Solution
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy.
How the standard developed1) The standards are developed to this .pdfapexcomputer54
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judgment as to whether or not there is
at least rough consensus to publish the edited document as an Internet standard.
9) A key factor that opinion within the IETF is whether or not there is running code.
Internet Society:
1) Founded 1991 by Internet Pioneers Founded 1991 by Internet Pioneers
2) International, not-for-profit, membership org. International, not-for-profit, membership org.
a) 80+ organisation organisation members members
b) 21,000+ individual members 21,000+ individual members
c) 75+ chapters, 10+ chapters forming
3) Organisation Organisation members fund activities in members fund activities in
a) Standards Standards
b) Education Education
c) Policy
Solution
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judg.
Hi!The chlorphylls having lower Rf values indicates that they inte.pdfapexcomputer54
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : )
Solution
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : ).
Determining the hybridization state requires that you know how many .pdfapexcomputer54
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents.
Solution
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents..
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. It consists of the letter "E", the number "5th", the word "Solution", and repeats the letter "E" and number "5th" again without any other details.
Command language is more beneficial than other input methods in foll.pdfapexcomputer54
Command language is more beneficial than other input methods in following cases
Some example applications are
Solution
Command language is more beneficial than other input methods in following cases
Some example applications are.
Clastic structure1)Roundingit is the degree of smoothing due to .pdfapexcomputer54
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diagenesis of
sediments.cementation occured primarily below the water table regardless of sedimentar grin
size present.
non clastic texture.
1)crystalline texture:it are visible and form an interlocking network.unlike igneious crystalline
textures.sedimentary crystallind textures are typically formed one material throught the entire
rock.
2)skeletal texture:it term used to describe the habit of enhedral to subhedral crystals in igneous
rock contains crystallogically oriented hallows and gaps.crystalls can bdescribed as re-entrant.the
voids with in skeletal crystals are filled with ground mass material.
3)oolite texture:it is a sedimentary rock formed by ooids,spherical grains composed of concentric
layers.oolites consisted of ooids of diameter 0.25-2mm rocks composed of ooids larger than
2mmare called pisolites.
Solution
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diage.
Balance in the Retained Earnings account immediately after Event 2 i.pdfapexcomputer54
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Solution
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000.
How to Add Chatter in the odoo 17 ERP ModuleCeline George
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
we will use tables of normal distribution ( Z distribution)P( z .pdfapexcomputer54
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587
Solution
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587.
The Northeast blackout of 2003 was a widespread power outage that oc.pdfapexcomputer54
The Northeast blackout of 2003 was a widespread power outage that occurred throughout parts
of the Northeasternand Midwestern United States and the Canadian province of Ontario on
Thursday, August 14, 2003, just after 4:10 p.m. All told, 50 million people lost power for up to
two days in the biggest blackout in North American history. The event contributed to at least 11
deaths and cost an estimated $6 billion. A surge of electricity to western New York and Canada
touched off a series of power failures and enforced blackouts that left parts of at least eight states
in the Northeast and the Midwest without electricity.Transmission system operators scattered
across some 300 control centers nationwide monitor voltage and current data from SCADA
(supervisory control and data acquisition) systems placed at transformers, generators and other
critical points.The widespread failures provoked the evacuation of office buildings, stranded
thousands of commuters. As circuit breakers tripped at generating stations from New York to
Michigan and into Canada, millions of people were instantly caught up in the largest blackout in
American history. In New York City, power was shut off by officials struggling to head off a
wider blackout. Cleveland and Detroit went dark, as did Toronto and sections of New Jersey,
Pennsylvania, Connecticut and Massachusetts. Officials worked into the night to put the grid
back in operation and restore electric service. Mayor Michael R. Bloomberg said that that the
power was back on in parts of Brooklyn, the Bronx and Queens by 11 p.m. -- but not Manhattan.
The blackout began just after the stock exchanges had closed for the day, a slow summer day of
relatively light trading, as thousands of workers were about to head home. Office workers who
were still at their desks watched their computer monitors blink off without warning on a hot and
hazy afternoon. Soon hospitals and government buildings were switching on backup generators
to keep essential equipment operating, and the police were evacuating people trapped in
elevators. Airports throughout the affected states suffered serious disruptions, including the three
major airports in the New York metropolitan region, but did not close. Thousands of subway
passengers in New York City had to be evacuated from tunnels, and commuter trains also came
to a halt. Gov.George E. Pataki said that 600 trains were stranded.The office of the Canadian
prime minister, Jean Chrétien, initially said the power problems were caused by lightning in New
York State but later retracted that. Canadian officials later expressed uncertainty about the exact
cause but continued to insist the problem began on the United States side of the border. The
Nuclear Regulatory Commission said that the seven nuclear plants in New York and New Jersey
and two in the Midwest had shut down automatically when the failure occurred.Telephone
service was disrupted, especially calls to and from cellular phones.Officials said the.
The chemical reaction responsible for formation of water2H2+ O2--.pdfapexcomputer54
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water.
Solution
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water..
SolutionDomain name system is used to determine the IP address of.pdfapexcomputer54
Solution
:
Domain name system is used to determine the IP address of the host/domain name. DNS is very
useful to every user where it helps retrieve the website IP address. Without DNS we cannot
retrieve the website IP address. This process is known as forward DNS resolution.
DNS should be registered by some companies. If any attacker has details then the dns will be
controlled by the attacker. when we are sending the e-mail the dns data will be used then it can
be catches anywhere in the network.
So it is very important that a DNS require reliable communications..
sample mean = population mean = 2.11standard deviation = popul.pdfapexcomputer54
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137
Solution
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137.
Polymerase chain reaction (PCR) is a process used to replicate DNA i.pdfapexcomputer54
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication starts at single
origin (ori C). It do not occur in PCR.
Prokaryotic DNA replication is followed by the three basic step:-
1. initiation, 2. elongation, and 3. termination.
The process of PCR is followed by the three basic step:-
1. Denaturation, 2.,Annealing, and 3. Primer Extension
1) the process of forming single stranded DNA, :-In PCR the DNA denatures at high tempreture,
at 95oC to form its single-stranded. During the prokaryotic DNA Replication, Ori C unwinds the
DNA by DNA B or helicase, and extension of ssDNA for copying during initiation step.
2) priming of the DNA polymerase,:-In PCR primer binds with the complementry strand of DNA
during the Annealing step. In prokaryotic replcation, primase form RNA primer extended by
DNA polymerase III,during the elongation step. Synthesis ofleading strand and lagging strand
takes place in this step. It do not occur in PCR.
3) the DNA polymerase itself:-In prokaryotic DNA replication DNA polymerase III synthesize
the segment of DNA for both leading and lagging strand. In PCR, DNA polymerase extends the
primer, Taq polymerase, is used in this process and extension is performed at 72oC.
4) the end product:- As a PCR two ds DNA is formed. And in prokaryotic DNA replication 2
circular DNA is formed , which remain interlinked and topoisomerase II make them seprate by
making cut over them.
Protien involed in prokaryotic DNA replication:-
Dna A protein-It initiate DNA replication by the binding with the Dna A boxes.
Dna helicase- it seprates the ds DNA.
Topoisomerase- From the replication fork, it removes positive supercoiling.
Single stranded binding protein-to prevent the reformation of ds DNA, it binds with single
stranded structure,
primase - synthesis the short RNA primer.
DNA polymerase III-In the leading and lagging strands, it synthesize the DNA.
DNA polymerase I- Removes RNA primer
DNA ligase- it joins the fragments.
Tus;-binds with ter sequence.
DNA replication different than PCR:--PCR generates large quantities of DNA from very small
amount of DNA, PCR is a INVITRO technique, while DNA replication occurs inside the iving
system, it is a invivo technique. The whole DNA is replicated, in DNA replication, while in PCR
the amplification of segment of DNA takes place.
Solution
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication st.
Glutamic acid and Lycine Hydrogen bonding could .pdfapexcomputer54
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring.
Solution
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring..
Need a clear pictureSolutionNeed a clear picture.pdfapexcomputer54
This very short document discusses needing a clear picture but does not provide any details on the topic or solution. It repeats the phrase "Need a clear picture" but does not elaborate on what is needed, why it is needed, or how any solution might help provide a clear picture.
This one sentence document contains the word "Li2S" repeated three times. It appears to be referring to lithium sulfide, a chemical compound with the formula Li2S, but provides no other context or information.
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = .pdfapexcomputer54
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1
Solution
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1.
let an=(-1)^(2n+1)and L=-1let us ahow that L=lim anindeed if e.pdfapexcomputer54
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1
Solution
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1.
D) more, vibrational A vibration happens in any .pdfapexcomputer54
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy
Solution
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy.
How the standard developed1) The standards are developed to this .pdfapexcomputer54
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judgment as to whether or not there is
at least rough consensus to publish the edited document as an Internet standard.
9) A key factor that opinion within the IETF is whether or not there is running code.
Internet Society:
1) Founded 1991 by Internet Pioneers Founded 1991 by Internet Pioneers
2) International, not-for-profit, membership org. International, not-for-profit, membership org.
a) 80+ organisation organisation members members
b) 21,000+ individual members 21,000+ individual members
c) 75+ chapters, 10+ chapters forming
3) Organisation Organisation members fund activities in members fund activities in
a) Standards Standards
b) Education Education
c) Policy
Solution
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judg.
Hi!The chlorphylls having lower Rf values indicates that they inte.pdfapexcomputer54
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : )
Solution
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : ).
Determining the hybridization state requires that you know how many .pdfapexcomputer54
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents.
Solution
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents..
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. It consists of the letter "E", the number "5th", the word "Solution", and repeats the letter "E" and number "5th" again without any other details.
Command language is more beneficial than other input methods in foll.pdfapexcomputer54
Command language is more beneficial than other input methods in following cases
Some example applications are
Solution
Command language is more beneficial than other input methods in following cases
Some example applications are.
Clastic structure1)Roundingit is the degree of smoothing due to .pdfapexcomputer54
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diagenesis of
sediments.cementation occured primarily below the water table regardless of sedimentar grin
size present.
non clastic texture.
1)crystalline texture:it are visible and form an interlocking network.unlike igneious crystalline
textures.sedimentary crystallind textures are typically formed one material throught the entire
rock.
2)skeletal texture:it term used to describe the habit of enhedral to subhedral crystals in igneous
rock contains crystallogically oriented hallows and gaps.crystalls can bdescribed as re-entrant.the
voids with in skeletal crystals are filled with ground mass material.
3)oolite texture:it is a sedimentary rock formed by ooids,spherical grains composed of concentric
layers.oolites consisted of ooids of diameter 0.25-2mm rocks composed of ooids larger than
2mmare called pisolites.
Solution
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diage.
Balance in the Retained Earnings account immediately after Event 2 i.pdfapexcomputer54
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Solution
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000.
How to Add Chatter in the odoo 17 ERP ModuleCeline George
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
Physiology and chemistry of skin and pigmentation, hairs, scalp, lips and nail, Cleansing cream, Lotions, Face powders, Face packs, Lipsticks, Bath products, soaps and baby product,
Preparation and standardization of the following : Tonic, Bleaches, Dentifrices and Mouth washes & Tooth Pastes, Cosmetics for Nails.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
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The simplified electron and muon model, Oscillating Spacetime: The Foundation...RitikBhardwaj56
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
How to Fix the Import Error in the Odoo 17Celine George
An import error occurs when a program fails to import a module or library, disrupting its execution. In languages like Python, this issue arises when the specified module cannot be found or accessed, hindering the program's functionality. Resolving import errors is crucial for maintaining smooth software operation and uninterrupted development processes.