Using gordon growth model Value of stock=Dividend paid(1+growt.pdf

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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.

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

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moles of NH3 = 3417 = 2 moles of O2 = 4032 = 1..pdf

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NO has got a strong bond because it has got an co.pdf

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Yearscost of machin7 Year MACRS rateannual depreciation = cost.pdf

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Years cost of machin 7 Year MACRS rate annual depreciation = cost of machine* MACRS rate 1 190000 0.143 27170 2 190000 0.245 46550 3 190000 0.175 33250 4 190000 0.125 23750 5 190000 0.089 16910 6 190000 0.089 16910 7 190000 0.089 16910 8 190000 0.045 8550 total 1 190000 Year EBIT less depreciation EBT less tax-30% EAT Add depreciation EATBD present value of EATBD = EATBD/1+r)^n r= 12% 1 58000 27170 30830 9249 21581 27170 48751 43527.68 2 58000 46550 11450 3435 8015 46550 54565 43498.88 3 58000 33250 24750 7425 17325 33250 50575 35998.29 4 29000 23750 5250 1575 3675 23750 27425 17429.08 5 29000 16910 12090 3627 8463 16910 25373 14397.32 6 29000 16910 12090 3627 8463 16910 25373 12854.75 7 29000 16910 12090 3627 8463 16910 25373 11477.46 8 29000 8550 20450 6135 14315 8550 22865 9234.79 9 29000 0 29000 8700 20300 0 20300 7320.384 10 29000 0 29000 8700 20300 0 20300 6536.057 sum of present value of EATBD 202274.7 less present value of cash outflow 190000 net present value 12274.69 Machine should be purchased as it results In positive NPV Years cost of machin 7 Year MACRS rate annual depreciation = cost of machine* MACRS rate 1 190000 0.143 27170 2 190000 0.245 46550 3 190000 0.175 33250 4 190000 0.125 23750 5 190000 0.089 16910 6 190000 0.089 16910 7 190000 0.089 16910 8 190000 0.045 8550 total 1 190000 Year EBIT less depreciation EBT less tax-30% EAT Add depreciation EATBD present value of EATBD = EATBD/1+r)^n r= 12% 1 58000 27170 30830 9249 21581 27170 48751 43527.68 2 58000 46550 11450 3435 8015 46550 54565 43498.88 3 58000 33250 24750 7425 17325 33250 50575 35998.29 4 29000 23750 5250 1575 3675 23750 27425 17429.08 5 29000 16910 12090 3627 8463 16910 25373 14397.32 6 29000 16910 12090 3627 8463 16910 25373 12854.75 7 29000 16910 12090 3627 8463 16910 25373 11477.46 8 29000 8550 20450 6135 14315 8550 22865 9234.79 9 29000 0 29000 8700 20300 0 20300 7320.384 10 29000 0 29000 8700 20300 0 20300 6536.057 sum of present value of EATBD 202274.7 less present value of cash outflow 190000 net present value 12274.69 Machine should be purchased as it results In positive NPV Solution Years cost of machin 7 Year MACRS rate annual depreciation = cost of machine* MACRS rate 1 190000 0.143 27170 2 190000 0.245 46550 3 190000 0.175 33250 4 190000 0.125 23750 5 190000 0.089 16910 6 190000 0.089 16910 7 190000 0.089 16910 8 190000 0.045 8550 total 1 190000 Year EBIT less depreciation EBT less tax-30% EAT Add depreciation EATBD present value of EATBD = EATBD/1+r)^n r= 12% 1 58000 27170 30830 9249 21581 27170 48751 43527.68 2 58000 46550 11450 3435 8015 46550 54565 43498.88 3 58000 33250 24750 7425 17325 33250 50575 35998.29 4 29000 23750 5250 1575 3675 23750 27425 17429.08 5 29000 16910 12090 3627 8463 16910 25373 14397.32 6 29000 16910 12090 3627 8463 16910 25373 12854.75 7 29000 16910 12090 3627 8463 16910 25373 11477.46 8 29000 8550 20450 6135 14315 8550 22865 9234.79 9 29000 0 29000 8700 20300 0 20300 7320.384 10 29000 0 29000 8700 20300 0 20300 6536.057 sum.

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moles of NH3 = 3417 = 2 moles of O2 = 4032 = 1..pdf

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Years cost of machin 7 Year MACRS rate annual depreciation = cost of machine* MACRS rate 1 190000 0.143 27170 2 190000 0.245 46550 3 190000 0.175 33250 4 190000 0.125 23750 5 190000 0.089 16910 6 190000 0.089 16910 7 190000 0.089 16910 8 190000 0.045 8550 total 1 190000 Year EBIT less depreciation EBT less tax-30% EAT Add depreciation EATBD present value of EATBD = EATBD/1+r)^n r= 12% 1 58000 27170 30830 9249 21581 27170 48751 43527.68 2 58000 46550 11450 3435 8015 46550 54565 43498.88 3 58000 33250 24750 7425 17325 33250 50575 35998.29 4 29000 23750 5250 1575 3675 23750 27425 17429.08 5 29000 16910 12090 3627 8463 16910 25373 14397.32 6 29000 16910 12090 3627 8463 16910 25373 12854.75 7 29000 16910 12090 3627 8463 16910 25373 11477.46 8 29000 8550 20450 6135 14315 8550 22865 9234.79 9 29000 0 29000 8700 20300 0 20300 7320.384 10 29000 0 29000 8700 20300 0 20300 6536.057 sum of present value of EATBD 202274.7 less present value of cash outflow 190000 net present value 12274.69 Machine should be purchased as it results In positive NPV Years cost of machin 7 Year MACRS rate annual depreciation = cost of machine* MACRS rate 1 190000 0.143 27170 2 190000 0.245 46550 3 190000 0.175 33250 4 190000 0.125 23750 5 190000 0.089 16910 6 190000 0.089 16910 7 190000 0.089 16910 8 190000 0.045 8550 total 1 190000 Year EBIT less depreciation EBT less tax-30% EAT Add depreciation EATBD present value of EATBD = EATBD/1+r)^n r= 12% 1 58000 27170 30830 9249 21581 27170 48751 43527.68 2 58000 46550 11450 3435 8015 46550 54565 43498.88 3 58000 33250 24750 7425 17325 33250 50575 35998.29 4 29000 23750 5250 1575 3675 23750 27425 17429.08 5 29000 16910 12090 3627 8463 16910 25373 14397.32 6 29000 16910 12090 3627 8463 16910 25373 12854.75 7 29000 16910 12090 3627 8463 16910 25373 11477.46 8 29000 8550 20450 6135 14315 8550 22865 9234.79 9 29000 0 29000 8700 20300 0 20300 7320.384 10 29000 0 29000 8700 20300 0 20300 6536.057 sum of present value of EATBD 202274.7 less present value of cash outflow 190000 net present value 12274.69 Machine should be purchased as it results In positive NPV Solution Years cost of machin 7 Year MACRS rate annual depreciation = cost of machine* MACRS rate 1 190000 0.143 27170 2 190000 0.245 46550 3 190000 0.175 33250 4 190000 0.125 23750 5 190000 0.089 16910 6 190000 0.089 16910 7 190000 0.089 16910 8 190000 0.045 8550 total 1 190000 Year EBIT less depreciation EBT less tax-30% EAT Add depreciation EATBD present value of EATBD = EATBD/1+r)^n r= 12% 1 58000 27170 30830 9249 21581 27170 48751 43527.68 2 58000 46550 11450 3435 8015 46550 54565 43498.88 3 58000 33250 24750 7425 17325 33250 50575 35998.29 4 29000 23750 5250 1575 3675 23750 27425 17429.08 5 29000 16910 12090 3627 8463 16910 25373 14397.32 6 29000 16910 12090 3627 8463 16910 25373 12854.75 7 29000 16910 12090 3627 8463 16910 25373 11477.46 8 29000 8550 20450 6135 14315 8550 22865 9234.79 9 29000 0 29000 8700 20300 0 20300 7320.384 10 29000 0 29000 8700 20300 0 20300 6536.057 sum.

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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..

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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.

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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.

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Import java.awt.*; Import acm.program.*; Import acm.graphics.*; public class Sierpinski extends GraphicsProgram { public void run() { GRect box = new GRect(20, 20, 242, 242); box.setFilled(true); add(box); drawGasket(20, 20, 243); } private void drawFigure(int x, int y, int side) { int sub = side / 3; GRect box = new GRect(x + sub, y + sub, sub - 1, sub - 1); box.setFilled(true); box.setColor(Color.WHITE); add(box); if (sub >= 3) { drawFigure(x, y, sub); drawFigure(x + sub, y, sub); drawFigure(x + 2 * sub, y, sub); drawFigure(x, y + sub, sub); drawFigure(x + 2 * sub, y + sub, sub); drawFigure(x, y + 2 * sub, sub); drawFigure(x + sub, y + 2 * sub, sub); drawFigure(x + 2 * sub, y + 2 * sub, sub); } } C.8 SQUARE in recursive method:- import java.awt.*; import java.awt.event.*; import javax.swing.*; public class SierpinskiCarpet extends JPanel implements ActionListener { public static void main(String[] args) { JFrame window = new JFrame(\"Sierpinski Carpet\"); window.setContentPane( new SierpinskiCarpet() ); window.pack(); window.setResizable(false); window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); Dimension screensize = Toolkit.getDefaultToolkit().getScreenSize(); int x = Math.max(10, (screensize.width - window.getWidth()) / 2); int y = Math.max(10, (screensize.height - window.getHeight()) / 2); window.setLocation(x,y); window.setVisible(true); } private int level; private JRadioButton[] levelButton; private JCheckBox[][] regionCheckbox; private Display display; private class Display extends JPanel { Display() { setBackground(Color.WHITE); setPreferredSize(new Dimension(729,729)); } protected void paintComponent(Graphics g) { super.paintComponent(g); drawCarpet(g,level,0,0,729); } } public SierpinskiCarpet() { setLayout(new BorderLayout(3,3)); setBackground(Color.GRAY); setBorder(BorderFactory.createLineBorder(Color.GRAY,3)); display = new Display(); add(display,BorderLayout.CENTER); Box controls = Box.createVerticalBox(); controls.setBorder(BorderFactory.createEmptyBorder(10, 10, 10, 10)); controls.setBackground(Color.WHITE); controls.setOpaque(true); add(controls, BorderLayout.EAST); controls.add(new JLabel(\"Controls\")); controls.add(Box.createVerticalStrut(30)); levelButton = new JRadioButton[7]; ButtonGroup grp = new ButtonGroup(); for (int i = 0; i < 7; i++) { levelButton[i] = new JRadioButton(\"Recursion Level \" + i); grp.add(levelButton[i]); levelButton[i].addActionListener(this);controls.add(levelButton[i]); } level = 4; levelButton[4].setSelected(true); regionCheckbox = new JCheckBox[3][3]; JPanel checks = new JPanel(); checks.setLayout(new GridLayout(3,3,5,5)); for (int r = 0; r < 3; r++) { for (int c = 0; c < 3; c++) { regionCheckbox[r][c] = new JCheckBox(); checks.add(regionCheckbox[r][c]); regionCheckbox[r][c].addActionListener(this); regionCheckbox[r][c].setSelected(true); } } checks.setMaximumSize(new Dimension(90,90)); checks.setAlignmentX(LEFT_ALIGNMENT); checks.setBackground(Color.WHITE); regionCheckbox[1][1].setSelec.

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How the standard developed1) The standards are developed to this .pdf

How the standard developed1) The standards are developed to this .pdf

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.pdf

Hi!The chlorphylls having lower Rf values indicates that they inte.pdf

Hi!The chlorphylls having lower Rf values indicates that they inte.pdf

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 .pdf

Determining the hybridization state requires that you know how many .pdf

Determining the hybridization state requires that you know how many .pdf

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..

E) 5thSolutionE) 5th.pdf

E) 5thSolutionE) 5th.pdf

E) 5thSolutionE) 5th.pdf

Command language is more beneficial than other input methods in foll.pdf

Command language is more beneficial than other input methods in foll.pdf

Command language is more beneficial than other input methods in foll.pdf

Clastic structure1)Roundingit is the degree of smoothing due to .pdf

Clastic structure1)Roundingit is the degree of smoothing due to .pdf

Clastic structure1)Roundingit is the degree of smoothing due to .pdf

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.pdf

Balance in the Retained Earnings account immediately after Event 2 i.pdf

Balance in the Retained Earnings account immediately after Event 2 i.pdf

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.

PCOS corelations and management through Ayurveda.

PCOS corelations and management through Ayurveda.

PCOS corelations and management through Ayurveda.

Dr. Shivangi Singh Parihar

C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx

C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx

C1 Rubenstein AP HuG xxxxxxxxxxxxxx.pptx

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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..

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How the standard developed1) The standards are developed to this .pdf

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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.pdf

Hi!The chlorphylls having lower Rf values indicates that they inte.pdf

Hi!The chlorphylls having lower Rf values indicates that they inte.pdf

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 .pdf

Determining the hybridization state requires that you know how many .pdf

Determining the hybridization state requires that you know how many .pdf

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..

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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.pdf

Balance in the Retained Earnings account immediately after Event 2 i.pdf

Balance in the Retained Earnings account immediately after Event 2 i.pdf

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.

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