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The Final Accounts of non-trading concerns consists of: 1. Receipts and Payments Account 2. Income and Expenditure Account, and 3. Balance Sheet. 1. Receipts and Payments Account: It is a Real Account. It is a consolidated summary of Cash Book. It is prepared at the end of the accounting period. All cash receipts are recorded on the debit side and all cash payments are recorded on the credit side. Cash Book consisting of entries of receipts and payments in a chronological order while the Receipts and payments is a summary of total cash receipts and cash payments. It starts with opening balance of Cash and Bank and ends with closing balance of Cash and Bank. It does not take into account outstanding amounts of receipts and payments. Receipts and Payments may be of Capi tal or Revenue nature; they may relate to the current or previous year or subsequent year; so long as they are actually received or paid, they must appear in this account. Features of Receipts and Payment Account, In Brief: 1. It starts with opening balance and ends with closing balance 2. It is the summary of cash and bank transactions. 3. Actual cash transactions are entered. 4. It includes capital as well as revenue items. 5. It follows cash system of accounting 6. It shows cash position and excludes all non-cash items. 7. It is a real account. 8. It does not take any income/expense outstanding at the beginning or at the end. 2. Income and Expenditure Account: It is a Nominal Account. It is in the form of Profit and Loss Account. It is concerned with only revenue items—expenses and incomes. It records all losses and expenses on its debit side and all incomes and gains on its credit side. Of the incomes and expenses of revenue nature, only the portion pertaining to the current year is shown in the Income and Expenditure Account i.e. amount relating to the previous year or future year are excluded. Again, the incomes and expenses of current year, whether received or not, must be shown. In other words, incomes and expenses have to be adjusted for both out-standing and pre- payments. All non-cash items, Depreciation, Bad Debts, Provision for Doubtful Debts etc. are taken into account. The difference between the debit side and the credit side is either surplus or deficit for the year concerned and the difference will be transferred to the Capital Fund (also called General Fund or Accumulated Fund) appearing in Balance Sheet. Features of Income and Expenditure Account, In Brief: 1. It is prepared in lieu of Profit and Loss Account. 2. It is a nominal account. 3. It is based on mercantile system of accounting. 4. There is no opening balance. 5. It ends with Surplus or Deficit. 6. It excludes all capital income and capital expenses. 7. It includes only revenue items. 8. It records all expenses whether paid or not, and all incomes whether received or not. 3. Balance Sheet: Balance Sheet in case of non-trading concern is prepared in the usual manner and consists of all liabilities and .

The Final Accounts of non-trading concerns consists of1. Receipts.pdf

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sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1. according to Hardy-Weinberg law frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2 q = square root of 0.1 = 0.316 then p = 1 - q = 1 - 0.316 = 0.684 p = frequency of healthy allele = 0.684 Frequency of healthy allele = 0.684 2.Ans. how many people in a population of 10,000 will have sickel cell anemia. sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia. 3.Ans. how many people in a population of 10,000 will be sickel cell carriers. sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers 4.Ans. mutations occurs and new allele introduced into the population. small population which causes genetic drift. the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained. 3. Alkaptonuria caused by recessive allele. survey revels that 5 in 50 people have Alkaptonuria a. if a population of 50 people if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1 q2 = 0.1 then q = square root of 0.1 = 0.316 q = 0.316 p = 1 - q =0.684 p = 0.684 b. Heterozygous to Alkaptonuria = 2pq 2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele C. Homozygous Dominant = AA =p2 AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant Solution sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1. according to Hardy-Weinberg law frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2 q = square root of 0.1 = 0.316 then p = 1 - q = 1 - 0.316 = 0.684 p = frequency of healthy allele = 0.684 Frequency of healthy allele = 0.684 2.Ans. how many people in a population of 10,000 will have sickel cell anemia. sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia. 3.Ans. how many people in a population of 10,000 will be sickel cell carriers. sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers 4.Ans. mutations occurs and new allele introduced into the population. small population which causes genetic drift. the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained. 3. Alkaptonuria caused by recessive allele. survey revels that 5 in 50 people have Alkaptonuria a. if a population of 50 people if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1 q2 = 0.1 then q = square root of 0.1 = 0.316 q = 0.316 p = 1 - q =0.684 p = 0.684 b. Heterozygous to Alkaptonuria = 2pq 2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele C. Homozygous Dominant = AA =p2 AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant.

sickel cell anemia is caused by recessive allele. frequency of sicke.pdf

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I think that (e) have only non-singular generalized inversesSol.pdf

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ConvertingSeconds.java import java.util.Scanner; public class ConvertingSeconds { public static void main(String[] args) { //Declaring variables int fromseconds,hours,minutes,seconds,toSeconds; //Scanner class Object is used to read the inputs entered by the user Scanner sc=new Scanner(System.in); //getting the number of seconds entered by the user System.out.print(\"Enter Total Number of Seconds :\"); fromseconds=sc.nextInt(); //calling the method secondTime() by passing the seconds as arguments String str=secondTime(fromseconds); //Converting the string to String array where \',\' is the delimeter String arr[]=str.split(\",\"); //Converting the String into integer hours=Integer.parseInt(arr[0]); minutes=Integer.parseInt(arr[1]); seconds=Integer.parseInt(arr[2]); //calling the method inSeconds by passing the hours,minutes,seconds as arguments toSeconds=inSeconds(hours,minutes,seconds); //Displaying the hours,minutes,seconds to seconds System.out.println(hours+\" hours \"+minutes+\" minutes \"+seconds+\" seconds corresponds to \"+toSeconds+\" seconds\"); } /*This method will converts the hours,minutes,seconds to seconds * Params:hours,minutes,seconds * Return:totalSeconds */ private static int inSeconds(int hours, int minutes, int seconds) { //Converting the hours,minutes and seconds to totalseconds int totSeconds=hours*3600+minutes*60+seconds; return totSeconds; } /*This method will converts the seconds to hours,minutes,seconds * Params:total Seconds * Return:String */ private static String secondTime(int totseconds) { //Declaring variable int secs=totseconds; //calculating the number of hours int hours=totseconds/3600; totseconds=totseconds-(hours*3600); //calculating the number of minutes int minutes=totseconds/60; totseconds=totseconds-(minutes*60); //Calculating the number of seconds int seconds=totseconds; //displaying the total seconds to hours,minutes,seconds System.out.println(secs+\" Seconds corresponds to \"+hours+\" hours \"+minutes+\" minutes \"+seconds+\" seconds\"); return hours+\",\"+minutes+\",\"+seconds; } } _____________________________________________ Output: Enter Total Number of Seconds :10000 10000 Seconds corresponds to 2 hours 46 minutes 40 seconds 2 hours 46 minutes 40 seconds corresponds to 10000 seconds _____________________________________________ Output1: Enter Total Number of Seconds :15000 15000 Seconds corresponds to 4 hours 10 minutes 0 seconds 4 hours 10 minutes 0 seconds corresponds to 15000 seconds _________________________________________Thank You Solution ConvertingSeconds.java import java.util.Scanner; public class ConvertingSeconds { public static void main(String[] args) { //Declaring variables int fromseconds,hours,minutes,seconds,toSeconds; //Scanner class Object is used to read the inputs entered by the user Scanner sc=new Scanner(System.in); //getting the number of seconds entered by the user System.out.print(\"Enter Total Number of Seconds :\"); fromseconds=sc.nextInt(); //calling the method seco.

ConvertingSeconds.javaimport java.util.Scanner;public class Conv.pdf

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Before Attempting to measure risk aversion, following two factors should be considered: 1 ) Estimating probabilities: It includes measuring the future events based on their possibility of occurrence. generally, the Probabilities are assigned based on past experiences. 2) Life tables and estimates: it includes the estimation of the amount of risk undertaken by the investors in their previous investment. By looking at their investment history and analyzing behavior, this table can be prepared. Solution Before Attempting to measure risk aversion, following two factors should be considered: 1 ) Estimating probabilities: It includes measuring the future events based on their possibility of occurrence. generally, the Probabilities are assigned based on past experiences. 2) Life tables and estimates: it includes the estimation of the amount of risk undertaken by the investors in their previous investment. By looking at their investment history and analyzing behavior, this table can be prepared..

Before Attempting to measure risk aversion, following two factors sh.pdf

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Answer: The general circumstance for retaining outside experts is where the auditor requires a special skill or knowledge outside of accounting or auditing, the auditors feel they do not have the level of expertise, or they may want an additional opinion. PCAOB interim Using the Work of a Specialist provides guidance and recognizes that auditors are not expected to be experts in all areas, auditors may encounter material matters which are complex or subjective, and an auditor may use the work of a specialist as evidential matter to evaluate financial statement assertions (PCAOB AU 336, 2010). In the case of auditing pensions, the auditors would most likely use an actuary as a specialist as the auditors would not be expected to have that level of expertise. The actuary would be knowledgeable of the entire pension development process and be able to more quickly comprehend the nuances of the actuarial assumptions and mechanics of the calculations. For example, some of the tasks that the actuary would be involved with the company on for pensions includes the measurement of pension obligations, assignment of plan obligations to time periods, development of a cost allocation procedure, development of a contribution allocation procedure, determination of types and levels of benefits, projection of pension obligations, projection of plan costs or contributions (AAA, 2010). Much of the financial impact of pensions is based on going from current known data (number employees, salaries, etc.) through a number of assumptions (how many years work, how many years collect a pension, mortality rates, what rate able to invest funds at, etc.) to develop the most likely funding needs over time to provide the retirement benefits as agreed to cover the pension expenses. Outside of determining the current employees, salaries, etc., almost everything else is an assumption. How do we make the most appropriate and best assumption to ensure having dollars available in the future? How much do we need to put aside now that will grow and cover the pension requirements? This is where the audit firm should employ an actuary to examine the actuarial-type assumptions made by the company on its employees and current retirees to see if they are reasonable assumptions given the company’s data. The actuary would be able to help examine the pension plan assets, the investments, the effective earning rates from the investments (in comparison to the discount rate being used), the projected pension expenses, and the resultant pension plan funding situation. The actuary’s review should be able to assist the auditor in identifying potential issue areas for the auditor’s further examination. Solution Answer: The general circumstance for retaining outside experts is where the auditor requires a special skill or knowledge outside of accounting or auditing, the auditors feel they do not have the level of expertise, or they may want an additional opinion. PCAOB interim Using the Work of a S.

Answer The general circumstance for retaining outside experts is wh.pdf

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Ans: Consolidated financial statements are the aggregated or combined figure of statements between a parent company and its subsidiaries. Consolidated balance sheet As on 31/12/2020 Assets P S Total Current assets: Cash 70,000 28,000 98,000 Accounts receivable 210,000 224,000 434,000 Inventory 252,000 140,000 392,000 532,000 392,000 924,000 Noncurrent assets: Land 140,000 -- 140,000 Equipment 7,000,000 3,780,000 10,780,000 Amortization, equipment (2,478,000) (1,736,000) (4,214,000) Investment in S 1,120,000 -- 1,120,000 5,782,000 2,044,000 7,826,000 Total assets 6,314,000 2,436,000 8,750,000 Liabilities and shareholders equity: Current liabilities: Accounts payable 630,000 280,000 910,000 Noncurrent liabilities: Loan payable 420,000 700,000 1,120,000 1,050,000 980,000 2,030,000 Shareholders equity: Share capital 1,680,000 420,000 2,100,000 Retained earnings 3,584,000 1,036,000 4,620,000 5,264,000 1,456,000 6,720,000 6,314,000 2,436,000 8,750,000 Consolidated income statement For the year ended 31/12/2020 Particulars P S Total Revenue: Sales 2,804,200 2,100,000 4,904,200 Royalties 210,000 -- 210,000 Dividends 100,800 -- 100,800 3,115,000 2,100,000 5,215,000 Expenses: Cost of sales 1,680,000 1,260,000 2,940,000 Other 784,000 575,400 1,359,400 2,464,000 1,835,400 4,299,400 Net income 651,000 264,600 915,600 Retained earnings statement For the year ended 31/12/2020 P S Total Opening retained earnings 3,353,000 897,400 4,250,400 Net income 651,000 264,600 915,600 Dividends declared (420,000) (126,000) (546,000) Closing retained earnings 3,584,000 1,036,000 4,620,000 Figures are added. P + S = Total. Assets P S Total Current assets: Cash 70,000 28,000 98,000 Accounts receivable 210,000 224,000 434,000 Inventory 252,000 140,000 392,000 532,000 392,000 924,000 Noncurrent assets: Land 140,000 -- 140,000 Equipment 7,000,000 3,780,000 10,780,000 Amortization, equipment (2,478,000) (1,736,000) (4,214,000) Investment in S 1,120,000 -- 1,120,000 5,782,000 2,044,000 7,826,000 Total assets 6,314,000 2,436,000 8,750,000 Liabilities and shareholders equity: Current liabilities: Accounts payable 630,000 280,000 910,000 Noncurrent liabilities: Loan payable 420,000 700,000 1,120,000 1,050,000 980,000 2,030,000 Shareholders equity: Share capital 1,680,000 420,000 2,100,000 Retained earnings 3,584,000 1,036,000 4,620,000 5,264,000 1,456,000 6,720,000 6,314,000 2,436,000 8,750,000 Solution Ans: Consolidated financial statements are the aggregated or combined figure of statements between a parent company and its subsidiaries. Consolidated balance sheet As on 31/12/2020 Assets P S Total Current assets: Cash 70,000 28,000 98,000 Accounts receivable 210,000 224,000 434,000 Inventory 252,000 140,000 392,000 532,000 392,000 924,000 Noncurrent assets: Land 140,000 -- 140,000 Equipment 7,000,000 3,780,000 10,780,000 Amortization, equipment (2,478,000) (1,736,000) (4,214,000) Investment in S 1,120,000 -- 1,120,000 5,782,000 2,044,000 7,826,000 Total assets 6,314,000 2,436,00.

AnsConsolidated financial statements are the aggregated or combin.pdf

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Answer b)S phaseS-phase is synthetic phase where semiconservative.pdf

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a)The NLS resides somewhere between amino acids 240 and 357. b)The.pdf

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A database is generally used for storing related, structured data, with well defined data formats, in an efficient manner for insert, update and/or retrieval (depending on application). On the other hand, a file system is a more unstructured data store for storing arbitrary, probably unrelated data. The file system is more general, and databases are built on top of the general data storage services provided by file systems. A Data Base Management System is a system software for easy, efficient and reliable data processing and management. It can be used for: Creation of a database. Retrieval of information from the database. Updating the database. Managing a database. It provides us with the many functionalities and is more advantageous than the traditional file system in many ways listed below: 1) Processing Queries and Object Management: In traditional file systems, we cannot store data in the form of objects. In practical-world applications, data is stored in objects and not files. So in a file system, some application software maps the data stored in files to objects so that can be used further. We can directly store data in the form of objects in a database management system. Application level code needs to be written to handle, store and scan through the data in a file system whereas a DBMS gives us the ability to query the database. 2) Controlling redundancy and inconsistency: Redundancy refers to repeated instances of the same data. A database system provides redundancy control whereas in a file system, same data may be stored multiple times. For example, if a student is studying two different educational programs in the same college, say ,Engineering and History, then his information such as the phone number and address may be stored multiple times, once in Engineering dept and the other in History dept. Therefore, it increases time taken to access and store data. This may also lead to inconsistent data states in both places. A DBMS uses data normalization to avoid redundancy and duplicates. 3) Efficient memory management and indexing: DBMS makes complex memory management easy to handle. In file systems, files are indexed in place of objects so query operations require entire file scans whereas in a DBMS , object indexing takes place efficiently through database schema based on any attribute of the data or a data-property. This helps in fast retrieval of data based on the indexed attribute. 4) Concurrency control and transaction management: Several applications allow user to simultaneously access data. This may lead to inconsistency in data in case files are used. Consider two withdrawal transactions X and Y in which an amount of 100 and 200 is withdrawn from an account A initially containing 1000. Now since these transactions are taking place simultaneously, different transactions may update the account differently. X reads 1000, debits 100, updates the account A to 900, whereas X also reads 1000, debits 200, updates A to 800. In bot.

A database is generally used for storing related, structured data, w.pdf

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9x^4y^10z^-2Solution9x^4y^10z^-2.pdf

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212+423 is same as 2 36+4 46 is same as 6 76 is same as 7 16...pdf

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1. Atoms: Atoms are structured by 3 main components. Protons, neutrons and electrons. The core of the atom consists of protons and neutrons. Protons are positively charged, and neutrons are neutral. Electrons circle the core of the atom, and form a \"cloud\" of electrons. Electrons are negatively charged. 2. Bonds: You have different elements such as Na, Cl, H, O etc. Several COMPOUNDS that you may have heard of include, NaCl and H2O. In order for these elements to form these compounds, bonds need to form between them. When bonds are made energy is released. 3. Periodic Table: The number of electrons that an atom of an element has is usually equal to it\'s atomic number. More importantly there is a trend that you can determine what the number of valence (outer most) electrons are. Group 1 has 1 valence electron, Group 2 has 2, Group 3 has 3, Group 8 has 0, Group 7 has 7, Group 6 has 6 (O, S, Se, Te), Group 5 has 5( N, P, As) and Group 4 has 4 (C,Si). 4. Electronegativity: This means how reactive a certain element is. If an element has lots of valence electrons it will have a high electronegativity because it wants that 1 electron, to fill its 8 compartments in its shell. The trend for electronegativity is that electronegativity increases as you go across the periodic table towards the right ->. As you go towards the right, you have more and more valence electrons, therefore more \"want\" for electrons to fill up the compartments. When you go (^) up the periodic table electronegativity also increases due to the distance the electrons are away from the nucleus (core) of the atom. 5. Polarity: Polarity just means you have 2 poles. Think of it like a magnet, you usually have a north pole and a south pole OR a positive pole and a negative pole. Different atoms have different charges. For example, H2O, consists of 2 hydrogens and 1 oxygen. Hydrogen is positively charged and oxygen is negatively charged. This is why water is said to be polar. It consists of both a positive and a negative. Water is also an example of hydrogen bonding, because Hydrogen is bonding with a large molecule, oxygen. Non-polar bonds are where there are no positives or negatives. An example would be a diatomic molecule, such as oxygen gas: O2. Methane: CH4 is also considered to be non-polar because its carbon atom covalently bonds with hydrogen atoms sharing almost completely equally. Hope this helps! :) Solution 1. Atoms: Atoms are structured by 3 main components. Protons, neutrons and electrons. The core of the atom consists of protons and neutrons. Protons are positively charged, and neutrons are neutral. Electrons circle the core of the atom, and form a \"cloud\" of electrons. Electrons are negatively charged. 2. Bonds: You have different elements such as Na, Cl, H, O etc. Several COMPOUNDS that you may have heard of include, NaCl and H2O. In order for these elements to form these compounds, bonds need to form between them. When bonds are made energy is released. 3. Perio.

1. Atoms Atoms are structured by 3 main components. Protons, ne.pdf

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1) importjavax.swing.* => package contains JFrame and JButton classes 2) javadoc comments begin with the /* symbol, and end with the */ symbol. 3) If you want to read in values from the user in the console window you must use a ____Scanner______ object. 4) float, int, long, and double are forms of ____C_____ data types. 5) An easy way to access every element of a Two dimensional array is to use _____2_nested_____ for loops. 6) To search the elements of an array, ___linear_____ search can always be used. 7) There are two types of method calls, void return methods and ___data____ return methods. 8.) import keyword is used to import built-in and user-defined packages into your java source file. key word that is used to bring in java.utils is ____Import______ 9. When an array name is passed to a method, it is passing the memory address, therefore it is passing the array by ____reference_____. because array name refers to the memory address of the array Solution 1) importjavax.swing.* => package contains JFrame and JButton classes 2) javadoc comments begin with the /* symbol, and end with the */ symbol. 3) If you want to read in values from the user in the console window you must use a ____Scanner______ object. 4) float, int, long, and double are forms of ____C_____ data types. 5) An easy way to access every element of a Two dimensional array is to use _____2_nested_____ for loops. 6) To search the elements of an array, ___linear_____ search can always be used. 7) There are two types of method calls, void return methods and ___data____ return methods. 8.) import keyword is used to import built-in and user-defined packages into your java source file. key word that is used to bring in java.utils is ____Import______ 9. When an array name is passed to a method, it is passing the memory address, therefore it is passing the array by ____reference_____. because array name refers to the memory address of the array.

1)importjavax.swing. = package contains JFrame and JButton class.pdf

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1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site is fomed at c with double bond O. 2)This electrophile formed attacks the position of sulphur. Because, the lone pair gets conjugated and forms Negative sites on both positions. But it attacks -Carbon with less congestion(presence of less bulky groups). 3) It combines to form given product aong with release of HCl. Solution 1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site is fomed at c with double bond O. 2)This electrophile formed attacks the position of sulphur. Because, the lone pair gets conjugated and forms Negative sites on both positions. But it attacks -Carbon with less congestion(presence of less bulky groups). 3) It combines to form given product aong with release of HCl..

1) SnCl4 takes cl- from acid halide to form SnCl5+. A positive site .pdf

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(a) NaCl is an ionic compound. The intermolecular forces are electro.pdf

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% Runs Newton\'s method to find roots of f(z) . It col function [a,n,TOL,N]=P3(a,n,TOL,N) A=zeros(n,n,\'uint8\'); f = inline(\'z^3-1\'); % Define f and f\'. fprime = inline(\'3*z^2\'); rts=roots([1 0 0 -1]) h=2*a/n; for j=1:n+1, % Try initial values with imaginary parts between y = -a + (j-1)*h; % -a and a for i=1:n+1, % and with real parts between x = -a + (i-1)*h; % -a and a. z = x + 1i*y; zk = z; kount = 0; % kount is the total number of iterations. conv1 = 0; % conv1,2,3 count iterations when approx soln is within conv2 = 0; % TOL of root1,2,3. conv3 = 0; while kount < 40 & conv1 < 5 & conv2 < 5 & conv3 < 5, kount = kount + 1; zk = zk - f(zk)/fprime(zk); % This is the Newton step. if abs(zk-root1) < TOL, % Check for convergence to root1. conv1 = conv1 + 1; else conv1 = 0; end; if abs(zk-root2) < TOL, % Check for convergence to root2. conv2 = conv2 + 1; else conv2 = 0; end; if abs(zk-root3) < TOL, % Check for convergence to root3. conv3 = conv3 + 1; else conv3 = 0; end; end; if conv1 >=3, M(j,i,1) = 255; end; % Converged to root 1. Color point green. if conv2 >=3, M(j,i,2) = 255; end; % Converged to root 2. Color point red. if conv3 >=3, M(j,i,3) = 255; end; % Converged to root 3. Color point blue. end; end; imshow(A) Solution % Runs Newton\'s method to find roots of f(z) . It col function [a,n,TOL,N]=P3(a,n,TOL,N) A=zeros(n,n,\'uint8\'); f = inline(\'z^3-1\'); % Define f and f\'. fprime = inline(\'3*z^2\'); rts=roots([1 0 0 -1]) h=2*a/n; for j=1:n+1, % Try initial values with imaginary parts between y = -a + (j-1)*h; % -a and a for i=1:n+1, % and with real parts between x = -a + (i-1)*h; % -a and a. z = x + 1i*y; zk = z; kount = 0; % kount is the total number of iterations. conv1 = 0; % conv1,2,3 count iterations when approx soln is within conv2 = 0; % TOL of root1,2,3. conv3 = 0; while kount < 40 & conv1 < 5 & conv2 < 5 & conv3 < 5, kount = kount + 1; zk = zk - f(zk)/fprime(zk); % This is the Newton step. if abs(zk-root1) < TOL, % Check for convergence to root1. conv1 = conv1 + 1; else conv1 = 0; end; if abs(zk-root2) < TOL, % Check for convergence to root2. conv2 = conv2 + 1; else conv2 = 0; end; if abs(zk-root3) < TOL, % Check for convergence to root3. conv3 = conv3 + 1; else conv3 = 0; end; end; if conv1 >=3, M(j,i,1) = 255; end; % Converged to root 1. Color point green. if conv2 >=3, M(j,i,2) = 255; end; % Converged to root 2. Color point red. if conv3 >=3, M(j,i,3) = 255; end; % Converged to root 3. Color point blue. end; end; imshow(A).

Runs Newtons method to find roots of f(z) . It col function [a.pdf

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The java Payroll that prompts user to enter hourly rate .pdf

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They represent the position of the other function.pdf

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