Let f be holomorphic on and inside a Jordan curve C and let f have m.pdf

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Let f be holomorphic on and inside a Jordan curve C and let f have maximum value M on C (such a number exists by the Extreme Value Theorem). Suppose that |f(z0)| = M for some z0 ins(C). Show that f is constant over ins(C). Solution By maximum modulus principle, if |f| attains the maximum value at the interior , it should be the constant function. Hence the result.

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Which of the following are true statementsI. sin2+cos2=1II. sec.pdf

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What scientific mechanism for evolution did Charles Darwin and Alfred Wallace propose in 1858? Describe how this concept works. What is the difference between microevolution and macroevolution? Solution 1. Charles Darwin during 1858 had proposed the Theory of Evolution by Natural Selection, where he has said that organisms change over time as a result of changes in heritable physical or behavioral traits. Changes that allow an organism to better adapt to its environment will help it survive and have more offspring. Alfred Wallace who is also called the Father of Biogeography, studied the geographical distribution of animal species and said the evolution is caused due to environmental pressures on varieties of species it forces different species to adapt themselves to the local conditions, leading to different populations in different locations. Both groups of scientists have said that natural selection is the single most and important factor in evolutionary changes seen in species. Scientifically, This concept works due to different phenomenon such as: a)Mutation- It is the Sudden change in the genetic makeup of an organism.And it is the driving force of evolution, which influences the population’s gene pool.The change in the genetic makeup of organism which favors the life of species makes it the fittest to survive! b)Genetic Drift-It can occur when a small group of individuals leaves a population and establishes a new one in a geographically isolated region.It occurs in a large population and helps the population to survive. c)Natural selection-This happens when populations of organisms are subjected to the environment. The fittest creatures are more likely to survive and pass their genes to their offspring, producing a population that is better adapted to the environment. 2.MicroEvolution-Microevolution is defined as changes in gene frequency in a population from one generation to the next.These small changes in species occur by recombining existing genetic material within the group of same species. Micro Evolution is caused due to Mutation, Migration or Gene Flow, Genetic Drift and Selection (natural and artificial). Macro Evolution: It refers to major evolutionary changes over time, the origin of new types of organisms from previously existing, but different, ancestral types.It is an evolution on a larger scale where we find the descent of many species from one common ancestor over billions of years.It is very difficult to see the Macro Evolution happening because it takes a longer time to happen. But Instead, we can reconstruct the history of life using all available evidence like geology, fossils, and living organisms..

What scientific mechanism for evolution did Charles Darwin and Alfred.pdf

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What is the main difference between doing business process analysis for ERP and other types of systems Solution The key differences in between the erp and other type of analysis can be explained by taking three basic concepts and bt these we can understand that and it is system implementation: In ERP systems it requires data conversations to estimate the analysis which contains seperate shells so thet it will be very helpful whereas in other systems there is no special software which is used it is designed manually and gone through the source code and not so conversions are required. Technological scope: In ERP systems it is homogenous environments and eith common data structures and tranfer overflows and it is data centric mainly whereas in other systems it is process centric and supports IT application and supports hetrogenous environments. Domain specific: In ERP domain is specific that is static which contains international sttings whereas in other systems it is quite opposite it does not contain domains and not having any international settings..

What is the main difference between doing business process analysis .pdf

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What are the genetic differences between a processed pseudogene and a mutated pseudogene and how do their typical locations relative to their functional paralogous genes differ? Solution Pseudogenes are the genes that have lost some part or entire of its original ability, of their real gene relative and thus they have a altered or no gene expression activity. Most commonly, pseudogenes are formed as a result of accumulation of multiple mutations in a single gene, whose gene product is not essential for survival. Pseudogenes may not be fully functional and may perform regulatory functions. Thus a pseudogene has its association with a functional gene, and both of them share a common ancestor. Processed pseudogenes are the result of retrotransposition, i.e, a part of the mRNA or hnRNA transcript from a gene spontaneously reverses back (transcribed again) into DNA and gets inserted into chromosomal DNA.Thus they are processed once and then reverses back to result in a pseudogene. As a result, random copies of genes are produced in vitro. When such pseudogenes are inserted into the genome, they usually contain a poly-A tail,they are without introns (introns are spliced out) and also lack the upstream promoters. Thus after transposition event, they become non-functional pseudogenes. Such pseudogenes are randomly inserted into the genome. On the other hand, mutated pseudogenes arise after multiple mutation events which prevent a gene from normal transcription and make them less functional or non-functional; like gene duplication event caused by homologous recombination (repititive sequences on misaligned chromosomes become non-functional) or one allele of gene having deleterious mutation resulting in loss of function, non-sense mutations, unequal cross overs and sister chromatid exchanges etc. In such pseudogenes, the location of genes remain same as their functional paralogous genes..

What are the genetic differences between a processed pseudogene and .pdf

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True or False? _____ Cells Composed of chemicals and structures, make up all living things, and vary in size and shape. _____ Cells Work together to allow for proper functioning of processes necessary for life. _____ Every cell in the human body contain Almost all human cells contain a nucleus, organelles, cytoplasm and a cell membrane. _____ The plasma membrane acts as a protective covering, maintains the cell’s integrity and allows it to survive. _____ Osmosis, filtration, facilitated diffusion, endocytosis and exocytosis are examples of passive transport mechanisms. _____ Diffusion is when a substance moves from high concentration to low concentration (down the concentration gradient). _____ Refering to osmotic pressure: water tends to travel across the plasma membrane from areas that have a high concentration of a solute to a lower concentration until the concentration is equal on both sides. _____Cystic Fibrosis and Diabetes Mellitus are examples of passive transport disorders. _____Ribosomes are the site for production of enzymes and other proteins needed for cell repair. _____Mitochondria is referred to as the “power house of the cell” and provides up to 95% of the body’s energy needs for cellular repair, movement and reproduction. _____Lysosomes have a pH of 7 and contain powerful enzymes that clean up waste from the cytoplasm. _____ Mitosis is the process of cellular reproduction and the process of sorting chromosomes so the new cell gets all the genetic material. _____Organs are groups of specialized cells that are similar in structure and that perform common functions. _____Skin, lining of the mouth, bladder, lungs, and blood vessels are examples of epithelial tissue. _____Simple epithelium tissues are a single layer of cells. _____A basement membrane is located directly beneath the cells of connective tissue and is a supporting non-cellular layer. _____Connective tissue Support the softer organs of the body against the forces of gravity. _____Muscle tissue consists of cells that are specialized to shorten and contract resulting in movement of some kind. _____Skeletal tissue and cardiac tissue are examples of connective tissue. _____Nervous tissue consists primarily of cells that are specialized for generating and transmitting electrical impulses. _____Serous, mucous, synovial and cutaneous are examples of tissue membranes. _____The Integumentary system consists of skin and all its accessory components (hair, nails, sweat and sebaceous glands). _____Intact skin is the best protection from most infections. _____Skin is the largest organ of the body. _____The epidermis layer of the skin is vascular and contains nerve cells. _____Cells of the epidermis level are constantly shedding replaced with new cells from the stratum basale. _____Melanocytes are specialized cells located deep in the epidermis and are responsible for skin color. _____Albinism is a condition in which individual has too much pigment in their skin, hair and eyes. ___.

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The picture above shows the life cycle of a moss. Mark all true statements about the structures indicated on the picture above. a. “B” is haploid b. “B” is part of a sporophyte c. “A” is haploid d. Cells that make up “C” and “B” are genetically identical e. “A” is part of a gametophyte f. “A” is part of a sporophyte g. “C” is the reproductive structure of the offspring of plant “B” (the next generation) h. “C” is part of a sporophyte i. “B” is part of a gametophyte j. “C” is the reproductive structure of plant “B” k. “C” is part of a gametophyte l. “A” is diploid m. “B” is diploid n. Cells that make up “A” and “B” are genetically identical a. “B” is haploid b. “B” is part of a sporophyte c. “A” is haploid d. Cells that make up “C” and “B” are genetically identical e. “A” is part of a gametophyte f. “A” is part of a sporophyte g. “C” is the reproductive structure of the offspring of plant “B” (the next generation) h. “C” is part of a sporophyte i. “B” is part of a gametophyte j. “C” is the reproductive structure of plant “B” k. “C” is part of a gametophyte l. “A” is diploid m. “B” is diploid n. Cells that make up “A” and “B” are genetically identical U Solution Some moss plants have both male and female plants as different , but others have both male and female parts in the same plant. At matured stage during wet conditions male plant releases sperm cells from antheridium, which fertilises the egg cell enclosed in female organs ( archegonia ) , a spore producing plant develops with capsule and lid. When spre capsule matured , it releases spores . Spores will germinate during wet conditions developed into moss plant . Then again the cycle repeats. Diagram explanation In the diagram on the left side labelled plant is spore producing plant with capsule . Capsule releasing spores Spores developed. Into separate female and male plants.( upper plant male plant , lower plant female plant) Sperm cells entry into the female organs ( archegonia) Fertilisation of egg Development of spore producing plant..

The picture above shows the life cycle of a moss. Mark all true stat.pdf

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The conversion of pyruvate to acetyl-CoA. A. requires the addition of CO2 B. occurs in the mitochondria of the cell and is irreversible. C. occurs in he nucleus of the cell. D. is needed for red blood cell production. Solution The conversion of pyruate to acetyl COA (B) occurs in the mitochondria of the cell and is irriversible. The process is carriedout with the help of an enzyme pyruate dehydrogenase, which is located in mitochondrial matrix. During the conversion process each pyruate molecule loses one carbon atom with the release of CO2..

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Suppose S is a set of n + 1 integers. Prove that there exist distinct a, b S such that a - b is a multiple of n. Solution There n distinct remainders modulo n ie 0,1,2,...,n-1 So let there be n boxes each corresponding to one of these remainders We need to put these n+1 integers into these n boxes So two of them must be in the same box ie two of them give the same remainder modulo n Let those two integers be a,b SO,a=b modulo n a-b=0 modulo n ie a-b is a multiple of n HEnce proved.

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Structural features of fungi Complete the following paragraph to describe the structural features of fungi. Solution The thallus of most fungi is known as a(n) mycelium. which is divided into a network of filaments called hyphae These hyphae increase the surface area, maximizing the absorption of nutrients. Some species of fungi have hyphae that are partitioned by walls of tissue called septae Fungal cells have cell walls comprised of chitin Similar to animals they store energy in the form of glycogen Fungi tend to be Nonmotile, although some do contain flagella at some point in their life cycle. Fungi reproduce sexually and asexually during their life cycles. Cells produced during the Asexual phase are called spores. These are haploid as they contain only one set of chromosomes.

Structural features of fungi Complete the following paragraph to desc.pdf

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State some successful predictions from Einstein\'s theory of special relativity Solution Special relativity implies a wide range of consequences, which have been experimentally verified, including length contraction, time dilation, relativistic mass, mass–energy equivalence, a universal speed limit and relativity of simultaneity.The theory is \"special\" in that it only applies in the special case where the curvature of spacetime due to gravity is negligible. In order to include gravity, Einstein formulated general relativity in 1915 Several experiments predating Einstein\'s 1905 paper are now interpreted as evidence for relativity.Particle accelerators routinely accelerate and measure the properties of particles moving at near the speed of light, where their behavior is completely consistent with relativity theory and inconsistent with the earlier Newtonian mechanics. These machines would simply not work if they were not engineered according to relativistic principles. In addition, a considerable number of modern experiments have been conducted to test special relativity. Some examples 1.) According to special relativity, the properties of particles moving approximately at thespeed of light significantly deviate from the predictions of Newtonian mechanics. For instance, the speed of lightcannot be reached by massive particles. Today, those relativistic expressions for particles close to the speed of light are routinely confirmed inundergraduate laboratories, and necessary in the design and theoretical evaluation of collision experiments inparticle accelerators. 2.) The Ives–Stilwell experiment tested the contribution of relativistic time dilation to the Doppler shift of light. The result was in agreement with the formula for the transverse Doppler effect, and was the first direct, quantitative confirmation of the time dilation factor. 3.) Time dilation of moving particles as predicted by special relativity can be measured in particle lifetime experiments. According to special relativity, the rate of clock C traveling between two synchronized laboratory clocks A and B is slowed with respect to the laboratory clock rates. 4.) The Kennedy–Thorndike experiment, first conducted in 1932, is a modified form of the Michelson–Morley experimental procedure, testing special relativity.The modification is to make one arm of the classical Michelson–Morley (MM) apparatus shorter than the other one. While the Michelson–Morley experiment showed that the speed of light is independent of the orientation of the apparatus, the Kennedy–Thorndike experiment showed that it is also independent of the velocity of the apparatus in different inertial frames 5.) Hughes–Drever experiments (also clock comparison-, clock anisotropy-, mass isotropy-, or energy isotropy experiments) are spectroscopic tests of the isotropy of mass and space.Unlike Michelson–Morley type experiments, Hughes–Drever experiments test the isotropy of the interactions of matter itself, that is, of .

State some successful predictions from Einsteins theory of special.pdf

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QUESTION 7 Forward and futures markets provide insurance or hedging a.pdf

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Problem 14-1A On January 1, 2017, Geffrey Corporation had the following stockholders\' equity accounts. Common Stock ($20 par value, 50,000 shares issued and outstanding) Paid-in Capital in Excess of Par-Common Stock Retained Earnings $1,000,000 190,000 550,000 During the year, the following transactions occurred. Feb. 1 Declared a $1 cash dividend per share to stockholders of record on February 15, payable March 1. Mar. 1 Paid the dividend declared in February. Apr. 1 Announced a 2-for-1 stock split. Prior to the split, the market price per share was $35 July 1 Declared a 10% stock dividend to stockholders of record on July 15 distributable uly 31 On uly 1, the market price of the stock was SA er share. 31 Dec. 1 31 Issued the shares for the stock dividend. Declared a $0.30 per share dividend to stockholders of record on December 15, payable January 5, 2018. Determined that net income for the year was $300,000 Solution a) Journal Entries Date Accounts and Titles Debit Credit Feb.1 Retained Earnings (50,000 X $1) $50,000.00 Dividends Payable $50,000.00 Mar.1 Dividends Payable $50,000.00 Cash $50,000.00 Apr.1 No Entry Two-for-one stock split increases number of shares to 100,000 = (50,000 X 2) and reduces par value to $10 per share Jul.1 Retained Earnings (100,000 x 10% x $14) $140,000.00 Common Stock Dividends Distributable (10,000 X $10) $100,000.00 Paid-in Capital in Excess of Par Value (10,000 X $4) $40,000.00 Jul.31 Common Stock Dividends Distributable $100,000.00 Common Stock $100,000.00 Dec. 1 Retained Earnings (110,000 X $.30) $33,000.00 Dividends Payable $33,000.00 Dec. 31 Income Summary $300,000.00 Retained Earnings $300,000.00 b) Geffrey Corporation Balance Sheet (Partial) December 31, 2017 Stockholders’ equity Paid- in capital Capital stock Common stock, $10 par value, 110,000 shares issued and outstanding $1,100,000.00 Additional paid-in capital In excess of par value ($190,000 + $40,000) $230,000.00 Total paid-in capital $1,330,000.00 Retained earnings ($300,000 + 33000 + 140000 + 50000) $523,000.00 Total stockholders’ equity $1,853,000.00.

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on the way to the cerebrum 15) Almost all somatic sensory signals pass through the A) cerebellum B) thalamus C) hypothalamus D) corpus callosum E) corticospinal tracts Solution A large number of pathways travel through the thalamus before reaching the cortex. All sensory signals except for olfaction also pass through the thalamus first. Why so? Because the thalamus is considered as the relay station. The thalamus is made up of a number of nuclei that possess functional specialization for dealing with particular type of information. This means that all the sensory information is first travelled to the thalamus which is then routed to a nucleus tailored to dealing with that type of sensory data. The information is then sent from the nucleus to the appropriate area in the cortex where it is further processed. In this way, the nuceli of the thalamus make sure to transfer the signal to the area which is specialized to handle that information for further processing..

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Please recreate a simple game of flappy birds using two classes and incorporate images and ez java The project must use EZ and incorporate at least 2 of the following concepts: Array Lists Finite State Machines File reading / parsing File writing 2D arrays Inheritance Private, public member variables and member functions Solution Java code for flappy bird are below: /* * Added frameRate(30) in because the program apparently started going 60fps when it was configured to run at 30 * */ frameRate(30); // variables for the width and space in the middle of pipes var pipeW = 40; var space = 100; var endW = pipeW + 10; var endH = 20; // space between pipes var pipeSpace = width / (4 - 1); // max and min height of pipes var pipeMaxY = 250; var pipeMinY = 100; // speed the pipe\'s move left var pipeSpeed = 4; // variables for bird\'s location and movement var birdX = 100; // x is constant var birdY = 200; var birdR = 10; var birdDY = 0; // var birdDX = 0; // PHYSICS var grav = 1.5; var jump = 0.15; var groundY = 350; var pipeStartLocation = 500; var pipes = [[pipeStartLocation, random(pipeMinY, pipeMaxY)], [pipeStartLocation + pipeSpace, random(pipeMinY, pipeMaxY)], [pipeStartLocation + pipeSpace*2, random(pipeMinY, pipeMaxY)], [pipeStartLocation + pipeSpace*3, random(pipeMinY, pipeMaxY)]]; noStroke(); var inGame = false; var inStart = true; var inGameOver = false; var score = 0; var gotFirstPoint = false; var drawBird = function() { pushMatrix(); translate(birdX, birdY); // angle the bird rotate(-40); rotate(max(-0, min(120, 20*birdDY))); // body stroke(0, 0, 0); strokeWeight(1); fill(255, 221, 0); ellipse(0, 0, birdR*2 - 2, birdR*1.6 - 2); // belly fill(255, 170, 0); arc(0, 0, birdR*2 - 2, birdR*1.6 - 2, 30, 150); // wing fill(254, 255, 178); pushMatrix(); rotate(sin(frameCount * 80) * 15); // flap rotation ellipse(-birdR*4/5, 0, birdR * 0.9, birdR * 0.6); popMatrix(); // eye fill(255, 255, 255); pushMatrix(); rotate(22); ellipse(birdR/2, -birdR/3, birdR, birdR * 0.8); popMatrix(); fill(0, 0, 0); ellipse(birdR*0.9, -birdR/20, birdR/5, birdR / 3); // lips fill(255, 0, 72); ellipse(birdR/2, birdR*3/5, birdR*0.8, birdR / 4); ellipse(birdR*0.6, birdR*2/5, birdR, birdR / 4); noStroke(); popMatrix(); }; var drawGround = function() { // dirt fill(214, 189, 145); rect(0, groundY, 400, 400 - groundY); // dark green grass fill(68, 143, 62); rect(0, groundY, 400, 5); stroke(82, 179, 85); // diagonal grass strokeWeight(3.0); for(var x = 0; x < 440; x += 10) { pushMatrix(); translate(x, groundY + 5); if(!inGameOver) { translate(-(pipeSpeed*frameCount % 10), 0); } rotate(225); line(0, 2, 0, 5); popMatrix(); } noStroke(); }; var drawPipe = function(x, y) { strokeWeight(4.0); // pipe body fill(65, 186, 79); stroke(0, 0, 0, 180); rect(x - pipeW / 2, 0, pipeW, y - space / 2); // top rect(x - pipeW / 2, y + space / 2, pipeW, 400 - y - space / 2 + 10); // bottom noStroke(); // pipe stripes colorMode(HSB); var bodyColW = 5; for(var c = x - pipeW / 2; c < x + pipeW / 2; .

Please recreate a simple game of flappy birds using two classes and .pdf

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My C proggram is having trouble in the switch in main. Also the a couple of funtion like delete and display tree are failing #include #include #include #include #include #include struct tree_node { int data; struct tree_node *left, *right; }; typedef struct nodeT { int key; struct nodeT *left, *right; int data; } nodeT, *treeT; //struct nodeT *t1, *t2; void DisplayTree(nodeT*t) { if (t != NULL) { DisplayTree(t->left); printf(\"%c \", t->key); DisplayTree(t->right); } } nodeT *FindNode(nodeT *t, int key){ if(t==NULL) return 0; if (key == t->key) return t; if(key < t->key){ return FindNode(t->left, key);; } else { return FindNode(t->right, key);; } } int max(nodeT *p){ nodeT *t, *tmp; while(tmp->right != NULL) tmp=tmp->right; return tmp->data; } int min(nodeT *t){ nodeT *tmp; while(tmp->left != NULL) tmp = tmp->left; return tmp->data; } void ModifyPreOrderWalk(nodeT *t, int h) { int i; if (t == NULL) return; for(i=0; i<-1;i++){ printf(\"| \"); } if(h>1)printf(\"+---\"); printf(\"%d\ “\", t->key); ModifyPreOrderWalk(t->left, h+1); ModifyPreOrderWalk(t->right, h+1); } void PrintTree(nodeT *t) { ModifyPreOrderWalk(t, 1); } void delete(nodeT **p){ nodeT *target, *imd_r, *plmd_r; target=*p; if (target->left==NULL && target->right==NULL) { *p=NULL; } else if (target->left == NULL) { *p=target->right; } else if (target->right == NULL) { *p=target->left; } else { plmd_r = target; imd_r = target->right; while( imd_r->left != NULL){ plmd_r = imd_r; imd_r = imd_r->left; } if(plmd_r == target) plmd_r->right = imd_r->right; else plmd_r->left = imd_r->right; imd_r->left = target->left; imd_r->right = target->right; *p = imd_r; } free(target); } void listInorder(nodeT *t){ if (t != NULL) { DisplayTree(t->left); printf(\"%d \", t->key); DisplayTree(t->right); } } void listPreorder(nodeT *t) { if (t != NULL) { printf(\"%d \", t->key); DisplayTree(t->left); DisplayTree(t->right); } } void listPostOrder(nodeT *t){ if (t != NULL) { DisplayTree(t->left); DisplayTree(t->right); printf(\"%d\", t->key); } } void InsertNode(nodeT **tptr, int key){ nodeT *t; nodeT *tmp; t=*tptr; if (t == NULL) { tmp= malloc(sizeof(nodeT)); tmp->key = key; tmp->left=tmp->right=NULL; *tptr=tmp; return; } if (key < t->key) { InsertNode (&t->left, key); } else { InsertNode(&t->right, key); } } int height(nodeT *t){ int lDepth = height(t->left); int rDepth = height(t->right); if(lDepth > rDepth) return(lDepth+1); else return(rDepth+1); } int sum(nodeT *p){ if (p == NULL) return 0; else return (p->data + sum(p->left) +sum(p->right) ); } int count(nodeT *t){ if(t == NULL) return 0; int i = 1+count(t->left)+count(t->right); return i; } /*nodeT *New(){ nodeT *tmp; tmp = (nodeT *)malloc(sizeof(nodeT)); if (tmp==NULL) return NULL; return tmp; }*/ int main(){ nodeT *node; nodeT *t=NULL, *findA; int f; int d, a, b, c, j, k, e, s, y; while (1) { printf(\"\ I - insert\ F - find node\ D - delete\ L - list in order\ R - list preorder\ P - list PorstOrder\ M - max\ N - min\ H - height\ C - count\ S - sum.

My C proggram is having trouble in the switch in main. Also the a co.pdf

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Need help on creating code using cart. The output has to show multiple lines of the users information. So far this is what I have: index.html: chapter8.jsp <%@page contentType=\"text/html\" pageEncoding=\"UTF-8\"%> JSP Page <%@ page import=\"java.io.*, business.*, data.*, list.*\" %> <% // get parameters from the request String firstName = request.getParameter(\"firstName\"); String lastName = request.getParameter(\"lastName\"); String email = request.getParameter(\"email\"); String phoneNumber = request.getParameter(\"phoneNumber\"); // get a relative file name // ServletContext sc = this.getServletContext(); // String path = sc.getRealPath(\"/WEB-INF/EmailList.txt\"); File tempdir = (File) this.getServletContext().getAttribute(\"javax.servlet.context.tempdir\"); String temppath = tempdir.toString(); String path = temppath + \"/EmailList.txt\"; // use regular Java objects User1 user = new User1(firstName, lastName, email, phoneNumber); UserIO.add(user, path); %> Thanks for joining our email list Here is the information that you entered:First NameLast NameEmailPhone Number<%= user.getFirstName() %><%= user.getLastName() %><%= user.getEmailAddress() %><%=user.getPhoneNumber() %> User1.java package business; public class User1 { private String firstName; private String lastName; private String email; private String phoneNumber; public User1() { firstName = \"\"; lastName = \"\"; email = \"\"; phoneNumber = \"\"; } public User1(String firstName, String lastName, String email, String phoneNumber) { this.firstName = firstName; this.lastName = lastName; this.email = email; this.phoneNumber = phoneNumber; } public void setFirstName(String firstName) { this.firstName = firstName; } public String getFirstName() { return firstName; } public void setLastName(String lastName) { this.lastName = lastName; } public String getLastName() { return lastName; } public void setEmailAddress(String email) { this.email = email; } public String getEmailAddress() { return email; } public void setPhoneNumber(String phoneNumber) { this.phoneNumber = phoneNumber; } public String getPhoneNumber() { return phoneNumber; } } UserIO.java package data; import java.io.IOException; import java.io.PrintWriter; import java.io.*; import business.User1; import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; public class UserIO extends HttpServlet { public static void add(User1 user, String filepath) throws IOException { File file = new File(filepath); PrintWriter out = new PrintWriter( new FileWriter(file, true)); out.println(user.getEmailAddress()+ \"|\" + user.getFirstName() + \"|\" + user.getLastName() + \"|\" + user.getPhoneNumber()); out.close(); } } *NEED HELP!* cart.java package list; import java.io.IOException; import java.io.PrintWriter; import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; .

Need help on creating code using cart. The output has to show multip.pdf

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Name key elements of Object-Oriented systems. Note : no plagiarism please Name key elements of Object-Oriented systems. Name key elements of Object-Oriented systems. Name key elements of Object-Oriented systems. Note : no plagiarism please Solution there are 3 basic elements of object oriented system.they are:- 1.inheritance 2.polymorphism 3.encapsulation Inheritance:- inheritance is a concept where a class or an object can get the properties of its parent or sibling class,for example let us consider A is a parent class and it has a method called read_me().and B is another class inherited from class A.so using inheritance B class also has access rigths or can explicitly use the read_me() method from A. polymorphism:- (poly-many) (morph-forms) polymorphism is an ability of an object to take on many forms.in general sense polymorphism means a parent class refering the child class object encapsulation:- encapsulation in oop is nothing but binding or grouping of methods and member function in a class.

Name key elements of Object-Oriented systems. Note .pdf

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Match the item on the left with the meaning on the right. Just show .pdf

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Linear algebra. Do both part i and part ii. I have found part i and wanted to check my work. Part ii im severly confused about. 3. Let a= | 3 (i) Compute the perpendicular projection of the vector b1on the line in the direction 2 of a. (ii) Write a formula for the distance of the point Xa to the point b. Show that this distance attains the minimum value when a_ p is the perpendicular projection found in (i). Solution Projection of b in the direction of a=(a dot b) =(1 3 -2) dot (0 1 2) =(1*0)+(3*1)+(-2*2) =0+3-4 =-1 a dot a=(1^2)+(3^2)+(-2^2) =1+9+4 =14 now [(a dot b)/a dot a]=-1/14 now {[(a dot b)/a dot a]}b=(-1/14)(0, 1, 2) =(0,-1/14,-2/14).

Linear algebra. Do both part i and part ii. I have found part i and .pdf

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Let A = {Leah, Dora, Mitch, Andy} which of the following are correct statements? D A Leah A A Mitch D A Mitch A Solution A={Leah, Dora, Mitch, Andy} Leah is belongs to A So correct statements is second statement ( Leah belongs to A ) and fourth statement ( D not belongs to A ) Remaining are in correct because D not in A ( first is wrong ), Mitch in A ( fifth is wrong) and A is a super set, so third is wrong..

Let A = {Leah, Dora, Mitch, Andy} which of the following are corr.pdf

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