Above is a trace depicting mechanical activity of frog heart. A stud.pdf

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Above is a trace depicting mechanical activity of frog heart. A student tapped the table when atria were contracting. Indicate what peaks 1 and 2 represent Tapping noise Solution Now any other stimulation to ventricle during diastole creates an extrasystole and it will be followed by a very long diastole which is called a compensatory pause.

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Why do viral particles, zymosan on fungi, endotoxin (LPS) from gram negative bacteria, bacterial DNA, and bacterial flagellin each induce the same proinflammatory response, and especially the production of IL-8, IL-12, IL-1, IL-6, and TNF-alpha? What is the link between this and c-reactive protein levels in the blood stream? Solution The immune system is comprised of two branches: innate and acquired immunity. The innate immune system is the first line of host defense against pathogens and is mediated by phagocytes including macrophages and dendritic cells (DCs). The innate immune response relies on recognition of evolutionarily conserved structures on pathogens, termed pathogen-associated molecular patterns (PAMPs), through a limited number of germ line-encoded pattern recognition receptors (PRRs), Upon PAMP recognition, PRRs present at the cell surface or intracellularly signal to the host the presence of infection and trigger proinflammatory and antimicrobial responses by activating a multitude of intracellular signaling pathways, including adaptor molecules, kinases, and transcription factors. Several classes of PRRs, including Toll-like receptors triggers functional maturation of dendritic cells and leads to initiation of antigen-specific adaptive immune responses. Macrophages activated at sites of tissue injury produce a number of cytokines into the bloodstream, most notable of which are the interleukins IL-1, IL-6 and IL-8, and TNF-, which induces hepatocytes to synthesize acute phase proteins (APP) including C- reactive protein. The level of CRP in blood plasma can rise as high as 1000-fold with inflammation. Thus the greatly raised level of c- reactive protein in the blood stream can be viewed as a strategy to overcome the poplulation of pathogens..

Why do viral particles, zymosan on fungi, endotoxin (LPS) from gram .pdf

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Which of the following used historical/comparative methods in their research? Select one: a. Durkheim, for Elementary Forms of Religion Life b. Marx, for Das Capital c. All of the above d. Weber, for Protestant Ethic. Solution d. Weber, for Protestant Ethic has used historical/comparative methods in their research. Comparative or historical research examines historical events in order to create explanations which are valid beyond a particular time and place..

Which of the following used historicalcomparative methods in their .pdf

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Which of following is not a class of the phylum platyhelminthes? Trematoda Turbellaria Nemertea Monogenea. Cestoda Which one of the following statements regarding the origin of metazoans has the most support? Ancient metazoans, similar to members of the phylum Placozoa, have been ruled out as ancestral metazoans. The choanoflagellates share features with the sponges, like having collars of surrounding a flagellum and being colonial. These shared features suggest a link to at least metazoan lineage. Metazoans are derived from many separate of unicellular organisms. Modern sponges have a genetic makeup that reflects their ancestral status. The traditional view of cnidarians relative to germ layers is that cnidaria possess two germ layers, ectoderm and endoderm. only one germ layer, the ectoderm, that differentiates into the two tissue layers. two germ layers, ectoderm and mesoderm. definitely all three germ layers including the mesoglea as it is derived from ectoderm. two germ layers, endoderm and mesoderm. Throughout most cnidarian groups, the basal or pedal disc serves to be the base of the tentacles. harbor sensory cells. generate buds. perform extracellular digestion. attach the cnidarian to the substrate. How are schistosomes unique, compared to other trematodes? They strike the rich and the poor, educated and ignorant, alike Infections are easy to cure by drugs surgery All species must cycle through humans to survive They have in their life do they leave the body of the bost larger separate male and female individuals, and the male is larger which of the following is a characteristic of stauromedusans? The planula swims initially and then becomes sessile The top of the polyp has nine extensions Polyps reproduce asexual They have a solitary polyp body They do not have a mouth If a sponge is fragmented and cells are dissociated from one another, the cells will reorganize their structure and function, and clumps of isolated cells will form a new. reproduce sexually. form spicules in the pattern of the cloth. die from being separated. Solution 14] nemertea is not a class of platyhelmenthes Answer = c) Nemertea Nemertea is a phylum not a class is phylum of invertebrate animals also known as ribbon worm or proboscis worm explaination = Phylum Platyhelhelminthes divided into four 4 classes 1] Turbellaria = include mainly free living marine species 2] Monogenea = these are ectoparasites mostly of fish 3] Trematoda = Are internal parasites of mollusca & humans 4] Cestoda = Are internal parasites of vertebrates.

Which of following is not a class of the phylum platyhelminthes Tre.pdf

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What is wrong with this code? Please fix. code: #include #include #define ROWS 3 #define COLS 3 void display(double[ROWS][COLS]); double main(){ double mat[ROWS][COLS] = { {1./2, 1, 0}, {0, 2./3, 0}, {-1./2, -1, 2./3} }; display(mat); return 0; } void display(double nums[ROWS][COLS]) { double rowNum, colNum; for (rowNum = 0.; rowNum < ROWS; rowNum++) { for (colNum = 0.; colNum < COLS; colNum++) printf(\"%1.1s\", nums[rowNum][colNum]); printf(\"\ \"); } } Solution #include #include #define ROWS 3 #define COLS 3 void display(double[ROWS][COLS]); int main(){ double mat[ROWS][COLS] = { {1./2, 1, 0}, {0, 2./3, 0}, {-1./2, -1, 2./3} }; display(mat); return 0; } void display(double nums[][COLS]) { //problem with declaring array subscripts with double, so changed to int int rowNum, colNum; for (rowNum = 0.; rowNum < ROWS; rowNum++) { for (colNum = 0.; colNum < COLS; colNum++) printf(\"%1.1f\\t\", nums[rowNum][colNum]); printf(\"\ \"); } }.

What is wrong with this code Please fix.code#include stdio.h.pdf

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What are two ways that meiosis could produce gametes that contain genetic material that is different from the parent cell? What are two ways that meiosis could produce gametes that contain genetic material that is different from the parent cell? What are two ways that meiosis could produce gametes that contain genetic material that is different from the parent cell? Solution Answer: The two ways that meiosis could produce gametes which differ genetically from its parent cells are as under: (1) During meiosis, homologous chromosomes undergo independent assortment. This means that the chromosomes associate in no way with one another. Thus, the gametes so formed may have any combination of paternal and maternal chromosomes. (2) Crossing over results in the exchange of genetic material between homologous chromosomes, resulting in recombinant chromosomes. It occurs during the pachytene stage of meiosis I. This process gives rise to new recombinant varieties and thus the gamete so produced differ genetically from the parentals..

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what is the process in society that made this change to advertising objective throughout history? what objective in the list of advertising objectives bears a qualitative change in it or represents a new level of diff Solution Over the years, the advertising objectives has changed with change in societal structure and its morality. Advertising is a process to attract customer towards a product or service and it takes the route to arouse various emotion of the target audience. As the societal value has changed over the years, so has changed the pattern of advertisement to lure the consumer. In earlier days, when advertisindg came into picture, the society used to be closely knitted and had joint families. Hence, it was possible to capture a large audience at a go and that too with a well directed common messaage that includes the societal value of being together. As time progressed, families broken apart and nuclear families arose in the scene-so advertisers started targeting individual people\'s choices and came up with 2-3 versions of a single advertisement that can attract different customers having a same choice but with different mentality. Moreover, the division of society into high income group, middle income group and lower income group made the advertiser think in different ways.The concept of niche-advertising crept in- advertisers started targeting people based on their income. Then came the era of internet that changed the advertising objective all together. Initially the case was to attract the customer towards a product now marketers try to present their product at a place and time when customers need it, Today consumer is busy, having various options to compare - so reaching to the customer at a right time and at a right place is very necessary-so came the practice of using direct mail communication, tracking customer search history to know his/her preferences and present him/her with the exact product they want, came the concept of testimony from satisfied customer to gain confidence of other customers. Qualitative changes over the years that represents a new level of difference: - Niche advertising- Targeting preferences of individual class - Personal Targeting through internet- helped to save conumer efforts and made the experience rich..

what is the process in society that made this change to advertising .pdf

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What is the best way to go about designing an Android App? Solution The best way to design an Android App is the Eclipse ADT. The Android Development Tools (ADT) with Eclipse IDE gives a powerful, intergrated environment in which the Android Application is built. ADT enlarge the capabilities of Eclipse to rapidly set up a new Android project, create an application UI, add packages based on the Android Framework API, debug the applications with the Android SDK tools, and even send signed or unsigned .apk files in order to share the application..

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Use the following word bank to complete the numbers 51-100. Acoelomate Arthropoda Alternation alveolus Book lungs cloaca cephalopoda chitin ditellim emulsification Conjugation companion filter feeding gametes lacteals Custaceans endotherm hemaphrodite closed circulation inferior vena cava Liver monotremes Operculum Pancreas Parazoa Protonephridium Plasmodium Phototrophs protein sino atrial node sporozoan pulmonary circuit Spinnerets shell spiral systole opposable thumb Stomata Kelps tracheids tracheal tube radial It.ventrides Radially trypanosomes monotremes dosed mycelium vessel elements rhizoids sieve tube elements This structure protects fish gills. This is the site in mammals where gases are exchanged with the environment This is the site of gas exchange in insects This vein collects blood from the lower body and carries it to the heart The common opening of the digestive system, the excretory system and the reproductive system is known as the Bile is produced in the Phloem, responsible for transporting nutrients in plants is composed of two types of cells and cells. These organs in spiders produce silk _In this kind of circulatory system blood is carried entirely in vessels: This organ is both and endocrine and exocrine gland This group includes platypus and spiny anteaters. Solution 51 - This structure protects fish gills OPERCULUM 52 - This is the site in mammals where gases are exchange with the environment LUNGS 53 - This is the site of gas exchange in insects TRACHEAL TUBE 54 - This vein collect blood from the lower body and carries it to the heart INFERIOR VENA CAVA 55 - The common opening of digestive system , excretory system and reproductive system is known as CLOACA 56 - Bile is produced in the LIVER 57 - Phloem , responsible for transporting nutrients in plants is composed of two types of cells SIEVE TUBE ELEMENTS AND COMPANION CELLS 59 - These organs in spiders produce silk SPINNERETS 60 - In these kind of circulatory system blood is carried entirely in vessels CLOSED CIRCULATORY SYSTEM 61 - This organ is both an endocrine and exocrine gland PANCREAS 62 - This group includes platypus and spiny anteaters MONOTREMES.

Use the following word bank to complete the numbers 51-100. Acoeloma.pdf

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Todo package hwk6; // This class contains the configuration of a type of apartment public class Apartment { int numOfUnits; // the number of apartments of this type Room[] rooms; // rooms in this type of apartment Apartment(int numOfUnits, Room[] rooms) { this.numOfUnits = numOfUnits; this.rooms = rooms; } // return an array of window orders for one unit of this type of apartment WindowOrder[] orderForOneUnit() { // TODO } // return an array of window orders for all units of this type of apartment WindowOrder[] totalOrder() { // TODO } // return text like: // // 15 apartments with (Living room: 5 (6 X 8 window)) (Master bedroom: 3 (4 X 6 window)) (Guest room: 2 (5 X 6 window)) public String toString() { // TODO } } class OneBedroom extends Apartment { OneBedroom(int numOfUnits) { super(numOfUnits, new Room[] { new LivingRoom(), new MasterBedroom() }); } } class TwoBedroom extends Apartment { TwoBedroom(int numOfUnits) { super(numOfUnits, new Room[] { new LivingRoom(), new MasterBedroom(), new GuestRoom() }); } } class ThreeBedroom extends Apartment { ThreeBedroom(int numOfUnits) { super(numOfUnits, new Room[] { new LivingRoom(), new MasterBedroom(), new GuestRoom(), new GuestRoom() }); } // return an array of window orders for all units of this type of apartment // // Notice we have two guest rooms and they have the same size of windows. // override the inherited method to merge the order for the two guest rooms since their windows have the same size @Override WindowOrder[] orderForOneUnit() { // TODO } } package hwk6; public class Building { Apartment[] apartments; public Building(Apartment[] apartments) { this.apartments= apartments; } // Return an array of window orders for all apartments in the building // Ensure that the orders for windows of the same sizes are merged. WindowOrder[] order() { // TODO } // return a string to represent all types of apartments in the building such as: // 20 apartments with (Living room: 5 (6 X 8 window))(Master bedroom: 3 (4 X 6 window)) // 15 apartments with (Living room: 5 (6 X 8 window))(Master bedroom: 3 (4 X 6 window))(Guest room: 2 (5 X 6 window)) // 10 apartments with (Living room: 5 (6 X 8 window))(Master bedroom: 3 (4 X 6 window))(Guest room: 2 (5 X 6 window))(Guest room: 2 (5 X 6 window)) // public String toString() { // TODO } } package hwk6; public class Hwk6 { public static void main(String[] args) { Apartment[] apartments = { new OneBedroom(20), new TwoBedroom(15), new ThreeBedroom(10) }; Building building = new Building(apartments); WindowOrder[] orders = building.order(); System.out.println(building); System.out.println(\"Window orders are: \"); for(WindowOrder order: orders) { System.out.println(order); } } } package hwk6; public class Room { Window window; int numOfWindows; Room(Window window, int numOfWindows) { this.window = window; this.numOfWindows = numOfWindows; } WindowOrder order() { return new WindowOrder(window, numOfWindows); } // Print text like: 5 (6 X 8 window) @Override publ.

Todopackage hwk6; This class contains the configuration of a t.pdf

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This question refers to the ABO blood type locus. Remember that the A and B alleles are codominant to each other and both are dominant to the O allele. Assume no epistatic interactions with the H locus. A man has blood type A. Mark all genotypes below that are possible for this man. Select one or more: AA AB AO BB BO OO Solution Hence both A and B are codominant and dominant over O, the possible genotype of man having blood group A are AA and AO. Explanation: As both A and B are codominant if both are present iegenotype AB results in AB blood group BB and BO results in B blood group (as there is no A in genotype and B is dominant over O) OO results in O bllod group(Absence of both A and B in genotype).

This question refers to the ABO blood type locus. Remember that the A.pdf

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This is testing Algorithm Writing For this assessment we will be combining while and if control structures together. (A):Write a pseudocode (pen and paper) algorithm that solves the ODE y\'= f(t, y) using Euler’s method. (B):Convert this algorithm into MATLAB syntax, as the function euler method . B1:Your inputs should be the start time t0, the end time tend, the function f, an initial condition y0 and an initial step size h. B2:If h × f(t, y) at any point exceeds 1, your function should halve the step size h. B3:Equally, if h × f(t, y) at any point drops below 0.01, your function should double the step size h. B4:You should also make sure that the final solution step doesn’t ‘overshoot’ tend. (i.e. change h during the final step to exactly reach tend). B5:Your MATLAB code should use the function header below, and be wellcommented with a sensible layout. (C)Include your pseudocode algorithm, as well as the MATLAB code in your Portfolio submission. Also include a discussion of when the code could break, or give incorrect outputs (you DO NOT need to design your code to avoid these). Solution Answer:) A) The derivative term in the first order ivp: y\' = f(t, y), y(t0) = y0 is approtimated by making use of Taylor series approtimation of the dependent variable y(t) at the point ti+1. That is y(ti+1) = y(ti+ Dt) = y(ti) + Dty\'(ti) + (Dt2 / 2)y\'\'(ti) + . . . = y(ti) + Dtf(ti, yi) + (Dt2 / 2)y\'\'(ti) + . . . (... y\'(ti) = f(ti, yi)) if the infinite series is truncated from the term Dt2 onwards, then y(ti+1) = y(ti) + Dt y\'(ti) (or) yi+1 = yi + Dt fi for all i That is, for i = 0, y1 = y0 + Dt f0 i = 1, y2 = y1 + Dt f1 ! i = n-1, yn = yn-1 + Dt fn-1 Since y0 and hence f0 are known (from initial condition) in the equation corresponding to i = 0, all the terms on the r.h.s are known. So y1 that is, y at t1 is calculated easily from this equation. Similarly once y1 is known, r.h.s of the equation corresponding to i = 1 is also known so y2 can be computed. As we proceed in the same way until i = n-1, yn can be obtained. This is an etplicit method because in any equation there is only one unknown which can be separated to the left side of the equation. Local truncation error : The error in the approtimation, that is the difference between the exact solution at ti+1 and the numerical solution yi+1 is called the local truncation error (assumed that yi+1 is calculated with exact arithmetic with out any round off error). Ti+1 = y(ti+1) - yi+1 = y(ti+1) - yi - Dtfi = h2/2 y\'(t) (by Taylor series & remainder theorem) where ti < t < ti+1. Hence the order of the local truncation error for Euler scheme is O(Dt2) as Dt -> 0.

This is testing Algorithm Writing For this assessment we will be c.pdf

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The rest of the headings are explailalUly sell 5) ABOVE THE CHART AT THE TOP OF THE PAGE... write the Characteristics that the KINGDOM PLANTAE has (All Plants have that the Green Algae/Charophytes do NOT) Ex: Antheridia. Solution Charophytes are considered ancestors to land plants as they share many common characteristics but some of the differences originated when plants started migrating from water to land or adapting to land environment. Differences are as follows, 1) Plants produce embryos and protect them in enclosed spaces 2) Plants have true roots, stems, leaves and adaptation to land has acquired waxy cuticles, stomata, and vasculature.

The rest of the headings are explailalUly sell 5) ABOVE THE CHART AT .pdf

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The hepatic portal system, pulmonary circulation and fetal circulation all vary from the “normal” pattern of blood flow. Be able to describe these special cases. Solution 1. HEPATIC PORTAL SYSTEM 2. PULMONARY CIRCULATION In Pulmonary circulation, deoxygenated blood from the right ventricle is carried to the air sacs of the lungs and returns oxygenated blood from the lungs to the left atrium. In hepatic portal circulation, blood is carried from capillaries of the gastrointestinal organs and spleen to sinusoids of the liver. The liver stores, modifies, or detoxifies some of the substances that have been absorbed by the gastrointestinal tract. The hepatic portal system includes the veins that drain blood from the pancreas, spleen, stomach, intestines, and gallbladder and transport it to the hepatic portal vein. The pulmonary and systemic circulations are different in several ways: a. Blood in the pulmonary circulation need not be pumped as far as blood in the systemic circulation b. Pulmonary arteries have larger diameters, thinner walls, and less elastic tissue; consequently, resistance to blood flow is very low and therefore less pressure is needed to move blood through the lungs. c. As a result, normal pulmonary capillary hydrostatic pressure is lower than average systemic capillary hydrostatic pressure. 3. FETAL CIRCULATION Featal circulation exists only in the fetus and consists of structures that permit exchange of substances between the developing fetus and its mother. i. The fetal circulation permits the fetus to obtain oxygen and nutrients from the maternal blood and eliminate its carbon dioxide and wastes into the maternal blood. ii. The exchange of substances between the fetal blood and maternal blood occurs through the placenta, which attaches to the navel (umbilicus) of the fetus by the umbilical cord. iii. Deoxygenated blood flows from the fetus to the placenta via two umbilical arteries. iv. Oxygenated blood returns from the placenta to the fetus via one umbilical vein which delivers its blood primarily to the ductus venosus(fetal blood vessel that joins aorta and pulmonary artery) that drains into the inferior vena cava. .

The hepatic portal system, pulmonary circulation and fetal circulati.pdf

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The expriment is (solution chemical and physical properties of ethanol in water) I need ur help on this experiment, provides me the Literture review of this topic, like the form what i attached on the the paper I would be thankful for that Thank u Literature Review Citation Authors Historically, the introduction of synthetic ion exchange resin Abrams I. M et is developed by two treatment expert and polymer chemist is al, 1997 and Dale B. A. Adams and E. L. Holmes on 1935 which is do the J. A. et al, 1992. reaction of formaldehyde with phenolic or amine derivatives to create phenol-formaldehyde structure by condensation of resin. From the condensate resin era in early 1940, with continue with addition polymer in ate 1950, and transforming into new adsorbents for varies of application. A class of polymer beads such a macroporous polymer beads Wheaton R. M et was contained a fixed well- developed porous structure in al, 2000 withered state, that having a wide range of applications such as water softening, dealkalization, purification, and many more. The benefit of the porous system is a sensitive function of the diameter of internal pores, their division, and morphology Sulfonation reaction can operate at monomer before Meyer M. et al. polymerization or during post-polymerization. It will affect 2013 the degree of sulfonation and the final ion exchange capacities of the sulfonated polymer. Most commonly during Solution Answer : Ethanol Ethanol, C2H5OH, (also called Ethyl Alcohol) is the second member of the aliphatic alcohol series. It is a clear colourless liquid with a pleasant smell. Except for alcoholic beverages, nearly all the ethanol used industrially is a mixture of 95% ethanol and 5% water, which is known simply as 95% alcohol. Although pure ethyl alcohol (known as absolute alcohol) is available, it is much more expensive and is used only when definitely required. Preparation Ethanol is prepared as 95% alcohol (i.e. a 95% solution of ethanol in water) by distillation of the solution which results from the fermentation of sugars. Manufacture There are two major industrial pathways to ethanol. Ethanol which is intended for industrial use is made by the first method, while ethanol intended for food use tends to be made by the second method. Reaction of Ethene with Steam Most of the ethanol used in industry is made, not by alcoholic fermentation, but by an addition reaction between ethene and steam. Alcoholic Fermentation The fermentation process takes three days and is carried out at a temperature between 250C and 300C. The ethanol is then obtained by fractional distillation..

The expriment is (solution chemical and physical properties of ethan.pdf

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The ABO and MN blood groups are given below for four sets of parents (1 to 4) and four children (a to d). Recall that the ABO blood group has three alleles, I^A, I^B, and i. The MN blood group has two codominant alleles, M and N. When all the genetic data is considered together and by the process of elimination you should be able to match each of the four children with the correct set of parents. Solution Beginning from the first set of parents, let us try to determine the genotype of all sets. Parents 1: The father can be IBIB or IBi and the mother can be IAIAor IAi. For the MN group, The father is NN and the mother is MN. So, their children can of A, B, AB or O blood type and of either MN or N blood group. Parents 2: The father can be IAIA or IAi and the mother can be IAIAor IAi. For the MN group, The father is NN and the mother is also NN. So, their children can of A or O blood type and of N blood group. Parents 3: The father can be IAIAor IAi and the mother is IAIB. For the MN group, both the parents are of MN type. So, their children can of A, B, or AB blood type and of either M, N or MN blood group. Parents 4: The father is ii and the mother is IAIAor IAi. For the MN group, both the parents are of MM type. So, their children can of A or O blood type and of M blood group. Amongst the given set of children only one has the N type blood group. So this child (a) definitely belongs to the parents who can have children with only N blood group i.e. Parents 2. For the remaining 3 cases (b, c and d), we want parents with possibility of giving children of M type blood for children (c) and (d). Parents 1 cannot give an offspring with M type blood group, so it has to be matched with child (b) where it can easily give AB and MN blood types. For child (d), Only parents 4 are a match because that is the only set of parents from the remaining parents which can have a child with O blood group. That leaves parents 3 to give birth to child (c) with A blood group and M blood type and it is possible. So, 1- b, 2- a, 3-c, 4-d.

The ABO and MN blood groups are given below for four sets of parents .pdf

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Table 19.1 Table 20.3 Construct an evolutionary history using the information in Table 19.1 in Chapter 19 and Table 20.3 Use names. Include only the 4 traits listed in these tables. You will need to lumpsome organisms of the groups of and some traits. The is a colonial ancestral protist derived is trait distinguishing from the animals, the the new) t at the end of the branches coming off the main slanted line. Recall that the progression of evolution (and time) is from bottom to top (outgroup) (Parazoa) groups here) (5 groups here) (2 animals multicellularity Solution Embryonic Development. Radial (Adults) Bilatetal (Larvae)PhylumOrganizationSymmetryGerm layers Embryonic Development.PoriferaCellAssymetrical2N/ACnidariaTissueRadial2N/APlatyhelminthsOrganBil ateral3ProtostomesNematodaOrganBilateral3protostomeMolluscaOrganBilateral (some are assymetrical)3protostomesAnnelidaOrganBilateral3ProtostomesArthropodaOrganBilateral3Proto stomesEchinodermataOrgan Radial (Adults) Bilatetal (Larvae)3DeuterostomesChordataOrganBilateral3Deuterostomes.

Table 19.1Table 20.3 Construct an evolutionary history.pdf

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Suppose memory for the address space (process image) is allocated contiguously using dynamic memory allocation. What extra hardware is needed to generate a physical address and why? Solution Memory management is the functionality of an operating system which handles or manages primary memory and moves processes back and forth between main memory and disk during execution. Its function i to track whenever some memory gets freed or unallocated .It then updates the status.There are 2 ways dynamic and static. dynamic routines of the library are stored on a disk in relocatable form and are loaded into memory only when they are needed by the program.A reference to the dynamic module is not used to link the library with the program when dynamic linking is used. Dynamic Link Libraries (DLL) in Windows and Shared Objects in Unix are examples of dynamic libraries.memory pool allocation is used. virtual memory may also be used.The Translation Lookaside Buffer (TLB ) is the main component of the processor responsible for virtual-memory..

Suppose memory for the address space (process image) is allocated co.pdf

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Select three methods in the ObjectList class to work through algorithmically. Describe any special cases or boundary conditions that might exist for each of the three methods selected. Be sure to draw pictures as you work through the methods. Remember that you cannot work on linked lists without drawing pictures. The homework will not be accepted without submitting the pictures! Write a Java method with the signature: public objectList intersect (objectList list1, objectList list2) that accepts two unordered linear linked lists and returns a third linear linked list whose nodes contains the intersection of the two original linear linked lists. Note that the intersection of two lists is a new list containing the elements that the two lists have in common, without modifying the two original lists. Describe any boundary conditions that might exist. Solution Node.java //Defines each node of a linked list public class Node{ int data; Node next; Node(int d) { data = d; next = null; } } ObjectList.java public class ObjectList{ public Node head; /* Utility function to print list */ public void printList() { Node temp = head; while(temp != null) { System.out.print(temp.data+\" \"); temp = temp.next; } System.out.println(); } /* Inserts a node at start of linked list */ public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* A utilty function that returns true if data is present in linked list else return false */ public boolean isPresent (int data){ Node t = this.head; while (t != null) { if (t.data == data) return true; t = t.next; } return false; } } Intersection.java //Implements the intersection method. class Intersection{ static public ObjectList intersect(ObjectList list1, ObjectList list2){ ObjectList result = new ObjectList(); //Node t1 = head1; Node t1 = list1.head; Node t2 = list2.head; // Traverse list1 and search each element of it in list2. // If the element is present in list 2, then insert the // element to result while (t1 != null){ if (list2.isPresent(t1.data)) result.push(t1.data); t1 = t1.next; } return result; } /* Driver program to test above functions */ public static void main(String args[]){ ObjectList llist1 = new ObjectList(); ObjectList llist2 = new ObjectList(); ObjectList unin = new ObjectList(); ObjectList intersecn = new ObjectList(); /*create a linked lits 10->15->5->20 */ llist1.push(20); llist1.push(4); llist1.push(15); llist1.push(10); /*create a linked lits 8->4->2->10 */ llist2.push(10); llist2.push(2); llist2.push(4); llist2.push(8); intersecn = intersect(llist1, llist2); intersecn.printList(); } }.

Select three methods in the ObjectList class to work through algori.pdf

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