True or falsea. List index values must be integers.b. Lists can.pdf

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True or false? a. List index values must be integers. b. Lists can change their size, getting larger or smaller. c. A function cannot return a list. d. All elements of a list are of the same type. e. Lists cannot contain strings as elements. f. A function cannot change the length of a list argument. Solution Below are True statements a. List index values must be integers. b. Lists can change their size, getting larger or smaller. f. A function cannot change the length of a list argument. Below are False statements A function cannot return a list. All elements of a list are of the same type. e. Lists cannot contain strings as elements..

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State true or false with reason (all from Complex Analysis):- (a) Power series has an antiderivative everywhere inside its disc of convergence (c) There does not exist any analytic function f defined on C\\ {0} such that f \' (z) = 1/z (d) If a mobius transformation T carries z1 and z2 into a same number w1 then either z1 = z2 or else T is a constant map Solution (a) Power series has an antiderivative everywhere inside its disc of convergence. (TRUE).

State true or false with reason (all from Complex Analysis)-(a) P.pdf

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send edited code Main 2 public class Main2 { public static void main(String[] args) { EZ.initialize(512, 512); EZ.addImage(\"metropolis.jpg\", 256, 256); Hero myHero = new Hero(100,390); //HeroPlus mySecondHero = new HeroPlus(100,390); while(true){ myHero.processStates(); //mySecondHero.processStates(); EZ.refreshScreen(); } } } Hero Plus public class HeroPlus { EZImage heroStand; EZImage heroJump; EZImage heroLand; EZImage heroHover; EZImage heroPunch; // PUNCH int posx = 0; int posy = 0; int heroState = STAND; static final int STAND = 1; static final int JUMP = 2; static final int LAND = 3; static final int HOVER = 4; static final int PUNCH = 5; static final int JUMPHEIGHT = 100; static final int PUNCHDISTANCE = 50; // PUNCH int punchCounter = 0; // PUNCH int jumpCounter = 0; HeroPlus(int x, int y) { posx = x; posy = y; heroStand = EZ.addImage(\"superstand.png\", posx, posy); heroLand = EZ.addImage(\"superland.png\", posx, posy); heroJump = EZ.addImage(\"superjump.png\", posx, posy); heroHover = EZ.addImage(\"superhover.png\", posx, posy); heroPunch = EZ.addImage(\"superpunch.png\", posx, posy); // PUNCH hideHero(); heroStand.show(); } void hideHero() { heroStand.hide(); heroLand.hide(); heroJump.hide(); heroHover.hide(); heroPunch.hide(); // PUNCH } void positionHero(int x, int y) { heroStand.translateTo(x, y); heroJump.translateTo(x, y); heroLand.translateTo(x, y); heroHover.translateTo(x, y); heroPunch.translateTo(x, y); // PUNCH } void processStates() { switch (heroState) { case STAND: if (EZInteraction.wasKeyPressed(\'j\')) { heroState = JUMP; jumpCounter = 0; hideHero(); heroJump.show(); } break; case JUMP: jumpCounter++; if (EZInteraction.wasKeyPressed(\'p\')) { punchCounter = 0; hideHero(); heroPunch.show(); heroState = PUNCH; } if (jumpCounter > JUMPHEIGHT) { heroState = LAND; hideHero(); heroLand.show(); } else { posy -= 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed(\'h\')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case PUNCH: // PUNCH punchCounter++; posx += 2; positionHero(posx, posy); if (punchCounter > PUNCHDISTANCE) { heroState = LAND; hideHero(); heroLand.show(); } break; case LAND: jumpCounter--; if (jumpCounter <= 0) { heroState = STAND; hideHero(); heroStand.show(); } else { posy += 2; positionHero(posx, posy); if (EZInteraction.wasKeyPressed(\'h\')) { heroState = HOVER; hideHero(); heroHover.show(); } } break; case HOVER: if (EZInteraction.wasKeyPressed(\'l\')) { heroState = LAND; hideHero(); heroLand.show(); } break; } } } Solution public class Main2 { public static void main(String[] args) { EZ.initialize(512, 512); EZ.addImage(\"metropolis.jpg\", 256, 256); // Hero myHero = new Hero(100,390); HeroPlus mySecondHero = new HeroPlus(100,390); while(true){ //myHero.processStates(); mySecondHero.processStates(); EZ.refreshScreen(); } } } class HeroPlus { EZImage heroStand; EZImage heroJump; EZImage heroLand; EZImage heroHover; EZImage heroPunch; // PUNCH int posx = 0; int posy = 0; int her.

send edited codeMain 2public class Main2 { public static voi.pdf

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Question 16 What is a common tactic hackers use to allow themselves re-access to a system? Keystroke capture Trap door Backdoor Rootkit 2.5 points Question 17 What types of viruses propagate through the Internet with no user intervention? Virus hoaxes Stealth viruses Worms Trojan Horses 2.5 points Question 18 A Web merchant wants to ensure that people cannot deny that they entered into a transaction, so they starting using digital signatures and encryption as part of their transaction processing. This is an example of what? Integrity Confidentiality Authenticity Non-repudiation 2.5 points Question 19 What is the main solution used to prevent unauthorized access to systems, data, and applications? Passwords User profiles Biometrics Firewalls 2.5 points Question 20 Which of the following refers to security solutions that protect access to systems, applications, or data by ensuring users or systems are authenticated and allowed to have such access? Physical access controls User Profiles Logical access controls System access controlsA. Keystroke captureB. Trap doorC. BackdoorD. Rootkit Solution 16. (D) Rootkit 17. (C) Worms 18. (D) Non-repudiation 19. (D) Firewalls 20. (D) System access controls.

Question 16What is a common tactic hackers use to allow themselves.pdf

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Qualified plan documents can be written to accommodate plan forfeitures in different ways. Which of the following statements is correct? A. Stock bonus plans always allocate plan forfeitures to participants’ accounts since it is a stock plan. B. Safe harbor 401(k) plans never allocate plan forfeitures of matching contributions to participants’ accounts. C. Defined benefit plans (but not cash balance plans) have two options to deal with forfeitures. D. Employers can use the forfeited funds to pay other expenses of the business outside the qualified plan. Solution.

Qualified plan documents can be written to accommodate plan forfeitu.pdf

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Proteins fold. How? Why? Solution ANS: Protein folding: Proteins are the molecules in the body which carries out many important functions, such as enzymes and antibodies. Before proteins carry out their function, they must first assemble themselves or fold. PROTEIN FOLDING MODELS: 1. Frame work model: In this the protein local elements which are stretches of amino acids liked by one another by covalently with peptide bonds. The region of the primary amino acids sequence in peptide bonds make different domains.in this domains some of regions are made with Alfa - helix and some are beta sheet domains. These primary elements form secondary structure independently of tertiary structure (kim and baldwin).these elements can diffuse until they collided, successfully adhering and coalescing to give the tertiary structure (diffus-collision model), (karplus and weaven). 2. The classic nucleation model: It postulates that some neighboring residues in the sequence would form native secondary structure that would act as a nucleus form which the native structure would propagate in a step wise manner. Thus the tertiary structure would form as necessary consequence of the secondary structure (wetlaufer). 3. Hydrophobic collapse model: It is hypothesized that a protein would collapse rapidly around its hydrophobic side chain and then arrange from restricted conformational space occupied by the intermediate. Here the secondary structure would be directed by native like tertiary structure (ptitsyn & kwajima). Leviathal’s paradox: It is formulated by Cyrus Leviathal. It explains, there are vast too many different possible conformations for a protein to fold by a random search. Structural organization of protein: 1. Primary Structure: the sequence of amino acids along with the di sulphide bonds. 2. SECONDARY STRUCTURE: Alfa Helices and Beta- Pleated sheets are most frequently occur in secondary structure in polypeptides. Both of these can case local folding of protein. The Alfa Helices in polypeptide made up of L- Alfa amino acids are right handed, the Alfa - Helices is stabilized by hydrogen bonds formed between oxygen and amide hydrogen’s of peptide bonds that are a part of linear sequence. Beta- Pleated sheets: 2 types A. Parallel: in this the adjacent hydrogen bonding segments are connected by cross over connection [Right handed]. B. Anti-Parallel: In this the adjacent hydrogen bonding segments are connected by HAIRPIN turns or reverse turn [Left Handed]. 3. Tertiary structure: It is arisen because of the folding of poly peptides upon itself. 4. Quaternary structure: Some proteins are made up of 2 or more polypeptides chains, the interaction between different poly peptide are subunits gives rise to the quaternary structure..

Proteins fold. How WhySolutionANSProtein folding Protein.pdf

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PLN 360/560 Environmental Impact Assessment Homework No. 4 1. An applicant has applied to the U.S. Army Corps of Engineers (USACE) for a Clean Water Act (CWA) Section 404 Permit to construct a marina at the mouth of Chaotic Creek on Bagel Bay on Lake Promise, a Water of the United States and Class \"B\" surface water in New York State. This site is located in the upper reach of a large commercial harbor used for deep-draft navigation. The project entails (1) construction of a breakwall; (2) dredging (and future maintenance dredging) of navigation entrance and creek channels; (3) installation of 50 docks with boat slips; and (4) construction of a boat launching ramp. During construction, it is estimated that a total of 60,000 cubic yards of material will be dredged and need to be managed, and about 2,500 cubic yards of fill will need to be placed in Lake Promise. Management alternatives for the dredged material include placement at an upland permitted municipal landfill, or discharge at a designated open- water site in the harbor. The site is an abandoned \"hole\" and since discharge of the dredged sediment is conceived to have benefits through reducing DDE bioaccumulation through capping and restoration of shallow water aquatic habitat, it is referred to as a habitat restoration site (HRS). The applicant had a consultant sample the sediments to be dredged for the marina as well as a lake reference area and HRS, and had the samples analyzed by a laboratory. The following is a summary of the test results that are to be submitted with the application: Contaminant Concentration in sediment sample of Concentration in Concentration in reference bulk HRS bulk marina sediment to be sediment sample sediment sample dredged Bulk Standard sediment elutriate (mg/kg)(mg/L) (mg/kg) (mg/kg) 13 0.02 Barium Cyanide Iron Mercury Dieldrin p.p\'- DDE Chlorobenzene 16 65 0.70 13,000 120 14 53 0.40 12,500 130 0.60 11,000 100 0.02 3x10-7 8x10 1.3x10 0.32 0.008 0.0034 0.12 1.3 38,000 Total organic 21,000 10,300 carbon (TOC) a. What type of CWA regulatory approvals (i.e., permits, etc.) would be required for this project? (2 pts.) Solution Approvals required for this project - Storm Water permit - Clean Water Act (CWA) permit coverage for discharge of storm water runoff from the construction site which are issued throught the US Environmental protection agency\'s NPDES (national pollutant discharge elimination system) or the state NPDES authority Permit under section 404 of the CWA issued by US Army Crops for Engineers Resource Conservation and Recovery Act (RCRA) permit for generators of hazardous wastes (The RCRA regulations list quantities of hazardous wastes that determine whether a generator is large or small.), Storage of hazardous wastes (The RCRA regulations specify the time hazardous wastes can be stored at a site. If the storage time is exceeded, a RCRA permit is required) and Transport, treatment, and disposal of hazardous wastes (a transporter should be a .

PLN 360560 Environmental Impact Assessment Homework No. 4 1. An appl.pdf

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one of the main design goals for the java programming language is security. Explain, in detail how java achieves this goal? Solution 1.Security became an essential piece of a java language from the starting day itself.Additionally it is an progressing design aim of the java community .An intrinsic security is observed ,hence the java language is in much now-a-days which is the major cause. 2.Every action is possible through Java virtual machine(JVM) .it is mainly to establish the attributes like trust,integrity,confidentiality..etc. 3.Java offers large compile perusal ,leads with secondary stage of examining .language attributes motivates the programmers closely to good programming recurents. especially the memory management representation is uncomplicated.it removes the total classes of programing mistakes which causes a continous trouble to the C and C++ programmers.moreover the objects are fabricated with a new operator.precised programmer defined pointer data types, pointer arthematic and self-activated garbage collector is absent. 4.java language is manipulated in distributed domains that tells that the security is most important including the security attributes which is intended in the java language and run time system. 5.In case of network habitats ,the applications in java language are secured from interruption with an unofficial code venturing to acheive alrady the scenes and occupy file systems..

one of the main design goals for the java programming language is se.pdf

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Meyer and colleagues performed three rounds of X-ray crystallography, each with a different elastase crystal. For each experiment, they positioned the elastase crystal at a different angle. Why might researchers have rotated elastase crystals to position them at different angles? Select one: a. Similar to when you get X-rayed in different positions, X-rays would produce different images depending on the angle and position of the elastase crystal. Researchers likely rotated elastase crystals to different angles to understand the conformation of this enzyme from all sides. b. There was no reason for the researchers to rotate the elastase crystals. It would have been sufficient to carry out one X-ray crystallography experiment on an elastase crystal. c. Rotating the elastase crystals to different angles likely affected the rate of the enzymatic reaction; at a different angle, the hexapeptide substrate could remain bound to elastase for a longer period of time. There was thus more time to get detailed X-ray images. Solution a.Similar to when you get X-rayed in different positions,X-rays would produce different images depending on the angle and the position of the elactase crystal.Researchers likely rotated elastase crystals to different angles and position of the elactase crystal.Researchers likely rotated elactase crystals to different angles to understand to confirmation of this enzyme from all sides..

Meyer and colleagues performed three rounds of X-ray crystallography.pdf

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Numerous diseases are caused by problems in DNA replication or repair. For example, Bloom Syndrome is caused by a mutation in one of the helicase enzymes involved in DNA replication. Select one of the enzymes involved in DNA replication or repair and find a disease related to a mutation in this enzyme (a few of these diseases are mentioned in your textbook). Then, in at least 3 pages, address the following topics: 1. What is the specific mutation? 2. What are the symptoms of the disease? 3. How was the relationship between the mutation and the disease discovered? 4. How does this mutation impact DNA replication and repair and how is this different from normal DNA replication? 5. How does this mutation impact transcription and translation and how is this different from normal transcription and translation? 6. How does this problem in DNA replication lead to one symptom of the disease? 7. Give two examples of how genetic engineering technology could be used to learn more about, detect earlier, or treat this disease? Be sure to include a description of what genetic engineering is. You should write this Assignment as an integrated paper that addresses all the questions. This Assignment should not be done in question and answer format. Be sure to use appropriate headings for each section of the paper. Solution Disease caused :Progressive external ophthalmoplegia (PEO) is caused by mutations in mitochondrial DNA. 1.Specific mutation:Human mitochondrial DNA replication takes place by two-subunitsof DNA polymerase gamma (polgamma).Mutations in the polymerase and exonuclease domains of the human polgamma is the cause behind the occurence of PEO. 2.Symptoms of the disease: 3.Discovery of the relationship between DNA replication and repair: The diseases was discovered the symptoms that the patients experience followed by a genetic test on muscle biopsy sample.This led to the conclusion that there occured several deletions inmitochondrial DNA.Research further revealed that there occured mutations in polgamma on chrmosome 15q25.In particular, three autosomal dominant mutations were revealed that altered the amino acids located within polymerase motif B of polgamma. 4.Impact of mutation on DNA replication: The mutation resulted in dysfunctioning of the polgamma and thus some regions of the mitochondrial DNA are deleted in comparision to normal replication where the entire mitochondrial DNA is replicated fully. 5.Impact on transcription and translation: Mitochondrial DNA encodes 13 protein subunits of multimeric oxidative phosphorylation, 2 ribosomal RNAs and all the translated RNAs that are required for the translation of its own messenger RNA.In presence of mutations, some regions of mitochindrial DNA is not replicated and this leads to impaired protein synthesis. 6.Problem in DNA replication lead to one symptom of the disease: Deletions in segments of mitochondrial DNA leads to defective function of oxidative phosphorylation due to impaired protein synth.

Numerous diseases are caused by problems in DNA replication or repai.pdf

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In the Mexican swordtail fish, the wild-type color is olive-green. Other color variants include golden and Take caution with ratio there is some reduced viability in this system. Matings involving fish of these three phenotypes are shown below: Explain the inheritance of these color phenotypes in the fish Solution At any rate, the mode of inheritance of albinism in the goldfish has not hitherto been known. Using mates of different varieties of the goldfish, scientists arrived at the conclusion that albinism of the goldfish is not a simple Mendelian character but it is represented by the double homozygote of two non-linked autosomal recessive genes since because the F2 generation shows 15:1 ratio for non-albino and albino types like we see in the case of duplicate genes. example: If the animal was A/a, because a is recessive and A is dominant, the animal\'s phenotype wouldn\'t be albino, but it would still carry the gene for albinism (since it has an \"a\"). Since it carries both \"A\" and \"a\", it is known as \"heterozygous\". If the animal had the A/A combination, its phenotype wouldn\'t be albino, and it wouldn\'t carry the gene for albinism (both genes being the same, it is called \"homozygous\" for \"A\"). If it were homozygous for \"a\" (i.e. a/a), the animal\'s phenotype would be albino. Since \"a\" is recessive, both alleles need to be \"a\" in order for albinism to be expressed in the phenotype. The genes for the pigment cells are inherited independently of one another, and there is no known linkage to any other genes. So, each pigment type is coded by two different genes, one on each of a pair of chromosomes. The phenotype of wild type animals is dark, non-melanoid, non- albino and non-axanthic. The golden colour (D/D a/a or D/d a/a). It has normal migration of pigment cells, but lacks melanophores, hence the yellow/gold appearance..

In the Mexican swordtail fish, the wild-type color is olive-green. Ot.pdf

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If we maintain a strict biological definition of “species”, H. sapiens and H. neanderthalensis were NOT separate species as they successfully interbred. What impact might this information have on assigning a date to the origin of humans? Are there any other implications? Solution As we know the common male ancester is usually much younger than the common female ancester.Moreover the Mitochondrial DNA which is present in the Neanderthals and it is absent in humans Suppose we have related populations A, B and C and we can have that A can interbreed with B, and also the B can interbreed with C,.At the same time ,A can not interbreed with C. Noe a ccording to the strict BSC, the A and B are in the same species and at the same time B and C are in the same species, but A and C are not.These are the implications here. It is impossible to know the date of origins of humans due to difference between sapiens and neanderthals.we can fix one date on which an old species was found and can assume it to be the date of origin..

If we maintain a strict biological definition of “species”, H. sapie.pdf

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If taxpayers Monica, Traci , and Phillip are responsible persons in the failure to withhold and pay over employment taxes but only Monica and Traci acted willfully which of the following statements is correct? A the IRs may assess the trust fund recovery penalty against Monica , Traci and Phillip, but can only collect it from Monica and Traci b. The IRS can only asses and collect a trust fund recovery penalty from Phillip c. The IRS cannot assess a trust find recovery penalty against Phillip d. the IRS may asses a trust fund recovery penalty against Monica , Traci, and Phillip Solution The tax IRs may assess a trust recovery penalty against monica, traci, and phillip, but can only collect it from monica and trici. The employer need to withhold taxes from its employees wages and pay them to IRS. Withheld taxes are been said as trust fund taxes. When these taxes are not paid by the employer intenstionally or not to get paid the taxes IRs checks for the responsible person and willful person and upon all IRs may assess for thr trust fund recovery. IRs may ask for to pay the penality by any of the responsible persons..

If taxpayers Monica, Traci , and Phillip are responsible persons in .pdf

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human biology scientific method project Step 1: Observe and Generalize ----common cold example hand washing name two more Observe different characteristics related to your topic of question, and then generalize about it. Take into consideration what is already known on the subject, varying opinions, past mistakes, and new aspects that you may have not foreseen. Summarize: Observation 1: Observation 2: Observation 3: Solution Answer: Observation 1: Common cold is caused by bacterial or viral infection and transmitted from one person to other through droplets from infected person. The person infected with common cold show protect his nose or mouth with steril mask, which covers mouth and nose. If nose or mouth is not covered by mask, droplets can transmit common cold to many people in surrounding environment. Observation 2: Common cold is widely transmitted very easily, if a person infected with common cold and he sneezes in public places which is having high population, several thousands of droplets can be transported through air at onetime, when person sneezes, so avoid moving in public places when common cold is rampant in specific area. Observation 3: Doctors who examine the persons suffering with common cold should be careful, the physician should use antibiotics or vaccines to prevent common cold infection..

human biologyscientific method projectStep 1 Observe and Genera.pdf

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I would appreciate help with these questions. Thank you. Question 1 How long is the typical default keepalive timer for TCP? 10 milliseconds 1 second 5 minutes 2 hours Question 2 What is the primary reason that large packets may not be more efficient in routing IP networks? The additional processing time associated with the cut-through nature of the IP router The additional overhead associated with a larger bandwidth connection The additional latency due to the store-and-forward nature of the IP router The TCP connection overhead required from hop-to-hop Question 3 What additional parameter is typically used to protect against wrapped sequence numbers on high-bandwidth connections? keepalive TTL timestamp DF Question 4 What does OID stand for? Object IDentifier Object Instance Data Organized Identifying Data Ordered Information Director Question 5 What is the typical port for an SNMP agent to use for managers to connect to? 80 161 22 1812 Question 6 Which of the following are notifications generated by the snmp agent on a router, rather than information requested by an snmp manager? get get-bulk get-next trap1. 10 milliseconds2. 1 second3. 5 minutes4. 2 hours Solution Answer : Question 1 How long is the typical default keepalive timer for TCP? Answer : 4. 2 hours 2 hours is the typical default keepAlive time for TCP Question 3 What additional parameter is typically used to protect against wrapped sequence numbers on high-bandwidth connections? Answer : 3.timestamp timestamp parameter is typically used to protect against wrapped sequence numbers on high- bandwidth connections Question 4 What does OID stand for? Answer : 1.Object IDentifier OID stand for Object IDentifier Question 5 What is the typical port for an SNMP agent to use for managers to connect to? Answer : 2.161 161 is the typical port for an SNMP agent to use for managers to connect to. Question 6 Which of the following are notifications generated by the snmp agent on a router, rather than information requested by an snmp manager? Answer : 4.trap Traps are notifications generated by the snmp agent on a router, rather than information requested by an snmp manager..

I would appreciate help with these questions. Thank you.Question 1.pdf

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For settlements, what are the major components of a soil settlment? What is the difference between between settlment in sands and clays, please describe the details. 2. Solution settlement have three components: 1. Immediate settlement, 2. Primary settlement and 3.secondary settlement. immediate settlement takes place immediately after the application of load. Primary settlement is the long term process which occurs after the application of load. secondary settlement depends on time. In sands, immediate settlement takes place and in clays primary settlement occurs..

For settlements, what are the major components of a soil settlment W.pdf

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Explain the heterogenous subunits of Rpb1 by answering these questions. What is the difference between IIo, IIa and IIb? Which enzymes convert IIa to IIo, IIo to IIa and IIa to IIb? Solution Rpb I is a subunit of DNA directed RNA polymerase II which is encoded by POLR2A GENE. IIa - stabilises TATA binding proteins (TBP ) and TAF binding ( TBP associated factors ) . This has one subunit. IIb - this GTF HAS 2 subunits. intermediate in the recruitment of RNA polymerase II. Influences selection of the start point of transcription. This protein would enter the pre initiation complex after TBP. This protein bridges the TATA bound TBP and polymerase. *There appears to be a mistype in IIo as none of the transcription factors are named as IIo, but yes there is a transcription factor as IId. Kindly leave in the comments if that is the same you want details about and feel free to ask any question..

Explain the heterogenous subunits of Rpb1 by answering these questio.pdf

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Drinking alcohol often makes the consumer need to urinate. This is because ethanol inhibits the production of anti-diuretic hormone. Anti-diuretic hormone inserts channel proteins into the kidneys that allow for water to pass through them. What are these channel proteins called? Glycoproteins Gated channels Lipophilic proteins Ion channels Aquaporins Solution Answer : The correct answer is E (Aquaporins) Information : Aquaporins are a family of integral proteins which provides channels for rapid movement of water molecules across all plasma membranes..

Drinking alcohol often makes the consumer need to urinate. This is be.pdf

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Drug release profile of acetaminophen (APAP) in cellulose excipient was observed by dissolving the tablet in the saturated solution of APAP. If the initial porosity of the tablet before dissolution is 2%, diffusion constants of excipient and APAP are 0.3 and 0.5, ?u=?v=1.5 and Solution INTRODUCTION: Ethylcellulose polymer is widely used in pharmaceutical coating, especially when producing modified release dosage forms. Drug release through barrier membrane systems is influenced by several process and formulation variables; however, our understanding of the influence of these variables is hindered by a lack of comprehensive models to guide our coating efforts. The purpose of this work was to study the influence of an aqueous or an organic coating process on drug release profile characteristics from an ethylcellulose barrier membrane coated multiparticulate system. Either ETHOCEL ethylcellulose (Standard 20 Premium, The Dow Chemical Company) or aqueous ethylcellulose dispersion (Surelease E-7-19040, Colorcon, USA) was used as the barrier membrane polymer. Hypromellose (HPMC, METHOCEL.

Drug release profile of acetaminophen (APAP) in cellulose excipient .pdf

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Di Paoloma University had a faculty reduction and lost 15 of its fa.pdf

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Define a relation R on the set Z of all integers by a R b if and only if 3a + b is a multiple of 4. Prove that R is an equivalence relation. Determine the equivalence classes associated with R. Solution 1. 3a+a=4a Hence, aRa for all integers a Hence, R is reflexive 2. Let, aRb So 3a+b= multiple of 4 3(3a+b)=3b+9a is multiple of 4 3b+a+8a is multiple of 4 Hence, 3b+a is multiple of 4 Hence,bRa So, R is symmetric 3. Let, aRb and bRc 3a+b is multiple of 4, 3b+c is multipe of 4 So, 3a+b+3b+c =3a+c+4b is multiple of 4 HEnce, 3a+c is multiple of 4 So, aRc Hence, R is equivalence relation 3a+b=4a-a+b Hence, 4a+(b-a) is multiple of 4 ie b-a is multipel of 4 Hence all elemenets of an equivalence class are one which give same remainder modulo 4 so there are 4 equivalence classes for each remainder modulo 4 ie [0],[1],[2],[3].

Define a relation R on the set Z of all integers by a R b if and only.pdf

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