Let L R^n rightarrow R^m be a linear transformation. Show that if n .pdf

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Let L: R^n rightarrow R^m be a linear transformation. Show that if n > m, then L must have a nontrivial kernel, i.e., dim(ker(L)) > 0. If n lessthanorequalto m does L have to be one-to-one? Solution For second proof, If T is one-one, it can\'t map a non-zero vector into 0, since it already maps 0 into 0. This means that its kernel can contain only 0. Suppose the kernel contains just the zero vector. To check if T is one-one, check if it can map two different vectors u and v in Rn into the same vector in Rm. If T(u) = T(v), then T(u – v) = T(u) – T(v) = 0. This means that the non-zero vectoru – v is in the kernel of T, which is impossible since we assumed the kernel contained just 0. Then T cannot map different vectors into the same vector, and must be one-one..

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select all that apply to ground substance Solution Lets us see what is ground substance. It is an amorphous gel-like or a transparent flud substance surrounding the cells,present in the connective tissue . The ground substance and fibers between cells in a connective tissue (most abundant tissues in the body-functions : binding and supporting; consists of relatively few cells in a generous matrix ) are surrounded and supported by matrix called as extracellular matrix. The Ground substance includes all the other components of the extracellular matrix excluding fibers (Eg: collagen and elastic fibers). Depending upon the tissue, the components of the ground substances vary . Ground substance is primarily composed of water, proteoglycans,glycosaminoglycans (most notably hyaluronan-hyaluronic acid), calcium and glycoproteins. In cytology, it may also refers to cytosol or protoplasm. The stroma is the tissue that forms the ground substance, base or the foundation, or the framework of an organ, as opposed to its functional parts (parenchyma)..

select all that apply to ground substanceSolutionLets us see w.pdf

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Relying on the above pathway for biosynthesis of tryptophan, fill in the table below to predict growth (+) or failure to grow (-) when mutants defective in each of the trp genes are placed on media supplemented with the compounds listed. Solution Answer: A mutant for a particular gene in the tryptophan pathway will only grow when it is supplemented with the required nutrient so that the biosynthetic pathway for the tryptophan synthesis is resumed. The mutant strain (auxotroph) will proliferate only when the medium is supplemented with the required substance leading to the completion of the tryptophan biosynthetic pathway. + indicates growth when a particular supplement is given. Mutants which have dysfunctional trpB, trpC, trpD or trpE can still grow if supplemented with any of Indole, tryptophan or Indole+tryptophan because in all of these cases the strain will get the required tryptophan necessary for its survival.SupplementMutanttrpAtrpBtrpCtrpDtrpENone-----Anthranilic acid---- +Indole-++++Tryptophan+++++Typtophan and indole+++++.

Relying on the above pathway for biosynthesis of tryptophan, fill i.pdf

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Question 7 (2 points) A researcher conducts a study on the realtionship between caffeine consumption and performance on a math test. What kind of study is this? Question 7 options: There is not enough information here to tell. Correlational Quasi-experimental Experimental Save Question 8 (2 points) Which researcher is most likely to engage in content analysis? Question 8 options: Dr. Williams, who studies transcripts of police interviews with suspects. Dr. McKenzie, who studies relationships between various personality traits. Dr. Perryman, who studies the social skills of indigenous people. Dr. Ramirez, who studies mathematical competence in adolescents. Save Question 9 (2 points) Which of the following would be more likely to be found in an empirical research article on bullying than in a popular magazine article on the same topic? Question 9 options: Opinions about why people bully from the point of view of the author of the article. Excerpts from actual interviews with bullies and/or victims. Summaries of the methods and results found by previous researchers on bullying accompanied by reference citations. Humerous anecdotes about how victims fantasize about getting back at their bullies. Save Question 10 (2 points) Dr. Simmons is interested in studying attachment behavior in infants. She trained her research assistants to recognize classic attachment behaviors such as fearful reactions to strangers and proximity seeking with the attachment figure. The research assistants then learned to code these behaivors while watching video tapes of infants and caregivers interacting while they waited in the doctor\'s waiting room. What type of research is going on here? Question 10 options: Quasi-experimental research Naturalistic observation Archival data collection Interrupted time-series designa) There is not enough information here to tell.b) Correlationalc) Quasi-experimentald) Experimental Solution b) Correlational a) Dr. Williams, who studies transcripts of police interviews with suspects. c) Summaries of the methods and results found by previous researchers on bullying accompanied by reference citations c) Archival data collection b) Correlational.

Question 7 (2 points)A researcher conducts a study on the realtion.pdf

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Primary difference between the income sections for form 1120 and 1120s Primary difference between the income sections for form 1120 and 1120s Solution Answer :- The primary difference between form 1120 and 1120s is that the form 1120 is filed by corporations on annual basis under internal revenue service (IRS) code whereas the form 1120s is filed by S corporation (corporation who have opted for \"S\" status) on annual basis under internal revenue service (IRS) code..

Primary difference between the income sections for form 1120 and .pdf

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Once it\'s readable i will be okay. Solution \"I will be okay in the end. If it\'s not okay, it\'s not the end.\" What are some incidences when this John Lennon quote was proved true in real life? Many inspiring stories of people\'s lives show it... Consider Abraham Lincoln. He met failure after failure throughout his life before he became the President of USA at 52. Thomas Edison... 10,000 failures before inventing the light bulb. Henry Ford... was broke at the age of 40. Conolel Sanders... broke at 65... knocked at 1000 doors before he could make his first sale of a chicken dish...founder of KFC.... Each and everytime in life , On all the small tasks and bigger challenges...its proven true. It may be preparing a cup of tea or coffee or Working on a home . The tea or coffee after two boils are perfectly okay for me but you may boil it more with extra milk. I work on decor with few frames of modern art and exclusive vases but you may go for plain walls. A waiter takes the order and serves and is okay with it...but another one wears a smile and is quite alert and won\'t let the glass of water to be empty because its his level of \"okay\". So each of us working and completing our daily task work on this principle. We try hard and reach our okay\'s. There\'s a variation in \"okay\" of all of us...that\'s it..

Once its readable i will be okay.SolutionI will be okay in.pdf

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Need the mutiple choice answered uslde .An)_can omunicate with a system inside a firewall system to be used to filter out undesirable traffic and prevent employees from accessing potentially hostile web sites Multiple-Choice Quiz e three types of event logs generated What are the three t t bw Windows Which of the following is not a capability of 6. Windows 2003 and Vista by A. Event, Process, and Security B. Application, User, and Security c. User, Event, and Security network-based IDS A. Can detect denial-of-service attacks B. Can decrypt and read encrypted traffic C. Can decode UDP and TCP packets D. Can be tuned to a particular network Application, System, and Security What are the two main types of intrusion detection systems? A. Network-based and host-based B. Signature-based and event-based C. Active and reactive D. Intelligent and passive environment An active IDS can: A. Respond to attacks with TCP resets B. Monitor for malicious activity C. A and B D. None of the above Honeypots are used to: , 7. s. What was the first commercial, network-based 8. IDS product? A. Stalker B. NetRanger C. IDES D. RealSecure What are the two main types of IDS signatures? 9. Egress filtering is used to detect spam that is: A. Network-based and file-based B. Context-based and content-based C Active and reactive A. Attract attackers by simulating systems with open network services B. Monitor network usage by employees C. Process alarms from other IDSs D. Attract customers to e-commerce sites A. Coming into an organization B. Sent from known spammers outside your organization C. Leaving an organization D. Sent to mailing lists in your organization Preventative intrusion detection systems: A. Are cheaper B. D. None of the above . Which of the following describes a passive, host based IDS? 10. A. Runs on the local system Are designed to stop malicious activity from occurring B. Does not C. Can l D. All of the above not interact with the traffic around it C. Can only monitor activity D. Were the first types of IDS ook at system event and error logs r 13:Indusion Detection Systems Detection Systems and Network Security Solution 1. What are the three types of event logs generated by Windows 2003 and Vista systems? D. Application, System, and Security 2. What are the two main types of intrusion detection systems? A. Network-based and host-based 3. What was the first commercial, network-based IDS product? B. NetRanger 4. What are the two main types of IDS signatures? B. Context-based and content-based 5. Which of the following describes a passive, host-based IDS? D. All of the above 6. Which of the following is not a capability of network-based IDS? B. Can decrypt and read encrypted traffic 7. An active IDS can: C. A and B 8. Honeypots are used to: A. Attract attackers by simulating systems with open network services 9. Egress filtering is used to detect spam that is: C. Leaving an organization 10. Preventative intrusion detection systems: B. Are designed to stop m.

Need the mutiple choice answered uslde .An)_can omunicate with a sys.pdf

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Need a strong biologist/geneticist to answer this P element question for Drosophila... PLEASE What do you know about P-elements in Drosophila? How might P-elements have entered Drosophila melanogaster? What do you know about P-element structure? How big are the native elements? How do P-elements move around the genome? Connect your ideas to the special P- elements used in this lab. What is the connection between transposition of mobile DNA elements and DNA replication? Why are P-elements important for biologists and Drosophila researchers in particular? What do you know about the Gene Disruption Project (GDP) in Drosophila? Solution Ans.) P element is a type of transposon found particularly in Drosophila and is used for mutagenesis and for the production of genetically modified flies that is extensively used in various genetic and medical researches. It has ability to initiate a phenotype known as hybrid dysgenesis. It encodes the protein P transposase and is flanked by terminal inverted repeats which are important for its mobility. The P element enters in Drosophila by the horizontal gene transfer method. It has a sequence of 4 exons with 3 introns. The absolute splicing of the intron produces the transposase enzyme, whereas the alternative partial splicing of intron 1 and 2 produce the P element repressor. The P element is a class II transposon and it is moved by DNA-based \"cut and paste\" mechanism. The P element encodes the transposase enzyme which recognizes the 31 bp terminal inverted repeats of the P element and catalyzes P element excision and re-insertion. P element is important for biologist and genetics because nowadays it is widely used to solve many puzzles of genetics through its use in fly transformation, insertional mutagenesis as well as reverse genetics. The Gene Disruption Project (GDP) is a public collection of mutant that is created by applying a non-targeted transposon mutagenesis strategy. Its main objective is to acquire a transposable element insertion in each gene and this collection is available online and new mutants may be requested while they are being balanced. These balanced stocks are available at the Bloomington Drosophila Stock Center..

Need a strong biologistgeneticist to answer this P element question.pdf

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Modify HuffmanTree.java and HuffmanNode.java to allow the user to select the value of n 9 to construct an n ary Huffman Tree. Once finished, verify the accuracy of your tree by running it for the English Alphabet for n = 3, 4, 5, 6, 7, 8, 9. For each value of n, compute the average codeword length. Write these in a table (n vs. ACW L(Cn)). / HuffmanTree.java import java.io.FileNotFoundException; import java.io.FileReader; import java.util.ArrayList; import java.util.Scanner; public class HuffmanTree { public static void main(String[] args){ Scanner in = new Scanner(System.in); System.out.print(\"Enter the file name: \"); String fileName = in.nextLine(); ArrayList nodes = new ArrayList<>(); ArrayList temp = new ArrayList<>(); try { Scanner fin = new Scanner(new FileReader(fileName)); String symbol; double prob; while(fin.hasNext()){ symbol = fin.next(); prob = fin.nextDouble(); nodes.add(findInsertIndex(nodes, prob), new HuffmanNode(symbol, prob)); } fin.close(); temp.addAll(nodes); HuffmanNode root = createHuffmanTree(temp); setCodeWords(root); double avg = 0; for (HuffmanNode node : nodes) { System.out.println(node.getS() + \": \" + node.getC() + \": \" + node.getL()); avg += node.getL(); } if(avg != 0) avg /= nodes.size(); System.out.println(\"Average code length is: \" + avg); } catch (FileNotFoundException e) { e.printStackTrace(); } } private static HuffmanNode createHuffmanTree(ArrayList nodes){ HuffmanNode root = null; if(nodes.size() > 0) { HuffmanNode node0, node1, node; while (nodes.size() > 1) { node0 = nodes.get(0); node1 = nodes.get(1); node = new HuffmanNode(null, node0.getP() + node1.getP()); node.setLeft(node0); node.setRight(node1); nodes.remove(0); nodes.remove(0); nodes.add(findInsertIndex(nodes, node.getP()), node); } root = nodes.get(0); } return root; } private static void setCodeWords(HuffmanNode root){ if(root != null){ setCodeWords(root, \"\"); } } private static void setCodeWords(HuffmanNode root, String codeWord){ if(root.getS() != null){ root.setC(codeWord); root.setL(codeWord.length()); } else{ setCodeWords(root.getLeft(), codeWord + \"0\"); setCodeWords(root.getRight(), codeWord + \"1\"); } } private static int findInsertIndex(ArrayList nodes, double prob){ for(int i = 0; i < nodes.size(); ++i){ if(prob < nodes.get(i).getP()){ return i; } } return nodes.size(); } } // HuffmanNode.java public class HuffmanNode { private String S; private double P; private String C; private int L; private HuffmanNode left, right; public HuffmanNode(String s, double p) { S = s; P = p; } public String getS() { return S; } public void setS(String s) { S = s; } public double getP() { return P; } public void setP(double p) { P = p; } public String getC() { return C; } public void setC(String c) { C = c; } public int getL() { return L; } public void setL(int l) { L = l; } public HuffmanNode getLeft() { return left; } public void setLeft(HuffmanNode left) { this.left = left; } public HuffmanNode getRight() { return right; } public void se.

Modify HuffmanTree.java and HuffmanNode.java to allow the user to se.pdf

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Listed are 2 of the core theories of evolution. For each, give a definition or brief explanation of the theory and 2 examples of the type of evidence we have that supports the theory. The theory of common descent – 2 examples of supporting evidence: The theory of natural selection – 2 examples of supporting evidence: Solution Evolution: Evolution is a process of change through time. It is the process by which modern organisms have descended from ancient organisms. In genetic terms, the change in the frequency of alleles in populations from one generation to another generation. (change over time). A. The theory of common descent: According to the principle of common descent, all species (living and extinct) are descended from ancient common ancestors. In common descent theory darwin explains, that organisms are descended from common ancestors that gradually change over time into entirely new organism. To create new species, many generations must pass on traits with slight modifications. Examples: 1. Archaeopteryx is between reptiles and birds. 2. Eustheopteron is an amphibious fish. 3. Seymouria is a reptile-like amphibian. 4. Therapsids were mammal-like reptiles. B. Darwin\'s Natural selection: According to the Darwin, It is the process that results in adaptation of an organism to its environment which means of selectivity reproducing changes in the genotype. In the process of evolution the natural selection is a basic mechanism, it is mainly happens due to the mutations, migrations and genetic drift. Examples: 1. The beaks of Darwin finches. The finches of the Galapagos islands were adopted to the island in different ways, but are believed to share a common ancestor. 2. Peppered moths and industrial pollution: Natural selection favored dark colored moths in areas of heavy pollution, while light colored moths are survived better in less polluted area. 3.The soapberry bugs have must mutated their beaks length genes, when they ran out of food. Beak size variation: In South Florida soapberry bugs feed on balloon vine tree with large fruits, longer beak present in them. In North Florida soapberry bugs feed on golden rain tree with small fruits, they have shorter beaks..

Listed are 2 of the core theories of evolution. For each, give a def.pdf

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Light is confined within the core of a simple clad optical fiber by.pdf

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: IS path vector routing algorithm closer to the distance vector routing algorithm or to the link state routing algorithm? Explain. Solution The path-vector routing algorithm is truly distance-vector routing using the finest path instead of the direct distance as the metric. Each node first create a forwarding table. Assuming it can only arrive at instant neighbors . The forwarding table is slowly better as path vectors turn up from the instant neighbors..

IS path vector routing algorithm closer to the distance vector rou.pdf

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a growth medium that can determine if a bacterium contains several d.pdf

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Identify the function of the circulatory system in fishes. Explain how it works. Solution The circulatory arrangement of fish is very straightforward. It comprises of a heart, blood, and veins. The heart of a fish is a basic strong structure that is situated behind (and underneath) the gills. It is encased by the pericardial layer or pericardium. The heart comprises of a chamber, a ventricle, a thin-walled structure known as sinus venosus, and a tube called bulbus arteriosus. The blood contains plasma (the liquid part) and platelets. The red platelets or the erythrocytes contain hemoglobin, a protein that conveys oxygen all through the body. The white platelets contain a basic part of the invulnerable framework. The thrombocytes perform capacities that are proportionate to the part of platelets in the human body, i.e. they help in blood coagulating. Veins convey the blood all through the body. While corridors convey oxygenated blood from the gills to whatever remains of the body, veins return deoxygenated blood from various parts of the body to the heart. The cardiovascular arrangement of a fish involve a heart, veins, corridors, blood, and fine vessels. The vessels are minute vessels that shape a system called a fine bed, where the blood vessel and venous blood get connected. Vessels have thin dividers that encourage dispersion, a procedure through which oxygen and different supplements from the blood vessel blood are moved into the cells. In the meantime, carbon dioxide and waste materials are moved into the vessels. Vessels with deoxygenated blood (contains carbon dioxide) deplete into little veins called venules, which thusly deplete into bigger veins. The veins convey the deoxygenated blood into the sinus venosus, which resembles a little gathering chamber. The sinus venosus has pacemaker cells that are in charge of starting constrictions, so that the blood is moved into the thin-walled chamber, which has not very many muscles. The chamber produces feeble compressions to push blood into the ventricle. The ventricle is a thick-walled structure with bunches of cardiovascular muscles. It produces enough weight to pump the blood all through the body. The ventricle pumps blood inside it into bulbus arteriosus, a little chamber with flexible segments. While bulbus arteriosus is the name of the chamber in teleosts (rayfinned, hard fish), the structure is known as conus arteriosus in elasmobranchs (angle with cartilaginous skeleton and placoid scales). Conus arteriosus has numerous valves and muscles, while bulbus arteriosus has no valves. The primary capacity of this structure is to diminish the beat weight created by the ventricle, with a specific end goal to maintain a strategic distance from harm to the thin-walled gills. Gills are the essential respiratory organs of fish. They encourage trade of gasses, i.e. retention of oxygen from water and end of carbon dioxide. Corridors convey the oxygenated blood (from the gills) all through the body..

Identify the function of the circulatory system in fishes. Explain h.pdf

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How did the discovery of three categories of petite mutations in yeast lead researchers to postulate extranuclear inheritance of colony size? Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive petites followed non-Mendelian patterns that were consistent with the involvement of nuclear inheritance. Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive petites followed non-Mendelian patterns that were consistent with the involvement of an extranuclear agent. Whereas suppressive petites exhibited Mendelian inheritance, both neutral and segregational petites followed non-Mendelian patterns that were consistent with the involvement of an extranuclear agent. Whereas suppressive petites exhibited Mendelian inheritance, both neutral and segregational petites followed non-Mendelian patterns that were consistent with the involvement of nuclear inheritance. Solution Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive petites followed non-Mendelian patterns that were consistent with the involvement of nuclear inheritance. --------- Extra nuclear inheritance means, inheritance patterns involving genetic material outside the nucleus, here the genome is residing in mitochondria or chloroplast. The segregation petite followed Mendelian type of inheritance (wild type and segregation when mated, segregated in 1:1 ratio), and if any mutation has occurred in the cell nucleus, it affected the protein production needed for mitochondrial function. Where as neutral petites, they did not have most of their mt DNA and but then it was mated with wild type resulted in all wild type 4: neutral 0. So, these are not following Mendelian type of inheritance. In case, of suppressive petites, when suppressive mutants mated with wild type resulted in All wild type 4: 0 all petite. So they are also not following Mendelian inheritance..

How did the discovery of three categories of petite mutations in yea.pdf

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A certain system can experience 2 difference types of defects. Let ,.pdf

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Find to three decimal places, the coordinate of point P lying in the 4th quadrant on the Cartesian plan such that the line tangent to the cosine function passes through the origin. X:[0,2Pi] Show all work and describe all steps. Solution t = cos x tangent passing through origin t\' = -sin x So, tangent y = mx+c y = -xsinx+c here, (0,0) is point on tangent, 0 = -0sin0 + c c = 0 So, y = -xsinx is tangent..

Find to three decimal places, the coordinate of point P lying in the.pdf

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Find the mobility (DoF) and Grashof condition of the mechanism below..pdf

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5. Discuss the pros and cons of American business being excluded by American law (Pres Kennedy issued an Executive Order) from conducting any form of business activity on the island of Cuba. Express your perspectives of the European Union and other first world nations (like Canada) steadfastly generating billion of euros (and other currencies) from trade with Cuba. Solution There are significant pros and cons of prohibition of conducting trade with Cuba. The pros of the prohibition are as follows 1. Cuba has not fulfilled the conditions of lifting the trade Ban which was set up by the US government. 2. The US government made various attempts to reduce the restrictions of trade. However this was met with aggression by the Cuban government. 3. There is no benefit to the US government from lifting the trade ban since there are hardly any private businesses in Cuba. 4. Reversing the embargo could be ruinous to the US image. The cons of the trade Ban are as follows 1. The rest of the world is in favour of reversing the trade ban and this ban reduces the US goodwill in the world. 2. The Ban is considered to be ineffective since it was created to make Cuba promote capitalism but has not made its objective. 3. The ban has an impact on the small and family Run enterprises in United States and is expensive for the nation. The main imports of Cuba are food, machinery and fuel. Its major exports are sugar, pharmaceuticals and tobacco. The main trading partners of Cuba are Venezuela and European Union. Tourism stemming from Canada has boosted the economy of Cuba. The European Union is Cubas main trade partner and also the biggest foreign investor in Cuba. Cuba will benefit from the European Union thematic program which covers issues such as support of internationalisation of small and medium enterprises as well as increasing the technical assistance in agriculture. The European Union is also exploring opportunities in fields of culture and health..

5. Discuss the pros and cons of American business being excluded by .pdf

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Explain the chemical basis of why proteins have such a large variety of structures but DNA has so few. Solution The DNA is a molecule that carries the genetic information from one generation to another. Structurally, DNA is made of 04 nitrogen containing nucleobases namely, Adenine (A), Guanine (G), Cytosine (C) and Thymine (T). These chemical bases pairs with each other, like A pairs with T with 02 Hydrogen Bonds and C pairs with G with 03 Hydrogen bonds to form units called Base Pairs. Each base is then attached to a 5 carbon sugar molecule and a phosphate group. Combined base, sugar and phosphate are called nucleotide. These nucleotides are arranged in two long strands to form a spiral structure called double helix. The double helix structure looks like ladder, the base pairs form the ‘ladders rugs’ and sugar with phosphate forms the vertical sidepieces of ladder. Only four different bases in nucleic acid but it contains a long backbone with million of bases. The appearance of nucleotides in nucleic acid codes for the information that is being carried out in the molecule. Protein is a very large biological molecule that is composed of one or more polypeptides that are folded into specific structures. Structurally, proteins are the polymers of 20 different Amino Acids (AA). The AA are the building block of the the proteins. Each AA consist of - carbon atom bonded specifically with carboxyl group (COO-), amino group (NH3+) and distinctive side chain. The specific chemical property of different AA side chains leads to the role of each AA in protein structure and function. The AA are joined together by peptide bonds between - amino group of 1 AA to - carboxyl group of second. Each polypeptides has 2 distinct ends, one terminates in -amino group (N- terminus) and other in -carboxyl group (C-terminus) attached to the same tetrahedron atom C. The proteins have large variety of structure is because, proteins have 20 different Amino Acids and each amino acid contains a vast range of functional groups including alcohol, thiols, thioesters, carboxylic acid, carboxamides and variety of basic groups. Proteins are complex structure of 20 AA, but there are only four bases in DNA (A,T, C,G). A minimum of three bases are required to code for at least 20 AA . The AA can be distinguished on the basis of their properties of side chains. Few contain non polar side chains that do not interact with water, other have uncharged but polar side chains. Some with charged side chain with basic groups. Unlike DNA which forms ladder like structure protein adopts three dimensional structures. The protein structure has four different levels. The primary structure is depicted by sequence of Amino Acid in polypeptide chain. The secondary structure depicts the transformation of AA sequence into and chains. The third level depicts the completely folded 3 Dimensional structure of protein. The quaternary structure consist of different interactions between different polypeptides.

Explain the chemical basis of why proteins have such a large variety .pdf

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Element# of protons# of neutrons# of electronsCarbon (atomic.pdf

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