(a).Let me show that H is a subgroup. Then you can make the identica.pdf
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(a).Let me show that H is a subgroup. Then you can make the identical argument and show that K is. e = (e, e) is in H let (x, e) be in H. G1 group so x^-1 in G1. Now (x^-1, e) = (x,e)^-1 since (x, e)(x^-1, e) = (xx^- 1, e) = (e, e) which is the identity in G let (x, e), (y, e) be in H. Then (x,e)(y,e) = (xy,ee) = (xy,e) is in H. Therefore H is a subgroup of G b) Let me show that KH=G. You can make the identical argument to show that HK=G. Let k =(e, y) be in K, h=(x, e) in H. Then kh = (e,y)(x,e) = (ex,ye)=(x,y) is in G, so KH subset of G Let (x,y) be in G. Then (e,y) is in K and (x,e) is in H, and (e,y)(x,e) = (x,y). So G is a subset of KH. Therefore G=KH Solution (a).Let me show that H is a subgroup. Then you can make the identical argument and show that K is. e = (e, e) is in H let (x, e) be in H. G1 group so x^-1 in G1. Now (x^-1, e) = (x,e)^-1 since (x, e)(x^-1, e) = (xx^- 1, e) = (e, e) which is the identity in G let (x, e), (y, e) be in H. Then (x,e)(y,e) = (xy,ee) = (xy,e) is in H. Therefore H is a subgroup of G b) Let me show that KH=G. You can make the identical argument to show that HK=G. Let k =(e, y) be in K, h=(x, e) in H. Then kh = (e,y)(x,e) = (ex,ye)=(x,y) is in G, so KH subset of G Let (x,y) be in G. Then (e,y) is in K and (x,e) is in H, and (e,y)(x,e) = (x,y). So G is a subset of KH. Therefore G=KH.
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International Journal of Engineering Research and Applications (IJERA) is an open access online peer reviewed international journal that publishes research and review articles in the fields of Computer Science, Neural Networks, Electrical Engineering, Software Engineering, Information Technology, Mechanical Engineering, Chemical Engineering, Plastic Engineering, Food Technology, Textile Engineering, Nano Technology & science, Power Electronics, Electronics & Communication Engineering, Computational mathematics, Image processing, Civil Engineering, Structural Engineering, Environmental Engineering, VLSI Testing & Low Power VLSI Design etc.
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International Journal of Engineering Research and Applications (IJERA) is an open access online peer reviewed international journal that publishes research and review articles in the fields of Computer Science, Neural Networks, Electrical Engineering, Software Engineering, Information Technology, Mechanical Engineering, Chemical Engineering, Plastic Engineering, Food Technology, Textile Engineering, Nano Technology & science, Power Electronics, Electronics & Communication Engineering, Computational mathematics, Image processing, Civil Engineering, Structural Engineering, Environmental Engineering, VLSI Testing & Low Power VLSI Design etc.
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A STUDY ON L-FUZZY NORMAL SUB L -GROUP
This paper contains some definitions and results of L-fuzzy normal sub l -group and its generalized characteristics.
A STUDY ON L-FUZZY NORMAL SUBl -GROUP
This paper contains some definitions and results of L-fuzzy normal sub l -group and its generalized
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A STUDY ON L-FUZZY NORMAL SUBl -GROUP
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This paper contains some definitions and results of L-fuzzy normal sub l -group and its generalized characteristics.
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ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BA...
ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BA...UniversitasGadjahMada
Chaatit, Mascioni, and Rosenthal de ned nite Baire index for a bounded real-valued function f on a separable metric space, denoted by i(f), and proved that for any bounded functions f and g of nite Baire index, i(h) i(f) + i(g), where h is any of the functions f + g, fg, f ˅g, f ^ g. In this paper, we prove that the result is optimal in the following sense : for each n; k < ω, there exist functions f; g such that i(f) = n, i(g) = k, and i(h) = i(f) + i(g).IRJET - Fuzzy Soft Hyperideals in Meet Hyperlattices
https://irjet.net/archives/V7/i3/IRJET-V7I3138.pdf
International Journal of Mathematics and Statistics Invention (IJMSI)
International Journal of Mathematics and Statistics Invention (IJMSI) is an international journal intended for professionals and researchers in all fields of computer science and electronics. IJMSI publishes research articles and reviews within the whole field Mathematics and Statistics, new teaching methods, assessment, validation and the impact of new technologies and it will continue to provide information on the latest trends and developments in this ever-expanding subject. The publications of papers are selected through double peer reviewed to ensure originality, relevance, and readability. The articles published in our journal can be accessed online.
TOPOLOGY 541Let M be a set with two members a and b. Define the fu.pdf
TOPOLOGY 541
Let M be a set with two members a and b. Define the function D : M × M as follows: D(a,a) =
D(b,b) = 0,D(a,b) = D(b,a) = r where r is a positive real number. Prove that (M, D) is a metric
space.
Solution
Definition 1.1 (Metric). A metric, or distance function, on a set X is a mapping d : X × X R such
that • d(x, y) 0 for all x, y X, and d(x, y) = 0 if and only if x = y. • d(x, y) = d(y, x) for all x, y
X. • d(x, z) d(x, y) + d(y, z) for all x, y, z X. We call (X, d) a metric space. Definition 1.2 (Open
ball). Let (X, d) be a metric space. For x X and > 0, the set Bd(x, ) defined by Bd(x, ) = {y X |
d(x, y) < } is callen an open ball in the set X. Definition 1.3 (Open sets in metric spaces). Let (X,
d) be a metric space and let U be any subset of X. Then U is called an open set in X if every
point of U is an interior point of U; that is, for any a U, there is an open ball B(a, ) such that B(a,
) U. Definition 1.4 (Properties of open sets). Let (X, d) be a metric space. • and X are open. •
The union of an arbitrary collection of open sets is open. • The intersection of a finite number of
open sets is open. Definition 1.5 (Closed set). A subset A of a metric space (X, d) is closed if it’s
complement X\\A is open in X. Definition 1.6 (Properties of closed sets). Let (X, d) be a metric
space. • and X are closed. • The union of an finite collection of closed sets is closed. • The
intersection of an arbitrary number of closed sets is closed. Definition 1.7 (Limit point of a
subset). Let (X, d) be a metric space and let A be a subset of X. Then a point x in X is a limit
point of A if every open ball B(x, ) contains at least one point of A. The set of all limit points of
A is called the derived set A . Definition 1.8 (Closure of a set). Let (X, d) be a metric space and
let A X. Then the set consisting of A and its limit points is called the closure of A, denoted A. A
= A A
Corollary 1.19. We note that A is dense in X if and only if for any x X and > 0, there is a point
a A such that d(x, a) < . 1.2. Subspaces. Definition 1.20 (Open sets in a subspace). Let (X, d) be
a metric space and (Y, dY ) be a metric subspace of (X, d). Let G be a subset of Y . Then G is
open in Y if and only if, for any x G, there is an open ball B(x, ) in X such that B(x, ) Y G A
subset H of Y is closed in Y if its complement G = Y \\H of H is open in Y . Theorem 1.21
(Open sets in a metric subspace). Let (Y, dY ) be a metric subspace of a metric space (X, d), and
let G Y . Then G is open in Y if and only if there exists an open subset U in X such that G = U
Y . 1.3. Convergence in a Metric Space. Definition 1.22 (Convergence). A sequence (xn) in a
metric space (X, d) is said to converge to a point x X if for any > 0, there exists N such that n >
N implies d(xn, x) < The point x is called a limit of the sequence (xn) Corollary 1.23. A
sequence (xn) in a metric space (X, d) is said to converge to a point x X if any open ball B(x, )
contains almost all xn. Theorem 1..
VideoGamepublic class VideoGam.pdf
/*****************VideoGame*****************/
public class VideoGame {
// instance variables
private String gameName;
private String gameConsole;
// Default constructor
public VideoGame() {
}
// parameterized constructor
public VideoGame(String gameName, String gameConsole) {
this.gameName = gameName;
this.gameConsole = gameConsole;
}
/* getters and setters */
public String getGameName() {
return gameName;
}
public void setGameName(String gameName) {
this.gameName = gameName;
}
public String getGameConsole() {
return gameConsole;
}
public void setGameConsole(String gameConsole) {
this.gameConsole = gameConsole;
}
}
/****************VideoGameManager**************/
import java.io.BufferedReader;
import java.io.FileOutputStream;
import java.io.FileReader;
import java.io.PrintWriter;
import java.util.LinkedList;
import java.util.Scanner;
public class VideoGameManager {
Scanner keyboard = new Scanner(System.in);
//array list
private LinkedList games;
private LinkedList currentResult;
//tab
private static final String delim = \"\\t\";
//Default constructor
public VideoGameManager(){
games = new LinkedList();
}
public void ReadGameFile(String filePath)
{
//This reconstructs a new instance of the VideGame Linked list.
//This is done to clear the Linked list
games = new LinkedList();
BufferedReader br = null;
try
{
//Create a new file Scanner
//Scanner fileScanner = new Scanner(new File(fileName));
br=new BufferedReader(new FileReader(filePath));
String fileLine=\"\";
//Reads each line in the file one-by-one
while((fileLine=br.readLine())!=null)
{
//That line is then split using the delimiter (\\t)
String[] splitStrings = fileLine.split(delim);
//If the newly created array is not 2 items in length then
//that line is not correctly formatted and should be ignored.
if(splitStrings.length != 2)
continue;
String gameName = splitStrings[0];//The first element is the game name
String console = splitStrings[1];//Next is the console
VideoGame newVideoGame = new VideoGame(gameName,console);
games.add(newVideoGame);//Added to the Linked list
}
br.close();
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
public void WriteToGameFile(String fileName, boolean append)
{
if(games == null)//if the file name is null then return
return;
try
{
//Creates the new instance of a print writer
PrintWriter fileWriter = new PrintWriter(new FileOutputStream(fileName,append));
for(VideoGame aVideoGame : games)
{
//Prints to the file
fileWriter.println(aVideoGame.getGameName()+delim+
aVideoGame.getGameConsole());
}
fileWriter.close();
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
//returns an array list of video games based on a search criterion (name and console)
public void PrintSearchCriteria(String nameInput, String consoleInput)
{
currentResult=new LinkedList();
for(VideoGame aVideoGame : games)
{
String gameName=aVideoGame.getGameName().toLowerCase();
String consoleName=aVideoGame.getGameConsole().toLowerCase();
if(gameName.contains(nameInput.to.
Use the order of operations to simplify the expression. Parenthese.pdf
Use the order of operations to simplify the expression.
Parentheses: 5
Solution
Use the order of operations to simplify the expression.
Parentheses: 5.
The RASopathies are a clinically defined group of medical genetic sy.pdf
The RASopathies are a clinically defined group of medical genetic syndromes caused by
germline mutations in genes that encode components or regulators of the Ras/mitogen-activated
protein kinase (MAPK) pathway (Raue, 2013). Further, since theRASopathies: developmental
syndromes of Ras/MAPK pathway results from duplication ofchromosomeband 12q24.11q24.23
it results in apparent ...X-linkedintellectual disability, (Michelson)
Solution
The RASopathies are a clinically defined group of medical genetic syndromes caused by
germline mutations in genes that encode components or regulators of the Ras/mitogen-activated
protein kinase (MAPK) pathway (Raue, 2013). Further, since theRASopathies: developmental
syndromes of Ras/MAPK pathway results from duplication ofchromosomeband 12q24.11q24.23
it results in apparent ...X-linkedintellectual disability, (Michelson).
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ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BA...
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Chaatit, Mascioni, and Rosenthal de ned nite Baire index for a bounded real-valued function f on a separable metric space, denoted by i(f), and proved that for any bounded functions f and g of nite Baire index, i(h) i(f) + i(g), where h is any of the functions f + g, fg, f ˅g, f ^ g. In this paper, we prove that the result is optimal in the following sense : for each n; k < ω, there exist functions f; g such that i(f) = n, i(g) = k, and i(h) = i(f) + i(g).IRJET - Fuzzy Soft Hyperideals in Meet Hyperlattices
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International Journal of Mathematics and Statistics Invention (IJMSI)
International Journal of Mathematics and Statistics Invention (IJMSI) is an international journal intended for professionals and researchers in all fields of computer science and electronics. IJMSI publishes research articles and reviews within the whole field Mathematics and Statistics, new teaching methods, assessment, validation and the impact of new technologies and it will continue to provide information on the latest trends and developments in this ever-expanding subject. The publications of papers are selected through double peer reviewed to ensure originality, relevance, and readability. The articles published in our journal can be accessed online.
TOPOLOGY 541Let M be a set with two members a and b. Define the fu.pdf
TOPOLOGY 541
Let M be a set with two members a and b. Define the function D : M × M as follows: D(a,a) =
D(b,b) = 0,D(a,b) = D(b,a) = r where r is a positive real number. Prove that (M, D) is a metric
space.
Solution
Definition 1.1 (Metric). A metric, or distance function, on a set X is a mapping d : X × X R such
that • d(x, y) 0 for all x, y X, and d(x, y) = 0 if and only if x = y. • d(x, y) = d(y, x) for all x, y
X. • d(x, z) d(x, y) + d(y, z) for all x, y, z X. We call (X, d) a metric space. Definition 1.2 (Open
ball). Let (X, d) be a metric space. For x X and > 0, the set Bd(x, ) defined by Bd(x, ) = {y X |
d(x, y) < } is callen an open ball in the set X. Definition 1.3 (Open sets in metric spaces). Let (X,
d) be a metric space and let U be any subset of X. Then U is called an open set in X if every
point of U is an interior point of U; that is, for any a U, there is an open ball B(a, ) such that B(a,
) U. Definition 1.4 (Properties of open sets). Let (X, d) be a metric space. • and X are open. •
The union of an arbitrary collection of open sets is open. • The intersection of a finite number of
open sets is open. Definition 1.5 (Closed set). A subset A of a metric space (X, d) is closed if it’s
complement X\\A is open in X. Definition 1.6 (Properties of closed sets). Let (X, d) be a metric
space. • and X are closed. • The union of an finite collection of closed sets is closed. • The
intersection of an arbitrary number of closed sets is closed. Definition 1.7 (Limit point of a
subset). Let (X, d) be a metric space and let A be a subset of X. Then a point x in X is a limit
point of A if every open ball B(x, ) contains at least one point of A. The set of all limit points of
A is called the derived set A . Definition 1.8 (Closure of a set). Let (X, d) be a metric space and
let A X. Then the set consisting of A and its limit points is called the closure of A, denoted A. A
= A A
Corollary 1.19. We note that A is dense in X if and only if for any x X and > 0, there is a point
a A such that d(x, a) < . 1.2. Subspaces. Definition 1.20 (Open sets in a subspace). Let (X, d) be
a metric space and (Y, dY ) be a metric subspace of (X, d). Let G be a subset of Y . Then G is
open in Y if and only if, for any x G, there is an open ball B(x, ) in X such that B(x, ) Y G A
subset H of Y is closed in Y if its complement G = Y \\H of H is open in Y . Theorem 1.21
(Open sets in a metric subspace). Let (Y, dY ) be a metric subspace of a metric space (X, d), and
let G Y . Then G is open in Y if and only if there exists an open subset U in X such that G = U
Y . 1.3. Convergence in a Metric Space. Definition 1.22 (Convergence). A sequence (xn) in a
metric space (X, d) is said to converge to a point x X if for any > 0, there exists N such that n >
N implies d(xn, x) < The point x is called a limit of the sequence (xn) Corollary 1.23. A
sequence (xn) in a metric space (X, d) is said to converge to a point x X if any open ball B(x, )
contains almost all xn. Theorem 1..
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VideoGamepublic class VideoGam.pdf
/*****************VideoGame*****************/
public class VideoGame {
// instance variables
private String gameName;
private String gameConsole;
// Default constructor
public VideoGame() {
}
// parameterized constructor
public VideoGame(String gameName, String gameConsole) {
this.gameName = gameName;
this.gameConsole = gameConsole;
}
/* getters and setters */
public String getGameName() {
return gameName;
}
public void setGameName(String gameName) {
this.gameName = gameName;
}
public String getGameConsole() {
return gameConsole;
}
public void setGameConsole(String gameConsole) {
this.gameConsole = gameConsole;
}
}
/****************VideoGameManager**************/
import java.io.BufferedReader;
import java.io.FileOutputStream;
import java.io.FileReader;
import java.io.PrintWriter;
import java.util.LinkedList;
import java.util.Scanner;
public class VideoGameManager {
Scanner keyboard = new Scanner(System.in);
//array list
private LinkedList games;
private LinkedList currentResult;
//tab
private static final String delim = \"\\t\";
//Default constructor
public VideoGameManager(){
games = new LinkedList();
}
public void ReadGameFile(String filePath)
{
//This reconstructs a new instance of the VideGame Linked list.
//This is done to clear the Linked list
games = new LinkedList();
BufferedReader br = null;
try
{
//Create a new file Scanner
//Scanner fileScanner = new Scanner(new File(fileName));
br=new BufferedReader(new FileReader(filePath));
String fileLine=\"\";
//Reads each line in the file one-by-one
while((fileLine=br.readLine())!=null)
{
//That line is then split using the delimiter (\\t)
String[] splitStrings = fileLine.split(delim);
//If the newly created array is not 2 items in length then
//that line is not correctly formatted and should be ignored.
if(splitStrings.length != 2)
continue;
String gameName = splitStrings[0];//The first element is the game name
String console = splitStrings[1];//Next is the console
VideoGame newVideoGame = new VideoGame(gameName,console);
games.add(newVideoGame);//Added to the Linked list
}
br.close();
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
public void WriteToGameFile(String fileName, boolean append)
{
if(games == null)//if the file name is null then return
return;
try
{
//Creates the new instance of a print writer
PrintWriter fileWriter = new PrintWriter(new FileOutputStream(fileName,append));
for(VideoGame aVideoGame : games)
{
//Prints to the file
fileWriter.println(aVideoGame.getGameName()+delim+
aVideoGame.getGameConsole());
}
fileWriter.close();
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
//returns an array list of video games based on a search criterion (name and console)
public void PrintSearchCriteria(String nameInput, String consoleInput)
{
currentResult=new LinkedList();
for(VideoGame aVideoGame : games)
{
String gameName=aVideoGame.getGameName().toLowerCase();
String consoleName=aVideoGame.getGameConsole().toLowerCase();
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Use the order of operations to simplify the expression. Parenthese.pdf
Use the order of operations to simplify the expression.
Parentheses: 5
Solution
Use the order of operations to simplify the expression.
Parentheses: 5.
The RASopathies are a clinically defined group of medical genetic sy.pdf
The RASopathies are a clinically defined group of medical genetic syndromes caused by
germline mutations in genes that encode components or regulators of the Ras/mitogen-activated
protein kinase (MAPK) pathway (Raue, 2013). Further, since theRASopathies: developmental
syndromes of Ras/MAPK pathway results from duplication ofchromosomeband 12q24.11q24.23
it results in apparent ...X-linkedintellectual disability, (Michelson)
Solution
The RASopathies are a clinically defined group of medical genetic syndromes caused by
germline mutations in genes that encode components or regulators of the Ras/mitogen-activated
protein kinase (MAPK) pathway (Raue, 2013). Further, since theRASopathies: developmental
syndromes of Ras/MAPK pathway results from duplication ofchromosomeband 12q24.11q24.23
it results in apparent ...X-linkedintellectual disability, (Michelson).
The Brocas area is located in the frontal part of the brain on the.pdf
The Broca\'s area is located in the frontal part of the brain on the left hemisphere. ... Thus, when
Broca\'s aphasia occurs, it can have a debilitating effect on the person\'s communication skills.
The damage forces people to speak in fragmented sentences that include only nouns and verbs.
The Wernicke\'s area is located in the temporal lobe on the left side of the brain and is
responsible for the comprehension of speech (Broca\'s area is related to the production of
speech). ... When this area of the brain is damaged, a disorder known as Wernicke\'s aphasia can
result
The ventral stream is associated with object recognition and form representation. Also described
as the \"what\" stream, it has strong connections to the medial temporal lobe(which stores long-
term memories), the limbic system (which controls emotions), and the dorsal stream (which
deals with object locations and motion).All the areas in the ventral stream are influenced by
extraretinal factors in addition to the nature of the stimulus in their receptive field. These factors
include attention, working memory, and stimulus salience. Thus the ventral stream does not
merely provide a description of the elements in the visual world—it also plays a crucial role in
judging the significance of these elements.Damage to the ventral stream can cause inability to
recognize faces or interpret facial expression
The function of the dorsal pathway is to map auditory sensory representations onto articulatory
motor representations. Hickok & Poeppel claim that the dorsal pathway is necessary because,
\"learning to speak is essentially a motor learning task. The primary input to this is sensory,
speech in particular. So, there must be a neural mechanism that both codes and maintains
instances of speech sounds, and can use these sensory traces to guide the tuning of speech
gestures so that the sounds are accurately reproduced.Conduction aphasia affects a subject\'s
ability to reproduce speech (typically by repetition), though it has no influence on the subject\'s
ability to comprehend spoken language. This shows that conduction aphasia must reflect an
impairment of the ventral pathway but instead of the dorsal pathway. Hickok and Poeppel found
that conduction aphasia can be the result of damage, particularly lesions, to the Spt (Sylvian
parietal temporal). This is shown by the Spt\'s involvement in acquiring new vocabulary, for
while experiments have shown that most conduction aphasiacs can repeat high-frequency, simple
words, their ability to repeat low-frequency, complex words is impaired.
Hippocampal damage can result in anterograde amnesia: loss of ability to form new memories,
although older memories may be safe. Thus, someone who sustains an injury to the hippocampus
may have good memory of his childhood and the years before the injury, but relatively little
memory for anything that happened since.
Solution
The Broca\'s area is located in the frontal part of the brain on the left hemispher.
the conflict of internet surveilance versus individuals and privacy .pdf
the conflict of internet surveilance versus individuals and privacy rights have been rage for years
and mjor internet and technology companies including microsoft , google ,ebay and intel are
seeking to update the electronic communication privacy act ie ECPA and the ECPA prevents
internet service providers ie ISP\'s from releasing personal information including names
,addresses , phone numbers , credit card and account details about their subscribers without the
court permission and protects the content of the files stored on individuals personal computers
with the advent of the cloud storage and ubiquitous mobile phone usage since ECPA was passed
in 1986 prosecutors and general public are at odds about whether police should be required to
obtain a search warrant to retrieve an individuals email communications or to identify the
location of smart phones or mobile devices so in this way and yes the congress should enact laws
requiring isp\'s to formally track their customers activities in attempt to aid police investigating
crimes , attorneys litigating divorce and insurance fraud cases and the entertainment industry in
locating illegally downloaded files like tracking the hacker so inorder to protect the information
from the hackers congress enacted these laws.
Solution
the conflict of internet surveilance versus individuals and privacy rights have been rage for years
and mjor internet and technology companies including microsoft , google ,ebay and intel are
seeking to update the electronic communication privacy act ie ECPA and the ECPA prevents
internet service providers ie ISP\'s from releasing personal information including names
,addresses , phone numbers , credit card and account details about their subscribers without the
court permission and protects the content of the files stored on individuals personal computers
with the advent of the cloud storage and ubiquitous mobile phone usage since ECPA was passed
in 1986 prosecutors and general public are at odds about whether police should be required to
obtain a search warrant to retrieve an individuals email communications or to identify the
location of smart phones or mobile devices so in this way and yes the congress should enact laws
requiring isp\'s to formally track their customers activities in attempt to aid police investigating
crimes , attorneys litigating divorce and insurance fraud cases and the entertainment industry in
locating illegally downloaded files like tracking the hacker so inorder to protect the information
from the hackers congress enacted these laws..
Question Tell me about yourselfAnswer We begin by introducing .pdf
Question: Tell me about yourself:
Answer : We begin by introducing ourselves and then goes to qualifications ,achievements and
our qualities , positives and negatives.Thus I would start as ,\" My name is Alisha Singh I am
simple honest and hard working girl I live in joint family my father is businessman I have
completed all my basic and elementary education with goods grades and performance . I am also
having work experience of 2.5 years . My work is always appreciated by my colleagues and my
seniors . I do my work with full faith and confidence I make sure that is no error in my
performance .Just like any human being i also have my strengths and weakness. I am very
conversant with any person I meet I am very fond of teaching since childhood, i have done
volunteering work of teaching students in NGOs wherein I have been loved by the kids.\"
1.Utilize your notes! Have a copy of your CV to hand, print the job specification, have the
company website open on your laptop, bring up the interviewer’s LinkedIn profile, etc. There is
no excuse for being under prepared!
2. Make notes throughout the interview, and use this information to ask questions at the end. If
you progress to the next stage, use this information to help you prepare.
3. You can be more flexible with times – participate from home / work / abroad, etc. However,
don’t commit to a telephone interview if you are likely to be rushed or interrupted.
4.No cost of travel. Screening for basic requirement is done more easily.
Cons of telephone interviews
1. It is more difficult to build rapport over the phone. It makes a big difference to how you
sound.
2. Telephone interviews are often shorter, so you have less time to sell yourself. Be concise as
you have a lot of ground to cover. A common starter for telephone interviews is “tell me your
‘story’ in 5 minutes” – stick to the time allocated, and make sure you’ve rehearsed this. If you
can’t articulate your experience in 5 minutes then you need to get practicing! This is one of the
most common reasons why candidates don’t get past a phone screen .
Face to Face Interview :
1. It is much easier to build rapport in person, so get your personality across!
2. It is also much easier to read how the interview is going – look at their body language, do they
look interested? Do they ask additional probing questions?
3. You can see the environment that you would be working in which gives you a good sense as to
whether this is the right opportunity for you.
Cons :
1. Beware of interviewer bias. First impressions count for a lot – be early (leave ample time to
account for traffic, train delays, etc.), be presentable (dress for the environment) and on behalf of
interviewers everywhere. PLEASE don’t have a limp handshake.
2.Cost of travel – trains are expensive, but remember that this could be your future job, so it’s
worth it, right? If you are attending too many and it’s getting too expensive,
3. You can’t refer to as many notes during a face-to-face in.
Pls post the question again, i cant see any.SolutionPls post t.pdf
Pls post the question again, i cant see any.
Solution
Pls post the question again, i cant see any..
Please find the summary belowContemporary geographical conditions.pdf
Please find the summary below:
Contemporary geographical conditions reveal alot about the primitive states of fauna and flora.
In this regard, a very discrete observation has been made with respect to diversion and evolution
of fauna and flora of Central America. During past few decades, it has been observed that a close
association occurs between the reproductive traits of flora and the nature of fauna found in
Central American lowland forest areas. Researchers have observed that in last few centuries,
some large herbivore species have become extinct. There extinction has lead to significant
decline in the seed-dispersal and cross-pollination efficacy of the local plants of this area.
However, inhabitation of humans and colonization has resulted into rearing of large cattles such
as horses in these areas. The rearing of these large herbivores has played a crucial role in
restoring the floral diversity in these areas because these herbivores help in seed-dispersal and
sexual reproduction of plants by grazing and movements. Conseqeuntly, these observations has
lead to recognize the impact and capability of restoration of local extinct/near-extinct floral
diversity just by merely introducing large/small herbivores. Thus, a comparative analysis could
be made between artifical methods of floral-diversity restoration as compared to natural and
environmentally favourable methods.
Solution
Please find the summary below:
Contemporary geographical conditions reveal alot about the primitive states of fauna and flora.
In this regard, a very discrete observation has been made with respect to diversion and evolution
of fauna and flora of Central America. During past few decades, it has been observed that a close
association occurs between the reproductive traits of flora and the nature of fauna found in
Central American lowland forest areas. Researchers have observed that in last few centuries,
some large herbivore species have become extinct. There extinction has lead to significant
decline in the seed-dispersal and cross-pollination efficacy of the local plants of this area.
However, inhabitation of humans and colonization has resulted into rearing of large cattles such
as horses in these areas. The rearing of these large herbivores has played a crucial role in
restoring the floral diversity in these areas because these herbivores help in seed-dispersal and
sexual reproduction of plants by grazing and movements. Conseqeuntly, these observations has
lead to recognize the impact and capability of restoration of local extinct/near-extinct floral
diversity just by merely introducing large/small herbivores. Thus, a comparative analysis could
be made between artifical methods of floral-diversity restoration as compared to natural and
environmentally favourable methods..
IFRS 13Fair Value Measurement applies to IFRSs that require or per.pdf
IFRS 13
Fair Value Measurement applies to IFRSs that require or permit fair value measurements or
disclosures and provides a single IFRS framework for measuring fair value and requires
disclosures about fair value measurement. The Standard defines fair value on the basis of an
\'exit price\' notion and uses a \'fair value hierarchy\', which results in a market-based, rather than
entity-specific, measurement.
IFRS 13 was originally issued in May 2011 and applies to annual periods beginning on or after 1
January 2013.
History of IFRS 13
Related Interpretations
Amendments under consideration by the IASB
In addition, the IASB has signalled an intention to conduct a post-implementation review of
IFRS 13, commencing in 2015.
Publications and resources
Summary of IFRS 13
Objective
IFRS 13: [IFRS 13:1]
IFRS 13 applies when another IFRS requires or permits fair value measurements or disclosures
about fair value measurements (and measurements, such as fair value less costs to sell, based on
fair value or disclosures about those measurements), except for: [IFRS 13:5-7]
Additional exemptions apply to the disclosures required by IFRS 13.
Key definitions
[IFRS 13:Appendix A]
The price that would be received to sell an asset or paid to transfer a liability in an orderly
transaction between market participants at the measurement date
A market in which transactions for the asset or liability take place with sufficient frequency and
volume to provide pricing information on an ongoing basis
The price that would be received to sell an asset or paid to transfer a liability
The use of a non-financial asset by market participants that would maximise the value of the
asset or the group of assets and liabilities (e.g. a business) within which the asset would be used
The market that maximises the amount that would be received to sell the asset or minimises the
amount that would be paid to transfer the liability, after taking into account transaction costs and
transport costs
The market with the greatest volume and level of activity for the asset or liability
Fair value hierarchy
Overview
IFRS 13 seeks to increase consistency and comparability in fair value measurements and related
disclosures through a \'fair value hierarchy\'. The hierarchy categorises the inputs used in
valuation techniques into three levels. The hierarchy gives the highest priority to (unadjusted)
quoted prices in active markets for identical assets or liabilities and the lowest priority to
unobservable inputs. [IFRS 13:72]
If the inputs used to measure fair value are categorised into different levels of the fair value
hierarchy, the fair value measurement is categorised in its entirety in the level of the lowest level
input that is significant to the entire measurement (based on the application of judgement). [IFRS
13:73]
Level 1 inputs
Level 1 inputs are quoted prices in active markets for identical assets or liabilities that the entity
can access at the measurement date. [IFRS 13:76]
A quoted ma.
Cyanobacteria does not compete with methanogens , rather it acts as .pdf
Cyanobacteria does not compete with methanogens , rather it acts as source of hydrogen for
methane formation. Halophiles and fermenters are non competitive with methanogens.
Solution
Cyanobacteria does not compete with methanogens , rather it acts as source of hydrogen for
methane formation. Halophiles and fermenters are non competitive with methanogens..
Correlation is the key in diversification of risk. If one asset does.pdf
Correlation is the key in diversification of risk. If one asset does well, the other does not.
Therefore, adding assets that do not behave like other assets in your portfolio is good and can
reduce risk. The two companies should have a correlation of less than 1.0. Even when we expand
the portfolio to many assets, correlation among assets remains the primary determinant of
portfolio risk. Lower correlations are associated with lower risk. Unfortunately, most assets have
high positive correlations. The challenge in diversifying risk is to find assets that have a
correlation that is much lower than +1.0.
For Diversified portfolios investor must have domestic stocks, domestic bonds, foreign stocks,
foreign bonds, real estate, cash, and other asset classes.The low correlations between stocks and
bonds are attractive for portfolio diversification. Similarly, including international securities in a
portfolio can also control portfolio risk
Solution
Correlation is the key in diversification of risk. If one asset does well, the other does not.
Therefore, adding assets that do not behave like other assets in your portfolio is good and can
reduce risk. The two companies should have a correlation of less than 1.0. Even when we expand
the portfolio to many assets, correlation among assets remains the primary determinant of
portfolio risk. Lower correlations are associated with lower risk. Unfortunately, most assets have
high positive correlations. The challenge in diversifying risk is to find assets that have a
correlation that is much lower than +1.0.
For Diversified portfolios investor must have domestic stocks, domestic bonds, foreign stocks,
foreign bonds, real estate, cash, and other asset classes.The low correlations between stocks and
bonds are attractive for portfolio diversification. Similarly, including international securities in a
portfolio can also control portfolio risk.
Ans.) Exosomes are layer enclosed vesicles effectively discharged in.pdf
Ans.) Exosomes are layer enclosed vesicles effectively discharged into the extracellular space,
whose substance reflects the physiological condition of the cells they begin from. These vesicles
are involved in cell-to-cell correspondence and exchange of naturally dynamic proteins, lipids,
and RNAs.
A ) Molecular composition of virus-modified exosomes- Exosomes are basically cytoplasm
encased in a lipid bilayer with uncovered at the outside of transmembrane proteins. Exosomes
may contain a wide range of biomolecules like proteins, sugars, lipids and furthermore nucleic
acid also. New sanitization strategies giving exceedingly immaculate arrangements of exosomes
have permitted the utilization of proteomic and atomic systems to comprehend the sub-atomic
piece of exosomes.
B) Both the virus-modified exosomes and typical eukaryotic cell membrane is made up of lipid
bilayer. The chemical composition of both the mentioned entities is almost same as both of them
are composed of proteins, sugars, lipids and nucleic acid. Exosomes are found inside the
eukaryotic cell, they are derived from the eukaryotic cell.
C) These virus-modified exosomes are used to target the RNA delivery, a new advancement in
the field of nanomedicine. Recent studies propose that a specific type of nano-sized lipid vesicles
called exosomes can consolidate and transport useful RNAs into target cells and may serve as an
appealing option. Confirmation is gathered that most pluricellular organism maintains exosome-
based interchanges by means of transfer between mRNA and miRNAs between the cells. We
found that viruses have discovered approaches to endeavor this correspondence. We talk about
disclosures in exosome biogenesis their physical properties, focusing on and transport
methodologies and how the information of exosome creation in virus tainted cells could impel
their entrance into clinical trials.
Solution
Ans.) Exosomes are layer enclosed vesicles effectively discharged into the extracellular space,
whose substance reflects the physiological condition of the cells they begin from. These vesicles
are involved in cell-to-cell correspondence and exchange of naturally dynamic proteins, lipids,
and RNAs.
A ) Molecular composition of virus-modified exosomes- Exosomes are basically cytoplasm
encased in a lipid bilayer with uncovered at the outside of transmembrane proteins. Exosomes
may contain a wide range of biomolecules like proteins, sugars, lipids and furthermore nucleic
acid also. New sanitization strategies giving exceedingly immaculate arrangements of exosomes
have permitted the utilization of proteomic and atomic systems to comprehend the sub-atomic
piece of exosomes.
B) Both the virus-modified exosomes and typical eukaryotic cell membrane is made up of lipid
bilayer. The chemical composition of both the mentioned entities is almost same as both of them
are composed of proteins, sugars, lipids and nucleic acid. Exosomes are found inside the
eukaryotic cell, they are derived.
The java program that display a menu of choices and p.pdf
/**
* The java program that display a menu of choices
* and prompt user choice and then call the corresponding
* method and then prints the results to console.
* */
//TextOperations.java
import java.util.Scanner;
public class TextOperations {
private static Scanner scanner=new Scanner(System.in);
public static void main(String[] args)
{
char ch;
String text;
while(true)
{
ch=printMenu();
switch (ch)
{
case \'c\':
System.out.println(\"Enter a sample text:\");
text=scanner.nextLine();
int numChars=getNumOfNonWSCharacters(text);
System.out.println(\"Number of characters : \"+numChars);
break;
case \'w\':
System.out.println(\"Enter a sample text:\");
text=scanner.nextLine();
int numWords=getNumOfWords(text);
System.out.println(\"Number of words:\"+numWords);
break;
case \'f\':
System.out.println(\"Enter a sample text:\");
text=scanner.nextLine();
System.out.println(\"Enter a word or phrase to be found:\");
String search=scanner.nextLine();
int count=findText(text,search);
System.out.println(\"more \"+search+\" instances:\"+count);
break;
case \'r\':
System.out.println(\"Enter a sample text:\");
text=scanner.nextLine();
String result=replaceExclamation(text);
System.out.println(\"Result : \"+result);
break;
case \'s\':
System.out.println(\"Enter a sample text:\");
text=scanner.nextLine();
String shortenText=shortenSpace(text);
System.out.println(shortenText);
break;
case \'q\':
System.out.println(\"Good bye!\");
System.exit(0);
}
}
}
/*
* The method that replaces 2 or more spaces
* witht single space
* */
private static String shortenSpace(String text) {
String temp=text;
return temp.trim().replaceAll(\" +\", \" \");
}
/**
* The method that takes text as input and
* replaces the exclamation with .
* */
private static String replaceExclamation(String text) {
String temp=text;
temp=temp.replace(\'!\', \'.\');
return temp;
}
/**
* The method that takes source text and search
* string and returns the number of search words
* find in the source text
* */
private static int findText(String text, String search) {
String temp=text.trim();
int count=0;
temp=temp.trim();
String[] arr=temp.split(\" \");
for (String string : arr)
{
if(string.contains(search))
count++;
}
return count;
}
/**
* The method that returns the number of words in a text
* */
private static int getNumOfWords(String text) {
String temp=text.trim();
temp=temp.trim();
String[] arr=temp.split(\" \");
return arr.length;
}
/**
* The method that returns the number of non-white
* space characters in a given string
* */
private static int getNumOfNonWSCharacters(String text) {
String temp=text.trim();
int countChars=0;
temp=temp.trim();
String[] arr=temp.split(\" \");
for (String string : arr)
{
countChars+=string.length();
}
return countChars;
}
/**
* The method prompts user to enter user choice and returns
* user choice
* */
private static char printMenu() {
char ch;
do
{
System.out.println(\"***MENU***\");
System.out.println(\"c - Number of non-whitespace characters \");
System.out.println(\.
ASSETS 2016 2015 Current assets Cash 3251 3407 Accouns re.pdf
ASSETS 2016 2015 Current assets: Cash 3251 3407 Accouns receivable 4777 5801
Inventory 12438 13802 Total current assets 20466 23010 Fixed assets: Net plant and
equipment 49749 74097 Total assets 70215 97107 LIABILITIES & OWNERS\' EQUITY
Current liabilities: Accounts payable 2143 2580 Other 88 105 Notes payable 1740 2096
3971 4781 Long term debt 13600 16360 Total debt 17571 21141 Owners\' equity:
Common stock and paid in surplus 37000 37000 Accumulated earnings 15644 38966 Total
owners\' equity 52644 75966 Total liabilities and owners\' equity 70215 97107 Current ratio =
Current assets/Current liabilities = 5.15 4.81 (20466/3971) (23010/4781) Quick ratio =
(Cash+Accounts receivable)/Current liabilities = 2.02 1.93 (3251+4777)/3971))
(3407+5801)/4781)) Cash ratio = (Cash+marketable securities)/Current liabilities 0.82 0.71
(3251/3971) (3407/4781) Total asset turnover = Sales/Average total assets =
186970/((70215+97107)/2)) = 2.23 Inventory turnover = Cost of goods sold/Average inventory
= 126003/((12438+13802)/2)) = 9.60 Receivables turnover = Sales/Average receivables =
186970/((4777+5801)/2)) = 35.35 Total debt ratio = Total debt/Total assets 0.25 0.22 Debt
equity ratio = Total debt/Total equity 0.33 0.28 Equity multiplier =Total assets/Total equity
1.33 1.28 Times interest earned = EBIT/Interest expense = 55614/1450 = 38.35 Cash
coverage ratio = (EBIT+Depreciation)/Interest expense = (55614+5353)/1450 = 42.05 Profit
margin = Net income/Sales = 35207/186970 = 18.83% Return on assets = Net income/Average
assets = 35207/((70215+97107)/2)) = 42.08% Return on equity = Net income/Average equity =
35207/((52644+75966))/2) = 54.75%
Solution
ASSETS 2016 2015 Current assets: Cash 3251 3407 Accouns receivable 4777 5801
Inventory 12438 13802 Total current assets 20466 23010 Fixed assets: Net plant and
equipment 49749 74097 Total assets 70215 97107 LIABILITIES & OWNERS\' EQUITY
Current liabilities: Accounts payable 2143 2580 Other 88 105 Notes payable 1740 2096
3971 4781 Long term debt 13600 16360 Total debt 17571 21141 Owners\' equity:
Common stock and paid in surplus 37000 37000 Accumulated earnings 15644 38966 Total
owners\' equity 52644 75966 Total liabilities and owners\' equity 70215 97107 Current ratio =
Current assets/Current liabilities = 5.15 4.81 (20466/3971) (23010/4781) Quick ratio =
(Cash+Accounts receivable)/Current liabilities = 2.02 1.93 (3251+4777)/3971))
(3407+5801)/4781)) Cash ratio = (Cash+marketable securities)/Current liabilities 0.82 0.71
(3251/3971) (3407/4781) Total asset turnover = Sales/Average total assets =
186970/((70215+97107)/2)) = 2.23 Inventory turnover = Cost of goods sold/Average inventory
= 126003/((12438+13802)/2)) = 9.60 Receivables turnover = Sales/Average receivables =
186970/((4777+5801)/2)) = 35.35 Total debt ratio = Total debt/Total assets 0.25 0.22 Debt
equity ratio = Total debt/Total equity 0.33 0.28 Equity multiplier =Total assets/Total equity
1.33 1.28 Times interest earned = EBIT/Interest expense = 55614/.
ACA (Affordable care Act) signed by Obama on 23 march 2010. .pdf
ACA (Affordable care Act) signed by Obama on 23 march 2010. Putting
Information for Consumers Online So that consumers can compare health insurance coverage
options and pick the coverage that works for them. Prohibiting Denying Coverage of Children
Based on PreExisting Conditions The health care law includes new rules to prevent insurance
companies from denying coverage to children under the age of 19 due to a pre-existing
condition. Prohibiting Insurance Companies from Rescinding Coverage In the past, insurance
companies could search for an error, or other technical mistake, on a customer\'s application and
use this error to deny payment for services when he or she got sick. But now this is illegal. After
media reports cited incidents of breast cancer patients losing coverage, insurance companies
agreed to end this practice immediately. Eliminating Lifetime Limits on Insurance Coverage
Insurance companies will be prohibited from imposing lifetime dollar limits on essential
benefits, like hospital stays. Regulating Annual Limits on Insurance Coverage Insurance
companies\' use of annual dollar limits on the amount of insurance coverage a patient may
receive will be restricted for new plans in the individual market and all group plans. In 2014, the
use of annual dollar limits on essential benefits like hospital stays will be banned for new plans
in the individual market and all group plans. Appealing Insurance Company Decisions
Provides consumers with a way to appeal coverage determinations or claims to their insurance
company, and establishes an external review process. Establishing Consumer Assistance
Programs in the States States that apply ACA receive federal grants to help set up or expand
independent offices to help consumers navigate the private health insurance system. These
programs help consumers file complaints and appeals; enroll in health coverage; and get
educated about their rights and responsibilities in group health plans or individual health
insurance policies. The programs will also collect data on the types of problems consumers have,
and file reports with the U.S. Department of Health and Human Services to identify trouble spots
that need further oversight. Improving Quality and lowering costs Both this head get
amended from time to time so that consumer receive best to best service. Increasing Access to
Affordable Care Hoe ACA Affects Reiumburshment Short Term Effects:
The most immediate expected effect of the ACA for providers is a sudden rise in patient
populations. Millions of Americans are expected to obtain coverage under the ACA
Payers are required to cover more than ever,under the ACA, individual and small group health
plans are required to cover 10 essential health benefits Long Term Effects:
Changing payment and care models,biggest changes in healthcare right now are the new fee-for-
value payment models that are replacing traditional fee-for-service programs Through
Medicare and Medicaid, the government has been .
we have, V1 T1= V2T2 Volume = V2 = 12.126 lit.pdf
we have, V1/ T1= V2/T2 Volume = V2 = 12.126 liter
Solution
we have, V1/ T1= V2/T2 Volume = V2 = 12.126 liter.
Should just be proportional .0951.28=x1 .074=.pdf
Should just be proportional: .095/1.28=x/1 .074=x
Solution
Should just be proportional: .095/1.28=x/1 .074=x.
only C represents the pair of resonance structure.pdf
only C represents the pair of resonance structures.. B is pair of tautomers...
Solution
only C represents the pair of resonance structures.. B is pair of tautomers....
pH =1.074 pH = - log(concentration of H+) = -log .pdf
pH =1.074 pH = - log(concentration of H+) = -log (8.43E-02) = 1.074
Solution
pH =1.074 pH = - log(concentration of H+) = -log (8.43E-02) = 1.074.
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the conflict of internet surveilance versus individuals and privacy .pdf
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Question Tell me about yourselfAnswer We begin by introducing .pdf
Question Tell me about yourselfAnswer We begin by introducing .pdf
Pls post the question again, i cant see any.SolutionPls post t.pdf
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Please find the summary belowContemporary geographical conditions.pdf
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only C represents the pair of resonance structure.pdf
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Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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(a).Let me show that H is a subgroup. Then you can make the identica.pdf
- 1. (a).Let me show that H is a subgroup. Then you can make the identical argument and show that K is. e = (e, e) is in H let (x, e) be in H. G1 group so x^-1 in G1. Now (x^-1, e) = (x,e)^-1 since (x, e)(x^-1, e) = (xx^- 1, e) = (e, e) which is the identity in G let (x, e), (y, e) be in H. Then (x,e)(y,e) = (xy,ee) = (xy,e) is in H. Therefore H is a subgroup of G b) Let me show that KH=G. You can make the identical argument to show that HK=G. Let k =(e, y) be in K, h=(x, e) in H. Then kh = (e,y)(x,e) = (ex,ye)=(x,y) is in G, so KH subset of G Let (x,y) be in G. Then (e,y) is in K and (x,e) is in H, and (e,y)(x,e) = (x,y). So G is a subset of KH. Therefore G=KH Solution (a).Let me show that H is a subgroup. Then you can make the identical argument and show that K is. e = (e, e) is in H let (x, e) be in H. G1 group so x^-1 in G1. Now (x^-1, e) = (x,e)^-1 since (x, e)(x^-1, e) = (xx^- 1, e) = (e, e) which is the identity in G let (x, e), (y, e) be in H. Then (x,e)(y,e) = (xy,ee) = (xy,e) is in H. Therefore H is a subgroup of G b) Let me show that KH=G. You can make the identical argument to show that HK=G. Let k =(e, y) be in K, h=(x, e) in H. Then kh = (e,y)(x,e) = (ex,ye)=(x,y) is in G, so KH subset of G Let (x,y) be in G. Then (e,y) is in K and (x,e) is in H, and (e,y)(x,e) = (x,y). So G is a subset of KH. Therefore G=KH