One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
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One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever, the tongue appears furred, with centrally coated and a red margin around. There are other illness also which shows coated tongue, but with different appearance. In malaria, the tongue is uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In typhoid, the patch is centrally located on the tongue surrounded by a red margin. During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur, which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip and edges and completely disappears by the thid week of infection, leaving a glazed, dry or smooth red tongue. By observing the coat pattern, the Doctor would have diagnosed the fever well in advance. Solution One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever, the tongue appears furred, with centrally coated and a red margin around. There are other illness also which shows coated tongue, but with different appearance. In malaria, the tongue is uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In typhoid, the patch is centrally located on the tongue surrounded by a red margin. During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur, which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip and edges and completely disappears by the thid week of infection, leaving a glazed, dry or smooth red tongue. By observing the coat pattern, the Doctor would have diagnosed the fever well in advance..
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both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:
Solution
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:.
Introduction One of the key goals for the Windows Subsystem for Li.pdf
Introduction
One of the key goals for the Windows Subsystem for Linux is to allow users to work with their
files as they would on Linux, while giving full interoperability with files the user already has on
their Windows machine. Unlike a virtual machine, where you have to use network shares or
other solutions to share files between the host and guest OS, WSL has direct access to all your
Windows drives to allow for easy interop.
Windows file systems differ substantially from Linux file systems, and this post looks into how
WSL bridges those two worlds.
File systems on Linux
Linux abstracts file systems operations through the Virtual File System (VFS), which provides
both an interface for user mode programs to interact with the file system (through system calls
such as open, read, chmod, stat, etc.) and an interface that file systems have to implement. This
allows multiple file systems to coexist, providing the same operations and semantics, with VFS
giving a single namespace view of all these file systems to the user.
File systems are mounted on different directories in this namespace. For example, on a typical
Linux system your hard drive may be mounted at the root, /, with directories such as /dev, /proc,
/sys, and /mnt/cdrom all mounting different file systems which may be on different devices.
Examples of file systems used on Linux include ext4, rfs, FAT, and others.
VFS implements the various system calls for file system operations by using a number of data
structures such as inodes, directory entries and files, and related callbacks that file systems must
implement.
Inodes
The inode is the central data structure used in VFS. It represents a file system object such as a
regular file, directory, symbolic link, etc. An inode contains information about the file type, size,
permissions, last modified time, and other attributes. For many common Linux disk file systems
such as ext4, the on-disk data structures used to represent file metadata directly correspond to the
inode structure used by the Linux kernel.
While an inode represents a file, it does not represent a file name. A single file may have
multiple names, or hard links, but only one inode.
File systems provide a lookup callback to VFS which is used to retrieve an inode for a particular
file, based on the parent inode and the child name. File systems must implement a number of
other inode operations such as chmod, stat, open, etc.
Directory entries
VFS uses a directory entry cache to represent your file system namespace. Directory entries only
exist in memory, and contain a pointer to the inode for the file. For example, if you have a path
like /home/user/foo, there is a directory entry for home, user, and foo, each with a pointer to an
inode. Directory entries are cached for fast lookup, but if an entry is not yet in the cache, the
inode lookup operation is used to retrieve the inode from the file system so a new directory entry
can be created.
File objects
When an inode is opened, .
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.JButton;
import javax.swing.JColorChooser;
import javax.swing.JFrame;
import javax.swing.JMenu;
import javax.swing.JMenuBar;
import javax.swing.JMenuItem;
import javax.swing.JToolBar;
public class DrawApp extends JFrame {
//declaring data fields
private Drawable selectedFig = null;
//default border color
private Color presentBorderColor = Color.BLACK;
private Color presentInteriorrColor = Color.WHITE;
private Drawable[] theFigKinds;
// Constructor
private DrawApp(String args[]) {
//title
super(\"Draw App\");
//frame size
setSize(650, 650);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
loadFigKinds(args);
JMenu drawingToolMenu = createDrawingToolMenu();
JToolBar drawingToolToolBar = createDrawingToolToolBar();
selectedFig = theFigKinds[0];
JMenu colorMenu = createColorMenu();
JMenuBar menuBar = new JMenuBar();
menuBar.add(colorMenu);
menuBar.add(drawingToolMenu);
setJMenuBar(menuBar);
getContentPane().add(drawingToolToolBar, BorderLayout.EAST);
getContentPane().add(new DrawPanel(this),
BorderLayout.CENTER);
}
// creating drawing tool menu
private JMenu createDrawingToolMenu() {
JMenu drawingToolMenu = new JMenu(\"Drawing Tool\");
for (int i = 0; i < theFigKinds.length; i++) {
// create a menu item for this figure kind
JMenuItem item = new JMenuItem(theFigKinds[i].getName());
// set the action listener
item.addActionListener(new ToolSelector(theFigKinds[i]));
// add the item to the menu
drawingToolMenu.add(item);
}
return drawingToolMenu;
}
// creating menu tool bar
JToolBar createDrawingToolToolBar() {
JToolBar drawingToolToolBar = new JToolBar(JToolBar.VERTICAL);
for (int i = 0; i < theFigKinds.length; i++) {
JButton aButton = new JButton(theFigKinds[i].getIcon(16));
aButton.addActionListener(new ToolSelector(theFigKinds[i]));
drawingToolToolBar.add(aButton);
}
return drawingToolToolBar;
}
//creating color menu
private JMenu createColorMenu() {
JMenu colorMenu = new JMenu(\"Colors\");
JMenuItem setInteriorColor
= new JMenuItem(\"Set Interior Color\");
setInteriorColor.addActionListener(new SetInteriorColor());
colorMenu.add(setInteriorColor);
JMenuItem setBorderColor = new JMenuItem(\"Set Border Color\");
setBorderColor.addActionListener(new SetBorderColor());
colorMenu.add(setBorderColor);
return colorMenu;
}
public Drawable getSelectedFig() {
return selectedFig;
}
public static void main(String args[]) {
DrawApp drawApp = new DrawApp(args);
drawApp.setVisible(true);
}
// Inner Classes
private class ToolSelector implements ActionListener {
// Data Fields
private Drawable desiredFig;
// Constructor
public ToolSelector(Drawable desiredFig) {
this.desiredFig = desiredFig;
}
// Methods
@Override
public void actionPerformed(ActionEvent e) {
selectedFig = desiredFig;
}
}
private class SetInteriorColor implements ActionListener {
@Override
public void actionPerformed(ActionEvent e) {
presentInteriorrColo.
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
More Related Content
More from anwarfoot
0 to -1 Solution 0 to -1 .pdf
This very short document contains a range from 0 to -1 and the word "Solution" but provides no other context or explanation. It is unclear what problem or question this brief statement is providing a solution for.
phosphate group and deoxyribose The groups are .pdf
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder.
Solution
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
Osmosis .pdf
Osmosis is the spontaneous net movement of solvent molecules through a selectively permeable membrane into a region of higher solute concentration in order to dilute the solute and equalize the solute concentrations on both sides of the membrane. The process of osmosis is important for many physiological processes in both plants and animals and is used in industrial applications such as desalination, reverse osmosis water purification, and separation of molecules by their molecular weight. Osmosis plays a key role in homeostasis of organisms by regulating water balance across biological membranes and the flow of nutrients and waste.
in case of solid oxygen the atoms of oxygen are s.pdf
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a
covalent bond between them so there is a dipole dipole bond
Solution
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a
covalent bond between them so there is a dipole dipole bond.
Yeast is a microorganism and doesnt itself rise.pdf
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes
Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes
the bread or cake \"rise\".
Solution
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes
Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes
the bread or cake \"rise\"..
First, lets analyze what each component of the .pdf
First, let\'s analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how
to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa +
log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just
determined. You can calculate pKa from the given Ka value. Thus, you should be able to
calculate pH.
Solution
First, let\'s analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now,.
when they reach equilibrium te cell potential is .pdf
when they reach equilibrium te cell potential is zero
Solution
when they reach equilibrium te cell potential is zero.
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s.pdf
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq)
[UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011
E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V
Solution
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq)
[UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011
E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V.
D) Not B) because Cl-Benzene bond develops a doub.pdf
D) Not B) because Cl-Benzene bond develops a double bond character due to
resonance
Solution
D) Not B) because Cl-Benzene bond develops a double bond character due to
resonance.
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
Yes
Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there
Solution
Yes
Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there.
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2
Solution
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2.
lower.. think about it.. take a straw draw a line.pdf
lower.. think about it.. take a straw draw a line on it and fill it with water, then look
at it from different angles
Solution
lower.. think about it.. take a straw draw a line on it and fill it with water, then look
at it from different angles.
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
r= 1+ sin theta
dr/d = cos = 0
Therefore
= /2
Solution
r= 1+ sin theta
dr/d = cos = 0
Therefore
= /2.
From left to right iodocyclopropane; 1-bromo-3-m.pdf
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2-
dimethylpropane;
Solution
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2-
dimethylpropane;.
For inorganic compounds Chemical properties remai.pdf
For inorganic compounds Chemical properties remain same but colour or physical
appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is
colourless.Colour change can be explained by theories of Co ordination compounds. In organic
compounds ; chemical properties also change as dehydration changes the chemical nature;
C2H5OH---> C2H4+H2O
Solution
For inorganic compounds Chemical properties remain same but colour or physical
appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is
colourless.Colour change can be explained by theories of Co ordination compounds. In organic
compounds ; chemical properties also change as dehydration changes the chemical nature;
C2H5OH---> C2H4+H2O.
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
n=4
l=3
we know that
L=h/2(l(l+1))
L=3.7*10-34
Solution
n=4
l=3
we know that
L=h/2(l(l+1))
L=3.7*10-34.
Over forty interstellar molecules have been found, ranging from simp.pdf
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules
with over a dozen different atoms, including a few polycyclic. Their significance lies in that
planets are formed from such nebulas( area of the cloud), so this tells us what may be found.
Solution
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules
with over a dozen different atoms, including a few polycyclic. Their significance lies in that
planets are formed from such nebulas( area of the cloud), so this tells us what may be found..
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:
Solution
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:.
Introduction One of the key goals for the Windows Subsystem for Li.pdf
Introduction
One of the key goals for the Windows Subsystem for Linux is to allow users to work with their
files as they would on Linux, while giving full interoperability with files the user already has on
their Windows machine. Unlike a virtual machine, where you have to use network shares or
other solutions to share files between the host and guest OS, WSL has direct access to all your
Windows drives to allow for easy interop.
Windows file systems differ substantially from Linux file systems, and this post looks into how
WSL bridges those two worlds.
File systems on Linux
Linux abstracts file systems operations through the Virtual File System (VFS), which provides
both an interface for user mode programs to interact with the file system (through system calls
such as open, read, chmod, stat, etc.) and an interface that file systems have to implement. This
allows multiple file systems to coexist, providing the same operations and semantics, with VFS
giving a single namespace view of all these file systems to the user.
File systems are mounted on different directories in this namespace. For example, on a typical
Linux system your hard drive may be mounted at the root, /, with directories such as /dev, /proc,
/sys, and /mnt/cdrom all mounting different file systems which may be on different devices.
Examples of file systems used on Linux include ext4, rfs, FAT, and others.
VFS implements the various system calls for file system operations by using a number of data
structures such as inodes, directory entries and files, and related callbacks that file systems must
implement.
Inodes
The inode is the central data structure used in VFS. It represents a file system object such as a
regular file, directory, symbolic link, etc. An inode contains information about the file type, size,
permissions, last modified time, and other attributes. For many common Linux disk file systems
such as ext4, the on-disk data structures used to represent file metadata directly correspond to the
inode structure used by the Linux kernel.
While an inode represents a file, it does not represent a file name. A single file may have
multiple names, or hard links, but only one inode.
File systems provide a lookup callback to VFS which is used to retrieve an inode for a particular
file, based on the parent inode and the child name. File systems must implement a number of
other inode operations such as chmod, stat, open, etc.
Directory entries
VFS uses a directory entry cache to represent your file system namespace. Directory entries only
exist in memory, and contain a pointer to the inode for the file. For example, if you have a path
like /home/user/foo, there is a directory entry for home, user, and foo, each with a pointer to an
inode. Directory entries are cached for fast lookup, but if an entry is not yet in the cache, the
inode lookup operation is used to retrieve the inode from the file system so a new directory entry
can be created.
File objects
When an inode is opened, .
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.JButton;
import javax.swing.JColorChooser;
import javax.swing.JFrame;
import javax.swing.JMenu;
import javax.swing.JMenuBar;
import javax.swing.JMenuItem;
import javax.swing.JToolBar;
public class DrawApp extends JFrame {
//declaring data fields
private Drawable selectedFig = null;
//default border color
private Color presentBorderColor = Color.BLACK;
private Color presentInteriorrColor = Color.WHITE;
private Drawable[] theFigKinds;
// Constructor
private DrawApp(String args[]) {
//title
super(\"Draw App\");
//frame size
setSize(650, 650);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
loadFigKinds(args);
JMenu drawingToolMenu = createDrawingToolMenu();
JToolBar drawingToolToolBar = createDrawingToolToolBar();
selectedFig = theFigKinds[0];
JMenu colorMenu = createColorMenu();
JMenuBar menuBar = new JMenuBar();
menuBar.add(colorMenu);
menuBar.add(drawingToolMenu);
setJMenuBar(menuBar);
getContentPane().add(drawingToolToolBar, BorderLayout.EAST);
getContentPane().add(new DrawPanel(this),
BorderLayout.CENTER);
}
// creating drawing tool menu
private JMenu createDrawingToolMenu() {
JMenu drawingToolMenu = new JMenu(\"Drawing Tool\");
for (int i = 0; i < theFigKinds.length; i++) {
// create a menu item for this figure kind
JMenuItem item = new JMenuItem(theFigKinds[i].getName());
// set the action listener
item.addActionListener(new ToolSelector(theFigKinds[i]));
// add the item to the menu
drawingToolMenu.add(item);
}
return drawingToolMenu;
}
// creating menu tool bar
JToolBar createDrawingToolToolBar() {
JToolBar drawingToolToolBar = new JToolBar(JToolBar.VERTICAL);
for (int i = 0; i < theFigKinds.length; i++) {
JButton aButton = new JButton(theFigKinds[i].getIcon(16));
aButton.addActionListener(new ToolSelector(theFigKinds[i]));
drawingToolToolBar.add(aButton);
}
return drawingToolToolBar;
}
//creating color menu
private JMenu createColorMenu() {
JMenu colorMenu = new JMenu(\"Colors\");
JMenuItem setInteriorColor
= new JMenuItem(\"Set Interior Color\");
setInteriorColor.addActionListener(new SetInteriorColor());
colorMenu.add(setInteriorColor);
JMenuItem setBorderColor = new JMenuItem(\"Set Border Color\");
setBorderColor.addActionListener(new SetBorderColor());
colorMenu.add(setBorderColor);
return colorMenu;
}
public Drawable getSelectedFig() {
return selectedFig;
}
public static void main(String args[]) {
DrawApp drawApp = new DrawApp(args);
drawApp.setVisible(true);
}
// Inner Classes
private class ToolSelector implements ActionListener {
// Data Fields
private Drawable desiredFig;
// Constructor
public ToolSelector(Drawable desiredFig) {
this.desiredFig = desiredFig;
}
// Methods
@Override
public void actionPerformed(ActionEvent e) {
selectedFig = desiredFig;
}
}
private class SetInteriorColor implements ActionListener {
@Override
public void actionPerformed(ActionEvent e) {
presentInteriorrColo.
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phosphate group and deoxyribose The groups are .pdf
phosphate group and deoxyribose The groups are .pdf
in case of solid oxygen the atoms of oxygen are s.pdf
in case of solid oxygen the atoms of oxygen are s.pdf
Yeast is a microorganism and doesnt itself rise.pdf
Yeast is a microorganism and doesnt itself rise.pdf
First, lets analyze what each component of the .pdf
First, lets analyze what each component of the .pdf
when they reach equilibrium te cell potential is .pdf
when they reach equilibrium te cell potential is .pdf
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D) Not B) because Cl-Benzene bond develops a doub.pdf
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We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
lower.. think about it.. take a straw draw a line.pdf
lower.. think about it.. take a straw draw a line.pdf
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r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
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From left to right iodocyclopropane; 1-bromo-3-m.pdf
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For inorganic compounds Chemical properties remai.pdf
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
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Over forty interstellar molecules have been found, ranging from simp.pdf
Over forty interstellar molecules have been found, ranging from simp.pdf
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Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
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Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
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Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
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Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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