the link is not working can u pls write questions.pdf
•
0 likes•2 views
the link is not working can u pls write questions dirctly here itself thanks Solution the link is not working can u pls write questions dirctly here itself thanks.
Report
Share
Report
Share
Download to read offline
Recommended
The metal will undergo oxidation forming metal ion and releasing .pdf
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas]
Solution
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas].
Static Keyword Static is a keyword in C++ used to give special chara.pdf
Static Keyword Static is a keyword in C++ used to give special characteristics to an element.
Static elements are allocated storage only once in a program lifetime in static storage area. And
they have a scope till the program lifetime. Static Keyword can be used with following, Static
variable in functions Static Class Objects Static member Variable in class Static Methods in class
Static variables inside Functions Static variables when used inside function are initialized only
once, and then they hold there value even through function calls. These static variables are stored
on static storage area , not in stack. void counter() { static int count=0; cout << count++; } int
main(0 { for(int i=0;i<5;i++) { counter(); } } Output : 0 1 2 3 4 Let\'s se the same program\'s
output without using static variable. void counter() { int count=0; cout << count++; } int main(0
{ for(int i=0;i<5;i++) { counter(); } } Output : 0 0 0 0 0 If we do not use static keyword, the
variable count, is reinitialized everytime when counter() function is called, and gets destroyed
each time when counter() functions ends. But, if we make it static, once initialized count will
have a scope till the end of main() function and it will carry its value through function calls too.
If you don\'t initialize a static variable, they are by default initialized to zero. Static class Objects
Static keyword works in the same way for class objects too. Objects declared static are allocated
storage in static storage area, and have scope till the end of program. Static objects are also
initialized using constructors like other normal objects. Assignment to zero, on using static
keyword is only for primitive datatypes, not for user defined datatypes. class Abc { int i; public:
Abc() { i=0; cout << \"constructor\"; } ~Abc() { cout << \"destructor\"; } }; void f() { static Abc
obj; } int main() { int x=0; if(x==0) { f(); } cout << \"END\"; } Output : constructor END
destructor You must be thinking, why was destructor not called upon the end of the scope of if
condition. This is because object was static, which has scope till the program lifetime, hence
destructor for this object was called when main() exits. Static data member in class Static data
members of class are those members which are shared by all the objects. Static data member has
a single piece of storage, and is not available as separate copy with each object, like other non-
static data members. Static member variables (data members) are not initialied using constructor,
because these are not dependent on object initialization. Also, it must be initialized explicitly,
always outside the class. If not initialized, Linker will give error. class X { static int i; public:
X(){}; }; int X::i=1; int main() { X obj; cout << obj.i; // prints value of i } Once the definition for
static data member is made, user cannot redefine it. Though, arithmetic operations can be
performed on it. Static Member Functions These functions work for the .
Sr2+ is most likely to substitute for Ca2+ becaus.pdf
Sr2+ is most likely to substitute for Ca2+ because they are in the same group in
Periodic table and has most common properties. Good Luck! Please rate.
Solution
Sr2+ is most likely to substitute for Ca2+ because they are in the same group in
Periodic table and has most common properties. Good Luck! Please rate..
may be that peak id due to presence of alkyl gro.pdf
may be that peak id due to presence of alkyl group and moisture,
Solution
may be that peak id due to presence of alkyl group and moisture,.
There should only have one singlet resonance for .pdf
There should only have one singlet resonance for 1,1,2,2-tetrachloroethane. There is
one hydrogen on each carbon. both of them have the exact same environment. Thus, it is shown
as 1 singlet resonance.
Solution
There should only have one singlet resonance for 1,1,2,2-tetrachloroethane. There is
one hydrogen on each carbon. both of them have the exact same environment. Thus, it is shown
as 1 singlet resonance..
The one have higher value of E(cell) is acting as.pdf
The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e---
>2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode
= 1.07- (-0.13) =1.20V
Solution
The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e---
>2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode
= 1.07- (-0.13) =1.20V.
Nicotine has a molecular formula of C10H14N2 .pdf
Nicotine has a molecular formula of C10H14N2
Solution
Nicotine has a molecular formula of C10H14N2.
Recommended
The metal will undergo oxidation forming metal ion and releasing .pdf
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas]
Solution
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas].
Static Keyword Static is a keyword in C++ used to give special chara.pdf
Static Keyword Static is a keyword in C++ used to give special characteristics to an element.
Static elements are allocated storage only once in a program lifetime in static storage area. And
they have a scope till the program lifetime. Static Keyword can be used with following, Static
variable in functions Static Class Objects Static member Variable in class Static Methods in class
Static variables inside Functions Static variables when used inside function are initialized only
once, and then they hold there value even through function calls. These static variables are stored
on static storage area , not in stack. void counter() { static int count=0; cout << count++; } int
main(0 { for(int i=0;i<5;i++) { counter(); } } Output : 0 1 2 3 4 Let\'s se the same program\'s
output without using static variable. void counter() { int count=0; cout << count++; } int main(0
{ for(int i=0;i<5;i++) { counter(); } } Output : 0 0 0 0 0 If we do not use static keyword, the
variable count, is reinitialized everytime when counter() function is called, and gets destroyed
each time when counter() functions ends. But, if we make it static, once initialized count will
have a scope till the end of main() function and it will carry its value through function calls too.
If you don\'t initialize a static variable, they are by default initialized to zero. Static class Objects
Static keyword works in the same way for class objects too. Objects declared static are allocated
storage in static storage area, and have scope till the end of program. Static objects are also
initialized using constructors like other normal objects. Assignment to zero, on using static
keyword is only for primitive datatypes, not for user defined datatypes. class Abc { int i; public:
Abc() { i=0; cout << \"constructor\"; } ~Abc() { cout << \"destructor\"; } }; void f() { static Abc
obj; } int main() { int x=0; if(x==0) { f(); } cout << \"END\"; } Output : constructor END
destructor You must be thinking, why was destructor not called upon the end of the scope of if
condition. This is because object was static, which has scope till the program lifetime, hence
destructor for this object was called when main() exits. Static data member in class Static data
members of class are those members which are shared by all the objects. Static data member has
a single piece of storage, and is not available as separate copy with each object, like other non-
static data members. Static member variables (data members) are not initialied using constructor,
because these are not dependent on object initialization. Also, it must be initialized explicitly,
always outside the class. If not initialized, Linker will give error. class X { static int i; public:
X(){}; }; int X::i=1; int main() { X obj; cout << obj.i; // prints value of i } Once the definition for
static data member is made, user cannot redefine it. Though, arithmetic operations can be
performed on it. Static Member Functions These functions work for the .
Sr2+ is most likely to substitute for Ca2+ becaus.pdf
Sr2+ is most likely to substitute for Ca2+ because they are in the same group in
Periodic table and has most common properties. Good Luck! Please rate.
Solution
Sr2+ is most likely to substitute for Ca2+ because they are in the same group in
Periodic table and has most common properties. Good Luck! Please rate..
may be that peak id due to presence of alkyl gro.pdf
may be that peak id due to presence of alkyl group and moisture,
Solution
may be that peak id due to presence of alkyl group and moisture,.
There should only have one singlet resonance for .pdf
There should only have one singlet resonance for 1,1,2,2-tetrachloroethane. There is
one hydrogen on each carbon. both of them have the exact same environment. Thus, it is shown
as 1 singlet resonance.
Solution
There should only have one singlet resonance for 1,1,2,2-tetrachloroethane. There is
one hydrogen on each carbon. both of them have the exact same environment. Thus, it is shown
as 1 singlet resonance..
The one have higher value of E(cell) is acting as.pdf
The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e---
>2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode
= 1.07- (-0.13) =1.20V
Solution
The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e---
>2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode
= 1.07- (-0.13) =1.20V.
Nicotine has a molecular formula of C10H14N2 .pdf
Nicotine has a molecular formula of C10H14N2
Solution
Nicotine has a molecular formula of C10H14N2.
HClO4 in aqueous medium ionizes as Hydrogen(+1)ca.pdf
HClO4 in aqueous medium ionizes as Hydrogen(+1)cation and Chlorate (-1) anion
Molar conc of Hydrogen cation =.245 Molar conc of chlorate anion = .245
Solution
HClO4 in aqueous medium ionizes as Hydrogen(+1)cation and Chlorate (-1) anion
Molar conc of Hydrogen cation =.245 Molar conc of chlorate anion = .245.
Which of the following would be a description of a system unitA c.pdf
Which of the following would be a description of a system unit?
A case and internal components used to process data into information
A computer’s main memory component
The module that interprets and carries out computer instructions
A data and information storage device
Ans: A case and internal components used to process data into information
Question 2
Select one answer.
True or false? A tower or minitower PC is a type of all-in-one desktop unit.
True
False
Ans: True
Question 3
Fill in the blanks by selecting one option from each menu.
A/an --expansion card internal drive bay processor chip is inserted on the motherboard and
provides more memory or controls a peripheral device such as a graphics adapter that supports
video games.
Ans: Expansion card
Question 4
Select one answer
What is a peripheral?
A high-speed storage location that temporarily holds data and instructions for future use
A component of the system board that controls the timing of computer activities
A slot on the motherboard that holds memory modules and helps speed computer processes
A device that connects to the system unit and is controlled by the computer’s processor
Ans: A device that connects to the system unit and is controlled by the computer’s processor
Question 5
Fill in the blanks by selecting one option from each menu.
If you often play the latest computer games on your PC, which of the following would you
consider critical components of the PC? Select Yes or No for each option.
The weight and size of your PC
--YesNo
Ans: No
Sophisticated graphics and sound adapter cards
--YesNo
Ans: Yes
Processor speed and amount of RAM
--YesNo
Ans: Yes
Question 1
Select one answer.
If you use your desktop computer only for doing schoolwork, which input and output device
combination would be best from the options below (considering the best combination of cost,
screen resolution, and keyboard design)?
A gaming keyboard and a 27\" LCD monitor
A foldable keyboard and a 15\" LCD monitor
A wireless keyboard and a 24\" LED monitor
A split-style keyboard and a 30\" LED monitor
Ans: A wireless keyboard and a 24\" LED monitor
Question 2
Select one answer.
You are considering using cloud storage as your primary system for saving your data. What
advantages does this storage media have over using your system’s hard drive or some other
external media (external hard drive, flash drive, R/W disks) for storing your data?
You can always get to your data from anywhere.
The data is more secure in cloud storage than on your own storage devices.
The cloud storage provider ensures that you have a backup of your data.
Cloud storage is definitely less expensive than the cost of a hard drive or external devices.
Ans: You can always get to your data from anywhere.
Question 3
Select one answer.
Which of the following can be considered an output device?
Microphone
Stylus
Wireless mouse
Speaker
Ans: Speaker
Question 4
Select one answer.
In this diagram of a simple home network, what appears to be missing?
.
Water is a polar inorganic solvent. Benzene is a nonpolar organic so.pdf
Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will
generally dissolve solutes with similar polarity (with certain exceptions).
CaCO3 - insoluble in water, insoluble in benzene
CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
NaNO3 - soluble in water, insoluble in benzene
NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
K2SO4 - soluble in water, insoluble in benzene
K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
Petroleum jelly - insoluble in water, soluble in benzene
Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but
soluble in benzene.
Butane - insoluble in water, soluble in benzene
Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene.
Acetone - soluble in water, soluble in benzene
Acetone is a polar organic compound. Thus it is soluble in both water and benzene.
Solution
Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will
generally dissolve solutes with similar polarity (with certain exceptions).
CaCO3 - insoluble in water, insoluble in benzene
CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
NaNO3 - soluble in water, insoluble in benzene
NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
K2SO4 - soluble in water, insoluble in benzene
K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
Petroleum jelly - insoluble in water, soluble in benzene
Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but
soluble in benzene.
Butane - insoluble in water, soluble in benzene
Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene.
Acetone - soluble in water, soluble in benzene
Acetone is a polar organic compound. Thus it is soluble in both water and benzene..
viruses which have single strande DNA in their genome come under cat.pdf
viruses which have single strande DNA in their genome come under category of paroviridae
which generally includes animal pathogens. These viruses when infect the host, the single
stranded DNA is first converted into double stranded DNA, and then using the host machinary
for transcription, ds DNA transcribes into viral mRNA.
For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host
and divided into initiation, elongation and termaination. For gene expression in viruses
transcription is generally not coupled with translation in ssDNA virus. The virus instructs
transcription machinary to make viral components in the host cell.
Solution
viruses which have single strande DNA in their genome come under category of paroviridae
which generally includes animal pathogens. These viruses when infect the host, the single
stranded DNA is first converted into double stranded DNA, and then using the host machinary
for transcription, ds DNA transcribes into viral mRNA.
For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host
and divided into initiation, elongation and termaination. For gene expression in viruses
transcription is generally not coupled with translation in ssDNA virus. The virus instructs
transcription machinary to make viral components in the host cell..
This word problem is about a triangle whose perimeter is 47 miles. S.pdf
This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the
length of each side, then there are 3 unknowns. So here\'s what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.
Solution
This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the
length of each side, then there are 3 unknowns. So here\'s what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined l.
CH4 + 2O2 -- CO2 + 2H2O note CH4 combustion is.pdf
CH4 + 2O2 --> CO2 + 2H2O note: CH4 combustion is a process to react with O2.
Solution
CH4 + 2O2 --> CO2 + 2H2O note: CH4 combustion is a process to react with O2..
Carbon 2 is where D and L differ for all sugars. .pdf
Carbon 2 is where D and L differ for all sugars. For D and L glucose they will have
the same chirality everywhere cept at 2.
Solution
Carbon 2 is where D and L differ for all sugars. For D and L glucose they will have
the same chirality everywhere cept at 2..
Program to print the Diamond Shape -#include stdio.h int ma.pdf
Program to print the Diamond Shape :-
#include
int main()
{
int i, j, n;
printf(\"Enter a positive int number greater than one to print the Diamond Shape : \");
scanf(\"%d\", &n);
//To Print the upper pyramid
for(i=1; i<=n; i++)
{
for(j=i; j=1; i--)
{
for(j=i; j<=n; j++)
{
printf(\" \");
}
for(j=2; j<(2*i-1); j++)
{
printf(\"*\");
}
printf(\"\ \");
}
return 0;
}
Solution
Program to print the Diamond Shape :-
#include
int main()
{
int i, j, n;
printf(\"Enter a positive int number greater than one to print the Diamond Shape : \");
scanf(\"%d\", &n);
//To Print the upper pyramid
for(i=1; i<=n; i++)
{
for(j=i; j=1; i--)
{
for(j=i; j<=n; j++)
{
printf(\" \");
}
for(j=2; j<(2*i-1); j++)
{
printf(\"*\");
}
printf(\"\ \");
}
return 0;
}.
Part D option 4 is answerUnless untill they are exposed by some me.pdf
Part D option 4 is answer
Unless untill they are exposed by some means we can not find them
PArt E option 1 is answer as they hypo... teraposed in rocks
Solution
Part D option 4 is answer
Unless untill they are exposed by some means we can not find them
PArt E option 1 is answer as they hypo... teraposed in rocks.
Isomers which have their atoms connected in the same sequence but di.pdf
Isomers which have their atoms connected in the same sequence but differ in the way the atoms
are oriented in space - i.e. the difference between two stereoisomers lies only in the three
dimensional arrangement of atoms.
since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and
arrangement is same in 2D space , they are stereo isomers
Solution
Isomers which have their atoms connected in the same sequence but differ in the way the atoms
are oriented in space - i.e. the difference between two stereoisomers lies only in the three
dimensional arrangement of atoms.
since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and
arrangement is same in 2D space , they are stereo isomers.
import java.util.Scanner;import java.text.DecimalFormat;import j.pdf
import java.util.Scanner;
import java.text.DecimalFormat;
import java.io.*;
public class SavingAccount //MUST match the file name!
{
public static void main(String[] args)throws IOException
{
//Create a decimal format for displaying dollars
DecimalFormat dollar = new DecimalFormat(\"#,###.##\");
//Constants
double depositSum = 0.0;
double withdrawalSum = 0.0;
double earnedInt = 0.0;
double startBalance = 500; //Starting balance
// Executables
System.out.println(\"This program solves Programming Challenge 6.11\");
System.out.println();
//Create Scanner object for keyboard input.
Scanner keyboard = new Scanner(System.in);
//Get interest rate
System.out.print(\"Enter the annual interest: \");
double testInterest = keyboard.nextDouble();
//Create an object that accept the starting balance and annual interest
SavingAccount1 account = new SavingAccount1(startBalance, testInterest);
//Open Deposit file.
File file = new File (\"BankDeposits.txt\");
Scanner inputFile = new Scanner(file);
//Read line in file
while (inputFile.hasNext());
{
//Read numbers
double num = inputFile.nextDouble();
//Add the numbers
depositSum += num;
}
//Deposit the file input.
account.deposit(depositSum);
//Close the file
inputFile.close();
//Open Withdrawal file
File file2 = new File(\"BankWithdrawal.txt\");
Scanner inputFile2 = new Scanner(file2);
//Read lines in file
while (inputFile2.hasNext());
{
//Read numbers
double num2 = inputFile2.nextDouble();
//Add the numbers
withdrawalSum += num2;
}
//Withdrawal the file input from account.
account.withdraw(withdrawalSum);
//Close the file
inputFile2.close();
//Add the monthly interest
account.addInt();
//Get amount of interest earned.
earnedInt += account.getInterest();
//Display the data
System.out.println(\"Account balance $\" + dollar.format(account.getBalance()));
System.out.println(\"Total interest earned $\" + dollar.format(account.getInterest()));
}
}//end class
public class SavingAccount1 //MUST match the file name!
{
public static void main(String[] args)
{
System.out.println(\"This program solves Programming Challenge 6.10\");
System.out.println();
}
//Fields
private double balance; //Account balance
private double annualInterest; //annual interest
private double monthInt; //monthly interest
private double earnedInt; //earned interest
double totalWithdraw;
double totalDeposit;
/**
018
This constructor sets the starting balance
019
and the annual interest at 0.0.
020
*/
public void Ward_Tassinda_SavingAccount1()
{
balance = 0.0;
annualInterest = 0.0;
}
/**
029
This constructor set the starting balance and the annual interest rate
030
to the value passed as an argument.
031
@param startBalance The starting balance.
032
*/
public void Ward_Tassinda_SavingAccount1(double startBalance, double interestRate)
{
balance = startBalance;
annualInterest = interestRate;
}
/**
041
This constructor sets the starting balance to
042
the value in the String argument.
043
@param str The starting balance, as a String.
044
*/
public void .
am sorry but i have no idea about what you are sa.pdf
am sorry but i have no idea about what you are saying
Solution
am sorry but i have no idea about what you are saying.
Points.java import java.util.ArrayList; import java.util.Scan.pdf
// Points.java
import java.util.ArrayList;
import java.util.Scanner;
public class Points{
double x;
double y;
Points(double xco, double yco){
x = xco;
y = yco;
}
}
// PolygonArea.java
import java.util.ArrayList;
import java.util.Scanner;
public class PolygonArea {
// funtion to determine area of polygon
public static double getArea(ArrayList coordinates) {
double sumxy = 0;
double sumyx = 0;
for (int i = 0; i < coordinates.size(); i++) {
int index = (i + 1) % coordinates.size();
Points point1 = coordinates.get(i);
Points point2 = coordinates.get(index);
sumxy += (point1.x * point2.y);
sumyx += (point1.y * point2.x);
}
double areaPolygon = 0.5 * (sumxy - sumyx);
if (areaPolygon > 0)
return areaPolygon;
else
return -areaPolygon;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print(\"Enter the number of the points: \");
int totalPoints = scan.nextInt();
System.out.print(\"Enter the coordinates of the points: \");
ArrayList coordinates = new ArrayList<>();
for (int i = 0; i < totalPoints; i++) {
coordinates.add(new Points(scan.nextDouble(), scan.nextDouble()));
}
System.out.println(\"The total area is \" + getArea(coordinates));
}
}
/*
output:
Enter the number of the points: 3
Enter the coordinates of the points: 0 0 0 3 4 0
The total area is 6.0
*/
Solution
// Points.java
import java.util.ArrayList;
import java.util.Scanner;
public class Points{
double x;
double y;
Points(double xco, double yco){
x = xco;
y = yco;
}
}
// PolygonArea.java
import java.util.ArrayList;
import java.util.Scanner;
public class PolygonArea {
// funtion to determine area of polygon
public static double getArea(ArrayList coordinates) {
double sumxy = 0;
double sumyx = 0;
for (int i = 0; i < coordinates.size(); i++) {
int index = (i + 1) % coordinates.size();
Points point1 = coordinates.get(i);
Points point2 = coordinates.get(index);
sumxy += (point1.x * point2.y);
sumyx += (point1.y * point2.x);
}
double areaPolygon = 0.5 * (sumxy - sumyx);
if (areaPolygon > 0)
return areaPolygon;
else
return -areaPolygon;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print(\"Enter the number of the points: \");
int totalPoints = scan.nextInt();
System.out.print(\"Enter the coordinates of the points: \");
ArrayList coordinates = new ArrayList<>();
for (int i = 0; i < totalPoints; i++) {
coordinates.add(new Points(scan.nextDouble(), scan.nextDouble()));
}
System.out.println(\"The total area is \" + getArea(coordinates));
}
}
/*
output:
Enter the number of the points: 3
Enter the coordinates of the points: 0 0 0 3 4 0
The total area is 6.0
*/.
General Instructions 1. The objective of this check list is to guid.pdf
General Instructions: 1. The objective of this check list is to guide the Internal Auditors in the
conduct of Internal Audit in a systematic manner so that all the significant points in finance &
accounts, procurement, contract & tenders, payment of bills of contractors, planning process,
flow of funds and staffing etc. are checked before finalization of Internal Audit Report 2. The
Internal Auditors should familiarize themselves with the Manual on Financial Management &
Procurement thoroughly before conducting Internal Audit. Para 100 to 105 of the Manual on
Financial Management & Procurement provides broad guidelines for Internal Audit of SSA and
Paras 51 to 99.7 of the Manual contains guidelines for maintenance of accounts . 3. Scope and
coverage of Internal Audit and report – Internal Audit of District Project Offices and sub -district
units selected on a percentage basis should be conducted so as to cover all districts and sub-
districts units at least once in 3 years . The Internal Audit report should indicate name of unit
covered, date of Audit, review of planning process to ascertain whether the activities under SSA
have been planned as per prescribed norms, review of financial systems to check the efficacy of
internal checks and controls, review of establishment matters and staffing, review of
Procurement and status of Statutory Audit, C& AG Audit and Internal Audit. 4 Finance &
Accounts – 4.1 To check whether 22 Registers( Annexure-I is attached for ready reference)
indicated in para 52 of FM&P Manual are maintained. 4.2 To verify about the number of savings
Bank Accounts maintained for SSA at SPO /DPO /Block & VEC Levels. The matter should be
reported if more then one savings bank accounts are maintained at any levels. The reasons for
maintenance of multiple Bank Accounts should also be mentioned in the Internal Audit Report
4.3 Cash book - To check cash book for ascertaining whether cash book is written daily as
envisaged in para 79 of FM&P Manual monthly on regular basis. The Internal Auditor must also
conduct physical verification of cash and report heavy cash balance if any. 4.4 To check whether
Bank Reconciliation is done monthly on regular basis or not 4.5 Payment of bills of contractors
Annexure-P To verify whether 4.5.1 Original bill duly signed by contractors is submitted. 4.5.2
Contractor has put his initials in all cuttings and corrections in the bill. 4.5.3 All Supporting
documents are attached with the bills. 4.5.4 The rates, security deposit, and deductions are as per
terms and conditions specified in the agreement 4.5.5 The variations in quantities and completion
period etc. have been authorized by the competent authority 4.5.6 Job completion certificate has
been processed by the dealing assistant of SSA. 4.5.7 Bills passed for payment are as per rules of
the society. 4.5.8 All terms and conditions of the contract are fulfilled before passing the bills.
4.5.9 Every final bill is checked in detail with measurement boo.
Computer security, additionally called cybersecurity or IT security,.pdf
Computer security, additionally called cybersecurity or IT security, is that the protection of
knowledge systems from thieving or harm to the hardware, the package, and to the knowledge on
them, moreover as from disruption or misdirection of the services they supply.[1]
It includes dominant physical access to the hardware, moreover as protective against damage that
will come back via network access, information and code injection,[2] and thanks to malpractice
by operators, whether or not intentional, accidental, or thanks to them being tricked into
deviating from secure procedures.[3]
The field is of growing importance thanks to the increasing reliance on laptop systems and
therefore the web in most societies,[4] wireless networks like Bluetooth and Wi-Fi - and
therefore the growth of \"smart\" devices, as well as smartphones, televisions and small devices
as a part of the net of Things.
A vulnerability may be a system susceptibleness or flaw. several vulnerabilities ar documented
within the Common Vulnerabilities and Exposures (CVE) information. AN exploitable
vulnerability is one that a minimum of one operating attack or \"exploit\" exists.[5]
To secure a automatic data processing system, it\'s vital to know the attacks that may be created
against it, and these threats will generally be classified into one among the classes below:
Backdoors[edit]
A backdoor in an exceedingly automatic data processing system, a cryptosystem or AN
algorithmic rule, is any secret methodology of bypassing traditional authentication or security
controls. they will exist for variety of reasons, as well as by original style or from poor
configuration. they will are accessorial by a licensed party to permit some legitimate access, or
by AN assailant for malicious reasons; however notwithstanding the motives for his or her
existence, they produce a vulnerability.
Denial-of-service attack[edit]
Denial of service attacks ar designed to create a machine or network resource unavailable to its
supposed users.[6] Attackers will deny service to individual victims, like by deliberately coming
into a wrong countersign enough consecutive times to cause the victim account to be fast, or they
will overload the capabilities of a machine or network and block all users directly. whereas a
network attack from one IP address may be blocked by adding a brand new firewall rule, several
sorts of Distributed denial of service (DDoS) attacks ar doable, wherever the attack comes from
an outsized variety of points – and defensive is far harder. Such attacks will originate from the
zombie computers of a botnet, however a spread of alternative techniques ar doable as well as
reflection and amplification attacks, wherever innocent systems ar fooled into causing traffic to
the victim.
Solution
Computer security, additionally called cybersecurity or IT security, is that the protection of
knowledge systems from thieving or harm to the hardware, the package, and to the knowledge on
them, moreover.
C. Nitrosomonas, Nitrosolobus, and NitrosovibrioNitrifiers are che.pdf
C. Nitrosomonas, Nitrosolobus, and Nitrosovibrio
Nitrifiers are chemolithotropic organisms which obtains energy by oxidating inorganic nitrogen
compounds. These three organisms convert Ammonia to Nitrites which inturn converted into
nitrate by other organisms such as Nitrobacter or Nitrospina.
False: Neisseria gonorrhoeae - responsible for causing sexually transmitted disease Gonorrhea
Neisseria meningitidis- responsible for causing meningitis and sepsis
Thiobacillus spp- It belongs to Hydrogenophilaceae family and mostly responsible for oxidizing
hydrogen.
Solution
C. Nitrosomonas, Nitrosolobus, and Nitrosovibrio
Nitrifiers are chemolithotropic organisms which obtains energy by oxidating inorganic nitrogen
compounds. These three organisms convert Ammonia to Nitrites which inturn converted into
nitrate by other organisms such as Nitrobacter or Nitrospina.
False: Neisseria gonorrhoeae - responsible for causing sexually transmitted disease Gonorrhea
Neisseria meningitidis- responsible for causing meningitis and sepsis
Thiobacillus spp- It belongs to Hydrogenophilaceae family and mostly responsible for oxidizing
hydrogen..
C is correct. Routing and logical addressing occur at the Internet l.pdf
C is correct. Routing and logical addressing occur at the Internet layer of the TCP/IP stack.
A, B, and D are incorrect. A is incorrect because the Network layer is an OSI model layer. B is
incorrect because Data Link is an OSI model layer. D is incorrect because framing, error
checking, and media access occur at the Network Access layer of the TCP/IP stack.
Solution
C is correct. Routing and logical addressing occur at the Internet layer of the TCP/IP stack.
A, B, and D are incorrect. A is incorrect because the Network layer is an OSI model layer. B is
incorrect because Data Link is an OSI model layer. D is incorrect because framing, error
checking, and media access occur at the Network Access layer of the TCP/IP stack..
AnswerThe gill and siphon withdrawal complex in Aplysia (a marine.pdf
Answer:
The gill and siphon withdrawal complex in Aplysia (a marine snail) is an involuntary defensive
reflex, which causes the marine snail to retract its siphon and gill when the animal is disturbed.
There are 3 mechanisms used to stimulate the process of involuntary defensive reflex as soon as
an external stimuli trigger its activity on the peripheral organs of mollusk, Aplysia. The three
mechanisms are habituation, dishabituation and sensitization. The central and pheripheral
neurons are involved in an involuntary defensive reflex of Aplysia as part of defense mechanism.
The neurobiological model is mediated by six central motor neurons in Aplysia with abdominal
ganglion to produce involuntary defensive reflex to stimulate gills & these neurons are L7,
LDG1, LDG2 and RDG used for larger gill contractions. Smaller gill involuntary defensive
reflex is mediate L9G1 and L9G2 whereas siphon is innervated by LDS1, LDS2, LDS3, RDS,
LBS1, LBS2 to mediate involuntary defensive reflex. Habituation and dishabituation is mediated
based on the external stimuli on the gills finally some reflexes are going to activate “ink glands”
to trigger release of ink when a predator arrived to its surrounds result in escaping from the
predator
Solution
Answer:
The gill and siphon withdrawal complex in Aplysia (a marine snail) is an involuntary defensive
reflex, which causes the marine snail to retract its siphon and gill when the animal is disturbed.
There are 3 mechanisms used to stimulate the process of involuntary defensive reflex as soon as
an external stimuli trigger its activity on the peripheral organs of mollusk, Aplysia. The three
mechanisms are habituation, dishabituation and sensitization. The central and pheripheral
neurons are involved in an involuntary defensive reflex of Aplysia as part of defense mechanism.
The neurobiological model is mediated by six central motor neurons in Aplysia with abdominal
ganglion to produce involuntary defensive reflex to stimulate gills & these neurons are L7,
LDG1, LDG2 and RDG used for larger gill contractions. Smaller gill involuntary defensive
reflex is mediate L9G1 and L9G2 whereas siphon is innervated by LDS1, LDS2, LDS3, RDS,
LBS1, LBS2 to mediate involuntary defensive reflex. Habituation and dishabituation is mediated
based on the external stimuli on the gills finally some reflexes are going to activate “ink glands”
to trigger release of ink when a predator arrived to its surrounds result in escaping from the
predator.
Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
More Related Content
More from KUNALHARCHANDANI1
HClO4 in aqueous medium ionizes as Hydrogen(+1)ca.pdf
HClO4 in aqueous medium ionizes as Hydrogen(+1)cation and Chlorate (-1) anion
Molar conc of Hydrogen cation =.245 Molar conc of chlorate anion = .245
Solution
HClO4 in aqueous medium ionizes as Hydrogen(+1)cation and Chlorate (-1) anion
Molar conc of Hydrogen cation =.245 Molar conc of chlorate anion = .245.
Which of the following would be a description of a system unitA c.pdf
Which of the following would be a description of a system unit?
A case and internal components used to process data into information
A computer’s main memory component
The module that interprets and carries out computer instructions
A data and information storage device
Ans: A case and internal components used to process data into information
Question 2
Select one answer.
True or false? A tower or minitower PC is a type of all-in-one desktop unit.
True
False
Ans: True
Question 3
Fill in the blanks by selecting one option from each menu.
A/an --expansion card internal drive bay processor chip is inserted on the motherboard and
provides more memory or controls a peripheral device such as a graphics adapter that supports
video games.
Ans: Expansion card
Question 4
Select one answer
What is a peripheral?
A high-speed storage location that temporarily holds data and instructions for future use
A component of the system board that controls the timing of computer activities
A slot on the motherboard that holds memory modules and helps speed computer processes
A device that connects to the system unit and is controlled by the computer’s processor
Ans: A device that connects to the system unit and is controlled by the computer’s processor
Question 5
Fill in the blanks by selecting one option from each menu.
If you often play the latest computer games on your PC, which of the following would you
consider critical components of the PC? Select Yes or No for each option.
The weight and size of your PC
--YesNo
Ans: No
Sophisticated graphics and sound adapter cards
--YesNo
Ans: Yes
Processor speed and amount of RAM
--YesNo
Ans: Yes
Question 1
Select one answer.
If you use your desktop computer only for doing schoolwork, which input and output device
combination would be best from the options below (considering the best combination of cost,
screen resolution, and keyboard design)?
A gaming keyboard and a 27\" LCD monitor
A foldable keyboard and a 15\" LCD monitor
A wireless keyboard and a 24\" LED monitor
A split-style keyboard and a 30\" LED monitor
Ans: A wireless keyboard and a 24\" LED monitor
Question 2
Select one answer.
You are considering using cloud storage as your primary system for saving your data. What
advantages does this storage media have over using your system’s hard drive or some other
external media (external hard drive, flash drive, R/W disks) for storing your data?
You can always get to your data from anywhere.
The data is more secure in cloud storage than on your own storage devices.
The cloud storage provider ensures that you have a backup of your data.
Cloud storage is definitely less expensive than the cost of a hard drive or external devices.
Ans: You can always get to your data from anywhere.
Question 3
Select one answer.
Which of the following can be considered an output device?
Microphone
Stylus
Wireless mouse
Speaker
Ans: Speaker
Question 4
Select one answer.
In this diagram of a simple home network, what appears to be missing?
.
Water is a polar inorganic solvent. Benzene is a nonpolar organic so.pdf
Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will
generally dissolve solutes with similar polarity (with certain exceptions).
CaCO3 - insoluble in water, insoluble in benzene
CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
NaNO3 - soluble in water, insoluble in benzene
NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
K2SO4 - soluble in water, insoluble in benzene
K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
Petroleum jelly - insoluble in water, soluble in benzene
Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but
soluble in benzene.
Butane - insoluble in water, soluble in benzene
Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene.
Acetone - soluble in water, soluble in benzene
Acetone is a polar organic compound. Thus it is soluble in both water and benzene.
Solution
Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will
generally dissolve solutes with similar polarity (with certain exceptions).
CaCO3 - insoluble in water, insoluble in benzene
CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
NaNO3 - soluble in water, insoluble in benzene
NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
K2SO4 - soluble in water, insoluble in benzene
K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
Petroleum jelly - insoluble in water, soluble in benzene
Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but
soluble in benzene.
Butane - insoluble in water, soluble in benzene
Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene.
Acetone - soluble in water, soluble in benzene
Acetone is a polar organic compound. Thus it is soluble in both water and benzene..
viruses which have single strande DNA in their genome come under cat.pdf
viruses which have single strande DNA in their genome come under category of paroviridae
which generally includes animal pathogens. These viruses when infect the host, the single
stranded DNA is first converted into double stranded DNA, and then using the host machinary
for transcription, ds DNA transcribes into viral mRNA.
For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host
and divided into initiation, elongation and termaination. For gene expression in viruses
transcription is generally not coupled with translation in ssDNA virus. The virus instructs
transcription machinary to make viral components in the host cell.
Solution
viruses which have single strande DNA in their genome come under category of paroviridae
which generally includes animal pathogens. These viruses when infect the host, the single
stranded DNA is first converted into double stranded DNA, and then using the host machinary
for transcription, ds DNA transcribes into viral mRNA.
For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host
and divided into initiation, elongation and termaination. For gene expression in viruses
transcription is generally not coupled with translation in ssDNA virus. The virus instructs
transcription machinary to make viral components in the host cell..
This word problem is about a triangle whose perimeter is 47 miles. S.pdf
This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the
length of each side, then there are 3 unknowns. So here\'s what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.
Solution
This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the
length of each side, then there are 3 unknowns. So here\'s what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined l.
CH4 + 2O2 -- CO2 + 2H2O note CH4 combustion is.pdf
CH4 + 2O2 --> CO2 + 2H2O note: CH4 combustion is a process to react with O2.
Solution
CH4 + 2O2 --> CO2 + 2H2O note: CH4 combustion is a process to react with O2..
Carbon 2 is where D and L differ for all sugars. .pdf
Carbon 2 is where D and L differ for all sugars. For D and L glucose they will have
the same chirality everywhere cept at 2.
Solution
Carbon 2 is where D and L differ for all sugars. For D and L glucose they will have
the same chirality everywhere cept at 2..
Program to print the Diamond Shape -#include stdio.h int ma.pdf
Program to print the Diamond Shape :-
#include
int main()
{
int i, j, n;
printf(\"Enter a positive int number greater than one to print the Diamond Shape : \");
scanf(\"%d\", &n);
//To Print the upper pyramid
for(i=1; i<=n; i++)
{
for(j=i; j=1; i--)
{
for(j=i; j<=n; j++)
{
printf(\" \");
}
for(j=2; j<(2*i-1); j++)
{
printf(\"*\");
}
printf(\"\ \");
}
return 0;
}
Solution
Program to print the Diamond Shape :-
#include
int main()
{
int i, j, n;
printf(\"Enter a positive int number greater than one to print the Diamond Shape : \");
scanf(\"%d\", &n);
//To Print the upper pyramid
for(i=1; i<=n; i++)
{
for(j=i; j=1; i--)
{
for(j=i; j<=n; j++)
{
printf(\" \");
}
for(j=2; j<(2*i-1); j++)
{
printf(\"*\");
}
printf(\"\ \");
}
return 0;
}.
Part D option 4 is answerUnless untill they are exposed by some me.pdf
Part D option 4 is answer
Unless untill they are exposed by some means we can not find them
PArt E option 1 is answer as they hypo... teraposed in rocks
Solution
Part D option 4 is answer
Unless untill they are exposed by some means we can not find them
PArt E option 1 is answer as they hypo... teraposed in rocks.
Isomers which have their atoms connected in the same sequence but di.pdf
Isomers which have their atoms connected in the same sequence but differ in the way the atoms
are oriented in space - i.e. the difference between two stereoisomers lies only in the three
dimensional arrangement of atoms.
since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and
arrangement is same in 2D space , they are stereo isomers
Solution
Isomers which have their atoms connected in the same sequence but differ in the way the atoms
are oriented in space - i.e. the difference between two stereoisomers lies only in the three
dimensional arrangement of atoms.
since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and
arrangement is same in 2D space , they are stereo isomers.
import java.util.Scanner;import java.text.DecimalFormat;import j.pdf
import java.util.Scanner;
import java.text.DecimalFormat;
import java.io.*;
public class SavingAccount //MUST match the file name!
{
public static void main(String[] args)throws IOException
{
//Create a decimal format for displaying dollars
DecimalFormat dollar = new DecimalFormat(\"#,###.##\");
//Constants
double depositSum = 0.0;
double withdrawalSum = 0.0;
double earnedInt = 0.0;
double startBalance = 500; //Starting balance
// Executables
System.out.println(\"This program solves Programming Challenge 6.11\");
System.out.println();
//Create Scanner object for keyboard input.
Scanner keyboard = new Scanner(System.in);
//Get interest rate
System.out.print(\"Enter the annual interest: \");
double testInterest = keyboard.nextDouble();
//Create an object that accept the starting balance and annual interest
SavingAccount1 account = new SavingAccount1(startBalance, testInterest);
//Open Deposit file.
File file = new File (\"BankDeposits.txt\");
Scanner inputFile = new Scanner(file);
//Read line in file
while (inputFile.hasNext());
{
//Read numbers
double num = inputFile.nextDouble();
//Add the numbers
depositSum += num;
}
//Deposit the file input.
account.deposit(depositSum);
//Close the file
inputFile.close();
//Open Withdrawal file
File file2 = new File(\"BankWithdrawal.txt\");
Scanner inputFile2 = new Scanner(file2);
//Read lines in file
while (inputFile2.hasNext());
{
//Read numbers
double num2 = inputFile2.nextDouble();
//Add the numbers
withdrawalSum += num2;
}
//Withdrawal the file input from account.
account.withdraw(withdrawalSum);
//Close the file
inputFile2.close();
//Add the monthly interest
account.addInt();
//Get amount of interest earned.
earnedInt += account.getInterest();
//Display the data
System.out.println(\"Account balance $\" + dollar.format(account.getBalance()));
System.out.println(\"Total interest earned $\" + dollar.format(account.getInterest()));
}
}//end class
public class SavingAccount1 //MUST match the file name!
{
public static void main(String[] args)
{
System.out.println(\"This program solves Programming Challenge 6.10\");
System.out.println();
}
//Fields
private double balance; //Account balance
private double annualInterest; //annual interest
private double monthInt; //monthly interest
private double earnedInt; //earned interest
double totalWithdraw;
double totalDeposit;
/**
018
This constructor sets the starting balance
019
and the annual interest at 0.0.
020
*/
public void Ward_Tassinda_SavingAccount1()
{
balance = 0.0;
annualInterest = 0.0;
}
/**
029
This constructor set the starting balance and the annual interest rate
030
to the value passed as an argument.
031
@param startBalance The starting balance.
032
*/
public void Ward_Tassinda_SavingAccount1(double startBalance, double interestRate)
{
balance = startBalance;
annualInterest = interestRate;
}
/**
041
This constructor sets the starting balance to
042
the value in the String argument.
043
@param str The starting balance, as a String.
044
*/
public void .
am sorry but i have no idea about what you are sa.pdf
am sorry but i have no idea about what you are saying
Solution
am sorry but i have no idea about what you are saying.
Points.java import java.util.ArrayList; import java.util.Scan.pdf
// Points.java
import java.util.ArrayList;
import java.util.Scanner;
public class Points{
double x;
double y;
Points(double xco, double yco){
x = xco;
y = yco;
}
}
// PolygonArea.java
import java.util.ArrayList;
import java.util.Scanner;
public class PolygonArea {
// funtion to determine area of polygon
public static double getArea(ArrayList coordinates) {
double sumxy = 0;
double sumyx = 0;
for (int i = 0; i < coordinates.size(); i++) {
int index = (i + 1) % coordinates.size();
Points point1 = coordinates.get(i);
Points point2 = coordinates.get(index);
sumxy += (point1.x * point2.y);
sumyx += (point1.y * point2.x);
}
double areaPolygon = 0.5 * (sumxy - sumyx);
if (areaPolygon > 0)
return areaPolygon;
else
return -areaPolygon;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print(\"Enter the number of the points: \");
int totalPoints = scan.nextInt();
System.out.print(\"Enter the coordinates of the points: \");
ArrayList coordinates = new ArrayList<>();
for (int i = 0; i < totalPoints; i++) {
coordinates.add(new Points(scan.nextDouble(), scan.nextDouble()));
}
System.out.println(\"The total area is \" + getArea(coordinates));
}
}
/*
output:
Enter the number of the points: 3
Enter the coordinates of the points: 0 0 0 3 4 0
The total area is 6.0
*/
Solution
// Points.java
import java.util.ArrayList;
import java.util.Scanner;
public class Points{
double x;
double y;
Points(double xco, double yco){
x = xco;
y = yco;
}
}
// PolygonArea.java
import java.util.ArrayList;
import java.util.Scanner;
public class PolygonArea {
// funtion to determine area of polygon
public static double getArea(ArrayList coordinates) {
double sumxy = 0;
double sumyx = 0;
for (int i = 0; i < coordinates.size(); i++) {
int index = (i + 1) % coordinates.size();
Points point1 = coordinates.get(i);
Points point2 = coordinates.get(index);
sumxy += (point1.x * point2.y);
sumyx += (point1.y * point2.x);
}
double areaPolygon = 0.5 * (sumxy - sumyx);
if (areaPolygon > 0)
return areaPolygon;
else
return -areaPolygon;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print(\"Enter the number of the points: \");
int totalPoints = scan.nextInt();
System.out.print(\"Enter the coordinates of the points: \");
ArrayList coordinates = new ArrayList<>();
for (int i = 0; i < totalPoints; i++) {
coordinates.add(new Points(scan.nextDouble(), scan.nextDouble()));
}
System.out.println(\"The total area is \" + getArea(coordinates));
}
}
/*
output:
Enter the number of the points: 3
Enter the coordinates of the points: 0 0 0 3 4 0
The total area is 6.0
*/.
General Instructions 1. The objective of this check list is to guid.pdf
General Instructions: 1. The objective of this check list is to guide the Internal Auditors in the
conduct of Internal Audit in a systematic manner so that all the significant points in finance &
accounts, procurement, contract & tenders, payment of bills of contractors, planning process,
flow of funds and staffing etc. are checked before finalization of Internal Audit Report 2. The
Internal Auditors should familiarize themselves with the Manual on Financial Management &
Procurement thoroughly before conducting Internal Audit. Para 100 to 105 of the Manual on
Financial Management & Procurement provides broad guidelines for Internal Audit of SSA and
Paras 51 to 99.7 of the Manual contains guidelines for maintenance of accounts . 3. Scope and
coverage of Internal Audit and report – Internal Audit of District Project Offices and sub -district
units selected on a percentage basis should be conducted so as to cover all districts and sub-
districts units at least once in 3 years . The Internal Audit report should indicate name of unit
covered, date of Audit, review of planning process to ascertain whether the activities under SSA
have been planned as per prescribed norms, review of financial systems to check the efficacy of
internal checks and controls, review of establishment matters and staffing, review of
Procurement and status of Statutory Audit, C& AG Audit and Internal Audit. 4 Finance &
Accounts – 4.1 To check whether 22 Registers( Annexure-I is attached for ready reference)
indicated in para 52 of FM&P Manual are maintained. 4.2 To verify about the number of savings
Bank Accounts maintained for SSA at SPO /DPO /Block & VEC Levels. The matter should be
reported if more then one savings bank accounts are maintained at any levels. The reasons for
maintenance of multiple Bank Accounts should also be mentioned in the Internal Audit Report
4.3 Cash book - To check cash book for ascertaining whether cash book is written daily as
envisaged in para 79 of FM&P Manual monthly on regular basis. The Internal Auditor must also
conduct physical verification of cash and report heavy cash balance if any. 4.4 To check whether
Bank Reconciliation is done monthly on regular basis or not 4.5 Payment of bills of contractors
Annexure-P To verify whether 4.5.1 Original bill duly signed by contractors is submitted. 4.5.2
Contractor has put his initials in all cuttings and corrections in the bill. 4.5.3 All Supporting
documents are attached with the bills. 4.5.4 The rates, security deposit, and deductions are as per
terms and conditions specified in the agreement 4.5.5 The variations in quantities and completion
period etc. have been authorized by the competent authority 4.5.6 Job completion certificate has
been processed by the dealing assistant of SSA. 4.5.7 Bills passed for payment are as per rules of
the society. 4.5.8 All terms and conditions of the contract are fulfilled before passing the bills.
4.5.9 Every final bill is checked in detail with measurement boo.
Computer security, additionally called cybersecurity or IT security,.pdf
Computer security, additionally called cybersecurity or IT security, is that the protection of
knowledge systems from thieving or harm to the hardware, the package, and to the knowledge on
them, moreover as from disruption or misdirection of the services they supply.[1]
It includes dominant physical access to the hardware, moreover as protective against damage that
will come back via network access, information and code injection,[2] and thanks to malpractice
by operators, whether or not intentional, accidental, or thanks to them being tricked into
deviating from secure procedures.[3]
The field is of growing importance thanks to the increasing reliance on laptop systems and
therefore the web in most societies,[4] wireless networks like Bluetooth and Wi-Fi - and
therefore the growth of \"smart\" devices, as well as smartphones, televisions and small devices
as a part of the net of Things.
A vulnerability may be a system susceptibleness or flaw. several vulnerabilities ar documented
within the Common Vulnerabilities and Exposures (CVE) information. AN exploitable
vulnerability is one that a minimum of one operating attack or \"exploit\" exists.[5]
To secure a automatic data processing system, it\'s vital to know the attacks that may be created
against it, and these threats will generally be classified into one among the classes below:
Backdoors[edit]
A backdoor in an exceedingly automatic data processing system, a cryptosystem or AN
algorithmic rule, is any secret methodology of bypassing traditional authentication or security
controls. they will exist for variety of reasons, as well as by original style or from poor
configuration. they will are accessorial by a licensed party to permit some legitimate access, or
by AN assailant for malicious reasons; however notwithstanding the motives for his or her
existence, they produce a vulnerability.
Denial-of-service attack[edit]
Denial of service attacks ar designed to create a machine or network resource unavailable to its
supposed users.[6] Attackers will deny service to individual victims, like by deliberately coming
into a wrong countersign enough consecutive times to cause the victim account to be fast, or they
will overload the capabilities of a machine or network and block all users directly. whereas a
network attack from one IP address may be blocked by adding a brand new firewall rule, several
sorts of Distributed denial of service (DDoS) attacks ar doable, wherever the attack comes from
an outsized variety of points – and defensive is far harder. Such attacks will originate from the
zombie computers of a botnet, however a spread of alternative techniques ar doable as well as
reflection and amplification attacks, wherever innocent systems ar fooled into causing traffic to
the victim.
Solution
Computer security, additionally called cybersecurity or IT security, is that the protection of
knowledge systems from thieving or harm to the hardware, the package, and to the knowledge on
them, moreover.
C. Nitrosomonas, Nitrosolobus, and NitrosovibrioNitrifiers are che.pdf
C. Nitrosomonas, Nitrosolobus, and Nitrosovibrio
Nitrifiers are chemolithotropic organisms which obtains energy by oxidating inorganic nitrogen
compounds. These three organisms convert Ammonia to Nitrites which inturn converted into
nitrate by other organisms such as Nitrobacter or Nitrospina.
False: Neisseria gonorrhoeae - responsible for causing sexually transmitted disease Gonorrhea
Neisseria meningitidis- responsible for causing meningitis and sepsis
Thiobacillus spp- It belongs to Hydrogenophilaceae family and mostly responsible for oxidizing
hydrogen.
Solution
C. Nitrosomonas, Nitrosolobus, and Nitrosovibrio
Nitrifiers are chemolithotropic organisms which obtains energy by oxidating inorganic nitrogen
compounds. These three organisms convert Ammonia to Nitrites which inturn converted into
nitrate by other organisms such as Nitrobacter or Nitrospina.
False: Neisseria gonorrhoeae - responsible for causing sexually transmitted disease Gonorrhea
Neisseria meningitidis- responsible for causing meningitis and sepsis
Thiobacillus spp- It belongs to Hydrogenophilaceae family and mostly responsible for oxidizing
hydrogen..
C is correct. Routing and logical addressing occur at the Internet l.pdf
C is correct. Routing and logical addressing occur at the Internet layer of the TCP/IP stack.
A, B, and D are incorrect. A is incorrect because the Network layer is an OSI model layer. B is
incorrect because Data Link is an OSI model layer. D is incorrect because framing, error
checking, and media access occur at the Network Access layer of the TCP/IP stack.
Solution
C is correct. Routing and logical addressing occur at the Internet layer of the TCP/IP stack.
A, B, and D are incorrect. A is incorrect because the Network layer is an OSI model layer. B is
incorrect because Data Link is an OSI model layer. D is incorrect because framing, error
checking, and media access occur at the Network Access layer of the TCP/IP stack..
AnswerThe gill and siphon withdrawal complex in Aplysia (a marine.pdf
Answer:
The gill and siphon withdrawal complex in Aplysia (a marine snail) is an involuntary defensive
reflex, which causes the marine snail to retract its siphon and gill when the animal is disturbed.
There are 3 mechanisms used to stimulate the process of involuntary defensive reflex as soon as
an external stimuli trigger its activity on the peripheral organs of mollusk, Aplysia. The three
mechanisms are habituation, dishabituation and sensitization. The central and pheripheral
neurons are involved in an involuntary defensive reflex of Aplysia as part of defense mechanism.
The neurobiological model is mediated by six central motor neurons in Aplysia with abdominal
ganglion to produce involuntary defensive reflex to stimulate gills & these neurons are L7,
LDG1, LDG2 and RDG used for larger gill contractions. Smaller gill involuntary defensive
reflex is mediate L9G1 and L9G2 whereas siphon is innervated by LDS1, LDS2, LDS3, RDS,
LBS1, LBS2 to mediate involuntary defensive reflex. Habituation and dishabituation is mediated
based on the external stimuli on the gills finally some reflexes are going to activate “ink glands”
to trigger release of ink when a predator arrived to its surrounds result in escaping from the
predator
Solution
Answer:
The gill and siphon withdrawal complex in Aplysia (a marine snail) is an involuntary defensive
reflex, which causes the marine snail to retract its siphon and gill when the animal is disturbed.
There are 3 mechanisms used to stimulate the process of involuntary defensive reflex as soon as
an external stimuli trigger its activity on the peripheral organs of mollusk, Aplysia. The three
mechanisms are habituation, dishabituation and sensitization. The central and pheripheral
neurons are involved in an involuntary defensive reflex of Aplysia as part of defense mechanism.
The neurobiological model is mediated by six central motor neurons in Aplysia with abdominal
ganglion to produce involuntary defensive reflex to stimulate gills & these neurons are L7,
LDG1, LDG2 and RDG used for larger gill contractions. Smaller gill involuntary defensive
reflex is mediate L9G1 and L9G2 whereas siphon is innervated by LDS1, LDS2, LDS3, RDS,
LBS1, LBS2 to mediate involuntary defensive reflex. Habituation and dishabituation is mediated
based on the external stimuli on the gills finally some reflexes are going to activate “ink glands”
to trigger release of ink when a predator arrived to its surrounds result in escaping from the
predator.
More from KUNALHARCHANDANI1 (20)
HClO4 in aqueous medium ionizes as Hydrogen(+1)ca.pdf
HClO4 in aqueous medium ionizes as Hydrogen(+1)ca.pdf
Which of the following would be a description of a system unitA c.pdf
Which of the following would be a description of a system unitA c.pdf
Water is a polar inorganic solvent. Benzene is a nonpolar organic so.pdf
Water is a polar inorganic solvent. Benzene is a nonpolar organic so.pdf
viruses which have single strande DNA in their genome come under cat.pdf
viruses which have single strande DNA in their genome come under cat.pdf
This word problem is about a triangle whose perimeter is 47 miles. S.pdf
This word problem is about a triangle whose perimeter is 47 miles. S.pdf
CH4 + 2O2 -- CO2 + 2H2O note CH4 combustion is.pdf
CH4 + 2O2 -- CO2 + 2H2O note CH4 combustion is.pdf
Carbon 2 is where D and L differ for all sugars. .pdf
Carbon 2 is where D and L differ for all sugars. .pdf
Program to print the Diamond Shape -#include stdio.h int ma.pdf
Program to print the Diamond Shape -#include stdio.h int ma.pdf
Part D option 4 is answerUnless untill they are exposed by some me.pdf
Part D option 4 is answerUnless untill they are exposed by some me.pdf
Isomers which have their atoms connected in the same sequence but di.pdf
Isomers which have their atoms connected in the same sequence but di.pdf
import java.util.Scanner;import java.text.DecimalFormat;import j.pdf
import java.util.Scanner;import java.text.DecimalFormat;import j.pdf
am sorry but i have no idea about what you are sa.pdf
am sorry but i have no idea about what you are sa.pdf
Points.java import java.util.ArrayList; import java.util.Scan.pdf
Points.java import java.util.ArrayList; import java.util.Scan.pdf
General Instructions 1. The objective of this check list is to guid.pdf
General Instructions 1. The objective of this check list is to guid.pdf
Computer security, additionally called cybersecurity or IT security,.pdf
Computer security, additionally called cybersecurity or IT security,.pdf
C. Nitrosomonas, Nitrosolobus, and NitrosovibrioNitrifiers are che.pdf
C. Nitrosomonas, Nitrosolobus, and NitrosovibrioNitrifiers are che.pdf
C is correct. Routing and logical addressing occur at the Internet l.pdf
C is correct. Routing and logical addressing occur at the Internet l.pdf
AnswerThe gill and siphon withdrawal complex in Aplysia (a marine.pdf
AnswerThe gill and siphon withdrawal complex in Aplysia (a marine.pdf
Recently uploaded
Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
special B.ed 2nd year old paper_20240531.pdf
Instagram:-
https://instagram.com/special_education_needs_01?igshid=YmMyMTA2M2Y=
WhatsApp:-
https://chat.whatsapp.com/JVakNIYlSV94x7bwunO3Dc
YouTube:-
https://youtube.com/@special_education_needs
Teligram :- https://t.me/special_education_needs
Slide Shere :-
https://www.slideshare.net/shabnambano20?utm_campaign=profiletracking&utm_medium=sssite&utm_source=ssslideview
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
678020731-Sumas-y-Restas-Para-Colorear.pdf
KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK
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/
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Recently uploaded (20)
Digital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and Research
2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Home assignment II on Spectroscopy 2024 Answers.pdf
Home assignment II on Spectroscopy 2024 Answers.pdf
"Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Unit 8 - Information and Communication Technology (Paper I).pdf
Unit 8 - Information and Communication Technology (Paper I).pdf
Instructions for Submissions thorugh G- Classroom.pptx
Instructions for Submissions thorugh G- Classroom.pptx
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI fund...
The French Revolution Class 9 Study Material pdf free download
The French Revolution Class 9 Study Material pdf free download
the link is not working can u pls write questions.pdf
- 1. the link is not working can u pls write questions dirctly here itself thanks Solution the link is not working can u pls write questions dirctly here itself thanks