Answer:
The method written in Java is as follows:
public static void classAttendance(){
Scanner input = new Scanner(System.in);
String[][] names = new String[10][10];
for(int i =0;i<10;i++){
for(int j =0;j<10;j++){
System.out.print("Student Name: "+(i+1)+" , "+(j+1)+": ");
names[i][j] = input.nextLine();
}
}
}
Explanation:
This defines the classAttendance() method
public static void classAttendance(){
Scanner input = new Scanner(System.in);
This declares the 2D array of 10 by 10 dimension as string
String[][] names = new String[10][10];
This iterates through the rows of the array
for(int i =0;i<10;i++){
This iterates through the columns of the array
for(int j =0;j<10;j++){
This prompts user for student name
System.out.print("Student Name: "+(i+1)+" , "+(j+1)+": ");
This gets the student name from the user
names[i][j] = input.nextLine();
}
}
The method ends here
}
<em>See attachment for complete program that include main method</em>
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
The answer is : Ka=0.000137
HA -> H+ + A-
0.10 x x
0.10-x +x +x
[H+]= 10^-2.44 = 0.00363
x= 0.00363
(0.00363)(0.00363)/(0.10-0.00363) = 0.000137
Ka=0.000137
