Answer:
String word = "George slew the dragon";
int pos = word.indexOf("dr");
String drWord = word.substring(pos, pos+4);
System.out.println(drWord);
Explanation:
Assuming dr is always there, we don't have to check the validity of 'pos'. Normally, you would!
Answer:
sqrt(area)
Explanation:
- Here sqrt is a short form of square root.
- area is a variable name having he area of square.
- When the function is called, it will give the square root of the value stored in variable area.
As we have to find the length of the diagonal, we must knew that as all sides of the square are same in length so are the diagonals. This means that each of the four sides of the square and it diagonals are equal in length.
So a square has: length=breadth=diagonal
As Area= length*breadth
√area = length (as length = diagonal length)
So √area = diagonal length
Answer:
loop
Explanation:
Loop is the one which is used to execute the specific statement again and again until the condition is true.
In the programming, there are 3 basic loop used.
1. for loop
<u>Syntax:</u>
for(initialization, condition, increment/decrement)
{
statement;
}
the above statement execute until the condition in the for loop true when it goes to false, the loop will terminate.
2. while loop
<u>Syntax:</u>
initialization;
while(condition)
{
statement;
increment/decrement;
}
it is work same as for loop and the increment/decrement can be write after or before the statement.
3. do while
syntax:
initialization;
do
{
statement;
increment/decrement;
}while(condition);
here, the statement execute first then, it check the condition is true or not.
so, if the condition is false it execute the statement one time. this is different with other loops.
Answer:
b. The names in the list should be in alphabetical order.
Explanation:
A binary search is an algorithm used for searching for an item in a list or array. The algorithm first sorts the data structure into order and then divides it into halves. If the searched item is less than the middle item in the list, then the algorithm searches for the target in the first half, else, in the second half. This reduces the time complexity of the search.
Answer:
Pretty sure it's concentration camps
