The skill thar is being displayed I thing is being outgoing and loyal
Answer: is a unified standard used when recording the sounds of instruments
Answer:
public class Main
{
// required method
public static void fizzBuzz(){
// looping through 1 to 100
for(int i = 1; i <= 100; i++){
//if number is evenly divisible by both 3 and 5
if(i%3 == 0 && i%5 == 0){
System.out.println("fiz buzz");
}
// if number is divisible by 3
else if (i%3 == 0){
System.out.println("fizz");
}
// if number is divisible by 5
else if (i%5 == 0){
System.out.println("buzz");
}
// if number is not divisible by both 3 and 5
else {
System.out.println(i);
}
}
}
// main method
public static void main(String[] args) {
//calling function
fizzBuzz();
}
}
Explanation:
Answer:
public class MagicSquare {
public static void main(String[] args) {
int[][] square = {
{ 8, 11, 14, 1},
{13, 2, 7,12},
{ 3, 16, 9, 6},
{10, 5, 4, 15}
};
System.out.printf("The square %s a magic square. %n",
(isMagicSquare(square) ? "is" : "is not"));
}
public static boolean isMagicSquare(int[][] square) {
if(square.length != square[0].length) {
return false;
}
int sum = 0;
for(int i = 0; i < square[0].length; ++i) {
sum += square[0][i];
}
int d1 = 0, d2 = 0;
for(int i = 0; i < square.length; ++i) {
int row_sum = 0;
int col_sum = 0;
for(int j = 0; j < square[0].length; ++j) {
if(i == j) {
d1 += square[i][j];
}
if(j == square.length-i-1) {
d2 += square[i][j];
}
row_sum += square[i][j];
col_sum += square[j][i];
}
if(row_sum != sum || col_sum != sum) {
return false;
}
}
return d1 == sum && d2 == sum;
}
}
Answer: Joystick is the only one that makes sense, a stylus and digitizer are both used on drawing tablets, not computer games.
