Answer:
A chart and a graph are kind of image that are indispensable and need added text to go with it.
Answer:
C. Practiced in front of a trusted adult..
Answer:
Option B is the correct answer.
Explanation:
- In the above code, the loop will execute only one time because the loop condition is false and it is the Do-While loop and the property of the Do-while loop is to execute on a single time if the loop condition is false.
- Then the statement "x*=20;" will execute one and gives the result 200 for x variable because this statement means "x=x*20".
- SO the 200 is the answer for the X variable which is described above and it is stated from option B. Hence it is the correct option while the other is not because--
- Option A states that the value is 10 but the value is 200.
- Option C states that this is an infinite loop but the loop is executed one time.
- Option D states that the loop will not be executed but the loop is executed one time
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;
}
}
