Answer:
It throws an error.
the public class needs a name.
like this:
public class G{ public static void main(String[] args) {
int x=5 , y = 10;
if (x>5 && y>=2) System.out.println("Class 1");
else if (x<14 || y>5) System.out.println(" Class 2");
else System.out.println(" Class 3"); }// end of main
}
if you give the class a name and format it, you get:
Class 2
Explanation:
Face you can talk to friends post picture's and see pictures of friends i am not really sure what is linkedln
1.)
<span>((i <= n) && (a[i] == 0)) || (((i >= n) && (a[i-1] == 0))) </span>
<span>The expression will be true IF the first part is true, or if the first part is false and the second part is true. This is because || uses "short circuit" evaluation. If the first term is true, then the second term is *never even evaluated*. </span>
<span>For || the expression is true if *either* part is true, and for && the expression is true only if *both* parts are true. </span>
<span>a.) (i <= n) || (i >= n) </span>
<span>This means that either, or both, of these terms is true. This isn't sufficient to make the original term true. </span>
<span>b.) (a[i] == 0) && (a[i-1] == 0) </span>
<span>This means that both of these terms are true. We substitute. </span>
<span>((i <= n) && true) || (((i >= n) && true)) </span>
<span>Remember that && is true only if both parts are true. So if you have x && true, then the truth depends entirely on x. Thus x && true is the same as just x. The above predicate reduces to: </span>
<span>(i <= n) || (i >= n) </span>
<span>This is clearly always true. </span>
Explanation:
Talk About The Benefits. of your service over other compines
Answer:
margins
Explanation:
the definition margins are the edges
