Answer:
Oct 30, 2018 — ... design? A) difficulty level B) level duration C) perspective D) player-adjusted time. ... Add answer+5 pts. Log in to add comment. jonathanwilson is waiting for your help. Add your answer and earn points. ... New questions in Computer Science. answer ... classified the computer on the basic of operations.Explanation:
Because the string is invalid or your source is not attached to your search engine.
Answer:
Variable Where Declared
In Sub1 :
A Sub1
Y Sub1
Z Sub1
X Main
In Sub2:
A Sub2
B Sub2
Z Sub2
Y Sub1
X Main
In Sub3 :
A Sub3
X Sub3
W Sub3
Y Main
Z Main
Explanation:
In static-scoped languages with nested subprograms, the declaration of a variable is checked with the subprogram, if it is not found, it check within the parent method that called it, it continue until it find a declaration, if no declaration is found, it display an error.
In Sub1, a, y, z is declared there while the declaration of x is found in main.
In Sub2, a, b, z is declared, declaration of y is found in sub1 and declaration of x is found in main.
In Sub3, a, x, w is declared while the declaration of y, z is from the main.
Answer:
Explanation:
The following program creates a function called region_Matches that takes in two strings as arguments as well as an int for starting point and an int for amount of characters to compare. Then it compares those characters in each of the words. If they match (ignoring case) then the function outputs True, else it ouputs False. The test cases compare the words "moving" and "loving", the first test case compares the words starting at point 0 which outputs false because m and l are different. Test case 2 ouputs True since it starts at point 1 which is o and o.
class Brainly {
public static void main(String[] args) {
String word1 = "moving";
String word2 = "loving";
boolean result = region_Matches(word1, word2, 0, 4);
boolean result2 = region_Matches(word1, word2, 1, 4);
System.out.println(result);
System.out.println(result2);
}
public static boolean region_Matches(String word1, String word2, int start, int numberOfChars) {
boolean same = true;
for (int x = 0; x < numberOfChars; x++) {
if (Character.toLowerCase(word1.charAt(start + x)) == Character.toLowerCase(word2.charAt(start + x))) {
continue;
} else {
same = false;
break;
}
}
return same;
}
}
Pie graph but idk if thats the answer