Which code segment results in "true" being returned if a number is even? Replace "MISSING CONDITION" with the correct code segment. function isEven(num){ if(MISSING CONDITION){ return true; } else { return false;
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- Option A - A variable num modulo 2 is used to check the num variable is even or not. The modulo operator is used to check the remainder value.
- Option B - Same num variable is checking with the value of modulo 0. The even number checks based on the value modulo 2, not 0 and it is not equivalent to 2. Numbers are interchanged in this condition.
- Option C - The modulo 1 is not checking for the even number; the condition should return 'true' as per the specification prompted.
- Option D - The equivalent number value is not 2 and then the modulo should be modulo 2, not modulo 1. It will return the 'false' and wrong checks.
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- Cyclops numbersdef is_cyclops(n):A nonnegative integer is said to be a cyclops number if it consists of an odd number of digits so that the middle (more poetically, the “eye”) digit is a zero, and all other digits of that number are nonzero. This function should determine whether its parameter integer n is a cyclops number, and return either True or False accordingly n Expected result 0 True 101 True 98053 True 777888999 False 1056 False 675409820 False#include <stdio.h>#include <stdlib.h> int cent50 = 0;int cent20 = 0;int cent10 = 0;int cent05 = 0; //Function definitionvoid calculateChange(int change) {if(change > 0) {if(change >= 50) {change -= 50;cent50++;} else if(change >= 20) {change -= 20;cent20++;} else if(change >= 10) {change -= 10;cent10++;} else if(change >= 05) {change -= 05;cent05++;}calculateChange(change);}} //Define the functionvoid printChange() { if(cent50)printf("\n50 Cents : %d coins", cent50); if(cent20)printf("\n20 Cents : %d coins", cent20); if(cent10)printf("\n10 Cents : %d coins", cent10); if(cent05)printf("\n05 Cents : %d coins", cent05); cent50 = 0;cent20 = 0;cent10 = 0;cent05 = 0; } //Function's definitionint TakeChange() { int change;printf("\nEnter the amount : ");scanf("%d", &change);return change; }//main functionint main() {//call the functionint change = TakeChange(); //use while-loop to repeatedly ask for input to the userwhile(change != -1){if((change %…#include <stdio.h>#include <stdlib.h> int cent50 = 0;int cent20 = 0;int cent10 = 0;int cent05 = 0; //Function definitionvoid calculateChange(int change) {if(change > 0) {if(change >= 50) {change -= 50;cent50++;} else if(change >= 20) {change -= 20;cent20++;} else if(change >= 10) {change -= 10;cent10++;} else if(change >= 05) {change -= 05;cent05++;}calculateChange(change);}} //Define the functionvoid printChange() { if(cent50)printf("\n50 Cents : %d coins", cent50); if(cent20)printf("\n20 Cents : %d coins", cent20); if(cent10)printf("\n10 Cents : %d coins", cent10); if(cent05)printf("\n05 Cents : %d coins", cent05);cent50 = 0;cent20 = 0;cent10 = 0;cent05 = 0; } //Function's definitionint TakeChange() { int change;printf("\nEnter the amount : ");scanf("%d", &change);return change; }//main functionint main() {//call the functionint change = TakeChange(); //use while-loop to repeatedly ask for input to the userwhile(change != -1){if((change %…
- #include <stdio.h>#include <stdlib.h> int cent50 = 0;int cent20 = 0;int cent10 = 0;int cent05 = 0; //Function definitionvoid calculateChange(int change) {if(change > 0) {if(change >= 50) {change -= 50;cent50++;} else if(change >= 20) {change -= 20;cent20++;} else if(change >= 10) {change -= 10;cent10++;} else if(change >= 05) {change -= 05;cent05++;}calculateChange(change);}} //Define the functionvoid printChange() { if(cent50)printf("\n50 Cents : %d coins", cent50); if(cent20)printf("\n20 Cents : %d coins", cent20); if(cent10)printf("\n10 Cents : %d coins", cent10); if(cent05)printf("\n05 Cents : %d coins", cent05);cent50 = 0;cent20 = 0;cent10 = 0;cent05 = 0; } //Function's definitionint TakeChange() { int change;printf("\nEnter the amount : ");scanf("%d", &change);return change; }//main functionint main() {//call the functionint change = TakeChange(); //use while-loop to repeatedly ask for input to the userwhile(change != -1){if((change %…int FindSmallestVal() { int num = 0, min = 0; // reads num until the num > 0 while (num <= 0) { cin >> num; // finds the min value in the min,num min = num < min ? num : min; } // returns min return min; }Sum of two squares def sum_of_two_squares(n): Some positive integers can be expressed as a sum of two squares of some positive integers greater than zero. For example, 74 = 49 + 25 = 72 + 52. This function should find and return a tuple of two positive integers whose squares together add up to n, or return None if the parameter n cannot be broken into a sum of two squares. To facilitate the automated testing, the returned tuple must present the larger of its two numbers first. Furthermore, if some integer can be broken down to a sum of squares in several ways, return the breakdown that maximizes the larger number. For example, the number 85 allows two such representations 72 + 62 and 92 + 22 , of which this function must return (9, 2). The technique of two approaching indices that start from the beginning and end of a sequence, respectively, and approach each other until they meet somewhere, used in the function two_summers in one of our class examples, is directly applicable to this…
- Sum of two squares def sum_of_two_squares(n): Some positive integers can be expressed as a sum of two squares of some positive integers greater than zero. For example, 74 = 49 + 25 = 72 + 52. This function should find and return a tuple of two positive integers whose squares together add up to n, or return None if the parameter n cannot be broken into a sum of two squares.To facilitate the automated testing, the returned tuple must present the larger of its two numbers first. Furthermore, if some integer can be broken down to a sum of squares in several ways, return the breakdown that maximizes the larger number. For example, the number 85 allows two such representations 72 + 62 and 92 + 22 , of which this function must return (9, 2).The technique of two approaching indices that start from the beginning and end of a sequence, respectively, and approach each other until they meet somewhere, used in the function two_summers in one of our class examples, is directly applicable to this…5- int fun() int a, b, c, result; cout>a>>b>>c; cout<<"The result is "; reutrn result; } The error is Your answerQuestion 1: break phrase Problem statement Breaking a string into two parts based on a delimiter has applications. For example, given an email address, breaking it based on the delimiter "@" gives you the two parts, the mail server's domain name and email username. Another example would be separating a phone number into the area code and the rest. Given a phrase of string and a delimiter string (shorter than the phrase but may be longer than length 1), write a C++ function named break_string to break the phrase into two parts and return the parts as a C++ std::pair object (left part goes to the "first" and right part goes to the "second"). Do the following Write your algorithm as code comments. I recommend to follow UMPIRE technique Implement your function
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