Answer:
Explanation:
Given
Required
The result when 
and 
Analyzing the given instruction
a*=(++a)/(6)+(b++3)
Single out and solve the expressions in bracket
(++a) = a -- When the ++ operator appears before an operand, it is called pre increment. meaning that the operation will be done before the operand will be incremented.
So: in this case: ++a = a
The operator, as used in that statement is the same as: b + 3.
So:
The above expression is calculated as:
So:
The answer is B Okay good luck and dont...mess it up
Hello!
My best guess is you would have to use mediocre satellites that float over for internet connection.
Answer:
The solution code is written in Java
- public static void checkCommonValues(int arr1[], int arr2[]){
- if(arr1.length < arr2.length){
- for(int i = 0; i < arr1.length; i++){
- for(int j = 0; j < arr2.length; j++){
- if(arr1[i] == arr2[j]){
- System.out.print(arr1[i] + " ");
- }
- }
- }
- }
- else{
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr2[i] == arr1[j]){
- System.out.print(arr2[i] + " ");
- }
- }
- }
- }
- }
Explanation:
The key idea of this method is to repeated get a value from the shorter array to check against the all the values from a longer array. If any comparison result in True, the program shall display the integer.
Based on this idea, an if-else condition is defined (Line 2). Outer loop will traverse through the shorter array (Line 3, 12) and the inner loop will traverse the longer array (Line 4, 13). Within the inner loop, there is another if condition to check if the current value is equal to any value in the longer array, if so, print the common value (Line 5-7, 14-16).
Answer:
(c) the dynamic type of reference will determine which of the methods to call.
Explanation:
Polymorphism in Object Oriented Programming typically means the same method name can cause different actions depending on which object it is invoked on. Polymorphism allows for dynamic binding in that method invocation is not bound to the method definition until the program executes.
So in the case of Animal superclass and Mammal subclass, both having a method called eat() with identical signatures and return types, depending on which reference, the correct method eat() will be called dynamically upon execution.
For example, if we have the following;
================================
<em>Mammal mammal = new Animal();</em>
<em>mammal.eat()</em>
================================
The eat() method that will be called is the one in the Mammal subclass.
However, if we have;
================================
<em>Animal animal = new Animal();</em>
<em>animal.eat()</em>
================================
The eat() method of the Animal superclass will be called.
