Answer Research is defined as the creation of new knowledge and/or the use of existing knowledge in a new and creative way so as to generate new concepts, methodologies and understandings
What is the question here?
Answer:
1100111111110100000110 = �
Binary -> UTF-16
3,407,110 in decimal form
Additionally, it also translates to the color green in hexadecimal.
Not really sure what you are trying to translate this in to though.
The kinds and ways to improve your diagnostic and troubleshooting skills are:
- Be Relax and never panic when you encounter it.
- Know everything about your computer.
- Look for solutions and clues and state them down.
- Find out the repeatability.
<h3>What is diagnostic and troubleshooting?</h3>
Diagnosing is known to be the act of finding out the root cause of any issue through an act of elimination but troubleshooting is known to be the act of fixing of the problem after diagnosis is said to have been carried out.
Therefore, The kinds and ways to improve your diagnostic and troubleshooting skills are:
- Be Relax and never panic when you encounter it.
- Know everything about your computer.
- Look for solutions and clues and state them down.
- Find out the repeatability.
Learn more about troubleshooting skills from
brainly.com/question/14983884
#SPJ1
Answer:
Explanation:
The following code is written in Java. It creates a function that takes in an ArrayList and an int parameter. Then it loops through the array and adds each element into a new array called newList but adds them the number of times that the numRepeats argument states. Output can be seen in the attached image below.
import java.util.ArrayList;
class Brainly{
public static void main(String[] args) {
ArrayList list = new ArrayList();
list.add('a');
list.add('b');
list.add('c');
ArrayList newList = repeatArrayList(list, 3);
System.out.println(newList);
}
public static ArrayList repeatArrayList(ArrayList list, int numRepeats) {
ArrayList newList = new ArrayList();
for (int x = 0; x < numRepeats; x++) {
for (int i = 0; i < list.size(); i++) {
newList.add(list.get(i));
}
}
return newList;
}
}