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vichka [17]
4 years ago
10

In order to use the Active Directory Recycle Bin, all DCs in the forest must be running at least what Windows Server operating s

ystem?
Computers and Technology
1 answer:
AleksAgata [21]4 years ago
4 0

Answer:

Windows Server 2008 R2

Explanation:

The Windows Server 2008 R2 is a window server operating system that was developed by Microsoft. The server was developed on the enhancement of the Window Server 2008. This type of window server came with different improvements and benefits (such as power consumption) over the previous window servers in the market.

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In which country did the World Cyber Games originate?
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The world cyber games originated in South Korea.
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4 years ago
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Complete the method, longestWord(), below that takes in an array of Strings and returns the longest String in the array. Note: y
beks73 [17]

Answer:

I am writing a JAVA program.

public class StringMethod {

public static String longestWord(String []words) { //method that takes an array  "word" of strings and returns the longest string in the array word

    int pos = 0; // contains index of the longest word of array words

    int longest = words[0].length(); //initially holds the length of the 1st element of words array

    for(int i=1; i< words.length; i++) { // iterates through each word of the array until the end of array

        if(words[i].length() > longest) { // if the length of word at i-th index of words[] is greater than the length of the word that is contained in longest variable

            pos = i; // assigns the index of the longest array element to pos

            longest = words[i].length();    } } //assigns the word in words[] array with the longest length to longest variable

   return words[pos]; } // returns the longest string in the array    

public static void main(String[] args) { //start of main() function body

String[] animals = {"cat", "horse", "elephant", "bird"}; //creates a String type array named animals which has 4 elements/words

   String longest_word= longestWord(animals); //calls longestWord method to find the longest string in the array animals and assign that longest string to longest_word.

  System.out.println(longest_word); }} //displays the longest word in output

Explanation:

Suppose we have the array of these elements: "cat", "horse", "elephant", "bird" These are four words in the array. Now lets see how the method longestWord() works:

int pos = 0; This variable is initialized to 0.

int longest = words[0].length();

longest holds the length of element at 0th index of words[] array. length() method is used to return the length of the first element words[0] array.

As the first element of array is "cat" so length of cat is 3. Hence longest = 3

for(int i=1; i< words.length; i++)

The loop has a variable i which is initialized to 1. The loop checks if the value of i is less than length of words[] array. The length of words array is 4 as there are 4 elements i.e. cat, horse, elephant and bird in this array. So the condition evaluates to true because i<words.length i.e. 1<4. The body of the loop will execute now. The body of loop contains an if condition if(words[i].length() > longest) which checks if the length of element of array at i-th index is greater than length of the element that is stored in longest variable. As i=1 so word[i] is word[1] which is the second element of words[] array. The second element is horse.

words[i].length()  means the length of i-th element of words[] array. So this means words[1].length() i.e. the length of element of words[] at 1st index. The length of horse is 5. So the statement if(words[i].length() > longest) is true because longest=3 and words[1].length() = 5 and 5>3. So the statements of if condition will execute now.

pos = i; this variable pos is used to hold the index position of the longest element in array. So pos = 1 as longest element computed so far is "horse".

longest = words[i].length(); This statement then assigns the length of horse to longest. So now the value of longest = 5

At second iteration i is now 2.  Loop condition evaluates to true because i<words.length i.e. 2<4. If condition is also true as i=2 so word[i] is word[2] which is elephant.

words[i].length() now means words[2].length() i.e. the length of element of words[] at 2nd index. The length of elephant is 8. So the statement if(words[i].length() > longest) is true because longest=5 and words[2].length() = 8 and 8>5. So the statements of if condition will execute now.

pos = i; now becomes pos = 2 as longest element computed so far is "elephant".

longest = words[i].length();  This statement then assigns the length of horse to longest. So now the value of longest = 8

At third iteration i is now 3.  The loop again checks if the value of i is less than length of words[] array.The condition evaluates to true because i<words.length i.e. 3<4. The body of the loop will execute now.  if(words[i].length() > longest) checks if the length of element of array at i-th index is greater than length of the element that is stored in longest variable. As i=3 so word[i] is word[3] which is the fourth element of words[] array. The fourth element is bird.

words[i].length() now means words[3].length() i.e. the length of element of words[] at 3rd index. The length of bird is 4. So the statement if(words[i].length() > longest) is false because longest=8 and words[2].length() = 4 and 4<8. So the statements of if condition will not execute now.

At fourth iteration i is now 4.  The loop again checks if the value of i is less than length of words[] array.The condition evaluates to false because i==words.length i.e. 4=4. So the loop breaks.

 return words[pos]; statement returns the index position of longest word in words[] array. pos = 2 So the array element at 2nd index is basically the third element of array words[] i.e. elephant. So the longest string in the array is elephant.

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3 years ago
Types of Hazards Mitigation Measures
Fynjy0 [20]

Answer:

auto mechanics

Explanation:

keeps dirt and grease off hands

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If your vehicle has fuel injection and the engine is cold,
olga nikolaevna [1]
D. Press the accelerator to the floor once and release it.
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4 years ago
If a priority queue is being implemented using an unordered list, what is the Big-O complexity of the Enqueue operation?
lidiya [134]

Answer:

<u>the answer is O(1)</u>

Explanation:

A priority queue is a type of queue whereby each of the elements are linked with a priority and the elements are served according to their priority.

The word enqueue means to add elements to the back of the queue

<em><u>For this question, while using an unordered list, O(1) is the big-O complexity of the enqueue operation. O(1) is a natural Choice for queues.</u></em>

6 0
3 years ago
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