Explanation:
the question does not have an answer probably what they have in common
Hello,
The answer is option A p<span>ay attention to URL endings, such as .gov and .com.
Reason:
Its A because its a very important guideline to make sure you look for the root words .gov and .com to make sure its professional. Its not option B because using search engines can cause viruses. Its also not option C because anything can look professional just by how the website looks and says.
If you need anymore help feel free to ask me!
Hope this helps!
~Nonportrit</span>
Complete Question:
LDAP is an industry standard employed by Microsoft, which enables IT departments to use a(n) __________ structure when creating user accounts and user groups.
(a)directory tree
(b)Organizational Unit (OU)
(c)forest
(d)file system
Answer:
(a) directory tree structure
Explanation:
Lightweight Directory Access Protocol (LDAP) is a special type of Directory Access Protocol (DAP) and it is a software protocol that allows the storage and retrieval of data objects (such as user accounts and user groups) that are arranged in an hierarchical directory structure. This hierarchical directory structure is called directory tree or directory information tree.
LDAP stores users and resources information of an organization. These information include usernames, passwords, email, human resource data e.t.c
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
The correct answer is d) all of the above
Explanation:
The size of the info is necessary to give the users heads up about the amount of data needed to access it, the type of the video is also important so the user can know the environment appropriate for viewing it and also tips for viewing will help users know the requirements needed to view it.
