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diamong [38]
3 years ago
5

How many assignments would you have failed without brainly? lol.

Computers and Technology
1 answer:
svetoff [14.1K]3 years ago
7 0
How many assignments would I have failed without brainy? Lol

ALL OF THEM
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Two DHCP servers, Server1 and Server2, are running Windows Server 2016. As the administrator, you create a scope called Scope1.
kumpel [21]

Answer:

The answer is "Option a"

Explanation:

Split-scope is also an easy and simple approach to deliver DHCP consistency and workload management into your system. Server 2008 R2 provides a convenient divide-scope guide which removes several operational efforts but can only be to use if all databases run on R2, and wrong choices can be described as follows:

  • In option b, It uses the Ip address for multicast, that's why it is wrong.
  • In option c, It is wrong because it uses a windows interface, that works on policies.  
  • In option d, It is wrong because it is an administrative feature.

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3 years ago
The acronym LAH stands for
Nina [5.8K]
D.Limited Access Highway
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3 years ago
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Nina visited an area that receives a large amount of precipitation during all twelve months and is typically warm year round. Th
slava [35]

Answer: Rainforest

Explanation:

A rainforest is an area that typically has tall, green trees and also has a high amount of rainfall. Rainforest can be found in the wet tropical uplands and lowlands. The largest rainforests can be seen in the Amazon River Basin.

Since the area visited by Nina has a large amount of precipitation during all twelve months, typically warm year round and has a large variety of organism diversity, then the biome thatcNina most likely visited is the rainforest.

7 0
3 years ago
1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

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3 years ago
A good website design combines which of the following elements? (select all that apply) powerful web server hardware components
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Sounds like it'd be all of the above.


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3 years ago
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