Why is my speedtest is very good but chrome is slow

Every call to a recursive function has its own code and its own set of ____ and local variables

777dan777 [17]

Answer: Parameters

Explanation:

Whenever a call to a recursive function is made, then the function has its own code and its own set of parameters with local variables. These parameters are within the scope of the recursive function. For example while finding the factorial of a number we are given the function with parameter such as int recursive(int n) where int n is a parameter passed into the function.

8 0

3 years ago

Write a program that has a conversation with the user. The program must ask for both strings and numbers as input. The program m

konstantin123 [22]

Answer:

yo im sorry eat my cookie

Explanation

doorkoeeworkwoeroewkrwerewrwe

5 0

3 years ago

In the output at the command line from a router that reads - ip ospf message-digest-key 1 md5 CISCO-123 - what does the CISCO-12

11111nata11111 [884]

Answer: a) it is used as a password.

Explanation:

In order to be able to exchange routing information , CISCO routers use an authentication process, so when they send a message (a HELLO message for instance) the router that receives the message confirms that the message is authentic.

There are several types of authentication in OSPF, null, plain and MD5.

The difference between them is as follows:

1) null, implies that no password is included in the packet header, it's the default mode.

2) plain: In this type of authentication, the password is sent transparently from a router to his neighbor, as a plain text which makes it vulnerable to attacks from packet inspection software, like Sniffer.

3) MD5, is a more secure mode, in which the password is not sent directly, but a hash built based on the packet contents and the password itself instead.

MD5 (Message Digest 5) is the name of the algoritm used to build the hash that receiver will compare with the sent by the sender in order to confirm authentication.

The chosen password, is placed at the end of the command, after the type of authentication (MD5 in this case), which in this case is "CISCO-123".

7 0

4 years ago

Contrast the following terms (provide examples): Partial dependency; transitive dependency

kiruha [24]

Answer:

Explanation:

Transitive dependency

In this case, we have three fields, where field 2 depends on field 1, and field three depends on field 2.

For example:

Date of birth --> age --> vote

Partial dependency

It is a partial functional dependency if the removal of any attribute Y from X, and the dependency always is valid

For example:

Course and student these tables have a partial dependency, but if we have the field registration date, this date will depend on the course and student completely, we must create another table with the field registration date to remove this complete dependency.

If we remove or update the table registration date, neither course nor student must not change.

5 0

3 years ago

⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val

timurjin [86]

Answer:

a) There are approximately 2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b) ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 = 4.824 × 10^9 bytes. Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a) There are approximately 2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b) ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 = 4.824 × 10^9 bytes. Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

8 0

3 years ago