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zhuklara [117]
2 years ago
3

A Transmission Control Protocol (TCP) connection is established and two devices ensure that they're speaking the same protocol.

What has occured?
A. Three-way handshake
B. Two-way handshake
C. Handshake
D. Four-way handshake
Computers and Technology
1 answer:
klasskru [66]2 years ago
7 0

Answer:

The correct option is;

A. Three-way handshake

Explanation:

For establishment of connection within Transmission Control Protocol, (T. C. P.), involves a three-way way handshake. Prior to attempting a server connection, the server to which connection is sought passively opens a port by listening at the port. Upon establishment of passive open, active open by the client can then be initiated by the client. A connection establishment requires a three-way handshake as follows;

1. The client sends a SYN to the server

2. The server responds by sending a SYN-ACK

3. The client further responds sending ACK back to the server.

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All of the following are specific security challenges that threaten corporate systems in a client/server environment except:
Fudgin [204]

Answer:

Option D i.e., radiation is the correct answer.

Explanation:

When any person challenges the specific security system for the purpose of stealing the resources from the corporate security system then they unable to threaten the radiation because it is the electromagnetic waves that affect the body or skin of any human being and it can be generated by those devices that have high-voltage.

8 0
2 years ago
9. Which of the following are true for all regular languages and all homomorphisms? (a) h (L1 ∪ L2)= h (L1) ∩ h (L2). (b) h (L1
mixer [17]

Answer:

h (L1 ∪ L2)= h (L1) ∩ h (L2).

This is true.

There will be w ∈  (L1 ∪ L2) for any s ∈ h (L1 ∪ L2) in such a way that s=h(w)

we can assume that w ∈ L1

So In this case h(w) ∈ L (S1). Hence s ∈ L(S1)

for any s ∈ h (L1) U h(L2)

We can assume that s ∈ L(S1)

There exists w ∈ L1 such that s= h(w)

In this case it is w ∈ L1 U L2 as well.

Hence , s ∈ h (L1 U L2)

Explanation

consider  = 0,1 and  = a,b and h(0) = a , h(1) = ab

(a) Consider L1 = 10,01 and L2 = 00,11

   Now L1 ∪ L2 = 00,01,10,11

   h (L1 ∪ L2) = h(00) , h(01) , h(10) , h(11) = h(0)h(0) , h(0)h(1) , h(1)h(0) , h(1)h(1)

  = aa, aab , aba , abab

Hence h (L1 ∪ L2) = aa, aab , aba , abab .

Here h (L1) = h(10) , h(01) = h(1)h(0) , h(0)h(1) = aba , aab

Hence h (L1) = aba , aab .

Here h (L2) = h(00) , h(11) = h(0)h(0), h(1)h(1) = aa, abab

Hence h(L2) = aa, abab.

Finally Hence , h (L1 ∪ L2)= h (L1) ∩ h (L2).

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