The most effective way to exploit TCP three-way handshake is TCP SYN flood.
- SYN flood attack is simply a kind of computer attack that begins with 3-way TCP handshake.
- TCP SYN flood is also called SYN flood and it is a type of distributed Denial of Service (DDoS) attack that uses every part of the normal TCP three-way handshake to eat up resources on the known server and render it non-functional
- With TCP SYN flood, a hacker can eat up resources of target's computer and makes it unresponsive.
From the above, we can therefore say that the most effective way to exploit TCP three-way handshake is TCP SYN flood.
Learn more from:
brainly.com/question/14739766
Answer:
C. ground antennas
Explanation:
AKA Satellite Dishes That Communicate Just Like Direct Tv dish It is focused by a bowl-shaped parabolic dish onto a device in the center called a "feed horn", which channels the signal to a "low-noise block down converter" (LNB) which filters out unwanted interference, and sometimes converts it to yet another frequency before amplifying it and sending it to the satellite receiver
Answer:
I believe it is a sperm cell.
Explanation:
Not a sex cell because that can apply to both female and male reproduction.
Answer:
The correct answer to the following question will be Option D.
Explanation:
- A collection of nested title scopes are defined by the class hierarchy, even though a collection of nested methodologies as well as blocks see in an everything else.
- From the all, the lexical role characterizes the connection between certain scopes of names - if method d would be decided to declare somewhere within procedure c, therefore the namespace within the namespace of c would be nested.
- The more common classes in such a family hierarchy were near the peak of the tree as well as more specific classes are towards the edges.
Other choices have no connection with the given situation. So Option D seems to be the right response.
If 29 bits of the 32 available addressing bits are used for the subnet, then only 3 bits giving 2^3=8 combinations remain for the host addresses.
In reality, the all zeros and all ones addresses are reserved. So, 8 addresses can exist, but 6 of those are available.
The way the question is formulated it seems the answer 8 is what they're after.
