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o-na [289]
3 years ago
6

Find the Thevenin Equivalent.

Engineering
2 answers:
Eva8 [605]3 years ago
8 0

Answer:

pake ayus nong pic

Explanation:

di ko gets ei

arlik [135]3 years ago
4 0

Answer:

hindi ko po gets

Explanation:

paulit po ng picture

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How many different powerball combinations are there
svlad2 [7]
Answer:





There 11,238,513 different Powerball
4 0
2 years ago
What do you think the top TWO game elements are that directly contribute to player immersion?
fredd [130]
Sound (like music and background sound) and detail. The music and background sound is what sets the mood for the game. In horror games, the sound is sometimes uneasy to make the player feel anxious, and when something chases you the music turns into fast-pace and making the player scared and feel that adrenaline. The detail also is a great factor. The more realistic a game is, the more it feels like real life and determines if the player will get a real reaction from the game.
6 0
3 years ago
This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2
amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

4 0
3 years ago
What are the two reasons for a clear cut
Inessa [10]

Answer:

to clear land for agriculture and settlement and to use or sell timber for lumber, paper products, or fuel.

3 0
3 years ago
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