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Shalnov [3]
3 years ago
6

What is the best way to improve an online search

Computers and Technology
2 answers:
Slav-nsk [51]3 years ago
6 0
Best way is to put your exact question! And do "advanced search"

Hope I helped and if you need more info or another question, then feel free to ask me!
Have a nice day!
Zepler [3.9K]3 years ago
6 0
Update your content regularly
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How many responses does a computer expect to receive when it broadcasts an ARP request?why?
Alla [95]

Answer: The response that is expected when it broadcast an ARP request is one or zero.

Explanation: ARP request means Address Resolution Protocol which is a protocol responsible for the mapping of the IP(Internet protocol)address of a system to the MAC(Media Access Control) layer. Only one response is received only if the IP address is present in the ARP otherwise if the IP address does not matches then no response is returned.Thus only one or zero response can be received when a ARP request is process.

5 0
3 years ago
Where does append add the new elements?
ZanzabumX [31]

Answer:

"Option 1: To the end of an array." is the correct answer.

Explanation:

The word "Append" means adding to the end of a document.

Appending means whatever new content is added, it is added at the end of the document or data structure.

Similarly,

Appending an array means adding new elements to the end of the array.

Hence,

"Option 1: To the end of an array." is the correct answer.

7 0
3 years ago
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Leviafan [203]
1. a.
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3. b
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6 0
4 years ago
Read 2 more answers
let m be a positive integer with n bit binary representation an-1 an-2 ... a1a0 with an-1=1 what are the smallest and largest va
Ahat [919]

Answer:

Explanation:

From the given information:

a_{n-1} , a_{n-2}...a_o in binary is:

a_{n-1}\times 2^{n-1}  + a_{n-2}}\times 2^{n-2}+ ...+a_o

So, the largest number posses all a_{n-1} , a_{n-2}...a_o  nonzero, however, the smallest number has a_{n-2} , a_{n-3}...a_o all zero.

∴

The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times

i.e.

(11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}

= \dfrac{1(2^n-1)}{2-1}

\mathbf{=2^n -1}

(1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}

\mathbf {= 2 ^{n-1}}

Hence, the smallest value is \mathbf{2^{n-1}} and the largest value is \mathbf{2^{n}-1}

3 0
2 years ago
I will give you brainliest!!!
Feliz [49]

Answer:

sure i dont mind subbing

Explanation:

7 0
3 years ago
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