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4 years ago

which option of the AutoCorrect tool enables you to add and delete words that do not follow abbreviation rules?

Computers and Technology

2 answers:

mixas84 [53]4 years ago

6 0

Answer:

The answer is the "Exceptions" option

Explanation:

The Exceptions tab helps you to add or remove exceptions to short forms and words with two initial capitals such as CDs. To add exceptions to the list of short forms, first type the short form in the Abbreviation box or Words with TWo INitial CApitals box. Then, click the New button. For example, let’s assume you want to add the word TVs to your list of exceptions. You have to type “TVs” (without quotes) in the Words with TWo INitial CApitals box. Then, click the New button. Make sure you check the AutoInclude box.

quester [9]4 years ago

3 0

The replace text as you type is an option in the AutoCorrect tool that enables the user to add or delete words that do not follow abbreviation rules. Moreover, this is a convenient way to use especially if you are dealing with long string of words that only needs abbreviation.

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The code int *p; declares p to be a(n) ____ variable. new

Pavel [41]

Int* p; defines a pointer variable.

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4 years ago

What is the difference between password protection and encryption? please answer quick, i cant pass this test for the life of me

sergiy2304 [10]

Answer:Password protection is like locking something in a safe-deposit. It means no one can get to the locked content without knowing the right combination. This method is used on separate documents, folders, and other data the computer's user may want to protect from other people who might have access to the device. The problem is, if someone interested in such content obtains the password or finds a way to open it without it, the content might be revealed despite the owner's efforts to keep it hidden. Unfortunately, there are a lot of ways hackers could obtain the password or hack in without it. For example, it could be obtained with the help of malware, or it might be guessed if the user chooses a weak password. Not to mention, when it comes to PDF documents, the passwords placed on them can be removed using the CMD window or specific.

Password encryption is a step up from password protection. The term can be a tad confusing because, in fact, you cannot encrypt the password itself. Instead, by setting up "password encryption" you are creating a password AND encrypting the contents of the file. In our example (see instructions below), the contents of the user's PDF document are not only password protected, but also encrypted. It is a process during which the content one wishes to keep secret is altered to make it unrecognizable. For example, if it is a text document, letters of each word might be shuffled with additional characters so the words would no longer make any sense. The reverse process is only available if the person who wants to decrypt this data can provide a specific decryption key or a password. In other words, even if the password is removed no one could read the hidden content as it still would need to be decrypted. Of course, it is important to realize you might be unable to retrieve it too if you lose the decryption key, aka, the password.

PLS MARK ME AS BRAINLIEST.

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2 years ago

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

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kozerog [31]

The answer is
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4 0

3 years ago

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Yuri [45]

And select copy from the file tab.

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3 years ago