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solong [7]
2 years ago
7

What is the next line?

Computers and Technology
1 answer:
lianna [129]2 years ago
4 0

Answer:

4

Explanation:got it right on edg.

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1. _______ is when two things happen at one time when using the Scratch program.
Alja [10]
1. Parallel Execution
3. Hope that a company calls you and offers you a position
4. Intranet
5. Percent
8 0
3 years ago
Libby wrote an email to her friend. She pressed Shift and the number key 2 together to enter the email address. Which symbol did
exis [7]

When Libby wrote the email to her friend, she typed the '@' symbol. When pressing SHIFT and 2 together, it pastes this aforementioned symbol. However, there isn't any specific name for the symbol. As a matter as fact, there are several names that this symbol goes by.

The most famous name this symbol is called is the "at symbol" or the "at sign". In terms of a formal name, "commercial at" would be a good one.

Here's an example with the symbol:

[email protected]

This is essentially telling the email server where to send your email. From this, they'll know it's located at brainly.com! It's pretty neat.

4 0
3 years ago
Read 2 more answers
C++
Luden [163]

Answer:

The program to this question as follows:

Program:

//header file iostream

#include<iostream> //including file for use basic function

//using name space

using namespace std;

//main method

int main() //defining main method

{

   int a[3][3]; //defining two dimension array

   int x,y,sum=0; //defining variables

   cout<<"Enter array elements: "<<endl; //message

   for(x=0;x<3;x++) // for row

   {

       for(y=0;y<3;y++) //for column

       {

          cin>>a[x][y]; //input values from user.

       }

       

   }

   //loop for calculting sum.

   for(x=0;x<3;x++)

   {

       for(y=0;y<3;y++)

       {

       sum=sum+a[x][y];//add all elements

       }

       

   }

   cout<<"Sum: "<<sum; //print sum.

   return 0;

}

Output:

Enter array elements:  

1

2

3

4

5

6

7

8

9

Sum: 45

Explanation:

In the above C++ programming language code first, a header file is included then the main method is declared, inside a main method 2D array that is "a[][]", and an integer variable is defined that are "i, j, and sum". In the next line for loop is used, this loop is used two times that can be described as follows:

  • The first time it is used for inserting elements from user ends.  
  • The second time, it uses the sum variable to add all array elements. and in the last print function that is "cout" is used for print sum variable value.
7 0
3 years ago
1. Define lexemes. Give an example of an lexeme in any programming language.<br>​
tresset_1 [31]

The correct answer is; " a unit of lexical meaning that underlies a set of words that are related through inflection."

Further Explanation:

The lexeme is part of the stream where the tokens are taken and identified from. When these streams become broken, there is always an error message. The lexeme is known as "the building blocks of language."

These are a very important part of computer programming. If any string or character is misplaced this can stop the entire program or operation.

One example of a lexeme is;

  • the symbols (, ), and -> are for the letter C.

Learn more about computer programming at brainly.com/question/13111093

#LearnwithBrainly

4 0
3 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

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

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

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

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

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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