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a_sh-v [17]
3 years ago
15

If she presses the left arrow , what will happened

Computers and Technology
1 answer:
Harman [31]3 years ago
4 0
The keyboard will move
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write a c++ program that reads from the user a positive integer (in a decimal representation), and prints its binary (base 2) re
Vesna [10]

Answer:

The program to this question can be described as follows:

Program:

#include <iostream> //defining header file

using namespace std;

int main() //defining main method

{

int x1,rem,i=1; //defining integer variable

long Num=0;// defining long variable

cout << "Input a decimal number: "; // print message

cin >> x1; //input value by user

while (x1!=0) //loop for calculate binary number

{

//calculating binary value    

rem= x1%2;

x1=x1/2;

Num =Num +rem*i;//holding calculate value

i=i*10;

}

cout <<Num;//print value

return 0;

}

Output:

Input a decimal number: 76

1001100

Explanation:

In the above code, four variable "x1, rem, i, and Num" is declared, in which three "x1, rem, and i" is an integer variable, and one "Num" is a long variable.

  • In the x1 variable, we take input by the user and use the while loop to calculate its binary number.
  • In the loop, first, we check x1 variable value is not equal to 0, inside we calculate it binary number that store in long "Num" variable, after calculating its binary number the print method "cout" is used to prints its value.  
5 0
3 years ago
In this exercise, first run the code as it is given to see the intended output. Then trace through each of the 3 variables to se
dlinn [17]

Answer:

The memory with variable names str1, str2, and str3 all have equal and the same value after the first if-statement.

Explanation:

The str1 was first assigned a null value while the str2 and str3 were assigned the string value "Karen" with the String class and directly respectively. On the first if-statement, the condition checks if the str1 is null and assigns the value of the variable str2 to str1, then the other conditional statement compares the values of all the string variables.

3 0
3 years ago
Research and describe a recent development in theater and film lighting and explain how the technology can be used. (Maybe limit
Hunter-Best [27]

Answer:

A computerized light dimmer is one helpful development in lighting equipment. Computerized light dimmers help to control the brightness of light. These dimmers lower or increase the intensity of light. I first noticed light dimmers in a scene from the movie Children of Men. In a war scene, innocent people are shown running helter-skelter to save their lives. Artificial lights are used in scene to create that dark appearance in broad daylight, which makes the scene look gloomy and scary. It helped to generate the feeling of uncertainty and suspense through the proper use of lights. Light dimmers help light designers to create the exact atmosphere that the director demands for a scene. Also, because of dimmers, a light designer need not use extra accessories to diffuse light or make the scene brighter.

Explanation:

This is Plato's sample answer so take pieces out.

6 0
3 years ago
Read 2 more answers
Two electronics technicians are discussing electrical quantities. Technician A says that resistance is an opposition to electric
inessss [21]
Neither are correct.

Resistance is opposition to CURRENT not power.
Technition B Is wrong about the voltage thingy.
6 0
3 years ago
Consider sending a 2400-byte datagram into a link that has an mtu of 700 bytes. suppose the original datagram is stamped with th
Feliz [49]

Explanation:

Let, DG is the datagram so, DG= 2400.

Let, FV is the Value of Fragment and F is the Flag and FO is the Fragmentation Offset.

Let, M is the MTU so, M=700.

Let, IP is the IP header so, IP= 20.

Let, id is the identification number so, id=422

Required numbers of the fragment = [\frac{DG-IP}{M-IP} ]

Insert values in the formula = [\frac{2400-20}{700-20} ]

Then,        = [\frac{2380}{680} ] = [3.5]

The generated numbers of the fragment is 4

  • If FV = 1 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=0 and F=1.
  • If FV = 2 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=85(85*8=680 bytes) and F=1.
  • If FV = 3 then, bytes in data field of DG= 720-20 = 680 and id=422 and FO=170(170*8=1360 bytes) and F=1.
  • If FV = 4 then, bytes in data field of DG= 2380-3(680) = 340 and id=422 and FO=255(255*8=2040 bytes) and F=0.

3 0
3 years ago
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