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posledela
3 years ago
8

Assume you have taken another square picture with the 25-megapixel digital camera discussed in the previous question. This time

around you decide to view the picture on your computer screen that can display 2000 pixels by 1000 pixels. What fraction of the image is viewable on the screen? Tnoo
Computers and Technology
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

8% of the picture

Explanation:

Given:

Square picture pixels = 25 MP

Pixels that can be displayed by the computer = 2000 pixels by 1000 pixels

or

Pixels that can be displayed by the computer = 2000000 pixels

Now,

The fraction of picture viewable on the screen = \frac{\textup{2MP}}{\textup{25MP}}

or

The fraction of picture viewable on the screen = 0.08

or

The fraction of picture viewable on the screen = 8% of the picture

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The ________ is the biggest power consumer on a mobile computing device.
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2 years ago
Write a C program that reads two hexadecimal values from the keyboard and then stores the two values into two variables of type
sattari [20]

Solution :

#include  $$

#include $$

#include $$

//Converts $\text{hex string}$ to binary string.

$\text{char}$ * hexadecimal$\text{To}$Binary(char* hexdec)

{

 

long $\text{int i}$ = 0;

char *string = $(\text{char}^ *) \ \text{malloc}$(sizeof(char) * 9);

while (hexdec[i]) {

//Simply assign binary string for each hex char.

switch (hexdec[i]) {

$\text{case '0'}:$

strcat(string, "0000");

break;

$\text{case '1'}:$

strcat(string, "0001");

break;

$\text{case '2'}:$

strcat(string, "0010");

break;

$\text{case '3'}:$

strcat(string, "0011");

break;

$\text{case '4'}:$

strcat(string, "0100");

break;

$\text{case '5'}:$

strcat(string, "0101");

break;

$\text{case '6'}:$

strcat(string, "0110");

break;

$\text{case '7'}:$

strcat(string, "0111");

break;

$\text{case '8'}:$

strcat(string, "1000");

break;

$\text{case '9'}:$

strcat(string, "1001");

break;

case 'A':

case 'a':

strcat(string, "1010");

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case 'B':

case 'b':

strcat(string, "1011");

break;

case 'C':

case 'c':

strcat(string, "1100");

break;

case 'D':

case 'd':

strcat(string, "1101");

break;

case 'E':

case 'e':

strcat(string, "1110");

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case 'F':

case 'f':

strcat(string, "1111");

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default:

printf("\nInvalid hexadecimal digit %c",

hexdec[i]);

string="-1" ;

}

i++;

}

return string;

}

 

int main()

{ //Take 2 strings

char *str1 =hexadecimalToBinary("FA") ;

char *str2 =hexadecimalToBinary("12") ;

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int p,n;

scanf("%d",&p);

scanf("%d",&n);

//keep j as length of str2

int j=strlen(str2),i;

//Now replace n digits after p of str1

for(i=0;i<n;i++){

str1[p+i]=str2[j-1-i];

}

//Now, i have used c library strtol

long ans = strtol(str1, NULL, 2);

//print result.

printf("%lx",ans);

return 0;

}

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3 years ago
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