Answer:
a) 255
b) 61602
c)
d)
Explanation:
You need to understand the decimal equivalent of hexadecimal numbers, <em>from 0 to 9 numbers are represented the same way, from 10 to 15 we use the alphabet, meaning 10 equals A in hexadecimal base, 11-B, 12-C, 13-D, 14-E, and 15-F.</em>
For your first exercise you'll enumerate the number's positions fromright to lef begining with 0:
a. F F
position 1 0
Now you'll multiply your hexadecimal number (using the decimal equivalent for your letters) for the base (16) elevated to the number of the position:
Finally, you'll add your results:
240+15=255
FF=255
b. F 0 A 2
position 3 2 1 0
F0A2=61602
c. F 1 0 0
position 3 2 1 0
0F100=61696
d. 1 0 0
position 2 1 0
100=256
I hope you find this information useful! Good luck!
☆ <u>Since, the computer accepts raw data as input and converts into information by means of data processing, it is called information processing machine (IPM).</u>
HOPE IT HELPS ❣️
Answer:
First by
Clicking
New Button (+) THEN click Vendor THEN click Credit
Secondly
Click Expenses Center then click New Transaction then finally click Vendor Credit
Answer:
Given
The above lines of code
Required
Rearrange.
The code is re-arrange d as follows;.
#include<iostream>
int main()
{
int userNum;
scanf("%d", &userNum);
if (userNum > 0)
{
printf("Positive.\n");
}
else
{
printf("Non-positive, converting to 1.\n");
userNum = 1;
printf("Final: %d\n", userNum);
}
return 0;
}
When rearranging lines of codes. one has to be mindful of the programming language, the syntax of the language and control structures in the code;
One should take note of the variable declarations and usage
See attachment for .cpp file
Answer:
The sum of all positive even values in arr
Explanation:
We have an array named arr holding int values
Inside the method mystery:
Two variables s1 and s2 are initialized as 0
A for loop is created iterating through the arr array. Inside the loop:
num is set to the ith position of the arr (num will hold the each value in arr)
Then, we have an if statement that checks if num is greater than 0 (if it is positive number) and if num mod 2 is equal to 0 (if it is an even number). If these conditions are satisfied, num will be added to the s1 (cumulative sum). If num is less than 0 (if it is a negative number), num will be added to the s2 (cumulative sum).
When the loop is done, the value of s1 and s2 is printed.
As you can see, s1 holds the sum of positive even values in the arr
