Answer:
The algorithm to find A is even or odd.
- input A.
- Check the remainder on diving by 2 by A%2.
- If remainder is 1 then A is odd Print(Odd).
- If remainder is 0 print(Even).
Explanation:
To check if the number is even or odd we use modulo operator(%).Which gives the remainder on dividing.So if we do this A%2 it will give the remainder that will come out on dividing the value of A by 2.
So if the remainder comes out is 1 then the number is odd and if the remainder is 0 then the number is odd.
Answer:
The other person is correct the answer is 7
Explanation:
I got it right on edge
Answer:
<u><em>C</em></u>
Explanation:
Flash drives usually have an input (E.G. one of those android charger connectors), so just plug that into your camera, and load the flash drive with the photos. Then go home and plug the USB on the flash drive into your PC. If there isn't already, a folder will be made in your "Pictures" folder on your PC, to which you can open the folder, and then your photos should be there. From there you can open them, print them, etc.
Hope this helped, and sorry if I didn't answer in time.
This represents Guy <span>Kawasaki Rule for PowerPoint Presentations. Here is what it stands for:
</span><span>10 slides are the optimal number to use in a presentation.
20 minutes is the longest amount of time that you should speak.
<span>30 point font is the smallest font that you should use on slides.
</span></span>
Answer:
The algorithm is as follows:
Input number
count = 0
while(number not equal 0)
number = number / 10
count = count + 1
end
Print count
Explanation:
This gets input for the integer number
Input number
This initializes count of digits to 0
count = 0
The following loop is repeated while number is not 0
while(number not equal 0)
This performs integer division of the number by 10; the resulting division is saved in variable number
number = number / 10
The count variable is incremented by 1
count = count + 1
The loop ends here
end
This prints the count of integers
Print count
<em>See attachment for flowchart</em>
