It appears on Document window tab
I believe it is 200 per revolution!
Answer:
10.5
Explanation:
1010.1
Handy method you can find searching google images, and I originally learned
from some guy teaching an online java course.
.1 --> 1 * 2^(-1) = 0.5
0 --> 0 * 2^(0) = 0
1 --> 1 * 2^(1) = 2
0 --> 0 * 2^(2) = 0
1 --> 1 * 2^(3) = 8
0.5 + 2 + 8 = 10.5
If for some reason it isn't very clear, just take the number, (x) and multiply it
by two to the power of the position it is in. (e.g. first number before decimal point is 0, second 1, etc).
Hi, you haven't provided the programing language, therefore, we will use python but you can extend it to any programing language by reading the code and the explanation.
Answer:
n1 = int(input("First numeber: "))
n2 = int(input("Second numeber: "))
for i in range(5):
r1 = n1%10
r2 = n2%10
print(r1+r2)
n1 = n1//10
n2 = n2//10
Explanation:
- First, we ask for the user input n1 and n2
- We create a for-loop to calculate the sum of each place-value of two numbers
- We obtain the last number in n1 by using the mod operator (%) an the number ten this way we can always take the last value, we make the same for n2
- Then we print the result of adding the last two numbers (place value)
- Finally, we get rid of the last value and overwrite n1 and n2 to continue with the process