(73)10 = (1 0 0 1 0 0 1)2
73 written in base 2 is... 1001001
10 base 2 = 2 base 10
11011 base 2 is equal to 27 in base 10 321 base 4 is equal to 57 in base 10 27+57=84
The number 11 in base 2 represents the binary value. To convert it to base 10, you calculate (1 \times 2^1 + 1 \times 2^0), which equals (2 + 1 = 3). Therefore, 11 in base 2 is equal to 3 in base 10.
(73)10 = (1 0 0 1 0 0 1)2
73 written in base 2 is... 1001001
10 base 2 = 2 base 10
( 1010 )2 = ( 10 )10
log(e)100 = log(10)100 / log(10)e = log(10)100 / log(10) 2.71828.... = 2/ 0.43429448... = 4.605170186..... (The answer). NB Note the change of log base to '10' However, on a calculator type in ;- 'ln' (NOT log). '100' '=' The answer shown os 4.605....
1010 base 2 = 10 base 10 1010 base 10 = 11 1111 0010 base 2
To subtract in base 2, we need to borrow from the next higher place value if necessary. In this case, when subtracting 11 from 101 in base 2, we need to borrow from the leftmost digit. So, 101 in base 2 is 5 in decimal, and 11 in base 2 is 3 in decimal. When subtracting 3 from 5 in decimal, we get 2 in decimal, which is 10 in base 2. Therefore, 101 base 2 minus 11 base 2 is 10 base 2.
20 to the base 10 = 1 Therefore 2 + 1 = 3
101001, base 10 = 11000101010001001, base 2
73/2 = 36.5 of them. 73/2 = 36.5 of them. 73/2 = 36.5 of them. 73/2 = 36.5 of them.
11011 base 2 is equal to 27 in base 10 321 base 4 is equal to 57 in base 10 27+57=84
Adding in base 2 is binary for computers. 10 = 2 because 10 means 1 of your base. The rules are no different.