Give the base-ten numeral for 101,010two and 912twelve
(0 * 2^0) + (1 * 2^1) + (0 * 2^2) + (1 * 2^3) + (0 * 2^4) + (1 * 2^5)
0 + 2 + 0 + 8 + 0 + 32 = 42
(2 * 12^0) + (1 * 12^1) + (9 * 12^2)
2 + 12 + 1296 = 1310
To convert a number from a different base to base-ten (decimal), we need to understand the positional value of each digit.
1. Let's start with 101,010 in base two (binary). Each digit's position represents a power of two: starting from the rightmost digit, the powers of two are 2^0 (1), 2^1 (2), 2^2 (4), 2^3 (8), 2^4 (16), and 2^5 (32).
To calculate the decimal equivalent, we multiply each digit (0 or 1) by its corresponding power of two and sum them up:
(1 × 2^5) + (0 × 2^4) + (1 × 2^3) + (0 × 2^2) + (1 × 2^1) + (0 × 2^0) = 32 + 0 + 8 + 0 + 2 + 0 = 42
Therefore, the base-ten numeral for 101,010 in base two is 42.
2. Moving on to 912 in base twelve. In base twelve, we have twelve different digits: 0-9 and A-B (representing 10 and 11, respectively). The positional values follow powers of twelve: starting from the rightmost digit, the powers of twelve are 12^0 (1), 12^1 (12), and 12^2 (144).
To calculate the decimal equivalent, we multiply each digit by its corresponding power of twelve and add them up:
(2 × 12^2) + (9 × 12^1) + (1 × 12^0) = 2 × 144 + 9 × 12 + 1 = 288 + 108 + 1 = 397
Hence, the base-ten numeral for 912 in base twelve is 397.