Any time you throw a zero in there, the answer will be zero.
Any value raised to the power 'zero'(0) equals '1'. Hence 2^(0) = 1 10 ^(0) = 1 Hence 2^(0) X 10^(0) = 1 x 1 = 1 the answer.
In expanded notation using powers of ten, 250,000 can be expressed as (2 \times 10^5 + 5 \times 10^4 + 0 \times 10^3 + 0 \times 10^2 + 0 \times 10^1 + 0 \times 10^0). Simplifying this, it becomes (2 \times 100,000 + 5 \times 10,000). Thus, the expanded form is (200,000 + 50,000).
527,519 = 500,000 + 20,000 + 500 + 10 + 9 5 4 2 1 ( 5 times 10 ) + ( 2 times 10 ) + ( 5 times 10 ) + (1 times 10 ) + 0 ( 9 times 10 ) 5 times 1,000,000 + 2 times 10,000 + 1 times 10 + 9 times 1
0 = 0/10, 1/2 = 5/10 and 1 = 10/10 so 6/10 is nearest to 5/10 or 1/2.
100 = (1 x 10^2) + (0 x 10^1) + (0 x 10^0) or 1 x 10^2
Any value raised to the power 'zero'(0) equals '1'. Hence 2^(0) = 1 10 ^(0) = 1 Hence 2^(0) X 10^(0) = 1 x 1 = 1 the answer.
In expanded notation using powers of ten, 250,000 can be expressed as (2 \times 10^5 + 5 \times 10^4 + 0 \times 10^3 + 0 \times 10^2 + 0 \times 10^1 + 0 \times 10^0). Simplifying this, it becomes (2 \times 100,000 + 5 \times 10,000). Thus, the expanded form is (200,000 + 50,000).
527,519 = 500,000 + 20,000 + 500 + 10 + 9 5 4 2 1 ( 5 times 10 ) + ( 2 times 10 ) + ( 5 times 10 ) + (1 times 10 ) + 0 ( 9 times 10 ) 5 times 1,000,000 + 2 times 10,000 + 1 times 10 + 9 times 1
100101 1 times 2^0 = 1 PLUS 0 times 2^1 = 0 PLUS 1 times 2^2 = 4 PLUS 0 times 2^3 = 0 PLUS 0 times 2^4 = 0 PLUS 1 times 2^5 = 32 EQUALS 37
(4.9 × 10^-2) × (9.8 × 10^2) = (4.9 × 9.8) × 10^(-2 + 2) = 48.02 × 10^0 = (4.802 × 10^1) × 10^0 = 4.802 × 10^(1 + 0) = 4.802 × 10^1
0 = 0/10, 1/2 = 5/10 and 1 = 10/10 so 6/10 is nearest to 5/10 or 1/2.
100 = (1 x 10^2) + (0 x 10^1) + (0 x 10^0) or 1 x 10^2
[(4 * 10^2) + (0 * 10^1) + (0 * 10^0)] + [(1 * 10^1) + (0 * 10^0)]
21000 = (2 x 10000) + (1 x 1000) + (0 x 100) + (0 x 10) + (0 x 1) OR (2 x 10^4) + (1 x 10^3) + (0 x 10^2) + (0 x 10^1) + (0 x 10^0)
(73)10 = (1 0 0 1 0 0 1)2
It is: 10 times 1/2 = 5
In the decimal system, the places are 10^0 (1), 10^1 (10), 10^2 (100), and 10^3 (1000) In the binary system, the places are 2^0 (1), 2^1 (2), 2^2 (4), and 2^3 (8) 1101 has one 1, one 4 and one 8. That makes 13, base ten.