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Every terminating decimal representation can be written as a decimal fraction, a fraction whose denominator is a power of 10 (e.g. 1.585 = 1585 / 1000 ); it may also be written as a ratio of the form k / 2 n ·5 m (e.g. 1.585 = 317 / 2 3 ·5 2 ).
Thus, in the above example, after an increase and decrease of x = 10 percent, the final amount, $198, was 10% of 10%, or 1%, less than the initial amount of $200. The net change is the same for a decrease of x percent, followed by an increase of x percent; the final amount is p (1 - 0.01 x )(1 + 0.01 x ) = p (1 − (0.01 x ) 2 ) .
In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30. As 120 ÷ 30 = 4, and 90 ÷ 30 = 3, one gets
If the ratio consists of only two values, it can be represented as a fraction, in particular as a decimal fraction. For example, older televisions have a 4:3 aspect ratio, which means that the width is 4/3 of the height (this can also be expressed as 1.33:1 or just 1.33 rounded to two decimal places). More recent widescreen TVs have a 16:9 ...
The only known powers of 2 with all digits even are 2 1 = 2, 2 2 = 4, 2 3 = 8, 2 6 = 64 and 2 11 = 2048. [12] The first 3 powers of 2 with all but last digit odd is 2 4 = 16, 2 5 = 32 and 2 9 = 512. The next such power of 2 of form 2 n should have n of at least 6 digits.
An example of a fraction that cannot be represented by a decimal expression (with a finite number of digits) is 1 / 3 , 3 not being a power of 10. More generally, a decimal with n digits after the separator (a point or comma) represents the fraction with denominator 10 n, whose numerator is the integer obtained by removing the separator.
Thus, 6.25 = 110.01 in binary, normalised to 1.1001 × 2 2 an even power so the paired bits of the mantissa are 01, while .625 = 0.101 in binary normalises to 1.01 × 2 −1 an odd power so the adjustment is to 10.1 × 2 −2 and the paired bits are 10. Notice that the low order bit of the power is echoed in the high order bit of the pairwise ...
In the division of 43 by 5, we have: 43 = 8 × 5 + 3, so 3 is the least positive remainder. We also have that: 43 = 9 × 5 − 2, and −2 is the least absolute remainder.