例句與造句

1. where the base 10 logarithm has been used in the final answer for ease of calculation.
2. And of course we like base 10 logarithms : 1, 10, 100, 1000.
3. This is 20 times the base 10 logarithm of the ratio of a given root-mean-square sound pressure to the reference sound pressure.
4. The base 10 logarithm is taken of this function, and a first order linear regression, linear fit, is performed on the resulting data set, that is
5. The DIN system was inspired by Scheiner's system, but the sensitivities were represented as the base 10 logarithm of the sensitivity multiplied by 10, similar to decibels.
6. It's difficult to find base 10 logarithm in a sentence. 用base 10 logarithm造句挺難的
7. For converting to and from base 10 logarithms and natural logarithms the natural logarithm of 10 ( 2.30259 ) and e ( 2.71828 ) are printed on the case.
8. It is approximately the negative of the base 10 logarithm of the molar concentration, measured in units of Pure water is neutral, at pH 7, being neither an acid nor a base.
9. Thus an increase of 20?( and not 19?as in Scheiner's system ) represented a hundredfold increase in sensitivity, and a difference of 3?was much closer to the base 10 logarithm of 2 ( 0.30103 & ):
10. The Nernst equation is frequently expressed in terms of base 10 logarithms ( " i . e . ", common logarithms ) rather than natural logarithms, in which case it is written, " for a cell at 25 癈 ":
11. Formally, " L AT " is 10 times the base 10 logarithm of the ratio of a root-mean-square A-weighted sound pressure during a stated time interval to the reference sound pressure and there is no time constant involved.
12. Originally the sensitivity was written as a fraction with'tenths'( for example " 18 / 10?DIN " ), where the resultant value 1.8 represented the relative base 10 logarithm of the speed .'Tenths'were later abandoned with DIN 4512 : 1957-11, and the example above would be written as " 18?DIN ".
13. Note that the log and exp functions can be used, as in your first equation, in place of the general y ^ x function, so you could write f ( x ) = fStart * 10 ^ ( log ( fEnd / fStart ) * x / steps ), where log is base 10 logarithm or the 10 ^ is replaced with exponentiation of the same base as the log.