it is the property of the liquids. all the liquids are incompressible. this is because the change in volume due to pressure is negligible for liquids. in liquids the molecules are almost compact and hence no space for compression.
There is no perfect incompressible fluid.You can compress anything with certain pressure.
You need to put the strings in an array, and then loop through the array to output the strings. Something like this would be a simple example: ---------------- var strings = ["s1","s2","s3"]; for ( var i in strings ) { document.write( strings[i] ); }
c strings are terminated by \0 character
The equals method in Strings is used to check if two strings are the same.
by notes, not chords when i enumerate the positions, that is the instruction on how to place your finger lower do = 2nd set of strings, 1st fret " re = 2nd set of strings, 3rd fret " mi = 3rd set of strings (open) or 2nd set of strings, 5th fret " fa = 3rd set of strings, 1st fret " so = 3rd set of strings, 3rd fret " la = 4th set of strings (open) or 3rd set of strings, 5th fret " ti = 4th set of strings, 2nd fret do = 4th set of strings, 3rd fret re = 5th set of strings (open) or 4th set of strings, 5th fret mi = 5th set of strings, 2nd fret fa = 5th set of strings, 3rd fret so = 6th set of strings (open) or 5th set of strings, 5th fret la = 6th set of strings, 2nd fret ti = 6th set of strings, 4th fret higher do = 6th set of strings, 5th fret " re = 6th set of strings, 7th fret " mi = 6th set of strings, 9th fret " fa = 6th set of strings, 10th fret " so = 6th set of strings, 12th fret " la = 6th set of strings, 14th fret " ti = 6th set of strings, 16th fret " do = 6th set of strings, 17th fret if a note is in # or sharp, move 1 fret to the right, if in b or flat, to the left
Yes, it is true that if a language is undecidable, then it must be infinite.
No, the class of undecidable languages is not closed under complementation.
Yes, the halting problem is undecidable, meaning that there is no algorithm that can determine whether a given program will halt or run indefinitely.
if it halts
The following are undecidable cfl problems: If A is a cfl - Does A = Sigma star? If A & B cfls - is A a contained within B?
Yes, it is true that determining whether a given context-free grammar generates a specific language is undecidable.
Yes, the problem of determining whether a given context-free grammar (CFG) is undecidable.
There are two main types of piano - upright, in which the strings are set vertically, and grand, in which the strings are set horizontally.
Using computers as an example, just whack it a few times until lights flash. You might discover a new 'undecidable' problem.
Undecidable languages are languages for which there is no algorithm that can determine whether a given input string is in the language or not. Examples of undecidable languages include the Halting Problem and the Post Correspondence Problem. Decidable languages, on the other hand, are languages for which there exists an algorithm that can determine whether a given input string is in the language or not. Examples of decidable languages include regular languages and context-free languages. The key difference between undecidable and decidable languages is that decidable languages have algorithms that can always provide a definite answer, while undecidable languages do not have such algorithms.
An example of an undecidable language is the Halting Problem, which involves determining whether a given program will eventually halt or run forever. This problem cannot be solved by any algorithm.