L is a decidable language because the Turing machine accepts it.
L is a recognizable language if TM M recognizes it.
<h3>How do you know if a language is decidable?</h3>
A language is said to be decidable only when there seems to exists a Turing machine that is said to accepts it,
Here, it tends to halts on all inputs, and then it answers "Yes" on words that is seen in the language and says "No" on words that are not found in the language. The same scenario applies to recognizable language.
So, L is a decidable language because the Turing machine accepts it.
L is a recognizable language if TM M recognizes it.
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<span>The RAID level 2 could give high
information exchange rates and would be straightforward contrasted with other
Raid levels. Be that as it may, it has a high cost and would need a high rate move
required with a specific end goal to legitimize this cost.</span>
Answer:
d) daco = new Banana;
Explanation:
Dynamically allocated variables have their memory allocated in the heap memory.We declare a dynamical variable like this:-
int *a=new int ;
It means a pointer a is created on the stack memory which hold the address of the block that hold the value of variable a in heap memory.
We already have the pointer daco. We just have to initialize with keyword new.
It will be like daco=new Banana; which matches the option d.
The answers are 1, 3, and 5.
Answer:
The correct pseudocode to find the records of all citizens over the age of 50 is IF(age > 50).
OR EACH item IN citzlist
{
WHILE(not end of citzlist)
{
IF(age > 50)
{
DISPLAY(name)
}
}
}
If this is run, it will bring out all the names of the citizen who are over the age of 50 in the list.
Explanation:
