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bazaltina [42]
3 years ago
15

A regular language can not be infinite. True False

Computers and Technology
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

False

Explanation:

A regular language can be represented by a regular expression. A regular language can be infinite. Let us consider a simple example of an infinite regular language:

a* is a regular language represented by a regular expression.

The languages matches all strings containing or more a's.

Clearly this is an infinite language.

Note that all finite languages are regular but all regular languages need not be finite.

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Which of the following calculates to 10?
nikdorinn [45]
The third one because u have to do parenthesis
7 0
3 years ago
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What is basically the only thing that computers (microprocessors and prgrammable devices) understand?
horsena [70]

Answer:

Binary Code

Explanation:

All microprocessors and programmable devices understand is Binary Code. These are various combinations of 0's and 1's which when placed together in a sequence represent a set of instructions that the microprocessor can read and understand to complete complex tasks. There are various other programming languages to program these tasks in an easier to read syntax for the programmers themselves but they simply take the written code and convert it into Binary before sending it to the microprocessor.

4 0
2 years ago
Difference between desktop, web and mobile apps
Lyrx [107]

there are many differences, but the most prominent are that they have different apis (aplication programing interface) and different architecture

also mobile apps tend to use less system resources and are more regulated by services like the google play store, opposed to the fairly unregulated uwp platform used by most windows machines.  


4 0
3 years ago
Consider the following code segment, which is intended to set the Boolean variable inRange to true if the integer value num is g
Natasha_Volkova [10]

Answer:

Option d  num = 50, min = 50, max = 50

Explanation:

Given the code segment:

  1. boolean isBigger;
  2. boolean isSmaller;
  3. boolean inRange;
  4. if (num < max)
  5.        {
  6.            isSmaller = true;
  7.        }
  8.        else {
  9.            isSmaller = false;
  10.        }
  11.        if (num > min)
  12.        {
  13.            isBigger = true;
  14.        }
  15.        else {
  16.            isBigger = false;
  17.        }
  18.        if (isBigger == isSmaller) {
  19.            inRange = true;
  20.        } else {
  21.            inRange = false;
  22.        }

If we have num = 50, min = 50 , max = 50,  the condition num < max will be evaluated to false and therefore isSmaller is set to false.

The condition num > min will be evaluated to false as well and therefore isBigger is set to false.

Since isSmaller and isBigger are both false and therefore isBigger == isSmaller will be evaluated to true and set the inRange = true. This has violated the statement that if the integer value num is greater than min value and less than max value, then only set inRange to true. This is because num = 50 is neither greater than min nor less than max, it is supposedly not in range according to the original intention of the code design.

8 0
3 years ago
What is the available vector for each resource and the need matrix for each thread. Note that an available vector depicts the qu
blsea [12.9K]

Answer:

If request granted then T₁ and T₂ are in deadlock.

Explanation:

See attached image

4 0
3 years ago
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