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Andreyy89
3 years ago
7

Machines may aid in making work easier over a longer distance. true or false.

Computers and Technology
1 answer:
Dmitry [639]3 years ago
4 0
This seems like more of an opinion based question, but since they added "over a longer distance" If I were you I would go with false.
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Could anyone please answer this?
Anastaziya [24]

Answer:

D

Explanation:

The answer is D because if you're looking for lightweight materials, you'll be dealing with chemistry, and chemistry is sience.

4 0
3 years ago
___________ system allows us to talk to any person in the world it any time​
GuDViN [60]
World wireless system
7 0
3 years ago
Read 2 more answers
Consider the following language:
VMariaS [17]

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.

Learn more about programming language from

brainly.com/question/16936315

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8 0
2 years ago
Plz plz plz plz plz plz plz plz plz plz plz pl plz plz plz plz plz plz plz plz plz plz plz plz olz plz plz
zvonat [6]

REM PROGRAM TO DISPLAY AVERAGE OF TWO NUMBERS

CLS

INPUT “ENTER FIRST NUMBER”; A

INPUT “ENTER SECOND NUMBER”; B

INPUT “ENTER THIRD NUMBER”; C

AV = (A + B + C) / 3

PRINT “AVERAGE OF THREE NUMBERS”; AV

END

hope this helpz

6 0
3 years ago
Write a calculator program that will allow only addition, subtraction, multiplication &amp; division. Have the
ozzi

num1 = float(input("Enter the first number: "))

num2 = float(input("Enter the second number: "))

operation = input("Which operation are you performing? (a/s/m/d) ")

if operation == "a":

   print("{} + {} = {}".format(num1, num2, num1+num2))

elif operation == "s":

   print("{} - {} = {}".format(num1, num2, num1-num2))

elif operation == "m":

   print("{} * {} = {}".format(num1, num2, num1*num2))

elif operation == "d":

   print("{} / {} = {}".format(num1, num2, num1/num2))

I hope this helps!

8 0
3 years ago
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