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Alex Ar [27]
3 years ago
11

Do you think tracking your screen time will reduce your usage? Why or Why not?

Computers and Technology
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

I dont think it will help much

Explanation:

Well, just tracking it will not necessarily make you stop. I would recommend putting app limits or downtime (if you have a mac). Also make a log. Write down how much time you need for different things, like 1 hour of free time, 5 hours of schoolwork. Also, make a schedule based off the things you put on your log.

I found doing this helped me reduce my time on screen and made me more productive.

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Window is a very popular operating system becouse of its _ environment .​
Inga [223]
Window is a vary popular operating system because of its runtime and compatible environment .
hope it help
4 0
3 years ago
Read 2 more answers
How important "saving" in every individual?​
photoshop1234 [79]

Answer:

The importance of saving money is simple: It allows you to enjoy greater security in your life. If you have cash set aside for emergencies, you have a fallback should something unexpected happen. And, if you have savings set aside for discretionary expenses, you may be able to take risks or try new things.

4 0
3 years ago
5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S
Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0
3 years ago
What corrective action should be taken on a printer that prints unknown characters??
Lynna [10]
Reinstall the driver. 
7 0
3 years ago
A programmer wants to write a procedure that calculates the net elevation - total number of feet a traveler goes up and down. Fo
ch4aika [34]

Answer:

Pseudocode

////////////////////////////////////////////////////////////////////////////////////////////////////////////

Integer netElevation(list of elements of type elevation - type and number)

<em>function open</em>

   Define running total = 0

   for each element from list

   <em>loop open</em>

       elevation type = element[i].type

      if (elevation type == Up)

           running total = running total + element[i].number

       else

           running total = running total - element[i].number

   <em>loop close</em>

   return running total

<em>function close</em>

3 0
3 years ago
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