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Luba_88 [7]
3 years ago
7

Because one memory location can be used repeatedly with different values, you can write program instructions once and then use t

hem for thousands of separate calculation. T/F
Computers and Technology
1 answer:
CaHeK987 [17]3 years ago
3 0

Answer:

True.

Explanation:

The statement written in the question is True.We can use one memory location and use it with different values.

For example:- When we are using a loop be it for,while or do-while.The counter that we use for iteration is one and we use that counter to run the loop.We are using a single memory location and we are updating the count in that memory location many times.

for(int i=0;i<1000;i++)

{

     //body.

}

We are using i's memory location and changing it 1000 times.

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2 years ago
Display all the natural numbers from 1 to 100 that are exactly divisible by 3 and 7 using FOR … NEXT. Without using Mod
Tcecarenko [31]

Answer:

FOR i% = 1 TO 100

 IF ((i%\3) = i%/3) AND ((i%\7) = i%/7)  THEN

   PRINT i%

 END IF

NEXT i%

Explanation:

Of course using MOD would be cleaner, but another way to check if a number is integer divisable is to compare the outcome of an integer division to the outcome of a floating-point division. If they are equal, the division is an integer division.

The program outputs:

21

42

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6 0
3 years ago
What will be displayed after code corresponding to the following pseudocode is run? Main Set OldPrice = 100 Set SalePrice = 70 C
Tatiana [17]

Answer:

A jacket that originally costs $ 100 is on sale today for $ 80                                    

Explanation:

Main : From here the execution of the program begins

Set OldPrice = 100  -> This line assigns 100 to the OldPrice variable

Set SalePrice = 70   -> This line assigns 70to the SalePrice variable

Call BigSale(OldPrice, SalePrice)  -> This line calls BigSale method by passing OldPrice and SalePrice to that method

Write "A jacket that originally costs $ ", OldPrice  -> This line prints/displays the line: "A jacket that originally costs $ " with the resultant value in OldPrice variable that is 100

Write "is on sale today for $ ", SalePrice  -> This line prints/displays the line: "is on sale today for $ " with the resultant value in SalePrice variable that is 80

End Program -> the main program ends

Subprogram BigSale(Cost, Sale As Ref)  -> this is a definition of BigSale method which has two parameters i.e. Cost and Sale. Note that the Sale is declared as reference type

Set Sale = Cost * .80  -> This line multiplies the value of Cost with 0.80 and assigns the result to Sale variable

Set Cost = Cost + 20  -> This line adds 20 to the value of Cost  and assigns the result to Cost variable

End Subprogram  -> the method ends

This is the example of call by reference. So when the method BigSale is called in Main by reference by passing argument SalePrice to it, then this call copies the reference of SalePrice argument into formal parameter Sale. Inside BigSale method the reference &Sale is used to access actual argument i.e. SalePrice which is used in BigSale(OldPrice, SalePrice) call. So any changes made to value of Sale will affect the value of SalePrice

So when the method BigSale is called two arguments are passed to it OldPrice argument and SalePrice is passed by reference.

The value of OldPrice is 100 and SalePrice is 70

Now when method BigSale is called, the reference &Sale is used to access actual argument SalePrice = 70

In the body of this method there are two statements:

Sale = Cost * .80;

Cost = Cost + 20;

So when these statement execute:

Sale = 100 * 0.80 = 80

Cost = 100 + 20 = 120

Any changes made to value of Sale will affect the value of SalePrice as it is passed by reference. So when the Write "A jacket that originally costs $ " + OldPrice Write "is on sale today for $ " + SalePrice statement executes, the value of OldPrice remains 100 same as it does not affect this passed argument, but SalePrice was passed by reference so the changes made to &Sale by statement in method BigSale i.e.  Sale = Cost * .80; has changed the value of SalePrice from 70 to 80 because Sale = 100 * 0.80 = 80. So the output produced is:

A jacket that originally costs $ 100 is on sale today for $ 80                              

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