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nlexa [21]
3 years ago
13

When using an online media source, students should ensure the information is

Computers and Technology
2 answers:
OLEGan [10]3 years ago
6 0
Viable and true. hope this helps!
vitfil [10]3 years ago
4 0
Relevant, credible, and valid.

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Write a program using integers userNum and x as input, and output userNum divided by x four times. Ex: If the input is: 2000 2
STatiana [176]

Answer:

// here is code in c++.

// include header

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// variables to read input

   int userNum,x;

   cout<<"enter the value of userNum and x :";

   // read the input from user

   cin>>userNum>>x;

   // divide the userNum with x 4 times

   for(int a=0;a<4;a++)

   {

       userNum=userNum/x;

       cout<<userNum<<" ";

   }

       cout<<endl;

return 0;

}

Explanation:

Declare two variables "userNum" and "x". Read the value of these. Run a for loop 4 time and divide the "userNum" with "x" and print  the value of "userNum".

<u>Output:</u>

enter the value of userNum and x :2000 2                                                                                  

1000 500 250 125  

3 0
2 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 is one effective way for employees to keep their skillsets current
iragen [17]
<span> Create a large personal network online</span>
4 0
3 years ago
You are comparing two properties for lease. The first, A, is all-inclusive at $4,000 per month. The second, B, is $2,300 per mon
Lemur [1.5K]
Take 4,000 times 12 to get 48,000. that’s how much A costs.

B is a little harder. you have to take (2,300+300)x12+(4*250). this equals $32,100

the correct answer would be B.
6 0
2 years ago
Katrina studied hard for a biology test about the anatomy of a frog, but she did poorly and is using the "glows and grows” strat
melomori [17]

Answer:

how the frog grows through its life

Explanation:

7 0
2 years ago
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