Question: 1. Addition 2 2. #include iostream 3. using namespace std 4. //the functions prototype declares the function 5. int addition (int, int) Copy addition2 6.

 1. Addition 2 2. #include iostream 3. using namespace std 4.
//the functions prototype declares the function 5. int addition (int, int) Copy
addition2 6. int main The scope of variables 7. declared within a
function nt Z is only inside the function. 9. nt a 5
It is impossible to use the 10 int b 3 11 //this

1. Addition 2 2. #include iostream 3. using namespace std 4. //the functions prototype declares the function 5. int addition (int, int) Copy addition2 6. int main The scope of variables 7. declared within a function nt Z is only inside the function. 9. nt a 5 It is impossible to use the 10 int b 3 11 //this is calling the function addition variables a, b or r 12 z addition (a,b) directly in function main 13 cout "The result is Z since they are variables 14 return 0 local to function addition. 15. It is impossible to use the 16. //addition function 17. int addition int aint b variable z directly within 18. function addition, since 19 nt r this was a variable local to 20 r-a+b the function main. 21 return (r); 22. Programming Logic ana Uesign, Devenin taltion

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