Answer:
int count =0;
for(int i=0;i<10;i++)
{
if(myArray[i]>=0)
{
count++;
}
}
cout<<"Number of positive integers is "<<count<<endl;
Explanation:
The above written loop is for counting positive integers in the myArray[].
For counting we have taken a count integer initialized with 0.On iterating over the array if the element is greater than or equal to 0 we consider it as positive and increasing the count.At the end printing the count.
Answer:
C, Or D.
Explanation:
<em>Because A speed enhancing hard drive Can store any type of quick file if you just click on it will load fast. Same thing goes for a hard drive but you have to transfer the file</em>
Im pretty sure you can but different schools have different rules so try contacting your guidence counselor and asking them
Answer:I love Python, very useful
Explanation:python is very easy and user friendly!
Answer:
No you can not tell that recursion is ever required to solve a problem.
Recursion is required when in the problem, the solution of the input depends on the solution of the subsets of the input.
Iteration is also another form of repetitive approach we follow to solve that kind of problems.
But the difference between recursion and iteration is :
- In recursion we call the function repeatedly to return the result to next level.
- In iteration certain bunch of instructions in a loop are executed until certain conditions met.
Explanation:
For example in the Fibonacci sequence problem, to find
, we need to compute
and
before that.
- In case of recursion we just call the method Fibonacci(n) repeatedly only changing the parameter Fibonacci(n-1), that calculates the value and return it.
Fibonacci(n)
1. if(n==0 or n==1)
2. return 1.
3.else
4. return( Fibonacci(n-1)+Fibonacci(n-1) )
- But in case of iteration we run a loop for i=2 to n, within which we add the value of current
and
to find the value of 
Fibonacci(n)
1. if(n<=2)
2. result = 1
3. else
4. result1 =1 and result2=1.
5. { result = result1 +result2.
6. result1= result2.
7. result2 = result.
8. }
9. output result.