Answer:view
Explanation:i just took the test
Answer:
There are two ways to print 1 to 1000
- Using Loops.
- Using Recursion.
Explanation:
Using loops
for(int i=1;i<=1000;i++)
{
cout<<i<<" ";
}
Using recursion
Remember you can implement recursion using a function only.
void print(int n)
{
if(n==0)
return;
print(n-1);
cout<<n<<" "';
}
you should pass 1000 as an argument to the function print.
Answer:
As many as we need
Explanation:
https://quizizz.com/admin/quiz/5e68ffb7ada381001bd58b3a/codehs-unit-2-lessons-13-to-19
Answer:
The Following are the solution to this question:
Explanation:
In Option a:
In the point (i) 
is transitive, which means it converts one action to others object because if 
indicates 
. It's true by definition, that becomes valid. But if 
, which implies 
. it's a very essential component. If 
. They 
will also be 
.
In point (ii), The value of 
is convergent since the 
. It means they should be dual a and b constant variable, therefore 
could only be valid for the constant variable, that is 
.
In Option b:
In this algorithm, the input size value is equal to 1 object, and the value of A is a polynomial-time complexity, which is similar to its outcome that is 
. It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop
. All internal loops operate on a total number of 
generations and therefore the final time complexity is .
Answer:
ExChunking is a good strategy for completing large assignments because it makes the work
less boring.
more thorough.
less difficult.
more manageable.planation:
Explanation: less difficult (C)
