To present something infront of a crowd
Answer:
Essentially additive editing and stringing “the good stuff” together, whereas subtractive is more about stringing all your raw footage together and “removing the bad stuff”
Additive editing feels confident and concerned with the pursuit of a specific, existing vision. And it’s faster. Subtractive editing feels like a deeper listening to what the footage is saying, and holding on to many potential permutations.
Explanation:
Addictive editing - creating a program from raw footage by starting by starting with nothing and adding selected components
Subtraction editing - creating a program by removing redundant or poor quality material from the original footage
Answer: This is a python code
def lightyear():
rate=3*100000000 //speed of light
seconds=365*24*60*60 //number of seconds in 1 year
return str((rate*seconds)/1000)+" km" //distance=speed x time
print(lightyear()) //will print value of light hear in kilometers
OUTPUT :
9460800000000.0 km
Explanation:
In the above code, there is a variable rate, which stores the speed of light, i.e. distance traveled by light in 1 second which is in meters. Another variable is seconds, which store the number of seconds in 1 year, which is no of days in 1 year multiplied by the number of hours in a day multiplied by the number of minutes in an hour multiplied by the number of seconds in a minute. Finally, distance is speed multiplied by time, so distance is printed in kilometers and to convert distance in kilometers it is divided by 1000.
Answer:
i think its C.
if its wrong i'm truly sorry
Explanation:
<u>Explanation:</u>
Hey there! you need not to panic about it ,your program didn't have Driver program i.e main program! the correct & working code is given below:
// C++ program to count even digits in a given number .
#include <iostream>
using namespace std;
// Function to count digits
int countEven(int n)
{
int even_count = 0;
while (n > 0)
{
int rem = n % 10;
if (rem % 2 == 0)
even_count++;
n = n / 10;
}
cout << "Even count : "
<< even_count;
if (even_count % 2 == 0 )
return 1;
else
return 0;
}
// Driver Code
int main()
{
int n;
std::cin >>n;
int t = countEven(n);
return 0;
}
