Don’t know what the answer is but do you, I have the test for that today
It takes places during surveying.
For example, let's say you want to review your own produced game application software, and found out whether it satisfy your consumer or not
Before you conduct the survey, you have to identify what aspect you want to find out from your consumers , such as : Does it have any bug ? Is it challenging enough ? is the graphic good enough ? etc
Answer:
There is no need to make an algorithm for this simple problem. Just add the two numbers by storing in two different variables as follows:
Let a,b be two numbers.
c=a+b;
print(c);
But, if you want to find the sum of more numbers, you can use any loop like for, while or do-while as follows:
Let a be the variable where the input numbers are stored.
while(f==1)
{
printf(“Enter number”);
scanf(“Take number into the variable a”);
sum=sum+a;
printf(“Do you want to enter more numbers? 1 for yes, 0 for no”);
scanf(“Take the input into the variable f”);
}
print(Sum)
Explanation:
hi there answer is given mar me as brainliest
Answer:
Changing the customer number on a record in the Orders table to a number that does not match a customer number in the customer table would violate referential integrity.
If deletes do not cascade, deleting a customer that has orders would violate referential integrity.
If deletes cascade, customer can be deleted and all orders for that customer will automatically be deleted.
Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]