Answer:
Following are the program in python language
def prob3_6(k): #function definition
c = 0 #variable declaration
while k != 1: #iterating the while loop
print(k) #print k
if k % 2 == 0:#check if condition
k= k // 2 #divisible by 2
else: #else condition
k = k * 3 + 1
c = c + 1
print(k) #print k
print c #print count
prob3_6(3)#function call
Output:
3
10
5
16
8
4
2
1
7
Explanation:
Following are the description of program
- Create a function "prob3_6" in this function we passing an integer parameter of type "int" named "k".
- Inside that function we declared a variable "c" of type "int" that is used for counting purpose .
- After that we iterated the while for print the value of when it is not equal to 1 .
- We check if condition when k gives 0 on modulus then k is divisible by 2 otherwise else block will be executed in the else part we multiply by 3 to k and add 1 to the k variable .
- Finally print "k" and "c"
Answer:
public class Main
{
public static void main(String[] args) {
System.out.println(min(3, -2, 7));
}
public static int min(int n1, int n2, int n3){
int smallest = Math.min(Math.min(n1, n2), n3);
return smallest;
}
}
Explanation:
*The code is in Java.
Create a method named min that takes three parameters, n1, n2, and n3
Inside the method:
Call the method Math.min() to find the smallest among n1 and n2. Then, pass the result of this method to Math.min() again with n3 to find the min among three of them and return it. Note that Math.min() returns the smallest number among two parameters.
In the main:
Call the method with parameters given in the example and print the result
Answer:
Click home tab, click conditional formatting, click new rule, use formula to determine
Answer:
Explanation:
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