Answer:
Obsolescencia programada es cuando un producto está diseñado deliberadamente para tener un tiempo de vida específico. ... Los productos dejan de funcionar al cabo de un tiempo, no porque estén estropeados, sino por que han sido diseñados para fallar al cabo de ese periodo.
Explanation:
espero y esto te pueda ayudar
Answer:
retupmoc
Explanation:
1.) Anwser will be retupmoc
because
public static String mysteryString(String s){
if(s.length() == 1){
return s;
}
else{
return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));
}
}
In this program input is "computer" . So the function mysteryString(String s) it does
return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));
so when it enters the first time ??s.substring(s.length() -1) and it will be give you 'r' then it calls the function recursively by reducing the string length by one . So next time it calls the mysteryString function with string "compute" and next time it calls return s.substring(s.length()-1)? + mysteryString(s.substring(0,s.length-1)) so this time it gives "e" and calls the function again recursively . It keeps on doing till it matched the base case.
so it returns "retupmoc".
Answer:
The program to this question can be described as follows:
Program:
#include <iostream> //defining header file
using namespace std;
int main() //defining main method
{
int x; //defining integer variable
for(x=0;x<=100;x++) //defining loop to count value from 0 to 100
{
if(x%7==0) //check value is divisable by 7
{
cout<<x<<endl; //print value
}
}
return 0;
}
Output:
please find the attachment.
Explanation:
In the above code, an integer variable x is declared, which is used in the for loop, in this loop variable "x" starts from 0 and ends when the value of x is less than and equal to 100.
- Inside the loop an, if block is used that defines a condition that is (i%7==0), it will check, that the value is divided by 7.
- In this loop, a print method is used, that prints its values.
