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**Answer:
**

The program to this question can be given as:

**Program:
**

#include<iostream> //header file

using namespace std;

int main() //main method

{

int x[10],i,largest = 0,second_largest=0,n; //variable

cout << "Enter Number of elements :"; //message

cin>>n;

cout << "Insert array elements :"; //message

for(i=0;i<n;i++) //insert elements in array

{

cin >>x[i];

}

//Finding Largest element

for(i=0;i<n;i++)

{

if (x[i]>largest)

{

largest = x[i];

}

}

//finding second largset element

for(i=0;i<n;i++)

{

if (x[i]>second_largest)

{

if(x[i]==largest)

{

continue; //Ignoring largest in order to get second largest

}

second_largest=x[i];

}

}

//print value

cout <<"Largest Number:"<<largest<<endl;

cout <<"Second Largest Number:"<<second_largest;

return 0;

}

**Output:
**

Enter Number of elements :5

Insert array elements :33

45

75

87

23

Largest Number:87

Second Largest Number:75

**Explanation:
**

In the above program firstly we define the header file then we define the main method in the main method we define the array and other variables. We first input the number for the size of the array. Then we insert array elements after inserting array elements we search the largest number and the second largest number in the array. To search the largest number in the array we use the loop. To search the first largest number we define a condition that array is greater than the largest number and store the value into the largest variable. Then we check the second largest number in the array for this we use two conditions that are array is greater than the second largest number in this we use another condition that is array is equal to the largest number. If the inner condition is true then it will move forward and end of an inner condition. In the outer condition, the value will be stored on the second_largest variable all the conditions will be done inner the loop. At the last we print values.

**Answer:**

D

**Explanation:**

Current tables such as the new iPads use solid state drives to store data.

**Answer:**

**// here is code in java.**

import java.util.*;

// class definition

class Solution

{

// main method of the class

public static void main (String[] args) throws java.lang.Exception

{

try{

// scanner object to read innput

Scanner s=new Scanner(System.in);

// variables

long min,years,days;

long temp;

System.out.print("Please enter minutes:");

// read minutes

min=s.nextLong();

// make a copy

temp=min;

// calculate days

days=min/1440;

// calculate years

years=days/365;

// calculate remaining days after years

days=days%365;

// print output

System.out.println(temp+" minutes is equal to "+years+" years and "+days+" days");

}catch(Exception ex){

return;}

}

}

**Explanation:**

Read the number of minutes from user and assign it to variable "minutes" of long long int type.Make a copy of input minutes.Then calculate total days by dividing the input minutes with 1440, because there is 1440 minutes in a day.Then find the year by dividing days with 365.Then find the remaining days and print the output.

** Output:**

** please enter the minutes:1000000000**

** 1000000000 minutes is equal to 1902 years and 214 days.**