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allsm [11]
3 years ago
13

I'm not sure how to do these. By the way, it has to be python.

Computers and Technology
1 answer:
frutty [35]3 years ago
6 0

Task 1:

float75 = float(75)

string75 = "75"

# you cannot add together a number and a string because a string has no inherent numerical value like a number does.

Task 2:

num = float(input("Enter a number"))

print(num**2)

Task 3:

num = int(input("Enter an integer: "))

print("When you divide "+str(num)+" by 7, the quotient is "+str(num//7)+" and the remainder is "+str(num%7)+".")

Task 4:

gigs = int(input("How many gigabytes does your flashdrive hold? "))

print("A flashdrive with "+str(gigs)+ " gigabyte(s) holds "+str(gigs*8589934592)+" bit(s).")

For task 4, you might have to change the number 8589934592 to something else. I'm not entirely sure how many bits are in a gigabyte. I hope this helps though.

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Advantages of using Unicode to represent data
Alona [7]
Global source and binary.
Support for mixed-script computing environments.
Improved cross-platform data interoperability through a common codeset.
Space-efficient encoding scheme for data storage.
Reduced time-to-market for localized products.
Expanded market access.
5 0
2 years ago
A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2 until it reaches its final angu
Nataly_w [17]
The first thing we are going to do is find the equation of motion:
 ωf = ωi + αt
 θ = ωi*t + 1/2αt^2
 Where:
 ωf = final angular velocity
 ωi = initial angular velocity
 α = Angular acceleration
 θ = Revolutions.
 t = time.
 We have then:
 ωf = (7200) * ((2 * pi) / 60) = 753.60 rad / s
 ωi = 0
 α = 190 rad / s2
 Clearing t:
 753.60 = 0 + 190*t
 t = 753.60 / 190
 t = 3.97 s
 Then, replacing the time:
 θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
 θ1 = 1494.51 rad
 For (10-3.97) s:
 θ2 = ωf * t
 θ2 = (753.60 rad / s) * (10-3.97) s
 θ2 = 4544,208 rad
 Number of final revolutions:
 θ1 + θ2 = (1494.51 rad + 4544.208 rad) * (180 / π)
 θ1 + θ2 = 961.57 rev
 Answer:
 the disk has made 961.57 rev 10.0 s after it starts up
3 0
3 years ago
Read 2 more answers
The ArrayList class contains a trim method that resizes the internal array to exactly the capacity. The trim method is intended
andreev551 [17]

Answer:

public void trimToSize() {

modCount++;

if (size < elementData.length) {

elementData = (size == 0)

? EMPTY_ELEMENTDATA

: Arrays.copyOf(elementData, size);

}

}

Now, the running time for copyOf() function is O(N) where N is the size/capacity of the ArrayList. Hence, the time complexity of trimToSize() function is O(N).

Hence, the running time of building an N-item ArrayList is O(N^2).

Please find the sample code below.

CODE

==================

import java.util.ArrayList;

public class Driver {

  public static void main(String[] args) throws Exception{

      int N = 100000;

      ArrayList<Integer> arr = new ArrayList<>(N);

     

      long startTime = System.currentTimeMillis();

      for(int i=0; i<N; i++) {

          arr.add(i);

          arr.trimToSize();

      }

      long endTime = System.currentTimeMillis();

      double time = (endTime - startTime)/1000;

      System.out.println("Total time taken = " + time + " seconds.");

  }

}

Explanation:

5 0
3 years ago
When she manages a software development project, Candace uses a program called __________, because it supports a number of progr
alexandr402 [8]

Answer:

PLS programming language supporting.

Explanation:

7 0
3 years ago
Write a java program called allDigitsOdd that returns whether every digit of a positive integer is odd. Return true if the numbe
Vlada [557]

Answer:

public class Digits

{

   public static boolean allDigitsOdd(int num)

   {

       boolean flag=true;

       int rem;

       while(num>0)

       {

           rem=num%10;

           num=num/10;

           if(rem%2==0)    // if a even digit found immediately breaks out of loop

           {

               flag=false;

               break;

           }

       }

       return flag;     //returns result

   }

   public static void main(String args[])

   {

       System.out.println(allDigitsOdd(1375));    //returns true as all are odd digits

   }

}

OUTPUT :

true

Explanation:

Above program has 2 static methods inside a class Digits. Logic behind above function is that a number is divided by 10 until it is less than 0. Each time its remainder by 0 is checked if even immediately breaks out of the loop.

4 0
3 years ago
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