c++ 2.30 LAB: Phone number breakdown Given a long long integer representing a 10-digit phone number, output the area code, prefi

Answer:

import random

random.seed(1,10) # Note: seed takes one argument, so this line has an error

a = random.randint(1, 10)

b = random.randint(1, 10)

print("Calculate: " + str(a) + " X " + str(b) + "?")

ans = int(input("Your answer: "))

if ans == a * b:

print("Well done!")

else:

print("That's wrong!")

Explanation:

Question:

The function is_positive should yield True if the supplied number is positive; otherwise, it gives None. Can you fill in the blanks accordingly?

def is_positive(number):

       if  _____ :

           return _____

Answer:

def is_positive(number):

   if (number > 0):

       return True

  else:

       return "None"

---------------------------------------------------------------------------------

Code Test and Sample Output:

print(is_positive(6))

>> True

print(is_positive(-7))

>> None

----------------------------------------------------------------------------------

Explanation:

This segment of code is composed in Python.

To explain it further, let's break down the content of each line;

Line 1: Creates a function identified as is_positive which takes in a variable number.i.e

def is_positive(number):

Line 2: Evaluates if the given number is positive; a number qualifies as positive if it exceeds zero. i.e

if (number > 0):

Line 3: Produces a boolean result of True when the input number is positive. i.e

return True

Line 4: Initiates the else block to be executed in case the input number isn’t positive. i.e

else:

Line 5: Returns the string "None" when the number is not positive. i.e

return "None"

When combined, this results in;

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

def is_positive(number):

   if (number > 0):

       return True

   else:

       return "None"

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

An instance test for the code provided an argument of 6 and -7, resulting in True and None respectively. i.e

print(is_positive(6))   = True

print(is_positive(-7))  = None

This problem is a personal reflection and cannot be solved without discussing it after the individual provides input.

a)

C++

#include<iostream>  

using namespace std;  

int factorial(int num)  {  

   if (num == 0)  

       return 1;  

   return num * factorial(num - 1);  }    

int main()  {  

   int integer;

   cout<<"Enter a non negative integer: ";

   cin>>integer;

   cout<< "Factorial of "<< integer<<" is "<< factorial(integer)<< endl;  }

Python:

def factorial(num):  

   if num == 0:  

       return 1

   return num * factorial(num-1)  

integer = int(input("Enter a non negative integer: "))  

print("Factorial of", integer, "is", factorial(integer))

b)

C++

#include <iostream>  

using namespace std;

double factorial(int number) {  

if (number == 0)  

 return 1;  

return number * factorial(number - 1); }  

double estimate_e(int num){

    double e = 1;

    for(int i = 1; i < num; i++)

     e = e + 1/factorial(i);

     cout<<"e: "<< e; }  

int main(){

int term;

cout<<"Enter a term to evaluate: ";

cin>>term;

estimate_e(term);}

Python:

def factorial(number):  

   if number == 0:  

       return 1

   return number * factorial(number-1)  

def estimate_e(term):

   if not term:

       return 0

   else:

       return (1 / factorial(term-1)) + estimate_e(term-1)

number = int(input("Enter how many terms to evaluate "))

print("e: ", estimate_e(number))

c)

C++

#include <iostream>

using namespace std;

int main(){

   float terms, sumSeries, series;

   int i, number;

   cout << " Input the value of x: ";

   cin >> number;

   cout << " Input number of terms: ";

   cin >> terms;

   sumSeries = 1;

   series = 1;

   for (i = 1; i < terms; i++)      {

       series = series * number / (float)i;

       sumSeries = sumSeries + series;     }

   cout << " The sum  is: " << sumSeries << endl;  }  

Python    

def ePowerx(number,terms):

   sumSeries = 1

   series =1

   for x in range(1,terms):

       series = series * number / x;

       sumSeries = sumSeries + series;

   return sumSeries    

num = int(input("Enter a number: "))

term=int(input("Enter a number: "))

print("e^x: ",ePowerx(num,term))

Explanation:

a)

The program includes a factorial method that takes a number as an argument and calculates its factorial using recursion. For instance, if number = 3

The base case occurs at  if (number == 0)

and the recursion is handled with return number * factorial(number - 1);  

With number = 3 not equaling zero, the function calls itself recursively to get the factorial of 3

return 3* factorial(3- 1);

3 * factorial(2)

3* [2* factorial(2- 1) ]

3 * 2* [ factorial(1)]

3 * 2 * [1* factorial(1- 1) ]

3 * 2 * 1* [factorial(0)]

At this point at factorial(0), the base condition is satisfied as number==0, so factorial(0) returns 1

The resulting output is:

3 * 2 * 1* 1

yielding 6

So, the final program output will be

Factorial of 3 is 6

b)

The estimate_e method takes a number, termed as num, which signifies the term to estimate the mathematical constant e

The for loop extends through each term. For example, if num is set to 3

Then the core statement:

e = e + 1/factorial(i);  

The preceding calculation works as:

e = 1 + 1/1! +1/2!

Since the term count is 3

Initially, e is set to 1

i is initialized at 1

Inserting this into the calculation gives us:

e = 1 + 1/factorial(1)

The factorial function computes and returns 1, as the factorial of 1 is 1. Thus,

e = 1 + 1/1

This results in e = 2

Proceeding to the next iteration, where i = 2 and e = 2, we calculate e = 2 + 1/factorial(2)

Thus, e = 2 + 1/2 results in e = 2.5

Following to the next iteration with i = 3, we have e = 3 + 1/factorial(3)

This yields e = 3 + 1/6 resulting in approximately e = 3.16666

Therefore, the output is:

c)

This program calculates the sum of a series based on the formula:

e^x = 1 + x/1! + x^2/2! + x^3/3! +...

The for loop iterates according to the number set for terms. Assuming x is 2, and the number of terms is set to 3, the series would read:

e^2 = 1 + 2/1! + 2^2/2!

In this setup: number = 2 and terms = 3

Initial values for series and sumSeries are both 1

Starting with i equal to 1, the update statement series = series * number / (float)i; applies as follows:series = 1 * 2 /1 results in series = 2

Then, for sumSeries, we have sumSeries = sumSeries + series; Outputs sumSeries as 1 + 2, yielding 3

Continuing to the next iteration: i=2, with series = 2 and sumSeries = 3, we recalculate as series = 2 * 2/2 imposing series = 2 again. Thus, we find: sumSeries = 3 + 2 giving a final sumSeries value of 5