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2 years ago

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please answer as soon as possible and don't write wrong answer .wrong answer will be reported and correct answer will be marked

as brainiest

Computers and Technology

1 answer:

LenaWriter [7]2 years ago

8 0

Answer:

a sprite

6 operator boolean blocks

11 operator reporter blocks

trojan/ trojan horse

a podcast

a container

the trash

Explanation:

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In this question, you will experimentally verify the sensitivity of using a precise Pi to the accuracy of computing area. You ne

zhannawk [14.2K]

Answer:

Follows are the code to this question:

import math as x #import math package

#option a

radius = 10#defining radius variable

print("radius = ", radius)#print radius value

realA = x.pi * radius * radius#calculate the area in realA variable

print("\nrealA = ", realA)#print realA value

#option b

a1 = 3.1 * radius * radius#calculate first area in a1 variable

print("Area 1= ", a1)#print Area

print("Percentage difference= ", ((realA - a1)/realA) * 100) #print difference

a2 = 3.14 * radius * radius#calculate first area in a2 variable

print("Area 2= ", a2)#print Area

print("Percentage difference= ", ((realA - a2)/realA) * 100)#print difference

a3 = 3.141 * radius * radius#calculate first area in a2 variable print("Area 3= ", a3)#print Area

print("Percentage difference= ", ((realA - a3)/realA) * 100) #print difference

Output:

please find the attached file.

Explanation:

In the given Python code, firstly we import the math package after importing the package a "radius" variable is defined, that holds a value 10, in the next step, a "realA" variable is defined that calculate the area value.

In the next step, the "a1, a2, and a3" variable is used, which holds three values, that is "3.1, 3.14, and 3.141", and use the print method to print its percentage difference value.

4 0

3 years ago

We piped the results of the Get-Process cmdlet to the Sort-Object cmdlet to sort them in descending order. Which property was us

Basile [38]

The answer is yes hope this helps did before

5 0

3 years ago

Write a program that uses while loops to perform the following steps:

Vladimir79 [104]

Answer:

Explanation:

//Include the required header files.

#include<iostream>

using namespace std;

//Define the main function.

int main()

{

//Define the variables.

int i, sum = 0, sqSum = 0, firstNum = 1, secondNum = 0;

char ch;

//Check for valid input.

while (!(firstNum < secondNum))

{

cout << "Enter starting number: ";

cin >> firstNum;

cout<<"Enter ending number(must be > startingNumber): ";

cin >> secondNum;

}

//Store first number in i

i = firstNum;

//Dispaly the number.

cout << "The odd numbers between " << firstNum

<< " and " << secondNum << " are:\n";

//Iterate between first and second number.

while (i <= secondNum)

{

//Check for even numbers.

//Store the sum

if (i % 2 == 0)

sum = sum + i;

//Print the odd numbers

//Evaluate the square of sum of odd number.

else

{

cout << i << " ";

sqSum = sqSum + i * i;

}

//Increase the value of i.

i++;

}

//Dispaly the sum of even numbers.

cout << "\n\nThe sum of the even numbers is:"

<< sum << endl << endl;

//Dispaly the sum of square of odd number.

cout << "The sum of squares the odd numbers is:"

<< sqSum << endl;

//Set i to 1.

i = 1;

//Dispaly the message.

cout << "\nNumber Square\n";

//Iterate and print number between 1 andd 10

//along with the sum.

while (i <= 10)

{

cout << " " << i << "\t " << i * i << endl;

i++;

}

//USe for visual studio.

system("pause");

//Return the value 0.

return 0;

}

Explanation:

The program code will perform the function listed below using loop.

Prompt the user to input two integers: firstNum and secondNum (firstNum must be less than secondNum). Output all odd numbers between firstNum and secondNum. Output the sum of all even numbers between firstNum and secondNum. Output the numbers and their squares between 1 and 10. Separate the numbers using any amount of spaces. Output the sum of the square of the odd numbers between firstNum and secondNum. Output all uppercase letters.

Hope this helps!

3 0

3 years ago

Express the worst case run time of these pseudo-code functions as summations. You do not need to simplify the summations. a) fun

ser-zykov [4K]

Answer:

The answer is "O(n2)"

Explanation:

The worst case is the method that requires so many steps if possible with compiled code sized n. It means the case is also the feature, that achieves an average amount of steps in n component entry information.

In the given code, The total of n integers lists is O(n), which is used in finding complexity.
Therefore, O(n)+O(n-1)+ .... +O(1)=O(n2) will also be a general complexity throughout the search and deletion of n minimum elements from the list.

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3 years ago

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Pani-rosa [81]

Answer:

¹²³⁴⁵⁶⁷⁸⁹

Explanation:

4 0

4 years ago