Question

WHERE DO I GO TO DO THIS AND WHAT DO I WRITE?????

Answer 1

You can just look up "python ide online" on google and paste this code:

n = -1

count = 0

while n < 0:

   n = int(input("We're checking to see if a number is prime or not! Enter a positive number: "))

if n % 2 == 0:

   if n == 2:

       print("{} is a prime number".format(n))

   else:

       print("{} is not a prime number".format(n))

else:

   for x in range(n, 1, -1):

       if n % x == 0:

           count += 1

   if count > 1 or n == 1:

       print("{} is not a prime number".format(n))

   else:

       print("{} is a prime number".format(n))

I've written some code that checks to see if a number entered by the user is a prime number or not.

Sorry, but I'm not too good with pseudocode plans and all that. I hope this helps.

Answer 2

Answer:

import math

print("Let's solve ax² + bx + c = 0")

a = int(float(input('Enter a value for a: ')))

b = int(float(input('Enter a value for b: ')))

c = int(float(input('Enter a value for c: ')))

D = b*b-4*a*c

if (D<0):

   print("Sorry, this equation has no solutions.")

elif (a == 0):

   if (b == 0):

       if (c == 0):

           print("Every value of x is a solution")

       else:

           print("Sorry, this equation has no solutions")

   else:

       x = -c/b

   print("The one solution is x={:.3g}".format(x))

elif (D==0):

   x = (-b + math.sqrt(D)) / (2*a)

   print("The one solution is x={:.3g}".format(x))

else:

   x1 = (-b + math.sqrt(D)) / (2*a)

   x2 = (-b - math.sqrt(D)) / (2*a)

   print("This equation has two solutions: x={:.3g} or x={:.3g}".format(x1, x2))

Explanation:

Above is another little program to use the quadratic formula.