The second factor as it did not make travel easy at all
Answer:
For question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!
Solution :
x = float_(input())
y = float_(input())
z = float_(input())
res1 = x**z
res2 = x**(y**z)
res3 = abs(x-y)
res4 = (x**z)**0.5
print('{:.2f} {:.2f} {:.2f} {:.2f}'.format(res1,res2,res3,res4))
Output is :
5.0
1.5
3.2
172.47 361.66 3.50 13.13