# 19/19

#1

def distance_from_zero(n):
return type(n) ==int or type(n) == float
print distance_from_zero(15)
print distance_from_zero(15.2)
print distance_from_zero("35")
print("\n")
print max(15,16,17,18)
print min(15,16,17,18)
print("\n")

print abs(15)
print abs(-15)

its return is true and not ten how do I fix this?

#2

Well... The issue here is that you are not using the function parameter anywhere in your function.

TRY:

``````def distance_from_zero(number):
try:
return abs(number)
except TypeError as error:
print("You failed to enter a number: %s" % error)``````

EDIT:

I meant you never used the parameter to work on, you did use it as a comparative.

EDIT:

@sarahbailey3

``````def distance_from_zero(p):
if type(p) in [int,float]:
return abs(p)
else:
return("Nope")``````

#3

it still says Your function seems to fail on input True when it returned '1' instead of 'Nope'

#4

This is what I put in and it worked.
Just don't copy and paste it since you wouldn't learn anything from doing that

``````def distance_from_zero(hi):
if type(hi) == int or type(hi) == float:
return abs(hi)
else:
return "Nope"``````

#5

If you want to use what Zeziba said do this:

``````def distance_from_zero(p):
if type(p) in [int,float]:
return abs(p)
else:
return "Nope"``````

#6

Try this mate

def distance_from_zero(n):
if type(n) == int or type(n) == float:
return abs(n)
else:
return "Nope"

#7

I tried your code just to see what would happen i tried your code and i still get a name error

#8

It's more simple than that:
def distance_from_zero(n):
if type(n) == int or type(n) == float:
return abs(n)
else:
return "Nope"
print distance_from_zero(14)
print distance_from_zero(-14.5)

Trying to use a Python Shell on your own computer when the things goes wrong.