Answer:
str1=str(input("Enter a character and phrase "))
count= 0
for i in range(0,len(str1)):
if str1[0]==str1[i]:
count+=1
count1=count-1
print(str1[0] + ' appears ' + 'this many times in ' + str1[1:] + ' is : ' + str(count1) )
B. abrasion. Abrasion is pretty much friction on the rocks. Yardangs don't really apply to this situation and I have never heard of a deflated rock:P Hope that helps:)
Answer:
It can be "ON" or "OFF". So it can store the numbers 1 and 0, but it certainly doesn't have the capacity to store a letter of the alphabet.
Explanation:
In terms of music taking one song and incorporating it into your own song
A. He has not saved the document yet.
reason being is that most documents today will change their title on what it is thats written on the paper unless you personally change it yourself
