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!
Answer:
Center Processing Unit ....
Answer:
a=input("Amount in pennies")
b=int(a)
dollars=0
dimes= 0
quarters=0
nickels=0
pennies = 0
dollars = int(b/100)
b= b- dollars *100
quarters=int(b/25)
b=b-quarters*25
dimes = int(b/10)
b = b -dimes*10
nickels=int(b/5)
b=b - nickels * 5
pennies = b
print(dollars)
print(dimes)
print(nickels)
print(pennies)
Explanation:
The required program is in answer section. Note, the amount is entered in pennies.
Answer:
Decomposition is when we break a problem down into smaller parts to make it easier to tackle.
Hope this helps I am in high school and I’m gonna work for Apple so I know a lot about computers and electronics
Answer:
int x;
indata.open("lottowins");
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata.close();
