Posted by Anonymous on Saturday, February 9, 2013 at 1:00pm.
I need help starting this, I don't want a plain answer just a what to do.
My best guess is to use the distributive
(a+b)(a+c)= a(b+c). But now that there is a 4th variable what do I do?
Boolean Algebra - Steve, Saturday, February 9, 2013 at 2:23pm
use it twice
= bc'(ab'+cd') + a'd(ab'+cd')
= abb'c' + bcc'd' + aa'b'd + a'cdd'
Now, since xx' = 0, the result is 0
Boolean Algebra - Anonymous, Sunday, February 10, 2013 at 1:01am
OOK, thank you, I see how you did that step. that makes perfect sense.
so simplified you get
Boolean Algebra - Steve, Sunday, February 10, 2013 at 6:54pm
no, the result is 0, since
abb'c = a(bb')c = a0c = 0
each term is 0
Answer This Question
More Related Questions
- Distributive property - How would I multiply these expression? 6(h-4) -3(x+8) -...
- Algebra - 1. Use the Distributive Property to simplify the expression. (-1) (4-c...
- Algebra - 1. Use the Distributive Property to simplify the expression. (-1)(4-c...
- Distributive Property Algebra - I have two questions I don't understand and I ...
- Boolean algebra - Simplify this function using Boolean algebra laws? I am stuck ...
- math - Write an algebraic expression that requires each of the following ...
- algebra 2 - It says to: Use the Distributive Property to show that the following...
- Algebra/Distributive Properties - Solve these fraction equations: 2/3 - 8/9 = c/...
- Math-Distributive Property - Use the distributive property to simplify: 3(m-1)-(...
- Math - okay so i feel like a dummie for not knowing this but i need the answer ...