I'd like to simplify the following expression "xy + x'z + yz": xy + x'z + yz = xy + z (x' +y) = (xy + z) (xy + x' + y) = (xy + z) (y (x + 1) + x'...

2014年8月27日 · I need to simplify xy + xz' + x'yz into xz' + yz. I know that these expressions are equal in truth value, but I'm not sure how to simplify the first to get the second. Here are the steps I can do:...

If xy+yz+xz=-10 and xyz=8, then \frac {1} {x}+\frac {1} {y}+\frac {1} {z} equals what?

2015年4月6日 · I'm trying to prove that XYZ + XYZ' + XY'Z + X'YZ = XY+ XZ + YZ. I have gotten so far i think. I don't know if I am on the right track but i keep getting stuck when i end up with this xy + xy’z + x’yz.... here is what i have done: F (X,Y,Z) = XYZ + XYZ'+XY'Z + X'YZ. XYZ + XYZ’ + ZXY’ + XYX’ (Commutative) XY (Z) + XY (Z’) + ZXY’ + ZYX’ (associative)

2021年5月4日 · So using K-maps I was able to simplify xy + (!x)z + yz to xy + (!x)z, and I double-checked that the truth tables are the same. I'm having trouble understanding how I would have used boolean algebra to get this result. I tried my usual tricks of adding zero, letting addition distribute over multiplication, etc., without avail. Any thoughts ...

As stated in the title, I'm trying to simplify the following expression: $xy + xy'z + x'yz'$ I've only gotten as far as step 3: $xy + xy'z + x'yz'$ $=x (y+y’z) + x’ (yz’)$ $=x (y+y’z)+x (y’+z)$ But...

XY + YZ .

2019年9月24日 · There are only 8 possibilities to look at. Both xz + yz′ +x′y′ x z + y z ′ + x ′ y ′ and xy +x′z′ +y′z x y + x ′ z ′ + y ′ z are correct, though they look different. If you've got one of them, it is correct.

2019年4月1日 · Applying this principle, we can rewrite your expression as: f = x'yz + xy'z + xyz' + xyz. That said, as the diagram clearly shows, you can simply take AND together each pair of inputs, and OR them together to get the same result. By doing this, you ensure that:

Enhanced with AI, our expert help has broken down your problem into an easy-to-learn solution you can count on. Here’s the best way to solve it. Distribute x in the first term of the expression: x (y z + y ′ z) + x y + x ′ y + x z = x y z + x y ′ z + x y + x ′ y + …