Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Boolean Logic: reduce to a minimum sum-of-product form:
X'Y'Z' + X'YZ + XYZ
X'Y'Z' + X'Y'Z + XY'Z + XYZ'
for the first one, I have it down to X'Y'Z' + YZ, but is there anything more I can do that I'm not seeing?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

|dw:1327917442864:dw|
X'Y'Z' + YZ
no other solution for sure ... you are welcome

- anonymous

2nd one = X'Y'+X(Y'Z+YZ') = X'Y'+Y'Z+YZ' = X'Y' + (Y xor Z)

- anonymous

Chris ... we have not x'y' ... please check your answer . I'm sure its not right answer

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

What do you mean by "we have not x'y'"? The (2nd) given formula was X'Y'Z' + X'Y'Z + XY'Z + XYZ' and there you can easily see, that the first two terms lead to X'Y'(Z'+Z) = X'Y'. Or maybe I misunderstood you?

- anonymous

look to original Boolean equ
X'Y'Z' + X'YZ + XYZ
X'Y'Z + XY'Z + XYZ' yours
do you have any difference before simplification

- anonymous

?

- anonymous

I'm all the time talking about the 2nd (second, the lower, the latter, the last) of the two expressions that were mentioned in the original post. (At least, as far as I can understand, the two lines with two expressions, one consisting of three, the other of four terms, are two independent expressions, each of he has to reduce.) Maybe I see a different posting, but for me the 2nd expressions reads: \[X'Y'Z' + X'Y'Z + XY'Z + XYZ'\].

- anonymous

in second term you consider y' while its not ..

- anonymous

any way .. logic is simple .. and simplification is easy . its just need to take it easy and step by step .. I'm sure that you can solve harder than this, I like your way of deffence because that is mean ... You are completely understand logic ... respect for you buddy .. see u around

Looking for something else?

Not the answer you are looking for? Search for more explanations.