The answers are as follows:
a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC
= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)
= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC
= B'(C' + C) + B(C' + C) = B' + B = 1
b) F(x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms
with x'1, which can be factored and removed as in (a). The remaining 2n1
product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms
with x'2, which and be factored to remove x2 and x'2, continue this
process until the last term is left and xn + x'n = 1
Answer:
No
Explanation:
You had the whole week for the essay but didn't do it and then turned it in late.
Hey there!
In Microsoft Access, you can click the Tab key then the Enter key on your keyboard to establish a new insertion point directly to the right of your current insertion point when entering data in datasheet view.
Hope this helped you out! :-)
