Solution:XOR: X+Y= XY’ + X’YDual of XOR:= (X +Y’)+(X’+Y)= XX’+XY +X’Y’ +YY’= XY + X’Y’ Complement of XOR (XNOR)= (X+Y)’= (XY’ + X’Y)’=(X’+Y)+(X +Y’)= XX’+ XY + X’Y’+YY’= XY + X’Y’
HOPE IT HELPS
Answer:
(A) No solution
(B) One solution
(C) One solution
(D) One solution
(E) No solution
Please tell me if this is incorrect. I hope this helps!
Step-by-step explanation:
Let 
Then 
or
This gives us or all integer multiples of 
Step-by-step explanation:
gradient = slope or several other words.
it describes how strongly a line (or tangent to a bent curve) is going up or down or ... if it is changing at all.
it is represented by the ratio
(y coordinate change / x coordinate change)
when going from one point on the line to another.
in our case, when going from A to B we have
x changes by -2k (from 3k to k).
y changes by -11 (from 8 to -3).
so, the gradient or slope is
-11/-2k = 3
11/2k = 3
11 = 3×2k = 6k
k = 11/6
A = (33/6, 8) = (11/2, 8)
B = (11/6, -3)
Answer:
3, in both a), b)
Step-by-step explanation:
a) The slope of the line tangent to the curve that passes through the point (2,-10) is equal to the derivative of p at x=2.
Using differentiation rules (power rule and sum rule), the derivative of p(x) for any x is 
. In particular, the value we are looking for is 
.
If you would like to compute the equation of the tangent line, we can use the point-slope equation to get 
b) The instantaneus rate of change is also equal to the derivative of P at the point x=2, that is, P'(2). This is equal to 
.
