Answer:Since the integers are closed under addition, . m C n C 1/ is an integer, and hence the last equation shows that x C y is even. Therefore, we Page 3 Appendix C. Answers for Exercises 539 have proven that if x and y are odd integers, then x C y is an even integer. P1 There exists an integer k m D 2k.
Step-by-step explanation:
Answer:
To prove:
X+Y.Z=(X+Y).(X+Z)
Taking R.H.S
= (X+Y).(X+Z)
By distributive law
= X.X+X.Z+X.Y+Y.Z --- (1)
From Boolean algebra
X.X = X
X.Y+X.Z = X.(Y+Z)
Using these in (1)
=X+X(Y+Z)+Y.Z
=X(1+(Y+Z)+Y.Z --- (2)
As we know (1+X) = 1
Then (2) becomes
=X.1+Y.Z
=X+Y.Z
Which is equal to R.H.S
Hence proved,
X+Y.Z=(X+Y).(X+Z)
Number = 125
Here, ∛125 = 5
so, in exponential form, it would be: 5³
So, your final answer is 5³
Hope this helps!
X is going to equal 33. 25+29+x/3=33
