Answer:
2
If R is a relation that is transitive and symmetric, then R is reflexive on dom(R)={a∣(∃b)aRb}: if a∈dom(R), then there is b such that aRb, thus bRa by symmetry, so aRa by transitivity.
Note that if R is symmetric, then dom(R)=range(R)={b∣(∃a)aRb}.
Hence, to get an example of a relation R on a set A that is transitive and symmetric but not reflexive (on A), there has to be some a∈A which is not R-related to any b∈A. There are many examples of this:
A={0,1} and R={(0,0)},
not reflexive on A because (1,1)∉R,
A={0,1,2} and R={(0,0),(0,1),(1,0),(1,1)},
not reflexive on A because (2,2)∉R.
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
Step-by-step explanation:
To solve these problems, just figure out the sequence, multiply, then add.
In the first problem, the terms progress by -3
So, the 5th term would be -3 x 4 + -19:
-3 x 4 = -12 -12 + -19 = -31 The 5th term is -31.
Do the same for the 9th and 100th term:
-3 x 8 = -24 - 24 + -19 = -43 The 9th term is -43.
-3 x 99 = -297 - 297 + -19 = -316 The 100th term is -316.
For the second problem, The numbers progress by + 10.
So let's do the same thing as last time!
The 5th term is 44.
The 9th term is 84.
The 100th term is 994.
In this last problem, the sequence <u><em>is</em></u> geometric.
How?
Each number is multiplied by -6.
-6 x -1 = 6
-6 x 6 = -36
-6 x -36 = 216
And the next number would be:
-6 x 216 = -1,296
well, take a peek at the cos(2θ) identities.
