Answer: What is your question?
Answer:
Step-by-step explanation:
Let A = R−{0}, the set of all nonzero real numbers, and consider the following relations on A × A.
Given that (a,b) R (c,d) if 
Or (a,b) R (c,d) if determinant
![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D0)
a) Reflexive:
We have (a,b) R (a,b) because ab-ab =0 Hence reflexive
b) Symmetric
(a,b) R (c,d) gives ad-bc =0
Or da-cb =0 or cb-da =0 Hence (c,d) R(a,b). Hence symmetric
Well, accordingly to the graph, it has a positive slope so we can eliminate b and d. So it is either a or c. As x increases, so does y. Y is almost not quite equal to 3x-2. In fact, it is greater, so I'd go with A. I'm pretty confident.
All of them are the correct answer.
Answer:
5
Step-by-step explanation:
I drew it out on a piece of paper. R is going up and down, through PQ, which is going left to right. If the ratio is 1:3, that means that for every 1 unit there is on one side, there are 3 units in the other. If R is -1 and P is -3, they are two numbers away from each other, which means that on the other side of R (to the right of R), there needs to be 6 numbers in between to make the 1:3 ratio. R is at -1, so then you would write the equation, -1+6=? The answer is 5.
-1+6=5 That is where Q would be
