Answer:
[see below]
Step-by-step explanation:
Quadrant One has a positive value for both x and y. (+ , +)
Quadrant Two has a negative x value. (- , +)
Quadrant Three has a negative value for both x and y. (- , -)
Quadrant Four has a negative y value. (+, -)
Therefore:
(2, -3) would be plotted in Q4. It does not lie on any axis.
(0, 8) would be on the positive y-axis. (0 is neither negative nor positive.)
(-1, -2) would be plotted in Q3. It does not lie on any axis.
(4, 7) would be plotted in Q1. It does not lie on any axis.
Answer:
Can you post image of the triangle
Step-by-step explanation:
Answer:
38% if you round it from 37.80.
Step-by-step explanation:
Answer:
Step-bThis is an interesting problem. To solve it I should find two irrational numbers r and s such that rs is rational.
I am not sure I am able to do that. However, I am confident that the following argument does solve the problem.
As we know, √2 is irrational. In particular, √2 is real and also positive. Then √2√2 is also real. Which means that it is either rational or irrational.
If it's rational, the problem is solved with r = √2 and s = √2.
Assume √2√2 is irrational. Let r = √2√2 and s = √2. Then rs = (√2√2)√2 = √2√22 = √22 = 2. Which is clearly rational.
Either way, we have a pair of irrational numbers r and s such that rs is rational. Or do we? If we do, which is that?
(There's an interesting related problem.)
Chris Reineke came up with an additional example. This one is more direct and constructive. Both log(4) and √10 are irrational. However
√10 log(4) = 10log(2) = 2.y-step explanation:
