Answer:
-0.5
Step-by-step explanation:
C. 1/5,2/5,3/5,4/5 and 1 since all will be on the same mark as 2/10,4/10,6/10,8/10 and 10/10.
d. there would be 10/10 for every 0 to 2.
Answer:
I'd say about 50/50 cuz say there is 6 red cards and 7 black cards then it depends on the luck of the draw
Step-by-step explanation:
ik my answer sucks but thats what i think
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:
Answer:
The slope is -1
Step-by-step explanation:
Anytime the x-intercept values have 0 in it, the y-intercept value which is below/above the x-values (Which in this case is -1) is the y-intercept.
