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 inequality 44x + 1230 ≥ 6000 can be used.
Step-by-step explanation:
Number of members = 44
Amount to raise = $6000
Amount already raised = $1230
Let,
x be the amount each member needs to raise.
44x + 1230 ≥ 6000
Therefore,
The inequality 44x + 1230 ≥ 6000 can be used.
Area = length x width.
24 = (x + 2)(x - 3) = x² - x - 6
x² - x - 30 = 0
(x + 5)(x - 6) = 0
We can eliminate the root x = - 5 because dimensions cannot be negative. So, the only valid answer is x = 6
The length = x + 2 = 6 + 2 = 8 in.
The width = x - 3 = 6 - 3 = 3 in.
Jim owns 48 cars.
4:3 as x:36
36/3 is 12, so the scale is 12.
x is 4*12 = 48