Question

Check please.

Answer 1

Solve:\[x^2 - 4x + \lfloor x\rfloor + 3 = 0\]I did this:\[x^2 - 4x + x - \{ x \} + 3 = 0\]\[x^2 - 3x + 3 = \{ x \}\]So I imposed the condition:\[0 \le x^2 - 3x + 3< 1 \]\[\Rightarrow (x-2)(x-1) < 0\]\[\Rightarrow x \in (1,2)\]So I basically have to solve in the interval \((1,2)\) the following:\[x^2 - 3x + 3 = x - 1\]\[\Rightarrow x^2 - 4x + 4 = 0\]\[\Rightarrow (x -2)^2 = 0 \]And so that means no solutions.

Answer 2

Why does WA return 2? That's obviously not a solution.

Answer 3

that looks nice! wolfram also says no solutions
http://www.wolframalpha.com/input/?i=solve+x%5E2-4x%2Bfloor%28x%29%2B3%3D0+over+reals

Answer 4

Weird.
http://www.wolframalpha.com/input/?i=x%5E2+-+4x+%2B+floor%28x%29+%2B3+%3D0

Answer 5

That's the first time I've seen something like that.

Answer 6

it does say "numerical solution" so that accounts for some error...

Answer 7

Oh, I got it!

Answer 8

https://brilliant.org/problems/can-you-draw-its-graph-1/?group=S1VvD8PL1YP8&ref_id=725593 Why is it not accepting zero as an answer?

Answer 9

try 2

Answer 10

What's \(P(P(\emptyset))\)?

Answer 11

Shouldn't that be \(\emptyset \) itself?

Answer 12

A : {}

P(A) : {{}}

P(P(A)) : {{}, {{}}}

Answer 13

2 is incorrect as well!

Answer 14

try 4

Answer 15

answer has to be a power of 2

Answer 16

Could you do that, please? :P

Answer 17

4 is correct

Answer 18

Yeah, I understand that.

Answer 19

That's funny. This question seems to test more of set theory.