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I hope so. I cannot see any mistakes so far.
I have no idea who is talking to whom anymore.
Thanks, Bob. I am pretty sure that you would've got it if you continued.
Hi bobbym
MTG? Magic: The Gathering? ![]()
Done. There might be mistakes though, so there may be more solutions.
I do have a method. I will post it here. As Bob would always say, watch this space.
Hi jamesbailey1990 and kdipakj
Welcome to the forum! ![]()
Hi she taco
Welcome to the forum! ![]()
Hi bobbym
Hi Bob
I am getting those answers, too.
Nessun problema.
Hi bobbym
Didn't 2 people already solve Brittle math 2?
Can you immediately tell me what they equal? I bet you can, that means you have seen them so much you have memorized them. Why wait?
7^3, no. 11^3 neither, but I can calculate that one pretty quickly in my head.
I agree on 6^3, though, because everyone who does even a little bit of combinatorics will see that number.
10^3 is pretty straightforward.
Well, I really don't think that's what he's looking for right now. ![]()
And, aren't you missing a term there?
I agree on 6^3. I'm not sure why you would need 7^3 and 11^3, though.
You design questions. Or rather, your subconscience does.
For a quadratic to always be positive you need:
For a quadratic to have positive roots, you need the conditions Agnishom posted, except, the second condition should be
Hi Agnishom
Hi bobbym
Hi Bob
I'm guessing he's referring to this one