Math Is Fun Forum

I thought there is 1 if the letter belongs to both set, and 0 if it belongs only to the set A. In this case. letters "a" and "e" both belong to sets A and B, so shouldn't there be 1 in: "If the letter is 'a', then the value is  0."?
Anyway, I understand that, but is this the way I have to describe it? Besides, is letter "y" a vowel too, in English?

And one more thing, thanx for your help

Thanx mate, that's the answer I needed to get

OMG I've never hear of that O_O
Thank you, I'll try to understand that and maybe use it later

thanx man, but I did it in another more complicated way.

x³-9-6x=0
(x³-27)+18-6x=0
(x-3)(x²+27x+27²)-6(x-3)=0
(x-3)(x²+27x+723)=0

x-3=0
x=3

and
x²+27x+723=0
D<0 no solutions

Now the answer is correct.

Could u explain me how you get this (x^2+3*x+3)?

yeah, maybe you're right.

oh, I knew I did it right, but just missed that slash

or you simply can see a formula (a-b)(a+b)=a² - b² ; as a result (x-1)(x+1)=x²-1²
so x(x-1)(x+1)=x(x²-1²)=x³-x

hey this is weird.
I understand everything till that 'thin moment'
first of all what the '^' mean? is it multiplying or what?
then where did you get √((2^2+√5^2)