Thread: A little maths problem for you

ok, commander. post the answer.
but show the method / steps you used.

x^2 = 80?

how did you get that?

very good clyde!!!!!!!!!

ur right!!! good for u!!!

clyde, do you want to post how you got it or do u want me to post how i got it?

I didn't want to post this...but ppl r putting a lot of pressure

and while clyde is posting the solution, here's another one...

Code:

A={1, 2,3, 5, 8, 13, 21, 34, 55}

Question : How many numbers between 3 & 89 cannot be written 
as the sum of two elements on the of the set

it's quite easy...

HINT : Do not use programming or try to make a list of all the possible sums of the {} A

At a glance I'd say 50. If I get a chance to look at it I'll check

I'll post it, I'll just have to write it in a way so u guys can understand

Originally posted by fyodor
At a glance I'd say 50. If I get a chance to look at it I'll check

Very close!

Originally posted by Driveway
Then commander, post how YOU got it.
Hur ry Up

I'm posting it now.....

Code:

[(x+9)^(1/3)] - [(x-9)^(1/3)] = 3

or, [(x+9)^(1/3)] = 3 + [(x-9)^(1/3)]

or, x+9 = 27 + [27(x-9)^(1/3)] + [9((x-9)^2)^(1/3)] + x - 9   // Cubing both sides

or, [27(x-9)^(1/3)] + [9((x-9)^2)^(1/3)] = -9

or, [3(x-9)^(1/3)] + [((x-9)^2)^(1/3)] = -1

or, [(x-9)^(1/3)] * [3 + (x-9)^(1/3)] = -1

since, [3 + (x-9)^(1/3)] = [(x+9)^(1/3)]                      // from ln 2

so, [(x-9)^(1/3)][(x-9)^(1/3)] = -1                           

or, (x-9)(x+9) = -1                                           // cube buth sides

or, x^2 - 81 = -1

or, x^2 = 80                                                  // BINGO !!!!!!!!!!

I found another problem...I just dont know who to solve it yet....I don't wanna post it b4 i solve it because then if someone asks for the answer, I won't be able to give it to them!