Thats what we are learning atm. in mathematic... dont find the rule for it. 27 ___ 3 n+3 = the n+3 is high. Any Ideas?
Well I do Maths and Further maths at AS level at the moment, but I don't really understand what you're asking us to do, or what "the n+3 is high" means...
that it. You now know what I mean? <a class="postlink" href="http://www.pic-upload.de/view-8792007/WTFORFL.jpg.html" onclick="window.open(this.href);return false;">http://www.pic-upload.de/view-8792007/WTFORFL.jpg.html</a>
So what you are asking is the limit of f(x)= 27 / 3^(n+3) with N going to infinity aka going "high"? Then lim f(x) = 0. Since the number is getting smaller and smaller because the denominator is getting bigger and bigger. Or did you just want to solve the equation like antou did?
I made an example from that I think I calculated it right... Example under it the right excercise <a class="postlink" href="http://www.pic-upload.de/view-8793334/math_sucks.jpg.html" onclick="window.open(this.href);return false;">http://www.pic-upload.de/view-8793334/m ... s.jpg.html</a>
Antou is right, 27/(3^(n+x)) = 3^3/(3^(n+x)) = 3^3 * 3^(-(n+x)) = 3^(3-n-x) And just pick a number for x You can't erase or solve n with just that given.
27/(3^(n+2)) = 3^3/(3^(n+x2)) = 3^3 * 3^(-(n+2)) = 3^(3-n-2) the x is a 2 not an x . But that is right so`?
Yea, that makes: 3^(1-n) (But if I'm not mistaken, you still don't get it, else you wouldn't have asked if that was right?)
nah i think i got it.. jsut didnt know what to first because it was 27/3 but if its 3^3/3 etc its clear for me thx . (wasnt sure if that was wanted from me thats why i asked )