For starters, , not 9.
Also how did you magically turn into , into and into ?
Hi all, I've gone through the 'read before' threads so I hope this is in the right section and that I've typed it out ok, I'm still working with pen and paper here
Basically I've just started to get back into mathematics and I'm doing a course externally. I've had a try at simplifying the problem below but all the rules are a very new to me and so I have no idea if I'm on the right track or not. Any feedback at all would be a huge help, Thanks.
(z + 1)^2 (3xy^2)^3 / (9y^2)^1/2 x^-3 (z + 1) ^-1
= (z + 1)^2 (9x^3y^6) / 3y x^-3 (z + 1)^-1
= (z + 1) ^2 9x^3y^6 1/3y^-1 x^3 (z + 1)
= (z + 1)^3 9x^9y^6 1/3y^-1
= (z + 1)^3 3x^9y^-6
I've typed it out as I've done the steps because I'd almost guarantee the final answer is wrong so like I said any feedback on what I've done right or wrong would be a huge help.
Thanks, cant believe I missed the first part considering how many times I've looked at it.
For the rest I thought that if something was to the power of -1 below the dividing line then it can be moved to the top (numerator section?) and would become a positive to the same power. I was trying to do the same for 3y, moving it above the line as 1/3y^-1?
(z + 1)^2 (27x^3y^6)
3y x^-3 (z + 1)^-1
be a correct first step? (sorry I couldn't find any suggestions about the best way to type out a problem). If it is ok I suppose I'm stuck there on how to simplify it further.
Yes it would be. And to typeset mathematics, we use LaTeX. There is an inbuilt compiler on this site and a subforum dedicated to learning how to use it and gaining assistance with it.
As for negative powers, , so whenever you see negative powers, you can assume there is division. With some manipulation, this would be equivalent to moving everything with a negative power from the top to the bottom with a positive power, or vice versa.