Melody

Sorry Max, I didn't see your answer there. :)

Sorry, the Latex is playing up.

its (-1,1]

Sometimes I just gotta laugh.   

\(\frac{3-z}{z+1} \geq 1 \\ \text{ Assuming that z is a real number}\\ z\ne-1\\ \frac{3-z}{z+1} \geq 1 \\~\\ If\;\;z>-1\;\;then\\ 3-z \geq z+1 \\ -2z\ge -2\\ z\le1\\ \therefore -1

or

\(If\;\;z<-1\;\;then\\ 3-z \leq z+1 \\ -2z\le -2\\ z\ge1\\ \therefore no\;\;soln \)

Hence

\(-1

I find it astonishing that this site in general would be specifically approved by a teacher and not the net in general.

there are many brilliant teaching resources in You Tube and on the net in general.  

Maybe it is a stranger danger thing idk.

Anyway, your explanation is reasonable.

Most of the time it is best to ignore Gingers tirades.

I might delete some of thiese posts later later for now I am just going to lock the question so that no more posts can be added.

Thanks for your answer Vinculum.    

Guest answerer clearly got out of bed on the wrong side today and I am not condoning his/her aggressive words.

But the suggestion to  "Google how to calculate hydronium ions" was most likely a good one.

If you learn to look things up on the net you can usually get more comprehensive answers and sometimes you learn more.

Often it is just knowing what words to use to get the answer you want that is the hard bit and guest has given you a good suggestion.

As for saying that googling is cheating, when you are here looking for an exact answer, that was a rather crazy thing to say.

You are welcome. 

Just use a calc to work out what those two are (one dec place will do) 

and then write down/ count the numbers between them.

There is no question here.

ok thanks for that, I thought I might have double counted.

Ok here is the go

the first bit of what I did was correct

  # W #[BR]# W #[BR]# W #  

If they both stay together then there are  6 places they can go

If they are seperated then there are  6C2 = 15 places they can go.

Which adds to 15+6=21 places

now lets look at a scenario

consider starting with

  # W #[BR]# W #[RB]# W #           

If I added 2 reds tied together they will all be different than any permutation in the first set, that is 6 ways

If 2 two added reds are to be added seperately then

 # W #[BR]# W R[BR]# W #  

will all be the same as

 # W #[BR]# W #[RB]R W # 

and there are 4 places where the second R can be put, so if I add15 in this second set I will have double counted by 4

So rather than 15 there will be 11 

The next set will have 4 less again and the same with the last set

So rather than it being

21+21+21+21

it will be

21+17+13+9 = 60