Melody

Here is a small learning unit for it.

what you have written is exactly the same as B.  No need to rearrange anything.

B just has one added set of brackets but they do not change the value at all.

I will give you 2 good hints.

You can come back with your thoughts if you need more help.

1)   let y=3^x

2)  \((a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4\)

Or you could do it without a calculator    

\(5*10^{-4}*3.5*120\\ =5*7*60*10^{-4}\\ =7*300*10^{-4}\\ =2100*10^{-4}\\ =0.21 \)

Thanks Tiggsy.

I have been playing, unsuccessfully with this one. 

Never would have thought of that.   :)

So it is!  I had read it wrong the whole time! 

thanks Alan :)

I do not understand what you assumed but there is often more than one way to interpret these questions.

I think you found a more complicated way to do so.

I think it is     \((\frac{1}{7})_{base 10}\;=\;(0.\bar{05})_{base6}\)

I used this video to show me how.

You are right.   There is NO maximum.

just replace the x in the g function with   2x^2-3x+4

and then sub 5 for x.

You tell us what you get.   

Hi Alan,

Could you show me 2 or more possible sets of draws where there is just one unseeded team hosting?

Because I think there is only 1.