Melody

Well there you go! Thanks Alan.

Now zegroes, I am not sure if your comment was serious or not. I will assume that it was serious.

I think all the answers are the same but you have been offered a few ways to remember it.  You just have to choose the one you like the best and forget about the others!

$$1+\frac{\sqrt3}{2}=\frac{2+\sqrt3}{2}$$

okay don't go away. I want to talk you through it.

With any whole number there is an invisable decimal point at the end.

You need to move it by jumping over the digits one at a time  until it is between the 6 and the 2.

How many digits (numbers) do you have to jump over?

okay bioschip why don't we do this together.

Where is the decimal point at the moment?

Where do you need to move it to?

How many digits do you have to jump over to get it there?

Do you know what I am talking about?

@@ End of Day Wrap - Wednesday 14/5/14   Sydney, Australia Time 21:30

hi all,

Another busy day! Great answers from Alan, CPhill, Heureka, problem, zegroes, Goldenaleaf, gamrgrrl156, and Kitty<3.  Thank you everyone.  You all do a great job.

Here are some interesting posts for the day.

Polar co-ordinates

Thanks Alan.

Maybe look here

http://www.mathsisfun.com/data/binomial-distribution.html

I will use the same symbols as the last answerer.

B=P(1+r)ⁿ 

P=70,000

r = interest rate per compounding period, written as a decimal

n=number of compounding periods in total. 

eg 5%per annum compounded yearly r=0.05

6% per annum compounded monthly r=0.06/12

-----------------------------------------------------------

If the money is compounded yearly then the first answer is correct

B=5000(1+0.09)⁹

$${\mathtt{5\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1.09}}\right)}^{{\mathtt{9}}} = {\mathtt{10\,859.466\: \!397\: \!211\: \!546\: \!945}}$$

Thats $10859.47

If the money is compounding monthly then this would be the calculation

B=5000(1+(0.09/12))^9*12

$${\mathtt{5\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{0.09}}}{{\mathtt{12}}}}\right)\right)}^{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{12}}\right)} = {\mathtt{11\,205.620\: \!861\: \!161\: \!185\: \!437\: \!5}}$$

That's $11205.62

etc

You can't solve that - for starters there is no equal signs.

you can simplify it by hand but I don't think that you can on any remotely normal calculator.

(i could be wrong)