Melody

Have you drawn it on a number plane?

The four-digit number “25AB” is divisible by nine, with A the tens digit and B the units digit. How many different such four-digit numbers could “25AB” represent?

Well,

The four digits must add to a number that is divisable by 9.

A+B can be no more than 18

2+5=7

A+B+7 can be no more than 25

so

A+B+7=9 or 18 

You can finish it. 

Why is it a 'challange questions'?

\(\cos(2\theta) = 2\cos(\theta)\\ cos^2\theta-\sin^2\theta = 2\cos(\theta)\\ cos^2\theta-(1-\cos^2\theta) = 2\cos(\theta)\\ 2cos^2\theta-1 = 2\cos\theta\\ 2cos^2\theta-2\cos\theta-1 = 0\\\)

Now let x=cos (theta)  and solve for x.

etc

YES I will second that. 

Thanks Tom...riddle

Hi MagicKitten,

I saw you ask how this forum should be used. (earlier today)

How you and and Tom worked together on this thread is EXACTLY how it should be used.

To bad the asker was not involved but it doesn't matter.  It was fabulous that you stepped in. :)

Thanks Max,

It is good to see you again :)

Thanks guest. I had forgotten how to do that.

So at the very least you have refreshed my memory.

3 \nabla 6 = \dfrac{3 + 6}{1 + 3 \times 6} = \dfrac{9}{19}.

Asker guest,

I repeat CalculatorUser's question.

Was this meant to be solved with the aid of a computer program OR were you meant to devise your own mathermatical technique.

Answering guest, I am pleased that you answered and that you showed your code. Thanks for that

Just sub in the numbers and then do the addition.

I will start you off

\(\displaystyle{f(x)=x^{(x+1)}(x+2)^{(x+3)}}\\ \displaystyle{f(0)=0^{(0+1)}(0+2)^{(0+3)}}\\ \displaystyle{f(0)=0^{1}*2^3}\\ \displaystyle{f(0)=0}\)

Yes I suppose there is an ettique to foruming that is slowly learned. 

But I think people need to learn to use words and phrases like 'please' and 'thank you' 

and

'Thanks but I do not understand this bit of your answer .....'

And they need to learn not to use terms  like  'quick', 'urgent',  'anwer now' etc.