Melody

Don't worry about it this time unless someone asks for it. 

Just don't do it again :)

Every now again I wll spend a long time answering a question and then when I go to post it the question has been removed, or changed without comment. It can be infuriating.

Please do not delete your questions.

Someone could have been in the process of answering it.

Plus someone else could have learned from that answer.

If you do not need an answer anymore just write a new post saying something like:

"Thanks to all those that look at my question but I do not need an answer any more.

I have worked it our for myself   "

\(\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\\ LHS\\ =\binom{8}{3} + \binom{8}{4} \\ =\frac{8!}{3!\;5!}+\frac{8!}{4!\;4!}\\ =\frac{9! }{3!\;5!*9}+\frac{9!}{4!\;4!*9}\\ =\frac{9! }{3!\;4!*5*9}+\frac{9!}{3!*4\;*4!*9}\\ =\frac{9!}{3!4!*9}\left( \frac{1}{5}+\frac{1}{4} \right)\\ =\frac{9!}{3!4!*9}\left( \frac{9}{20} \right)\\ =\frac{9!}{3!4!}\left( \frac{1}{4*5} \right)\\ =\frac{9!}{4!5!}\\ =\binom{9}{4}\quad or \quad \binom{9}{5}\)

So the largest value of n is 5

These are all concave up parabolas.

they are parabolas becasue the highest power of x (the degree) for each one is 2

They are all concave up becasue the number in front of x^2 (the leading coefficient) is positve.

The higher the leading coefficient the steeper, or narrower, the parabola.

Now you can work if out for yourself.

Thanks Alan.

The derivative is the gradient of the tangent to the curve at any given point.

Hi Alan,

How did you work out that  b must be greater or equal to 2-2a  ?

\(\frac{(cot\alpha cot\beta -1)}{(cot\alpha +cot\beta )}\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{sin\alpha sin\beta}\div \left(\frac{cos\alpha}{sin\alpha}+\frac{cos\beta}{sin\beta}\right)\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{sin\alpha sin\beta}\div \frac{2cos\alpha sin\beta}{sin\alpha sin\beta}\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{2cos\alpha sin\beta}\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{2cos\alpha sin\beta}\\ =\frac{cos(\alpha+\beta)}{sin(\alpha+\beta)}\\ =cot(\alpha+\beta)\)

Could you please edit your question to get rid of all the irrelevant $ signs.

Then it will be easier to read and someone might attempt an answer.

شاید یک دهه یا دو ساله ام خرید کنم!