Melody

I will admit that I do not understand the situation you have described.

But I have also, at times, been disappointed by your behaviour.

I know you do interact with people sometimes but here is an example of what I have also seen you do (or rather not do)

Alan gave you a great answer here and you did not even comment on it or give him any points.   

Find the coefficient of y^4 in the expansion of (2y - 5)^7. 

The general term in the expansion will be

\(\binom{7}{n}(-5)^n(2y)^{7-n}\)

if I rearrange  this I get

\(\binom{7}{n}(-5)^n(2)^{7-n}(y)^{7-n}\\ \)

you want y^4

so  7-n=4    gives     n=3

sub n=3 into the term and you will have your answer

\(\binom{7}{n}(-5)^n(2)^{7-n}(y)^{7-n}\\ n=3\\ \binom{7}{3}(-5)^3(2)^{7-3}(y)^{7-3}\\ \)

You can finish it.

Thanks for the feedback.  :)

You can take this one step further.

For any three dimensional object:

If you expand the lengths by a factor of y ,  [so a side of lenth 4 is epanded to length 4y]

then the area of any surface will become y^2 times bigger

and the volume will enlarge by    y^3

This is useful for many physics applications too.

A certain game is won when one player gets 3 points ahead. Two players A and B are playing the game and at a particular point, A is 1 point ahead. Each player has an equal probability of winning each point. What is the probability that A wins the game?

We do not care about the number of players we only care about the number of points that A is in front.

So draw a number line.

It goes from -3 to 3   (a total of 6 units)

If the difference gets to -3 then A loses, if it gets to +3 A wins.

A starts at +1

A must go a 2 units to the right to win and 4 units to the left to lose.

So the liklihood of a win is 4/6   and the likelihood of a loss is 2/6

So he has a 2/3 chance of winning.

This is not a formal solution.  I don't know how to solve fomally but I believe this is effectively a speed test so intuition is of utmost importance.

ok thanks for responding.  Did you get anything from it?

I quit 3/4 of the way through because I already know that stuff.

It was not the most exciting presentation but the content was good.

Thanks for your response.

I am still interested in your question. 

I would have liked you to explain it.  So I at least understood the question.

It seems like this is only a part of a question.

I know that the gradient of T' is  tan(theta)

so I know that the gradient of OP is  -cot(theta)

Which points are fixed?

Can you describe the situation better please?

It will have a real solution so long as  b^2-4ac is greater or equal to 0

\(b^2-4ac\ge0\\ 25-8c\ge0\\ -8c\ge-25\\ 8c\le25\\ c\le 3.125\)

Try watching this.

I ahve watch 3/4 of it and it looks really good.

Remember that If 2 figures are just different by a scaling factor, then they are similar figures.