Melody

That is great :)

Thanks Max,

It is really great to have you back here again :)

Hi Kakashi,

I like people to interact with their answerers.

If the answerer gives a partial answer, it may be enough to get the asker thinking in the right direction.

That way, askers get help and still have to think thourgh at least some of the problem for themselves. 

They are more likely to learn that way. 

If they look at the answer, think about it for a while and still want more help they can ask for clarifications.

Or for more help.

That is how I encourage answerers to think.  

I want this to be a teaching/help site.

Not a cheat site.

Most times it is not really necessary to give the final answer. 

Here is a start

This is a cross section of half the fulcrum with the chopped off cone sitting on top.

Find x using similar triangles.

Please do not answer in full max.  Let the guest do some of the thinking, they might learn that way :/

Thanks BuilderBoi,  you are correct of course.

I just wanted to show people how to do it with a contour map.

I really like contour maps, in this case it was probably as easy either way but sometimes contour maps make probablility questions MUCH easier.

The sample space is  5*5 = 25    (area)

The desirable area is  2*4+ 1*3 = 8+3 = 11

So  P(positive product) = 11/25

Try a couple of times over a couple of days and if it still does not work then report it.

  There is a report a problem link at the bottom of the page.

Make sure you include all info that could be relevant. Like the country you are in, your browser etc.

I cannot promiss this will make a difference but problems have no chance of being fixed if they are not reported properly.

I hope this problem fixes itself for you 

No one else has been complaining about this in the immediate past.

Nice find Kakashi :)

It is against Web2.0 policy to answer aops questions.    Sorry,

It is called  'Art of Problems Solving' becasue they want you to learn to problem solve - without outside help.

They provide a notice board with some hints.

They do not want you to get outside help.

It is not  called  'Art of getting people to help on the net'.   

I am not trying to be harsh but AoPS have shown their frustration enough times.

I can most definitely understand their point of view.

please give us the link to the original post.

It should be answered on the original.

You do not nedd to repeat the question,  Just give the link and ask for more help.

Lets see.

$\displaystyle  \int_1^2 \frac{(x+3)^2}{3} dx\\ =\displaystyle  \frac{1}{3}\int_1^2 (x^2+6x+9) dx\\ =\displaystyle  \frac{1}{3}* \left[\frac{x^3}{3}+\frac{6x^2}{2}+9x\right] _1^2\\ =\displaystyle  \frac{1}{3}* \left[(\frac{2^3}{3}+\frac{6*2^2}{2}+9*2)-(\frac{1^3}{3}+\frac{6*1^2}{2}+9*1)\right] \\ =\displaystyle  \frac{1}{3}* \left[(\frac{8}{3}+\frac{24}{2}+18)-(\frac{1}{3}+\frac{6}{2}+9)\right] \\ =\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+\frac{18}{2}+9)\right]\\ =\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+18)\right]\\ =\displaystyle   \left[(\frac{7}{9}+6)\right]\\~\\ =6\frac{7}{9}$

LaTex:

\displaystyle  \int_1^2 \frac{(x+3)^2}{3} dx\\
=\displaystyle  \frac{1}{3}\int_1^2 (x^2+6x+9) dx\\
=\displaystyle  \frac{1}{3}* \left[\frac{x^3}{3}+\frac{6x^2}{2}+9x\right] _1^2\\
=\displaystyle  \frac{1}{3}* \left[(\frac{2^3}{3}+\frac{6*2^2}{2}+9*2)-(\frac{1^3}{3}+\frac{6*1^2}{2}+9*1)\right] \\
=\displaystyle  \frac{1}{3}* \left[(\frac{8}{3}+\frac{24}{2}+18)-(\frac{1}{3}+\frac{6}{2}+9)\right] \\
=\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+\frac{18}{2}+9)\right]\\
=\displaystyle  \frac{1}{3}* \left[(\frac{7}{3}+18)\right]\\
=\displaystyle   \left[(\frac{7}{9}+6)\right]\\~\\
=6\frac{7}{9}