Melody

I found it

Hi asdf,

If you want to draw attention to an old question please just paste in the address of the original question (preferable in a link so people can hyperlink straight to it.)

You can also explain why  you want people to take a look but the answer should continue on the original thread so you do not need to copy in the question.

Can you find the address of the original question and post it into a link please.

If you can't find it then let me know. :)

What is the value of 

\( \text{b+c if }\;\;x^2+bx+c>0\qquad \text{only when }\;\;x\in (-\infty, -2)\cup(3,\infty)?\)

Well that means that  y cannot not be  positive for  x in [-2,3]

So the roots are x=-2 and x=3 and it is concave up

so the expression is

\((x+2)(x-3) = x^2-x-6 \)

You can do the addition. 

Thanks Alan. 

That is interesting. It looked so complicated.  

Hi Gibson:   

Before I start answering exact questions I want to work out what is happening. - I should have done it just by drawing the displacement graph. It is actually easier that way. I've done it both ways though.  :)

The biker starts 1 mile north of Pidgeon lake.  (the writing says so)

At first the biker is travelling north at 11mph.  

       after 1 hours he is 12km north of Pidgeon lake

At 1 hour he stops and changes direction and heads off south at 10mps 

      so he travels 2 hours south at 10mph that is 20miles south

      he was 12miles noth to start with so now he is 8 south

At the end of the 3rd hour, he stops and changes direction again and heads north at 9mph

After the end of 4 hours (one hour later) he is exhausted and he stops.

    In this last hour he has travelled 9miles north.

    He was 8 miles south in the beginning so now he is 1 mile north.

    So he is back to where he started at the beginning of the 4 hours.

It is actually EASIER to see this if you do the distance vs time graph FIRST.

I'll try and show you how.

It is much easier if you line the graphs up one under the other so that the times are in the same relative position.

You should be able to answer the questions now.

It is undefined  :)

Not so stupid really - questions such as this have been asked many times.

Thanks Alan :)

I did that, I just didn't think to finish it with a run. 

I was looking for some other way to do it. 

But I guess I could do it in EXCEL easily enough. :/

Thanks Alan :)

No it wouldn't

I think this is your actual question.......

Alan is telling you the question that your friend has effectively answered, which is a different question.  Some confusion here. :)

Anyway....

\((2x^3y)^6\\ =(2\times x^3\times y)^6\\ =2^6\times(x^3)^6\times y^6\\ =64\times x^{18}\times y^6\\ =64 x^{18} y^6\\ \)