Melody

Almost Geno :)

But there are 2 R's so I think that the answer is 

$$\frac{6!}{2!}=\frac{6!}{2}$$

$${\frac{{\mathtt{6}}{!}}{{\mathtt{2}}}} = {\mathtt{360}}$$   ways

Hi Chris and Sabi,

I want to try this one too :)

The parabola at the top is concave up so it is the one with the positive leading coefficient.

so you want

$$\\4<(\frac{1}{2}x^2+7)-(\frac{-1}{4}x^2+3x)<7\\\\
4<(\frac{2x^2}{4}+7)-(\frac{-x^2}{4}+3x)<7\\\\
4<\frac{2x^2}{4}+7+\frac{x^2}{4}-3x<7\\\\
4<\frac{3x^2}{4}+7-3x<7\\\\
-3<\frac{3x^2}{4}-3x<0\\\\
-12<3x^2-12x<0\\\\
-4 -4 x^2-4x>-4\qquad and \quad x(x-4)<0\\\\
x^2-4x+4>0\qquad and \quad x(x-4)<0\\\\
(x-2)^2>0\qquad and \quad x(x-4)<0\\\\$$

$$y=(x-2)^2$$    is a concave up parabola with a double root at x=2 so it is going to be positive for all valures of y except  x=2

$$y=x(x-4)$$  is a concave up parabola with roots at x=0 and x=4 . It will be BELOW the xaxis (negative) between the 2 roots.      Negative for   $$0

These will both be true when    $$0      EXCEPT  when x=2

This is identical to CPhill's answer so by the law of averages we are probably both correct.   LOL

Yep, I like your logic Chris :))

On the web2.0calc it is really easy.

Here is an example

Just make sure that the top right corner of the calc is set to equations :))

You just solve the 2 equations simultaneously :)

I have just added this post to the "reference material" sticky topic

I have entered it twice

Once for solving inequalities and once for graphing polynomials :))

That's ok Chris, none of us can answer all the questions in great detail.  We answer what we can with as much time as we can afford to donate at that point in time.

Any correct answer is better than none, I am sure that Sabi appreciated your help :)

I like answering questions like this - it is one of my pet subject areas :))

Oh, thanks Sabi - and you are very welcome :)

Hi Sabi,

(x-2)(x²-1)>0

(x-2)(x-1)(x+1)>0

consider y=(x-2)(x-1)(x+1)

The roots will be 2,1 and -1.

3 is the highest power of x   (It is degree 3)  

This means that the graph will have 3 (or less, see my notes for the last question) directions.  

(Since there are 3 distinct roots this one has to have 3 directions )

I do not need to do the expansion to see that the coefficient of x³ is +1.

Since it is positive, one of the ends of the graph will be in the TOP RIGHT CORNER

Always start your sketch on the right side.

Above the x axis if it is a positive leading coefficient

and below the x axis if it is a negative leading coefficient.

This is what it must look like

It can be seen from the graph that y is positive  when  -1<x<1  and when x>2

Any more questions the just ask :))

It is not, it is undefined.