Melody

the prob that the first one drawn is  J Q or K of hearts  is    3/52

the prob that the 2nd one drawn is  J Q or K of hdiamonds  is    3/52

the prob that the 3rd one drawn is  J Q or K of spades  is    3/52

the prob that the 4th one drawn is  J Q or K of clubs  is    3/52

These can be in any order so I would have to multiply by 4!

So I think the answer is    \(\frac{3^4}{52^4}*4!\)

6=2*3

8=2*2*2

27=3*3*3

smallest common muliple is   3*3*3*2*2*2

Thanks Builderboi  

Please do not 'make people's lives easier' by answering over me Builderboi.

If guest had have got back with some suggestion of doing some work for themselves then by all means, help some more.

But I asked thei guest to come back and show that they had done something for themselves and I do not want other answerers answering over me.

Note:

If I ever answer over you and you  take exception to what I have done then by all means tell me that you are not happy.

I do not expect this to be a one way street.

Anyway, yoiu cannot just cancel out x+3 because sometimes it is negative and somteimes it is positive.

This is what I would do...

\(\frac{dy}{dx}=\frac{2x}{3y}\\ \frac{d^2y}{dx^2}=\frac{(3y)(2)-(3\frac{dy}{dx})(2x)}{(3y)^2}\\ \frac{d^2y}{dx^2}=\frac{(y)(2)-(\frac{dy}{dx})(2x)}{3(y)^2}\\ \frac{d^2y}{dx^2}=\frac{2y-(\frac{2x}{3y})(2x)}{3y^2}\\ \frac{d^2y}{dx^2}=\frac{2y}{3y^2}-\frac{4x^2}{9y^3}\\ \frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\ \)

OR

\(2x^2 - 3y^2 = 4\\ 4x-6y\frac{dy}{dx}=0\\ 4-6[y(\frac{d^2y}{dx^2})+(\frac{dy}{dx}\cdot \frac{dy}{dx})=0\\ 4-6[y(\frac{d^2y}{dx^2})+\frac{2x}{3y}\cdot \frac{2x}{3y})=0\\ 4-6[y(\frac{d^2y}{dx^2})+\frac{4x^2}{9y^2}]=0\\ 4-6y(\frac{d^2y}{dx^2})-\frac{6*4x^2}{9y^2}=0\\ 3y(\frac{d^2y}{dx^2})=2-\frac{4x^2}{3y^2}\\ 3y(\frac{d^2y}{dx^2})=\frac{6y^2-4x^2}{3y^2}\\ \frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\\)

LaTex

\frac{dy}{dx}=\frac{2x}{3y}\\
\frac{d^2y}{dx^2}=\frac{(3y)(2)-(3\frac{dy}{dx})(2x)}{(3y)^2}\\
\frac{d^2y}{dx^2}=\frac{(y)(2)-(\frac{dy}{dx})(2x)}{3(y)^2}\\
\frac{d^2y}{dx^2}=\frac{2y-(\frac{2x}{3y})(2x)}{3y^2}\\
\frac{d^2y}{dx^2}=\frac{2y}{3y^2}-\frac{4x^2}{9y^3}\\
\frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\

what do you think?

How could you start it?

I have already answered another part of this question so look at what I did.

You are welsome to get back with your thoughts.

Assume all the marbels are different.  

the number of ways that three blues can be chosen is   6C3 

the number of ways that three white can be chosen is    5C3

the number of ways that three red can be chosen is      4C3

the number of ways any 3 can be chosen is   18C3

Can you finish it?

 (2x+10)(x+3)<(3x+9)(x-8).

So what have you done towards solving this???

I might start by expanding both sides.

Do something yourself and get back to us if you still have problems.

101 = 1.01e2 = 101

101^2 = 10201

101^3 = 1030301

101^9 = 1093685272684360901

101^10 = 110462212541120451001

We seem to have a pattern here, based on the pattern   the last 3 digits are most likely  201

You question is poorly presented and makes no sense.

what has a cone got to do with anything?