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Cal please give Helpbot time and incentive to work it out for him/her self.

It is good that you are thinking, and your hint was good, I just do not want helpbot to be given that much help yet.

Yes but that is not the question Cal  AND  I want Helpbot to answer and not just be totally spoonfed.

Just put in 108 fr 'y' in your equation ...then solve for 't' using quadratic formul or factoring.... throw out any negative values for t

108 = -6t² + 51 t

-6t²+51t - 108 = 0     now use quadratic formula or factoring to find the values of 't'

The total number of ways would be binomial (16 5)

Let's set the equation to

-6t^2+51t=108

Try if you can solve the equation :D

If you can't, I'll try to help!

Have you read the question or tried to think it through?

It says HOW MANY

Why would that be a fraction?

Thanks... got a notification that someone answered!

I'll use the second method....

Would it be 7C5/16C5?

Possibilities/Total?

At the end of the year, the Math Club decided to hold an election for which 5 equal officer positions were available. However, 16 candidates were nominated, of whom 7 were past officers. Of all possible elections of the officers, how many will have at least 1 of the past officers?

Method1:

9 new members and 7 that have been members before.   16 altogether

5 positions.

they can all come from the past members   that is 7C5 ways

You can have one from the new and 4 from the old.  That is  9C1*7C4

etc

OR

method 2

there are 16C5 possible selections

9C5 of those would have no old member

the rest would have at least one.

The second method is easier.

You work it out from here and then show us your working/logic.

AC=?? in

Back square is  9km x 5 km =......

triangular ends are 3-4-5 rt triangles and there ae TWO  so     2   x 1/2 x 3 x 4 =.......

top rectangle    9 x 3  = .......

front rctangle    9 x 4 = ........                  ADD all of the REDs

I expect some people still have a little fun here.

I hoexpect that there are some friendships, both public and private.

I do not follow what is happening so much any more.

Is this possible?????!!

It is good that you have responded but I have an observation.

When you say "that is correct" it makes me think 2 things.

1. It sounds condescending. 

2. It sounds like the only thing you are interested in is the final answer.

How about now? :(

Thanks for sharing the memories Rosala,

You have brought back many memories.

Rosala and zegroes used to have us all in stitches (from laughing) on this forum.

The other characters are all well remembered by me as well. They were all great.

We used to have a lot of fun here.

I've answered to EP's answer 

Thanks, that is correct!

Sorry about that, I wasn't online :(

I'll take a look at the others

I was about to answer, but I looked at your record first.

You give your self points but you never have given an answerer any point.

You have not responded to your answerers in any way, other than "That is incorrect"

EP gave you a full answer that Cal responded to.  You didn't though.

so I am not inspired to help you.

We can create two inequalities from this...

x-2<3x

3x+4<13

Try solving these 

Nevermind, I do remember how to solve this. 

First let's cube \(x-\frac{1}{x}=2\)

\((x-\frac{1}{x})^3=2^3\)

\(x^3-3x+\frac{3}{x}-\frac{1}{x^3}=8\)

\(x^3-\frac{1}{x^3}=8+3(x-\frac{1}{x})\)

\(x^3-\frac{1}{x^3}=8+3(2)\)

\(x^3-\frac{1}{x^3}=\boxed{14}\)

I think EP answered that already :D

Wow, I think that's right, EP.... I wanted to answer this too, but had no idea how!

Let's hear what HB says :)