The table is 30 inches wide and 180 inches long. The width of the 2 and 3 rectangles is 12 inches. The sides of the 4 rectangle are 15 inches and the top and bottom are 12 inches. The 1 rectangle is twice as wide as the 2

3. What are the dimensions of the 1 rectangle?

4. What are the measurements of one of the 5 rectangles?

5. You are playing shuffleboard in P.E. class. To score points, your disc must land in a box on the other side of the table and you are awarded the number of points that is in that box. What is the probability of scoring 1 point?

6. What is the probability of scoring 2 points?

7. What is the probability of scoring 3 points?

8. What is the probability of scoring 4 point?

9. What is the probability of scoring 5 point?

10. What is the probability of scoring at all (any number of points)?

]]>Secondly in the following formula

why did we divide through by and not or ?

Well, you *can* divide it by

1. I think to do this, you'd have to work through all the possibilities. geogebra as I mentioned before would help.

2. What is meant by a 'broken' line?

Bob

]]>Think of a vector as being a 'journey' ** from A to B. If you go A to B; then B to C; then C to A; then you end up where you started so your vector for the three journeys is zero.

Here's an example ( I'll use horizontal rather than the usual vertical to save time) :

(1,0) + (0,1) + (-1,0) + (0,-1) take you round the sides of a square and these add to (0,0).

What you describe in your post is a journey around the perimeter of a regular polygon. Once you've gone all the way round, you are back to the start.

http://www.mathsisfun.com/algebra/vectors.html

Bob

** vectors are used for other things than 'journeys', but it's a good place to start.

]]>1+(sinx+cosx)(cosx-sinx)=1+(cosx+sinx)(cosx-sinx)=1+cos*squared*x-sin*squared*x=cos*squared*+sin*squared*x+cos*squared*x-sin*squared*x=2cos*squared*x

For the fifth question:

(cosx+sinxtanx)/tanx=(cosx/tanx)+sinx=(cos*squared*x/sinx)+(sin*squared*x/sinx)=cos*squared*x+sin*squaredx/sinx=1/sinx

I responded to this question in your previous thread -- did you see it?

]]>I see there are other answers to this questions like this, but as far as I saw not for exactly what I am seeking.

Networking event: The goal is for each person to meet as many people as possible;

They may not repeat tables within day one nor may they repeat within day two - I know this is possible. However, they do not want the attendees to see the same people at the tables over the entire two day event = 7 rotations. Would you tell me if this is possible and share the breakdown? I have done it with a randomize function once and a pivot table...but both times table guests were repeated.

This is the breakdown...

We anticipate 30 participants; 5 tables (6 at a table)

Day One:

Must have three rotations - they may never repeat the same table twice (on this day) nor do they want them to see the same people at a table (however I think with a group of 30 this may be impossible...?)

Day Two:

Must have four rotations - they may never repeat the same table twice (on this day) nor do they want them to see the same people at a table.

Looking forward to your expert advise.

Thank you.

In the first the vertex is A and the angles ABC and ACB are the base angles of a triangle and are bisected.

Let's suppose ABC = 2x and ACB = 2y, and that the bisectors cross at 90.

Then, in triangle BDE we have angles of x, 90 and y and so in triangle CDE angle BEC = x.

So angle BEA = 180 - x

and in triangle ABE angle A = 0.

So in this case the bisectors cannot cross at 90.

In my second diagram the vertex is at G and angles FGH and FGJ are bisected.

If FGH = 2x and FGJ = 2y, then 2x + 2y + 2x + 2y = 360, => x+y = 90, and the angles are supplementary.

For my third diagram the angles meet at a point and share a common line. If the angles are 2x and 2y and we know that x+y = 90, then 2x + 2y = 180 and the angles are supplementary.

??????

Maybe I should adapt the third so that the angles do not share a common line. Then the angles would not be supplementary. I leave this fourth diagram to the interested reader.

Bob

]]>Bob

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