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Good work!

Yes...... that is what I would do.     $/(can/week)  is the units   (don't forget units in your math answers.....technically, w/out units your answer is incorrect!)      

Thanx for your help GA !   I am still getting 're-acquainted' with this stuff.......can you backcheck my answer for the 5 Repubs and 5 Dems sitting around a table???    G'Day ...     ~EP

Solution

For any given one of the 200 points, we can find exactly one square with that point as one of the vertices---that point, the point diametrically opposite it, and the endpoints of the diameter that is perpendicular to the diameter formed by connecting the first two points. Each square then accounts for 4 vertices, so there are \(200/4=\boxed{50}\) squares.

Here is my solution, EP. Correct me if I'm wrong.....

Choose any 5 consecutive seats in which to place the Democrats -- it doesn't matter which 5 consecutive seats that we choose, since we can rotate the table. Then there are 5! ways to place the Democrats in their seats, and 5! ways to place the Republicans in their seats, for a total of 5!*5!=14,400 arrangements.

As a block of 5 and 5 there are 10 ways to sit around the table   10.

  Now , WITHIN  EACH block there is 5! ways to sit

     AND for EACH of the 5! ways to sit in one block there are 5! ways to sit in the OTHER block....

so   10 x 5! x 5! = 144,000 ways

      (I am still learning this stuff like you.....as GA can attest!  So this answer may be incorrect!)     ~EP

Yes, that's correct! Good job, EP!

Very nice solution, CPhill! Great job! 

You’re right! I withdraw my critical assessment.  

The knot does equate to a straight line; changing from the circular permutation P=(n-1) to the linear P=n!

The caveat for reflections (/2) being equivalent for a linear permutation seems unusual, but there isn’t a reason you couldn’t do it. When I scratch my left ear while looking in a mirror I see me scratching my right ear in Alice’s universe, but were both content.

See the answer here: https://web2.0calc.com/questions/a-pencil-eraser-and-notebook-cost-a-total-of-s1-00-if

Here's my take  :

Fix the chief in any position

We have two ways to arrange the mother and wife......either the wife is on the left of the chief and the mother on the right, or vice-versa

And the other  5 members of the tribe can be arranged in 5! ways

So...the total possible arrangements are   2 * 5!  =  2 * 120    =  240

GA.....I'm still confused I guess.  The knot in the bracelet makes it essentially 'knot' (pun) a circle but a straight line configuration......the beads cannot rotate...they are in a straight line.    So I (think) my naswer is correct, not coincidental.  Maybe we are viewing the question differently in our minds' eyes?

Your answer matches because of coincidence, not because of logical analyses.   

You said the passengers for the bus are standing in a line, that’s not a circle. If they were standing in a circle, and the bus door prevented them from rotating, then your solution is wrong because you have to count the bus door as a passenger, and you didn’t do that. 

Your methodology solves a string of beads in a line, not attached at the end, or a bracelet (ends attached) and each rotation counting as a unique arrangement –this isn’t normally done in circle combinations.  

Each square requires 4 points.....then you can rotate it by 1 point....then 1 point again.....

you can rotate 49 times before you start to repeat the square pattern.....the original square plus 49 more = 50 squares.

You're welcome!  Hope we didn't confuse you as much as I often am. 

Thanks, EP and GingerAle! 

P (3,2)   means that  we are choosing any 2 things out of 3 and then permuting (giving all the possible orders  of the things selected )

So...suppose the math books  are  A, B and C

The possible selections we can make is   

{ A, B}   { B , C}  or { A, C}    (1)

But  we can also arrange these as  { B, A}  {C, B}  or   {C, A}.......so we have a total of 6 possible arrangements

Note that   (1)  is  C(3,2) = 3  =   the number of possible sets

But  P(3,2)   =  6  =  the number of posssible arrangements of these sets

[ This stuff is hard for everyone  !!!  ]

Hmmmmm...... the question stated there is a knot (bus door essentially) so the beads CANNOT rotate......so I got 12 too.

10,000(1+.045)^2  - 10,000 = $920.25 interest after two years.

Hi, I don't understand how you got the p(3,2) can you explain it to me? (Thank you! I'm really struggling with this topic.)

The wheel makes one revolution in 10 sec.....it is at 4 ft  at 7.5 sec  and again at 4 ft, 10 seconds later

So...the first statement is false

At  0 sec.....the loading height is 34 ft....but the lowest height is at 4 ft

So....the second statement isn't true, either

The third statement is true

The last statement is also true....the  max height is 64 and the minimum is 4 ft

So....the radius is  ( 64 - 4) / 2   =  60/2  =  30 ft

Unless the people line up in a circle, it’s not the same.

For a bracelet without a knot, there are (4-1)! / 2 arrangements. One (1) is subtracted to fix the position of a bead, because the beads can be rotated such that one particular bead can be placed in a start position without rearranging them. The division by two (2) is because the bracelet can be flipped.

A bracelet with a knot prevents rotation. Treat the knot as an additional bead, fixed in position, giving (5-1)! / 2 = 12 arrangements. 

GA

First one :

The graph is horizontally compressed by a factor of 2 and shifts up 1 unit.

Here are the graphs :  https://www.desmos.com/calculator/umiqrxhmie

Second One

2. Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. The tide measures 5.15 ft at midnight, rises to a high of 10.2 ft, falls to a low of 0.1 ft, and then rises to 5.15 ft by the next midnight.

In the form   y  =  Asin (Bx) + C

C  =  5.15

The amplitude, A, is  [ 10.2 - .1] / 2  =  5.05

And we can find B  as

2pi / 24   =   pi/ 12

So....the function  is  y  =  5.05sin ((pi/12) x )  + 5.15

Here's the graph : https://www.desmos.com/calculator/7k1p2hksdu

Second one:

First one: