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There are 4 children, which means there are 4! = 4 x 3 x 2 x 1 = 24 ways to seat the children.

There are also 4 adults, which means there are, again, 4! = 4 x 3 x 2 x 1 = 24 ways to seat the adults.

Since the children needs to sit next to each other, and all the adults needs to be next to each other, we can view all the children as 1 "object"and all the adults as 1 "object".

Now we need to find out how many ways we can place the childrens next to the adults.

children = C

adults = A

It can be:       CA                or                 AC

Since the problem said that two seatings are considered the same if one can be rotated to form the other, that means there is actually only ONE way to place them next to each other, since rotating AC gives us CA.

Finally, since we can seat the children in 24 different ways, and we can seat the adults in 24 different ways, that gives us 24 x 24 = 576 ways to seat the PEOPLES.

Sub the equation of the line into the equation of the  circle

(x - 4)^2 + (3x - 2 -5)^2  = 68

(x -4)^2  + ( 3x - 7)^2  = 68

x^2 - 8x + 16 + 9x^2 -42x + 49   = 68

10x^2 - 50x - 3   = 0

Using the Quadratic Formula the solutions  for  x  are

x = (5/2) + sqrt (131/5) / 2

x = (5/2) - sqrt (131/5) / 2

Adding  these values and  dividing  by 2 gives us  the  x coordinate of the  midpoint  =  5/2

The y values  are    3 (5/2 + sqrt (131/5)  / 2)  - 2    =   15/2  + 3sqrt (131/5)/2 - 2

And  3(5/2 - sqrt (131/5) /2) -2 =  15/2 - 3sqrt (131/5)/2 - 2

Adding these and  dividing by 2 gives us   the y coordinate of the  midpoint = 11/2

Midpoint = (5/2, 11/2)

(a° b)(3)   means  a (b(3))

(b ° a ) (3)   means b (a (3))

b(3) =  1                                    a(3) = 8

a (b (3)) = a(1)  = 7                    b(a(3)) = b (8) = 11

So

(a ° b)(3)  + (b ° a)(3) =   7  +  11  =   18

Triangle PUQ similar to Triangle TUS

ST =  (5) / (2 + 5)  * 20  =  100/7

So   ST   / QP    =   (100/7) / 20  =  100 /140  =  10/14 =  5/7

Therefore altitude of  PUQ  = QV

 And altitude of TUS  =   VR

Therefore    VR / QV =  5/7

Therefore     QV   =  (7/12) QR

And triangles  QUV and  QSR are similar  

Therefore    UV = (7/12)SR =  (7/12) (20)  =  140/12  = 35 / 3   

Draw OX = sqrt (3)  = OP 

Since QR  is a tangent  meeting the circle at X , then angle OXR = 90

And since the triangle is equilateral, then angle PRQ  = 60

So triangle ORX   is a 30 -60 -90 right triangle

So XR  =  OX /sqrt 3  =   sqrt (3) / (sqrt (3)    = 1

OR = 2

PR  = OR + OP  =   2 + sqrt (3)  =  QR

QR  -  XR  = QX  

2 + sqrt 3  -  1  =  QX  =      1  + sqrt 3

( f ° g) x = (g ° f) x

3(3x + c) - 4 = 3(3x - 4) + c

9x + 3c - 4 = 9x - 12 + c

3c - 4 = -12 + c

2c  =  -8

c = -8 / 2  =    -4

[PR^2 - QR^2 - PQ^2 ] / [- 2 (QR * PQ] = cos PQR

[23^2 -25^2 - 18^2 ] / [ -2 (25 * 18) ] = cos PQR  = 7/15

PM^2  =   QM^2 + QP^2  - 2 (QM * QP) * cos PQR

PM^2 = 12.5^2 + 18^2 - 2(12.5 * 18)* (7/15)

PM = sqrt (270.25) ≈  16.44

f (f(x)) =  p ( px + q) + q =  p^2x + pq + q

f(f(f(x)))  =   p ( p^2x + pq+ q) + q =   p^3x + p^2q + pq + q

So

p^3x  + p^2q + pq + q =  64x - 63x -105 + 278

p^3x  + p^2q + pq  + q = x + 173

Equating terms

p^3  = 1        so  p  =1

And

q ( p^2 + p + 1)  =  173

q  (1^2  + 1 + 1)   = 173

q (3) = 173

q =173/3

p + q =   1 + 173/3  =   176 / 3

\($x^3 y^2 - x^2 y^2 - xy^4 + xy^3 \ge 0 and 0 \le x \le y$\)

This area (in the blue)  is unbounded 

See here :  https://www.desmos.com/calculator/hvtmevyg4f

Let O  be the center of the square

Let M  be the top right vertex of the  square

Let N be the rightmost vertex of the hexagon  and P  be the top right vertex of the  hexagon

Triangle  PON is equilateral  so   PN =  ON  =  1

OM  =  (1/2) diagonal of square =   S /sqrt 2       where S is the side of the square 

Angle  OMN  = 75°     Angle MON =  45°    Angle MNO = 60° 

Using the Law of Sines

sin (OMN) / 1  =  sin MNO / (S/sqrt 2)

sin (75) /1 =  sin (60)/ (S/ sqrt 2)

S / sqrt (2) =  sin 60 / sin 75

S =    sqrt (2) sin (60) / sin (75) 

S = sqrt (2) (sqrt (3) / 2) /  ( sin (45 + 30)

S = sqrt (6) / (2 (sin 45cos 30  + cos 45 sin 30) )

S =  sqrt (6) / ( 2 ( sqrt (2)/2 * sqrt (3)/2  + sqrt(2/2 )(1/2) )

S =  sqrt (6) / (2 (sqrt (6) /4  + sqrt (2)  / 4)) 

S = sqrt (6) / [ (sqrt (6) + sqrt (2)) / 2

S = 2 sqrt (6) / (sqrt 6 + sqrt 2) 

S = 2 sqrt (6) ( sqrt 6 - sqrt 2) / ( 6-2)

S =  (12 - 2sqrt (12)) /4

S =  3 - 2sqrt (4) sqrt (3) / 4

S =  3 - 2*2 * sqrt (3) / 4

S = 3 - 4sqrt (3) / 4

S =  3 - sqrt (3)  ≈  1.27

Here's a pic  (AM =  1/2 the side of the square ≈  .63 )

This problem is very similar: 

https://www.quora.com/A-point-P-is-given-on-the-circumference-of-a-circle-of-radius-R-The-chord-QR-is-parallel-to-the-tangent-at-P-What-is-the-maximum-possible-area-of-a-triangle-PQR

See if that helps you

Thank you so much!

Can you help me solve the other problem I posted about the 52 cards?

1a  + 2  = 2b + 1

1 =  2b - a                                            

Let a =  5 , b = 3         

12₅ =  21₃

1(5) + 2 =   2(3) + 1

7  =  7

The way you solved it was really easy to understand! Wow!

\(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1 \)

This is just the perimemter  of a decagon

The length of one side can be  found as     radius/2 ( -1 + sqrt (5))

The perimeter  is    10 (1/2) ( -1 + sqrt 5)  =    5 ( -1 + sqrt 5) ≈  6.18

This link might help: 

https://www.quora.com/Three-balls-are-drawn-without-replacement-from-a-bag-containing-3-black-5-purple-and-6-yellow-balls-Given-that-the-drawn-balls-are-all-the-same-colour-what-is-the-probability-they-are-purple

The area is 1/4 of the circle's total area

60pi =   (1/4) pi r^2           divide out pi      multipy through by 4

240  =  r^2

sqrt (240) =  r   = 4sqrt (15)

4 x 2  rectangle at the center = 8

Two triangles at top / bottom with bases of 2 and heights of 1  =  2

Two triangle at left / right  with bases of 5 and  heights of 2  =  10

Total area =   8 + 2 + 10  =   20

(5/3)x   + 12/7   = y

5/3  is a  terminating decimal   while   12/7  is  not......so   it will not pass through any lattice points 

I also believe that there is  no minimum distance to any lattice point.....

The minimum distances will approach zero but they never converge to any particular value 

Here's what I think :

This will be a right triangle with legs = 2  and hypotenuse = 2sqrt (2)

Minimum perimeter =   2 + 2 + 2sqrt (2) =   4 + 2sqt (2)  

The region is  unbounded

See here :  https://www.desmos.com/calculator/m5ajeuqdn4

f(x) = 2x^2 + 22x - 76

2x^2  + 22x - 76    =     a

2x^2 + 22x  - 76 - a  =  0

2x^2 + 22x - (76 + a) = 0

Take the discriminant and set to 0

(-22)^2  - 4 (2) [ -76 - a ] = 0

484 + 608 + 8a  =0

1092  = -8a

a = 1092/ -8  =   -136.5

(The value of x  =  5)

(In other words, f(5)  = -136.5  )