CPhill

We can "anchor" any shell at either an outward-pointing or inward-pointing point

And for each of these, we have 9! ways to place the other shells

So....the total arrangements are  2 * 9!  = 725760 arrangements

The two digit primes < 40 are

11, 13, 17, 19, 23, 29, 31, 37

Any two of these eight can be selected  =  C(8,2)  = 28

And each of these can be arranged in two ways

So...the total number possible house  numbers  is  2 * 28  =  56

Possible  number of different selections for each home team  =  4 different home teams * C(7,3)  = 140

And for the other three teams, the  possible number of selections at each home site are 3* (7, 2) = 63

So...the number number of possible different tournament committees  =   140 * 63  = 8820

Nice, heureka  !!!

\(f(a) = \frac{1}{1-a}$, find the product $f^{-1}(a) \times a \times f(a)$. (Assume $a \neq 0$ and $a \neq 1$.)\)

Write     y  =  1  / ( 1 - a)        swap   a and y    and we want to get y by itself

a  = 1 / ( 1 - y)       multiply  both sides  by ( 1 - y)

a ( 1 - y)   =  1       divide both sides by  a

1 - y  =   1 /a       rearrange as

1 - 1 / a  = y          get a common denominator on the left side

(a - 1)  /  a  = y = f^-1(a)       and this is the  inverse

So

 f^-1 (a)  *  a  =     (a - 1) / a  * a     =  (a -1)

And this  product multiplied  by  f(a)  =   [ (a -1) ]  *  [ 1 / (1 - a) ]    =   [ (a -1) ]  * [ -  1 (a - 1) ]    =   -1

f

1.  \(r(\theta) = \frac{1}{1-\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)? \)

r(30)  =  1 / (1 -30)  = 1/-29  = -1/29

r(r(30))  =  1  / (1 - (- 1/29))   = 1 / (29/29  + 1/29)  =  1 / (30/29)  =  29/30

r (r((r(30)))  = 1  / ( 1  - 29/30)  = 1 / ( 30/30 - 29/30)  =  1  / (1/30)  = 30

r (r(r(r(30))))  = 1 / (1 - ( 30) )  =   1/ (-29) = -1/29

r (r (r(r(r(30)))))  = 1 / ( 1  -  (-1/29) )  = ( 29/29 + 1/29)  =  1 / (30/29) = 29/30

r (r (r (r(r(r(30)))))) =  1 / (1 - (29/30) )  =  1 / ( 30/30 - 29/30)  =  1/(1/30)   =  30

Note that....

angle  A = angle H

angle D  = angle  E

angle C  = angle F

and  

angle B  =  angle  G

So

ABCD ~ HGFE

One place to start is to  look at the divisors of 162  and see which ones are possible legs for  "primitive" Pythagorean Triple  right triangles

The  divisors  of  162  are   1 | 2 | 3 | 6 | 9 | 18 | 27 | 54 | 81 | 162

  3   will  work  because   3-4-5  is a primitive  triple

We need to divide  162  by each factor....and then  multiply  both 4 and 5 by this factor to find one triple

So

162/3  =   54

So  54 * 4  = 216

And 54 * 5 = 270

So    162 -  216 - 270  is a   "triple"

And

9  will work  because

9 - 40 - 41 is a primitive triple

So 162/9  = 18

And  18 * 40  = 720

And 18 * 41  =  738

So  162 - 720 - 738  is a  "triple'

There are two more which are a lttle hard to find...the next is

27 - 364 - 365

So 162/27  = 6

And 364 * 6  = 2184

And 365 * 6  = 2190

So 162 - 2184  -2190   is  a "triple'

The last  is 162 - 6560- 6562

So....there are 4    which have one leg  = 162

-0.1x^2  - 0.6x  + 0.4    =  0      

First off....the easiest thing to  do  is to multiply  everything through  by  -10....this gives us

x^2  + 6x - 4  = 0        take  1/2 of 6  = 3....square it  =  9....add and subtract it

x^2   + 6x  + 9  - 4  - 9  = 0        factor the first three terms and simplify  the rest

(x + 3)^2  - 13  = 0       add 13 to both sides

(x + 3)^2  = 13           take both roots

x + 3  = ±√13       subtract 3  from  both sides

x  = ±√13  - 3    or    -3 ±√13   if you prefer!!!!

\(f(x)=\frac{2x-7}{\sqrt{x^2-5x+6}} \)

We only need to worry about the denominator, since all real numbers will "work" for the numerator

Note that  the function inside the square root in the denominator must be  >  0  

So...let's find this

x^2 - 5x  + 6  >  0     factor

(x - 3) ( x - 2)  >  0

Setting each factor to 0  and solving for x produces  x  = 2   and  x  = 3

So....we have   three  possible  intervals  for solutions

(-inf, 2)   and   [2, 3]   and (3, infinity)

Note that  any  value of x  in the middle interval will make the inequality  false

So.....the domain  is       x < 2  ∪  x > 3

Here's a graph that will confirm this :

https://www.desmos.com/calculator/f0cw1ybcym