CPhill

We  have  a  "low/ high"  situation....so.....this graph has a horizontal asymptote at  y  = 0

So....we need to solve this

  2x - 6

______________  =   0

x^3 -7x^2 -2x + 6

Only  an x value  in the numerator can make this =   0

So

2x  - 6   = 0       add 6 to both sides

2x  = 6  divide both sides  by  2

x =   3

So....it crosses its horizontal asymptote  at  (3, 0)

See the graph here :

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

LOL!!!....I dunno  about that.....sometimes....I  make  the "easy" ones look "hard"

The winner  will  play

log 1024  / log 4  =  5   games

For convenience....let there  be  100  children who go to  the doctor.....so...

100 * .70  =    70 have fevers

70 * .30  =  21 have  fevers  and rash

100  - 70  = 30  have no fever

30 * .20  =  6  have  no fever  but  have a rash

21  + 6  =   27  have  rashes

So

6  have  a rash  but  no fever

_______________________  =   6  / 27   =   2 / 9

27    have a rash

EDIT.....thanks, Geno forspotting my earlier error   !!!

All correct.....good job  !!!!

This  will  follow  the principle  that  the angle  of incidence of the light beam  =  the  angle  of  reflection  of the light beam

We can solve this by using similar triangles such  that.....

1.8 /  2.75  =   height  of tower  / 500         cross-multiply

500 * 1.8 / 2.75  = height of tower  ≈  327.3  m

Proof

arctan( 373.3 / 500)  = 33.2°   =  angle of incidence

arctan   ( 1.8/ 2.75)  = 33.2°  = angle of reflection

We  have  that

pq  = p + q       (1)          

p - q  = 7  ⇒  p   =  7 + q    (2)

Sub  (2)  into (1)  and we have  that

(7 + q)q  =  7 + 2q       simplify

7q  + q^2  = 7 + 2q      rearrange

q^2  + 5q  -  7  =  0   complete  the  square  on q

q^2  + 5q   + (25/4)  =  7  + 25/4

(q + 5/2)^2  =   53/4      take the positive root

q + 5/2  =    √53/2

q =   √53/2 - 5/2    =    [√53 - 5 ] / 2

q^2  =    [ 53 - 10√53 + 25]  / 4   =   [ 78 - 10√53  ] / 4    =   [39 - 5√53] / 2

So....p =  √53/2  - 5/2  + 7    =  √53/2  + 9/2   =   [ √53  + 9 ]  / 2

p^2   =  [ 53  + 18√53  + 81] / 4   =  [ 134 +18√53 ] / 4  =  [ 67 + 9√53  ] / 2

Simplifying  the given  expression  we   have  that

(pq)^2                 (p + q)^2         

_________ =     ________ 

p^2 + q^2           p^2 + q^2

p + q  =   √53 + 2

(p+ q)^2   = (  53  + 4√53  + 4)   =  (57 + 4√53)

p^2  + q^2    =    [106  + 4√53 ] / 2   =  53  + 2√53

So  we  have  

(57 + 4√53 )        (53  - 2√53)             2597  + 98√53       

___________     ___________  =    ______________

( 53 + 2√53)        (53  - 2√53)               2597

So

a +  b  +  c   +  d  = 

2597  + 98 + 53   +  2597   =    5345

Nice, heureka  !!!

Yep....I screwed  this one up...  ........EP's  answer is correct  !!!!

The  midline  is  the   same  as  with  the  "regular"  sine curve....we  would have to have a plus or minus added to  the function to affect the  normal midline of  y  =  0 

The only  thing tricky about  this one  is that  it  is  a  "compressed" sine  curve wth  a  "phase shift"

Note that  we  can write  this  in  the following  manner

f(x)  =  sin (2pi ) ( x  - 1/4)

This  says  that  there  are  2pi  periods  in  2pi  rads....so   .....every period = 1 (rad)

Also.....the  regualr  sine curve  is   shifted  1/4  rads  to the right

Look at the regular  sine  curve  and this function on the same graph :

https://www.desmos.com/calculator/itkg2m4xrj

Note  that  the point   (.25,0)  "starts" the  shifted curve  and the point (1.25, 0)  completes  the period......

Also note that  the  curve  is shifted (1/4) rad to the right  of the regular curve