CPhill

It will have one real root  when k =  (5/2)^2   =  25/4 

This is as large as k can be for any real roots

Just multiply the dimensions

(1/2) (3/4)^6  (4/3)^7  =

(1/2) (3/4)^6 * ( 4/3)^6 * (4/3)  =

(1/2) (1) (4/3)  =

4 / 6   = 

2 / 3

(36 - 24)  = 12  =  those taking math only

(27 - 24) =   3  =   those taking physics only

Total taking neither  =

50 - ( 12 + 24 + 3)  =

50 -  ( 39)  = 

11

Draw DF  and CG

Really....the only  trick here is to find  DF = sqrt [ 4^2 + 4^2 ] = sqrt (32) = 4sqrt 2 = CG

Area of BHKA  =  [ BA *  (BC + DF + GH) ]   =  [ 4 * ( 4 + 4sqrt 2 + 4) ]  =  [ 4 ( 8 + 4sqrt 2 )  = 

32 + 16sqrt 2  

Area of rectangle DFGC  =  FG * DF =   4 * 4sqrt 2 =  16sqrt 2

Area of right triangle DFE  = (1/2) * DE * FE =  (1/2) * 4^2 =  8

Area of  DEFGC =  16sqrt 2  - 8  

Area of   polygon  =

32 +16sqrt 2 + 16sqrt 2  - 8  =

24 + 32sqrt 2

Radius of circle = (1/2) side length of square  =  25m

Area of circle = watered area =   pi (25)^2  ≈  1963 m^2

[(19.95 + 2.95) / 200] *   100   =  11.45  cents   (round to 11 cents)

Note that

40 =  AB * BC

40 = 10 * BC

BC = 4 = AD

Draw MO  perpendicular to DC

And MO = BC

Area of rectangle  AMOD =  AD * AM =  4 * 5  = 20

Area of right triangle  MON = (1/2)(BC) (ON)  =  (1/2)(4)(3) =  6

Area of   MBCN =  40 -  20  - 6  =   14

What the guest said is true

See the graphs here  : https://www.desmos.com/calculator/1iouv5eqqu

Note that they do not intersect  .......( there is no solution )

Interior angle of a regular n-gon =    (N - 2)  * 180   /  N

(10-2) (180)  / 10  -  (6-2) (180) / 6  =

180 [ 8/10  - 4/6] =

180 [ 4/5 - 2/3]  =

180 [  2 / 15 ] =

24°  greater