CPhill

Note that       1/ ( a - 1)  +  1/(b -1)    can  be  simplified to

[ (a + b)  - 2 ]   / ( ab  - (a + b) + 1 )       .......    (1)

Simplify the equation as

3x^2 - 11x  - 7  =  0

By Vieta   the sum of the  roots  (a + b)  =  11/3

And the  product of  the  roots  ab   = -7/3

Subbing into (1) we have

[ 11/3 - 2 ]  /  [ -7/3 - ( 11/3) + 1 ]  =

[  5/3 ] / [ -5 ] =

-5/15  =

-1/3

Angle C =  180  - 40 - 40  = 100°

Angles  ADC  and BEC  = 90°   (since AD and BE are altitudes)

We have quadrilateral  CEHD where the sum of  its  interior angles  = 360

So angle EHD =   360 - 90 - 90  - 100  =  80°

And angle AHB is a  vertical angle to angle EHD so it also measures 80°

Let BP  = x   and DP =  5 - x

By the Intersecting Chords Theorem we  have

DP * PB  = AP * PC

(5 - x) * x =  8 * 2

5x - x^2  =  16       rearrange  as

x^2 - 5x + 16 =  0       

We can see that this has no real solution for  x  because the  discriminant is  negative.....

A                               B

                                 4                               

                                 P

                                 1                       

D      4       Q     1     C

Let  A = (0, 5)

P = (5 , 1)

Q = (4,0)

AP  = sqrt [  5^2 + 4^2 ] =  sqrt (41)

AQ = sqrt [ 4^2 + 5^2 ] = sqrt (41)

PQ = sqrt [ 1^2 + 1^2)  = sqrt (2)

Using the Law of Cosines

PQ^2  =  AQ^2 + AP^2 - 2 ( AQ * AP) cos (PAQ)

 2  = 41 + 41  - 2 ( sqrt 41 * sqrt 41) cos (PAQ)

[ 2 - 41 -41 ]  / [ -2 * sqrt 41 * sqrt 41 ] =  cos PAQ

-80 / [- 2 * 41]  = cos PAQ

80 / 82  = 40 / 41  =  cos PAQ

sin PAQ   =  sqrt  [ 41^2  - 40^2 ] / 41  =  sqrt (81) / 41  =  9 / 41

-.05x^3  + 3.25x^2 - 10x + 20 =  34.24

Easiest to solve this  by a  graph :

In the 5th year....

The point  (a, -4/5)  is  on the  graph of sin x

So

 a  is a third quadrant angle  so sin a =  -4/5   and   cos a =  -3/5

And   (a, c)  is on the graph of  sin (2x)

So

sin (2a)  =  2sin a  cos a  =     2 (-4/5) (-3/5)  = 2 (12/25) = 24/25  = c

Likewise

(b , 5/13)  is on sin x

sin b = 5/13

b is a first quadrant angle so cos b =  12/13

So

sin (2b)  = 2 sin b  cos b  =   2 (5/13) (12/13)  =  120 / 169  =  d

tan (A + B) =   [ tan A + tan B ]  / [ 1 - tan A tanB ] =   [ 5/2 + 1/2] / [ 1 - (5/2)(1/2) ] = 3 / (-1/4)  =  -12

We can solve for tan G thusly :

tan (B + G)  = [ tan B + tan G ] / [ 1 - tanB tanG ] 

     3/2 =   [   1/2 + tan G ] / [ 1 - (1/2) tan G ]

     (3/2)  [ 1 - (1/2) tan G ]   =  1/2 + tan G

      (3/2) - (3/4)tanG = 1/2+ tanG

     (3/2) - (1/2)  = tan G + (3/4)tan G

           1 =   ( 1 + 3/4) tan G

           1 =  (7/4) tan G

            tan G  =  4/7

tan ( A + B + G)  =   [ (tan A + tan B + tan G)  - tanA tanB tan G ]  /  [ 1 - ( tanAtanB + tanAtanG + tanBtanG) ]

                                [ (5/2 + 1/2 + 4/7 )  - (5/2)(1/2)(4/7) ] / [ 1 - ( 5/4 + 20/14 + 4/14) ] =

                                [  (3 + 4/7 ) - ( 20/ 28) ] / [ 1 - ( 5/4 + 12/7 ) ]  =

                                [   25/7 - 5/7 ]  / [  -55/28 ]  =

                                [ 20/7 ] / [ -55/28 ] = 

                                  -(20 * 28) / ( 7 *55)  = 

                                   -20 * 4  / 55  =

                                     -80  / 55  =

                                      -16 / 11

I'm not too sure about the formula for the triple sum, but I think that it's correct !!!

tan (A + B) =   [ tan A + tan B ]  / [ 1 - tan A tanB ] =   [ 5/2 + 1/2] / [ 1 - (5/2)(1/2) ] = 3 / (-1/4)  =  -12

tan (B + G)  = [ tan B + tan G ] / [ 1 - tanB tanG ] 

     3/2 =   [   1/2 + tan G ] / [ 1 - (1/2) tan G ]

     (3/2)  [ 1 - (1/2) tan G ]   =  1/2 + tan G

      (3/2) - (3/4)tanG = 1/2+ tanG

     (3/2) - (1/2)  = tan G + (3/4)tan G

           1 =   ( 1 + 3/4) tan G

           1 =  (7/4) tan G

            tan G  =  4/7

tan ( A + B + G)  =   [ (tan A + tan B + tan G)  - tanA tanB tan G ]  /  [ 1 - ( tanAtanB + tanAtanG + tanBtanG) ]

                                [ (5/2 + 1/2 + 4/7 )  - (5/2)(1/2)(4/7) ] / [ 1 - ( 5/4 + 20/14 + 4/14) ] =

                                [  (3 + 4/7 ) - ( 20/ 28) ] / [ 1 - ( 5/4 + 12/7 ) ]  =

                                [   25/7 - 5/7 ]  / [  -55/28 ]  =

                                [ 20/7 ] / [ -55/28 ] = 

                                  -(20 * 28) / ( 7 *55)  = 

                                   -20 * 4  / 55  =

                                     -80  / 55  =

                                      -16 / 11

Just add the tips  and then divide this sum by 7.....that will give you the mean  (average)

I don't believe that an integer  exists to  make this prime

Simplifying  we have

l 6x^2 - 75x  + 15 l

l 3 ( 2x^2  - 25x + 5) l

3 l 2x^2 -25x + 5 l

The only way this  can  be  prime is  if the expression   in the absolute  value  bars  is  either = 1  or  = -1

2x^2 - 25x + 5  = 1

2x^2  - 25x + 4 =  0 

Using the Q formula  x is   not an integer  because  the discrminant = 25^2 - 4(2)*(4)  = 593 which is  not  a perfect square

Likewise

2x^2 - 25x + 5 =   -1

2x^2 - 25x + 6  =  0

The discriminant = 25^2 - 4* 2 * 5  =   585  which is also not a  perfect square