CPhill

Thanks, James

Another  method

We  have  the relationship that

BD / AD  = AD  /  DC

BD / 20  =  20 / 16           cross-multiply

16 BD  =  20 * 20

16 BD  =  400

BD  =  400  /16     =   25

-8r <  -12  + 2r           rearrange   as

12 <  2r + 8r

12 <  10r

12/10 < r

6/5 <  r       which also  means that    r  >  6/5

Good job, James.....exactly correct   !!!!

First.....find  the altitude, A,   as....  sqrt  (12^2  - 8^2)  = sqrt  [ 144 - 64 ] = sqrt (80) =  4sqrt 5

And  we  have  this relationship

x / A =  A  / 8

x / ( 4sqrt 5)  =  (4sqrt 5)  /  8        cross-multiply

8x  =  (4 sqrt 5)^2

8x  =  80

x  =  80 / 8   =   10

Area  =  pi  *  radius ^2  *  ( 40 / 360)   =

pi * 4^2  *  ( 40 / 360)  = 

16 pi  * ( 1/ 9)   =

16pi / 9    cm^2

Unit price  =

11.43 /  9     = 

$1.27

Note  that we can write  this as

C = 2n^(1/4)  [  20n^(1/8)  + 8n^(1/3)  ] 

C =  2 * 20* n^(1/4 + 1/8)    +  2* 8 * n^(1/4 + 1/3)   

C  =  40 n^(3/8)  + 16 n^(7/12) 

C  =  40 ⁸√ (n³)   +  16 ¹²√ (n⁷)

a = 3

b = 7

Let A  be your current age

(4/5)A  >  (3/4) (A + 1)

(4/5)A - (3/4)A  > 3/4

(1/20)A  > 3/4

A > (3/4)(20)

A >  15

And

(4/5)A  >  (5/6) (A - 2)

(4/5)A  >  (5/6)A  - 5/3

5/3  >  (5/6)A - (4/5)A

5/3  >  (1/30)A

(5/3) ( 30)  > A

50  > A

15  < A  < 50    where A is an integer

Area  =  (1/2) height  * ( sum of base lengths)

38  =  (1/2) (4)  ( 10  + x)

38 = 2  ( 10 + x)

38 / 2   =  10 +  x

19  =  10 + x

19 - 10  = x   =  9 =   the other base length

In every  hundred, we will  have  20  integers   that satisfy  this

So    20 (hundreds) * 20 ( occurnces in each hundred)  =  400

And we  have three more occurences  from 2001 - 2012

So

400  + 3   =

403  occurences