CPhill

Using the square of the distance formula = r^2....let the center = (x,y)

So we have

(x -22)^2  + (y-15)^2 = (x -22)^2 + (y + 29)^2   simplify as

(y -15)^2  = (  y - 29)^2

y^2 -30y + 225  =  y^2 -58y + 841

28y =  616

y = 616 / 28 =  22

So...using  (22,15)  and (-25,0)   we can  find x  as

(x -22)^2  + ( 22-15)^2  = (x + 25)^2  + ( 22 - 0)^2

(x -22)^2  + 49  = ( x + 25)^2  + 484

x^2 - 44x + 484  + 49  =  x^2 + 50x + 625 + 484

-44x  + 49  =  50x + 625

- 94x  =  576

x = 576 / -94 = -288/47

Coordinates are  ( -288/47 , 22)

Find the area using Heron's Formula

A =  sqrt [ 18.5 (3.5) (6.5) (8.5) ]    ≈  59.8

The shortest altitude is drawn to the longest side

59.8 =  (1/2) (15) (altitude)

59.8 / [ 15/2 ] =  altitude   ≈  7.97  

         B

     6           8

A                  C 30

We have a SSA  situation......there may be 1 triangle formed, 2 triangles formed or no triangles formed

Law of Sines

sin A / 8  =  sin 30 / 6

sin A / 8 = (1/2) / 6

sinA = (8/6) (1/2) = 2/3

arcsin (2/3) ≈  41.81°    or    180 - 41.81  = 138.19

So   angle B  =  180  - 30  - 41.81   =  108.19°

or    angle B =  180 - 30 - 138.19  =    11.81°

If B =  108.19°   area of ABC =  (1/2) (8)(6)sin (108.19°)     ≈  22.8

If B  =  11.81°  area of ABC  = (1/2) (8)(6) sin ( 11.81°)  ≈  4.9

    B

30    30

90C      D         A30

            12

BC = 12/sqrt 3  =  4sqrt 3

BA = 24/sqrt 3  = 8 sqrt 3

Since BD is  a bisector

BC / BA =   DC / DA

[4sqrt 3]/ [8/sqrt 3]  = DC /DA

4/8  =  DC/ DA

1/2 = DC / DA

Therefore   DA  =   2 / [1 + 2 ] AC   =  (2/3)AC = (2/3) 12 = 8

[ABD] = (1/2) (BA) ( DA)  sin (30)  =    (1/2) (8sqrt 3)(8) (1/2)  =  16 sqrt 3

https://web2.0calc.com/questions/right-triangles_35

A  = 6(1/2) s^2  sqrt (3) / 2

3/2 = (3/2)   sqrt (3)  s^2

1 = sqrt (3)  s^2

1/sqrt 3  = s^2

s =  sqrt (1 /sqrt 3)  =  1 / ⁴√3 

I like your polynomial  approach  !!!!!

At least 5 heads means the same as no more than 4 tails

0 tails = C(12,0) =  1 possible sequence

1 tail  = C(12,1) =    12 possible sequences

2 tails =  C(12,,2)  =   66 possible sequences

3 tails  = C(12, 3)  = 220 possible sequences

4 tails  = C (12,4)  =  495 possible sequences

At least 5 heads =   sum of 12rh row of Pascal's Triangle - ( 1 + 12 + 66 + 220 + 495)  =

2^12 - ( 1 + 12 + 66 + 220 + 495)  = 

3302 possible sequences

Factor 2000  =  2^4 * 5^3 =  1* 4^2 * 5^3 =   1 * 4 * 4 * 5 * 5 * 5

Possible identifiable sequences  =    7!  / [ 2! * 3! ]  =    420

ab -3a + 4b  = 131        take the product of the coefficients on a,b.....add this product to  both sides

ab - 3a + 4b + (4 * - 3)  = 131 + (4 * -3)

ab - 3a + 4b - 12  =  119            factor the left side

(a + 4) ( b - 3)  = 119

Factors of  119  are    1   7    17     119

Using   17 and  7

( 13 + 4) ( 10 - 3)   →  l a - b l  =  l13 - 10 l =  3