CPhill

mn = m + 11n  - 18

mn - m  =  11n - 18

m ( n -1)  = 11 n  - 18

m =  (11n - 18) / ( n -1)

m will not be < 12  as n approaches  - inf

m  will not be > 10  as  n  approaches inf

m       n

4        2

10      8

18      0

12     -6

(a + 2) / (a + 3)  = b / 7

7 (a + 2) = b(a + 3)

7a + 14  = ab + 3b

7a - ab  =  3b  - 14

3b + ab  = 7a + 14

b ( 3 + a)  = 7a -+14

b =  (7a +14) / (3 + a)

a        b

-10    8

-4    14

-2      0

4       6

x^2  + 3x + p

p =  [ (coefficient on x)  / 2 ] ^2 =  [ 3/2]^2 =  9/4

x^2  + 3x + 9/4   =

(x + 3/2)^2

Law of Cosines

AE^2 = AF^2  + FE^2  - 2 (AF * FE) cos (60°)

AE^2  = 2^2 + 5^2  - 2(2* 5) (1/2)

AE^2  = 19

AE = sqrt (19)

x^3 ( x^2 + tx)   =   x^5 + tx^4

This has no x^2 term anyway

18 / 16  = r^3

9 / 8  = r^3

r =  (9/8)^(1/3)

24th term  = 16 * (9/8)^(1/3) ≈   16.64

Here :   https://web2.0calc.com/questions/circles_146

Let D be the center of the  circle and let its radius = r

Let E be the center of one of the small semi-circles

We can construct right triangle BDE  with EBD  = 90°

BE = 1

BD = 2 - r

ED =  1 + r

BE^2 + BD^2  = ED^2

1^1 + (2 -r)^2  = (1 + r)^2

1 + r^2 - 4r + 4  = r^2 + 2r + 1

6r =  4

r = 4/6  = 2/3

I = (4,r)   where r is the radius of the incircle

To find r

[ ABC] = (1/2)(8 + 5 + 5) r

(1/2) 8 * 3 = (1/2) (18) r

r = 4/3

l = (4, 4/3)

ID = 4/3 

AI =  3 - (4/3)  = 5/3

Triangles  AMN  and ABC are similar such that  AI /  AD =   = (5/3) / 3  = 5/9

[AMN ] = (5/9)^2   [ ABC  ] =   (5/9)^2 (1/2) (8)(3)  = (25 / 81 ) * 12  = 300 / 81 =   100 / 27