CPhill

                      A

      3                                5

B                          D                      C

                      7

Law of Cosines

BD^2 = AB^2 + AC^2 - 2(AB * AC) cos BAC

7^2  = 3^2  + 5^2  - 2 (3 * 5) cos BAC

cos BAC  = [ 7^2 - 3^2 - 5^2 ] / [ -2 (3 * 5) ]  =  -1/2

arccos (-1/2)  =  measure of BAC  =  120°

Since AD is an angle bisector

BAD  = 60°

cos (BAD)   = cos (60°)  =  1/2

x + y = 10    →   y  = 10 - x

x^2  + (10-x)^2 = 64

2x^2 - 20x + 100 = 64

2x^2 -20x + 36 =  0

x^2 -10x + 18  =  0 

x^2 - 10x  =   -18

x^2 -10x + 25  = -18 + 25

(x - 5)^2  = 7            take both roots

x - 5 = sqrt 7            or            x  - 5 = -sqrt 7

x = 5 + sqrt 7                          x  = 5 - sqrt 7

(x , y)  = (5 + sqrt 7 , 5 -sqrt 7 )   or ( 5 -sqrt 7 , 5 + sqrt 7)

10 * 10 =  5 * x

100 = 5x

x = 100 / 5   =  20 min

10* 10  = 15 * x

100 / 15  =  x  =   20 / 3  min =   6 + 2/3 min  = 6 min 40 sec

Simplify as

-15x^2 -kx + 53 = 0

This will have no real solutions when the discriminant is < 0

k^2 - (4)(53)(-15)  < 0

k^2 + 3180 < 0

k^2 + 3180  is always >  0.....so no value of "k" will give us non-real solutions

Simplifying we get

x^2  - 28x + 392

The  minimum  value occurs at  x =   28 / [ 2 * 1 ] =  14

And  the minimum value is

(14)^2 -28(14) + 392   =  196

(x - 4)^2 + (y -5)^2  =  68

This is a circle  with a center at (4,5)

A perpepdicular line to the given one passing through the center amd intersecting the given line  will give us the midpoint of chord AB 

Equation of perpendicular line

y =(-1/3)(x - 4) + 5        set this equal to  the given line

3x -2 = (-1/3)(x - 4) + 5       mulriply through by 3

9x -6  = - (x - 4) + 15

10x = 25

x = 25/10  = 2.5

y = 3(2.5) -2  = 5.5

Midpoint =   (2.5 , 5.5)

x = -4     y =  2

-4 / ( -4 - 2*2)  =  -4 / (-4 -4)  = -4 / -8   =  1 / 2

x^2 + 7x + 12 ≤ 0

This is true on  [ -4, -3 ]

The max value for   x^2 + 2x + 1  =  (-4)^2 + 2(-4) + 1 =   9

1 + 2 + 3 + ....+ 200  =  [ 1 + 200] * 100 = 

20100 =

201 * 100  =

67 * 3 * 5^2 * 2^2 

67 is the greatest prime divisor