CPhill

               B

           45   45

A   45     D          45   C

Triangles ABD and CBD are congruent....BD is the altitude of both

AD = BD  = CD  =  6

[ ABD ]   = (1/2) AD * BD  =  (1/2) (6)(6)  =  18

Angle of depression =  arctan ( 100  / 250) ≈  21.8°

https://web2.0calc.com/questions/right-triangles_33 

Circumradius =  2/sqrt 3  * (1/2) a  =  a /sqrt 3

Inradius =  1/sqrt 3 *  (1/2)a  =   a / ( 2sqrt 3) =  a / sqrt 12

ab^4  = 48   (1)

ab^2 =   4    (2)

Divide (1) by (2)

b^2  = 12

b =  +/-  sqrt 12

Using (2)

a ( 12)  = 4

a = 1/3

(a,b)  =  (1/3 , sqrt 12) , (1/3, -sqrt 12)

No solution

If

q - 2r  = -2 + q

Then

-2r = -2

r =1      (1)

If r =1

Then 

p + r = 9 + p

r = 9

Which contradicts  (1)

https://web2.0calc.com/questions/semicircle_5

\(9^x - 2 \cdot 3^x + 1 \)

Take the derivative and  set to 0

ln 9 (9^x)  - 2ln3* 3^x  = 0

ln 9  ( 9^x)  - ln 3^2 * 3^x  = 0

ln9  (9^x)  - ln 9 ( 3^x)  = 0   divide out ln 9

9^x  - 3^x  =  0

9^x  = 3^x

(9/3)^x  = 1

3^x  = 1

x = 0

y= 9^(0)  - 2*3^(0) + 1 =

1 - 2 + 1 = 

  0 =  is the min

                  A

         7           17        

                12

B              M                 C

BM = CM

Law of Cosines

Note : cos CMA =  - cos BMA

We have these two equations

7^2 = 12^2 + BM^2  - 2(12 * BM) cos (BMA)

17^2  - 12^2  + BM^2 - 2(12 *BM) cos (CMA)     

7^2 = 12^2 + BM^2  - 2(12 * BM) cos (BMA)

17^2  = 12^2  + BM^2 + 2(12 *BM) cos (BMA)    add these

338 = 288 + 2BM^2

50 = 2BM^2

25 = BM^2

BM = 5 = CM

BC = 10

However, this triangle is  impossible because  BC + AB  must be >  AC