CPhill

Simplify  as

mn - 7m + 11n

m ( n - 7)  +  11(n - 7)  =   (m + 11) ( n - 7)

mn  - 7m + 11n - 77

https://web2.0calc.com/questions/help-pls_105

11*2 - 2(7i) + 11i    - 7i^2  = 

22 - 14i + 11i  + 7   =

29  -  3i

A

N       

C        M           B

AC^2  + [ (BC)/2 ] ^2  = 1

(AC/2)^2  + BC^2  =  1

AC^2  + BC^2/4  =  1

AC^2/4  + BC^2  = 1

4AC^2 + BC^2  =  4

AC^2/4  +  BC^2  =  1          subtract these

(15/4)AC^2  =  3

AC^2  = 12/15  = 4/5

(12/15)/2 + BC^2 =  1

(2/5) + BC^2  =1

BC^2  = 4/5

AB =  sqrt  [ AC^2 + BC^2 ]   =  sqrt [  4/5 + 4/5 ]  = sqrt [ 8/5]  =  (2/5)sqrt (10) 

13  - 6i

          P

      5            8

Q         M            R

            11

Law of Cosines

PR^2  = QP^2 + QR^2  - 2(QP * QR)cos(PQR)

8^2   = 5^2 + 11^2  - 2(5 * 11) cos (PQR)

64  = 25 + 121 - 110 cos (PQR)

[ 64 - 25 - 121 ]  / [ -110]  = 41/55 cos (PQR)

Law of Cosines again

PM^2  = QM^2  + PQ^2 - 2 (QM * PQ)cos (PQR)

PM^2  = (5.5)^2  + 5^2  - 2(6.5 * 5) (41/55)

PM^2  = 14.25

PM = sqrt [14.25  ]  ≈   3.77

Circumradius =  product of the sides /  [4 * Area]

Since the circumradius of each triangle is the same, then the areas are the same

So this means that  we can  solve this :    

[ AB * AX * BX[  = [ AC * AX * CX]

[AB * 5 * 5 ]  = [ AC * 5 * 4 ]

25AB = 20AC

AB  = (20 / 25)AC

AB = (4/5)AC

-cos ( AXB)  =  cos (AXC)

Law of  Cosines

AB^2 =  BX^2 + AX^2  - 2(BX * AX) cos (AXB)

AC^2 = CX^2 + AX^2 - 2(CX * AX)cos (AXC)

[(4/5)AC]^2  = 5^2  + 5^2  - 2(25) cos (AXB)

AC^2 = 4^2 + 5^2  - 2(20) (-cos AXB)

(16/25)AC^2  = 50 - 50cos (AXB)

AC^2  = 41 + 40cos (AXB)

cos (AXB)  = cos(AXB)....so......

[ (16/25)AC^2  - 50 ] / [-50 ]   =  [AC^2 - 41] / 40

40  [ (16/25)AC^2  - 50 ] = -50 [ AC^2 - 41]

(25.6)AC^2  - 2000  = -50AC^2 + 2050

(75.6)AC^2 =  4050

AC^2  = 4050 / 75.6

AC =  sqrt [ 4050 / 75.6 ]  ≈ 7.32

AB = (4/5)7.32  =  5.86

Semiperimeter  ≈   [ AB + AC +BC ]  /  2 =   [ 5.86 + 7.32 + 9  ]  /   2   = 11.088

[ABC]   = sqrt [ 11.088 * ( 11.088 - 5.86) (11.088 - 7.32) (11.088 - 9) ] ≈  21.36

1  1  3  1   1
-4 2  5  -4  2
k  0  1  k   0

Determinant   =     0

2 + 5k + 0  - [ 6k  + 0 - 4] =   0

2 + 5k  - 6k + 4  = 0

6 - k   =   0 

k = 6

https://web2.0calc.com/questions/for-a-certain-value-of-nbsp-k-nbsp-the-system_1

                   C

          8           4.5

A                          B

             M

Angle ACB =  90

AB = sqrt [ 8^2 + 4.5^2]  = sqrt (84.25 )     ......radius =  sqrt (84.25)  /  2

Call the center of the  circle O

OC = OA =  sqrt (84.25)  / 2

Law of  Cosines

AC^2  =  OC^2 + OA^2  -  2(OC * OA)  cos (COA)

8^2  = 84.25 / 4  + 84.25 / 4  -  2 ( 84.25 / 4) cos (COA)

64  = 42.125 - 42.125cos (COA)

[ 64 - 42.125 ] / [-42.125 ] = cos (COA)  = -175 / 337

arccos (-175/337)  ≈ 121.28°

This is the measure of central angle COA

CMA is an inscribed angle  intercepting the  same  arc as COA

So....its measure ≈  121.28  /  2  ≈  60.64°