CPhill

THX!!!!

Multiply the  first equation by -3

-3x - 3y  - 9z   =   -30

Add this to the  second equation and we get  that

-7x -4z = -22

Mutltiply the  last equation  by  -4

-4kx - 4z  =  -12

The  system will have no solutions when

-4k  =  -7

k =  -7/-4

k  = 7/4

We  need  to  find  the  height of the attachment  above  the  ground= H

sin 35  =  H  /  50

50* sin 35  =  H

When  the  120 ft  guy  wire  is attached....we  can find  the  angle  as

sin A  =   H  / 120

sin A  =  (50*sin 35)  / 120

To find  the  angle....take  the  arcsin

arcsin   [ (50 * sin 35)  / 120   ]  ≈ A ≈  13.8° 

66  is  the  sum of the first  n positive integers....so  we  have

N ( N + 1)   / 2    = 66

N ( N + 1)  = 132

N^2  + N  - 132   =   0

(N + 12) ( N -11)  = 0

Since N  is positive....N - 11   = 0  .....N  =11

It will be full after  the 11th stop

Eh....no big deal....easy to  make  mistakes  on here.......(I do it quite often  !!!!)

3800  -  (3/5)(3800)  -  210   =

3800  -  2280  -  210   =

$1310   left

2x^2   - 6x  + 5 =  -4

2x^2  - 6x  =   -9                 complete  the square on x  

2(x^2 - 3x  + 9/4)   = -9  + 9/2

2 ( x - 3/2)^2  =   -18/2  + 9/2

2 ( x - 3/2)^2   =  - 9/2

(x - 3/2)^2  =  -9/4          take both roots

x - 3/2  =   (3/2)i           or       x  - 3/2  =  -(3/2)i

x = (3/2)  + (3/2)i         or        x =  3/2  -(3/2)i

m =  n  = 3/2

mn   =  (3/2)^2  = 9/4

1.  (7 + 1/4) (4)   =   29

     (5  + 1/2) (4)  =   22

The first  train had 7 more  riders  than the 3rd

2.   3/4  of  a row  was empty.......(3/4) (4)  = 3  more riders needed  to completely fill the train

3.    8 - ( 5 + 1/2)  = 8 - ( 11/2)  =  (16/2)  - (11/2)   =  5/2  =  2 1/2  rows were empty

(5/2) (4)  =  20 / 2  = 10 more  rideres needed  to completely fill the train

Impressive, EP!!!!!

We're  missing something here.......