CPhill

I'm not totally confident about the  "width" answer, EP

I have a feeling that I missed something  .....

I just gave it my best shot   !!!!

Well.....there's nothing particularly magical about this

For instance

sqrt (12)  =  2sqrt (3)

But

2 tan (60°)   also =  2 sqrt (3)

We could find any number of these "strange" coincidences  !!!

The vertex  is  ( 2,1)  = (h, k)

Note that if we  have the form

y = a(x -h)^2 + k         substituting the vertex coordinates into this we have

y = a(x - 2)^2  +  1

Since  the point  (0, 5)  is on the graph we have that

5  = a( 0 - 2)^2  + 1

5 = a(-2)^2 + 1

4 = 4a

a = 1

So we have this

y = 1 ( x - 2)^2  + 1

y = 1x^2 - 4x + 4 + 1

y = 1x^2 - 4x + 5

a  = 1

b = -4

c =5

a + b + c  = 

1 - 4 + 5 =

2

Note that  pure water  contains   0 concentration of brine....so....

1L = 1000 g

110g/L  =  110g/1000g =  11% concentraion =.11

200g /L  = 200g/1000g  =  20% concentration  = .20

So

6 ( .11)  - x* (0) =  (6 - x) (.20)      where x  is the amount of water to be  boiled off  in L

6 (.11)  = 1.2 - .2x

.66 = 1.2 - .2x      rearrange as

.2x  = 1.2 - .66

.2x = .54

x = 2.7  L 

sqrt (50)  = sqrt (25 * 2)  =  5sqrt (2)

10 sin (45)  =  10 [sqrt (2) / 2 ]  = (10/2) * sqrt (2)  = 5 sqrt (2)    !!!

Note... sin (45)  = cos (45)

Since each pass of the  mower cuts a 2 foot swath.....the length  of the field  must  be

11*2 *2  + 14 *2* 2  =  (25)*4  =  100  ft = L

On the first  pass the square footage cut = 

2* (W) * 2   + 2 (L - 4)*2  =  

4W + 4(96)  = 

4 (W + 96) =

4W  + 384

On the 2nd pass  the square footage cut = 

2 (W - 4)*2  +2(L -8)*2 =

4W - 16 + 4(92) = 

4(W + 76)=

4W  + 352=

4W + 384 - 32 (1)

On the 3rd pass the square footage cut =

2(W - 8)*2 + 4(L - 12) =

4W - 32 + 4 * 88 =

4W + 320 =

4W + 384 - 32(2)

.....

So.....on the first 11 passes....the square footage cut  is

4W   +  sum [ 384 - 32n] for n = 0 to n =10 

11*4W + 2464

44W + 2464

.....

For the remaining 14 passes....

On the 11th  pass the square footage cut =

4W  - 32 (11)

......

On the 24th pass the square footage cut  =

4W  - 32 (24)

So....on the final 14 passes the square footage cut  =

14*4W  + sum [ 384 - 32n] for n = 11 to n =24

56W - 2464

And the  square footage of the first 11 passes = square footage of the remaining 14 passes...so.....

44W + 2464  = 56W - 2464   simplify

12W  = 4928

W =  [410 + 2/3 ] ft

We need to solve this

f(g(x))  =  g (f (x))

2(3x + c) + 7    = 3(2x + 7)  + c          simplify

6x + 2c + 7  =  6x + 21  + c      subtract  6x  from both sides

2c + 7  =  21 + c         subtract c , 7  from both sides

c  =  14

-548

720 - 548  =   172°

-532

720 -548  =  188°

-188

360 -188  = 172°

8     =    8°

532

532 - 360  = 172°

We  have  a 30-60-90  right triangle

Side  opposite 60° angle =  15

Side opposite of 30°  angle =  15/√3

Side opposite the 90° angle  = 2 times the side opposite the 30° angle = 2*15 / √3  =  30 / √3 = the hypotenuse

Assume this is

tan (-2pi/3)

_________ -  sec (-pi)         ???   .....if  so, we have ....

sin (7pi/4)

-2pi/3  =  -120° = (360 - 120)°  = 240° = 60° in Q3  =  tan (60)  = √3

7pi/4 * 180/pi  =  7 *(180/4)  =  7 * 45  =  315° =  45° in Q4

So    sin (-45°) = -√2/2

sec (-pi)  =  sec (pi)  = -1

So we have

  √3

_____ -  (-1)  =

  √2/2

2√3

___  +  1   =

 √2

2√3 * √2

_______  +  1     =

 √2 * √2

2√6

___  +    1   =

  2

√6  +  1