CPhill

Notice  that   4sqrt (1/2)  =    ±  4 (1) / sqrt (2)  =  4/sqrt (2)

Multiply  top/bottom  by  sqrt (2)   and we get

 ±  4sqrt (2) / 2 =   ±  2 sqrt (2)  =  b

Sorry that I didn't simplify it  more  !!!

The vendor  sells the jugs containing  13 , 15, 16, 21 and 22 gallons

This totals  87  gallons

1/3  of this =  29 gallons  which means that he sells the jugs containing 13 and 16 gallons to Randy

The remaining 2/3 = 58 gallons which means that he sells the jugs containing 15, 21 and 22 gallons to Lisa 

He  keeps the  remaining  jug which contains 21 gallons of  blue punch

x^2 + y^2/4  = 1      using implicit differentiation  we have that

2x + (1/2)yy'  = 0

y'  =  -2x / (1/2 y)

y'  =  -4x/ y

Yhe slope of the line  = 2

So....we want to find where

-4x / y =  2

-4x  = 2y

-2x  =  y

Sub this back into the equation of the ellipse for y

x^2   +  (-2x)^2/4  = 1

x^2  + 4x^2/4  = 1

2x^2  =  1

x^2  =   ±√(1/2)

When  x = √(1/2),  y =  -2√(1/2)

When x = -√(1/2) ,  y = 2√(1/2)

Subbing into the equation of the line

 -2√(1/2)  = 2√(1/2)  + b   →  b =  - 4√(1/2)

And

2√(1/2)  = -2√(1/2)  + b →  b = 4√(1/2)

See the graph here  :  https://www.desmos.com/calculator/7ifp0ajbri

Let the diagonal  length of  both  =  D

In the rectangle...  the side  opp the 30° angle  = D/2 and .the side  opp the 60° angle  =    D√3/2   and

The area of the rectangle  =  D^2 [√3/4]

In the square......the side length =  D/√2

Area of square  = (D / √2)^2  = D^2/2

R =  D^2 [ √3/4]  / [D^2 / 2 ]   =  (√3/2)

R^2 =  3/4  

EDIT TO CORRECT A PRIOR MISTAKE

The slope of this segment  is  [ 9 - 3 ] / [ 7 - 1 ]  =  6/6  = 1

The perpendicular line to this segment will have a  negative reciprocal slope   =  -1

Since this line is a perpendicular bisector, we need to find the  mdpoint of the segment.....this is given by

(  [ 1+ 7] / 2 , [ 3 + 9] / 2  )    =   (8/2 , 12/2 )  =  (4, 6)

The equation of the perpendicular line is

y = -1 ( x -4)  + 6

y = -x + 4 + 6      add x to both sides  and simplify

x + y  = 10

So...b  =  10

OK, parpar1.....

We  are looking for the  minimum (squared) distance  from the origin to the line   3x + y  =15

Let this point  be   (x  , 15 - 3x )

D^2  =  ( x- 0)^2  +  ( 15  -3x-0) ^2

D^2  =  x^2  + ( 15-3x)^2

D^2  = x^2  + 225 - 90x + 9x^2      take the derivative and set to 0

D'^2  =  2x  -90 + 18x   = 0

20x - 90  = 0

20x  =90

x= 9/2

y = 15 - 3(9/2)  = 30/2 - 27/2  =  3/2

So x^2  + y^2   is a minimum  when  x= 9/2  and y =3/2

The minimum  is   (9/2)^2  + (3/2)^2   =   [ 81 + 9 ] / 4  =   90 / 4 =  45/2  =  22.5

1

(a)  the populauion  consists  of  the guests staying at the hotel

(b)  the poulation  consists of the  teachers  at the school

(c)  the population consists of  the customers of the business

a = 7   b  =  5    c  = 5

The circumradius length is  given by:

     abc

_________________________________________  =

√[ (a + b + c) ( a + b - c) (a + c - b ) ( b + c - a)]

7* 5 * 5

________________________________________ =

√ [ ( 7 + 5 + 5)  ( 7) ( 7)  ( 3) ] 

     175

_____________   =

√[ (17) (49) (3) ]  

     175

_______________  =

  7 √51

25 / √51

Using the Law of Cosines, we can  calculate  angle BAC

7^2  - 5^2  - 5^2

_____________   = cos BAC

     -2(5)(5)

1 / 50  =cos BAC

arccos (1/50) =  BAC  = 88.85°

And  since BAC  is an inscribed angle, then central angle BOC  is twice this  = 177.7°

So the area of   BOC =  (1/2) ( 25/ √71)^2 sin (177.7)  ≈   .177  units^2 

Look at the  graph ,here :  https://www.desmos.com/calculator/grfdjntjey

It appears that just  G and I  satisfy the inequalities