CPhill

2.  Connect  ( 2, 1)  and (3, -3)....this creates a rectangle  with vertices (2,1) , (3, -3), (-5, -5) 

and (-6, -1)

The distance from  (2,1) to (3, -3)  forms one side of this rectangle  = √17 units

The distance  from (-6, -1) to (2,1)  forms the other side of this rectangle  = 2√17 units

So....the area of this rectangle is  √17 * 2√17  =  2 * 17  =  34 units^2   (1)

Connect (2, 1)  and (3, 1)......this forms a triangle at the bottom right with vertices  (2,1), (3,1) 

and (3, -3)

The base of this triangle  is  (2,1) to (3,1)  = 1 unit

And the height of this triangle is (3, 1)  to (3, -3)  =  4 units

So...the area of this triangle is  (1/2) (1) (4)  =  2 units^2   (2)

Connect (2,3)  and (3,3)....this creates a rectangle with vertices  (2,3) (3,3), (2,1) and (3,1)

The height of this rectangle is  (3,1) to (3, 3)  = 2 units

And the widthof this rectangle is from (2,1) to (3,1)  = 1 unit

So...the area of the rectangle is  2 * 1  = 2 units^2  (3)

Finally....we have a triangle with the vertices (2,3), (3,3)  and (2,7)

The base of this triangle is from (2,3) to (3,3)  = 1 unit

And the height of this triangle is from   (2,3) to (2,7) =  4 units

So....the area of this triangle  is  (1/2) (1) (4)  =  2 units^2   (4)

So  the total area =   (1)  + (2)  + (3) + (4)  =  [34 + 2 + 2 + 2]  =

40 units^2

1.     Draw VR....this creates triangle VMR  and rectangle VRDE

Let VR be the base of triangle VMR  = 6 units   and the height drawn from M  to the base  = 3 units

So...the area of this triangle is  (1/2)base* height  = (1/2) 6 *3  =  (1/2) 18  =

9 units^2   (1)

And rectangle VRDE has area VR * VE   =  6 * 5  =  30 units^2  (2)

So.....adding (1) and (2)  gives  the total area  of  39 units^2 

(x, y)  ⇒ reflected about  x axis  ⇒  ( x, -y)

⇒  rotated 90°  clockwise  about the origin  ⇒  ( -x , - y)

⇒  reflected in the y axis  ⇒  ( x, -y)

O will be the center of the octagon

Draw OB, OC  and OD

Note that   area OAB  =  area  ODC =  area ODE

And area OBM  =  1/2 area OAB

And area OCM  = area OBM  =  1/2 area OAB

So

area  EDCMO  =   area ODE + area ODC  + area OCM

Substituting, we have that

area EDCMO  =  area OAB + area OAB + 1/2area OAB   =  2.5 areaOAB

And area ABMO  =  area OAB + area OBM

Substituting, we have that

area ABMO  =  area OAB + 1/2 area OAB   =   1.5 area OAB

So

[ ABMO] / [ EDCMO]  =

[ 1.5 area OAB ] / [ 2.5 area OAB ]  =

[1.5] / [ 2.5]  =

3 / 5 

4y + 1x  + 40  =  0     rearrange as

4y  =  -1x - 40          divide both sides by 4

y  = (-1/4)x - 10

The slope of this line is  -1/4   ....a perpendicular line will have a negative reciprocal slope = 4

So...using the point-slope form, we have

y  = 4 ( x - 1/2)  + 0       simplify

y = 4x  -  2

Here's the graph of both lines :  https://www.desmos.com/calculator/xzeqsobhzr

y =  2x + 3y  =  24   ???

Is there a typo here  ??

[ 6n^2 + 5 ]  / [3n^2]

We have  a   Same Degree / Same Degree situation

The limit  as n approaches infinity in  Same / Same   is simply the ratio of the coefficients on the highest powers of n in the numerator / denominator...so...the limit  is  6 / 3  =  2

Correct  !!!

f(0)  = f(1) = f(2) = f(3)   = 1

And f(4)  = 0

In the form

ax^4 + bx^3 + cx^2 + dx + e        

If f(0)  = 1, then  e  = 1

And we have this system

a + b + c + d  =  0

16a + 8b + 4c + 2d  = 0

81a + 27b  + 9c + 3d  =  0

256a + 64b + 16c + 4d  = -1 

The solution to this system is

a = -1/24     b  =  1/4      c  =  -11/24   d = 1/4

So...the polynomial  is

(-1/24)x^4  + (1/4)x^3 - (11/24)x^2 +  (1/4)x  + 1

And   f(5)  =  - 4

Each dish will hold

[ 2. 5 lbs ] /  3  =   25/30 lbs.   =   (5/6)lb

And one bowl will serve 10 people....so....each person will have a serving size  of

( 5 /6)  / 10   lbs    =   5/60 lbs   =  1/12  lbs