CPhill

2* compact 2-door  = compact 4-door     (1)

3* compact 2-door  = SUV 2-door          (2)

compact 4-door = SUV 4 door                (3)

compact 2-door  + SUV 2-door  +  compact 4-door  + SUV 4-door  = 40       (4)

Sub (1), (2) and (3)  into (4)

Note that using  (1) and (3).....2compact 2-door = SUV 4-door....so.....

compact 2-door + 3compact 2-door + 2*compact 2-door + 2*compact 2-door  = 40

Simplifying....we have that

8 *compact 2-door  =40       divide both sides by  8

compact 2-door  =  40/8  =    5

Proof

There are  5 compact 2-door

There are 10 compact 4-door

There are 15 SUV 2-door

There are 10 SUV 4-door

y = 4cosθ  +  49 / cosθ

y = 4cosθ  +  49secθ      take the derivative

y'  = -4sinθ  + 49secθtanθ

y'= -4sinθ  +  49sinθ /cos^2θ     set this to 0

-4sinθ  + 49sinθ /cos^θ  =  0

49sinθ / cos^2θ  =   4sinθ

49sinθ = 4sinθcos^2θ

49sinθ - 4sinθcos^2  =  0

sinθ  (49 - 4cos^2θ)  =  0

sinθ (7 - 2cosθ)(7 + 2cosθ)  =  0

So  either

sinθ  = 0            7  - 2cosθ =  0                                 7 + 2cosθ = 0

θ = 0                  7  = 2cosθ                                          2cosθ  = -7

                           cosθ = 7/2  ( impossible)                     cosθ = -7/2   (impossible)

So....the minimum  occurs when θ  =  0....

And the minimum  =   4cos(0)  + 49/cos(0)  =  4   + 49/1  =   53

As this graph shows  : https://www.desmos.com/calculator/42hzhstsx6

Let the numbers  be  X, Y  and Z  in that order

And we have that

X + 4  = Z    ⇒  Z  - 4  = X        (1)

Y = 3Z + 5      (2)

X + Y  + Z  = 16      (3)

Sub (1)  and (2) into (3)   and we have that

[Z - 4] + [3Z + 5]  + Z  = 16   simplify

5Z + 1  =16       subtract 1 from each side

5Z  = 15       divide both sides by 5

Z  =3

So....the first number, X  = Z - 4   = 3  - 4   = -1

Let the twins' current ages  =  A

Let the current age  of the other brother =  A^3

And next year  we have that

A^3 + 1   =  ( A + 1)^2

A^3 + 1  = A^2 + 2A + 1

A^3 - A^2  - 2A  =  0

A(A^2 - A - 2)  =  0

One solution is that  A  =0    (reject)

And   using the second factor we have that

A^2 - A  - 2  = 0

(A - 2) ( A + 1)  = 0

Setting both factors to 0 and solving for A  we have that

A = 2     or  A = -1   (reject)

So

A= 2        and the non-twin  = 2^3 =  8

So.....the sum of the ages  =

2 + 2  + 8  = 

12

a

Net change  =  r(12)  - r(6)  =

[ 6 - 24/12]  - [ 6 - 24/6]  =

[6 - 2 ]  - [ 6 - 4]  =

4  -  2   =

2

b

Avg change  =  [ r(12) - r(6) / [ 12 - 6]  =    2 / [ 12 - 6 ]   =   2 / 6   =   1 / 3

See the following image :

Let the radius of the large circle  = R

Due to  symmetry, QF = EF   = R - 7

And PF  = R - 10

And FB  =  R

And by the intersecting chord theorem in the smaller circle

(QF)(EF)  = (PF)(FB)

(R - 7) (R - 7)  =(R- 10)(R)

(R - 7) (R -7) = R^2 - 10R

R^2 - 14R + 49 = R^2 - 10r     rearrange  as

49  = 4R

R =  49/4  =  12.25 =  the  radius of the larger circle

C'mon, dude....I just made a typo......get a grip  !!!!

a.

Net change =  r(12)  - r(6)  =

[ 6 - 24*12  ]  - [ 6 - 24*6 ]  =

-24 ( 12)  +  24(6)   =    -144

b.

Avg rate of change =    Net Change  / [ 12  - 6]   =   -144/6   =  -24