CPhill

My equations should have been : 

New Car

A =  350 [ (1 + .0279/12)^48 - 1 ] / [(.0279/12) ( 1 + .0279/12)^48 ]  = $15,879

Used Car

A  = 350 [ (1 + .0329/12)^48 - 1 ] / [(.0329/12) ( 1 + .0329/12)^48 ]  = $15, 721.34

3x^2  - 6x  - 2        factor out the 3

3 (x^2  - 2x  - 2/3)      take 1/2 of 2, square it  = 4...add and subtract it within  the parentheses

3(x^2  - 2x + 4  - 2/3  - 4)    factor the first three terms in parentheses ...simplify the rest

3 [ (x - 2)^2  -14/3 ]    distribute the 3 across the terms in the brackets

3 ( x- 2 )^2  - 14

a  = 3     h = 2   and  k = -14

And their sum is  -9

[AD  -  AC]  =  [30  - 16 ] = 14  = CD

[BD  - CD ]  = [20  - 14]  = 6   =   BC

The directrix is  above the vertex, so this parabola turns downward

The form is

-4p [ y - k ] = [x - h]^2

(h, k)  is the vertex = (-4, -6 )     

And p is the distance between the vertex  and the directrix  given by 3 -  k  = 3 - (-6)  = 9

So we have

-4(9) [ y - (-6) ]  = [ x -  (-4) ] ^2

-36 [ y +  6]  = [ x + 4 ] ^2      divide both sides by -36

y  + 6  =  -(1/36) (x + 4)^2

y  = -(1/36) ( x + 4)^2  - 6

We have 9 "whole" cars and 2 "half "  cars....but 2 "halves" gives us another whole one

So...we have 10 whole cars

And Coupeville manufactures 2 + 1/2 0f them  = 2.5

So....the fraction  made in Coupeville   =  2.5  / 10    =  1/4    ⇒   B

These aren't that hard to identify......

We  have the form

a (b)^x

When   "b"  >  0   but  <  1.....we have decay

When "b"  > 1     ...we have  growth

[The "a"  doesn't  matter  ]

So.....in the first part

h(x)   = (0.2)^x     and    f(x)  = 4(0.3)^x      are decay functions

And...in the second part

h(x)  = (1.9)^x   and   d(x)  = 2(4.7)^x     are growth  functions

I believe the last answer is correct...if we remove the outliers  [ lowest and highest data values] the median is not affected

Example

16   20   22   23   24  27   55      median  is   23

20  22  23   24   27       median is still 23 

f(x)   = -x^2 -2x + 8

This is a parabola  that turns "downward"  because thw leading coefficient is negative

To find the x coordinate of the vertex...

In the form    ax^2  + bx  + c

The x coordinate of the vertex is    -b / [ 2a ]

So....in our function,  b = (-2)   and  a = (-1)

So...  -b / [ 2a]  = 2 / [ 2 * -1 ]   =  2 / -2   = -1

To find the associated y value for the function....put  -1  back into it and evaluate

- (-1)^2  -2(-1) + 8 =  -1 + 2 + 8  = 9

So...(-1,9)  is the "high point"  of the function

So....this function will increase  from  (-infinity, to -1)

To find out where it tis positive on this interval...we can  find the  root on this side  [ where it crosses the x axis]

So...we want to solve

-x^2 - 2x  + 8  =0   multiply through by -1

x^2 + 2x  - 8  =  0     factor

(x + 4) (x - 2)  =0

Set each factor to 0  and solve for x and we get   x = - 4   or  x  = 2

So....this tells us that the function is positive and increasing from  

-4 < x  <  -1    ⇒   B

Here is the graph :  https://www.desmos.com/calculator/lmmko24qwo

I'm going with  "D"

ABC  is a right triangle  with  AC the hypotenuse  = 5

Its area  = (1/2) product of the leg lenghts =   (1/2) (3 * 4)   = 6

Since its base is  5.....its altitude is

6  = (1/2) AC * altitude

12  = 5 * altitude

altitude  = 12/5

And triangle WBZ is similar to triangle ABC

So

altitude of WBZ  / base of WBZ  = altitude of ABC  / base of ABC

(12/5  - S ] / S  = (12/5) / 5      where S  = the side of the square

(12/5 - S) * 5 = (12/5)S

12 - 5S  = (12/5)S

12 =  5S + (12/5)S

12 = S [ 5 + 12/5 ]

12 = S [  37/5 ] 

!2 [ 5/37]  = S

60/37    =  S