CPhill

2. Triangles PBA and PCD are similar

Since CD = 2AB, then the scale factor of triangle PCD to triangle PBA = 2

So....the area of triangle PCD  = 

area of triangle PBA * scale factor^2 =

10 * 2^2 =

10 * 4   =

40 units^2

Hit the equal sign twice....

I like Desmos......pretty easy to use....

The greatest integer on the "odd" side of the street is

3 + 6(20 - 1) =  3 +  6(19) = 3 + 114 = 117

2 houses have one digit each 3, 9

3 houses will have three digits each 105, 111, 117

So...15 houses will have two digits each 

So...for the "odd" side he collects  2(1) + 15(2) + 3(3) = 2 + 30 + 9 = $ 41   

For the "even" side, the greatest integer is   4 + 6(20-1) = 4 + 6(19) = 118

1 house has one digit

4 houses have three digits each [ 100, 106, 112, 118]

So.....15 houses wil have two digits each

So....for the "even" side he collects 1(1) + 15(2) + 4(3) = 1 + 30 + 12 = $43

So....the total he collects is    $ [ 41 + 43 ]  =  $ 84

This is an exponential function.......

Here is the graph.....https://www.desmos.com/calculator/0vbgmdeqfv

Nicely done, Melody !!!!

Using the answer to your other question, the equation of the hyperbola was found to be

(y - 1)^2      -   (x - 0)^2

________      ________ =   1

      4                 1

The center is   (0, 1)  = (h, k)

a = sqrt (4)  =  2  

The vertices  are given by

(h, k+a)   and  (h, k - a) =

(0, 1 + 2)  and   (0, 1 - 2) =

(0, 3)   and (0, -1)

-4x^2 +y^2 -2y = 3      rewrite as

y^2 - 2y - 4x^2 = 3        complete the square on y

y^2 - 2y + 1 - 4x^2 = 3 + 1

(y - 1)^2  -   4x^2 =   4         divide both sides by 4

(y - 1)^2     -   x^2  

______         ____  =      1

   4                  1

In this form,   a = 2  , b = 1     h = 0    k = 1

The equation of the asymptote with the positive slope is 

y =  (a/b) ( x - h) + k        ....so...

y = (2/1) ( x - 0) + 1

y = 2x + 1

Here's the graph :   https://www.desmos.com/calculator/tenipf6zfq

A must be a right angle

AB = 1

AC = sqrt (3)

BC = 2

Let A = (0,0)  B = (1, 0)  and C = (0, sqrt (3) )

The midpoint of BC is  [ (1 + 0)/2 , ( sqrt(3) + 0 ) / 2 ]  =   (1/2, sqrt (3)/2 )

Call this point, D

The distance from A to D is

sqrt [   (1/2)^2 + (sqrt (3)/ 2)^2]  =  sqrt [  1/4 + 3/4] = sqrt (1) = 1

And the distance from A to G is 2/3 of AD =   2/3 units

No prob.....!!!!