CPhill

Maeve's rate to Tanzi's rate   =  (1/3) bottle : 1 bottle  =  1  :  3

So  we can divide (1/3) of the bottle inro 4 equal parts

Maeve will use  (1/4)(1/3)  = 1/12  of the bottle  and at the same time, Tanzi wil use (3/4)(1/3) = 1/4 of the bottle

So Maeve  should pour (1/4)(1200)  = 300 ml into Tanzi's bottle

Check :  There are  (1/3)(1200)  = 400 ml left in Maeve's bottle

She will use (1/12) (1200) = 100 ml   at the same time that Tanzi will use 300 ml

So   100 : 300  =  1 : 3

A

              X

B       12            C

AB =  12sqrt 3

AC =  24

sin (angle A)  = sin (30)  =  (1/2)

Angle ABX = 45

Angle BXA = 180 - 30 - 45  =  105

Law of Sines

BX  /sin (30)  = AB / sin (105)

BX  /  sin (30) = 12 sqrt 3  / sin (105)

BX =  12sqrt (3) sin (30)  /sin (105)  = 12sqrt (6) / (sqrt (3) + 1)

[BXA ]  =  (1/2)AB*BX *sin (45)  =   (1/2) (12sqrt 3)(12sqrt6) / (sqrt (3)+ 1) *sqrt (2) / 2  =

216  / (1 + sqrt 3)  =   216 ( 1 -sqrt 3)  / [ 1 - 3]  =     216 [ sqrt (3) - 1] / 2  =  108 (sqrt (3) - 1)

Here :  https://web2.0calc.com/questions/circles_146

Here : https://web2.0calc.com/questions/semicircles_5

Here :   https://web2.0calc.com/questions/square_32/edit-2?

\(0.\overline{abcdef} = \frac{1}{7} \)

           .142857

7    [ 10          ]

          7

________

         30

         28

        ____

            20

            14

           ____

              60 

              56

            ____

                40

                 35

                ____

                    50

                    49

                   ___

Note something interesting

2/7 = .285714 (repeating)

3/7 = .428571 (repeating)

4/7 = .571428 (repeating)

5/7 = .714285 (repeating)

6/7 = .857142 (repeating)

Note that each fraction  contains  the same repeating digits (just in  different orders)

\(0.\overline{296841} \)

=  296841 / 999999

Draw altitude XW

XW = sqrt [ XZ^2 - (YZ/2)^2 ]  = sqrt  [10^2 - 8^2] = sqrt [36 ] = 6

Let V be the point of tangency between the triangle and the circle

Draw WV  = the radius of the semi-circle

We have two right triangles XVW and ZVW  with angles XVW and ZVW =  right angles

Let  XV  = x    and let ZV  =(10-x)

So

WV^2  = XW^2 - XV^2  =  6^2  - x^2       (1)

WV^2  = ( YZ/2)^2 - ZV^2  = 8^2 - (10-x)^2    (2)

Equate (1) and (2)

6^2 - x^2   = 8^2  - (10 -x)^2

36 - x^2  = 64 - x^2 + 20x - 100

72 = 20x

x =72/20  =  18/5

So

WV^2  =  6^2 - (18/5)^2  =  576/25  = r^2

Area  of semi-circle =  (1/2)r^2 *pi =     (1/2)(576/25)pi  =  (288/25)  pi

Let d be the distance between the center of the circle and UV

Let r be the radius of a circle

We can form  a right triangle such that

( UV /2)^2 + d^2   = r^2

(19)^2 + d^2  = r^2     (1)

And let the distance between the center of the  circle and YZ =  d + 6

We can form another right triangle such that

(YZ / 2)^2 + (d + 6)^2  = r^2

(11)^2 + (d + 6)^2  = r^2   (2)

Equate (1) , (2)

19^2 + d^2  = 11^2 + (d + 6)^2

361 + d^2 = 121 + d^2 + 12d + 36

361 = 157 + 12d

204  = 12d

d = 204 / 12 =  102 / 6  =  17

So  using (1)

361 +17^2  = r^2

361 + 289  = r^2

650 = r^2

Let the distance between the center of the circle and WX  = d + 3  = 17 + 3 =   20

And we can from  another right triangle such that

(WX/2)^2 + (d + 3)^2   = r^2

(WX / 2)^2  + 20^2  = 650

(WX / 2)^2  = 650 - 400

(WX / 2)^2  = 250

WX^2 / 4  = 250

WX^2  = 1000

WX = sqrt (1000) = 10sqrt (10)  ≈ 31.62

We can write this as the sum of two series

First series  =   1*3^(-1)  + 1*3^(-3) + 1*3^(-5) + ......

The sum of this series is   [ 1*3^(-1) ] / [ 1 - 3^(-2) ]  =  (1/3) / [ 8/9] = 9/24 = 3/8

Second series = 2*3^(-2) + 2*3^(-4) + 2*3^(-6) + ......

The sum of this series is  [  2*3^(-2) ]  / [ 1 - 3^(-2)] =  (2/9) / (8/9) =  1/4

The sum is   (3/8) + (1/4)  =  [12 + 8 ] / 32  = 20 /32  = 5/8 = . 12 (repeating) base 3