CPhill

By the exteriror angle theorem, 3y = 2y + x  ⇒ 3y - 2y  = x ⇒  y  = x

And in triangle PQR, since PQ  = RQ, then angle PQR  = angle PRQ  = x

So

3y  + x  +  x =  180

And since  y = x

3x + x + x  = 180

5x  = 180

x = 36

So 3x  = 3y  =  3(36)   = 108°

Not only Happy Birthday.....but....I might add that hectictar is a real asset to our site!!!

Her answers  are always clear and concise....

Since the line  has the form  y  = -1x   + b....the slope is -1

This means OP  = OQ  and QR  = SR

If    [QRS ] / [QOP ]  = 9/25.....the scale factor is the square root of this =  3/5  

This means that    QR : QO = 3 : 5

So...there are 8 equal parts to OR  and OQ  is 5 of them

So...the x coordinate of Q = the y coordinate of P  since OP  = OQ

And this coordinate is   ( 5 / [ 3 + 5 ] ) * 4 =  5/8 * 4  =  20/8  = 5/2 = 2.5

So...b = the y intercept P  = 5/2  =  2.5

Proof : [ QRS]  = (1/2)(QR) (SR)  = (1/2) (1.5)(1.5)  = (1/2)(2.25)

[QOP] =  (1/2) (OP)(OQ)  = (1/2)(2.5)(2.5)  = (1/2)(6.25)

[ QRS] / [ QOP ] =   2.25 / 6.25  =    9 / 25

See the graph, here :  https://www.desmos.com/calculator/m1ufgopozx

See...my  corrected answer under your orginal post, Lightning...

Call the value  before Monday's loss, V

After Monday...the stock is worth .90V

After Tuesday, the stock is worth  .80 (.90V)  = .72V

So....it loses  1  - .72  =    28% of its value

There are  12 * 60   =  720  possible minutes displayed

From 10 - 2....  4hrs * 60 mintes per hour   = 240  minutes display  a "1"

Every hour between 2 and 10.....the following minutes display at least one "1"

10 , 11, 12, 13, 14, 15, 16 ,17 , 18, 19

01, 21, 31, 41, 51

So...that's  15 occurrences per hour  * 8 hrs  =  120 additional minutes

So...the total number of minutes displaying  at least one "1"  = 240 + 120  = 360

So

360 / 720    = 1/2    !!!

DC  is parallel to AB

And FG  = 3  and AB  = 6

Therefore.....FG splits the sides of triangle AEB  in a 1: 1 ratio

This also  means that the altitude of triangle EFG  = BC  = 3

So...the altitude of triangle AEB is twice this = 6

So...the area of triangle AEB  =  1/2 * AB *  2 BC =   1/2 * 6 * 6  =  18 units^2

We can use symmetry to solve this

The area  of  the 60 degree sector on the right is given by  pi *12^2 * 1/6  = 144/6 pi =

24 pi  units^2

The area  of the  triangle on the right =  (1/2)(12^2)sqrt (3)/ 2 =  36sqrt3  units^2

So...half of the shaded area =  the differences in these areas

So....the whole shaded area  = twice this  =

2 [ 24 pi - 36sqrt 3 ]  =  [48pi - 72sqrt3 units^2] ≈  26.09 units^2

Call O the center of the circle.....draw radius OC perpendicular  to AB, and connect OB  and OA

In triangle OBC...angle OBC  = 30°  and angle BOC  = 60°

So....BC  = 6√3  cm

And in triangle OAC  = 15“  and angle COA  = 75°

Using the law of sines

CA / sin 75  = OC / sin 15

CA /  sin (30 + 45)  = 6 / sin (45 -30)

CA  =  6 sin (30 + 45)  / sin (45 - 30)

CA = 6 [ sin 30cos 45 + sin 45cos30 ] / [ sin 45cos 30 - sin 30 cos 45 ]

CA = 6  [ 1/2 * √2/2  + √2/2 * √3/2 ]  / [ √2/2 * √3 / 2 * 1/2 * √2/2 ]

CA = 6 [ √2/4 + √6/4 ] / [ √6 / 4  - √2/4]

CA = 6 [ √6 + √2 ] / [√6  - √2]

CA  = 6 [ 6 + 2√12 + 2 ] / [6 - 2]

CA = 6 [8 + 4√3 ] / [ 4]

CA = (3/2) [ 8 + 4√3]  =  3 [ 4 + 2√3]  = 12 + 6√3  cm

So....AB = BC + CA  =  6√3  + 12 + 6√3  =   12 + 12√3  cm 

(2/3)a  - 1/6  = 1/3

(2/3)a  =  1/3 + 1/6

(2/3)a  = 2/6 + 1/6

(2/3)a = 3/6

(2/3)a  =  1/2  multiply both sides by 3/2

a  = (3/2) (1/2)   =  3/4