CPhill

Call the angle A

Note  that   tan A  =  opposite side / adjacent side  =   10/14

So...to  find A  take the tan inverse   (arctan)

arctan (10/14)   ≈  35.53°  =   36°

Let n be the number of months when they will have the same amount

So  we have

145  +  10n  = 20  + 15n        subtract  10n , 20  from both sides

125  =  5n           divide both sides by 5

25 =  n

They will have the same amount  after 25 months

The total  number  of dominoes in this set    =  (18 + 1) (18 + 2)  / 2  =  19 * 20 / 2  =  190

There will 19  dominoes  that  are "doubles"  [  0,0    1,1   ......  18,18  ] 

So....the probability of selecting a double   is  just

19 / 190   =   

1 / 10

8 x 8 x 1  =   64  ft^3

6 x 4 x 1  =   24 ft^3

So....we have this proportion

48 / 64 =  x / 24        where x is the number of bags we need

3 / 4  =  x  /24             multiply   both sides by 24

18  =  x

We need 18 bags

Connect AD

Then triangle ADB  is right  with AB = 2 ,  AD  =1  and   BD  = √3

And in  triangle AEB, angle  AEB  = 120° 

And using the law of sines

EB / sin 30  = AB /sin 120

EB /(1/2)  = 2 / (√3/2)

2EB  = 4/√3

EB  = 2/√3  =  (2/3)√3

So   DB  - EB  =  DE  =  √3 - (2/3)√3  =  √3/3

And by symmetry DE = CE

And angle DEC  =120°

So the area of triangle  DEC =  (1/2) DE^2 * sin (120°)  =  (1/2) (√3/3)^2 * (√3/2)  =  √3/12    (1)

Next....let F  be  the  midpoint  of AB  

Connect   DF and CF

Angle DFC  =  60°

DF = CF  = 1

So the area  of sector DFC  = (1/2) (DF) (CF) * pi/3  = (1/2) (1)^2 * (pi/3)  = pi/6    (2)

And the area  of triangle DFC  = (1/2)(1) sin (60°)  = √3/4    (3)

So....the green  area =   (1) + [ (2) - (3) ]  =

√3/12  +  [ pi/6  - √3/4 ]   = 

pi/6  + √3/12  - 3√3/12  =

2pi /12 + √3 /12  - 3√3/12  =

[ 2pi   - 2√3 ]             pi  - √3

__________  =     _________ ≈   .235  units^2

      12                          6

See the following graph  :   https://www.desmos.com/calculator/gh92tovzqd

Let the number of children's tickets sold  = x

Let the number of adult tickets sold  = y

Five possibilities  are

(x , y )  =  

(0, 350)

(30, 320)

(52, 298)

(75, 275)

(300, 50)

P (orange on first draw  and  purple  on second)  =  (8/12 ) * (4/11)  = (2/3) * ( 4/11)  = 8 / 33

Here's my best attempt

We  have   C (20, 3)   = 1140  possible different sets of 3 integers

(a)  the sum will be even when   we have

even + even + even        or

even + odd +  odd

In the first case  we want to choose  any  3  of the 10  even numbers  =  C (10, 3)  = 120

In the second  case we want to  choose  any 1 of 10 evens  = C (10, 1)  and  any 2 of the 10 odds = C(10,2) = 45 

So  the probability  is

[  C (10,3)  +  C(10,1) * C(10,2) ]               120  + 10 * 45            570

____________________________  =      _____________  =   ______  =   .5

         C (20, 3)                                                1140                      1140

(b)  the only way we can have an odd product is when all three  integers are odd.....there  are C (10,3)  = 120 sets of these

So.....the probability that we have an even product =

1   -   120  / 1140  =

[ 1140  -120 ]  / 1140  =

1020 / 1140  = 

17  /19

No prob....glad to help   !!!!

By Vieta's Theorem

The sum of  the roots =  -b / -2

So

(3 + sqrt 5) + (3 -sqrt 5)  =  -b/-2

6  = -b / -2       multiply both sides by  -2

-b=  -12

b =12

Also  the product of the  roots  =   c / -2

So

(3 + sqrt 5) (3 - sqrt 5)  =  c / -2

9  - 5  = c / -2

4 =  c  /-2     multiply both sides  by -2

-8 = c

The parabola is

y=  -2x^2  + 12x  - 8

The x coordinate of the  vertex =  -b /[2a]  =     -12/ [ 2 * -2]  = -12 / -4  =  3

And the y coordinate of the  vertex is   -2(3)^2  + 12 (3)  - 8   =    -18 + 36 - 8 =  10

So....the vertex  is   ( 3, 10)