CPhill

Here's the second one

Angle AEB= Angle DEC

Angle ABE = Angle DCE 

So.....triangle AEB is similar to Triangle DEC

So 

DC / AB = 12/18 = 2/3

So

DE /  AE =  2 / 3   ⇒    AE = (3/2)DE

Which implies that

DE / AE  =  EC / EB

2 / 3  =   EC / EB

EC = (2/3)EB

EB + DE = 19

EC + AE = 21

EB + DE =  19

(2/3)EB  + (3/2)DE = 21

EB + DE  = 19

-EB - (9/4)DE = -63/2

-(5/4)DE = -25/2

DE = (25/2) (4/5) = 10

So

AE = (3/2)DE =  15

And

EB + DE = 19

EB + 10 = 19

EB = 9

So

EC = (2/3)EB

EC = (2/3)(9) = 6

And

[ BEC] = (1/2)(EB) (EC) sin BEC  =  (1/2) (9)(6) sin BEC

[ AED ] = (1/2)(DE) (AE) sinAED = (1/2) (10)( 15) sin AED

And sin BEC = sin AED

So

[BEC ] / [ AED] =   (9)(6) / [ 10 * 15]  =   54 / 150  =   9 /  25

(-3) + 8 =    5

(0) + 8   =    8

(5) + 8 =   13

LOL!!!

AB =  BC = CD = DA                                                 given

BD = BD                                                                    reflexive property

Triangle ADB congruent to Triangle CDB                 SSS congruency

Assuming that some boxes could remain empty.....

For the first one....let the number of balls = k   and the number of boxes = n

So...we have

C( k + n - 1 , n - 1)   = C( 6 + 3 - 1 , 3 - 1)  = C (8, 2)  = 28 ways

For the second one.....we have the following partitions of 5 indistinguishable balls in 3 indistinguishable  boxes

5, 0, 0

4, 1, 0

3, 2, 0

3, 1, 1

2, 2, 1   =  5 ways

K. Cannot be determined

We need at least 3 pieces of info to solve a triangle

Only knowing two side lengths isn't enough

C. gross income - (required deductions + optional deductions)

We have the form

y = Asin (Bx)  + D

A = the amplitude =  6

B =  2pi / period

The period = 4

So

B =  2pi / 4  =    pi/2

D = the shift up/down.....since the resting position is y = 0....we have no shift

The equation for the displacement from the resting position is

y =    6 sin [  (pi/2) t  ]

Here is the graph :  https://www.desmos.com/calculator/jn8ipp0b2g

x = y^4      (1)      and     x + y^2 = 1      (2)

Sub   (1)  into (2)   and we have that

y^4 + y^2 = 1         

Let  y^2  =  a      and we have

a^2 + a   =   1        complete the square on a

a^2 + a + 1/4   =  1 + 1/4       factor the left....simplify the right

(a + 1/2)^2  =  5/4           take both roots

a + 1/2   =   ±√5  /  2       subtract   1/2 from both sides

a =   ( - 1 ±√5 ) / 2

Since  y^2  must be positive, then  a  must be   ( √ 5 - 1 ) / 2    =  y^2      (3)

Using   x + y^2 = 1

x =  1 - y^2

x = 1 -  [  ( √ 5 - 1 ) / 2   ]  =    ( 3 - √ 5 ) / 2

These curves intersect at the same x values

So...the distance between the y values is the distance between the intersection points 

So.....using (3)

y =  √  [ ( √ 5 - 1 ) / 2 ]       or y =  - √ [ ( √ 5 - 1 ) / 2 ] 

So... the distance between these points is just

√  [ ( √ 5 - 1 ) / 2 ]   -  [  - √ [ ( √ 5 - 1 ) / 2 ]  ]  =

2 √  [ ( √ 5 - 1 ) / 2 ]   =

√ [ 4 ( √5 - 1 ) / 2 ]  =

√ [ 2√ 5 - 2 ]  =

√ [ - 2 + 2√ 5 ]   ⇒  (u, v)   =  (-2, 2)

No prob....!!!