CPhill

A, D and  E  are all the same.....they are just rotations of the same congruent right triangle...

First one :

Look at triangles  A, D  and E

Each  is a right triangle  with  leg legths  of 2 and 3

This insures  that their  hypotenuses are also  the same......so....by  they are congruent by  Leg-Leg  congruency

We only need to solve this

22/9  = (2a + 3)(2/3)  - a^2

22/9  =  (4/3)a + 2  - a^2     rearrange as

a^2  - (4/3)a  + 22/9 - 2  =  0      

a^2  - (4/3)a  + 4/9  =  0      factor as

(a - 2/3)^2  =  0

a  = 2/3

The line  is

y = (13/3)x - 4/9

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

OK, THX....I appreciate  that    !!!!!

We have  that

m + m^2  =  15/4

And

m * m^2  =  a^3

m^3  = a^3

This  implies  that  m  =  a

So

a + a^2  =  15/4

a^2  +  a  =  15/4        complete the square on a

a^2 + a + 1/4  =  15/4  + 1/4

(a + 1/2)^2  =  4     take both roots

a + 1/2  =  ±√4

a + 1/2  =  2             or             a + 1/2   =    - 2

     a  = 3/2             or              a  = - 5 / 2

( -1 - 8) i   +  (5 - - 6) j  =

-9 i  +  11  j

We have the  form

y  =  Asin (Bx )  + C

A   = the amplitude  =  2

C = the vertical shift  =  - 3

B  is  the  hardest to calculate

B  =   2pi  / period

B =   2pi  / 4

B =  pi/2

So....our function is  

y = 2sin[ (pi/2) x ]   -  3

Here's the graph  :  https://www.desmos.com/calculator/h2aglcdrbw

a)  Volume  =  (40 * 20 *  24) cm^3       =  19200 cm^3

b)  1L   =  1000cm^3

So  2L  = 2000 cm^3

So.....we can find how much the water wil rise  by  solving  this

(20 * 40  * h)  =   2000

800 *  h  =  2000     divide both sides by 800

h  =  2000/800  =  2.5 cm  =    height  that the water will rise

The sum   of  the roots  =  (2 - m)

So

a + b =  2m     square  both sides of this

a^2 + 2ab + b^2  = 4m^2      (1)

The product  of  the roots  = -( m + 3)

So

ab  = -(m + 3)

2ab =  -2(m + 3)      (2)

Sub (2)  into (1)

a^2  - 2 (m + 3)  + b^2   =  4m^2

a^2  + b^2  =  4m^2  + 2(m + 3) 

Note  that  the function will be maxiimized  when  (7/3)x + 1/6   =  (3pi/2)

So....the  max  temp   =  -6sin 3(pi/2) - 5  =   -6(-1)  - 5   =    1° F

And  the function will be  minimized whex  (7/3)x + 1/6 =  pi/2

So....the min temp  is   -6sin(pi2) - 5   =  -6(1)  - 5  =   -11° F

So....the temp falls  from 1°F to -11°F