CPhill

If   8 +  sqrt (3)  is a root, then so is its  conjugate,  8  -  sqrt (3)

Y

           28

24           

X                Z

XZ  =  sqrt   [  YZ^2  -  XY^2  ] =  sqrt [ 28^2  - 24^2  ]  =  sqrt (208)  =  sqrt (16 * 13)  =   4sqrt (13)

tan Y    =   XZ /  XY =    4sqrt(13) / 24   =      sqrt (13)  /  6

Angle ACB  =  180  -  60  - 45   =  75

And if AT  bisects  CAB  ....then angles CAT  and  BAT  =  30

Using  the Law of Sines

CT  =  24 sin (30°) / sin (75°)  =    12/sin (75°)   = 12csc(75°)

TB  =  24 sin (30°)  / sin (45°)  =   12 / sin (45°)   =  12csc (45°)

So    CB   =   CT  +  TB   =     12   ( csc (75°)  +  csc (45°) )    ≈  29.39   

Mmmmm.....haven't done one like this before

A  graph  seems the  best  way to go, here :  https://www.desmos.com/calculator/w1pkainlsv

The  solution point   is  (x , y)   =   (8 , -2)

p(q(x))  means we  are putting  q  into  p ....so....

(2x^3  + 1)^2   =    4x^6  +  4x^3   + 1

Adding   this to r(x)   gives  us

7x^4  + 4x^6  - x^3   - 6x  - 1

If a, b , c  are consecutive  integers   a  =  2    c  =  4    and b   = 3

So....the  area  =   (1/2) (a + c) (b)  = (1/2)(6)(3)    =   9  units^2

1. A teacher arranged her books on a shelf. She has 4 astronomy books, 3 Biology books, and 2 Chemistry books. How many ways can she arrange these books on the shelf if the books from each subject must be together?

The  number of ways  =    arrangements in each topic * arrangements of topics =   4!  * 3!  * 2!  *  3!  = 

1728  arrangements

2. How many ways can David, Sam, Emma, and Lanie stand in a line if David and Sam refuse to stand next to each other?

Number of   ways  without restrictions  =  4!  =   24

Number  of ways  if David and Sam DO  stand next to each other   =  3 places they can occupy * 2 arrangements in each place   =   6

So.....number of  ways  if  they refuse to stand next to each other  =  24  -  6   =   18 ways

(2x  + 1/2)^6

We  are looking  for   the  third  term

This  is  given by

C(6, 2)  (2x)^4  (1/2)^2  =

15 *  ( 16x^4)  ( 1/4)  =

15 *  4x^4  =

60 x^4

If BAC  =  23

Then   QAD  =  90 - 23   =  67

And QDA  =  23

So

QDA  +  QAD  =    23  +  67   =   90  (degrees)

Note that if DE = 3EC

Then    3EC  + EC  =  20

4EC =  20

EC =  20 / 4    =  5

So  DE =  3 * 5  =  15

Note that riangles DFE  and  BFA  are similar 

And  DE / BA   =    15/20  = 3/4

Which also means  that    DF / BF  =   3/4

So    BF / BD    =  BF / (BF + FD) =  4  / (4 + 3)  =     4/7

And  triangle  FGB is similar to triangle DCB

So

FG / DC     =     BF / BD

FG /  20  =   (4/7)

FG  =  20 (4/7)   =   80/7