CPhill

Thanks, Cal   !!!!

Those  are  going to  involve  drawing graphs......this is a little time-consuming....I'll see if I can do this in a little while, because I have other questions to answer.....sorry!!!

In right triangle  ABD, AD  = sqrt [ BD^2  - BA^2] =  sqrt [ 15^2  -  12^2]  =sqrt [ 225 -144] = sqrt [81  = 9

So...its area = (1/2)product of the legs  =(1/2) BA * AD =  (1/2)12 * 9  =  54

And  in right triangle BDC, DC = sqrt [ BC^2  - BD^2] = sqrt  [ 17^2  - 15^2] = sqrt [ 289 - 225] = sqrt [ 64] = 8

So....the  area of triangle  BDC  =(1/2) (8)(15)  = (1/2) (120) = 60

So....the  area of ABCD =   54 + 60 =   114

(c)   f(2x - 8)    just  takes  the points that we  found in (a)  and shifts them to the  RIGHT by 8  units

So

(-2,4)   becomes   (-2 + 8, 4)  =   (6,4)    so  c   = 6

And

(2, -4) becomes  ( 2 + 8, -4)  =   (10, -4)   so d =  10

I know  a couple of these

(a)  f(2x)    makes the  graph 1/2 as wide as the  original

So note  that   the points   (4, -4)  and  (-4, 4)   are on the original graph

So   ( a, 4)   on f(2x)  gives us  ( -2, 4)   so   a =  -2

And  (b, -4)  on f(2x)  gives  us  (2, -4)  so  b =  2

Note  that  DG is just  the  hypotenuse of a right triangle  with legs  DO  and OG

So...its length  =  sqrt [ DO^2  + OG^2 ]  =  sqrt [ 24^2  + 10^2] = sqrt  [576 + 100] = sqrt [ 676] = 26 

1.2n - 4.4 < 5.2   add 4.4 to both sides

1.2n <  9.6     divide  both sides  by  1.2

n <  8

So.....the positive integers  hat satisfy this are

1,2,3,4,5,6,7

And  their sum =   (7)(8) / 2 =  56 / 2   =  28

Note  that  the  altitude of the  right  triangle on the left =  sqrt (5^2  -1^2)=  sqrt (25 -1)  = sqrt (24)

And  ?   is the hypotenuse of the right  triangle o the right  with  legs of sqrt (24)  and 5

So

? =   sqrt [ 5^2  + (sqrt (24))^2 ]   =  sqrt  [ 25  + 24 ] = sqrt [49]   = 7

Right  triangle  ACD  is a  5 -12 -13  Pythagorean Triple

So AD  = 12

Since AB  = 15.....then   BD  =sqrt  [ AB^2 - AD^2] = sqrt ( 15^2 - 12^2 ] = sqrt  [225 - 144] = sqrt [81] = 9

So....the  area  of triangle  ADB  = (1/2)(BD)(AD)  = (1/2)(9)(12)  = 54    (1)

And  the area of triangle ACD = (1/2)(5)(12) = 30    (2)

So the area of triangle ACB = (1)  - (2)  =  54  - 30  =  24

\(\frac{1}{3}<{\frac{x}{5}}<{\frac{5}{8}} \)

Take  this by parts

1/3  <  x/5                              cross-multiply           x/5   <   5/8

5 < 3x    divide  both sides  by 3                           8x  <  5 * 5

5/3 < x                                                                    8x  <   25       divide both sides  by 8                                                            

  x > 5/3                                                                    x < 25/8

So the solution is

5/3  <  x  <  25/8