CPhill

h (h (t) )   means   that we are putting the function itself  in for t

So we have

h(h(t)) =

-2 ( -2t - 1)  - 1   =

4t + 2  - 1     =

4t + 1

We have 3 equations that give us possible values for  x

!st....assume  that  these two sides are equal

3x + 62 = 7x + 30

   62 - 30  = 7x - 3x

       32  = 4x

         x =  8                        and the perimeter  =  3(8)+ 62 + 7(8)+ 30 + 5(8) + 60    = 272

Next assume that these two sides are equal

3x + 62  = 5x + 60

62 - 60  = 5x - 3x

2  = 2x

x = 1                               The perimeter  =  3(1) + 62  + 7(1)  + 30  + 5(1)  + 60 =   167

Finally, assume  that

7x + 30 = 5x + 60

7 - 5x  = 60 -30

2x = 30

x = 15....    it's obvious that this  value of x will lead to the greatest perimeter

So....the perimeter  is the least  when  3x + 62 = 5x + 60  and the perimeter  = 167

x^2  + kx -10x + 36 + 12x  - 8     combine like terms

x^2 + (12 - 10 + k)x + 28

x^2  + ( 2 + k)x  + 28

If this can be written as the square  of a binomial  the discrminant  =  0

So

(2 + k)^2  - 4 (1) (28)   = 0

(2 + k)^2  =  112

2 + k  =  ±√112   = ± 4√ 7

k =  ± 4√7   - 2

The sum of all possible k's =   (4√ 7  - 2)  + ( -4 √ 7 -2 )  =  -4 

Let the number of textbooks = T

And let the number of workbooks = W

If the price of a workbook  was $6, then the price of a textbook  = $6 ( 1 + 3/5)  = $6 ( 8/5)  = $9.60

35% of the books sold were textbooks so 65% were workbooks

Let the total number of books sold =  S

So.....the textbooks = .35S    and the  workbooks  =  .65S

Putting all of this together......

Number of workbooks sold * cost each - Number of textbooks sold * cost each  =  $108

          .65S(6)   -  .35S ( 9.6) =  108         simplify

           3.90S -  3.36S  = 108

               .54S  = 108

                  108 / .54 =  S =   200  books total

True

f(x)  when  x = 3   = f(3)

Here's the same problem solved by heureka with only PQ = 7 instead of 6

https://web2.0calc.com/questions/pls-help-i-don-t-know-how-to-do-this

Let E be the midpoint of chord  CD

Then OE is  perpendicular to  DC

And OED  forms a right triangle with DE = 15 , OE  = 12

And by the Pythagorean Theorem, OD is a radius =   sqrt  ( DE^2  + OE^2 )  = sqrt ( 15^2 + 12^2 )  =

sqrt  [ 225 + 144]   =   sqrt  (369)  ≈ 19.21

See here, JM

Your intuition was correct

Let p = 1

Note that triangles DEC and ABC are similar

And triangle DEF is just a reflection of triangle DEC

And the scale of   DEC : ABC  =  1/2

So.....the scale of DEF : ABC   =  1/2

i.e......each successive triangle  will be similar  with a scale factor of 1/2 of its predecessor 

Each successive triangle  will  have  sides of 6p , 3p , 1.5p    and so on     (each will have 1/2 the side length of its predecessor )

And the succesive  heights follow the same pattern   (3sqrt 3)p , ( 1.5 sqrt 3) p , ( .75 sqrt 3)p  .....etc. 

The center of the circle will be on the  y axis.....let's call this point (0 , a)

The distance from this point to  (2,4)   and to (0,7)  will be the radius

So  we  can equate distances   (using the square of the distance formula on  both sides of the  equation )

( 0 - 0 )^2  +  ( 7 -a)^2  =  ( 2- 0)^2  + ( 4 - a)^2          simplify

     a2  - 14a  + 49   =  4  + a^2  - 8a  + 16 

         -14a +  49  =  -8a + 20

           49 -20  =  14a - 8a

               29  =   6a

                a = 29/6 

So the center is ( 0, 29/6)  and the radius =  7 - 29/6   =   42/6 - 29/6 =   13/6  

The radius  =   7 - 29/6  =   42/6 - 29/6 =  13/6

\(f(\sqrt{x + 1}) = \frac{1}{x + 1}\)

Note that   :     f (  sqrt (3 + 1) )  = f ( sqrt ( 4) )  =    f (2)

So

                        1                1

f(2)  =            _____  =   ____ 

                      3  + 1           4