CPhill

Let your the current age of your son =  N

Then your current age =  9N

And in one year we have this relationship

9N + 1  =   6 ( N + 1)        simplify

9N + 1  = 6N + 6

9N - 6N  =  6 - 1

3N  = 5

N =  5/3    =   1 + 2/3  yrs old  =   1yr  and  8 months old =  your son's age now

Cars and trucks are   ( 1 + 6) / (1 + 6 + 4)  =  7/11 of the total number, N, of  vevicles

So...

(7/11) N  =  42

N =   42 ( 11/7)  =  66   vehicles in total

So....there are   66 - 42  =  24 motorcycles

t^2 - 9t - 36  + t^2 - t + 24       simplify

2t^2 -10t -12

We have the form    at^2  + bt  + c

The  value of  t  that produces the smallest value  is  - (b)/ (2a)  =        - (-10) / (2 *2)  =  10  /4  =   2.5

x^2 - 15 < 2x + 7x - 40      rearrange as

x^2 - 9x + 25  <    0      (1)

Notice the graph here : https://www.desmos.com/calculator/5gdvkizm7m

The graph is always   above the x axis  (always > 0 )  so no x values make (1)  true.

Ignoring the denominators (since they are the same)

1  = -a^3

-1  = a^3       take the cube root of both sides

a =  -1

5N/  ( N + 7)   =  -8

5N  =  -8 (N + 7)

5N  = -8N  - 56

13N  =  -56

N =  -56 / 13

(3x + 12) (x + 4)  =  (4x^2 + 8x  + 8x^2) / (2x)               assuming that  x   is not 0

3x^2  +24x + 48  =  (4x) ( x + 2 + 2x) / ( 2x)

3x^2 + 24x + 48  = 2 ( 3x + 2)

3x^2 + 24x + 48  = 6x  + 4

3x^2 + 18x + 44  = 0

Using the Q formula

              -18  ± √ (18^2  - 4 * 3 * 44)       -18  ± √ ( -204)         -18  ± 2√51 i

x =         ______________________   = ____________  =  ____________ =   -3  ± √51 i / 3

                         2 * 3                                        6                             6

Intersecting Chord Theorem :

AQ  * BQ = CQ * DQ

6  * 15  = CQ * DQ

Let CQ = x    and DQ = 38- x....so we have

6 * 15 = x  (38 - x)

90 =  38x - x^2            rearrange as

x^2 - 38x  + 90  =  0     complete the square on x

x^2  - 38x    =   -90

x^2 - 38x + 361  =  -90 + 361

(x - 19)^2   = 271          take the negative root

x - 19 = -sqrt (271)

x =  -sqrt (271) + 19 ≈  2.54 =  minimum length of CQ

By the tangent - secant theorem

XY^2 =  AX * XB

9^2  =  16 * XB

81  = 16 * XB

81 / 16  = XB

So

BA =  AX - XB 

BA  = 16 - 81/16

BA =  (256 - 81) / 16

BA =  175 / 16

F = freshmen at first     T = total students at first

Initially

F / T  = .40   ⇒     F =  .40T      (1)

After 15 more Freshmen arrive  we have that

[F + 15] / [ T + 15 ] =  .60       (2)

Sub (1) into (2)

[ .40T  + 15 ] / [ T + 15]  =  .60        multiply both sides by  T + 15

.40 T + 15  =  .60 [ T + 15 ]

.40T  + 15 =  .60T  + 9

15 - 9  =  .60T - .40T

6   = .20T

6 / .20  = T  =   30

So....after the 15 arrive.....there are now    30 + 15 =  45  students at the dance