CPhill

3^(2c + 1)  = 28*3^c

3^(2c) * 3^1  =  28*3^c

(3^c)^2 * 3  = 28 *3^c      divide  both sids by  3^c

3^c * 3  =  28

3^c  =  28/3        take the log of each side

log 3^c =  log (28/3)

c * log 3  =  log (28/3)

c =  log (28/3)  / log (3)

4x + 5 = 19

4x  = 19 - 5

4x   = 14

So

[14 + 17 ]  /  [ 14 + 1 ] = 

31  /  15          { as pogfrog found  !!!!.....thx, pogfrog  !!!!  }

(10 *8 * 6 * 4 *2)  / (9 *7 *5 * 3 *1)  =

10/5 * 6/3  * 2/1 *  ( 8/9) ( 4/7)  =

2^3  *  32 / 63  =

256 / 63

Let f(x) =  (x - 1)^2  - 2

Then

f ( 2x - 3)  =  [ (2x - 3) - 1]^2  - 2   =

[ 2x - 4 ]^2  - 2  =

[ 2 ( x - 2)]^2  - 2  =

4 ( x - 2)^2 - 2

The vertex is    ( 2 , -2)

See here :  https://www.desmos.com/calculator/yehdvvbzxe

Assuming without replacement   the probability is

6  / 14  *    5 / 13     =    3/7 *5/13   =   15 / 91

Welcome aboard   !!!!

Thx, Supersonic Man 12   !!!!!

That's a great answer !!!!

Let the total games    =  12 + 2 + 1 + 3m  =    15 + 3m      where m is the number of weeks after the fifth one

And the number of wins  =  2 + 2m

So  we wish that

[2 + 2m ]  /  [ 15 + 3m ]  =     .50

2 + 2m =   .50  [ 15 + 3m ] 

2 + 2m =  7.5  + 1.5m

2m - 1.5m  = 7.5  - 2

.5m    =  5.5

m =  5.5  / .5   =    11 weeks  after the 5th one

So....you actually  need to play 16 weeks

See below......!!!!

x = -1     and  x =  -5 are roots so

y =  a ( x +1) ( x + 5)          and (0,5)  is a  point on the graph......so.....

5 =  a ( 0+ 1) ( 0 + 5)

5  =  a (5)

a = 1

So we have

y = ( x + 1) ( x + 5)

y = x^2  + 6x  +  5

100(1)  + 10(6)  +  5  =    165