CPhill

Let a  =  1

So

( x + 3) ( x - 8)    =  0

x^2 -5x - 24  =  0

b  = -5

[P.S. -   the value of "a" = 1  is  arbitrary.......other values of  b are possible  with other choices of "a"  ]

\($\frac14 < \frac{x}{5} < \frac23$\)

1/4 <  x /  5        and       x / 5 <  2/3

5/4 <  x                          x <  10/3

So.....the possible integer  values for x  are   2 , 3

See BuilderBoi's  excellent answer,  here :  https://web2.0calc.com/questions/quadratics_57

The altitude of this triangle  drawn from  the right angle  to AB  =  6 + 6sqrt (2)

We have this relationship

(AB/ 2) /  (6 + 6sqrt (2))  =    (6 + 6sqrt (2))  / (AB  / 2 )      simplify

(AB/2)^2  =   (6 + 6sqrt (2) )^2

AB^2  / 4   =  ( 6 + 6sqrt (2) )^2

AB^2  =  4 ( 6 + 6sqrt (2))^2      take the square root of both sides

AB  = 2 ( 6 + 6sqrt (2)) =   12 ( 1 + sqrt (2) )

We have      (75 - 3)   / 3  +   1 =    72/3  + 1 =  24 + 1  =    25  terms

Sum  =  

[ first  term  + last term ]   *  ( number of terms)   /  2   =

[ 75 + 3 ]  *  ( 25 / 2)  =

 ( [ 78 ]   / 2 )  *  25   = 

39 * 25   =

975

Let q be the number of weeks after the 5th week

We  have  this  equation

total wins                        [ 2  +  2q ]

_________     =             __________  =    .50    

total games                    [ 15  + 3q ]

Where n  is the  number of weeks  after  the  5th week

Multiply  both sides  by   15 + 3q  and we have that

2 + 2q   =   .50 [15 + 3q ]  

2 + 2q  =  7.5  + 1.5q      subtract  1.5n, 2  from both sides

.5q = 5.5      divide  both sides  by  .5

q =  11

So..it will take   5 + q =  n =   5 + 11 =   16 weeks to reach a  50%  winning percentage

Proof:

After  n =  16  weeks  we  have   2 + 2(11)  =   24 wins

And  15 + 3(11)  =  48  games will have been played

So

24  / 48    =   .50

Let a and b be the roots of the quadratic 2x^2 - 8x + 7 = x^2 - 5x + 3.  Compute a^3*b + a*b^3.

Note that

a^3b + ab^3  =  ab ( a^2 + b^2)

Rearrange  equation  as     x^2 - 3x  + 4  = 0

By Vieta

Sum of roots  = -(-3) /1  = 3     and product of roots  = 4/1 =  4

So

a + b  = 3     square both sides                            and             ab  = 4    so    2ab =  8

a^2  + 2ab + b^2 =  9

a^2 + b^2   =   9 - 2ab

a^2 + b^2  =  9  - 2(8) =    -7

So

a^3b + ab^3  =  ab ( a^2 + b^2) =   4 ( -7) =    -28

\( $10^n = 10^{-5}\times \sqrt{\frac{10^{13}}{0.01}}$\)

10^n  = 10^(-5)  *    sqrt ( 10^13 / (10^-2) )

10^n  = 10^(-5)  * sqrt (10^15) 

10^n   = 10^(-5)  *  (10^15)^(1/2)

10^n = 10^(-5)  *  10^(7.5)

10^n  =  10^(-5 + 7.5)

10^n  = 10^(2.5)

n = 2.5

(2x)  / ( 3x^2  - 6x  - 14)

We have a lower  power polynomial  over a higher power polynomial......this sets up a horizontal asymptote at y = 0

The vertical  asymptotes  will be where    3x^2 - 6x - 14  = 0

By Vieta   the sum of the asymptotes  =   the sum of the roots of this poynomial  =   a  + b  =    - (-6) / 3   = 2

So

a +  b +  c   =   0 +  2   = 2

W =  W

L = 5W + 7

LW  =  area

W ( 5W + 7)  =  16

5W^2 + 7W   = 16

5W^2 + 7W - 16  = 0

                -7   +  sqrt  [ 7^2  - 4*5* -16 ]              -7  +  sqrt [ 369]               

W =     ___________________________ =   ________________ ≈ 1.22 in 

                           2(5)                                                 10