CPhill

a)  Basketball only =  (1/5) (200 )  = 40

b)   1/5   = 20%       1/4  = 25%

So   100  -  20   -20   - 25  - 13  =   22%  like Football only

The number who like Football   =   (22 + 13) (200)  =  (35)(200)   = 70

9x^2 + sx  +  4                 

Take the positive root of  9x^2 and  both roots of 4   = 3x  and   +/- 2

(3x  + 2)^2   =

9x^2  + 12x +  4

s = 12

Also

(3x -2)^2  =  

9x^2  - 12x + 4

s =  -12

If we stated with 900 and each contained an equal number at the end, then each box contained 300 at the end

Let a,b,c  be the number in each box to start with

b started with b

And after the transfers

b + 18  - 5  = 300

b + 13  = 300

b = 300 - 13  =  287

c started  with     900 - (a + b)

And after the transfers 

(2/3) [ 900 -(a + b) + 5 ]  = 300

905 - a - b =  450

905 - a - 287  =  450

618 - a = 450

a = 618 - 450  =  168

c =  900 - 168 - 287 = 445

So

a started with 168

b started with  287

c started with  445

Let Ben begin with N sweets

Then Calvin had  4N sweets

And Andy had   840  - N  - 4N  =  840   - 5N sweets

When Andy and Ben each give away (1/2)  of theirs we have

N / 2  +  (840 - 5N)/2 +  4N =   620

N / 2  + 420 - (5/2)N + 4N  =  620

420 + 2N  =  620

2N  = 620  - 420

2N =  200

N = 100 =  the number that Ben had at first

In triangle AIB ,

(1/2)A  + (1/2)B  + angle AIB  =180

(1/2) angleA + (1/2) angleB  + 103  = 180

(1/2)angleA + (1/2) angleB =  180  -103

(1/2) ( angleA + angle B)  = 77       multiply through by 2

angleA + angle B   =  154

Angle C =   180  -  154  =   26°

We're adding 18 + 25 + 16 =   59

The radius =  sqrt (59)

Rearrange as

4x^2 +x^2  -12x + 2x  + k

5x^2  - 10x +  k 

The x coordinate of the  vertex of this  parabola   is     -(-10) / (2 *5)  =  10 / 10  = 1

So...we want

5(1)^2  - 10 (1)  +  k  =  - 1

5 -10 + k  =  -1

-5 + k = -1

k =  -1 + 5   =   4

x^2 -10x + y^2 + 8y =  18       complete the square on x, y

x^2 -10x + 25  + y^2 + 8y + 16    = 18 + 25 + 16

(x - 5)^2   + (y + 4)^2  = 59

The radius  =   sqrt (59)  units

Note that we can write

a(10)^7  + b(10)^6 + c (10)^5  + d (10)^4  + a(10)^3 + b(10^2) + c(10) + d = 

a (10^7 + 10^3)  + b(10^6 + 10^2)  + c(10^5 + 10)  +  d(10^4 + 1)  =

a ( 10^3) (10^4 + 1)  + b (10^2) (10^4 +1)  + c(10)(10^4 + 1) + d (10^4 + 1)  =

(10^4 + 1)  [ a (10^3 )  + b (10^2) + c (10)  +  d  ]

The   gcd =  ( 10^4 + 1 )  = 10001

-t^2  + 8t - 4 - 3t^2       simplify

-4t^2  +  8t  -  4      (1)

The  t that maximizes this  =  -8 / (2 * -4)  =    1

Putting this back into  (1)  the max value is

-4(1)^2  + 8(1)  - 4  =   0