CPhill

(x + 2) ( x - 5) < 0

This will be true whenever    x  = (-2, 5)

f (x)  =  1/x + x/[ 1 + 1/x ] =   1/x + x / ( [x + 1]  /x )  =  1/x + x^2 / (x + 1)

f(-2)  =  -1/2  + (-2)^2 / (-1)   =  -4 -1/2 = -9/2

f (f (-2)) =  f (-9/2)  =  1/ (-9/2)  + (-9/2)^2 / ( -9/2 + 1)  = 

-2/9  +  ( 81/4) /( -7/2)  =

-2/9  -  (81/4) (2/7)   =

-2/9  - 162 / 28   =

-2/9  - 81/14  =

-757 / 126

2x^2 - 7x + 2  = 0

The sum of the roots =  7/2

The product of the roots  =  2/2  = 1

1/ (a - 1)  + 1/ (b -1)  =

(a - 1) + (b -1)

___________  =

(a - 1) (b -1)

(a + b) - 2

____________   =

ab - (a + b) + 1

(7/2) - 2

____________   =

1 - (7/2) + 1

  (3/2)

______  =      -1

 - (3/2)

(x - 30) = 2 ( x  - 49)

x -30  = 2x - 98

x =  98 -  30

x = 68

Each had 68 at first

He can score any number  ending in 0 (starting with 10)  with an even number (starting at 2)  of  "5" flips 

He can score any number ending in 1  (starting with 11) with an  odd number of "5" flips and one "6" flip

He can score  any number ending with 2 (starting with 12)  with an two "6" flips  and any even number  (starting at 0) of "5"  flips 

He can score any number ending in 3  (starting wih 23) with an 3 flips of "6" and an odd number  of "5" flips   (He cannot make  13)

He can score any number ending in 4 (starting with 24)  =  four "6" flips  and an even number (starting at 0) of "5" flips   (he cannot make 14)

He can score any number ending in 5  with  any  number of "5" flips starting at 1 

He can score any number ending in 6   (starting with 6) with one "6" flip and an even number (starting at 0) of "5" flips 

He can score any  number ending in 7 (starting at 17) with  two "6" flips and  an odd number of "5" flips 

He can score any number ending in 8  (starting at 18)  with 3 flips of 6 and any number of even flips (starting at 0) of "5" 

He can score any number ending in 9 (starting with 29) with 4 flips of "6" and any odd number of "5" flips

(He cannot make 19)

So....

The greatest integer that is  impossible for him to flip  is  19

3 (24/2)  + 2 (24/3)  =

36  + 18  =

54 cords

(6a - 2b)^2  + (3a + 2b)^2  = 90

No integers possible for a, b

tan 27°  =  34 / JK

JK    =  34 / tan (27°)

Area = (1/2) product of the  legs = (1/2) (JK) (KL)  =     (1/2) ( 34 /tan (27°) ) ( 34)   ≈  1134.4  unts^2

Linus finishes  at  9:50

So   Jenny must run for  50 minutes  =  5/6 hr

So

D  =  R * T

10 km =  R  * ( 5/6) hr

10 km /  (5/6) hr = R   = 12 km /  hr  = rate Jenny must run

10 ^(2n)  < 25000000  take the log of  both sides

log 10^(2n)  < log ( 25000000)

(2n)  < log (25000000)  / log 10                    (log 10 =  1)

2n < log (25000000)

n<  log (25000000) / 2   ≈   3.698   →    n <  ≈ 3.698

Similarly we would have

log (25000000) < 10^( 2n + 1)

[ log (25000000) - 1] / 2  < n  →     n >   ≈  3.19

The integer value of n =  3