CPhill

Let J be the number that Uncle Joe started with

And we have that

J ( .50) (.50) =  14

J * .25  = 14

J =  14 /  .25  =    56

Proof

When  Peter ate 1/2 of these 56.... 28 remain

When Jack  ate 1/2 of these 28....  14 remain

[  ( x + 2y)  -  z)^8

The constant will be from  the  expansion of

C( 8 , 2) [(x + 2y)  ]^6 * z^2 =

C(8,2) * C( 6,3) (x^3) (2y)^3 * z^2 =

28 * 20 * 2^3 * x^3 * y^3 * z^2  = 

28 * 20 * 8 * x^3 y^3 z^2  =

4480   x^3 y^3 z^2

See here :   https://web2.0calc.com/questions/function_67

   Elements  Containing                                     Number of subsets                              

    9 ,1    and  1 other possible element                2  =   sum of row 1  of Pascal's Triangle

    8, 2   and  2 other possible elements               4  =   sum of row 2

    7, 3  and 3 other possible elements                 8  =   sum of  row 3

    6, 4 and  4 other possible elements               16  =   sum of row 4 

                                                                           __

                                                                           30

Each set contains  two specific elements  and the the number of total  number of subsets with these = 

( 2) ^(possible  number of other elements) 

\( y = $\frac{1}{2}x^2$ - 9 \)

Call the point we are looking  for (c , c^2/2 - 9 )

Using the square of the  distance  formula  we have

D^2  =  ( c - 0)^2  + ( c/^2 / 2  - 9  - 0  )^2

D^2  = c^2  + c^4/4 - 9c^2 + 81

D^2  =  c^4 /4 - 8c^2 +  81           take the derivative

D^2 '   =   c^3 -16c            set to 0

c^3 - 16c  = 0

c ( c^2 - 16)  = 0

The second factor  set  = 0   is what we want

c^2  = 16

Two vaues of c  will  minimize the distance c = 4 or c  =-4......choose  c = 4

And y  = 4^2 / 2 - 9  =    -1

The point that minimizes the distance is  ( 4 , -1) 

a^2  =  4^2 + (-1)^2  =    17

x + sqrt (x - 2)  =  4

sqrt (x - 2)  = 4  - x             square both sides

x - 2 =  x^2 - 8x + 16         simplify as

x^2 - 9x +  18 =   0

(x - 3) ( x - 6)  =  0

x - 3 =  0                          x - 6 =  0

x =  3                               x =  6       (reject this solution)

x = 3  is the solution

3x^2 -24x + 55 - x^2 - 49      simplify as

2x^2 - 24x  -  6

2 ( x^2 - 12x - 3)       complete the square on  x inside the parentheses

2 ( x^2 - 12x + 36 - 3 - 36)

2 (x - 6)^2    + 2 (-39)

2 (x - 6)^2  -  78

a +  d +  e     =     2 - 6  - 78  =    -82

4 galllons of the current  mixture is  ammonia  (80% of 5)

We need 3 gallons of ammonia at the  end

Let x be the number of gallons drained off and.8x the amount of  ammonia drained off

(4 - .8x) = .6 *5

4 - .8x  = 3

4 - 3  =  .8x

1 = .8x

1/.8  = x = 1.25

We need to drian off  1.25 gal

Proof

4 gallons of  ammmonia to start with

When we drain away 1.25  gallons, 80%  of this is ammonia.....so   4 - .8(1.25)  = 3 gallons of ammonia are left

Adding back 1.25 gallons  of  pure water  = 5 gallons  and 3 of  this is  ammonia...so....3/5=  .60  =  60% 

And 1.25 gallons =     5 qts

f ( x + 1)  = f ( 1 + 1)  = f(2)

So x = 1

So

f(2)  =  1/1    = 1

When x changes by  3, y changes by   (13 - a)

From  4 to 13   x changes by 9  so  y changes by  3 (13 - a)=   39 - 3a

And  notice that

39 - 3a =  b   rearrange   as

3a + b =  39