CPhill

Assuming that we can have empty boxes

Let k  = the number of balls

Let n = the number of boxes

The total ways  =

C ( k + n - 1 , n - 1 )    =

C ( 6 + 3 - 1 , 3 - 1 )  =

C ( 8 , 2 )  =

28 ways

EDIT

In the other problem here : https://web2.0calc.com/questions/help-please_8228

Sue's age was found to be 12

So....Sid is 18

OK.....Mr. "SHADOW"

              3x      +  5

x + 4  [   3x^2  + 17 x   + 25  ]

              3x^2  + 12x

             _________________

                          5x   +  25 

                          5x  +  20

                         ___________  

                                     5

So.....     3x + 5    R   5 /  [ x + 4 ]

Second one

             5x   - 1

x - 8    [ 5x^2 -  41x + 3      ]

             5x^2 - 40x

            ______________

                        -1x   + 3

                        -1x   + 8

                      __________

                                -5

So...      5x - 1   R   -5 / [ x - 8 ]

f (f(x) ) =

sqrt [  sqrt (x - 3) - 3  ]

sqrt (x - 3)  - 3     must be  ≥ 0

So...

sqrt ( x - 3) - 3 ≥ 0

sqrt (x - 3) ≥ 3        square both sides

x - 3 ≥ 9

x ≥ 12

So.....12 is the smallest real number in the domain

r [  c1     c2                      c3                         c4     ....                          ]

           r*c1                 r * [  r*c1 + c2]      r [  r * [ r*c1 + c2] + c3 ]        ]

     ____________________________________________________

     c1    r*c1 + c2     r * [ r*c1 + c2] + c3     r [  [ r*c1 + c2] + c3 ] + c4       .....etc

OK....using your formula from last week...we have

1/3 + 1/4  =  1/ Total Time Working Together

4/12  + 3/12 =  1 / T

7 /12  = 1 / T       [ remember that we can write ]

12 / 7  =  T / 1

12/ 7  = T     

T = 12/7 hrs ≈     1.7 hrs  =    1 + .7(60)  =   1 hr 42 min

x - y - z = 0

2x + 3y + z = 17

z + 2y = 7

No set way to do this GM.....but here is an approach....the idea is to get to two equations and two unknowns

Using the 3rd equation.....we can write    z = 7 - 2y

Sub this into the first two equations for z

x - y - [ 7 - 2y ]  =  0     ⇒    x - y - 7 + 2y = 0   ⇒    x + y    = 7     (4)

2x + 3y + [ 7 - 2y ] = 17 ⇒   2x + y  + 7 = 17 ⇒    2x + y = 10      (5)

We can rewrite (4)  as  y = 7 - x

Sub this into (5)

2x + ( 7 - x) = 0

x + 7 = 10

x = 3

So...y =  7 - x =   7 - 3  = 4

And

z = 7 - 2(y)   =  7 - 2(4)  = - 1

So   { x, y, z }      =      { 3, 4 , -1 }

a)  We can make up whatever we want

For the first....a possibility is    ( x - 3)^5 (x - 2)^2

[ We have repeated real zeroes, but the problem doesn't say that every zero has to be unique ]

The graph of this is here :   https://www.desmos.com/calculator/td0qq1sakf

b)  We can have

(x  - 3) ( x + 3) ( x - 4) ( x + 4) (x + 6) ( x^2 + 1)

The graph of this is here : https://www.desmos.com/calculator/0f8x7e1irp

c) We can have

(x^2 + 1) ( x^2 + 3)

Graph : https://www.desmos.com/calculator/sqj7sk0rcf

d) We can have

(x^2 + 1) (x + 3) ( x - 3) ( x + 2) ( x - 2)

Graph : https://www.desmos.com/calculator/nyabj6mmtz

I hope these links work, GM !!!

 (x-4)(x-8) = 12

Expand the left side

x^2 - 12x + 32 =  12        subtract 12 from both sides

x^2 - 12x + 20  =   0

Factor

(x  - 10 ) ( x - 2) = 0

Setting each factor to 0  and solving for x produces

x = 10    or  x  = 2