CPhill

Here's a pic  (not to scale) :

We have  16  possible outcomes  [2 for each coin = 2^4  = 16 ]

The penny and dime  can both show tails...and there will be 2 * 2  = 4 outcomes for the other two coins

or

The penny and dme  can both show heads...and there will be 2 * 2  = 4 outcomes for the other two coins

So....P ( penny and dime have the same results ) =   is  [ 4  + 4  ]  / 16  =  8 / 16   =  1 / 2

g (2x)  =  ³√ [  (2x + 3) / 4 ]    (1)  

2 (g(x) ) =  2 * ³ √  [ (x + 3 ) / 4 ]    (2)

Let  us  cube  (1) , (2)  and set them equal...so we have...

(2x + 3) / 4  =  8 (x + 3) / 4

(2x + 3)  = 8(x + 3)

2x +  3  = 8x + 24

-21  = 6x

x  = -21/6  =  -7/2

This one was a little difficult...!!!

Note : angle AEB  = angle DEC

and angle CDB  = angle DBA

So...by AA congruency....ΔBEA ~ ΔDEC  →  DC /AB  = DE / BE

Area of  ΔDBA  = area of  ΔADE  + area of ΔABE  = 20 + 50  = 70

So  area of Δ DBA  = (1/2)sin DBA(AB * DB)  = 70

So...sin DBA = 140 / ( AB * DB)

And sin (DBA) = sin CDB

So area of  ΔCDB  = (1/2) sin (DBA)(DB * DC)  = (1/2) (140 / [(AB * DB] ) (DB * DC) =

70 (DC/AB)  = 70 ( DE /BE)

Now

Area ΔAEB  = (1/2)sin AEB ( AE * BE) = 50   (1)

Area Δ ADE  = (1/2) sin DEA (AE * DE)  = 20  (2)

But  angle AEB  is  supplemental to angle DEA  so thier sines are equal

So  sin AEB  = sinDEA

Rearranging (1), (2)  we have that

sin AEB  = 100 / [ AE * BE]      =     sin DEA = 40 / [ AE * DE ]

 So

100 / BE   = 40 / DE  →   DE / BE  = 40/100  = 2/5

But area of ΔCDB  was found to be  70 (DE / BE) = 70 (2/5)  =  28

So...area of trapezoid ABCD  = area of ΔDBA  + area of ΔCDB  =

                                                               70        +        28             =      98

I get a slightly different result from Daisy

Reasoning :

Let us just choose any two black balls on the first five draws  [so, by default, the other three will be white ]

First possibility :

B B W W W  =  (5/10) (4/9) (5/8) ( 4/7) (3/6)  = 

(5 * 5 * 4 * 4 * 3)  /( 10 * 9 * 8 * 7 * 6)

Second  possibility : 

B W B W  W  = (5/10) (5/9) (4/8) (4/7) (3/6)  =

( 5 * 5 * 4 * 4 * 3) / (10 * 9 * 8 * 7 * 6)

Third possibilty 

B W W B W = ( 5/10) (5/9)(4/8)(4/7)(3/6) =

( 5 * 5 * 4 * 4 * 3) / (10 *9 * 8 * 7 * 6)

etc......

Out of any 5 positions, the black balls can occur  in any 2 of them...so we have  C(5,2)   = 10 possible arrangements  and each of these has the probabillity  (5 * 5 * 4 * 4 * 3) / (10 * 9 * 8 * 7 * 6) associated with it

So...P( 2 black in 5 drraws)  = P (3 white in 5 draws)   =

10 * ( 5 * 5 * 4 * 4 * 3) / ( 10 * 9 * 8 * 7 * 6)  =   25 /63  = 50 / 126

Daisy is correct !!!!...[ the sum of each row  is also 15...if the missing number is 6  ]

After 10 minutes...the last square will have side lengths of 1 + 10(2) = 21 cm

The area of the last square = 21^2  = 441 cm^2

The area of the first square  is 1^2  = 1 cm^2

The difference in their areas  is   441   -1    =  440

P(all white on first 5 draws)  =

5/11 * 4 / 10 * 3/ 9 * 2/ 8 * 1/7  =

5 / 11  * 2 / 5  * 1/3 * 1/4 * 1/7 =

(2 /11) *  (1/12) * (1/7)  =

2 / 924  =

1 / 462

(5ab)^(2/3)  =

(5^2 * a^2 * b^2) ^(1/3) =

³√ [ 25 a^2 b^2 ]

³√ [ 27 x^6 y^3 ]  =

( 3^3 * x^6 * y^3)^(1/3) =    [use the fact that   (a^m)^n  =  a^(m*n)

(3 * x^2 * y)

5√ [12xm^3 ]  =

5 √12 * √x * √m^3  =

5 * √4 * √3 * √x * √m^3 =

5 * 2 * √3 * √x * m√m  =

5 * 2 * m √ [ 3 * x * m ] =

10m √ [3xm ]

7√28  - 5√63   =

7√[4 * 7] - 5√[9 * 7 ]  =

7 * √4 *√7 -  5 *√9 * √7  =

7 * 2 * √7  - 5 * 3 * √7  =

14√7 - 15√7 =

-√7

2         √7                2√7

__  *   ___   =         ___

√7       √7                  7