CPhill

Nice, heureka!!!....one question....what is the  intercept theorem???.....I've seen you refer to this several times in the past.....

Let the number of brown counters  = B....so....the number of purple counters is 4B

So....the total number of counters is 5B

So.....

The probability of taking a brown one on the 1st draw is   B / [B + 4B]  = B / 5B  = 1/5

The probability that a brown one is taken on the second draw is  [ B -1]  / [ 5B - 1]

So....the total probability is

[1/5 ]  * [ B - 1 ] / [ 5B - 1]  = 17/ 445

[ B- 1] / [ 5B - 1]  = 17/89   cross multiply

89 [ B - 1 ]  =  17 [ 5B - 1]

89B - 89  = 85B - 17       add/subtract  89, 85B to/from  both sides

4B = 72     divide both sides by 4

B = 18  =  the original number of brown  counters

4B  = 72  =   the original number of purple counters

Expand and we have that

x^2 - (p + q)x + pq  = r^2 - (p+ q)r  + pq

x^2  - (p + q)k - r^2 + (p + q)r   = 0

The sum of the roots  =  (p + q) / 1   = p + q

The product of the roots  is [ -r^2  + (p + q)r] / 1  =  -r^2  + (p + q)r

Since  r  is one solution

Let  s   be the other   and we have

So

r + s  = p + q

s  = p + q  - r

Verify  that

r * s  =    r ( p + q - r)   =  -r^2  + (p + q)r

Let R be the intersection of the medians

DR  =1/3  of DP  = 6

ER  =1/3  of EQ  = 8

So....since the medians are perpendicular....then  DE  is the hypotenuse of right triangle DRE  

So....call S  the midpoint  of DE...so  DS  = 5

So.... SF  is another median

The sine of RDE  = 8/10  = 4/5

cos RDE   = 3/5

So the distance  from  R to S  can be found using the Law of Cosines  as

RS^2  = DS^2 + DR^2 -2(DS * DR)cos(RDE)

RS^2  = 5*2 + 6^2  - 2(30)cosRDE

RS^2  = 61 - 60(3/5)

RS^2  = 61  - 36

RS^2  = 25

RS  = 5

But  RS  = 1/3  of SF....so SF  = 15

We need to find  sin DSR

Using the Law of Sines we have

sinDSR / DR  = sinRDE/ RS

sinDSR/6  = sinRDE / 5

sinDSR / 6   = (4/5) / 5

sinDSR / 6  = 4/25

sinDSR  = 24/25

And the cos DSR   = √ [1  - (24/25)^2 ]  = √ [625 - 576 ] / 25  = √49 / 25 =  7/25

So...we can find DF with the Law of Cosines as

DF^2  = DS^2  + SF^2  - 2(DS*SF)cos(DSR)

DF^2  = 5^2  + 15^2  - 2(5*15)(7/25)

DF^2  = 250  - 2*3*7

DF  = 250 - 42

DF^2  = 208

DF = √208  = 4√13

OK

We need to  find the difference between the x coordinates of A and B= (5 -2)  =3

Since  AC  = 3AB

3 times this difference plus the x  coordinate of A  will give us the x coordinate of C

So    3(5-2)  + 2  =   3*3  + 2  = 11 

Similarly.......3 times the difference between the y coordinates of A  and B  plus the y coordinate of A  will give us the y coordinate of C....so we have

3 ( 9 - 3)  +  3   =  3*6  + 3  =  21

So....the coordinates of C   are   (11, 21)

The  volume  of the 8 steel balls  =

8 * (4/3)pi    =    32/3 pi

So....to find the radius, r, of the large ball, we have

32/3 pi  =  (4/3)pi (r)^3

(32/ 3) (3/4)  = r^3

32 / 4  =  r^3

8  =  r^3

r  = 2

antilog  3.2

We have

log 3.2  = x     ⇒  x  ≈   .505

In exponential form we have

10^.505  =  3.2

b  =10

x  = .505

n  = N  = 3.2

Ln N  =  3

In  exponential form, we have

e^3  =   N

b  =  e

x  = 3

n  = N  

antilog  ≈ 20.09

Since we can use any scale factor for the square that we desire, we can derive an algebraic solution

Let  A  = (0, 0)

B  =  (0, a)

C  =   (a, a)

D  = (a, 0)

E  =  (a/2, a)

F  = (a, a/2)

AB  = a

The slope  of AE  is  (a)  / ( a/2 - 0 )  =  a / (a/2)  =  2

The slope of BF  is  ( a - a/2)   / ( 0 - a)  =  -(1/2)a  / a  =  -1/2 

The equation  of  the line containing segment   AE  is    y  = 2x

The equation  of the line containing segment FB  y = (-1/2)x  + a

Set these equal  to find the x coordinate of G

2x  = (-1/2)x + a

(5/2)x  = a

x  = (2/5)a

And the y coordinate  is y  = 2[(2/5)a]  =  4/5a

So....G  =  (2/5 a, 4/5 a)

And DG  =  √ [ (2/5 a - a)^2  + (4/5 a)^2 ]  = √ [ ( 3/5 a)^2  + (4/5  a)^2 ] =

a√[ 3^2  + 4^2  ] / 5  =   a √25 / 5  = (5 / 5 )  a  = a   =  AB

C  =  [  2 + ( 5 -2)*3  ,  3 + (9 - 3)*3 ]  =  [ 2 + 9, 3 + 18 ]  =  ( 11, 21 )