CPhill

2

Because BY is an  angle bisector

BC  / CY  =  AB / AY

BC / 8  =  12   / 10

BC  =  8 * 12   /  10      =  9.6

*****

Surface area  =   b^2  + 4 * ( b^2 /3)

432 = (7/3)b^2

b^2 = 432 ( 3/7)  = 1296 / 7

b = sqrt (1296/ 7)  = 36 /sqrt 7

To find the  slant  height, s, of the pyramid

area of face = 1/2 b * s

b^2 / 3 =  (1/2) 36 / sqrt 7 *  s

1296 / (7*3)  = 18 / sqrt 7 * s

s = 1296 / 21 * sqrt 7 / 18 =  24 / sqrt 7

To find the height, h, of the pyramid by the Pythagorean Theorem

h  =  sqrt  ( s^2  - (b/2)^2 )

h =  sqrt [ (24/sqrt 7)^2  - (18/sqrt 7)^2 ]

h =   6

Volume of pyramid = 

(1/3)  base area * height  =

(1/3) (b^2) * h

(1/3) (1296  / 7) * 6  =  2592 / 7 units^3   ≈ 370.29  units^3  

****

  Avg of nth row of Pascal's Triangle =  2^n  / ( n + 1)

So

2^n / (n + 1)   = 2

2^n  = 2(n + 1)

n = 3

                        1                            Avg  = 1

                    1      1                        Avg =  1

                  1     2     1                    Avg  = 4/3

                1   3     3     1                 Avg  =  2

n =  3

5005

5115

5225

5335

5445

5555

5665

5775

5885

5995

B - 50 - (B - 50)(1/3)   =  (3/7)B

B -  50  - (1/3)B+ 50/3   = (3/7)B

(2/3)B -100/3  = (3/7)B

(2/3 - 3/7)B   = 100/3

(5/21)B = 100/3

B  =  (100/3) ( 21/5)   = (100 / 5) ( 21 / 3)  =    20 * 7  =    140  = total that he made

D + J + K   = 720     ⇒    J + K =  720  -  D      (1)

Derrick used .20D and had  .80D left

John had  J - 80 left

Kenneth had  K - .40D   left

.8D + J - 80 + K - .4D   = 448    simplify

.4D + J + K  =  448 + 80        ,...... sub in (1)

.4D + (720 -D)   =  528

-.6D + =  528 - 720

-.6D  = -192

D =  -192  /( -.6D)   =  320   = number Derrick had at first

Call the total time = T

First room =  (1/5)T

Second room = (T - 1/5 T) (1/3)   =  (4/5)T (1/3)  =  (4/15)T

Third room =  8

So

(1/5)T + (4/15)T + 8  = T

(7/15)T + 8 = T

8 = T - (7/15)T

8 = (8/15)T

T = 8 ( 15/8)  =   15 hours

Call the original number of sweets = S

Call the sweets he gave to his friends on Thursday = T

(S - T) = sweets left after Thursday   

Sweets given on Thursday = (3/5) of what was left

T = (3/5)(S-T)    simplify

(8/5)T = (3/5)S

8T = 3S

T = (3/8)S        (1)

Sweets  given on Friday  = 130

Sweets given over 2 days = 4 more the (9/16) of what he started with

T + 130 =  (9/16)S + 4

T + 126 = (9/16)S           (2)

Sub (1) into (2)

(3/8)S + 126 = (9/16)S

126 = (9/16 - 3/8)S

126 = (9/16 - 6/16)S

126 =  (3/16)S

S = 126 * 16/3 =   672  = number of sweets Edi had at first

Sweets given after 2 days  = 672 - [ (9/16)(672) + 4 ]  =  (7/16)(672) - 4  =  290

Sweets left after 2 days  = 672 - 290  =  382