CPhill

Correct-A-Mundo, CTG   !!!!!

A "Best Answer "....

THX, Zekken.....I did not see the other posting

BUT...we all  agree

81 !!!!!!!!!!

I think  that  you  must mean that  the AREA  =  5103 m^2   [  perimeter is a linear measure ]

So

Area  = W * L

5103  = W * 63  divide both sides by 63

81 (m) = W

a₁  = 3(1)  + 1    =  4

a₂  = 3(2) + 1  =    7 =     a₁  + 3

Note that we are adding 3 to every previous term to get the  next term

So

a₁  = 4     and the rule  is   a_n  =  a_n-1 + 3

Impressive work guys    !!!!

Call the roots a, b

The  sum of  the  roots  =   -4/3

So

a + b = -4/3      square both sides  of this

a^2 + 2ab + b^2  = 16/9     (1)

The product of the roots =  12/3  = 4

So

ab = 4    

And

2ab  =8      (2)

Sub (2)  into (1)  and we have that

a^2  + 8  + b^2  =  16/9

a^2 + b^2  =  16/9  - 8

a^2  + b^2 =  16/9 - 72/9

a^2  + b^2   =  [ 16 - 72 ] / 9

a^2 + b^2 =  -56 /9

See  the following image

We can find angle BAC  with the Law of Cosines

We have

BC^2  = AB^2 + AC^2  - 2 ( AB * AC)  cos(BAC)

9^2  = 5^2  + 8^2  - 2 (5*8)cos(BAC)

[ 9^2  - 5^2 - 8^2]

______________  = cos BAC =   1/10

[ -2 * 5 * 8 ]

So using the Law of Cosines, again we  have  that

DE  = √[ AE^2 + AD^2  - 2 (AE * AD) cos (BAC)]

DE  =   √[ 10^2  + 16^2  - 2 (10*16)(1/10)  ] 

DE  = √ [ 100 + 256  - 32]

DE = √324  =  18

1. How many ways are there to put  balls into  boxes, given that the balls are not distinguished but the boxes are?

This  one is easier  than the first one.....again, assuming no restrictions  [ we can have emplty boxes] 

The  "formula"  is

C ( k + n - 1 , n - 1)     where k = the number of balls  and  n  =  number of boxes

1. How many ways are there to put 4 balls into 3 boxes, given that the balls can all be distinguished but the boxes are not distinguished? (Thus, for example, putting all the balls in the first box is counted as the same outcome as putting all the balls in the second box.)

Balls  distiguishable   Boxes not

Let  k be the  number of  balls  and  n  be  the  number of  boxes

Assuming that we  can have empty boxes......this is a little  hard to calculate,  EW

I'm not too sure about this....but....I believe that it  uses something  called "Stirling Numbers of the Second Kind "

We  have  this

S(4,1)  + S(4,2)  + S(4,3)

  1  +   6   +     7  =      14 ways  

Note  that if  his  average  was  85  he must  have  85 * 5   = 425  total points

So  his score on the  fifth test  was

425 - ( 81 + 93 + 76 + 78)    =   97