CPhill

6  [     1 / [ 3^2 - 1 ]   +   1 / [ 5^2 - 1]   +  1/[7^2  - 1] +   1/ [ 9^2 - 1 ] +....... ]   =

6 [     1 /  8    +   1 / 24    +     1/48  +  1/80  +   ..... ]    =

6  [   1 / [ 2*4]   +  1/ [ 4*6]  +  1/[ 6*8] + 1/ [8*10] +  ....... ]   =

6   [ (1/4) ]  [   1 / [1 * 2]   +  1/ [2 *3]   + 1/ [ 3*4]  + 1 / [ 4*5] +   .....]

Note that            1/ [ n * (n + 1) ]   =         1 / n   -  1/[ n + 1]

So we can write

(6/4)   [   ( 1/1 - 1/2) + (1/2 - 1/3)  + (1/3 - 1/4) + (1/4 - 1/5)  + .......] =

(3/2)  [   1   + ( 1/2 -1/2) + (1/3 - 1/3)   + (1/4 - 1/4)  + ........ +  (-1/n) ]  = 

(3/2) [ 1  -   1/n ]         as n ⇒  infinity,   1/n   ⇒   0

So we have

(3/2 )  [ 1 - 0]  =

(3/2) (1)   =

3/2

Consider...if the side of the square is 7.3 m....the area   = 7.3^2  = 53.29 m^2

So......this is equal to the area of one of the circular tables

So.....

53.29  = pi * r^2

53.29/pi =  r^2       take the square root of both sides

r ≈  4.1 m

I'm wondering if this  was meant in the original problem instead of  a circumference of 4.1  m  ???

Circumference  =  2 pi * r

4.1  =  2 pi * r       divide both sides by 2pi

4.1 / [ 2pi]  = r

So....the area of one table is just the area of a circle  =  pi  r^2  =   pi *  [  4.1 / [2pi]  ]^2  =

 [ 4.1 / 2] ^2 / pi    m^2

And this is equal to the area of the square

So....the side of this square is just the square root of this  =  [ 4.1/2] / pi^(1/2) ≈  1.16 m

4) With a tailwind, a small plane traveled 420 miles in one hour and 30 min. The return trip took 30 min longer. Find the wind speed and airspeed of the plane...

This one is similar to (3)

1hr 30 min  = 1.5 hrs.....so

(R + W) * 1.5  =  420     (1)

(R-W )  *2  =   420    ⇒ 2R - 2W  = 420   ⇒  R - W  = 210  ⇒  R  = 210 + W   (2)

Sub (2) into (1)   for R  and we have that

(210 + W + W) * 1.5   = 420

(210 + 2W) * 1.5  =  420

315 + 3W  =  420     subtract   315 from each side

3W  =  105        divide both sides by 3

W  = 35   mph

And using (1) ,  R  =  210 + 35  =  245 mph

3) With a headwind, a plane traveled 840 miles northward in 2 hours. With the same wind as a tail wind, the return trip took 1 hour and 45 mineutes. Find the plane's airspeed and the wind speed.

Note....1hr 45 min  =  1.75 hrs

Let the speed of the plane in still air be S

Let the wind speed  be W

Against the wind the plane's rate is  (R - W)  ....the wind slows the plane's normal speed

With the wind the plane's rate is (R + W).....the wind increases the plane's normal speed

So.....  Rate * Time  =  Distance   and we have that

(R -W) * 2  =  840   ⇒  2R - 2W  =  840  ⇒   R - W  = 420  ⇒   R  = 420 + W    (1)

(R + W) *1.75  = 840  ⇒  1.75 R + 1.75W  = 840    (2)

Sub (1)  into (2)   for R  and we have that

1.75(420 + W) + 1.75W  = 840

735 + 1.75W + 1.75W  = 840

735 + 3.5W  = 840       subtract 735 from both sides

3.5W  =  105       divide both sides by  3.5

W  = 30  mph = wind speed

And using (1) the plane's speed  is  420 + 30  =  450 mph

THX, HiylinLink.......!!!

2. How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable but the boxes are?

Again....assuming that we can have empty boxes.....the number of ways of distributing k  identical balls into n  distinguishable boxes is given by :

C ( k +n - 1 , n - 1)   =  C (6 +3 - 1, 3 - 1 )  = C (8,2)  =  28 ways

1. How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable and neither are the boxes?

Assuming that  we can have empty boxes.....this boils down to the number of ways that we partition  6  identical balls into 3 identical boxes

These are

6, 0, 0

5, 1, 0

4, 2, 0

4, 1, 1

3, 3, 0

3, 2, 1

2, 2, 2

So......7 ways

Note that quadrilaterals EAHD and  BFCG  must  be  3/4 of the area of the square

But the area of  triangle EAH and  triangle FCG  are 1/2 the area of each of the above quadrilaterals

So.......their combined area  must be   (1/2) (3/4)  =  3/8  units^2

So [ FEGH]  +[ EAH ]  +[ FCG]   = [FACH]   =  1/4  + 3/8   =   2/8 + 3/8  =   5/8 units^2