CPhill

Thanks, EP.....I hadn't thought about that method   !!!!

Never done one like this before.....but....I believe that the rotation of this will produce two cones

The  volume of an octahedron is given by

V  = √2/3  * edge length^3

243* 3 / √2  =  edge length^3

729/√2  = edge length ^3

9 / (2)^(1/6)  = edge length 

The distance from any vertex to its non-adjacent vertex  = √2 edge  = 9√2 / {2)^(1/6)  = 9∛2

The radius of each cone will be (1/2) of this  =  4.5∛2

And this will also be the height of each cone

So....the volume of the 3D figure will be  the volume of the two cones =

2 [ (1/3) pi * radius^2 * height ]  =

(2pi)  ( 4.5∛2)^2 (4.5∛2) / 3  =

2pi  [ 4.5]^3 * 2 / 3  =

(4pi] * [91.125] / 3  =

4pi (243/8)  =

(243/ 2) pi units^3

We have orthogonal vectors when their dot products  = 0

So

(2, 1) dot (-3, 6)  =  -6 +6  =  0    orthogonal

(2, 1) dot (  1, 2 )  = 2 + 2   = 4    not  orthogonal

(2, 1) dot ( -2, - 3)  = -4 + -3  =  -7     not orthogonal

(2, 1) dot ( 1, - 2)  = 2 + -2  =  0       orthogonal

[ I'm assuming that your are familiar with dot products ....]

Here's your answer with the Shoelace Theorem

(-3, 2)       (6 , - 2)        (3, 5)

 (x1, y1)    (x2, y2)      (x3, y3)

The area is given by :

abs [ x1y2 + x2y3 + x3y1  - x1y3 - x2y1 - x3y2 ]

_________________________________________   =

                              2

abs [ (-3*-2) + (6*5) + (3*2) - (-3^5) - (6*2) - (3 * -2) ]

_____________________________________________   =

                             2

abs [ 6  + 30 + 6 - (-15)  - 12 - (-6) ]

_______________________________

                       2

abs [ 63 - 12 ] / 2  

51 / 2  =

25.5 units^2

Thanks for turning me on to this  !!!!.....I want to look at it a little more...I believe that it just involves a vector cross-product ......

I'm not familiar with this.....I checked my answer with WolframAlpha and it syncs up....

I see  that the Shoelace Theorem is on the internet....I'll look at it, and....if I can....I'll give you the solution using it..

The rule for divisibility by 8 is that, if the last three digits are divisible by 8, then the whole number is divisible by 8

The total number of outcomes is  6^8

Notice that  the smallest that the last three digits can be  =  111

And the largest that they can be  = 666

The number of integers between these two numbers that are divisible by 8  =

floor [ (666 - 111) / 8 ]  = 69

So....the number of outcomes that are divisible by 8  =  (6)^5 * 69

So the probability  =   (6)^5 * 69  /  (6)^8   =   69 / 6^3   =   69 / 216   =  23 / 72

We can use something known as "Pick's Theorem" here

The area is given by :

Number of boundary lattice points on the triangle/ 2   + number of interior lattice points   -  1

Where a lattice point is one with integer coordinates 

Number of boundary lattice points  =  5  

Number of interior lattice points = 24

So

Total area  =    5/2 + 24 - 1  =   5/2 + 23  =   2.5 + 23  =

25.5  units^2

Here's the pic :

2.There are 100 students in South High School. South High School offers only Chinese and Spanish. There are 8 more students in Chinese than in Spanish, and every student takes at least one language. If 37 students take only Spanish, then how many take both languages?

These always give me some trouble....!!!!

Let N be the number that take both

37 + N   =  number who take Spanish

37 + N + 8  = 45 + N  = number who take Chinese

So

Number ( Spanish + Chinese)  = Number(Spanish) + Number (Chinese)  - Number (Both)    = 100

(37 + N)  +  (45 + N )  - N   =  100 

37 + 45 + N = 100

82 + N  = 100

N = 18  =  number that take both  !!!!

So

37 take only Spahish , 18 take both and 45 take only Chinese  = 100

5.In the Gregorian calendar, every year which is divisible by  is a leap year, except for years which are divisible by 100; those years are only leap years if they're divisible by 400.(This may seem complicated, but the calendar is carefully designed to keep the average number of days per year very close to the number of days in one complete orbit of the Earth.)Assuming we keep using the Gregorian calendar, how many leap years will there be between 2001 and 2999?

We have   998 years

floor [ 998/4 ]  = 249  possible leap years

But  the years 2200, 2300, 2500, 2600, 2700 and 2900 do not qualify because they are not evenly divisible by 400

So

249 - 6  = 

243 leap years 

4.There are 40 animals at a petting zoo, all of which are goats or rabbits, and each animal is either an adult or a child. There are twice as many adult rabbits as adult goats, and three times as many child goats as adult goats. If there are an equal amount of adult rabbits as child rabbits, how many adult goats are there at the petting zoo?

AR = Adult rabbits

AG = Adult goats

CR = Child rabbits

CG = Child goats

2AG = AR            

3AG = CG            

AR = CR               

AR + AG + CR + CG  = 40        substitute

2AG  + AG + AR+ CG  = 40

2AG + AG + 2AG + 3AG  = 40

8AG  =  40         

AG = 5