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Okay I will try to explain this as best as I can. 

If the members were not allowed to hold two positions, the answer would be 25*24*23.

This is because there are 25 people who can run for the first spot, and 24 people who can run for the second, because

the first spot would be taken, and there will only be 24 people availible for the second. Same goes with the 23. (I hope I am not making this confusing.) 

But because we can't do that, we have to try something else. If a member can hold two spots, the number of spots will be 

25*25*24. This accomendates for two spots and allows a member to have two positions. 25*25*24 is equal to 15,000

So there are 15,000 combinations.

This problem is pretty simple if we think about it in a different way.

We see the we have 1 - 2 + 3 - 4 +... + 2005 - 2006 + 2007 - 2008 + 2009 - 2010.

It looks hard, but all we have to see are the -1.

1-2  is equal to -1, 3-4 is equal to -1, and so on.

Now, all we have to do is figure out how many -1 s we have.

the biggest number is 2010, and if we divide that by 2, (because every two numbers is a -1) we get 1005.

From this, we can conclude that the answer is  -1*1005=-1005

If EF is equal to FG, you know that the triangle is an isosceles. If BEF is equal to 130 degrees, EFG is also 130. You know that the whole triangle has 180 degrees, and the big angle is 130. 

180-130=50

50/2=25

so x is equal to 25

Hope its right and im not bad at math :)

The width of the rectangles is 50% its length. Given that the perimeter of 3 such rectangles is 90 cm, find the length of 1 rectangle.

Let l be the length

Let w be the width

w=1/2 l (1)

2(l+w) * 3 = 90 (2)

Simplifying (2): 2l+2w=30

l+w=15 (3)

(1) & (2) : 3/2 l = 15

1/2 l = 5

L = 10

Length is 10.

Enjoy you ungrateful guest.

Honestly, I think @mathisopandcool's solution is nice and clear enough. @guest, it is not his/her fault if YOU can't understand a thing. Why are you being rude?

did you ever get an answer

There are a maximum of 240 intersection points.

The domain of G(x) has width 4.

Consider the geometric series 4 + 20/a + 100/a^2 + .... If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?

Using the geometric series formula, we have: 4/(1 - 5/a)

Simplifying we have: 4a/(a-5)

4a/(a-5) = k^2

4a = ak^2 - 5k^2

4a/k^2 = a-5

k^2 will always stay positive, so a-5 = 1 (least square)

a=6

Let  the   number of blue  gems  =  B

Let  the number of   green gems  = 2B

Let  the  number of red  gems = B + 13

So

5B  + 1*2B  + 20 ( B + 13)  =  2150   simplify

27B + 260 =  2150

27B  = 2150 - 260

27B  = 1890

B =  1890 / 27  =  70   blue gems

2*70  =  140  green gems

70 + 13  =  83  red gems

Evaluate the infinite geometric series: 1 + 2/7 + 4/49 + 8/343 + ...

Sum: (first term)/(1 - rate of increase)

First term = 1

Rate of increase = 2/7

1/(1-2/7)

1/(5/7)

7/5

Leonhard should end up with 40 dollars.

Add the equations  and we have

(a + b + c )x  + (a + b + c)y  + (a + b + c) =  6x + 6y +  6

Equating terms....(a + b + c)   =   6

The ball will hit the ground in 0.32 seconds.

Let  the  nested roots = a

So  a =  8

Square  both sides  and  we  get  that

x +  a    = 64

x +   8   =  64

x = 64  - 8  

x = 56

The answer is 8.

The drone flew a distance of 2 + sqrt(2).

The common difference between  terms = 6

So

(2006 - 2000) * 6  =

4 * 6   =

24

x + 2y  =  4     

xy = -8

Square both sides of the  the first equation

x^2  + 4xy + 4y^2  =    16

x^2  + 4(-8)  + 4y^2  =  16

x^2  + 4y^2  -32  =  16

x^2 + 4y^2  =  16 + 32

x^2  + 4y^2  =  48

AP = 8/3.

OK, viola....

We have  this system

a + c  =  23

a + d  =  20

b + e =  24

e + c =   21

b + d  = 22       add  these equations together

2 ( a + b + c + d + e)   =  20 + 21 + 22 + 23 + 24

2 ( a + b + c + d + e)  =   110

a + b + c + d + e  =  110 / 2

a + b + c + d + e   =  55 

TYSM! YOUR ALL CORRECT! 

Well, I just figured it out myself o_o
thanks anyways cphill!

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18  such  numbers