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There are 36 possible seating arrangements.

Here’s a Translation: Too dumb to cheat!

This question is written in standard U.S. English. A similar question (with different countries) is posted below.

I have just looked at you post record.

It seems that so far you have never responded to an answer.

Perhaps this is why it took you five days to get an answer here.

You are expected to respond to your answerers.

I worked this out algebraically

Let the little squares be x*x   and the  big ones are   y*y   so the long ones are   x*y

Ths sum of the areas is 

 \(10x^2+9xy+2y^2\\ \)

I want two numbers that multiply to 10*2=20 and add to  9   Those numbers are  4 and 5

\(10x^2+9xy+2y^2\\ =10x^2+5xy+4xy+2y^2\\ =5x(2x+y)+2y(2x+y)\\ =(5x+2y)(2x+y)\\\)

So one side is    5x+2y   long and the other is   2x+y   long

Now it is quite easy.  If you have problems then assign a number to x and to y.   

I chose 1 and 7 but any numbers that have only 1 as a factor will do.

People ususally find it easier to deal with real numbers instead of with pronumerals

hmm....... i am a little bit sus that this is homework from ur teacher...

Let the center of the  large circle  be   =  (7,R)

Then the  center of   one of the  small circles   is  (11, 11)

So  the square of the distance   between  these two points  is   (R + 3)^2

So

(7-11)^2  + (11  - R)^2  =   (R + 3)^2

4^2  + R^2   -  22R + 121 =   R^2  + 6R  + 9     simplify

137 - 22R  =  6R + 9

137  - 9  =  28R

128  = 28R

128/28 =  R  =  32/7

There is 5 options of where to put the A. There is 4 options of where to put the D, because the A takes up one spot. FOr the remaining 3 slots, there is \(3^3 = 27 \) ways to arange them, so the answer is \(27 \times 20 = \color{red}540 \) ways.

ok, that explains the confusion.  :)

You have not asked a question Manuel.     

Looks like it has been translated a couple of times.  Who knows what the original was...

Not BilderBoi     and     not me either

AGD is 90 because it's given that BGD is a right angle, so the AGD is also a right angle considering the fact that AB is a straight line, which it is.

y=-2x^2+12x-23

a=-2 b=12 c=-23

Use formula -b/2a: -12/-4=3

x=3

Plug in 3 as x: y=-2(3)^2+12(3)-23

y=-5

(3,-5)

Jack made 2 1/3 dozen chocolate cookies,

1 dozen peanut cookies

1 1/2 dozen oatmeal cookies.

1 dozen = 12 cookies

Therfore, 2 1/3 * 12 = 7/3 * 12

                                = 84/3

                                = 28 chocolate cookies

                      1 * 12 = 12 peanut cookies

                1 1/2 * 12 = 3/2 * 12

                                = 36/2

                                = 18 oatmeal cookies

Find the greatest common denominator;

2 | 28  12  18

   |-----------------

     14   6   9

He could fill 2 bags with the same 

number of each cookies in each bag.

Peter saved = 12 one dollar bills

After he paid he left with 1/3 of them

         = 1/3 of 12

         = 4 dollar bills

6. $12 start --> $2 left = 5/6th money

          Alexander spent $10 at the

          mall and has $2 left.
5. Jordan spent $24

      he spent less than 1/2 of his money

         24 / (1/3) = 72    72 - 24 = 48

    Jordan spent $24 at the mall

    and has $48 kept.

7. Justin = $24 left   (1/2) yes 11/12 no 1/12 no

    24 / (1/2) = 48

    Justin spent $24 at the mall

    and has $24 left.

2. Benjamin spent $8 (1/12) yes 11/12 no

         8 / (1/12) = 96     96 - 8 = $88

    Benjamin spent $8 at the mall

    and has $88 left.

4. 11/12

    $60 start

   $60 * 11/12 = $55     60 - 55 = $5

    Zachary spent $55 at the mall

    and has $5 left.

Let Tom have = x strawberries

    Peter have =  y straberries

When Tom gave 1/3

      Tom = x - x/3 = 2x/3

      Peter = y + x/3

When Peter gave 1/4

    Peter = (y + x/3)3/4 = 3y/4 + x/4

      Tom = 2x/3 + (y + x/3) 1/3 = 2x/3 + y/4 + x/12

  When Tom gave 1/5

     Tom = (3x/4 + y/4)4/5 = (3x + y)/5
   Peter = ((3y + x)/4) + (3x/4 + y/4)1/5 => (4y + 2x)/5

     Tom =  (3x + y)/5 = 32

             => (1) --- 3x + y + 160

                         3x + y = 160

                    3(x + 2y = 140)

      3x + y = 160

      3x + 6y = 420

-------------------------

      0 - 5y = -260

             y = 52

Peter = (4y + 2x)/5 = 56

=> 4y + 2x = 280 (2)

     2y + x = 140

3x + (52) = 160

3x = 108

x = 36

The girls can be chosen in \({10 \choose 2}\) ways. The boys can be chosen in \(8 \choose 2\) ways. of the 4 members, there is 4 ways to choose a leader.

So... the answer is \({10 \choose 2} \times {8 \choose 2} \times 4 = 5040\) ways.

What's the problem?

What kind of a questions is this? Whoever wrote this problem should feel ashamed of themselves.