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Hi, I’m not really sure how you got the answer. Can you further explain it and/or your techniques to get the answer?

Let's get an equation for the line....we have that

y  = -2(x - 3) + 5         simplify

y = -2x + 6 + 5

y  =  -2x + 11        (1)

We can let  (3, 5)  be the center of a circle   with a radius  of  AB

So  we have the circular  equation

(x  - 3)*2  + ( y - 5)^2  =  (6√5)^2      sub (1)  into (2)  for  y   and we have that

(x - 3)^2  + ( -2x + 11 -  5)^2  =   180      simplify

(x -3)^2  + ( 6 - 2x)^2  = 180

x^2 - 6x + 9  + 4x^2 - 24x + 36   = 180

5x^2 - 30x + 45  = 180

5x^2 -30x - 135  =  0         divide through by 5

x^2  - 6x  - 27   =   0      factor

( x - 9) ( x + 3)  = 0

Set  both factors to 0 and solve for x and we have that

x =  9        and   x   =   -3

Using the equation of the line

y  = -2(9 -3) + 5              y  = -2(-3 - 3) + 5

y  = -12 + 5                    y  =  12 + 5

y  =  - 7                           y  = 17

So....the two possible endpoints for  B  are   ( -3, 17)   and  ( 9, -7)

Here's a graph   :    https://www.desmos.com/calculator/viyvhipurq

EP is correct

We have 2 * 2 * 2  =   2^3  =  8 possibilities

Only  two of these  are either all boys or all girls

So

[ 8 - 2  ] / 8   =

6 / 8  =

3 / 4

....do they taste like chicken? 

I believe this has been answered before.....but I'll try anew:

1=boy  0=girl

possibities

111

110

101

011

100

010

001

000      6 out of 8 include at least 1 boy 1 girl    6/8 = 3/4

Second one

The incenter  is the point where the angle  bisectors meet

Therefore  BE  is an angle bisector.....so we have the relationship that

AB / AE  =  BC  / EC

12 / 9  =  BC  / 12    cross-multiply

12 * 12  =  BC * 9

12 * 12 / 9  = BC

144 / 9  =  BC   =   16

The inradius  is the height of three smaller triangles with the sum of their bases = the perimeter of the large triangle

So.....the area of the large triangle is

(1/2) (perimeter) ( 1)  =

(1/2) (12) * 1  =

6 units^2

Then why don't you try working it out, instead of trying to cheat on your homework?

Let  A  = (0, 0)

Let B  =  (0, 3)

Let C  =  (4, 0)

We can find the incenter  thusly  :

[ Length of side opposite A * xcoordinate of A + Length of side opposite B * x coordinate of B + Length of side opposite C*xcoordinate of C ] ,  [ Length of side opposite A * ycoordinate of A + Length of side opposite B * y coordinate of B + Length of side opposite C*ycoordinate of C ]

 ( [(5(0)  + 0 (4) + 4(3)] / 12  , [5(0)  + 3(4)  + 0(3) ] / 12  )  =

(1 , 1 )

In a right triangle.....the circumcenter is located at the midpoint of the hypotenuse....this point is

(2, 3/2)

So....the distance between the incenter and the circumcenter is

sqrt  [ ( 2 - 1)^2  + (3/2 - 1)^2  ]  =

sqrt [ 1*2  +  (1/2)^2  ] =

sqrt [ 1 + 1/4  ]  =

sqrt  [ 5/4]   =   sqrt  (5)  / 2

Don't you think that you should interact with answering people, when the answer is wrong? What is it that you want? Can you clarify it? Is it ONLY the sum of 10 digits 1234567890 rearranged? Can you repeat ANY digit such as 1122345678??. Clarify your question PLEASE !! Thank you.

We can  take this absolute difference

  l  13 - -3  l  = l 16  l   =  16

I do.  My first cat was named "Bosco the Chocolate Bear". 

Are you the Taylor Swift?  I like her too. 

_.

I understand now, thank youuu 

Me   !!!!!

(Dogs, too )

Given  that the  equation through AB  has  the equation 5y + 2x  = 10

Then A  has the coordinates  (0, 2)

And B has the coordinates  (5, 0)

And the slope  of the line through  AB =  [ 2 -0 ] / [ 0 - 5 ]  = -2/5

So....the slope of the line through AD  is   5/2

So....the equation of the line through  AD is

y = (5/2) x +  2

2y - 5x  = 4

Since D lies on the x axis, then we have that

2(0) - 5x  = 4

x = -4/5

So....the distance  AD  =  sqrt [ (0 - - 4/5)^2  + (2 - 0)^2 ] = sqrt [16/25 + 4]  = sqrt [116] / 5

And the distance AB  = sqrt [ 5^2 + 2^2 ]  =  sqrt [ 29]

So.....the area of the rectangle  = AD * AB  =

sqrt (116 ) sqrt (29)   / 5  =  sqrt (3364)  / 5  =   58/5  units^2  =  11.6 units^2

see no one has anything to say anymore. 

oh k

I would guess that it prevents spam

ok thanks

when you post a question or answer there should be an image icon in the toolbar, click that then upload image choose the file and click I'm not a robot and it should appear.

why do you have to enter a more detailed answer if what you post is too short

EP's solution/approach are perfectly valid.....here's one more way....

Let  M  be the center  of  circle ADB

Draw DB

The area between chord DB  and the arc DB  is given by

[ pi/4  - 1/2 ]    (1)

Using symmetry......the shaded area  =

Area of circle containing arc DFE  - 4 times  (1)  =

pi  - 4 [ pi/4  - 1/2 ]  =

pi  - pi  +  2  =

2 units^2

i tryed to post an image but it didnt work???????!!!!!!!!

Well we are not doing it for you even if you cant

Someone usually does.