CPhill

x + y = 0

y = -x

Sub this into the first equation

-x = x^2

x^2 + x = 0

x ( x + 1) =  0

Set both factors to 0  and  solve for  x

x = 0    and x = -1

When x = 0  y = 0

When x = -1 , y = 1

(0,0) and (-1, 1)

Distance between  these  =  sqrt [ (-1)^2 + 1^2 ]  = sqrt (2)

AB =  [ 16  14   18 ]   *  C [ 0  ]   =    [ -10  ] 

            21  29   13              -2            -45

                                            1

A = [4 3 ]   *   BC  =  [ 5  ]    =     [ -10 ]     

       1 5                     -10               -45 

Each exterior angle   of the octagon  =   360 / 8  =  45°

We can constuct   2 isosceles right triangles....one at the  bottom right and  one  at the  bottom left

The one at the  bottom right   has a hypotenuse of 4

So  the legs are  4/sqrt (2)

The one at the bottom left has a hypotenuse of  2

So the legs are 2/sqrt (2) 

And by symmetry, we can constuct the  same triangles at the  top left = the one at the  bottom right and at the top right = the one at the bottom left

This  forms a rectangle with sides of   1 + 4/sqrt 2  + 2/sqrt/ 2 = 1 + 6/sqrt 2 

And 3 + 4sqrt (2) + 2/sqrt 2 =  3 + 6/sqrt 2

The area of this  rectangle  =  (1 + 6/sqrt 2) ( 3 + 6/sqrt 2) =  3 + 24/sqrt 2 + 18  = 21 + 24/sqrt 2 = 21 + 12sqrt (2)

From this we  must  subtract the areas of the  4 isosceles  triangles

These are equal  to   (4/sqrt2)^2  + (2/sqrt 2)^2 =  8 + 2 = 10

So....the total area of the  octagon =  21 + 12sqrt (2) - 10  =   11 + 12sqrt (2)

Just  add the like elements of   A and B

I'll  get you started

[ -5    6 ] 

 -10   2

  __ __

  __ __

See if you can  finish it.....     

No real number makes the  denominator = 0   since x^2 + 4 is always positive

So....the domain is  all real numbers

The triangle will be isosceles  with two equal legs

So

L^2 + L^2  =  6^2

2L^2 = 36

L^2  = 36/2

L^2   =18

Area =  (1/2)  (product of the  legs)  = (1/2) (L * L ) =  (1/2) (L^2) =  (1/2)(18)  = 9 in^2

Sub the first equation   into the  second  and we  have

y^4 + y^2 = 1 - y^4

2y^4 + y^2  - 1  =   0       factor this

(2y^2 -1) (y^2 + 1) = 0

The second factor has no real  solution......set the first factor  = 0

So

2y^2 - 1  =  0 

2y^2  = 1

y^2 =  1/2

y = 1/sqrt (2)     or  y = -1/sqrt (2)

And x =  y^4  =   (1 / sqrt (2))^4 =  1/4

The distance between  these points  is  just     1/sqrt (2)  -  -1/sqrt (2)  = 2/sqrt (2) =  sqrt (2) =  0 + 1sqrt (2)

(u,v)  = (0, 1)

LOL!!!!......can't argue with that !!!!

Inradius = [ ABC ] / [semiperimeter of ABC]

ABC is a right triagle so  the area =  (1/2)(product of the legs)  = (1/2) (5 * 12) =  30

Semi-perimeter  =  [ 5 + 12 + 13 ] / 2  =  15

Inradius =  (30) / ( 15)  = 2

Because ID is parallel to AC, then AE = AC - ID = 12 - 2  = 10

Draw AI.....this  bisects angle A

And  IE = 2

And triangle AIE is right because angle AEI  = 90°

So....the area of AEI = (1/2)(2)(10) =  10

And by   SAS, triangle  AFI is congruent to triangle AEI....so its area also = 10

So  [ AEIF ]  =   10+  10  =   20

25^(1/2) - log ₅ sqrt (5)  = 

sqrt (25)  - log ₅  5^(1/2)  =

5  -  (1/2) log ₅ 5          {  log ₅ 5  =  1 }

5 - (1/2)(1)  =

5 - 1/2 = 

9/2