CPhill

                                  B

          A  45                D              45   C

AB = CB

Angle B  = 90°

Angle ABD = 45

So angle ADB = angle CDB  = 90

So  triangles ABD and CBD are congruent  by HL

AD = CD  =  6

BD = 6

Area  =  (1/2)AD * BD =   (1/2) (6)(6)  = 18

A

     15

            X

      15

B                    C

AB =  sqrt (15^2 + 15^2)  = 15sqrt (2)

Triangles AXB and ABC  are similar

BX / AX =  BC / AB

15 / 15  = BC / 15sqrt (2)

1  = BC / 15sqrt (2)

15sqrt (2)  =  BC

We   find AD as follows

BD^2 = AD^2  + AB^2 - 2(AD * AB) cos (60°)

6^2 = AD^2 + 5^2 - 2 (AD * 5) (1/2)

36 -25 = AD^2 -5AD

11 = AD^2 - 5AD

AD^2 - 5AD  =  11

AD^2 - 5AD  + 25/4  = 11 + 25/4

(AD - 5/2)^2  =  [ 69 / 4]

AD - 5/2  =  sqrt (69) / 2

AD  = [ 5 + sqrt (69) ] / 2

The height =  (5/2)sqrt (3) =  5sqrt (3)/2 

Area  =  AD * height  =      [5sqrt (3)] / 2  *   [ 5 + sqrt (69)] / 2  =   25sqrt(3) / 4  + 5sqrt [ 207] / 4  =

25sqrt (3) / 4 + 15sqrt (23) / 4 =

(5/4) [ 5sqrt (3)  + 3 sqrt (23) ] ≈  28.8

sqrt [ 4x + 1 ]  / sqrt [ -4x - 1]

4x + 1 ≥ 0               -4x - 1 >  0

4x ≥ - 1                   -4x > 1            (divide by -4  and  change the  direction of the inequality sign )

x ≥ -1/4                      x <  - 1/4

There are no intersecting intervals....thus.....no values of  x that  make this well-defined

y = log ₂ (2x)

y = log ₄ (16 + x)

log ₂ (2x)  =  log ₄ (16 + x)         {change of base)

log (2x) / log 2  =  log (16 + x) / log 4

log (2x) / log 2  =  log ( 16 + x) / log 2^2

log (2x) / log 2  = log (16 + x) / [2 * log 2]

log (2x)  = log (16 + x) / 2

2 log (2x) = log (16 + x)

log (2x)^2  = log (16 + x)

4x^2  = 16 + x

4x^2 - x - 16  =  0

x must be positive.....so......

x =   [ 1 + sqrt ( 1 + 256 ) ] / 8  =   [ 1 + sqrt (257) ]  / 8

\(\log_2 (\log_3 x) = 2 \log_2 (\log_{27} x) \)

We can write

log₂ (log ₃x) = log₂ (log₂₇x)^2

The log bases are the same, so

log₃x = (log ₂₇ x)^2

log₃ x =  (log ₂₇ x) (log ₂₇ x)               {apply change-of-base theroem}

log x / log 3 =  ( log x / log 27) (log x / log 27)      {assume x not equal to 1} {divide out log x}

1/ log 3  =  (log x/ log 27) ( 1 /log 27)

log 27 / log 3  =  log x / log 27

(log 27)^2 / log 3 = log x

(log 3^3)^2   / log 3  = log x

(3 log 3)^2 / (log 3)  =  log x

9 (log 3)^2 / (log 3)  = log x

9 log 3  = log x

log 3^9  = log x

x = 3^9 =  19683

Let (x,y)  be the center

Since all radiuses are equal  the distance from the center to any of the  points  is equal

So  we can solve this  using (-7,10) and (5,4)

(x + 7)^2  + ( y -10)^2  =  (x - 5)^2  + ( y - 4)^2        simplify

x^2 + 14x + 49 + y^2 - 20y + 100  = x^2 -10x + 25 + y^2 -8y + 16

24x  - 12y  =  -108    ⇒     2x - y  = - 9   ⇒  y= 2x + 9      (1) 

And using  (-7,10) and (15, -10)

(x + 7)^2  + (y-10)^2  = (x -15)^2 + ( y +10)^2

x^2 + 14x + 49 + y^2 - 20y + 100  = x^2 -30x + 225 + y^2 + 20y + 100

44x - 40y  = 176

11x - 10y = 44       (2)

Sub (1)  into (2)

11x - 10 ( 2x + 9) =  44

11x -20x - 90  = 44

-9x = 134

x= -134/9

y = 2(-134/9) + 9  =  -187/9

Center of first circle = (0,0) =  P

Center of second  circle = (1,1)  = Q

Can you take it  from here  ???

Simplify as

x^2 - 14x + y^2 + 20y = -17           complete the square on x, y

x^2 - 14x + 49  + y^2 + 20y + 100  = -17 + 49 + 100

( x - 7)^2  + ( y + 10)^2  = 132

The center is (7, -10)   and the radius =  sqrt (132)  =  = sqrt (4 * 33)  = 2sqrt (33)

Let the height of the upper trapezoid = x    and the height of the lower trapezoid = y

Since the areas are equal

(1/2) (x) ( 1 + EF)  = (1/2)y (2 + EF)

x ( 1 + EF)  = y(2 + EF)

x/y = (2 + EF) / ( 1 + EF)

height = x + y =  (2 + EF) + (1 + EF)  =    3 + 2EF

And twice the area of the top trapezoid =  the area of the whole trapezoid

2 ( 1/2) x (1 + EF)  =  (1/2)(x + y)(1 + 2)

x (1 + EF)  = (3/2)(x + y)

(2 + EF) ( 1 + EF)  =  (3/2)(3 + 2EF) 

2 + 3EF + EF^2  = (3/2)(3 + 2EF)

EF^2 + 3EF + 2 = 3EF + 9/2

EF^2 = 9/2 - 2

EF^2  = 5/2

EF = sqrt (5/2) ≈  1.58