CPhill

x = 5^m   y  = 5^n

1)  5^(7m + 1) =    5^(7m) * 5   =  (5^m)^7 * 5  =  (x)^7 * 5  =   5x^7

2) 25 ^ (m + n)  =  (5^2)^(m + n)  =  (5^2)^m * (5^2)^n  = (5^m)^2 * (5^n)^2  =  x^2* y^2

3) ( 1 / 125)^n  =  (1/ 5^3)^n  =  (5^(-3 ) )^n  = ( 5^n)^(-3)  =  y^(-3) 

Here's my attempt at this  tricky problem

Note that  angle ABC  is inscribed in the  first circumcircle

Let the circumcenter   =  M  = O1

Then  angle   AMC is a  central angle intercepting the same arc  as angle ABC  so its measure =  120°

And triangle AMC  is isoceles  with AM = CM

And angles  MAC and MCA =   (180 - 120) / 2  =    30°

Using some trig  we can find circumradius  AM  as

AM / sin 30°   =   12 / sin 120°

AM / (1/2)  =  12  / ( [sqrt 3 ] / 2 ) 

AM  =  12/sqrt 3  =   4sqrt 3

Now triangle   ACD  is obtuse  and its circumcenter will be exterior to side AD

Angle   ADC will be inscribed in  the circumcircle of this triangle

Let N be the circumcenter O2

Then angle  ANC  is a central angle intercepting the same  arc as angle ADC so its measure = 60°

And triangle ANC  is isosoceles with AN = CN

So  angles   CAN and ACN   = (180 - 60) / 2  =  60°

So triangle ANC  is equilateral  with circumradius   AN  =   12

Let  A =  (0,0)

Then  M =  O1  has the coordinates   (   4sqrt (3) cos 30° , 4sqrt (3) sin 30°)  =  (6 , 2sqrt (3) )

And  N =  O2  has the coordinates   ( 12 cos (-60°) , 12 sin (-60°) )  =   ( 6  , -6sqrt (3) )

Assuming that    O1 O2    is the distance between the  circumcenters we have  that

O1 O2  =   2sqrt (3)   -  (-6sqrt (3) )   =    8 sqrt (3)

THX, Vinculum......this one is a little  tricky  !!!

Let    a,b, c, d   be in increasing order

Using some logic....we know that 

a + b  = 10       (1)

a + c =  18       (2)

c + d =  31      (3)

b + d =  23      (4)

What we don't know is  that if

a + d =  19   or   a + d  =  22  or   b + c = 19  or   b + c =   22

To see which is true    subtract   (4)  from (1)    and we  get 

a - d  =  -13        (5)

Now let us assume that   a + d  =  22      (6)

Add (5) and (6)  and we get that

2a = 9             but a here is not an integer

So....it must be that

a + d = 19

So

a - d = -13

a + d = 19          add these

2a  = 6

So.....using our equations.....

a = 3

b = 7

c = 15

d = 16

The vertex is   (1, -4)

The axis of  symmetry   is   :  x = x coordinate of the  vertex

So

axis of symmetry   is      x  = 1

You beat me to it, BuilderBoi  !!!!!   {LOL!!!}

Let A  = (0,0)   D = (0,1)  N= (.5, 1)   M  = (1,.5)

The slope of the line sebment AN  =  [1 -0] / [ .5 - 0 ]  =  2

The equation of the line containing this segment is

y = 2x

The slope of the line segment DM =  [ .5 - 1 ] / [ 1 - 0 ]  =   -.5 =  -1/2

The equation of the line containing this segment is

y = (-1/2)(x - 1) + .5

y = (-1/2)x + 1

Setting the equations = and solving for x will give us the x coordinate of O = the height of ADO

2x = (-1/2)x + 1

2.5x = 1

x = 1/2.5  = 10/25  =  .4

So....the area of ADO  =  (1/2) AD * .4  =   (1/2) (1) (.4)  =  .2  units

Here's a pic :

x^2 + y^2 <= 4x + 6y + 13 - 2x + 8y + 15

Rewrite  as

x^2 -2x + y^2 - 14y ≤ 28     complete the square on  x and y

x^2 -2x + 1  + y^2 -14y + 49 ≤  28 + 1 + 49

(x - 1)^2 + (y - 7)^2 ≤  78

This is a  circle  centered at (1, 7)  with a radius^2  of 78

The area lies within the circle and   =    78pi  units^2

Welcome aboard, Atlas  !!!!!

When the ball hits the ground the  height   = 0

Multiplying the original equation through by -1   and  setting this to 0  we have

16t^2  + 18t  - 405  = 0

Use the Q Formula

t  =  (  -18  ± sqrt  [ 18^2  + 4*16 *405 ] )  /  [ 2 * 16 ]  =    [  -18 ±  162 ] / 32

Take the positive root

t =  [ -18 + 162 ]  /32   =    144 / 32   =    4.5 sec