CPhill

Find the center of the circle.....call this  " O"

From O.....draw  bisector OG  to chord AB   and bisector OH  to chord EF 

But, by Euclid,  a radial line bisecting a chord does so at right angles

And since AB  = EF = CD.....then equal chords are equal distances from the center of a circle

Now....connect  O  and  P

So.....OG  = OH,  OP  = OP   and angles OGP  and OHP  are right

So, by HL,  have two  cogruent  right triangles formed, ΔOGP  and ΔOHP

But  angle GPO   =  angle HPO....so  angle P  is bisected

By similar reasoning, we can draw bisector  OI  to chord  CD  and  also connect  QO

This will form  two congruent right triangles - Δ OGQ and ΔOIQ

And angle GQO  =  angle IQO, so Q is bisected

And since PO  and QO are angle bisectors of triangle  QPR  that  meet  at O....then ,by definition, the center of the circle is the incenter of triangle PQR 

Note that  m angle PAD  =  1/2 m arc AD  =  m angle DBA

And

m angle PCD  =  1/2 m  arc   DC   =  m  angle DBC

And PA  = PC....so....

sin PAD / PD  =      sin DBA / PD

sin PCD / PD    =    sin DBC/ PD

sin PDA/ PC  = sin PAD / PD   =  sin DBA / PD

sin PDC / PC  =  sinPCD / PD  =    sin DBC / PD

PC / PD  =  sin PDA / sin DBA

PC / PD  = sin PDC / sin DBC

So.....

sin PDA / sin DBA   =  sin PDC/ sin DBC

Note  sin PDA = sin ADB     because they are supplemental   ...and..

sin PDC  = sin BDC  for the same reason....so....

sin ADB / sin DBA  =  sinBDC / sin DBC     (A)

sin ADB / AB  =  sin DBA / DA

sinBDC / BC  =  sin DBC / DC

sin ADB / sin DBA  =  AB /DA

sin  BDC / sin DBC  = BC / DC

From (A)    we must have, by substitution, that

AB / DA  =  BC / DC   ⇒   

AB * DC  =  BC * DA ⇒

AB * CD   = BC * DA

A = True

B  = False

C = True

D = True

E = Not strictly true....it looks like a litte more than $110....but, I suupose  that  "about"  makes this True

Written in the form

y  =  a (b)^x   

Whenever  b > 1.....we have a growth function

To  find the percent change, we can write 

2.4  - 1  =  1.4     ⇒  140%  

-5pi / 4  =   -225°   =  (360 - 225)°  =  135°   =  a  45° reference angle in the second quadrant =

  ( -√2 / 2 , √2 / 2 )

sin (-theta)  =  - sin (theta)

So

-sin (theta)  = -3/5      multiply both sides by -1

sin (theta)  = 3/5

And  cos(-theta)  =  cos(theta)

So......  the sine , tan > 0        this must be a first quadrant angle

So....cos (theta)  =

√ [ 1  -  sin^2 (theta) ]  =    √ [ 1 - (3/5)^2 ]  =  √ [ 1 - 9/25]  =

√ [ 25 - 9 ]  / 5  =     √ 16 /  5      =  4/5

tan (pi + x)  =

tan x  +  tan pi

_____________   =

1  -  tan x tan pi

(2/5) + 0

__________     =

 1  -  (2/5)*0

2/5

Power Property of Logs

Product Property of Logs

Equality  Property of Logs

These are alwas a little difficult

-4y^2 + 3xy - 12y + 9x            factor by grouping

y ( -4y + 3x)  + 3 (-4x + 3x)

Take out the commmon factor and then write the rest in parentheses

(-4y + 3x) ( y + 3)   =

(3x - 4y) ( y + 3)

We have a 30-60-90  right triangle

The side across from the 30° angle  = 5   cm  =  the base of the triangle

The side across from the 60° angle  is   √3   times this  =  5 √3  cm = height of the triangle

So.....the area is    (base) (height)  / 2  =

(5) * (5 √3)  /  2  =

(25/2) √3    cm^2    .....  this is exact  !!!