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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

Wow!   And THAT was a LOT of work !

Don't worry! It happens to all of us! 

You're welcome, MT !

Thanks, EP! You're amazing!

Thanks so much, EP! 

He has three WHOLE bottles left    500 x 3 =1500 ml  PLUS  25% of this amount left in the jug...

500 x 3  +  .25 (500 x 3) = 1875 ml  = 1.875 liters

(and .75 x 1500 = 1.125 l in his stomach !)

By the Pythagorean identity...

sin²(-θ) + cos²(-θ)   =   1

                                             Plug in  (-3/5)  for  sin(-θ)

(-3/5)² + cos²(-θ)   =   1

9/25   +  cos²(-θ)   =   1

                                             Subtract  9/25  from both sides of the equation.

cos²(-θ)   =   1  -  9/25

cos²(-θ)   =   16/25

                                             Take the  ±  square root of both sides.

cos(-θ)   =   ±√[ 16/25 ]

cos(-θ)   =   ± 4 / 5

Since  sin(-θ)  is negative,  -θ  must lie in Quadrant III or Quadrant IV.

Since  tan θ  is positive,  θ  must lie in Quadrant I or III, and so  -θ  must lie in Quadrant II or IV.

So  -θ  must lie in Quadrant IV, and  cos(-θ)  must be positive.

cos(-θ)   =   4 / 5

@Guest, It's a math website......lock this

YOU are correct X !!!  Thanx for fixin' it   ! 

Please do not do that.   This is a MATH site, not a religion site...AND there are many people on this forum with religions different than yours.

18 posts make for  17  spaces     17 x 2 = 34 ft

20 posts make 19 spaces             19 x 2 = 38

38 x 34 = 1292 sq ft

I see a minor error in ElectricPavlov's solution. I think this is simply a copying error.

After multiplying by the LCM of the denominator on both sides, all existing fractions disappear.

5(x+2)-4(x-1)=300

5x+10-4x+4=300

x+14=300

x=286

I like to call this error a "flow-through" error. 

amount after  0  years  =  N

amount after  5.27 years  =  (1/2)N

amount after  2 * 5.27 years  =  (1/2)(1/2)N

amount after  3 * 5.27 years  =  (1/2)(1/2)(1/2)N  =  (1/2)³N

amount after  x * 5.27 years  =  (1/2)^xN

x * 5.27  =  t

x  =  t / 5.27

amount after  t  years  =  (1/2)^{t / 5.27}N

amount after  t  years  =  \((\frac12)^{\frac{t}{5.27}}N\)

\(A=(\frac12)^{\frac{t}{5.27}}N\\ A=N(\frac12)^{\frac{t}{5.27}}\)

If the initial amount is  100, then...

\( A=100(\frac12)^{\frac{t}{5.27}}\)

\(\quad\sqrt{48x^4y^3} \\ =\\ \quad\sqrt{2\cdot2\cdot2\cdot2\cdot3\cdot x^4\cdot y^3} \\ =\\ \quad\sqrt{2^4\cdot3\cdot x^4\cdot y^3} \\ =\\ \quad\sqrt{2^4\cdot3\cdot x^4\cdot y^2\cdot y} \\ =\\ \quad\sqrt{2^4\cdot x^4\cdot y^2\cdot 3\cdot y} \\ =\)

\(\quad\sqrt{2^4}\cdot \sqrt{x^4}\cdot \sqrt{y^2}\cdot \sqrt{3\cdot y}\\ =\\ \quad2^2\cdot x^2\cdot y\cdot\sqrt{3\cdot y}\\ =\\ \quad4x^2y\sqrt{3 y} \)          Since  x  and  y  are positive, we can say...

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

population after  0  hours  =  50

population after  18  hours  =  2 * 50

population after  2*18 hours  =  2 * 2 * 50

population after  3*18 hours  =  2 * 2 * 2 * 50  =  2^3 * 50

population after  x*18 hours  =  2^x * 50

x*18  =  h

x  =  h/18

population after  h  hours  =  2^(h/18) * 50

population after  60  hours  =  2^(60/18) * 50   ≈  504

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

Sin (x ) and  sin (pi + x) have the same BUT negative values

cos (x ) and  cos (pi+x)  have the same values but NEGATIVE

so  tanx  =  sin x / cos x

      tan (x+pi) = -sinx / -cosx  = sinx/cos x  = 2/5

Multily everything by 20   to get

5(x+4) - 4 (x-1) = 300    expand

5x + 20 - 4x + 4 = 300   combine like terms

x + 24 = 300                  subtract 24 from both sides

x = 276

tan (pi + x)  =

tan x  +  tan pi

_____________   =

1  -  tan x tan pi

(2/5) + 0

__________     =

 1  -  (2/5)*0

2/5

1/4 + 1/5 = 1/t, solve for t

The LCD of 4, 5 = 20

[(1*5) + (1*4) ] / 20 = 1/t

[ 5 + 4 ] / 20 = 1/t

 9 / 20 =1/t   cross multiply

9t = 20         divide both sides by 9

t = 20 / 9