CPhill

The slant height of the cone is = BC

The volume of the cone is given by

432pi  = pi/3 * r^2 * height

432pi  = pi / 3 * 12^2  * height

432  = 144/3 * height      multiply both sides by 3/144

3 * 432  / 144  = height 

9 cm  = height

Now...the slant height of the cone  = √ [ r^2 +  h^2 ] = √[12^2 + 9^2] =√225  =15 cm = BC

So...the lateral surface area of the cone = the shaded area of the circle  =

pi * radius * slant height  =     pi * 12 * 15  =   180 pi  cm^2

And we can set up the following relationship

Total circumference of the circle / Total area of the circle = 

Total circumference of shaded area / Total area of shaded area

[2*pi * BC] /[ pi * BC^2]  =  [2pi *  BC * ( D /360)]  / [ 180pi ]

Where D is the number of degrees in the arc of the shaded sector

[2 * pi * 15]  /[ pi * 15^2 ] =[ 2 * pi * 15] * ( D / 360)] /[180pi]

1 / 15^2  =   (D / 360) / 180

1 / 15^2  = D / 64800

64800 / 225  = D = 288°

So...the number of degrees in ABC  = 360  - 288   =  72°

That is pretty cool  !!!

We have

f (a+ h)  = 3(a + h)^2 - 5  =   3 [ a^2 + 2ah + h^2  ] - 5    =

3a^2 + 6ah + 3h^2  - 5

f(a)  =  3(a)^2  -  5   =   3a^2  - 5

So

[ f(a + h)  - f(a) ] / h  =

[  (3a^2 + 6ah + 3h^2 - 5)  -  (3a^2 - 5)  ]   / h  =

[6ah + 3h^2] / h  =

h [ 6a  + 3h ] / h  =

6a  + 3h

Note  that if we let  h approach 0,  we get   6a  =  the derivative  of  3x^2 - 5   when x = a ]

A  = 14790

r = .069  / 12   =  23 / 4000

n = 48

P  =  14790 *(23/4000) *  [ 1 + 23/4000]^(48) ] / [ ( 1 + 23/4000)^48  - 1 ] = 353.48 

So..."C"

Nice, hectictar  !!!

The 110° angle  will subtend a major arc of  twice its measure  = 220°

So...the remaining minor arc subtended by its endpoints = 360 -  220  = 140° 

The  105° angle will subtend a major arc of twice its measure = 210°

So...the remaining minor arc subtended by its endpoints  = 360 - 210  = 150°

So....the major arc intercepted by  the enpoints of  α  =  150 + 140  = 290°

And its measure is  1/2  of this   =  1/2 * 290   = 145°

A   =  P  [ (1 + r)^n  - 1 ]

         _______________     

          r  ( 1 + r)^n

Where P  = 725    r  = .0825/12  = 11/1600    n  = 48 months

So  we  have

A = 725 [ (1 + 11/1600)^(48)  - 1 ]

      __________________________    ≈  29,555.11  ⇒  "B" seems the closest

         [ (11/1600) * ( 1 + 11/1600)^48

Using Omi's diagram  and the exterior angle theorem, we have that

y  = x + 48      (1)

z   = y + x        (2)

Sub  (1)  into ( 2) for y   and we have

z  = ( x + 48 ) + x     

z  = 2x + 48

So..in the triangle on the right  

48 + x + z  = 180

48 + x + (2x + 48) = 180

96 + 3x  = 180

3x  = 180 - 96

3x  = 84

x = 28° =  angle "B"

Let the bottom left vertex of the equilateral triangle be (0,0)

In a 30-60-90 right triangle, the side  opposite the right angle is [side opposite the 30° angle * 2 ] =  [6 *2] = 12  = side of equilateral triangle

The side of the the right triangle opposite the 60° angle is  [side opposite the 30° * √3 ] = [ 6√3 ]  = √108

So....the top right vertex of the right triangle  is   ( 12 , √108 )

So...the distance  that this vertex is  from ( 0 , 0)  is given by  :

√ [ 12^2  + (√108)^2  ]  =

√ [ 144 + 108  ]  =

√ [252]  =

√  [ 4 * 9 * 7 ] =

2*3 √7  =

6√7  units

 Δ EDC   ~  Δ  EBA

So

EC  / DC  = EA / BA

EC / 8  = 5 / 4       multiply both sides by 8

EC  =  8 *  ( 5 /4)  

EC =  40 / 4  

EC  = 10