CPhill

Sector Area  =  (1/2)r^2 * theta (in rads)

So we have :

(1/2) (10^2) (pi/5)  =

(1/2) (100/5) pi =

(1/2) (20) pi  =

10 pi  units^2 

Here's a few :

1. 

A(2)  - A(1)  =

[ 3  - 2(2)^2 + 4^2 ]  - [ 3 - 2(1)^2 + 4^1]

[ 3 - 8 + 16]  -  [ 3 - 2 + 4]

[ 11 ]  - [ 5]

6

3.     [ 3x - 7]  / [ x + 1]

We can put any x into the function  except  x  =  -1

So....the domain is  (-infinity , -1) U (-1, infinity )

4. [ 3x - 7 ] / [ x + 1 ]

Since we have a same power polynomial/ same power polynomial  we will have a horizontal asymptote   at    [ 3x ] / x   =    3

So....the range  is   (-infinity , 3) U (3 , infinity )

5.  

3  =  [ 2 - 3a ]  /  [ 5 - 2a ]     multiply both sides by 5 - 2x

3 [ 5 - 2a]  = 2 - 3a

15 - 6a  = 2 - 3a      add 6a to both sides, subtract 2 from each side

13 = 3a  divide both sides by 

13/3  =   a

Let's solve this with a graph, GameMaster

Look here :    https://www.desmos.com/calculator/gt2fmgddhr

Note that 164 billion was reached after about 4.978 years ≈ 5 years  =  about 1995

Thanks, coracat....!!!!!

3)A prism and a cone have the same base area and the same height. The volume of the prism is 1. What is the volume of the cone?

The base area of the cone  =  pi * r^2

But...this is also the base area of the prism

So.....the volume of the prism  =    base area * height  = 1

So

1  = base area * height

1  = pi (r^2) * height

1/ [ pi (r^2)]  =  height

So....the volume of the cone is

(1/3)[ pi * r^2]  [ 1 / [ pi * ^2 ]  =     1/3 

The area  is

(1/2) (3.2) (4.7) sin (62)  ≈ 6.64  ft^2

We want to solve this :

a)

7x^2 - 4x + 392  =  1904         subtract  1904 from both sides

7x^2 - 4x - 1512   =  0

Solving this for x gives = 14.985 yrs  ≈  15 years  (about the year  2000)

b)

Here's the graph :

https://www.desmos.com/calculator/13gj3kx7vy 

a)  The function is :

h  = -16t^2  + 24t + 16.4

b) 

13.2  = -16t^2 + 24t + 16.4        subtract 13.2 from each side

0  = -16t + 24t + 3.2     multipl through by - 1

16t^2 - 24t - 3.2  =  0       divide through by 3.2

5t^2 - 7.5t - 1  =  0

Solving this for t   we get ≈  1.6  seconds

1.

  A  =  180  - 74   =   106°

Note that if  the circumference of the larger circle is 10pi feet more....the track must be 5 feet wide

Proof....let the  radius of the smaller circle  = r

Then....let the radius of the larger circle be  r + 5

So.... circumference of the smaller circle is  2pi * r

And the circumference of the larger circle  is  2pi ( r + 5)  = 2pi * r +  10pi

Subtracting the smaller circumference from the larger, we get

[ 2 pi * r  + 10 pi ]  - [ 2 pi * r ]  =    10pi