CPhill

The radius of the circle  can be found as

4pi = pi*r^2   ⇒  r    = 2

(1/2) of the side, m,  of the triangle can be found as

2 / sin(30)  =  m / sin( 60)

2sin(60) / sin( 30)  = m

4 * sin 60    =

4 *√3/2  =  m  = 2√3

So.....the side of the triangle is  twice this = 4√3  = √48

And the area  is 

(1/2)( √48)^2 sin 60  =

24 * √3 / 2

12√3   cm^2

5. The first  answer is correct, NSS....to see why this is so....note that we have

 2(1) + 2(2) + 2(3) + 2(4) + ........=

2  + 4  +   6  + 8  + .....

6.  Note that the first term is

5(3)  = 15

And the last term is  

5(8)  = 40

And the number of terms is  [ 8 - 3] + 1   = 6

So....the sum is

 [ First term + Last term ] *  number of terms / 2   =

[ 15 + 40 ] * 6 / 2   =

55 * 3   =

165

The equation of BA  is

y = -2 (x - 4) + 4

y = -2x + 8 + 4

y   = -2x + 12

So..."B"  =  -2(0) + 12  =  12

And "A"  can be found as

0  = -2x + 12

2x  = 12

x = 6

So..."A"  =  (6,0)

So.....the area of the shaded quadrilateral   =  

Area of triangle OBA  +  area of triangle EAC  =

{Note that EAC has a base of 2  and a height of 4 }

(1/2)(12*6)  + (1/2)(4*2)    =

36 + 4   =

40 units^2 

2^2001 * 5^1950

______________  =

     4^27

2^2001 * 5^1950

_____________    =

        (2^2)^27

2^2001 * 5^1950

_____________    =

       2^54

2^1947 * 5^1950 =

2^ 1947 * 5^1947 * 5^3  =

(2*5)^1947  * 5^3  =

10^1947 * 125  =

125 x 10^1947 =

1.25 x 10^1949

So...... 1950 digits

3.  We have that

18 * [  r ]^2  = 8          divide both sides by 18 

 r^2  =  8/18  = 4/9

r^2  = 4/9

r = √[ 4/ 9]   = 2/3

So

The length of the second arc is

18 (2/3)  =  12 ft

4.  ​A little tricky, NSS

We have that

1250 * r^2  = 50     divide both sides by 1250

r^2  =  50/1250  = 1/25

r = √(1/25)  = 1/5

So.....the second term is

1250 (1/5)  =   250

Note that the total amount of money, A, doesn't change.....just the distribution of it does

So....this is a "zero-sum" game....the total amts won + total amts  lost   = 0

7 : 6 : 5

Means that  the first person  started with (7/18)A

The second (6/18)A  =  (1/3)A

The third (5/18)A

In the end   ⇒   6 : 5 : 4

The first person ends up  with (6/15)A  = ( 2/5)A

The second ends  up  with (5/15)A  = (1/3)A

The last ends  up wth (4/15)A

Note that the second person breaks even...he starts with (1/3)A  and ends up with the same amount

The first person must have won the $12  because  (2/5)A > (7/18)A

And the difference between (2/5A - 7/18A)  = 1/90A  =  $12

So.... $12  is 1/90 of the total amount  = $1080

So....the first person must have started with (7/18) * 1080  = $420

And he ends up with (2/5) * 1080  =  $432

So.....he makes $12

{Which means that since the second person breaks even....the third person must have lost the $12 }

1.  No...in a geometric series.the ratio between the successive term to the previous term must be constant ....note that

-4 / 2   =   -2

But

-16 / - 4  =  4    so....not geometric

2. Fifth term  

The common ratio  is  36/120  =  3/10

So....the fifth term is

First Term * (Common Ratio)^n-1   =

120 (3/10)^5-1  =

120 (3/10)⁴   = 

 0.972

LOL!!!!

Can't answer the second one....but....for the first

The vertex  is (0,0)

This parabola opens to the left

The general form  for this type of parabola with the vertex is at the origin  is

-4px  = y^2         where p is the distance between the focus and the vertex

So....we have

x =  -4y^2        divide both sides by -4

(-1/4)x  = y^2

So.......this implies that  (-1/4) = -4p  ⇒  p = 1/16

So.....the focus in this situation is given by

(0 - p , 0)  =  (0 - 1/16 , 0 )  =  (-1/16, 0 )

And the directrix  is give by:    x = 1/16

Here's a graph :  https://www.desmos.com/calculator/z66mdaucdo 

Note that when a = 2

P(x  ≤ 2)   =  the probability of this area  =  1/2  of the area of the figure  =  0.5

So....a  = 2