CPhill

r =  3/-5  =  -3/5

Sum  =   first term  / [ 1 - r ]     =  -5 / [ 1 - -3/5 ]  =    -5 / [ 8/5 ] =  -25 / 8

DX = 5

See the following :

Area  =    12^2 *sin (150°)  = 144 * (1/2)  =  72

OK

r = - 80 / 120  =  -2/3

Sum of 1st n terms of a geometric series = first term * [ 1 - r^n ] / [1 - r ]   =

120 *  [ 1 - (-2/3)^8 ] /  (1 - (-2/3) )

Taking it  by parts

[1 - (-2/3)^8  ] =    

[1 - 256/6561 ]  =

[ 6561 - 256 ] / 6561    =

[6305 / 6561 ]

And  ( 1 - -2/3 )  = ( 1 + 2/3)  = (5/3)

So we have

120 / ( 5/3)  *  ( 6305 /6561)  =

72 * (6305 / 6561)   =  453960 /  6561  =    [50440 * 9  ]  /  [ 729 * 9 ]  =  50440 / 729

Let d  be the distance from the center of the  circle to UV

Let the radius of the circle be r

So....by the Pythagorean Theorem

d^2  +  (UV /2)^2  =  r^2

d^2 + (3)^2  =r^2

d^2 + 9  = r^2     (1)

Let the distance between the center of the circle and YZ = d + 2

By the Pytagorean Theorem

(d +2)^2 + (YZ/2)^2  = r^2

(d + 2)^2 + (4/2)^2  = r^2

(d + 2)^2  + 2^2 = r^2

(d + 2)^2  + 4  = r^2     (2)

Equating (1) and (2)

d^2 + 9 = (d + 2)^2  + 4

d^2 + 9  =d^2 + 4d + 4 +4

1 = 4d

d = (1/4)

And

d^2 + 9  =r^2

(1/4) + 9 = r^2

(37/9) = r^2

To find (1/2) the length of WX

(d + 1)^2  + (WX/2)^2  = r^2

(1/4+ 1)^2  + (WX/2)^2  = 37/9

25/16  + (WX/2)^2  = 37/9

WX^2 / 4   =  37/9  - 25/16

WX^2 / 4  = 367/144

WX^2 =  367/36

WX = sqrt (367) / 6 ≈  3.19

Let x be the distance he steps back....so....

To find his  original  distance, d,  from the tree  we have

tan (68°) = 85/d

d = 85  / tan (68°)   =  34.34 ft

So

tan (41°)  =  85 / ( 34.34 + x)

(34.34 + x)  =  85  /tan (41°)

34.34 + x =  97.78

x = 97.78  - 34.34     ≈  63.44 ft

r =   -80/120  = -2/3

Sum of 1st  8 terms  

S₈  =  first term  *  [ 1 - r^8 ] / [ 1  - r ]  =

120*  [  1 - (-2/3)^8 ] / [ 1 - (-2/3) ]  =   50440 /   729 

2(n+1)!               2^(2n)

_________   *   ______   =

2^(2 (n +1))          2n!

2 (n + 1)!         2^(2n)

_______  *    _________   =

 2n!                 2^(2n +2)

n  *       2 ^(2n - (2n + 2)   =

n  *  2 ^(-2)  =

n / 2^2  = 

n / 4   = (1/4) n 

(1/4) * lim      [ n ]      >   1            {this series diverges } 

         n →inf

tan (-2pi / 3)  =  tan ( -120°) = tan (240°) =  tan of a 60° angle in quadrant 3  = sqrt (3)

sin (7pi/4)  

(7pi/4) - 2pi  = (7pi/4 - 8pi/4)  = -pi/4

sin (-pi/4)  = - 1/sqrt (2)

sec (-pi)  = sec (pi)  =  -1

So we have

sqrt (3)  / ( -1/sqrt 2)  -  (-1)   =

-sqrt (6) + 1

 f(0) = 5, f(2) = 8, f(3)=13, and f(7) = -5

We have the form

ax^3  + bx^2 + cx + d

Since f(0)  =  5    then d  = 5

We have this system

a(2)^3  + b(2)^2  + c(2) + 5 = 8

a(3)^3  + b(3)^2  + c(3) + 5 = 13

a(7)^3  + b(7)^2  +   c(7)  + 5 =  -5        simplify

8a + 4b  + 2c  =  3

27a + 9b + 3c  = 8

343a +49b + 7c = -10

I'm assuming that you  know how to  solve such a system

a  = -46/105  =  -92/210

b = 47/14 =   705/210

c = -727/210

Sum of  coefficients  =   [-92 + 705 - 727 ] / 210  =  -114 / 210  =  -57 / 105