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r =  14/16 = 7/8

32nd term  =   24 (7/8)^8  = 17294403 / 2097152

We need to solve this

1000*(9/10)^(n-1) =  1

(.9)^(n -1)  =   1/1000       take the log of  both sides

log (.9)^(n-1) = log (1/1000)

And we can write

(n-1) log (.9)  = log (1/1000)

n -1  =  log (1/1000) / log (.9)

n -1   = 65.56

n = 66.56

The 66th term is  1000(.9)(66-1)  = 1.061

The 67th term is  1000(.9)(67-1) = .955

So  66 terms are > 1

Call the center of the  small circle  M

Let N be the point of tangency between the small circle and OA

Let r be the radius of the small circle

Triangle  OMN  is  right

OM = 3 - r

MN = r

ON = 2 

OM^2 = MN^2 + ON^2

(3 -r)^2  = r^2 + 2^2

r^2 - 6r + 9  = r^2 + 4

6r  = 5

r = 5/6

r =  (32/3)  / 16  =  (2/3)

Since  -1 ≤ r ≤ 1     this series will converge  to  [  16 ] / [ 1 -2/3] = 16 / (1/3)  = 48

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

Also tell me how you solved it.

I don't know if this is the classroom approved method. 

A geometric series is formed by multiplying a term by a number to find the next term. 

So, if we call that multiplyer number x, then a series would look like this. 

1st number times x  =  2nd number  

2nd number times x  =  3rd number  

3rd number times x  = 4th number  

4th number times x  =  5th number  

By this we can see that 1st number times x⁴  =  5th number  

Now that we have a principle established, let's substitute    

the specific numbers given in the above problem   

48 times x⁴  =  243/16    

                                                      48x⁴  =  243/16   

                                                          x⁴  =  243/(16)(48)  

                                                          x  =  (243/768)^1/4   

Using my scientific calculator    

to take the fourth root                        x  =  0.75  

The series is    48,  36,  27,  20.25,  15.1875  

Note that 15.1875 is the same as 243/16  

_.

Let's first find the side lengths of PAQ with the given information:

The square's side length is 4 units

Side AQ= 1^2+4^2= 17 so    AQ= sqrt17

Side PQ= 2^2+3^2=13 so    PQ= sqrt13

Side AP= 2^2+4^2=20 so     AP= 2sqrt5

Now let's use SOHCAHTOA:

We're going to do the opposite over hypotenuse, depending on which angle you are using, that should give you your answer .

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

Do you mind explaining the process of how you got it I am kind of confused?

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

\(f(x)\cdot{g(x)} = (-7x^4 - 3x^2 + 2)(2x^8-3x^7+5x^6+3x^5-11x^4+3x^3-18x^2+17x-5)\)

Leading term is the term with the highest degree:\(-7x^4\cdot{2x^8}=-14x^{12}\)

Thus, the leading coefficient is -14

The (highest) degree of the function is 12

We know that the constant term must stem from the product of the constants of the original function. 2 * -5 = -10

Coefficient of x^6 would be -3x^2 * 3x^3 + 2*5x^6 + -7x^4 *-18x^2 = 7*18 - 9 + 10 = 127

Hypotenuse = 3sqrt(15)

Shorter Leg = 3 sqrt(6)

By pythagorean theorem, Longer Leg = 9

Let the point between the hypotenuse and the shorter leg be A. We know that point A must have the largest acute angle in the triangle by construction.

Using any trig function (preferably a easy-calculation one), such as sinA, we can use the inverse function of it to find the angle.

sinA = opposite/hypotenuse = 9/3sqrt(15) = 3sqrt(15)/15 = sqrt(15)/5

Then angle A = arcsin(sqrt(15)/5) = 0.886 radians = 50.8 degrees.

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

Let the unit circle's center be P, and the circle with radius 4, Q (the center). 

AP = 1 because AP is the radius of the unit circle. BQ = 4 with the same logic. With a translation (mapping of points) to AB, P becomes point A, and Q is moved along the line BQ by 1. Let these points be P' = A, and Q'. PAB is a 90 degree angle because the radius is always perpendicular to the tangent, and QBA is 90 degrees. Thus P'BQ' is a 90 degree triangle with P'BQ' = 90 degrees, BQ' = 3, and P'Q' = 5. By pythagorean theorem, AB = 4.

Let the midpoint of YZ = point M. Let the intersection points from M to XY and XZ be points P and Q respectively.

Let the radius of the semicircle be r, and also keep in mind that point M is also the center of the semicircle.

XMZ is a 45 - 45 - 90 because XM bisects angle X and is perpendicular to YZ, so XM = 10/sqrt(2) = 5sqrt(2)

Note how MP = MQ, which create 45 - 45 - 90 triangles: XPM and XQM. Thus, MP = MQ = r = 5sqrt(2)/sqrt(2) = 5

A semicircle is half the area of a circle: circle has area \(\pi r^2 = \pi 5^2 = 25\pi\), so the area of the semicircle = \(25\pi/2\)

Say the center of the inscribed circle is C      let the intersection of line OP that intersects minor arc AB be point P

and also let the radius of the inscribed circle be "r", a variable.

We know that PC = r, and that CO bisects angle O. Additionally, CO is the diagonal of a square with side length r, because the perpendicular from C to AO = r, and creates a 45 - 45 - 90 triangle. As a result, CO = \(r\sqrt{2}\)

Summing the line segments, \(PO = r + r\sqrt{2} = 1\) and to solve this equation:

\(r + r\sqrt{2} = r(1 + \sqrt{2}) = 1\)

\(r = {1\over{1 + \sqrt{2}}} = {1(1 - \sqrt{2})\over{(1)^2 - (\sqrt{2})^2}}= {1 - \sqrt{2}\over{-1}} = \sqrt{2} - 1\) - note, the expression after the second equation sign is deduced via difference of squares.

and if you need a decimal answer for sqrt(2) - 1 = 0.414