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f(17) =  17 + 30 =  47

f (f(17))  =  f (47)  =  47 + 13  =   60

f (f ( f (17) ) )   =  f (60)  =  60 + 13  =    73

r = 1/2

Sum of first  n terms of an infinite series  = 

first term  [ 1  - (r)^n ] / [ 1 - r ] = 

(1/4) [ 1 - (1/2)^n ] / [ 1 -1/2 ]  =   15/32

(1/4) [ 1 - (1/2)^n ] / (1/2)   =15/32

(1/2) [ 1 - (1/2)^n ]  =  15/32

1 - (1/2)^n  =  30/32

1- (1/2)^n  =  15/ 16

(1/2)^n  = 1 -15/16

(1/2)^n = 1 /16

n = 4

The denominator  cannot  =  0

Note that   l x^2 + 3x + 4  l   =  l (x -1) ( x + 4) l

And   l x^2 + 9x -10 l =  l (x -1) ( x + 10) l

So  we have

l (x -1) ( x + 4) l +  l (x -1) ( x + 10 l  =

( l x - 1 l  )  (  l (x + 4) l + l x + 10 l )

The only value of x that makes  this 0   is when x   = 1

So .....x =1   is not in the  domain

n ==19

GCD{13*19 + 8}, {5*19 + 7} =={255,  102}==51 - maximum possible value of the GCD.

a = the slope =    [ -17-5 ] / [ 8 -4]  =  -22 / 4  =  -11/2

b = the y intercept

So  using (4,5)  the equation of the line is

y = ( -11/2)(x - 4)  + 5

y = (-11/2)x  + 22 + 5

y = (-11/2)x + 27

b = 27

a - b  = 

-11/2  -27   =

-11/2  - 54/2  =

- 65 / 2

We have a 45-45-90  right triangle

XY  = XZ  =  12sqrt (2)

1/a  + 1/b =   1 / 34

Let  z = 34

Let a = z + m

Let b = z + n

And let's  assume that   a ≥ b

So we have

1/ ( z + m)   +  1/ ( z + n)  =  1  / z           simplify

(2z + m + n)  / [ (z + m) (z + n) ] =   1 / z       cross-multiply

z ( 2z + m + n)   =  (z + m) ( z + n)

2z^2  + mz + nz  = z^2 + mz + nz +  mn

z^2  =  mn

34^2  =  mn

1156  = mn

Factors of  1156   =  1 | 2 | 4 | 17 | 34 | 68 | 289 | 578 | 1156 

So we have

  m * n            a = z + m              b = z + n

1156 * 1          1190                         35

578 * 2              612                        36

289 * 4              323                        38

68 *   17            102                        51 

34  *  34              68                        68

x^2 + 11x - 5 + 5x  =

x^2 + 16x - 5

We have the form    ax^2 + bx + c

The x that will  give the minimum =  - b / [2a ]  =     -16 / ( 2 * 1)  =  -8

-(3 -c)  ( c + 2(3-c) + 5c - 7)       =

-(3 -c)  ( c +   6 - 2c + 5c - 7)  =

-(3 -c) ( 4c - 1)  =

(c - 3) (4c -1)  =

4c^2 - 12c -1c + 3  =

4c^2 -13c  + 3

Sum of coefficients  =  4 -13   =  -9

Find the slope  between the given points

[-2 - 7 ] / [-3 - -1 ] =    -9 / -2  =  9/2

Since we have a line.....the slope bewtween  any  two points is the same

So

[y - 7]   / [ -6 - -1 ] =   9/2

[y - 7] / -5  = 9/2

y - 7  = -5 (9/2)

y - 7  = -45/2

y = -45/2 + 7  

y = -31/2

We can solve this as follows

1/5 =2 / (x + 1)

x + 1  =  2  / (1/5)

x + 1  =  10

x = 9 

So

f^-1( 1/5)  =  9

Proof      

f(9)  =1/5    so    (9,1/5) is on the graph of f

f^-1 reverses  the coordinates    so   (1/5,  9)  is on the graph of the inverse

Since there are 360^o in a circle and a 90^o sector has been removed, 270^o remain.

Therefore, you want 270^o / 360^o, or 3/4^ths of the circumference (plus the two radii).

The formula for the circumference of a circle is:  2·pi·radius.

[ (3/4) · (2·pi·20) ] + [ 2·20 ]  =  the perimeter of the shaded region

Your  " - 171" should read - 71

Your sequence looks like this:

88 , 85 , 82 , 79 , 76 , 73 , 70 , 67 , 64 , 61 , 58 , 55 , 52 , 49 , 46 , 43 , 40 , 37 , 34 , 31 , 28 , 25 , 22 , 19 , 16 , 13 , 10 , 7 , 4 , 1 , -2 , -5 , -8 , -11 , -14 , -17 , -20 , -23 , -26 , -29 , -32 , -35 , -38 , -41 , -44 , -47 , -50 , -53 , -56 , -59 , -62 , -65 , -68 , -71 , -74 , >>Number of terms = 55>>Total Sum == 385

So,  -71 would be the 54th term.

$f^{-1}(1/5) = \frac{4}{3}$

I assumed " shorter than 3 units":

I will assume two cut points  x  and  y,  each chosen using a uniform distribution of reals from  [0,5].  Let us consider the square in  2 -space with corners at  (0,0)  and  (5,5)  and find the areas within that square for which the longest piece has length  ≥3. 

That occurs when both  x  and  y  are  ≤2.  That corresponds to a  2×2  square in the lower left. Similarly, when both are  ≥3,  we have an area  4  square in the upper right.

It also occurs when  x−y≥3.  That corresponds to a triangle in the lower right bounded by the line from  (3,0)  to  (5,2).  It has area  2.  For  y−x≥3,  we have an area  2  triangle at the upper left.

Those four areas are disjoint and add up to total area of  12.  In the area  13  space that remains, no piece has length  ≥3. 

Answer:  13/25

"What is the probability that all three resulting pieces are shorter than  units?"  How many units?

That is wrong. Does anyone else have an answer to this?

9 + 27d = - 840

d = - -849/27

9 + 18d = 23rd term = -557

The probability is 9/25.

3 + x             1 + x

____  =       ______                cross-multiply

5 + x             4 + x

(3 + x) ( 4 + x)  = (5 + x) ( 1 + x)          simplify

x^2 + 7x  + 12   =  x^2 + 6x + 5          subtract x^2 from both sides

7x + 12  = 6x + 5   

7x - 6x  =  5 - 12

1x = -7

x = -7

Actually , -171 will not appear

The difference between successive terms = -3

So....the  nth term is given by

88  - 3 (n -1) =   

91 - 3n

But note  that   no interger  n  can produce  -171    because

-171  =  91 - 3n

-171 - 91   =  -3n

-262 = -3n

-262  / -3  =  n   =   87 1/3   which is not an integer

(n-2) * 180 =

(21-2)(180) =

19(180) = ____________ degrees total

No prob  !!!!

Note that the first few terms of the sequence are :

1, 3 , 7 , 15 , 31 , 63 , 127  ..... 

The  terms are   2^n    - 1     where  n  =  the nth term 

a₂₃ =  p =   2^23 - 1  =   8388607

Great! Thanks CPhill