CPhill

17b =   (1b + 7)

(1b + 7)^2  =  b^2 + 14b + 49

211b  = 2b^2 + 1b + 1

b^2 + 14b + 49  = 2b^2 + 1b + 1

b^2 - 13b - 48 = 0

(b - 16) ( b + 3)  = 0

b - 16  = 0

b = 16

\(4+ { \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{4}}}} \)

Work from the bottom-up

4 + 1/4 =  17/4

1/(17/4) = 4/17

4 + 4/17  = 72/17

1/(72/17)  = 17/72

4 + 17/72  =  305 / 72

2.57 repeating

[257 - 2 ] / [990 ] =  255 / 990 = 17 / 66

11012 / 100000   =   2753 / 25000

12 people can  count 1/4 box in  one hour

So 1 person  can count  1/ 4 * 1/12  =    1/48 box in  1 hr

So

30 people can count   30 * (1/48)  = 30/48 =  5/8  of a box in  1  hour

So

30  / (5/8) =   48 hrs

1/x - 1/(xy) - 1 / (xyz)  =  1/3

Let x  =1

1 - 1/3 =  1/(y) + 1/(yz)

2/3  = 1/y + 1/(yz)

2/3 = [ z + 1] / [ yz]

2yz = 3z + 3

2yz - 3z  = 3

z (2y - 3)  =  3

  If y = 2  then z  = 3

x,y,z = ( 1,2,3)

Check

1/1 - 1/2 - 1/6  =   1/3

1/977  =  .001........

2^4 * 3 =   48

4

1^2 =  1

2^2 =  4

P

  x =   3 + 2sin (2 *-pi6)  =  3 + 2 sin (-pi/3)  = 3 + 2  *(-sqrt 3 / 2) = 3 - sqrt 3

  y =    4 + 2cos (2 *-pi/6)  =  4 + 2 cos (-pi/3)  =   4 + 2(1/2) =  5

Q

x = 3 + 2sin (2*0)  = 3 + 2(0) =  3

y = 4 + 2cos (2*0) =  4 + 2 (1) = 6

R

x= 3 + 2sin (2 *pi/3) =  3 + 2 sin (2pi/3) = 3 + 2 (sqrt 3 / 2) =  3 + sqrt 3

y = 4 + 2cos(2 *pi/3) = 4 + 2 (-1/2) = 3

S

x = 3 + 2sin (2 *pi/2)  = 3 + 2 (0) =  3

y = 4 + 2cos (2 *pi/2)  = 4 + 2(-1) = 2

Excellent, NTS !!!!

{Without the Totient Function , this would be very tedious....  Euler was a pretty smart guy......!!! }

https://web2.0calc.com/questions/probability_27845

Simplifying, we have

 sqrt [ ( sqrt 5 - sqrt 2)^2 ] + sqrt [ (sqrt 5 + sqrt 2)^2 ]

___________________________________________   =

                sqrt [ 5 - 2 ]

[ sqrt 5 - sqrt 2  ] + [ sqrt 5 + sqrt 2 ]

____________________________     =

                 sqrt 3

2sqrt 5

______  =

  sqrt 3

(2/3) sqrt 15

f(x) =  x^3 -6x^2 + 2x 

f (f(x)) =  [ x^3 - 6x^2 + 2x]^3 - 6 [ x^3 -6x^2 + 2x]^2 + 2[ x^3 - 6x^2 + 2x] =

x^9 - 18 x^8 + 114 x^7 - 294 x^6 + 300 x^5 - 312 x^4 + 154 x^3 - 36 x^2 + 4 x

So

x^3 -6x^2 + 2x  = x^9 - 18 x^8 + 114 x^7 - 294 x^6 + 300 x^5 - 312 x^4 + 154 x^3 - 36 x^2 + 4 x

x^9 - 18 x^8 + 114 x^7 - 294 x^6 + 300 x^5 - 312 x^4 + 153 x^3 - 30 x^2 + 2 x =  0

With a little help from WolframAlpha :

Solutions are :

                    C

            E

A              F          D          B

       7            2

Let DB = x

Since EF parallel to CD....Triangle AEF is similar to triangle ACD

AE/AC = AF/AD

AE/AC = 7/9

Since ED parallel to BC......Triangle AED is similar to triangle ACB

AE/AC = AD/AB

AE/AC = 9/ (9 + x)

So

7/9 = 9 / (9+x)

7(9+x) = 9*9

63 + 7x  = 81

7x = 18

x = 18/7  = DB

Using the Law of Cosines twice :

[BC^2 - BA^2 - AC^2] / [ -2 BA * AC]  = cos DAE

[8^2 - 3^2 - 7^2 ] / [ -2 * 3 * 7 ] =  cos DAE = -1/7

And

DE^2  = DA^2 + AE^2 - [2* DA * AE] cos ( DAE)

DE^2  = 21^2 + 9^2  - [ 2* 21 * 9] (-1/7)

DE^2  =  522 + 54

DE^2 = 576

DE = sqrt 576  =  24

191 is prime

phi (191)  =  (191 - 1)  =  190

1=1             5 = 5            9 = 3^2             13 =13 

2=2             6 = 2 * 3      10 = 2*5            14= 2*7

3=3             7 = 7            11=11                15 = 3*5

4=2^2         8 = 2^3         12 = 2^2 * 3      16 = 2^4

LCM = 2^4 * 3^2 * 5 * 7 * 11 * 13  = 720720 = M

No. of Divisors  =  5*3*2*2*2*2 = 240