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Not a geometric sum

r between the first two terms = 9/100

r between the 2nd and  3rd term =  9/10000

Let

A= (a,b)

B=(0,0)

M = (c,0)

D = [ (a +c)/2 , b/2)

E = (a/2,b/2)

DE is  parallel to BM   because the slope of  both segments  =  0

BM = c

DE = c/2

So DE = (1/2)  of BM

Angle DET = Angle BMT

Angle DTE  = Angle BTM

So triangle EDT  is similar to  triangle MBT

Since DE =(1/2) of BM.....then  ET = (1/2)MT

Since triangles DET and DMT are under the same  height and ET = (1/2)TM  then [DET] = (1/2)[DMT ] = (1/2)(20) = 10

And DT = (1/2)BT  ⇒  2DT = BT

And since triangles DET and BET  are under the same height, then  [BET] = 2 [ DET] = 2 (10)   = 20

1 + \frac{1}{5} + \frac{2}{5} + \frac{3}{5} + \frac{2}{25} + \frac{3}{25} + \frac{4}{25}   

           1         2          3           2         3           4  

  1 + –––– + –––– + –––– + –––– + –––– + ––––   

           5         5          5          25       25         25            

  You have to get everything with a common denominator.  

    25        5         10         15         2          3         4  

  –––– + –––– + –––– + –––– + –––– + –––– + ––––  

    25       25         25        25        25        25        25            

    64  

  ––––   =   2 & 14 / 25    

    25   

_.

Numbers    2       7       12           17             22

Sum               2        9           21           38             60

Let R =(0,5)    ,  O = (2,0)    M = (1,0)

The line through RQ  has a slope of    (-5/2)

The line through PL is perpendicular to the first line so it has a slope of 2/5   and it  passes through (0,0) so its equation is   y= (2/5)x       (1)

The line through  RM  has a slope of  -5   and passes through  (0,5)

So its equation is   y =-5x + 5         (2)

Equatiing (1) and (2)  to find the x coordinate of  F

-5x + 5  =(2/5)x                muliply hrough by 5

-25x + 25 = 2x 

25 = 27x

x = 25 / 27        and  y =  (2/5)(25/27)  =  10/27

PF =  sqrt  [ (25/27)^2 + ( 10/27)^2 ]   =  sqrt [ 25^2 + 10^2] / 27  = sqrt [ 725 ] / 27 =  

sqrt [ 25 * 29 ] /27  =  (5/27)sqrt (29)

x^3  + y^3  =  162 - (x^2 + y^2)

(x + y) (x^2 - xy + y^2)  = 162 - (x^2 + y^2

6( x^2 + y^2)  - 6xy = 162 - (x^2 + y^2)

7 (x^2 + y^2 + 2xy)  -20xy = 162

7 ( x + y)^2  - 20xy = 162

7 (6)^2  - 20xy  =162

90  = 20xy

9/2 = xy

(9/2)/x = y

x + (9/2)/x  = 6

x^2 + (9/2)  = 6x

x^2 - 6x   =  -9/2

x^2 - 6x + 9  = -9/2 + 9

(x - 3)^2  = 9/2

x -3  = sqrt(9/2)         or      x - 3  = -sqrt (9/2)

x = 3 + 3/sqrt (2)                x =  3 - 3/sqrt (2)

(x, y)  = ( 3 + 3/sqrt2 , 3- 3/sqrt 2)  and  ( 3 - 3/sqrt2 , 3 + 3/sqrt 2)

x^2  + (-4)^2  = 20

x^2 + 16 =  20

x^2  = 4

x= 2      or         x  =   -2

y = -4               y =    -4

AB =   2- (-2)  =  4

a1 + a3 =  1    ⇒  a1 + (a1 + 2d)  = 1  ⇒  2a1 + 2d  = 1           (1)

a2 + a4  = 2   ⇒  (a1 + d) +  (a1 + 3d)  = 2 ⇒  2a1 + 4d =  2    (2)

Subtract (1) from (2)

2d  = 1

d = 1/2

Using (1)

2a1 + 2(1/2)  = 1

2a1 + 1 = 1

2a1  = 0

a1 = 0       a2 =1/2    a3 = 1    a4 = 3/2

x^2 - 4x + y^2 - 2y  = 0

x^2 - 4x + 4  + y^2 - 2y + 1  =  5

(x - 2)^2  + (y - 1)^2  =  5

This is a  circle  centered at  (2, 1)  with a radius of sqrt (5)

x max  =  2 + sqrt (5)

The height of the trapezoid is 3 times the height of the  equilateral triangle

Area of equilateral triangle = (1/2)base * height  = (1/2) (2) (H) =  H

Bases  of trapezoid = 2 , 8......height =  3H

Area of trapezoid=   (1/2) (3H)( 2 + 8)  =   15H

Ratio of areas =   H / ( 15H)  =   1 / 15

\(\frac{x}{a} + \frac{y}{b} = 1 \)

Slope of line  = [ -5 - 4 ] / [ -1 - 5 ]   =  -9/-6  =  3/2

Equation of line   y  =  (3/2)(x - 5)  + 4

y = (3/2)x - 7/2      multiply through by -(2/7)

-(2/7)y = - (3/7)x + 1

(3/7)x - (2/7)y  =   1         and  we  can  write

 x/ (7/3)  + y / (-7/2)  = 1

a = (7/3)

4 *Area of ADE  =  (1/2) AD * DE = 

Area of quadrilateral ABCE  = (1/2)(AD) (AB + EC)

4 (1/2) AD * DE  = (1/2) (AD) (AB + EC)

4DE = AB + EC

4DE = (DE + EC) + EC

3 DE = 2 EC

DE / EC  =  2 / 3

y= x

y =2x

We can  find  m thusly 

  m  =  tan [ (arctan(slope of one  line) + arctan(slope of second line) / 2  ]  

  =      tan ([ arctan (2)   + arctan(1)] /  2)   ≈ 1.387

Since AB  = AC and BAC = 30

Then angle  ACB = (180 -30) / 2  = 75

And since BAC = 30.....then  minor arc BC = 60...so  angle BCE = (1/2)(60) = 30

So angle ACD  =180  - BCE - ACB   = 180  - 75 - 30 =  75°

Semiperimeter of top triangle =  19/2 = 9.5

Semiperimeter of bottom triangle  = 20/2 = 10

Area =  sqrt [ 9.5 ( 9.5 - 9) (9.5 - 7) (9.5 -3) ] + sqrt [ 10 (10 - 9) (10 -6) (10 - 5) ] =

sqrt [ 9.5 * .5 * 2.5 * 6.5 ]  + sqrt [ 10 * 4 * 5  ] =

sqrt [ 77.1875 ] +  sqrt [ 200]  ≈  22.93

WY  (by graph) ≈  5.34

Draw IZ  and YI

Angle AZY = 44°

So   minor arc YZ = 2 * angle AZY = 2 (44)  =   88°

And central angle YIZ intercepts this arc....so  its measure = 88°

This is AoPS homework.  Do not post AoPS homework.

Let 12 o'clock represent  0°....let the hands advance in a  positive angular  direction

At  3 o'clock   the hour  hand is  at 90°   and the  minute  hand is  at 0°

The hour hand  moves   15°   more   in  30 minutes.....so it is at 105°   at 3;30

And the minute hand is  at180°    at 3:30

So....the acute angle is   180 - 105  =   75°

(90 + d) / (60 + d)  = r , (165 + d) / (90 + d)  = r

Setting equal  for r  and simplifying

(90 + d)^2  = (60 + d) ( 165 + d)

d^2 + 180d + 8100 =  9900 + 225d + d^2

-45d = 1800

d = 1800 / -45  =  -40

r = [ 90 - 40 ] / [ 60 - 40 ] =  50 / 20   =  5/2

This is AoPS homework.  Do not post AoPS homework.

(1000) (.9)^(n - 1)  =  1

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

log .9^n  = log (1/1000)

(n-1) log (.9)  =  log (1/10000

n -1 = log (1/1000) / log (.9) ≈  65.56

n = 65.56 + 1 =   66.56

66 terms will be > 1

If Alice receives $n$ dollars from Bob, then she will have $7$ times as much money as Bob. 

A + n  = 7 ( B - n)        (1)

If, on the other hand, she gives $n$ dollars to Bob, then she will have $2$ times as much money as Bob

A - n  =  2 ( B + n)      (2)

Subtract  (2)  from (1)

2n = 5B - 9n

11n = 5B

n = (5/11)B

Using (1)

A + (5/11)B  = 7 (B - (5/11)B)

A + (5/11)B  = 7B - (35/11)B

A = 7B - (40/11)B

A =  (37/11)B

A / B  =  37  / 11

pi =  180°

On the unit circle,  the terminal point of this angle  can  be represented  by (x , y)  =  (-1. 0)

tan (pi)  =   y /  x  =   0   / -1  =   0

We have one choice for A and three for R. After that, no matter which R we choose, we'll have 3 C's to choose from. After that, we'll have three H's to choose from. Therefore, the answer is 3^3 = 27. :D