CPhill

121=   11^2

So

11^2 * 11  = 11^3   =   1331  =  smallest  perdect cube divisible by 121

Hint :  Multiply  top / bottom  by     [ 3^(2/3)  + 2^(2/3) + 6^(1/3)  ]

We have two equations

2x - 1  = x + 2               - ( 2x - 1) =  x + 2

x = 3                               -2x + 1  = x + 2

                                           -1  =  3x

                                               x =  -1/3

Sum =   3   - 1/3   =  9/3 - 1/3  =   8/3

sqrt  [ 1^2  + 7^2  ]  =  sqrt [50]  =   the farthest from  the origin = (-1,7)

xy^2 = 6   →   x =  6/y^2      (1)

x^2 y^6  =  72xy      →   xy^5  =  72    (2)

Sub  (1)  into (2)

(6/y^2) * y^5  =  72

6 * y^3  =  72

y^3  =  72/6 = 12

y = (12)^(1/3)

x=  6 / [ 12^(1/3) ] ^2 =  6 / (12)^(2/3)

xy =  6/ (12^(2/3) * 12^(1/3)  =  6 / (12^(1/3) =    [6 ] / [ 2^(1/3) * 6^(1/3) ]   = (6)^(2/3) / (2^(1/3)  =

3^(2/3)  * 2^(2/3)  / 2^(1/3)  =

(9)^(1/3) * 2^(1/3)  = 

∛9 * ∛2 =

∛18

F is also the midpoint of AD

Equation of line  containing   segment  AE  is

y = 2x

Equation of line containing segment BF  is

y = (-1/2)x + 1

Setting these equal  to get the x coordinate of the intersection of the segments

2x = (-1/2)x + 1

(5/2)x = 1

x = (2/5)

y =  2 (2/5)  = 4/5

Call the intersection pt   (2/5, 4/5)  = G

The area of right triangle DEF =  (1/2) (1) (1)   =1/2

Area of right  triangle AGB   = (1/2)(2)(4/5) = 4/5

Area of right triangle AFG =  (1/2)(2/5)(1)  =  1/5

Area of  ECBF =

Area of ABCD - Area of DEF - Area of AGB   - Area of AFG   =

2^2 - (1/2)  - (4/5) - (1/5)  =

4 - 1/2 - 1  =

2.5 cm^2

1 : 2 : 3 :  x

  x  /  [ 1 + 2 + 3 + x  ]  * 64

x  =

2

6

10

18

26

42

58

Number divisible by 2 =   floor [ 280  /2] =  140

Number divisible by 3 = floor [ 280/3 ]  = 93

Number divisible by  5 =  floor [ 280 / 5]   =  56

Number divisible by 2* 3  =   floor [ 280 / 6 ] = 46

Number divisible by 2*5  =  floor [ 280 / 10 ] = 28

Number divisible by 3*5  = floor [ 280 / 15] = 18

Number divisible  by  2 * 3 * 5  = floor  [ 280 / 30  ] = 9

N ( divisible by 2 or 3 or 5)  = 

140 + 93 + 56  - 46 - 28 - 18 - 9  =  188

Number NOT divisible by  2, 3 or 5  =    280 - 188  =   92

Slope of second line =  (-2/-3)  =    (2/3)

So....since the lines are parallel the slopes are the same......so....

[ 17  - - 9 ] /  [ j - 2 ]   =  2/3

26  =   (2/3) [ j -2]        

26 (3/2)  =  j - 2

39  =  j - 2

39 + 2  = j

41   =  j

(7^2 + 24^2)*4*(5^2*10^2)*5/((75^2*100^2)*6)  =

[625 * 4 * (5^2 * 5^2 * 2^2) * 5]  / [ 3^2 * 5^4 * 5^4 * 2^4 * 6] =

(  5^4 * 5^2 * 5^2 * 5  * 4 * 4 ) /  [ 5^4* 5^4 * 3^2 * 16 * 6 ]  =

( 5^9 )  ] / [ 5^8 * 54 ] /  =

5 / 54  =   5 ( 1/54)

n = 1 / 54