CPhill

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See  EP's  answer  below

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Let the area of the smaller polygon be  A

The scale  factor  of the larger polygon to  to the smaller polygon  = 18/12 =  3/2

So....the  area  of the larger polygon =  (scale factor)^2 * A =   (3/2)^2 * A =   (9/4)A

So......we have that

A  + (9/4)A   =  65

(13/4)A  = 65       multiply both sides by   4/13

A =  65 (4/13)  =   20

So......the area of the  larger polygon =  (9/4) * 20   =   45

See the following image  :

Let  C = (1,0)    A  = (-4, 3√3)     B  = (4, 3√3 )

Construct right triangle CHE

And  angle ACD  = angle BAC

So  we can also construct right triangle AIF  with AI = CH

And by ASA, right triangles CHE  and AIF  are congruent

So  F has  the x coordinate value of  -3

And by symmetry, we can note that the x coordinate of  G  = 3

So  FG  =  6

[  65 + 68 + 66  + 80  +  R  ]  /  5 =    70       where R  = Friday's run

[ 279  + R ]  / 5  =  70              multiply both sides by 5

279 + R  =  350         subtract  279 from both sides

71 km  = R

The function is positive on the intervals where it lies above the x axis....

These  are

(-4, 0)    and  (6, 8 ]

Note :  1.25kg  = 1250g

[ 850 + 1250 + 950   + W  ]  / 4 =    975

[3050 + W  ]  / 4  = 975                       multiply both sides by 4

3050 + W  = 3900        subtract  3050 from both sides

W =  850 g  =   weight of 4th bag

⁴√[4096a⁸b²⁰ ]  =  ⁴ √[ 2¹²a⁸b²⁰ ]  =   (2^12 )^(1/4)  * (a^8)^(1/4)  * (b^20)^(1/4)  =  2^3 * a^2 * b^5  =

8a²b⁵

√ [ 72 x¹² y ⁴ ] =  √ [ 36 * 2 ] *(x^12)^(1/2) * (y^4)^(1/2)  =   √2  * √36  * x⁶ * y²  =

√2  *  6  x⁶y²   =

6x⁶y² √2

Note that  we can  write

    1                                             1

______  - x^3 - y^3       as     ________  - ( x ^3 + y^3)     

x^3y^3                                      x^3y^3

xy ( x + y)  =  1       ⇒  x^2y + y^2x  = 1   ⇒  [ 3x^2y + 3y^2x ] = 3

x + y  =  1 / [xy]           cube both sides

x^3  + 3x^2y + 3y^2x  + y^3  =   1 / [x^3y^3 ]

x^3  + [ 3x^2y + 3y^2x ] + y^3  =  1 / {x^3y^3]

x^3 +   3   + y^3    =  1 / [ x^3y^3]

3  =   1 / [ x^3y^3 ] - (x^3 + y^3]

We can find the common difference as

76  -100          -  24

_______  =       ___    =   -  8

     3                   3

So  we need to solve this

100 -  8(n - 1)  =  0

100 - 8n + 8  =0

108  - 8n  = 0

54 - 4n  =  0

54  = 4n

54/4   = n

27/2  = n

13.5  = n

The  14th term will be < 0   ......so

100 - 8 (14 - 1)  =

100 - 8 * 13  =

100 - 104   =

-4