CPhill

We have

6 [ 1 - 5^(5-1) ] / [ 1 - .5]  + 6  =  17.25 ft

I get 

3,5,7,13,31,61

3 * 5 * 7  =  105

13 + 31 + 61   = 105

There may be something better .....

See the following image :

ED  = r

Angle HEF  = 120°     so angle DEF = 60°

And angle EDF  = 90°

And angle EFD = 30°

So triangle EDF is a 30 - 60 -90 right triangle  with ED = r, DF = √3 r  and EF = 2r

The area  of the smaller  equilateral triangle  = 3(1/2)(ED)^2* sin (120°)  =  3(1/2)(r)^2(√3 / 2) 

The area of the larger equilateral triangle = 3(1/2) (EF)^2  * sin (120°) = 3(1/2) (2r)^2 (√3 / 2)

So we have that

(3√3)/4 * [ (2r)^2 - r^2]  = 36

(3√3)/4 * [ 4r^2 - r^2 ] = 36        multiply through by 4

3√3 [ 3r^2]  =  144        

9√3 r^2  =  144

r^2  =  144 / (9√3)

r^2  =  16/ √3

r =   4 / ⁴√3   cm  ≈  3.04 cm

Whenever the hyperbola intersects the x axis  [like the one we did ]

The equation of the asymptotes  is

y  = ± (b/a) ( x - h) + k       where (h, k) is the center of the hyperbola

If the hyperbola intersects the y axis we have the form

(y - k)^2        ( x - h)^2

______    -   ________   = 1

 a^2                  b^2

And the equation of the asymptotes is

y  =  ± (a/b) (x - h) +  k     where (h, k)  is the center

BTW....here's a resource that I find helpful :

https://www.purplemath.com/modules/index.htm

Last one

x^2         y^2

___  +   ____  =   1   

25          16

The " + "  between the terms signals an ellipse

Like the hyperbola before, this is centered  at the origin = (0, 0)

The major axis is along x  and the minor along y

a^2  = 25       b^2  = 16

a = 5     b  =  4

The vertices here are located at   ( 5, 0) , (-5, 0), (0, 4) and (0, -4)

The foci are located on the major axis and  are given by  ( ±√[ a^2 - b^2 ] , 0) =

( ±√[ 25 - 16 ] , 0 )   =   (± √9 , 0 )   =  (3, 0) and (-3, 0)

Here's a graph :  https://www.desmos.com/calculator/j579zix3zu

Second one

x^2          y^2

___   -     ____   =  1

 25             1

We have the form

(x - h)^2           (y -k)^2

______   -      ________   =       1         

    a^2                  b^2

This is a hyperbola intersecting the x axis  with a center  at (h, k)  = (0, 0)

a = 5      and b  = 1

Tve vertices   lie    along the x axis  and  are given by  (a, 0)  and (-a, 0)  = (5, 0) and (-5, 0) 

The foci  are given by   ( √[ a^2 + b^2], 0 )    and (- √[a^2 + b^2], 0 )   =

( √ [25 + 1] , 0 )   =  (√26, 0)       and  ( - √26, 0)

The asypmtotes are given by :

y = ± (b/a) (x - h) + k     =   

y  = ± (1/5) ( x - 0) + 0    = 

y =  ± (1/5)x 

Here's a graph:  https://www.desmos.com/calculator/aabdp29eok

First one

-16x  = y^2      this is a parabola  opening to the left.....we have the form

4p ( x - h)  =  (y - k)^2

4p ( x - 0)  = (y - 0)^2

The vertex is at  (h, k)  = (0,0)

We can find "p" [ the distance between the focus and the vertex ] as

4p = - 16       divide both sides by 4

p = - 4

The focus is given by

(0 + p, 0)  =  (0 + (-4), 0)  =  (-4, 0)

The directrix is given by

x =  - (p) 

x= - (-4)

x  = 4

Here's a graph :  https://www.desmos.com/calculator/cdaauapqz1

Nice, sinclairdragon.....you approached it from the other "end"  !!!!

Probably easier ways to do this....but...let the first term  = a

So we have

a , (2a + 3) , (4a + 9) , (8a + 21), (16a + 45), (32a + 93), (64a + 189)

So

64a + 189  = 125

64 a  =  -64       divide both sides by -64

a  =  - 1

So  ....the sequence is

-1, 1, 5, 13, 29, 61, 125