CPhill

tan x  = 15 / 8

To find the angle x, take the arctan  of 15/8

arctan (15/8)  ≈  61.93°

Lets work with this, first

3                  1

____    -     ___

a^2             3

Cross-multiply ans keep the same sign as between the fractions

3*3 - 1*a^2   =  9 - a^2

Take the product of the denominators 

3a^2

So   we have

( 9 - a^2) 

________

    3a^2

So...puitting this together, we have

[ a^2 / 9 ]  ÷  [ ( 9 - a^2) / ( 3a^2) ]

Invert the second fraction and multiply

a^2               3a^2                      3a^4

___   *         ______    =           _______   

9                  9 - a^2                  9(9 - a^2)

Cancel  the 3 in the numerator and the 9 in the denominator  and we are left with

a^4                        a^4

_______     =      ________

3(9 - a^2)           27 - 3a^2

No prob, Jenny.....!!!

This is kind of counter-intuitive....but......

The  "1/4"   means that we have a horzontal stretch of the original  function  by a factor of 4.....  [ the reciprocal of the fraction ]

In general  we have   y = log (b x)

When b > 0 but  < 1.....we have  a horizontal stretch of y = log (x)

When b > 1......we have a horizontal "compression"   of y = log(x)....   [ not, "comprehension" ]

So... y = log (4x)    has a compression  and  the compression  factor is a reciprocal of 4  =  1/4

Look at the graphs, here : https://www.desmos.com/calculator/tbguagxjhp

Remember that  a^m  x a ^n   =  a^(m + n)

So.....the numerator becomes

3^8  x 3^4   = 3^(8 +4)  =  3^12

And the denominator becomes

3^ 2  x 3^8  =  3^(2 + 8)  =  3^(10)

So  we have

3^12

_____        (1)

3^10

And

a^m

____    =     a^(m - n)

a^n

So (1)  evaluates to    3^(12- 10)   =  3^2  

We have the following "formula"

A  = P ( 1 + r)^t

Where

A = 800

P = 500

r = 1.5%  = .015

And we are looking for t

So we have

800  = 500 ( 1 + .015)^t       divide both sides by 500

(800/500)   =  (1 + .015)^t

(8/5) = (1.015)^t        take the log of both sides

log (8/5)  =log (1.015)^t        and we can write

log(8/5)  = t * log (1.015)     divide both sides by  log(1.015)

log(8/5) /log (1.015)  = t ≈  31. 57  yrs  =  32 years (to the nearest year)

Thanks, hashtagmath.......

 Your answers are well-presented......!!!

Answered here :   https://web2.0calc.com/questions/geometry-problem_21#r1

There may be a much easier way to do this, but we can find the volume of the frustum by creating a cross-sectional kite .......the sphere will  have a circular cross-section and the frustum will have a cross-sectional trapezoid 

First......since the sphere is tangent to the frustum at both the top and bottom bases, the frustum height  is  =  6

So.....we can create a 6-8-10  Pythagorean right triangle DGF  as shown in the illustration - with one leg of 6 - (GF)  - one leg of 8 - (DG) -  and the slant height - (DF)  -forming the hypotenuse........the cosine of  angle FDG  ( at the bottom right vertex of the frustum ) =   8/10   =  4/5

Now......the radii of the circle will form kite  ABCD  

Connect AC......angle ABC will be supplementary to angle FDG......so its cosine  = -4/5

And because DA and DC are tangents drawn from the same point D, they are equal

And we can use the Law of Cosines to find AC

Note: AB  and BC are radii of the circle

AC^2 =  AB^2 + BC^2  - 2(AB)(BC)(cos ABC)

AC^2  =  3^2 + 3^2  - 2(3)(3) (-4/5)

AC^2 = 18  + 72/5

AC^2  =  [ 90 + 72] / 5

AC^2  =  [ 162 / 5 ]

And Using the Law of Cosines once more, we can find the lenths of DA and DC.....let their lengths = x......and we have

AC^2  =   x^2 + x^2  -  2 (x^2)cos(CDA)

[ 162 / 5 ]  =  2x^2 - 2x^2 [ 4/5 ]

[162/ 5 ]  = 2x^2 [ 1 - 4/5]

[ 162 / 5 ] = 2x^2 [ 1/5]

162 = 2x^2

x^2  = 81

x = 9  =   DA  =  DC

But DA  =  radius  of the bottom base of the frustum

And since DC = 9.....then CF  = FD - DC  =  10 - 9   =  1

But CF = FE since they are tangents drawn from the point F

And FE  =  radius of the top base of the frustum

So.....using the fomula for the volume of a frustum, we have

V  = pi * GF * ( DA^2 + DA*FE^2 + FE^2 ) / 3  

Where 

GF  = frustum height

DA =  radius of bottom base

FE = radius of top base

So we have

pi * ( 6) *  ( 9^2  + 9 * 1 + 1^2 ) / 3  =

pi (6)  ( 81 + 9 + 1 ) / 3    =

pi (6) (91) / 3  =

2  * 91 pi  =

182 pi  units^3

B

2

X

1

A     1     Y         2               C

Note that   BX  = 2AX

And  YC  =  2AY

So we have that

(3AX)^2 + AY^2  = BY^2   ⇒   9AX^2 + AY^2  = 256  ⇒  AY^2  = 256 - 9AX^2    (1)

And

AX^2  + (3AY)^2  = 28^2  ⇒  AX^2  + 9AY^2  = 784  (2)

Subbing (1)  into (2)  we have

AX^2 + 9(256 - 9AX^2)  = 784

AX^2 + 2304 - 81AX^2  = 784

-80AX^2  = -1520

AX^2  =  -152 / - 8

AX^2 = 19

AX = sqrt(19)

So....BA  =  3AX  =  3sqrt(19)

And

AX^2 + 9AY^2  = 784

19 + 9AY^2  = 784

9AY^2  = 765

AY^2 =  85

AY  =  sqrt (85)

So AC  = 3sqrt (85)

So  BC  =  sqrt  [ BA^2 + AC^2 ] = 

sqrt [ 9 * 19 + 9 * 85 ] =

sqrt [ 9 ( 19 + 85)  ]  =

3sqrt (104) =

3sqrt ( 26 * 4)  =

2 * 3 sqrt (26)  =

6 sqrt (26)