CPhill

Here is the equation we need to solve

160 = 80( 1.05)^t    where t is the number of days to double the height

divide both sides by 80

2 =  ( 1.05)^t       take the log of both sides

log 2 = log ( 1.05)^t     and, by a log property, we can write

log 2 = t log (1.05)     divide both sides by log (1.05)

log 2 / log 1.05  = t ≈  14.2 days   

Note that the original height is immaterial....it would take this many days to double the height no matter how tall its original height 

The new width will be (8/5) * 3  =  24/5 =  4 + 4/5 inches

5(5n+9) - 6(4n+10)  =

25n + 45 - 24n - 60  =   [ combine like terms]

n - 15

f(x)=  [4x+1] /  [x-5]  and g(x)=  [5x-1] /  [x+4]

There is a simple test to determine if these are inverses  if  f [ g(x)]  =  g [ f(x) ] = x, then they are inverses....however......we only need one counter-example to show that they are not

If  (x, y)   is on the function f(x), then the reverse coordinates are on  g(x)....

Let's put  x = 0   into the first function.....then f(0)  =  -1/5 ...then the point (0, -1/5)  is on the first graph

Then ( -1/5, 0)   should be on the second  .....but   g(-1/5)  = -10/19 .....so  the coordinates are not reversed and this indicates that the functions are not inverses of each other....!!!

Guest....look at what the progressive sums would be if r = -8

We are evaluating  ∑ 9(-8)^n-1      for n = 1 to infinity

n = 1    sum = 9

n = 2    sum = -68

n = 3  sum =  513

n = 4  sum  = -4095

n = 5  sum = 32,769........and so on

The progressive sums diverge ........!!!!

f(x) = x^2 + 2   for  f(x), write y

y = x^2 + 2        get x by itself

y - 2  = x^2        take the positive and negative square roots of both sides

± √ [y - 2]   = x     "swap" x and y

± √ [x - 2]   = y       for y, write f^-1(x)

± √ [x - 2]   =   f^-1(x)

If we restict the original function to the domain ( -inf, 0), then  - √ [x - 2]   is the inverse function

If we restrict the original function to the domain (0, inf) then + √ [x - 2]   is the inverse function

Note : "0"   can belong to either  of the original domains  and either of the inverse domains

Mmmm.......I not certain that the "guest's" answer is correct...if "r"   is  > 1  or < -1, the series diverges and has  no sum

Note:

9 / [ 1 - ( - 1/2)]  = 6

9 / [ 1 - (- 4/5 )]  = 5

So....we are trying to find some "r"   between  -1 and 1 such that

9 / [ 1 - r) ]  = 4       →   r =  -5/4 ......this series will diverge

And

9/ [ 1 - r ]  = 3  →  r = -2.........and this series will diverge, as well

Seemingly, for the positive integer sums of less than 5, "r"  will  be less than -1....and any such series will diverge

So...it appears that the smallest integer that can be the sum of an infinite series whose first term is 9 is produced when r = -4/5.....and that sum is 5

Circumference = 2*pi*r

Differentiating this with respect to time, we have

dC / dt  = 2*pi *dr / dt

2.5 cm/s =  2*pi * dr/dt     divide both sides by 2*pi

[ 2.5 ]/ [2 *pi] cm/s  = dr/dt   =  [ 5 / (4pi) ] cm/s

3x + 15  =

3 (x + 5)

19² = 361