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a)

f(x)   =   3x - 2       Instead of  " f(x) "  let's say  y .

y  =  3x - 2            Now let's solve this equation for  x .

                             Add  2  to both sides.

y + 2  =  3x

                             Divide both sides by  3 .

(y + 2)/3  =  x

                             So the inverse function is....

f^-1(x)  =  (x + 2)/3

b)   I guess that this means find  (g o f)(x) .

(g o f)(x)  =  g( f(x) )

(g o f)(x)  =  g( 3x - 2 )

(g o f)(x)  =  10/( 3x - 2 ) + 2

$x = {-2m \pm \sqrt{(2m)^2-4(m^2+2m+3)} \over 2} = {-2m \pm \sqrt{4m^2-4m^2-8m-12} \over 2}= -m \pm \sqrt{-2m-3}$

$r = -m + \sqrt{-2m-3} \\~\\ r^2=(-m + \sqrt{-2m-3})(-m + \sqrt{-2m-3}) \\~\\ r^2=m^2-2m\sqrt{-2m-3}+(-2m-3)$

and

$s = -m-\sqrt{-2m-3} \\~\\ s^2=(-m-\sqrt{-2m-3})(-m-\sqrt{-2m-3}) \\~\\ s^2=m^2+2m\sqrt{-2m-3}+(-2m-3)$

$r^2+s^2=m^2-2m\sqrt{-2m-3}+(-2m-3)+m^2+2m\sqrt{-2m-3}+(-2m-3) \\~\\ r^2+s^2=2m^2+2(-2m-3) \\~\\ r^2+s^2=2m^2-4m-6$

The minimum value of  r^2 + s^2  occurs when  m   =   4/( 2 * 2 )   =   1

When m = 1 ,   r² + s²   =   2 - 4 - 6   =   -8

https://math.stackexchange.com/questions/1897276/log-4n-40-sqrt3-log-3n-45-find-n3

{YOU ARE NOT ONLY A PHONY BUT A FLAGRANT PLAGIARIST AND A PATHOLOGICAL LIAR!!  SHAME ON YOU!!]. 

a) In order to factor using this method, let's try and identify the GCF first. 9 is the greatest common factor between 36 and 81. a^4 is the greatest common factor between the a's, and b^10 is the factor for the b's. Let's factor it out!
 

b) The beginning binomial is a difference of squares to begin with, so it is possible to start with this first! 

Well, these are the two techniques. 

Find the inverses

y =  3 - 4x                                         y   =   3 + 4x                      

y - 3  = - 4x                                       y - 3  =  4x

3 - y  = 4x                                         [ y - 3 ] / 4  = x

[3 - y ] / 4  = x                                   [ x -  3 ] / 4  =  the inverse

[3 - x] / 4  = the inverse

So..the sum of these inverses is :

[ 3 - x] / 4   +  [ x - 3] / 4  =

[ 3 - 3 - x + x ] / 4   =    0 / 4   =   0

4) volume of sphere: 4/3 * 3.14 * 10^3

volume of sphere: 4/3 * 3.14 * 1000

volume of sphere: 4186.67

The first one    f(x) = y = 3-4x    solve for 'x'

y-3 = -4x

3-y = 4x

(3-y)/4 = x      now switch the x's and y's

f^-1 (x) =  (3-x)/4

The second one  y=3+4x

                            (y-3)/4 = x

f^-1 (x) =  (x-3)/4

Now add the two f^-1  (x) in red       (3-x)/4 + (x-3)/4    =  (x-x +3-3)/4 = 0

6) 45cm^2 * 2 = 90cm^2 new area

3. 

It's easiest to simplify the numerator and denominator separately

1/x^2  -  2/y   =   [ y - 2x^2] / [ x^2y]

5/x  - 6/y^2  =  [ 5y^2  - 6x ] / [ xy^2]

So we have

( [ y - 2x^2] / [ x^2y] )  / (  [ 5y^2  - 6 ] / [ xy^2] )

Invert the second fraction and multiply

y  - 2x^2              xy^2       

______     *      __________ 

  x^2y                  5y^2 - 6 

y - 2x^2             xy^2

_______   *  ________

5y^2 - 6            x^2y

y - 2x^2             y

______   *      ___

5y^2 - 6            x

y^2  - 2x^2y

__________

5xy^2 - 6x

7) 3 : 2

3^3 : 2^3

9 : 6

5) 42 * 6.5 = 273 new perimeter

Ah, yes...I see that, now.....thanks for the the correction!!!!

Here's the graph, Sam

https://www.desmos.com/calculator/v1v3mj3tqk

Picking the point (0,4)  shows that this area is correct -  it makes both equations true

Write this as

y  =  24 [1 + 3e^(-1.3x)]^(-1)

The derivative of this function is

y '  =     -24[ 1 + 3e^(-1.3x)]^(-2) ( -3.9e^(-1.3x))

Looking at the graph of the derivative here : 

https://www.desmos.com/calculator/sunmexdhiz

We see that the max growth rate  occurs at  about  x  = .845

The graph of the function here : 

https://www.desmos.com/calculator/pkls30n5mn

Shows that   the y  value associated with this x value  = 12

So....the max growth rate occurs  at ≈  (.85, 12 )

CPhill: The question says "The three-digit positive integer  has a ones digit of 3". Doesn't that  mean that the 3rd digit is fixed at 3? E.G. 103, 113, 123, 133, 143, 153 163, 173, 183, 193......and so on for a total of 9 x 10 = 90 3-digit numbers ALL ending in 3. And there would be: 90/3 =30 numbers out the 90 divisible by 3. Or 30/90 =1/3.

8) This one will require the formula that yields the volume of a cylinder. $V_{\text{cylinder}}=\pi r^2h$. We can manipulate this formula so that we can find any missing information such as the height, in this case. 

9) If the height of the un-consumed soup was 8 centimeters tall and 3-centimeters-worth of soup is consumed, then 5 centimeters of soup remains. We already know the radius of this soup can (that I assume is cylinder-shaped despite not being explicitly stated), so we can determine the volume.

10) If the town park enlarges its area by a factor of 5, then both dimensions of the park are affected by this scale factor. For example, if we assume, for the sake of understanding, that the park is perfectly rectangular with dimensions 3yd by 100yd, then both dimensions (the length and the width) would be affected by this scale factor. This means that, on area, the scale factor actually affects the area by its square, or 25 in this case. 

$300\text{yd}^2*25=7500\text{yd}^2$

Deleted....

For this to be divisible by 3, the sum of the first two digits must be divisible by 3

There will be   (99-12)/ 3 + 1  = 30  such numbers

And there are (999-100) + 1  =  900 three digit numbers....so

The probability that one ends in  "3"  and is divisible by 3 is just

30 / 900      =    1/30

The problem is easier than I first thought. 

By the given information, we know that there are three right-angled triangles in the diagram. We know that $m\angle AEB=m\angle BEC=m\angle CED= 60^{\circ}$. We can use this information to determine that every right triangle is also a 30-60-90 triangle. We also know that $AE=24$.

A 30-60-90 triangle is a special kind of right triangle where the ratio of the side lengths are $1:\sqrt{3}:2$. $\overline{AE}$ is the longest side length because it is the hypotenuse of the largest right triangle in the diagram. $\overline{BE}$ is the shortest side length of $\triangle ABE$ because this side is opposite the smallest angle. We mentioned earlier what the ratio of the side lengths are, so we can determine the length of $\overline{BE}$ without doing anything too computationally demanding. 

Of course, the ultimate goal is to figure out the length of $\overline{CE}$. If you look at $\triangle BCE$, carefully, you will notice that we are in an identical situation to when we solved for $BE$. Notice that $\overline{BE}$ is the hypotenuse of this triangle, and $\overline{CE}$ is the shortest side length since it is opposite the 30º angle. We can use the same $1:\sqrt{3}:2$ relationship of the side lengths to find the missing length.

Thank you so much!

First one:

The formula for a sphere is 4/3 pi r^3

 if scale UP 'r' by a factor of 4.5     r^3   will yield a volume  (4.5)^3 larger  = 91.13 times larger

Second question:  

 The formula for the volume of a cone is  pi r^2  h/3

         if we scale it DOWN by 1/2      the   r^2 will be  1/4   and the h/3 will become  1/2 (h)/3

     so the volume will be ruduced to 1/4 x 1/2 = 1/8 of the original   1/8 (360) = 45 m^3

1.

$y=\frac{20}{1+9e^{3x}} \\~\\ y(1 + 9e^{3x})=20 \\~\\ 1 + 9e^{3x}=\frac{20}{y} \\~\\ 9e^{3x}=\frac{20}{y}-1 \\~\\ 9e^{3x}=\frac{20-y}{y} \\~\\ e^{3x}=\frac{20-y}{9y} \\~\\ 3x=\ln(\frac{20-y}{9y}) \\~\\ x=\frac{\ln(\frac{20-y}{9y})}{3}$

There is an asymptote when   9y  =  0   which is when   y = 0

There is an asymptote when   $\frac{20-y}{9y}$  =  0   which is when   y = 20

sin (x) = 0     where ?    at   x = 0 degrees

sin (0 degrees) =0

arcsin(sin (0 degrees)) = arcsin (0)
0 degrees = arcsin (0)

Guest troll visits some strange websites.

Do you know where   cos(x) = -1  ?    This is true at   pi radians (or 180 degrees ----same thing)

    so    cos (pi) = -1

    now arcos both sides

        arcos ( cos(pi) )  = arcos (-1)

                     pi = arcos (-1)

When you use the sin function, you 'undo' it when you use the sin inverse (also called the arcsin) function     so   if you start with theta

    then you use the sine function ,  sin (theta)     then    arcsin ( sin theta) = theta

Your equation:   sin(theta) = x/10      now    arcsin both sides

       arcsin(sin (theta))  = arcsin  (x/10)

                 theta = arcsin (x/10)