Alan

Do you mean y(x) = √(25x-75) + 3.  If so here is a part of its graph:

It goes off to infinity to the right, but doesn't go further left than x = 3 because the term under the square root sign would be negative there. 

I don't know what is meant by "make it easy to graph using a translation".

The function g/f is (x-2)/(3x-6).  This is valid for all real numbers except x = 2, because when x =  2 we have (x-2)/(3x-6) = 0/0, which is undefined.  So the domain is all numbers except x = 2.

We start by giving a label, say x, to the sister's age (you can think of x as a label on a box inside which is the number representing her age).  We have to use the information we are given in order to calculate x (the number in the box).

We are told that the age of Carlos is 2*x + 5 and that this equals 13, so the equation is

2*x + 5 = 13.

Subtract 5 from both sides of this equation

2*x = 8

Divide both sides by 2

x = 4

So his sister is 4 (this is the number in the box labelled x) 

You are right Melody; but what if the centre of dilation was, say (2,3)?  In this case you would need to halve the difference between the Px-values and 2, and the difference between the Py-values and 3, and then add these halved differences back on to 2 and 3 respectively to get the P'x and P'y coordinates.  Rom gave the general procedure first, then started to apply it to the particular data set of the question. 

You have made use of the relation x^a*b = (x^a)^b.  But this has a limited range of applicability. In particular it doesn't apply when x is negative and b is a fraction.  If you insist on applying it outside of its range of applicability you will get inconsistent mathematics (as your example illustrates!).

If you mean how do you plot a function involving log(x) then, select the "create graph" button at the bottom right of the calculator.  Type your function into the box near the top left, and type your desired x and y limits in the boxes below the graph.  See the example below:

 However, as far as I'm aware, you can't choose to have log scales for the axes, if that's what you wanted.

Indeed, the axes don't show a numerical scale.  This graphing capability is severely limited!

Fantastic picture!  I'm not sure I could bring myself to eat such pie should I ever get the chance!

Melody said: I am trying to upload a picture but I haven't worked out how to do it yet!

Here is a bullet point list of the steps:

1.  Create your picture and save it on your computer as an image file (perhaps using a facility in the software used to generate the picture, or using the Windows snipping tool, or some other such tool).

2.  Within the Answer box select the icon to the far right of the Formatting ribbon (looks like a picture of a mountain range?).

3. At the top of the window that appears is a button for selecting the Tiny Pic Upload Service. Click and this and another window appears in which you can select Choose file.  

4. Select Choose file and browse to your saved image.

5.  Select it and click on Upload Now.  The file will be uploaded and a URL will appear in a box near the bottom of the window (if you've a big picture you might have to scroll down to see this).

6.  Select the URL and copy it.  There is a button that says Click to copy, however, this doesn't always seem to work, so I copy using Ctrl_C.

7.  Close the Tiny Pic window and paste the URL into the Image URL box (fill in the other boxes as appropriate if you like)

8.  Close the image upload window and you should find your picture in the Answer box. 

You could also approach this using conservation of energy.  If we assume air resistance is negligible then the potential energy (PE) at the top of the tower gets turned into kinetic energy (KE) at the bottom.

PE = m*g*h   where m = mass, g = gravitational acceleration (I usually use 9.81m/s², but I'll use 9.8m/s² here because that's what Chris used), h = tower height (40m)

KE = (1/2)*m*v² where v is speed when the ball hits the ground.

So:

(1/2)*m*v² = m*9.8*40

The mass cancels, multiply through by 2 and take the square root of both sides to get:

v = √(2*9.8*40) m/s

$${\mathtt{v}} = {\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{40}}}} = {\mathtt{v}} = {\mathtt{28}}$$ m/s