Welcome to the forum.

So work out the top, and work out the bottom, and simplify to a single fraction.

Hope that helps,

Bob

]]>Could you tell me where I can find more info on this SketchPad application please?

Geometer's Sketchpad: http://www.dynamicgeometry.com/

GeoGebra: www.geogebra.org/

]]>It provides a graphical editor for the formula which is rather handsome. The graphs are also looking a bit nicer for my taste than with Wolfram Alpha....

]]>Also that is not the equation you gave in post #664. In post #664 you gave this equation

log(x+1) - 2logx^2=1

That is the equation I said does not have a root of 1 / 3.

Now you are giving this equation.

EbenezerSon wrote:

This is it;

log(x^2 + 1) - 2logx = 1

They are not the same.

]]>If you are looking for a nice introduction into the topic p-values, significance and confidence levels I can recommend you the following article:

https://www.plotgraphs.com/blog/statist … igma-mean/

It breaks the topic down to a understandable level I thin :-)

]]>]]>

You need to supply a more detailed algorithm.

eg.

Find minimum values

If only one, state its index.

If more than one, what?

(a) state the one in the middle

(b) state the one nearest to index 4

(c) state ..........?

Bob

]]>Hope that helped and welcome to the forum.

]]>I've not seen this way of defining point symmetry before so I've had to do a bit of experimenting. When I come across a tricky bit of math, I try to simplify the numbers first, and also pick an example for what I'm trying to investigate.

For symmetry, I wouldn't choose a trig. function because the graphs have too many lines and points of symmetry, so its impossible to see what's going on.

I chose y = x^3, because it only has point symmetry around (0,0)

You'll find it is worth using the function grapher at

http://www.mathsisfun.com/data/function … c1=(1+x)^3

I've set this up with the function y = (1+x)^3. Try some values:

x = -3 , y = -8

x = -2 , y = -1

x = -1 , y = 0

x = 0 , y = 1

x = 1 , y = 8

You can see from this that the graph had point symmetry around (-1,0)

Try changing that '1' into other numbers and you'll see that, with a = -1

And f(a-x) = -f(a+x)

Now let's try adding a constant.

y = x^3 + 2

You'll see that this shifts the graph up by 2, so the symmetry point is moved to (0,2)

Then

Here I've put square brackets around the function and introduced an additional '2' (either + 2 or -2) to balance the equation

so in general

You don't need the square brackets here, but I thought it would help you to compare with the line above if I left them in.

This function as point symmetry around (0,b)

Finally put both bits together:

So y = (1+x)^3 + 2

x = -3 , y = -6

x = -2 , y = 1

x = -1 , y = 2

x = 0 , y = 3

x = 1 , y = 10

Try it on the grapher as well.

You'll see the point symmetry is now around (-1,2)

Again you can vary the value of the constant.

So if I put in square brackets for the function again:

y = [(1+x)^3 + 2]

a = -1 and b = 2

and

so in general

with the point of symmetry (-a,b)

Bob

]]>If the price increase one week is p pence, find expressions in terms of p for:

-the number N of items sold weekly;

-the production cost £C;

-the revenue £R from the sales.

-Show that the profit £P is given by

P = 250p – 10200 – p^2

]]>