...I didn't think that would work properly - they are not usually that compatible.

I copy everything between, and including, begin{document} and end{document}

I haven't used any esoteric features so I don't know what the limitations of this are!

I didn't think that would work properly - they are not usually that compatible.

Do copy in everything or just the main command parts?  I'll have to try that.

Thanks.

To dispaly properly we really need some version of align! I'll leave him alone at the moment but when other problems lessen I will ask Andre about it.  Maybe by then I (or you) will have worked it out. 

5x+10x-4x

+10x  +  -4x  = +6x

+6x  +  +5x  = +11x

Answer is +11x

$${\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{y}} = {\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{y}}$$

I haven't used either of begin{align} or begin{equarray}, I'm afraid.  

I suspect that my problems are more likely to be of my own making in the main as I'm still very much a beginner with LaTeX.  I find it easier to use Texmaker and then copy and paste into the LaTex Formula box than write it directly here.   

$${\frac{{\mathtt{12\,345\,678\,910.109\: \!876\: \!543\: \!21}}}{{\mathtt{13\,579.246\: \!81}}}} = {\mathtt{909\,157.855\: \!575\: \!487\: \!293\: \!408\: \!3}}$$

$${\frac{{\mathtt{4}}}{{\mathtt{5}}}} = {\mathtt{0.8}}$$

8+8=16

I I I I I I I I + I I I I I I I I I I = I I I I I I I I I I I I I I I I

hi Alan,

I have had to split up LaTex before.  It seems to happen when the output is long.

Today and I think yesterday the \frac was not working properly.  

I'll try it now to show you

1-\frac{3}{(x+5)^2}

displays as - NOW IT WORKS!  It wouldn't work when i was anwering Stu's question. 

$$1-\frac{3}{(x+5)^2}$$

Have you managed to get begin{align} or begin{equarray} (I used to use this one all the time and now I am not even sure that the command is correct)  to work?  I have really missed not having those.

Thanks.

Energy required = 220*1000*4.18 J

$${\mathtt{Energyrequired}} = {\mathtt{220}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}{\mathtt{\,\times\,}}{\mathtt{4.18}} = {\mathtt{Energyrequired}} = {\mathtt{919\,600}}$$ Joules

Power available = 1.84*746 J/s

$${\mathtt{Poweravailable}} = {\mathtt{1.84}}{\mathtt{\,\times\,}}{\mathtt{746}} = {\mathtt{Poweravailable}} = {\mathtt{1\,372.64}}$$ Joules/second

Time = Energy required/(60*Power available) minutes

$${\mathtt{Time}} = {\frac{{\mathtt{919\,600}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{1\,372.64}}\right)}} = {\mathtt{Time}} = {\mathtt{11.165\: \!831\: \!293\: \!468\: \!547\: \!2}}$$minutes

...Anyone else had problems today?

I had some problems with the LaTeX insertion in  

It wouldn't take all the LaTeX in one go so I had to split it into two parts.  The second part then wouldn't format properly (the four equations that are left-hand justified should all be centred).  

@@ End of Day Wrap - Tuesday 29/4/14   Sydney, Australia Time 20:20

Hi everyone,

Lots of great answers were delivered today by Rom, Alan, CPhill, Colyn, DavidQD, Kitty<3, Professor, gabbylona, Anonymous4338 and other anonymous people.  Thank you all.

The frac command in the latex was not working properly which was all very irritating.  These things are sent to try me/us.

Kitty<3 wrote a 'how do you do fractions on this calculator' post.  She put it on the end of the 'Information pages worth keeping or developing' sticky thread.  We may change its location but this is the address at present. Thanks Kitty. This is great.

This question from Stu has promoted interest.  Alan is trying to show us how to insert an image, sounds complicated but I guess it is easy when you know how to do it.

These ones also promoted some interest.

This one was too much for me. Thanks Alan.

http://web2.0calc.com/questions/how-do-you-integrate-e--x-2

and a bit of fun with this one 

As I type this I am having problems.  It keeps going into strange mode. Anyone else had problems today?

That's it - I hope I haven't included too many threads.  I don't get all that many feedbacks from my wraps.  I know people like me writing them but I am not sure what you want me to include.

Enjoy the rest of your Monday/Tuesday 

Melody 

melodymathforum@gmail.com

Thanks David.  Interesting.  You live and learn!

Thanks David 

Incidentally, to import the graph I used the Tiny Pic upload service now available with the imported image facility (icon on the far right of the formatting ribbon).  I created the graph in a piece of software on my PC; used the Windows snipping tool to copy and save it as a .png file; uploaded it using the Tiny Pic option in insert image; copied the url and pasted that into the Image URL box in insert image.  

Do you mean

$${\frac{{\mathtt{10}}}{\left({\sqrt{{\mathtt{22}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}} = {\mathtt{1.150\: \!692\: \!933\: \!039\: \!049\: \!3}}$$

or 

$${\frac{{\mathtt{10}}}{{\sqrt{{\mathtt{22}}}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\mathtt{6.132\: \!007\: \!163\: \!556\: \!104\: \!3}}$$

or something else?

Alan's answer is good but I'll give you another answer Stu - I got called away before.  Sorry.

The LaTex fraction display is not  working properly so my presentation is not a good as I wanted.

$$f(x)=x-2ln(x+1)\:\:\: 0\le x\le2\\

f'(x)=1- 2/(x+1)\\

f'(x)=1-2(x+1)^{-1}\\

f''(x)=+2(x+1)^{-2}\\

f''(x)=2/(x+1)^2$$

Since the (x+1) is squared, the second derivative is always positive and this means that any turning point will be a minimum.  Which in turn means that the two end points will be higher.

Now we need to find any turning points   f'(x)=0

$$0=1-2/(x+1)\\

2/(x+1)=1\\

2 = x+1\\

x=1\\

f(1)=1-2ln(1+1) = 1-2ln2 \approx -0.386$$

So (1,-0.386) is the minumum

Now to find the maximum - you need to look at the end points.

f(0)=0-2ln(0+1) = 0

f(2)=2-2ln(2+1) = 2-2ln3 = -0.197

So the maximum is 0 at x=0 and the minimum is 1-2ln2 at x=1

Just type this into the calculator on this site and press Enter:

$${{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{177}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{836}} = {\mathtt{0}} = \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{4}}\\
{\mathtt{x}} = {\mathtt{19}}\\
{\mathtt{x}} = -{\mathtt{11}}\\
\end{array} \right\}$$

This obviously factors into (x-4)(x-19)(x+11)=0

1.  Look at a graph of your function between the limits of 0 and 2:

You can see that there is a single minimum in the range.

2. To find the turning points on a curve you need to find the places where the tangents are zero.  A zero tangent means that a line through that point is horizontal.  You can see from the curve that a tangent line at the minimum of the curve will be horizontal.

3. The tangent line to a curve at any point has the same gradient as the curve at that point.  The gradient is obtained by differentiating the function with respect to x. So if the curve is described by f(x) = x - 2ln(x+1) you have to differentiate this to find the gradient.  Differentiating the first term, x, with respect to x is just 1.  Differentiating the second term with respect to x gives 2/(x+1).  Overall then the gradient of the curve at any point is given by gradient(x) = 1-2/(x+1).

4.  We want the point where this gradient is zero, so we set 1-2/(x+1) = 0 and find the value of x that makes this true.  Here we can see that x = 1 is required since 1-2/(1+1) is indeed zero.

5.  The value of f(x) at this point is obtained by putting x = 1 into the expression for f(x) -  this will give us the value of the minimum of f(x):  1-2ln(1+1)  or 1-2ln(2) = -0.386...

6.  You can see that there isn't a turning point maximum in the range from x = 0 to x = 2, but the maximum value of the function in this range clearly happens when x = 0.  The value of the function here is f(0) = 0 - 2ln(0+1) = -2ln(1) = 0.

Hope this helps a bit more. 

I don't know how standard it is - It is what I was taught and so it is what I teach.

 (I was lucky, I had a couple of really good maths teachers when I was at school, most of what they taught me was correct.  Now I pass that learning on.  It may not all be flawless) 

I dont get it. But I need to find the global minima and global maxima for this equation. What was the minima and x value, and same for maxima? 

What steps are used to get the answers?

Hi bioschip,

I just thought I would add a litte

2 angles that add to 90 degrees are called complementary angles

2 angles that add to 180 degrees are called supplementary angles.

If you can remember the 2 names it is easy to remember which is which because

2 90s make 180 and

2 c's make an s (one is back to front)

does that make sense?

... I always use a little supertext c to represent radians.

Interesting! I've never seen that done anywhere; is it standard in Australian schools?  I'm much more used to assuming that radians are meant unless the degree symbol is used, or you know you are in degree mode (with a calculator, say)

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