All Answers 358509 Answers

Excellent Creeperface2000!  

I don't think that I understand the question!

Thank you Alan,

This is where I want it.  As more bits and peices get added it would be good if it gets cleaned up into a neater thread but for now this is great.

-----------------------------

More limits

$$\small{\text{$\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{(x+4)(x-2)}{(x-4)(x-6)} =\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{x^2+2x-8}{x^2-10x+24}=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}}{ \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}}=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ 1 + \dfrac{2}{x}-\dfrac{8}{x^2}}{ 1 - \dfrac{10}{x}+\dfrac{24}{x^2}}= 9\cdot \dfrac{ 1 }{ 1 }=9 $}}\\\\$$

\small{\text{$\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{(x+4)(x-2)}{(x-4)(x-6)}

=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{x^2+2x-8}{x^2-10x+24}

=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ \dfrac{x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{8}{x^2}}{ \dfrac{x^2}{x^2}-\dfrac{10x}{x^2}+\dfrac{24}{x^2}}

=\lim \limits_{x \rightarrow \infty } 9\cdot \dfrac{ 1 + \dfrac{2}{x}-\dfrac{8}{x^2}}{ 1 - \dfrac{10}{x}+\dfrac{24}{x^2}}= 9\cdot \dfrac{ 1 }{ 1 }=9 $}}\\\\

With regard to problem 2, I've realised that everytime a member answers a question, the one line summary that appears in the Answer (and on the Question) page contains a link to their profile.  The one line summary is text only - it doesn't even cope with superscripts and subscripts for example.  If the first line in the reply is something that can't be represented by text - a LaTeX insert, for example, a blank appears on the summary line and just the profile link is left.  

I've discovered how to put labels on a summation sign correctly.  For example:

\begin{document}

Starting with:
$$\frac{3x+h}{x+h}$$

Limit as h tends to zero:
$$\lim_{h\rightarrow0}\frac{3x+h}{x+h}$$

\end{document}

produces

$$Starting with:
$$\frac{3x+h}{x+h}$$

Limit as h tends to zero:
$$\lim_{h\rightarrow0}\frac{3x+h}{x+h}$$$$

However, if you leave out the begin{} and end{} and the $$ signs you get (after ignoring the text and first expression):

$$\lim_{h\rightarrow0}\frac{3x+h}{x+h}$$

I don't know if you want this in this thread Melody.  If not I'll delete it and create a separate thread.

create graph

Just enter this directly into the site calculator;

$${\frac{{\left({\mathtt{27.2}}{\mathtt{\,-\,}}{\mathtt{54}}\right)}^{{\mathtt{2}}}}{{\mathtt{54}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\left({\mathtt{44.6}}{\mathtt{\,-\,}}{\mathtt{54}}\right)}^{{\mathtt{2}}}}{{\mathtt{54}}}} = {\mathtt{14.937\: \!037\: \!037\: \!037\: \!037}}$$

5/12 of the boys do not wear glasses

1/4 are girls so 3/4 are boys

4/9 of the boys wear glasses so 5/9 of the boys do not wear glasses.

$$\frac{3}{4}\times\frac{5}{9}=\frac{5}{12}$$

so $$\frac{5}{12}$$  of the students are boys who do not wear glasses.  Just like Alan told you 

Let the number of pupils be n.  There are n/4 girls, hence 3n/4 boys.  

(4/9)*3n/4 = n/3 boys wear spectacles, so 3n/4 - n/3 = 9n/12 - 4n/12 = 5n/12 boys don't.  So the boys who don't wear spectacles, expressed as a fraction of the number of pupils in the class is 5/12.

I think there needs to be a little more:

A straight line that touches a curve at a point, but if extended does not cross it at that point.

Otherwise a radial line that stops at the curve, for example, would meet the criteria in the other posts.

Hi Kitty,

I have added this post to the sticky topic "Information pages worth keeping or developing"

I will use it every time I get a question like this - or maybe I will condense it properly so they don't have to open lots.

Would you like to do that for me on a new post?  You don't have to it was just a thought.

Thank you

Melody.

NEWEST AND BEST ANSWER - 13/5/14  (it includes the fraction calc and the answer below)

Fractions on the calculator

COOL !    THANKS ROM!

I LIKED THE QUESTION TOO!

It is a little more than that.

It is a line that touches a curve a a point. It does not cut the curve at that point - it just touches it. 

3ab   

I am told I have to enter more detail.  How about 3ab is the answer.

3ab [(3*a)*(b)]

A line which touches a circle at only one point.

Hi hello this is a cool web thanks guys

Thanks Alan  

Is it? 

Compare it with athletics; more specifically, with the high jump.  Is the high jump hard? If you are trying to break the Olympic record the answer is yes; even if you can do it, it’s hard! On the other hand, if the bar is 2cm off the ground it’s easy for any able-bodied person.  Somewhere in between it starts to get hard for everyone, though the point at which it does so differs for everyone.  We each have a “personal best” height.  Often by working hard at it we can improve our personal best, and there is a lot of satisfaction in doing so.

I think of mathematics in a similar way.  If you are trying to prove Fermat’s last theorem it is hard (it has now been proven, of course, by Andrew Wiles – and he spent years on it before succeeding!).  On the other hand, at the level of 2 + 2 = 4, most of us find that very easy.  Somewhere  in between everyone starts to struggle, no matter how clever they are.  We all have a “personal best”, different for each of us, where we find things hard (actually, because mathematics is much more diverse than the high jump, each of us probably has several different personal bests, depending on what piece of mathematics we are dealing with!).  As in the case of the high jump, we can often improve our personal best by putting in some effort and working at it.  There is enormous satisfaction in being able to understand something today, that you didn't understand yesterday, as a result of your own efforts.

Here’s my (admittedly oversimplified) summary equation

Hard + Effort = Easy

Click on the words "create graph" at the bottom right of the calculator.  Enter the expression you want to be graphed (can only have a single variable name and that must be x). Enter the x and y limits that you want to be covered in the boxes below the graph.  Control of anything else about the graph is very limited though.

This "equality" was first developed by a famous Indian mathematician called Ramanjuan.  It arises from something called analytic continuation and the Riemann equation (very advanced mathematics - way over my head).

Oops!