I can, but it will clutter up everything. Just mouse over the Latex and the code window pops up. Select and copy that code and paste it in an editor to play with it.

It’s as easy as

  $${\pi}$$

Since

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)} = {\mathtt{cosxcosy}}{\mathtt{\,-\,}}{\mathtt{sinxsiny}}$$

and 2x=x + x

you have

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)} = {\mathtt{cosxcosx}}{\mathtt{\,-\,}}{\mathtt{sinxsinx}}$$

that is

cos(2x)=(cosx)^2-(sinx)^2

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

$${\frac{{\mathtt{6}}}{{\mathtt{7}}}} = {\mathtt{0.857\: \!1}}$$ (4 d.c.p.)

0.8 < 0.8571 

We can also solve this by making the denominators the same. 

$${\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)}} = {\frac{{\mathtt{28}}}{{\mathtt{35}}}}$$

$${\frac{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}{\left({\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}} = {\frac{{\mathtt{30}}}{{\mathtt{35}}}}$$

28<30

$$\left({\frac{{\mathtt{4}}}{{\mathtt{9}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{5}}}{{\mathtt{7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right) = {\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{2}}\right)}} = {\frac{{\mathtt{20}}}{{\mathtt{126}}}} = {\frac{{\mathtt{10}}}{{\mathtt{63}}}}$$

You can also check it on your calculator.

Is there a way to write fractions without using brackets?

I have started a new thread here

Tues 14/4/15

FTJ means "For the Juniors"

Mon 13/4/15

FTJ means "For the Juniors"

@@ End of Day Wrap   Mon 13/4/15    Sydney, Australia    Time 3:40 pm   ♪ ♫

Hi everyone,

I am early today, how about that :)

Toay our great answers were  delivered by Bertie, Geno3141, CPhill, Alan, Mellie and Nausiated.  Thank you.     I think I might have missed some.  If I missed you I am  sorry.   

Interest posts:

FTJ means "For the Juniors"

Sorry Mathsisbae, 

you misunderstood my request.   (Private message)

I wanted you to make a post here to ask me to explain more . 

I did not want you to make another post that was the same - we don't do that.  :)

okay you want me to expain some more. :)

The perimeter of the square is  80p+19 and that is in cms

There are 4 sides and they are all the same.  when you add up all the sides you will have the perimeter.

So on side must be a quarter of the perimeter.

$$\\side=\frac{1}{4}\times (80p+19)\\\\
side=\frac{80p+19}{4}\;\;cm$$

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

If you are still not sure think about what you would do with something simpler

Say the perimeter is    20cm

then the side would have to be  20/4 = 5cm   BECAUSE  5+5+5+5=20cm

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

This method of replacing something complicated with something easy to work out how to do it is really handy.

I still do this regularly.  :)

Does that help ?

Most of the time it is nice to have you about Nauseated.  

Are  you sure we can't have your autograph  ??     

Thanks MathsGod1,

It is a good thread, it has not been added to for a long time because NinjaDeva can no longer do it and I don't have time.  It is a pity but i do include a little more threads in the "reference Material" to make up a little. 

Ninja used to help me a lot - I continue to be very appreciative of all the effort he put in :)

Wow! 13 points! It looks like I do have a fan club.

Some of my snarky posts might have received minus 13 points, except those are blanked, either by CPhill, or by Melody, who makes Scooby Snacks out of them.

Anyway, thanks to my fans. . . . Sorry, no autographs please.

Thank you Alan and Bertie  :)

I also found the language of this post and the error in it very confusing.   :)

(47784-11570) * 0.27

Thanks Nauseated,

Could you please include the code for each bit.  

Maybe separate it up so we can see the code then the output, then the next bit of code and the output.

It would be great it you did that for us.  :))

$$\\f(x) =\frac{ ( x^7 + 4x^3 ) }{ x^4}\\\\
f(x) = x^3 + 4x^{-1} \\\\
f'(x) = 3x^2-4x^{-2} \\\\
f'(x) = 3x^2-\frac{4}{x^{2}} \\\\$$

10C8(3/5)^7(2/5)^2

you can enter it into the web2 calc like this.

nCr(10,8)*(3/5)^7*(2/5)^2

(8x-5) + (2x-3)

=8x-5+2x-3

=8x+2x    -5-3

= 10x   -8

I see that you are on the forum Mathisbae,  Have a think about it and if you don't understand I can explain some more :)

$$\Displaystyle \text {Basic form} \\

\Displaystyle x^{3^4}\\

\Displaystyle \text {This format will work for more complex forms}\\

\Displaystyle \begin{Large} {X^{2^4^{^6^}^{^8^}^{^{Etc}}}\end{Large}\\\\

\text {Here are some other presentation forms that may be useful.}\\$$

$$\noindent \sqrt[\leftroot{-2}\uproot{4} 7]9876 \leftarrow \Text${Use noindent to prevent pesky indentation}\\\

\noindent \sqrt[\leftroot{-2}\uproot{4} 7]{9876}\leftarrow \Text${Note the root function covers all the numbers here. Compare the codes to see why} \\\\

\Text{Here's a method to nest "n-root" and align the "n" using "leftroot" and "uproot"}\\\\\

\sqrt[\leftroot{-2}\uproot{4} 7]{\sqrt[\leftroot{-2}\uproot{4} 5]{ (\sqrt[\leftroot{-2}\uproot{2} 9]{x^2})^3}} = Z \\\

\Text{Note the subtle difference in the exponent of 3}\\\

\sqrt[\leftroot{-2}\uproot{4} 7]{\sqrt[\leftroot{-2}\uproot{4} 5]{ (\sqrt[\leftroot{-2}\uproot{2} 9]{x^2})^{^3}}} = Z \\\


\Text{One great thing about writing great LaTex presentations: You can be dumber than a box of rocks and still look intelligent --at least for awhile. I have been getting away with it for years . . . .}$$

(80p+19) /4   cm

100% +8% = 108% = 1.08

Use the change of base rule

$$\\x=log_8(23)\\\\
x=\frac{log23}{log8}$$

you can have it in whatever base you want - the answer will always be the same :)

With the web2 calc you don't even have to change the base

log(23,8)             $${{log}}_{{\mathtt{8}}}{\left({\mathtt{23}}\right)} = {\mathtt{1.507\: \!853\: \!985\: \!352\: \!337\: \!8}}$$

y=-5-5x^2 at (-2,-25)

Always remember that the fisrst derivative gives the gradient of the tangent to the curve!

y'=0-10x= -10x

x=-2    y'=+20

so the gradient is 20 and a point on the tangent is (-2,-25)

Now can you finish it?