Melody

No problems here, it is all really good MG   

When you are trying to learn mathematics try not to care excessively about Latex.  Don't let it get in the way of your maths learning.   :)

Oh, if I write too much LaTex it does do that.

I cut some of my lines until it is working properly.  Then I 'ok' it.

Then I open a new Latex box and paste the cut code in there.  

So I may end up with 2 or 3 consecutive lots of LaTex coding.

It ususally only happens when the code is very long though :/

I might try looking for units that you can do elsewhere and then I can tie my teaching in with it.  It is VERY important that you get grounded properly in basic arithmetic.  Then you will be able to fly ahead.  :)

Thanks Alan,

I doubt I would have got that on my own EVEN if I had been a good enother forensic mathematician to work out the question in the first place.   LOL

I put this into the Great Answers to Learn from Sticky Thread :)

$${\frac{{\mathtt{8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{40}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{70}}^\circ\right)}}} = {\mathtt{5.472\: \!322\: \!293\: \!214\: \!088\: \!1}}$$

Well I don't know about best but here is a few (not altogether negative) things that happened.

1)  Nicholas Copernicus  1473 -1543  Polish astronomer lived.  Regarded as the founder of modern astronomy.

2) Williams Shakespeare was born 1564 and died 1616

 He is widely acknowledge as the greatest writer in the English language.

3)  Maori's settled in New Zealand around 1000AD

4) Christopher Columbus discovered the new world in 1492

5) Queen Elizabeth I was coronated Queen of England at the age of 25 on 17th Nov 1558.  

She ruled for 45 years until she died in 1603

Thanks MG and thanks for being patient :)

Q1 and 2 have the correct answers but I will talk about them.

Q3 is not correct but I do not worry about that too much yet.  :)

Q1  Your answer is right but you working was shaky 

$$\frac{35}{21}$$      If you divide this by 7 then the value would change.

BUT if you divide it by1 then the value will be the same.   for example

$$\\8\div 1=8\\\\
\frac{3}{4}\div 1 = \frac{3}{4}\\\\
$Now any number divided by itself = 1$\\\\
\frac{5}{5}=1 \qquad \frac{12}{12}=1 \qquad \frac{-6}{-6}=1\\\\\\
\frac{35}{21}\div 7 \qquad $will change value BUT $\\\\
\frac{35\div 7}{21 \div 7} \qquad $will keep the same value$\\\\
=\frac{5}{3}
=1\frac{2}{3}$$

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

Q2     I have spread this answer out a real lot so that you can see better what is happening

$$\frac{16}{12}\times \frac{15}{10}\\\\
=\frac{16}{4*3}\times \frac{5*3}{10}\\\\
=\frac{16}{4*\not{3}^1}\times \frac{5*\not{3}^1}{10}\\\\
=\frac{16}{4}\times \frac{5}{10}\\\\
=\frac{16}{4}\times \frac{5*1}{5*2}\\\\
=\frac{16}{4}\times \frac{\not{5}*1}{\not{5}*2}\\\\
=\frac{16}{4}\times \frac{1}{2}\\\\
=\frac{4*4}{4*1}\times \frac{1}{2}\\\\
=\frac{\not{4}*4}{\not{4}*1}\times \frac{1}{2}\\\\
=\frac{4}{1}\times \frac{1}{2}\\\\
=\frac{2*2}{1}\times \frac{1}{2*1}\\\\$$

$$\\=\frac{\not{2}*2}{1}\times \frac{1}{\not{2}*1}\\\\
=\frac{2}{1}\times \frac{1}{1}\\\\
=2\times 1\\\\
=2$$

What I have done in question 2 is called cancelling :)

I'll do a few more for you 

Simlify

$$\\1) \;\;\frac{10}{12} \qquad$the common factor here is 2 so I divide the top and bottom by 2$\\
\frac{\not{10^5}}{\not{12}^6}=\frac{5}{6}\\\\
$I have divided top and bottom by 2 but i didn't actually write the $\div 2 $ there, I just did the cancelling.$\\\\
$Your turn$\\\\
$Simplify these with cancelling$\\\\
2)\;\;\frac{15}{10}\\\\
3)\;\;\frac{49}{56}\\\\
4)\;\;\frac{8}{12}\\\\$$

If you use LaTex     \not{8}^4    is what I might use, it would look like this       $$\not{8}^4$$

I know you want to do higher level maths than this but you need a really good grounding in basic maths and then algebra will be easy, so please be patient  :)

$$\\\sqrt3tan^2x=tanx\\
\sqrt3tan^2x-tanx=0\\
tanx(\sqrt3tanx-1)=0\\
tanx=0 \quad or \quad \sqrt3tanx-1=0\\
tanx=0 \quad or \quad tanx=\frac{1}{\sqrt3}\\
-180^0,\;\;0^0,\;\;180^0,\quad or \quad -150^0,\;\;30^0\\
$so all the solutions are$\\
x=-180^0,\;\;-150^0,\;\;0^0,\;\;30^0,\;\;180^0$$

Use the calc on the home page please.

@@ End of Day Wrap   Wed 27/5/15   Sydney, Australia Time   7:15pm     ♪ ♫

Hi everyone,

Our great answers today were provided by Heureka, Cphill, alan, Civonamzuk, Geno3141, Rosala, TitaniumRome, DarkBlaze347, Zacismyname, Brodudedoodebrodude, radix and thejamesmachine.  Thanks all. 

Thurs 28/5/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts:

FTJ means 'For the juniors' 

Thanks Brodude for that great explanation.  :)

I had not even noticed your first answer there.