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@@ End of Day Wrap :       Mon 4/8/14      Sydney, Australia     Time 10:35 pm        ♬

Hi all,             

Our great answers today were provided by Alan, CPhill, NinjaDevo, Rosala and Sasini.  Thanks  

Now for some interest threads:

1) A little LaTex lesson  

Just use the on-site calculator.....it will even "reduce" this for you....magic!!!

I just want to point out that I DID NOT give myself 3 points on my last post.

I am assuming that it was Chris but whoever it was, thank you for your support.  

Thanks Ninja, I was thinking the same thing!

Thanks Rosala, I thought that your "Treasure hunt = Answer hunt" analagy was quite clever so I gave you (and Chris and Ninja) thumbs up.  

I think that Chris should give himself thumbs up too 

Inverse sine is the same as arc sine.

So you use 2nd  then   asin

How many quarters?

Whos john venn?

Alan is right!

But I thought I would point out that all these triangles would be similar figures.  

And, since the sides would all be in the same ratio then the length of the sides could all be expressed using just one variable.

I am thinking as I am writing.

Let the sides of one of these similar figures  be 1, a and b and let the given angles be <A, <B and <C

Using the Sine rule.

$$\begin{array}{rll}
\frac{a}{SinA}&=&\frac{1}{SinC}\\\\
a&=&\frac{SinA}{SinC}\\\\
\end{array}\\
Also\\\\
\begin{array}{rll}
\frac{b}{SinB}&=&\frac{1}{SinC}\\\\
b&=&\frac{SinB}{SinC}\\\\
\end{array}\\\\
\mbox{So the sides are in the ratio } \quad \frac{SinA}{SinC}\;:\;\frac{SinB}{SinC}\;:\;1\\\\
\mbox{So the sides are equal to } \quad \frac{kSinA}{SinC}\;,\;\frac{kSinB}{SinC}\;,\;k\qquad \mbox{Where k}\in Z \\\\$$

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

Here is the code for people trying to learn LaTex. 

\begin{array}{rll}

\frac{a}{SinA}&=&\frac{1}{SinC}\\\\

a&=&\frac{SinA}{SinC}\\\\ \

end{array}\\

Also\\\\

\begin{array}{rll}

\frac{b}{SinB}&=&\frac{1}{SinC}\\\\

b&=&\frac{SinB}{SinC}\\\\

\end{array}\\\\

\mbox{So the sides are in the ratio } \quad \frac{SinA}{SinC}\;:\;\frac{SinB}{SinC}\;:\;1\\\\

\mbox{So the sides are equal to } \quad \frac{kSinA}{SinC}\;,\;\frac{kSinB}{SinC}\;,\;k\qquad \mbox{Where k}\in Z \\\\

Great answer Ninja,

Grreat use of LaTex too!  (I answered your latex question for you as well)

I'm going to do it a different way (and I will practice some LaTex while I am at it)

3y-5x=-13

 $$\\If\; x=0\;\; then \;\;3y=-13 \;\;so \;\;y=\frac{-13}{3}=-4\frac{1}{3}\;\;(0,-4\frac{1}{3})\\\\
If\; y=0\;\; then \;\;-5x=-13 \;\;so \;\;x=\frac{13}{5}=2\frac{3}{5}\;\;\;(2\frac{3}{5},0)\\\\
\begin{array}{|c|c|c|}\hline
$x$&0&2\frac{3}{5}\\\hline
$y$&-4\frac{1}{3}&0\\\hline
\end{array}\\$$

FOR PEOPLE LEARNING LATEX

This is the code that I have used.

\\If\; x=0\;\; then \;\;3y=-13 \;\;so \;\;y=\frac{-13}{3}=-4\frac{1}{3}\;\;(0,-4\frac{1}{3})\\\\

If\; y=0\;\; then \;\;-5x=-13 \;\;so \;\;x=\frac{13}{5}=2\frac{3}{5}\;\;\;(2\frac{3}{5},0)\\\\

\begin{array}{|c|c|c|}\hline

$x$&0&2\frac{3}{5}\\\hline

$y$&-4\frac{1}{3}&0\\\hline

\end{array}\\

I drew the picture with GeoGebra

Hi Ninja, my knowledge is not great - this is about all that I know 

SPACING

67

6\;7

6\:7

6\quad7

6\qquad7

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

$$\\67\\
6\:7\\
6\;7\\
6\quad7\\
6\qquad7\\$$

Here is another 'learning LaTex' page from today.

135.2857142857142857

The quotient is the result of dividing one number by another, so your quotient is  422.0875 (which is 337.67/0.8)

What do you want to do with this? 

Expand it as (4479*777 + 4479*x)/11 = (3480183 + 44798x)/11 ?

Separate the terms (3480183/11) + (44798/11)x ?

Approximate in decimal form  316380.3 + 4072.5x ?

...?

The calculator here doesn't automatically round to a specific number of decimal places.  However, it should be easy to do this by hand.  For example:

2.71828 becomes 2.718

3.14159 becomes 3.142

etc.

It is unclear what you are asking for here.  Do you mean

(i) 4268*log(943)*7

(ii) 4268*log(943*7)

(iii) log(4268)*943*7

(iv) ...?

Percent means out of 100, so 1 in every 4 means (1/4)*100 = 25% 

√(7 - y) = 6

Square both sides:

7 - y = 36

Add y to both sides:

7 = 36 + y

Subtract 36 from both sides:

-29 = y   or just y = -29.

Check in the original equation:

√(7 - (-29)) = √(7+29) = √36 = 6  ok.

You can't, unless you have at least one side length!  There are an infinite number of similar triangles with the same three angles - they are just different sizes.

Because there are 60' in one degree and 3600" in one degree you can enter your angle as (48 + 30/60 + 24/3600)°. 

Hey guys,

I had another question about LaTeX. 
How do you put a single space between words in LaTeX? I know you can do /quad to put 4 spaces between text in LaTeX, but is there a way to put just one space, like when writing a sentence? 

No that's alright CPhill.

Your answers is plain, simple and to the point.

Mine has a little more information that needed probably. I gave you a point for what it's worth. :)

Thanks for the thumbs up Chris