Melody

or maybe we have

.08=0.75x/1-.75x

$${\mathtt{0.08}} = {\frac{{\mathtt{0.75}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{1}}}}{\mathtt{\,-\,}}{\mathtt{0.75}}{\mathtt{\,\times\,}}{\mathtt{x}} \Rightarrow \left\{ \begin{array}{l}\end{array} \right\}$$

The empy parentheses indicate that this is an empty set.   There are NO solutions.

It is very nice to meet you Adrienne.  

This is a question that crops up from time to time.

To me if I am asked for "the difference between A and B" I interprete that as an absolute value.

so |A-B|=|B-A|=big one - little one.

However, I think other mathematicians may disagree with me.   

Great explanation anon.  

not ususally.

$$\\2-x^{0.5}=4-x\\
-x^{0.5}=2-x\\
x-2=\sqrt{x} \\
$square both sides$\\
x^2-4x+4=x\\
x^2-5x+4=0\\
(x-4)(x-1)=0\\
x=4\;\;or\;\;x=1$$

check when x=4

2-2=4-4  true

check when x=1

2-1=4-1 not true  

So this is only true when x=4

I did this a little differently (the answer is the same although Chris's is not quite finished.)

The co in cosine stands for complement of sine.  Complementary angles add to 90 degrees.

So 

$$\boxed{sin\theta=cos(90-\theta)}\\\\$$

also, 140 degrees is in the 2nd quadrant.  Sine is positive in the 2nd quad.  so

$$\\sin140^0=sin(180-140)=sin40\\\\
sin40=cos50\\
so \\
sin140=cos50$$

This is the same as CPhill's answer.

Now we know that sin50=p=p/1

Draw a right angled triangle and label on of the acute angles as 50 degrees.

The opposite side is p

The hypotenuse is 1

So using pythagoras' Theorum the adjacent side must be     $$\sqrt{1-p^2}$$

$$\\cos50=\frac{adj}{hyp}=\frac{\sqrt{1-p^2}}{1}\\\\
cos50^0=\sqrt{1-p^2}$$

a CONE is a circular pyramid.  It has no triangles.  How about that.   LOL

Where???????

@@ End of Day Wrap        Mon 6/10/14 Sydney,     Australia Time 9:45pm      ♬

Hi all, 

It was a quiet day today.  As always there were great answers from Alan, CPhill, radio and Adrienne.  Thanks you all.

CPhill often produces great mathematics answers but he also has a skill which I am sadly lacking.

He is a forensic mathematician.  He can read between the lines and work out what the intended question is.

I sometimes get stuborn and answer the question asked because I want people to learn to present their questions properly.  They can always repost with their intended question.

But even though this is sometimes true I am really not gifted at fathoming what an intended question is.  So with this one I really did not 'see' what the intended question was.  On re-reading the question I can see that I was at fault this time.  I needed to spend more time thinking about the actual wording of the question.

Chris read it properly and answered the intended question.  Sometimes the logic is more obscure than this, Chris really is quite gifted at reading between the mathematical lines.  It is an admirable gift.  Thank you Chris.