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I have no idea what that notation means ??

This is much better written as 

$$0.8     or it can be expressed as     domain (0.8,0.9)

what do you want me to do with it?

I have to think about that one anon - Thanks 

@@ End of Day Wrap     Fri 9 /1/15     Sydney, Australia       Time 9:55pm       ♪ ♫

Hi all,

Lots of great answers the last 2 days from Heureka, CPhill, Geno3141, Oboeduck, Shaniab29544, Nadvera, usiu96, YourAssassin, Hussein101, Alan, Tetration, VampireYuki and Sangyun.  A big thankyou to each of you. 

Interest posts:

1) Using Gyazo (there are other similar prgrams available if you prefer) - Good info for all levels

Sat 10/1/14

1) Limit 

Heueka just wrtote some electricity formulas with LaTex - a few new symbols I think.  Thanks Heureka.

Thank you Heureka  

I am going to reference this from the LaTex page too.  Thanks again :)

if Voltage source is 12, resistance is 6800 and voltage 1 is 8 what is R2

$$\small{\text{series circuit
}}$\\$
\small{\text{$
R=6800\ \Omega \quad U = 12\ V\quad U_1 = 8\ V\quad R_2=\ ? \ \Omega
$
}}$\\$
\small{\text{
I.
$
\quad I=\frac{R_1}{U_1}=\frac{R_2}{U_2} \quad \Rightarrow \quad \boxed{R_2=\frac{U_2}{U_1}*R_1}
$
}}$\\\\$
\small{\text{
II.
$
\quad R=R_1+R_2 \quad \Rightarrow \quad \boxed{R_1=R-R_2}
$
}}$\\\\$
\small{\text{
III.
$
\quad U=U_1+U_2 \quad \Rightarrow \quad \boxed{U_2=U-U_1}
$
}}$\\\\$
\small{\text{
we substitute $U_2=U-U_1 $ and $R_1=R-R_2$
}}$\\\\$
\small{\text{
$
R_2=\left(\frac{U-U_1}{U_1}\right) *(R-R_2)\quad \Rightarrow \quad \boxed{\textcolor[rgb]{1,0,0}{ R_2 = \left( 1-\frac{U_1}{U} \right) *R } } $
}} $\\\\$
\small{\text{
$
R_2 = \left( 1-\frac{8\ V}{12\ V} \right) * 6800 \ \Omega = \left( 1-\frac{2}{3} \right) * 6800 \ \Omega =\frac{1}{3}* 6800 \ \Omega =2266.\overline{6} \ \Omega
$
}} $\\\\$
\small{\text{
$
\textcolor[rgb]{1,0,0}{R_2 = 2266.\overline{6} \ \Omega }
$
}}$$

$${\frac{{\mathtt{9\,999}}}{{\mathtt{60}}}} = {\frac{{\mathtt{3\,333}}}{{\mathtt{20}}}} = {\mathtt{166.65}}$$

166.65 hours

-800x10^-3

=-800.      x    10^-3        There is an invisable decimal point after the 800

Now the index (of 10) is negative so this is going to be a little number.

Move the decimal point 3 places in the direction to make the number smaller.

=-.8          I jumped the point over 0, over 0, and over 8 that is the 3 jumps.

now put a zero infront of the point so that you can see it easily.

= -0.8

Thank you Chris for discovering this for us  

This Wikipaedia site that Chris has referred us too is interesting.

 x=1/2, -2      what is p-q? 

$$x_1 = \frac{1}{2} \ and \ x_2=-2$$

$$2x^2-px+q=0 \quad | \quad :2 \\
x^2\underbrace{-\frac{p}{2}}_{P}x\underbrace{+\frac{q}{2}}_{Q}=0\\
x^2+ Px+Q=0 \\\\
\[[Vieta\]]\ : -(x_1+x_2) = P = -\frac{p}{2} \ and \ x_1*x_2 = Q = \frac{q}{2} \\\\
\small{\text{
$
p-q = -2P-2Q = (-2)[ -(x_1+x_2) ] - 2x_1*x_2 = 2(x_1+x_2-x_1*x_2) =2[\frac{1}{2}-2-\frac{1}{2}*(-2)]=2(\frac{1}{2}-2+1)=-1
$
}}
\\
\textcolor[rgb]{1,0,0}{p-q=-1}$$

put the numbers in order - then the MIDDLE one is the MEDIAN.

Here is a site for you

cos B cannot equal 42.25 because cosine must be between -1 and +1

$$\frac{-20}{15}$$

Divide top and bottom by 5

$$=\frac{-4}{3}$$

$$=-(\frac{3}{3}+\frac{1}{3})$$

$$=-(1+\frac{1}{3})$$

$$=-1\frac{1}{3}$$

p-1/-3=-4 

p+(-1/-3)=-4

a neg divided by a neg = a pos

so this is better written as 

p+(1/3)=-4

subtract 1/3 from both sides

p= -4-(1/3)

p=-(4+1/3)

$$p=-4\frac{1}{3}$$

You cannot solve for anything - there is no equal sign  

What numbers can you use.

Can you explain a little better please?

Yes

It has to be a right angled triangle

A little more on this strange graph.....

Apparently, if this can be factored into this form....(x + y) (x - y) = 0, we have a graph of intersecting lines...let's see...

2x^2 - 3y^2 = 5xy 

2x^2 - 5xy - 3y^2 = 0

(2x + y) (x - 3y)  = 0

Since this "reducible" to this form, this is a degenerate conic that will form two intersecting lines.

The best Scooby snacks need a napkin and a mop. LOL ROSF (sticky floor).

This is the most unexpected graph ever.  Mmm one for the interest posts definitely !

 Thanks for bringing it to my attention Chris