All Answers 358087 Answers

Where is the working Titanium, how can anon learn anything from what you have written?

Maybe anon can learn some rounding I guess. :/

TitaniumRome

IT IS NOT ZERO - IT IS UNDEFINED - YOU CANNOT DIVIDE BY 0

LOL  Thanks Titanium - I am definitely female :)

I am not sure if it makes a difference but I want to see what the board actually looks like.  

I think that it does make a difference.  :/

I am not particularly good with the calculator but what exactly do you want to work out?

I downloaded this graphing program Alan.  It looks VERY impressive but i have not really played with it yet. 

Melody....I only vaguely remembered this from Calculus.....I had to do some "reviewing," too.  I don't know if this is taught much now in many places. The algebra gets pretty "sticky".......

WOW, That rock has sure given Sisyphus some muscles!

Why doesn't he just crack it open to find the Roman zero?

I guess his brawn is bigger than his brain !

I tried to tell Sisyphus to quit fooling with that stupid rock....!!!!

Oh, rotate the Axes  at least the question makes some degree of sense.

I will have to study your answer Chris ://

I would have had NO idea how do do this one!

Very impressive Chris    

you can just type it into the input box.   The | is above the backslash on your keyboard :)

$${\left|{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{3}}\right|} = {\mathtt{10}}$$

So what was the work done on Alan?  Cause I can see the sweat rolling off that man. :/

doesn't exist

the number must be between -1 and +1

THERE IS A FRACTION BUTTON

http://web2.0calc.com/questions/how-do-i-type-in-fractions

use   asin  it is the same thing

there are too many questions, we cannot answer them all.

If you are still waiting for an answer to your own question you are best just to guide us to that one :)

Heureka beat me!     hi Heureka 

Do you want to know what shape it is?   Looks like a circle to me.   

$$\\36x^2+36y^2-24x-12y=31\\
36x^2-24x+36y^2-12y=31\\
36(x^2-\frac{2x}{3}+y^2-\frac{1y}{3})=31\\
(x^2-\frac{2x}{3})+(y^2-\frac{1y}{3})=\frac{31}{36}\\
(x^2-\frac{2x}{3}+\frac{1}{9})+(y^2-\frac{1y}{3}+\frac{1}{36})=\frac{31}{36}+\frac{1}{9}+\frac{1}{36}\\
(x-\frac{1}{3})^2+(y-\frac{1}{6})^2=1\\$$

$$\\Circle:\qquad centre\;(\frac{1}{3},\frac{1}{6})\qquad radius=1\;unit$$

I think this is supposed to be :

2x^2-72xy+23y^2-80x-60y-125=0

We need to first calculate this:

cot2Θ = [A - C]  / B   = [2 - 23]/[-72]  = [-21]/[-72]  = [7/24]

And, from trig, we have that cos2Θ  = 7/25

And using trig again, we can caclulate  sinΘ  and cosΘ, thusly

 sinΘ  = √[(1 - cos2Θ)/2 ] = √[(1 - 7/25)/2] = √[9/25] = 3/5

cosΘ =  √[(1 + cos2Θ)/2 ] = √[(1 + 7/25)/2] = √[16/25] = 4/5

Now....we need to make the following substitutions

x = cosΘ X  -  sinΘ Y      and y =  sinΘ X + cosΘ Y    ....  or, just .......

x = (4/5)X - (3/5)Y       and  y = (3/5)X + (4/5)Y

So we have

2[(4/5)X - (3/5)Y]^2-72[ (4/5)X - (3/5)Y][(3/5)X + (4/5)Y]+23[(3/5)X + (4/5)Y]^2-80[ (4/5)X - (3/5)Y]-60[ (3/5)X + (4/5)Y -125=0-80[ (4/5)X - (3/5)Y]  

And term by term, we have.......

[(32/25)X^2  - (48/25)XY + (18/25)Y^2] +

[(-864/25)X^2 - (504/25)XY + (864/25)Y^2] +

[(207/25)X^2 +(552/25)XY + (368/25)Y^2]  +

[48Y  - 64X]  +

[-36X  - 48Y ] -

125    =  0

Simplifying the above, we get

50Y^2  - 25X^2 -100X  - 125 = 0     divide through by 25

2Y^2 - x^2 - 4x - 5 = 0   complete the square on x  

2Y^2 - [x^2 + 4x + 4  - 4] = 5

2Y^2 - (x +2)^2 = 1

And we have the recognizable equation of a hyperbola.

Here's a picture of both the original  graph with the xy term and the "normal" graph without rotation...

36x^2+36y^2-24x-12y=31

\(\begin{array}{rcll} 36x^2+36y^2-24x-12y &=& 31 \\ (36x^2-24x)+(36y^2-12y) &=& 31 \quad & | \quad : 36 \\\\ (x^2-\frac{2}{3}x)+(y^2-\frac{1}{3}y) &=& \frac{31}{36} \\ \\ (x-\frac{2}{6})^2-\frac{4}{36}+(y-\frac{1}{6})^2-\frac{1}{36} &=& \frac{31}{36} \\ \\ (x-\frac{1}{3})^2 +(y-\frac{1}{6})^2&=& \frac{31}{36} +\frac{4}{36}+\frac{1}{36} \\ \\ (x-\frac{1}{3})^2 +(y-\frac{1}{6})^2&=& 1 \\ \\ \end{array}\)

36X^2+36y^2-24x-12y=31

This is a circle with center

\( \left(\dfrac{1}{3} ,\; \dfrac{1}{6}\right)\)

and the radius = 1

Thanks Geno.

That is a really good website anon 

I can see why you find it so helpful :)

Thanks for sharing.

Axially through simply refers to a situation in which the current is flowing in the direction of the long red arrow (see diagram below).  Radially outward refers to a situation in which the current is flowing in the direction of the short blue arrow.

The direction current flows will be determined by the potential difference in a particular case.  If there is a potential difference between the ends then current will flow axially; if there is a potential difference between an inner and an outer curved surface (as in the original problem) then current will flow radially.

.

That’s because Christianity had a 600-year head start, and the greatest power advantage during the dark ages.

Maybe you are not religious, but you are at least a little biased because you just recommended Buddhism over the other two (and 15 other major religions that weren’t mentioned at all).

That is an impressive 4