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Thanky you Chris   

Is there  supposed to be a second hat in there?

Best "non-math" answer of the day  (IMHO) :

http://web2.0calc.com/questions/help-asap-i-have-a-question#r106014

I'm confused.

What is the question?

Everyone's "ticked off" now !!!

Awe - Don't I get a tick too?  

9x^2+4y^2+36x-24y+36=0

OK...hang  on !!

Let's rewrite this as

9x^2 + 36x + 4y^2 -24y = -36      Now factoring out the "9" and the "4" we have

9(x^2 + 4x +     )    + 4(y^2 - 6y +     )    = -36      (A)

Now, take half of "4", square it  and add the result (4) into the first parentheses

9(x^2 + 4x +  4   )    + 4(y^2 - 6y +     )    = -36

And take half of the -6 in the second parentheses, square it and add the result (9) into the second parentheses

9(x^2 + 4x +  4   )    + 4(y^2 - 6y +   9  )    = -36

Notice, on the left side we've actually added 9(4) and 4(9) = 72 to the original expression (A)

So, I need to add this to the right side, too.  And we get

9(x^2 + 4x +  4   )    + 4(y^2 - 6y +   9  )    = -36 + 72    And we have

9(x^2 + 4x +  4   )    + 4(y^2 - 6y +   9  )    =  36

Now divide everything on both sides by 36  This gives

(x^2 + 4x +  4 ) / 4    + (y^2 - 6y +   9) / 9    =   1

Now, factor the stuff in both sets of parentheses   ......     We get  ......

(x + 2)^2 / 4  +  (y - 3)^2 /  9  = 1

And we have an ellipse centered at  (-2, 3) with a minor (horizonta)l axis of 4 and a major (vertical) axis of 6.

And that's it !!!

Lets take a look.

9x^2+4y^2+36x-24y+36=0

$$9x^2+36x+4y^2-24y=-36$$

$$9(x^2+4x)+4(y^2-6y)=-36$$

$$9(x^2+4x+4)+4(y^2-6y+9)=-36+(9*4)+(4*9)\\

9(x+2)^2+4(y-3)^2=-36+36+36\\

9(x+2)^2+4(y-3)^2=36\\

\mbox{Divide throughout by 36}\\$$

$$\frac{x+2}{4}+\frac{y-3}{9}=1$$

That looks like an ellipse to me - centred at (-2,3)

this is what it looks like

https://www.wolframalpha.com/input/?i=%283%28x%2B2%29%29%5E2%2B%282%28y-3%29%29%5E2%3D36

2x² = x² - 1

First, subtract x² from both sides.

x²= -1

Then take the square root of both sides.

so $${\mathtt{x}} = {\sqrt{-{\mathtt{1}}}}$$ and $${\mathtt{x}} = {\mathtt{\,-\,}}{\sqrt{-{\mathtt{1}}}}$$

But in real numbers, you cannot have the square root of a negative number. So instead, there is this number, $${i}$$ , the imaginary number.

$${i}$$ by definition is the square root of -1

So x = i or x = -i

negative b plus or minus the square root of b squared minus 4ac, all divided by 2a

Nice pic, kitty<3 !!!

Thumbs Up !!

We can't help you right now....but we've found someone who might be able to.......

And as CPhill said earlier, too

It would certainly help if you gave an example along with your question, but no worries:)

There are quite a few ways to solve for a side in a right triangle, and one of the most common is the Pythagorean Theorem: 

 $${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$

a and b are legs of the triangle (the two shorter sides) and c is the hypotenuse (longer, diagonal side)

If you have the measure of an angle, you can use trigonometry to solve for an unknown side (using sine, cosine, tangent)

It really depends on what info you have about the triangle. . .

It depends upon which side is "unknown"

If it's the hypoteneuse, then we have

Hypoteneuse = √(Leg₁² + Leg₂²)

If it's a leg, then we have

Missing Leg = √(Hypoteneuse² - Known Leg ²)

I have to admit, too....I never heard of "imperial minims," either !!

These sites should be of great help to stu.

http://www.mouseprogram.com/mousepractice/

After mastering the basics of this, you should be able to navigate this site without major problems. This is as easy as pushing buttons on an elevator to take you to the level you wish to go.

Hope this helps.

Remedial Site Navigation Research Team.

Conversion web page

Thurs 15th May

Who has heard of a minim before - not I

660,000,000 km³

From this web page:

What are "Imperial minims"?

I just looked them up

Anything multiplied by zero is zero.

82*0=0

Yes I know.

The search function doesn't work very well yet.  There are still a number of issues that Andre Massow has to work through.  

Many features do work well.  Andre is working hard and he site is constantly being improved but it will take time.  Sorry.

I have no great advise for your learning/study techniques.  Even if I had a technique that suited me well (which I don't) it still probably would not work for you. Everyone learns differently.   

If you click on the blue "Answers" tab you should see the latest posts (answers).

They may - or may  not - coincide with the latest questions under the blue "Questions" tab.

I can only give you study advice that worked for me....practice a lot of problems. One more thing....if you know the proceedure, don't spend too much time on that particular problem. I found that most of my problems on tests came from material I was uncomfortable with, so that's where you should spend most of your study time.

sin ÷ cos