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Thanks for that answer, Bertie.....I went through the algebra as you suggested.......that's an interesting result to note that the product of both lines gives us the rotated conic equation.....I suspect that this might just be a coincidence, albeit a pretty fascinating one ....

Two questions......

The normal rotated conic usually contains an x and a y term, but I noticed that you omit them in your assumed equation........why is this???......

Also.....

Why do we let X and Y approach infinity??  ....Is this just to eliminate the c/X^2  term so we can solve the quadratic that follows??  I have seen Alan use this technique in solving another hyperbola problem......I was never able to wrap my head around it.....!!!!

All in all....this is a  really neat problem......

BTW......graphing this in Desmos just produces a pair of intersecting lines  .....https://www.desmos.com/calculator/quxgucllfk     ....????

Fri 22/1/16  

1)  NEW FEATURE ADDED        Thanks    Dragonslayer554   and   Admin    

     http://web2.0calc.com/questions/new-feature-added

2)  I thought this was funny :)  Thanks Rarinstraw1195, CPhill and Admin   

      http://web2.0calc.com/questions/i-find-it-odd

3)   How do you multiply a whole number with a fraction.  (For the juniors) Great answer from Gibsonj338   

     http://web2.0calc.com/questions/question_74695

4)   Dealing with negative indices (just the basics)    Thanks Guest and Melody

      http://web2.0calc.com/questions/help_86079

5)  Can you find a smaller number with 13 factors ?   For the juniors     Thanks rarinstraw1195 and guest   :)

     http://web2.0calc.com/questions/factors_4

6)  Function of a function.   Yhanks CPhill,

     http://web2.0calc.com/questions/if-v-x-sin-x-and-h-x-e-x-find-h-v-x

7)  Limit Question - hole in half a parabola     Great answer from guest, improved presentation from Melody.

     http://web2.0calc.com/questions/hi-i-need-help-again-c

8)  Find the height of the rhombus.    Thanks Heureka

     http://web2.0calc.com/questions/rhombus_1

9)  An old conics question that Solveit, to her credit, has persisted with. 

    https://www.desmos.com/calculator/ws0izufvrt

10)    This conics question should go up in our Hall of Fame.  We must get one of those!

                  Thanks Solveit, Melody, Chris and Alan and new thanks to Heureka and Bertie.   

         http://web2.0calc.com/questions/write-the-equation-of-hyperbole-which-is-tangent-with-the-straights-5x-6y-16-0-and-13x-10y-48-0_1#r15

LOL, I'm glad you enjoyed them, I'll get you some new ones later...:)

Yes, I expect that you do :))

You're quite welcome Melody!

archeological digging

Yes, I suppose I do have to dig quite a lot to find quality posts for u to enjoy... JJ

All of the water from a full cylinder tank is drained into a rectangular aquarium. The cylinder has a height of 50 cm and a diameter of 30 cm. How deep is the water in the aquarium, if it has a rectangular base measuring 40 cm by 20 cm?

A cylinder with radius r units and height h units

has a volume of V cubic units given by 

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} V_{\text{cylinder tank}} &=& \pi \cdot r^2 \cdot h \\ &=& \pi \cdot 30^2 \cdot 50 \ cm^3 \\ &=& 45000 \cdot \pi \ cm^3 \\ &=& 141371.669412\ cm^3 \\ \end{array} ~}\\\\ \end{array}\\\)

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} V_{\text{aquarium}} &=& a\cdot b \cdot h \\ &=& 40\ cm \cdot 20\ cm \cdot h \\ &=& 800\ cm^2 \cdot h \\ \end{array} ~}\\\\ \end{array}\\\)

\(\begin{array}{lcll} \boxed{~ \begin{array}{lcll} V_{\text{cylinder tank}} &=& V_{\text{aquarium}} \\ 141371.669412\ cm^3 &=& 800\ cm^2 \cdot h_{\text{aquarium}} \\ h_{\text{aquarium}} &=& \frac{141371.669412\ cm^3}{800\ cm^2} \\ h_{\text{aquarium}} &=& 176.714586764\ cm \end{array} ~} \end{array}\)

The water in the aquarium is 176.7 cm deep.

Or in meters: 1.767 m

@@ What is Happening?  [Wrap4]   Thurs 21/1/16   Sydney,  Australia Time 5:35pm   ♪ ♫

Hi everyone, it has been busy!!

Answer credits are awarded to CPhill, Alan, Heureka, Bertie, Omi67, asinus, BreeDeBoss, OneDirection, rarinstraw1195, Will85237, Solveit, Coldplay, Dragonlance, xvxvxv, InjustaGod and Wondergirl1257.  Thanks everyone.  

Belated Happy Birthday to our wonderful storyteller, Dragonlance.  We hope that you had a great day  

Interest Posts: 

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.

Communications and Fun:

1)  A forum history trip, complete with lots of comedy.  Coldplay has been doing some archeological digging.

     Thanks so much Coldplay, these old posts are great!

Very good effort CPhill, but the answer in my book is different by a few thousands dollars. This might be up in "heureka's alley!". I shall get the "shortcut" from our teacher tomorrow.

Thanks I got the same answer!!

\(\sqrt{x+\sqrt{x-2}}=2 ===>> {(x+\sqrt{x-2})}^{1/2}=2 ==>>> (x+\sqrt{x-2})=4 ===>>>4-x =\sqrt{x-2}\)

\(4-x=\sqrt{x-2} ==>>>{(4-x)}^{2}=x-2==>>{4}^{2}-2*4*x+{x}^{2}=x-2\) 

\(16-8x+{x}^{2}=x-2====>>18-9x+{x}^{2}=0\)

Applying Bhaskara we gonna have x=6 or x=3

And 

\(\sqrt{3+\sqrt{3-2}}=2\)

X=3

OK....assuming end of year deposits, we have

1(1.05)^99 + 2(1.05)^98  + 3(1.05)^97 +  ....+  99(1.05)^1 + 100(1.05)^0

n = 100

∑      (n)(1.05)^(100-n)  ≈  $52,810.53

n = 1

P.S. The deposits are made at the end of each year.

Thanks Bertie, 

I have only looked at your post superficially. I look forward to finding the time to study it and fill in the gaps. :)

CPhill told me privately that he was too :)

I have not finished with this one yet  ://     

Yeah, Melody....I see that I'm probably wrong.....I'm just trying to figure out my mistake....do you see it???   ....did I "double-count" somehow???

[ My answer seems  "logical"......LOL!!!! ]

thank you so much!

Thanks, Melody....!!!

I really like that too.  I shall have to try and remember it.

Moronic Condescening Idiot Disorder.   MCID

That does fit most of Nauseated's symptoms!   MCID   MCID   MCID ....

That is me trying to ram it into my brain :))))       Hahahahaha   

\(\sqrt{5^9}\)

Assuming that the deposits are made at the beginning of each year ....we have....

1(1.05)^100  + 2(1.05)^99  + 3(1.05)^98  +   ...... +  99(1.05)^2  + 100(1.05)

n = 99

∑         (n + 1)(1.05)^(100-n)       =  about $55,451.05

n = 0

Maybe heureka knows a shortcut ???

See, that's what I didn't understand! I though it was everything being multiplied. You are correct then guest. It is -7. Sorry to worry you! I feel like such an idiot now! 

See, I wouldn't believe that I am, but it all depends on someone's POV.

Nice work Chris :))

In how many ways can we seat 7 people in a round table with a certain 3 people not in consecutive order?

I do not understand the question ://

What does it mean that the 3 people are not in consecutive order?

I will take it to mean that all 3 of these people cannot sit together.  I think CPhill has interpreted it the same way.

2 CAN sit together but not all 3

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

ok, as CPhill said there are 3!=6 different ways these three can sit together - that is ignoring everyone else.

Mmm   Altogether there are 6! = 720 ways to sit 7 people at a table (rotations are counted as the same)  

So CPhill's answer can't be right.  Sorry Chris.   

My nexty task is to try and work out how many ways the group can be seated with the 3 siting next to each other.

Now if I tie the tree together and treat them like one person then there are 4 real people plus the triplet and these can be seated in 4! ways = 24 ways.

But there are 6 ways the triplets can be tired so that is  24*6=144 ways the three can be together

so

The number of ways that they can be seated if those three are not all together must be 720-144 = 576 ways

That is what I think anyway LOL

I believe this is supposed to  be  x √(x+(√x-2))=2

x √(x+(√x-2))= 2   divide both sides by x

√(x+(√x-2))  = 2/x  squre both sides

x + √x  - 2  =  4/x^2

√x  = 4/x^2  - x  + 2   

√x  = [ 4 - x^3  + 2x^2] / x^2      multiply both sides by x^2

x^(5/2)  = 4 - x^3 + 2x^2

x^3 + x^(5/2) -2x^2 - 4  = 0     

WolframAlpha shows the solution to this as   x ≈ 1.83436