Melody

Thanks guest, that is a great answer.  I just wanted to think about it a little. 

There are an infinite number of answers.

(a)^3 + (b)^3 + (c)^3 = 33

what is a, b and C?

If a=b=c then

\(3*(a^3)=33\\ a^3=11\\ a=\sqrt[3]{11}\\ a=b=c=\sqrt[3]{11}\)

Here is another answer.

(a)^3 + (b)^3 + (c)^3 = 33

\(a=1,\;\;b=2,\;\;c=\sqrt[3]{24}\)

0=26sin60t + (1/2)(-9.8)t^2

0=(26sin60)t + (1/2)(-9.8)t^2     It is better with the extra brackets.

0=t [(26sin60)+ (1/2)(-9.8)t ]

t=0       or      26sin60 = (1/2)(9.8)t 

      \(t=0\qquad or \qquad 26*\frac{\sqrt3}{2} = \frac{(9.8)\;t}{2} \\ t=0\qquad or \qquad 26\sqrt3 = (9.8)\;t\\ t=0\qquad or \qquad \frac{26\sqrt3}{9.8} = \;t\\\)

And you can finish it.         

@@ What is Happening?  [Wrap4]   Sun 15/11/15   Sydney, Australia Time 12:50am (really Monday)  ♪ ♫

Hello everyone,

I trust you have all enjoyed your weekend.  

We had soem great answers today from Alan, Sarafoyiti123 and Bertie.  Thanks you.

Repeat request:

In the post below where I have invited people to discuss their views of the forum.  All sensible thoughts will be greatly appreciated.

Thanks you Bertie,

We have to get you back on this forum properly!  I nearly missed your post!

I will take a really good look at what you have said and tuck your post away in a safe spor that i can find later. :)

If you are in the mood it would be good if you post a modular question every now and again so that I, and maybe others, can practice.  

For some reason modular arithmetic and diophantine equations have caught my imagination for the moment.  Probably because they do not seem that difficult but I have only had very recent experience with them.

Anyway, thanks, much appreciated :))

sqrt(7-2*sqrt(3))+sqrt(3

\(\sqrt{7-2\sqrt3}+\sqrt3\)

What do you want me to do with this?

I can use the forum calce to approximate it I guess.

sqrt(7-2*sqrt(3))+sqrt(3) = 3.6124492719540845

Hi Mellie,

I love answering your questions. I would of answered this one earlier but it got lost. I thought I saw it but then I could not find it.     

Points S and T are on side CD of rectangle ABCD such that AS and AT trisect angle DAB . If CT=3 and DS=6, then what is the area of ABCD?

Here is the diagram. I answered it just  from a rough sketch but i wanted to check it was correct so I draw it to scale. :)

 DS is given as 6 units and TC is given as 3 units, ST is not given.

The LaTex function is not working so I will have to do it by hand.

tan 30 = DS/H = 6/H

1/sqrt3 = 6/H

H=6sqrt3

tan60 = DT/H

sqrt3 = (6+ST)/6sqrt3

6*3=6+ST

18=6+ST

12=ST

so

DC=6+12+3 = 21

Area = AD * DC

Area = 6sqrt3 * 21

Area = 126 sqrt3      units squared.

That was a great question - thanks Mellie.

The most recent Nauseated one was a physics answer. Not intended to be funny and I doubt i would be able to find it.

Here is an old one that you will appreciate. :)             It certainly brought  smile to my face. :)

Sorry Sara,

I misread your question.   

3 ✖️ n ✖️ n ✖️ n ✖️ d ➗ w =

\(\frac{3n^3d}{w}\)

What is zero times infinity

It is undefined.