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I just want to play with this one for a moment.  (This has been edited)

I don't think Chris's is right I think it is 1/3 of a different cube which i could find but I want to try a different way.

Let the side lenght of the cube be 1.

Inside the cube is the tranglular required trianglular pyramid but

the space surrounding this consist of 3(no 4 as alan has pointed out) congruent right triangular pyramids

The volume for each of these is

$$V=1/3 \mbox{ x area of base x height}\\
V=\frac{1}{3}\times (1/2 * 1*1)*1\\
V=\frac{1}{6}\qquad u^3\\$$

So for all three added together the volume is 1/6*4 = 2/3  u³

The volume of the whole cube is 1u³

Volume of the tetrahedron (triangular pyramid) = 1-2/3 = (1/3)u³

Which is one third the volume of the whole cube.

Did I make a mistake?  I can't find one. (yes I did there were 4 not 3)

I gave you another 3 points.

Thank you for your excellent answer NinjaDivo 

I have included the address of this  answer in the Sticky note: "Information pages worth keeping or developing"

It was CPhill's suggestion but I thought it was a god idea too!

I'm sorry Bertie, but do you mind writing it out?

I know it is a lot to ask, but I'm completely lost at the part where you start taking (14-L) out and adding rows and such. I think I don't remember enough rules about computing the determinant to be able to follow you by your explanation.

(And yes, I am not allowed to use a calculator on my exam).

It is a good idea to think about a number line when you deal with negative and positive numbers.

What is the biggest number 14 is, it is furthest over towards the right.

(14 is really +14 the positive sign is just invisable but it really is there!)

Any number to the left of 14 is smaller than 14

So, ALL the negative numbers are smaller than aLL the positive numbers!

Well, put them on the forum and we will see.  

10 to the power of 16 is 1 followed by 16 zeros.  You do not need a calculator for powers of 10.  They are easy!

10^7 = 1 with 7 zeros = 10 000 000

etc     

Tealftlove, you are obviously capable of inserting pictures - Your instructions keep giving us a different question from the one you seem to want answered.  It is very confusing.  

I walk around in a fog of confusion ALL THE TIME.  I do not need you to make the fog more dense!! lol

This is number 10

This is me!

Thank you CPhill and Problem,

First the queston in the heading is different from the question in the body of the post so CPHill and Problem have answered different questions.

ANONYMOUS - Please be much more careful next time you make a post.

CPhill's answer is much better because Problem's answer is an estimate.  It is not an exact value for the question.

ALSO,

The second solution that Problem has given is wrong. 

$$\sqrt{3}-\frac{4}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}(1+\sqrt{3})}{1+\sqrt{3}}-\frac{4}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}+3-4}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}-1}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}-1}{1+\sqrt{3}}\times\frac{1-\sqrt{3}}{1-\sqrt{3}}\\\\
=\frac{-(\sqrt{3}-1)^2}{1-3}\\\\
=\frac{-(3-2\sqrt{3}+1)}{-2}\\\\
=\frac{-(4-2\sqrt{3})}{-2}\\\\
=2-\sqrt{3}\\\\$$

Some fine german lessons here!

ACTUALLY

1) If the square root is written into the question then, by convention, the answer is POSITIVE.

$$\sqrt{4}=+2$$

2) However, if the question is

$$x^2=4$$

then this is how you solve it

$$\sqrt{x^2}=\pm\sqrt{4}\\

x=\:\pm\; 2$$

This is because the square root was not a part of the question.  It was introduced as a part of the solution.

So this time there are 2 answers, +2 and -2

The question is jibberish hence there is no mathematical answer.

BUT WAIT, what is that i hear.

THE ANSWER IS ....... 42

Oh Sorry Lord from far future time.  I have erred. Please leave me with my head.  

One headless moderator is more than any forum doth need.

Your humble servant.

Lady Guinevere.

It's when you are reduced to threats and witchcraft that I KNOW I am in the lead!!!!!

Quiet, whippersnapper!!! Things are what we say they are !!!

I'll get you, my pretty...and your little dog, too!!

See if you can answer your own question....are 48 hrs. 2 days??

How could you find out??  If 1 day = 24 hrs., what would you need to do to find out how many hrs. are in 2 days??

Think about it a little........

No such thing.......whatever you think might be the highest....I can always add 1 to it !!!

There ARE some well-known "Large" numbers, however....here's one.....

I'm assuming you want to find the volume of a tetrahedron inscribed in a cube....I don't know much about this, but there is a way it can be done.....(see below)....the volume of this is just 1/3 the volume of the cube = s^3/3  where s is the side of the cube....

There may be other answerers who could give you more help, but if not, here's a page that will probably tell you more.....

thank you Chris.

This is an equation relating to the Gateway Arch in St.Louis.......I don't know if the poster wanted an answer or was just curious as to what it was (I wouldn't know how to give one, anyway).....but I guess if he/she were interested, they could look at this page:

This one makes no sense to me  and since no one else has answered it i am assuming  it makes no sense to anyone else either!

Yes I know functions and relations have domains and ranges but this has no equal sign.

Techically speaking - does it really have a domain and a range.  To me it is just an expression.

(I know an equal sign and a y or f(n) is implied BUT it is not really there!)

I know I am being padentic but mathematics is a padentic field of study!

I think you have some problem while typing

(11.0-12.3)^2+(10.7-12.3)^2+(13.1-12.3)^2+(13.9-12.3)^2+(12.9-12.3)^2+(12.5-12.3)^2+(11.9-12.3)^2+(12.4-12.3)^2

=(-1.3)^2+(-1.6)^2+(0.8)^2+(1.6)^2+(0.6)^2+(0.2)^2+(-0.4)^2+(0.1)^2

=1.69+2.56+0.64+2.56+0.36+0.04+0.16+0.1

=8.11

The volume of clay is given by volume of the outer hemisphere less volume of the inner hemisphere =  (2/3)*pi (8.2^3 - 7.7^3) ≈ 198.6cm³ = pi^3 * k

Dividing this by pi^3, k ≈ 6.41cm³

I'm assuming that this topic relates to the volume of spheres ???

The volume of a hemisphere is given by

$$volume=\frac{2}{3}\pi r^3$$

where r is the radius.

The desired volume of clay will be the difference between the outer and inner hemispherical volumes, so:

$$k\pi^3=\frac{2}{3}\pi8.2^3-\frac{2}{3}\pi7.7^3 = \frac{2}{3}\pi(8.2^3-7.7^3)$$ 

The exact value of k is therefore

$$k=\frac{2}{3}\frac{(8.2^3-7.7^3)}{\pi^2}$$

Numerically, the approximate value is:

$${\mathtt{k}} = {\frac{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({{\mathtt{8.2}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{{\mathtt{7.7}}}^{{\mathtt{3}}}\right)}{{{\mathtt{\pi}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{k}} = {\mathtt{6.405\: \!862\: \!967\: \!147\: \!401\: \!7}}$$

or k ≈ 6.41 (it will have units of cm³)

I would say this topic is Geometry.