Math Is Fun Forum

That is a tough one for pre-algebra if you haven't learned the laws of rational exponents. Here's another way to look at it:

If you know the rules of rational exponents, then:

This part you can use some reasoning for. "k" should be equal to its cube root * its cube root * its cube root, right?
So if the problem above shows that cube root k * cube root k * -2 = k, then that must mean -2 is the cube root of k.
That then means (-2)³ = k, or -2*-2*-2 = -6 = k.

EDIT: As per below, (-2)³ is, in fact, -8, not -6.

I believe it's related to the number of posts you've made, John.

Good question. I don't know the even-square algorithm

Edit: I'd probably fall back on 17 simultaneous equations

Classic one, Katie:

Two from me, the first written by JRR Tolkein, the second is a bit silly:

1.
It cannot be seen, cannot be felt
Cannot be heard, cannot be smelt.
It lies behind stars and under hills,
And empty holes it fills.
It comes first and follows after,
Ends life, kills laughter.

2.
Used my men a thousand times
Used by the world only twice
What's it good for? Absolutely nothing.

Unless you're the only person who speaks it:
http://en.wikipedia.org/wiki/Amurdag_language

You're using every odd number, right? Here you go:

093    115    137    159    001    023    045    067    089
113    135    157    017    021    043    065    087    091
133    155    015    019    041    063    085    107    111
153    013    035    039    061    083    105    109    131
011    033    037    059    081    103    125    129    151
031    053    057    079    101    123    127    149    009
051    055    077    099    121    143    147    007    029
071    075    097    119    141    145    005    027    049
073    095    117    139    161    003    025    047    069

It all adds up to 729.

There's an algorithm I know for doing this and it worked for the odd numbers luckily.

1. The first number goes in the middle cell of the top row.
2. Place the next number 1 right and 1 up from that number (the order of this is important).
     2.a If you're at the right-most cell when you try to move right, go to the left most cell of the same row instead.
     2.b If you're at the top most cell when you try to move up, then go to the bottom of the current column instead.
3. If the destination cell is already populated, then fill in the cell 1 down from the last one you filled in instead.
4. Repeat until you've filled in the whole thing.

It should work for any magic square with odd-dimensions.

I suspect that he's thinking a christmas present must be wrapped as the attached, which is the way I wrap presents. His assumption is that those four flaps (two on the visible side and two more on the opposite side) should have a height of C as well. Which would mean the area of those triangles is 4x0.5CC = 2CC.

Now that I think about it... that wouldn't work. We already know the ideal width for the paper is A+C, meaning that the height of those flaps should be HALF of C.

So his equation should ACTUALLY be:
Area=2ab+2ac+2bc+(4)(0.5)(0.5c)c
A = 2(ab+ac+bc)+c²

That's amusing. They must have thought the "c2" was a typo and removed it .

I would have loved to eavesdrop on the conversation when they asked how he arrived at this magic formula and (to cover for himself) he started going into how he used advanced differential calculus to find the ideal solution for three variables.

"Gimme my commission now, kthxbai!"

Nice one. Far more elegant than my number juggling.

Jane's got sabers and Null's got chainsaws

Daniel

That sounds right. Just think of it this way:

1/(x²-1) is pretty much the same as 1/(some numbers). Once you work out the possibilities for the 'some numbers' then you should be able to analyze what possible values 1/(x²-1) can give you.

Here 'some numbers' is defined as x²-1, where x is any Real number greater than 1. So we know it can be anything from just over zero to as big as you want and that it increases continuously throughout.

So when you try the range of values from x²-1 you see 1/(a very small number) gives you a very large number. When you try 1/(a very large number) you get a small, positive number.

Just go ahead and post your question, Youssef. We'll be happy to help.

Hmm... something else: