CPhill

As an addendum.....apparently......our calculator never heard of this, either...LOL!!!

If you key in  !N, where N is some positive integer........nothing.....!!!!

√3 tan x + 2 sin x = 0   rewrite as

√3sin x / cos x + 2 sin x = 0  factor

sin x (√3 / cos x + 2 ) = 0       

And sin x  = 0   at  -180, 0 and 180 on the given interval

For he second factor, we have 

(√3 / cos x + 2 )  = 0       rewrite as  ( √3sec x + 2 ) = 0

So we have

sec x = -2 / √3   and this would be where the cos x = -√3 / 2

And this would occur at -150 and 150 on the given interval

Here's a graph of the solutions....

So.....by Alan's reasoning ....the general solution would be for one die to have six faces labeled  n, n+1, n+ 2, n+3, n + 4  and n+ 5  for n ≥ 1. And the other die would have three faces labeled  -n + 1  and three labeled -n + 7

Yep.... about 12.24°

Be aware that it also might be a 3rd quadrant angle because the tangent is also positve there....i.e., 192.24°

SKOAL.....!!!!

Uh....sorry, Anonymous...unless you have devised a new system of Mathematics, the square root of 7 is about 2.646.....

The proof is this, Melody.....it's in the way that "e" is defined...

e = lim as n →∞  (1 + 1/n)^n

Notice, that Anonymous is letting n, the number of compounding periods per year, grow larger and larger......if we theoretically make "n" "huge," then the number of compoundings per year ≈ "continuous" because the time intervals of compounding → 0........thus, we have an "infinite" number of compoundings per year of extremely short duration each....!!!!

Yeah.....I love it!!! (them)

I think he means this, Melody...

Let's suppose that we have the ordered elements of a set, {A,B,C}

Then, the possible ways that none of them can be in their "natural" positions is just

{B,C,A} and {C,A,B}  = 2 "derangements"

And this equals.....according to heureka's formula....

3! ( 1 - 1/1! + 1/2!+ 1/3!) =

3! ( 1/2! + 1/3!) =

3! [3 - 1]/ 6=

6 (2/6] =

6[1/3] =

2 "derangements"

Hey.....I've been "deranged" my whole life.....now .....I've found the "formula" as to why....!!!

I befriended a very friendly wolf and he really likes me. I decided to take him home to live with me and be my hunting companion. Before I had my wolf-dog, Wolfie, I used to bring back home approximately h pounds of meat from hunting in the forest. Now that Wolfie is with me, the ratio of how much meat I brought home before I found Wolfie to after I found him is m:b. I almost always eat approximately 1/8 of the meat we bring home now. Wolfie eats about w of what I eat out of the rest of the meat that we have. We eat 3 meals a day of about the same amounts. Each pound of meat we have left over is sold to merchants for (on average) for $d. How much money do we get selling the meat we have left over daily now that I have Wolfie?

The amount of meat now brought home is h(b/m)lbs. = (hb/m)lbs.

The amount consumed by you is (1/8) of this =[hb/(8m)] lbs. ...so there's (7/8)(hb/m) lbs. left

And Wolfie consumes w of what you eat...and "of" means "multiply".... so he eats → w[hb/(8m)]lbs

So, the amount left over is (7/8)(hb/m) - (1/8)(w)(hb/m) = [(hb)/(8m)](7 - w)lbs

So....the amount of money you make is just what you're paid per pound times the number of lbs. left over =

$(d)[(hb)/(8m)](7 - w)

{The 3 meals per day is a "red herring" ....an irrelevant fact.....}