xkcd

Hmmm your A level students aren't likely to like it (speaking as an AS student). Year 7 might like it but after year 9 they will probably be too jaded for it. That said it we like sweets and will put up with (and maybe even participate in) anything if there is a high probability of them :) I'm teac...

Sadistic Humor wrote:Achievement Unlocked: Not looking like a prostitute.

It would almost be worth the stern costernation of administration to put that into writing. Also, sad that it's come to that though.

I'm a high school math teacher, and in the interest of promoting student involvement and participation, I want to implement an achievement system similar to the ones used on Xbox Live or in World of Warcraft. (They'll spend hours working on those, but can't spare 15 mins for extra help...) For the s...

I clearly see now that fixing either the spawn time or the speed and varying the other has the same effect on the problem. I didn't get quite that meaning when I read your earlier post.

We are told the speed the dots are spawned is a constant 1/sec. This is independent of the speed of our elevator. If you'll allow me a discrete example to illustrate why I'm having trouble with this idea... Imagine n cash registers that begin empty. Every minute a customer arrives at an empty regist...

One dot is added at 0. Then dots are added, two at a time, at 0.9 and 1.0, every 0.1 seconds. This just happens, randomly... It is never in the interests of the elevator to head down to the 0, if the goal is to reduce the average wait time of dots after they are spawned. Yeah, but the P(X>.9) is on...

Since the points are truly random, I think the best method is to just travel back and forth from 0,1 over and over. The worst case scenario is a wait time of 2/v, but the average should be no worse than 1/v and I'd wager most wait times are less. I just don't have the time to try and nail down an ac...

How has noone pointed to these? For all your non-orientable drinking requirements.

Here's how this one goes: /snip What is r, the polynomial you get from dividing f by g. You obviously can't use the division theorem, and f is not irreducible as far as I can see. If, as you assert in the problem, the answer is a polynomial, then we know g divides evenly into f. That should simplif...

Token wrote:Change of variables: y=e^x, solve for y as a quadratic, then take log for x.

Since e^x is never zero I multiplied through the equation by it, then did the u substitution and solved the quadratic. Double checked on graphing calculator.

Based on the receipts in my wallet, you round to the nearest cent, and if the subtotal ends in 0.10, 0.30, 0.50, 0.70, or 0.90, you round up.... I have a lot of receipts in my wallet.... Incidentally, at the Greenhouse Cafe, a cup of Tazo tea is $1.39, a bubble tea is $4.25, and a tomato basil mozz...

Assuming they each ordered the same thing and tax was not included in the price -- there were 3 of them, the total for the tea and sandwich was $6.01.

$6.01 * 3 * 1.05 = $18.94

And this is what I've done so far: A = 2 \int 1 + {1 \over 3}y ds Now, using \int f(x,y) ds = \int f(x(t),y(t)) \sqrt {[x'(t)]^2 + [y'(t)]^2} dt We get 2 \int_0^{\pi \over 2} 1 + 10sin^3t \sqrt{(-90cos^2tsint)^2+(90sin^2tcost)^...

[math]2 \int_0^{\pi \over 2} 1 + 10sin^3t{90sintcost} dt[/math]
Applied a trig identity:
[math]2 \int_0^{\pi \over 2} 1 + 10sin^3t{45 sin 2t} dt[/math]


Instead of using a trig identity can't that be simplified further to:
[math]2 \int_0^{\pi \over 2} 1 + 900sin^4t{cost} dt[/math]

So, I'm looking for a periodic function f:\mathbb{R}\rightarrow \{0,1\} . Here's the caveat. I want the function to have some sort of variable periodicity -- say I want to generate a pulse every 1.5 seconds, but say at random the next pulse might come at t+.75 seconds. Here's the second caveat. It ...

Well, yes, but it's no longer a set. Sets don't have repeated elements. It's pretty bad notation on her part, but it's not worth arguing the point. (Seriously though, sets don't have repeated elements.) Sets don't have repeated elements, or a set with repetitions is equivalent to one with repetitio...

SlyReaper wrote:

GreedyAlgorithm wrote:The turn comes 7, the river comes 7. Now do you think it's less likely your opponent holds a 7?

Slaps down a pair of sevens on the table. Six of a kind! I win!

No... he has the uh, royal sampler.

My new College Algebra teacher told me "that is not always the case." Life lesson: Smile and nod, and use her definition of a function in her class . People in power aren't always right, but they can really mess up your GPA. Or understand that at a lower level a simplified definition was ...

OK, I don't know how to do the markups to make the pretty math symbols, but I think I can explain it without. Let the left circle be the unit circle centered at the origin and the right circle be a unit circle with center (a,0). These circles intersect at x=a/2. Consider now only the semicir...

It is easy to see that one can go through the entire deck: Each ace can add 4 cards to the pile, each king can add 3, each queen can add 2, and each jack can add 1 for a total of 40 cards. Adding in the 16 face cards makes a total of 56 possible cards dealt. Why do you add in the 16 "face"...

Let's see, there's 52 C 7 ways to deal 7 cards. Of those 36 C 7 have no face cards, so 6% of the time you don't even drink... so sad. I think the easiest way to solve this would be to run a simulation, there are so many test cases that the math would get long and boring before I arrived at a solutio...

It is easy to see that one can go through the entire deck: Each ace can add 4 cards to the pile, each king can add 3, each queen can add 2, and each jack can add 1 for a total of 40 cards. Adding in the 16 face cards makes a total of 56 possible cards dealt. Why do you add in the 16 "face"...

As a general rule the absolute value is not differentiable because of the sharp point at zero... the limit of dy/dx is different from the left and right. If you break it up into a piecewise function and do the two halves seperately, then it usually works out ok. Disclaimer - Teaching HS has left me ...

If you're saying you want a homeomorphism between Im(c) ∪ Im(d) and the union of two lines with the same endpoints, you won't find it, since the former might actually have more (distinct) intersections than the latter. OK, but if it isn't possible for the former to have FEWER interesections than th...

Is it necessary to prove that c and d are homeomorphic to lines with the same endpoints, or is that agreeably proven by someone else with a lot more education than me? Homeomorphism isn't the word you're looking for; you want homotopic. (The homeomorphism is obvious: take the inverse of one paramet...

OK, as before feel free to correct me anywhere and everywhere. Is it necessary to prove that c and d are homeomorphic to lines with the same endpoints, or is that agreeably proven by someone else with a lot more education than me? If so the slope of the image of c is bounded by the interval (-1,1) a...

Well, as I've read the discussion it seemed to be, "oh we can't use the intermediate value theorem because we don't have functions". If as you say I've restated the original problem successfully with a function, then we can use IVT and the result should be trivial for someone with a little...

OK, if we're still trying to solve the original problem... Let x(t) be the horizontal displacement of c inside the square; x(0)=0 and there exists a t c such that x(t c )=1 Similarly, let y(t) be the vertical displacement of d inside the square. Now x(t) necessarily takes on every value from [0,1] e...