We have a winner! Wooo!
Your work leaves a bit to be desired, and I used a different condition (ax=0 instead of ay=g, so everything I had was worked out in x, not y!), but your answer for the magnitude of N is correct, so I trust that you got the right answer. (2) follows from (3) in any case, so I’ll give it to you
But you were so close! And it’s not so bad, just a couple of bits you need to tack onto N.
What do you mean? We’re all enrolled at the university of Huck. We have our own programing class lead by @NavyFish and co. We have a screenshot class by @Hutchings and now we have a suffering class by these folks here.
Edit: On that note, I wish I had time to actually search for materials to be able to do this type of math solving/proofing. But I’ll have to stick to simple carpenters math (Which oddly enough is the most confusing shit in the world sometimes.)
If people would like more, I’m full of problems like these. Feedback first, though. Are they too tough? Too easy? Any preferences for subject material? As y’all might have guessed, I was pretty infatuated with classical mechanics, but I can be flexible. Probability problems are another favorite of mine, and I can probably dig up some neat abstract algebra problems or something.
I can also provide a worked solution to the quadratic slope problem if people desire… Tomorrow. Finals studying tonight.
Guys, I need a little help in Probability. I have this problem: From a deck of cards (52) we remove the face cards (J, Q, K) with probability 1/2 (each one). After that we draw 5 cards from the remaining ones. If X is the number of aces in the hand then calculate E[X]. Thanks for anyone who can solve it
(I tried to solve it assuming that X is a hypergeometric variable conditioned by Y, a binomial variable with parameters n=12 (all the face cards) and p = 1/2. Then i calculate E[X] = Sum(E[X|Y=y]P(Y=y)) where E[X|Y=y]= 5*4/(52-y), the expected value of the hypergeometric variable and P(Y=y) is the
density of Y in y. Well that formula is pretty ugly, i calculated it and the value was near 10/23, one of the possible answers to the question in my exam simulation, but not equal to… so i think there must be another way to do this!)
Given two values, p and q, such that pq = 0, either p, q, or both p and q must be zero. This applies to functions as well.
That means, on the interval (a,b), either f(x) = 0, or df/dx = 0, or both. And since f(x) = 0 is a constant function, and df/dx = 0 requires a constant function, all roads lead to f(x) = const.
Math comes back with a vengeance. Just kidding, but I stumbled upon this little “problem” on a news site (the articles headline was something along the lines of ‘The net goes crazy over trying to find all the triangles. Can you find them all?’, so I didn’t bother actually reading it, just took a look at the picture which showed the triangle). I wasn’t interested in drawing in every triangle possible, but I wanted to know, how many triangles there are so I did some calculations and thought it would be fun to share this with you. So, what do you think?
How many triangles are there? (and more importantly: how did you reach your conclusion?)
Let’s assume that any shape that is constructed using 3 straight lines that intersect with each other counts as a triangle, no matter what other (if any at all) shape(s) is/are inside of that shape.
I count 10. The base triangle, then the 2 put together to make the base. Then 2 more making the top left of those 2(the equilateral). Then the top 3 double inner, 1 in the middle, 1 on the bottom left. That makes 10.
My triangles are defined as triangles…where the sum of the angles is 180 degrees and have 3 sides.
Ooh 6 extra. Wish I could see those. I believe there is, this reminds me of graph theory in discrete. Theres definitely a formula where you count the edges and vertices and get some sort of answer.
