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 Mar24 answered $\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial y}=1$ Mar23 comment Combinatorics Question, drawing from a pot I read the question again and it seems d) is impossible, if we interpret it like c). So yes, you are right. Mar23 comment Combinatorics Question, drawing from a pot I've edited the answer. d) and e) are the same. It is possible to use combinatorics, but you'll have to deal with conditional probabilities and probably end up with some sum, but you will not get a nice formula. Mar23 revised Combinatorics Question, drawing from a pot added 137 characters in body Mar23 answered Combinatorics Question, drawing from a pot Mar22 comment Find the radius in dependency of a (the side length) The new picture is still not enough to correct your problem. Can you tell what exactly your professor told you? Mar22 comment Find the radius in dependency of a (the side length) When I said not dependent I meant $r$ cannot be expressed in terms of $a$. You can make $a$ as big as you want without affecting $r$. Mar22 awarded Commentator Mar22 comment Find the radius in dependency of a (the side length) Looking at picture, $r$ is not dependent of $a$. Mar20 revised Probability triangle question? added 189 characters in body Mar20 answered Probability triangle question? Mar20 awarded Citizen Patrol Mar20 comment Finding the probability that one of the given independent events happens As I understood, the question didn't say "one and only one of events should happen". Mar20 comment Chain rule and conditional probability What do you mean by "chain rule for conditional probability"? Mar20 comment Finding the probability that one of the given independent events happens Are those events intersecting? Mar20 comment Finding a second linearly independent solution to a differential equation I think you have a typo there, because $x(t)=t$ is not a solution. Mar20 answered Weird factoring out of a derivative Mar19 comment Functions satisfying $f(m+f(n)) = f(m) + n$ @Biswanath, see the answer below, it explains why you can conclude f(0)=0. Mar19 comment Functions satisfying $f(m+f(n)) = f(m) + n$ Taking $n=0$ gives us $f(m)=f(m+f(0))$ or $f(0)=0$. Next, taking $m=0$ we'll have $f(f(n))=n$ Mar18 awarded Supporter