Recent content by Rikudo

I found something that discussed the relationship between addition of probabilities and number of ways. Here is what it says:Suppose the event can happen in two ways which cannot concur and let ##\frac {a_1}{b_1}##, ##\frac {a_2}{b_2}## be the chances of the happening of the event in these two...

##\displaystyle P(\text{white})=P(\text{box 1 AND white})\,\text{OR}\,P(\text{box 2 AND white})=\frac{1}{2}\cdot\frac{20}{40}+\frac{1}{2}\cdot\frac{10}{15}=\frac{7}{12}## That is the correct answer from the other thread. From the denominator of this final result, we can conclude that the...

But, this is the definition of probability itself. So, isn't that should be true for any cases?

https://www.physicsforums.com/threads/choosing-a-ball-at-random-from-a-randomly-selected-box.1034377/ First of all, I would like to point out that this is the same exact question from what is being discussed in the thread above. In that thread, the problem is solved by adding the probability...

So, to find ##k##, should I make use of the quotient instead of the remainder? The quotient will be : ##\frac n k + \frac {k-1} {k²} ##Here is what I thought initially: At first glance, the only possible value of ##k## is 1. Otherwise, the second term will become a fraction (consequently, the...

So, ##n\, |\, (p − 1)## implies ##p = nk + 1## and ##p ≥ n + 1##. Clearly, ## p \,|\, n^3 − 1## implies either : ##p \,|\, n − 1 ## (which is impossible, because p cannot be less than ##n-1##) or ##p \,|\,n^2 + n + 1##. Now, our main focus is ##p\, |\,n^2 + n + 1##. Since ##p = nk + 1##...

Why can't we just substitute ##(1+x)^{1/x}## as ##e?## then, this equation will become: $$\lim _{x\to 0} \frac{f(x)-f(0)}{x} = \lim _{x\to 0} \frac {e- e}{x}$$ Since x is near zero, then we can make it like this: $$\lim _{x\to 0} \frac{f(x)-f(0)}{x} = \lim _{x\to 0} \frac{0}{x}$$ So, the...

Ah. I see.. Now I managed to prove it. There are some things that is bothering me,though.. 1. Why my method in post 1 does not work? 2. I tried a different way to transform the term ##ln(1+x)##, but I only get the first term of the maclaurin series. Here is what I did: let ##y= ln(1+x)##...

Ok. so, I tried to differentiate the ##(1+x)^{1/x}##, and using the rule, I get: $$\lim _{x\to 0} \frac{x}{(1+x)^{1/x} - e} = \frac{1}{\frac{x(1+x)^{-1+\frac{1}{x}}\,\,-\,\,(1+x)^{\frac 1 x}\,\,ln(1+x)}{x^2}}$$ Then, all I need to do is to substitute the value of ##x## as zero, right? The...

Right now, I am trying to prove this : I tried to use this identity to solve it: Then, the limit will become ##\frac {x}{e-e}## However, the result is still ##\frac 0 0 ## Could you please give me hints to solve this problem?

I have a different way in solving the problem, but strangely, the result is different from that written in the solution manual. My method: Firstly, we will solve the ##AB=A## equation $$AB=A$$ $$B=A^{−1}A$$ $$B=I$$ where ## I## is an identity matrix Similarly, we can solve ##BA=B## using the...

Umm... well, it is is true, but what is the relation of it with the area of ring? Why, though? We can't use ##ds## as its height since it is not perpendicular to the bases.

Yes. But, ##ds## is not the height of the trapezoid. So, in my opinion, we are not supposed to multiply this with the ##2πa sin\theta## to get the area.

Why the area of the thin rings are ##2πasin\theta \, ds##? (a is the radius of the hollow sphere) If we look from a little bit different way, the ring can be viewed as a thin trapezoid that has the same base length ( ##2πa sin\theta##), and the legs are ## ds##. The angle between the leg and...

$$\int -mg \mu d(l) = \frac 1 2 m(v²-u²)$$ Where v is the speed of the lower block just before the 2nd collision. $$v = \sqrt {u² - 2g\mu l_0}$$ So, there are two answers: 1. ## v = 0## if ##u² ≤2g\mu l_0## 2. ## v = \sqrt {u² - 2g\mu l_0}## if ##u² ≥2g\mu l_0## Do I also need to think of...