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• Today, 08:55
Setting a function, in an object, "private" means that it can only be seen or used by that object. Setting a constructor "private" means that object...
2 replies | 25 view(s)
• Today, 08:51
In what order are you doing these transformations?
1 replies | 11 view(s)
• Today, 05:06
That argument does not work, because the function $f(x) = 1 + x^n$ depends on $n$. In fact, there is a whole sequence of functions $f_n(x) = 1 +... 4 replies | 114 view(s) • Yesterday, 20:06 MarkFL replied to a thread Related Rates in Calculus Yes, the length of his shadow is increasing at a rate of 4/3 ft/s, and the tip of his shadow is moving at a rate of 10/3 ft/s. :D 8 replies | 96 view(s) • Yesterday, 16:29 MarkFL replied to a thread Related Rates in Calculus I got 4/3 ft/s where you got 3 ft/s. You have the line: \d{y}{t}=\frac{3}{2}\d{x}{t} Now, this implies: \d{x}{t}=\frac{2}{3}\d{y}{t} 8 replies | 96 view(s) • Yesterday, 16:20 Nicely done, Chris! (Yes) This is the solution I found elsewhere: We are given to evaluate: \lim_{n\to\infty}\left(\sqrt{\int_0^1... 4 replies | 114 view(s) • Yesterday, 15:18 For$0\leqslant x\leqslant1$,$1+x^n \leqslant2$. Therefore \int_0^1 \left(1+x^n\right)^n\,dx \leqslant 2^n and hence \sqrt{\int_0^1... 4 replies | 114 view(s) • Yesterday, 15:15 MarkFL replied to a thread Related Rates in Calculus Suppose we wanted to work the problem in general terms. Please refer to the following diagram:$P$is the height of the pole,$M$is the... 8 replies | 96 view(s) • Yesterday, 13:16 HallsofIvy replied to a thread Related Rates in Calculus You are correct that this involves two similar triangles. The shadow itself extends from the man in the direction away from the light. The larger... 8 replies | 96 view(s) • Yesterday, 10:30 Here is this week's POTW: ----- Right triangle$ABC$has right angle at$C$and$\angle BAC =\theta$; the point$D$is chosen on$AB$so that... 0 replies | 37 view(s) • Yesterday, 10:26 Congratulations to Opalg for his correct solution, which follows. Also, honorable mention to Kiwi. Let$\mathcal P$be the vector space of all... 1 replies | 89 view(s) • February 26th, 2017, 23:47 Compute the following limit: \lim_{n\to\infty}\left(\sqrt{\int_0^1 \left(1+x^n\right)^n\,dx}\right) 4 replies | 114 view(s) • February 26th, 2017, 18:51 Okay, 3, 4, and 5 are equilibrium values because: when y= 3, y'= (3- 3)(3- 4)(3- 5)= 0 when y= 4, y'= (4- 3)(4- 4)(4- 5)= 0 when y= 5, y'= (5-... 4 replies | 78 view(s) • February 26th, 2017, 09:33 Rereading the problem statement, I think we can deduce the following: $$a_{11}=a_{22}=a_{33}=0$$ since it is given that the particle makes a... 4 replies | 98 view(s) • February 26th, 2017, 06:31$m>r$holds because by definition of$r$,$m=rd$and by assumption$d>1$. But I don't see where$s>1$is used in the proof. It can happen, for... 1 replies | 41 view(s) • February 26th, 2017, 06:20 If$=$, then$x$can be canceled and we have$i=j$. 1 replies | 42 view(s) • February 26th, 2017, 02:44 I am reading Anderson and Feil - A First Course in Abstract Algebra. I am currently focused on Ch. 8: Integral Domains and Fields ... I need... 1 replies | 42 view(s) • February 26th, 2017, 02:02 I am reading Anderson and Feil - A First Course in Abstract Algebra. I am currently focused on Ch. 8: Integral Domains and Fields ... I need... 1 replies | 41 view(s) • February 26th, 2017, 01:48 We could likely model this with a function of the form (which comes from Newton's Law Of Cooling): f(t)=c_1e^{-kt}+350 Since we know$f(0)=70$,... 1 replies | 77 view(s) • February 26th, 2017, 01:01 My solution: Consider the objective function: F(A,B,C)=\sin(A)+\sin(B)+\sin(C)-\cos(A)-\cos(B)-\cos(C) Subject to the constraint: ... 3 replies | 114 view(s) • February 25th, 2017, 23:35 Let's look at the behavior of the solutions near the equilibrium solution$y=3$. Suppose we have: y(0)=2.9 Then we see: y'<0 And for: 4 replies | 78 view(s) • February 25th, 2017, 18:55 Hi jerryd! Welcome to MHB! ;) It appears we're applying the angle difference identity for the cosine:$$\cos(\alpha - \beta) = \cos\alpha... 2 replies | 62 view(s) • February 25th, 2017, 18:35 Yep. And more precisely, surjective means that for every y there has to be at least one x. 7 replies | 116 view(s) • February 25th, 2017, 16:22 Okay. For the record, it seems that$P(A)\setminus B$contains elements that are integers, but$P(A)$is supposed to only contain sets doesn't it?... 7 replies | 116 view(s) • February 25th, 2017, 15:32 Hey FallArk! (Smile) If I understand correctly, we're talking about something like this: \begin{tikzpicture} %preamble... 7 replies | 116 view(s) • February 25th, 2017, 10:22 Hi Umar! ;) If I'm not mistaken$a_{31}$is the probability that the quantum state shifts from state 1 to state 3? (Wondering) Then$a_{21}$is... 4 replies | 98 view(s) • February 25th, 2017, 09:15 Why would you "assume" that? Are you clear on what an equilibrium solution is? An "equilibrium" position is one that does not change, such as a... 4 replies | 78 view(s) • February 25th, 2017, 08:09$\cos72^\circ$, "using geometry": \coordinate (A) at (90:3) ; \coordinate (B) at (18:3) ; \coordinate (C) at (306:3) ; \coordinate (D) at... 7 replies | 119 view(s) • February 25th, 2017, 07:25 As for evaluating$\displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } \end{align*}\$, we need to look at a regular pentagram. If we...
7 replies | 119 view(s)
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### 24 Visitor Messages

1. Thanks Fantini
2. Haha. \widehat's cool. Also check out \widetilde. But these sometimes get ugly.
3. Yep. Corrected now.
4. Thanks for the thanks. For what it's worth, I have already let Jameson know that an option to turn off or disable the social media buttons would be a good idea in my opinion.
5. Hey, good catch...that is in fact what I meant to type, but my darn keyboard seems to have a mind of its own at times. Thanks for letting me know so I could fix it.
6. Yes thanks Fantini, edited
7. Not at all...it's GOOD to think about these things.
8. Fantini, have I seen you on SE before?
9. Fantini - Hey, thanks! Yeah, the colors go well together, I think. Bach was the greatest!
10. Hey Fantini . I reported the thread so the Mods in green can decide where to put it. I don't mess with topics I'm not very familiar with haha.
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#### February 6th, 2017

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