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• Today, 11:02
To follow up, we can use the derivative to find the time rate of change of the angular displacement, which is typically denoted by $\omega$. ...
2 replies | 45 view(s)
• Today, 09:23
Ackbach replied to a thread vector algebra in Geometry
Well, here's another hint: $\mathbf{a}-\mathbf{b}=\mathbf{a}+(-\mathbf{b})$. So, comparing $\mathbf{a}+\mathbf{b}$ with $\mathbf{a}+(-\mathbf{b})$...
3 replies | 46 view(s)
• Today, 09:19
Ackbach replied to a thread AlphaZero Crushes Chess in Chat Room
Very interesting! Thanks for posting!
7 replies | 347 view(s)
• Yesterday, 13:06
Ackbach replied to a thread vector algebra in Geometry
(b) If you write out the formula for the dot product, what do you get? Now compare that with the RHS. What can you conclude? (c) Again, I would...
3 replies | 46 view(s)
• Yesterday, 06:59
Hello and welcome to MHB! (Wave) We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions....
2 replies | 45 view(s)
• Yesterday, 06:16
Yes. Sorry. I meant sine. Fixed in my previous post.
4 replies | 77 view(s)
• Yesterday, 02:52
Thanks GJA ... Appreciate your help ... Peter
3 replies | 48 view(s)
• Yesterday, 01:55
MarkFL replied to a thread Domain...2 in Pre-Calculus
No factor in the denominator can be zero, and nothing under a square root radical can be negative...so this leads to: t-1\ne0 0<t What do...
1 replies | 28 view(s)
• Yesterday, 01:47
I would write: y=\frac{2x}{x-1}=\frac{2x-2+2}{x-1}=\frac{2(x-1)+2}{x-1}=2+\frac{2}{x-1} We see this will have a horizontal asymptote at $y=2$,...
1 replies | 30 view(s)
• Yesterday, 01:31
Let's look at a definition: |u|=\begin{cases}u, & 0\le u \\ -u, & u<0 \\ \end{cases} Can you see that we must have: 0\le|u| ?
2 replies | 48 view(s)
• March 19th, 2018, 22:43
I am reading "The Basics of Abstract Algebra" by Paul E. Bland ... ... I am currently focused on Chapter 3: Sets with Two Binary Operations:...
3 replies | 48 view(s)
• March 19th, 2018, 22:30
Yes, and as Wilmer was pointing out, you would likely want to use the binomial theorem. If you raise both sides to the 5th power, you get: ...
10 replies | 107 view(s)
• March 19th, 2018, 21:56
You're being asked to find an area, and in essence, you're doing so by adding up a bunch of vertical lines, the length of which are determined by the...
5 replies | 54 view(s)
• March 19th, 2018, 21:48
To follow up, we get: \d{x}{t}=\frac{\d{y}{t}\left(x-12y^2\right)}{9x^2-y} Plugging in the given values, we find: ...
4 replies | 52 view(s)
• March 19th, 2018, 21:07
That's already included in the "top curve minus the bottom curve." :)
5 replies | 54 view(s)
• March 19th, 2018, 17:00
Hi karthikS, welcome to MHB! Can it be that you mean the angle of the line with the X-Z plane? Respectively the angle with the Y-Z plane?...
4 replies | 77 view(s)
• March 19th, 2018, 16:40
Hi Alexis87, It looks all correct to me. For clarity I would just add a final sentence saying that therefore f(x+c) is reducible, which is a...
1 replies | 42 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?
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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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