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• Today, 10:41
Here is this week's POTW: A charge $-q_1$ is located at the origin of a $y$-axis and a charge $-q_2$ is located at $y=d$. At what point along...
0 replies | 9 view(s)
• Today, 10:40
Congratulations to the following members for their correct submission: kaliprasad My solution is as follows: For the entire system...
1 replies | 50 view(s)
• Today, 09:13
Wilmer replied to a thread Right triangle. in Geometry
(x-6)^2 + (y-5)^2 = 17 x^2 + (y-12)^2 = 68 Expand: x^2 - 12x + y^2 - 10y = -44 x^2 - 00x + y^2 - 24y = -76 Ahhh...ain't that cute! 2...
19 replies | 193 view(s)
• Yesterday, 15:42
Hello and welcome to MHB! :D We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This...
1 replies | 59 view(s)
• Yesterday, 09:35
Wilmer replied to a thread Right triangle. in Geometry
YIKES! I feel guilty for starting this thread :)
19 replies | 193 view(s)
• April 24th, 2018, 21:44
Wilmer replied to a thread Right triangle. in Geometry
Nope Mr.Fly: Sangaku translation: calculation tablet) are Japanese geometrical problems or theorems on wooden tablets which were placed as offerings...
19 replies | 193 view(s)
• April 23rd, 2018, 15:50
Wilmer replied to a thread Right triangle. in Geometry
GOOD idea...never thought of that!!
19 replies | 193 view(s)
• April 23rd, 2018, 13:39
Wilmer replied to a thread Right triangle. in Geometry
I'm also lazy!!
19 replies | 193 view(s)
• April 23rd, 2018, 13:32
Wilmer replied to a thread Right triangle. in Geometry
19 replies | 193 view(s)
• April 23rd, 2018, 13:16
Wilmer replied to a thread Right triangle. in Geometry
Nice. How about "without using trigonometry"? :)
19 replies | 193 view(s)
• April 23rd, 2018, 12:09
Wilmer replied to a thread Right triangle. in Geometry
Agree. And the 2 solutions for C are (2,4) and (38/5,44/5). BUT I was trying to solve without involving a circle. (guess I should have said...
19 replies | 193 view(s)
• April 23rd, 2018, 06:56
Wilmer started a thread Right triangle. in Geometry
Right triangle ABC. BC = a = sqrt(17) AC = b = sqrt(68) AB = c = sqrt(85) A's coordinates: 0,12 B's coordinates: 6,5 What's EASIEST way...
19 replies | 193 view(s)
• April 22nd, 2018, 10:17
MarkFL replied to a thread Olinguito in Introductions
Hello, and welcome to MHB! (Wave)
8 replies | 106 view(s)
• April 21st, 2018, 14:15
There was no inconvenience, except to you trying to post and to those trying to read the thread. It was bad coding on my part, done before I knew the...
19 replies | 522 view(s)
• April 20th, 2018, 22:16
Update - Version 1.3.1: Bug fix for usernames with special characters.
2 replies | 224 view(s)
• April 20th, 2018, 19:16
I have added code to that external script so a single quote in a username won't throw an error, and I edited the post so that only the username is...
19 replies | 522 view(s)
• April 20th, 2018, 12:00
Here is this week's POTW: A small bug is placed between two blocks of masses $m_1$ and $m_2$ ($m_1>m_2$) on a frictionless table. A horizontal...
1 replies | 50 view(s)
• April 20th, 2018, 11:59
No one answered this week's problem, and my solution is as follows: A formula for power $P$ we can apply here is: P=\frac{1}{2}\mu\omega^2A^2v...
1 replies | 75 view(s)
• April 16th, 2018, 17:36
Google both terms and you should be ok...
2 replies | 55 view(s)
• April 15th, 2018, 20:01
Awesome work, as always Chris! (Yes) Using your post, where the base case and induction statement have been given, then the induction step would...
6 replies | 133 view(s)
• April 15th, 2018, 12:02
Here are the choices: A.) \frac{1}{2} B.) \frac{1}{2}-\frac{2^{2006}}{3^{2^{2006}}-1} C.) \frac{1}{2}-\frac{2^{2005}}{3^{2^{2005}}-1} D.)...
6 replies | 133 view(s)
• April 14th, 2018, 10:56
The actual sum he's given is: S=\sum_{k=0}^{2006}\left(\frac{2^k}{3^{2^k}+1}\right) And he's given several choices for the result. I was hoping...
6 replies | 133 view(s)
• April 14th, 2018, 10:10
Hello MHB! (Wave) A young man in high school I know has been essentially tasked with finding the following partial sum: ...
6 replies | 133 view(s)
• April 12th, 2018, 11:47
MarkFL replied to a thread Hello from Pan in Introductions
Hello, and welcome to MHB panpan! (Wave)
6 replies | 94 view(s)
• April 12th, 2018, 11:15
Hello, MHB Community! (Wave) anemone has asked me to stand in for her for a few weeks, so please be gentle. (Bigsmile) Here is this week's...
1 replies | 75 view(s)
• April 12th, 2018, 11:11
Hello MHB Community! (Wave) I am going to stand in for anemone for a few weeks. Congratulations to the following for their correct submissions:...
1 replies | 136 view(s)
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