meta user
network profile
| bio | website | |
|---|---|---|
| location | Israel | |
| age | ||
| visits | member for | 2 years, 1 month |
| seen | Feb 2 at 0:20 | |
| stats | profile views | 9 |
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Sep 20 |
revised |
Transforming triangular coordinate system from angle to another added 17 characters in body |
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Sep 20 |
asked | Transforming triangular coordinate system from angle to another |
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Sep 2 |
comment |
Creating a special vector from two vectors Thats working great, thanks. |
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Sep 2 |
accepted | Creating a special vector from two vectors |
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Sep 2 |
comment |
Creating a special vector from two vectors @J. M. added an image. |
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Sep 2 |
revised |
Creating a special vector from two vectors added 150 characters in body |
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Sep 2 |
comment |
Creating a special vector from two vectors @J. M. its not perpendicular to them. the vector that connects the two endpoints $\vec {BD}$ is perpendicular to $\vec {AB}$ and same for the other vector. |
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Sep 2 |
asked | Creating a special vector from two vectors |
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Aug 31 |
awarded | Editor |
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Aug 31 |
comment |
Calculating the area of a special hexagon @Glen Wheeler: you are right, I forgot a constraint, see the edit. |
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Aug 31 |
comment |
Calculating the area of a special hexagon @Mark Bennet: you are right, I forgot a constraint, see the edit. |
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Aug 31 |
revised |
Calculating the area of a special hexagon added 139 characters in body |
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Aug 31 |
asked | Calculating the area of a special hexagon |
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Jun 13 |
accepted | dot product negative angle |
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Jun 13 |
comment |
dot product negative angle yes, I've been able it to that. didn't know cross product is such an easy operation. |
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Jun 13 |
comment |
dot product negative angle Isn't there an easier way? |
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Jun 13 |
asked | dot product negative angle |
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Apr 14 |
accepted | Finding the first term of a geometric series by the sum and $n$ |
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Apr 14 |
awarded | Student |
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Apr 14 |
comment |
Finding the first term of a geometric series by the sum and $n$ actually by the link Raskolnikov gave its $S = a \cdot \frac{r^{n+1}-1}{r-1}$ because my series terminates in $n$ not $n-1$ |
