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geometry · This list is for general discussions of geometry and related branches of mathematics, including proofs, constructions, trigonom
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Messages 453 - 482 of 1084   Oldest  |  < Older  |  Newer >  |  Newest
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453
Hey i was just wondering if someone can teach me a little geometry.Please.Thanks....
vektor3000
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Aug 8, 2005
7:25 am
454
This is my firt time to join this group. I need any information about spatial reasoning, geometry reaoning, how to know and how to improve this ability. For...
teresia_sri
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Aug 18, 2005
7:01 am
455
Are you studiyng Geometry right now?Is it the basic parts or the more harder ones?Im interseted.But i only now the basic stuff. ... ...
israel Campos
vektor3000
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Aug 19, 2005
1:41 am
456
Can anybody in this group help me to find about measurement of spatial ability and how to measure it? Please, I need your. help. This is my address:...
teresia_sri
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Aug 24, 2005
7:07 am
457
Hi I have a line from point P1(x1,y2) to P2(x2,y2). I want to draw perpendiculars to P1 and P2. How can I compute every point belonging to any of the...
Adi
ady_tiganus
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Sep 2, 2005
11:39 pm
458
P1 and P2 lie along the line y=y2, since the abscissae are different but ordinates are the same. The lines perpendicular to the two points are x=x1 and x=x2,...
Carl Adkins
carlito1942
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Sep 3, 2005
3:58 am
459
... different but ordinates are the same. The lines perpendicular to the two points are x=x1 and x=x2, respectively. ... points belonging to a line...
Adi
ady_tiganus
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Sep 6, 2005
1:38 am
460
First, determine the slope of line P1P2, which is (difference of ordinates)/(difference of abscissae)=(y2-y1)/(x2-x1). A line perpendicular to this line will...
Carl Adkins
carlito1942
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Sep 6, 2005
4:18 am
461
Hi I have 2 lines:AB and CD. How can i calculate the cobb angle? Thanks...
Adi
ady_tiganus
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Sep 13, 2005
1:14 am
462
You could go to a medical site and ask!!! Carl Adkins ... From: Adi To: geometry@yahoogroups.com Sent: Monday, September 12, 2005 8:14 PM Subject: [geometry]...
Carl Adkins
carlito1942
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Sep 13, 2005
3:45 am
463
Enter http://www.fpnotebook.com/ORT339.htm Luiz - PY1LL ... From: "Adi" <ady_tiganus@...> To: <geometry@yahoogroups.com> Sent: Monday, September 12, 2005...
Luiz Amaral
py1ll
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Sep 13, 2005
12:06 pm
464
Hi, I'm trying to help a young friend in 4th grade with a geometry problem. The problem asks him to draw a quadrangle that has 2 pairs of equal sides but is...
Tim Blackwood
tblackwood1
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Sep 13, 2005
8:54 pm
465
Draw a figure (like a lozenge) with 2 equal sides forming an angle and the other equal sides forming the opposite angle. Luiz Amaral ... From: "Tim Blackwood"...
Luiz Amaral
py1ll
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Sep 14, 2005
12:33 am
466
Hello Luiz, Thank you. :) However, I'm not entirely certain what this means. I've got a picture of a lozenge in my mind and, using your description, it could...
Timothy Blackwood
tblackwood1
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Sep 14, 2005
2:29 am
467
A quadrangle (same as quadrilateral) is a four sided figure, as you probably already know. Draw a straight line for the base. Using a compass with a given...
Carl Adkins
carlito1942
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Sep 14, 2005
3:36 am
468
Hi Tim ... As an alternative to the kite, if you start with a rectangle ABCD then ABDC is a quadrangle looking like a bow-tie and with alternate sides having...
Adrian Rossiter
adrianrossiter
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Sep 14, 2005
6:58 am
469
If the opposite sides are equal, the figure is a parallelogram, but see the figure below (it is the best I can draw with text characters). Is it a solution for...
Luiz Amaral
py1ll
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Sep 14, 2005
11:51 am
470
Correction: both pairs of opposite sides must be equal to yield a parallelogram. Carl Adkins ... From: Luiz Amaral To: geometry@yahoogroups.com Sent:...
Carl Adkins
carlito1942
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Sep 14, 2005
6:27 pm
471
... Draw one triangle ABC with sides of length AB = 9 cm, AC = 12 cm and BC = 15 cm. Now take the mirror image over the BC side, creating point A' such that...
Alberto Monteiro
albmont
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Sep 14, 2005
9:08 pm
472
Hie Everybody, I just read this amazing book called ‘Vedic Mathematics Made Easy’ by Dhaval Bathia (Jaico Publishing House). It has some mind-blowing...
Shahrukh Sengupta
shahrukh_sen...
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Sep 20, 2005
9:41 am
473
Hi I have a line determined by P1(x1,y1) and P2(x2,y2). I also have a point P3(x3,y3). I need to find the formula of the distance from P3 to the line, and also...
Adi
ady_tiganus
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Oct 2, 2005
8:07 pm
474
Hi Adi ... You could work out the equations of the two lines and find their intersection point as simultaneous equations. There is also a more general solution...
Adrian Rossiter
adrianrossiter
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Oct 3, 2005
7:39 am
475
Hi I have a Point P1. From P1 I draw a line to a point P2 and a line to a point P3. P2P1 and P3P1 forms a semi-closed figure. I need to know if a given point G...
Adi
ady_tiganus
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Oct 10, 2005
8:45 pm
476
It would take a lot of time and space to explain how to solve this problem. If you have a text on Analytic Geometry, please consult the section on "normal...
Carl Adkins
carlito1942
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Oct 20, 2005
4:35 am
477
Hi, I'm new to the list. I was wondering if someone could tell me what the areas of the three overlapping circles (below) are called. Do they have names? ...
efjhome@...
efj_geo
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Nov 9, 2005
12:11 pm
478
Hi Friends, We are HR Consultants based in Pune. We at the moment have some interesting opportunities going on in some of our client companies wherein they are...
Cinque
cinque_2004
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Nov 29, 2005
12:41 pm
479
Is there a geometric proof for the sum of two cosines? (i.e. prove the identity cos(A) + cos(B) = 2cos1/2(A+B)cos1/2(A-B) geometrically)...
pollom01
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Dec 9, 2005
5:31 am
480
Interesting question!!! Geometric proofs exist for cos(A+B)=cos(A)*cos(B)-sin(A)*sin(B) and cos(A-B)=cos(A)*cos(B)+sin(A)*sin(B). Adding these two equations...
Carl Adkins
carlito1942
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Dec 9, 2005
6:27 am
481
Thanks Carl. Yeah, I knew that the substitutions lead to the sum-to-product identities, but was wondering about a geometric proof. I know you can use the idea...
Mike Pollock
pollom01
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Dec 10, 2005
1:21 am
482
Damn, it has sure been a while... I just realized today that it's been 12 months since I last went out with a special lady, but I am pretty happy to be...
diva.bensley6707@...
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Dec 10, 2005
10:54 am
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