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S1
Indago dalībnieks
Indago dalībnieks


Joined: 31 Jul 2005
Posts: 219
Location: Jelgava

PostPosted: Sun Aug 13, 2006 12:39 pm    Post subject:

vai varētu palūtgt Cx aprēķinu, ja Cy ir zināms, laikam bez tā neiztikt...
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bubu
Indago Uzvarētājs
Indago Uzvarētājs


Joined: 23 Mar 2004
Posts: 3223
Location: Riga

PostPosted: Sun Aug 13, 2006 1:56 pm    Post subject:

Kamoon, tas taču kādā piektajā klasē jau jāmāk:
cy = k*cx + m
cy-m = k*cx
cx = (cy-m)/k

Ja taisnes ir vertikālas (ax=bx), tad cx = ax = bx un nevajag neko dalīt/reizināt.
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S1
Indago dalībnieks
Indago dalībnieks


Joined: 31 Jul 2005
Posts: 219
Location: Jelgava

PostPosted: Sun Aug 13, 2006 3:11 pm    Post subject:

bubu wrote:
Kamoon, tas taču kādā piektajā klasē jau jāmāk:
cy = k*cx + m
cy-m = k*cx
cx = (cy-m)/k

ko lai saku, ...alus vainīgs, jā tieši tā Laughing
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