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Can somedbody help me with problem 3F8, (b)? I can't understand the answer given...
 one year ago
 one year ago
Can somedbody help me with problem 3F8, (b)? I can't understand the answer given...
 one year ago
 one year ago

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NoelGrecoBest ResponseYou've already chosen the best response.0
There are several somebodies out here who would be glad to help if you included a link to the problem itself.
 one year ago

tribsantosBest ResponseYou've already chosen the best response.0
Sure, sorry about that: the problem is in this pdf file
 one year ago

StaceyBest ResponseYou've already chosen the best response.1
dw:1359705070117:dw Basically we end up with OB=y and AQ=x
 one year ago

StaceyBest ResponseYou've already chosen the best response.1
The point O is (0, 2y) and the point Q is (2x, 0), giving the slope of the line to be y/x.
 one year ago

tribsantosBest ResponseYou've already chosen the best response.0
Thank you very much for your answer. I still don't know if I got it, though. What I understood from the problem is that I should find a curve such that for whatever point P in the curve that I chose, if I took the tangent at that point, P would be at exactly the midpoint of the segment of that line that lies in the first quadrant. But looking at your curve, and the one from the solution given, it seems to me that if I move up (down), the upper part of the line would be shorter (longer) than the lower part. Did I get the question wrong? Thank you!
 one year ago

StaceyBest ResponseYou've already chosen the best response.1
dw:1359951384469:dw You seem to understand the problem correctly. The upper portion of my curve is definitely off, but I did manage to show a tangent for the lower section that should illustrate the concept.
 one year ago

StaceyBest ResponseYou've already chosen the best response.1
dw:1359952181692:dw This graph might illustrate it better.
 one year ago

tribsantosBest ResponseYou've already chosen the best response.0
Thank you very much
 one year ago
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