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Someone gave this to me as a response to an earlier questiuon but i need the steps he took broken down a little but more.

interpolate and extrapolate.

at 10,000,000 revolutions, (interpolate)
[(80-85)/(21939300-6271600)]=
[(x-85)/(10,000,000-6271600)]
-5/15667700 = (x-85)/3728400
x-85 = -1.189836415
x = 83.81016359

at 100,000,000 revolutions, (extrapolate)
[70-75)/(87358800-62070100)]=
[(x-70)/(100,000,000-87358800)
-5/25288700 = (x-70)/12641200
x-70 = -2.499377192
x = 67.50062281

2007-06-20 06:32:35 · 1 answers · asked by Spencer D 3 in Science & Mathematics Mathematics

i am a little rusty with my math can you explain how this part was worked out?

5/15667700 = (x-85)/3728400
x-85 = -1.189836415
x = 83.81016359

2007-06-20 07:10:54 · update #1

1 answers

ok. the answerer assumed that there is a linear relationship between the revoltions and the stress. so they took two points (21939300, 80) and (6271600, 85) and created a ratio on the left of the slope, expressed as run over rise (not sure why that ratio. normally is rise over run, but as a ratio it doesn't matter). then they assumed that the ratio would be the same between one of those points and the desired value. so they expressed the anticipated run as x-85 and the rise as 10,000,000 (desired value) - 6271600.
same thing for the extrapolation.
personally I would have graphed the points and estimated a best fit line, either linear or first power of x.
but there's your answer.

2007-06-20 07:03:27 · answer #1 · answered by Piglet O 6 · 0 0

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