A mildly relevant question
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- Xisiqomelir
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A mildly relevant question
I know everyone ends up with y = mx + b, but which do you normally start with?
- Mitth`raw`nuruodo
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I use y=mx+b because it's just.. easier, I guess. Unless you dont know the y-intercept, in which case you have to go with the 2nd one... I think.
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I am an old fan of y = mx +b.
But for some higher level statistics and regressions I generally turn and use the alternative. You just need 2 point to determine the slope, then take that and one point to generate the y-value.
But for some higher level statistics and regressions I generally turn and use the alternative. You just need 2 point to determine the slope, then take that and one point to generate the y-value.
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I've always quite liked the latter, even though the former was more use, the second one did me well in the Pure Maths exams at AS level.
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y = F + vx
F = Fixed Costs
vx = Variable costs.
This is the equation used in cost accounting to describe cost behavior, and the one I most often use.
F = Fixed Costs
vx = Variable costs.
This is the equation used in cost accounting to describe cost behavior, and the one I most often use.
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y = mx + b was the first I learned. Then progressed to ax + cy = b after that. -shrugs-
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y = ax + b is more convenient, but I more often have to start with y - y1 = m(x - x1)
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Bah, always use parametric symmetry.
(x-a)/A=(y-b)/B=(z-c)/C=...
Works in any euclidean geometry.
R(3) through infinity.
(x-a)/A=(y-b)/B=(z-c)/C=...
Works in any euclidean geometry.
R(3) through infinity.
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I got taught y = mx + c (or in this case y = mx + b)
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