3 Matching Annotations
- Jun 2024
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math.stackexchange.com math.stackexchange.com
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g(L(t))=(D−C)×(L(t)−C)=0(D−C)×((B−A)⋅t+A−C)=0t=(C−A)×(D−C)(D−C)×(B−A)
This is wrong. The correct solution is a negative fraction that is $$ t = −\frac{(C−A)×(D−C)}{(D−C)×(B−A)} = \frac{(C−A)×(D−C)}{(B−A)×(D−C)} $$
because $$ g(L(t)) = (D−C)×(L(t)−C) \ = (D−C)×((B−A)⋅t + A − C)\ = ((D−C)×(B−A))⋅t + (D−C)×(A−C) = 0 \ \implies ((D−C)×(B−A))⋅t = −(D−C)×(A−C) \ \implies t = −\frac{(C−A)×(D−C)}{(D−C)×(B−A)} $$
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- Dec 2023
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docs.omnivore.app docs.omnivore.app
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saved:2022-04-21.._
Use saved:2022-04-21..* instead
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published:_..2020-01-01
Use published:*..2020-01-01 instead
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