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**Unformatted text preview: **E/∂ C = 2(C − 2D − 4) + 2(C − D − 2) + 2(C + 1) + 2(C + D) + 2(C + 2D) = 10C − 10
and
∂ E/∂ D = 2(C − 2D − 4)(−2)+2(C − D − 2)(−1)+0+2(C + D)(1)+2(C +2D)(2) = 20D +20
so we need to solve 5C = 5, 10D = −10 so C = 1 and D = −1. The line of best ﬁt is 1 − t. 26) We are trying solve C + D = 0, C + E = 1, C − D = 3 and C − E = 4, that is to 1 1
1
0
0
1
0
11
1
1
400
1 1...

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