calc16 - 2 = 1 in the above, the first and last terms of...

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Notice that we have y 1 (3,−1) = −cos(3_) = 1, y 2 (3,−1) = −sin(3_) = 0 and y 3 (3,−1) = ln 10. Thus plugging in x 1 = 3 and x
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Unformatted text preview: 2 = 1 in the above, the first and last terms of the sum become 0, and we are left with y 2 3 _x 2 cos(_x 1 ) =...
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This note was uploaded on 02/11/2010 for the course MA 141 taught by Professor Zeck during the Spring '10 term at Quebec Trois-Rivieres.

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