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March 1, 2015

March 1, 2015

Posted by **Jake** on Thursday, March 29, 2012 at 8:23pm.

I did this so far,

2k-1=5

6/2

k=3

- Calculus -
**Steve**, Friday, March 30, 2012 at 11:26amwhat's the problem?

(k,2k-1,3)•(k,5,-4) = k*k + (2k-1)(5) + 3(-4)

so, we need

k^2 + 10k - 5 - 12 = 7

k^2 + 10k - 24 = 0

(k-2)(k+12) = 0

so, k=2 or -12

k=2: (2,3,3)•(2,5,-4) = 4+15-12 = 7

k=-12: (-12,-25,3)•(-12,5,-4) = 144-125-12 = 7

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