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March 3, 2015

March 3, 2015

Posted by **Unknown** on Friday, November 30, 2012 at 1:21pm.

27. h(x)= |x-3| ; x=0,1,2,3,4,5,6

h(x); |0-3| =3

h(x); |1-3|=2

h(x); |2-3|=1

h(x); |4-3|=1

h(x); |5-3|=2

h(x); |6-3|=3

- Algebra2 -
**Steve**, Friday, November 30, 2012 at 2:31pmcorrect, but you left out h(3) = 0

rather than h(x);|1-3| it is more customary to write

h(1) = |1-3| = 2

since h(x) = |x-3|

and you just replace all the x's with 1's

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