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Math Help - h(31)=?

  1. #1
    s3a
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    h(31)=?

    "The period of function h is 15. If h (1) = 0 and h(-6) = 3, calculate the value of:

    a) h(31)
    b) h(9)
    c) h(-14)
    d) d(-51)"

    I just need to see the work for a) at least to understand what this problem is all about so please show me how to do a). Thanks in advance!
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  2. #2
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    Hello, s3a!

    You're expected to know what a periodic function is.


    The period of function h is 15.

    If h (1) = 0 and h(\text{-}6) = 3, calculate the values of:

    . . (a)\;h(31) \qquad(b)\;h(9)\qquad(c)\;h(\text{-}14)\qquad(d)\;h(\text{-}51)

    The function is periodic ... and "repeats itself" every 15 units.


    \text{So if }h(1) = 0\text{, then: }\;h(16) = 0,\;\;\underbrace{h(31) = 0}_{(a)},\;\;h(46) = 0,\;\; \hdots

    \text{Moving "backwards": }\;h(1) = 0,\;\;\underbrace{h(\text{-}14) = 0}_{(c)},\;\;h(\text{-}29)=0,\;\;\hdots


    Get the idea?

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  3. #3
    s3a
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    Quote Originally Posted by Soroban View Post
    Hello, s3a!

    You're expected to know what a periodic function is.



    The function is periodic ... and "repeats itself" every 15 units.


    \text{So if }h(1) = 0\text{, then: }\;h(16) = 0,\;\;\underbrace{h(31) = 0}_{(a)},\;\;h(46) = 0,\;\; \hdots

    \text{Moving "backwards": }\;h(1) = 0,\;\;\underbrace{h(\text{-}14) = 0}_{(c)},\;\;h(\text{-}29)=0,\;\;\hdots


    Get the idea?

    -14+15=1 not 0 for the backward one. I get the one on top though.
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  4. #4
    o_O
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    Yes: {\color{blue}-14 + 15} = 1

    That is why: h(-14) = h \Big({\color{blue}-14 + 15} \Big) = h(1) = 0
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