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anonymous

  • one year ago

which sets of ordered pairs represent functions from a to b , A={u,v,w} and B={-2,-1,0,1,2}

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  1. anonymous
    • one year ago
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    @Nnesha

  2. anonymous
    • one year ago
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    @Loser66 @Luigi0210

  3. anonymous
    • one year ago
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    @mathmate

  4. anonymous
    • one year ago
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    can you please help @mathmate

  5. mathmate
    • one year ago
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    There are many possible answers. If you have answer choices, choose the choices where there is no repetition in the set A, i.e. u, v, w appear at most once in the set of ordered pairs. This is because a function can only have one single value of the image given a value of the preimage.

  6. anonymous
    • one year ago
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    the choices are (a) {(v,-1) , (u,2) , (w,0) , (u,-2)} (b) (u,-2) , (v,2) , (w,1) (c) (u,2) , (v,0) , (w,2)

  7. anonymous
    • one year ago
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    @mathmate

  8. mathmate
    • one year ago
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    Please try to understand my previous post, and reread if necessary. You will have no problem making your choices. :)

  9. anonymous
    • one year ago
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    thank you i got it

  10. anonymous
    • one year ago
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    i had missed out one of the answer choices and it happened to be the right one

  11. mathmate
    • one year ago
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    If we examine each of the answers choices, (a) {(v,-1) , (u,2) , (w,0) , (u,-2)} (b) (u,-2) , (v,2) , (w,1) (c) (u,2) , (v,0) , (w,2) For the relation {(v,-1) , (u,2) , (w,0) , (u,-2)} Ask yourself, what would be the value of f(v)? It is clear that f(v)=-1 because the ordered pair (v,-1) tells us that. However, what is the value of f(u)? There are two ordered pairs, (u,2), and (u,-2). We do not know which one is true. A relation that contains ambiguities like this is NOT considered a function. So this relation is not a function. For both of the following relations, (u,-2) , (v,2) , (w,1) (u,2) , (v,0) , (w,2) the ambiguity does not exist, given any value of u, v or w, we can find a UNIQUE value of the image, so both are functions.

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