For what value(s) of k will the relation not be a function? #1: A = {(k^2, 16), (4k, 32)} #2: B = {(k^2-5k, 10), (k+7, 4)}

Question

For what value(s) of k will the relation not be a function?
#1: A = {(k^2, 16), (4k, 32)}
#2: B = {(k^2-5k, 10), (k+7, 4)}
#3: R = {(k^3-5k^2+3k, -5), (-k, 4)}
#4: S = {(|k+1|+2, 4), (8, 7)}
Will give 100 points and Brainliest!

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Euphemia 4 years 2021-08-23T07:03:35+00:00 1 Answers 149 views 0

Answers ( )

  1. linhdan
    -1
    2021-08-23T07:05:34+00:00
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    Answer:

    • See below

    Step-by-step explanation:

    The relation is not a function when x -value is same but y- values are different.

    #1

    • k^2 = 4k
    • k = 0, k = 4

    #2

    • k^2 – 5k = k + 7
    • k^2 – 6k = 7
    • k^2 – 6k + 9 = 16
    • k – 3 = 4 ⇒ k = 7
    • k – 3 = -4 ⇒ k = -1

    #3

    • k^3 – 5k^2 + 3k = -k
    • k^3 – 5k^2 + 4k = 0
    • k(k^2 – 5k + 4) = 0
    • k = 0, k = 1, k = 4

    #4

    • |k + 1| + 2 = 8
    • |k + 1| = 6
    • k = 5, k = -7

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