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sineoko [7]
4 years ago
13

Help plzz and how?? please don't take to long

Mathematics
1 answer:
Sergio039 [100]4 years ago
8 0
The edison marble question is 225 blue


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For what value(s) of k will the relation not be a function?
kolbaska11 [484]

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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If C = 6π = πd, then the diameter, d, must be 6.

7 0
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