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24 October, 01:44

Can somebody solve this out? Identify Two additional vValues for x and y in a direct variation relationship where y=11 when x=8.

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Answers (2)
  1. 24 October, 02:20
    0
    You can multiply both values, x and y by the same value and it still holds true.

    y=22 when x=16

    y=44 when x=32

    y=55 when x=40
  2. 24 October, 04:42
    0
    Direct variation: y=kx

    Ask ourselves: what is k in this particular relationship? Substitute:

    y=kx

    11=8k

    k=8/11

    Two additional values for x and y; we can just throw in random numbers for x and see what y becomes. I'm going to use 11 and 22.

    y = (8/11) x

    y = (8/11) 11

    y = 8

    (11,8)

    y = (8/11) x

    y = (8/11) 22

    y = (8/11) 2*11

    y = 8*2

    y = 16

    (22,16)
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