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samalami Group Title

Pre Calc: If y varies inversely as the cube root of x, and y = 12 when x = 8, find y when x = 1. y = ? xoxo

  • 2 years ago
  • 2 years ago

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  1. Jonask Group Title
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    \[y=\frac{k}{x^3}\] put(8,12) then find k

    • 2 years ago
  2. Jonask Group Title
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    after since you have k put x=1 to get y

    • 2 years ago
  3. samalami Group Title
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    @Jonask so y=12?

    • 2 years ago
  4. Jonask Group Title
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    did you get k first

    • 2 years ago
  5. zepdrix Group Title
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    cube ROOT of x, might wanna fix that real quick @Jonask ! :D

    • 2 years ago
  6. Jonask Group Title
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    yes\[y=\frac{ k }{\sqrt[3]{x} }\] thanks @zepdrix

    • 2 years ago
  7. samalami Group Title
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    Ifound 12=k/8^3 (y=k/x^3 with substitutions) ~12=1644/8^3

    • 2 years ago
  8. Jonask Group Title
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    @samalami the formular had mistake

    • 2 years ago
  9. samalami Group Title
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    ohhhhhh!

    • 2 years ago
  10. samalami Group Title
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    then k is 24

    • 2 years ago
  11. samalami Group Title
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    @Jonask ??

    • 2 years ago
  12. samalami Group Title
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    thanks for the clarification @zepdrix :)

    • 2 years ago
  13. zepdrix Group Title
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    k=24? yah that sounds right :D

    • 2 years ago
  14. Jonask Group Title
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    so now find y when x=1\[y=\frac{24}{\sqrt[3]{1}}\]

    • 2 years ago
  15. samalami Group Title
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    y would equal 24 @Jonask

    • 2 years ago
  16. samalami Group Title
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    @zepdrix does this mean y=24 is the answer to the entire question?

    • 2 years ago
  17. zepdrix Group Title
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    Hmm yah that sounds right :D Yayyy sammy! \c:/

    • 2 years ago
  18. samalami Group Title
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    @zepdrix much thanks! xoxo

    • 2 years ago
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