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samalami

  • 3 years ago

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

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

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

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

  4. Jonask
    • 3 years ago
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    did you get k first

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

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

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

  8. Jonask
    • 3 years ago
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    @samalami the formular had mistake

  9. samalami
    • 3 years ago
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    ohhhhhh!

  10. samalami
    • 3 years ago
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    then k is 24

  11. samalami
    • 3 years ago
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    @Jonask ??

  12. samalami
    • 3 years ago
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    thanks for the clarification @zepdrix :)

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

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

  15. samalami
    • 3 years ago
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    y would equal 24 @Jonask

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

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

  18. samalami
    • 3 years ago
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    @zepdrix much thanks! xoxo

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