anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[y=\frac{k}{x^3}\] put(8,12) then find k
anonymous
  • anonymous
after since you have k put x=1 to get y
anonymous
  • anonymous
@Jonask so y=12?

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anonymous
  • anonymous
did you get k first
zepdrix
  • zepdrix
cube ROOT of x, might wanna fix that real quick @Jonask ! :D
anonymous
  • anonymous
yes\[y=\frac{ k }{\sqrt[3]{x} }\] thanks @zepdrix
anonymous
  • anonymous
Ifound 12=k/8^3 (y=k/x^3 with substitutions) ~12=1644/8^3
anonymous
  • anonymous
@samalami the formular had mistake
anonymous
  • anonymous
ohhhhhh!
anonymous
  • anonymous
then k is 24
anonymous
  • anonymous
anonymous
  • anonymous
thanks for the clarification @zepdrix :)
zepdrix
  • zepdrix
k=24? yah that sounds right :D
anonymous
  • anonymous
so now find y when x=1\[y=\frac{24}{\sqrt[3]{1}}\]
anonymous
  • anonymous
y would equal 24 @Jonask
anonymous
  • anonymous
@zepdrix does this mean y=24 is the answer to the entire question?
zepdrix
  • zepdrix
Hmm yah that sounds right :D Yayyy sammy! \c:/
anonymous
  • anonymous
@zepdrix much thanks! xoxo

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