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