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Mathematics 12 Online
OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

after since you have k put x=1 to get y

OpenStudy (anonymous):

@Jonask so y=12?

OpenStudy (anonymous):

did you get k first

zepdrix (zepdrix):

cube ROOT of x, might wanna fix that real quick @Jonask ! :D

OpenStudy (anonymous):

yes\[y=\frac{ k }{\sqrt[3]{x} }\] thanks @zepdrix

OpenStudy (anonymous):

Ifound 12=k/8^3 (y=k/x^3 with substitutions) ~12=1644/8^3

OpenStudy (anonymous):

@samalami the formular had mistake

OpenStudy (anonymous):

ohhhhhh!

OpenStudy (anonymous):

then k is 24

OpenStudy (anonymous):

@Jonask ??

OpenStudy (anonymous):

thanks for the clarification @zepdrix :)

zepdrix (zepdrix):

k=24? yah that sounds right :D

OpenStudy (anonymous):

so now find y when x=1\[y=\frac{24}{\sqrt[3]{1}}\]

OpenStudy (anonymous):

y would equal 24 @Jonask

OpenStudy (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:/

OpenStudy (anonymous):

@zepdrix much thanks! xoxo

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