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

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

samalami
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@Jonask so y=12?

Jonask
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did you get k first

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

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

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

Jonask
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@samalami the formular had mistake

samalami
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ohhhhhh!

samalami
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then k is 24

samalami
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@Jonask ??

samalami
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thanks for the clarification @zepdrix :)

zepdrix
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k=24? yah that sounds right :D

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

samalami
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y would equal 24 @Jonask

samalami
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@zepdrix does this mean y=24 is the answer to the entire question?

zepdrix
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Hmm yah that sounds right :D
Yayyy sammy! \c:/

samalami
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@zepdrix much thanks! xoxo