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March 30, 2015

March 30, 2015

Posted by **Shauna** on Tuesday, November 22, 2011 at 2:32pm.

- Algebra II -
**Steve**, Tuesday, November 22, 2011 at 10:52pmy = k * √w/x^3

The new y, y' will be

y' = k * √3w/(x/2)^3

= k * √3 * √w / (x^3/8)

= k * √w/x^3 * [√3 * 8]

= y * 8√3

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