Question 199482
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If *[tex \Large y] varies directly as *[tex \Large x] and inversely as *[tex \Large z^2], then you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \frac{kx}{z^2}]


So if *[tex \Large y = 3] when *[tex \Large x = 2] and *[tex \Large z = 4], then you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2k}{4^2} = 3]


Solve for *[tex \Large k]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2k}{16} = 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2k = 48]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k = 24]


Now you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \frac{24x}{z^2}]


And if you want to know the value of *[tex \Large x] when *[tex \Large y = 9] and *[tex \Large z = 4], then just make the substitutions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9 = \frac{24x}{4^2}]


And solve for *[tex \Large x].


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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