Question 189494
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For your first question, neither of your answers are correct.


If <i>y</i> varies inversely as <i>x</i>, then you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y = \frac{k}{x}]


where <i>k</i> is called the constant of proportionality.  The first task is to determine the constant of proportionality based on the given values of <i>x</i> and <i>y</i>.  To do this, substitute the given values and then solve for <i>k</i>.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  8 = \frac{k}{2}]


Multiply both sides by 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k = 16]


Now you can create the specific relationship for your problem by substituting this newly discovered value for <i>k</i>:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \frac{16}{x}]


And finally, if you need to know the value of <i>x</i> when <i>y</i> = 17, you need to substitute this value in place of <i>y</i> and solve for <i>x</i>:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17 = \frac{16}{x}]


Multiply both sides of the equation by *[tex \Large \frac{x}{17}]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{16}{17}]


And that's your answer.


The other two problems are done the same way, except that a direct variation looks like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y = kx]


And in your third problem, you have to set up two equations and solve them both.


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