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vegas14
y varies inversely as twice x. When x = 4, y = 1. Find y when x = 5.
y varies inversely as twice x will be something like \[y = \frac{k}{2x}\] does that help? or do you need more hints?
good day vegas14, this is an example of a proportionality problem. what you need to do is read the problem word for word, each word will be giving you hint as to how to approach the problem. I will list the possible cases in this type of problem, I hope that it will help you. caseI: if the problem states that "y varies directly to x" it means that y is directly proportional to x "but", it will be proportional only if you multiply a proportionality constant "k" \[y \alpha x\] this equation (eq1) symbolizes the direct proportionality of y to x, to be able to replace the proportionality symbol alpha to an equal (=) sign you must multiply x to a proportionality constant which is "k". therefore transforming the equation to, \[y=k(x)\]
caseII: (this is the case of your problem) when the problem states that "y varies inversely as x" it means that your equation would be like this, \[y \alpha \frac{ 1 }{ x }\] from this equation you should again have to replace the proportionality symbol "alpha" to an equal (=) sign. transforming the equation to be,\[y=k(\frac{ 1 }{ x })\] which if simplified will become \[y=\frac{ k }{ x }\]
of course problems may vary in this cases, for example "it may state twice as x therefore you should use "2x" or thrice as x use "3x" or may be as the square of x use \[x^{2}\] it will now depend on your own analysis of the problem.
i really wanted to solve the problem for you, but it might be best if you do it on your own, just to test yourself if you really understand the logic behind =) but just to give you more tips: 1) Apply CaseII 2) Find the proportionality constant "k" ---- just this I will not give you which on the given you will use to solve for "k" =) 3) Solve for y(new) I hope this posts will help you. will check your answers soon. . . =)