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y varies inversely as twice x. when x=9, y=2. find y when x=3
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\[(y) \alpha \frac{1}{2x}\] Thus, \[y=k \frac{1}{2x}\] where k is an arbitrary constant. When x=9, y=2 2=k/(2*9) k=36 When x=3, y= (36) /(2*3) y=6
its polite to at least give people a thank you for their help. medals are meant as a means to give thanks as well; but not a dire requirement :)
i was comparing what I had with what you had to make sure I understood the problem and how the answer was derived. Thank you for your help!!!
what is the solution set for Ix-6I+3<10
|x-6| +3 <10 |x-6| < 7 Breaking the abs sign, x-6 <7 or x-6>-7
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thanks
Np!
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