Gamom3 Posted January 21, 2009 Share Posted January 21, 2009 Ds is working on Direct and Inverse Variations He pretty much understands what to do, but he came upon a problem that has confused him/us. He is using Chalkdust pre-algebra 7.4 Q 37 which is a Physics question Here's the Q: For a constant temperature, the pressure(P) of a gas varies inversely as the volume(V). If the pressure is 25 lb/in^2 when the volume is 400ft^3, find the pressure when the volume is 150ft^3. Round to the nearest hundredth. He thought that the first mentioned would go first..which is what he has been doing the whole chapter. So he was writing P=k/V but the answer key did V=k/P Can someone tell me why? Thanks Quote Link to comment Share on other sites More sharing options...
kiana Posted January 21, 2009 Share Posted January 21, 2009 I'm not completely sure why the answer key started it that way, but the two equations are equivalent and his method is not wrong. Quote Link to comment Share on other sites More sharing options...
Gamom3 Posted January 21, 2009 Author Share Posted January 21, 2009 Thats what was throwing us off. Thanks!! Quote Link to comment Share on other sites More sharing options...
fractalgal Posted January 22, 2009 Share Posted January 22, 2009 Ds is working on Direct and Inverse VariationsHe pretty much understands what to do, but he came upon a problem that has confused him/us. He is using Chalkdust pre-algebra 7.4 Q 37 which is a Physics question Here's the Q: For a constant temperature, the pressure(P) of a gas varies inversely as the volume(V). If the pressure is 25 lb/in^2 when the volume is 400ft^3, find the pressure when the volume is 150ft^3. Round to the nearest hundredth. He thought that the first mentioned would go first..which is what he has been doing the whole chapter. So he was writing P=k/V but the answer key did V=k/P Can someone tell me why? Thanks When written V=k/P, it is implied that the volume (V) is the dependent variable and the pressure (P) is the independent variable if this is thought of in terms of a function. Quote Link to comment Share on other sites More sharing options...
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