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anonymous
 4 years ago
After t hours, Liza's distance from home, in miles, is given by D(t) = 138 + 40(t3) what is the practical interpretation of the constants 3 and 138?
anonymous
 4 years ago
After t hours, Liza's distance from home, in miles, is given by D(t) = 138 + 40(t3) what is the practical interpretation of the constants 3 and 138?

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sasogeek
 4 years ago
Best ResponseYou've already chosen the best response.1i'm not sure if this is right but here's what i think and i'll let someone confirm or provide the correct answer... from y=mx+c I believe that 138 is the distance she was away from home after 3 seconds.

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0thought she was that far (138) at the beginning.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2no IsTim  sasogeek is right (except for the units of time).

sasogeek
 4 years ago
Best ResponseYou've already chosen the best response.140*3 would change the distance when t=0

IsTim
 4 years ago
Best ResponseYou've already chosen the best response.0Sigh. IsTim is wrong again.

sasogeek
 4 years ago
Best ResponseYou've already chosen the best response.1but IsTim has learnt now :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help guys
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