SOLUTION: After t hours on a particular day, a freight train is s(t)=−17.98t+16.732t^2-0.78t^3 miles due east of its starting point (for 0≤t≤18).
a) Where is the train (bo
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-> SOLUTION: After t hours on a particular day, a freight train is s(t)=−17.98t+16.732t^2-0.78t^3 miles due east of its starting point (for 0≤t≤18).
a) Where is the train (bo
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Question 970764: After t hours on a particular day, a freight train is s(t)=−17.98t+16.732t^2-0.78t^3 miles due east of its starting point (for 0≤t≤18).
a) Where is the train (both distance and direction from starting point) after 9 hours? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! After t hours on a particular day, a freight train is
s(t)= -17.98t + 16.732t^2 - 0.78t^3 miles due east of its starting point (for 0≤t≤18).
a) Where is the train (both distance and direction from starting point) after 9 hours?
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s(t)= -17.98t + 16.732t^2 - 0.78t^3
Sub 9 for t
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s(9)= -17.98*9 + 16.732*9^2 - 0.78*9^3
Just arithmetic.
