SOLUTION: R(p)=-2p^2+2000p for what range of prices will revenue exceed $400,000

Question 1187836: R(p)=-2p^2+2000p
for what range of prices will revenue exceed $400,000
Answer by ikleyn(53765) (Show Source): You can put this solution on YOUR website!
.


To answer this question, solve the quadratic equation


    -2p^2 + 2000p = 400000.


Your answer will be the set of positive real ( ? integer ? ) numbers in between the roots.

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