Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 10 - Section 10.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 625: 49

Answer

15,000 discs, 3.75

Work Step by Step

Demand: $p=-.01x^2 - .2x+9$ Supply: $p=.01x^2-.01x+3$ $-.01x^2-.2x+9=.01x^2-.1x+3$ $-.01x^2-.2x+9+.01x^2+.2x-9=.01x^2-.1x+3+.01x^2+.2x-9$ $0 = .02x^2+.1x-6$ $0*50 =(.02x^2+.1x-6)*100$ $0 = x^2+5x-300$ $(x-15)(x+20)=0$ $x-15=0$ $x-15+15=0+15$ $x=15$ $x+20=0$ $x+20-20=0-20$ $x=-20$ We can’t have a negative quantity, so $x=15$. $p=.01x^2-.01x+3$ $x=15$ $p=.01x^2-.1x+3$ $p=.01*15^2-.1*15+3$ $p=.01*225-1.5+3$ $p=2.25+1.5$ $p=3.75$
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