Q:Solving the given demand equation for x in terms of p, p= -0.02x+400 (o< or = x < or =20000) text book answer: x=f(p) = -50p + 20000 I got x = -p/0.02 + 20000 I don't understand how is it 50p. Thanks
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Originally Posted by hovermet Q:Solving the given demand equation for x in terms of p, p= -0.02x+400 (o< or = x < or =20000) text book answer: x=f(p) = -50p + 20000 I got x = -p/0.02 + 20000 I don't understand how is it 50p. Thanks note that ...
Originally Posted by hovermet Q:Solving the given demand equation for x in terms of p, p= -0.02x+400 (o< or = x < or =20000) text book answer: x=f(p) = -50p + 20000 I got x = -p/0.02 + 20000 I don't understand how is it 50p. Thanks 1/(0.02) = 50
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