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 $\displaystyle p = -0.02x + 400$ note that $\displaystyle 0.02 = \frac{2}{100} = \frac{1}{50}$ ... $\displaystyle p = -\frac{1}{50} x + 400$ $\displaystyle p - 400 = -\frac{1}{50} x$ $\displaystyle -50p + 20000 = x$
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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