A farmer F_1 has a land in the shape of a tri | vedclass

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(C) The vertices of the triangle are $P(0,0)$,$Q(1,1)$,and $R(2,0)$.The area of $	riangle PQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} =

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$4$

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(C) The vertices of the triangle are $P(0,0)$,$Q(1,1)$,and $R(2,0)$.The area of $	riangle PQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes PR 	imes 	ext{ordinate of } Q = \frac{1}{2} 	imes 2 	imes 1 = 1 	ext{ unit}^2$.The side $PQ$ lies on the line $y = x$ for $x \in [0, 1]$.The area taken away by farmer $F_2$ is the area between the line $y = x$ and the curve $y = x^n$ from $x = 0$ to $x = 1$.Area $= \int_0^1 (x - x^n) dx = \left[ \frac{x^2}{2} - \frac{x^{n+1}}{n+1} ight]_0^1 = \frac{1}{2} - \frac{1}{n+1}$.Given that this area is $30\%$ of the area of $	riangle PQR$,we have:$\frac{1}{2} - \frac{1}{n+1} = \frac{30}{100} 	imes 1 = \frac{3}{10}$.$\frac{1}{n+1} = \frac{1}{2} - \frac{3}{10} = \frac{5-3}{10} = \frac{2}{10} = \frac{1}{5}$.Therefore,$n + 1 = 5$,which gives $n = 4$.

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