The area of the region {(x, y) in R^2: y geq | vedclass

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(C) The given region is bounded by $y \geq \sqrt{|x+3|}$ and $5y \leq x+9 \leq 15$.From $5y \leq x+9 \leq 15$,we get $y \leq \frac{x+9}{5}$ and $-6 \l

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$\frac{3}{2}$

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(C) The given region is bounded by $y \geq \sqrt{|x+3|}$ and $5y \leq x+9 \leq 15$.From $5y \leq x+9 \leq 15$,we get $y \leq \frac{x+9}{5}$ and $-6 \leq x \leq 6$.The intersection points of $y = \sqrt{x+3}$ and $y = \frac{x+9}{5}$ are found by solving $25(x+3) = (x+9)^2$,which gives $x^2 - 7x - 6 = 0$ (not integer roots) or checking the graph provided.Based on the graph,the region is bounded by the line $y = \frac{x+9}{5}$ above and the curves $y = \sqrt{-x-3}$ (for $x \in [-4, -3]$) and $y = \sqrt{x+3}$ (for $x \in [-3, 1]$) below.The area is given by $\int_{-4}^{1} \frac{x+9}{5} dx - \int_{-4}^{-3} \sqrt{-x-3} dx - \int_{-3}^{1} \sqrt{x+3} dx$.Calculating the integral of the line: $\int_{-4}^{1} \frac{x+9}{5} dx = \frac{1}{5} [\frac{x^2}{2} + 9x]_{-4}^{1} = \frac{1}{5} [(\frac{1}{2} + 9) - (8 - 36)] = \frac{1}{5} [9.5 + 28] = \frac{37.5}{5} = 7.5 = \frac{15}{2}$.Calculating the area under the curves: $\int_{-4}^{-3} \sqrt{-(x+3)} dx = [-\frac{2}{3}(-(x+3))^{3/2}]_{-4}^{-3} = 0 - (-\frac{2}{3}(1)) = \frac{2}{3}$.$\int_{-3}^{1} \sqrt{x+3} dx = [\frac{2}{3}(x+3)^{3/2}]_{-3}^{1} = \frac{2}{3}(4^{3/2} - 0) = \frac{2}{3}(8) = \frac{16}{3}$.Required Area $= \frac{15}{2} - \frac{2}{3} - \frac{16}{3} = \frac{15}{2} - \frac{18}{3} = \frac{15}{2} - 6 = \frac{3}{2}$.

The area of the region $\{(x, y) \in R^2: y \geq \sqrt{|x+3|}, 5y \leq x+9 \leq 15\}$ is equal to

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