The area of the region bounded by y=||x-3|-4| | vedclass

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(C) To find the area bounded by $y=||x-3|-4|-5$ and the $X$-axis,we first determine the $X$-intercepts by setting $y=0$:$||x-3|-4|-5 = 0$$||x-3|-4| =

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$49$

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(C) To find the area bounded by $y=||x-3|-4|-5$ and the $X$-axis,we first determine the $X$-intercepts by setting $y=0$:$||x-3|-4|-5 = 0$$||x-3|-4| = 5$$|x-3|-4 = 5$ or $|x-3|-4 = -5$$|x-3| = 9$ or $|x-3| = -1$ (impossible)$x-3 = 9$ or $x-3 = -9$$x = 12$ or $x = -6$The function $y=||x-3|-4|-5$ is a $W$-shaped graph. The vertices are at $(-6, 0)$,$(-1, -5)$,$(3, -1)$,$(7, -5)$,and $(12, 0)$.The area consists of two triangles and two trapezoids formed between the graph and the $X$-axis:$1$. Triangle with vertices $(-6, 0)$,$(-1, 0)$,and $(-1, -5)$: Area $= \frac{1}{2} 	imes 5 	imes 5 = 12.5$$2$. Trapezoid with vertices $(-1, 0)$,$(3, 0)$,$(3, -1)$,and $(-1, -5)$: Area $= \frac{1}{2} 	imes (5 + 1) 	imes 4 = 12$$3$. Trapezoid with vertices $(3, 0)$,$(7, 0)$,$(7, -5)$,and $(3, -1)$: Area $= \frac{1}{2} 	imes (1 + 5) 	imes 4 = 12$$4$. Triangle with vertices $(7, 0)$,$(12, 0)$,and $(7, -5)$: Area $= \frac{1}{2} 	imes 5 	imes 5 = 12.5$Total Area $= 12.5 + 12 + 12 + 12.5 = 49$ square units.

The area of the region bounded by $y=||x-3|-4|-5$ and the $X$-axis is

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