The shaded region shown in the figure is give | vedclass

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(A) The vertices of the shaded triangular region are $A(0, 14)$,$B(5, 0)$,and $C(19, 14)$.$1$. The line passing through $A(0, 14)$ and $B(5, 0)$ has t

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$14x + 5y \geq 70; y \leq 14$ and $x - y \leq 5$

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(A) The vertices of the shaded triangular region are $A(0, 14)$,$B(5, 0)$,and $C(19, 14)$.$1$. The line passing through $A(0, 14)$ and $B(5, 0)$ has the equation $\frac{x}{5} + \frac{y}{14} = 1$,which simplifies to $14x + 5y = 70$. Since the shaded region lies above this line (e.g.,testing point $(5, 14)$ gives $14(5) + 5(14) = 70 + 70 = 140 > 70$),the inequality is $14x + 5y \geq 70$.$2$. The line passing through $A(0, 14)$ and $C(19, 14)$ is the horizontal line $y = 14$. Since the shaded region lies below this line,the inequality is $y \leq 14$.$3$. The line passing through $B(5, 0)$ and $C(19, 14)$ has a slope $m = \frac{14 - 0}{19 - 5} = \frac{14}{14} = 1$. Using the point-slope form $y - 0 = 1(x - 5)$,we get $y = x - 5$,or $x - y = 5$. Since the shaded region lies to the left of this line (e.g.,testing point $(5, 14)$ gives $5 - 14 = -9 Thus,the correct set of inequations is $14x + 5y \geq 70$,$y \leq 14$,and $x - y \leq 5$.

The shaded region shown in the figure is given by the inequations:

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