The set of values that beta can assume so tha | vedclass

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(A) The triangle is formed by the intersection of the lines $L_1: 3x+y+2=0$,$L_2: 2x-3y+5=0$,and $L_3: x+4y-14=0$.To find the range of $\beta$ for the

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$\left[\frac{5}{3}, \frac{7}{2}\right]$

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(A) The triangle is formed by the intersection of the lines $L_1: 3x+y+2=0$,$L_2: 2x-3y+5=0$,and $L_3: x+4y-14=0$.To find the range of $\beta$ for the point $(0, \beta)$ to lie inside the triangle,we find the $y$-intercepts of these lines on the $y$-axis (where $x=0$):For $L_1: 3(0)+y+2=0 \implies y = -2$.For $L_2: 2(0)-3y+5=0 \implies y = 5/3$.For $L_3: 0+4y-14=0 \implies y = 14/4 = 7/2$.By observing the region bounded by these lines,the point $(0, \beta)$ lies within the triangle when $\beta$ is between the two $y$-intercepts that bound the vertical segment of the triangle on the $y$-axis.From the graph,the $y$-intercepts are $-2$,$5/3$,and $7/2$. The segment on the $y$-axis inside the triangle lies between $y = 5/3$ and $y = 7/2$.Thus,the set of values for $\beta$ is $\left[\frac{5}{3}, \frac{7}{2}\right]$.

The set of values that $\beta$ can assume so that the point $(0, \beta)$ lies on or inside the triangle formed by the lines $3x+y+2=0$,$2x-3y+5=0$,and $x+4y-14=0$ is

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