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Ann [662]
3 years ago
6

Which graph best represents the feasibility region for the system shown above? ( pics are in this)

Mathematics
2 answers:
zalisa [80]3 years ago
5 0

Consider the system of inequalities

\left\{ \begin{array}{l}  y\ge 2 \\  x\le 6 \\  y\le 3x+2 \\  y\le -x+10. \end{array}\right.

1. Plot all lines that are determined by equalities (see attached diagram)

\left\{ \begin{array}{l}  y=2 \text{ (red line)} \\  x= 6  \text{ (blue line)}\\  y=3x+2 \text{ (green line)} \\  y= -x+10  \text{ (orange line)}. \end{array}\right.

2. Determine which bounded part of the plane you should select:

  1. y\ge 2 means that you should take points with y-coordinates greater than or equal to 2 (top part of the coordinate plane that was formed by the red line);
  2. x\le 6 means that you should take points with x-coordinates less than or equal to 6 (left part of the coordinate plane that was formed by the blue line);
  3. for y\le 3x+2 you can check where the origin is placed. Since 0\le 3\cdot 0+2, the origin belongs to the needed part and you have to take the right part of the coordinate plane that was formed by green line.
  4. for y\le -x+10 you can check where the origin is placed. Since 0\le -1\cdot 0+10, the origin belongs to the needed part and you have to take the bottom part of the coordinate plane that was formed by orange line.

3. According to the previous explanations, the shaded region is as in A diagram.

Answer: correct choice is A.


IRINA_888 [86]3 years ago
3 0
The first graph is the correct answer.
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given abc with a(-4 -2), b(4,4), and c(18,-8). write the equation of the line that contains the altitude that passes through b i
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check the picture below.

so red line of BD is perpendicular to AC, hmmmm let's firstly find the slope of AC, bearing in mind that perpendicular lines have <u>negative reciprocal</u> slopes.


\bf A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{18}~,~\stackrel{y_2}{-8}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-8-(-2)}{18-(-4)}\implies \cfrac{-8+2}{18+4} \\\\\\ \cfrac{-6}{22}\implies -\cfrac{3}{11} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{11}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{11}{3}}\qquad \stackrel{negative~reciprocal}{\cfrac{11}{3}}}\impliedby \textit{BD's slope}


so we're really looking  for the equation of a line whose slope is 11/3 and runs through B(4,4).  Keeping in mind that

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient


\bf B(\stackrel{x_1}{4}~,~\stackrel{y_1}{4})~\hspace{10em} slope = m\implies \cfrac{11}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=\cfrac{11}{3}(x-4)


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