7x + 9y = - 27_________eq1
3x - y = 3__________eq2
Multiply eq 2 by 9 and then add the resulting eq with eq1.
eq2 x 9 gives us 27x - 9y = 27___________eq3
add eq3 with eq1 we get.
34x = 0.
where x=0/34
x=0.
substitute x into any of the above equations to get y.
substitute x into eq1. we get:
7(0) + 9y =-27
9y = -27
y= -27/9
y = -3.
<span>Constraints (in slope-intercept form)
x≥0,
y≥0,
y≤1/3x+3,
y</span>≤ 5 - x
The vertices are the points of intersection between the constraints, or the outer bounds of the area that agrees with the constraints.
We know that x≥0 and y≥0, so there is one vertex at (0,0)
We find the other vertex on the y-axis, plug in 0 for x in the function:
y <span>≤ 1/3x+3
y </span><span>≤1/3(0)+3
y = 3.
There is another vertex at (0,3)
Find where the 2 inequalities intersect by setting them equal to each other
(1/3x+3) = 5-x Simplify Simplify Simplify
x = 3/2
Plugging in 3/2 into y = 5-x: 10/2 - 3/2 = 7/2
y=7/2
There is another vertex at (3/2, 7/2)
There is a final vertex where the line y=5-x crosses the x axis:
0 = 5 -x , x = 5
The final vertex is at point (5, 0)
Therefore, the vertices are:
(0,0), (0,3), (3/2, 7/2), (5, 0)
We want to maximize C = 6x - 4y.
Of all the vertices, we want the one with the largest x and smallest y. We might have to plug in a few to see which gives the greatest C value, but in this case, it's not necessary.
The point (5,0) has the largest x value of all vertices and lowest y value.
Maximum of the function:
C = 6(5) - 4(0)
C = 30</span>
Answer:
D
Step-by-step explanation:
21% were absent, if you divide 146 by 680 and multiply by 100, then round to the nearest whole number, you get 21
