<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>
8d = -56
Divide both sies by 8:-
d = -56/8
d = -7 (answer)
Answer:
The rate of change will be $25
Step-by-step explanation:
Since she started from $45 we have to count again so on the chart it goes 5, 10, 15, 20, and 25!
<h2 />
The point P has coordinates (x,y) = (-2,6) so x = -2 and y = 6
Replace x and y with those values into the rule given
So,
(x,y) ---> (x-2, y-16)
turns into
(-2,6) ---> (-2-2, 6-16) = (-4,-10)
P = (-2,6)
P ' = (-4,-10)
The answer is -10 because your teacher just wants the y coordinate of point P'
It is B. the differences are added together
