<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>
A polygon<span> has as many interior </span>angles<span> as sides. An equilateral triangle has three equal 60 </span>degree angles<span>. The </span>sum<span> of the </span>angles<span> of this and any triangle is </span>180 degrees<span>.The </span>sum<span> of the four interior </span>angles<span> of a square is 360 </span>degrees<span>, which is the same for any quadrilateral.</span>
Answer:
55860
Step-by-step explanation:
This way super dupper easy
You cannot solve for x but you can factor
to solve in ax^2+bx+c form you must find
b=x+y
a times c=x times y
so solve
3 times 3=9
what 2 number multily to get 9 and add to get 10
9=x times y
the numbers are 9 and 1 so
3x^2+1x+9x+3
(3x^2+1x)+(9x+3)
factor
(x)(3x+1)+(3)(3x+1)
reverse distribute
ab+ac=a(b+c)
(x)(3x+1)+(3)(3x+1)=(x+3)(3x+1)
factored out form is (x+3)(3x+1)
