<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:
Step-by-step explanation:
roots of a complex number is given by DeMoivre's formula.
![\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}](https://tex.z-dn.net/?f=%5Csf%20%5Cboxed%7B%5Cbf%20r%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5Cleft%5BCos%20%5Cdfrac%7B%5Ctheta%20%2B%202%5Cpi%20k%7D%7Bn%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%5Cright%5D%7D)
Here, k lies between 0 and (n -1) ; n is the exponent.
a = -1 and b = √3
n = 4
For k = 0,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%20%2B0%7D%7B4%7D%2BiSin%20%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%2B0%7D%7B4%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cz%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%20-%5Cpi%20%20%7D%7B12%7D%2BiSin%20%20%5C%20%5Cdfrac%7B-%5Cpi%7D%7B12%7D%5Cright%5D%5C%5C%5C%5C%5C%5Cz%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5B-Cos%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D-i%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D%5Cright%5D)
For k =1,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%5Cright%5D)
For k =2,
![z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 3,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 4,
![\sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%5Cright%5D)
Distributing same key numbers
I hope this helps you
-15+6= -9
-9+6= -3
-3+6=3
add 6
