Radius = half of diameter
16 mm/2 = 8 mm
Solution: 8 mm
You must graph the two given lines and find their point of intersection. It is (10,5). Shade the areas BELOW each of the 2 lines. One line intersects the y-axis at (0,15) and the other line intersects the x-axis at (12.5,0)
Thus, you have an area defined by the vertices given above.
What to do next? Steal the coordinates of each point and subst. them into the given objective function P = 15x + 20y. For exampel, for (12.5,0), the value of P is 15(12.5) + 20(0) = 187.5.
Find P for each of the remaining 3 vertices.
The largest value of P is the answer to this question.
Answer:
The point ( 0,-1) is a solution.
Step-by-step explanation:
y<= (x+1) (x+5)
y>= (x-1)^2 -6
See attached graph.
To prove algebraically, put the point (0,-1) into both equations and see if it is true for both of them.
y<= (x+1) (x+5)
-1 <= (0+1) (0+5)
-1<= 1*5
-1 <=5
True
y>= (x-1)^2 -6
-1 >= (-1)^2 -6
-1 >= 1-6
-1 >=-5
True
The point is a solution
By de Moivre's theorem,

![\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B4%5D%7B%281%20-%20i%29%5E2%7D%20%3D%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi%282%5Cpi%20k-%5Cpi%2F2%29%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi%284k-1%29%5Cpi%2F8%7D)
where
. The fourth roots of
are then
![k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}](https://tex.z-dn.net/?f=k%20%3D%200%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7B-i%5Cpi%2F8%7D)
![k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}](https://tex.z-dn.net/?f=k%20%3D%201%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi3%5Cpi%2F8%7D)
![k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}](https://tex.z-dn.net/?f=k%20%3D%202%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi7%5Cpi%2F8%7D)
![k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}](https://tex.z-dn.net/?f=k%20%3D%203%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi11%5Cpi%2F8%7D)
or more simply
![\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cpm%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7B-i%5Cpi%2F8%7D%20%5Ctext%7B%20and%20%7D%20%5Cpm%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi3%5Cpi%2F8%7D%7D)
We can go on to put these in rectangular form. Recall


Then




and the roots are equivalently
![\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cpm%5Csqrt%5B4%5D%7B2%7D%5Cleft%28%5Csqrt%7B%5Cdfrac12%20%2B%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%20-%20i%5Csqrt%7B%5Cdfrac12%20-%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%5Cright%29%20%5Ctext%7B%20and%20%7D%20%5Cpm%5Csqrt%5B4%5D%7B2%7D%5Cleft%28%5Csqrt%7B%5Cdfrac12%20%2B%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%20%2B%20i%20%5Csqrt%7B%5Cdfrac12%20-%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%5Cright%29%7D)
Answer:
30 degrees
Step-by-step explanation:
A right angle is 90 degrees. Subtract 30 and you get 60. Then divide it by two because the leftover is 2x.