rad 89 = 9.43398
bcuz: 9.43398^2 = 88.9 which rounds to 89
The answer is D I think, cause sum means plus and product means times
Answer: complex equations has n number of solutions, been n the equation degree. In this case:
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi11%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi101%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi191%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi281%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi78%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi168%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi258%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi348%2C75%C2%B0%7D)
Step-by-step explanation:
I start with a variable substitution:

Then:

Solving the quadratic equation:


Replacing for the original variable:
![Z=\sqrt[4]{0,5+0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5%2B0%2C5i%7D)
or ![Z=\sqrt[4]{0,5-0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5-0%2C5i%7D)
Remembering that complex numbers can be written as:

Using this:

Solving for the modulus and the angle:
![Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.](https://tex.z-dn.net/?f=Z%3D%5Cleft%20%5C%7B%20%7B%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi45%7D%7D%20%7D%20%5Catop%20%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi-45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi-45%7D%7D%20%7D%7D%20%5Cright.)
The possible angle respond to:

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"
In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º
Obtaining 8 different angles, therefore 8 different solutions.
Answer:
x<7
Step-by-step explanation:
65-8x>9
-8x>9-65
-8x>-56
x<7
Solution:
<u>A few changes were made:</u>
<u>New equation:</u>
- 4 + 0.3 + 0.09 = 4.00 + 0.30 + 0.09
<u>Solving the equation:</u>
- 4.00 + 0.30 + 0.09
- => 4.39 (Refer to image for work)
Correct option is B.