The answer is c, or 3b+6!!
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:
Five more than a number
Step-by-step explanation:
Answer:
<em>20 sq yd</em>
Step-by-step explanation:
To find the total amount of sq yd's of paint needed, you have to find the area.
The area of a triangle is: A = 1/2 bh
We know:
b = base = 8
h = height = 5
Plug in and solve:
A = 1/2 (8) (5)
A = 1/2 (40)
A = 20
Answer:
A. simpson's paradox
Step-by-step explanation:
The Simpson's paradox was named after Edward Simpson, the person who described this paradox for the first time in 1951. In this paradox, you find two contrary patterns. For example, a positive and a negative correlation, depending on how data is analyzed. The differences in the analyses are how data are grouped. This paradox is observed often in social researches. Most of the times, results are affected by the sample on each group or additional information related to the data.
