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:
D. 85.7%
Step-by-step explanation:
its figuring out the percentage decrease
so you would go
35-5= 30
30/35 x 100 = 85.7
Answer:
See below
Step-by-step explanation:
Out of 25 games, they won 14
Fraction won = won / total = 14 / 25
Percent won = won /total * 100 % = 14/25 * 100 = .56 * 100 % = 56 %
To find the number of games lost
25-14 = 11
Fraction won = won / total = 11 / 25
Percent won = won /total * 100 % = 11/25 * 100 = .44 * 100 % = 44 %
To check your work, the percent total should be 100
56+44 = 100
First you would need to find out how many 9 payments of 58$ would be and that is 522. Then you would need to subtract the 522 and the 480 and your answer would be 42$ so the interest would be 42$
Rewriting the left hand side,
csc²t - cost sec t
= (1/sin²t)-(cost)(1/cost)
= 1/sin²t - 1
= 1/sin²t - sin²t/sin²t
= (1-sin²t)/sin²t
= cos²t/sin²t
= cot²t
