Given a complex number in the form:
![z= \rho [\cos \theta + i \sin \theta]](https://tex.z-dn.net/?f=z%3D%20%5Crho%20%5B%5Ccos%20%5Ctheta%20%2B%20i%20%5Csin%20%5Ctheta%5D)
The nth-power of this number,

, can be calculated as follows:
- the modulus of

is equal to the nth-power of the modulus of z, while the angle of

is equal to n multiplied the angle of z, so:
![z^n = \rho^n [\cos n\theta + i \sin n\theta ]](https://tex.z-dn.net/?f=z%5En%20%3D%20%5Crho%5En%20%5B%5Ccos%20n%5Ctheta%20%2B%20i%20%5Csin%20n%5Ctheta%20%5D)
In our case, n=3, so

is equal to
![z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20%5Crho%5E3%20%5B%5Ccos%203%20%5Ctheta%20%2B%20i%20%5Csin%203%20%5Ctheta%20%5D%20%3D%20%285%5E3%29%20%5B%5Ccos%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%2B%20i%20%5Csin%20%283%20%5Ccdot%20330%5E%7B%5Ccirc%7D%29%20%5D)
(1)
And since

and both sine and cosine are periodic in

, (1) becomes
Answer:
y - 4 = 3(x - 1)
Step-by-step explanation:
It looks like you have slope 3 and a point (1, 4)
so it should be y - 4 = 3(x - 1)
which is your first choice: y minus 4 = 3 (x minus 1)
Answer:
3 1/3
Step-by-step explanation:
Make the fractions into improper fractions:
2 1/2 = 5/2
1 1/3 = 4/3
5/2 × 4/3 =
5×4/2×3 =
20/6 =
10/3 =
3 1/3