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
![z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]](https://tex.z-dn.net/?f=z%5E3%20%3D%20125%20%5B%5Ccos%20270%5E%7B%5Ccirc%7D%20%2B%20i%20%5Csin%20270%5E%7B%5Ccirc%7D%20%5D)
.75x +1.29 = 7.00
x = the amount of bagels he wants.
For the 51-deg angle, y is the opposite leg, and 12 is the hypotenuse.
The trig ratio that relates the opposite leg and the hypotenuse is the sine.
sin A = opp/hyp
sin 51 deg = y/12
y = 12 * sin 51 deg
y = 9.3
Answer:
xisans
Step-by-step explanation:
ty z×ynrkkkdiod
C = 8.5x + 6y
Rearrange the equation by making the side with x, be on the Left Hand Side (LHS)
8.5x + 6y = C
Subtract 6y from both sides of the equation
8.5x + 6y - 6y = C - 6y
8.5x = C - 6y
Divide both sides of the equation by 8.5
(8.5x/8.5) = (C - 6y)/8.5
x = (C - 6y)/8.5
I hope this helped.
