13. Find AC. PLEASEE I NEED HELP!!!!!
2 answers:
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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)
The nth-power of this number,
- the modulus of
In our case, n=3, so
And since
and both sine and cosine are periodic in
Answer:
Step-by-step explanation:
let number of passion fruits=x
number of dragon fruits=x+10% of x
=x+0.1 x=(1+0.1)x=1.1 x
total number of fruits=x+1.1 x=(1+1.1)x=2.1 x
1.1x+60% of 1.1x+x+9% of x=2.1x+75
1.1 x+0.6 of 1.1 x+x+0.09 of x=2.1x+75
1.1x+0.66 x+x+0.09 of x =2.1x+75
2.1 x+0.75x-2.1 x=75
0.75 x=75