<span>we have that
the cube roots of 27(cos 330° + i sin 330°) will be
</span>∛[27(cos 330° + i sin 330°)]
we know that
e<span>^(ix)=cos x + isinx
therefore
</span>∛[27(cos 330° + i sin 330°)]------> ∛[27(e^(i330°))]-----> 3∛[(e^(i110°)³)]
3∛[(e^(i110°)³)]--------> 3e^(i110°)-------------> 3[cos 110° + i sin 110°]
z1=3[cos 110° + i sin 110°]
cube root in complex number, divide angle by 3
360nº/3 = 120nº --> add 120º for z2 angle, again for z3
<span>therefore
</span>
z2=3[cos ((110°+120°) + i sin (110°+120°)]------ > 3[cos 230° + i sin 230°]
z3=3[cos (230°+120°) + i sin (230°+120°)]--------> 3[cos 350° + i sin 350°]
<span>
the answer is
</span>z1=3[cos 110° + i sin 110°]<span>
</span>z2=3[cos 230° + i sin 230°]
z3=3[cos 350° + i sin 350°]<span>
</span>
Answer:y=-4/5x+4
Step-by-step explanation:
I search and found different answers but the nearest is 24
(2x - 1)(x + 2y - 3) =
2x(x + 2y - 3) + (-1)(x + 2y - 3) =
2x^2 + 4xy - 6x - x - 2y + 3 =
2x^2 + 4xy - 7x - 2y + 3
Answer:
16
Step-by-step explanation:
Let's represent this with an equation, where x is the first number. The next number must be x+2, because it has to be even, not x+1, and the rest continue the same way:
x+(x+2)+(x+4)+(x+6)+(x+8) = 100
Solve for x.
5x+20 = 100
5x = 80
<u>x = 16</u>
