<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: B. 2 = 3x + 10x2
Step-by-step explanation:
This is the concept of quadratic equations; We required to find the type of equation that can be solved using the model that has been used to solve the equation such that the answer is:
[-3+-sqrt(3^2+4(10)(2))]/(2(10))
The formual that was applied here was a quadratic formula given by:
x=[-b+\-sqrt(b^2-4ac)]/2a
whereby from the our substituted values above,
a=10,b=3 and c=-2
such that the quadratic equation will be:
10x^2+3x-2
Answer:
Correct option is
B
90
∘
,90
∘
,90
∘
Let AB and CD be two lines Intersecting at O, such that, ∠AOD=90
∘
Now, ∠AOD=∠COB=90
∘
(Vertically opposite angles)
⟹∠AOD+∠DOB=180
o
(Angles on a straight line)
⟹90+∠DOB=180
o
∠DOB=90
∘
∠DOB=∠AOC=90
∘
(Vertically opposite angles)
Thus, all angles are 90
∘
.
