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Keith_Richards [23]
3 years ago
9

How can you solve a simpler problem to help you find the solution

Mathematics
2 answers:
sveta [45]3 years ago
5 0
Ya you just need to round the number
irina1246 [14]3 years ago
4 0
Just round the number
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Precal help if you could could you answer both. Me finishing this helps me graduate.
oee [108]

Exercise 1:

The easiest way to compute powers of complex numbers is to write them in the form

z = \rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))

In this form, you have

z^n = \rho^n e^{in\theta} = \rho^n(\cos(n\theta)+i\sin(n\theta))

The magnitude of the number is given by

z=a+bi \implies \rho = \sqrt{a^2+b^2}

So, we have

z=-1+\sqrt{3}i \implies \rho = \sqrt{1+3}=2

As for the angle, we have

z=a+bi \implies \theta = \text{atan2}\dfrac{b}{a}

So, we have

z=-1+\sqrt{3}i \implies \theta = \text{atan2}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\dfrac{2\pi}{3}

Finally,

z=-1-\sqrt{3}i = 2\left(\cos\left( -\dfrac{2\pi}{3}\right) + i\sin\left( -\dfrac{2\pi}{3}\right)\right) \implies z^6 = 2^6\left(\cos\left( -4\pi\right) + i\sin\left(-4\pi\right)\right) = 64

Exercise 2:

You simply have to compute the trigonometric function:

\cos\left(-\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2},\quad \sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}

So, we have

z = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2}\right) = 1-i

7 0
3 years ago
Help asap will mark branniest thx
Paha777 [63]

Answer:

(\pi)r ^{2} h \div 3

Volume = 3.14 × 5² ×6 ÷ 3

volume = 157 cm³ B

6 0
3 years ago
In the problem above what comes after distribution property?
faltersainse [42]

Answer:

A

Step-by-step explanation:

Id say A Im sorry If im wrong though, this is what I would put

4 0
3 years ago
Read 2 more answers
Which of these is NOT a terminating decimal?<br> A) 1.88<br> B) 1.875 <br> C) 0.88<br> D) 0.8
slamgirl [31]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
−r3≤6<br> The solution is
lawyer [7]

Answer:

r>=-3

Step-by-step explanation:

4 0
3 years ago
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