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noname [10]
3 years ago
11

Is 3, 1, 1/3, 1/9,... a geometric sequence?

Mathematics
1 answer:
barxatty [35]3 years ago
7 0
a_1;\ a_2;\ a_3;\ a_3;\ ...;\ a_n;\ a_{n+1};...\ is\ a\ geometric\ sequence\\if\ a_{n+1}:a_n=const.\\-------------------\\a_1=3;\ a_2=1;\ a_3=\dfrac{1}{3};\ a_4=\dfrac{1}{9}\\\\a_4:a_3=\dfrac{1}{9}:\dfrac{1}{3}=\dfrac{1}{9}\cdot\dfrac{3}{1}=\dfrac{1}{3}\\\\a_3:a_2=\dfrac{1}{3}:1=\dfrac{1}{3}\\\\a_2:a_1=1:3=\dfrac{1}{3}\\\\Answer:Yes.\ It's\ the\ geometric\ sequence.
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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

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Answer:

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If carbon dioxide, methane, and water vapor were not in the atmosphere, then
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<h2>Answer:</h2>

D. The Earth's surface would be much colder than it is.

<h3>Step-by-step explanation:</h3>

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3 0
2 years ago
What is 4/20 + 11/20 in its simplest form​
vivado [14]

Answer:

Step-by-step explanation:

4/20 + 11/20 = 15/20

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