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slavikrds [6]
3 years ago
5

Lines a and b are parallel. What is the measure of ∠3 if ∠6 measures 84°?

Mathematics
1 answer:
Gennadij [26K]3 years ago
7 0
We need to know wat it looks like
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Ganezh [65]
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What is the square root of -2i?
Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of z=0-2i. First, convert from Cartesian to polar form:

r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2

\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}

z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})

Next, use the formula \displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr] where \displaystyle k=0,1,2,...\:,n-1 to find the square roots:

<u>When k=1</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]

\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)

\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i

<u>When k=0</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)

\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i

Thus, the square roots of -2i are 1-i and -1+i

4 0
2 years ago
Hello can someone help me with this please <br> thank you
Aleks04 [339]

y = 2 - x

y(-3) = 2 - (-3) = 2 + 3 = 5

y(0) = 2 - 0 = 2

y(5) = 2 - 5 = -3

8 0
2 years ago
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Find the 10th term of 3, 9 /2​ , 27/4
Amanda [17]

Answer:

Step-by-step explanation:

the nth term of a geometric sequence is

∙

x

a

n

=

a

r

n

−

1

where a is the first term and r the common ratio

r

=

a

2

a

1

=

a

3

a

2

=

...

...

=

a

n

a

n

−

1

here  

a

=

3

and  

r

=

9

3

=

27

9

=

3

 

⇒

a

10

=

3

×

3

9

=

3

10

=

59049

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(20 POINTS) HELP ASAP AND USE IMAGE TO HELP!!
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Hey are use desmos I don’t know if I can help you also watway
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