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jeyben [28]
3 years ago
7

Can any one help me with this?

Mathematics
2 answers:
Mumz [18]3 years ago
6 0

Answer: to #3 is no it is not a function

Step-by-step explanation:

erik [133]3 years ago
5 0
10. x= -1

3x + 7 =11x +15
-3x -3x

7=8x + 15
-15. -15

-8=8x
—. —
8. 8

X=-1
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3 0
3 years ago
2 to the power of 15<br> Is it correct to say like this?
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yes you can.

Step-by-step explanation:

...

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2 years ago
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Solve 6x + ( 14x - 5 ) + ( 17 - 3x ) , - ( 2 - x ) -3 ( 6 + 8x ) -12, ( 4x - 9 ) + 8 ( 2x + 3 ) -7x
gregori [183]

Answer:

Step-by-step explanation:

6x + ( 14x - 5 ) + ( 17 - 3x )

= 17x + 12

- ( 2 - x ) -3 ( 6 + 8x ) -12,

= -23x - 32

( 4x - 9 ) + 8 ( 2x + 3 ) -7x

= 13x +15

5 0
3 years ago
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
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