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3 years ago

Solve equation 2x+9=4x-23

Mathematics

2 answers:

Sladkaya [172]3 years ago

7 0

Answer:=16

Step-by-step explanation:

hope this helps:)

ser-zykov [4K]3 years ago

4 0

Answer:

x = 16

Step-by-step explanation:

2x + 9 = 4x - 23

+23 +23

2x + 32 = 4x

-2x -2x

32 = 2x

divide both sides by 2

x = 16

Hope this Helps!!!

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3 years ago

For the month of March in a certain city, 57% of the days are cloudy. Also in the month of March in the same city, 55% of the da

oksian1 [2.3K]

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Step-by-step explanation:

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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:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

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

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

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

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

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

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

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

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2 years ago