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3 years ago

10

3y - 21 = 12x Help please

1 answer:

Irina18 [472]3 years ago

3 0

Simply
3Y+-21=12X

Re-order the terms
-21+3Y=12X

(Solving for Y variable)

Move all terms containing y to the left, all other terms to the right.

Add 21 to each side of the equation
-21+21+3Y=21+12X

Combine like terms: -21+21=0
0+3Y=21+12X
3Y=21+12X

Divide each side by 3
Y=7+4X

Simplifying
Y=7+4x
I hope this helps!! :)

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3 years ago

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3 years ago

Which expression is a cube root of -1+i√3?

Tpy6a [65]

Answer:

<em>The correct option is C.</em>

Step-by-step explanation:

<u>Root Of Complex Numbers</u>

If a complex number is expressed in polar form as

$Z=(r,\theta)$

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$\displaystyle Z_1=\left(\sqrt[3]{r},\frac{\theta}{3}\right)$

$\displaystyle Z_2=\left(\sqrt[3]{r},\frac{\theta}{3}+120^o\right)$

$\displaystyle Z_3=\left(\sqrt[3]{r},\frac{\theta}{3}+240^o\right)$

We are given the complex number in rectangular components

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It's located in the second quadrant, so

$\theta=120^o$

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$Z=(2,120^o)$

Its cubic roots are

$\displaystyle Z_1=\left(\sqrt[3]{2},\frac{120^o}{3}\right)=\left(\sqrt[3]{2},40^o\right)$

$\displaystyle Z_2=\left(\sqrt[3]{2},40^o+120^o\right)=\left(\sqrt[3]{2},160^o\right)$

$\displaystyle Z_3=\left(\sqrt[3]{2},40^o+240^o\right)=\left(\sqrt[3]{2},280^o\right)$

Converting the first solution to rectangular coordinates

$z_1=\sqrt[3]{2}(\ cos40^o+i\ sin40^o)$

The correct option is C.

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4 years ago