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3 years ago

Math question help plsss

Mathematics

1 answer:

Tomtit [17]3 years ago

3 0

Answer would be C
Point A is at (1,1) then point B is three units down and point C is three over from point B :)

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4 is the answer to that

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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

$Z=(r,\theta)$

Then the cubic roots of Z are

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

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

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

We are given the complex number in rectangular components

$Z=-1+i\sqrt{3}$

Converting to polar form

$r=\sqrt{(-1)^2+(\sqrt{3})^2}=2$

$\displaystyle tan\theta=\frac{\sqrt{3}}{-1}=-\sqrt{3}$

It's located in the second quadrant, so

$\theta=120^o$

The number if polar form is

$Z=(2,120^o)$

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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Step-by-step explanation:

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here a = 3

area of ∆ = 3.9 in^2

area of circle

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3 years ago

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maks197457 [2]

thirty-five thousand, three hundred twenty-six.

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The base of a solid is bounded by y=√(x), y=0, x=2, and x=6. Its cross-sections, taken perpendicular to the x-axis, are squares.

Gre4nikov [31]

Answer:

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Step-by-step explanation:

Volume of the solid

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pi × [6²/2 - 2²/2]

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Since the solid is only above the x-axis half of this volume would be generated:

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