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3 years ago

9

It due today answer this correctly and I mark you Brainliest look at the 2 picture it 5 question easy right, if you don’t answer

and put I don’t know I will report you...

1 answer:

Anon25 [30]3 years ago

3 0

Answer:

1.=40

2=8

3=-16

Step-by-step explanation:

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ANSWER ASAP<br><br> x+4.5<br><br> x= 1.2<br><br> A: 3.3<br> B: 6.7<br> C: 5.7

swat32

Answer:

C.5.7

Step-by-step explanation:

plug in the x value then add 4.5 to 1.2 and booyyah

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3 years ago

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Lelu [443]

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4 years ago

Which trigonometric functions have a domain of [–1, 1]?

Annette [7]

Answer:

The answer is A. y = arcsinx and y = arccosx

Step-by-step explanation:

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3 years ago

Carla had $ 5.50 more than Rhonda. Rhonda had $ 5.00. How much does Carla have in total? A $10.00 B $10.50 C $9.50 D $5.50

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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.

8 0

3 years ago