9r-5r+2+r ????????? Pls help me

Mathematics

1 answer:

Oksanka [162]2 years ago

6 0

Answer:

5r + 2

Step-by-step explanation:

9r-5r+2+r=5r+2

\mathrm{Group\:like\:terms}

=9r-5r+r+2

\mathrm{Add\:similar\:elements:}\:9r-5r+r=5r

=5r+2

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Which expression is a fourth root of -1+isqrt3?

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Answer:

Step-by-step explanation:

$\sf n^{th}$ roots of a complex number is given by DeMoivre's formula.

$\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}$

Here, k lies between 0 and (n -1) ; n is the exponent.

$\sf -1 + i\sqrt{3}$

a = -1 and b = √3

$\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}$

$\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}$

$\sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})$

$\sf = \dfrac{-\pi }{3}$

n = 4

For k = 0,

$\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]$