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GenaCL600 [577]
2 years ago
15

A fifth-degree polynomial equation with rational coefficients has the roots 3, 8i, and 7- radical 5. Which are also roots of the

polynomial equation?

Mathematics
1 answer:
Sliva [168]2 years ago
8 0

Using <u>complex numbers and radicals</u>, it is found that the correct option is C.

When a complex number is a root of a polynomial function, it's conjugate also has to be.

  • Hence, since 8i is a root of the function, -8i is also a root.

The same is valid for radicals, as they are in the \pm format.

  • Since 7 - \sqrt{5} is a root, 7 + \sqrt{5} is also a root.

Hence, option C is correct.

For more on <u>complex numbers and radicals</u>, you can check brainly.com/question/25173944

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Right triangle since it has a right angle = 90°

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2 years ago
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garik1379 [7]

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3 0
3 years ago
PLZ HELP
Alex

Answer:

The correct option is C ) 109.5°

Therefore,

m\angle E=109.5\°

Step-by-step explanation:

Given:

In Triangle DEF

d = 10

e = 18

f = 12

To Find

angle E = ?

Solution:

In Triangle DEF , Cosine Rule says

\cos E=\dfrac{f^{2}+d^{2}-e^{2}}{2fd}

Substituting the values we get

\cos E=\dfrac{12^{2}+10^{2}-18^{2}}{2\times 12\times 10}

\cos E=\dfrac{-80}{240}=-\dfrac{1}{3}

Therefore,

\angle E=\cos^{-1}(-0.3333)

m\angle E=109.5\°       ............As it is in Second Quadrant

Therefore,

m\angle E=109.5\°

4 0
3 years ago
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