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Fittoniya [83]
3 years ago
8

a polynomial function has -5

" align="absmiddle" class="latex-formula"> as a root. which of the following must also be a root of the function?
Mathematics
1 answer:
Goshia [24]3 years ago
8 0

Answer:

5 \sqrt{3}i

Step-by-step explanation:

Assuming one root of a polynomial function is

- 5 \sqrt{3} i

Then the other root is

5 \sqrt{3}i

The reason is that , complex roots of polynomial functions occur in pairs.

One complex root is the conjugate of the other.

This means that, if

x -  yi

is a root, then

x  +  yi

is also a root.

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Answer: x = -3

Step-by-step explanation:

i simply divided by -7, in order to get x by itself.

-7/-7x = 21/-7

that cancels out the -7 on the left side, and leaves us with -3 on the right side!

x = -3

6 0
3 years ago
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Integration of ∫(cos3x+3sinx)dx ​
Murljashka [212]

Answer:

\boxed{\pink{\tt I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C}}

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\  dx

  • Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
  • Now , Rewrite using du and u .

\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I =  \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C }}}}}

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3 years ago
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Veseljchak [2.6K]

Answer:

A.

Step-by-step explanation:

Standard form is as follows...

AX^2 + BX + C

- Hope that helped! Please let me know if you need further explanation.

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Answer:

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Answer:

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