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dolphi86 [110]
2 years ago
9

Multiplicity of F(x)= 4x^3+19x^2-41x+9

Mathematics
1 answer:
Evgen [1.6K]2 years ago
6 0

The multiplicity of the function is 1 for each term after factorization.

<h3>What is polynomial?</h3>

Polynomial is the combination of variables and constants systematically with "n" number of power in ascending or descending order.

\rm a_1x+a_2x^2+a_3x^3+a_4x^4..........a_nx^n

We have a polynomial function:

\rm F(x)= 4x^3+19x^2-41x+9

After factorization:

\rm F(x) = \left(4x-1\right)\left(x^2+5x-9\right)=0

\rm F(x) =(4x-1)^1(\:x-\frac{-5+\sqrt{61}}{2})^1(\:x+\frac{-5-\sqrt{61}}{2})^1

The multiplicity of the function is 1 for each term.

Thus, the multiplicity of the function is 1 for each term after factorization.

Learn more about Polynomial here:

brainly.com/question/17822016

#SPJ1

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Which toy pack cost of 2.50 per toy ?(brainlist will be given out)!!
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Answer: Yo-yo

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What is the solution set?
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Simplify the expression the root of negative sixteen all over the quantity of three minus three i plus the quantity of one minus
viva [34]

I think the question is simplify

\dfrac{\sqrt{-16}}{3 - 3i} + 1-2i

= \dfrac{4\sqrt{-1}}{3(1-i)}+ 1-2i

= \dfrac{4}{3}\dfrac{i}{1-i} \cdot \dfrac{1+i}{1+i} + 1-2i


= \dfrac{4}{3} \dfrac{i-1}{2} + 1-2i

= \dfrac{2}{3} i - \dfrac{2}{3} + 1-2i

= \dfrac{1}{3} - \dfrac{4}{3} i

Answer: (1/3) - (4/3)i

I might be getting the question wrong. It might be

\dfrac{\sqrt{-16}}{3 - 3i + (1-2i)}

Let's do this one too.

= \dfrac{4i}{4 - 5i} \cdot \dfrac{4+5i}{4+5i}

= \dfrac{16i - 20}{4^2 + 5^2}

=-\dfrac{20}{41} + \dfrac{16}{41} i


4 0
3 years ago
Can someone please explain what I did wrong here? Thank you!
WARRIOR [948]

Answer:

P(Tamika)=\frac{1}{3} \times\frac{1}{3} =\frac{1}{9} \\P(Jayden)=\frac{1}{3} ;\\P=P(Tamika)\timesP(Jayden)=\frac{1}{9}\times\frac{1}{3}=\frac{1}{27}

4 0
3 years ago
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