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Akimi4 [234]
4 years ago
8

Which algebraic expression is a polynomial with a degree of 2?

Mathematics
2 answers:
Sauron [17]4 years ago
4 0
6x^2 - 6x + 5 <== this is ur polynomial with a degree of 2

the one u have marked is not a polynomial because it has a variable in the square root...a polynomial cannot have that


jeka944 years ago
4 0

Answer:

D) 6x² − 6x + 5 has degree 2.

Step-by-step explanation:

Given : Polynomial .

To find : Which algebraic expression is a polynomial with a degree of 2.

Solution: We have given polynomial

Degree : The highest power of the polynomial is called degree.

We can see from the given polynomials two polynomial have 2 power

first is  10x^{2} -\sqrt{x}.

Second is 6x² − 6x + 5.

But the first term have root so, it is not a polynomial.

So  6x² − 6x + 5 has degree 2

Therefore, D) 6x² − 6x + 5 has degree 2.

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Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
  • N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3

3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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