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Sagot :
The given polynomials can be rewritten in simplified form by dividing by
a power of the variable of the polynomial.
Responses:
- [tex]\dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}\ \Rightarrow \ \underline{ d. \ 1}[/tex]
- [tex]\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9} \Rightarrow \underline{a. \ \dfrac{1}{x^4}}[/tex]
- [tex]\dfrac{7 \cdot x +2}{4 \cdot x + x^3} \Rightarrow \underline{ \ b. \ \dfrac{7}{x^2}}[/tex]
- [tex]\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5} \Rightarrow \underline{ \ c. \ \dfrac{2}{x^3}}[/tex]
Which method are used to approximate fractional polynomials?
The possible functions based on a similar question posted online are;
[tex]\dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}[/tex]
[tex]\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9}[/tex]
[tex]\dfrac{7 \cdot x +2}{4 \cdot x + x^3}[/tex]
[tex]\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5}[/tex]
The options are;
[tex]a. \hspace{0.15 cm}\dfrac{1}{x^4}[/tex]
[tex]b. \hspace{0.15 cm}\dfrac{7}{x^2}[/tex]
[tex]c. \hspace{0.15 cm}\dfrac{2}{x^3}[/tex]
d. 1
- [tex]\mathbf{\dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}}[/tex]
[tex]\mathbf{\dfrac{\frac{2}{ x^4}+\frac{x 3 \cdot x^2}{ x^4} + \frac{ x^4}{ x^4} }{\frac{x}{ x^4} + \frac{3 \cdot x^3 }{x^4} +\frac{x^4}{x^4} } }\approx \dfrac{0 + 0 + 1}{0 + 0 + 1} \approx 1[/tex]
Therefore;
[tex]The \ match \ for \ \dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}\ is \ d. \ 1[/tex]
- [tex]\mathbf{\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9}}[/tex]
[tex]\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9} = \mathbf{ \dfrac{\frac{5 \cdot x}{x^5} + \frac{2 \cdot x^2}{x^5} +\frac{x^5}{x^5} }{\frac{5 \cdot x}{x^5} + \frac{4 \cdot x^2}{x^5} + \frac{x^9}{x^5} }} \approx \dfrac{0 + 0 + 1}{0 + 0 + x^4} \approx \dfrac{1}{x^4}[/tex]
[tex]The \ match \ for \ \dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9} \ is \ a. \ \dfrac{1}{x^4}[/tex]
- [tex]\mathbf{\dfrac{7 \cdot x +2}{4 \cdot x + x^3}}[/tex]
[tex]\dfrac{7 \cdot x +2}{4 \cdot x + x^3} = \dfrac{7 \cdot x }{4 \cdot x + x^3}+ \dfrac{ 2}{4 \cdot x + x^3} = \mathbf{ \dfrac{7 }{4 + x^2}+ \dfrac{ 2}{4 \cdot x + x^3}} \approx \dfrac{7}{x^2}[/tex]
Therefore;
[tex]The \ match \ for \ \dfrac{7 \cdot x +2}{4 \cdot x + x^3} \ is \ b. \ \dfrac{7}{x^2}[/tex]
- [tex]\mathbf{\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5}}[/tex]
[tex]\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5} =\mathbf{ \dfrac{\frac{4}{x^2} + \frac{2 \cdot x^2 }{x^2} }{\frac{5 \cdot x}{x^2} + \frac{4 \cdot x^3}{x^2} +\frac{x^5}{x^2}}} \approx \dfrac{0 +2}{0 +4 \cdot x +x^3} \approx \dfrac{2}{x^3}[/tex]
[tex]The \ match \ for \ \dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5} \ is \ c. \ \dfrac{2}{x^3}[/tex]
Learn more about polynomials here:
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