Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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:
https://brainly.com/question/1528517
https://brainly.com/question/12006853
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
