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2 years ago

Self-testing is an effective strategy to

SAT

2 answers:

umka2103 [35]2 years ago

4 0

Answer:

A. review information you've already read.

Explanation:

You can use an analogy towards studying for example. You review the information you already used or have been taught previously, and test yourself again. Hope I helped!

eimsori [14]2 years ago

3 0

Answer:

A. review information you have already read

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2 years ago

Match the following functions with functions that exhibit a similar behavior for large values of $x. $.

Delvig [45]

The given polynomials can be rewritten in simplified form by dividing by

a power of the variable of the polynomial.

Responses:

$\dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}\ \Rightarrow \ \underline{ d. \ 1}$

$\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9} \Rightarrow \underline{a. \ \dfrac{1}{x^4}}$

$\dfrac{7 \cdot x +2}{4 \cdot x + x^3} \Rightarrow \underline{ \ b. \ \dfrac{7}{x^2}}$

$\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5} \Rightarrow \underline{ \ c. \ \dfrac{2}{x^3}}$

<h3>Which method are used to approximate fractional polynomials?</h3>

The possible functions based on a similar question posted online are;

$\dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}$

$\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9}$

$\dfrac{7 \cdot x +2}{4 \cdot x + x^3}$

$\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5}$

The options are;

$a. \hspace{0.15 cm}\dfrac{1}{x^4}$ <em />

$b. \hspace{0.15 cm}\dfrac{7}{x^2}$

$c. \hspace{0.15 cm}\dfrac{2}{x^3}$

d. 1

$\mathbf{\dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}}$

$\mathbf{\dfrac{\frac{2}{ x^4}+\frac{x 3 \cdot x^2}{ x^4} + \frac{ x^4}{ x^4} }{\frac{x}{ x^4} + \frac{3 \cdot x^3 }{x^4} +\frac{x^4}{x^4} } }\approx \dfrac{0 + 0 + 1}{0 + 0 + 1} \approx 1$

Therefore;

$The \ match \ for \ \dfrac{2 + 3 \cdot x^2 + x^4}{x + 3 \cdot x^3 + x^4}\ is \ d. \ 1$

$\mathbf{\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9}}$

$\dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9} = \mathbf{ \dfrac{\frac{5 \cdot x}{x^5} + \frac{2 \cdot x^2}{x^5} +\frac{x^5}{x^5} }{\frac{5 \cdot x}{x^5} + \frac{4 \cdot x^2}{x^5} + \frac{x^9}{x^5} }} \approx \dfrac{0 + 0 + 1}{0 + 0 + x^4} \approx \dfrac{1}{x^4}$

$The \ match \ for \ \dfrac{5 \cdot x + 2 \cdot x^2 + x^5}{5 \cdot x + 4 \cdot x^2 + x^9} \ is \ a. \ \dfrac{1}{x^4}$

$\mathbf{\dfrac{7 \cdot x +2}{4 \cdot x + x^3}}$

$\dfrac{7 \cdot x +2}{4 \cdot x + x^3} = \dfrac{7 \cdot x }{4 \cdot x + x^3}+ \dfrac{ 2}{4 \cdot x + x^3} = \mathbf{ \dfrac{7 }{4 + x^2}+ \dfrac{ 2}{4 \cdot x + x^3}} \approx \dfrac{7}{x^2}$

Therefore;

$The \ match \ for \ \dfrac{7 \cdot x +2}{4 \cdot x + x^3} \ is \ b. \ \dfrac{7}{x^2}$

$\mathbf{\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5}}$

$\dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3 + x^5} =\mathbf{ \dfrac{\frac{4}{x^2} + \frac{2 \cdot x^2 }{x^2} }{\frac{5 \cdot x}{x^2} + \frac{4 \cdot x^3}{x^2} +\frac{x^5}{x^2}}} \approx \dfrac{0 +2}{0 +4 \cdot x +x^3} \approx \dfrac{2}{x^3}$