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shusha [124]
2 years ago
5

Homes in a new home development have been built with a floor plan that doesn't appeal to today's buyers. In a cost approach valu

ation of the houses, what sort of depreciation will the appraiser apply
SAT
1 answer:
Novay_Z [31]2 years ago
6 0

The kind of depreciation that the appraiser will apply in this case is called "functional obsolescence".

<h3>What is "functional obsolescence"?</h3><h3 />

The kind of loss of value that is attributable to loss of appeal is what tis called Functional obsolescence depreciation method.

It can be curable or incurable. curable means that by giving the property a face life, it re-acquires value.

Learn more about depreciation at;
brainly.com/question/1203926
#SPJ1

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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

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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}

The \ match \ for \ \dfrac{4 + 2 \cdot x^2 }{5 \cdot x + 4 \cdot x^3  + x^5} \ is \ c. \  \dfrac{2}{x^3}

Learn more about polynomials here:

brainly.com/question/1528517

brainly.com/question/12006853

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