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3 years ago

Determine the total annual FICA tax for an annual salary of $38,480. A. $294. 37 b. $717. 96 c. $2,385. 76 d. $2,943. 72.

1 answer:

7 0

Answer:
d. $2,943.72 is the total annual FICA tax for an annual salary of $38,480.

Explanation:
6.2% of the annual salary for the Social Security Tax, and 1.45% of the annual salary for the Medicare Tax. In total, FICA takes 7.65% of your annual salary as tax. $38,480 x 7.65% = $2,943.72.

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3 years ago

An executor should be willing and reliable because brainly

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<h3>What is the role of an executor?</h3>

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2 years ago

Simplify 16^x = 0.5

gayaneshka [121]

Answer:

Explanation:

Convert the improper fraction 16/3 to a mixed number. Divide 16 by 3: 16 ÷ 3 = 5 with remainder of 1; The whole number result is 5; The remainder is 1. With 1 as the numerator and 3 as the denominator

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3 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;

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

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

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