In 1978, the Social Security and Medicare rate combined was 6.05%, up to $17,700 earned. Express the Social Security tax s(x) for 1978 as a piecewise function. Ten years later, the percent had increased to 7.51% and the maximum taxable income had increased to $45,000. Express the Social Security tax s(x) for 1988 as a piecewise function. If a person earned $50,000 in 1978, and $50,000 in 1988, what was the difference in the Social Security and Medicare taxes paid?

Step-by-step explanation:

From the given question:

a) The piecewise function for the Social Security tax s(x) is expressed as:

s(x) = x * 6.05% when x < $17,700

The 6.05% of the given amount is:

= (6.05/100) * 17700

= 1070.85

s(x) = 1070.85 when x ≥ $17,700

b) Ten years later, i.e. 1988; the percentage had increased to 7.51% & maximum taxable income also increased to $45000

The piecewise function for the Social Security tax s(x) is expressed as:

s(x) = x * 7.51% when x < $45,000

The 7.51% of the given amount is:

= (7.51/100) * 45000

= 3379.5

s(x) = 3379.5  when x ≥ $45,500

c) The difference in the Social Security and Medicare taxes paid is as follows:

In 1978, s(x) = 1070.85 when x ≥ $17,700

In 1988, s(x) = 3379.5  when x ≥ $45,500

The difference in the Social Security and Medicare taxes paid  = $3379.5 - $1070.85

The difference in the Social Security and Medicare taxes paid = $2308.65

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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