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Makovka662 [10]
3 years ago
9

The sales tax on a $20 hammer is 7 percent, or $1.40. why is this tax a bigger burden for josh, who has a $15,000 income, than f

or aaron, who has a $150,000 income?
Mathematics
2 answers:
AlekseyPX3 years ago
7 0
The tax represents a larger proportion of josh's income
Bond [772]3 years ago
3 0

Answer:

The ratio of the sales tax to Josh income is greater than the sales tax to aaron income (As the income of the Josh is less as compared to the arron income) this shows the tax is bigger burden for the Josh as compared to the aaron .

Step-by-step explanation:

As given

The sales tax on a $20 hammer is 7 percent, or $1.40.

As income of the  josh is 15000 and the income of the aaron is $150000 .

Now the ratio of the sales tax to the josh income .

\frac{Sales\ tax}{Josh\ income} = \frac{1.40}{15000}

\frac{Sales\ tax}{Josh\ income} = \frac{14}{15000\times 10}

\frac{Sales\ tax}{Josh\ income} = \frac{7}{75000}

In decimal form

\frac{Sales\ tax}{Josh\ income} = 0.000093\ (Approx)

Now the ratio of the  sales tax to the aaron income .

\frac{Sales\ tax}{aaron\ income} = \frac{1.40}{150000}

\frac{Sales\ tax}{aaron\ income} = \frac{14}{150000\times 10}

\frac{Sales\ tax}{aaron\ income} = \frac{7}{750000}

In decimal form

\frac{Sales\ tax}{aaron\ income} = 0.0000093\ (Approx)

Thus as the ratio of the sales tax to Josh income is greater than the sales tax to aaron income (As the income of the Josh is less as compared to the arron income) this shows the tax is bigger burden for the Josh as compared to the aaron .

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Step-by-step explanation:

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where a, b, and c are real numbers, and a \ \neq \ 0.

The standard form or the vertex form of a quadratic function is, however, a little different from the general form. To get the standard form from the general form, we need to use the "complete the square" method.

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                          f(x) \ = \ a\left(x \ + \ \displaystyle\frac{b}{2a}\right)^{2} \ + \ c \ - \ a\left(\displaystyle\frac{b^{2}}{4a^{2}}\right) \\ \\ \\ f(x) \ = \ a\left(x \ + \ \displaystyle\frac{b}{2a}\right)^{2} \ + \ c \ - \ \displaystyle\frac{b^{2}}{4a}

Let

                                         h \ = \ -\displaystyle\frac{b}{2a}     and     k \ = \ c \ - \ \displaystyle\frac{b^{2}}{4a},

then the expression reduces into

                                              f(x) \ = \ a \left(x \ - \ h\right)^{2} \ + \ k,

where the point (<em>h</em>, <em>k</em>) are the coordinates for the vertex of the quadratic function.

There are two different methods to approach this question. First, we consider the general form of the quadratic function, it is observed that has a y-intercept at the point \left(0, \ 2\right), so

                                            f(0) \ = \ -2 \\ \\ \\ f(0) \ = \ a(0)^{2} \ + \ b(0) + c \\ \\ \\ c = \ -2.

Additionally, it is pointed that two distinct points (-1, \ -3) and (-4, \ 6) lies on the quadratic graph, hence

                                       f(-1) \ = \ -3 \\ \\ \\ f(-1) \ = \ a(-1)^{2} \ + \ b(-1) \ -2 \\ \\ \\ \-\hspace{0.36cm} -3 \ = \ a \ - \ b \ -2 \\ \\ \\ \-\hspace{0.3} a \ - \ b \ = \ -1 \ \ \ \ \ \ $-----$ \ (1)

and

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Substitute a \ = \ 1 into equation (1),

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\rule{12.5cm}{0.02cm}

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Answer:

<h3>415 mm</h3>

Step-by-step explanation:

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