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Alexxandr [17]
3 years ago
6

Brenton’s weekly pay, P(h) , in dollars, is a function of the number of hours he works, h. He gets paid $20 per hour for the fir

st 40 hours he works in a week. For any hours above that, he is paid overtime at $30 per hour. He is not permitted to work more than 60 hours in a week. Which set describes the domain of P(h)? {h| 0 ≤ h ≤ 40} {h| 0 ≤ h ≤ 60} {P(h)| 0 ≤ P(h) ≤ 1,400} {P(h)| 0 ≤ P(h) ≤ 1,800}
Mathematics
2 answers:
Irina18 [472]3 years ago
8 0

Answer: soooooooo.......

That is, if you work more than 40 hours a week, your hourly wage for the extra hours is 1.5 times your ... Write a piece wise function that gives your weekly pay "P" in terms of the number "H" of hours you work. b. How much will you get paid if you work 45 hours?

hope this helps

HACTEHA [7]3 years ago
7 0

Answer:

The first set would be,  {h| 0 ≤ h ≤ 40} then it would come up to {P(h)| 0 ≤ P(h) ≤ 1,400} {P(h)| 0 ≤ P(h) ≤ 1,800} so it would leave you with 25$\



P(h)? {h| 0 ≤ h ≤ 40} {h| 0 ≤ h ≤ 60} {P(h)| 0 ≤ P(h) ≤ 1,400} {P(h)| 0 ≤ P(h) ≤ 1,800}    

$30 per hour + 40 hours he works in a week.

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1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.
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Answer:

Step-by-step explanation:

1.

To write the form of the partial fraction decomposition of the rational expression:

We have:

\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}

2.

Using partial fraction decomposition to find the definite integral of:

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx

By using the long division method; we have:

x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }

                  - 2x^3 -16x^2-40x

                 <u>                                         </u>

                                            x+ 20

So;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}

By using partial fraction decomposition:

\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}

                         = \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now;  we have to relate like terms on both sides; we have:

A + B = 1   ;   2A - 10 B = 20

By solvong the expressions above; we have:

A = \dfrac{5}{2}     B =  \dfrac{3}{2}

Now;

\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}

Thus;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}

Now; the integral is:

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3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

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