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2 years ago

Someone please help Me algebra 2

Mathematics

1 answer:

Bumek [7]2 years ago

5 0

Answer:

Table I fill-in: 1.404

Table II fill-in: 1.661 (instead of the 2 that's there)

Step-by-step explanation:

The functions shown in Table I and II are inverses of each other. This means that the x's for Table I are the y's in Table II. And the y's in Table I are the x's in Table II.

So look at Table II, there is a coordinate pair

(7, 1.404). Switch the x and y and you get (1.404,7) That's your answer to fill in on Table I.

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2 years ago

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2 years ago

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:

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

We have:

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

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:

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

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$x+ 20$

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:

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

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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;

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

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

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