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VashaNatasha [74]
3 years ago
7

Can you plz help me on this i will give you brainlist

Mathematics
1 answer:
lukranit [14]3 years ago
4 0

1: Whole Numbers: (Do not include Negative)

- 15

- 134

Rational Numbers:

- 15 (15/1) (You can change 15 into a rational by putting it over 1)

- 134 (134/1)

- -1/3

- 2.4

- -18 (-18/1)

2: Least to Greatest

0.09, 1/9, 9/10, 91%

3: Greatest to Least

5/4, 1 1/5, 100%, 10.5

<em>I hope this helps, and Happy Holidays! :)</em>

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For all values of x, which expression is equivalent to 2x + 5 − x + 3x + x − 2?
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Answer:

5x + 3

Step-by-step explanation:

1) Collect like terms.

(2x − x + 3x + x) + (5 − 2)

2) Simplify

5 x + 3

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3 years ago
What is three fourths divided by eight thirdteenths?
xz_007 [3.2K]
\frac{3}{4} , \frac{8}{13} is \frac{39}{32}

\frac{3}{4}  *  \frac{13}{8} = ? , multiple by the reciprocal of the second fraction to get the answer.
3 0
3 years ago
1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.
german

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:

\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx

\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}

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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3 years ago
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Answer:

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