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3 years ago

100 points! Look at the picture attached. I will mark brainliest!

Mathematics

2 answers:

Pepsi [2]3 years ago

4 0

Answer:

1432

Step-by-step explanation:

a = by/2. then add all areas together to get final answer.

MakcuM [25]3 years ago

4 0

Answer:

x > 6
6 < x < 62

Step-by-step explanation:

<h3>Question 5</h3>

The sides containing the given angles 37 and 84 are of same length:

one side is marked same, the second side is common.

The sides opposite angles of 37 and 84 are proportional to those angles, therefore:

0< x + 14 < 2(3x - 8)

The first part:

x > - 14

The second part:

x + 14 < 6x - 16
5x > 30
x > 6

Combination of the two is:

x > 6

<h3>Question 6</h3>

The triangles are isosceles, because:

The left triangle has two equal angles, so the opposite sides are congruent
The common side with the right triangle is same as marked side, so the triangle on the right is also isosceles.

The angles between equal sides

Left triangle:

180° - 2*70° = 40°

Right triangle:

180° - 2*66° = 48°

Since 40° < 48°, the opposite sides will have same inequality due to being proportional to angles:

3x - 18 < 2(x + 22)
3x - 18 < 2x + 44
3x - 2x < 44 + 18
x < 62

Same time we know the side must be a positive number:

3x - 18 > 0
3x > 18
x > 6

So the possible values of x are:

6 < x < 62 or x = (6, 62)

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Answer:

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|x|=5

x=±5

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#11
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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:

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:

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

$- 2x^3 -16x^2-40x$

$x+ 20$

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.