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OlgaM077 [116]
3 years ago
8

HELPP

Mathematics
1 answer:
nasty-shy [4]3 years ago
4 0

Answer:

\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1} = 2 \cdot x^2 - 9 \cdot x + 7 \  Remainder \ (x - 2)

Step-by-step explanation:

Part I

The problem can be expressed as follows;

The dividend is 4·x⁴ - 5·x³ + 2·x² - x + 5

The divisor is x² + x + 1

\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1}

Part II

The number of times x² goes into the larest term, 4·x⁴ = 4·x² times

2·x² - 9·x + 7

\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1}

<u>4·x⁴ + 4·x³ + 4·x²</u>

-9·x³ - 2·x² - x + 5

<u>-9·x³ - 9·x² - 9·x</u>

7·x² + 8·x + 5

<u>7·x² + 7·x + 7</u>

x - 2

Therefore, we have;

\dfrac{4 \cdot x^4 - 5\cdot x^3 + 2 \cdot x^2 - x + 5}{x^2 + x + 1} = 2 \cdot x^2 - 9 \cdot x + 7 \  Remainder \ (x - 2)

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Which pair of angles in the figure below are vertical angles? Check all that apply
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Step-by-step explanation:

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What is the side length of the eectagkes
dedylja [7]

1. 5 in and 1/3 in: Area = 5/3 in^2

2. 5 in and 4/3 in: Area= 20/3 in^2

3. 5/2 in and 4/3 in: Area=10/3 in^2

4. 7/6 in and 6/7 in: Area = 1 in^2

Step-by-step explanation:

<u>1. 5 in and 1/3 in</u>

Here,

Let\\l=5\\w=\frac{1}{3}\\Area=l*w\\=5*\frac{1}{3}\\=\frac{5}{3}\ in^2

<u>2. 5 in and 4/3 in</u>

Here,

l= 5\ in\\w=\frac{4}{3}\ in\\Area=l*w\\=5*\frac{4}{3}\\=\frac{20}{3}\ in^2

<u>3. 5/2 in and 4/3 in</u>

let\\l=\frac{5}{2}\\w=\frac{4}{3}\\Area= \frac{5}{2}*\frac{4}{3}\\=\frac{20}{6}=\frac{10}{3}\ in^2

<u>4. 7/6 in and 6/7 in</u>

<u>Let</u>

l=\frac{6}{7}\\w=\frac{7}{6}\\Area=l*w\\=\frac{6}{7}*\frac{7}{6}\\=1\ in^2

<u>Hence,</u>

1. 5 in and 1/3 in: Area = 5/3 in^2

2. 5 in and 4/3 in: Area= 20/3 in^2

3. 5/2 in and 4/3 in: Area=10/3 in^2

4. 7/6 in and 6/7 in: Area = 1 in^2

Keywords: Rectangle, Area

Learn more about rectangles at:

  • brainly.com/question/10703930
  • brainly.com/question/10772025

#LearnwithBrainly

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