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LekaFEV [45]
2 years ago
11

Can someone help me with this? pleasealguien me puede ayudar con esto? porfavorr

Mathematics
1 answer:
soldi70 [24.7K]2 years ago
8 0

See attachment for the number line that compares the numbers 5.3, 5 1/5 and 5

<h3>How to plot the numbers on the number line?</h3>

The numbers are given as:

5.3, 5 1/5 and 5

We start by representing the numbers as decimals (i.e. real numbers).

So, we have:

5.3, 5.2 and 5

Reorder the numbers in ascending number (i.e. from least to greatest)

So, we have

5, 5.2 and 5.3

Next, we plot the numbers on a number line using a scale of 0.1

This can be visualized as

5        5.2      5.3

See attachment for the number line that compares the numbers 5.3, 5 1/5 and 5

Read more about number line at

brainly.com/question/24644930

#SPJ1

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Answer:

314 in²  (nearest whole number)

Step-by-step explanation:

<u>Radius of a regular polygon</u>: The distance from the <u>center</u> of the polygon to any vertex.  The radius of a hexagon is equal to the length of one side.

Therefore, from inspection of the given diagram:

  • radius = 11 in  ⇒  side length = 11 in

To find the area of a regular polygon, we first need to calculate the apothem.   The <u>apothem</u> is the line drawn from the center of the polygon to the midpoint of one of its sides.

\textsf{Length of apothem (a)}=\dfrac{s}{2 \tan\left(\frac{180^{\circ}}{n}\right)}

where:

  • s = length of one side
  • n = number of sides

Given:

  • s = 11 in
  • n = 6

Substitute the given values into the formula and solve for a:

\implies \textsf{a}=\dfrac{11}{2 \tan\left(\frac{180^{\circ}}{6}\right)}=\dfrac{11\sqrt{3}}{2}

<u>Area of a Regular Polygon</u>

\textsf{A}=\dfrac{n\:s\:a}{2}

where:

  • n = number of sides
  • s = length of one side
  • a = apothem

Given:

  • n = 6
  • s = 11
  • \textsf{a}=\dfrac{11\sqrt{3}}{2}

Substitute the given values into the formula and solve for A:

\implies \sf A=\dfrac{6 \cdot 11 \cdot \dfrac{11\sqrt{3}}{2}}{2}

\implies \sf A=314.3672216...

\implies \sf A=314\:\:in^2\:\:(nearest\:whole\:number)

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2 years ago
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