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lara [203]
3 years ago
12

find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. show all of your work.​

Mathematics
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

The surface area of right regular hexagonal pyramid = 82.222 cm³

Step-by-step explanation:

Given as , for regular hexagonal pyramid :

The of base side = 3 cm

The slant heights = 6 cm

Now ,

The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times a^{2} + 3 a \sqrt{h^{2}+ 3\times \frac{a^{2}}{4}}

Where a is the base side

And h is the slant height

So, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 3^{2} + 3 \times 3 \sqrt{6^{2}+ 3\times \frac{3^{2}}{4}}

Or, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 9 + 9 \sqrt{36+ 3\times \frac{9}{4}}

Or,  The surface area of right regular hexagonal pyramid = 23.38 + 9 × \frac{171}{4}

∴  The surface area of right regular hexagonal pyramid = 23.38 + 9 × 6.538

I.e The surface area of right regular hexagonal pyramid = 23.38 + 58.842

So,  The surface area of right regular hexagonal pyramid = 82.222 cm³ Answer

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Ginnie plans to paint a wall. She measures its height and width. She finds that she needs enough paint to cover 8 square meters.
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Answer:

The 8 square meters represent the product of the height by the width of the wall and therefore, its area.

Step-by-step explanation:

Ginnie is going to paint a wall and measures the height and the width, the walls are usually in the form of a rectangle or square. To find the area of both we need to multiply the height by the width (in the case of the square both are the same) and this will give us the total amount of paint that we need to paint the wall.

In this example, Ginnie finds that she needs enough paint to cover 8 square meters, therefore, these 8 square meters represent the product of the height by the width (that we don't know but it doesn't matter) of the wall and therefore, its area.

7 0
3 years ago
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m_a_m_a [10]

Answer:

  • hemisphere volume: 262 m³
  • cylinder volume: 942 m³
  • composite figure volume: 1204 m³

Step-by-step explanation:

A. The formula for the volume of a hemisphere is ...

  V = (2/3)πr³

For a radius of 5 m, the volume is ...

  V = (2/3)π(5 m)³ = 250π/3 m³ ≈ 261.799 m³

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B. The formula for the volume of a cylinder is ...

  V = πr²h

For a radius of 5 m and a height of 12 m, the volume is ...

  V = π(5 m)²(12 m) = 300π m³ ≈ 942.478 m³

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C. Then the total volume is ...

  V = hemisphere volume + cylinder volume

  V = 261.799 m³ +942.478 m³ = 1204.277 m³

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Rounded to the nearest integer, the volumes are ...

  • hemisphere volume: 262 m³
  • cylinder volume: 942 m³
  • composite figure volume: 1204 m³

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As a rule, you only want to round the final answers. Here, the numbers are such that rounding the intermediate values still gives the correct final answer. That is not always the case.

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Problem 35

The bar graph is shown below

You simply draw various rectangles such that the heights represent the frequency of each animal type.

Eg: there are 20 elephants, so the "elephant" bar is 20 units tall.

You can make the bar graph by hand, or use spreadsheet software. I recommend going with software (if you can).

=======================================================

Problem 36

1 book = 20 mm thickness

5 books = 5*20 = 100 mm thickness

1 paper = 0.016 mm thickness

5 papers = 5*0.016 = 0.08 mm thickness

total thickness = 100+0.08 = 100.08 mm

<h3>Answer: 100.08 mm</h3>

=======================================================

Problem 37

121/11 = 11

121 = 11*11

If we say 11+11+11+...+11, and have 11 copies of these values added, then we get to a sum of 121

This is probably the easiest way to get the answer assuming repeated values are allowed.

You can stop here if your teacher allows you to use repeated values. If not, then move onto the next section.

-----------

If your teacher requires you to add 11 <u>different</u> numbers, then you can follow this procedure

  1. Write out eleven copies of "11" in a sequence
  2. Subtract 2 from the first "11" (to get 9) and add it to the last copy of "11" (to get 13)
  3. Subtract 4 from the second "11" (to get 7) and add it to the second to last copy of "11" (to get 15)
  4. Subtract 6 from the third "11" (to get 5) and add it to the third to last copy of "11" (to get 17)
  5. Subtract 8 from the fourth "11" (to get 3) and add it to the fourth to last copy of "11" (to get 19)
  6. Finally, subtract 10 from the fifth "11" (to get 1) and add it to the fifth to last copy of "11" (to get 21)

After carefully following those steps, you'll get this sequence (in the exact order shown):

{9, 7, 5, 3, 1, 11, 21, 19, 17, 15, 13}

There are three properties to notice of this sequence

  1. It's composed of two decreasing arithmetic sequences {9, 7, 5, 3, 1} and {21, 19, 17, 15, 13}
  2. The 11 in the middle hasn't been changed from the original sequence of nothing but "11"s.
  3. You should find that the terms of this new sequence add to 121.

That sequence we got sorts to {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}

and we can say 1+3+5+7+9+11+13+15+17+19+21 = 121

<h3>Answer: 1+3+5+7+9+11+13+15+17+19+21 = 121</h3>

5 0
2 years ago
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