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jekas [21]
3 years ago
15

Unit 5: Given the coordinates of the vertices of a quadrilateral, determine whether it is a square, a rectangle, or a

Mathematics
1 answer:
Aleks [24]3 years ago
5 0

Answer:

jfifjugswfshgfuebdjh uy7ferhjbdchuyz8hudbjchuy8o hcjndfm eciuojnm

Step-by-step explanation:

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The axis of symetry, or A.o.S, intersects a parabola at its vertex. The answer is x=1
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While researching Facts on the planet earth Kate learned that a true earth day is about 23.93 hours long how many hours are in 2
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335.02

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the hardcover version of a book weighs 7 ounces while its paperback version weighs 5 ounces. Forty-five copies of the book weigh
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Keywords:

<em>System of equations, variables, hardcover version, paperback version, books </em>

For this case we must construct a system of two equations with two variables. Let "h" be the number of hardcover version books, and let "p" be the number of paperback version books. If the hardcover version of a book weighs 7 ounces and the paperback version weighs 5 ounces, to reach a total of 249 ounces we have:

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p = 45-h

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x = 5 (45-h)

Option D

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What is all of the surface area and volume of this Castle? Find the surface area and volume of all the figures below, then out o
motikmotik

Answer:

Step-by-step explanation:

There are a few formulas that are useful for this:

  • lateral area of a pyramid or cone: LA = 1/2·Ph, where P is the perimeter and h is the slant height
  • lateral area of a cylinder: LA = π·dh, where d is the diameter and h is the height
  • area of a rectangle: A = lw, where l is the length and w is the width
  • volume of a cone or pyramid: V = 1/3·Bh, where B is the area of the base and h is the height
  • volume of a cylinder or prism: V = Bh, where B is the area of the base and h is the height

You will notice that for lateral area purposes, a pyramid or cone is equivalent to a prism or cylinder of height equal to half the slant height. And for volume purposes, the volume of a pyramid or cone is equal to the volume of a prism or cylinder with the same base area and 1/3 the height.

Since the measurements are given in cm, we will use cm for linear dimensions, cm^2 for area, and cm^3 for volume.

___

The heights of the cones at the top of the towers can be found from the Pythagorean theorem.

  (slant height)^2 = (height)^2 + (radius)^2

  height = √((slant height)^2 - (radius)^2) = √(10^2 -5^2) = √75 = 5√3

The heights of the pyramids can be found the same way.

  height = √(13^2 -2^2) = √165

___

<u>Area</u>

The total area of the castle will be ...

  total castle area = castle lateral area + castle base area

These pieces of the total area are made up of sums of their own:

  castle lateral area = cone lateral area + pyramid lateral area + cylinder lateral area + cutout prism lateral area

and ...

  castle base area = cylinder base area + cutout prism base area

So, the pieces of area we need to find are ...

  • cone lateral area (2 identical cones)
  • pyramid lateral area (2 identical pyramids)
  • cylinder lateral area (3 cylinders, of which 2 are the same)
  • cutout prism lateral area
  • cylinder base area (3 cylinders of which 2 are the same)
  • cutout prism base area

Here we go ...

Based on the above discussion, we can add 1/2 the slant height of the cone to the height of the cylinder and figure the lateral area of both at once:

  area of one cone and cylinder = π·10·(18 +10/2) = 230π

  area of cylinder with no cone = top area + lateral area = π·1^2 +π·2·16 = 33π

  area of one pyramid = 4·4·(13/2) = 52

The cutout prism outside face area is equivalent to the product of its base perimeter and its height, less the area of the rectangular cutouts at the top of the front and back, plus the area of the inside faces (both vertical and horizontal).

  outside face area = 2((23+4)·11 -3·(23-8)) = 2(297 -45) = 504

  inside face area = (3 +(23-8) +3)·4 = 84

So the lateral area of the castle is ...

  castle lateral area = 2(230π + 52) +33π + 504 + 84 = 493π +692

  ≈ 2240.805 . . . . cm^2

The castle base area is the area of the 23×4 rectangle plus the areas of the three cylinder bases:

  cylinder base area = 2(π·5^2) + π·1^2 = 51π

  prism base area = 23·4 = 92

  castle base area = 51π + 92 ≈ 252.221 . . . . cm^2

Total castle area = (2240.805 +252.221) cm^2 ≈ 2493.0 cm^2

___

<u>Volume</u>

The total castle volume will be ...

  total castle volume = castle cylinder volume + castle cone volume + castle pyramid volume + cutout prism volume

As we discussed above, we can combine the cone and cylinder volumes by using 1/3 the height of the cone.

  volume of one castle cylinder and cone = π(5^2)(18 + (5√3)/3)

  = 450π +125π/√3 ≈ 1640.442 . . . . cm^3

 volume of flat-top cylinder = π·1^2·16 = 16π ≈ 50.265 . . . . cm^3

The volume of one pyramid is ...

  (1/2)4^2·√165 = 8√165 ≈ 102.762 . . . . cm^3

The volume of the entire (non-cut-out) castle prism is the product of its base area and height:

  non-cutout prism volume = (23·4)·11 = 1012 . . . . cm^3

The volume of the cutout is similarly the product of its dimensions:

  cutout volume = (23 -8)·4·3 = 180 . . . . cm^3

so, the volume of the cutout prism is ...

  cutout prism volume = non-cutout prism volume - cutout volume

  = 1012 -180 = 832 . . . .  cm^3

Then the total castle volume is ...

  total castle volume = 2·(volume of one cylinder and cone) + (volume of flat-top cylinder) +2·(volume of one pyramid) +(cutout prism volume)

  = 2(1640.442) + 50.265 +2(102.762) +832 ≈ 4368.7 . . . . cm^3

4 0
3 years ago
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