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Mars2501 [29]
3 years ago
14

A 208-yard-long road is divided into 16 parts of equal length. Mr.Ward paints a 4-yard-long strip in each part. How long is the

unpainted strip of each part of the road?
Mathematics
2 answers:
Andreas93 [3]3 years ago
5 0

Answer:

Length of unpainted strip of each part of the road = 9 yard

Step-by-step explanation:

Total length of the road = 208 yard

When it is divided into 16 equal parts,

length of each part = \frac{208}{16}

                                 = 13 yard

Mr.Ward paints 4 yard long strips in each part .That is he does not completely paint the part but paints only a small section of each part.

Length of unpainted strip of each part of the road = length of each strip - length of strip painted by Mr.Ward

Length of unpainted strip of each part of the road = 13 - 4

Length of unpainted strip of each part of the road = 9 yard

Allushta [10]3 years ago
4 0

Answer:

In each part, unpainted strip is 9 yard long.

Step-by-step explanation:

The total strip is of given length 208 yard long.

It is divided into 16 parts, so the length of each strip is,

= \frac{208}{16} = 13 yard .

Now it is given that,  <em>Mr.Ward paints a 4-yard-long strip in each part,</em>

so the unpainted part of strip length is given by,

13 - 4 = 9 yard.

<em>Thus, the unpainted part of strip is 9 yard long.</em>

You might be interested in
How would you calculate the area and perimeter of 3yd 6yd 5yd 10yds
spayn [35]
Okay, start with the smaller square, you know the side length is 3 and that the width is 5, multiply this together:
3 * 5 = 15

That's the smaller left side of this shape, now the bigger side, divide the 10 yards at the bottom by 2 to properly represent the other side then multiply this by the 6 over there.
5 * 6 = 30

Conclusion for Area: After adding the results the answer should be 45.

The perimeter is solely adding all the sides together. You have a 10, two 5's, one 6, and two 3's. Add this together:

10 + 5 + 5 + 6 + 3 + 3 = 32

Conclusion for Perimeter: So the perimeter would be 32.

I hope this helps in someway, have a great rest of your day! ^ ^
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3 years ago
What part of the expression 4(x+3) is a factor
makkiz [27]
4,x,3,and (x+3)
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3 years ago
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

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A town has three roads that form an obtuse triangle. James wants to set up a refreshment stand so that the shortest distance fro
GarryVolchara [31]
He should set up the refreshment stand on the incenter of the obtuse triangle. The incenter of a triangle is described as the intersection between the angle bisectors of a triangle. The inradius are the line segments from the incenter of the triangle to each of the three sides of the triangle which are all equal. The inradius is depicted as the radius of an inscribed circle in the triangle. Therefore, the shortest equal distance from his stand to each road is C. on the incenter.
4 0
3 years ago
What is the area , in square meters of the shaded part of the rectangle below
Licemer1 [7]

Answer:

123.5 m^2

Step-by-step explanation:

The area of the whole rectangle is 9 x 19 = 171 m^2

The area of the region that isn't shaded is (5 x 19)/2 = 47.5 m^2 because the height of the triangle is 5 (9 - 4) and the area of a triangle is

(height x length)/2

171 - 47.5 = 123.5 m^2

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3 years ago
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