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mina [271]
3 years ago
14

Mr. Osterhout is putting trim around the edge of a circular merry-go-round that has a diameter of 15 feet. How much trim does he

need to buy to the nearest tenth?
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0
Given:
circular merry-go-round that has a diameter of 15 feet.

Find: How much trim does he need to buy to put around the edge of the merry-go-round?

We need to find the circumference of the merry-go-round to get the measurement of the trim needed.

Circumference = π d
π = 3.14
d = diameter = 15 feet

Circumference = 3.14 * 15 feet
Circumference = 47.10 feet.

Mr. Osterhout needs to buy 47.10 feet of trim to put around the circular merry-go-round.
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Lorico [155]

Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

<em>Is the transformation a rigid motion?</em>

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

6 0
3 years ago
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