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tester [92]
2 years ago
5

1. ADEA has vertices D(8,4), E(2, 6), and F(3, 1). What are the vertices of the image after a dilation with a scale factor of 5

using the origin as the center of dilation? TO​
Mathematics
1 answer:
lubasha [3.4K]2 years ago
3 0

Answer:

The vertices of the image are D' (40, 20), E' (10, 30), F' (15, 5)

Step-by-step explanation:

If the point (x, y) dilated by a scale factor k using the origin as a center of dilation, then its image is (kx, ky)

Let us use this rule to solve our question

∵ The vertices of ΔDEF are D (8, 4), E (2, 6), and F (3, 1)

∵ ΔDEF dilated by a scale factor 5

∴ k = 5

∵ The origin as the center of dilation

→ That means, multiply the coordinates of every point by 5

∴ D' = (8 × 5, 4 × 5)

∴ D' = (40, 20)

∴ E' = (2 × 5, 6 × 5)

∴ E' = (10, 30)

∴ F' = (3 × 5, 1 × 5)

∴ D' = (15, 5)

∴ The vertices of the image are D' (40, 20), E' (10, 30), F' (15, 5)

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Identify the area of segment MNO to the nearest hundredth. HELP PLEASE!! I don't understand it!
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Answer:

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Step-by-step explanation:

I don't understand it, either.

Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.

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The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.

The area of a segment is given by the formula ...

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Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...

  A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²

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3 0
3 years ago
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<span>log 5 7 =1.2090619551</span>

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