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liq [111]
3 years ago
14

A need help Dilations-He Said, She Said (12/10)

Mathematics
1 answer:
Elodia [21]3 years ago
6 0

Answer:

We conclude that the original triangle ΔDEF is dilated by a scale factor of 5/2 to get the image triangle ΔD'E'F'.

Therefore, 'Cárl' is correct. Hence, option 'a' is correct.

Step-by-step explanation:

Checking the location of the vertices of the original triangle ΔDEF

  • D(-2, 4)
  • E(2, 4)
  • F(0, 2)

Checking the location of the dilated vertices of the image triangle ΔD'E'F'

  • D'(-5, 10)
  • E'(5, 10)
  • F'(0, 5)

Notice that if we multiply the vertices of the original triangle ΔDEF by 5/2, we get the correct image vertices of image triangleΔD'E'F'

i.e.

D(-2, 4) → D'(5/2 (-2), 5/2 (4))  → D'(-5, 10)

E(2, 4) → E'(5/2 (2), 5/2(4))  → D'(5, 10)

F(0, 2) → E'(5/2 (0), 5/2(2))  → F'(0, 5)

Thus, we conclude that the original triangle ΔDEF is dilated by a scale factor of 5/2 to get the image triangle ΔD'E'F'.

Therefore, 'Cárl' is correct. Hence, option 'a' is correct.

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The surface area of the composite figure is 714 sq. in.

<h3>What is formula of surface area of cube ?</h3>

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Let, length of the cuboid = l unit

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∴ Surface area of cube = (6×4²) sq. in. = 96 sq. in.

Now, given, length of cuboid = 15 in.

Width of the cuboid = 10 in. & height of the cuboid = 7 in.

∴ Surface area of the cuboid = 2{(15×10)+(15×7)+(10×7)} sq. in.

                                                = 2{150+105+70} sq. in.

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Here, we have to find the surface area of the composite figure.

We have to subtract the surface area from both of cube & cuboid, where they are attached.In that case, we have to subtract (4×4) from both.

So, we have to subtract 2(4×4) sq. in. = 32 sq. in.

∴ The surface area of the composite figure = (650+96-32) sq. in.

                                                                        = 714 sq. in.

Learn more about composite figure here :

brainly.com/question/27915109

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