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Svetradugi [14.3K]
3 years ago
11

Point (2, 4) is reflected across the x-axis. What are the coordinates of its image?

Mathematics
1 answer:
klio [65]3 years ago
5 0

Answer:

The coordinates of its image are (2, -4)

Step-by-step explanation:

<em>Let us revise the rules of reflection across the axes</em>

  • If the point (x, y) reflected across the x-axis, then its image is (x, -y), the rule of reflection is rx-axis (x, y) → (x, -y)
  • If the point (x, y) reflected across the y-axis, then its image is (-x, y), the rule of reflection is ry-axis (x, y) → (-x, y)

∵ The point (2, 4) is reflected across the x-axis

→ By using the first rule above rx-axis (x, y) → (x, -y)

∴ Change the sign of its y-coordinate

∴ Its image is (2, -4)

∴ The coordinates of its image are (2, -4)

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3 years ago
5x - 6 = -16 (Show work)
zysi [14]
  • <em>Answer:</em>

<em>x = - 2</em>

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<em>Hi there ! </em>

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Read 2 more answers
How do I solve this, please help
Bas_tet [7]

9514 1404 393

Answer:

  • EF = DE = 44
  • FG = DG = 36
  • FH = DF = 31

Step-by-step explanation:

Since EH is the perpendicular bisector of DF, ∆DEF is isosceles and sides DE and EF have the same length.

  DE = EF

  (9x -1) = (7x +9)

  2x = 10 . . . . . . . add 1-7x

  x = 5 . . . . . . . . . divide by 2

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Similarly, marked sides GD and GF are the same length, so ...

  GD = GF

  (10y -4) = (7y +8)

  3y = 12 . . . . . . . . . . add 4-7y

  y = 4 . . . . . . . . . divide by 3

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Now, we have what we need to calculate the side lengths.

  EF = 7x+9 = 7·5 +9 = 44

  DE = 9x-1 = 9·4 -1 = 44

  FG = 7y+8 = 7·4 +8 = 36

  DG = 10y-4 = 10·4 -4 = 36

  FH = 3x+4y = 3·5 +4·4 = 31

  DF = FH = 31

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