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umka2103 [35]
3 years ago
12

Solve the equation for x by graphing.

Mathematics
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer:

A

Step-by-step explanation:

You literally just have to use a calc or solve it by yourself. Doesn't need a graph.

-2x + 3 = 3x - 2 (-3 and -x create positive)

-5x + 3 = -2 (subtract 3x from both sides)

-5x = -5

x= 1

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What property -3(m + 8) = -3m - 24
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The vertices of xyz are x (1,-4)
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Answer:

5. The vertices of ΔX'Y'Z' are (-3, -7), (-6, -4), (-1, -2)

6. The vertices of ΔX'Y'Z' are (6, -7), (3, -4), (8, -2)

Step-by-step explanation:

If the point (x, y) translated by T → (h, k), then its image is (x + h, y + k)

#5

In ΔXYZ

∵ X = (1, -4), Y = (-2, -1), Z = (3, 1)

∵ T → (-4, -3)

∴ h = -4 and k = -3

→ Use the rule above to find the image of the vertices of the Δ

∵ X' = (1 + -4, -4 + -3)

∴ X' = (-3, -7)

∵ Y' = (-2 + -4, -1 + -3)

∴ Y' = (-6, -4)

∵ Z' = (3 + -4, 1 + -3)

∴ Z' = (-1, -2)

∴ The vertices of ΔX'Y'Z' are (-3, -7), (-6, -4), (-1, -2)

#6

In ΔXYZ

∵ X = (1, -4), Y = (-2, -1), Z = (3, 1)

∵ T → (5, -3)

∴ h = 5 and k = -3

→ Use the rule above to find the image of the vertices of the Δ

∵ X' = (1 + 5, -4 + -3)

∴ X' = (6, -7)

∵ Y' = (-2 + 5, -1 + -3)

∴ Y' = (3, -4)

∵ Z' = (3 + 5, 1 + -3)

∴ Z' = (8, -2)

∴ The vertices of ΔX'Y'Z' are (6, -7), (3, -4), (8, -2)

6 0
3 years ago
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