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Lady_Fox [76]
3 years ago
8

The vertices of xyz are x (1,-4)

Mathematics
1 answer:
dexar [7]3 years ago
6 0

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)

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The solution to given system of equations are (x,y) = (\frac{4}{25} , \frac{123}{25})

<em><u>Solution:</u></em>

Given that we have to find solution to the system of equations

<em><u>Given equations are:</u></em>

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<em><u>Substitute eqn 2 in eqn 1</u></em>

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\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

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Answer:

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Step-by-step

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