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Mashcka [7]
3 years ago
14

Which expression represents the sixth term in the binomial expansion of (2a - 3b)10?

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
8 0

<u>Answer</u>

2,737,152a⁵b⁵


<u>Explanation</u>

(2a - 3b)¹⁰ = (2a)¹⁰ - 10(2a)⁹(3b) + 47(2a)⁸(3b)² - 144(2a)⁷(3b)³ - 212(2a)⁶(3b)⁴ + 352(2a)⁵(3b)⁵

The 6th term in this expression is 352(2a)⁵(3b)⁵

352(2a)⁵(3b)⁵ = 352 × 2⁵a⁵ × 3⁵b⁵

                         = 352 × 32 ×243 ×a⁵b⁵

                         = 2,737,152a⁵b⁵

kaheart [24]3 years ago
3 0

Answer:

<h2>A. 10C5 (2a)^5 (-3b)^5 </h2>

Step-by-step explanation:

<h3><em>hope this helps!</em></h3>

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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)

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In ΔXYZ

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

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∵ X' = (1 + -4, -4 + -3)

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∴ Y' = (-6, -4)

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#6

In ΔXYZ

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

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

∵ X' = (1 + 5, -4 + -3)

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∵ 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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