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2 years ago

Need help on this! thanks :)

Mathematics

2 answers:

Natasha2012 [34]2 years ago

4 0

Answer:

-2,4

Step-by-step explanation:

dedylja [7]2 years ago

3 0

Answer:

1/3

Step-by-step explanation:

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Find the coefficient of the x^3 in the expansion of (2x-9)^5

il63 [147K]

Use the binomial theorem:

$\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}$

The x ³ terms occurs for 5 - k = 3, or k = 2, and its coefficient would be

$\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}$

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Step-by-step explanation:

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3 years ago

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Dafna11 [192]

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