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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%282x-9%29%5E5%20%3D%20%5Csum_%7Bk%3D0%7D%5E5%20%5Cbinom5k%20%282x%29%5E%7B5-k%7D%28-9%29%5Ek%20%3D%20%5Csum_%7Bk%3D0%7D%5E5%20%5Cfrac%7B5%21%7D%7Bk%21%285-k%29%21%7D%202%5E5%20%5Cleft%28-%5Cfrac92%5Cright%29%5Ek%20x%5E%7B5-k%7D)
The <em>x</em> ³ terms occurs for 5 - <em>k</em> = 3, or <em>k</em> = 2, and its coefficient would be
![\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}](https://tex.z-dn.net/?f=%5Cdfrac%7B5%21%7D%7B2%21%285-2%29%21%7D%202%5E5%20%5Cleft%28-%5Cdfrac92%5Cright%29%5E2%20%3D%20%5Cboxed%7B6480%7D)
Answer:
José has 49 used stamps
Step-by-step explanation:
7 used stamps for every 4 new stamps
To solve this type of problem, we can use a ratio
7 : 4 ratio
He has 28 new stamps
7 : 4 = x : 28
Ratios can be written as fractions
Cross multiply
4x = 7(28) = 196
Divide by 4 on both sides
x = 49
Hope this helps :)
Answer:
its c
Step-by-step explanation:
It 5 hope it helps you have a good night