Answer:
65625/4(x^5)(y²)
Step-by-step explanation:
Using binomial expansion
Formula: (n k) (a^k)(b ^(n-k))
Where (n k) represents n combination of k (nCk)
From the question k = 5 (i.e. 5th term)
n = 7 (power of expression)
a = 5x
b = -y/2
....................
Solving nCk
n = 7
k = 5
nCk = 7C5
= 7!/(5!2!) ------ Expand Expression
=7 * 6 * 5! /(5! * 2*1)
= 7*6/2
= 21 ------
.........................
Solving (a^k) (b^(n-k))
a = 5x
b = -y/2
k = 5
n = 7
Substituting these values in the expression
(5x)^5 * (-y/2)^(7-5)
= (3125x^5) * (-y/2)²
= 3125x^5 * y²/4
= (3125x^5)(y²)/4
------------------------------------
Multiplying the two expression above
21 * (3125x^5)(y²)/4
= 65625/4(x^5)(y²)
Since 9 is a factor of an unknown number, that means that that number has to be a multiple of 9. If 9 is the factor of something then that something HAS to be a multiple of 9.
Answer:
14
Step-by-step explanation:
add the stuff together
