Step-by-step answer:
We are looking at the coefficient of the 22nd term of (x+y)^25.
Following the sequence, first term is x^0y^25, second term is x^1y^24, third term is x^2y^23...and so on, 22nd term is x^21y^4.
The twenty-second term of (x+y)^25 is given by the binomial theorem as
( 25!/(21!4!) ) x^21*y^4
=25*24*23*22/4! x^21y^4
= 12650 x^21 y^4
The coefficient required is therefore 12650, for a binomial with unit valued coefficients.
For other binomials, substitute the values for x and y and expand accordingly.
Question would have been more clearly stated if the actual binomial was given, as commented above.
Answer:
A
Step-by-step explanation:
1.) The sum(addition) of 21 and 5 times(multiplication) a number f is(=) 61.
f = unknown number/variable [So 21 plus 5f(5 times f) equals 61]
21 + 5f = 61 [21(one-time) + 5f(number x variable) = 61(total)]
2.) Seventeen more(addition) than seven times(multiplication) a number j is(=) 87.
j = unknown number/variable [So 17 plus 7j(7 times j) equals 87]
17 + 7j = 87
3.) n = number of calls
18 + 0.05n = 50.50
[Company charges $18 plus five cents per call(n), and the total charge was $50.50]
4.) s = the number of students
40 + 30s = 220
[Tutor charges $40 plus $30 per student(s), and the total charge was $220]
Answer:
if this is a fraction, then negative 5 over a⁶
Answer:
w+4 is bigger!
Step-by-step explanation:
0 is worth nothing
