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:
80% that is what you got is this a test question or no?
Step-by-step explanation:
This question doesn't make sence to me could you explain
Answer is 3 seconds
When the bullet reaches the ground, ground being x in graph (and here its s which is = 0)
s = -16t^2 + 48t
s = 0, solve for t
0 = -16t^2 + 48t
0 = t ( -16t + 48)
0 = 16t ( - t + 3)
now you have two equation
0 = 16t and 0 = -t +3 ( you can look at the graph line touches x twice)
0 = 16 t
0 = t ( you know its false, because time = 0)
You are left with
0 = -t + 3
t = 3
It takes 3 seconds for the bullet to return to the ground.
// Hope this helps.
