What will be the value of the 22nd coefficient of a binomial raised to the power 25

Step-by-step

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.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.

5

step-by-step explanation: